"Science, Faculty of"@en . "Mathematics, Department of"@en . "DSpace"@en . "UBCV"@en . "Moyles, Iain"@en . "2015-06-04T16:51:28Z"@en . "2015"@en . "Doctor of Philosophy - PhD"@en . "University of British Columbia"@en . "In this thesis we present an analysis of the Gierer-Meinhardt model with saturation (GMS) on various curve geometries in \u00E2\u0084\u009D\u00C2\u00B2. We derive a boundary fitted coordinate framework which translates an asymptotic two-component differential equation into a single component reaction diffusion equation with singular interface conditions. We create a numerical method that generalizes the solution of such a system to arbitrary two-dimensional curves and show how it extends to other models with singularity properties that are related to the Laplace operator. This numerical method is based on integrating logarithmic singularities which we handle by the method of product integration where logarithmic singularities are handled analytically with numerically interpolated densities. In parallel with the generalized numerical method, we present some analytical solutions to the GMS model on a circular and slightly perturbed circular curve geometry. We see that for the regular circle, saturation leads to a hysteresis effect for two dynamically stable branches of equilibrium radii. For the near circle we show that there are two distinct perturbations, one resulting from the introduction of a angular dependent radius, and one caused by Fourier mode interactions which causes a vertical shift to the solution. We perform a linear stability analysis to the true circle solution and show that there are two classes of eigenvalues leading to breakup or zigzag instabilities. For the breakup instabilities we show that the saturation parameter can completely stabilize perturbations that we show are always unstable without saturation and for the zigzag instabilities we show that the eigenvalues are given by the near circle curve normal velocity. The breakup analysis is based on the reduction of an implicit non-local eigenvalue problem (NLEP) to a root finding problem. We derive conditions for which this eigenvalue problem can be made explicit and use it to analyze a stripe and ring geometry. This formulation allows us to classify certain technical properties of NLEPs such as instability bands and a Hopf bifurcation condition analytically."@en . "https://circle.library.ubc.ca/rest/handle/2429/53715?expand=metadata"@en . "Hybrid Asymptotic-NumericalAnalysis of Pattern FormationProblemsbyIain MoylesB.Sc., The University of Ontario Institute of Technology, 2009M.Sc., The University of British Columbia, 2011A THESIS SUBMITTED IN PARTIAL FULFILLMENT OFTHE REQUIREMENTS FOR THE DEGREE OFDOCTOR OF PHILOSOPHYinThe Faculty of Graduate and Postdoctoral Studies(Mathematics)THE UNIVERSITY OF BRITISH COLUMBIA(Vancouver)June 2015c\u00C2\u00A9 Iain Moyles 2015AbstractIn this thesis we present an analysis of the Gierer-Meinhardt model withsaturation (GMS) on various curve geometries in R2. We derive a boundaryfitted coordinate framework which translates an asymptotic two-componentdifferential equation into a single component reaction diffusion equation withsingular interface conditions. We create a numerical method that generalizesthe solution of such a system to arbitrary two-dimensional curves and showhow it extends to other models with singularity properties that are relatedto the Laplace operator. This numerical method is based on integrating log-arithmic singularities which we handle by the method of product integrationwhere logarithmic singularities are handled analytically with numerically in-terpolated densities. In parallel with the generalized numerical method, wepresent some analytical solutions to the GMS model on a circular and slightlyperturbed circular curve geometry. We see that for the regular circle, sat-uration leads to a hysteresis effect for two dynamically stable branches ofequilibrium radii. For the near circle we show that there are two distinctperturbations, one resulting from the introduction of a angular dependentradius, and one caused by Fourier mode interactions which causes a verticalshift to the solution. We perform a linear stability analysis to the true circlesolution and show that there are two classes of eigenvalues leading to breakupiiAbstractor zigzag instabilities. For the breakup instabilities we show that the sat-uration parameter can completely stabilize perturbations that we show arealways unstable without saturation and for the zigzag instabilities we showthat the eigenvalues are given by the near circle curve normal velocity. Thebreakup analysis is based on the reduction of an implicit non-local eigen-value problem (NLEP) to a root finding problem. We derive conditions forwhich this eigenvalue problem can be made explicit and use it to analyze astripe and ring geometry. This formulation allows us to classify certain tech-nical properties of NLEPs such as instability bands and a Hopf bifurcationcondition analytically.iiiPrefaceThe work in Chapter 4 has been submitted for application in [53] along withmy supervisor Dr. Michael Ward and my colleague Wang Hung Tse. Themajor contribution of Wang Hung Tse in [53] is on the modelling of urbancrime which is not presented in this thesis. Dr. Ward was the supervisingauthor and was primarily responsible for conceptualizing the generalized ex-plicit NLEP formulation as outlined in section 4.1 of Chapter 4 which wasmotivated from his work in [57]. My main contribution of this work was inthe extension of this framework to the models presented in sections 4.2 and4.3 of Chapter 4. All numerical simulations presented in [53] were performedby me and also appear in Chapter 5 Figures 5.1, 5.2, 5.3, 5.4, 5.5, and 5.6.The work in Chapter 6 has been submitted for application in [54] along withmy supervisor Dr. Brian Wetton. I was the lead investigator of this project,responsible for deriving the numerical method, performing all of the analysis,and conducting the numerical experiments. Dr. Wetton was the supervisoryauthor and was involved in project design, conception, and manuscript revi-sion.The remaining work in Chapter 2 and Chapter 3, of which I was the leadivPrefaceinvestigator, was original and is in preparation for a manuscript. I am re-sponsible for the thesis manuscript composition with revision support fromDr. Ward and Dr. Wetton.vTable of ContentsAbstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiPreface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ivTable of Contents . . . . . . . . . . . . . . . . . . . . . . . . . . viList of Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . xList of Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . xiiAcknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . xxviDedication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xxix1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 Main Contribution and Summary of Previous Work . . . . 61.2 Thesis Outline . . . . . . . . . . . . . . . . . . . . . . . . 102 General Curve Formulation and Quasi-Steady Solutions 142.1 Choosing a Coordinate System . . . . . . . . . . . . . . . 142.2 Asymptotic Expansion of Steady-State . . . . . . . . . . . 172.2.1 Global Inhibitor Sharp Interface Limit . . . . . . . 23viTable of Contents2.3 Quasi-Steady State Profiles for the Gierer-Meinhardt Model 262.3.1 Inhibitor Problem on a Circular Curve . . . . . . . 312.3.2 Inhibitor Problem on a Near Circular Curve . . . . 483 Linear Stability of Ring Solutions to Breakup and ZigzagModes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 683.1 Linear Stability Formulation . . . . . . . . . . . . . . . . 683.2 Eigenvalues Associated with \u00CE\u00A60 Even . . . . . . . . . . . . 763.2.1 Removing Saturation: The case b = 0 . . . . . . . 783.2.2 Real Eigenvalues . . . . . . . . . . . . . . . . . . . 793.2.3 Real Eigenvalues: m = O(1) . . . . . . . . . . . . 843.2.4 Real Eigenvalues: m\u001D O(1) . . . . . . . . . . . . 923.2.5 Real Eigenvalue Summary . . . . . . . . . . . . . . 943.2.6 Complex Eigenvalues . . . . . . . . . . . . . . . . 943.2.7 Complex Eigenvalues: 0 \u00E2\u0089\u00A4 m < mb\u00E2\u0088\u0092 . . . . . . . . 1003.2.8 Complex Eigenvalues: m > mb\u00E2\u0088\u0092 . . . . . . . . . . 1013.2.9 Eigenvalue Summary . . . . . . . . . . . . . . . . . 1023.2.10 Numerical Computation of Eigenvalues . . . . . . 1043.2.11 Computing Eigenvalues, \u00CF\u0084 6= 0 . . . . . . . . . . . 1063.2.12 Adding Saturation . . . . . . . . . . . . . . . . . . 1153.3 Eigenvalues Associated with \u00CE\u00A60 Odd . . . . . . . . . . . . 1193.3.1 Global Inhibitor Eigenvalue Problem . . . . . . . . 1334 Classification of Explicitly Solvable Non-Local EigenvalueProblems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1414.1 Explicit Non-Local Eigenvalue Formulation . . . . . . . . 141viiTable of Contents4.2 Explicit Stability Formulation for the Gierer-Meinhardt Modelon a Stripe . . . . . . . . . . . . . . . . . . . . . . . . . . 1484.2.1 Explicit Stripe Eigenvalues, \u00CF\u0084 = 0 . . . . . . . . . 1544.2.2 Explicit Stripe Eigenvalues, \u00CF\u0084 > 0 . . . . . . . . . 1594.3 Explicit Stability Formulation for the Gierer-Meinhardt Modelon a Ring . . . . . . . . . . . . . . . . . . . . . . . . . . . 1745 Full Numerical Simulations of the Gierer-Meinhardt Model1835.1 Stripe Numerical Experiments . . . . . . . . . . . . . . . 1855.2 Ring Numerical Experiments . . . . . . . . . . . . . . . . 1935.2.1 Explicit Formulation . . . . . . . . . . . . . . . . . 1945.2.2 Non-Explicit Formulation . . . . . . . . . . . . . . 1956 Solving the Gierer-Meinhardt Problem for Arbitrary Curvesin Two Dimensions . . . . . . . . . . . . . . . . . . . . . . . 2106.1 Layer Potential Formulation . . . . . . . . . . . . . . . . . 2116.1.1 Incorporating Neumann Boundary Conditions . . . 2146.1.2 Scaled Arclength parametrization . . . . . . . . . . 2166.1.3 Curve Dynamics . . . . . . . . . . . . . . . . . . . 2186.1.4 Normal Velocity Condition . . . . . . . . . . . . . 2196.1.5 Singular Integration . . . . . . . . . . . . . . . . . 2206.2 Numerical Formulation of Curve Motion Problem . . . . . 2246.2.1 Discretizing Integrals . . . . . . . . . . . . . . . . 2266.2.2 Numerical Equations . . . . . . . . . . . . . . . . . 2356.3 Solving the GMS Model . . . . . . . . . . . . . . . . . . . 244viiiTable of Contents6.3.1 Including Saturation and Computing Homoclinic Or-bits . . . . . . . . . . . . . . . . . . . . . . . . . . 2446.3.2 GMS Results . . . . . . . . . . . . . . . . . . . . . 2477 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2667.1 Future Work and Open Problems . . . . . . . . . . . . . . 273Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277AppendixA Derivation of Boundary Properties for Single Layered Po-tentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 288ixList of Tables3.1 Comparison for \u00CF\u0084 = 0 and b = 0 of numerical and asymptoticcomputations of mb\u00E2\u0088\u0092 and mb+ for a variety of exponent sets,\u000F, R, and r0 with D = 1 for all. The (n) refers to numericcomputations of (3.16) using eigs in Matlab. mb\u00E2\u0088\u0092(a) is com-puted via Newton\u00E2\u0080\u0099s method on (3.33), mb+(a1) is computedvia (3.41) while mb+(a2) is computed via (3.43). . . . . . . 1064.1 Asymptotic and numerical comparison of the neutral stabilitypoints mb\u00E2\u0088\u0092 , mb+ , and the dominant wave mode mdom. Thenumerical values (n) are obtained from Figure 4.1 and theasymptotic approximations (a) are obtained from (4.28) formb\u00E2\u0088\u0092 , (4.30) for mb+ , and (4.34) for mdom . . . . . . . . . . 1596.1 Numerical-analytic comparison of integrating\u00E2\u0088\u00AB 10 cos(\u00CF\u0083) log |\u00CF\u0083|d\u00CF\u0083using the product integration method with linear interpola-tion. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2306.2 Numerical-analytic comparison of integrating\u00E2\u0088\u00AB 10 cos(\u00CF\u0083) log |\u00CF\u0083|d\u00CF\u0083using the product integration method with quadratic interpo-lation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231xList of Tables6.3 The global truncation error for solving the MS problem withconcentric circles R1 = 1 (top table), R2 = 2 (bottom ta-ble) solving to T = 0.0469. We define xerr as the error inthe x-component of the curve position. The error in the y-component is the same and omitted. Verr and Lerr are theerrors in the normal velocity and curve length respectively.The rat suffix for each indicates the ratio of successive errorsto the previous one. The convergence is O(\u00E2\u0088\u0086\u00CF\u00832 log \u00E2\u0088\u0086\u00CF\u0083) asexpected. The CPU timings reflect the computation of bothcurves and does not included anything that can be precom-puted such as the singular scalar logarithmic integrals. . . 2416.4 The global truncation error for solving the GMS problem ona circle of radius r0 = 0.5 with R = 1, r0 = 1/2, D = 1,exponent set (2, 1, 2, 0) and saturation \u00CF\u0083\u00CB\u0086 = 10 solving to T =0.0469. We define xerr as the error in the x-component ofthe curve position. The error in the y-component is the sameand omitted. Verr and Lerr are the errors in the normalvelocity and curve length respectively. The rat suffix for eachindicates the ratio of successive errors to the previous one. Theconvergence is O(\u00E2\u0088\u0086\u00CF\u00832 log \u00E2\u0088\u0086\u00CF\u0083) as expected. The CPU timingsreflect the computation of both curves and does not includedanything that can be precomputed such as the singular scalarlogarithmic integrals. . . . . . . . . . . . . . . . . . . . . . 255xiList of Figures2.1 The boundary fitted coordinate system for some curve in R2.The normal points inward relative to the moving curve whichis parametrized by unit arclength. . . . . . . . . . . . . . . 152.2 Solutions to (2.22) for different values of b. Here we chooseL\u00CB\u0086 = 20 as a sufficient representation of infinity. Note we cansolve the equation on [0, L\u00CB\u0086] and use symmetry because thefunctions are even. . . . . . . . . . . . . . . . . . . . . . . 292.3 Numerical computation of the b derivative of A for o = 1 too = 6. Here we see that the derivative is always positive andeach value of o is bounded from below by the previous values.The integral diverges as b approaches bc from the left. . . . 342.4 Phase portrait of (2.43) for D = 1, b = 0, exponent set(2, 1, 2, 0), and various values of R. . . . . . . . . . . . . . 362.5 Phase portrait of (2.43) for D = 1, b = 0, exponent set(2, 2, 2, 0), and various values of R. . . . . . . . . . . . . . 37xiiList of Figures2.6 Bifurcation diagram to (2.43) for different values of the expo-nent q and b = 0. The differential equation undergoes a saddlenode bifurcation when R = 3.6220 (for q = 1) and R = 1.4296(for q = 2). The larger of the equilibrium r0 values belong tothe stable branch. The red dashed curve represents an asymp-totic approximation to the lower radius. . . . . . . . . . . 392.7 Modified saturation parameter b as a function of r0 for varioussaturation values \u00CF\u0083. Here we take D = 1, R = 1 and exponentset (2, 1, 2, 0). . . . . . . . . . . . . . . . . . . . . . . . . . 412.8 Growth of H\u00CB\u0086 versus \u00E2\u0088\u0092 log(r0) for R = 1 and when \u00CF\u0083 = 5. . 422.9 Right-hand side to (2.43) for various saturation values, \u00CF\u0083 andboundary values R. The exponent set here is (2, 1, 2, 0) andD = 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 432.10 Right-hand side to (2.43) for various saturation values, \u00CF\u0083 andboundary values R. The exponent set here is (2, 2, 2, 0) andD = 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 442.11 Bifurcation diagram to (2.43) for exponent set (2, 1, 2, 0) anddifferent values of \u00CF\u0083. The dashed curve represents an asymp-totic approximation for r0 \u001C 1. The smallest and highestequilibrium values are stable while there is an unstable tran-sition branch in the middle. . . . . . . . . . . . . . . . . . 45xiiiList of Figures2.12 Asymptotic corrections compared to numeric simulations ofthe curve inhibitor value U0 and the corresponding satura-tion value b from solving (2.32) for a perturbed circle withradius (2.52) and h(\u00CE\u00B8) = cos(6\u00CE\u00B8). Here we take exponent set(2, 1, 2, 0), R = 1, D = 1, r0 = 0.5, \u00CF\u0083 = 10, and \u00CE\u00B5 = 0.01. . 662.13 Asymptotic corrections compared to numeric simulations ofthe curve velocity V0 from solving (2.32) for a perturbed circlewith radius (2.52) and h(\u00CE\u00B8) = cos(6\u00CE\u00B8). Here we take exponentset (2, 1, 2, 0), R = 1, D = 1, r0 = 0.5, \u00CF\u0083 = 10, and \u00CE\u00B5 = 0.01. 673.1 Computation of f(\u00C2\u00B5) from (3.19b) for various o with m = 0,r0 = 0.5, and \u000F = 0.025. We set m = 0 solely to satisfy\u00C2\u00B5 = \u00CE\u00BB and deal with a single variable. The properties (3.27)derived analytically from o = 2 or o = 3 still hold for variousexponents. . . . . . . . . . . . . . . . . . . . . . . . . . . . 813.2 Numerical computation of f(\u00C2\u00B5) when m = 0, r0 = 0.5, \u000F =0.025, and o = 2 along with the asymptotic expression (3.28)demonstrating the simple pole at \u00C2\u00B5 = \u00CE\u00BD0. . . . . . . . . . . 843.3 Numerical computation of the derivative of (3.30) with respectto m. For a given value of m, R, and \u00CE\u00B8\u00CE\u00BB, we compute theorder derivative of (3.30) over r0 \u00E2\u0088\u0088 [0, R] and then take themaximum value over that interval. The figure shows eachmaximal value of the derivative as a function of \u00CE\u00B8\u00CE\u00BB for variousvalues of R. . . . . . . . . . . . . . . . . . . . . . . . . . . 87xivList of Figures3.4 Computation of fR and fI from (3.47) for various o with m =0, r0 = 0.5, and \u000F = 0.025. We set m = 0 solely to satisfydealing with a single variable \u00CE\u00BBI . The properties (3.54) derivedanalytically from o = 2 or o = 3 in Proposition 3.1 and 3.2 of[79] still hold for various exponents. . . . . . . . . . . . . . 993.5 Numerical computation of the largest real part of the eigen-value \u00CE\u00BB in (3.16) for the case \u00CF\u0084 = 0 and b = 0 using eigs inMatlab. The blue solid curves are where the largest eigenvalueis negative while the red dashed curves are where it is positive.In all experiments D = 1. . . . . . . . . . . . . . . . . . . 1053.6 Numerical computation for b = 0 of the largest real part of theeigenvalue \u00CE\u00BB in (3.16) using Newton\u00E2\u0080\u0099s method on (3.59). Thesolid curves are where the largest eigenvalue has negative realpart while the dashed curves are where it has positive realpart. In all experiments (2, q, o, s) = (2, 1, 2, 0), \u000F = 0.025,D = 1, R = 1, and r0 = 0.5. . . . . . . . . . . . . . . . . . 1113.7 Numerical computation of eigenvalues near m = mb\u00E2\u0088\u0092 for \u00CF\u0084 >\u00CF\u0084 \u00E2\u0088\u0097mb\u00E2\u0088\u0092 and \u00CF\u0084 < \u00CF\u0084\u00E2\u0088\u0097mb\u00E2\u0088\u0092 . In all experiments (2, q, o, s) = (2, 1, 2, 0),\u000F = 0.025, D = 1, R = 1, and r0 = 0.5. . . . . . . . . . . . 1123.8 Imaginary part of the eigenvalues with largest real part com-puted using Newton\u00E2\u0080\u0099s method on (3.59). The black x markpoints where Re(\u00CE\u00BB) = 0. In all experiments (2, q, o, s) =(2, 1, 2, 0), \u000F = 0.025, D = 1, R = 1, and r0 = 0.5. . . . . . 1143.9 Computation of b dependent eigenvalues to L0b given by (3.7). 116xvList of Figures3.10 Computation of eigenvalues for b 6= 0 and \u00CF\u0084 = 0. In all cases(2, q, o, s) = (2, 1, 2, 0), \u000F = 0.025, R = 1, r0 = 0.5, and D = 1 1173.11 Computation of eigenvalues for b = 0.2 and \u00CF\u0084 6= 0. In all cases(2, q, o, s) = (2, 1, 2, 0), \u000F = 0.025, R = 1, r0 = 0.5, and D = 1 1184.1 Eigenvalues for \u00CF\u0084 = 0 computed from (4.27) versus m for\u000F = 0.05, and s = 0 for several values of l. The curves fromhighest maximum to smallest maximum are l = 0, l = 0.5,and l =\u00E2\u0088\u009E respectively. . . . . . . . . . . . . . . . . . . . . 1584.2 Plot of \u00CE\u00BB versus m for q = 1 (solid curve) and q = 2 (dashedcurve). The parameters here are s = 0, l =\u00E2\u0088\u009E, and \u00CF\u0084 = 2. . 1704.3 Plot of \u00CF\u0084Hm and \u00CE\u00BBIH from (4.52) for m in 0 < m < mb\u00E2\u0088\u0092 andl =\u00E2\u0088\u009E. The parameter values are s = 0 and \u000F = 0.05 while thesolid and dashed curves are for q = 1 and q = 2 respectively. 1714.4 Plot of eigenvalues \u00CE\u00BB versus m for l = 0.1 (solid curve) andl = 0.8 (dashed curve). The parameter values are q = 1, s = 0,\u000F = 0.05, and \u00CF\u0084 = 2. . . . . . . . . . . . . . . . . . . . . . 1744.5 Bifurcation diagram for (4.54) for q = 2. A saddle-node bifur-cation occurs when l = 3.622. The larger of the equilibriumr0 values belong to the stable branch. . . . . . . . . . . . . 1764.6 Plot of (4.54) for q = 1 and various values of l. We alwayshave that dr0/dT < 0 and therefore there are no equilibriumring radii when q = 1. . . . . . . . . . . . . . . . . . . . . . 1774.7 Eigenvalues \u00CE\u00BB versus m for q = 2, s = 0, \u000F = 0.05, l = 5, and\u00CF\u0084 = 0 using (4.56). The solid curve is for r0 = 1.08 while thedashed curve is for r0 = 2.56. . . . . . . . . . . . . . . . . 180xviList of Figures4.8 Eigenvalues \u00CE\u00BB versus m for q = 2, s = 0, \u000F = 0.05, l = 5,and \u00CF\u0084 = 6. The lighter curve is for r0 = 1.08 while the heavycurve is for r = 2.56. We plot both for 0 < m < 40 sincelarge m behaviour is not very impacted by increasing \u00CF\u0084 andwill be represented by Figure 4.7. The positive eigenvalues arein dash while the negative eigenvalues are in solid. . . . . . 1825.1 Experiment 1: Contour plot of the solution v to (5.1a) withstripe geometry at four times with with exponent set (3, 1, 3, 0).The parameter values are \u000F0 = 0.05, D0 = 1, \u00CF\u0084 = 0.1, andd0 = 2. This corresponds to \u000F = 0.05, l = 1/2, \u00CF\u0084 = 0.1, andd = 2 in (4.27) of Chapter 4. . . . . . . . . . . . . . . . . . 1875.2 Experiment 1: Discrete Fourier transform of the solution v to(5.1a) with stripe geometry at four times with with exponentset (3, 1, 3, 0). The parameter values are \u000F0 = 0.05, D0 = 1,\u00CF\u0084 = 0.1, and d0 = 2. The upper left plot shows the amplitudesfrom the Fourier transform while the upper right plot displaysthe phase. Dominant modes are defined as any modes thathave an amplitude within 95% of the largest amplitude mode.The bottom graphic in each panel shows an inverse Fouriertransform of a solution comprised of only the most dominantmode. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188xviiList of Figures5.3 Experiment 2: Contour plot of the solution v to (5.1a) withstripe geometry at four times with with exponent set (3, 1, 3, 0).The parameter values are \u000F0 = 0.05, D0 = 1, \u00CF\u0084 = 0.1, andd0 = 3. This corresponds to \u000F = 0.05, l = 1/2, \u00CF\u0084 = 0.1, andd = 3 in (4.27) of Chapter 4. . . . . . . . . . . . . . . . . . 1895.4 Experiment 2: Discrete Fourier transform of the solution v to(5.1a) with stripe geometry at four times with with exponentset (3, 1, 3, 0). The parameter values are \u000F0 = 0.05, D0 = 1,\u00CF\u0084 = 0.1, and d0 = 3. The upper left plot shows the amplitudesfrom the Fourier transform while the upper right plot displaysthe phase. Dominant modes are defined as any modes thathave an amplitude within 95% of the largest amplitude mode.The bottom graphic in each panel shows an inverse Fouriertransform of a solution comprised of only the most dominantmode. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1905.5 Experiment 3: Contour plot of the solution v to (5.1a) withstripe geometry at four times with with exponent set (3, 1, 3, 0).The parameter values are \u000F0 = 0.05, D0 = 0.1, \u00CF\u0084 = 0.1,and d0 = 2. This corresponds to \u000F = 0.05\u00E2\u0088\u009A10 \u00E2\u0089\u0088 0.1581,l =\u00E2\u0088\u009A10/2 \u00E2\u0089\u0088 1.58, \u00CF\u0084 = 0.1, and d = 2\u00E2\u0088\u009A10 \u00E2\u0089\u0088 6.32 in (4.27) ofChapter 4. . . . . . . . . . . . . . . . . . . . . . . . . . . . 192xviiiList of Figures5.6 Experiment 3: Discrete Fourier transform of the solution v to(5.1a) with stripe geometry at four times with with exponentset (3, 1, 3, 0). The parameter values are \u000F0 = 0.05, D0 = 0.1,\u00CF\u0084 = 0.1, and d0 = 2. The upper left plot shows the amplitudesfrom the Fourier transform while the upper right plot displaysthe phase. Dominant modes are defined as any modes thathave an amplitude within 95% of the largest amplitude mode.The bottom graphic in each panel shows an inverse Fouriertransform of a solution comprised of only the most dominantmode. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1935.7 Experiment 4: Contour and Fourier transform plot of thesolution v to (5.1a) with ring geometry with exponent set(3, 2, 3, 0). The parameter values are \u000F0 = 0.01, D0 = 0.04,and \u00CF\u0084 = 0.1. This corresponds to \u000F = 0.05, and l = 5 in (4.56)of Chapter 4. . . . . . . . . . . . . . . . . . . . . . . . . . 1955.8 Experiment 5: Contour plot of the solution v to (5.1a) withring geometry at four times with with exponent set (2, 1, 2, 0).The parameter values are \u000F0 = 0.025, D0 = 1, and \u00CF\u0084 = 0.1.This corresponds to \u000F = 0.025, R = 1, and \u00CF\u0084 = 0.1 in thenumerical computation of (3.16) of Chapter 3. . . . . . . . 197xixList of Figures5.9 Experiment 5: Discrete Fourier transform of the solution v to(5.1a) with ring geometry at four times with exponent set(2, 1, 2, 0). The parameter values are \u000F0 = 0.05, D0 = 1,and \u00CF\u0084 = 0.1. The upper left plot shows the amplitudes fromthe Fourier transform while the upper right plot displays thephase. Dominant modes are defined as any modes that havean amplitude within 95% of the largest amplitude mode. Thebottom graphic in each panel shows an inverse Fourier trans-form of a solution comprised of only the most dominant mode. 1985.10 Experiment 6: Contour plot of the solution v to (5.1a) withring geometry at four times with with exponent set (2, 1, 2, 0).The parameter values are \u000F0 = 0.025, D0 = 1, \u00CF\u0084 = 0.1, and\u00CF\u0083 = 25. This corresponds to \u000F = 0.025, R = 1, and \u00CF\u0084 = 0.1 inthe numerical computation of (3.16) of Chapter 3. . . . . . 2005.11 Experiment 6: Discrete Fourier transform of the solution vto (5.1a) with ring geometry at four times with exponent set(2, 1, 2, 0). The parameter values are \u000F0 = 0.025, D0 = 1,\u00CF\u0084 = 0.1, and \u00CF\u0083 = 25. The upper left plot shows the amplitudesfrom the Fourier transform while the upper right plot displaysthe phase. Dominant modes are defined as any modes thathave an amplitude within 95% of the largest amplitude mode.The bottom graphic in each panel shows an inverse Fouriertransform of a solution comprised of only the most dominantmode. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201xxList of Figures5.12 Experiment 7: Contour plot of the solution v to (5.1a) withring geometry at four times with with exponent set (2, 1, 2, 0).The parameter values are \u000F0 = 0.025, D0 = 1, \u00CF\u0084 = 0.1, and\u00CF\u0083 = 950. This corresponds to \u000F = 0.025, R = 1, and \u00CF\u0084 = 0.1in the numerical computation of (3.16) of Chapter 3. . . . 2035.13 Experiment 7: Discrete Fourier transform of the solution vto (5.1a) with ring geometry at four times with exponent set(2, 1, 2, 0). The parameter values are \u000F0 = 0.05, D0 = 1, \u00CF\u0084 =0.1, and \u00CF\u0083 = 950. The upper left plot shows the amplitudesfrom the Fourier transform while the upper right plot displaysthe phase. Dominant modes are defined as any modes thathave an amplitude within 95% of the largest amplitude mode.The bottom graphic in each panel shows an inverse Fouriertransform of a solution comprised of only the most dominantmode. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2045.14 Eigenvalues of (3.16) for exponent set (2, 1, 2, 0), \u000F = 0.05,\u00CF\u0084 = 0, and R = 1 with r0 = 0.213 and \u00CF\u0083 = 950. . . . . . . 2055.15 Experiment 8: Contour plot of the solution v to (5.1a) withring geometry at four times with with exponent set (2, 1, 2, 0).The parameter values are \u000F0 = 0.025, D0 = 1, \u00CF\u0084 = 0.1, and\u00CF\u0083 = 950. This corresponds to \u000F = 0.025, R = 1, and \u00CF\u0084 = 0.1.We take as an initial radius r0 = 0.5 + 0.02 cos(6\u00CE\u00B8). . . . . 206xxiList of Figures5.16 Experiment 9: Contour plot of the solution v to (5.1a) withring geometry at four times with with exponent set (2, 1, 2, 0).The parameter values are \u000F0 = 0.01, D0 = 0.01, \u00CF\u0084 = 0.1, and\u00CF\u0083 = 5910. This corresponds to \u000F = 0.1, R = 10, and \u00CF\u0084 = 0.1.We take as an initial radius r0 = 0.5 + 0.02 cos(6\u00CE\u00B8). . . . . 2086.1 This shows the value of integrating (6.38) with \u00E2\u0088\u0086\u00CF\u0083 = 0.02 forall the possible discrete values of \u00CF\u0083\u00E2\u0088\u0097 \u00E2\u0088\u0088 [0, 1). The blue solidcurve represents the technique used in the integral splitting(6.26) where singularities within a full period on either side ofthe true singularity are removed while the red dashed curverepresents removing only the true singular value. . . . . . . 2326.2 The plot of F (\u00CF\u0083\u00E2\u0088\u0097) as defined in (6.39). The function has anabsolute maximum at \u00CF\u0083\u00E2\u0088\u0097 = 0.5 and and absolute minimum at\u00CF\u0083\u00E2\u0088\u0097 = 0. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2346.3 The solution to the Mullins-Sekerka problem for concentriccircles with an outer radius R2 = 2 and inner radius R1 = 1.The solid blue curve is the numeric solution at t = 0 and thered dashed curve is the numeric solution at t = 0.2. The hollowcircles are the analytic solution as computed with (6.42) and(6.43) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 240xxiiList of Figures6.4 Evolution of non concentric circles with MS. The first circle iscentered at (\u00E2\u0088\u00921, 0) with radius R1 = 1 and the second circleis centered at (6, 6) with radius R2 = 2. The initial curve isin a blue solid line while the final curve at time t = 1.5 (\u00E2\u0088\u0086t =1 \u00C3\u0097 10\u00E2\u0088\u00922) is in a red dashed line. As time evolves, an effectknown as Ostwald ripening occurs [62] which favours growthof larger objects at the expense of shrinking small objects. 2426.5 Evolution of an ellipse to MS with major axis 3 and minor axis1. The initial curve is in a blue solid line while the final curveat time t = 2 (\u00E2\u0088\u0086t = 1\u00C3\u009710\u00E2\u0088\u00922) is in a red dashed line. The curvebecomes more circular as time evolves which is a consequenceof the area preserving and length shrinking property of theMS model [84]. . . . . . . . . . . . . . . . . . . . . . . . . 2436.6 Circle evolution under the GMS model with \u00CF\u0083\u00CB\u0086 = 0, R = 1,r0 = 1/2, D = 1, exponent set (2, 1, 2, 0) and time step 1 \u00C3\u009710\u00E2\u0088\u00923. The lines represent the numerical solution while thecircles represent the analytic solution computed using (2.42).The outer black line represents the boundary curve r = R = 1. 2496.7 U0 values computed numerically (solid line) and analytically(dashed-line) using (2.40) for the GMS model with \u00CF\u0083\u00CB\u0086 = 0,R = 1, r0 = 1/2, D = 1, and exponent set (2, 1, 2, 0). . . . 2506.8 Slope-field for circle evolution using the GMS model with R =4, D = 1, exponent set (2, 2, 2, 0) and \u00CF\u0083\u00CB\u0086 = 0. There is anunstable equilibrium at r0/R \u00E2\u0089\u0088 0.044 (r0 \u00E2\u0089\u0088 0.176) and a stableequilibrium at r0/R \u00E2\u0089\u0088 0.69 (r0 \u00E2\u0089\u0088 2.76). . . . . . . . . . . . 251xxiiiList of Figures6.9 Circle evolution using the GMS model with R = 4, D = 1,exponent set (2, 2, 2, 0), and \u00CF\u0083\u00CB\u0086 = 0. The boundary curve atR = 4 has been omitted. . . . . . . . . . . . . . . . . . . . 2516.10 Circle evolution under the GMS model with \u00CF\u0083\u00CB\u0086 = 10, R =1, r0 = 1/2, D = 1, exponent set (2, 1, 2, 0) and time step1\u00C3\u0097 10\u00E2\u0088\u00923. The lines represent the numerical solution while thecircles represent the analytic solution computed using (2.42).The outer black line represents the boundary curve r = R = 1. 2536.11 U0 values computed numerically (solid line) and analytically(dashed-line) using (2.40) for the GMS model with \u00CF\u0083\u00CB\u0086 = 10,R = 1, r0 = 1/2, D = 1, and exponent set (2, 1, 2, 0). Since\u00CF\u0083\u00CB\u0086 6= 0 a Newton\u00E2\u0080\u0099s method was used to solve the analytic value. 2546.12 Circle evolution under the GMS model with R = 1, r0 =1/2, D = 1, exponent set (2, 1, 2, 0) and time step 1 \u00C3\u0097 10\u00E2\u0088\u00922.The lines represent the numerical solution for different valuesof the saturation parameter \u00CF\u0083\u00CB\u0086 (0,10,30,50) at t = 0.1. Theboundary curve R = 1 has been omitted to more clearly showthe separate curves. . . . . . . . . . . . . . . . . . . . . . . 2546.13 Initial circle U0 formulation under the GMS model with R =4, r0 = 2, D = 1, and exponent set (2, 1, 2, 0). The bluesolid curve is the computed U0 solution from the numericalinterface problem with an initial guess of cos(3\u00CF\u0083) while thered dashed curve is the convergent solution to (2.51) by usingthe computed solution as an initial guess. . . . . . . . . . . 257xxivList of Figures6.14 Perturbation of a circle with perturbed radius r = 1/2 +0.1 cos(6\u00CE\u00B8) using the GMS model with R = 1, D = 1, ex-ponent set (2, 1, 2, 0), and \u00CF\u0083\u00CB\u0086 = 10. The boundary curve atR = 1 has been omitted. . . . . . . . . . . . . . . . . . . . 2596.15 Perturbation of a circle with perturbed radius r = 1/2 +0.3 cos(2\u00CE\u00B8) using the GMS model with R = 1, D = 1, ex-ponent set (2, 1, 2, 0), and \u00CF\u0083\u00CB\u0086 = 10. The boundary curve atR = 1 has been omitted. . . . . . . . . . . . . . . . . . . . 2606.16 Ellipse with major axis a = 1/2 and minor axis b = 1/4 usingthe GMS model with R = 1, D = 1, exponent set (2, 1, 2, 0),and \u00CF\u0083\u00CB\u0086 = 10. The boundary curve at R = 1 has been omitted. 2616.17 Circle evolution using the GMS model with R = 4, D = 1,exponent set (2, 1, 2, 0), and \u00CF\u0083\u00CB\u0086 = 10. . . . . . . . . . . . . . 2626.18 Non-concentric circle evolution with centre [\u00E2\u0088\u00921, 2] and radiusr0 = 1/2 using the GMS model with R = 4, D = 1, andexponent set (2, 1, 2, 0). The boundary curve at R = 4 hasbeen omitted. . . . . . . . . . . . . . . . . . . . . . . . . . 2636.19 Perturbation of a circle with perturbed radius r = 5+0.2 cos(6\u00CE\u00B8)using the GMS model with R = 10, D = 1, exponent set(2, 1, 2, 0), and \u00CF\u0083\u00CB\u0086 = 5. The boundary curve at R = 10 hasbeen omitted. . . . . . . . . . . . . . . . . . . . . . . . . . 264xxvAcknowledgementsThe completion of a thesis, particularly for a PhD, is at least as much atrial in endurance as it is an intelectual pusuit. Therefore, it is impossibleto achieve any success without a vartiey of support in a number of places. Iwant to take the time to acknowledge the important support I have receivedfor which I am eternally grateful.First and foremost, I need to thank my supervisors, Brian Wetton andMichael Ward for their patience and support while completing this thesis.Each has provided technical and moral support in their own ways through-out the entire process. I am particularly grateful for the support they offeredthrough encouraging and funding my participation at a variety of conferenceswhere I could present my work and network with the mathematical commu-nity.I am extremely grateful to the financial support I received through the VanierCanada Graduate Scholarship program. This relieved a lot of financial bur-den and allowed me to engage fully in my project. I was also supported bya four-year fellowship which activated at the completion of the Vanier awardand am grateful for the UBC tuition waiver applied during my studies.xxviAcknowledgementsIn terms of moral support, I could not say enough to appropriately capturethe love and understanding I have received from my wonderful fiance\u00C2\u00B4e, Sonia,to whom this thesis is dedicated. Through every long night and missed event,her patience and belief in me has helped me get through this entire process.She encourages me to pursue my dreams, but also keeps me grounded andreminds me of the truly important things in life.I am grateful for the support of my mom, grandma, and family back home,as well as to Aida, Joe, and my new family in Vancouver. In particular, I\u00E2\u0080\u0099dlike to acknowledge my mom for her unwaivering support throughout my en-tire life. From a young age, I had a scientific curiosity brewing inside of me,and she did everything she could to nurture my insatiable appetite for knowl-edge. I owe all of my commitment to making the world a better place throughcommunity engagement to the example she set in our community growing up.There are too many friends that have provided support of various amountsto mention them all here, but I will isolate a few very significant people.Firstly, I need to thank Kai, Fred, and Chen for our numerous pub outingsduring my time at UBC. They were the perfect relaxation needed after longperiod of stress. In particular I\u00E2\u0080\u0099m grateful to Kai for our lengthy discussions,commiserating together through the stressful and frustrating moments, ofwhich there are many, while completing my PhD. Thanks to Mike, Bern-hard, Carmen, Kyle, and more for, among other things, great conversation,great sushi, and great laughs! Thank you to Devin and Devon for friendshipxxviiAcknowledgementsand support. The ferry ride to visit them on Vancouver island, always clearsmy mind and relaxes me. I will always be grateful to the continued supportfrom my physics company at UOIT: Mike, Ryan, and Matt, who still inspireme to great things. In particular, I owe a lot of thanks to Matt for being anacademic role model. His success in Europe gave me the self-confidence tostart the next phase of my career there.Finally, I need to thank the little ones in my life, my nieces and nephews:Addison, Zackary, Emma, and Max for continuing to sharpen what may bethe most important tool for any scientist, my imagination.xxviiifor SoniaxxixChapter 1IntroductionPattern formation is the observation of orderly outcomes arising from com-mon attributes in a particular system. Patterns exist across all branchesof science and occurs on many magnitudes of scale from cell division at themicroscopic level to dune formations in the desert. The history of pattern for-mation is rich and was developed by groups of scientists with interdisciplinaryknowledge in mathematics, chemistry, biology, and physics. The origins ofpattern formation in science are traced to the study of oscillation in chemicalreactions [69]. One of the earliest papers to address this issue mathemati-cally is credited to Lotka [48] in which he solved differential equations thatcould represent a chemical system and showed that damped oscillation so-lutions occur. However, he conceded that, as of the time of writing (1910),he was unaware of any real chemical reactions which could be explained bysuch models. Lotka\u00E2\u0080\u0099s work was eventually extended to what became knownas the Lotka-Voltera equations (cf. [49],[16]), a general competition modelwhich has most frequently been used to study predator-prey relationships.One significant criticism of this type of model to general pattern formationwas that it was more representative of a mechanical equation, such as that ofa pendulum, where the final solutions were heavily dependent on initial data[69]. Later, a theoretical framework was presented which allowed for this1Chapter 1. Introductionrequired structural stability through limit cycle oscillations. These modelsbecame referred to as Brusselator models (cf. [75],[16]), although at the timethere was still little evidence of actual chemical reactions that had any sortof oscillatory behaviour. In the many years that followed however, chem-ical reactions were discovered such as the Briggs-Rauscher (cf. [10], [61])and perhaps most famously, the Belousov-Zhabotinsky reaction (cf. [8], [83],[19]), which can be described with the Brusselator framework. In the contextof limit cycles, Schnakenberg (cf. [71], [16]) refined the general ideas of theBrusselator model into a set of required conditions for limit cycles to form.Tangent to the study of chemical reactions was the study of patterns in thecontext of fluid dynamics, such as the Rayleigh-Be\u00C2\u00B4nard convection patterns(cf. [21], [13]). Unlike the difficulty in experimentally confirming theoreti-cal models of patterns in chemical reactions, the Rayleigh-Be\u00C2\u00B4nard instabilitywas, experimentally, very well described [69].Modern understanding of biological pattern formation is mostly attributedto the seminal paper by Turing [74] in which he showed that diffusion, amechanism typically associated with stability, could be a destabilizing mech-anism in a two-compartment system. His paper showcases several types ofsolutions including oscillations and travelling waves. The breadth of patternsdiscussed from this work was a catalyst in the rediscovery and correlation ofliterature in pattern formation which stimulated the research field [69]. Inparticular, many of Turing\u00E2\u0080\u0099s spatiotemporal patterns were found in papers inpopulation dynamics (cf. [17], [37]). In the context of this thesis, which is ahybrid analytical and numerical analysis of pattern formation, Turing\u00E2\u0080\u0099s paper2Chapter 1. Introductiondemonstrates extra significance because it was the first paper to incorporatenumerically computed solutions of reaction-diffusion equations alongside an-alytical results with Turing himself contributing heavily to the building ofthe computer used [69].Since Turing\u00E2\u0080\u0099s pioneering work, pattern formation problems have been stud-ied in a variety of contexts including animal spotting [63], sea-shell formation[67], urban crime analysis [41], and animal aggregation [12]. A common at-tribute of most pattern formation problems is that pattern initiation is gen-erally attributed to very complex dynamical systems involving positive feed-back loops, self-reinforcing conditions, and antagonistic tendencies [67]. Theparticular focus on pattern formation in a biological context has interestedscientists because of the large scale where cascades of chemical reactions andbiological processes, often beginning with a single cell, eventually develop intocomplex structures that are necessary to support life in organisms. While itmay seem a near impossible endeavour to understand the formation of thesecomplex structures, many of them, as a first approximation, can be taken tobe independent from one another. For example, the legs of most amphibiansdevelop regularly even when placed in ectopic positions indicating that thegrowth is primarily due to the influence of local variables (with respect tothe larger organism as a whole) [67].Following the work of Turing, it was postulated that more complex bio-logical pattern formation requires two conditions in order to persist: localself-enhancement and long-range inhibition (cf. [20], [72]). One such model,3Chapter 1. Introductionpresented in [20], is called the Gierer and Meinhardt (GM) model. The au-thors postulated a coupled partial differential equation model to describemorphogen activation and inhibition as it related to head formation in hy-dra. However, this model later proved to be useful in describing patterns insuch things as the formation of embryonic axes, leaf formation at the tip ofa growing shoot, and shell patterns on mollusks [67]. It is of interest to notethat, when postulating their model, Gierer and Meinhardt were unfamiliarwith the work of Turing and only became aware of it when it was mentionedby one of their article reviewers [51]. This model has a rich history of analysis(cf. [39], [80], [33], [46], [32], [15], [79] among many others) and will be thefocus of this thesis. A general non-dimensional GM model can be writtendown asvt = Dv\u00E2\u0088\u0086v \u00E2\u0088\u0092 v +vpuq(1.1a)\u00CF\u0084ut = Du\u00E2\u0088\u0086u\u00E2\u0088\u0092 u+vous(1.1b)with v an activator and u an inhibitor. The coefficients Di are the respectivediffusivities, \u00CF\u0084 is the inhibitor time constant, and the exponents (p, q, o, s)satisfy [20],p > 1, q > 0, o > 0, s \u00E2\u0089\u00A5 0, p\u00E2\u0088\u0092 1q Dc for somecritical diffusivity that depends on the outer disk radius and exponent set.In the presence of saturation a ring equilibrium does exist for all diffusivityvalues and initiates a hysteresis effect with an unstable branch of equilibriumbetween two stable branches. This is a phenomena that has not been ob-served in previous steady-state analysis of the GM or related models. Awayfrom the equilibria values, the dynamics of the ring radii follow a differentialequation which persists on an O(\u000F\u00E2\u0088\u00922) time scale. As such, in contrast tothe stripe problem, the steady-state is quasi time dependent and thereforethe linear stability analysis formulation involves an application of the WKBmethod (or alternatively the multiple time scales method) [9] for handling81.1. Main Contribution and Summary of Previous Workthe eigenvalue problems. Typically, this slow-time evolution analysis ariseswith the dynamic evolution of a critical bifurcation parameter [76], but israrely needed for a linear stability formulation.Using a boundary fitted coordinate system, we are able to pose the GMmodel in an arbitrary two-dimensional framework. We use this model to findthe existence of saturated near-ring solutions. We believe this is one of thefirst attempts to analytically treat a non-standard, non-symmetric geometrywith this model. The generalization of the model to the boundary fittedcoordinates, makes it non-self-adjoint and, as such, intractable to analysisin general geometries. However, it is easily recast as a singular interfaceproblem which can be analyzed and computed numerically using the methodof layer potentials. We formulate a numerical method to solve generalizedsingular interface problems, one of which is the saturated GM model. Thesesingular interface problems are in contrast to pattern formation models de-veloped in [23] which reduce to non-singular curve evolution equations whichare more tractable for computation than the models we present. Aside fromsolving the GM model, we show the generality of our numerical method byapplying it to the well studied Mullins-Sekerka problem [55]. The methodinvolves singular properties of the Laplace operator and related operators,such as the modified Helmholtz operator in the GM model. Recent work (cf.[68]) has studied the modified Helmholtz operator using a layer potentialformulation as well. However, the focus of that work was for solutions inall of space based on a set of static singular interfaces. Since we are gener-ally interested solely on the evolution of curves, we have designed our method91.2. Thesis Outlineto focus only on the tracking of curves subject to a set of dynamic conditions.Finally, for both stripes and rings using general exponent sets for (1.3), thestability analysis involves understanding a non-self-adjoint, non local eigen-value problem (NLEP). The complex structure of the eigenvalue problemgenerally leads to bounding arguments for eigenvalues such as the resultsin [39] and what we present in Chapter 3. However, for specific choices ofexponent sets, the eigenvalue problems can be formed in an explicitly solv-able way. These results were first discovered in [57] and we generalize theconditions for which an NLEP can be explicitly solvable, leading to a newstability classification for a previously unreported exponent set.1.2 Thesis OutlineWe present the material of this thesis as follows. In Chapter 2 we formulatethe general boundary fitted coordinate Laplace operator for use in solvingthe Gierer-Meinhardt model and reduce it to a singular interface probleminvolving a flux jump condition across some (possibly disconnected) curve\u00CE\u0093 and a normal velocity condition. We then restrict this problem to twogeometries. In 2.3.1, we first consider the asymptotic construction of a ringsolution on some circular domain \u00E2\u0084\u00A6ring including the extension of the normalvelocity condition to a dynamic differential equation for the ring radius. Weuse this to analyze the ring radii equilibrium and conclude that the equilib-rium structure is drastically different with and without saturation where asaddle node bifurcation occurs for the former and a hysteresis effect for the101.2. Thesis Outlinelatter. We also show the existence of ring solutions for which the inhibitoris non-radially symmetric. The determination of these solutions relies onstudying a root finding problem of a Fourier transform decomposition andis extremely preferential to the radially symmetric solution root. However,this root finding problem can be used to verify non-radially symmetric solu-tions that are found with the numerical study of Chapter 6. Next in 2.3.2,we utilize the boundary fitted coordinate framework to analyze the quasi-steady solutions of a near circular solution where the ring radius r = r(\u00CE\u00B8).The same model could be derived from a polar coordinate framework butthe jump and velocity conditions would be less natural to implement. Theboundary-fitted framework extracts both of these conditions regardless of theunderlying geometry. We show, using a Fourier analysis, that the first or-der corrections introduce sinusoidal perturbations to both the curve inhibitorvalue and radial velocity but that a second order correction is required toaccount for vertical shifts that may occur. This shifting is due to the inter-action of Fourier modes at higher order. Furthermore, we show that whenthe base radius of a slightly perturbed ring radius is small enough, the ve-locity corrections are in phase with the curve perturbation and therefore actto stabilize the curve toward a circle. This is due to the curvature being astabilizing term in the normal velocity equation.In Chapter 3 we return to the pure radially symmetric ring problem, andperform a linear stability analysis for arbitrary initial ring radius r0, notnecessarily in equilibrium. Because of this, the problem is not separable intime and we rely on a multiple scales argument through the WKB method111.2. Thesis Outlineto formulate the stability problem. We then derive an NLEP that exists fora certain class of even functions as activator perturbations. We show thatwithout saturation there is, for all \u00CF\u0084 , an unstable branch of real eigenvaluesthat leads to breakup instabilities. However, with the inclusion of satura-tion, we show that breakup patterns can be stabilized. We verify all of thesestability bands in 3.2.10, where we discretize the eigenvalue problem andsolve it numerically. This requires a robust Newton solve due to the highnonlinearity in the eigenvalue. We perform several numerical experiments tocompare asymptotic and numeric computations of upper and lower stabilitybounds. In 3.3 we consider a second class of instabilities where the activatorperturbation is odd. These eigenvalues are perturbations of \u00CE\u00BB = 0 and, as weshow, are of order \u000F2. The analysis of these eigenvalues is intimately relatedto the asymptotic construction of the activator and inhibitor solutions in theinner region near the ring.In Chapter 4 we introduce the notion of an explicitly solvable NLEP andprovide a general framework to classify an NLEP as such. We then ap-ply this framework to a specific exponent set for the GM model in 4.2 fora rectangular domain and in 4.3 for a circular domain. This exponent setyields the same conclusions as the non-explicit case in Chapter 3 but in away that is more tractable to analysis. Furthermore, in the case of a stripe,the simplicity of the Green\u00E2\u0080\u0099s function allows us to determine the stabilityboundary, dominant mode, and Hopf bifurcation analytically which is gen-erally not obtainable in the classic NLEP formulation. For the ring in whichthe Green\u00E2\u0080\u0099s function involves modified Bessel functions, we can recast the121.2. Thesis Outlineproblem in a way that is amendable to the framework for general NLEPs ona ring in Chapter 3. In this decomposition, we see that the explicit formu-lation has the effect of extracting the singular behaviour of the eigenvaluepole, a crucial component of the NLEP analysis. Chapter 5 is devoted to fullnumerical computations of (1.3) in order to verify the existence of stripe andring solutions, as well as to verify stability results. This includes performinga number of numerical experiments which verify predicted dominant wave-modes for breakup instabilities, the stabilizing effect of saturation, and thestability effects of a circle to zig-zag instabilities.Finally, in Chapter 6 we derive a numerical scheme to solve the singularinterface limit of (1.3) derived in Chapter 2 for any curve geometry in R2.This involves the use of a layer potential formulation to handle the curvesingularities and normal interface velocity. We formulate the problem overM possibly disjoint curves using a scaled arclength formulation in 6.1.2. In6.1.5, we discuss methods of handling the singular logarithmic integrals anddiscretize them using a combined Lagrange interpolation, trapezoid methodin 6.2.1. In order to validate our method we solve a different, but relatedproblem known as the Mullins-Sekerka problem in 6.2.2. We show that theerror converges in the standard way with chosen finite difference and timestepping schemes. In 6.3 we solve the saturated GM model for a variety ofinitial curves including circles, perturbed circles, ellipses in concentric andnon-concentric initializations. We use the analytical results of Chapter 2to confirm the numerical errors are consistent with the chosen discretizingschemes.13Chapter 2General Curve Formulation andQuasi-Steady SolutionsConsider a general reaction diffusion equation for an activator v and inhibitoru of the formvt = \u000F2\u00E2\u0088\u0086v \u00E2\u0088\u0092 v + g(u, v) (2.1a)\u00CF\u0084ut = D\u00E2\u0088\u0086u\u00E2\u0088\u0092 u+1\u000Ff(u, v) (2.1b)on a domain \u00E2\u0084\u00A6 \u00E2\u008A\u0082 R2 subject to Neumann boundary conditions on \u00E2\u0088\u0082\u00E2\u0084\u00A6. \u00CF\u0084is an effective time scale delay between the activator and inhibitor while Dand \u000F2 are the diffusivities of the inhibitor and activator respectively. As wasdiscussed in Chapter 1, we consider D = O(1) while \u000F\u001C 1 which defines thesemi-strong regime.2.1 Choosing a Coordinate SystemBefore proceeding to specific geometries, we will derive a general global prob-lem for the inhibitor based on activator localization on arbitrary curves \u00CE\u0093.To do this, we will consider a boundary fitted coordinate system (cf. Figure142.1. Choosing a Coordinate System2.1, [22], [35], [27]) wherex \u00E2\u0089\u00A1 \u00CE\u00B31(s) + \u00CE\u00B7n\u00CB\u0086x, y \u00E2\u0089\u00A1 \u00CE\u00B32(s) + \u00CE\u00B7n\u00CB\u0086y.Here s is the arclength of the curve with \u00E3\u0080\u0088\u00CE\u00B31(s), \u00CE\u00B32(s)\u00E3\u0080\u0089 being the parametriza-tion of \u00CE\u0093 and \u00CE\u00B7 is the signed normal distance from the curve with normaln\u00CB\u0086 = \u00E3\u0080\u0088n\u00CB\u0086x, n\u00CB\u0086y\u00E3\u0080\u0089. We consider the normal to be the inward pointing normalrelative to \u00CE\u0093 and so \u00CE\u00B7 > 0 denotes inside the curve for a single curve.\u00CE\u0093 = \u00E3\u0080\u0088\u00CE\u00B31(s), \u00CE\u00B32(s)\u00E3\u0080\u0089t\u00CB\u0086n\u00CB\u00861Figure 2.1: The boundary fitted coordinate system for some curve in R2. Thenormal points inward relative to the moving curve which is parametrized byunit arclength.We need to determine the Laplace operator using these boundary fitted coor-dinates. If we keep in mind that the curve is being parametrized by arclength152.1. Choosing a Coordinate Systemthen the unit tangent and normal vectors satisfyt\u00CB\u0086 \u00E2\u0089\u00A1 \u00E3\u0080\u0088\u00CE\u00B3\u00E2\u0080\u00B21, \u00CE\u00B3\u00E2\u0080\u00B22\u00E3\u0080\u0089, n\u00CB\u0086 \u00E2\u0089\u00A1 \u00E3\u0080\u0088\u00E2\u0088\u0092\u00CE\u00B3\u00E2\u0080\u00B22, \u00CE\u00B3\u00E2\u0080\u00B21\u00E3\u0080\u0089;t\u00CB\u0086\u00E2\u0080\u00B2 = \u00CE\u00BAn\u00CB\u0086, n\u00CB\u0086\u00E2\u0080\u00B2 = \u00E2\u0088\u0092\u00CE\u00BAt\u00CB\u0086where prime denotes differentiation with respect to s and \u00CE\u00BA is the signed cur-vature (positive for convex curves). Using this information we can computethe (s, \u00CE\u00B7) derivatives for (x, y) to get the Jacobian,J =\u00EF\u00A3\u00AE\u00EF\u00A3\u00B0xs x\u00CE\u00B7ys y\u00CE\u00B7\u00EF\u00A3\u00B9\u00EF\u00A3\u00BB =\u00EF\u00A3\u00AE\u00EF\u00A3\u00B0(1\u00E2\u0088\u0092 \u00CE\u00BA\u00CE\u00B7)\u00CE\u00B3\u00E2\u0080\u00B21 \u00E2\u0088\u0092\u00CE\u00B3\u00E2\u0080\u00B22(1\u00E2\u0088\u0092 \u00CE\u00BA\u00CE\u00B7)\u00CE\u00B3\u00E2\u0080\u00B22 \u00CE\u00B3\u00E2\u0080\u00B21\u00EF\u00A3\u00B9\u00EF\u00A3\u00BB ; det J = 1\u00E2\u0088\u0092 \u00CE\u00BA\u00CE\u00B7by noting that |t\u00CB\u0086| =\u00E2\u0088\u009A\u00CE\u00B3\u00E2\u0080\u00B221 + \u00CE\u00B3\u00E2\u0080\u00B222 = 1. Therefore, by the inverse functiontheorem,\u00EF\u00A3\u00AE\u00EF\u00A3\u00B0sx sy\u00CE\u00B7x \u00CE\u00B7y\u00EF\u00A3\u00B9\u00EF\u00A3\u00BB =\u00EF\u00A3\u00AE\u00EF\u00A3\u00B0\u00CE\u00B3\u00E2\u0080\u00B21(1\u00E2\u0088\u0092\u00CE\u00BA\u00CE\u00B7)\u00CE\u00B3\u00E2\u0080\u00B22(1\u00E2\u0088\u0092\u00CE\u00BA\u00CE\u00B7)\u00E2\u0088\u0092\u00CE\u00B3\u00E2\u0080\u00B22 \u00CE\u00B3\u00E2\u0080\u00B21.\u00EF\u00A3\u00B9\u00EF\u00A3\u00BB (2.2)Taking higher order derivatives and using the chain rule we can ascertainthat the Laplace operator in the new coordinate system can be written as\u00E2\u0088\u0086 = \u00E2\u0088\u0082xx + \u00E2\u0088\u0082yy = \u00E2\u0088\u0082\u00CE\u00B7\u00CE\u00B7 \u00E2\u0088\u0092\u00CE\u00BA(1\u00E2\u0088\u0092 \u00CE\u00BA\u00CE\u00B7)\u00E2\u0088\u0082\u00CE\u00B7 +1(1\u00E2\u0088\u0092 \u00CE\u00BA\u00CE\u00B7)\u00E2\u0088\u0082s(\u00E2\u0088\u0082s(1\u00E2\u0088\u0092 \u00CE\u00BA\u00CE\u00B7)). (2.3)It is worth noting that there is a slight issue with this formulation and thatis the singularity (1 \u00E2\u0088\u0092 \u00CE\u00BA\u00CE\u00B7) in the Laplace operator. The curvature at apoint s is the inverse of the radius of the osculating circle tangent to thecurve at s which is centered at some x \u00E2\u0088\u0088 R2. The singularity arises if the162.2. Asymptotic Expansion of Steady-Statenormal distance from the curve is equal to this osculating radius. Since pointsinfinitesimally close to s will also lie on the osculating circle then they wouldbe the same distance \u00CE\u00B7 away from the centre of the circle x. Therefore xis no longer uniquely defined by a single (s, \u00CE\u00B7) coordinate which means itcannot be a coordinate system for all of R2. However, the specific use of thiscoordinate transformation will be in analysis of the activator problem where\u00CE\u00B7 \u001C 1 in which case the coordinate system is uniquely defined as long as thecurvature is not sufficiently large.2.2 Asymptotic Expansion of Steady-StateWe begin by determining a quasi steady-state solution to (2.1) in the bound-ary fitted coordinates. We define quasi-steady solutions such that the onlytime dependence should come from a potential motion of the curve \u00CE\u0093, i.e.for some slow time scale T = a(\u000F)t, we have that the boundary fitted coordi-nates satisfy (\u00CE\u00B7, s) = (\u00CE\u00B7(T ), s(T )). Quasi-steady solutions are typical whenconsidering front motion problems (cf. [65], [31], [11]). We rescale our timederivative,\u00E2\u0088\u0082\u00E2\u0088\u0082t\u00E2\u0086\u0092 a\u00CE\u00B7\u00CB\u0099 \u00E2\u0088\u0082\u00E2\u0088\u0082\u00CE\u00B7+ as\u00CB\u0099\u00E2\u0088\u0082\u00E2\u0088\u0082swhere the dot indicates differentiation with respect to T . We will beginby considering an inner region near the front where v is localized. Havingdefined the boundary fitted coordinate system, it is easy to see that an O(\u000F)region near \u00CE\u0093 can be defined with the local distance coordinate \u00CE\u00B7\u00CB\u0086 = \u00CE\u00B7\u000F . We172.2. Asymptotic Expansion of Steady-Statewill also define the activator and inhibitor variable near the front,u\u00CB\u009C(\u00CE\u00B7\u00CB\u0086, s) = u(\u000F\u00CE\u00B7\u00CB\u0086, s), v\u00CB\u009C(\u00CE\u00B7\u00CB\u0086, s) = v(\u000F\u00CE\u00B7\u00CB\u0086, s).In the local coordinate frame, the system (2.1) becomes,a\u000F\u00CE\u00B7\u00CB\u0099v\u00CB\u009C\u00CE\u00B7\u00CB\u0086 + as\u00CB\u0099v\u00CB\u009Cs =v\u00CB\u009C\u00CE\u00B7\u00CB\u0086\u00CE\u00B7\u00CB\u0086 \u00E2\u0088\u0092\u000F\u00CE\u00BA(1\u00E2\u0088\u0092 \u000F\u00CE\u00BA\u00CE\u00B7\u00CB\u0086) v\u00CB\u009C\u00CE\u00B7\u00CB\u0086+\u000F2(1\u00E2\u0088\u0092 \u000F\u00CE\u00BA\u00CE\u00B7\u00CB\u0086)\u00E2\u0088\u0082\u00E2\u0088\u0082s(v\u00CB\u009Cs(1\u00E2\u0088\u0092 \u000F\u00CE\u00BA\u00CE\u00B7\u00CB\u0086))\u00E2\u0088\u0092 v\u00CB\u009C + g(u\u00CB\u009C, v\u00CB\u009C) (2.4a)\u000Fa\u00CF\u0084 \u00CE\u00B7\u00CB\u0099u\u00CB\u009C\u00CE\u00B7\u00CB\u0086 + \u000F2a\u00CF\u0084 s\u00CB\u0099u\u00CB\u009Cs =Du\u00CB\u009C\u00CE\u00B7\u00CB\u0086\u00CE\u00B7\u00CB\u0086 \u00E2\u0088\u0092\u000FD\u00CE\u00BA(1\u00E2\u0088\u0092 \u000F\u00CE\u00BA\u00CE\u00B7\u00CB\u0086) u\u00CB\u009C\u00CE\u00B7\u00CB\u0086+\u000F2D(1\u00E2\u0088\u0092 \u000F\u00CE\u00BA\u00CE\u00B7\u00CB\u0086)\u00E2\u0088\u0082\u00E2\u0088\u0082s(u\u00CB\u009Cs(1\u00E2\u0088\u0092 \u000F\u00CE\u00BA\u00CE\u00B7\u00CB\u0086))\u00E2\u0088\u0092 \u000F2u\u00CB\u009C+ \u000Ff(u\u00CB\u009C, v\u00CB\u009C). (2.4b)We expand the inner solutions as follows:v\u00CB\u009C \u00E2\u0088\u00BC v\u00CB\u009C0+\u000Fv\u00CB\u009C1+. . . u\u00CB\u009C \u00E2\u0088\u00BC u\u00CB\u009C0+\u000Fu\u00CB\u009C1+. . . , \u00CE\u00BA \u00E2\u0088\u00BC \u00CE\u00BA0+\u000F\u00CE\u00BA1+. . . , \u00CE\u00B7\u00CB\u0099 \u00E2\u0088\u00BC \u00CE\u00B7\u00CB\u00990+\u000F\u00CE\u00B7\u00CB\u00991+. . .and note that since we want the curve motion to be on a sub-order one timescale, it is most natural to take a = \u000F2. To first order we get,v\u00CB\u009C0\u00CE\u00B7\u00CB\u0086\u00CE\u00B7\u00CB\u0086 \u00E2\u0088\u0092 v\u00CB\u009C0 + g(u\u00CB\u009C0, v\u00CB\u009C0) = 0 (2.5a)u\u00CB\u009C0\u00CE\u00B7\u00CB\u0086\u00CE\u00B7\u00CB\u0086 = 0. (2.5b)We can solve (2.5b) to get,u\u00CB\u009C0 = A\u00CE\u00B7\u00CB\u0086 +B182.2. Asymptotic Expansion of Steady-Statebut since we expect the global solution for the inhibitor to be O(1), matchingwould required that A = 0. Therefore we have thatu\u00CB\u009C0 = U0(s, T )where we explicitly note the possible dependence on the arclength and timescaleT . We now draw our attention to (2.5a) which is supplemented by far-fieldconditions decaying to zero so that the solution is entirely localized. Sincethe problem exhibits translational invariance, we will impose that v\u00CB\u009C0\u00CE\u00B7\u00CB\u0086(0) = 0as a front centering condition. We consider the following Lemma for theexistence of a homoclinic orbit solution,Lemma 2.2.0.1 Consider the problemwyy \u00E2\u0088\u0092 w + f(w) = 0, \u00E2\u0088\u0092\u00E2\u0088\u009E < y <\u00E2\u0088\u009E, w \u00E2\u0086\u0092 0 as |y| \u00E2\u0086\u0092 \u00E2\u0088\u009E,w\u00E2\u0080\u00B2(0) = 0, wm = w(0) > 0 (2.6)and assume f(w) is C2 smooth on w > 0 with f(0) = 0 and f \u00E2\u0080\u00B2(0) < \u00E2\u0088\u00921. Ifwe define Q(w) \u00E2\u0089\u00A1 f(w)\u00E2\u0088\u0092w then a unique, positive, homoclinic orbit solutionexists when1. Q(0) = 0, Q\u00E2\u0080\u00B2(0) < 02. For s > 0 Q(s) = 0, Q\u00E2\u0080\u00B2(s) > 0; Q(w) < 0, for 0 < w < s3. Q(w) > 0 for s < w < wm with wm satisfying\u00E2\u0088\u00AB wm0 Q(w) dw = 0.The requirements on f(w) are needed so that w decays exponentially in thefar-field. When f \u00E2\u0080\u00B2(0) is finite (such as the case when f(w) = wp) then it192.2. Asymptotic Expansion of Steady-Stateis easy to see the exponential decay condition holds. In an example wheref \u00E2\u0080\u00B2(0) is not finite such as f(w) = w logw we can solve (2.6) exactly and showthat w \u00E2\u0086\u0092 exp(\u00E2\u0088\u0092y2) as y tends to infinity which has an even faster decaythan the finite case. The conditions on Q(w) can be proven by taking afirst integral of (2.6) and using the front centering condition w\u00E2\u0080\u00B2(0) = 0. Ifwe take w = v\u00CB\u009C0 and f(w) = g(U0, v\u00CB\u009C0) then Lemma 2.2.0.1 gives the condi-tions for (2.5a) to have homoclinic orbit solutions. See [39], [33], and [43] asexamples of where this formulation is used to form homoclinic orbit solutions.Completely separate from the homoclinic orbit solution, we note that g(u, v)does not depend explicitly on the space parameter and so the solutions to(2.5a) can be written as a superposition of an even and odd function [30].Since the homoclinic orbit satisfies positivity, this must be the even solutionand so formally we say that v\u00CB\u009C0 is the even homoclinic orbit solution to (2.5a).A corollary to the even homoclinic orbit solution is that the odd solution to(2.5a) necessarily blows up as |\u00CE\u00B7\u00CB\u0086| \u00E2\u0086\u0092 \u00E2\u0088\u009E.Continuing the expansion of (2.4) we have at O(\u000F),Lv\u00CB\u009C1 = \u00CE\u00BA0v\u00CB\u009C0\u00CE\u00B7\u00CB\u0086 \u00E2\u0088\u0092 gu(U0, v\u00CB\u009C0)u\u00CB\u009C1 + v\u00CB\u009C0\u00CE\u00B7\u00CB\u0086\u00CE\u00B7\u00CB\u00990 (2.7a)u\u00CB\u009C1\u00CE\u00B7\u00CB\u0086\u00CE\u00B7\u00CB\u0086 = \u00E2\u0088\u00921Df(U0, v\u00CB\u009C0) (2.7b)whereLv\u00CB\u009C1 = v\u00CB\u009C1\u00CE\u00B7\u00CB\u0086\u00CE\u00B7\u00CB\u0086 \u00E2\u0088\u0092 v\u00CB\u009C1 + gv(U0, v\u00CB\u009C0)v\u00CB\u009C1.202.2. Asymptotic Expansion of Steady-StateConsider the problem for the homoclinic orbit (2.5a) and differentiate,(v\u00CB\u009C0\u00CE\u00B7\u00CB\u0086)\u00CE\u00B7\u00CB\u0086\u00CE\u00B7\u00CB\u0086 \u00E2\u0088\u0092 v\u00CB\u009C0\u00CE\u00B7\u00CB\u0086 + gv(U0, v\u00CB\u009C0)v\u00CB\u009C0\u00CE\u00B7\u00CB\u0086 = Lv\u00CB\u009C0\u00CE\u00B7\u00CB\u0086 = 0. (2.8)Here we see that v\u00CB\u009C0\u00CE\u00B7\u00CB\u0086 is a homogeneous solution to (2.7a) and therefore wewill require an orthogonality condition with the source terms. This conditionis\u00CE\u00BA0\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ev\u00CB\u009C20\u00CE\u00B7\u00CB\u0086 d\u00CE\u00B7\u00CB\u0086 + \u00CE\u00B7\u00CB\u00990\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ev\u00CB\u009C20\u00CE\u00B7\u00CB\u0086 d\u00CE\u00B7\u00CB\u0086 \u00E2\u0088\u0092\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Egu(U0, v\u00CB\u009C0)v\u00CB\u009C0\u00CE\u00B7\u00CB\u0086u\u00CB\u009C1 d\u00CE\u00B7\u00CB\u0086\u00EF\u00B8\u00B8 \u00EF\u00B8\u00B7\u00EF\u00B8\u00B7 \u00EF\u00B8\u00B8I= 0. (2.9)If we defineG \u00E2\u0089\u00A1\u00E2\u0088\u00AB v\u00CB\u009C00gu(U0, x) dx (2.10)then we can simplify the final integral and use integration by parts to getI = \u00E2\u0088\u0092\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009EGu\u00CB\u009C1\u00CE\u00B7\u00CB\u0086 d\u00CE\u00B7\u00CB\u0086by noting that since v\u00CB\u009C0 is even then G(\u00E2\u0088\u0092\u00E2\u0088\u009E) = G(\u00E2\u0088\u009E) = 0. We now define anew function,G\u00CB\u0086(\u00CE\u00B7\u00CB\u0086) \u00E2\u0089\u00A1\u00E2\u0088\u00AB \u00CE\u00B7\u00CB\u00860G(x) dx (2.11)so that integrating I by parts once more, we haveI = \u00E2\u0088\u0092 u\u00CB\u009C1\u00CE\u00B7\u00CB\u0086G\u00CB\u0086\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009E+\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009EG\u00CB\u0086u\u00CB\u009C1\u00CE\u00B7\u00CB\u0086\u00CE\u00B7\u00CB\u0086 d\u00CE\u00B7\u00CB\u0086.212.2. Asymptotic Expansion of Steady-StateAgain, since v\u00CB\u009C0 is even then G is as well, which ensures G\u00CB\u0086 is an odd function,so thatI = \u00E2\u0088\u0092G\u00CB\u0086(\u00E2\u0088\u009E) (u\u00CB\u009C1\u00CE\u00B7\u00CB\u0086(\u00E2\u0088\u009E) + u\u00CB\u009C1\u00CE\u00B7\u00CB\u0086(\u00E2\u0088\u0092\u00E2\u0088\u009E))\u00E2\u0088\u00921D\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009EG\u00CB\u0086f(U0, v\u00CB\u009C0) d\u00CE\u00B7\u00CB\u0086, (2.12)where we have simplified the last integral by using (2.7b). However, since v\u00CB\u009C0is even then so to is f(U0, v\u00CB\u009C0) so the final integrand is odd and hence vanishesover the domain. Finally then we can write the solvability condition (2.9) as\u00CE\u00B7\u00CB\u00990 = \u00E2\u0088\u0092\u00CE\u00BA0 \u00E2\u0088\u0092G\u00CB\u0086(\u00E2\u0088\u009E)\u00E2\u0088\u00AB\u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009E v\u00CB\u009C20\u00CE\u00B7\u00CB\u0086 d\u00CE\u00B7\u00CB\u0086(u\u00CB\u009C1\u00CE\u00B7\u00CB\u0086(\u00E2\u0088\u009E) + u\u00CB\u009C1\u00CE\u00B7\u00CB\u0086(\u00E2\u0088\u0092\u00E2\u0088\u009E)) (2.13)which prescribes the leading order velocity of the curve. We can also relatethe difference in u\u00CB\u009C1\u00CE\u00B7\u00CB\u0086 by integrating (2.7b),(u\u00CB\u009C1\u00CE\u00B7\u00CB\u0086(\u00E2\u0088\u009E)\u00E2\u0088\u0092 u\u00CB\u009C1\u00CE\u00B7\u00CB\u0086(\u00E2\u0088\u0092\u00E2\u0088\u009E)) = \u00E2\u0088\u00921D\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ef(U0, v\u00CB\u009C0) d\u00CE\u00B7\u00CB\u0086. (2.14)Finally, we will consider the expansion at O(\u000F2) as this will be needed whenanalyzing the spectrum of the linearization for a radial geometry,Lv\u00CB\u009C2 =\u00CE\u00BA0v\u00CB\u009C1\u00CE\u00B7\u00CB\u0086 + \u00CE\u00BA1v\u00CB\u009C0\u00CE\u00B7\u00CB\u0086 + \u00CE\u00BA20\u00CE\u00B7\u00CB\u0086v\u00CB\u009C0\u00CE\u00B7\u00CB\u0086 + \u00CE\u00B7\u00CB\u00990v\u00CB\u009C1\u00CE\u00B7\u00CB\u0086 + \u00CE\u00B7\u00CB\u00991v\u00CB\u009C0\u00CE\u00B7\u00CB\u0086 + s\u00CB\u00990v\u00CB\u009C0s \u00E2\u0088\u0092 v\u00CB\u009C0ss\u00E2\u0088\u0092 12guu(U0, v\u00CB\u009C0)u\u00CB\u009C21 \u00E2\u0088\u0092 guv(U0, v\u00CB\u009C0)u\u00CB\u009C1v\u00CB\u009C1 \u00E2\u0088\u009212gvv(U0, v\u00CB\u009C0)v\u00CB\u009C21 \u00E2\u0088\u0092 gu(U0, v\u00CB\u009C0)u\u00CB\u009C2,(2.15a)u\u00CB\u009C2\u00CE\u00B7\u00CB\u0086\u00CE\u00B7\u00CB\u0086 =1DU0 + \u00CE\u00BA0u\u00CB\u009C1\u00CE\u00B7\u00CB\u0086 \u00E2\u0088\u00921Dfu(U0, v\u00CB\u009C0)u\u00CB\u009C1 \u00E2\u0088\u00921Dfv(U0, v\u00CB\u009C0)v\u00CB\u009C1. (2.15b)Once again, this will have an orthogonality relationship with v\u00CB\u009C0\u00CE\u00B7\u00CB\u0086 producinga condition for the velocity correction \u00CE\u00B7\u00CB\u00991 but we do not derive this here as222.2. Asymptotic Expansion of Steady-Statewe will consider a singularity limit where we take \u000F to zero and hence thiscorrection will occur at higher order.2.2.1 Global Inhibitor Sharp Interface LimitTurning our attention to the global region where \u00CE\u00B7 = O(1), we have thatv \u00E2\u0089\u00A1 0 (to within exponential order) and so we only need to consider theproblem for the inhibitor u:D\u00E2\u0088\u0086u\u00E2\u0088\u0092 u+ 1\u000Ff(u, v) = 0 (2.16)where we assume that f(u, 0) = 0 so that the inhibitor solution does notexperience global blowup as \u000F\u00E2\u0086\u0092 0. Therefore then, the only contribution tothe reaction term is very close to the curve. In fact, since f(u, 0) = 0 and vdecays super-linearly to zero we have that,f(u, v)\u000F=f(u, v\u00CB\u009C(\u00CE\u00B7\u000F))\u000F=\u000F\u00E2\u0086\u00920\u00EF\u00A3\u00B1\u00EF\u00A3\u00B4\u00EF\u00A3\u00B2\u00EF\u00A3\u00B4\u00EF\u00A3\u00B3\u00E2\u0088\u009E, \u00CE\u00B7 = 00, elseand so (cf. [33])lim\u000F\u00E2\u0086\u00920f(u, v)\u000F= A\u00CE\u00B4(\u00CE\u00B7)with \u00CE\u00B4(\u00CE\u00B7) the Dirac mass centered at \u00CE\u00B7 = 0. To find A, we integrate over asmall domain including zero and scale to the inner coordinate,lim\u000F\u00E2\u0086\u00920\u00E2\u0088\u00AB 0+0\u00E2\u0088\u0092f(u(\u00CE\u00B7), v(\u00CE\u00B7))\u000Fd\u00CE\u00B7 = lim\u000F\u00E2\u0086\u00920\u00E2\u0088\u00AB 0+/\u000F0\u00E2\u0088\u0092/\u000Ff(u\u00CB\u009C(\u00CE\u00B7\u00CB\u0086), v\u00CB\u009C(\u00CE\u00B7\u00CB\u0086)) d\u00CE\u00B7\u00CB\u0086 = A.232.2. Asymptotic Expansion of Steady-StateExpanding out f(u\u00CB\u009C, v\u00CB\u009C),f(u\u00CB\u009C, v\u00CB\u009C) = f(U0, v\u00CB\u009C0) + \u000F (fu(U0, v\u00CB\u009C0)u\u00CB\u009C1 + fv(U0, v\u00CB\u009C0)v\u00CB\u009C1) + . . . =\u000F\u00E2\u0086\u00920f(U0, v\u00CB\u009C0)so thatA =\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ef(U0, v\u00CB\u009C0) d\u00CE\u00B7\u00CB\u0086,and finally (2.16) becomesD\u00E2\u0088\u0086u\u00E2\u0088\u0092 u = \u00E2\u0088\u0092(\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ef(U0, v\u00CB\u009C0) d\u00CE\u00B7\u00CB\u0086)\u00CE\u00B4(\u00CE\u00B7). (2.17)We refer to this as the sharp interface limit of (2.16) since we have capturedall of the asymptotic structure via a singularity at the interface. By usingthe sharp interface limit, we do not need to expand u in powers of \u000F as (2.17)captures the entire global problem. Upon solving u we will need to match tothe inner region viau(\u000F\u00CE\u00B7\u00CB\u0086, s) \u00E2\u0088\u00BC u(0, s) + \u000F\u00CE\u00B7\u00CB\u0086 \u00E2\u0088\u0082u\u00E2\u0088\u0082\u00CE\u00B7\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00CE\u00B7=0\u00C2\u00B1+ \u00C2\u00B7 \u00C2\u00B7 \u00C2\u00B7 = u\u00CB\u009C0(\u00C2\u00B1\u00E2\u0088\u009E) + \u000Fu\u00CB\u009C1(\u00C2\u00B1\u00E2\u0088\u009E),where the \u00C2\u00B1 indicates approaching the curve from either side of \u00CE\u00B7 = 0. Uponperforming the matching we have u(0, s) = U0 andu\u00CB\u009C1\u00CE\u00B7\u00CB\u0086(\u00C2\u00B1\u00E2\u0088\u009E) =\u00E2\u0088\u0082u\u00E2\u0088\u0082\u00CE\u00B7\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00CE\u00B7=0\u00C2\u00B1. (2.18)Using this with (2.14) we have that[\u00E2\u0088\u0082u\u00E2\u0088\u0082n]\u00CE\u00B7=0= \u00E2\u0088\u0092 1D\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ef(U0, v\u00CB\u009C0) d\u00CE\u00B7\u00CB\u0086,242.2. Asymptotic Expansion of Steady-Statewhere [\u00C2\u00B7]\u00CE\u00B7=a indicates a jump from \u00CE\u00B7 = a+ to \u00CE\u00B7 = a\u00E2\u0088\u0092. This is the equiva-lent singularity structure to match the Dirac measure in (2.17). Combiningeverything we can write the problem for u asD\u00E2\u0088\u0086u\u00E2\u0088\u0092 u = 0, x \u00E2\u0088\u0088 \u00E2\u0084\u00A6 (2.19a)\u00E2\u0088\u0082u\u00E2\u0088\u0082n= 0, x \u00E2\u0088\u0088 \u00E2\u0088\u0082\u00E2\u0084\u00A6 (2.19b)u = U0(s), x \u00E2\u0088\u0088 \u00CE\u0093 (2.19c)[\u00E2\u0088\u0082u\u00E2\u0088\u0082n]\u00CE\u0093= \u00E2\u0088\u0092 1D\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ef(U0, v\u00CB\u009C0) d\u00CE\u00B7\u00CB\u0086, x \u00E2\u0088\u0088 \u00CE\u0093. (2.19d)This quasi-steady problem is subject to a normal velocity V given by (2.13)which we can rewrite using the outer coordinates asV0 = \u00CE\u00BA0 +H(\u00E2\u0088\u0082u\u00E2\u0088\u0082n\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00CE\u00B7=0++\u00E2\u0088\u0082u\u00E2\u0088\u0082n\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00CE\u00B7=0\u00E2\u0088\u0092)(2.19e)where we have definedH \u00E2\u0089\u00A1 G\u00CB\u0086(\u00E2\u0088\u009E)\u00E2\u0088\u00AB\u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009E v\u00CB\u009C20\u00CE\u00B7\u00CB\u0086 d\u00CE\u00B7\u00CB\u0086. (2.20)Note that if V is the normal velocity measured with respect to the originthen \u00CE\u00B7\u00CB\u0099 = \u00E2\u0088\u0092V . This is because for a single curve, if \u00CE\u00B7\u00CB\u0099 > 0 then pointsinside the curve are increasing their distance from the curve, i.e. the curveis expanding in space. However, we have defined the inward pointing normalto be positive and so V > 0 means that the curve shrinks, hence V = \u00E2\u0088\u0092\u00CE\u00B7\u00CB\u0099.The system (2.19) is what we will solve for quasi-steady state profiles.252.3. Quasi-Steady State Profiles for the Gierer-Meinhardt Model2.3 Quasi-Steady State Profiles for theGierer-Meinhardt ModelMoving forward, we will consider the saturated Gierer-Meinhardt (GMS)model (cf. [20], [39], [82]),g(u, v) =v2uq(1 + \u00CF\u0083v2), f(u, v) =vous(2.21)for some exponent set (2, q, o, s) and saturation parameter \u00CF\u0083 > 0. Notice thatthese functions obey the required properties of f(u, v) and g(u, v) outlinedabove. With this formulation then we can write v\u00CB\u009C0 = Uq0w and recast (2.5a)for w withw\u00CE\u00B7\u00CB\u0086\u00CE\u00B7\u00CB\u0086 \u00E2\u0088\u0092 w +w21 + bw2= 0, w\u00E2\u0080\u00B2(0) = 0, lim|\u00CE\u00B7\u00CB\u0086|\u00E2\u0086\u0092\u00E2\u0088\u009Ew = 0, (2.22)whereb = U2q0 \u00CF\u0083 > 0 (2.23)is a modified saturation parameter. Notice if \u00CF\u0083 = b = 0 then we can analyt-ically satisfy the conditions of Lemma 2.2.0.1 and get that the positive evenhomoclinic orbit solution to (2.22) isw(\u00CE\u00B7\u00CB\u0086) =32sech 2(\u00CE\u00B7\u00CB\u00862). (2.24)262.3. Quasi-Steady State Profiles for the Gierer-Meinhardt ModelIf b 6= 0 we can no longer obtain an analytic solution but we can investigatethe criteria on b for which a homoclinic solution exists via Lemma 2.2.0.1and compute it numerically. This was considered in [39] for a stripe and wereproduce the analysis here. LetQ(w) =w21 + bw2\u00E2\u0088\u0092 wand look for rest-points of the differential equation when Q(w) = 0. We seethat w = 0 is a root for all b and the other two roots are given by,w\u00C2\u00B1 =1\u00C2\u00B1\u00E2\u0088\u009A1\u00E2\u0088\u0092 4b2b(2.25)where we notice that real roots fail to exist if b > 1/4. If b = 1/4 then we havethat the two roots degenerate to a single root w\u00E2\u0080\u00A0 = 2. If we classify the rootsof (2.25) by linear theory, we have that Q\u00E2\u0080\u00B2(0) < 0 for all b which classifies itas a saddle point as was required via condition 1 in Lemma 2.2.0.1. Asidefrom w\u00E2\u0088\u0097 = 0, we haveQ\u00E2\u0080\u00B2(w\u00E2\u0088\u0097) =2\u00E2\u0088\u0092 w\u00E2\u0088\u0097w\u00E2\u0088\u0097which is zero when w\u00E2\u0088\u0097 = w\u00E2\u0080\u00A0 owing to the degeneracy of the root. Thereforethis root cannot be classified by linear theory but we have that Q\u00E2\u0080\u00B2\u00E2\u0080\u00B2(w\u00E2\u0080\u00A0) =\u00E2\u0088\u00921/2 and therefore for b = 1/4 since w = 0 and w = w\u00E2\u0080\u00A0 are the only rootsthen Q(w) < 0 for all w. Therefore we cannot satisfy condition 2 or 3 inLemma 2.2.0.1 and no homoclinic orbit exists at this value. For 0 \u00E2\u0089\u00A4 b < 1/4,since w+ > w\u00E2\u0080\u00A0 (w\u00E2\u0088\u0092 < w\u00E2\u0080\u00A0) then Q\u00E2\u0080\u00B2(w+) < 0 (Q\u00E2\u0080\u00B2(w\u00E2\u0088\u0092) > 0) and thereforew = w\u00E2\u0088\u0092 is a center while w = w+ is another saddle point. Therefore, in272.3. Quasi-Steady State Profiles for the Gierer-Meinhardt Modelterms of condition 2 in Lemma 2.2.0.1, s = w\u00E2\u0088\u0092. To satisfy condition 3, werequire\u00E2\u0088\u00AB w(0)0Q(u) du = 0. (2.26)We already concluded that no homoclinic orbit exists when b = 1/4 but asb decreases from this value, causing w = w\u00E2\u0080\u00A0 to degenerate into w+ and w\u00E2\u0088\u0092then by condition 3 of Lemma 2.2.0.1, a homoclinic orbit will begin to existat the moment when w(0) = w+. We can determine this critical b = bc valueby numerically solving (2.26) with w+ given by (2.25). We conclude thatbc = 0.2113763204, w+(bc) = 3.295208658. (2.27)Notice that at this point exactly, the maximum value w(0) = w+ is also asaddle point of the phase space and therefore we actually have a heteroclinicorbit here connecting w = 0 to w = w+. However for b < bc then w(0) < w+and therefore, the homoclinic orbit exists on 0 \u00E2\u0089\u00A4 b < bc. Having chosen a bvalue in the acceptable range then we can compute solutions to (2.22) usinga standard finite difference solver on a truncated domain [0, L\u00CB\u0086] (see section6.3). Examples of homoclinic orbits for different b values are in Figure 2.2.282.3. Quasi-Steady State Profiles for the Gierer-Meinhardt Model\u00E2\u0088\u009220 \u00E2\u0088\u009215 \u00E2\u0088\u009210 \u00E2\u0088\u00925 0 5 10 15 2000.20.40.60.811.21.41.6 Homoclinic orbit for b= 0\u00CF\u0081w(a) b = 0\u00E2\u0088\u009220 \u00E2\u0088\u009215 \u00E2\u0088\u009210 \u00E2\u0088\u00925 0 5 10 15 2000.20.40.60.811.21.41.61.82 Homoclinic orbit for b= 0.12807\u00CF\u0081w(b) b = 0.12807\u00E2\u0088\u009220 \u00E2\u0088\u009215 \u00E2\u0088\u009210 \u00E2\u0088\u00925 0 5 10 15 2000.511.522.5 Homoclinic orbit for b= 0.19211\u00CF\u0081w(c) b = 0.19211\u00E2\u0088\u009220 \u00E2\u0088\u009215 \u00E2\u0088\u009210 \u00E2\u0088\u00925 0 5 10 15 2000.511.522.533.5 Homoclinic orbit for b= 0.21132\u00CF\u0081w(d) b = 0.21132Figure 2.2: Solutions to (2.22) for different values of b. Here we choose L\u00CB\u0086 = 20as a sufficient representation of infinity. Note we can solve the equation on[0, L\u00CB\u0086] and use symmetry because the functions are even.The homoclinic orbit problem holds for any curve \u00CE\u0093 and hence the only effectof the geometry of the curve is in the inhibitor problem u. Separating thehomoclinic, we can write (2.19) in a more tractable way. Firstly we have that\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ef(U0, v\u00CB\u009C0) d\u00CE\u00B7\u00CB\u0086 = U\u00CE\u00B20\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ewo d\u00CE\u00B7\u00CB\u0086,\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ev\u00CB\u009C20\u00CE\u00B7\u00CB\u0086 d\u00CE\u00B7\u00CB\u0086 = U2q0\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew2\u00CE\u00B7\u00CB\u0086 d\u00CE\u00B7\u00CB\u0086; (2.28a)\u00CE\u00B2 =qo\u00E2\u0088\u0092 s. (2.28b)Furthermore, we can actually simplify G\u00CB\u0086 in (2.11) which we first write using292.3. Quasi-Steady State Profiles for the Gierer-Meinhardt Modelthe homoclinic asG\u00CB\u0086(\u00E2\u0088\u009E) = \u00E2\u0088\u0092qU2q\u00E2\u0088\u009210\u00E2\u0088\u00AB \u00E2\u0088\u009E0\u00E2\u0088\u00AB w(\u00CE\u00B7\u00CB\u0086)0v21 + bv2dv d\u00CE\u00B7\u00CB\u0086 = qU2q\u00E2\u0088\u009210\u00E2\u0088\u00AB \u00E2\u0088\u009E0W(w) d\u00CE\u00B7\u00CB\u0086where W is defined byW \u00E2\u0089\u00A1\u00E2\u0088\u00AB w0v2(1 + bv2)dv. (2.29)To simplify this, consider the homoclinic orbit problem (2.22), multiply byw\u00CE\u00B7\u00CB\u0086, and integrate to getw2\u00CE\u00B7\u00CB\u0086 \u00E2\u0088\u0092 w2 + 2W(w) = 0.If we integrate again, we get\u00E2\u0088\u00AB \u00E2\u0088\u009E0W(w) d\u00CE\u00B7\u00CB\u0086 = 14(\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew2 d\u00CE\u00B7\u00CB\u0086 \u00E2\u0088\u0092\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew2\u00CE\u00B7\u00CB\u0086 d\u00CE\u00B7\u00CB\u0086)(2.30)where we have exploited thatW(w) is an even function. Finally then we canuse this expression for G\u00CB\u0086 with (2.28a) in H to write (2.20) asH = \u00E2\u0088\u0092 q4U0(\u00E2\u0088\u00AB\u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew2 d\u00CE\u00B7\u00CB\u0086\u00E2\u0088\u00AB\u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew2\u00CE\u00B7\u00CB\u0086 d\u00CE\u00B7\u00CB\u0086\u00E2\u0088\u0092 1)\u00E2\u0089\u00A1 \u00E2\u0088\u0092 q4U0H\u00CB\u0086. (2.31)This form is convenient because we avoid the integral with respect to w andthe spatial integrals can easily be computed from an analytic or numericallycomputed homoclinic. This form is also useful because it holds for any func-tion g(u, v) with perhaps slightly different leading constants which makes itmore universal. Furthermore, as long as g(u, v) is such that U0 can easily be302.3. Quasi-Steady State Profiles for the Gierer-Meinhardt Modelextracted from v\u00CB\u009C0 then we explicitly remove the U0 dependence in a tractableway. With this simplification we now have the inhibitor problem isD\u00E2\u0088\u0086u\u00E2\u0088\u0092 u = 0, x \u00E2\u0088\u0088 \u00E2\u0084\u00A6 (2.32a)\u00E2\u0088\u0082u\u00E2\u0088\u0082n= 0, x \u00E2\u0088\u0088 \u00E2\u0088\u0082\u00E2\u0084\u00A6 (2.32b)u = U0(s), x \u00E2\u0088\u0088 \u00CE\u0093 (2.32c)[\u00E2\u0088\u0082u\u00E2\u0088\u0082n]\u00CE\u0093= \u00E2\u0088\u0092 1DU\u00CE\u00B20A, (2.32d)V0 = \u00CE\u00BA0 \u00E2\u0088\u0092q4U0H\u00CB\u0086(\u00E2\u0088\u0082u\u00E2\u0088\u0082n\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00CE\u00B7=0++\u00E2\u0088\u0082u\u00E2\u0088\u0082n\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00CE\u00B7=0\u00E2\u0088\u0092), (2.32e)where we have definedA =\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ewo d\u00CE\u00B7\u00CB\u0086. (2.33)2.3.1 Inhibitor Problem on a Circular CurveWe start by considering the curve \u00CE\u0093 to be a circle of radius r0 inside a circulardomain 0 \u00E2\u0089\u00A4 r \u00E2\u0089\u00A4 R. In this case, the inward normal is n\u00CB\u0086 = \u00E2\u0088\u0092r\u00CB\u0086 anddudn\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00CE\u00B7\u00CB\u0086=0\u00C2\u00B1= \u00E2\u0088\u0092ur(r\u00E2\u0088\u00930 )where we note that \u00CE\u00B7\u00CB\u0086+ is slightly on the inside of the curve which from theradial coordinate perspective is r\u00E2\u0088\u00920 and the opposite is true for \u00CE\u00B7\u00CB\u0086\u00E2\u0088\u0092. Usinga polar coordinate system, we have, for this geometry, that the inhibitor312.3. Quasi-Steady State Profiles for the Gierer-Meinhardt Modelproblem (2.32) reduces to1r\u00E2\u0088\u0082\u00E2\u0088\u0082r(r\u00E2\u0088\u0082u\u00E2\u0088\u0082r)+1Dr2\u00E2\u0088\u00822u\u00E2\u0088\u0082\u00CE\u00B82\u00E2\u0088\u0092 1Du = 0, 0 \u00E2\u0089\u00A4 r \u00E2\u0089\u00A4 R, r 6= r0, (2.34a)\u00E2\u0088\u0082u\u00E2\u0088\u0082r= 0, r = R (2.34b)u = U0(\u00CE\u00B8), r = r0 (2.34c)[\u00E2\u0088\u0082u\u00E2\u0088\u0082r]r0= \u00E2\u0088\u0092 1DU\u00CE\u00B20A, (2.34d)dr0dt= \u00E2\u0088\u0092 1r0\u00E2\u0088\u0092 q4U0H\u00CB\u0086(\u00E2\u0088\u0082u\u00E2\u0088\u0082r\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3r=r+0+\u00E2\u0088\u0082u\u00E2\u0088\u0082r\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00CE\u00B7=r\u00E2\u0088\u00920).(2.34e)Here we have noted that the curvature \u00CE\u00BA0 and normal velocity V are, \u00CE\u00BA0 = 1r0 ,dr0dt = \u00E2\u0088\u0092V respectively.Radially Symmetric SolutionWe will begin by considering U0 a constant and denote this as the radiallysymmetric ring solution, the details of which are similar to [45]. We canimmediately write the bounded solution to (2.34a) satisfying ur = 0 on r = Rand u bounded as r \u00E2\u0086\u0092 0+ asu(r) =\u00EF\u00A3\u00B1\u00EF\u00A3\u00B4\u00EF\u00A3\u00B2\u00EF\u00A3\u00B4\u00EF\u00A3\u00B3AI0(r\u00E2\u0088\u009AD), 0 \u00E2\u0089\u00A4 r \u00E2\u0089\u00A4 r0E(\u00CE\u00B10I0(r\u00E2\u0088\u009AD)+K0(r\u00E2\u0088\u009AD)), r0 \u00E2\u0089\u00A4 r \u00E2\u0089\u00A4 R, \u00CE\u00B10 = \u00E2\u0088\u0092K0(R\u00E2\u0088\u009AD)\u00E2\u0080\u00B2I0(R\u00E2\u0088\u009AD)\u00E2\u0080\u00B2where I0 and K0 are the zeroth order modified Bessel functions and primeindicates differentiation with respect to r. To avoid certain chain rule ex-pressions, we will adopt the following notation for the location of the prime322.3. Quasi-Steady State Profiles for the Gierer-Meinhardt Modelthroughout:f \u00E2\u0080\u00B2(u(r)) =dfdu\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3u=u(r), f(u(r))\u00E2\u0080\u00B2 =df(u(r))dr= f \u00E2\u0080\u00B2(u(r))dudr. (2.35)Enforcing continuity and the jump condition (2.34d) at r0 we haveu(r) =r0DU\u00CE\u00B20AG0(r; r0) (2.36)where we have used the Wronskian relationship,W (r) = I0(r\u00E2\u0088\u009AD)K0(r\u00E2\u0088\u009AD)\u00E2\u0080\u00B2\u00E2\u0088\u0092K0(r\u00E2\u0088\u009AD)I0(r\u00E2\u0088\u009AD)\u00E2\u0080\u00B2= \u00E2\u0088\u00921r. (2.37)Here G0(r; r0) is the Green\u00E2\u0080\u0099s functionG0(r; r0) =\u00EF\u00A3\u00B1\u00EF\u00A3\u00B4\u00EF\u00A3\u00B2\u00EF\u00A3\u00B4\u00EF\u00A3\u00B3J0,1(r)J0,2(r0), 0 \u00E2\u0089\u00A4 r \u00E2\u0089\u00A4 r0J0,1(r0)J0,2(r), r0 \u00E2\u0089\u00A4 r \u00E2\u0089\u00A4 R(2.38)whereJ0,1(r) = I0(r\u00E2\u0088\u009AD), J0,2(r) = \u00CE\u00B10I0(r\u00E2\u0088\u009AD)+K0(r\u00E2\u0088\u009AD). (2.39)We can determine the value of U0 by solving u(r0) = U0 to getU0 =(Dr0AG0(r0; r0))1/(\u00CE\u00B2\u00E2\u0088\u00921). (2.40)For b = 0, this is an explicit expression for U0, However, for b 6= 0 then Adepends on b and hence U0 as well. However, this can be solved with an332.3. Quasi-Steady State Profiles for the Gierer-Meinhardt Modeliterative technique such as Newton\u00E2\u0080\u0099s Method. Regardless we can use (2.23)to define (cf. [39])G\u00CB\u009C(b) \u00E2\u0089\u00A1 bA2q\u00CE\u00B2\u00E2\u0088\u00921 =(Dr0G0(r0; r0)) 2q\u00CE\u00B2\u00E2\u0088\u00921\u00CF\u0083 (2.41)and numerically (Figure 2.3) we see that dAdb > 0. Therefore,dG\u00CB\u009Cdb > 0 and sofor each \u00CF\u0083 there is a unique b and vice-versa. An elegant proof in appendixB of [82] shows analytically that dG\u00CB\u009Cdb > 0 when o = 2. The monotonicity ofG\u00CB\u009C guarantees that there is a unique root U0 to find with the Newton solve of(2.40).0 0.05 0.1 0.15 0.2 0.25010002000300040005000600070008000900010000bdA/db o=1o=2o=3o=4o=5o=6Figure 2.3: Numerical computation of the b derivative of A for o = 1 too = 6. Here we see that the derivative is always positive and each value ofo is bounded from below by the previous values. The integral diverges as bapproaches bc from the left.342.3. Quasi-Steady State Profiles for the Gierer-Meinhardt ModelFinally, we can determine the normal velocity of the ring solution. We canre-write (2.36) asu(r) =U0G0(r0; r0)G0(r; r0) (2.42)which we substitute into (2.34e) to getdr0dT= \u00E2\u0088\u0092 1r0\u00E2\u0088\u0092 q4H\u00CB\u0086(J \u00E2\u0080\u00B20,1(r0)J0,1(r0)+J \u00E2\u0080\u00B20,2(r0)J0,2(r0)). (2.43)Rather than consider varying R and D separately, typically (cf. [39]) theouter radius is set to R = 1 in an absolute geometry frame and then it isrescaled to an effective domain with a unit diffusion coefficient. We note thatthis can be achieved in the current formulation by setting D = 1 everywhereand then replacingR =1\u00E2\u0088\u009AD= `where this D or ` is to be varied. Unless otherwise stated, we will adopt thisformulation moving forward. It is worth noting that when the saturation isnot zero, the problems are not entirely equivalent as the diffusivity does notproperly scale in (2.41). We begin by considering no saturation (\u00CF\u0083 = b = 0)where Figure 2.4 shows (2.43) versus r0 for exponent set (2, 1, 2, 0) and variousvalues of R = `.352.3. Quasi-Steady State Profiles for the Gierer-Meinhardt Model0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1\u00E2\u0088\u00922\u00E2\u0088\u00921.5\u00E2\u0088\u00921\u00E2\u0088\u00920.500.51r0/Rdr0/dt `=1`=4`=10`=20Figure 2.4: Phase portrait of (2.43) for D = 1, b = 0, exponent set (2, 1, 2, 0),and various values of R.Notice that (2.43) does not depend on the exponents o or s and so onlyvarying q can make a difference. The plot for q = 2 is shown in Figure 2.5which overall does not show any qualitative difference to q = 1.362.3. Quasi-Steady State Profiles for the Gierer-Meinhardt Model0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1\u00E2\u0088\u00922\u00E2\u0088\u00921.5\u00E2\u0088\u00921\u00E2\u0088\u00920.500.511.52r0/Rdr0/dt `=1`=4`=10`=20Figure 2.5: Phase portrait of (2.43) for D = 1, b = 0, exponent set (2, 2, 2, 0),and various values of R.In both Figure 2.4 and 2.5, we have that for small values of R = ` thereare no equilibrium values to (2.43) and that transitions to a stable and un-stable equilibrium as ` increases. Therefore, (2.43) undergoes a saddle-nodebifurcation. We compute this bifurcation curve numerically by using a New-ton\u00E2\u0080\u0099s method continuation on r0 starting from the smallest root r0s. We canapproximate these roots asymptotically by definingF (r) =(J \u00E2\u0080\u00B20,1(r)J0,1(r)+J \u00E2\u0080\u00B20,2(r)J0,2(r))=I1 (r)I0 (r)+\u00CE\u00B10I1 (r)\u00E2\u0088\u0092K1 (r)\u00CE\u00B10I0 (r) +K0 (r)(2.44)372.3. Quasi-Steady State Profiles for the Gierer-Meinhardt Modeland then for r \u001C 1 we haveF (r) \u00E2\u0088\u00BC \u00E2\u0088\u0092 1(\u00CE\u00B10 + log(2)\u00E2\u0088\u0092 log(r)\u00E2\u0088\u0092 \u00CE\u00B3)rwith \u00CE\u00B3 \u00E2\u0089\u0088 0.5772156649 the Euler-Mascheroni constant. Therefore from(2.43),dr0dT\u00E2\u0088\u00BC(qH\u00CB\u00864(\u00CE\u00B10 + log(2)\u00E2\u0088\u0092 log(r)\u00E2\u0088\u0092 \u00CE\u00B3)\u00E2\u0088\u0092 1)1r0+O(r0). (2.45)and we can approximate r0s by setting the expression in brackets to zero,r0s \u00E2\u0089\u0088 exp(\u00CE\u00B10 + log(2)\u00E2\u0088\u0092 \u00CE\u00B3 \u00E2\u0088\u0092qH\u00CB\u00864). (2.46)As R tends to infinity, \u00CE\u00B10 tends to zero and so there is a limiting small radius,r0s \u00E2\u0089\u0088 1.1229 exp(\u00E2\u0088\u0092q) (2.47)where we have noted that when \u00CF\u0083 = 0 then H\u00CB\u0086 = 4. In Figures 2.4 and 2.5we can see this limiting small stationary point radius being approached as` increases. For the continuation method then we take R sufficiently large(here we choose R = 10) and take as an initial guess r0 = r0s and iterate untilconvergence. We then increase r0 and find the corresponding R that createsan equilibrium value. We plot the \u00CF\u0083 = 0 bifurcation diagram in Figure 2.6.Notice that for q = 1 the asymptotic approximation is not very accurate forthe lower root but from Figure 2.4 we see that atR = 10 the small equilibriumradius is r0s \u00E2\u0089\u0088 1 which is sufficiently far from the small r0 asymptotic regime382.3. Quasi-Steady State Profiles for the Gierer-Meinhardt Modelthat terms of O(r0) can no longer be neglected in (2.45). However, since weare mostly interested using the asymptotic approximation for initializing thenumerical algorithm, the accuracy is not of critical importance.0 2 4 6 8 100246810Rr 0 NumericAsymptotic(a) q = 10 2 4 6 8 100246810Rr 0 NumericAsymptotic(b) q = 2Figure 2.6: Bifurcation diagram to (2.43) for different values of the exponentq and b = 0. The differential equation undergoes a saddle node bifurcationwhen R = 3.6220 (for q = 1) and R = 1.4296 (for q = 2). The larger ofthe equilibrium r0 values belong to the stable branch. The red dashed curverepresents an asymptotic approximation to the lower radius.The case with no saturation was mentioned as an extension of work for392.3. Quasi-Steady State Profiles for the Gierer-Meinhardt Modelring solutions with the Grey-Scott model [45]. However, the analysis didnot consider saturation which we now present. The addition of saturationrequires a more delicate approach since the saturation value \u00CF\u0083 is the staticparameter and not b which varies as U0 varies. Therefore, H\u00CB\u0086 in (2.43) is nolonger static when b 6= 0 and to solve (2.43) we must first solve (2.40) fora given r0 and then compute b (and hence H\u00CB\u0086) with (2.23). We solve theproblem (2.40) for U0 numerically using Newton\u00E2\u0080\u0099s method. However, we doso in a special way that stabilizes b. The details of this are presented in6.3.2 and are omitted here but we indicate that the reason is to prevent bfrom exceeding its maximum value artificially (i.e. as in the intermediateNewton\u00E2\u0080\u0099s method steps). Something we notice immediately from (2.41) isthat the denominator tends to zero as r0 tends to zero and thus for any\u00CF\u0083 > 0 we have that A diverges or that b tends to the critical value as r0tends to 0. This is demonstrated numerically in Figure 2.7.402.3. Quasi-Steady State Profiles for the Gierer-Meinhardt Model0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 100.050.10.150.20.25r0b \u00CF\u0083=0.01\u00CF\u0083=0.5\u00CF\u0083=2\u00CF\u0083=5\u00CF\u0083=10\u00CF\u0083=25Figure 2.7: Modified saturation parameter b as a function of r0 for varioussaturation values \u00CF\u0083. Here we take D = 1, R = 1 and exponent set (2, 1, 2, 0).When \u00CF\u0083 = 0 we had that as r0 tended to zero then F (r0) from (2.44) tendedto zero as well since H\u00CB\u0086 was fixed. Therefore dr0/dt tended to negative infinityas evidenced by Figures 2.4 and 2.5. However, when \u00CF\u0083 6= 0 then since b tendsto the critical value as r0 tends to zero then H\u00CB\u0086 tends to infinity. While itcan be hard to analyze the growth of H\u00CB\u0086 analytically, from Figure 2.8, we canconjecture that it grows faster than log r0 and so therefore we actually havethat F (r0) (and hence dr0/dt) tends to positive infinity as r0 tends to zero.412.3. Quasi-Steady State Profiles for the Gierer-Meinhardt Model0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10510152025303540r0 H\u00CB\u0086\u00E2\u0088\u0092 log(r0)Figure 2.8: Growth of H\u00CB\u0086 versus \u00E2\u0088\u0092 log(r0) for R = 1 and when \u00CF\u0083 = 5.We plot (2.43) in Figure 2.9 for saturation values \u00CF\u0083 = 0.5, 5, 10, and 25 forexponent set (2, 1, 2, 0) and various values of R = ` (with D = 1). Figure(2.10) repeats the experiment but with the exponent set (2, 2, 2, 0) and ex-cludes \u00CF\u0083 = 25 since the single root is so close to r0 = R for R large that it isdifficult to compute.422.3. Quasi-Steady State Profiles for the Gierer-Meinhardt Model0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1\u00E2\u0088\u00925\u00E2\u0088\u00924\u00E2\u0088\u00923\u00E2\u0088\u00922\u00E2\u0088\u00921012345r0/Rdr 0/dt `=1`=4`=10`=20(a) \u00CF\u0083 = 0.50 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1\u00E2\u0088\u00925\u00E2\u0088\u00924\u00E2\u0088\u00923\u00E2\u0088\u00922\u00E2\u0088\u00921012345r0/Rdr 0/dt `=1`=4`=10`=20(b) \u00CF\u0083 = 50 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1\u00E2\u0088\u00925\u00E2\u0088\u00924\u00E2\u0088\u00923\u00E2\u0088\u00922\u00E2\u0088\u00921012345r0/Rdr 0/dt `=1`=4`=10`=20(c) \u00CF\u0083 = 100 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1\u00E2\u0088\u00925\u00E2\u0088\u00924\u00E2\u0088\u00923\u00E2\u0088\u00922\u00E2\u0088\u00921012345r0/Rdr 0/dt `=1`=4`=10`=20(d) \u00CF\u0083 = 25Figure 2.9: Right-hand side to (2.43) for various saturation values, \u00CF\u0083 andboundary values R. The exponent set here is (2, 1, 2, 0) and D = 1.432.3. Quasi-Steady State Profiles for the Gierer-Meinhardt Model0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1\u00E2\u0088\u00925\u00E2\u0088\u00924\u00E2\u0088\u00923\u00E2\u0088\u00922\u00E2\u0088\u00921012345r0/Rdr 0/dt `=1`=4`=10`=20(a) \u00CF\u0083 = 0.50 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1\u00E2\u0088\u00925\u00E2\u0088\u00924\u00E2\u0088\u00923\u00E2\u0088\u00922\u00E2\u0088\u00921012345r0/Rdr 0/dt `=1`=4`=10`=20(b) \u00CF\u0083 = 50 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1\u00E2\u0088\u00925\u00E2\u0088\u00924\u00E2\u0088\u00923\u00E2\u0088\u00922\u00E2\u0088\u00921012345r0/Rdr 0/dt `=1`=4`=10`=20(c) \u00CF\u0083 = 10Figure 2.10: Right-hand side to (2.43) for various saturation values, \u00CF\u0083 andboundary values R. The exponent set here is (2, 2, 2, 0) and D = 1.The addition of saturation drastically alters the root structure to (2.43). If\u00CF\u0083 \u001C 1 then the behaviour should be similar to \u00CF\u0083 = 0 since, from Figure 2.7,b very quickly tends to zero. This means that very quickly, for R not toolarge, dr0/dt goes negative and there must be a root very close to r0 = 0 as isevidenced in Figure 2.9a when \u00CF\u0083 = 0.5. Since dr0/dt starts positive, this rootis necessarily stable and exists prior to R = Rc, the saddle-node bifurcationpoint when \u00CF\u0083 = 0. Therefore, the effect of the saturation is to add an extrastability branch emanating from r0 = 0. Since these stable roots are very442.3. Quasi-Steady State Profiles for the Gierer-Meinhardt Modelsmall, they can quickly cause instability in Newton algorithms where b isnear the critical value. Furthermore, since dr0/dt starts at positive infinity,we can no longer guarantee the existence of a small root for R \u001D 1 andtherefore cannot initialize around something analogous to (2.47). However,as r0 tends to R we have that b tends to zero and so we still expect thelarger stable root near r0 = R. As such, we begin the bifurcation solverby numerically searching for a root near r0 = R and follow a downwardcontinuation in r0. Figure 2.11 shows the bifurcation diagram for \u00CF\u0083 = 10and \u00CF\u0083 = 8 with exponent set (2, 1, 2, 0) along with the small r0 asymptoticexpression computed from (2.46) where now this has to also be handled witha Newton solve since H\u00CB\u0086 = H\u00CB\u0086(r0).1 2 3 4 5 6 700.511.522.533.5Rr 0 NumericAsymptotic(a) \u00CF\u0083 = 101 2 3 4 5 6 700.511.522.533.5Rr 0 NumericAsymptotic(b) \u00CF\u0083 = 8Figure 2.11: Bifurcation diagram to (2.43) for exponent set (2, 1, 2, 0) anddifferent values of \u00CF\u0083. The dashed curve represents an asymptotic approxi-mation for r0 \u001C 1. The smallest and highest equilibrium values are stablewhile there is an unstable transition branch in the middle.Non-Radially Symmetric SolutionsStandard techniques guarantee that solutions to the modified Helmholtzproblem are unique for prescribed Dirichlet or Neumann boundary data (cf.452.3. Quasi-Steady State Profiles for the Gierer-Meinhardt Model[68]) and thus one would expect that the radially symmetric solution to (2.34)is unique. However, since the boundary data U0 is an unknown of the prob-lem, we can no longer prescribe uniqueness and we briefly demonstrate thatby finding solutions for which U0 is a periodic, non-constant, function. Evenwhile removing the restriction that U0 be constant, (2.34a) is still separableand therefore we can perform a Fourier eigenfunction expansion,u(r, \u00CE\u00B8) =\u00E2\u0088\u009E\u00E2\u0088\u0091n=\u00E2\u0088\u0092\u00E2\u0088\u009EUn(r) exp(in\u00CE\u00B8), U0 =\u00E2\u0088\u009E\u00E2\u0088\u0091n=\u00E2\u0088\u0092\u00E2\u0088\u009Ean exp(in\u00CE\u00B8),U\u00CE\u00B20 =\u00E2\u0088\u009E\u00E2\u0088\u0091n=\u00E2\u0088\u0092\u00E2\u0088\u009Efn exp(in\u00CE\u00B8)where we treat the U\u00CE\u00B20 term separately just for simplicity. Using this expan-sion in (2.34) we get for each eigenmode n,1rddr(rdUndr)\u00E2\u0088\u0092 n2Unr2\u00E2\u0088\u0092 UnD= 0, (2.48a)dUndr\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3r=R= 0, (2.48b)Un(r0) = an, (2.48c)[dUndr]r=r0= \u00E2\u0088\u0092 1DfnA. (2.48d)This is the n > 0 analogue of the radially symmetric case and so if we definethe functionsJn,1(r) = In(r\u00E2\u0088\u009AD), Jn,2(r) = \u00CE\u00B1nIn(r\u00E2\u0088\u009AD)+Kn(r\u00E2\u0088\u009AD)(2.49a)462.3. Quasi-Steady State Profiles for the Gierer-Meinhardt Modelwhere\u00CE\u00B1n = \u00E2\u0088\u0092Kn(R\u00E2\u0088\u009AD)\u00E2\u0080\u00B2In(R\u00E2\u0088\u009AD)\u00E2\u0080\u00B2 =\u00EF\u00A3\u00AB\u00EF\u00A3\u00ADKn+1(R\u00E2\u0088\u009AD)\u00E2\u0088\u0092 n\u00E2\u0088\u009ADR Kn(R\u00E2\u0088\u009AD)n\u00E2\u0088\u009ADR In(R\u00E2\u0088\u009AD)+ In+1(R\u00E2\u0088\u009AD)\u00EF\u00A3\u00B6\u00EF\u00A3\u00B8 (2.49b)then the solution can be written asUn(r) =r0DfnAG0,n(r; r0)with G0,n(r; r0), the Green\u00E2\u0080\u0099s functionG0,n(r; r0) =\u00EF\u00A3\u00B1\u00EF\u00A3\u00B4\u00EF\u00A3\u00B2\u00EF\u00A3\u00B4\u00EF\u00A3\u00B3Jn,1(r)Jn,2(r0), 0 \u00E2\u0089\u00A4 r \u00E2\u0089\u00A4 r0Jn,1(r0)Jn,2(r), r0 \u00E2\u0089\u00A4 r \u00E2\u0089\u00A4 R. (2.50)Using the Dirichlet value at r0 we have a condition to solve for U0,an =r0DAG0,n(r0; r0)fn.Notice that for radial symmetry, an = fn = 0 for n 6= 0 and f0 = a\u00CE\u00B20 leadingto the form for U0 in (2.40). We can approximate the Fourier coefficients cnof a discrete vector g using the discrete Fourier transform,cn \u00E2\u0089\u00881NN\u00E2\u0088\u0091j=1exp(\u00E2\u0088\u00922piin(j \u00E2\u0088\u0092 1)N)gj,where here N is the number of discrete wavemodes to consider. We can writethis asc =1NFg,472.3. Quasi-Steady State Profiles for the Gierer-Meinhardt Modelfor a Fourier transform matrix F whereFn,j = exp(\u00E2\u0088\u00922piin(j \u00E2\u0088\u0092 1)N).Using this, we can get the Fourier coefficients we desire by numerically solv-ing,FU0 = diag(Y)FU\u00CE\u00B20 , (2.51)for fixed r0 where Y is a vector with entries,Yn =r0DAG0,n(r0; r0).Computing solutions to (2.51) using Newton iterations starting at arbitraryinitial data tends to converge to the radially symmetric solution which is notsurprising since the decay of Fourier coefficients for increasing n will alwayshave the constant term dominate regardless. We will therefore return to thisformulation in 6.3.2 when we compute solutions on arbitrary domains to usethis as a verification of non-radially symmetric solutions.2.3.2 Inhibitor Problem on a Near Circular CurveNow consider the curve \u00CE\u0093 to be a perturbed circle given byr = r0 + \u00CE\u00B5h(\u00CE\u00B8), \u00CE\u00B5\u001C 1 (2.52)482.3. Quasi-Steady State Profiles for the Gierer-Meinhardt Modelstill inside a global circular domain 0 \u00E2\u0089\u00A4 r \u00E2\u0089\u00A4 R. We will not consider theequilibria of such a geometry because dynamics may distort it beyond a near-circle and we will therefore consider the inhibitor and velocity perturbationfor any initial base ring radius 0 < r0 < R. Recall that in deriving thesingular limit inhibitor problem (2.32), we took the limit as \u000F tends to zeroand so for our machinery to work with this perturbed circle geometry, werequire \u000F\u001C \u00CE\u00B5\u001C 1. We can write the inner normal to \u00CE\u0093 asn\u00CB\u0086 =\u00E3\u0080\u0088\u00E2\u0088\u00921, \u00CE\u00B5h\u00E2\u0080\u00B2(\u00CE\u00B8)r0+\u00CE\u00B5h(\u00CE\u00B8)\u00E3\u0080\u0089\u00E2\u0088\u009A1 + \u00CE\u00B52h\u00E2\u0080\u00B2(\u00CE\u00B8)2(r0+\u00CE\u00B5h(\u00CE\u00B8))2,where prime here indicates differentiation with respect to \u00CE\u00B8. We can use thisnormal vector to write\u00E2\u0088\u0082u\u00E2\u0088\u0082n=\u00E2\u0088\u0092ur + \u00CE\u00B5h\u00E2\u0080\u00B2(\u00CE\u00B8)(r0+\u00CE\u00B5h(\u00CE\u00B8))2u\u00CE\u00B8\u00E2\u0088\u009A1 + \u00CE\u00B52h\u00E2\u0080\u00B2(\u00CE\u00B8)2(r0+\u00CE\u00B5h(\u00CE\u00B8))2.We consider solving (2.32) with a formal expansionu(r, \u00CE\u00B8) = u0(r) + \u00CE\u00B5u1(r, \u00CE\u00B8) + \u00CE\u00B52u2(r, \u00CE\u00B8) + . . .where we explicitly note that we want to consider a perturbation from theradially symmetric inhibitor solution hence why u0 is independent of \u00CE\u00B8. Using492.3. Quasi-Steady State Profiles for the Gierer-Meinhardt Modelthe condition (2.32c) we haveU0 = u(r0 + \u00CE\u00B5h(\u00CE\u00B8)) \u00E2\u0088\u00BC u0(r0) + \u00CE\u00B5(u0r(r)h(\u00CE\u00B8) + u1(r))r0+ \u00CE\u00B52(12u0rr(r)h(\u00CE\u00B8)2 + u1r(r)h(\u00CE\u00B8) + u2(r))r0= U00 + \u00CE\u00B5U01(\u00CE\u00B8) + \u00CE\u00B52U02(\u00CE\u00B8). (2.53)Here we have put the evaluation at r = r0 outside the brackets since individ-ually, each term may be discontinuous but together, they must be continuoussince U0 has a fixed value. Using our asymptotic expansion we can write thenormal derivative asdudn\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00CE\u00B7\u00CB\u0086=0\u00C2\u00B1\u00E2\u0088\u00BC \u00E2\u0088\u0092 u0r(r\u00E2\u0088\u00930 ) + \u00CE\u00B5(\u00E2\u0088\u0092u0rr(r\u00E2\u0088\u00930 )h(\u00CE\u00B8)\u00E2\u0088\u0092 u1r(r\u00E2\u0088\u00930 ))+ \u00CE\u00B52(\u00E2\u0088\u009212u0rrr(r\u00E2\u0088\u00930 )h(\u00CE\u00B8)2 \u00E2\u0088\u0092 u1rr(r\u00E2\u0088\u00930 )h(\u00CE\u00B8) +u1\u00CE\u00B8(r\u00E2\u0088\u00930 )h\u00E2\u0080\u00B2(\u00CE\u00B8)r20+u0r(r\u00E2\u0088\u00930 )h\u00E2\u0080\u00B2(\u00CE\u00B8)22r20\u00E2\u0088\u0092 u2r(r\u00E2\u0088\u00930 ))(2.54)where we recall that because we are using the inner normal, \u00CE\u00B7\u00CB\u0086 = 0\u00C2\u00B1 corre-sponds to r = r\u00E2\u0088\u00930 . Finally we use (2.53) to writeU\u00CE\u00B20 \u00E2\u0088\u00BC U\u00CE\u00B200 + \u00CE\u00B5\u00CE\u00B2U\u00CE\u00B2\u00E2\u0088\u0092100 U01 + \u00CE\u00B52\u00CE\u00B2U\u00CE\u00B2\u00E2\u0088\u00922002((\u00CE\u00B2 \u00E2\u0088\u0092 1)U201 + 2U00U02). (2.55)In the presence of saturation, both A and H\u00CB\u0086 depend on the curve inhibitorvalue U0 and so we will also need to consider an expansion of the effective502.3. Quasi-Steady State Profiles for the Gierer-Meinhardt Modelsaturation parameter b,b = U2q0 \u00CF\u0083 \u00E2\u0088\u00BC U2q00\u00CF\u0083 + \u00CE\u00B52qU2q\u00E2\u0088\u0092100 U01\u00CF\u0083 + \u00CE\u00B52qU2q\u00E2\u0088\u0092200((2q \u00E2\u0088\u0092 1)U201 + 2U00U02)\u00CF\u0083= b0 + \u00CE\u00B5b1 + \u00CE\u00B52b2, (2.56)and then writeA \u00E2\u0088\u00BC A0 + \u00CE\u00B5dA0dbb1 + \u00CE\u00B52(dA0dbb2 +12d2A0db2b21), (2.57a)H\u00CB\u0086 \u00E2\u0088\u00BC H\u00CB\u00860 + \u00CE\u00B5dH\u00CB\u00860dbb1 + \u00CE\u00B52(dH\u00CB\u00860dbb2 +12d2H\u00CB\u00860db2b21), (2.57b)where the zero subscript indicates evaluation at b = b0, the unperturbedsaturation parameter. We will begin by looking at corrections to the inhibitorvalue before analyzing the corrections to the curve front velocity. The leadingorder problem for u0 is given precisely by the radially symmetric version of(2.34) in section 2.3.1 with solution (2.36) and curve value U00 given by (2.40)and as such we continue directly to the problem at O(\u00CE\u00B5):1r\u00E2\u0088\u0082\u00E2\u0088\u0082r(r\u00E2\u0088\u0082u1\u00E2\u0088\u0082r)+1r2\u00E2\u0088\u00822u1\u00E2\u0088\u0082\u00CE\u00B82\u00E2\u0088\u0092 1Du1 = 0, r 6= r0, (2.58a)\u00E2\u0088\u0082u1\u00E2\u0088\u0082r\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3r=R= 0, (2.58b)[u1]r0 = \u00E2\u0088\u0092h(\u00CE\u00B8)[du0dr]r0, (2.58c)[\u00E2\u0088\u0082u1\u00E2\u0088\u0082r]r0= \u00E2\u0088\u0092h(\u00CE\u00B8)[d2u0dr2]r0\u00E2\u0088\u0092 A0DU\u00CE\u00B2\u00E2\u0088\u0092100 A\u00C2\u00AF0U01,(2.58d)512.3. Quasi-Steady State Profiles for the Gierer-Meinhardt Modelwhere we have definedA\u00C2\u00AF0 = \u00CE\u00B2 +2qb0A0dA0db. (2.59)The continuity condition (2.58c) comes from utilizing (2.53) and that glob-ally, the inhibitor value U0 must be continuous across the front. The deriva-tive condition (2.58d) comes from (2.54), (2.55), and (2.56) where we haveexpressed b1 in terms of b0. We can rewrite (2.58d) by noticing that[d2u0dr2]r0=A\u00C2\u00AF0r0[d2G0dr2]r0= A\u00C2\u00AF0r0 (J0,1(r0)J0,2(r0)\u00E2\u0080\u00B2\u00E2\u0080\u00B2 \u00E2\u0088\u0092 J0,1(r0)\u00E2\u0080\u00B2\u00E2\u0080\u00B2J0,2(r0))=A\u00C2\u00AF0r0(K0(r0\u00E2\u0088\u009AD)\u00E2\u0080\u00B2\u00E2\u0080\u00B2I0(r0\u00E2\u0088\u009AD)\u00E2\u0088\u0092K0(r0\u00E2\u0088\u009AD)I0(r0\u00E2\u0088\u009AD)\u00E2\u0080\u00B2\u00E2\u0080\u00B2)where we defineA\u00C2\u00AF0 =A0DU\u00CE\u00B200, (2.60)G0(r; r0) by (2.38) and J0,1 and J0,2 by (2.39). If we use the Wronskian(2.37) and differentiate we get,W \u00E2\u0080\u00B2(r) = K0(r\u00E2\u0088\u009AD)\u00E2\u0080\u00B2\u00E2\u0080\u00B2I0(r\u00E2\u0088\u009AD)\u00E2\u0088\u0092K0(r\u00E2\u0088\u009AD)I0(r\u00E2\u0088\u009AD)\u00E2\u0080\u00B2\u00E2\u0080\u00B2=1r2.Therefore,[d2u0dr2]r0=A\u00C2\u00AF0r0(2.61)522.3. Quasi-Steady State Profiles for the Gierer-Meinhardt Modeland the derivative jump condition (2.58d) becomes[\u00E2\u0088\u0082u1\u00E2\u0088\u0082r]r0= \u00E2\u0088\u0092h(\u00CE\u00B8)A\u00C2\u00AF0r0\u00E2\u0088\u0092 A\u00C2\u00AF0U00A\u00C2\u00AF0U01. (2.62)Since the problem (2.58) is linear in the \u00CE\u00B8 dependence, we can perform aneigenfunction expansion,u1(r, \u00CE\u00B8) =\u00E2\u0088\u009E\u00E2\u0088\u0091n=\u00E2\u0088\u0092\u00E2\u0088\u009EV1n(r) exp(in\u00CE\u00B8), U01(\u00CE\u00B8) =\u00E2\u0088\u009E\u00E2\u0088\u0091n=\u00E2\u0088\u0092\u00E2\u0088\u009EU1n exp(in\u00CE\u00B8),h(\u00CE\u00B8) =\u00E2\u0088\u009E\u00E2\u0088\u0091n=\u00E2\u0088\u0092\u00E2\u0088\u009EHn exp(in\u00CE\u00B8)for integer eigenmodes, n. Upon this expansion, (2.58) becomes1rddr(rdV1ndr)\u00E2\u0088\u0092 n2r2V1n \u00E2\u0088\u00921DV1n = 0, r 6= r0, (2.63a)dV1ndr\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3r=R= 0, (2.63b)[V1n]r0 = HnA\u00C2\u00AF0, (2.63c)[dV1ndr]r0= \u00E2\u0088\u0092HnA\u00C2\u00AF0r0\u00E2\u0088\u0092 A\u00C2\u00AF0U00A\u00C2\u00AF0U1n. (2.63d)This problem is very similar to the non-radially symmetric case for the purecircle geometry and so if we define Jn,1, Jn,2, and \u00CE\u00B1n by (2.49) then we canwriteV1n =\u00EF\u00A3\u00B1\u00EF\u00A3\u00B4\u00EF\u00A3\u00B2\u00EF\u00A3\u00B4\u00EF\u00A3\u00B3AJn,1(r), 0 \u00E2\u0089\u00A4 r < r0BJn,2(r), r0 < r \u00E2\u0089\u00A4 R532.3. Quasi-Steady State Profiles for the Gierer-Meinhardt Modelwhere the boundary derivative condition (2.63b) has been used. Using (2.63c)and (2.63d) we have\u00EF\u00A3\u00AE\u00EF\u00A3\u00B0\u00E2\u0088\u0092Jn,1(r0) Jn,2(r0)\u00E2\u0088\u0092J \u00E2\u0080\u00B2n,1(r0) J \u00E2\u0080\u00B2n,2(r0)\u00EF\u00A3\u00B9\u00EF\u00A3\u00BB\u00EF\u00B8\u00B8 \u00EF\u00B8\u00B7\u00EF\u00B8\u00B7 \u00EF\u00B8\u00B8=M\u00EF\u00A3\u00AE\u00EF\u00A3\u00B0AB\u00EF\u00A3\u00B9\u00EF\u00A3\u00BB =\u00EF\u00A3\u00AE\u00EF\u00A3\u00B0HnA\u00C2\u00AF0\u00E2\u0088\u0092HnA\u00C2\u00AF0r0 \u00E2\u0088\u0092A\u00C2\u00AF0U00 A\u00C2\u00AF0U1n\u00EF\u00A3\u00B9\u00EF\u00A3\u00BB =\u00EF\u00A3\u00AE\u00EF\u00A3\u00B0A\u00C2\u00AF1nB\u00C2\u00AF1n\u00EF\u00A3\u00B9\u00EF\u00A3\u00BB .Using the Wronskian relationship (2.37), which holds for any n, we have thatdetM = 1/r0 and therefore\u00EF\u00A3\u00AE\u00EF\u00A3\u00B0AB\u00EF\u00A3\u00B9\u00EF\u00A3\u00BB =\u00EF\u00A3\u00AE\u00EF\u00A3\u00B0A\u00C2\u00AF1nr0J \u00E2\u0080\u00B2n,2(r0)\u00E2\u0088\u0092 B\u00C2\u00AF1nr0Jn,2(r0)A\u00C2\u00AF1nr0J \u00E2\u0080\u00B2n,1(r0)\u00E2\u0088\u0092 B\u00C2\u00AF1nr0Jn,1(r0)\u00EF\u00A3\u00B9\u00EF\u00A3\u00BB .If we define G0,n(r; r0) as in (2.50) and G1,n viaG1,n(r; r0) =\u00EF\u00A3\u00B1\u00EF\u00A3\u00B4\u00EF\u00A3\u00B2\u00EF\u00A3\u00B4\u00EF\u00A3\u00B3Jn,1(r)J \u00E2\u0080\u00B2n,2(r0), 0 \u00E2\u0089\u00A4 r < r0J \u00E2\u0080\u00B2n,1(r0)Jn,2(r), r0 < r \u00E2\u0089\u00A4 R. (2.64)then we can writeV1n(r) = A\u00C2\u00AF1nr0G1,n(r; r0)\u00E2\u0088\u0092 B\u00C2\u00AF1nr0G0,n(r; r0). (2.65)Now we can determine U1n via (2.53),U1n = V1n +Hndu0dr,542.3. Quasi-Steady State Profiles for the Gierer-Meinhardt Modelevaluated from either side of r = r0. However, since this must be continuousacross r0 we can actually compute it using the average value,U1n =A\u00C2\u00AF1nr0(\u00E3\u0080\u0088G1,n(r; r0)\u00E3\u0080\u0089r0 +\u00E2\u008C\u00A9dG0dr\u00E2\u008C\u00AAr0)+(A\u00C2\u00AF1n +A\u00C2\u00AF0r0U00A\u00C2\u00AF0U1n)G0;n(r0, r0),where \u00E3\u0080\u0088\u00C2\u00B7\u00E3\u0080\u0089a indicates the average value across r = a. If we notice thatA\u00C2\u00AF0r0 =U00G0(r0; r0),then we can solve for U1n and getU1n =HnU00G0(r0; r0)(1\u00E2\u0088\u0092 A\u00C2\u00AF0G0;n(r0; r0)G0(r0; r0))\u00E2\u0088\u00921(G0,n(r0; r0)r0+\u00E2\u008C\u00A9G1,n(r, r0) +dG0dr\u00E2\u008C\u00AAr0).(2.66)Typically, we consider the perturbation h(\u00CE\u00B8) to be a finite combination ofsinusoidal modes and if this is the case then the correction at O(\u00CE\u00B5) willadd components in each of those modes. However, unless an n = 0 mode isexplicitly part of h(\u00CE\u00B8), the perturbation at this order can not describe verticalshifting as there is no cross-mode influence. Therefore, we will consider theexpansion at O(\u00CE\u00B52) to account for this shifting since mode interactions occur552.3. Quasi-Steady State Profiles for the Gierer-Meinhardt Modelat this level through quadratic terms. The problem at O(\u00CE\u00B52) is1r\u00E2\u0088\u0082\u00E2\u0088\u0082r(r\u00E2\u0088\u0082u2\u00E2\u0088\u0082r)+1r2\u00E2\u0088\u00822u2\u00E2\u0088\u0082\u00CE\u00B82\u00E2\u0088\u0092 1Du2 = 0, r 6= r0 (2.67a)\u00E2\u0088\u0082u2\u00E2\u0088\u0082r\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3r=R= 0, (2.67b)[u2]r0 = \u00E2\u0088\u0092h(\u00CE\u00B8)[\u00E2\u0088\u0082u1\u00E2\u0088\u0082r]r0\u00E2\u0088\u0092 h(\u00CE\u00B8)22[d2u0dr2]r0,(2.67c)[\u00E2\u0088\u0082u2\u00E2\u0088\u0082r]r0=12h\u00E2\u0080\u00B2(\u00CE\u00B8)2r20[du0dr]r0+h\u00E2\u0080\u00B2(\u00CE\u00B8)r20[\u00E2\u0088\u0082u1\u00E2\u0088\u0082\u00CE\u00B8]r0\u00E2\u0088\u0092 h(\u00CE\u00B8)[\u00E2\u0088\u00822u1\u00E2\u0088\u0082r2]r0\u00E2\u0088\u0092 12h(\u00CE\u00B8)2[d3u0dr3]r0\u00E2\u0088\u0092 A\u00C2\u00AF0U200(A\u00C2\u00AF1U201 + A\u00C2\u00AF0U00U02), (2.67d)where we defineA\u00C2\u00AF1 =12\u00CE\u00B2(\u00CE\u00B2 \u00E2\u0088\u0092 1) + qb0(2q \u00E2\u0088\u0092 1)A0dA0db+2q2b20A0d2A0db2.As with the O(\u00CE\u00B5) expansion, this is generated by appropriately substitutingthe expansions from (2.53), (2.54), (2.55), and the corrections to b have beenexpressed using (2.56). We can rewrite the continuity condition (2.67c) using(2.61) and (2.62) to get[u2]r0 = h(\u00CE\u00B8)2 A\u00C2\u00AF02r0+ h(\u00CE\u00B8)A\u00C2\u00AF0A\u00C2\u00AF0U00U01. (2.68)562.3. Quasi-Steady State Profiles for the Gierer-Meinhardt ModelNext we turn our attention to the jump condition (2.67d). First we noticeusing (2.63c) that[\u00E2\u0088\u0082u1\u00E2\u0088\u0082\u00CE\u00B8]r0=\u00E2\u0088\u009E\u00E2\u0088\u0091n=\u00E2\u0088\u0092\u00E2\u0088\u009Ein [V1n]r0 exp(in\u00CE\u00B8) = A\u00C2\u00AF0\u00E2\u0088\u009E\u00E2\u0088\u0091n=\u00E2\u0088\u0092\u00E2\u0088\u009EinHn exp(in\u00CE\u00B8) = A\u00C2\u00AF0h\u00E2\u0080\u00B2(\u00CE\u00B8).Next, we need to determine[\u00E2\u0088\u00822u1\u00E2\u0088\u0082r2]r0=\u00E2\u0088\u009E\u00E2\u0088\u0091n=\u00E2\u0088\u0092\u00E2\u0088\u009E(A\u00C2\u00AF1nr0[d2G1,ndr2]r0\u00E2\u0088\u0092 B\u00C2\u00AF1nr0[d2G0,ndr2]r0)exp(in\u00CE\u00B8)[d3u0dr3]r0= A\u00C2\u00AF0r0[d3G0dr3]r0.Since the Wronskian relationship (2.37) also holds for n 6= 0 we haveW0,n(r) = In(r\u00E2\u0088\u009AD)Kn(r\u00E2\u0088\u009AD)\u00E2\u0080\u00B2\u00E2\u0088\u0092Kn(r\u00E2\u0088\u009AD)In(r\u00E2\u0088\u009AD)\u00E2\u0080\u00B2= \u00E2\u0088\u00921r(2.69)which we can differentiate to getW \u00E2\u0080\u00B20,n(r) =In(r\u00E2\u0088\u009AD)Kn(r\u00E2\u0088\u009AD)\u00E2\u0080\u00B2\u00E2\u0080\u00B2\u00E2\u0088\u0092Kn(r\u00E2\u0088\u009AD)In(r\u00E2\u0088\u009AD)\u00E2\u0080\u00B2\u00E2\u0080\u00B2=[d2G0,ndr2]r=1r20,572.3. Quasi-Steady State Profiles for the Gierer-Meinhardt Modeljust as in the n = 0 case. If we differentiate this again,W \u00E2\u0080\u00B2\u00E2\u0080\u00B20,n(r) = In(r\u00E2\u0088\u009AD)Kn(r\u00E2\u0088\u009AD)\u00E2\u0080\u00B2\u00E2\u0080\u00B2\u00E2\u0080\u00B2\u00E2\u0088\u0092Kn(r\u00E2\u0088\u009AD)In(r\u00E2\u0088\u009AD)\u00E2\u0080\u00B2\u00E2\u0080\u00B2\u00E2\u0080\u00B2+ In(r\u00E2\u0088\u009AD)\u00E2\u0080\u00B2Kn(r\u00E2\u0088\u009AD)\u00E2\u0080\u00B2\u00E2\u0080\u00B2\u00E2\u0088\u0092Kn(r\u00E2\u0088\u009AD)\u00E2\u0080\u00B2In(r\u00E2\u0088\u009AD)\u00E2\u0080\u00B2\u00E2\u0080\u00B2=[d3G0,ndr3]r+ In(r\u00E2\u0088\u009AD)\u00E2\u0080\u00B2Kn(r\u00E2\u0088\u009AD)\u00E2\u0080\u00B2\u00E2\u0080\u00B2\u00E2\u0088\u0092Kn(r\u00E2\u0088\u009AD)\u00E2\u0080\u00B2In(r\u00E2\u0088\u009AD)\u00E2\u0080\u00B2\u00E2\u0080\u00B2= \u00E2\u0088\u0092 2r30+W1,n(r0),where we define the new Wronskian,W1,n(r) = In(r\u00E2\u0088\u009AD)\u00E2\u0080\u00B2Kn(r\u00E2\u0088\u009AD)\u00E2\u0080\u00B2\u00E2\u0080\u00B2\u00E2\u0088\u0092Kn(r\u00E2\u0088\u009AD)\u00E2\u0080\u00B2In(r\u00E2\u0088\u009AD)\u00E2\u0080\u00B2\u00E2\u0080\u00B2. (2.70)Furthermore,[d2G1,ndr2]r0= In(r0\u00E2\u0088\u009AD)\u00E2\u0080\u00B2Kn(r0\u00E2\u0088\u009AD)\u00E2\u0080\u00B2\u00E2\u0080\u00B2\u00E2\u0088\u0092Kn(r0\u00E2\u0088\u009AD)\u00E2\u0080\u00B2In(r0\u00E2\u0088\u009AD)\u00E2\u0080\u00B2\u00E2\u0080\u00B2,and therefore[d3G0dr3]r0= \u00E2\u0088\u0092 2r30\u00E2\u0088\u0092W1,0(r0),[d2G1,ndr2]r0= W1,n(r0).To determine the Wronskian (2.70), consider that y = u0 = In(r\u00E2\u0088\u009AD)andy = v0 = Kn(r\u00E2\u0088\u009AD)satisfyd2ydr2+1rdydr\u00E2\u0088\u0092(n2r2+1D)y = 0582.3. Quasi-Steady State Profiles for the Gierer-Meinhardt Modeland differentiate this expression. Therefore u = In(r\u00E2\u0088\u009AD)\u00E2\u0080\u00B2and v = Kn(r\u00E2\u0088\u009AD)\u00E2\u0080\u00B2satisfyd2udr2+1rdudr\u00E2\u0088\u0092(n2 + 1r2+1D)u+2n2r3u0 = 0,d2vdr2+1rdvdr\u00E2\u0088\u0092(n2 + 1r2+1D)v +2n2r3v0 = 0respectively. We define W1,n(r) = u\u00E2\u0080\u00B2v \u00E2\u0088\u0092 uv\u00E2\u0080\u00B2 and so multiplying the firstexpression by v and the second by u we havedW1,ndr+W1,nr\u00E2\u0088\u0092 2n2r3W0,n(r) =dW1,ndr+W1,nr+2n2r4= 0.Solving for this Wronskian, we getW1,n(r) =n2r3+1Dr, (2.71)where the 1/Dr term is determined by looking at the small r asymptotics ofW1,n(r) for n = 0 (since the expression must hold for all n). We thereforehave that[\u00E2\u0088\u00822u1\u00E2\u0088\u0082r2]r0=\u00E2\u0088\u009E\u00E2\u0088\u0091n=\u00E2\u0088\u0092\u00E2\u0088\u009E(A\u00C2\u00AF1n(n2r2+1D)\u00E2\u0088\u0092 B\u00C2\u00AF1nr0)exp(in\u00CE\u00B8),[d3u0dr3]r0= \u00E2\u0088\u0092A\u00C2\u00AF0(2r20+1D).We can simplify the first expression,[\u00E2\u0088\u00822u1\u00E2\u0088\u0082r2]r0=A\u00C2\u00AF0r20(h(\u00CE\u00B8)\u00E2\u0088\u0092 h\u00E2\u0080\u00B2\u00E2\u0080\u00B2(\u00CE\u00B8)) + A\u00C2\u00AF0Dh(\u00CE\u00B8) +A\u00C2\u00AF0A\u00C2\u00AF0U00r0U01592.3. Quasi-Steady State Profiles for the Gierer-Meinhardt Modeland finally write[\u00E2\u0088\u0082u2\u00E2\u0088\u0082r]r0=A\u00C2\u00AF0h\u00E2\u0080\u00B2(\u00CE\u00B8)22r20+A\u00C2\u00AF0r20h(\u00CE\u00B8)h\u00E2\u0080\u00B2\u00E2\u0080\u00B2(\u00CE\u00B8)\u00E2\u0088\u0092 A\u00C2\u00AF02Dh(\u00CE\u00B8)2\u00E2\u0088\u0092 A\u00C2\u00AF0A\u00C2\u00AF0U00r0h(\u00CE\u00B8)U01 \u00E2\u0088\u0092A\u00C2\u00AF0U200(A\u00C2\u00AF1U201 + A\u00C2\u00AF0U00U02). (2.72)We wish to perform an eigenfunction expansion on this problem but we mustdelicately handle the product of infinite sums that occurs. We will define theproduct in the following way,Definition 2.3.2.1 Assume two functions f(\u00CE\u00B8) and g(\u00CE\u00B8) have a fourier se-ries given byf(\u00CE\u00B8) \u00E2\u0088\u00BC\u00E2\u0088\u009E\u00E2\u0088\u0091n=\u00E2\u0088\u0092\u00E2\u0088\u009Ean exp(in\u00CE\u00B8), g(\u00CE\u00B8) \u00E2\u0088\u00BC\u00E2\u0088\u009E\u00E2\u0088\u0091m=\u00E2\u0088\u0092\u00E2\u0088\u009Ebm exp(im\u00CE\u00B8)and that there exists some N and M such that |an| = 0 when |n| > N and|bm| = 0 when |m| > M . If this is the case then we can define the product ofthese functions asf(\u00CE\u00B8)g(\u00CE\u00B8) =\u00E2\u0088\u009E\u00E2\u0088\u0091m=\u00E2\u0088\u0092\u00E2\u0088\u009E\u00E2\u0088\u009E\u00E2\u0088\u0091n=\u00E2\u0088\u0092\u00E2\u0088\u009Eanbm exp(i(n+m)\u00CE\u00B8).This definition states that when a Fourier series terminates then we canuse the finite series product. This is in contrast to the Cauchy-producttypically used for infinite sums. Since we expect the perturbation h(\u00CE\u00B8) iscomposed of finite sinusoidal modes, the Fourier series will terminate. Thisallows us to use Definition 2.3.2.1 for quadratic products of h(\u00CE\u00B8) at O(\u00CE\u00B52).Using definition 2.3.2.1 for handling series products, we can perform the602.3. Quasi-Steady State Profiles for the Gierer-Meinhardt Modeleigenfunction expansion asu2(r, \u00CE\u00B8) =\u00E2\u0088\u009E\u00E2\u0088\u0091k=\u00E2\u0088\u0092\u00E2\u0088\u009EV2k(r) exp(ik\u00CE\u00B8), U02(\u00CE\u00B8) =\u00E2\u0088\u009E\u00E2\u0088\u0091k=\u00E2\u0088\u0092\u00E2\u0088\u009EU2k exp(ik\u00CE\u00B8),and the base problem is identical to (2.63) so we can immediately writeV2k =\u00EF\u00A3\u00B1\u00EF\u00A3\u00B4\u00EF\u00A3\u00B2\u00EF\u00A3\u00B4\u00EF\u00A3\u00B3AJk,1(r), 0 \u00E2\u0089\u00A4 r < r0BJk,2(r), r0 < r \u00E2\u0089\u00A4 R.The continuity and derivative jump conditions for each mode k become[V2k]r0 =A\u00C2\u00AF02r0\u00E2\u0088\u0091n\u00E2\u0088\u0088NHnHk\u00E2\u0088\u0092n +A\u00C2\u00AF0A\u00C2\u00AF0U00\u00E2\u0088\u0091n\u00E2\u0088\u0088NHnU1k\u00E2\u0088\u0092n = A\u00C2\u00AF2k, (2.73a)[dV2kdr]r0=\u00E2\u0088\u0092 A\u00C2\u00AF02r20\u00E2\u0088\u0091n\u00E2\u0088\u0088Nn(k \u00E2\u0088\u0092 n)HnHk\u00E2\u0088\u0092n \u00E2\u0088\u0092A\u00C2\u00AF0r20\u00E2\u0088\u0091n\u00E2\u0088\u0088N(k \u00E2\u0088\u0092 n)2HnHk\u00E2\u0088\u0092n\u00E2\u0088\u0092 A\u00C2\u00AF02D\u00E2\u0088\u0091n\u00E2\u0088\u0088NHnHk\u00E2\u0088\u0092n \u00E2\u0088\u0092A\u00C2\u00AF0A\u00C2\u00AF0U00r0\u00E2\u0088\u0091n\u00E2\u0088\u0088NHnU1k\u00E2\u0088\u0092n\u00E2\u0088\u0092 A\u00C2\u00AF0U200(A\u00C2\u00AF1\u00E2\u0088\u0091n\u00E2\u0088\u0088NU1nU1k\u00E2\u0088\u0092n + A\u00C2\u00AF0U00U2k)= B\u00C2\u00AF2k. (2.73b)Here the sum is over a set N which contains the integer modes n that producevalid integer k modes. For example, if h(\u00CE\u00B8) = cos(6\u00CE\u00B8) then there are twomodes at O(\u00CE\u00B5) of n = \u00E2\u0088\u00926 and n = 6. Various quadratic combinations ofthese modes leads to k = \u00E2\u0088\u009212, k = 0, and k = 12 as the only possible modesthat can occur at O(\u00CE\u00B52). Therefore, if k = \u00E2\u0088\u009212 then N = {\u00E2\u0088\u00926} since it isonly through this mode that frequencies exp(\u00E2\u0088\u009212i\u00CE\u00B8) can occur but for k = 0then the set N = {\u00E2\u0088\u00926, 6} since these two modes lead to terms of frequency 1.612.3. Quasi-Steady State Profiles for the Gierer-Meinhardt ModelAs another example, consider h(\u00CE\u00B8) = cos(3\u00CE\u00B8) + cos(6\u00CE\u00B8) which has admissiblemodes at O(\u00CE\u00B52) of k = 0,\u00C2\u00B13,\u00C2\u00B16,\u00C2\u00B19,\u00C2\u00B112 and here for k = 0, the set isN = {\u00C2\u00B13,\u00C2\u00B16} while for k = 9, the set is N = {3, 6}. In similar fashion to(2.63) at O(\u00CE\u00B5), we can use (2.73) to write,V2k(r) = A\u00C2\u00AF2kr0G1,k(r; r0)\u00E2\u0088\u0092 B\u00C2\u00AF2kr0G0,k(r; r0). (2.74)As with U1n, we can solve U2k by taking the average value of (2.53) at O(\u00CE\u00B52)U2k =A\u00C2\u00AF0r02d2G0dr2\u00E2\u0088\u0091n\u00E2\u0088\u0088NHnHk\u00E2\u0088\u0092n + r0\u00E2\u0088\u0091n\u00E2\u0088\u0088NHn(A\u00C2\u00AF1k\u00E2\u0088\u0092ndG1,k\u00E2\u0088\u0092ndr\u00E2\u0088\u0092 B\u00C2\u00AF1k\u00E2\u0088\u0092ndG0,k\u00E2\u0088\u0092ndr)+ r0A\u00C2\u00AF2kG1,k \u00E2\u0088\u0092 r0B\u00C2\u00AF2kG0,k.If we defineB\u00CB\u009C2k = B\u00C2\u00AF2k +A\u00C2\u00AF0A\u00C2\u00AF0U00U2kthen we can solveU2k =(1\u00E2\u0088\u0092 A\u00C2\u00AF0G0,k(r0; r0)G0(r0; r0))\u00E2\u0088\u00921(A\u00C2\u00AF0r02\u00E2\u008C\u00A9d2G0dr2\u00E2\u008C\u00AAr0\u00E2\u0088\u0091n\u00E2\u0088\u0088NHnHk\u00E2\u0088\u0092n+r0\u00E2\u0088\u0091n\u00E2\u0088\u0088NHn(A\u00C2\u00AF1k\u00E2\u0088\u0092n\u00E2\u008C\u00A9dG1,k\u00E2\u0088\u0092ndr\u00E2\u008C\u00AAr0\u00E2\u0088\u0092 B\u00C2\u00AF1k\u00E2\u0088\u0092n\u00E2\u008C\u00A9dG0,k\u00E2\u0088\u0092ndr\u00E2\u008C\u00AAr0)+r0A\u00C2\u00AF2k \u00E3\u0080\u0088G1,k\u00E3\u0080\u0089r0 \u00E2\u0088\u0092 r0B\u00CB\u009C2kG0,k(r0; r0)). (2.75)Thus we have completely solved the inhibitor problem up to O(\u00CE\u00B52). We willpresent an example verifying the asymptotic calculations in section 2.3.2 butfirst will consider velocity corrections.622.3. Quasi-Steady State Profiles for the Gierer-Meinhardt ModelVelocity CorrectionWe will now furnish corrections to the velocity magnitude (2.32e) which werewrite asV0 = \u00CE\u00BA0 \u00E2\u0088\u0092q2U0H\u00CB\u0086\u00E2\u008C\u00A9dudn\u00E2\u008C\u00AA\u00CE\u00B7=0.The process of determining this asymptotically is much simpler than deter-mining the inhibitor value because it can be computed explicitly in termsof already known quantities. First we note that in polar coordinates we canwrite the curvature as (cf [25]),\u00CE\u00BA0 =\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3dt\u00CB\u0086ds\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3 =r2 + 2r\u00E2\u0080\u00B22 \u00E2\u0088\u0092 rr\u00E2\u0080\u00B2\u00E2\u0080\u00B2(r2 + r\u00E2\u0080\u00B22)3/2where t\u00CB\u0086 is the unit tangent vector and once again prime is differentiationwith respect to s. Using the near-circle radius (2.52) we have that\u00CE\u00BA0 \u00E2\u0088\u00BC1r0\u00E2\u0088\u0092 \u00CE\u00B5(h\u00E2\u0080\u00B2\u00E2\u0080\u00B2(\u00CE\u00B8) + h(\u00CE\u00B8)r20)+ \u00CE\u00B52(4h(\u00CE\u00B8)h\u00E2\u0080\u00B2\u00E2\u0080\u00B2(\u00CE\u00B8) + h\u00E2\u0080\u00B2(\u00CE\u00B8)2 + 2h(\u00CE\u00B8)22r30). (2.76)We can make a velocity expansion as followsV0 \u00E2\u0088\u00BC V00 + \u00CE\u00B5V01 + \u00CE\u00B52V02and to leading order, the velocity is exactly that which was derived for theradially symmetric case (2.43). Using (2.53), (2.54), (2.76), and (2.57b) we632.3. Quasi-Steady State Profiles for the Gierer-Meinhardt Modelcan get the velocity correction at O(\u00CE\u00B5) isV01 =\u00E2\u0088\u0092(h\u00E2\u0080\u00B2\u00E2\u0080\u00B2(\u00CE\u00B8) + h(\u00CE\u00B8)r20)\u00E2\u0088\u0092 qH\u00CB\u008602U200(1\u00E2\u0088\u0092 2qb0H\u00CB\u00860dH\u00CB\u00860db)\u00E2\u008C\u00A9du0dr\u00E2\u008C\u00AAr0U01+qH\u00CB\u008602U00(h(\u00CE\u00B8)\u00E2\u008C\u00A9d2u0dr2\u00E2\u008C\u00AAr0+\u00E2\u008C\u00A9\u00E2\u0088\u0082u1\u00E2\u0088\u0082r\u00E2\u008C\u00AAr0)(2.77)All of the necessary values to compute V1 have been obtained and can there-fore be directly substituted. Notice that if r0 \u001C 1 then to leading orderV01 \u00E2\u0088\u00BC \u00E2\u0088\u0092(h\u00E2\u0080\u00B2\u00E2\u0080\u00B2(\u00CE\u00B8) + h(\u00CE\u00B8)r20),and for typical h(\u00CE\u00B8) = cos(m\u00CE\u00B8) thenV01 \u00E2\u0088\u00BC(m2 \u00E2\u0088\u0092 1r20)cos(m\u00CE\u00B8),which is in phase with h(\u00CE\u00B8) and so the velocity is positive where the perturbedradius is larger than the base radius r0 and negative where the radius issmaller. Therefore, since the inward normal is positive, this has the effect ofcircularizing the curve. If r0 \u001D 1 then it is possible for the curve velocity tobe negative allowing the perturbation to grow but this does not necessarilymean that the pattern destabilizes. Continuing in the expansion, we can get642.3. Quasi-Steady State Profiles for the Gierer-Meinhardt Modelthat the velocity at O(\u00CE\u00B52) isV02 =(4h(\u00CE\u00B8)h\u00E2\u0080\u00B2\u00E2\u0080\u00B2(\u00CE\u00B8) + h\u00E2\u0080\u00B2(\u00CE\u00B8)2 + 2h(\u00CE\u00B8)22r30)+q2((U201U200\u00E2\u0088\u0092 U02U200)H\u00CB\u00860 \u00E2\u0088\u0092U01U200dH\u00CB\u00860dbb1+1U00(12d2H\u00CB\u00860db2b21 +dH\u00CB\u00860dbb2))\u00E2\u008C\u00A9du0dr\u00E2\u008C\u00AAr0+q2U200(U00dH\u00CB\u00860dbb1 \u00E2\u0088\u0092 U01H\u00CB\u00860)(h(\u00CE\u00B8)\u00E2\u008C\u00A9d2u0dr2\u00E2\u008C\u00AAr0+\u00E2\u008C\u00A9\u00E2\u0088\u0082u1\u00E2\u0088\u0082r\u00E2\u008C\u00AAr0)\u00E2\u0088\u0092 qH\u00CB\u008602U00(\u00E2\u0088\u009212h(\u00CE\u00B8)2\u00E2\u008C\u00A9d3u0dr3\u00E2\u008C\u00AAr0\u00E2\u0088\u0092h(\u00CE\u00B8)\u00E2\u008C\u00A9\u00E2\u0088\u00822u1\u00E2\u0088\u0082r2\u00E2\u008C\u00AAr0+h\u00E2\u0080\u00B2(\u00CE\u00B8)r20\u00E2\u008C\u00A9\u00E2\u0088\u0082u1\u00E2\u0088\u0082\u00CE\u00B8\u00E2\u008C\u00AAr0+h\u00E2\u0080\u00B2(\u00CE\u00B8)22r20\u00E2\u008C\u00A9du0dr\u00E2\u008C\u00AAr0\u00E2\u0088\u0092\u00E2\u008C\u00A9\u00E2\u0088\u00822u2\u00E2\u0088\u0082r2\u00E2\u008C\u00AAr0)(2.78)which is also explicitly known in terms of previously computed values.Numerical Validation of the Asymptotic TheoryIn Chapter 6, we discuss and derive a method for solving the full problem(2.32) for arbitrary curves of which a near circle could be chosen. Therefore,while we omit the numerical details here, we can compare our asymptoticcorrections to the full numerically computed simulations. To ensure thatthe errors we make are asymptotic and not numeric, we choose a propercomputational resolution that is significantly smaller than our choice of \u00CE\u00B5(taken here to be 0.01). In all of our simulations we take \u00CF\u0083 = 10, R = 1,D = 1, exponent set (2, 1, 2, 0), and r0 = 0.5 and we consider the near circleperturbation h(\u00CE\u00B8) = cos(6\u00CE\u00B8). Figure 2.12 shows the corrections at each orderof \u00CE\u00B5 for U0 and b.652.3. Quasi-Steady State Profiles for the Gierer-Meinhardt Model0 1 2 3 4 5 60.09730.09740.09750.09760.09770.09780.09790.0980.09810.0982\u00CE\u00B8U 0 NumericU00U00 + \u00CE\u00B5U01U00 + \u00CE\u00B5U01 + \u00CE\u00B52U02(a) U0 corrections0 1 2 3 4 5 60.09460.09480.0950.09520.09540.09560.09580.0960.09620.0964\u00CE\u00B8b Numericb0b0 + \u00CE\u00B5b1b0 + \u00CE\u00B5b1 + \u00CE\u00B52b2(b) b correctionsFigure 2.12: Asymptotic corrections compared to numeric simulations ofthe curve inhibitor value U0 and the corresponding saturation value b fromsolving (2.32) for a perturbed circle with radius (2.52) and h(\u00CE\u00B8) = cos(6\u00CE\u00B8).Here we take exponent set (2, 1, 2, 0), R = 1, D = 1, r0 = 0.5, \u00CF\u0083 = 10, and\u00CE\u00B5 = 0.01.We see that indeed the correction at O(\u00CE\u00B5) introduces the sinusoidal pertur-bation only whereas the correction at O(\u00CE\u00B52) allows for the vertical shift cor-rection. Figure 2.13 shows the corrections at each order of \u00CE\u00B5 for the velocityV0. In Figure 2.13b we zoom in to better show the asymptotic alignment.662.3. Quasi-Steady State Profiles for the Gierer-Meinhardt Model0 1 2 3 4 5 600.511.522.53\u00CE\u00B8V 0 NumericV00V00 + \u00CE\u00B5V01V00 + \u00CE\u00B5V01 + \u00CE\u00B52V02(a) Velocity corrections0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 100.511.522.53\u00CE\u00B8V 0 NumericV00V00 + \u00CE\u00B5V01V00 + \u00CE\u00B5V01 + \u00CE\u00B52V02(b) Velocity corrections (zoom)Figure 2.13: Asymptotic corrections compared to numeric simulations of thecurve velocity V0 from solving (2.32) for a perturbed circle with radius (2.52)and h(\u00CE\u00B8) = cos(6\u00CE\u00B8). Here we take exponent set (2, 1, 2, 0), R = 1, D = 1,r0 = 0.5, \u00CF\u0083 = 10, and \u00CE\u00B5 = 0.01.67Chapter 3Linear Stability of RingSolutions to Breakup andZigzag ModesWe will now consider the linear stability analysis to the problem of the ra-dially symmetric inhibitor value produce by the activator localized on a ringof radius r0. The base problem for this isvt = \u000F2 1r\u00E2\u0088\u0082\u00E2\u0088\u0082r(r\u00E2\u0088\u0082v\u00E2\u0088\u0082r)+\u000F2r2\u00E2\u0088\u00822v\u00E2\u0088\u0082\u00CE\u00B82\u00E2\u0088\u0092 v + v2uq(1 + \u00CF\u0083v2)(3.1a)\u00CF\u0084ut = D1r\u00E2\u0088\u0082\u00E2\u0088\u0082r(r\u00E2\u0088\u0082u\u00E2\u0088\u0082r)+Dr2\u00E2\u0088\u00822u\u00E2\u0088\u0082\u00CE\u00B82\u00E2\u0088\u0092 u+ 1\u000Fvous(3.1b)which leads to a quasi-steady state for the activator v = U q0w where w is ahomoclinic orbit obtained by solving (2.22) and U0 is given by (2.40). Thequasi-steady state for the inhibitor u is then given by (2.36).3.1 Linear Stability FormulationTo perform a linear stability analysis of this steady state we first note thatsince the activator is locally confined to a ring of radius r0 then we can define683.1. Linear Stability Formulationan inner radius\u00CF\u0081 =r \u00E2\u0088\u0092 r0(T )\u000Fand we expect that the perturbation will also be localized entirely aroundthis radius. Secondly, we note that the linearization of the full equation (3.1)is linear in \u00CE\u00B8 and so we can perform a Fourier expansion as follows:v \u00E2\u0088\u00BC v\u00CB\u009Ce(r \u00E2\u0088\u0092 r0\u000F, T)+ \u00CF\u0086(r \u00E2\u0088\u0092 r0\u000F, t)exp(im\u00CE\u00B8) (3.2a)u \u00E2\u0088\u00BC ue(r, T ) +M (r, t) exp(im\u00CE\u00B8) (3.2b)where ve is the radial geometry form of the homoclinic orbit discussed insection 2.3 and ue is the steady-state from section 2.3.1. Continuity in \u00CE\u00B8dictates that m is an integer on (\u00E2\u0088\u0092\u00E2\u0088\u009E,\u00E2\u0088\u009E), but in what follows we will considerm to be a continuous parameter bearing in mind that all results will need tobe rounded down to the nearest integer mode. Furthermore, we will considerit to be a positive parameter since the eigenfunctions for m < 0 are thesame as for m > 0 and so we just need to be aware that all results needto accompany the complex conjugate mode. Note that in our expansion(3.2), we do not perform a standard Laplace expansion in time because thebase state is actually dependent on the slow time T = \u000F2t. Substituting our693.1. Linear Stability Formulationexpansion (3.2) into (3.1) we get\u000F2\u00CF\u0086T \u00E2\u0088\u0092 \u000F\u00CF\u0086\u00CF\u0081dr0dT=1r0 + \u000F\u00CF\u0081\u00E2\u0088\u0082\u00E2\u0088\u0082\u00CF\u0081((r0 + \u000F\u00CF\u0081)\u00CF\u0086\u00CF\u0081) (3.3a)\u00E2\u0088\u0092 \u000F2m2(r0 + \u000F\u00CF\u0081)2\u00CF\u0086\u00E2\u0088\u0092 \u00CF\u0086\u00E2\u0088\u0092 qv\u00CB\u009C2eu\u00CB\u009Cq+1e (1 + \u00CF\u0083v\u00CB\u009C2e)M(r0 + \u000F\u00CF\u0081)+2v\u00CB\u009Ceu\u00CB\u009Ce(1 + \u00CF\u0083v\u00CB\u009C2e)2\u00CF\u0086, (3.3b)\u000F2\u00CF\u0084MT =Dr\u00E2\u0088\u0082\u00E2\u0088\u0082r(rMr)\u00E2\u0088\u0092Dm2r2M \u00E2\u0088\u0092M + 1\u000Fovo\u00E2\u0088\u00921euse\u00CF\u0086(r \u00E2\u0088\u0092 r0\u000F)\u00E2\u0088\u0092 1\u000Fsvoeus+1eM (3.3c)where we have scaled to the inner coordinate for the activator problem andhave explicitly removed the slow-time dependence on r0. Furthermore, wehave rewritten the time dependence on \u00CF\u0086 and M using the long time scale.We now consider a WKB ansatz (cf. [9]) for each eigenfunction,\u00CF\u0086(\u00CF\u0081, T ) = \u00CE\u00A6(\u00CF\u0081, T ) exp(\u00CF\u0095(T )\u000F2), M(r, T ) = N(r, T ) exp(\u00CF\u0095(T )\u000F2). (3.4)Here we assume that the amplitude can vary with the radial coordinate butnot the phase and we choose the phase function to be the same for eacheigenfunction. The derivation of (3.4) can also be done via an applicationof multiple time scales (cf. [27]). We use this formulation in (3.3) to get703.1. Linear Stability Formulation(dividing out the exponential function)\u000F2\u00CE\u00A6T + \u00CE\u00A6\u00CF\u0095T \u00E2\u0088\u0092 \u000F\u00CE\u00A6\u00CF\u0081dr0dT=1r0 + \u000F\u00CF\u0081\u00E2\u0088\u0082\u00E2\u0088\u0082\u00CF\u0081((r0 + \u000F\u00CF\u0081)\u00CE\u00A6\u00CF\u0081)\u00E2\u0088\u0092 \u000F2m2(r0 + \u000F\u00CF\u0081)2\u00CE\u00A6\u00E2\u0088\u0092 \u00CE\u00A6\u00E2\u0088\u0092 qv\u00CB\u009C2eu\u00CB\u009Cq+1e (1 + \u00CF\u0083v\u00CB\u009C2e)N(r0 + \u000F\u00CF\u0081)+2v\u00CB\u009Ceu\u00CB\u009Ce(1 + \u00CF\u0083v\u00CB\u009C2e)2\u00CE\u00A6, (3.5a)\u000F2\u00CF\u0084NT + \u00CF\u0084N\u00CF\u0095T =Dr(rNr)r \u00E2\u0088\u0092Dm2r2N \u00E2\u0088\u0092N + 1\u000Fovo\u00E2\u0088\u00921euse\u00CE\u00A6(r \u00E2\u0088\u0092 r0\u000F)\u00E2\u0088\u0092 1\u000Fsvoeus+1eN. (3.5b)Since N is O(1) to leading order, we seek a natural asymptotic expansion ofthe inner activator function and phase,\u00CE\u00A6 \u00E2\u0088\u00BC \u00CE\u00A60 + \u000F\u00CE\u00A61 + . . . \u00CF\u0095 \u00E2\u0088\u00BC \u00CF\u00950 + \u000F\u00CF\u00951 + . . .to finally get the leading order problem\u00CE\u00A60\u00CF\u00950T =\u00CE\u00A60\u00CF\u0081\u00CF\u0081 \u00E2\u0088\u0092\u000F2m2r20\u00CE\u00A60 \u00E2\u0088\u0092 \u00CE\u00A60 +2w(1 + bw2)2\u00CE\u00A60 \u00E2\u0088\u0092qU q\u00E2\u0088\u009210 w2(1 + bw2)N(r0), (3.6a)\u00CF\u0084N\u00CF\u00950T =Dr(rNr)r \u00E2\u0088\u0092Dm2r2N \u00E2\u0088\u0092N + 1\u000Fovo\u00E2\u0088\u00921euse\u00CE\u00A60(r \u00E2\u0088\u0092 r0\u000F)\u00E2\u0088\u0092 1\u000Fsvoeus+1eN,(3.6b)where we have used in (3.6a) that to leading order u\u00CB\u009Ce \u00E2\u0088\u00BC U0 and v\u00CB\u009Ce \u00E2\u0088\u00BC U q0w.Note that this formulation assumes that N(r0) is O(1) to leading order andwe will analyze when this is so. We also include terms that are O(\u000F2m2)713.1. Linear Stability Formulationbecause if m\u001D 1 then this can be an O(1) term. If we define,L0b\u00CE\u00A60 = \u00CE\u00A60\u00CF\u0081\u00CF\u0081 \u00E2\u0088\u0092 \u00CE\u00A60 +2w(1 + bw2)2\u00CE\u00A60, (3.7)then we can write (3.6a) asL0b\u00CE\u00A60 \u00E2\u0088\u0092 qU q\u00E2\u0088\u009210w2(1 + bw2)N(r0) =(\u00CF\u00950T +\u000F2m2r20)\u00CE\u00A60, lim|\u00CF\u0081|\u00E2\u0086\u0092\u00E2\u0088\u009E\u00CE\u00A6 = 0, (3.8)and so we can think of \u00CF\u00950T = \u00CE\u00BB as the eigenvalues of (3.8). In (3.8) wedon\u00E2\u0080\u0099t define the normalization of the eigenfunctions but rather leave thisdiscussion to section 3.2.11 where we compute them numerically. If we solvedthe eigenvalues of (3.8) we would have\u00CF\u0086(\u00CF\u0081, T ) \u00E2\u0088\u00BC \u00CE\u00A60 exp(\u00E2\u0088\u00AB T0\u00CE\u00BB(s) ds/\u000F2), (3.9)where we have assumed that \u00CF\u00950(0) = 0 without loss of generality. Noticethat if the eigenvalues were not time-dependent we would get the standardexp(\u00CE\u00BBt) from the Laplace expansion in a linear stability analysis. This slightdifference can actually lead to delays in stability or instability if an eigenvaluechanges sign over the long-time domain. This behaviour is similar to what isresponsible for parameter delayed bifurcations in certain dynamical systemsmodels (cf. [76], [50], [28]). We will now turn our attention to actuallydetermining the eigenvalues of (3.8). To analyze this eigenvalue problem weneed to first solve (3.6b) which requires looking at the apparent singularterm,1\u000Fovo\u00E2\u0088\u00921euse\u00CE\u00A6(r \u00E2\u0088\u0092 r0\u000F)=ov\u00CB\u009Ce( r\u00E2\u0088\u0092r0\u000F)o\u00E2\u0088\u00921u\u00CB\u009Ce( r\u00E2\u0088\u0092r0\u000F)s \u00CE\u00A6 =\u000F\u00E2\u0086\u00920A\u00CE\u00B4(r \u00E2\u0088\u0092 r0).723.1. Linear Stability FormulationTo find A, we follow a procedure similar to section 2.2.1 and integrate thisexpression around the singularity r0 to getA = oU\u00CE\u00B2\u00E2\u0088\u0092q0\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ewo\u00E2\u0088\u00921\u00CE\u00A60 d\u00CF\u0081.In a similar fashion we have that1\u000Fsvoeus+1eN(r) =\u000F\u00E2\u0086\u00920sU\u00CE\u00B2\u00E2\u0088\u009210 N(r0)A\u00CE\u00B4(r \u00E2\u0088\u0092 r0), A =\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ewo d\u00CF\u0081,and so we can write (3.6b) as1r(rNr)r \u00E2\u0088\u0092m2r2N \u00E2\u0088\u0092 \u00CE\u00B82\u00CE\u00BBN=(sU\u00CE\u00B2\u00E2\u0088\u009210DN(r0)A\u00E2\u0088\u0092oU\u00CE\u00B2\u00E2\u0088\u0092q0D\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ewo\u00E2\u0088\u00921\u00CE\u00A60 d\u00CF\u0081)\u00CE\u00B4(r \u00E2\u0088\u0092 r0), (3.10)where\u00CE\u00B8\u00CE\u00BB =\u00E2\u0088\u009A1 + \u00CF\u0084\u00CE\u00BBD. (3.11)Like for the non-radially symmetric ring solution (2.48) or the near circularring solution (2.63) from section 2.3.1, (3.10) is the n > 0 analogue to theradially symmetric problem (eqn:circprob) and the solution technique imme-diately mimics that of section sec:radialsym. As such, we omit the detailshere but simply write downN(r) = r0(oU\u00CE\u00B2\u00E2\u0088\u0092q0D\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ewo\u00E2\u0088\u00921\u00CE\u00A60 d\u00CF\u0081\u00E2\u0088\u0092sU\u00CE\u00B2\u00E2\u0088\u009210DN(r0)A)G\u00C2\u00AF0,m(r; r0), (3.12)733.1. Linear Stability FormulationwhereG\u00C2\u00AF0,m(r; r0) =\u00EF\u00A3\u00B1\u00EF\u00A3\u00B4\u00EF\u00A3\u00B2\u00EF\u00A3\u00B4\u00EF\u00A3\u00B3J\u00C2\u00AFm,1(\u00CE\u00B8\u00CE\u00BBr)J\u00C2\u00AFm,2(\u00CE\u00B8\u00CE\u00BBr0), 0 \u00E2\u0089\u00A4 r \u00E2\u0089\u00A4 r0J\u00C2\u00AFm,1(\u00CE\u00B8\u00CE\u00BBr0)J\u00C2\u00AFm,2(\u00CE\u00B8\u00CE\u00BBr), r0 \u00E2\u0089\u00A4 r \u00E2\u0089\u00A4 R,withJ\u00C2\u00AFm,1(\u00CE\u00B8\u00CE\u00BBr) = Im (\u00CE\u00B8\u00CE\u00BBr) , (3.13a)J\u00C2\u00AFm,2(\u00CE\u00B8\u00CE\u00BBr) = \u00CE\u00B1\u00C2\u00AFmIm (\u00CE\u00B8\u00CE\u00BBr) +Km (\u00CE\u00B8\u00CE\u00BBr) , (3.13b)and \u00CE\u00B1\u00C2\u00AFm given by\u00CE\u00B1\u00C2\u00AFm = \u00E2\u0088\u0092Km (\u00CE\u00B8\u00CE\u00BBR)\u00E2\u0080\u00B2Im (\u00CE\u00B8\u00CE\u00BBR)\u00E2\u0080\u00B2 =(Km+1 (\u00CE\u00B8\u00CE\u00BBR)\u00E2\u0088\u0092 m\u00CE\u00B8\u00CE\u00BBRKm (\u00CE\u00B8\u00CE\u00BBR)m\u00CE\u00B8\u00CE\u00BBRIm (\u00CE\u00B8\u00CE\u00BBR) + Im+1 (\u00CE\u00B8\u00CE\u00BBR)). (3.13c)If we notice thatAU\u00CE\u00B2\u00E2\u0088\u009210 r0D=1G0(r0; r0),with G0 defined by (2.38) then we can write (3.12) asN(r) =(oU1\u00E2\u0088\u0092q0G0(r0; r0)A\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ewo\u00E2\u0088\u00921\u00CE\u00A60 d\u00CF\u0081\u00E2\u0088\u0092sG0(r0; r0)N(r0))G\u00C2\u00AF0,m(r; r0).(3.14)By evaluating this expression at r = r0, we solve for N(r0) to getN(r0) = oU1\u00E2\u0088\u0092q0(s+J0,1(r0)J0,2(r0)J\u00C2\u00AFm,1(\u00CE\u00B8\u00CE\u00BBr0)J\u00C2\u00AFm,2(\u00CE\u00B8\u00CE\u00BBr0))\u00E2\u0088\u00921 \u00E2\u0088\u00AB\u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ewo\u00E2\u0088\u00921\u00CE\u00A60 d\u00CF\u0081A (3.15)743.1. Linear Stability Formulationwith J0,1 and J0,2 defined by (2.39). There is something significant to noteabout N(r0) and that is the integral involving \u00CE\u00A60. The structure of (3.8)dictates that \u00CE\u00A60 will admit both an even and an odd solution (cf. [30]).Since w is even then if \u00CE\u00A60 is odd, N(r0) vanishes and the assumption thatit is O(1) to leading order fails. If \u00CE\u00A60 is even and has multiple nodal pointsthen N(r0) may still vanish. In this case, the eigenvalue of (3.8) is part ofthe spectrum of (3.7) for which we adapt the following Lemma of [7]:Lemma 3.1.0.1 Consider the problemwyy \u00E2\u0088\u0092 w + f(w) = 0, wy(0) = 0, w \u00E2\u0086\u0092 0 as |y| \u00E2\u0086\u0092 \u00E2\u0088\u009E, w(0) > 0and assume this has a homoclinic orbit solution via Lemma 2.2.0.1. Theassociated linearized operatorL\u00CE\u00A6 = \u00CE\u00A6yy \u00E2\u0088\u0092 \u00CE\u00A6 + f \u00E2\u0080\u00B2(w)\u00CE\u00A6 = \u00CE\u00BB\u00CE\u00A6has a discrete simple positive eigenvalue \u00CE\u00BB0 > 0 associated to a positive eigen-function \u00CE\u00A60. There is also a discrete eigenvalue \u00CE\u00BB1 = 0 with the eigenfunction\u00CE\u00A61 = w\u00E2\u0080\u00B2. Furthermore when f \u00E2\u0080\u00B2(0) is finite, a continuous spectrum exists onRe(\u00CE\u00BB) \u00E2\u0089\u00A4 \u00E2\u0088\u00921 + f \u00E2\u0080\u00B2(0) < 0 with Im(\u00CE\u00BB) = 0.A corollary to Lemma 3.1.0.1 is that if other discrete eigenvalues \u00CE\u00BBj, j > 1exist then \u00E2\u0088\u00921 + f \u00E2\u0080\u00B2(0) < \u00CE\u00BBj < 0. See [14] for other discretely computedeigenvalues with f(w) = wp. Since the principal eigenvalue is the only onefor which Re(\u00CE\u00BB) > 0 and its eigenfunction is even and of one sign, instabilitycan only occur for even eigenfunctions where N(r0) does not vanish.753.2. Eigenvalues Associated with \u00CE\u00A60 Even3.2 Eigenvalues Associated with \u00CE\u00A60 EvenSince the curve is defined as the point where the activator reaches its maxi-mum (defined to be at r = r0 for the steady-state) then if the eigenfunctionis even, this will affect the amplitude of the maximal value but not thelocation (the derivative at r0 still vanishes). Therefore, we consider eveneigenfunctions to correspond with amplitude or break-up instabilities. Usingthe expression (3.15), we define\u00CF\u0087m = qo(s+J0,1(r0)J0,2(r0)J\u00C2\u00AFm,1(\u00CE\u00B8\u00CE\u00BBr0)J\u00C2\u00AFm,2(\u00CE\u00B8\u00CE\u00BBr0))\u00E2\u0088\u00921,so that we can write (3.8) asL0b\u00CE\u00A60 \u00E2\u0088\u0092\u00CF\u0087mAw2(1 + bw2)\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ewo\u00E2\u0088\u00921\u00CE\u00A60 d\u00CF\u0081 =(\u00CE\u00BB+\u000F2m2r20)\u00CE\u00A60, (3.16)subject to far-field decay conditions. We call (3.16) the non-local eigenvalueproblem (NLEP) for \u00CE\u00BB and the study of NLEPs has a rich history of study(cf. [39], [45], [46], [81]). The non-local feature of the eigenvalue problemis common in pattern formation problems as a measure of the long-rangeinhibitor effect in the semi-strong regime. Aside from being non-local, NLEPssuch as (3.16) are also non-self-adjoint and as such are notoriously difficultfor finding conditions for which the eigenvalues satisfy Re(\u00CE\u00BB) < 0. If wedefine the quantities\u00C2\u00B5 = \u00CE\u00BB+\u000F2m2r20, A(\u00CE\u00A60) =\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ewo\u00E2\u0088\u00921\u00CE\u00A60 d\u00CF\u0081 (3.17)763.2. Eigenvalues Associated with \u00CE\u00A60 Eventhen we can write (3.16) asL0b\u00CE\u00A60 \u00E2\u0088\u0092\u00CF\u0087mAw2(1 + bw2)A(\u00CE\u00A60) = \u00C2\u00B5\u00CE\u00A60.We notice that L0b is a self-adjoint operator and with this in mind, multiplythe eigenvalue problem with a function \u00CF\u0088, satisfying the same boundaryconditions as \u00CE\u00A60, and integrate\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009E(L0b\u00CE\u00A6\u00E2\u0088\u0092 \u00C2\u00B5\u00CE\u00A60)\u00CF\u0088 d\u00CF\u0081 =\u00CF\u0087mA\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew2(1 + bw2)A(\u00CE\u00A60)\u00CF\u0088 d\u00CF\u0081.Since L0b is self-adjoint,\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009E(L0b\u00CE\u00A60 \u00E2\u0088\u0092 \u00C2\u00B5\u00CE\u00A60)\u00CF\u0088 d\u00CF\u0081 =\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009E(L0b\u00CF\u0088 \u00E2\u0088\u0092 \u00C2\u00B5\u00CF\u0088)\u00CE\u00A60 d\u00CF\u0081,and so\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009E(L0b\u00CF\u0088 \u00E2\u0088\u0092 \u00C2\u00B5\u00CF\u0088)\u00CE\u00A60 \u00E2\u0088\u0092\u00CF\u0087mAw2(1 + bw2)A(\u00CE\u00A60)\u00CF\u0088 d\u00CF\u0081 = 0.Define \u00CF\u0088 such that(L0b \u00E2\u0088\u0092 \u00C2\u00B5)\u00CF\u0088 =w2(1 + bw2), (3.18)so that\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew2(1 + bw2)(\u00CE\u00A60 \u00E2\u0088\u0092\u00CF\u0087mA A(\u00CE\u00A60)\u00CF\u0088)d\u00CF\u0081 = 0.Since everything outside the brackets in the integrand is positive, the integralcan only vanish if\u00CE\u00A60 =\u00CF\u0087mA A(\u00CE\u00A60)\u00CF\u0088.773.2. Eigenvalues Associated with \u00CE\u00A60 EvenWe can use this to write (3.16) as\u00CF\u0087mA A(\u00CE\u00A60)L0b\u00CF\u0088 \u00E2\u0088\u0092\u00CF\u0087mA A(\u00CE\u00A60)\u00CF\u0087mAw2(1 + bw2)\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ewo\u00E2\u0088\u00921\u00CF\u0088 d\u00CF\u0081 =\u00CF\u0087mA A(\u00CE\u00A60)\u00C2\u00B5\u00CF\u0088.Dividing through by \u00CF\u0087mA A(\u00CE\u00A6) and using (3.18) we get,1\u00CF\u0087m\u00E2\u0088\u0092\u00E2\u0088\u00AB\u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ewo\u00E2\u0088\u00921\u00CF\u0088 d\u00CF\u0081A = 0.We define Cm(\u00CE\u00BB) and f(\u00C2\u00B5) by,Cm(\u00CE\u00BB) \u00E2\u0089\u00A11\u00CF\u0087m=1qo(s+J0,1(r0)J0,2(r0)J\u00C2\u00AFm,1(\u00CE\u00B8\u00CE\u00BBr0)J\u00C2\u00AFm,2(\u00CE\u00B8\u00CE\u00BBr0)), (3.19a)f(\u00C2\u00B5) \u00E2\u0089\u00A1\u00E2\u0088\u00AB\u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ewo\u00E2\u0088\u00921\u00CF\u0088 d\u00CF\u0081A , (3.19b)and so the eigenvalue problem (3.16) becomes a root finding problemgm(\u00CE\u00BB) \u00E2\u0089\u00A1 Cm(\u00CE\u00BB)\u00E2\u0088\u0092 f(\u00C2\u00B5) = 0, (3.20)subject to (3.18) where \u00C2\u00B5 is given by (3.17). Note that we are only interestedin roots satisfying Re(\u00CE\u00BB) > 0 which correspond to unstable eigenvalues.3.2.1 Removing Saturation: The case b = 0We begin by considering \u00CF\u0083 = 0 (hence b = 0) as this will allow us to recoversome analytic properties for the even eigenfunctions for which the nonlocal783.2. Eigenvalues Associated with \u00CE\u00A60 Eventerm in (3.16) does not vanish. First when b = 0, we write (3.7) asL0\u00CE\u00A60 = L00\u00CE\u00A60 = \u00CE\u00A60\u00CF\u0081\u00CF\u0081 \u00E2\u0088\u0092 \u00CE\u00A60 + 2w\u00CE\u00A60, (3.21)so that the NLEP (3.16) simplifies toL0\u00CE\u00A60 \u00E2\u0088\u0092\u00CF\u0087mA w2\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ewo\u00E2\u0088\u00921\u00CE\u00A60 d\u00CF\u0081 = \u00C2\u00B5\u00CE\u00A60. (3.22)For this limiting case, the root finding problem (3.20) remains unchangedexcept that now the function \u00CF\u0088 from (3.19b) satisfies(L0 \u00E2\u0088\u0092 \u00C2\u00B5)\u00CF\u0088 = w2, (3.23)instead of (3.18). We will decompose the process of determining unstableeigenvalues by considering real and complex eigenvalues separately.3.2.2 Real EigenvaluesIf \u00CE\u00BB is purely real then the root finding problem remains completely un-changed. When b = 0 then we can determine the discrete eigenvaluesL0\u00CE\u00A8 = \u00CE\u00BD\u00CE\u00A8 using Lemma 3.1.0.1 explicitly (cf. [14], [47]). The continuousspectrum exists on Re(\u00CE\u00BD) < \u00E2\u0088\u00921 and the discrete eigenvalues and eigenfunc-tions satisfy,\u00CE\u00BD0 =54, \u00CE\u00A80 = w3/2; \u00CE\u00BD1 = 0, \u00CE\u00A81 = w\u00E2\u0080\u00B2; \u00CE\u00BD2 = \u00E2\u0088\u009234, \u00CE\u00A82 =(1\u00E2\u0088\u0092 56w)w1/2.(3.24)793.2. Eigenvalues Associated with \u00CE\u00A60 EvenThe principal eigenfunction \u00CE\u00A80, is a solution to the homogeneous problem(3.23) when \u00C2\u00B5 = \u00CE\u00BD0 and since the operator is self-adjoint, there will only bea solution \u00CF\u0088 if the compatibility condition,\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew2\u00CE\u00A8 d\u00CF\u0081 = 0,is satisfied which is impossible because both functions are even and hencetheir product is as well. Therefore we immediately understand that f , asdefined in (3.19b), will have a vertical asymptote at \u00C2\u00B5 = \u00CE\u00BD0. Properties off(\u00C2\u00B5) in (3.19b) have been analyzed in [79] where the asymptotic structure off(\u00C2\u00B5) for \u00C2\u00B5\u001C 1 is given byf(\u00C2\u00B5) \u00E2\u0088\u00BC 1 +(1\u00E2\u0088\u0092 12o)\u00C2\u00B5+ \u00CE\u00BAc\u00C2\u00B52 + . . . (3.25)with\u00CE\u00BAc =\u00E2\u0088\u00AB\u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ewo\u00E2\u0088\u00921\u00CF\u0088c d\u00CF\u0081A , \u00CF\u0088c = L\u00E2\u0088\u009230 (w2). (3.26)In general, \u00CF\u0088c needs to be computed numerically. However, in Proposition3.1 of [79], an explicit analytic representation is given for the case o = 2 ando = 3 (based on the exponent 2 in (2, q, o, s)). With these same cases, thefollowing global properties of f(\u00C2\u00B5) are also provided in Proposition 3.5:f \u00E2\u0080\u00B2(\u00C2\u00B5) > 0, \u00C2\u00B5 \u00E2\u0088\u0088 [0, \u00CE\u00BD0) (3.27a)f \u00E2\u0080\u00B2\u00E2\u0080\u00B2(\u00C2\u00B5) > 0, \u00C2\u00B5 \u00E2\u0088\u0088 [0, \u00CE\u00BD0) (3.27b)f(\u00C2\u00B5) < 0, \u00C2\u00B5 \u00E2\u0088\u0088 (\u00CE\u00BD0,\u00E2\u0088\u009E). (3.27c)803.2. Eigenvalues Associated with \u00CE\u00A60 EvenThese properties rely on easily being able to compute or bound the integralsin (3.19b) and solve (3.23) along with derivatives of eigenfunctions. However,in the absence of analytic solutions, we can solve (3.23) and (3.19b) numeri-cally and we conjecture that (3.27) holds for any exponent o. In Figure 3.1,we plot f(\u00C2\u00B5) for o = 1, 4, and 5.0 0.5 1 1.5 2 2.5 3 3.5 4\u00E2\u0088\u009210\u00E2\u0088\u00928\u00E2\u0088\u00926\u00E2\u0088\u00924\u00E2\u0088\u009220246810\u00C2\u00B5f(a) o = 10 0.5 1 1.5 2 2.5 3 3.5 4\u00E2\u0088\u009210\u00E2\u0088\u00928\u00E2\u0088\u00926\u00E2\u0088\u00924\u00E2\u0088\u009220246810\u00C2\u00B5f(b) o = 40 0.5 1 1.5 2 2.5 3 3.5 4\u00E2\u0088\u009210\u00E2\u0088\u00928\u00E2\u0088\u00926\u00E2\u0088\u00924\u00E2\u0088\u009220246810\u00C2\u00B5f(c) o = 5Figure 3.1: Computation of f(\u00C2\u00B5) from (3.19b) for various o with m = 0,r0 = 0.5, and \u000F = 0.025. We set m = 0 solely to satisfy \u00C2\u00B5 = \u00CE\u00BB and deal witha single variable. The properties (3.27) derived analytically from o = 2 oro = 3 still hold for various exponents.To understand the singularity properties of the asymptote to (3.19b) at \u00C2\u00B5 =813.2. Eigenvalues Associated with \u00CE\u00A60 Even\u00CE\u00BD0 we consider a small parameter \u00CE\u00B4 such that\u00C2\u00B5 = \u00CE\u00BD0 + \u00CE\u00B4\u00C2\u00B51, \u00CF\u0088 = \u00CF\u0089(\u00CE\u00B4)(\u00CF\u00880 + \u00CE\u00B4\u00CF\u00881),with \u00CF\u0089(\u00CE\u00B4) to be determined and substitute this into (3.23). The source ofthis asymptote, as previously stated, is that the homogeneous solution isincompatible with the non-homogeneous term. Near \u00CE\u00BD0, we can only achievesolution compatibility if the right hand-side of (3.23) is subdominant to \u00CF\u00880.If we expand our substitution,L0\u00CF\u00880 \u00E2\u0088\u0092 \u00CE\u00BD0\u00CF\u00880 + \u00CE\u00B4(L0\u00CF\u00881 \u00E2\u0088\u0092 \u00CE\u00BD0\u00CF\u00881 \u00E2\u0088\u0092 \u00CE\u00BD1\u00CF\u00880) +O(\u00CE\u00B42) =w2\u00CF\u0089,then a reasonable balance is to take \u00CF\u0089 = 1/\u00CE\u00B4. The leading order problemthen results in \u00CF\u00880 = \u00CE\u00A80, the eigenfunction for \u00CE\u00BD0. At next order the problemis,L0\u00CF\u00881 \u00E2\u0088\u0092 \u00CE\u00BD0\u00CF\u00881 = w2 + \u00C2\u00B51\u00CF\u00880.We notice that \u00CF\u00881 = \u00CE\u00A80 is a solution to the homogeneous problem and so, inorder for a solution to exist, we must have that the compatibility condition,\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew2\u00CE\u00A80 d\u00CF\u0081+ \u00C2\u00B51\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009E\u00CE\u00A820 d\u00CF\u0081 = 0,is satisfied. This yields that\u00C2\u00B51 = \u00E2\u0088\u0092\u00E2\u0088\u00AB\u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew2\u00CE\u00A80 d\u00CF\u0081\u00E2\u0088\u00AB\u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009E\u00CE\u00A820 d\u00CF\u0081.823.2. Eigenvalues Associated with \u00CE\u00A60 EvenIf we rewrite\u00CE\u00B4 =\u00C2\u00B5\u00E2\u0088\u0092 \u00CE\u00BD0\u00C2\u00B51,then we have that to leading order,\u00CF\u0088 \u00E2\u0088\u00BC \u00C2\u00B51\u00C2\u00B5\u00E2\u0088\u0092 \u00CE\u00BD0\u00CE\u00A80,for \u00C2\u00B5 near \u00CE\u00BD0. With this we can compute,f(\u00C2\u00B5) \u00E2\u0088\u00BC \u00C2\u00B51\u00C2\u00B5\u00E2\u0088\u0092 \u00CE\u00BD0\u00E2\u0088\u00AB\u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ewo\u00E2\u0088\u00921\u00CE\u00A80 d\u00CF\u0081\u00E2\u0088\u00AB\u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ewo d\u00CF\u0081, \u00C2\u00B5 \u00E2\u0089\u0088 \u00CE\u00BD0. (3.28)Therefore we have that there is a simple pole to f(\u00C2\u00B5) at \u00C2\u00B5 = \u00CE\u00BD0 satisfyingf \u00E2\u0086\u0092 \u00E2\u0088\u009E as \u00C2\u00B5 \u00E2\u0086\u0092 \u00CE\u00BD\u00E2\u0088\u00920 and f \u00E2\u0086\u0092 \u00E2\u0088\u0092\u00E2\u0088\u009E as \u00C2\u00B5 \u00E2\u0086\u0092 \u00CE\u00BD+0 . In Figure 3.2 we show f(\u00C2\u00B5)computed numerically for m = 0, r0 = 0.5, \u000F = 0.025, and o = 2 and overlaythe asymptotic approximation (3.28). The asymptotic agreement extendsquite well beyond the asymptote and generally describes the entire function.833.2. Eigenvalues Associated with \u00CE\u00A60 Even0 0.5 1 1.5 2 2.5 3 3.5 4\u00E2\u0088\u009210\u00E2\u0088\u00928\u00E2\u0088\u00926\u00E2\u0088\u00924\u00E2\u0088\u009220246810\u00C2\u00B5f numericasymptoticFigure 3.2: Numerical computation of f(\u00C2\u00B5) when m = 0, r0 = 0.5, \u000F = 0.025,and o = 2 along with the asymptotic expression (3.28) demonstrating thesimple pole at \u00C2\u00B5 = \u00CE\u00BD0.Returning to the root problem g given by (3.20), the structure can dependsignificantly on the magnitude of m and specifically centers around the term\u000F2m2/r20 being significant or not. As such we will independently investigatethese cases.3.2.3 Real Eigenvalues: m = O(1)Consider m = O(1) so that for \u000F\u001C 1,\u00C2\u00B5 = \u00CE\u00BB+\u000F2m2r20\u00E2\u0089\u0088 \u00CE\u00BB.843.2. Eigenvalues Associated with \u00CE\u00A60 EvenFirst note that if \u00CE\u00BB > \u00CE\u00BD0 then since Cm(\u00CE\u00BB) > 0 and f(\u00CE\u00BB) < 0 from (3.27) thenwe conclude that gm(\u00CE\u00BB) > 0 when \u00CE\u00BB > \u00CE\u00BD0. Therefore, to leading order, alleigenvalues to (3.22) satisfy \u00CE\u00BB < \u00CE\u00BD0. We begin by finding the neutral stabilitypoint \u00CE\u00BB = 0. From the asymptotic expansion of f(\u00C2\u00B5) given by (3.25), we havethat f(0) = 1 and sogm(0) = Cm(0)\u00E2\u0088\u0092 1.Furthermore, when \u00CE\u00BB = 0 then \u00CE\u00B8\u00CE\u00BB = 1/\u00E2\u0088\u009AD and when m = 0 we have thatJ\u00C2\u00AFm,i(\u00CE\u00B8\u00CE\u00BBr) = J0,i(r) so from (3.19a), C0(0) = s/qo and so since s + 1 < qo,g0(0) < 0. To understand the existence of roots we must look at the Cmfunction in (3.19a) and differentiate it with respect to order,\u00E2\u0088\u0082Cm\u00E2\u0088\u0082m= \u00E2\u0088\u0092 C\u00C2\u00AF0(J\u00C2\u00AFm,1(\u00CE\u00B8\u00CE\u00BBr0)J\u00C2\u00AFm,2(\u00CE\u00B8\u00CE\u00BBr0))2\u00E2\u0088\u0082\u00E2\u0088\u0082m(J\u00C2\u00AFm,1(\u00CE\u00B8\u00CE\u00BBr0)J\u00C2\u00AFm,2(\u00CE\u00B8\u00CE\u00BBr0)),whereC\u00C2\u00AF0 =J0,1(r0)J0,2(r0)qo> 0.Order derivative expressions have been formulated (cf. [1]) but are relativelyintractable for general m in terms of obtaining sign estimates. However, theydo simplify for m = 0 to,\u00E2\u0088\u0082\u00E2\u0088\u0082mIm (z)\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3m=0= \u00E2\u0088\u0092K0 (z) ,\u00E2\u0088\u0082\u00E2\u0088\u0082mKm (z)\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3m=0= 0,and so in this case\u00E2\u0088\u0082Jm\u00E2\u0088\u0082m\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3m=0= \u00E2\u0088\u00922\u00CE\u00B1\u00C2\u00AF0I0 (\u00CE\u00B8\u00CE\u00BBr0)K0 (\u00CE\u00B8\u00CE\u00BBr0)\u00E2\u0088\u0092K0 (\u00CE\u00B8\u00CE\u00BBr0)2 + I0 (\u00CE\u00B8\u00CE\u00BBr0)2\u00E2\u0088\u0082\u00CE\u00B1\u00C2\u00AFm\u00E2\u0088\u0082m\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3m=0,(3.29)853.2. Eigenvalues Associated with \u00CE\u00A60 EvenwhereJm(r0) = J\u00C2\u00AFm,1(\u00CE\u00B8\u00CE\u00BBr0)J\u00C2\u00AFm,2(\u00CE\u00B8\u00CE\u00BBr0). (3.30)To understand this better we require the order derivative of \u00CE\u00B1\u00C2\u00AF0,d\u00CE\u00B1\u00C2\u00AFmdm\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3m=0=\u00E2\u0088\u0082Km(\u00CE\u00B8\u00CE\u00BBR)\u00E2\u0088\u0082m\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3m=1\u00E2\u0088\u0092K0 (\u00CE\u00B8\u00CE\u00BBR)I1 (\u00CE\u00B8\u00CE\u00BBR)\u00E2\u0088\u0092K1 (\u00CE\u00B8\u00CE\u00BBR)(\u00E2\u0088\u0082Im(\u00CE\u00B8\u00CE\u00BBR)\u00E2\u0088\u0082m\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3m=1+ I0 (\u00CE\u00B8\u00CE\u00BBR))I1 (\u00CE\u00B8\u00CE\u00BBR)2 .Once again from [1] we can compute that\u00E2\u0088\u0082Im (z)\u00E2\u0088\u0082m\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3m=1= K1 (z)\u00E2\u0088\u0092I0 (z)z,\u00E2\u0088\u0082Km (z)\u00E2\u0088\u0082m\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3m=1=K1 (z)z,and using the Wronskian relationship (2.37) we can finally write thatd\u00CE\u00B1\u00C2\u00AFmdm\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3m=0=1(\u00CE\u00B8\u00CE\u00BBR)2I1 (\u00CE\u00B8\u00CE\u00BBR)2 (1\u00E2\u0088\u0092 \u00CE\u00B8\u00CE\u00BBR)\u00E2\u0088\u0092 \u00CE\u00B1\u00C2\u00AF20.For \u00CE\u00B8\u00CE\u00BBR > 1, this expression is negative and for \u00CE\u00B8\u00CE\u00BBR < 1, if we use the smallargument asymptotics for the modified Bessel functions [1], we haved\u00CE\u00B1\u00C2\u00AFmdm\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3m=0\u00E2\u0088\u00BC \u00E2\u0088\u0092 14(\u00CE\u00B8\u00CE\u00BBR)3,and therefore the order derivative of \u00CE\u00B1\u00C2\u00AF0 is always negative so (3.29) is alwaysnegative and therefore\u00E2\u0088\u0082Cm\u00E2\u0088\u0082m\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3m=0> 0.863.2. Eigenvalues Associated with \u00CE\u00A60 EvenWe expect this still to hold for m near zero and in fact, numerically, we canconfirm positivity on Cm holds for all m with r0 \u00E2\u0088\u0088 [0, R] via Figure 3.3 whichplots the maximum of the order derivative of (3.30) for r0 \u00E2\u0088\u0088 [0, R] versus \u00CE\u00B8\u00CE\u00BBfor various values of R. This derivative is always negative and tending tozero so therefore we have that \u00E2\u0088\u0082Cm\u00E2\u0088\u0082m > 0 for all m.5 10 15 20 25\u00E2\u0088\u00920.03\u00E2\u0088\u00920.02\u00E2\u0088\u00920.0100.010.020.03\u00CE\u00B8\u00CE\u00BBmax r 0\u00E2\u0088\u0088[0,R]\u00E2\u0088\u0082J \u00E2\u0088\u0082m m=5m=10m=20m=50(a) R = 0.15 10 15 20 25\u00E2\u0088\u00920.03\u00E2\u0088\u00920.02\u00E2\u0088\u00920.0100.010.020.03\u00CE\u00B8\u00CE\u00BBmax r 0\u00E2\u0088\u0088[0,R]\u00E2\u0088\u0082J \u00E2\u0088\u0082m m=5m=10m=20m=50(b) R = 15 10 15 20 25\u00E2\u0088\u00926\u00E2\u0088\u00924\u00E2\u0088\u009220246 x 10\u00E2\u0088\u00923\u00CE\u00B8\u00CE\u00BBmax r 0\u00E2\u0088\u0088[0,R]\u00E2\u0088\u0082J \u00E2\u0088\u0082m m=5m=10m=20m=50(c) R = 10Figure 3.3: Numerical computation of the derivative of (3.30) with respectto m. For a given value of m, R, and \u00CE\u00B8\u00CE\u00BB, we compute the order derivative of(3.30) over r0 \u00E2\u0088\u0088 [0, R] and then take the maximum value over that interval.The figure shows each maximal value of the derivative as a function of \u00CE\u00B8\u00CE\u00BB forvarious values of R.Since the derivative is always positive, we can have at most one root to Cm.873.2. Eigenvalues Associated with \u00CE\u00A60 EvenIf we consider m large then we can use the small argument asymptotics forIm (x) and Km (x) which are valid for x \u001C\u00E2\u0088\u009Am+ 1 [1]. These expansionsare,Im (x) \u00E2\u0088\u00BC1\u00CE\u0093(m+ 1)(x2)m, Km (x) \u00E2\u0088\u00BC\u00CE\u0093(m)2(2x)m(3.31)and so thereforeCm(\u00CE\u00BB) \u00E2\u0089\u0088m\u001D1C\u00C2\u00AF0Jm(r0)\u00E2\u0088\u00BC C\u00C2\u00AF02m1 +( r0R)2m \u001D 1. (3.32)This in combination with C0(0) < 0 anddCmdm > 0 shows there exists exactlyone point where Cm(0) = 1 and hence a single value m = mb\u00E2\u0088\u0092 such thatgmb\u00E2\u0088\u0092 (0) = 0. (3.33)Now, just because there is a value of m for which a neutral stability point oc-curs, this does not mean that \u00CE\u00BB = 0 is the largest eigenvalue when m = mb\u00E2\u0088\u0092 .To investigate this, we need to consider the \u00CE\u00BB derivative and second deriva-tive for Cm. In a similar fashion to what we did with the order derivative,we can conclude numerically that\u00E2\u0088\u0082Cm\u00E2\u0088\u0082\u00CE\u00BB> 0,\u00E2\u0088\u00822Cm\u00E2\u0088\u0082\u00CE\u00BB2< 0, (3.34)when \u00CF\u0084 6= 0. When \u00CF\u0084 = 0 then Cm(\u00CE\u00BB) = Cm(0) since the eigenvalue onlyoccurs in a product with \u00CF\u0084 and therefore dCmd\u00CE\u00BB = 0 when \u00CF\u0084 = 0. Using these883.2. Eigenvalues Associated with \u00CE\u00A60 Evenexpressions along with (3.27) we have that\u00E2\u0088\u00822gm\u00E2\u0088\u0082\u00CE\u00BB2=\u00E2\u0088\u00822Cm\u00E2\u0088\u0082\u00CE\u00BB2\u00E2\u0088\u0092 \u00E2\u0088\u00822f\u00E2\u0088\u0082\u00CE\u00BB2< 0. (3.35)Since this expression holds for all m, it must hold at m = mb\u00E2\u0088\u0092 and so gmb\u00E2\u0088\u0092can have at most 2 roots of which we already know \u00CE\u00BB = 0 is one. As \u00CE\u00BB tendsto \u00CE\u00BD0 from the left then gmb\u00E2\u0088\u0092 tends to negative infinity. Ifdgmb\u00E2\u0088\u0092d\u00CE\u00BB < 0 when\u00CE\u00BB = 0 then the only possible way there can exist a second root and havethe far-field behaviour be satisfied is if there are at least three roots to gmb\u00E2\u0088\u0092which violates the maximal root condition. Conversely, if the derivative ispositive when \u00CE\u00BB = 0 then there must be a second crossing to achieve thefar-field behaviour and therefore a second, larger, root must exist. When\u00CF\u0084 = 0, we know that dCmd\u00CE\u00BB = 0 and so by (3.27)dgmb\u00E2\u0088\u0092d\u00CE\u00BB\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00CF\u0084=0< 0, (3.36)always. If \u00CF\u0084 \u001D 1 then \u00CE\u00B8\u00CE\u00BB \u001D 1 and we can use the large argument asymptoticexpansions for the Bessel functions [1],Im (x) \u00E2\u0088\u00BCexp(x)\u00E2\u0088\u009A2pix, Km (x) \u00E2\u0088\u00BC\u00E2\u0088\u009Api2xexp(\u00E2\u0088\u0092x)to writeCm(\u00CE\u00BB) \u00E2\u0088\u00BC 2C\u00C2\u00AF0r0\u00CE\u00B8\u00CE\u00BB1(1 + exp(\u00E2\u0088\u00922\u00CE\u00B8\u00CE\u00BB(R\u00E2\u0088\u0092 r0)))\u00E2\u0089\u0088 2C\u00C2\u00AF0r0\u00CE\u00B8\u00CE\u00BB \u001D 1 \u00CF\u0084 \u001D 1.(3.37)893.2. Eigenvalues Associated with \u00CE\u00A60 EvenConsider \u00CE\u00BB small such that f \u00E2\u0088\u00BC 1 via (3.25) but large enough (i.e. \u00CE\u00BB >O(1/\u00CF\u0084)) so that for \u00CF\u0084 \u001D 1gm \u00E2\u0088\u00BC Cm \u001D 1.Specifically then, when m = mb\u00E2\u0088\u0092 , and if \u00CF\u0084 is large, we havedgmb\u00E2\u0088\u0092d\u00CE\u00BB\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00CF\u0084\u001D1> 0for some range of \u00CE\u00BB and therefore this must be true when \u00CE\u00BB = 0 otherwise thetwo-root condition will be violated. Therefore, there must exist some criticalvalue \u00CF\u0084 = \u00CF\u0084 \u00E2\u0088\u0097mb\u00E2\u0088\u0092 which satisfiesdgmb\u00E2\u0088\u0092d\u00CE\u00BB(0)\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00CF\u0084=\u00CF\u0084\u00E2\u0088\u0097mb\u00E2\u0088\u0092= 0 (3.38)such that a new root \u00CE\u00BB\u00CF\u0084\u00E2\u0088\u0097mb\u00E2\u0088\u0092> 0 is created and persists for \u00CF\u0084 > \u00CF\u0084 \u00E2\u0088\u0097mb\u00E2\u0088\u0092 . We canfind this point by using a numerical root-finding algorithm to solve (3.33) and(3.38). For example using Newton\u00E2\u0080\u0099s method with r0 = 0.5, R = 1, D = 1,and exponent set (2, 1, 2, 0), we get,mb\u00E2\u0088\u0092 \u00E2\u0089\u0088 0.4003 \u00CF\u0084 \u00E2\u0088\u0097mb\u00E2\u0088\u0092 \u00E2\u0089\u0088 1.8376. (3.39)While we solve this problem numerically, it is still an asymptotic approxima-tion as we are ignoring the terms \u000F2m2/r20.Now that we have investigated the neutral stability point, we will turn ourattention to the intervals that m = mb\u00E2\u0088\u0092 creates. Firstly, we will look at903.2. Eigenvalues Associated with \u00CE\u00A60 Evenm \u00E2\u0088\u0088 [0,mb\u00E2\u0088\u0092). When m = mb\u00E2\u0088\u0092 we have that Cmb\u00E2\u0088\u0092 = 1 and sincedCmdm > 0 forall m then for 0 \u00E2\u0089\u00A4 m < mb\u00E2\u0088\u0092 we must have Cm(0) < 1. Since f(0) = 1 thengm(0) < 0. When \u00CF\u0084 = 0 then recall Cm(\u00CE\u00BB) = Cm(0) and therefore via (3.36),there are no positive \u00CE\u00BB for m \u00E2\u0088\u0088 [0,mb\u00E2\u0088\u0092) when \u00CF\u0084 = 0. When \u00CF\u0084 is really largethen (3.37) holds for Cm and by a similar logic as when m = mb\u00E2\u0088\u0092 there is arange of \u00CE\u00BB for which gm(\u00CE\u00BB) > 0. Furthermore, we still have that gm \u00E2\u0086\u0092 \u00E2\u0088\u0092\u00E2\u0088\u009Eas \u00CE\u00BB \u00E2\u0086\u0092 \u00CE\u00BD\u00E2\u0088\u00920 , and so combining this information there are at least two rootsto gm for \u00CF\u0084 \u001D 1 for m \u00E2\u0088\u0088 [0,mb\u00E2\u0088\u0092). Using (3.35) we can restrict this furtherand conclude that there are exactly two roots when \u00CF\u0084 \u001D 1. Of course, thismeans there must exist some \u00CF\u0084 = \u00CF\u0084 \u00E2\u0088\u0097m such that there is exactly one positiveroot \u00CE\u00BB\u00CF\u0084\u00E2\u0088\u0097m which will occur whengm(\u00CE\u00BB\u00CF\u0084\u00E2\u0088\u0097m) =\u00E2\u0088\u0082gm\u00E2\u0088\u0082\u00CE\u00BB\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00CE\u00BB\u00CF\u0084\u00E2\u0088\u0097m= 0.Note that since the only neutral stability point occurs at m = mb\u00E2\u0088\u0092 , thesepositive roots cannot transition through \u00CE\u00BB = 0 and must become real via thecomplex plane.Finally, we consider m \u00E2\u0088\u0088 (mb\u00E2\u0088\u0092 ,\u00E2\u0088\u009E), bearing in mind that we are only asymp-totically considering m = O(1). On this region, Cm(0) > Cmb\u00E2\u0088\u0092 (0) = 1 andso gm(0) > 1. When \u00CF\u0084 = 0, then once again (3.36) holds and by the far-fieldbehaviour of gm, there must be exactly one root to gm. When \u00CF\u0084 6= 0 then(3.35) states that there can be at most one critical point to gm. Regardlessof the sign of dgmd\u00CE\u00BB (0), there will be exactly one root to gm as otherwise thefar-field behaviour dictates the single critical point condition will be violated.913.2. Eigenvalues Associated with \u00CE\u00A60 Even3.2.4 Real Eigenvalues: m\u001D O(1)We now turn our attention to the case when m is large enough that \u00C2\u00B5 \u00E2\u0089\u0088 \u00CE\u00BB isno longer valid as an approximation to (3.17). This approximation ceases tobe valid when m = O(\u000F\u00E2\u0088\u00921) and so we definem\u00CB\u009C = \u000Fm.Since m \u001D 1, we can use (3.32) as an approximation to Cm which meansthat \u00CF\u0087m = O(\u000F\u00E2\u0088\u00921)\u001C 1, and so to leading order (3.22) becomesL0\u00CE\u00A60 =(\u00CE\u00BB+m\u00CB\u009C2r20)\u00CE\u00A60. (3.40)We once again start by looking for a neutral stability point and so setting\u00CE\u00BB = 0, we know from Lemma 3.1.0.1 that \u00CE\u00BD0 in (3.24) is the only eigenvalue,eigenfunction pair with Re(\u00CE\u00BB) > 0 to L0, and so we have a neutral stabilitypoint m\u00CB\u009C = m\u00CB\u009Cb+ which satisfiesm\u00CB\u009Cb+ = r0\u00E2\u0088\u009A\u00CE\u00BD0. (3.41)We can find a correction to this by expandingm\u00CB\u009C2b+ \u00E2\u0088\u00BC r20\u00CE\u00BD0 + \u000Fm\u00CB\u009C1, \u00CE\u00A60 \u00E2\u0088\u00BC \u00CE\u00A80 + \u000F\u00CE\u00A601,and after substituting into (3.22), \u00CE\u00A601 and m\u00CB\u009C1 satisfyL0\u00CE\u00A601 \u00E2\u0088\u0092 \u00CE\u00BD0\u00CE\u00A601 =w22C\u00C2\u00AF0r0\u00E2\u0088\u009A\u00CE\u00BD0A\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ewo\u00E2\u0088\u00921\u00CE\u00A80 d\u00CF\u0081+m\u00CB\u009C1r20\u00CE\u00A80. (3.42)923.2. Eigenvalues Associated with \u00CE\u00A60 EvenHere we have used (3.32) for large m asymptotics to Cm (and hence \u00CF\u0087m) andhave taken that since r0 < R,( r0R)2mtends to zero as m\u00E2\u0086\u0092\u00E2\u0088\u009E. The solution\u00CE\u00A601 = \u00CE\u00A80 is a homogeneous solution to (3.42) and so we require a solvabilitycondition for the right-hand side. From this we get m\u00CB\u009C1 satisfiesm\u00CB\u009C1 = \u00E2\u0088\u0092r02C\u00C2\u00AF0\u00E2\u0088\u009A\u00CE\u00BD0B,whereB =\u00E2\u0088\u00AB\u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew2\u00CE\u00A80 d\u00CF\u0081\u00E2\u0088\u00AB\u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ewo\u00E2\u0088\u00921\u00CE\u00A80 d\u00CF\u0081A\u00E2\u0088\u00AB\u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009E\u00CE\u00A820 d\u00CF\u0081.Therefore, overall we have,mb+ \u00E2\u0088\u00BC1\u000F\u00E2\u0088\u009Ar20\u00CE\u00BD0 + \u000Fm\u00CB\u009C1 =r0\u00E2\u0088\u009A\u00CE\u00BD0\u000F\u00E2\u0088\u0092 B4C\u00C2\u00AF0\u00CE\u00BD0. (3.43)Now, unlike the case when m = mb\u00E2\u0088\u0092 , \u00CE\u00BB = 0 is the only eigenvalue whenm = mb+ . This is because if \u00CE\u00BB > 0 then \u00C2\u00B5 > \u00CE\u00BD0 and there is no solution to(3.40). In fact, this conclusion holds on the interval m > mb+ because for all\u00CE\u00BB \u00E2\u0089\u00A5 0, \u00C2\u00B5 > \u00CE\u00BD0 always. Now we consider m\u00CB\u009C < m\u00CB\u009Cb+ and first note that by thesame logic just discussed, if \u00CE\u00BB > \u00CE\u00BD0\u00E2\u0088\u0092 m\u00CB\u009C then there are no solutions to (3.40).We therefore restrict our attention to \u00CE\u00BB < \u00CE\u00BD0 \u00E2\u0088\u0092 m\u00CB\u009C. From (3.27) we havef(0) = 1, and from (3.32) we have Cm(0) \u001D 1 so gm(0) \u001D 1. Once againvia (3.35), we have that gm can have at most one critical point and sincegm \u00E2\u0086\u0092 \u00E2\u0088\u0092\u00E2\u0088\u009E as \u00CE\u00BB\u00E2\u0086\u0092 \u00CE\u00BD\u00E2\u0088\u00920 then there is exactly one root to gm on m\u00CB\u009C \u00E2\u0088\u0088 [0, m\u00CB\u009Cb+).Asymptotically, this root is given by,\u00CE\u00BB = \u00CE\u00BD0 \u00E2\u0088\u0092m\u00CB\u009C2r20. (3.44)933.2. Eigenvalues Associated with \u00CE\u00A60 Even3.2.5 Real Eigenvalue SummaryWe have now analyzed all the possible cases for real eigenvalues and wesummarize them here. For 0 \u00E2\u0089\u00A4 m < mb\u00E2\u0088\u0092 with mb\u00E2\u0088\u0092 given asymptoticallyby the numerical solution of (3.33) and (3.38), we have that there are noreal, positive eigenvalues when \u00CF\u0084 is sufficiently small and two positive realeigenvalues when \u00CF\u0084 is sufficiently large. Furthermore, there exists some \u00CF\u0084 =\u00CF\u0084 \u00E2\u0088\u0097m where there is exactly one real eigenvalue on this interval. When m = mb\u00E2\u0088\u0092then \u00CE\u00BB = 0 is the largest eigenvalue for 0 \u00E2\u0089\u00A4 \u00CF\u0084 < \u00CF\u0084 \u00E2\u0088\u0097mb\u00E2\u0088\u0092 with \u00CF\u0084\u00E2\u0088\u0097mb\u00E2\u0088\u0092 satisfying(3.38). When \u00CF\u0084 > \u00CF\u0084 \u00E2\u0088\u0097mb\u00E2\u0088\u0092 , there exists a non-zero positive real eigenvalue as thelargest eigenvalue. When mb\u00E2\u0088\u0092 < m < mb+ with mb+ given asymptoticallyby (3.43) then there is exactly one real eigenvalue for all values of \u00CF\u0084 . Whenm = mb+ , \u00CE\u00BB = 0 is the largest eigenvalue and when m > mb+ all the realeigenvalues are strictly negative.3.2.6 Complex EigenvaluesWe now consider the possibility that \u00CE\u00BB is complex. Note that the possibilityof complex eigenvalues is due to the full operator in (3.22) being non self-adjoint. We start by letting\u00CE\u00BB = \u00CE\u00BBR + i\u00CE\u00BBI , \u00CF\u0088 = \u00CF\u0088R + i\u00CF\u0088I (3.45)943.2. Eigenvalues Associated with \u00CE\u00A60 Evenwhich we substitute into (3.23) to getL0\u00CF\u0088R =(\u00CE\u00BBR +\u000F2m2r20)\u00CF\u0088R \u00E2\u0088\u0092 \u00CE\u00BBI\u00CF\u0088I + w2, (3.46a)L0\u00CF\u0088I =(\u00CE\u00BBR +\u000F2m2r20)\u00CF\u0088I + \u00CE\u00BBI\u00CF\u0088R. (3.46b)We then definefR =\u00E2\u0088\u00AB\u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ewo\u00E2\u0088\u00921\u00CF\u0088R d\u00CF\u0081A , fI =\u00E2\u0088\u00AB\u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ewo\u00E2\u0088\u00921\u00CF\u0088I d\u00CF\u0081A , (3.47)and gm = gmR + igmI withgmR = Re(Cm(\u00CE\u00BB))\u00E2\u0088\u0092 fR, gmI = Im(Cm(\u00CE\u00BB))\u00E2\u0088\u0092 fI , (3.48)and Cm(\u00CE\u00BB) still defined by (3.19a). To determine the number of roots to gm,we will use the Nyquist criteria [70] which says that the change in argumentof a function is related to the number of zeros, N0, and the number of poles,Np, inside a given closed contour, \u00CE\u0093, via[arg f(x)]\u00CE\u0093 = 2pi(N0(f)\u00E2\u0088\u0092Np(f)). (3.49)We take as a contour \u00CE\u0093 = \u00CE\u0093I \u00E2\u0088\u00AA \u00CE\u0093K with\u00CE\u0093K :{\u00CE\u00BB = K exp(it)|t \u00E2\u0088\u0088[\u00E2\u0088\u0092pi2,pi2]}, \u00CE\u0093I : \u00E2\u0088\u0092Ki \u00E2\u0089\u00A4 \u00CE\u00BB \u00E2\u0089\u00A4 Ki,traversed counter-clockwise and we consider what happens as K tends toinfinity while assuming that \u00CF\u0084 is chosen so that gm has no roots on the953.2. Eigenvalues Associated with \u00CE\u00A60 Evenimaginary axis. As K tends to infinity, so does \u00C2\u00B5, and from (3.23) we havethat \u00CF\u0088 tends to zero. Therefore fR and fI tend to zero as well as \u00C2\u00B5 \u00E2\u0086\u0092 \u00E2\u0088\u009E.Also since |\u00CE\u00BB| \u001D 1 then (3.37) holds andCm(\u00CE\u00BB) \u00E2\u0088\u00BC 2C\u00C2\u00AF0r0\u00E2\u0088\u009AK\u00CF\u0084Dexp(it2)(3.50)where we have chosen the principal value for the square root by taking thebranch cut along the negative real axis. We are interested in the change inargument of gm as we traverse \u00CE\u0093K in the counter clockwise direction. Whent = \u00E2\u0088\u0092pi/2 then from (3.50),arg gm = argCm = \u00E2\u0088\u0092pi4,and conversely when t = pi/2,arg gm = argCm =pi4,and so[arg gm]\u00CE\u0093R =pi2.For the change in argument along \u00CE\u0093I , since Cm(\u00CE\u00BB) is holomorphic and realvalued when \u00CE\u00BB is real then Cm(\u00CE\u00BB\u00C2\u00AF) = Cm(\u00CE\u00BB). This similarly holds for f andso we can write[arg gm]\u00CE\u0093I = 2[arg gm]\u00CE\u0093+I963.2. Eigenvalues Associated with \u00CE\u00A60 Evenwhere \u00CE\u0093+I is the positive imaginary axis traversed from infinity to zero. Wetherefore have from (3.49) thatN0(gm) =14+Np(gm) +1pi[arg gm]\u00CE\u0093+I .We are already aware from (3.28) that f(\u00C2\u00B5) (and hence gm) has a simple polewhen \u00C2\u00B5 = \u00CE\u00BD0. However, we also determined that if m > mb+ then \u00C2\u00B5 > \u00CE\u00BD0,and so the pole only exists inside the contour on m \u00E2\u0088\u0088 [0,mb+). Therefore,we have the number of roots to gm is given byN0(gm) =\u00EF\u00A3\u00B1\u00EF\u00A3\u00B4\u00EF\u00A3\u00B2\u00EF\u00A3\u00B4\u00EF\u00A3\u00B354 +1pi [arg gm]\u00CE\u0093+I , m < mb+14 +1pi [arg gm]\u00CE\u0093+I , m > mb+. (3.51)At the start of \u00CE\u0093+I , coming in from infinity, where K \u001D 1, we can once againuse (3.37) and writeCm(\u00CE\u00BB) \u00E2\u0088\u00BC 2C\u00C2\u00AF0r0\u00E2\u0088\u009AiK\u00CF\u0084D,whereRe(Cm) = Im(Cm) \u00E2\u0088\u00BC\u00E2\u0088\u009A2C\u00C2\u00AF0r0\u00E2\u0088\u009AK\u00CF\u0084D. (3.52)As we already discussed, \u00CF\u0088 \u00E2\u0086\u0092 0 as \u00CE\u00BB\u00E2\u0086\u0092\u00E2\u0088\u009E and so for K \u001D 1,arg gm = argCm = arctan(Im(Cm)Re(Cm))=pi4.Near the end of \u00CE\u0093+I , traversing towards the origin, we have K \u001C 1 and sowe can use (3.32) which, even though derived for m \u001D 1, was based off of973.2. Eigenvalues Associated with \u00CE\u00A60 Evensmall argument asymptotics and so will be valid for all m with K \u001C 1. Inthis case we haveRe(Cm) =2C\u00C2\u00AF0m1 +( r0R)m , Im(Cm) = 0.From Proposition 3.1 and 3.2 of [79], we have that for \u00CE\u00BBI \u001C 1 and m = O(1),fR(\u00CE\u00BBI) \u00E2\u0088\u00BC 1\u00E2\u0088\u0092 \u00CE\u00BAc\u00CE\u00BB2I +O(\u00CE\u00BB4I), (3.53a)fI(\u00CE\u00BBI) \u00E2\u0088\u00BC(1\u00E2\u0088\u0092 12o)\u00CE\u00BBI +O(\u00CE\u00BB3I) (3.53b)where \u00CE\u00BAc is defined by (3.26). If m\u001D O(1) then for \u00CE\u00BBI \u001C 1 (3.46b) simplifiestoL0\u00CF\u0088I =\u000F2m2r20\u00CF\u0088Iwhich only has a non-trivial solution at m = mb+ but we are not consideringthis point since it places the pole on the contour. Therefore, regardless ofm, we have fI = 0 for \u00CE\u00BBI \u001C 1 and gmI = 0. For gmR, recall that we alreadydetermined gm(0) < 0 on 0 \u00E2\u0089\u00A4 m < mb\u00E2\u0088\u0092 and gm(0) > 0 on m > mb\u00E2\u0088\u0092 sotherefore,arg gm = arctan(gmIgmR)=\u00EF\u00A3\u00B1\u00EF\u00A3\u00B4\u00EF\u00A3\u00B2\u00EF\u00A3\u00B4\u00EF\u00A3\u00B3pi, m \u00E2\u0088\u0088 [0,mb\u00E2\u0088\u0092)0, m \u00E2\u0088\u0088 (mb\u00E2\u0088\u0092 ,\u00E2\u0088\u009E) \mb+,near \u00CE\u00BBI = 0 of \u00CE\u0093+I . All that we are left to do now is determine the pathof gm as it changes its global argument and to do that, we will need a new983.2. Eigenvalues Associated with \u00CE\u00A60 Evenset of properties for fR and fI . Particularly we require that for \u00CE\u00BBI > 0 andm = O(1),\u00E2\u0088\u0082fR\u00E2\u0088\u0082\u00CE\u00BBI< 0, fI(\u00CE\u00BBI) > 0, (3.54)which for o = 2 has been proven explicitly in Proposition 3.1 and 3.2 of[79]. However, as we did for (3.27), we can conjecture numerically that theseproperties hold for any o. In Figure 3.4, we verify (3.27) for o = 1, 4, and 5.0 1 2 3 4 5 6 7 8 9 1000.20.40.60.811.21.4\u00CE\u00BBIf fRfI(a) o = 10 1 2 3 4 5 6 7 8 9 1000.20.40.60.811.21.4\u00CE\u00BBIf fRfI(b) o = 40 1 2 3 4 5 6 7 8 9 1000.20.40.60.811.21.4\u00CE\u00BBIf fRfI(c) o = 5Figure 3.4: Computation of fR and fI from (3.47) for various o with m = 0,r0 = 0.5, and \u000F = 0.025. We set m = 0 solely to satisfy dealing with a singlevariable \u00CE\u00BBI . The properties (3.54) derived analytically from o = 2 or o = 3in Proposition 3.1 and 3.2 of [79] still hold for various exponents.993.2. Eigenvalues Associated with \u00CE\u00A60 Even3.2.7 Complex Eigenvalues: 0 \u00E2\u0089\u00A4 m < mb\u00E2\u0088\u0092On this range we know that the argument must transition from pi/4 to pi andso we will determine the path to get here by looking at crossings along theimaginary axis where gmR = 0 of which at least one must exist. From (3.34)and (3.54) we havedgmRd\u00CE\u00BBI=dd\u00CE\u00BBIRe(Cm)\u00E2\u0088\u0092dfRd\u00CE\u00BBI> 0 (3.55)for all \u00CE\u00BBI and so by the mean value theorem there is a unique crossing of theimaginary axis. To determine which branch gets crossed, we need to considerthe sign of gmI at the crossing. Recall, if \u00CF\u0084 = 0 then Cm(\u00CE\u00BB) = Cm(0) and sofor \u00CF\u0084 = 0,Re(Cm) = Cm(0), Im(Cm) = 0,and so from (3.54),gmI = Im(Cm)\u00E2\u0088\u0092 fI = \u00E2\u0088\u0092fI < 0,and so the negative imaginary axis is crossed. Therefore,[arg gm]\u00CE\u0093+I = \u00E2\u0088\u00925pi4, \u00CF\u0084 \u001C 1.1003.2. Eigenvalues Associated with \u00CE\u00A60 EvenWhen \u00CF\u0084 \u001D 1 then from (3.37), we can use (3.52) for the real and imaginaryparts of Cm. Furthermore, since fI does not have a vertical asymptote thengmI \u00E2\u0089\u0088 Im(Cm)\u001D 1and so for \u00CF\u0084 \u001D 1, the positive imaginary axis is crossed and[arg gm]\u00CE\u0093+I =3pi4, \u00CF\u0084 \u001D 1.Therefore on 0 \u00E2\u0089\u00A4 m \u00E2\u0089\u00A4 mb\u00E2\u0088\u0092 we have from (3.51)N0(gm) =\u00EF\u00A3\u00B1\u00EF\u00A3\u00B4\u00EF\u00A3\u00B2\u00EF\u00A3\u00B4\u00EF\u00A3\u00B30, \u00CF\u0084 \u001C 12, \u00CF\u0084 \u001D 1.Therefore, for \u00CF\u0084 small, there are no complex eigenvalues with positive realpart which eventually transition to two complex eigenvalues with positivereal part. These eigenvalues cannot transition through \u00CE\u00BB = 0 since, if \u00CE\u00BB = 0for one value of \u00CF\u0084 , it is an eigenvalue for all values of \u00CF\u0084 and therefore, theeigenvalues must cross through the imaginary axis at some value \u00CF\u0084 = \u00CF\u0084Hmwhere a Hopf bifurcation occurs. We know from analyzing the real eigenval-ues that, eventually for \u00CF\u0084 large enough, these two complex eigenvalues mustbecome purely real and therefore \u00CF\u0084Hm < \u00CF\u0084\u00E2\u0088\u0097m.3.2.8 Complex Eigenvalues: m > mb\u00E2\u0088\u0092On this region, the argument transitions from pi/4 to 0 and so there mustbe zero or an even number of crossings through the imaginary axis. With1013.2. Eigenvalues Associated with \u00CE\u00A60 Evenm = O(1), we concluded in the previous section that via (3.55), there canexist at most one crossing through the imaginary axis and therefore, onm > mb\u00E2\u0088\u0092 with m = O(1), we must have zero crossings. This tells us that[arg gm]\u00CE\u0093+I = \u00E2\u0088\u0092pi4. (3.56)For m = O(\u000F\u00E2\u0088\u00921), we have that (3.32) holds for Cm \u001D 1 and since, unlike thecase for purely real eigenvalues, fR does not have a vertical asymptote form 6= mb+ thengmR \u00E2\u0089\u0088 Re(Cm) \u00E2\u0088\u00BC O(m)\u001D 1.Therefore gmR > 0 always and cannot cross the imaginary axis so (3.56)holds as well. Therefore, via (3.51), we haveN0(gm) =\u00EF\u00A3\u00B1\u00EF\u00A3\u00B4\u00EF\u00A3\u00B2\u00EF\u00A3\u00B4\u00EF\u00A3\u00B31, mb\u00E2\u0088\u0092 < m < mb+0, m > mb+.Since we have already determined that a real eigenvalue exists on m \u00E2\u0088\u0088(mb\u00E2\u0088\u0092 ,mb+), this must be the only positive eigenvalue for this range.3.2.9 Eigenvalue SummaryWe are now finally in a position to classify the entire spectrum of eigenvaluesto (3.16) for \u00CE\u00A60 even and \u00CF\u0083 = 0 via the following principal result:Principal Result 3.2.9.1 Eigenvalue Classification:On 0 < m < mb\u00E2\u0088\u0092:There are no eigenvalues with positive real part when \u00CF\u0084 is sufficiently small1023.2. Eigenvalues Associated with \u00CE\u00A60 Evenand two complex eigenvalues with positive real part when \u00CF\u0084 is sufficientlylarge. These eigenvalues undergo a Hopf bifurcation when \u00CF\u0084 = \u00CF\u0084Hm and as \u00CF\u0084increases further, these eigenvalues coincide as a single real eigenvalue when\u00CF\u0084 = \u00CF\u0084 \u00E2\u0088\u0097m before splitting on the real axis for \u00CF\u0084 > \u00CF\u0084\u00E2\u0088\u0097m.If m = mb\u00E2\u0088\u0092:The eigenvalue \u00CE\u00BB = 0 persists for all \u00CF\u0084 and when \u00CF\u0084 < \u00CF\u0084 \u00E2\u0088\u0097mb\u00E2\u0088\u0092 then \u00CE\u00BB = 0 is thelargest eigenvalue. If \u00CF\u0084 > \u00CF\u0084 \u00E2\u0088\u0097mb\u00E2\u0088\u0092 then there exists some \u00CE\u00BB\u00CF\u0084 > 0 that is purelyreal and positive.On mb\u00E2\u0088\u0092 < m < mb+ :There is exactly one real positive eigenvalue on this interval for all \u00CF\u0084 .On m > mb+:There are no eigenvalues with positive real part on this interval for all \u00CF\u0084 .The condition that there is always at least one unstable eigenvalue for all r0and all values of \u00CF\u0084 allows us to classify these eigenvalues as being an instabilityon an O(1) time scale. This is because if mb\u00E2\u0088\u0092 < m < mb+ then \u00CE\u00BB(0) > 0 sowhen we use condition (3.9), the integral diverges immediately. For an initialradius r0, unstable modes near the neutral stability points could be delayed orprevented. Specifically, for mb+ satisfied by (3.41) asymptotically, the modedecreases with increasing radius. Therefore modes that are initially unstablemay stabilize depending if secondary bifurcations occur before the patterncan stabilize or not. This is irrelevant however, since there will always bea band of unstable modes for random perturbations to amplify. However, if1033.2. Eigenvalues Associated with \u00CE\u00A60 Eventhe ring radius decreases then the upper instability mode increases and thiswould lead to previously stable modes becoming unstable.3.2.10 Numerical Computation of EigenvaluesWe will now verify some of our analytic conclusions by numerically computingthe full non-local eigenvalue problem (3.16) for any b or \u00CF\u0084 as desired. To dothis, we can discretize L0b with a standard second order finite differencescheme and in order to discretize the non-local part involving the integral of\u00CE\u00A60, we first truncate to a finite domain\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ewo\u00E2\u0088\u00921\u00CE\u00A6 d\u00CF\u0081 \u00E2\u0089\u0088\u00E2\u0088\u00AB L\u00E2\u0088\u0092Lwo\u00E2\u0088\u00921\u00CE\u00A6 d\u00CF\u0081.for some L to be chosen so that as L increases the change in the integral isbelow some tolerance. We disretize over N + 1 points withxk = \u00E2\u0088\u0092L+ kh, k = 0 . . . N, h =2LN.With this in mind we write the integral as\u00E2\u0088\u00AB L\u00E2\u0088\u0092Lwo\u00E2\u0088\u00921\u00CE\u00A6 d\u00CF\u0081 =N\u00E2\u0088\u00921\u00E2\u0088\u0091k=0\u00E2\u0088\u00AB \u00E2\u0088\u0092L+(k+1)h\u00E2\u0088\u0092L+khwo\u00E2\u0088\u00921\u00CE\u00A6 d\u00CF\u0081and discretize each integral with the trapezoid method [6]. Finally, we sup-plement far-field conditions that \u00CE\u00A60 = 0 at x = \u00C2\u00B1L.We will begin computations by setting \u00CF\u0083 = b = 0 since this is what wehave previously analyzed analytically. Furthermore, when \u00CF\u0084 = 0, \u00CF\u0087m is in-1043.2. Eigenvalues Associated with \u00CE\u00A60 Evendependent of \u00CE\u00BB and the eigenvalue problem becomes completely linear. Assuch eigenvalues can easily be computed with a standard eigenvalue packagesuch as eigs in Matlab. Figure 3.5 plots max{Re(\u00CE\u00BB)} as a function of mwhen \u00CF\u0084 = 0 for various parameters computed in Matlab.0 5 10 15 20 25\u00E2\u0088\u00921\u00E2\u0088\u00920.500.511.5mmaxRe(\u00CE\u00BB)(a) (2, q, o, s) = (2, 1, 2, 0), \u000F = 0.025,R = 1, r0 = 0.50 5 10 15 20 25 30\u00E2\u0088\u00921\u00E2\u0088\u00920.500.511.5mmaxRe(\u00CE\u00BB)(b) (2, q, o, s) = (2, 2, 3, 0), \u000F = 0.025,R = 1, r0 = 0.50 5 10 15 20 25 30\u00E2\u0088\u00921\u00E2\u0088\u00920.500.511.5mmaxRe(\u00CE\u00BB)(c) (2, q, o, s) = (2, 4, 3, 3), \u000F = 0.025,R = 1, r0 = 0.50 50 100 150\u00E2\u0088\u00921\u00E2\u0088\u00920.500.511.5mmaxRe(\u00CE\u00BB)(d) (2, q, o, s) = (2, 1, 2, 0), \u000F = 0.025,R =\u00E2\u0088\u009A10, r0 = 9/\u00E2\u0088\u009A10Figure 3.5: Numerical computation of the largest real part of the eigenvalue\u00CE\u00BB in (3.16) for the case \u00CF\u0084 = 0 and b = 0 using eigs in Matlab. The bluesolid curves are where the largest eigenvalue is negative while the red dashedcurves are where it is positive. In all experiments D = 1.In all cases, Principal Result 3.2.9.1 holds and there is a spectrum of realpositive eigenvalues. Using the eigenvalue solver we can also numerically1053.2. Eigenvalues Associated with \u00CE\u00A60 Evenobtain mb\u00E2\u0088\u0092 and mb+ and compare them to the asymptotic approximations(3.33) for mb\u00E2\u0088\u0092 and (3.43) for mb+ . This is presented in Table 3.1 for a varietyof cases, and shows an excellent agreement with the analytical theory. Manyof the experimental cases have been chosen to mimic a similar eigenvaluediscussion for stripe solutions considered in [39].(2, q, r, s) \u000F R = ` r0 mb\u00E2\u0088\u0092(n) mb\u00E2\u0088\u0092(a) mb+(n) mb+(a1) mb+(a2)(2,1,2,0) 0.1 1 0.5 0.4033 0.4003 5.3542 5.5902 5.3671(2,1,2,0) 0.05 1 0.5 0.4008 0.4003 10.9528 11.1803 10.9573(2,1,2,0) 0.025 1 0.5 0.3996 0.4003 22.1389 22.3607 22.1376(2,1,2,0) 0.025 1/\u00E2\u0088\u009A10 1/2\u00E2\u0088\u009A10 0.0485 0.0485 7.0482 7.0711 7.0469(2,1,2,0) 0.025\u00E2\u0088\u009A10\u00E2\u0088\u009A10/2 2.3362 2.3373 69.3076 70.7107 69.3350(2,1,2,0) 0.025\u00E2\u0088\u009A10 2/\u00E2\u0088\u009A10 0.9756 0.9745 27.6761 28.2843 27.6900(2,1,2,0) 0.025\u00E2\u0088\u009A10 9/\u00E2\u0088\u009A10 3.4737 3.4771 125.7202 127.2792 125.7232(2,1,3,0) 0.025 1 0.5 0.6974 0.6986 22.0085 22.3607 22.0122(2,2,3,0) 0.025 1 0.5 1.4307 1.4251 21.6334 22.3607 21.6637(2,4,3,3) 0.025 1 0.5 2.1107 2.0866 20.8712 22.3607 20.9667Table 3.1: Comparison for \u00CF\u0084 = 0 and b = 0 of numerical and asymptoticcomputations of mb\u00E2\u0088\u0092 and mb+ for a variety of exponent sets, \u000F, R, and r0with D = 1 for all. The (n) refers to numeric computations of (3.16) usingeigs in Matlab. mb\u00E2\u0088\u0092(a) is computed via Newton\u00E2\u0080\u0099s method on (3.33), mb+(a1)is computed via (3.41) while mb+(a2) is computed via (3.43).3.2.11 Computing Eigenvalues, \u00CF\u0084 6= 0When \u00CF\u0084 6= 0 then \u00CF\u0087m in (3.16) is a function of \u00CE\u00BB and the eigenvalue problemis more intricate as \u00CF\u0087m depends nonlinearly on the product \u00CF\u0084\u00CE\u00BB. For this case,we must solveT (\u00CE\u00BB)\u00CE\u00A6 = 0,where \u00CE\u00A6 is the discrete eigenvector and T (\u00CE\u00BB) are the discretized coefficientsfor (3.16). In order to enforce uniqueness of eigenvector solutions, we require1063.2. Eigenvalues Associated with \u00CE\u00A60 Evensome normalizing constraint,vT\u00CE\u00A6 = 1.We can write the problem as a block systemF(\u00CE\u00BB,\u00CE\u00A6) =\u00EF\u00A3\u00AE\u00EF\u00A3\u00B0T (\u00CE\u00BB)\u00CE\u00A6vT\u00CE\u00A6\u00E2\u0088\u0092 1\u00EF\u00A3\u00B9\u00EF\u00A3\u00BB = 0and hence our eigenvalues and eigenvectors are roots to F which we can solvewith Newton\u00E2\u0080\u0099s Method. We can write the Jacobian to the problem as,J(\u00CE\u00BB,\u00CE\u00A6) =\u00EF\u00A3\u00AE\u00EF\u00A3\u00B0T\u00CE\u00BB(\u00CE\u00BB)\u00CE\u00A6 T (\u00CE\u00BB)0 vT\u00EF\u00A3\u00B9\u00EF\u00A3\u00BBand so the Newton problem is,J(\u00CE\u00BBk,\u00CE\u00A6k)\u00EF\u00A3\u00AE\u00EF\u00A3\u00B0\u00CE\u00BBk+1 \u00E2\u0088\u0092 \u00CE\u00BBk\u00CE\u00A6k+1 \u00E2\u0088\u0092\u00CE\u00A6k\u00EF\u00A3\u00B9\u00EF\u00A3\u00BB = \u00E2\u0088\u0092F (\u00CE\u00BBk,\u00CE\u00A6k), (3.57)where k indicates the iteration count. We note that the normalizing vectorcan also change with each iterate and so v = vk. If we write out the firstequation from the matrix multiplication we get(\u00CE\u00BBk+1 \u00E2\u0088\u0092 \u00CE\u00BBk)T\u00CE\u00BB(\u00CE\u00BBk)\u00CE\u00A6k + T (\u00CE\u00BBk)\u00CE\u00A6k+1 \u00E2\u0088\u0092 T (\u00CE\u00BBk)\u00CE\u00A6k = \u00E2\u0088\u0092T (\u00CE\u00BBk)\u00CE\u00A6k,1073.2. Eigenvalues Associated with \u00CE\u00A60 Evenwhich can simplify to\u00CE\u00A6k+1 = \u00E2\u0088\u0092uk(\u00CE\u00BBk+1 \u00E2\u0088\u0092 \u00CE\u00BBk), (3.58)where we defineuk = T (\u00CE\u00BBk)\u00E2\u0088\u00921T\u00CE\u00BB(\u00CE\u00BBk)\u00CE\u00A6k.If we left-multiply by vTk then\u00CE\u00BBk+1 = \u00CE\u00BBk \u00E2\u0088\u0092vTk \u00CE\u00A6k+1vTk uk,which still appears to depend on the solution \u00CE\u00A6k+1. However, if we expandout the second equation in (3.57) then we getvTk \u00CE\u00A6k+1 = 1and so\u00CE\u00BBk+1 = \u00CE\u00BBk \u00E2\u0088\u00921vTk uk. (3.59)This determines \u00CE\u00BBk+1 and we could use (3.58) to determine \u00CE\u00A6k+1 but sinceit is just uk up to a constant we instead renormalize via,\u00CE\u00A6k+1 =ukvTk uk,1083.2. Eigenvalues Associated with \u00CE\u00A60 Evenwhich completes the problem. For simplicity we take,vk =uk\u00E2\u0088\u00921|uk\u00E2\u0088\u00921|2to normalize. There are several practical issues to deal with when implement-ing Newton\u00E2\u0080\u0099s method for this problem because there are several eigenvaluesthat exist (including the continuous spectrum that exists below \u00CE\u00BB = 1). Weare only interested in capturing the eigenvalues at each m for a given \u00CF\u0084 thathave the largest positive real part. To accomplish this we take as initial datathe \u00CF\u0084 = 0 case where the problem is linear and the largest eigenvalues are eas-ily computed. We choose an m = m\u00E2\u0088\u0097 value in the interval mb\u00E2\u0088\u0092 < m < mb+(typically the average point) because from Principal Result 3.2.9.1, whenb = 0 there is only one positive root in this interval and therefore, the New-ton solve will be more robust in this region. Having chosen m\u00E2\u0088\u0097, we computethe Newton solve by slowly varying \u00CF\u0084 and finding eigenvalues at m = m\u00E2\u0088\u0097until a desired final \u00CF\u0084 value is reached. At this point, we begin anotherNewton solve, fixing \u00CF\u0084 , for m < m\u00E2\u0088\u0097 and m > m\u00E2\u0088\u0097 until we have traversedthe entire spectrum of m, or until we enter the continuous spectrum. Weallow the stepsize on m to vary dynamically, getting smaller when there areconvergence issues and getting larger when roots are found in relatively fewiterations. On m > m\u00E2\u0088\u0097, there is very little trouble because once m > mb+we know that there are no eigenvalues with positive real part for all \u00CF\u0084 and sowe can terminate the Newton iteration quickly after this point. However, onm < m\u00E2\u0088\u0097, if \u00CF\u0084 satisfies \u00CF\u0084Hm < \u00CF\u0084 < \u00CF\u0084\u00E2\u0088\u0097m, the Newton iteration has some stabilityissues. This is due to the existence of complex eigenvalues with positive realpart on this interval of \u00CF\u0084 . On this region, there are still real eigenvalues for1093.2. Eigenvalues Associated with \u00CE\u00A60 Evenm \u00E2\u0089\u00A5 mb\u00E2\u0088\u0092 and so there is a point where eigenvalues transition from beingreal to complex. Since the Newton algorithm at this stage is based on usingthe previous m values as initial data, the stability issues arise from tryingto find complex solutions with real initial data. Therefore, if the numberof attempted iterations exceeds a certain threshold within this region of m,the code modifies the initial guess to include an imaginary component ofmagnitude on order with the selected tolerance and the algorithm continues.One last issue of practicality that occurs when solving the problem is whenthe eigenvalues get close to zero (m = mb\u00E2\u0088\u0092 and m = mb+). The problem isthat \u00CE\u00BB = 0 is always an eigenvalue of (3.16) with eigenfunction \u00CE\u00A6 = w\u00E2\u0080\u00B2 (seeLemma 3.1.0.1). Notice this eigenfunction is odd and so this zero eigenvalueis ignored in the present analysis of even eigenfunctions but it still existsand can be captured by the algorithm. Therefore, we have a checking crite-rion that the solved eigenfunction at some m = m1 is not orthogonal to theeigenfunction of the previous m = m0 value. If \u00CE\u00A6Tm1\u00CE\u00A6m0 is less than somepreset threshold then we discard the solution, reduce the step size in m, andreinitialize the algorithm with the values at m0. In Figure 3.6 we plot thenumerically computed eigenvalues using Newton\u00E2\u0080\u0099s method as outlined abovefor various values of \u00CF\u0084 with (p, q, o, s) = (2, 1, 2, 0), \u000F = 0.025, D = 1, R = 1,and r0 = 0.5.1103.2. Eigenvalues Associated with \u00CE\u00A60 Even0 5 10 15 20 25\u00E2\u0088\u00921\u00E2\u0088\u00920.500.511.5mmaxRe(\u00CE\u00BB)(a) \u00CF\u0084 = 0.40 5 10 15 20 25\u00E2\u0088\u00921\u00E2\u0088\u00920.500.511.5mmaxRe(\u00CE\u00BB)(b) \u00CF\u0084 = 10 5 10 15 20 25\u00E2\u0088\u00921\u00E2\u0088\u00920.500.511.5mmaxRe(\u00CE\u00BB)(c) \u00CF\u0084 = 20 5 10 15 20 25\u00E2\u0088\u00921\u00E2\u0088\u00920.500.511.5mmaxRe(\u00CE\u00BB)(d) \u00CF\u0084 = 30Figure 3.6: Numerical computation for b = 0 of the largest real part of theeigenvalue \u00CE\u00BB in (3.16) using Newton\u00E2\u0080\u0099s method on (3.59). The solid curves arewhere the largest eigenvalue has negative real part while the dashed curvesare where it has positive real part. In all experiments (2, q, o, s) = (2, 1, 2, 0),\u000F = 0.025, D = 1, R = 1, and r0 = 0.5.When \u00CF\u0084 \u001C 1 as in Figure 3.6a then the plot is almost indistinguishable fromthe linear case of \u00CF\u0084 = 0 (Figure 3.5a) as should be expected. This helps verifythat the Newton algorithm is working properly. In Figure 3.6c when \u00CF\u0084 = 2 wesee that at m = mb\u00E2\u0088\u0092 the zero eigenvalue is no longer the largest eigenvalue.In (3.39) we estimated that \u00CF\u0084 \u00E2\u0088\u0097mb\u00E2\u0088\u0092 = 1.8376 and in this case \u00CF\u0084 > \u00CF\u0084\u00E2\u0088\u0097mb\u00E2\u0088\u0092 . Figure3.7 verifies our predicted \u00CF\u0084 \u00E2\u0088\u0097mb\u00E2\u0088\u0092 value as in Figure 3.7a-3.7b \u00CF\u0084 = 1.82 < \u00CF\u0084\u00E2\u0088\u0097mb\u00E2\u0088\u00921113.2. Eigenvalues Associated with \u00CE\u00A60 Evenand we indeed still see at m = mb\u00E2\u0088\u0092 that \u00CE\u00BB = 0 as the largest eigenvalue. InFigure 3.7c-3.7d, \u00CF\u0084 = 1.84 > \u00CF\u0084 \u00E2\u0088\u0097mb\u00E2\u0088\u0092 and at m = mb\u00E2\u0088\u0092 there is an eigenvaluelarger than 0.0 5 10 15 20 25\u00E2\u0088\u00921\u00E2\u0088\u00920.500.511.5mmaxRe(\u00CE\u00BB)(a) \u00CF\u0084 = 1.820.3 0.35 0.4 0.45 0.5\u00E2\u0088\u00920.01\u00E2\u0088\u00920.008\u00E2\u0088\u00920.006\u00E2\u0088\u00920.004\u00E2\u0088\u00920.00200.0020.0040.0060.0080.01mmaxRe(\u00CE\u00BB)(b) \u00CF\u0084 = 1.82 (zoom)0 5 10 15 20 25\u00E2\u0088\u00921\u00E2\u0088\u00920.500.511.5mmaxRe(\u00CE\u00BB)(c) \u00CF\u0084 = 1.840.3 0.35 0.4 0.45 0.5\u00E2\u0088\u00920.01\u00E2\u0088\u00920.008\u00E2\u0088\u00920.006\u00E2\u0088\u00920.004\u00E2\u0088\u00920.00200.0020.0040.0060.0080.01mmaxRe(\u00CE\u00BB)(d) \u00CF\u0084 = 1.84 (zoom)Figure 3.7: Numerical computation of eigenvalues near m = mb\u00E2\u0088\u0092 for \u00CF\u0084 > \u00CF\u0084 \u00E2\u0088\u0097mb\u00E2\u0088\u0092and \u00CF\u0084 < \u00CF\u0084 \u00E2\u0088\u0097mb\u00E2\u0088\u0092 . In all experiments (2, q, o, s) = (2, 1, 2, 0), \u000F = 0.025, D = 1,R = 1, and r0 = 0.5.In Figure 3.7b it may appear that there is a second neutral eigenvalue nearm = mb\u00E2\u0088\u0092 as there is certainly another eigenvalue where Re(\u00CE\u00BB) = 0. However,as we mentioned in doing the analytical analysis, there should only be onetruly neutral stability point. To emphasize this, we plot the imaginary part1123.2. Eigenvalues Associated with \u00CE\u00A60 Evenof the eigenvalues for a range of \u00CF\u0084 in Figure 3.8. For \u00CF\u0084 = 1.83, we seethat indeed the second eigenvalue with Re(\u00CE\u00BB) = 0 has a non-zero imaginarypart and thus there is a Hopf bifurcation at this m value for this particularvalue of \u00CF\u0084 . We see that when \u00CF\u0084 = 1.84 > \u00CF\u0084 \u00E2\u0088\u0097mb\u00E2\u0088\u0092 and \u00CE\u00BB = 0 is no longerthe biggest eigenvalue that the eigenvalues on m < mb\u00E2\u0088\u0092 are complex andtherefore \u00CF\u0084 \u00E2\u0088\u0097m > 1.84. However, when \u00CF\u0084 = 30 we that all of the eigenvalue arepurely real so \u00CF\u0084 \u00E2\u0088\u0097m < 30.1133.2. Eigenvalues Associated with \u00CE\u00A60 Even0 0.2 0.4 0.6 0.8 1\u00E2\u0088\u00921\u00E2\u0088\u00920.500.511.5mIm(maxRe(\u00CE\u00BB))(a) \u00CF\u0084 = 0.40 0.2 0.4 0.6 0.8 1\u00E2\u0088\u00921\u00E2\u0088\u00920.500.511.5mIm(maxRe(\u00CE\u00BB))(b) \u00CF\u0084 = 10 0.2 0.4 0.6 0.8 1\u00E2\u0088\u00921\u00E2\u0088\u00920.500.511.5mIm(maxRe(\u00CE\u00BB))(c) \u00CF\u0084 = 1.820 0.2 0.4 0.6 0.8 1\u00E2\u0088\u00921\u00E2\u0088\u00920.500.511.5mIm(maxRe(\u00CE\u00BB))(d) \u00CF\u0084 = 1.840 0.2 0.4 0.6 0.8 1\u00E2\u0088\u00921\u00E2\u0088\u00920.500.511.5mIm(maxRe(\u00CE\u00BB))(e) \u00CF\u0084 = 30Figure 3.8: Imaginary part of the eigenvalues with largest real part computedusing Newton\u00E2\u0080\u0099s method on (3.59). The black x mark points where Re(\u00CE\u00BB) = 0.In all experiments (2, q, o, s) = (2, 1, 2, 0), \u000F = 0.025, D = 1, R = 1, andr0 = 0.5.1143.2. Eigenvalues Associated with \u00CE\u00A60 Even3.2.12 Adding SaturationWe now turn our attention to the case when there is saturation and b 6= 0.Fortunately, the Newton algorithm for (3.59) holds for any value of b andso we can easily consider this case numerically. One thing that does holdtrue analytically for b 6= 0 is the leading result on mb+ of section 3.2.4 whenm \u001D 1. There we have that the non-local part in (3.16) is negligible toleading order and therefore the only contribution with saturation is throughthe operator L0b (3.7). This operator is still classified by Lemma 3.1.0.1 andso there must still be a single positive eigenvalue with an eigenfunction thathas no nodal lines when a homoclinic orbit exists. If we let \u00CE\u00BD0(b) be thislargest discrete eigenvalue to b 6= 0 then the upper bound neutral stabilitywave mode is given by,mb+ \u00E2\u0088\u00BCr0\u00E2\u0088\u009A\u00CE\u00BD0(b)\u000F.In section 2.3 we have that the homoclinic orbit solution w(\u00CE\u00B7\u00CB\u0086) only exists upto a critical value b < bc with bc given by (2.27). When b = bc we saw thatw became a heteroclinic orbit and therefore in this instance w\u00E2\u0080\u00B2 is even andhas no nodal lines. Therefore when b = bc, we must have that the largesteigenvalue is the zero eigenvalue and thereforelimb\u00E2\u0086\u0092b\u00E2\u0088\u0092cmb+ = 0.We can compute the eigenvalues of L0b numerically using eigs which we plotin Figure 3.9 and we indeed notice that the largest eigenvalue goes to zeroas b approaches the critical value.1153.2. Eigenvalues Associated with \u00CE\u00A60 Even0 0.05 0.1 0.15 0.2 0.2500.20.40.60.811.21.4bmax\u00CE\u00BBFigure 3.9: Computation of b dependent eigenvalues to L0b given by (3.7).We demonstrate the spectrum of the NLEP (3.16) becoming entirely negativein Figure 3.10 where we take (2, q, o, s) = (2, 1, 2, 0), \u000F = 0.025, R = 1, r0 =0.5, and D = 1. In this figure, we have plotted the numerically computedeigenvalues to (3.16) for \u00CF\u0084 = 0 and various values of b. It can be seen thatby the time b = 0.2 all of the eigenvalues satisfy Re(\u00CE\u00BB) < 0.1163.2. Eigenvalues Associated with \u00CE\u00A60 Even0 5 10 15 20 25\u00E2\u0088\u00921\u00E2\u0088\u00920.500.511.5mmaxRe(\u00CE\u00BB) b =0b =0.01b =0.05b =0.1b =0.15b =0.2Figure 3.10: Computation of eigenvalues for b 6= 0 and \u00CF\u0084 = 0. In all cases(2, q, o, s) = (2, 1, 2, 0), \u000F = 0.025, R = 1, r0 = 0.5, and D = 1Finally, in Figure 3.11 we plot the eigenvalues for b = 0.2, (2, q, o, s) =(2, 1, 2, 0), \u000F = 0.025, R = 1, r0 = 0.5, and D = 1 for various values of \u00CF\u0084 .1173.2. Eigenvalues Associated with \u00CE\u00A60 Even0 5 10 15 20 25\u00E2\u0088\u00921\u00E2\u0088\u00920.8\u00E2\u0088\u00920.6\u00E2\u0088\u00920.4\u00E2\u0088\u00920.200.2mmaxRe(\u00CE\u00BB) \u00CF\u0084 =0.4\u00CF\u0084 =2\u00CF\u0084 =5\u00CF\u0084 =10\u00CF\u0084 =30\u00CF\u0084 =300\u00CF\u0084 =3000Figure 3.11: Computation of eigenvalues for b = 0.2 and \u00CF\u0084 6= 0. In all cases(2, q, o, s) = (2, 1, 2, 0), \u000F = 0.025, R = 1, r0 = 0.5, and D = 1Unsurprisingly, since we saw gm \u001D 1 for \u00CF\u0084 \u001D 1, we have that there is athreshold of \u00CF\u0084 where eigenvalues begin to enter Re(\u00CE\u00BB) > 0. However, unliker0, and b, \u00CF\u0084 is a static parameter and so for a fixed \u00CF\u0084 , there will exist a rangeof b for which all of the eigenvalues satisfy Re(\u00CE\u00BB) < 0. There is some delicacyrequired here since as we saw in section 2.3.1, b depends intimately on r0 andgenerally decreases as r0 increases. Therefore, in the dynamic transition ofcircle radii, bands of instability can arise. Fortunately due to the delay effectfrom (3.9), as long as these instability bands are transient, it is possible tostabilize everything. As is noted in Figure 2.7, the rate at which b decreasesas r0 increases gets slower as a function of saturation and so it is possible to1183.3. Eigenvalues Associated with \u00CE\u00A60 Oddfind a saturation value \u00CF\u0083\u00E2\u0088\u0097 such that for \u00CF\u0083 > \u00CF\u0083\u00E2\u0088\u0097, the spectrum of the NLEP(3.16) satisfies Re(\u00CE\u00BB) < 0 for most values of r0.3.3 Eigenvalues Associated with \u00CE\u00A60 OddWe will now consider the second class of eigenfunctions for which \u00CE\u00A60 is odd.Odd functions do not obtain their maximum values at r = r0 and so thisclass of functions will affect the position of the curve where maximal acti-vator occurs. Therefore, we consider odd eigenfunctions to correspond withtranslation or zig-zag instabilities. As we already discussed, when \u00CE\u00A60 is oddthen from (3.15), we have N(r0) = 0 to leading order, and so for \u00CE\u00A60 theleading order problem from (3.8) isL0b\u00CE\u00A60 =(\u00CE\u00BB+\u000F2m2r20)\u00CE\u00A60 = \u00C2\u00B5\u00CE\u00A60.For \u00CE\u00BB \u00E2\u0089\u00A5 0 then \u00C2\u00B5 \u00E2\u0089\u00A5 0 and from [14] and section 3.2.12, we have that the onlypossibility for \u00CE\u00A60 to be odd is if \u00C2\u00B5 = 0 and \u00CE\u00A60 = w\u00E2\u0080\u00B2 where w is the homoclinicorbit solution (2.22) and prime indicates differentiation with respect to \u00CF\u0081. Wecould once again analyze for m = O(1) and m\u001D O(1) but for the latter casenote that \u00C2\u00B5 = 0 implies\u00CE\u00BB = \u00E2\u0088\u0092\u000F2m2r20< 0and so there are no unstable roots for m\u001D 1 with an odd eigenfunction. Assuch we will strictly consider m = O(1) moving forward. If we consider theexpansion (3.5), the most natural scaling for the inhibitor function is \u000F and1193.3. Eigenvalues Associated with \u00CE\u00A60 Oddtherefore, to leading order,\u00CE\u00A60 \u00E2\u0088\u00BC w\u00E2\u0080\u00B2, \u00CE\u00BB = \u00CF\u0089(\u000F)\u00CE\u00BB\u00CB\u0086, N(r) = \u000FN\u00CB\u0086with \u00CF\u0089 \u001C 1 to be determined. We let N\u00CB\u009C(\u00CF\u0081) = N\u00CB\u0086(r0 + \u000F\u00CF\u0081) and make thefollowing expansions,\u00CE\u00A6 \u00E2\u0088\u00BC w\u00E2\u0080\u00B2 + \u000F\u00CE\u00A61 + \u000F2\u00CE\u00A62 + . . . , N\u00CB\u009C(\u00CF\u0081) \u00E2\u0088\u00BC N\u00CB\u009C0 + \u000FN\u00CB\u009C1 + . . . .We also consider the asymptotic expansion of the equilibrium solution fromsections 2.2 and 2.3,v\u00CB\u009Ce \u00E2\u0088\u00BC U q0w + \u000Fv\u00CB\u009C1 + \u000F2v\u00CB\u009C2 + . . . , u\u00CB\u009Ce \u00E2\u0088\u00BC U0 + \u000Fu\u00CB\u009C1 + \u000F2u\u00CB\u009C2 + . . . .We have already satisfied the leading order problem and so we consider theproblem at O(\u000F) for \u00CE\u00A61 (by expanding (3.5a)) which becomes,L0b\u00CE\u00A61 = \u00E2\u0088\u0092\u00CE\u00A60\u00CF\u0081r0\u00E2\u0088\u0092 \u00CE\u00A60\u00CF\u0081dr0dT\u00E2\u0088\u0092 a1\u00CE\u00A60 +qU q\u00E2\u0088\u009210(1 + bw2)w2N\u00CB\u009C0 + \u00CE\u00BB\u00CB\u0086\u00CE\u00A60, (3.60)where we have defineda1 \u00E2\u0089\u00A12U0(1 + bw2)2((1\u00E2\u0088\u0092 3bw2)U q\u00E2\u0088\u009210 (1 + bw2)v\u00CB\u009C1 \u00E2\u0088\u0092 qwu\u00CB\u009C1), (3.61)1203.3. Eigenvalues Associated with \u00CE\u00A60 Oddand, at least temporarily, consider \u00CF\u0089(\u000F) = \u000F. Next, consider the problem forv\u00CB\u009C1 and u\u00CB\u009C1 given by (2.7) which, in radial coordinates, isL0bv\u00CB\u009C1 = \u00E2\u0088\u0092U q0r0w\u00E2\u0080\u00B2 + qU q\u00E2\u0088\u009210w2(1 + bw2)u\u00CB\u009C1 \u00E2\u0088\u0092 U q0dr0dTw\u00E2\u0080\u00B2, (3.62a)u\u00CB\u009C\u00E2\u0080\u00B2\u00E2\u0080\u00B21 = \u00E2\u0088\u00921DU\u00CE\u00B20 wo. (3.62b)Upon differentiating (3.62a), we obtainL0bv\u00CB\u009C\u00E2\u0080\u00B21 + 2(1\u00E2\u0088\u0092 3bw2)(1 + bw2)3w\u00E2\u0080\u00B2v\u00CB\u009C1 =\u00E2\u0088\u0092U q0r0w\u00E2\u0080\u00B2\u00E2\u0080\u00B2 + qU q\u00E2\u0088\u0092102w(1 + bw2)2w\u00E2\u0080\u00B2u\u00CB\u009C1+ qU q\u00E2\u0088\u009210w2(1 + bw2)u\u00CB\u009C\u00E2\u0080\u00B21 \u00E2\u0088\u0092 U q0dr0dTw\u00E2\u0080\u00B2\u00E2\u0080\u00B2, (3.63)where we note that(L0by)\u00E2\u0080\u00B2 = L0by\u00E2\u0080\u00B2 + 2(1\u00E2\u0088\u0092 3bw2)(1 + bw2)3w\u00E2\u0080\u00B2y. (3.64)We can rearrange (3.63) to geta1w\u00E2\u0080\u00B2 = \u00E2\u0088\u0092 1U q0L0bv\u00CB\u009C\u00E2\u0080\u00B21 \u00E2\u0088\u0092w\u00E2\u0080\u00B2\u00E2\u0080\u00B2r0+qU0w2(1 + bw2)u\u00CB\u009C\u00E2\u0080\u00B21 \u00E2\u0088\u0092dr0dTw\u00E2\u0080\u00B2\u00E2\u0080\u00B2, (3.65)so that (3.60) becomesL0b\u00CE\u00A61 =1U q0L0bv\u00CB\u009C\u00E2\u0080\u00B21 +qU0w2(1 + bw2)(U q0 N\u00CB\u009C0 \u00E2\u0088\u0092 u\u00CB\u009C\u00E2\u0080\u00B21) + \u00CE\u00BB\u00CB\u0086w\u00E2\u0080\u00B2, (3.66)1213.3. Eigenvalues Associated with \u00CE\u00A60 Oddwhere we have used that \u00CE\u00A60 = w\u00E2\u0080\u00B2. We know \u00CE\u00A61 = w\u00E2\u0080\u00B2 is a homogeneous solu-tion to (3.66) and so we must satisfy the following compatibility condition:1U q0\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew\u00E2\u0080\u00B2L0bv\u00CB\u009C\u00E2\u0080\u00B21 d\u00CF\u0081+qU0\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew2(1 + bw2)(U q0 N\u00CB\u009C0 \u00E2\u0088\u0092 u\u00CB\u009C\u00E2\u0080\u00B21)w\u00E2\u0080\u00B2 d\u00CF\u0081+ \u00CE\u00BB\u00CB\u0086\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew\u00E2\u0080\u00B22 d\u00CF\u0081 = 0. (3.67)Now since L0b is self-adjoint and w\u00E2\u0080\u00B2 a homogeneous solution then,\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew\u00E2\u0080\u00B2L0bv\u00CB\u009C\u00E2\u0080\u00B21 d\u00CF\u0081 =\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ev\u00CB\u009C1L0bw\u00E2\u0080\u00B2 d\u00CF\u0081 = 0. (3.68)In this way, (3.67) reduces to\u00E2\u0088\u0092 qU0\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew2(1 + bw2)(U q0 N\u00CB\u009C0 \u00E2\u0088\u0092 u\u00CB\u009C\u00E2\u0080\u00B21)w\u00E2\u0080\u00B2 d\u00CF\u0081 = \u00CE\u00BB\u00CB\u0086\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew\u00E2\u0080\u00B22 d\u00CF\u0081. (3.69)We now wish to understand the behaviour of the function F\u00CB\u009C defined by,F\u00CB\u009C0 \u00E2\u0089\u00A1 U q0 N\u00CB\u009C0 \u00E2\u0088\u0092 u\u00CB\u009C\u00E2\u0080\u00B21,which is easier to obtain if we differentiate,F\u00CB\u009C \u00E2\u0080\u00B20 = U q0 N\u00CB\u009C \u00E2\u0080\u00B20 \u00E2\u0088\u0092 u\u00CB\u009C\u00E2\u0080\u00B2\u00E2\u0080\u00B21.First from (3.62b) we have thatu\u00CB\u009C\u00E2\u0080\u00B2\u00E2\u0080\u00B21 = \u00E2\u0088\u0092U\u00CE\u00B20Dwo,1223.3. Eigenvalues Associated with \u00CE\u00A60 Oddand for N\u00CB\u009C0 we turn to the expansion (3.5b) in inner coordinates:\u000F4\u00CF\u0084N\u00CB\u009CT + \u000F2\u00CF\u0084N\u00CB\u009C\u00CF\u0089(\u000F)\u00CE\u00BB\u00CB\u0086 =D(r0 + \u000F\u00CF\u0081)((r0 + \u000F\u00CF\u0081)N\u00CB\u009C\u00E2\u0080\u00B2)\u00E2\u0080\u00B2 \u00E2\u0088\u0092 \u000F2 Dm2(r0 + \u000F\u00CF\u0081)2N\u00CB\u009C \u00E2\u0088\u0092 \u000F2N\u00CB\u009C+ov\u00CB\u009Co\u00E2\u0088\u00921eu\u00CB\u009Ces\u00CE\u00A6\u00E2\u0088\u0092 \u000F sv\u00CB\u009Coeu\u00CB\u009Cs+1eN\u00CB\u009C , (3.70)which to leading order satisfies,N\u00CB\u009C \u00E2\u0080\u00B2\u00E2\u0080\u00B20 = \u00E2\u0088\u0092oU\u00CE\u00B2\u00E2\u0088\u0092q0Dwo\u00E2\u0088\u00921w\u00E2\u0080\u00B2.If we multiply this by U q0 and integrate we haveU q0 N\u00CB\u009C\u00E2\u0080\u00B20 = \u00E2\u0088\u0092U\u00CE\u00B20Dwo,and therefore F\u00CB\u009C \u00E2\u0080\u00B20 = 0 so F\u00CB\u009C0 is a constant. We already know that the derivativeof ue in the global region is discontinuous and therefore F\u00CB\u009C0 being constantimplies that N\u00CB\u0086 must be discontinuous as well but in such a way that theappropriate sum is continuous and F\u00CB\u009C0 is defined. Having F\u00CB\u009C0 be constantmakes the integrand on the left-hand side of (3.69) an odd function and sothe integral vanishes. Therefore we conclude that \u00CF\u0089(\u000F)\u001C \u000F. Next we consider\u00CF\u0089(\u000F) = \u000F2 and by expanding (3.5a) to the proper order we get the problemfor \u00CE\u00A62:L0b\u00CE\u00A62 =qUq\u00E2\u0088\u009210w2(1 + bw2)N\u00CB\u009C1 + a\u00C2\u00AF2Uq0 N\u00CB\u009C0 \u00E2\u0088\u0092 a1\u00CE\u00A61 \u00E2\u0088\u0092 a2w\u00E2\u0080\u00B2 \u00E2\u0088\u0092\u00CE\u00A6\u00E2\u0080\u00B21r0+\u00CF\u0081w\u00E2\u0080\u00B2\u00E2\u0080\u00B2r20\u00E2\u0088\u0092 dr01dTw\u00E2\u0080\u00B2\u00E2\u0080\u00B2 \u00E2\u0088\u0092 dr00dT\u00CE\u00A6\u00E2\u0080\u00B21 +(\u00CE\u00BB\u00CB\u0086+m2r20)w\u00E2\u0080\u00B2, (3.71)1233.3. Eigenvalues Associated with \u00CE\u00A60 Oddwhere we define a2 and a\u00C2\u00AF2 by,a2 \u00E2\u0089\u00A12U0(1 + bw2)2((1\u00E2\u0088\u0092 3bw2)U q\u00E2\u0088\u009210 (1 + bw2)v\u00CB\u009C2 \u00E2\u0088\u0092 qwu\u00CB\u009C2)+q(q + 1)U20w(1 + bw2)2u\u00CB\u009C21\u00E2\u0088\u0092 2qU q+10(1\u00E2\u0088\u0092 3bw2)(1 + bw2)3u\u00CB\u009C1v\u00CB\u009C1 \u00E2\u0088\u009212bwU2q0(1\u00E2\u0088\u0092 bw2)(1 + bw2)4v\u00CB\u009C21 (3.72a)a\u00C2\u00AF2 \u00E2\u0089\u00A12qU q+10w(1 + bw2)2v\u00CB\u009C1 \u00E2\u0088\u0092q(q + 1)U20w2(1 + bw2)u\u00CB\u009C1. (3.72b)In (3.71) we replace r0 = r00 + \u000Fr01 where r00 is the value previously usedfor r0 (i.e. it satisfies (2.43)) and r01 is added as a correction to satisfy anorthogonality condition of the base-state at O(\u000F2). It may seem erroneousto introduce a radial correction, r01, without any consideration to matchingconditions from previous quasi steady-state analysis. However, for the outerproblem, we are interested in the singular limit solution where all variables areO(1) and matching terms generated by radial corrections furnish conditionssmaller than this order. Therefore, whenever global matching is concerned,we will consider r0 + \u000F\u00CF\u0081 \u00E2\u0089\u0088 r00 + \u000F\u00CF\u0081. As we did with the O(\u000F) problem, weconsider the problem for v\u00CB\u009C2 and u\u00CB\u009C2 given by (2.15a) and (2.15b) respectivelywhich, in radial coordinates, isL0bv\u00CB\u009C2 =\u00E2\u0088\u0092v\u00CB\u009C\u00E2\u0080\u00B21r00+\u00CF\u0081U q0r200w\u00E2\u0080\u00B2 \u00E2\u0088\u0092 dr00dTv\u00CB\u009C\u00E2\u0080\u00B21 \u00E2\u0088\u0092dr01dTU q0w\u00E2\u0080\u00B2 \u00E2\u0088\u0092 Uq\u00E2\u0088\u009220 q(q + 1)2w2(1 + bw2)u\u00CB\u009C21+2qU0w(1 + bw2)2u\u00CB\u009C1v\u00CB\u009C1 \u00E2\u0088\u00921U q0(1\u00E2\u0088\u0092 3bw2)(1 + bw2)3v\u00CB\u009C21 + Uq\u00E2\u0088\u009210 qw2(1 + bw2)u\u00CB\u009C2,(3.73a)u\u00CB\u009C\u00E2\u0080\u00B2\u00E2\u0080\u00B22 =U0D\u00E2\u0088\u0092 u\u00CB\u009C\u00E2\u0080\u00B21r00+1DU\u00CE\u00B2\u00E2\u0088\u009210 swou\u00CB\u009C1 \u00E2\u0088\u00921DU\u00CE\u00B2\u00E2\u0088\u0092q0 owo\u00E2\u0088\u00921v\u00CB\u009C1. (3.73b)1243.3. Eigenvalues Associated with \u00CE\u00A60 OddAs before, we differentiate (3.73a) and use (3.64) to geta2w\u00E2\u0080\u00B2 = \u00E2\u0088\u0092 1U q0L0bv\u00CB\u009C\u00E2\u0080\u00B22 \u00E2\u0088\u0092v\u00CB\u009C\u00E2\u0080\u00B2\u00E2\u0080\u00B21U q0r00+w\u00E2\u0080\u00B2r200+\u00CF\u0081w\u00E2\u0080\u00B2\u00E2\u0080\u00B2r200\u00E2\u0088\u0092 1U q0dr00dTv\u00CB\u009C\u00E2\u0080\u00B2\u00E2\u0080\u00B21\u00E2\u0088\u0092 dr01dTw\u00E2\u0080\u00B2\u00E2\u0080\u00B2 + a\u00C2\u00AF2u\u00CB\u009C\u00E2\u0080\u00B21 \u00E2\u0088\u0092a1U q0v\u00CB\u009C\u00E2\u0080\u00B21 +qU0w2(1 + bw2)u\u00CB\u009C\u00E2\u0080\u00B22. (3.74)Upon substituting (3.74) into (3.71) and simplifying we getL0b\u00CE\u00A62 =qU0w2(1 + bw2)F\u00CB\u009C1 + a\u00C2\u00AF2F\u00CB\u009C0 \u00E2\u0088\u0092 a1\u00CE\u00A61 +1U q0L0bv\u00CB\u009C\u00E2\u0080\u00B22 +v\u00CB\u009C\u00E2\u0080\u00B2\u00E2\u0080\u00B21U q0r00\u00E2\u0088\u0092 w\u00E2\u0080\u00B2r200+1U q0dr00dTv\u00CB\u009C\u00E2\u0080\u00B2\u00E2\u0080\u00B21 +a1U q0v\u00CB\u009C\u00E2\u0080\u00B21 \u00E2\u0088\u0092\u00CE\u00A6\u00E2\u0080\u00B21r00\u00E2\u0088\u0092 dr00dT\u00CE\u00A6\u00E2\u0080\u00B21 +(\u00CE\u00BB\u00CB\u0086+m2r200)w\u00E2\u0080\u00B2, (3.75)where we define,F\u00CB\u009C1 = U q0 N\u00CB\u009C1 \u00E2\u0088\u0092 u\u00CB\u009C\u00E2\u0080\u00B22. (3.76)To simplify further we return to the problem for \u00CE\u00A61 given by (3.66), whichwe can simplify asL0b(\u00CE\u00A61 \u00E2\u0088\u00921U q0v\u00CB\u009C\u00E2\u0080\u00B21)=qF\u00CB\u009C0U0w2(1 + bw2). (3.77)Now, considering the even eigenfunction problem discussed in section 3.2,when \u00C2\u00B5 = 0 there is a function \u00CF\u0088 in (3.18) which satisfiesL0b\u00CF\u0088 =w2(1 + bw2), (3.78)1253.3. Eigenvalues Associated with \u00CE\u00A60 Oddand so we can write (3.77) asL0b(\u00CE\u00A61 \u00E2\u0088\u00921U q0v\u00CB\u009C\u00E2\u0080\u00B21 \u00E2\u0088\u0092qF\u00CB\u009C0U0\u00CF\u0088)= 0.Therefore, the function being operated on must be some multiple of the nulleigenfunction w\u00E2\u0080\u00B2 which we can take to be zero without loss of generality, andwe have that,\u00CE\u00A61 =1U q0v\u00CB\u009C\u00E2\u0080\u00B21 +qF\u00CB\u009C0U0\u00CF\u0088. (3.79)We can actually determine \u00CF\u0088 analytically by first noticing that when b = 0,we can directly verify that \u00CF\u0088 = w. With this in mind, we consider how theL0b operator acts on w even with saturation:L0bw = w\u00E2\u0080\u00B2\u00E2\u0080\u00B2 \u00E2\u0088\u0092 w + 2 w2(1 + bw2)2=w2(1 + bw2)2\u00E2\u0088\u0092 b w4(1 + bw2)2(3.80)where we have used (2.22) to simplify the second derivative term. We can geta better understanding of the last term in this expression by differentiatingthe homoclinic orbit problem (2.22) with respect to the saturation parameterb,w\u00E2\u0080\u00B2\u00E2\u0080\u00B2b \u00E2\u0088\u0092 wb + 2wwb(1 + bw2)2\u00E2\u0088\u0092 w4(1 + bw2)2= 0which impliesL0bwb =w4(1 + bw2)21263.3. Eigenvalues Associated with \u00CE\u00A60 Oddand this is precisely the last term in (3.80). With this in mind we attempt asolution for \u00CF\u0088 of the form,\u00CF\u0088 = w + cwb,with c to be determined. Substituting this into (3.78) we have,w2(1 + bw2)= L0b(w + cwb) =w2(1 + bw2)2+ (c\u00E2\u0088\u0092 b) w4(1 + bw2)2=w2(1 + bw2)(1 + (c\u00E2\u0088\u0092 b)w2)(1 + bw2)which will hold if c = 2b and thus we have\u00CF\u0088 = w + 2bwb. (3.81)Substituting (3.79) into (3.75) we getL0b\u00CE\u00A62 =qU0w2(1 + bw2)F\u00CB\u009C1 +(a\u00C2\u00AF2 \u00E2\u0088\u0092qU0(a1\u00CF\u0088 +1r00\u00CF\u0088\u00E2\u0080\u00B2 +dr00dT\u00CF\u0088\u00E2\u0080\u00B2))F\u00CB\u009C0+1U q0L0bv\u00CB\u009C\u00E2\u0080\u00B22 \u00E2\u0088\u0092w\u00E2\u0080\u00B2r200+(\u00CE\u00BB\u00CB\u0086+m2r200)w\u00E2\u0080\u00B2.1273.3. Eigenvalues Associated with \u00CE\u00A60 OddOnce again, \u00CE\u00A62 = w\u00E2\u0080\u00B2 is a homogeneous solution to this and so we need tosatisfy the solvability condition,qU0\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew2(1 + bw2)w\u00E2\u0080\u00B2F\u00CB\u009C1 d\u00CF\u0081\u00EF\u00B8\u00B8 \u00EF\u00B8\u00B7\u00EF\u00B8\u00B7 \u00EF\u00B8\u00B8I1+ F\u00CB\u009C0\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ea\u00C2\u00AF2w\u00E2\u0080\u00B2 \u00E2\u0088\u0092 qU0(a1\u00CF\u0088w\u00E2\u0080\u00B2 +1r00\u00CF\u0088\u00E2\u0080\u00B2w\u00E2\u0080\u00B2 +dr00dT\u00CF\u0088\u00E2\u0080\u00B2w\u00E2\u0080\u00B2)d\u00CF\u0081\u00EF\u00B8\u00B8 \u00EF\u00B8\u00B7\u00EF\u00B8\u00B7 \u00EF\u00B8\u00B8I2+(\u00CE\u00BB\u00CB\u0086+m2 \u00E2\u0088\u0092 1r200)\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew\u00E2\u0080\u00B22 d\u00CF\u0081 = 0, (3.82)where we have once again used (3.68) to remove the L0bv\u00CB\u009C\u00E2\u0080\u00B22 term. First considerthe integral I2 in (3.82). We can use (3.65) to write,\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ea1\u00CF\u0088w\u00E2\u0080\u00B2 d\u00CF\u0081 =\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009E\u00E2\u0088\u0092 1U q0\u00CF\u0088L0bv\u00CB\u009C\u00E2\u0080\u00B21 \u00E2\u0088\u0092w\u00E2\u0080\u00B2\u00E2\u0080\u00B2\u00CF\u0088r00+qU0\u00CF\u0088u\u00CB\u009C\u00E2\u0080\u00B21w2(1 + bw2)\u00E2\u0088\u0092 dr00dTw\u00E2\u0080\u00B2\u00E2\u0080\u00B2\u00CF\u0088 d\u00CF\u0081=\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009E\u00E2\u0088\u0092 1U q0\u00CF\u0088L0bv\u00CB\u009C\u00E2\u0080\u00B21 +w\u00E2\u0080\u00B2\u00CF\u0088\u00E2\u0080\u00B2r00+qU0\u00CF\u0088u\u00CB\u009C\u00E2\u0080\u00B21w2(1 + bw2)+dr00dTw\u00E2\u0080\u00B2\u00CF\u0088\u00E2\u0080\u00B2 d\u00CF\u0081,(3.83)where we have integrated by parts on the second and last term. We also have\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ea\u00C2\u00AF2w\u00E2\u0080\u00B2 d\u00CF\u0081 =2qU q+10\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Eww\u00E2\u0080\u00B2(1 + bw2)2v\u00CB\u009C1 d\u00CF\u0081\u00E2\u0088\u0092q(q + 1)U20\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew2w\u00E2\u0080\u00B2(1 + bw2)u\u00CB\u009C1 d\u00CF\u0081=qU q+10\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ev\u00CB\u009C1(L0b\u00CF\u0088)\u00E2\u0080\u00B2 d\u00CF\u0081\u00E2\u0088\u0092 q(q + 1)U20\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009EW \u00E2\u0080\u00B2u\u00CB\u009C1 d\u00CF\u0081,1283.3. Eigenvalues Associated with \u00CE\u00A60 Oddwhere we have differentiated (3.18) to get the simplification in the first inte-gral and W is given by (2.29). Integrating each term by parts\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ea\u00C2\u00AF2w\u00E2\u0080\u00B2 d\u00CF\u0081 = \u00E2\u0088\u0092 qU q+10\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ev\u00CB\u009C\u00E2\u0080\u00B21L0b\u00CF\u0088 d\u00CF\u0081+q(q + 1)2U20H\u00CB\u0086\u00E2\u008C\u00A9duedr\u00E2\u008C\u00AAr00\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew\u00E2\u0080\u00B22 d\u00CF\u0081where the last term was simplified using (2.12), (2.18), and that\u00E2\u0088\u00AB\u00E2\u0088\u009E0 W d\u00CF\u0081\u00E2\u0088\u00AB\u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew\u00E2\u0080\u00B22 d\u00CF\u0081=H\u00CB\u00864. (3.84)Using the self-adjoint property of L0b we can finalize once more to obtain\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ea\u00C2\u00AF2w\u00E2\u0080\u00B2 d\u00CF\u0081 = \u00E2\u0088\u0092 qU q+10\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009E\u00CF\u0088L0bv\u00CB\u009C\u00E2\u0080\u00B21 d\u00CF\u0081+q(q + 1)2U20H\u00CB\u0086\u00E2\u008C\u00A9duedr\u00E2\u008C\u00AAr00\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew\u00E2\u0080\u00B22 d\u00CF\u0081.(3.85)Combining (3.83) and (3.85) we can simplify I2 in (3.82) asI2 =q(q + 1)2U20H\u00CB\u0086\u00E2\u008C\u00A9duedr\u00E2\u008C\u00AAr00\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew\u00E2\u0080\u00B22 d\u00CF\u0081\u00E2\u0088\u0092 q2U20\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009E\u00CF\u0088u\u00CB\u009C\u00E2\u0080\u00B21w2(1 + bw2)d\u00CF\u0081\u00E2\u0088\u0092 2qU0(1r00+dr00dT)\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew\u00E2\u0080\u00B2\u00CF\u0088\u00E2\u0080\u00B2 d\u00CF\u0081.Using (2.34e) we can re-write the last term and we haveI2 =q(q + 1)2U20H\u00CB\u0086\u00E2\u008C\u00A9duedr\u00E2\u008C\u00AAr00\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew\u00E2\u0080\u00B22 d\u00CF\u0081\u00E2\u0088\u0092 q2U20\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009E\u00CF\u0088u\u00CB\u009C\u00E2\u0080\u00B21w2(1 + bw2)d\u00CF\u0081+q2U20H\u00CB\u0086\u00E2\u008C\u00A9duedr\u00E2\u008C\u00AAr00\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew\u00E2\u0080\u00B2\u00CF\u0088\u00E2\u0080\u00B2 d\u00CF\u0081.1293.3. Eigenvalues Associated with \u00CE\u00A60 OddFinally we substitute \u00CF\u0088 from (3.81) to getI2 =q2U20H\u00CB\u0086\u00E2\u008C\u00A9duedr\u00E2\u008C\u00AAr00\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew\u00E2\u0080\u00B22 d\u00CF\u0081+3q22U20H\u00CB\u0086\u00E2\u008C\u00A9duedr\u00E2\u008C\u00AAr00\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew\u00E2\u0080\u00B22 d\u00CF\u0081+q2U20\u00EF\u00A3\u00AB\u00EF\u00A3\u00AC\u00EF\u00A3\u00AC\u00EF\u00A3\u00AC\u00EF\u00A3\u00AD2bH\u00CB\u0086\u00E2\u008C\u00A9duedr\u00E2\u008C\u00AA\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew\u00E2\u0080\u00B2w\u00E2\u0080\u00B2b d\u00CF\u0081\u00E2\u0088\u0092\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew3(1 + bw2)u\u00CB\u009C\u00E2\u0080\u00B21 d\u00CF\u0081\u00EF\u00B8\u00B8 \u00EF\u00B8\u00B7\u00EF\u00B8\u00B7 \u00EF\u00B8\u00B8I3\u00E2\u0088\u00922b\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew2wb(1 + bw2)u\u00CB\u009C\u00E2\u0080\u00B21 d\u00CF\u0081\u00EF\u00B8\u00B8 \u00EF\u00B8\u00B7\u00EF\u00B8\u00B7 \u00EF\u00B8\u00B8I4\u00EF\u00A3\u00B6\u00EF\u00A3\u00B7\u00EF\u00A3\u00B7\u00EF\u00A3\u00B7\u00EF\u00A3\u00B8.Integrate I4 by parts to getI4 = 2\u00E2\u0088\u00AB \u00E2\u0088\u009E0w2wb(1 + bw2)d\u00CF\u0081\u00E2\u008C\u00A9duedr\u00E2\u008C\u00AAr00\u00E2\u0088\u0092\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009E(\u00E2\u0088\u00AB \u00CF\u00810w2wb(1 + bw2)dx)u\u00CB\u009C\u00E2\u0080\u00B2\u00E2\u0080\u00B21 d\u00CF\u0081,but the last integral vanishes since u\u00CB\u009C\u00E2\u0080\u00B2\u00E2\u0080\u00B21 is an even function. For the integralI3, we also integrate by parts to split the integral,I3 = \u00E2\u0088\u00923\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew2w\u00E2\u0080\u00B2(1 + bw2)u\u00CB\u009C1 d\u00CF\u0081+ 2b\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew4w\u00E2\u0080\u00B2(1 + bw2)2u\u00CB\u009C1 d\u00CF\u0081,but then integrate by parts on each separate integral again, twice, transfer-ring the derivative back to the u\u00CB\u009C1 term. After doing this, and once againremoving integrals involving u\u00CB\u009C\u00E2\u0080\u00B2\u00E2\u0080\u00B21, we getI3 =(6\u00E2\u0088\u00AB \u00E2\u0088\u009E0W d\u00CF\u0081\u00E2\u0088\u0092 4b\u00E2\u0088\u00AB \u00E2\u0088\u009E0\u00E2\u0088\u00AB w0v4(1 + bv2)2dv d\u00CF\u0081)\u00E2\u008C\u00A9duedr\u00E2\u008C\u00AAr00=(32H\u00CB\u0086\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew\u00E2\u0080\u00B22 d\u00CF\u0081\u00E2\u0088\u0092 4b\u00E2\u0088\u00AB \u00E2\u0088\u009E0\u00E2\u0088\u00AB w0v4(1 + bv2)2dv d\u00CF\u0081)\u00E2\u008C\u00A9duedr\u00E2\u008C\u00AAr00,1303.3. Eigenvalues Associated with \u00CE\u00A60 Oddwhere we have used (3.84) to simplify. Transferring everything together backinto I2 we haveI2 =q2U20H\u00CB\u0086\u00E2\u008C\u00A9duedr\u00E2\u008C\u00AAr00\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew\u00E2\u0080\u00B22 d\u00CF\u0081\u00E2\u0088\u0092 bq2U20\u00E2\u008C\u00A9duedr\u00E2\u008C\u00AAr00(4\u00E2\u0088\u00AB \u00E2\u0088\u009E0w2wb(1 + bw2)d\u00CF\u0081+4\u00E2\u0088\u00AB \u00E2\u0088\u009E0\u00E2\u0088\u00AB w0v4(1 + bv2)2dv d\u00CF\u0081\u00E2\u0088\u0092 2H\u00CB\u0086\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew\u00E2\u0080\u00B2w\u00E2\u0080\u00B2b d\u00CF\u0081).We can actually simplify this if we differentiate H\u00CB\u0086 in (3.84) with respect to band notice that this is exactly the last term in the expression. Therefore wehave,I2 =q2U20H\u00CB\u0086\u00E2\u008C\u00A9duedr\u00E2\u008C\u00AAr00\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew\u00E2\u0080\u00B22 d\u00CF\u0081(1\u00E2\u0088\u0092 2bqH\u00CB\u0086dH\u00CB\u0086db). (3.86)Finally, we consider I1 in (3.82) which we integrate by parts to getI1 =\u00E2\u0088\u0092H\u00CB\u00864\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew\u00E2\u0080\u00B22 d\u00CF\u0081(F\u00CB\u009C \u00E2\u0080\u00B21(\u00E2\u0088\u009E) + F\u00CB\u009C \u00E2\u0080\u00B21(\u00E2\u0088\u0092\u00E2\u0088\u009E))+\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009E\u00E2\u0088\u00AB \u00CF\u00810(W(x) dx) F\u00CB\u009C \u00E2\u0080\u00B2\u00E2\u0080\u00B21 d\u00CF\u0081.(3.87)To simplify this we need to understand the behaviour of F\u00CB\u009C1 given by (3.76)and specifically, its second derivative,F\u00CB\u009C \u00E2\u0080\u00B2\u00E2\u0080\u00B21 = U q0 N\u00CB\u009C \u00E2\u0080\u00B2\u00E2\u0080\u00B21 \u00E2\u0088\u0092 u\u00CB\u009C\u00E2\u0080\u00B2\u00E2\u0080\u00B2\u00E2\u0080\u00B22 .1313.3. Eigenvalues Associated with \u00CE\u00A60 OddTherefore, we need to consider the problem for N\u00CB\u009C1 which we obtain by ex-panding (3.70) to O(\u000F2),N\u00CB\u009C \u00E2\u0080\u00B2\u00E2\u0080\u00B21 =\u00E2\u0088\u0092N\u00CB\u009C \u00E2\u0080\u00B20r00+sU\u00CE\u00B2\u00E2\u0088\u009210DwoN\u00CB\u009C0 \u00E2\u0088\u0092oU\u00CE\u00B2\u00E2\u0088\u0092q0Dwo\u00E2\u0088\u00921(v\u00CB\u009C\u00E2\u0080\u00B21U q0+qF\u00CB\u009C0U0\u00CF\u0088)+osU\u00CE\u00B2\u00E2\u0088\u0092q\u00E2\u0088\u009210Dwo\u00E2\u0088\u00921w\u00E2\u0080\u00B2u\u00CB\u009C1 \u00E2\u0088\u0092o(o\u00E2\u0088\u0092 1)U\u00CE\u00B2\u00E2\u0088\u00922q0Dwo\u00E2\u0088\u00922w\u00E2\u0080\u00B2v\u00CB\u009C1. (3.88)If we multiply this by U q0 and subtract u\u00CB\u009C\u00E2\u0080\u00B2\u00E2\u0080\u00B2\u00E2\u0080\u00B22 by differentiating (3.73b) we getF\u00CB\u009C \u00E2\u0080\u00B2\u00E2\u0080\u00B21 =U\u00CE\u00B2\u00E2\u0088\u009210Dwo\u00E2\u0088\u00921F\u00CB\u009C0 (sw \u00E2\u0088\u0092 o\u00CF\u0088) (3.89)which is even and so, sinceW is also even, the integrand in the second integralin (3.87), is odd and hence, the integral vanishes. Rather than attempt anduse F\u00CB\u009C \u00E2\u0080\u00B21 to simplify I1 we will instead consider the global F problem and relatethem viaF(r00 + \u000F\u00CF\u0081) \u00E2\u0088\u00BC F(r00) + \u000F\u00CF\u0081dFdr\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3r00+ \u00C2\u00B7 \u00C2\u00B7 \u00C2\u00B7 = F\u00CB\u009C0 + \u000FF\u00CB\u009C1, (3.90)where we note that sided limits apply when functions are not continuous.Using (3.90), we can write (3.87) asI1 =\u00E2\u0088\u0092H\u00CB\u00862\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew\u00E2\u0080\u00B22 d\u00CF\u0081(U q0\u00E2\u008C\u00A9dN\u00CB\u0086dr\u00E2\u008C\u00AAr00\u00E2\u0088\u0092\u00E2\u008C\u00A9d2uedr2\u00E2\u008C\u00AAr00). (3.91)Finally, returning to (3.82) and substituting I1 and I2, we see that everythingdoes not vanish, and so we require eigenvalues at this order to satisfy theorthogonality condition. Hence \u00CF\u0089(\u000F) = \u000F2 and the eigenvalues, to leading1323.3. Eigenvalues Associated with \u00CE\u00A60 Oddorder, are given by\u00CE\u00BB\u00CB\u0086 =1\u00E2\u0088\u0092m2r200\u00E2\u0088\u0092 q2U20H\u00CB\u0086(1\u00E2\u0088\u0092 2bqH\u00CB\u0086dH\u00CB\u0086db)\u00E2\u008C\u00A9duedr\u00E2\u008C\u00AAr00F(r00)+qH\u00CB\u00862U0(U q0\u00E2\u008C\u00A9dN\u00CB\u0086dr\u00E2\u008C\u00AAr00\u00E2\u0088\u0092\u00E2\u008C\u00A9d2uedr2\u00E2\u008C\u00AAr00). (3.92)The fact that the eigenvalues exist to leading order at O(\u000F2) is why we canclassify them as long-time instabilities because even positive values will onlybegin to cause relative instability in (3.9) when T = O(1) which correspondsto regular time t = O(\u000F\u00E2\u0088\u00922). Since N\u00CB\u0086 is not continuous, but F(r00) = F\u00CB\u009C0is, we will define F(r00) by the average value as we did when discussing theperturbed circle problem in section (2.3.2) and therefore,F(r00) = U q0\u00E2\u008C\u00A9N\u00CB\u0086\u00E2\u008C\u00AAr00\u00E2\u0088\u0092\u00E2\u008C\u00A9duedr\u00E2\u008C\u00AAr00.We now need to consider the global problem for N\u00CB\u0086 .3.3.1 Global Inhibitor Eigenvalue ProblemSubstitute N = \u000FN\u00CB\u0086 into (3.6b) and simplify to get the base problem for N\u00CB\u0086 is1r(rN\u00CB\u0086r)r \u00E2\u0088\u0092m2r2N\u00CB\u0086 \u00E2\u0088\u0092 1DN\u00CB\u0086 +1\u000F2oDvo\u00E2\u0088\u00921euse\u00CE\u00A6(r \u00E2\u0088\u0092 r0\u000F)\u00E2\u0088\u0092 1\u000FsDvoeus+1eN\u00CB\u0086\u00EF\u00B8\u00B8 \u00EF\u00B8\u00B7\u00EF\u00B8\u00B7 \u00EF\u00B8\u00B8S1= 0, (3.93)where we do not use \u00CE\u00B8\u00CE\u00BB here as we did for even eigenfunctions because \u00CE\u00BB =O(\u000F2). Focusing solely on the singular terms S1 and expanding to all singular1333.3. Eigenvalues Associated with \u00CE\u00A60 Oddpowers of \u000F, we have,S1 =oU\u00CE\u00B2\u00E2\u0088\u0092q0D\u000F2wo\u00E2\u0088\u00921(\u00C2\u00B7)w\u00E2\u0080\u00B2(\u00C2\u00B7) + oU\u00CE\u00B2\u00E2\u0088\u0092q0D\u000Fwo\u00E2\u0088\u00921(\u00C2\u00B7)\u00CE\u00A61(\u00C2\u00B7)\u00E2\u0088\u0092osU\u00CE\u00B2\u00E2\u0088\u0092q\u00E2\u0088\u009210D\u000Fwo\u00E2\u0088\u00921(\u00C2\u00B7)u\u00CB\u009C1(\u00C2\u00B7)w\u00E2\u0080\u00B2(\u00C2\u00B7)+o(o\u00E2\u0088\u0092 1)U\u00CE\u00B2\u00E2\u0088\u00922q0D\u000Fwo\u00E2\u0088\u00922(\u00C2\u00B7)v\u00CB\u009C1(\u00C2\u00B7)w\u00E2\u0080\u00B2(\u00C2\u00B7)\u00E2\u0088\u0092sU\u00CE\u00B2\u00E2\u0088\u009210D\u000Fwo(\u00C2\u00B7)N\u00CB\u009C0(\u00C2\u00B7), (3.94)where the dot indicates evaluation at (r \u00E2\u0088\u0092 r0)\u000F\u00E2\u0088\u00921, i.e. the inner functionsevaluated in the outer region. In a manner similar to that as we discussed insection 3.1, when deriving the singularity behaviour for the even eigenfunc-tions functions, we havewo( r\u00E2\u0088\u0092r0\u000F)\u000F=\u000F\u00E2\u0086\u00920A\u00CE\u00B4(r \u00E2\u0088\u0092 r0),where \u00CE\u00B4 is the Dirac measure and A is again given by (2.33). Differentiatingthis expression we geto\u000F2wo\u00E2\u0088\u00921(r \u00E2\u0088\u0092 r0\u000F)w\u00E2\u0080\u00B2(r \u00E2\u0088\u0092 r0\u000F)= A\u00CE\u00B4\u00E2\u0080\u00B2(r \u00E2\u0088\u0092 r0),and the first term of S1 in (3.94) is precisely a multiple of this. If we associateinverse powers of \u000F as leading to Dirac singularities then we expect thisbehaviour for the remaining terms in (3.94) and soS1 =U\u00CE\u00B2\u00E2\u0088\u0092q0DA\u00CE\u00B4\u00E2\u0080\u00B2(r \u00E2\u0088\u0092 r0) + A\u00CE\u00B4(r \u00E2\u0088\u0092 r0),1343.3. Eigenvalues Associated with \u00CE\u00A60 Oddwhere we can determine A by integrating all but the first term in (3.94) overthe entire inner domain. This yields thatA =\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009EoU\u00CE\u00B2\u00E2\u0088\u00922q0Dwo\u00E2\u0088\u00921v\u00CB\u009C1 +oU\u00CE\u00B2\u00E2\u0088\u0092q\u00E2\u0088\u009210DqF(r0)wo\u00E2\u0088\u00921\u00CF\u0088 \u00E2\u0088\u0092osU\u00CE\u00B2\u00E2\u0088\u0092q\u00E2\u0088\u009210Dwo\u00E2\u0088\u00921u\u00CB\u009C1w\u00E2\u0080\u00B2+o(o\u00E2\u0088\u0092 1)U\u00CE\u00B2\u00E2\u0088\u00922q0Dwo\u00E2\u0088\u00922v\u00CB\u009C1w\u00E2\u0080\u00B2 \u00E2\u0088\u0092 sU\u00CE\u00B2\u00E2\u0088\u009210DwoN\u00CB\u009C0 d\u00CF\u0081,where we have substituted (3.79) for \u00CE\u00A61. We can integrate the third andfourth term by parts and upon simplifying we haveA =U\u00CE\u00B2\u00E2\u0088\u0092q\u00E2\u0088\u009210DF(r0)\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009E(qowo\u00E2\u0088\u00921\u00CF\u0088 \u00E2\u0088\u0092 swo)d\u00CF\u0081.Substituting \u00CF\u0088 using (3.81) and recognizing thatdAdb=\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Eowo\u00E2\u0088\u00921wb d\u00CF\u0081,we finally have thatA =U\u00CE\u00B2\u00E2\u0088\u0092q\u00E2\u0088\u009210DF(r0)\u00CE\u00B2(A+ 2qbdAdb),where \u00CE\u00B2 is given by (2.28b). Putting everything together we have,S1 =U\u00CE\u00B2\u00E2\u0088\u0092q0DA\u00CE\u00B4\u00E2\u0080\u00B2(r \u00E2\u0088\u0092 r0) +U\u00CE\u00B2\u00E2\u0088\u0092q\u00E2\u0088\u009210DF(r0)A(\u00CE\u00B2 +2qbAdAdb)\u00CE\u00B4(r \u00E2\u0088\u0092 r0), (3.95)1353.3. Eigenvalues Associated with \u00CE\u00A60 Oddand therefore the global problem for N\u00CB\u0086 is (by once again taking r0 \u00E2\u0089\u0088 r00),1r(rN\u00CB\u0086r)r \u00E2\u0088\u0092m2r2N\u00CB\u0086 \u00E2\u0088\u0092 1DN\u00CB\u0086 =\u00E2\u0088\u0092 U\u00CE\u00B2\u00E2\u0088\u0092q0DA\u00CE\u00B4\u00E2\u0080\u00B2(r \u00E2\u0088\u0092 r00)\u00E2\u0088\u0092 U\u00CE\u00B2\u00E2\u0088\u0092q\u00E2\u0088\u009210DF(r00)A(\u00CE\u00B2 +2qbAdAdb)\u00CE\u00B4(r \u00E2\u0088\u0092 r00),which we can rewrite using constants A\u00C2\u00AF0 (given by (2.60)) and A\u00C2\u00AF0 (given by(2.59)) used in section 2.3.2 to get:1r(rN\u00CB\u0086r)r \u00E2\u0088\u0092m2r2N\u00CB\u0086 \u00E2\u0088\u0092 1DN\u00CB\u0086 = \u00E2\u0088\u0092 A\u00C2\u00AF0U q0\u00CE\u00B4\u00E2\u0080\u00B2(r \u00E2\u0088\u0092 r00)\u00E2\u0088\u0092A\u00C2\u00AF0A\u00C2\u00AF0U q+10F(r00)\u00CE\u00B4(r \u00E2\u0088\u0092 r00). (3.96)It is more convenient if we turn this into a problem with homogeneous sourcesaway from the interface r 6= r00 supplemented by jump conditions acrossthe interface. To do this we first multiply by r and integrate over a smalldomain containing r00. Since the differential equation has a dipole sourceterm then N\u00CB\u0086 will have jump discontinuities, but otherwise will be continuousand therefore, by the integral mean value theorem, the second and thirdterms on the left-side of (3.96) will vanish over the integral. Therefore wehave that[dN\u00CB\u0086dr]r00=A\u00C2\u00AF0U q0r00\u00E2\u0088\u0092 A\u00C2\u00AF0A\u00C2\u00AF0U q+10F(r00) =A\u00C2\u00AF0U q0r00\u00E2\u0088\u0092 A\u00C2\u00AF0A\u00C2\u00AF0U q+10U q0\u00E2\u008C\u00A9N\u00CB\u0086\u00E2\u008C\u00AAr00+A\u00C2\u00AF0A\u00C2\u00AF0U q+10\u00E2\u008C\u00A9duedr\u00E2\u008C\u00AAr00. (3.97a)1363.3. Eigenvalues Associated with \u00CE\u00A60 OddTo get the jump in the function itself, we first compute the indefinite integralof (3.96), ignoring constants of integration,rN\u00CB\u0086r \u00E2\u0088\u0092m2\u00E2\u0088\u00AB1rN\u00CB\u0086 dr \u00E2\u0088\u0092 1D\u00E2\u0088\u00ABN\u00CB\u0086 dr =\u00E2\u0088\u0092 A\u00C2\u00AF0U q0r00\u00CE\u00B4(r \u00E2\u0088\u0092 r00) +A\u00C2\u00AF0U q0H(r \u00E2\u0088\u0092 r00)\u00E2\u0088\u0092 A\u00C2\u00AF0A\u00C2\u00AF0U q+10r00F(r00)H(r \u00E2\u0088\u0092 r00)with H the Heaviside function. We then divide by r, and integrate thisexpression over a small domain centered around r00 to get[N\u00CB\u0086 ]r00 = \u00E2\u0088\u0092A\u00C2\u00AF0U q0, (3.97b)where once again appropriate continuous terms have been neglected as theyvanish over the region of integration. We could now solve the problem for N\u00CB\u0086subject to the jump conditions (3.97), but we have seen a similar problemto this in section 2.3.2 when we looked at V1n, the perturbed circle steady-state problem at O(\u00CE\u00B5), given by (2.63). In fact if we add the jump conditions(2.63d) to (3.97a) and (2.63c) to (3.97b), letting n = m, and define a functionZ = HnU q0 N\u00CB\u0086 + V1n,1373.3. Eigenvalues Associated with \u00CE\u00A60 Oddthen we have the following problem for Z,1r(rZr)r \u00E2\u0088\u0092n2r2Z \u00E2\u0088\u0092 1DZ = 0, r 6= r00dZdr\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3r=r00= 0,[Z]r00 = 0,[dZdr]r00= \u00E2\u0088\u0092A\u00C2\u00AF0A\u00C2\u00AF0U0Z(r00).Solving this problem we haveZ(r) = r00A\u00C2\u00AF0A\u00C2\u00AF0U0Z(r00)G0;n(r; r00)with G0,n given by (2.50). Evaluating at r = r00 leads to the conclusionthat Z(r00) = 0 is the only possibility for arbitrary r00 and therefore Z isidentically zero and thus,N\u00CB\u0086 = \u00E2\u0088\u0092 1HnUq0V1n. (3.98)Therefore our perturbed inhibitor eigenfunction is already determined by thesolution to V1n. This allows us to write the small eigenvalues (3.92) as\u00CE\u00BB\u00CB\u0086 =1\u00E2\u0088\u0092m2r200+q2HmU20H\u00CB\u0086(1\u00E2\u0088\u0092 2bqH\u00CB\u0086dH\u00CB\u0086db)\u00E2\u008C\u00A9duedr\u00E2\u008C\u00AAr00(\u00E3\u0080\u0088V1m\u00E3\u0080\u0089r00 +Hm\u00E2\u008C\u00A9duedr\u00E2\u008C\u00AAr00)\u00E2\u0088\u0092 qH\u00CB\u00862HmU0(\u00E2\u008C\u00A9dV1mdr\u00E2\u008C\u00AAr00+Hm\u00E2\u008C\u00A9d2uedr2\u00E2\u008C\u00AAr00). (3.99)1383.3. Eigenvalues Associated with \u00CE\u00A60 OddIt is rather unsurprising that there is such an intimate relationship betweenthe problem for N\u00CB\u0086 and the perturbed circle since the odd eigenfunctionsdisplace the activator curve sinusoidally and thus exactly deform the circleinto a near-circle, at least initially. The problems are, in fact, even moreintimately related than just through the N\u00CB\u0086 function. Recall that we definethe concentrated activator curve at r = r0 + \u000F\u00CF\u0081\u00E2\u0088\u0097(\u00CE\u00B8) where \u00CF\u0081\u00E2\u0088\u0097 is defined as thelocation of the maximum of the activator and hencev\u00CB\u009C(\u00CF\u0081\u00E2\u0088\u0097)\u00E2\u0080\u00B2 = v\u00CB\u009C\u00E2\u0080\u00B2e(\u00CF\u0081\u00E2\u0088\u0097) + \u00CE\u00A6\u00E2\u0080\u00B2(\u00CF\u0081\u00E2\u0088\u0097) exp(im\u00CE\u00B8 +\u00E2\u0088\u00AB T0\u00CE\u00BB\u00CB\u0086(s) ds)= 0. (3.100)The radial velocity of the curve will be given bydrdT=dr0dT+ \u000Fd\u00CF\u0081\u00E2\u0088\u0097dTand differentiating (3.100) with respect to T , we have0 =(v\u00CB\u009C\u00E2\u0080\u00B2\u00E2\u0080\u00B2e (\u00CF\u0081\u00E2\u0088\u0097) + \u00CE\u00A6\u00E2\u0080\u00B2\u00E2\u0080\u00B2(\u00CF\u0081\u00E2\u0088\u0097) exp(im\u00CE\u00B8 +\u00E2\u0088\u00AB T0\u00CE\u00BB\u00CB\u0086(s) ds))d\u00CF\u0081\u00E2\u0088\u0097dT+ \u00CE\u00A6\u00E2\u0080\u00B2(\u00CF\u0081\u00E2\u0088\u0097) exp(im\u00CE\u00B8 +\u00E2\u0088\u00AB T0\u00CE\u00BB\u00CB\u0086(s) ds)\u00CE\u00BB\u00CB\u0086(T ).Using (3.100) we can simplify and writed\u00CF\u0081\u00E2\u0088\u0097dT=v\u00CB\u009C\u00E2\u0080\u00B2e(\u00CF\u0081\u00E2\u0088\u0097)v\u00CB\u009C\u00E2\u0080\u00B2\u00E2\u0080\u00B2(\u00CF\u0081\u00E2\u0088\u0097)\u00CE\u00BB\u00CB\u0086(T ) (3.101)and thereforedrdT=dr0dT+ \u000Fv\u00CB\u009C\u00E2\u0080\u00B2e(\u00CF\u0081\u00E2\u0088\u0097)v\u00CB\u009C\u00E2\u0080\u00B2\u00E2\u0080\u00B2(\u00CF\u0081\u00E2\u0088\u0097)\u00CE\u00BB\u00CB\u0086(T ).1393.3. Eigenvalues Associated with \u00CE\u00A60 OddIf we think of the coefficient on \u00CE\u00BB\u00CB\u0086 playing to role of \u00CE\u00B5 in the near circleperturbation then \u00E2\u0088\u0092\u00CE\u00BB\u00CB\u0086 should be the first velocity correction we obtained.Indeed if we compare the negative of (3.99) with (2.77) then if we takeh(\u00CE\u00B8) = exp(im\u00CE\u00B8)we see perfect agreement. Notice this required choice of h(\u00CE\u00B8) to balancecomes from the eigenfunction expansion of the linear stability problem (3.2).The coefficients Hm in the Fourier expansion of h(\u00CE\u00B8) were arbitrary and, ifdesired, could be extracted from the coefficient in front of \u00CE\u00BB\u00CB\u0086.140Chapter 4Classification of ExplicitlySolvable Non-Local EigenvalueProblems4.1 Explicit Non-Local EigenvalueFormulationIn Chapter, 3 we considered the stability of the Gierer-Meinhardt model on aring to breakup instabilities. This led to the derivation of a non-self-adjoint,non-local eigenvalue problem (3.16) which we analyzed by recasting it as aroot-finding problem. This allowed us to determine regions of stability butdid not lend itself well to determining eigenvalues explicitly. In this chapterwe consider scenarios for which the explicit determination of these eigenvaluescan be obtained. A similar analysis in what follows was done in [57] for themodel,vt = \u000F2vxx \u00E2\u0088\u0092 v + a(u)v2r\u00E2\u0088\u00923,\u00CF\u0084ut = uxx + (ub \u00E2\u0088\u0092 u) +1\u000Fb(u)vr,1414.1. Explicit Non-Local Eigenvalue Formulationfor certain conditions on the constants r, ub and the functions a(u) and b(u).To setup the theory, we generally consider a class of problems for which linearstability analysis produces a non-local eigenvalue problem of the formL0\u00CE\u00A6\u00E2\u0088\u0092 \u00CF\u0087(\u00CE\u00BB)h(w)\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Eg(w)\u00CE\u00A6 dy = \u00CE\u00BB\u00CE\u00A6, \u00E2\u0088\u0092\u00E2\u0088\u009E < y <\u00E2\u0088\u009E;\u00CE\u00A6\u00E2\u0086\u0092 0 as |y| \u00E2\u0086\u0092 \u00E2\u0088\u009E, (4.1)where ,L0\u00CE\u00A6 = \u00CE\u00A6yy \u00E2\u0088\u0092 \u00CE\u00A6 + f \u00E2\u0080\u00B2(w)\u00CE\u00A6, (4.2)with prime indicating differentiation with respect to w. \u00CF\u0087(\u00CE\u00BB) is a transcenden-tal function of the eigenvalue parameter \u00CE\u00BB and w(y) is the unique homoclinicofwyy \u00E2\u0088\u0092 w + f(w) = 0, \u00E2\u0088\u0092\u00E2\u0088\u009E < y <\u00E2\u0088\u009E;w \u00E2\u0086\u0092 0 as |y| \u00E2\u0086\u0092 \u00E2\u0088\u009E, wy(0) = 0, w(0) > 0. (4.3)This homoclinic orbit exists for certain functions f(w) that satisfy Lemma2.2.0.1 in section 2.2. For example, this criteria was used in section 2.3 tohelp establish the critical saturation parameter b for the saturated Gierer-Meinhardt model. Along with restrictions on f(w) we also require in (4.1)1424.1. Explicit Non-Local Eigenvalue Formulationthatg(0) = 0, g(w) > 0 for w > 0, g(w) is C1 as w \u00E2\u0086\u0092 0+;h(0) = 0, h(w) is C1 as w \u00E2\u0086\u0092 0+. (4.4)The conditions on g(w) and h(w) are such that the integral in (4.1) vanishesas \u00CE\u00A6\u00E2\u0086\u0092 0 and that \u00CE\u00A6 has exponentially decaying solutions when y is asymp-totically large and Re(\u00CE\u00BB) > 0. In order to form (4.1) in an explicitly solvableway, we will exploit the eigenvalue structure of the L0 operator which wediscussed in 3.2.2 but will repeat here. Assuming f(w) holds such that ahomoclinic orbit solution to (4.3) exists then Lemma 3.1.0.1 provides the de-tails of the eigenvalue spectrum to L0\u00CF\u0088 = \u00CE\u00BD\u00CF\u0088. Specifically, it admits a simplediscrete eigenvalue \u00CE\u00BD0 with eigenfunction \u00CF\u00880 of one sign and an eigenvalue\u00CE\u00BD1 = 0 with \u00CF\u00881 = w\u00E2\u0080\u00B2. Consider a choice of g(w) in (4.1) which satisfies (4.4),and is such thatL0g(w) = \u00CF\u0083g(w), (4.5)for some \u00CF\u0083 > 0. Since there can only be one positive eigenvalue to L0, wemust have that \u00CF\u0083 = \u00CE\u00BD0 and g(w) = \u00CF\u00880. If we multiply (4.1) by g(w) andintegrate over the entire domain then\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Eg(w)L0\u00CE\u00A6 dy \u00E2\u0088\u0092 \u00CF\u0087(\u00CE\u00BB)\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Eh(w)g(w) dy\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Eg(w)\u00CE\u00A6 dy = \u00CE\u00BB\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Eg(w)\u00CE\u00A6 dy.(4.6)1434.1. Explicit Non-Local Eigenvalue FormulationWhile the entire NLEP is not self adjoint, the operator L0 given by (4.2) isand so we can integrate by parts the first term in (4.6) to simplify,\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Eg(w)\u00CE\u00A6(\u00CF\u0083 \u00E2\u0088\u0092 \u00CF\u0087(\u00CE\u00BB)\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Eh(w)g(w) dy \u00E2\u0088\u0092 \u00CE\u00BB)dy = 0,where we have used (4.5) to simplify. Assuming that\u00E2\u0088\u00AB\u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009E g(w)\u00CE\u00A6 dy 6= 0 thenwe have an explicit relationship for \u00CE\u00BB,\u00CE\u00BB = \u00CF\u0083 \u00E2\u0088\u0092 \u00CF\u0087(\u00CE\u00BB)\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Eh(w)g(w) dy, (4.7)where this integral converges because of the decay behaviour on h(w) andg(w). Before continuing, consider the case\u00E2\u0088\u00AB\u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009E g(w)\u00CE\u00A6 dy = 0, where (4.1)reduces toL0\u00CE\u00A6 = \u00CE\u00BB\u00CE\u00A6,i.e. \u00CE\u00A6 is an eigenfunction of the operator L0. The only positive solution tothis is for \u00CE\u00BB = \u00CE\u00BD0 but this would require \u00CE\u00A6 = g(w) and we would not be ableto satisfy\u00E2\u0088\u00AB\u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009E g(w)\u00CE\u00A6 dy = 0. Therefore, any unstable eigenvalue to (4.1)needs to satisfy (4.7).We now need to consider the appropriate f(w) in (4.3) that produces a g(w)for (4.5) and ultimately allow (4.7) to hold. If we recognize thatg(w)yy = g\u00E2\u0080\u00B2\u00E2\u0080\u00B2(w)w2y + g\u00E2\u0080\u00B2(w)wyy1444.1. Explicit Non-Local Eigenvalue Formulationthen we can write (4.5) as,g\u00E2\u0080\u00B2\u00E2\u0080\u00B2(w)w2y + g\u00E2\u0080\u00B2(w)wyy + f\u00E2\u0080\u00B2(w)g(w) = (1 + \u00CF\u0083)g(w). (4.8)To remove wyy we can use (4.3) and to remove w2y multiply that same equationand integrate over the entire domain to getw2y \u00E2\u0088\u0092 w2 + 2W = 0, W =\u00E2\u0088\u00AB w0f(s) ds.Substituting into (4.8) we get,g\u00E2\u0080\u00B2\u00E2\u0080\u00B2(w)w2 \u00E2\u0088\u0092 2g\u00E2\u0080\u00B2\u00E2\u0080\u00B2(w)W + g\u00E2\u0080\u00B2(w)w \u00E2\u0088\u0092 g\u00E2\u0080\u00B2(w)f(w) + f \u00E2\u0080\u00B2(w)g(w) = (1 + \u00CF\u0083)g(w).If we integrate by parts and use that g(0) = f(0) = 0 we can simplify to get(w2 \u00E2\u0088\u0092 2W)g\u00E2\u0080\u00B2(w) = \u00CE\u00A3(w)\u00E2\u0088\u0092 f(w)g(w); (4.9a)\u00CE\u00A3(w) \u00E2\u0089\u00A1\u00E2\u0088\u00AB w0\u00CE\u00BE(s) ds, \u00CE\u00BE(s) \u00E2\u0089\u00A1 sg\u00E2\u0080\u00B2(s) + (\u00CF\u0083 + 1)g(s). (4.9b)If we differentiate (4.9a) with respect to w then2(w \u00E2\u0088\u0092 f(w))g\u00E2\u0080\u00B2(w) + (w2 \u00E2\u0088\u0092 2W)g\u00E2\u0080\u00B2\u00E2\u0080\u00B2(w) = \u00CE\u00A3\u00E2\u0080\u00B2(w)\u00E2\u0088\u0092 (f(w)g(w))\u00E2\u0080\u00B2.Using (4.9a) we can simplify to2wg\u00E2\u0080\u00B2(w)\u00E2\u0088\u0092 2f(w)g\u00E2\u0080\u00B2(w) + \u00CE\u00A3(w)\u00E2\u0088\u0092 f(w)g(w)g\u00E2\u0080\u00B2(w)g\u00E2\u0080\u00B2\u00E2\u0080\u00B2(w) = \u00CE\u00A3\u00E2\u0080\u00B2(w)\u00E2\u0088\u0092 (f(w)g(w))\u00E2\u0080\u00B2.1454.1. Explicit Non-Local Eigenvalue FormulationDividing by g\u00E2\u0080\u00B2(w) and rearranging this expression we havef \u00E2\u0080\u00B2(w)g(w)g\u00E2\u0080\u00B2(w)\u00E2\u0088\u0092 f(w)g\u00E2\u0080\u00B2(w)g\u00E2\u0080\u00B2(w)\u00E2\u0088\u0092 f(w)g\u00E2\u0080\u00B2\u00E2\u0080\u00B2(w)g\u00E2\u0080\u00B2(w)2=\u00CE\u00A3\u00E2\u0080\u00B2(w)g\u00E2\u0080\u00B2(w)\u00E2\u0088\u0092 \u00CE\u00A3(w)g\u00E2\u0080\u00B2\u00E2\u0080\u00B2(w)g\u00E2\u0080\u00B2(w)2\u00E2\u0088\u0092 2w. (4.10)We can recognize the expression on the left of (4.10) as,f \u00E2\u0080\u00B2(w)g(w)g\u00E2\u0080\u00B2(w)\u00E2\u0088\u0092 f(w)g\u00E2\u0080\u00B2(w)g\u00E2\u0080\u00B2(w)\u00E2\u0088\u0092 f(w)g\u00E2\u0080\u00B2\u00E2\u0080\u00B2(w)g\u00E2\u0080\u00B2(w)2= g2(fgg\u00E2\u0080\u00B2)\u00E2\u0080\u00B2,and the first two expressions on the right of (4.10) as,\u00CE\u00A3\u00E2\u0080\u00B2(w)g\u00E2\u0080\u00B2(w)\u00E2\u0088\u0092 \u00CE\u00A3(w)g\u00E2\u0080\u00B2\u00E2\u0080\u00B2(w)g\u00E2\u0080\u00B2(w)2=(\u00CE\u00A3(w)g\u00E2\u0080\u00B2(w))\u00E2\u0080\u00B2,so that finally, for a given g(w), we can compute f(w) as the solution to(f(w)g(w)g\u00E2\u0080\u00B2(w))\u00E2\u0080\u00B2=1g(w)2((\u00CE\u00A3(w)g\u00E2\u0080\u00B2(w))\u00E2\u0080\u00B2\u00E2\u0088\u0092 2w). (4.11)It is important to note that for a given g(w), while this formula providesthe necessary f(w), there is no guarantee that the resulting f(w) will sat-isfy the homoclinic orbit criteria in Lemma 2.2.0.1. Many common NLEPsfrom reaction diffusion problems (cf. [39], [40], [45], [46], [81], [57]) involveg(w) with algebraic powers in w. As such we will present two general casesg(w) = w and g(w) = w\u00CE\u00B1 for \u00CE\u00B1 > 1 for which this theory can easily be usedto determine f(w).First consider g(w) = w so that from (4.9b),\u00CE\u00BE(w) = (\u00CF\u0083 + 2)w, \u00CE\u00A3 =(\u00CF\u0083 + 2)w22.1464.1. Explicit Non-Local Eigenvalue FormulationFrom (4.11) we have(f(w)w)\u00E2\u0080\u00B2=1w2(((\u00CF\u0083 + 2)w22)\u00E2\u0088\u0092 2w)=\u00CF\u0083w,and therefore,f(w) = \u00CF\u0083w logw + Aw,where we can set A = 0 without loss of generality. Next consider g(w) = w\u00CE\u00B1for \u00CE\u00B1 > 1. In this case,\u00CE\u00BE(w) = (\u00CF\u0083 + 1 + \u00CE\u00B1)w\u00CE\u00B1, \u00CE\u00A3 =(\u00CF\u0083 + 1 + \u00CE\u00B1)w\u00CE\u00B1+1\u00CE\u00B1 + 1,and hence from (4.11),(f(w)\u00CE\u00B1w2\u00CE\u00B1\u00E2\u0088\u00921)\u00E2\u0080\u00B2=1w2\u00CE\u00B1(((\u00CF\u0083 + 1 + \u00CE\u00B1)w2\u00CE\u00B1(\u00CE\u00B1 + 1))\u00E2\u0080\u00B2\u00E2\u0088\u0092 2w)=2(\u00CF\u0083 + 1 + \u00CE\u00B1)\u00CE\u00B1(\u00CE\u00B1 + 1)w1\u00E2\u0088\u00922\u00CE\u00B1.Integrating this yields thatf(w) =(\u00CF\u0083 + 1\u00E2\u0088\u0092 \u00CE\u00B12)(1\u00E2\u0088\u0092 \u00CE\u00B12) w + Aw2\u00CE\u00B1\u00E2\u0088\u00921,for arbitrary A which we set to be one without loss of generality. Since thisholds for any \u00CF\u0083, we can avoid the term that is linear in w by choosing,\u00CF\u0083 = \u00CE\u00B12 \u00E2\u0088\u0092 1, (4.12)so that f(w) = w2\u00CE\u00B1\u00E2\u0088\u00921. We will now focus the rest of this chapter on studyingthe unsaturated Gierer-Meinhardt model as an example using the general1474.2. Explicit Stability Formulation for the Gierer-Meinhardt Model on a Stripeexplicit NLEP framework.4.2 Explicit Stability Formulation for theGierer-Meinhardt Model on a StripeIn contrast to Chapter 2, we will now focus solely on the unsaturated versionof the Gierer-Meinhardt model,vt = \u000F2\u00E2\u0088\u0086v \u00E2\u0088\u0092 v + vpuq, (4.13a)\u00CF\u0084ut = \u00E2\u0088\u0086u\u00E2\u0088\u0092 u+vo\u000Fus, (4.13b)and for the time being consider a rectangular domain with homogeneousNeumann conditions on the boundary. The rectangular domain \u00E2\u0084\u00A6 is definedby\u00E2\u0084\u00A6 \u00E2\u0089\u00A1 {(x1, x2)| \u00E2\u0088\u0092 l < x1 < l, 0 < x2 < d}for some length l and width d. We have taken the diffusivity on (4.13b)to be unity since, as per the discussion in section 2.3.1, the diffusivity canbe absorbed into the length scale. We want to use the explicitly solvableframework with g(w) = w\u00CE\u00B1, \u00CE\u00B1 > 1, and use the simplified condition on \u00CF\u0083given by (4.12). The f(w) in (4.3) is related to the exponent p in (4.13)and since we need the exponent set (p, q, o, s) to be integers, in order to use(4.12), we require p to be odd and specifically we take p = 3. Furthermoreas we saw for the ring derivation in 3.1, g(w) = w\u00CE\u00B1 is intimately tied to theexponent set o via \u00CE\u00B1 = o \u00E2\u0088\u0092 1, and by choosing p = 3, this specifies that1484.2. Explicit Stability Formulation for the Gierer-Meinhardt Model on a Stripe\u00CE\u00B1 = 2 and so we need o = 3. Therefore, by making our choice for p, we willconsider the specific variant of (4.13) to be:vt = \u000F2\u00E2\u0088\u0086v \u00E2\u0088\u0092 v + v3uq, (4.14a)\u00CF\u0084ut = \u00E2\u0088\u0086u\u00E2\u0088\u0092 u+v3\u000Fus, (4.14b)with q > 0 and s \u00E2\u0089\u00A5 0. We consider the activator v localized on an O(\u000F)region around the midpoint x1 = 0 for all x2 (hence the classification of astripe) and define an inner coordinate variable y = x/\u000F. The derivation ofthis stripe solution is very similar in approach to that of the ring in section2.3.1 and was presented in entirety in [39]. We therefore omit the detailshere and summarize the results. The equilibrium activator denoted ve(x1)and the equilibrium inhibitor denoted by ue(x1) are given byve(x1) \u00E2\u0088\u00BC U\u00CE\u00B30w(x1\u000F); ue(x1) \u00E2\u0088\u00BC U0Gl(x1)Gl(0), (4.15)where w(y) =\u00E2\u0088\u009A2 sech y is the unique homoclinic orbit solution that satisfiesw\u00E2\u0080\u00B2\u00E2\u0080\u00B2 \u00E2\u0088\u0092 w + w3 = 0, \u00E2\u0088\u0092\u00E2\u0088\u009E < y <\u00E2\u0088\u009E;w \u00E2\u0086\u0092 0 as |y| \u00E2\u0086\u0092 \u00E2\u0088\u009E , w(0) > 0, w\u00E2\u0080\u00B2(0) = 0. (4.16)We define the constants U0 and \u00CE\u00B3 byU \u00CE\u00B60 \u00E2\u0089\u00A11b\u00CB\u009CGl(0); b\u00CB\u009C \u00E2\u0089\u00A1\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew3 dy =\u00E2\u0088\u009A2pi; \u00CE\u00B6 \u00E2\u0089\u00A1 3q2\u00E2\u0088\u0092 (s+ 1) > 0; \u00CE\u00B3 \u00E2\u0089\u00A1 q2.(4.17)1494.2. Explicit Stability Formulation for the Gierer-Meinhardt Model on a StripeThe Green\u00E2\u0080\u0099s function Gl(x1) in (4.15) satisfiesGlx1x1 \u00E2\u0088\u0092Gl = \u00E2\u0088\u0092\u00CE\u00B4(x1), |x1| \u00E2\u0089\u00A4 l; Glx1(\u00C2\u00B1l) = 0,with \u00CE\u00B4(\u00C2\u00B7) the Dirac measure. This has solutionGl(x1) =cosh(l \u00E2\u0088\u0092 |x1|)2 sinh(l), (4.18)and therefore Gl(0) = 1/2 coth(l).We are now in a position to perform a linear stability analysis on this basestate. Since, unlike the ring solution in 2.3.1, the stripe always starts in itsequilibrium position, there are no long-time dynamics to consider and so wecan make the appropriate usual formulationv = ve + \u00CF\u0086(x1) exp(imx2 + \u00CE\u00BBt), u = ue + \u00CE\u00B7(x1) exp(imx2 + \u00CE\u00BBt); m =kpid,(4.19)where k is an integer. The restriction on m comes from the homogeneousNeumann conditions on x2 = 0 and x2 = d. These conditions also necessi-tate that we need to take the true solution as only the real part of (4.19).However, in what follows we treat m as a continuous variable and are justmindful that true unstable modes must satisfy the discrete condition. Alter-natively one can think of the operator eigenvalue problem having continuouseigenvalues for which we only sample the discrete ones to capture the appro-priate boundary conditions. Substituting (4.19) into (4.14), we obtain the1504.2. Explicit Stability Formulation for the Gierer-Meinhardt Model on a Stripeeigenvalue problem\u000F2\u00CF\u0086x1x1 \u00E2\u0088\u0092 \u00CF\u0086+3v2euqe\u00CF\u0086\u00E2\u0088\u0092 qv3euq+1e\u00CE\u00B7 = (\u00CE\u00BB+ \u000F2m2)\u00CF\u0086, |x1| \u00E2\u0089\u00A4 l; \u00CF\u0086x1(\u00C2\u00B1l) = 0,(4.20a)\u00CE\u00B7x1x1 \u00E2\u0088\u0092 (1 + \u00CF\u0084\u00CE\u00BB+m2)\u00CE\u00B7 = \u00E2\u0088\u00923ve2\u000Fuse\u00CF\u0086+sv3e\u000Fus+1e\u00CE\u00B7, |x1| \u00E2\u0089\u00A4 l; \u00CE\u00B7x1(\u00C2\u00B1l) = 0.(4.20b)Once again, as in 3.2, we do not ignore the terms \u000F2m2 in the event thathigh frequency modes lead to instability behaviour. Like the base state,ve, the activator perturbation \u00CF\u0086 will be entirely localized and so we take\u00CF\u0086(x1) \u00E2\u0088\u00BC \u00CE\u00A6(\u000F\u00E2\u0088\u00921x1). The leading order problem for \u00CE\u00A6 from (4.20a) is\u00CE\u00A6yy \u00E2\u0088\u0092 \u00CE\u00A6 + 3w2\u00CE\u00A6\u00E2\u0088\u0092 qU q/2\u00E2\u0088\u009210 w3\u00CE\u00B7(0) = (\u00CE\u00BB+ \u000F2m2)\u00CE\u00A6, \u00E2\u0088\u0092\u00E2\u0088\u009E < y <\u00E2\u0088\u009E,\u00CE\u00A6\u00E2\u0086\u0092 0 as |y| \u00E2\u0086\u0092 \u00E2\u0088\u009E. (4.21)Here \u00CE\u00B7(0) comes from the outer problem (4.20b) because it is not singularlyperturbed. The simplification of (4.20b) involves using Dirac measures forterms of O(\u000F\u00E2\u0088\u00921) (see 2.2.1) and therefore,3v2e\u000Fuse\u00CF\u0086 =3U2\u00CE\u00B3\u00E2\u0088\u0092s0\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew2\u00CE\u00A6 dy\u00CE\u00B4(x1) =3U1\u00E2\u0088\u0092\u00CE\u00B30b\u00CB\u009CGl(0)\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew2\u00CE\u00A6 dy\u00CE\u00B4(x1),sv3e\u000Fus+1e\u00CE\u00B7 =sU \u00CE\u00B60 \u00CE\u00B7(0)\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew3 dy\u00CE\u00B4(x1) =s\u00CE\u00B7(0)Gl(0)\u00CE\u00B4(x1),1514.2. Explicit Stability Formulation for the Gierer-Meinhardt Model on a Stripewhere we have simplified using (4.17). Therefore, \u00CE\u00B7(x1) satisfies\u00CE\u00B7x1x1 \u00E2\u0088\u0092 \u00CE\u00B82\u00CE\u00BB\u00CE\u00B7 =(sGl(0)\u00CE\u00B7(0)\u00E2\u0088\u0092 3U1\u00E2\u0088\u0092\u00CE\u00B30b\u00CB\u009CGl(0)\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew2\u00CE\u00A6 dy)\u00CE\u00B4(x1),|x1| \u00E2\u0089\u00A4 l; \u00CE\u00B7x1(\u00C2\u00B1l) = 0, (4.22)where \u00CE\u00B8\u00CE\u00BB \u00E2\u0089\u00A1\u00E2\u0088\u009A1 +m2 + \u00CF\u0084\u00CE\u00BB is the principal value of the square root. Bythis we mean the branch cut taken along the negative real axis such thatRe(\u00CE\u00BB) \u00E2\u0089\u00A4 \u00E2\u0088\u00921\u00E2\u0088\u0092m2\u00CF\u0084 which implies \u00CE\u00B7(0) is analytic in Re(\u00CE\u00BB) \u00E2\u0089\u00A5 0. If we considerthe Green\u00E2\u0080\u0099s function problemG\u00CE\u00BBx1x1 \u00E2\u0088\u0092 \u00CE\u00B82\u00CE\u00BBG\u00CE\u00BB = \u00E2\u0088\u0092\u00CE\u00B4(x1), |x1| \u00E2\u0089\u00A4 l, G\u00CE\u00BBx1(\u00C2\u00B1l) = 0,which has solution,G\u00CE\u00BB(x1) =cosh(\u00CE\u00B8\u00CE\u00BB(l \u00E2\u0088\u0092 |x1|))2\u00CE\u00B8\u00CE\u00BB sinh(\u00CE\u00B8\u00CE\u00BBl)(4.23)then we can write\u00CE\u00B7(x1) =(3U1\u00E2\u0088\u0092\u00CE\u00B30b\u00CB\u009CGl(0)\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew2\u00CE\u00A6 dy \u00E2\u0088\u0092 sGl(0)\u00CE\u00B7(0))G\u00CE\u00BB(x1).Using (4.18) and (4.23) we can solve for \u00CE\u00B7(0) to yield that\u00CE\u00B7(0) =3U1\u00E2\u0088\u0092\u00CE\u00B30b\u00CB\u009C\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew2\u00CE\u00A6 dy[s+\u00CE\u00B8\u00CE\u00BB tanh(\u00CE\u00B8\u00CE\u00BBl)tanh(l)]\u00E2\u0088\u00921. (4.24)This expression is non-zero as long as\u00E2\u0088\u00AB\u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew2\u00CE\u00A6 dy 6= 0 which as per thediscussion in section 4.1 holds for any unstable eigenvalue Re(\u00CE\u00BB) > 0. This1524.2. Explicit Stability Formulation for the Gierer-Meinhardt Model on a Stripemeans that \u00CE\u00A6 is even and as with section 3.2, this corresponds to breakupor amplitude instabilities. We substitute (4.24) into (4.21) to get our NLEPfor breakup instabilities of the stripe asL0\u00CE\u00A6\u00E2\u0088\u0092 \u00CF\u0087w3\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew2\u00CE\u00A6 dy = (\u00CE\u00BB+ \u000F2m2)\u00CE\u00A6, \u00E2\u0088\u0092\u00E2\u0088\u009E < y <\u00E2\u0088\u009E;\u00CE\u00A6\u00E2\u0086\u0092 0 as |y| \u00E2\u0086\u0092 \u00E2\u0088\u009E, (4.25a)\u00CF\u0087 \u00E2\u0089\u00A1 3qb\u00CB\u009C[s+\u00CE\u00B8\u00CE\u00BB tanh(\u00CE\u00B8\u00CE\u00BBl)tanh(l)]\u00E2\u0088\u00921, (4.25b)with L0 defined by (4.2). In terms of the explicit formulation discussed insection 4.1, by choosing p = o = 3 in (4.14), we prescribed \u00CE\u00B1 = 2 (g(w) = w2)and \u00CF\u0083 = 3 so thatL0w2 = 3w2.By comparing (4.25a) to (4.7) we take\u00CE\u00BB = \u00CE\u00BB+ \u000F2m2, h(w) = w3to get that (4.25a) becomes\u00CE\u00BB = 3\u00E2\u0088\u0092 \u000F2m2 \u00E2\u0088\u0092 9q2[s+\u00CE\u00B8\u00CE\u00BB tanh(\u00CE\u00B8\u00CE\u00BBl)tanh(l)]\u00E2\u0088\u00921, (4.26)where we have simplified that\u00E2\u0088\u00AB\u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew5 dyb\u00CB\u009C=32.We will now analyze (4.26) for \u00CF\u0084 = 0 and \u00CF\u0084 > 0.1534.2. Explicit Stability Formulation for the Gierer-Meinhardt Model on a Stripe4.2.1 Explicit Stripe Eigenvalues, \u00CF\u0084 = 0When \u00CF\u0084 = 0, we have that (4.26) becomes\u00CE\u00BB = 3\u00E2\u0088\u0092 \u000F2m2 \u00E2\u0088\u0092 9q2[s+\u00E2\u0088\u009A1 +m2 tanh(\u00E2\u0088\u009A1 +m2l)tanh(l)]\u00E2\u0088\u00921. (4.27)We begin by finding neutral stability points \u00CE\u00BB = 0. If m is O(1) then toleading order the neutral stability point (4.27) becomes0 = 3\u00E2\u0088\u0092 9q2[3q2+ \u00CE\u00BA(z)]\u00E2\u0088\u00921,wherez =\u00E2\u0088\u009Am2 + 1, z \u00E2\u0089\u00A5 1; \u00CE\u00BA(z) = z tanh(zl)tanh(l)\u00E2\u0088\u0092 (\u00CE\u00B6 + 1),with \u00CE\u00B6 from (4.17). This expression simplifies so that the neutral stabilitypoint is a root of \u00CE\u00BA(z). Now \u00CE\u00BA(1) = \u00E2\u0088\u0092\u00CE\u00B6 < 0 and \u00CE\u00BA(z) \u00E2\u0086\u0092 \u00E2\u0088\u009E as z \u00E2\u0086\u0092 \u00E2\u0088\u009E sothere is at least one root to \u00CE\u00BA(z). Furthermore,\u00CE\u00BA\u00E2\u0080\u00B2(z) =tanh(zl)tanh(l)+ zlsech 2(zl)tanh(l)> 0and so there is a unique root z\u00E2\u0088\u0092 > 1 to \u00CE\u00BA(z) and hence a unique neutralstability modemb\u00E2\u0088\u0092 =\u00E2\u0088\u009Az2\u00E2\u0088\u0092 \u00E2\u0088\u0092 1. (4.28)1544.2. Explicit Stability Formulation for the Gierer-Meinhardt Model on a StripeIf z < z\u00E2\u0088\u0092 then \u00CE\u00BB < 0 and so there are no unstable eigenvalues on 0 < m 0 is z > z\u00E2\u0088\u0092. If we now consider m = O(\u000F) anddefine m = \u000F\u00E2\u0088\u00921m\u00CB\u009C then (4.27) becomes\u00CE\u00BB = 3\u00E2\u0088\u0092 m\u00CB\u009C2 \u00E2\u0088\u0092 \u000F9q2[\u000Fs+\u00E2\u0088\u009Am\u00CB\u009C2 + \u000F2 tanh(\u00E2\u0088\u009Am\u00CB\u009C2 + \u000F2l\u000F\u00E2\u0088\u00921)tanh(l)]\u00E2\u0088\u00921. (4.29)If we search for the neutral stability point m\u00CB\u009C+b and expand m\u00CB\u009C+2b \u00E2\u0088\u00BC m\u00CB\u009C20 + \u000Fm\u00CB\u009C21then we get m\u00CB\u009C0 =\u00E2\u0088\u009A3 and for m\u00CB\u009C1 we havem\u00CB\u009C21 = \u00E2\u0088\u00929q2tanh(l)m\u00CB\u009C0,and so the upper neutral stability point ismb+ \u00E2\u0088\u00BC1\u000F\u00E2\u0088\u009Am\u00CB\u009C20 + \u000Fm\u00CB\u009C21 =\u00E2\u0088\u009A3\u000F\u00E2\u0088\u0092 3q4tanh(l). (4.30)If m > mb+ (m < mb+) then \u00CE\u00BB < 0 (\u00CE\u00BB > 0) and therefore unstable eigenvaluesexist on mb\u00E2\u0088\u0092 < m < mb+ . Next we seek to estimate the dominant wave modemdom where \u00CE\u00BB achieves its maximum. When m is an O(1) or O(\u000F\u00E2\u0088\u00921) numberwe have that to leading order \u00CE\u00BB is monotonic and therefore the dominantmode must occur at some intermediate scaling in \u000F. Therefore we let,m = \u000F\u00E2\u0088\u0092am\u00CB\u0086, 0 < a < 1,1554.2. Explicit Stability Formulation for the Gierer-Meinhardt Model on a Stripeand substitute into (4.27),\u00CE\u00BB = 3\u00E2\u0088\u0092 \u000F2\u00E2\u0088\u00922am\u00CB\u00862 \u00E2\u0088\u0092 \u000Fa9q2[\u000Fas+\u00E2\u0088\u009Am\u00CB\u00862 + \u000F2a tanh(\u00E2\u0088\u009Am\u00CB\u00862 + \u000F2a\u000F\u00E2\u0088\u0092al)tanh(l)]\u00E2\u0088\u00921.(4.31)The only hope for non-monotonic \u00CE\u00BB is if there is balance in the second andthird term and so we take a = 2/3. Differentiating (4.31) with respect to m\u00CB\u0086and setting to zero we getd\u00CE\u00BBdm\u00CB\u0086=\u00E2\u0088\u0092 2m\u00CB\u0086+ 9q2 tanh(l)[\u000F2/3s+\u00E2\u0088\u009Am\u00CB\u00862 + \u000F4/3 tanh(\u00E2\u0088\u009Am\u00CB\u00862 + \u000F4/3\u000F\u00E2\u0088\u00922/3l)tanh(l)]\u00E2\u0088\u00922(m\u00CB\u0086 tanh(\u00E2\u0088\u009Am\u00CB\u00862 + \u000F4/3\u000F\u00E2\u0088\u00922/3l)\u00E2\u0088\u009Am\u00CB\u00862 + \u000F4/3+ \u000F\u00E2\u0088\u00922/3lm\u00CB\u0086 sech 2(\u00E2\u0088\u009Am\u00CB\u00862 + \u000F4/3\u000F\u00E2\u0088\u00922/3l))= 0. (4.32)Substituting m\u00CB\u0086 = m\u00CB\u00860 + \u000F2/3m\u00CB\u00861 where the m\u00CB\u00861 correction comes from s intro-ducing terms of O(\u000F2/3), we get that to leading order\u00E2\u0088\u00922m\u00CB\u00860 +9q tanh(l)2m\u00CB\u008620= 0,and som\u00CB\u00860 =(9q tanh(l)4)1/3. (4.33)1564.2. Explicit Stability Formulation for the Gierer-Meinhardt Model on a StripeFor the m\u00CB\u00861 problem, both tanh(\u000F\u00E2\u0088\u00922/3) and sech (\u000F\u00E2\u0088\u00922/3) terms will remaintheir saturated values up to exponential order in \u000F and therefore\u00E2\u0088\u00922m\u00CB\u00861 \u00E2\u0088\u00929q tanh(l)m\u00CB\u00861m\u00CB\u00860\u00E2\u0088\u0092 9q tanh(l)2sm\u00CB\u008630\u00E2\u0089\u0088 0,and using (4.33),m\u00CB\u00861 \u00E2\u0089\u0088 \u00E2\u0088\u009223s tanh(l).Combining everything, we have,mdom \u00E2\u0088\u00BC \u000F\u00E2\u0088\u00922/3(9 tanh(l)4q)1/3\u00E2\u0088\u0092 2s tanh(l)3. (4.34)If one were to consider the effect of the walls negligible (taking l\u00E2\u0086\u0092\u00E2\u0088\u009E) then(4.27) becomes\u00CE\u00BB = 3\u00E2\u0088\u0092 \u000F2m2 \u00E2\u0088\u0092 9q2(s+\u00E2\u0088\u009A1 +m2),and we have thatmb\u00E2\u0088\u0092 =\u00E2\u0088\u009A\u00CE\u00B62 + 2\u00CE\u00B6, (4.35a)mb+ =\u00E2\u0088\u009A3/\u000F\u00E2\u0088\u0092 3q/4, (4.35b)mdom =\u000F\u00E2\u0088\u00922/3(9q/4)1/3 \u00E2\u0088\u0092 2s/3. (4.35c)This limit will allows us to obtain more tractable analytical results when weconsider \u00CF\u0084 > 0.We now plot (4.27) in Figure 4.1 with \u000F = 0.05, and s = 0 for q = 1 andq = 2. In each case we plot l = 0.5, l = 1, and l = \u00E2\u0088\u009E. In Table 4.1 we list1574.2. Explicit Stability Formulation for the Gierer-Meinhardt Model on a Stripethe numeric evaluation (denoted n) of mb\u00E2\u0088\u0092 , mb+ and mdom along with theasymptotic approximations (denoted a) given by (4.28) for mb\u00E2\u0088\u0092 , (4.30) formb+ , and (4.34) for mdom. The agreement for both q = 1 and q = 2 for allvalues of l agrees favourably with the analytic results.0 5 10 15 20 25 30 35\u00E2\u0088\u00920.500.511.522.53m\u00CE\u00BB l = 0.5l = 1l =\u00E2\u0088\u009E(a) q = 10 5 10 15 20 25 30 35\u00E2\u0088\u00920.500.511.522.53m\u00CE\u00BB l = 0.5l = 1l =\u00E2\u0088\u009E(b) q = 2Figure 4.1: Eigenvalues for \u00CF\u0084 = 0 computed from (4.27) versus m for \u000F = 0.05,and s = 0 for several values of l. The curves from highest maximum tosmallest maximum are l = 0, l = 0.5, and l =\u00E2\u0088\u009E respectively.1584.2. Explicit Stability Formulation for the Gierer-Meinhardt Model on a Stripeq = 1l mb\u00E2\u0088\u0092(n) mb\u00E2\u0088\u0092(a) mb+(n) mb+(a) mdom(n) mdom(a)0.5 0.75 0.75 34.29 34.29 7.42 7.461 0.86 0.86 34.06 34.07 8.76 8.82\u00E2\u0088\u009E 1.12 1.12 33.87 33.89 9.60 9.65q = 2l mb\u00E2\u0088\u0092(n) mb\u00E2\u0088\u0092(a) mb+(n) mb+(a) mdom(n) mdom(a)0.5 1.60 1.60 33.93 33.95 9.36 9.401 2.11 2.10 33.44 33.50 11.07 11.11\u00E2\u0088\u009E 2.85 2.83 33.03 33.14 12.13 12.16Table 4.1: Asymptotic and numerical comparison of the neutral stabilitypoints mb\u00E2\u0088\u0092 , mb+ , and the dominant wave mode mdom. The numerical values(n) are obtained from Figure 4.1 and the asymptotic approximations (a) areobtained from (4.28) for mb\u00E2\u0088\u0092 , (4.30) for mb+ , and (4.34) for mdomPredicting mdom allows us to approximate the number of spots Nspots thatwould occur in a breakup pattern whereNspots =\u00E2\u008C\u008Amdomd2pi\u00E2\u008C\u008Bwhere b\u00C2\u00B7c is the floor function, rounding down to the nearest integer and dis the rectangle width.4.2.2 Explicit Stripe Eigenvalues, \u00CF\u0084 > 0We now consider (4.26) for \u00CF\u0084 > 0, which becomes a transcendental equationfor \u00CE\u00BB because of the presence of \u00CE\u00B8\u00CE\u00BB. For simplicity, we consider the scenariodiscussed in section 4.2.1 where the sidewalls have no effect (l =\u00E2\u0088\u009E). In this1594.2. Explicit Stability Formulation for the Gierer-Meinhardt Model on a Stripecase (4.26) becomes\u00CE\u00BB = 3\u00E2\u0088\u0092 \u000F2m2 \u00E2\u0088\u0092 9q2(s+\u00E2\u0088\u009A1 +m2\u00E2\u0088\u009A1 + \u00CF\u0084\u00CB\u0086\u00CE\u00BB), \u00CF\u0084\u00CB\u0086 \u00E2\u0089\u00A1 \u00CF\u00841 +m2. (4.36)While we can make little progress analytically on prescribing values for theroots, we would like to understand the qualitative features of the roots aswell as how many exist. Therefore, we attempt to isolate \u00CE\u00BB separately from(4.36) and write,\u00CE\u00BB\u00E2\u0088\u0092 3 + \u000F2m2 = \u00E2\u0088\u0092 9q\u00E2\u0088\u009A1 +m2[2s\u00E2\u0088\u009A1 +m2+ 2\u00E2\u0088\u009A1 + \u00CF\u0084\u00CB\u0086\u00CE\u00BB]\u00E2\u0088\u00921.If we defined0 = \u00E2\u0088\u00922s\u00E2\u0088\u009A1 +m2\u00E2\u0089\u00A4 0, (4.37a)d1 = \u00E2\u0088\u00929q\u00E2\u0088\u009A1 +m2< 0, (4.37b)\u00CE\u00B2 \u00E2\u0089\u00A1 3\u00E2\u0088\u0092 \u000F2m2, (4.37c)then we can write (4.36) as a root finding problemF(\u00CE\u00BB) = 2\u00E2\u0088\u009A1 + \u00CF\u0084\u00CB\u0086\u00CE\u00BB\u00E2\u0088\u0092 G(\u00CE\u00BB), G(\u00CE\u00BB) \u00E2\u0089\u00A1 d0 \u00E2\u0088\u0092d1\u00CE\u00B2 \u00E2\u0088\u0092 \u00CE\u00BB. (4.38)We will look for the roots to (4.38) that satisfy Re(\u00CE\u00BB) > 0 by using theargument principle around the same closed contour \u00CE\u0093 = \u00CE\u0093I\u00E2\u0088\u00AA\u00CE\u0093K from section1604.2. Explicit Stability Formulation for the Gierer-Meinhardt Model on a Stripe3.2.6 with\u00CE\u0093K :{\u00CE\u00BB = K exp(it)|t \u00E2\u0088\u0088[\u00E2\u0088\u0092pi2,pi2]}\u00CE\u0093I : \u00E2\u0088\u0092Ki \u00E2\u0089\u00A4 \u00CE\u00BB \u00E2\u0089\u00A4 Kiwhich we traverse counter-clockwise and take the limit as K tends to infinity.When \u00CF\u0084 \u001D 1, F(\u00CE\u00BB) \u00E2\u0088\u00BC 2\u00E2\u0088\u009A\u00CF\u0084\u00CB\u0086\u00CE\u00BB and so on \u00CE\u0093K when t = \u00E2\u0088\u0092pi/2, argF = \u00E2\u0088\u0092pi/4and when t = pi/2, argF = pi/4. Therefore, the change in argument of F(\u00CE\u00BB)over \u00CE\u0093K is pi/2. When \u00CE\u00B2 < 0 (m >\u00E2\u0088\u009A3/\u000F) then for all Re(\u00CE\u00BB) > 0, G(\u00CE\u00BB) isanalytic but when \u00CE\u00B2 > 0 (0 < m <\u00E2\u0088\u009A3/\u000F) then there is a simple pole at\u00CE\u00BB = \u00CE\u00B2. Finally, since F is analytic (aside from \u00CE\u00BB = \u00CE\u00B2) and real valued when\u00CE\u00BB is real-valued then F(\u00CE\u00BB) = F(\u00CE\u00BB\u00C2\u00AF). Combining all of this together with theargument principle in a way similar to section 3.2.6, we have that the numberof roots J to F(\u00CE\u00BB) are given byJ =14+ h(\u00CE\u00B2) +1pi[argF ]\u00CE\u0093+I , where h(\u00CE\u00B2) =\u00EF\u00A3\u00B1\u00EF\u00A3\u00B4\u00EF\u00A3\u00B2\u00EF\u00A3\u00B4\u00EF\u00A3\u00B31, \u00CE\u00B2 > 0,0, \u00CE\u00B2 < 0, (4.39)and \u00CE\u0093+I is the positive imaginary axis i\u00CE\u00BBI with 0 < \u00CE\u00BBI <\u00E2\u0088\u009E traversed down-wards and [argF ] indicates the change in argument of F over the contour.When \u00CE\u00BB = i\u00CE\u00BBI , \u00CE\u00BBI > 0 we haveG(i\u00CE\u00BBI) = d0 \u00E2\u0088\u0092\u00CE\u00B2d1\u00CE\u00B22 + \u00CE\u00BB2I\u00E2\u0088\u0092 i \u00CE\u00BBId1\u00CE\u00B22 + \u00CE\u00BB2Iand2\u00E2\u0088\u009A1 + i\u00CF\u0084\u00CB\u0086\u00CE\u00BBI = 2(1 + \u00CF\u0084\u00CB\u00862\u00CE\u00BB2I)1/4 exp(iarctan(\u00CF\u0084\u00CB\u0086\u00CE\u00BBI)2).1614.2. Explicit Stability Formulation for the Gierer-Meinhardt Model on a StripeUsing the trig identitiescos2 \u00CE\u00B8 =1 + cos \u00CE\u00B82, sin2 \u00CE\u00B8 =1\u00E2\u0088\u0092 cos \u00CE\u00B82,we can write2\u00E2\u0088\u009A1 + i\u00CF\u0084\u00CB\u0086\u00CE\u00BBI = K+(\u00CF\u0084\u00CB\u0086\u00CE\u00BBI) + iK\u00E2\u0088\u0092(\u00CF\u0084\u00CB\u0086\u00CE\u00BBI), K\u00C2\u00B1(z) =\u00E2\u0088\u009A2(\u00E2\u0088\u009A1 + z2 \u00C2\u00B1 1)1/2,and therefore we can decompose F(i\u00CE\u00BBI) = FR(\u00CE\u00BBI) + iFI(\u00CE\u00BBI) withFR(\u00CE\u00BBI) \u00E2\u0089\u00A1 K+(\u00CF\u0084\u00CB\u0086\u00CE\u00BBI)\u00E2\u0088\u0092 d0 +\u00CE\u00B2d1\u00CE\u00B22 + \u00CE\u00BB2I, FI(\u00CE\u00BBI) \u00E2\u0089\u00A1 K\u00E2\u0088\u0092(\u00CF\u0084\u00CB\u0086\u00CE\u00BBI) +\u00CE\u00BBId1\u00CE\u00B22 + \u00CE\u00BB2I.(4.40)For \u00CE\u00BBI \u001D 1 then since \u00CF\u0084 > 0, FR \u00E2\u0088\u00BC FI \u00E2\u0088\u00BC\u00E2\u0088\u009A2\u00CF\u0084\u00CB\u0086\u00CE\u00BBI and therefore arg(F(i\u00CE\u00BBI))\u00E2\u0086\u0092pi/4 as \u00CE\u00BBI \u00E2\u0086\u0092 \u00E2\u0088\u009E. At \u00CE\u00BBI = 0, we have that FR(0) = 2\u00E2\u0088\u0092 G(0) and FI(0) = 0,so when \u00CE\u00BB = 0 is a root of F(\u00CE\u00BB) = 0 which occurs at mb\u00E2\u0088\u0092 and mb+ definedby (4.35a) and (4.35b) respectively then G(0) = 2. When m is O(1),G(0) = \u00CE\u00B6 + 12\u00E2\u0088\u009A1 +m2and so dG(0)/dm < 0. When m is O(\u000F\u00E2\u0088\u00921) thendG(0)dm\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3m=mb+= \u00E2\u0088\u0092 2mb+\u00E2\u0088\u009A1 +mb+2+18q\u000F2mb+\u00E2\u0088\u009A1 +mb+2(3\u00E2\u0088\u0092 \u000F2mb+2)\u00E2\u0088\u00BC 83q> 0.1624.2. Explicit Stability Formulation for the Gierer-Meinhardt Model on a StripeFinally, if m >\u00E2\u0088\u009A3/\u000F, G(0) < 0. Combining everything together we havethatG(0) < 2, mb\u00E2\u0088\u0092 < m < mb+ \u00E2\u0088\u00AAm >\u00E2\u0088\u009A3\u000F(4.41a)G(0) > 2, 0 < m < mb\u00E2\u0088\u0092 \u00E2\u0088\u00AAmb+ < m <\u00E2\u0088\u009A3\u000F. (4.41b)We are now in a position to classify the roots J to F(\u00CE\u00BB). If m >\u00E2\u0088\u009A3/\u000F then\u00CE\u00B2 < 0 and since d0 \u00E2\u0089\u00A4 0 and d1 < 0 then by (4.40), FR(\u00CE\u00BBI) > 0 for all \u00CE\u00BBI .Furthermore FR(0) > 0 and so the argument along \u00CE\u0093+I changes from pi/4 to0. Since FR(\u00CE\u00BBI) > 0 always then it cannot enter the negative real plane sowe must have [arg(F)]\u00CE\u0093+I = \u00E2\u0088\u0092pi/4. From (4.39) this means that J = 0 on thisregion. Next consider mb+ < m <\u00E2\u0088\u009A3/\u000F. On this interval \u00CE\u00B2 > 0 and from(4.41), FR(0) < 0 so the argument changes from pi/4 to pi. We can calculatethatdFRd\u00CE\u00BBI= \u00CF\u0084\u00CB\u0086K\u00E2\u0080\u00B2(\u00CF\u0084\u00CB\u0086\u00CE\u00BBI)\u00E2\u0088\u00922\u00CE\u00BBI\u00CE\u00B2d1(\u00CE\u00B22 + \u00CE\u00BB2I)2 , (4.42)which for mb+ < m <\u00E2\u0088\u009A3/\u000F is positive and therefore since FR(\u00CE\u00BBI) \u00E2\u0086\u0092 \u00E2\u0088\u009Eas \u00CE\u00BBI \u00E2\u0086\u0092 \u00E2\u0088\u009E then there is a unique root \u00CE\u00BB\u00E2\u0088\u0097I to FR. We can find this rootasymptotically by letting m =\u00E2\u0088\u009A3/\u000F+m1 with\u00E2\u0088\u00923q/4 < m1 < 0 and therefore\u00CE\u00B2 = 3 \u00E2\u0088\u0092 \u000F2m2 \u00E2\u0088\u00BC \u00E2\u0088\u00922\u00E2\u0088\u009A3m1\u000F. Substituting into (4.40) we get that to leadingorder FR(\u00CE\u00BB\u00E2\u0088\u0097I) = K+(\u00CF\u0084\u00CB\u0086\u00CE\u00BB\u00E2\u0088\u0097I) = 0 for which there is no solution. Therefore welet \u00CE\u00BB\u00E2\u0088\u0097I = \u000F\u00CE\u00BB\u00E2\u0088\u0097I0 and repeat the asymptotic expansion in (4.40) to get that toleading order,2 +18qm112m21 + \u00CE\u00BB\u00E2\u0088\u00972I0= 0,1634.2. Explicit Stability Formulation for the Gierer-Meinhardt Model on a Stripeand so we get \u00CE\u00BB\u00E2\u0088\u0097I0 =\u00E2\u0088\u009A\u00E2\u0088\u00923m1(4m1 + 3q) which exists for all m1 since m1 >\u00E2\u0088\u00923q/4. Using this root in FI from (4.40) we have that,FI(\u000F\u00CE\u00BB\u00E2\u0088\u0097I0) = K\u00E2\u0088\u0092(\u000F\u00CF\u0084\u00CB\u0086\u00CE\u00BB\u00E2\u0088\u0097I0) +\u000Fd1\u00CE\u00BB\u00E2\u0088\u0097I0\u00CE\u00B22 + \u000F2\u00CE\u00BB\u00E2\u0088\u00972I0\u00E2\u0088\u00BC \u00E2\u0088\u0092 9q\u00CE\u00BB\u00E2\u0088\u0097I0\u00E2\u0088\u009A3(12m21 + \u00CE\u00BB2I0)< 0.Since FI(\u00CE\u00BB\u00E2\u0088\u0097I) < 0 then [argF ]\u00CE\u0093+I = \u00E2\u0088\u00925pi/4 and from (4.39), J = 0. Next weconsider mb\u00E2\u0088\u0092 < m < mb+ where from (4.41), FR(0) > 0 and since \u00CE\u00B2 > 0and d1 < 0 then by (4.42), dFR/d\u00CE\u00BBI > 0. Therefore [argF ]\u00CE\u0093+I = \u00E2\u0088\u0092pi/4 andfrom (4.39) J = 1. This root is actually on the positive real axis and satisfies0 < \u00CE\u00BB < \u00CE\u00B2. To see this we consider (4.38) with 0 < F(0) < 2 and F \u00E2\u0086\u0092 \u00E2\u0088\u0092\u00E2\u0088\u009Eas \u00CE\u00BB\u00E2\u0086\u0092 \u00CE\u00B2 and therefore F has at least one root on 0 < \u00CE\u00BB < \u00CE\u00B2. If \u00CE\u00BB > \u00CE\u00B2 thenG(\u00CE\u00BB) < 0 and F > 0 for all \u00CE\u00BB. We have that\u00E2\u0088\u009A1 + \u00CF\u0084\u00CE\u00BB is an increasing concavefunction as well as G \u00E2\u0080\u00B2(\u00CE\u00BB) > 0 and G \u00E2\u0080\u00B2\u00E2\u0080\u00B2(\u00CE\u00BB) > 0 so that F \u00E2\u0080\u00B2\u00E2\u0080\u00B2(\u00CE\u00BB) < 0. ThereforeF(\u00CE\u00BB) can have at most one critical point and regardless of the sign of F \u00E2\u0080\u00B2(0),this can only be satisfied if F has exactly one root. Therefore there is exactlyone real root on mb\u00E2\u0088\u0092 < m < mb+ . Finally we consider 0 < m < mb\u00E2\u0088\u0092 , whereFR(0) < 0 and so the argument goes from pi/4 to pi. Once again dFR/d\u00CE\u00BBI > 0and FR \u00E2\u0086\u0092\u00E2\u0088\u009E as \u00CE\u00BBI \u00E2\u0086\u0092\u00E2\u0088\u009E so there is exactly one root to FR on this interval.Therefore either [argF ]\u00CE\u0093+I = \u00E2\u0088\u00925pi/4 and J = 0 or [argF ]\u00CE\u0093+I = 3pi/4 andJ = 2. If \u00CF\u0084\u00CB\u0086 = 0 then from (4.38), F(0) < 0 and F \u00E2\u0080\u00B2(\u00CE\u00BB) = \u00E2\u0088\u0092G \u00E2\u0080\u00B2(\u00CE\u00BB) < 0 sothere are no roots with positive real part. This should continue to be truefor \u00CF\u0084\u00CB\u0086 \u001C 1 and therefore J = 0 for \u00CF\u0084\u00CB\u0086 \u001C 1 on 0 < m < mb\u00E2\u0088\u0092 . If \u00CF\u0084\u00CB\u0086 \u001D 1 thenF(0) < 0 and F \u00E2\u0086\u0092 \u00E2\u0088\u0092\u00E2\u0088\u009E as \u00CE\u00BB \u00E2\u0086\u0092 \u00CE\u00B2\u00E2\u0088\u0092 but for \u00CE\u00BB \u001D O(\u00CF\u0084\u00CB\u0086\u00E2\u0088\u00921), F \u00E2\u0089\u0088 2\u00E2\u0088\u009A\u00CF\u0084\u00CB\u0086\u00CE\u00BB \u001D 1and since F \u00E2\u0080\u00B2\u00E2\u0080\u00B2(\u00CE\u00BB) < 0 there are exactly two roots with real positive part for\u00CF\u0084\u00CB\u0086 \u001D 1. Notice that eigenvalues cannot enter the real plane through \u00CE\u00BB = 01644.2. Explicit Stability Formulation for the Gierer-Meinhardt Model on a Stripesince \u00CE\u00BB = 0 is an eigenvalue for all \u00CF\u0084\u00CB\u0086 and therefore this shows that there mustexist a Hopf bifurcation at some \u00CF\u0084\u00CB\u0086 = \u00CF\u0084\u00CB\u0086Hm when 0 < m < mb\u00E2\u0088\u0092 . We summarizethe eigenvalue conclusions as follows:Principal Result 4.2.2.1 Let J denote the number of roots in Re(\u00CE\u00BB) > 0of F(\u00CE\u00BB) = 0 in (4.38). Then, for any \u00CF\u0084\u00CB\u0086 > 0,J =0, m > mb+J =1, mb\u00E2\u0088\u0092 < m < mb+J =0 or J = 2, 0 < m < mb\u00E2\u0088\u0092 .On mb\u00E2\u0088\u0092 < m < mb+ the root is on the positive real axis in the interval0 < \u00CE\u00BB < \u00CE\u00B2 and for 0 < m < mb\u00E2\u0088\u0092, there are no roots with positive realpart for \u00CF\u0084\u00CB\u0086 \u001C 1 and two roots with positive real part when \u00CF\u0084 \u001D 1. This lastcondition proves the existence of a Hopf bifurcation at some \u00CF\u0084\u00CB\u0086 = \u00CF\u0084\u00CB\u0086Hm .Explicit Representation for Hopf BifurcationWe now turn our attention to evaluating \u00CF\u0084\u00CB\u0086 = \u00CF\u0084\u00CB\u0086Hm and showing it is unique.If a Hopf bifurcation occurs then i\u00CE\u00BBI is a root to (4.38) and so from (4.40)we need to set FR = FI = 0. If we do this we get\u00E2\u0088\u009A2(\u00E2\u0088\u009Aa+ 1)1/2 = d0 \u00E2\u0088\u0092d1\u00CE\u00B2\u00CE\u00B8,\u00E2\u0088\u009A2(\u00E2\u0088\u009Aa\u00E2\u0088\u0092 1)1/2 = \u00E2\u0088\u0092d1\u00CE\u00BBI\u00CE\u00B8;a \u00E2\u0089\u00A1 1 + \u00CF\u0084\u00CB\u0086 2\u00CE\u00BB2I , \u00CE\u00B8 \u00E2\u0089\u00A1 \u00CE\u00B22 + \u00CE\u00BB2I . (4.43)1654.2. Explicit Stability Formulation for the Gierer-Meinhardt Model on a StripeBy dividing these expressions we get(\u00E2\u0088\u009Aa+ 1)1/2(\u00E2\u0088\u009Aa\u00E2\u0088\u0092 1)1/2 = \u00E2\u0088\u0092d0\u00CE\u00B8d1\u00CE\u00BBI+\u00CE\u00B2\u00CE\u00BBI,which we can simplify to get\u00E2\u0088\u009Aa+ 1\u00E2\u0088\u009Aa\u00E2\u0088\u0092 1 =1\u00CE\u00BBIA, A = \u00CE\u00B2 \u00E2\u0088\u0092 d0\u00CE\u00B8d1. (4.44)Using\u00E2\u0088\u009Aa\u00E2\u0088\u0092 1 = \u00CF\u0084\u00CB\u0086\u00CE\u00BBI and the first expression in (4.43) we have,\u00CF\u0084\u00CB\u0086 =d212\u00CE\u00B82A. (4.45)Next we determine an equation for \u00CE\u00B8. To do this we square and subtract thefirst two expressions of (4.43),2\u00E2\u0088\u009Aa+ 2\u00E2\u0088\u0092 2\u00E2\u0088\u009Aa+ 2 = 4 =(d0 \u00E2\u0088\u0092d1\u00CE\u00B2\u00CE\u00B82)2\u00E2\u0088\u0092 d21\u00CE\u00BB2I\u00CE\u00B82. (4.46)We can use that \u00CE\u00BB2I = \u00CE\u00B8\u00E2\u0088\u0092\u00CE\u00B22 in (4.46) to get that \u00CE\u00B8 is the root to the quadraticequation M(\u00CE\u00B8) given byM(\u00CE\u00B8) = (d20 \u00E2\u0088\u0092 4)\u00CE\u00B82 \u00E2\u0088\u0092 (2\u00CE\u00B2d0d1 + d21)\u00CE\u00B8 + 2d21\u00CE\u00B22 = 0. (4.47)In order for a solution \u00CE\u00BBI =\u00E2\u0088\u009A\u00CE\u00B8 \u00E2\u0088\u0092 \u00CE\u00B22 to exist we require \u00CE\u00B8 > \u00CE\u00B22. A Hopfbifurcation only occurs on 0 < m < mb\u00E2\u0088\u0092 and we will only consider thisinterval for M(\u00CE\u00B8). First we notice that,M(\u00CE\u00B22) = \u00CE\u00B24(d20 \u00E2\u0088\u00922d0d1\u00CE\u00B2+d21\u00CE\u00B22\u00E2\u0088\u0092 4)= \u00CE\u00B24(G(0)2 \u00E2\u0088\u0092 4) > 0,1664.2. Explicit Stability Formulation for the Gierer-Meinhardt Model on a Stripesince by (4.41), G(0) > 2 on this interval. If d0 = \u00E2\u0088\u00922 then the roots degen-erate and we get\u00CE\u00B81 =2\u00CE\u00B22d1d1 \u00E2\u0088\u0092 4\u00CE\u00B2(4.48)as the unique root to (4.47). At this root we have from (4.44) thatA = \u00CE\u00B2 +4\u00CE\u00B22d1 \u00E2\u0088\u0092 4\u00CE\u00B2=\u00CE\u00B2d1d1 \u00E2\u0088\u0092 4\u00CE\u00B2> 0since d1 < 0 and therefore in this case there is a unique \u00CF\u0084\u00CB\u0086 = \u00CF\u0084\u00CB\u0086Hm > 0 andunique Hopf-bifurcation eigenvalue\u00E2\u0088\u009A\u00CE\u00B81 \u00E2\u0088\u0092 \u00CE\u00B22. We next consider the tworemaining cases \u00E2\u0088\u00922 < d0 < 0 and d0 < \u00E2\u0088\u00922. If \u00E2\u0088\u00922 < d0 < 0 then M(\u00CE\u00B8) from(4.47) tends to negative infinity as \u00CE\u00B8 \u00E2\u0086\u0092 \u00C2\u00B1\u00E2\u0088\u009E and therefore since M(\u00CE\u00B22) > 0,by the intermediate value theorem, there are two roots to (4.47), one root\u00CE\u00B82\u00E2\u0088\u0092 in \u00CE\u00B22 < \u00CE\u00B82\u00E2\u0088\u0092 < \u00E2\u0088\u009E and another root \u00CE\u00B82+ in \u00E2\u0088\u0092\u00E2\u0088\u009E < \u00CE\u00B82+ < \u00CE\u00B22. Only \u00CE\u00B82\u00E2\u0088\u0092given by\u00CE\u00B82\u00E2\u0088\u0092 =2\u00CE\u00B2d0d1 + d212(d20 \u00E2\u0088\u0092 4)+\u00E2\u0088\u009A(2\u00CE\u00B2d0d1 + d21)2 \u00E2\u0088\u0092 8d21\u00CE\u00B22(d20 \u00E2\u0088\u0092 4)2(4\u00E2\u0088\u0092 d20), (4.49)is a valid solution to M(\u00CE\u00B8) that satisfies \u00CE\u00B8 > \u00CE\u00B22 and so we discount the secondroot. We now need to make sure that for this root, A > 0 so that \u00CF\u0084\u00CB\u0086Hm = \u00CF\u0084\u00CB\u0086 > 0.First, define \u00CE\u00B8c = d1\u00CE\u00B2/d0 as the unique root to (4.44) when A = 0. Sinced0 < 0 and d1 < 0 then dA/d\u00CE\u00B8 < 0 and so all we need to show to have A > 0when \u00CE\u00B8 = \u00CE\u00B82\u00E2\u0088\u0092 is that \u00CE\u00B82\u00E2\u0088\u0092 < \u00CE\u00B8c. If we evaluate M(\u00CE\u00B8c) we getM(\u00CE\u00B8c) =d21\u00CE\u00B22d0(d0 \u00E2\u0088\u0092d1\u00CE\u00B2)\u00E2\u0088\u0092 4d21\u00CE\u00B22d20=d21\u00CE\u00B22d0(G(0)\u00E2\u0088\u0092 4d0)< 0, (4.50)1674.2. Explicit Stability Formulation for the Gierer-Meinhardt Model on a Stripesince G(0) > 2. This inequality on G also implies thatd1d0> \u00CE\u00B2(1\u00E2\u0088\u0092 2d0)> \u00CE\u00B2and therefore that \u00CE\u00B8c > \u00CE\u00B22. This means that \u00CE\u00B8c \u00E2\u0088\u0088 (\u00CE\u00B22,\u00E2\u0088\u009E) and since \u00CE\u00B82\u00E2\u0088\u0092 isthe unique root to M(\u00CE\u00B8) in this interval then M(\u00CE\u00B8c) < 0 implies \u00CE\u00B82\u00E2\u0088\u0092 < \u00CE\u00B8c andtherefore A > 0 and \u00CF\u0084\u00CB\u0086Hm > 0 when \u00CE\u00B8 = \u00CE\u00B82\u00E2\u0088\u0092. Finally we consider the intervald0 < \u00E2\u0088\u00922. On this interval we still have M(\u00CE\u00B22) > 0 but now M(\u00CE\u00B8) \u00E2\u0086\u0092 +\u00E2\u0088\u009Eas \u00CE\u00B8 \u00E2\u0086\u0092 \u00C2\u00B1\u00E2\u0088\u009E. We still have from (4.50) that M(\u00CE\u00B8c) < 0 and so for d0 < \u00E2\u0088\u00922both roots to M(\u00CE\u00B8), \u00CE\u00B83\u00C2\u00B1 exist on \u00CE\u00B8 \u00E2\u0088\u0088 (\u00CE\u00B22,\u00E2\u0088\u009E). In order to satisfy A > 0 wetake the smaller root \u00CE\u00B83\u00E2\u0088\u0092 < \u00CE\u00B8c also given by (4.49) which leads to the uniqueHopf bifurcation \u00CF\u0084\u00CB\u0086 = \u00CF\u0084\u00CB\u0086Hm > 0.We now briefly consider the case when s = 0 and d0 = 0 identically. Inthis scenario, the roots to (4.47) simplify to\u00CE\u00B80 = c0 +\u00E2\u0088\u009Ac1; c0 = \u00E2\u0088\u0092d218, c1 =d4164+d21\u00CE\u00B222.We also have that when d0 = 0 that A = \u00CE\u00B2 and so we can write\u00CF\u0084\u00CB\u0086 =d21\u00CE\u00B22(1c0 +\u00E2\u0088\u009Ac1)2=d21\u00CE\u00B22c20 \u00E2\u0088\u0092 2c0\u00E2\u0088\u009Ac1 + c1(c20 \u00E2\u0088\u0092 c1)2.We can compute that,(c20 \u00E2\u0088\u0092 c1)2 =d41\u00CE\u00B244,1684.2. Explicit Stability Formulation for the Gierer-Meinhardt Model on a Stripeand so\u00CF\u0084\u00CB\u0086 =d2116\u00CE\u00B23+12\u00CE\u00B23\u00E2\u0088\u009Ad4164+d21\u00CE\u00B222+1\u00CE\u00B2=1\u00CE\u00B2(1 +c22+ c\u00E2\u0088\u009A1 +c24), c =d1\u00E2\u0088\u009A8\u00CE\u00B2= \u00E2\u0088\u0092 9q2\u00E2\u0088\u009A2(1 +m2)(3\u00E2\u0088\u0092 \u000F2m2).(4.51)We now summarize the results for the Hopf bifurcation.Principal Result 4.2.2.2 For 0 < m < mb\u00E2\u0088\u0092, there exists a unique value\u00CF\u0084Hm = (1 + m2)\u00CF\u0084\u00CB\u0086Hm > 0 for which \u00CE\u00BB = \u00CE\u00BBI is a root of F(i\u00CE\u00BBI) given by (4.38).The Hopf bifurcation point \u00CF\u0084\u00CB\u0086Hm and \u00CE\u00BBIH are given by\u00CF\u0084\u00CB\u0086Hm =d212\u00CE\u00B82A, \u00CE\u00BBIH =\u00E2\u0088\u009A\u00CE\u00B8 \u00E2\u0088\u0092 \u00CE\u00B22, A \u00E2\u0089\u00A1 \u00CE\u00B2 \u00E2\u0088\u0092 d0\u00CE\u00B8d1. (4.52)Here, \u00CE\u00B8 > \u00CE\u00B22 is the smallest root of (4.47) given by (4.48) if d0 = \u00E2\u0088\u00922 and(4.49) if d0 6= \u00E2\u0088\u00922. When s = 0 we can explicitly compute \u00CF\u0084\u00CB\u0086Hm via (4.51).As with \u00CF\u0084 = 0, we plot the eigenvalues of largest real part for \u00CF\u0084 6= 0 in Figure4.2 for l = \u00E2\u0088\u009E, \u000F = 0.05, \u00CF\u0084 = 2, and s = 0 for q = 1 and q = 2. ComparingFigure 4.2 to Figure 4.1, we have that for m \u001D 1 the curves have similarbehaviour when \u00CF\u0084 = 0 and \u00CF\u0084 6= 0. This is because \u00CE\u00B8\u00CE\u00BB =\u00E2\u0088\u009A1 +m2 + \u00CF\u0084\u00CE\u00BB \u00E2\u0089\u0088\u00E2\u0088\u009A1 +m2 for m \u001D 1 and \u00CF\u0084 = O(1) like in the case \u00CF\u0084 = 2 chosen here.Therefore changing \u00CF\u0084 should only have a noticeable affect near the lowerthreshold m = mb\u00E2\u0088\u0092 given by (4.35a).1694.2. Explicit Stability Formulation for the Gierer-Meinhardt Model on a Stripe0 5 10 15 20 25 30 35\u00E2\u0088\u00920.500.511.522.53m\u00CE\u00BB q = 1q = 2Figure 4.2: Plot of \u00CE\u00BB versus m for q = 1 (solid curve) and q = 2 (dashedcurve). The parameters here are s = 0, l =\u00E2\u0088\u009E, and \u00CF\u0084 = 2.For q = 2 in Figure 4.2 we have that \u00CE\u00BB \u00E2\u0086\u0092 0 as m \u00E2\u0086\u0092 mb\u00E2\u0088\u0092 but this is not sofor q = 1. We can understand this behaviour if we plot \u00CF\u0084Hm and \u00CE\u00BBIH from(4.52) which is in Figure 4.3.1704.2. Explicit Stability Formulation for the Gierer-Meinhardt Model on a Stripe0 0.5 1 1.5 2 2.5 301234567m\u00CF\u0084H m q = 1q = 2(a)0 0.5 1 1.5 2 2.5 300.511.522.5m\u00CE\u00BB IH q = 1q = 2(b)Figure 4.3: Plot of \u00CF\u0084Hm and \u00CE\u00BBIH from (4.52) for m in 0 < m < mb\u00E2\u0088\u0092 andl = \u00E2\u0088\u009E. The parameter values are s = 0 and \u000F = 0.05 while the solid anddashed curves are for q = 1 and q = 2 respectively.For q = 2, \u00CF\u0084Hm > 2 = \u00CF\u0084 in Figure 4.3a and so no Hopf bifurcation occurs for0 < m < mb\u00E2\u0088\u0092 . Furthermore, since the eigenvalues enter from the Re(\u00CE\u00BB) < 0complex plane then all of the eigenvalues are stable as evidenced in Figure4.2. For q = 1 however, \u00CF\u0084 > \u00CF\u0084Hm for all m in 0 < m < mb\u00E2\u0088\u0092 and so the eigen-values should have all transitioned through a Hopf bifurcation and entered1714.2. Explicit Stability Formulation for the Gierer-Meinhardt Model on a Stripethe Re(\u00CE\u00BB) > 0 complex plane. As such, all of the eigenvalues are unstable asevidenced in Figure 4.2.Finite lWe now briefly consider the case where l is finite. In this case, we use (4.26)to replace (4.38) with Fl(\u00CE\u00BB) and Gl(\u00CE\u00BB) defined byFl \u00E2\u0089\u00A1 2\u00E2\u0088\u009A1 + \u00CF\u0084\u00CB\u0086\u00CE\u00BB(tanh(l\u00E2\u0088\u009A1 +m2\u00E2\u0088\u009A1 + \u00CF\u0084\u00CB\u0086\u00CE\u00BB)tanh(l\u00E2\u0088\u009A1 +m2))\u00E2\u0088\u0092 Gl(\u00CE\u00BB), Gl(\u00CE\u00BB) \u00E2\u0089\u00A1 d0l \u00E2\u0088\u0092d1l\u00CE\u00B2 \u00E2\u0088\u0092 \u00CE\u00BB,(4.53)with d0l \u00E2\u0089\u00A4 0, d1l < 0 defined byd0l = \u00E2\u0088\u00922s\u00E2\u0088\u009A1 +m2tanh ltanh(l\u00E2\u0088\u009A1 +m2), d10 = \u00E2\u0088\u00929q\u00E2\u0088\u009A1 +m2tanh ltanh(l\u00E2\u0088\u009A1 +m2),and \u00CE\u00B2 still by (4.37c). We have that m = mb\u00E2\u0088\u0092 defined by (4.28) and m = mb+defined by (4.30) are the two values of m for which \u00CE\u00BB = 0 is a root to Fland as such these are the two values for which Gl(0) = 2. Therefore, we caneasily extend (4.41) for Gl(0) with the new values of mb\u00E2\u0088\u0092 and mb+ . To findthe roots we can still use (4.39) since Fl still has a pole at \u00CE\u00BB = \u00CE\u00B2 and stillchanges argument by pi/2 over \u00CE\u0093K . We define FRl(\u00CE\u00BBI) \u00E2\u0089\u00A1 Re(Fl(i\u00CE\u00BBi)) byFRl = CRl(\u00CE\u00BBI)\u00E2\u0088\u0092 d0l +d1l\u00CE\u00B2\u00CE\u00B22 + \u00CE\u00BB2I,CRl(\u00CE\u00BBI) \u00E2\u0089\u00A1 Re(2\u00E2\u0088\u009A1 + i\u00CF\u0084\u00CB\u0086\u00CE\u00BBItanh(l\u00E2\u0088\u009A1 +m2\u00E2\u0088\u009A1 + i\u00CF\u0084\u00CB\u0086\u00CE\u00BBI)tanh(l\u00E2\u0088\u009A1 +m2)),1724.2. Explicit Stability Formulation for the Gierer-Meinhardt Model on a Stripeand similarly,FIl = CIl(\u00CE\u00BBI) +d1l\u00CE\u00BBI\u00CE\u00B22 + \u00CE\u00BB2I,CIl(\u00CE\u00BBI) \u00E2\u0089\u00A1 Im(2\u00E2\u0088\u009A1 + i\u00CF\u0084\u00CB\u0086\u00CE\u00BBItanh(l\u00E2\u0088\u009A1 +m2\u00E2\u0088\u009A1 + i\u00CF\u0084\u00CB\u0086\u00CE\u00BBI)tanh(l\u00E2\u0088\u009A1 +m2)).It was shown in section 3 of [79] that CRl(\u00CE\u00BBI) is a monotonic increasingfunction of \u00CE\u00BB and therefore, since d1l < 0, dFRl(\u00CE\u00BBI)/d\u00CE\u00BBI > 0. We cantherefore easily retrieve the argument change for each case of m and makethe same conclusion for J in (4.39) as we did for l = \u00E2\u0088\u009E. Because of thiswe have that Principal Result 4.2.2.1 holds for any l in 0 < l < \u00E2\u0088\u009E. Wecan compute roots to (4.26) for any l using Newton\u00E2\u0080\u0099s method. We plot anexample of this for for q = 1 and \u00CF\u0084 = 2 in Figure 4.4.1734.3. Explicit Stability Formulation for the Gierer-Meinhardt Model on a Ring0 5 10 15 20 25 30 35\u00E2\u0088\u00920.500.511.522.53m\u00CE\u00BB l = 0.1l = 0.8Figure 4.4: Plot of eigenvalues \u00CE\u00BB versus m for l = 0.1 (solid curve) and l = 0.8(dashed curve). The parameter values are q = 1, s = 0, \u000F = 0.05, and \u00CF\u0084 = 2.It is worth mentioning that while Principal Result 4.2.2.1 holds, we cannotguarantee that the Hopf bifurcation for m on 0 < m < mb\u00E2\u0088\u0092 is unique and assuch that Principal Result 4.2.2.2 holds.4.3 Explicit Stability Formulation for theGierer-Meinhardt Model on a RingWe now consider the unsaturated Gierer-Meinhardt model (4.13) in a radialdomain\u00E2\u0084\u00A6 \u00E2\u0089\u00A1 {(r, \u00CE\u00B8)|0 \u00E2\u0089\u00A4 r \u00E2\u0089\u00A4 l, 0 \u00E2\u0089\u00A4 \u00CE\u00B8 < 2pi}, r = |x|1744.3. Explicit Stability Formulation for the Gierer-Meinhardt Model on a Ringwith the activator concentrated on a ring radius r = r0. The solution to thiswas constructed for p = 2 in 2.3.1 and the extension to p = 3 is straightfor-ward so the details are omitted here. We have thatv = U\u00CE\u00B30w(r \u00E2\u0088\u0092 r0\u000F), u = U0Gl(r; r0)Gl(r0; r0), \u00CE\u00B3 \u00E2\u0089\u00A1 q/2,where w is still given by (4.16). Here the Green\u00E2\u0080\u0099s function Gl(r; r0) is givenby (2.38) where R is replaced by l and the constant U0 satisfiesU \u00CE\u00B60 =(1r0\u00E2\u0088\u009A2piJ0,1(r0)J0,2(r0)), \u00CE\u00B6 \u00E2\u0089\u00A1 32q \u00E2\u0088\u0092 (s+ 1),with J0,i defined by (2.39). We have the dynamic condition for r0 given by(2.43) whereH\u00CB\u0086 =\u00E2\u0088\u00AB\u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew2 dy\u00E2\u0088\u00AB\u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew2y dy\u00E2\u0088\u0092 1 = 2, y = \u000F\u00E2\u0088\u00921(r \u00E2\u0088\u0092 r0),and thereforedr0dT= \u00E2\u0088\u0092 1r0\u00E2\u0088\u0092 q2(J \u00E2\u0080\u00B20,1(r0)J0,1(r0)+J \u00E2\u0080\u00B20,2(r0)J0,2(r0)), T \u00E2\u0089\u00A1 \u000F\u00E2\u0088\u00922t. (4.54)For the purposes of this section, we will consider the stability around trueequilibrium dr0/dT = 0 and like section 2.3.1, equilibrium radii will not existfor all values of l and there is a saddle node bifurcation. Figure 4.5 showsthe bifurcation diagram for p = 3 and q = 2.1754.3. Explicit Stability Formulation for the Gierer-Meinhardt Model on a Ring3 4 5 6 7 8 9 10012345678910lr 0Figure 4.5: Bifurcation diagram for (4.54) for q = 2. A saddle-node bifurca-tion occurs when l = 3.622. The larger of the equilibrium r0 values belongto the stable branch.Now, unlike Figure 2.6, when p = 3 there are no roots to (4.54) when q = 2as evidenced in Figure 4.6 and therefore we will not consider this case.1764.3. Explicit Stability Formulation for the Gierer-Meinhardt Model on a Ring0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1\u00E2\u0088\u00922\u00E2\u0088\u00921.5\u00E2\u0088\u00921\u00E2\u0088\u00920.500.51r0/ldr 0/dt `=1`=4`=10`=20Figure 4.6: Plot of (4.54) for q = 1 and various values of l. We alwayshave that dr0/dT < 0 and therefore there are no equilibrium ring radii whenq = 1.We now turn our attention to the linear stability problem,v = ve+\u00CE\u00A6(r \u00E2\u0088\u0092 r0\u000F)exp(im\u00CE\u00B8+\u00CE\u00BBt), u = ue+\u00CE\u00B7(r) exp(im\u00CE\u00B8+\u00CE\u00BBt), m \u00E2\u0088\u0088 Nwhere we note that since we are considering true equilibrium positions r0then we do not have a slow time dependence on the steady-state. As withthe stripe we will only consider \u00CE\u00A6 even for which Re(\u00CE\u00BB) > 0. The details ofthe derivation of the NLEP for p = 3 are similar to the p = 2 case in 3.1 andas such we omit the details here. The nonlocal eigenvalue problem (3.16)1774.3. Explicit Stability Formulation for the Gierer-Meinhardt Model on a RingbecomesL0\u00CE\u00A6\u00E2\u0088\u0092 \u00CF\u0087w3\u00E2\u0088\u00AB\u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew2\u00CE\u00A6 dy\u00E2\u0088\u00AB\u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew3 dy=(\u00CE\u00BB+\u000F2m2r20)\u00CE\u00A6;\u00CF\u0087 \u00E2\u0089\u00A1 3q(s+J0,1(r0)J0,2(r0)J\u00C2\u00AFm,1(\u00CE\u00B8\u00CE\u00BBr0)J\u00C2\u00AFm,2(\u00CE\u00B8\u00CE\u00BBr0))\u00E2\u0088\u00921, \u00CE\u00B8\u00CE\u00BB =\u00E2\u0088\u009A1 + \u00CF\u0084\u00CE\u00BB, (4.55)where J0,i is given by (2.39) and J\u00C2\u00AFm,i by (3.13). The operator L0 is given by(4.2) with f(w) = w3. Since the explicitly solveable criteria from section 4.1is not geometry specific, we still have L0w2 = 3w2 and therefore comparingto (4.7), (4.55) becomes\u00CE\u00BB = 3\u00E2\u0088\u0092 \u000F2m2r20\u00E2\u0088\u0092 9q2(s+J0,1(r0)J0,2(r0)J\u00C2\u00AFm,1(\u00CE\u00B8\u00CE\u00BBr0)J\u00C2\u00AFm,2(\u00CE\u00B8\u00CE\u00BBr0))\u00E2\u0088\u00921, (4.56)where we have used that\u00E2\u0088\u00AB\u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew5 dy = 3/2\u00E2\u0088\u00AB\u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew3 dy. Once again when\u00CF\u0084 = 0, we have that (4.56) becomes an explicit expression for \u00CE\u00BB. We start byfinding neutral stability points \u00CE\u00BB = 0. When m O(1) then for \u00CE\u00BB = 0 (4.56)becomesM = \u00CE\u00B6 + 1\u00E2\u0088\u0092 J0,1(r0)J0,2(r0)J\u00C2\u00AFm,1(r0)J\u00C2\u00AFm,2(r0)= 0. (4.57)M(0) = \u00CE\u00B6 > 0 and when m \u001D 1 we can use (3.31) for larger order asymp-totics to the modified Bessel functions to get thatJ\u00C2\u00AF1,m(r0)J\u00C2\u00AF2,m(r0) \u00E2\u0088\u00BC12m(1 + (r0/l)2m) \u00E2\u0089\u0088 12m, r0 > 1 (4.58)and soM\u00E2\u0086\u0092\u00E2\u0088\u0092\u00E2\u0088\u009E asm\u00E2\u0086\u0092\u00E2\u0088\u009E. Furthermore, from section 3.2, J\u00C2\u00AFm,1(r0)J\u00C2\u00AFm,2(r0)is a monotonic decreasing function of m and so therefore M is a monotonic1784.3. Explicit Stability Formulation for the Gierer-Meinhardt Model on a Ringdecreasing function as well. All of this combines to prove there is a uniqueroot m = mb\u00E2\u0088\u0092 to (4.57) such that \u00CE\u00BB = 0 when \u00CF\u0084 = 0. If \u00CF\u0084 = 0 then we canrewrite (4.56) as\u00CE\u00BB = \u00E2\u0088\u00923M(s+J0,1(r0)J0,2(r0)J\u00C2\u00AFm,1(r0)J\u00C2\u00AFm,2(r0))\u00E2\u0088\u00921and if m < mb\u00E2\u0088\u0092 ,M > 0 and hence \u00CE\u00BB < 0 for m in 0 < m < mb\u00E2\u0088\u0092 . Conversely,\u00CE\u00BB > 0 if m = O(1) in m > mb\u00E2\u0088\u0092 . When m = \u000F\u00E2\u0088\u00921m\u00CB\u009C with m\u00CB\u009C = O(1) then using(4.58) we have that (4.56) becomes0 \u00E2\u0089\u0088 3\u00E2\u0088\u0092 m\u00CB\u009C2r20\u00E2\u0088\u0092 9q2(s+2m\u00CB\u009CJ0,1(r0)J0,2(r0)\u000F)\u00E2\u0088\u00921\u00E2\u0088\u00BC 3\u00E2\u0088\u0092 m\u00CB\u009C2r20\u00E2\u0088\u0092 \u000F 9q4m\u00CB\u009CJ0,1(r0)J0,2(r0)(4.59)and so if we expand m\u00CB\u009C = m\u00CB\u009C0 + \u000Fm\u00CB\u009C1 then m\u00CB\u009C0 =\u00E2\u0088\u009A3r0 and2m\u00CB\u009C0m\u00CB\u009C1r20= \u00E2\u0088\u0092 9q4m\u00CB\u009C0J0,1(r0)J0,2(r0)and som\u00CB\u009C1 = \u00E2\u0088\u00923q8J0,1(r0)J0,2(r0).Therefore there exists m = mb+ given bymb+ \u00E2\u0088\u00BC1\u000F\u00E2\u0088\u009A3r0 \u00E2\u0088\u00923q8J0,1(r0)J0,2(r0),such that \u00CE\u00BB = 0 in (4.56). When \u00CF\u0084 = 0, if m > mb+ then \u00CE\u00BB < 0 and ifm = O(\u000F\u00E2\u0088\u00921) m < mb+ then \u00CE\u00BB > 0. Combining this with what we concluded1794.3. Explicit Stability Formulation for the Gierer-Meinhardt Model on a Ringin m = O(1), for m in mb\u00E2\u0088\u0092 < m < mb+ , Re(\u00CE\u00BB) > 0. We plot the eigenvaluesin Figure 4.7 for \u00CF\u0084 = 0 using (4.56) for q = 2 and l = 5. We plot theeigenvalues for the two equilibrium radii r0 \u00E2\u0089\u0088 1.08 and r0 \u00E2\u0089\u0088 2.56 from Figure4.5.0 10 20 30 40 50 60 70 80 90\u00E2\u0088\u00920.500.511.522.53m\u00CE\u00BB r0 = 1.08r0 = 2.56Figure 4.7: Eigenvalues \u00CE\u00BB versus m for q = 2, s = 0, \u000F = 0.05, l = 5, and\u00CF\u0084 = 0 using (4.56). The solid curve is for r0 = 1.08 while the dashed curveis for r0 = 2.56.When \u00CF\u0084 6= 0 then similar to section 4.2.2 we can defineR(\u00CE\u00BB) \u00E2\u0089\u00A1 Cm(\u00CE\u00BB)\u00E2\u0088\u0092 f(\u00C2\u00B5), \u00C2\u00B5 = \u00CE\u00BB+\u000F2m2r20,Cm(\u00CE\u00BB) =13q(s+J0,1(r0)J0,2(r0)J\u00C2\u00AFm,1(\u00CE\u00B8\u00CE\u00BBr0)J\u00C2\u00AFm,2(\u00CE\u00B8\u00CE\u00BBr0)), f(\u00C2\u00B5) = \u00E2\u0088\u0092 32(\u00C2\u00B5\u00E2\u0088\u0092 3) , (4.60)and then eigenvalues of (4.56) become roots to (4.60). We make the split inthis way because then Cm(\u00CE\u00BB) is exactly that given by (3.19a) in section 3.2.1804.3. Explicit Stability Formulation for the Gierer-Meinhardt Model on a RingWe see then the effect of the explicit formulation is to make f(\u00C2\u00B5) an explicitfunction in terms of \u00CE\u00BB. From (4.60) we can show directly thatf \u00E2\u0080\u00B2(\u00C2\u00B5) > 0, \u00C2\u00B5 \u00E2\u0088\u0088 [0, 3)f \u00E2\u0080\u00B2\u00E2\u0080\u00B2(\u00C2\u00B5) > 0, \u00C2\u00B5 \u00E2\u0088\u0088 [0, 3)f(\u00C2\u00B5) < 0, \u00C2\u00B5 \u00E2\u0088\u0088 (3,\u00E2\u0088\u009E)and therefore all of the analysis of section (3.2) holds and we can immediatelyconclude Principal Result 3.2.9.1 holds, i.e. that there exists a range ofunstable real eigenvalues for m in mb\u00E2\u0088\u0092 < m < mb+ and that on 0 < m < mb\u00E2\u0088\u0092 ,there exists a Hopf bifurcation as \u00CF\u0084 is increased from zero. To compute theeigenvalues for \u00CF\u0084 > 0 we can use the algorithm outlined in section 3.2.10. Asan example we plot the eigenvalues for q = 2, s = 0, l = 5, \u000F = 0.05 andr0 = 1.08 and r0 = 2.56 in Figure 4.8 for \u00CF\u0084 = 6.1814.3. Explicit Stability Formulation for the Gierer-Meinhardt Model on a Ring0 5 10 15 20 25 30 35 4000.511.522.5mmaxRe(\u00CE\u00BB)Figure 4.8: Eigenvalues \u00CE\u00BB versus m for q = 2, s = 0, \u000F = 0.05, l = 5, and\u00CF\u0084 = 6. The lighter curve is for r0 = 1.08 while the heavy curve is for r = 2.56.We plot both for 0 < m < 40 since large m behaviour is not very impacted byincreasing \u00CF\u0084 and will be represented by Figure 4.7. The positive eigenvaluesare in dash while the negative eigenvalues are in solid.182Chapter 5Full Numerical Simulations ofthe Gierer-Meinhardt ModelWe now wish to verify some of the analysis we predicted by computing fullnumerical simulations ofvt = \u000F20\u00E2\u0088\u0086v \u00E2\u0088\u0092 v +vpuqg(v), (5.1a)\u00CF\u0084ut = D0\u00E2\u0088\u0086u\u00E2\u0088\u0092 u+vous(5.1b)on some domain \u00E2\u0084\u00A6, where g(v) = 1 if saturation is not considered and g(v) =1+\u00CF\u0083v2 with \u00CF\u0083 the saturation parameter if p = 2 and saturation is considered.We want to perform computations on both a stripe and a ring. If we performcomputations on a stripe then the domain is\u00E2\u0084\u00A6stripe \u00E2\u0089\u00A1 {(x, y)|0 \u00E2\u0089\u00A4 x \u00E2\u0089\u00A4 1, 0 \u00E2\u0089\u00A4 y \u00E2\u0089\u00A4 d0}subject to Neumann conditions on \u00E2\u0088\u0082\u00E2\u0084\u00A6. If we perform computations on a ringthen the domain is\u00E2\u0084\u00A6ring \u00E2\u0089\u00A1 {(r, \u00CE\u00B8)|0 \u00E2\u0089\u00A4 r \u00E2\u0089\u00A4 1, 0 \u00E2\u0089\u00A4 \u00CE\u00B8 \u00E2\u0089\u00A4 2pi}183Chapter 5. Full Numerical Simulations of the Gierer-Meinhardt Modelsubject to periodicity in \u00CE\u00B8, Neumann conditions at r = 1 and a compatibilitycondition \u00E2\u0088\u0082v/\u00E2\u0088\u0082r = \u00E2\u0088\u0082u/\u00E2\u0088\u0082r = 0 at r = 0. For analysis purposes, in otherchapters, we often rescaled the diffusion coefficient into the length scale butthis is impractical for computations where it is more natural to computeon fixed-length domains. Therefore, for comparison between numerical andanalytic work for the stripe we takeD = 1, \u000F =\u000F0\u00E2\u0088\u009AD0, d =d0\u00E2\u0088\u009AD0, l =12\u00E2\u0088\u009AD0, (5.2a)where these quantities are used in section 4.2. Similarly for the ring we takeD = 1, \u000F =\u000F0\u00E2\u0088\u009AD0, R = l =1\u00E2\u0088\u009AD0, (5.2b)where these quantities are used in Chapter 2, 3, and section 4.3.For the Laplace operator in each scenario we use a cell-centered discretizationon a uniformly spaced 400\u00C3\u0097400 rectangular grid in both space variables. Oneof the main advantages of this discretization over a node based discretizationin this context is that we avoid evaluating at the singularity r = 0 directly.For derivative conditions at the endpoints we use a ghost point formulation[73]. The time-stepping is done via the adaptive step method ode15s inMatlab with the Jacobian supplied. For the case of radial coordinates, af-ter computing on an (r, \u00CE\u00B8) grid we conformal map to the circle via Matlab\u00E2\u0080\u0099spol2cart function.In order to stimulate breakup instabilities we will consider random perturba-1845.1. Stripe Numerical Experimentstions sampled uniformly from [\u00E2\u0088\u0092\u00CE\u00B4, \u00CE\u00B4] with \u00CE\u00B4 = 0.001 from a base homoclinicorbit. In all of the numerical experiments in which we predict spot breakuppatterns, we expect that the most dominant Fourier mode will dictate thenumber of spots that appears in the pattern. However, in both geometries,we concluded that there were neutral stability points at modes m = mb\u00E2\u0088\u0092 andm = mb+ such that there is a range of real unstable eigenvalues (in the ab-sence of saturation) for which Re(\u00CE\u00BB) > 0 on mb\u00E2\u0088\u0092 < m < mb+ . Therefore, dueto the random nature of the perturbations, modes other than the maximalmode may dominate the stability pattern. On a long enough time-scale, wewould still expect the dominant mode to be observed but all of the analysisis based upon linear stability and once a breakup occurs, secondary dynam-ics may initiate. Furthermore, since random amplitude perturbations do notfavour positive or negative amplitudes, this will induce a phase correction.To alleviate these issues, when looking for spot breakup, we will perform afull discrete Fourier transform on the solution and then isolate modes thatare within 95% of the maximal mode (which we call dominant modes). Thisis designed to represent the most dominant wavemode interference. We willalso plot the maximal mode and compare that to the predicted number ofspots. Since we are not concerned with translational effects, we will filter them = 0 mode.5.1 Stripe Numerical ExperimentsWe begin by presenting some numerical experiments to coincide with theexplicit stripe formulation in 4.2. In this scenario Fourier perturbations are1855.1. Stripe Numerical Experimentsof the form exp(ix2kpi/d) with k \u00E2\u0088\u0088 N. We initialize the code with the steady-state (4.15) from section 4.2 subject to the random amplitude perturbationpreviously discussed. For the first experiment, we consider exponent set(3, 1, 3, 0), \u000F0 = 0.05, D0 = 1, \u00CF\u0084 = 0.1, and d0 = 2 which correspond to\u000F = 0.05 l = 1/2 in (4.27). We note that this is meant to represent the\u00CF\u0084 = 0 case but we take \u00CF\u0084 small and finite to allow full dynamics to occur. Weconsidered this case analytically in Figure 4.1 and in Table 4.1, we computedthat mdom = 7.42, which corresponds to kdom = 4, where we have roundeddown to the nearest integer. The number of spots is given byNspot =mdomd2pi=kdom2, (5.3)which leads to N = 2 spots in this case. Note that the extra factor of 2 comesfrom the fact that minima do not produce spots. The numerical results ofexperiment 1 are in Figure 5.1 for times t = 0 (a), t = 2.64 (b), t = 3.16(c), and t = 5 (d). As time progresses it does indeed appear that a two spotpattern is dominant. However we can see this more clearly by looking atthe Fourier transform in Figure 5.2 where indeed the most dominant modeproduces a two spot pattern.1865.1. Stripe Numerical Experiments(a) t = 0 (b) t = 2.64(c) t = 3.16 (d) t = 5Figure 5.1: Experiment 1: Contour plot of the solution v to (5.1a) with stripegeometry at four times with with exponent set (3, 1, 3, 0). The parametervalues are \u000F0 = 0.05, D0 = 1, \u00CF\u0084 = 0.1, and d0 = 2. This corresponds to\u000F = 0.05, l = 1/2, \u00CF\u0084 = 0.1, and d = 2 in (4.27) of Chapter 4.1875.1. Stripe Numerical Experiments0 20 4000.0050.010.0150.020.025 kdom=34kAmplitude0 20 40\u00E2\u0088\u00924\u00E2\u0088\u00922024 t: 0kPhase0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2\u00E2\u0088\u00921\u00E2\u0088\u00920.500.51 x 10\u00E2\u0088\u00924y(a)0 20 4000.511.522.5 kdom=4 8kAmplitude0 20 40\u00E2\u0088\u00924\u00E2\u0088\u00922024 t: 2.64kPhase0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2\u00E2\u0088\u00920.01\u00E2\u0088\u00920.00500.0050.01y(b)0 20 400246810 kdom=4 8kAmplitude0 20 40\u00E2\u0088\u00924\u00E2\u0088\u00922024 t: 3.16kPhase0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2\u00E2\u0088\u00920.04\u00E2\u0088\u00920.0200.020.04y(c)0 20 40020406080 kdom=2 4kAmplitude0 20 40\u00E2\u0088\u00924\u00E2\u0088\u00922024 t: 5kPhase0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2\u00E2\u0088\u00920.2\u00E2\u0088\u00920.100.10.2y(d)Figure 5.2: Experiment 1: Discrete Fourier transform of the solution v to(5.1a) with stripe geometry at four times with with exponent set (3, 1, 3, 0).The parameter values are \u000F0 = 0.05, D0 = 1, \u00CF\u0084 = 0.1, and d0 = 2. Theupper left plot shows the amplitudes from the Fourier transform while theupper right plot displays the phase. Dominant modes are defined as anymodes that have an amplitude within 95% of the largest amplitude mode.The bottom graphic in each panel shows an inverse Fourier transform of asolution comprised of only the most dominant mode.For experiment 2 we consider the same parameters as experiment 1 but withd0 = 3. This produces the same dominant mode mdom = 7.42 as this isindependent of the width of the rectangle but it does affect the number ofspots (5.3), where now we predict N = 3. The numerical results are in Figure5.3 for times t = 0 (a), t = 2.42 (b), t = 2.64 (c), and t = 5 (d).1885.1. Stripe Numerical Experiments(a) t = 0 (b) t = 2.42(c) t = 2.64 (d) t = 5Figure 5.3: Experiment 2: Contour plot of the solution v to (5.1a) with stripegeometry at four times with with exponent set (3, 1, 3, 0). The parametervalues are \u000F0 = 0.05, D0 = 1, \u00CF\u0084 = 0.1, and d0 = 3. This corresponds to\u000F = 0.05, l = 1/2, \u00CF\u0084 = 0.1, and d = 3 in (4.27) of Chapter 4.In Figure 5.4 we plot the Fourier transform results of experiment 2 wherein fact it is a 4 spot pattern that emerges which is not surprising since inactuality we expect N = 3.54 and so anything between a three and fourspot pattern matches well. If we compare Figures 5.2 and 5.4 then we seethat the dominant modes competing for instability are more spread out inexperiment 2 which is a side effect of taking d0 larger. From Figure 4.1 in4.2.1, the dominant modes within 95% of the maximum satisfy 5 < m < 141895.1. Stripe Numerical Experimentswhich hold for all d but this results in 3 < k < 8 for experiment 1 and4 < k < 13 for experiment 2. Therefore, the clustering of competitive modesis more spread out as d0 increases and the most dominant mode has a betterchance of surviving through most random perturbations.0 20 4000.0050.010.0150.020.025 kdom=34kAmplitude0 20 40\u00E2\u0088\u00924\u00E2\u0088\u00922024 t: 0kPhase0 0.5 1 1.5 2 2.5 3\u00E2\u0088\u00921\u00E2\u0088\u00920.500.51 x 10\u00E2\u0088\u00924y(a)0 20 4000.511.52 kdom=8kAmplitude0 20 40\u00E2\u0088\u00924\u00E2\u0088\u00922024 t: 2.42kPhase0 0.5 1 1.5 2 2.5 3\u00E2\u0088\u0092505 x 10\u00E2\u0088\u00923y(b)0 10 20 30 4001234 kdom=8kAmplitude0 10 20 30 40\u00E2\u0088\u00924\u00E2\u0088\u00922024 t: 2.64kPhase0 0.5 1 1.5 2 2.5 3\u00E2\u0088\u00920.01\u00E2\u0088\u00920.00500.0050.01y(c)0 20 400204060 kdom=4kAmplitude0 20 40\u00E2\u0088\u00924\u00E2\u0088\u00922024 t: 5kPhase0 0.5 1 1.5 2 2.5 3\u00E2\u0088\u00920.2\u00E2\u0088\u00920.100.10.2y(d)Figure 5.4: Experiment 2: Discrete Fourier transform of the solution v to(5.1a) with stripe geometry at four times with with exponent set (3, 1, 3, 0).The parameter values are \u000F0 = 0.05, D0 = 1, \u00CF\u0084 = 0.1, and d0 = 3. Theupper left plot shows the amplitudes from the Fourier transform while theupper right plot displays the phase. Dominant modes are defined as anymodes that have an amplitude within 95% of the largest amplitude mode.The bottom graphic in each panel shows an inverse Fourier transform of asolution comprised of only the most dominant mode.For experiment 3 we repeat experiment 1 but with D0 = 0.1 which affects1905.1. Stripe Numerical Experimentsl d, and \u000F via (5.2a). We plot the numerical results of this experiment inFigure 5.5 for times t = 0 (a), t = 2.64 (b), t = 4.34 (c), and t = 5 (d).If we compute the eigenvalues using (4.27) from 4.2.1 then we have thatmdom \u00E2\u0089\u0088 4.24 and therefore from (5.3) with d = 2\u00E2\u0088\u009A10, we predict N = 4spots. If we were to plot the eigenvalues for this case, the instability bandforms on 1.01 < m < 10.19 which is much narrower than the band for l = 1/2.This narrowed clustering means that the dominant mode should emerge moreprominently as there are fewer modes to compete with. Furthermore, as wesaw from Figure 4.1 as l increases the magnitude of the largest eigenvaluedecreases. This means that for this value of l, the breakup instability shouldtake longer to form compared to experiments 1 and 2. Indeed by lookingat Figure 5.6 all of these predictions are verified as the four spot patternemerges quite distinctly without much competition and the prominence ofthe pattern is not noticed until t = 4.34 which is in contrast to t = 3.16 andt = 2.64 for experiments 1 and 2 respectively.1915.1. Stripe Numerical Experiments(a) t = 0 (b) t = 2.64(c) t = 4.34 (d) t = 5Figure 5.5: Experiment 3: Contour plot of the solution v to (5.1a) with stripegeometry at four times with with exponent set (3, 1, 3, 0). The parametervalues are \u000F0 = 0.05, D0 = 0.1, \u00CF\u0084 = 0.1, and d0 = 2. This corresponds to\u000F = 0.05\u00E2\u0088\u009A10 \u00E2\u0089\u0088 0.1581, l =\u00E2\u0088\u009A10/2 \u00E2\u0089\u0088 1.58, \u00CF\u0084 = 0.1, and d = 2\u00E2\u0088\u009A10 \u00E2\u0089\u0088 6.32 in(4.27) of Chapter 4.1925.2. Ring Numerical Experiments0 20 4000.0050.010.0150.020.025 kdom=34kAmplitude0 20 40\u00E2\u0088\u00924\u00E2\u0088\u00922024 t: 0kPhase0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2\u00E2\u0088\u00921\u00E2\u0088\u00920.500.51 x 10\u00E2\u0088\u00924y(a)0 20 4000.10.20.30.4 kdom=8kAmplitude0 20 40\u00E2\u0088\u00924\u00E2\u0088\u00922024 t: 2.64kPhase0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2\u00E2\u0088\u00921\u00E2\u0088\u00920.500.51 x 10\u00E2\u0088\u00923y(b)0 10 20 30 4002468 kdom=8kAmplitude0 10 20 30 40\u00E2\u0088\u00924\u00E2\u0088\u00922024 t: 4.34kPhase0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2\u00E2\u0088\u00920.02\u00E2\u0088\u00920.0100.010.02y(c)0 20 4005101520 kdom=8kAmplitude0 20 40\u00E2\u0088\u00924\u00E2\u0088\u00922024 t: 5kPhase0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2\u00E2\u0088\u00920.0500.05y(d)Figure 5.6: Experiment 3: Discrete Fourier transform of the solution v to(5.1a) with stripe geometry at four times with with exponent set (3, 1, 3, 0).The parameter values are \u000F0 = 0.05, D0 = 0.1, \u00CF\u0084 = 0.1, and d0 = 2. Theupper left plot shows the amplitudes from the Fourier transform while theupper right plot displays the phase. Dominant modes are defined as anymodes that have an amplitude within 95% of the largest amplitude mode.The bottom graphic in each panel shows an inverse Fourier transform of asolution comprised of only the most dominant mode.5.2 Ring Numerical ExperimentsWe will now consider numerical experiments for the ring geometry. Withthis geometry, Fourier perturbations are of the form exp(im\u00CE\u00B8) for m \u00E2\u0088\u0088 Z but1935.2. Ring Numerical Experimentswe only consider the positive integers keeping in mind the complex conjugatemodes also exist. In this case, unlike (5.3), the number of spots Nspot = mdomexactly.5.2.1 Explicit FormulationWe will begin by considering the explicit formulation as in section 4.2. Recallfrom 4.3 that steady-states only exist for q = 2 and so we will consider theexponent set (3, 2, 3, 0). We will consider an experiment that corresponds toFigure 4.7 in 4.3 and therefore, we will take l = 5, \u000F = 0.05, and \u00CF\u0084 = 0.1.This corresponds to taking D0 = 0.04 and \u000F0 = 0.01. We will take r0 to bethe equilibrium values at l = 5 which are r0 = 1.08 and r0 = 2.56. Whenscaled to be on r \u00E2\u0088\u0088 [0, 1], this corresponds to r0 = 0.216 and r0 = 0.512respectively. Using (4.56) from 4.3, for r0 = 0.216, we expect an instabilityband on 2 < m < 35 with a dominant mode mdom = 12, and hence 12spots. For r0 = 0.512 we expect an instability band 7 < m < 84 with adominant mode mdom = 30, and hence 30 spots. The numerical results anddiscrete Fourier transforms are shown in Figure 5.7 where the computationsterminate once the breakup pattern has emerged. For r0 = 0.216 a 14 spotpattern emerges and for r0 = 0.512 a 28 spot pattern emerges. Both resultsare very close to the predicted dominant spot pattern but are not perfectagain owing to the wide range of unstable bands. To ensure the accuracy ofthe mode prediction we ran several random seed perturbations and saw thatthe average unstable modes were indeed 12 and 30 spots respectively.1945.2. Ring Numerical Experiments(a) r0 = 0.2160 10 20010203040 mdom=14mAmplitude0 10 20\u00E2\u0088\u00924\u00E2\u0088\u00922024 t: 5mPhase0 1 2 3 4 5 6 7\u00E2\u0088\u00920.1\u00E2\u0088\u00920.0500.050.1$\u00CE\u00B8$r0= 0.21571(b) r0 = 0.216(c) r0 = 0.5120 50 100012345 mdom=28mAmplitude0 50 100\u00E2\u0088\u00924\u00E2\u0088\u00922024 t: 3mPhase0 1 2 3 4 5 6 7\u00E2\u0088\u00920.02\u00E2\u0088\u00920.0100.010.02$\u00CE\u00B8$r0= 0.51247(d) r0 = 0.512Figure 5.7: Experiment 4: Contour and Fourier transform plot of the solutionv to (5.1a) with ring geometry with exponent set (3, 2, 3, 0). The parametervalues are \u000F0 = 0.01, D0 = 0.04, and \u00CF\u0084 = 0.1. This corresponds to \u000F = 0.05,and l = 5 in (4.56) of Chapter 4.5.2.2 Non-Explicit FormulationWe now consider numerical experiments for the non-explicit ring geometryconsidered in Chapter 2 and Chapter 3. For simplicity we will consider theexponent set (2, 1, 2, 0) moving forward but emphasize that unlike the explicitcase, general exponents q, o, and s could be taken and, in fact, we only requirep = 2 when considering saturation. For these experiments, we will also beginat a radius that is not necessarily the equilibrium radius and as such the ring1955.2. Ring Numerical Experimentsradius could change dynamically. We numerically compute the ring radiusas the r position which produces the largest value of v when \u00CE\u00B8 = 0. We notethat this definition may not be correct for a breakup pattern if the spotsseparate in non-radially symmetric ways. We will begin by demonstratingthe breakup pattern expected in Figure 3.5a of 3.2.10. Here we take D0 = 1and \u000F0 = 0.025, which corresponds to \u000F = 0.025 and R = 1. We also taker0 = 0.5, which is not the equilibrium radius. However, since we anticipatebreakup on an O(1) timescale versus motion on a O(\u000F\u00E2\u0088\u00922) timescale the radiusshould stay fairly static. Indeed in Figure 5.8 we plot the numerical resultsfor this experiment for times t = 0 (a), t = 5.32 (b), t = 6.32 (c), and t = 10(d) and the numeric ring radius stays essentially static. From Figure 3.5awe predict mdom = 4.80 and hence a four spot breakup pattern which isevidenced in Figure 5.8. Although the m = 6 mode is most dominant in theFourier transform Figure 5.9, it is very balanced with the m = 4 mode. InFigure 3.5a, the magnitude of the positive eigenvalues are near and below\u00CE\u00BB = 1 which explains the slow breakup instability in Figure 5.8.1965.2. Ring Numerical Experiments(a) t = 0, r0 = 0.5 (b) t = 5.32, r0 = 0.495(c) t = 6.32, r0 = 0.493 (d) t = 10, r0 = 0.493Figure 5.8: Experiment 5: Contour plot of the solution v to (5.1a) with ringgeometry at four times with with exponent set (2, 1, 2, 0). The parametervalues are \u000F0 = 0.025, D0 = 1, and \u00CF\u0084 = 0.1. This corresponds to \u000F = 0.025,R = 1, and \u00CF\u0084 = 0.1 in the numerical computation of (3.16) of Chapter 3.1975.2. Ring Numerical Experiments0 10 2000.0050.010.015 mdom=13 18mAmplitude0 10 20\u00E2\u0088\u00924\u00E2\u0088\u00922024 t: 0mPhase0 1 2 3 4 5 6 7\u00E2\u0088\u00924\u00E2\u0088\u00922024 x 10\u00E2\u0088\u00925\u00CE\u00B8r0= 0.5(a)0 10 2000.511.52 mdom=4 6mAmplitude0 10 20\u00E2\u0088\u00924\u00E2\u0088\u00922024 t: 5.26mPhase0 1 2 3 4 5 6 7\u00E2\u0088\u0092505 x 10\u00E2\u0088\u00923\u00CE\u00B8r0= 0.49501(b)0 5 10 15 200246 mdom=4 6mAmplitude0 5 10 15 20\u00E2\u0088\u00924\u00E2\u0088\u00922024 t: 6.32mPhase0 1 2 3 4 5 6 7\u00E2\u0088\u00920.02\u00E2\u0088\u00920.0100.010.02\u00CE\u00B8r0= 0.49252(c)0 10 201020304050 mdom=1 2 3 4 5 6mAmplitude0 10 20\u00E2\u0088\u00924\u00E2\u0088\u00922024 t: 10mPhase0 1 2 3 4 5 6 7\u00E2\u0088\u00920.2\u00E2\u0088\u00920.100.10.2\u00CE\u00B8r0= 0.49252(d)Figure 5.9: Experiment 5: Discrete Fourier transform of the solution v to(5.1a) with ring geometry at four times with exponent set (2, 1, 2, 0). Theparameter values are \u000F0 = 0.05, D0 = 1, and \u00CF\u0084 = 0.1. The upper left plotshows the amplitudes from the Fourier transform while the upper right plotdisplays the phase. Dominant modes are defined as any modes that have anamplitude within 95% of the largest amplitude mode. The bottom graphicin each panel shows an inverse Fourier transform of a solution comprised ofonly the most dominant mode.Adding SaturationWe will now consider the numerical simulations where saturation is included.This has the most significant impact on the dynamics because as we con-cluded in section 3.2.12, adding saturation can stabilize breakup patterns.1985.2. Ring Numerical ExperimentsAdding saturation affects the steady-state activator profile but for initial-izing the code, we still use the steady-state with saturation equal to zero.Therefore, in our simulations where saturation is added, we expect that therewill be a transient period where the homoclinic fattens (see Figure 2.2 fromsection 2.3) before any secondary dynamics occurs. For all numerical exper-iments unless otherwise stated we are considering D0 = 1, \u000F0 = \u000F = 0.025,R = 1 and r0 = 0.5. We begin by taking the saturation \u00CF\u0083 = 25. From (2.41),this corresponds to b = 0.1356 which from Figure 3.10 in section 3.2.12 doesnot stabilize the pattern. However, it does significantly shrink the instabilityband and the magnitude of the eigenvalues satisfying Re(\u00CE\u00BB) > 0. We there-fore expect that in this scenario, the instability should take longer to formand the dominant mode (mdom = 4.68) should be more pronounced whencompared to Figure 5.8 which uses the same parameters but \u00CF\u0083 = 0. Theresults for this experiment are in Figure 5.10 with the Fourier Transform inFigure 5.11 for times t = 10 (a), t = 26.2 (b), t = 37.3 (c), and t = 50 (d). Incontrast to Figure 5.8 at t = 10 when the curve had not only dissociated intospots, but secondary instabilities had reduced it to a single spot, for Figure5.10, instabilities have not even begun to form at this time. However,the ringis thicker when compared to the zero saturation case, and since the breakuppattern takes longer to occur we see the dynamics of the ring structure since,at t = 10, it has shrunk from its original position of r0 = 0.5 to r0 = 0.47.From Figure 2.9, this shrinking behaviour is predicted since r0 is greater thanthe equilibrium value.1995.2. Ring Numerical Experiments(a) t = 10, r0 = 0.478 (b) t = 26.2, r0 = 0.438(c) t = 37.3, r0 = 0.358 (d) t = 50, r0 = 0.243Figure 5.10: Experiment 6: Contour plot of the solution v to (5.1a) with ringgeometry at four times with with exponent set (2, 1, 2, 0). The parametervalues are \u000F0 = 0.025, D0 = 1, \u00CF\u0084 = 0.1, and \u00CF\u0083 = 25. This corresponds to\u000F = 0.025, R = 1, and \u00CF\u0084 = 0.1 in the numerical computation of (3.16) ofChapter 3.2005.2. Ring Numerical Experiments0 10 2000.050.10.150.2 mdom=4mAmplitude0 10 20\u00E2\u0088\u00924\u00E2\u0088\u00922024 t: 10mPhase0 1 2 3 4 5 6 7\u00E2\u0088\u00924\u00E2\u0088\u00922024 x 10\u00E2\u0088\u00924\u00CE\u00B8r0= 0.47756(a)0 10 2005101520 mdom=4mAmplitude0 10 20\u00E2\u0088\u00924\u00E2\u0088\u00922024 t: 26.2mPhase0 1 2 3 4 5 6 7\u00E2\u0088\u00920.0500.05\u00CE\u00B8r0= 0.43766(b)0 10 200510152025 mdom=4mAmplitude0 10 20\u00E2\u0088\u00924\u00E2\u0088\u00922024 t: 37.3mPhase0 1 2 3 4 5 6 7\u00E2\u0088\u00920.1\u00E2\u0088\u00920.0500.050.1\u00CE\u00B8r0= 0.35786(c)0 10 200510152025 mdom=4mAmplitude0 10 20\u00E2\u0088\u00924\u00E2\u0088\u00922024 t: 50mPhase0 1 2 3 4 5 6 7\u00E2\u0088\u00920.1\u00E2\u0088\u00920.0500.050.1\u00CE\u00B8r0= 0.24314(d)Figure 5.11: Experiment 6: Discrete Fourier transform of the solution v to(5.1a) with ring geometry at four times with exponent set (2, 1, 2, 0). Theparameter values are \u000F0 = 0.025, D0 = 1, \u00CF\u0084 = 0.1, and \u00CF\u0083 = 25. The upper leftplot shows the amplitudes from the Fourier transform while the upper rightplot displays the phase. Dominant modes are defined as any modes that havean amplitude within 95% of the largest amplitude mode. The bottom graphicin each panel shows an inverse Fourier transform of a solution comprised ofonly the most dominant mode.As time goes on, eventually the dominant four spot pattern emerges andthen the spot structure continues to shrink as a ring. This spot collocationdynamic behaviour was previously observed and analyzed for the Schnaken-burg model in [44].2015.2. Ring Numerical ExperimentsNext we consider the same parameter set but for \u00CF\u0083 = 950. For this sat-uration value and r0 = 0.5 we have that b = 0.2010 which by Figure 3.10does stabilize all of the breakup instability modes. Therefore we expect thereto be an initial transient period where the ring fattens from the effect of sat-uration but where the radius remains relatively static. Afterwards, the ringshould remain stable and shrink according to (2.43). During this dynamicprocess, the width of the ring should also increase because the value of b,even for fixed \u00CF\u0083, is intimately tied to r0 and increases as r0 decreases. Indeedall of this behaviour is noted in Figure 5.12 for times t = 0 (a), t = 3.48 (b),t = 12.2 (c), and t = 20 (d) where the ring does not breakup. By looking atthe Fourier transform plot in Figure 5.13 we see that while m = 4 remainsthe dominant integer mode, the amplitudes are decreasing over time and wedo truly have a stabilizing pattern. By the end of the simulation, while stillstable, the m = 1 mode has become dominant and if we look at Figure 5.14this is exactly what we see is the dominant mode for r0 = 0.213 and \u00CF\u0083 = 950computed using the techniques of section 3.2.10.2025.2. Ring Numerical Experiments(a) t = 0, r0 = 0.5 (b) t = 3.48, r0 = 0.478(c) t = 12.2, r0 = 0.365 (d) t = 20, r0 = 0.213Figure 5.12: Experiment 7: Contour plot of the solution v to (5.1a) with ringgeometry at four times with with exponent set (2, 1, 2, 0). The parametervalues are \u000F0 = 0.025, D0 = 1, \u00CF\u0084 = 0.1, and \u00CF\u0083 = 950. This corresponds to\u000F = 0.025, R = 1, and \u00CF\u0084 = 0.1 in the numerical computation of (3.16) ofChapter 3.2035.2. Ring Numerical Experiments0 10 2000.0050.010.015 mdom=13 18mAmplitude0 10 20\u00E2\u0088\u00924\u00E2\u0088\u00922024 t: 0mPhase0 1 2 3 4 5 6 7\u00E2\u0088\u00924\u00E2\u0088\u00922024 x 10\u00E2\u0088\u00925\u00CE\u00B8r0= 0.5(a)0 5 10 15 2001234 x 10\u00E2\u0088\u00924 mdom=4mAmplitude0 5 10 15 20\u00E2\u0088\u00924\u00E2\u0088\u00922024 t: 3.48mPhase0 1 2 3 4 5 6 7\u00E2\u0088\u00921\u00E2\u0088\u00920.500.51 x 10\u00E2\u0088\u00926\u00CE\u00B8r0= 0.47756(b)0 5 10 15 2002468 x 10\u00E2\u0088\u00924 mdom=1mAmplitude0 5 10 15 20\u00E2\u0088\u00924\u00E2\u0088\u00922024 t: 12.2mPhase0 1 2 3 4 5 6 7\u00E2\u0088\u00922\u00E2\u0088\u00921012 x 10\u00E2\u0088\u00926\u00CE\u00B8r0= 0.36534(c)0 5 10 15 2001234 x 10\u00E2\u0088\u00923 mdom=1mAmplitude0 5 10 15 20\u00E2\u0088\u00924\u00E2\u0088\u00922024 t: 20mPhase0 1 2 3 4 5 6 7\u00E2\u0088\u00921\u00E2\u0088\u00920.500.51 x 10\u00E2\u0088\u00925\u00CE\u00B8r0= 0.21322(d)Figure 5.13: Experiment 7: Discrete Fourier transform of the solution v to(5.1a) with ring geometry at four times with exponent set (2, 1, 2, 0). Theparameter values are \u000F0 = 0.05, D0 = 1, \u00CF\u0084 = 0.1, and \u00CF\u0083 = 950. The upper leftplot shows the amplitudes from the Fourier transform while the upper rightplot displays the phase. Dominant modes are defined as any modes that havean amplitude within 95% of the largest amplitude mode. The bottom graphicin each panel shows an inverse Fourier transform of a solution comprised ofonly the most dominant mode.2045.2. Ring Numerical Experiments0 1 2 3 4 5 6\u00E2\u0088\u00920.5\u00E2\u0088\u00920.4\u00E2\u0088\u00920.3\u00E2\u0088\u00920.2\u00E2\u0088\u00920.100.10.20.30.40.5mmax(Re(\u00CE\u00BB))Figure 5.14: Eigenvalues of (3.16) for exponent set (2, 1, 2, 0), \u000F = 0.05, \u00CF\u0084 = 0,and R = 1 with r0 = 0.213 and \u00CF\u0083 = 950.We now consider the perturbations that lead to zig-zag instabilities. In sec-tion 2.3.2, we determined that the velocity corrections for a near circular per-turbation are in phase with the perturbation itself when r0 is small enough.Therefore, we expect that given an initial near circular perturbation withD = O(1), this curve should either grow or shrink by (2.43) and slowly cir-cularize via (2.77). In Figure 5.15, taken at times t = 0 (a), t = 4.92 (b),t = 15.76 (c), and t = 20 (d) we take all of the previous parameters except westart with an initial curve radius r0 = 0.5+0.02 cos(6\u00CE\u00B8). We still initialize thecurve with a homoclinic orbit as if there were no saturation. The homoclinicinitially fattens due to the presence of saturation, but as the curve continuesto evolve it indeed circularizes.2055.2. Ring Numerical Experiments(a) t = 0 (b) t = 4.92(c) t = 15.76 (d) t = 20Figure 5.15: Experiment 8: Contour plot of the solution v to (5.1a) with ringgeometry at four times with with exponent set (2, 1, 2, 0). The parametervalues are \u000F0 = 0.025, D0 = 1, \u00CF\u0084 = 0.1, and \u00CF\u0083 = 950. This correspondsto \u000F = 0.025, R = 1, and \u00CF\u0084 = 0.1. We take as an initial radius r0 =0.5 + 0.02 cos(6\u00CE\u00B8).Conversely, if r0 \u001D 1 then we could no longer guarantee that near circularperturbations would circularize. To demonstrate this, we compute a simula-tion for R = 10 and r0 = 5 in physical space which translates to parametersD0 = 0.01, and r0 = 0.5 in computational space. Furthermore, we take\u000F0 = 0.01 (\u000F = 0.1) to avoid the weak interaction regime where D = O(\u000F2)and different analysis is required (cf. [39]). While the rescaling in (5.2b)is equivarient in the spatial scales of the problem, it does impact the value2065.2. Ring Numerical Experimentsof U0 on the curve and hence the saturation required to achieve a given bvalue by (2.41). We wish to use the same b value as in Figure 5.12 and5.15 which was b = 0.210. To achieve this level of saturation for the currentparameter set we need to take \u00CF\u0083 = 5910. As in Figure 5.15, we initializethe curve with a perturbed radius r0 = 0.5 + 0.02 cos(6\u00CE\u00B8). Figure 5.16 showsthe results of this simulation at times t = 0 (a), t = 15 (b), t = 31 (c), andt = 50 (d) . We have that the curve initially fattens due to the presence ofsaturation and then begins to accentuate the small angular perturbations,overall lengthening the curve by the end of the simulation. Even though thiscurve destabilizes the circular solution it does not itself undergo any breakupinstabilities as it evolves.2075.2. Ring Numerical Experiments(a) t = 0 (b) t = 15(c) t = 31 (d) t = 50Figure 5.16: Experiment 9: Contour plot of the solution v to (5.1a) with ringgeometry at four times with with exponent set (2, 1, 2, 0). The parametervalues are \u000F0 = 0.01, D0 = 0.01, \u00CF\u0084 = 0.1, and \u00CF\u0083 = 5910. This correspondsto \u000F = 0.1, R = 10, and \u00CF\u0084 = 0.1. We take as an initial radius r0 = 0.5 +0.02 cos(6\u00CE\u00B8).The delicacy of manipulating \u000F0 in correspondence with D0 to avoid theweak interaction regime and the affect of \u00CF\u0083 increasing for a desired b whenthe domain length increases, is the side-effect of computing on a fixed domainlength. Therefore, it can be difficult to capture some of the rich dynamicsthat occur on an O(\u000F\u00E2\u0088\u00922) timescale for large curves or domains. Fortunately,the techniques developed in Chapter 6 for solving dynamics on arbitrarycurves are designed specifically to capture this behaviour and the issues that2085.2. Ring Numerical Experimentsarise from full numerical simulations underscores the need for this specializedcomputational infrastructure.209Chapter 6Solving the Gierer-MeinhardtProblem for Arbitrary Curvesin Two DimensionsWe consider the singular interface limit inhibitor problem (2.19) derived in2.2.1,D\u00E2\u0088\u0086u\u00E2\u0088\u0092 u = 0, x \u00E2\u0088\u0088 \u00E2\u0084\u00A6\u00E2\u0088\u0082u\u00E2\u0088\u0082n= 0, x \u00E2\u0088\u0088 \u00E2\u0088\u0082\u00E2\u0084\u00A6u = U0(s), x \u00E2\u0088\u0088 \u00CE\u0093[\u00E2\u0088\u0082u\u00E2\u0088\u0082n]\u00CE\u0093= \u00E2\u0088\u0092 1D\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ef(U0, v\u00CB\u009C0) d\u00CE\u00B7\u00CB\u0086, x \u00E2\u0088\u0088 \u00CE\u0093subject toV0 = \u00CE\u00BA0 +H(\u00E2\u0088\u0082u\u00E2\u0088\u0082n\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00CE\u00B7=0++\u00E2\u0088\u0082u\u00E2\u0088\u0082n\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00CE\u00B7=0\u00E2\u0088\u0092),with H given by (2.20).2106.1. Layer Potential Formulation6.1 Layer Potential FormulationBefore considering the dynamic portion of the curve, we will first look atsolving the problem of a static closed curve and then incorporate the veloc-ity condition into the calculation. For either case, we will use the methodof Layer potentials [34], which originates in the field of electrostatics. Themotivation comes from solving Laplace\u00E2\u0080\u0099s equation subject to closed curvesand surfaces at fixed potentials or electric fluxes. The electric potential atany point in space is written as an integral of the charge potential differ-ence between a set of delta sources on the curve and the desired point inspace. Mathematically, the method is similar to the Green\u00E2\u0080\u0099s function formu-lation. Typically, when solving these problems on non-standard geometries,the method of images is used to modify the Green\u00E2\u0080\u0099s function and use thecurve data as a source. The method of layer potentials does the opposite ofthis by using the original free space Green\u00E2\u0080\u0099s function (or its derivative) and,instead, considers the source density to be unknown. In order to proceedthen, we need the free space Green\u00E2\u0080\u0099s function or fundamental solution to themodified Helmholtz equation:D\u00E2\u0088\u0086q\u00CE\u00A8\u00E2\u0088\u0092\u00CE\u00A8 = \u00E2\u0088\u0092\u00CE\u00B4(p\u00E2\u0088\u0092 q), p, q \u00E2\u0088\u0088 R2where the subscript indicates differentiation with respect to q variables. Im-posing that the solution decay at infinity allow us to write the fundamentalsolution as\u00CE\u00A8(p, q) =12piDK0( |p\u00E2\u0088\u0092 q|\u00E2\u0088\u009AD), (6.1)2116.1. Layer Potential Formulationwith K0 the second modified Bessel function of zeroth order. Using the layerpotential formulation we need to find the continuous source density \u00CF\u0086 on \u00CE\u0093such that at any point x in space,u(x) =\u00E2\u0088\u00AB\u00CE\u0093\u00CE\u00A8(x, q)\u00CF\u0086(q) dqs. (6.2)With the integral in this form, it is often called the single, or monopole,layer potential (as opposed to the double, or dipole, layer potential thatuses the derivative of \u00CE\u00A8 [34]). The subscript s indicates that the integral iswith respect to the arclength of \u00CE\u0093 in the q variables. The method of layerpotentials, having been created for electrostatic problems, has a rich historyof investigation for the Laplace operator (cf. [34], [36], [18]) and so it is usefulto exploit those results. The fundamental solution to Laplace\u00E2\u0080\u0099s equation,\u00E2\u0088\u0086q\u00CE\u00A6 = \u00E2\u0088\u0092\u00CE\u00B4(p\u00E2\u0088\u0092 q)is\u00CE\u00A6 = \u00E2\u0088\u0092 12pilog |p\u00E2\u0088\u0092 q|. (6.3)The following property holds for the Laplace single layer potential with sourcedensity \u00CF\u0086 for some curve \u00CE\u0093 with Dirichlet boundary conditions u = f(x), x \u00E2\u0088\u0088\u00CE\u0093:f(x) =\u00E2\u0088\u00AB\u00CE\u0093\u00CE\u00A6(x, q)\u00CF\u0086(q) dqs. x \u00E2\u0088\u0088 \u00CE\u0093.2126.1. Layer Potential FormulationFurthermore, the limiting values of the normal derivative with the positivenormal on \u00CE\u0093 taken as the interior normal of a single curve arelim\u00CE\u00B1\u00E2\u0086\u00920+\u00E2\u0088\u0082u\u00E2\u0088\u0082nx(x+ \u00CE\u00B1n\u00CB\u0086xi) =\u00E2\u0088\u00AB\u00CE\u0093\u00E2\u0088\u0082\u00CE\u00A6\u00E2\u0088\u0082nx(x, q)\u00CF\u0086(q) dqs \u00E2\u0088\u0092\u00CF\u0086(x)2, (6.4a)lim\u00CE\u00B1\u00E2\u0086\u00920+\u00E2\u0088\u0082u\u00E2\u0088\u0082nx(x\u00E2\u0088\u0092 \u00CE\u00B1n\u00CB\u0086xi) =\u00E2\u0088\u00AB\u00CE\u0093\u00E2\u0088\u0082\u00CE\u00A6\u00E2\u0088\u0082nx(x, q)\u00CF\u0086(q) dqs +\u00CF\u0086(x)2, (6.4b)where we define normal derivatives with respect to a coordinate system a as,\u00E2\u0088\u0082u\u00E2\u0088\u0082na= \u00E2\u0088\u0087au \u00C2\u00B7 n\u00CB\u0086a.Subtracting (6.4b) from (6.4a) we get[\u00E2\u0088\u0082u\u00E2\u0088\u0082n]= \u00E2\u0088\u0092 (n\u00CB\u0086 \u00C2\u00B7 n\u00CB\u0086i)\u00CF\u0086(x), x \u00E2\u0088\u0088 \u00CE\u0093.The subscript i on the normal in (6.4) indicates that we are explicitly con-sidering \u00CE\u00B1 > 0 to be traversing the inner normal. It is important to makethis distinction if \u00CE\u0093 has multiple curves since n\u00CB\u0086 = \u00C2\u00B1n\u00CB\u0086i, depending on curveorientation. For a derivation of the properties of the Laplace operator, seeAppendix A. The Dirichlet and Neumann jump properties are inherentlytied to the logarithmic singularity as p\u00E2\u0086\u0092 q in \u00CE\u00A6(p, q). If we look at the samelimit for the fundamental solution \u00CE\u00A8(p, q) we have [1],\u00CE\u00A8(p, q) \u00E2\u0088\u00BCp\u00E2\u0086\u0092q1D\u00CE\u00A6(p, q) +R(|p\u00E2\u0088\u0092 q|)with R a differentiable function. Therefore the singularity structure betweenthe two problems is the same and we can immediately write down the prop-2136.1. Layer Potential Formulationerties for the modified Helmholtz problem (2.19):U0 =\u00E2\u0088\u00AB\u00CE\u0093\u00CE\u00A8(x, q)\u00CF\u0086(q) dqs, x \u00E2\u0088\u0088 \u00CE\u0093 (6.5a)[\u00E2\u0088\u0082u\u00E2\u0088\u0082n]\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00CE\u0093= \u00E2\u0088\u0092(n\u00CB\u0086 \u00C2\u00B7 n\u00CB\u0086i)\u00CF\u0086(x)D, x \u00E2\u0088\u0088 \u00CE\u0093 (6.5b)where the extra factor of D comes from the singularity relationship between\u00CE\u00A6 and \u00CE\u00A8.6.1.1 Incorporating Neumann Boundary ConditionsWe have thus far considered how to solve the problem (2.19) for a closed curve\u00CE\u0093 on an unbounded domain. However, the problem we want to ultimatelysolve has a closed domain with Neumann boundary conditions. This is aneasy feature to include in the layer potential formulation since the curve \u00CE\u0093does not have to be a single closed curve, and as such, we can define \u00E2\u0088\u0082\u00E2\u0084\u00A6 to beone of these curves. We can therefore write the new layer potential problemasu(x) =\u00E2\u0088\u00AB\u00CE\u0093\u00CE\u00A8(x, q)\u00CF\u0086(q) dqs +\u00E2\u0088\u00AB\u00E2\u0088\u0082\u00E2\u0084\u00A6\u00CE\u00A8(x, q)\u00CF\u0086b(q) dqs. (6.6)We introduce a new density \u00CF\u0086b here for the integral on the domain boundaryand split the integral in this way so that \u00CF\u0086b belongs to the, always static,curve \u00E2\u0088\u0082\u00E2\u0084\u00A6 and any dynamic curves belong to the set \u00CE\u0093. By replacing (6.2)2146.1. Layer Potential Formulationwith (6.6), we can define the new jump in normal derivative for x \u00E2\u0088\u0088 \u00CE\u0093:lim\u00CE\u00B1\u00E2\u0086\u00920+\u00E2\u0088\u0082u\u00E2\u0088\u0082nx(x+ \u00CE\u00B1n\u00CB\u0086xi) =\u00E2\u0088\u00AB\u00CE\u0093\u00E2\u0088\u0082\u00CE\u00A8\u00E2\u0088\u0082nx(x, q)\u00CF\u0086(q) dqs \u00E2\u0088\u0092\u00CF\u0086(x)2D+\u00E2\u0088\u00AB\u00E2\u0088\u0082\u00E2\u0084\u00A6\u00E2\u0088\u0082\u00CE\u00A8\u00E2\u0088\u0082nx(x, q)\u00CF\u0086b(q) dqs (6.7a)lim\u00CE\u00B1\u00E2\u0086\u00920+\u00E2\u0088\u0082u\u00E2\u0088\u0082nx(x\u00E2\u0088\u0092 \u00CE\u00B1n\u00CB\u0086xi) =\u00E2\u0088\u00AB\u00CE\u0093\u00E2\u0088\u0082\u00CE\u00A8\u00E2\u0088\u0082nx(x, q)\u00CF\u0086(q) dqs +\u00CF\u0086(x)2D+\u00E2\u0088\u00AB\u00E2\u0088\u0082\u00E2\u0084\u00A6\u00E2\u0088\u0082\u00CE\u00A8\u00E2\u0088\u0082nx(x, q)\u00CF\u0086b(q) dqs (6.7b)where we note that substitution is applicable in the last integral for x \u00E2\u0088\u0088 \u00CE\u0093and q \u00E2\u0088\u0088 \u00E2\u0088\u0082\u00E2\u0084\u00A6 since then, \u00CE\u00A8(p, q) is no where singular. It is now straightforwardto see that (6.5) becomesU0 =\u00E2\u0088\u00AB\u00CE\u0093\u00CE\u00A8(x, q)\u00CF\u0086(q) dqs +\u00E2\u0088\u00AB\u00E2\u0088\u0082\u00E2\u0084\u00A6\u00CE\u00A8(x, q)\u00CF\u0086b(q) dqs, x \u00E2\u0088\u0088 \u00CE\u0093 (6.8a)[\u00E2\u0088\u0082u\u00E2\u0088\u0082n]\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00CE\u0093= \u00E2\u0088\u0092(n\u00CB\u0086 \u00C2\u00B7 n\u00CB\u0086i)\u00CF\u0086(x)D, x \u00E2\u0088\u0088 \u00CE\u0093. (6.8b)To incorporate the condition \u00E2\u0088\u0082u\u00E2\u0088\u0082n = 0 on \u00E2\u0088\u0082\u00E2\u0084\u00A6, we once again look at thenormal jump formula. This boundary condition states that as we approachthe boundary from exterior to \u00E2\u0084\u00A6, the flux must vanish and thereforelim\u00CE\u00B1\u00E2\u0086\u00920+\u00E2\u0088\u0082u\u00E2\u0088\u0082n(x+ \u00CE\u00B1n\u00CB\u0086\u00E2\u0084\u00A6e) = 0, x \u00E2\u0088\u0088 \u00E2\u0088\u0082\u00E2\u0084\u00A6,2156.1. Layer Potential Formulationwhere n\u00CB\u0086\u00E2\u0084\u00A6e is the normal from \u00E2\u0088\u0082\u00E2\u0084\u00A6 pointing to the exterior of \u00E2\u0084\u00A6. Since we areconsidering the exterior normal and n\u00CB\u0086\u00E2\u0084\u00A6e = \u00E2\u0088\u0092n\u00CB\u0086\u00E2\u0084\u00A6i then we use (6.4b) and get0 =\u00E2\u0088\u00AB\u00CE\u0093\u00E2\u0088\u0082\u00CE\u00A8\u00E2\u0088\u0082n\u00E2\u0084\u00A6(x, q)\u00CF\u0086(q) dqs +\u00CF\u0086b(x)2D+\u00E2\u0088\u00AB\u00E2\u0088\u0082\u00E2\u0084\u00A6\u00E2\u0088\u0082\u00CE\u00A8\u00E2\u0088\u0082n\u00E2\u0084\u00A6(x, q)\u00CF\u0086b(q) dqs, x \u00E2\u0088\u0088 \u00E2\u0088\u0082\u00E2\u0084\u00A6.(6.9)6.1.2 Scaled Arclength parametrizationIn order to compute the integrals, we will use a scaled arclength parametriza-tion. First consider that \u00CE\u0093 is composed of M distinct closed curves,\u00CE\u0093 =M\u00E2\u008B\u0083j=1\u00CE\u0093j,and write (6.6) asu(x) =M\u00E2\u0088\u0091j=1\u00E2\u0088\u00AB\u00CE\u0093j\u00CE\u00A8(x, qj)\u00CF\u0086(qj) dqjs +\u00E2\u0088\u00AB\u00E2\u0088\u0082\u00E2\u0084\u00A6\u00CE\u00A8(x, q)\u00CF\u0086b(q) dqs, x \u00E2\u0088\u0088 \u00E2\u0084\u00A6 \u00E2\u0088\u00AA \u00CE\u0093.(6.10)Define the scaled arclength coordinate as \u00CF\u0083 = sj/Lj, where Lj is the lengthof curve j. In order to not confuse this with the saturation parameter \u00CF\u0083 fromthe previous chapters, we will now reserve \u00CF\u0083\u00CB\u0086 for the saturation parameter.Let the parametrization of \u00CE\u0093j be written aszj(\u00CF\u0083) = \u00E3\u0080\u0088z1j(\u00CF\u0083), z2j(\u00CF\u0083)\u00E3\u0080\u0089,2166.1. Layer Potential Formulationand the parametrization of \u00E2\u0088\u0082\u00E2\u0084\u00A6 aszb(\u00CF\u0083) = \u00E3\u0080\u0088z1b(\u00CF\u0083), z2b(\u00CF\u0083)\u00E3\u0080\u0089.With this in mind, we can write (6.10) for x \u00E2\u0088\u0088 \u00E2\u0084\u00A6 \u00E2\u0088\u00AA \u00CE\u0093 asu(x) =M\u00E2\u0088\u0091j=1Lj\u00E2\u0088\u00AB 10\u00CE\u00A8(x, zj(\u00CF\u0083))\u00CF\u0086j(\u00CF\u0083) d\u00CF\u0083 + L\u00E2\u0088\u0082\u00E2\u0084\u00A6\u00E2\u0088\u00AB 10\u00CE\u00A8(x, zb(\u00CF\u0083))\u00CF\u0086b(\u00CF\u0083) d\u00CF\u0083, (6.11)where \u00CF\u0086j = \u00CF\u0086(zj(\u00CF\u0083)). We can write the Dirichlet condition (6.8a) asU0(\u00CF\u0083\u00E2\u0088\u0097) =Lm\u00E2\u0088\u00AB 10\u00CE\u00A8(zm(\u00CF\u0083\u00E2\u0088\u0097), zm(\u00CF\u0083))\u00CF\u0086m(\u00CF\u0083) d\u00CF\u0083+M\u00E2\u0088\u0091j=1j 6=mLj\u00E2\u0088\u00AB 10\u00CE\u00A8(zm(\u00CF\u0083\u00E2\u0088\u0097), zj(\u00CF\u0083))\u00CF\u0086j(\u00CF\u0083) d\u00CF\u0083+ L\u00E2\u0088\u0082\u00E2\u0084\u00A6\u00E2\u0088\u00AB 10\u00CE\u00A8(zm(\u00CF\u0083\u00E2\u0088\u0097), zb(\u00CF\u0083))\u00CF\u0086b(\u00CF\u0083) d\u00CF\u0083, (6.12)where we have isolated the integral with the singularity separately from thesum. We can write the jump condition (6.8b) as\u00CF\u0086m(\u00CF\u0083\u00E2\u0088\u0097) = (n\u00CB\u0086 \u00C2\u00B7 n\u00CB\u0086i)\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ef(U0(\u00CF\u0083\u00E2\u0088\u0097), v\u00CB\u009C0) d\u00CE\u00B7\u00CB\u0086 (6.13)2176.1. Layer Potential Formulationwhere we have simplified the jump in the normal derivative using (2.19d).Finally, we can write the Neumann boundary condition (6.9) asM\u00E2\u0088\u0091j=1Lj\u00E2\u0088\u00AB 10\u00E2\u0088\u0082\u00CE\u00A8\u00E2\u0088\u0082n\u00E2\u0084\u00A6(zb(\u00CF\u0083\u00E2\u0088\u0097), zj(\u00CF\u0083))\u00CF\u0086j(\u00CF\u0083) d\u00CF\u0083 +\u00CF\u0086b(\u00CF\u0083\u00E2\u0088\u0097)2D+ L\u00E2\u0088\u0082\u00E2\u0084\u00A6\u00E2\u0088\u00AB 10\u00E2\u0088\u0082\u00CE\u00A8\u00E2\u0088\u0082n\u00E2\u0084\u00A6(zb(\u00CF\u0083\u00E2\u0088\u0097), zb(\u00CF\u0083))\u00CF\u0086b(\u00CF\u0083) d\u00CF\u0083 = 0. (6.14)A Note on UniquenessWe briefly comment on the uniqueness of the boundary value problem. Whenthe Dirichlet condition U0 is prescribed then it can be shown that the solutionto (6.12) is unique [68]. However, since the Dirichlet value on the curve isan unknown in the system, the guarantee of uniqueness no longer appliesand solution bifurcations may occur. We will demonstrate an example ofnon-unique solutions when we consider the GMS model in section 6.3.6.1.3 Curve DynamicsThe motion for the curve will be dictated by the normal velocity conditiongiven by (2.19e). However, in using the scaled arclength formulation, if asolution exists then the same solution would exist for any rotation of thecurve. To remedy this, we will introduce a tangential velocity but give itzero mean to prevent a rotation from an initial configuration. With theseconsiderations in mind for each \u00CF\u0083\u00E2\u0088\u0097 \u00E2\u0088\u0088 [0, 1), we have that the motion on curvem is given bydzmdt= V n\u00CB\u0086 + Vtt\u00CB\u0086, (6.15)2186.1. Layer Potential Formulationand that\u00E2\u0088\u00AB 10Vt d\u00CF\u0083 =\u00E2\u0088\u00AB 10dzmdt\u00C2\u00B7 t\u00CB\u0086 d\u00CF\u0083 = 0. (6.16)By prescribing the normal velocity and enforcing equal arclength (|zm\u00CF\u0083 | = L)for all time then the mean value condition is sufficient to implicitly impose atangential velocity [64]. Note that this is in contrast to imposing a tangentialvelocity that guarantees a given parametrization is equal arclength [84].6.1.4 Normal Velocity ConditionWe now want to incorporate the velocity condition given by (2.19e) into thelayer potential formulation. This turns out to be extremely straightforwardby using the jump conditions (6.7),lim\u00CE\u00B1\u00E2\u0086\u00920+\u00E2\u0088\u0082u\u00E2\u0088\u0082nx(x+ \u00CE\u00B1n\u00CB\u0086xi) + lim\u00CE\u00B1\u00E2\u0086\u00920+\u00E2\u0088\u0082u\u00E2\u0088\u0082nx(x\u00E2\u0088\u0092 \u00CE\u00B1n\u00CB\u0086xi)= 2\u00E2\u0088\u00AB\u00CE\u0093\u00E2\u0088\u0082\u00CE\u00A8\u00E2\u0088\u0082nx(x, q)\u00CF\u0086(q) dqs + 2\u00E2\u0088\u00AB\u00E2\u0088\u0082\u00E2\u0084\u00A6\u00E2\u0088\u0082\u00CE\u00A8\u00E2\u0088\u0082nx(x, q)\u00CF\u0086b(q) dqs.In (2.19e) we need to add the derivative contributions of x+\u00CE\u00B1n\u00CB\u0086 with n\u00CB\u0086 = \u00C2\u00B1n\u00CB\u0086i.However, this only affects the sign on the non-integral term in (6.7) and will2196.1. Layer Potential Formulationalways vanish when added. Therefore, we have thatV (\u00CF\u0083\u00E2\u0088\u0097) = \u00CE\u00BA+ 2H(U0)(Lm\u00E2\u0088\u00AB 10\u00E2\u0088\u0082\u00CE\u00A8\u00E2\u0088\u0082nz\u00E2\u0088\u0097m(zm(\u00CF\u0083\u00E2\u0088\u0097), zm(\u00CF\u0083))\u00CF\u0086m(\u00CF\u0083) d\u00CF\u0083 +M\u00E2\u0088\u0091j=1j 6=mLj\u00E2\u0088\u00AB 10\u00E2\u0088\u0082\u00CE\u00A8\u00E2\u0088\u0082nz\u00E2\u0088\u0097m(zm(\u00CF\u0083\u00E2\u0088\u0097), zj(\u00CF\u0083))\u00CF\u0086j(\u00CF\u0083) d\u00CF\u0083+L\u00E2\u0088\u0082\u00E2\u0084\u00A6\u00E2\u0088\u00AB 10\u00E2\u0088\u0082\u00CE\u00A8\u00E2\u0088\u0082nz\u00E2\u0088\u0097m(zm(\u00CF\u0083\u00E2\u0088\u0097), zb(\u00CF\u0083))\u00CF\u0086b(\u00CF\u0083) d\u00CF\u0083), (6.17)where we have defined, z\u00E2\u0088\u0097m = zm(\u00CF\u0083\u00E2\u0088\u0097).6.1.5 Singular IntegrationIf we attempt to solve (6.12), (6.13), (6.14), and (6.17) using standard numer-ical techniques, there will be an issue when some of the integrands becomesingular. First consider the integral\u00E2\u0088\u00AB 10\u00CE\u00A8(z(\u00CF\u0083\u00E2\u0088\u0097), z(\u00CF\u0083))\u00CF\u0086(\u00CF\u0083) d\u00CF\u0083 =12piD\u00E2\u0088\u00AB 10K0(\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3z(\u00CF\u0083)\u00E2\u0088\u0092 z(\u00CF\u0083\u00E2\u0088\u0097)\u00E2\u0088\u009AD\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3)\u00CF\u0086(\u00CF\u0083) d\u00CF\u0083, (6.18)which is singular when \u00CF\u0083 = \u00CF\u0083\u00E2\u0088\u0097. To determine the nature of the singularity,we perform an asymptotic expansion as \u00CF\u0083 \u00E2\u0089\u0088 \u00CF\u0083\u00E2\u0088\u0097,z(\u00CF\u0083) \u00E2\u0088\u00BC z(\u00CF\u0083\u00E2\u0088\u0097) + L(\u00CF\u0083 \u00E2\u0088\u0092 \u00CF\u0083\u00E2\u0088\u0097)t\u00CB\u0086z\u00E2\u0088\u0097 +L2\u00CE\u00BA2(\u00CF\u0083 \u00E2\u0088\u0092 \u00CF\u0083\u00E2\u0088\u0097)2n\u00CB\u0086z\u00E2\u0088\u0097 +O(\u00CF\u0083 \u00E2\u0088\u0092 \u00CF\u0083\u00E2\u0088\u0097)3 (6.19)2206.1. Layer Potential Formulationwith t\u00CB\u0086z\u00E2\u0088\u0097 and n\u00CB\u0086z\u00E2\u0088\u0097 , the tangent and normal vectors at z(\u00CF\u0083\u00E2\u0088\u0097) respectively. Herewe have used that\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3dzd\u00CF\u0083\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3 = L,\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3d2zd\u00CF\u00832\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3= L2\u00CE\u00BA,with \u00CE\u00BA the curvature at z(\u00CF\u0083\u00E2\u0088\u0097) and L the length. We can compute that theasymptotic norm is|z(\u00CF\u0083)\u00E2\u0088\u0092 z(\u00CF\u0083\u00E2\u0088\u0097)| = L|\u00CF\u0083 \u00E2\u0088\u0092 \u00CF\u0083\u00E2\u0088\u0097|+O(\u00CF\u0083 \u00E2\u0088\u0092 \u00CF\u0083\u00E2\u0088\u0097)3, (6.20)and therefore have that the Bessel function for \u00CF\u0083 \u00E2\u0088\u0092 \u00CF\u0083\u00E2\u0088\u0097 \u001C 1 has the form [1],K0(\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3z(\u00CF\u0083)\u00E2\u0088\u0092 z(\u00CF\u0083\u00E2\u0088\u0097)\u00E2\u0088\u009AD\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3)\u00E2\u0088\u00BC log(2\u00E2\u0088\u009ADL)\u00E2\u0088\u0092 \u00CE\u00B3 \u00E2\u0088\u0092 log |\u00CF\u0083 \u00E2\u0088\u0092 \u00CF\u0083\u00E2\u0088\u0097|+O(\u00CF\u0083 \u00E2\u0088\u0092 \u00CF\u0083\u00E2\u0088\u0097)2,where \u00CE\u00B3 is the Euler-Mascheroni constant. To alleviate the logarithmic sin-gularity, we will add and subtract the log term to (6.18), to get\u00E2\u0088\u00AB 10\u00CE\u00A8(z(\u00CF\u0083\u00E2\u0088\u0097), z(\u00CF\u0083))\u00CF\u0086(\u00CF\u0083) d\u00CF\u0083=12piD(\u00E2\u0088\u00AB 10K0(\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3z(\u00CF\u0083)\u00E2\u0088\u0092 z(\u00CF\u0083\u00E2\u0088\u0097)\u00E2\u0088\u009AD\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3)+ log |\u00CF\u0083 \u00E2\u0088\u0092 \u00CF\u0083\u00E2\u0088\u0097|)\u00CF\u0086(\u00CF\u0083) d\u00CF\u0083\u00E2\u0088\u0092 12piD\u00E2\u0088\u00AB 10log |\u00CF\u0083 \u00E2\u0088\u0092 \u00CF\u0083\u00E2\u0088\u0097|\u00CF\u0086(\u00CF\u0083) d\u00CF\u0083, (6.21)where the integral with the Bessel function is now non-singular \u00E2\u0088\u0080\u00CF\u0083 \u00E2\u0088\u0088 [0, 1].Due to the periodicity of the curve, there is a slight issue with this formulationand that is that \u00CF\u0083\u00E2\u0088\u0097 = 0 also induces a singularity at \u00CF\u0083 = 1. Furthermore, as\u00CF\u0083\u00E2\u0088\u0097 gets close to zero, it begins to notice the effects of a singularity at \u00CF\u0083 = \u00E2\u0088\u009212216.1. Layer Potential Formulationas well, though we classify this as a weak singularity since it is outside of thedomain of \u00CF\u0083. Since \u00CF\u0083\u00E2\u0088\u0097 \u00E2\u0088\u0088 [0, 1), there is no effect from the weak singularityat \u00CF\u0083 = 1. To remedy the effects of the periodicity inducing singularities, wewill remove the singularities that are a full period from \u00CF\u0083\u00E2\u0088\u0097 on either side andwrite (6.21) as\u00E2\u0088\u00AB 10\u00CE\u00A8(z(\u00CF\u0083\u00E2\u0088\u0097), z(\u00CF\u0083))\u00CF\u0086(\u00CF\u0083) d\u00CF\u0083 =12piD(\u00E2\u0088\u00AB 10K0(\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3z(\u00CF\u0083)\u00E2\u0088\u0092 z(\u00CF\u0083\u00E2\u0088\u0097)\u00E2\u0088\u009AD\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3)+ L(\u00CF\u0083, \u00CF\u0083\u00E2\u0088\u0097))\u00CF\u0086(\u00CF\u0083) d\u00CF\u0083\u00E2\u0088\u0092 12piD\u00E2\u0088\u00AB 10log |\u00CF\u0083 \u00E2\u0088\u0092 \u00CF\u0083\u00E2\u0088\u0097|\u00CF\u0086(\u00CF\u0083) d\u00CF\u0083 \u00E2\u0088\u0092 12piD\u00E2\u0088\u00AB 10log |\u00CF\u0083 \u00E2\u0088\u0092 (\u00CF\u0083\u00E2\u0088\u0097 + 1)|\u00CF\u0086(\u00CF\u0083) d\u00CF\u0083\u00E2\u0088\u0092 12piD\u00E2\u0088\u00AB 10log |\u00CF\u0083 \u00E2\u0088\u0092 (\u00CF\u0083\u00E2\u0088\u0097 \u00E2\u0088\u0092 1)|\u00CF\u0086(\u00CF\u0083) d\u00CF\u0083, (6.22)whereL(\u00CF\u0083, \u00CF\u0083\u00E2\u0088\u0097) = log |\u00CF\u0083 \u00E2\u0088\u0092 \u00CF\u0083\u00E2\u0088\u0097|+ log |\u00CF\u0083 \u00E2\u0088\u0092 (\u00CF\u0083\u00E2\u0088\u0097 + 1)|+ log |\u00CF\u0083 \u00E2\u0088\u0092 (\u00CF\u0083\u00E2\u0088\u0097 \u00E2\u0088\u0092 1)|. (6.23)Considering the singularity properties of the Bessel functions we will avoidevaluating at the singularity directly by defining a function K0 asK0(z(\u00CF\u0083), z(\u00CF\u0083\u00E2\u0088\u0097)) =\u00EF\u00A3\u00B1\u00EF\u00A3\u00B4\u00EF\u00A3\u00B4\u00EF\u00A3\u00B4\u00EF\u00A3\u00B4\u00EF\u00A3\u00B4\u00EF\u00A3\u00B2\u00EF\u00A3\u00B4\u00EF\u00A3\u00B4\u00EF\u00A3\u00B4\u00EF\u00A3\u00B4\u00EF\u00A3\u00B4\u00EF\u00A3\u00B3K0(\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3z(\u00CF\u0083)\u00E2\u0088\u0092z(\u00CF\u0083\u00E2\u0088\u0097)\u00E2\u0088\u009AD\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3)+ L(\u00CF\u0083, \u00CF\u0083\u00E2\u0088\u0097), \u00CF\u0083 6= \u00CF\u0083\u00E2\u0088\u0097log\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A32\u00E2\u0088\u009AD(\u00CF\u0083\u00E2\u0088\u0092(\u00CF\u0083\u00E2\u0088\u0097+1))(\u00CF\u0083\u00E2\u0088\u0092(\u00CF\u0083\u00E2\u0088\u0097\u00E2\u0088\u00921))L\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u0092 \u00CE\u00B3, \u00CF\u0083 = \u00CF\u0083\u00E2\u0088\u0097 6= 0log\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A32\u00E2\u0088\u009AD(\u00CF\u0083+1)L\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u0092 \u00CE\u00B3, \u00CF\u0083\u00E2\u0088\u0097 = 0, \u00CF\u0083 = 0, 1.(6.24)2226.1. Layer Potential FormulationNotice that using the asymptotic form is valid at \u00CF\u0083 = \u00CF\u0083\u00E2\u0088\u0097 because the errorterms vanish exactly. Finally, also defineS =\u00E2\u0088\u00AB 10log |\u00CF\u0083 \u00E2\u0088\u0092 \u00CF\u0083\u00E2\u0088\u0097|\u00CF\u0086(\u00CF\u0083) d\u00CF\u0083 +\u00E2\u0088\u00AB 10log |\u00CF\u0083 \u00E2\u0088\u0092 (\u00CF\u0083\u00E2\u0088\u0097 + 1)|\u00CF\u0086(\u00CF\u0083) d\u00CF\u0083+\u00E2\u0088\u00AB 10log |\u00CF\u0083 \u00E2\u0088\u0092 (\u00CF\u0083\u00E2\u0088\u0097 \u00E2\u0088\u0092 1)|\u00CF\u0086(\u00CF\u0083) d\u00CF\u0083, (6.25)so that (6.22) becomes\u00E2\u0088\u00AB 10\u00CE\u00A8(z(\u00CF\u0083\u00E2\u0088\u0097), z(\u00CF\u0083))\u00CF\u0086(\u00CF\u0083) d\u00CF\u0083 =12piD(\u00E2\u0088\u00AB 10K0(z(\u00CF\u0083), z(\u00CF\u0083\u00E2\u0088\u0097))\u00CF\u0086(\u00CF\u0083) d\u00CF\u0083 \u00E2\u0088\u0092 S).(6.26)Next consider the integral,\u00E2\u0088\u00AB 10\u00E2\u0088\u0082\u00CE\u00A8\u00E2\u0088\u0082nz\u00E2\u0088\u0097(z(\u00CF\u0083\u00E2\u0088\u0097), z(\u00CF\u0083))\u00CF\u0086(\u00CF\u0083) d\u00CF\u0083 =12piD3/2\u00E2\u0088\u00AB 10K1(\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3z(\u00CF\u0083)\u00E2\u0088\u0092 z(\u00CF\u0083\u00E2\u0088\u0097)\u00E2\u0088\u009AD\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3)(z(\u00CF\u0083)\u00E2\u0088\u0092 z(\u00CF\u0083\u00E2\u0088\u0097)) \u00C2\u00B7 n\u00CB\u0086z\u00E2\u0088\u0097|z(\u00CF\u0083)\u00E2\u0088\u0092 z(\u00CF\u0083\u00E2\u0088\u0097)| \u00CF\u0086(\u00CF\u0083) d\u00CF\u0083, (6.27)where K1 is the second order modified Bessel function of the second kind.Using the expansion for z(\u00CF\u0083) near \u00CF\u0083\u00E2\u0088\u0097 from (6.19) we get,K1(\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3z(\u00CF\u0083)\u00E2\u0088\u0092 z(\u00CF\u0083\u00E2\u0088\u0097)\u00E2\u0088\u009AD\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3)\u00E2\u0088\u00BC\u00E2\u0088\u009ADL1|\u00CF\u0083 \u00E2\u0088\u0092 \u00CF\u0083\u00E2\u0088\u0097| +O(\u00CF\u0083 \u00E2\u0088\u0092 \u00CF\u0083\u00E2\u0088\u0097)(z(\u00CF\u0083)\u00E2\u0088\u0092 z(\u00CF\u0083\u00E2\u0088\u0097)) \u00C2\u00B7 n\u00CB\u0086z\u00E2\u0088\u0097 \u00E2\u0088\u00BCL2\u00CE\u00BA2(\u00CF\u0083 \u00E2\u0088\u0092 \u00CF\u0083\u00E2\u0088\u0097)2 +O(\u00CF\u0083 \u00E2\u0088\u0092 \u00CF\u0083\u00E2\u0088\u0097)4,and along with the expansion of the norm (6.20) we haveK1(\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3z(\u00CF\u0083)\u00E2\u0088\u0092 z(\u00CF\u0083\u00E2\u0088\u0097)\u00E2\u0088\u009AD\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3)(z(\u00CF\u0083)\u00E2\u0088\u0092 z(\u00CF\u0083\u00E2\u0088\u0097)) \u00C2\u00B7 n\u00CB\u0086z\u00E2\u0088\u0097|z(\u00CF\u0083)\u00E2\u0088\u0092 z(\u00CF\u0083\u00E2\u0088\u0097)| \u00E2\u0088\u00BC\u00E2\u0088\u009AD\u00CE\u00BA2+O(\u00CF\u0083 \u00E2\u0088\u0092 \u00CF\u0083\u00E2\u0088\u0097).2236.2. Numerical Formulation of Curve Motion ProblemTherefore, we see that unlike the integral involving the zeroth order Besselfunction, K1 is not singular at \u00CF\u0083 = \u00CF\u0083\u00E2\u0088\u0097 and so we do not have to alter theintegral by removing any singularities. However, to avoid the numericaldifficulties of evaluating directly at \u00CF\u0083 = \u00CF\u0083\u00E2\u0088\u0097, we define the following functionK1(z(\u00CF\u0083), z(\u00CF\u0083\u00E2\u0088\u0097)) =\u00EF\u00A3\u00B1\u00EF\u00A3\u00B4\u00EF\u00A3\u00B2\u00EF\u00A3\u00B4\u00EF\u00A3\u00B31\u00E2\u0088\u009ADK1(\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3z(\u00CF\u0083)\u00E2\u0088\u0092z(\u00CF\u0083\u00E2\u0088\u0097)\u00E2\u0088\u009AD\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3)(z(\u00CF\u0083)\u00E2\u0088\u0092z(\u00CF\u0083\u00E2\u0088\u0097))\u00C2\u00B7n\u00CB\u0086z\u00E2\u0088\u0097|z(\u00CF\u0083)\u00E2\u0088\u0092z(\u00CF\u0083\u00E2\u0088\u0097)| , \u00CF\u0083 6= \u00CF\u0083\u00E2\u0088\u0097\u00CE\u00BA2 , \u00CF\u0083 = \u00CF\u0083\u00E2\u0088\u0097, (6.28)where we note that if \u00CF\u0083\u00E2\u0088\u0097 = 0 then due to the periodicity we evaluate thesecond branch if \u00CF\u0083 = 0 or \u00CF\u0083 = 1. With (6.28) we have that (6.27) becomes\u00E2\u0088\u00AB 10\u00E2\u0088\u0082\u00CE\u00A8\u00E2\u0088\u0082nz\u00E2\u0088\u0097(z(\u00CF\u0083\u00E2\u0088\u0097), z(\u00CF\u0083))\u00CF\u0086(\u00CF\u0083) d\u00CF\u0083 =12piD\u00E2\u0088\u00AB 10K1(z(\u00CF\u0083), z(\u00CF\u0083\u00E2\u0088\u0097))\u00CF\u0086(\u00CF\u0083) d\u00CF\u0083. (6.29)We are now ready to discuss the numerical solution of this problem.6.2 Numerical Formulation of Curve MotionProblemConsider a uniform discretization of \u00CF\u0083, \u00CF\u0083i = i\u00E2\u0088\u0086\u00CF\u0083 with \u00E2\u0088\u0086\u00CF\u0083 = 1N where N arethe chosen number of grid points. Define zij = zj(\u00CF\u0083i) as the discretized curvepositions. Using a standard centered difference discretization we havedzjd\u00CF\u0083\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00CF\u0083=\u00CF\u0083i=zi+1,j \u00E2\u0088\u0092 zi\u00E2\u0088\u00921,j2\u00E2\u0088\u0086\u00CF\u0083+O(\u00E2\u0088\u0086\u00CF\u00832),d2zjd\u00CF\u00832\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00CF\u0083=\u00CF\u0083i=zi+1,j \u00E2\u0088\u0092 2zi,j + zi\u00E2\u0088\u00921,j\u00E2\u0088\u0086\u00CF\u00832+O(\u00E2\u0088\u0086\u00CF\u00832),2246.2. Numerical Formulation of Curve Motion Problemwhich we can use to define the unit tangent, normal vectors, and curvatureup to O(\u00E2\u0088\u0086\u00CF\u00832)t\u00CB\u0086i,j =zj\u00CF\u0083(\u00CF\u0083i)Lj=zi+1,j \u00E2\u0088\u0092 zi\u00E2\u0088\u00921,j2Lj\u00E2\u0088\u0086\u00CF\u0083, (6.30)n\u00CB\u0086i,j = \u00E3\u0080\u0088\u00E2\u0088\u0092t\u00CB\u0086i,jy , t\u00CB\u0086i,jx\u00E3\u0080\u0089, (6.31)\u00CE\u00BAi,j =zj\u00CF\u0083\u00CF\u0083(\u00CF\u0083i) \u00C2\u00B7 n\u00CB\u0086i,jL2j=12\u00E2\u0088\u0086\u00CF\u00833L3j(zi+1,j \u00E2\u0088\u0092 2zi,j + zi\u00E2\u0088\u00921,j) \u00C2\u00B7 \u00E3\u0080\u0088\u00E2\u0088\u0092t\u00CB\u0086i,jy , t\u00CB\u0086i,jx\u00E3\u0080\u0089. (6.32)Here the x and y subscripts indicate the first and second components of thevector respectively. For the time discretization with a time step \u00E2\u0088\u0086t, we willconsider the implicit Backward Euler method so that for each curve m, (6.15)becomes(zk+1m \u00E2\u0088\u0092 zkm) \u00C2\u00B7 n\u00CB\u0086k+1 = \u00E2\u0088\u0086tV k+1,for the normal component where the subscript k indicates t = k\u00E2\u0088\u0086t and\u00E2\u0088\u00AB 10(zk+1m \u00E2\u0088\u0092 zkm) \u00C2\u00B7 t\u00CB\u0086k+1 d\u00CF\u0083 = 0,for the tangential component where we have used the zero mean condition(6.16). This is closed by prescribing the equal arclength parametrization,\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3dzk+1md\u00CF\u0083\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3 = Lm, (6.33)which we can discretize using (6.20) to get|zk+1i+1,m \u00E2\u0088\u0092 zk+1i,m | = Lm\u00E2\u0088\u0086\u00CF\u0083. (6.34)2256.2. Numerical Formulation of Curve Motion Problem6.2.1 Discretizing IntegralsBy doing the integral splitting in section 6.1.5 we are left to discretize regularand singular integrals. The regular integrals can be discretized using any ofthe standard interpolating methods which will all be of the form,\u00E2\u0088\u00AB 10f(\u00CF\u0083) d\u00CF\u0083 =N\u00E2\u0088\u0091i=0\u00CE\u00B1ifi\u00E2\u0088\u0086\u00CF\u0083,where fi = f(\u00CF\u0083i) and \u00CE\u00B1i are the quadrature weights. Unless otherwise spec-ified, since our scheme is accurate to O(\u00E2\u0088\u0086\u00CF\u00832) from the finite difference dis-cretization, we will consider the quadrature weights to be those that comefrom the trapezoid rule [6],\u00CE\u00B1i =\u00EF\u00A3\u00B1\u00EF\u00A3\u00B4\u00EF\u00A3\u00B2\u00EF\u00A3\u00B4\u00EF\u00A3\u00B312 , i = 0, i = N1, else.Generally due to periodicity in z, the integral can be written as a sum overN points however, we generalize because some of the integrals (such as thosein (6.24)) depend on \u00CF\u0083 explicitly and are not periodic.Discretizing Singular IntegralsIt is a slightly more delicate issue to consider the discretization of singularintegrals. Fortunately, due to the integral splitting in section 6.1.5, we onlyhave to consider integrals of the form,\u00E2\u0088\u00AB 10f(\u00CF\u0083) log |\u00CF\u0083 \u00E2\u0088\u0092 (\u00CF\u0083\u00E2\u0088\u0097 + a)| d\u00CF\u0083, (6.35)2266.2. Numerical Formulation of Curve Motion Problemwhere \u00CF\u0083\u00E2\u0088\u0097 \u00E2\u0088\u0088 [0, 1) is one of the discretized gridpoints (\u00CF\u0083\u00E2\u0088\u0097 = j\u00E2\u0088\u0086\u00CF\u0083) and a =\u00E2\u0088\u00921, 0, 1 is the shifted singularity being removed. The goal is to develop adiscretization so that we can write (6.35) as\u00E2\u0088\u00AB 10f(\u00CF\u0083) log |\u00CF\u0083 \u00E2\u0088\u0092 (\u00CF\u0083\u00E2\u0088\u0097 + a)| d\u00CF\u0083 =N\u00E2\u0088\u0091i=0wifi (6.36)for some weights wi. High order quadrature techniques have been devel-oped to handle singularities of logarithmic type [5]. They rely on a weightedtrapezoid rule method that adds gridpoints near the singularity as necessaryto counterbalance the singular behaviour. This technique will not be em-ployed here and instead we consider product integration [6] which allows usto continue to use our uniform spatial discretization. It is worth noting thatthe singular integral decomposition means that the singular integrals dependonly on \u00CF\u0083 and not on the specific curves themselves, and therefore higher or-der quadrature methods could easily be supplemented here if desired withoutimpacting the rest of the formulation significantly. The idea of the productintegration is to locally interpolate on f(\u00CF\u0083) using Lagrange interpolation andanalytically perform the polynomial-logarithmic integration that results. Westart by writing,\u00E2\u0088\u00AB 10f(\u00CF\u0083) log |\u00CF\u0083 \u00E2\u0088\u0092 \u00CF\u0083\u00E2\u0088\u0097| d\u00CF\u0083 =N\u00E2\u0088\u0091k=1\u00E2\u0088\u00AB k\u00E2\u0088\u0086\u00CF\u0083(k\u00E2\u0088\u00921)\u00E2\u0088\u0086\u00CF\u0083f(\u00CF\u0083) log |\u00CF\u0083 \u00E2\u0088\u0092 j\u00E2\u0088\u0086\u00CF\u0083| d\u00CF\u0083,where we are explicitly considering the case a = 0 which we will generalizelater. In what follows we consider the case of f(\u00CF\u0083) having a linear interpo-lation but the mechanism easily extends for higher order interpolants. The2276.2. Numerical Formulation of Curve Motion Problemlocal linear Lagrange interpolation of f(\u00CF\u0083) on [(k \u00E2\u0088\u0092 1)\u00E2\u0088\u0086\u00CF\u0083, k\u00E2\u0088\u0086\u00CF\u0083] isf \u00E2\u0088\u00BC (\u00CF\u0083 \u00E2\u0088\u0092 (k \u00E2\u0088\u0092 1)\u00E2\u0088\u0086\u00CF\u0083)fk \u00E2\u0088\u0092 (\u00CF\u0083 \u00E2\u0088\u0092 k\u00E2\u0088\u0086\u00CF\u0083)fk\u00E2\u0088\u00921\u00E2\u0088\u0086\u00CF\u0083,and so,\u00E2\u0088\u00AB 10f(\u00CF\u0083) log |\u00CF\u0083 \u00E2\u0088\u0092 \u00CF\u0083\u00E2\u0088\u0097| d\u00CF\u0083 =N\u00E2\u0088\u0091k=1fk\u00E2\u0088\u0086\u00CF\u0083\u00E2\u0088\u00AB k\u00E2\u0088\u0086\u00CF\u0083(k\u00E2\u0088\u00921)\u00E2\u0088\u0086\u00CF\u0083(\u00CF\u0083 \u00E2\u0088\u0092 (k \u00E2\u0088\u0092 1)\u00E2\u0088\u0086\u00CF\u0083) log |\u00CF\u0083 \u00E2\u0088\u0092 j\u00E2\u0088\u0086\u00CF\u0083| d\u00CF\u0083+fk\u00E2\u0088\u00921\u00E2\u0088\u0086\u00CF\u0083\u00E2\u0088\u00AB k\u00E2\u0088\u0086\u00CF\u0083(k\u00E2\u0088\u00921)\u00E2\u0088\u0086\u00CF\u0083(k\u00E2\u0088\u0086\u00CF\u0083 \u00E2\u0088\u0092 \u00CF\u0083) log |\u00CF\u0083 \u00E2\u0088\u0092 j\u00E2\u0088\u0086\u00CF\u0083| d\u00CF\u0083. (6.37)Consider evaluating the first integral in this expression by letting u = (\u00CF\u0083 \u00E2\u0088\u0092(k \u00E2\u0088\u0092 1)\u00E2\u0088\u0086\u00CF\u0083)/\u00E2\u0088\u0086\u00CF\u0083,1\u00E2\u0088\u0086\u00CF\u0083\u00E2\u0088\u00AB k\u00E2\u0088\u0086\u00CF\u0083(k\u00E2\u0088\u00921)\u00E2\u0088\u0086\u00CF\u0083(\u00CF\u0083 \u00E2\u0088\u0092 (k \u00E2\u0088\u0092 1)\u00E2\u0088\u0086\u00CF\u0083) log |\u00CF\u0083 \u00E2\u0088\u0092 j\u00E2\u0088\u0086\u00CF\u0083| d\u00CF\u0083= \u00E2\u0088\u0086\u00CF\u0083(log \u00E2\u0088\u0086\u00CF\u0083\u00E2\u0088\u00AB 10u du+\u00E2\u0088\u00AB 10u log |u+ (k \u00E2\u0088\u0092 1)\u00E2\u0088\u0092 j| du)=\u00E2\u0088\u0086\u00CF\u0083 log \u00E2\u0088\u0086\u00CF\u00832+ \u00E2\u0088\u0086\u00CF\u0083\u00CF\u00881,j(k \u00E2\u0088\u0092 1),where\u00CF\u00881,j(y) =\u00E2\u0088\u00AB 10u log |u+ y \u00E2\u0088\u0092 j| du,is computed analytically for any y and j. Using the same transformation onthe second integral in (6.37) we get1\u00E2\u0088\u0086\u00CF\u0083\u00E2\u0088\u00AB k\u00E2\u0088\u0086\u00CF\u0083(k\u00E2\u0088\u00921)\u00E2\u0088\u0086\u00CF\u0083(k\u00E2\u0088\u0086\u00CF\u0083 \u00E2\u0088\u0092 \u00CF\u0083) log |\u00CF\u0083 \u00E2\u0088\u0092 j\u00E2\u0088\u0086\u00CF\u0083| d\u00CF\u0083 = \u00E2\u0088\u0086\u00CF\u0083 log \u00E2\u0088\u0086\u00CF\u00832+ \u00E2\u0088\u0086\u00CF\u0083\u00CF\u00882,j(k \u00E2\u0088\u0092 1),2286.2. Numerical Formulation of Curve Motion Problemwhere\u00CF\u00882,j(y) =\u00E2\u0088\u00AB 10(1\u00E2\u0088\u0092 u) log |u+ y \u00E2\u0088\u0092 j| du.Using these functions we can define the following weights:w0,j =\u00E2\u0088\u0086\u00CF\u0083 log \u00E2\u0088\u0086\u00CF\u00832+ \u00E2\u0088\u0086\u00CF\u0083\u00CF\u00882,j(0),wN,j =\u00E2\u0088\u0086\u00CF\u0083 log \u00E2\u0088\u0086\u00CF\u00832+ \u00E2\u0088\u0086\u00CF\u0083\u00CF\u00881,j(N \u00E2\u0088\u0092 1),wi,j = \u00E2\u0088\u0086\u00CF\u0083 log \u00E2\u0088\u0086\u00CF\u0083 + \u00E2\u0088\u0086\u00CF\u0083(\u00CF\u00881,j(i\u00E2\u0088\u0092 1) + \u00CF\u00882,j(i)), i 6= 0, Nand can evaluate the discretized integral as a sum using (6.36). Notice thatthe computational cost for computing (6.35) in this way is not of great sig-nificance because the weights can be precomputed and thus, the method hasthe same local computational cost as integrating a non-singular function.The errors for product integration are of the same order as the equivalentmethods for non-singular integrals [6], i.e. the product integration rule usinglinear interpolation here is O(\u00E2\u0088\u0086\u00CF\u00832) in line with the Trapezoid rule for regularintegrals. If instead of a = 0, we have a = 1 (a = \u00E2\u0088\u00921) in (6.35) we define thepositive (negative) complementary functions:\u00CF\u0088c\u00C2\u00B11,j(y) =\u00E2\u0088\u00AB 10u log |u+ y \u00E2\u0088\u0092 j \u00E2\u0088\u0093N | du,\u00CF\u0088c\u00C2\u00B12,j(y) =\u00E2\u0088\u00AB 10(1\u00E2\u0088\u0092 u) log |u+ y \u00E2\u0088\u0092 j \u00E2\u0088\u0093N | du,2296.2. Numerical Formulation of Curve Motion Problemand the positive (negative) complementary weights:wc\u00C2\u00B10,j =\u00E2\u0088\u0086\u00CF\u0083 log \u00E2\u0088\u0086\u00CF\u00832+ \u00E2\u0088\u0086\u00CF\u0083\u00CF\u0088c\u00C2\u00B12,j(0),wc\u00C2\u00B1N,j =\u00E2\u0088\u0086\u00CF\u0083 log \u00E2\u0088\u0086\u00CF\u00832+ \u00E2\u0088\u0086\u00CF\u0083\u00CF\u0088c\u00C2\u00B11,j(N \u00E2\u0088\u0092 1),wc\u00C2\u00B1i,j = \u00E2\u0088\u0086\u00CF\u0083 log \u00E2\u0088\u0086\u00CF\u0083 + \u00E2\u0088\u0086\u00CF\u0083(\u00CF\u0088c\u00C2\u00B11,j(k \u00E2\u0088\u0092 1) + \u00CF\u0088c\u00C2\u00B12,j(k)), i 6= 0, N.To demonstrate the error accuracy of the method we will consider the case\u00E2\u0088\u00AB 10 cos(\u00CF\u0083) log |\u00CF\u0083| d\u00CF\u0083. Analytically we can integrate by parts to get that\u00E2\u0088\u00AB 10cos(\u00CF\u0083) log |\u00CF\u0083| d\u00CF\u0083 = \u00E2\u0088\u0092\u00E2\u0088\u00AB 10sin(\u00CF\u0083)\u00CF\u0083d\u00CF\u0083 = \u00E2\u0088\u0092Si(1).From [1] we can write this as a series,\u00E2\u0088\u0092\u00E2\u0088\u009E\u00E2\u0088\u0091k=0(\u00E2\u0088\u00921)k\u00E2\u0088\u00921(2k \u00E2\u0088\u0092 1)(2k \u00E2\u0088\u0092 1)! \u00E2\u0089\u0088 \u00E2\u0088\u00920.94608307036718301494where we obtained the approximation using 100 terms. We will use this asthe numerical value for comparison in our integrals. Table 6.1 shows theresults for various values of \u00E2\u0088\u0086\u00CF\u0083 along with the error ratio demonstrating the\u00E2\u0088\u0086\u00CF\u00832 convergence.\u00E2\u0088\u0086\u00CF\u0083 Error Ratio0.1 7.73\u00C3\u0097 10\u00E2\u0088\u00924 \u00E2\u0080\u00940.05 1.95\u00C3\u0097 10\u00E2\u0088\u00924 3.960.025 4.90\u00C3\u0097 10\u00E2\u0088\u00925 3.980.0125 1.22\u00C3\u0097 10\u00E2\u0088\u00925 3.990.00625 3.08\u00C3\u0097 10\u00E2\u0088\u00926 4.00Table 6.1: Numerical-analytic comparison of integrating\u00E2\u0088\u00AB 10 cos(\u00CF\u0083) log |\u00CF\u0083|d\u00CF\u0083using the product integration method with linear interpolation.2306.2. Numerical Formulation of Curve Motion ProblemWithout going through the details of defining the functions, we performthe same integration as in Table 6.1 using quadratic interpolation insteadof linear interpolation with the errors and ratios in Table 6.2. Notice thatthe ratio shows an O(\u00E2\u0088\u0086\u00CF\u00834) reduction as would be expected with a standardSimpsons integration rule [6].\u00E2\u0088\u0086\u00CF\u0083 Error Ratio0.1 1.70\u00C3\u0097 10\u00E2\u0088\u00926 \u00E2\u0080\u00940.05 1.02\u00C3\u0097 10\u00E2\u0088\u00927 16.570.025 6.27\u00C3\u0097 10\u00E2\u0088\u00929 16.300.0125 3.89\u00C3\u0097 10\u00E2\u0088\u009210 16.150.00625 2.42\u00C3\u0097 10\u00E2\u0088\u009211 16.04Table 6.2: Numerical-analytic comparison of integrating\u00E2\u0088\u00AB 10 cos(\u00CF\u0083) log |\u00CF\u0083|d\u00CF\u0083using the product integration method with quadratic interpolation.Finally, to demonstrate the importance of multiple singularity removal in theintegral splitting (6.26), consider a unit circle,z = \u00E3\u0080\u0088cos(2pi\u00CF\u0083), sin(2pi\u00CF\u0083)\u00E3\u0080\u0089,and integrate\u00E2\u0088\u00AB 10K0 (|z(\u00CF\u0083)\u00E2\u0088\u0092 z(\u00CF\u0083\u00E2\u0088\u0097)|) d\u00CF\u0083 (6.38)for a range of \u00CF\u0083\u00E2\u0088\u0097 \u00E2\u0088\u0088 [0, 1). Though analytic results are not available, thesymmetry of the circle indicates that the value of this integral should be thesame for any of the singularities. Figure 6.1 shows the value of integrating(6.38) for every \u00CF\u0083\u00E2\u0088\u0097 = k\u00E2\u0088\u0086\u00CF\u0083, k \u00E2\u0088\u0088 [0, N \u00E2\u0088\u0092 1], \u00E2\u0088\u0086\u00CF\u0083 = 0.02. The solid line isthe result where both of the singularities, a full period away on either side,2316.2. Numerical Formulation of Curve Motion Problemare removed and the dashed line is the value when only the actual singularvalue is removed. While the results are fairly constant in the middle, forthe dashed-line, there are errors greater than the quadrature error near theendpoints because the other singularities are being felt due to the periodicity.These artifacts are nearly removed for the solid curve.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10.5330.53320.53340.53360.53380.5340.53420.53440.53460.53480.535\u00CF\u0083*integral valueFigure 6.1: This shows the value of integrating (6.38) with \u00E2\u0088\u0086\u00CF\u0083 = 0.02 forall the possible discrete values of \u00CF\u0083\u00E2\u0088\u0097 \u00E2\u0088\u0088 [0, 1). The blue solid curve representsthe technique used in the integral splitting (6.26) where singularities withina full period on either side of the true singularity are removed while the reddashed curve represents removing only the true singular value.It may seem suspicious that we do not recover a constant exactly for the solidcurve since splitting the integral should not change its periodicity and, indeedthat is the case. However, the sole logarithmic integrals are handled exactlywhile the logarithmic terms that couple with the Bessel functions are handled2326.2. Numerical Formulation of Curve Motion Problemthrough the trapezoid rule (or some other quadrature rule) and hence willhave numeric error. This error is the source of breaking the periodicity inthe result. If we look at the asymptotic error expansion for the trapezoidrule through the Euler-Maclaurin formula [6] we have,E\u00E2\u0088\u0086\u00CF\u0083(\u00CF\u0083\u00E2\u0088\u0097) =\u00E2\u0088\u00AB 10K0(z(\u00CF\u0083), z(\u00CF\u0083\u00E2\u0088\u0097)) d\u00CF\u0083 \u00E2\u0088\u0092 T\u00E2\u0088\u0086\u00CF\u0083(\u00CF\u0083\u00E2\u0088\u0097) \u00E2\u0088\u00BC\u00E2\u0088\u0086\u00CF\u0083212F (\u00CF\u0083\u00E2\u0088\u0097) +O(\u00E2\u0088\u0086\u00CF\u00834),where K0(z(\u00CF\u0083), z(\u00CF\u0083\u00E2\u0088\u0097)) is given by (6.24), T\u00E2\u0088\u0086\u00CF\u0083 is the trapezoid approximationto the integral, andF (\u00CF\u0083\u00E2\u0088\u0097) = K0\u00CF\u0083(z(0), z(\u00CF\u0083\u00E2\u0088\u0097))\u00E2\u0088\u0092K0\u00CF\u0083(z(1), z(\u00CF\u0083\u00E2\u0088\u0097)). (6.39)F (\u00CF\u0083\u00E2\u0088\u0097) has an absolute maximum at \u00CF\u0083\u00E2\u0088\u0097 = 12 and absolute minimum at \u00CF\u0083\u00E2\u0088\u0097 = 0by recalling that 0 \u00E2\u0089\u00A4 \u00CF\u0083\u00E2\u0088\u0097 < 1 (see Figure 6.2).2336.2. Numerical Formulation of Curve Motion Problem0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1\u00E2\u0088\u00921.5\u00E2\u0088\u00921\u00E2\u0088\u00920.50\u00CF\u0083*F(\u00CF\u0083*)Figure 6.2: The plot of F (\u00CF\u0083\u00E2\u0088\u0097) as defined in (6.39). The function has anabsolute maximum at \u00CF\u0083\u00E2\u0088\u0097 = 0.5 and and absolute minimum at \u00CF\u0083\u00E2\u0088\u0097 = 0.If we consider the difference in F (\u00CF\u0083\u00E2\u0088\u0097) at its maximum and minimum valueswe haveFmax \u00E2\u0088\u0092 Fmin =16,and therefore,\u00E2\u0088\u0086maxE\u00E2\u0088\u0086\u00CF\u0083 =\u00E2\u0088\u0086\u00CF\u0083212(Fmax \u00E2\u0088\u0092 Fmin) =\u00E2\u0088\u0086\u00CF\u0083272\u001C \u00E2\u0088\u0086\u00CF\u00832.The error introduced by losing the periodicity is smaller than the quadratureerror and therefore of little significance. This procedure could be extendedusing Richardson extrapolation to account for non-periodic errors in higherorder quadrature schemes as well.2346.2. Numerical Formulation of Curve Motion Problem6.2.2 Numerical EquationsHaving discretized the integrals, we can now write the full system of equa-tions that we need to solve numerically. The normal and tangential velocityequations and the equal arclength parametrization arezk+1i,m \u00C2\u00B7 n\u00CB\u0086k+1i = zki,m \u00C2\u00B7 n\u00CB\u0086k+1i + \u00E2\u0088\u0086tV k+1i,m , (6.40a)N\u00E2\u0088\u0091i=0\u00CE\u00B1i(zk+1i,m \u00E2\u0088\u0092 zki,m) \u00C2\u00B7 t\u00CB\u0086k+1i \u00E2\u0088\u0086\u00CF\u0083 = 0, (6.40b)|zk+1i+1,m \u00E2\u0088\u0092 zk+1i,m | = Lm\u00E2\u0088\u0086\u00CF\u0083. (6.40c)The equations that prescribe the density \u00CF\u0086 and the value U0 become,Uk+10i,m =Lm2piDN\u00E2\u0088\u0091l=0(\u00CE\u00B1lK0(zk+1l,m , zk+1i,m )\u00CF\u0086k+1l,m \u00E2\u0088\u0086\u00CF\u0083 \u00E2\u0088\u0092Wl,i\u00CF\u0086k+1l,m)+M\u00E2\u0088\u0091j=1j 6=mLj2piDN\u00E2\u0088\u0091l=0\u00CE\u00B1lK0(\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3zk+1l,j \u00E2\u0088\u0092 zk+1i,m\u00E2\u0088\u009AD\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3)\u00CF\u0086k+1l,j \u00E2\u0088\u0086\u00CF\u0083+L\u00E2\u0088\u0082\u00E2\u0084\u00A62piDN\u00E2\u0088\u0091l=0\u00CE\u00B1lK0(\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3zbl \u00E2\u0088\u0092 zk+1i,m\u00E2\u0088\u009AD\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3)\u00CF\u0086k+1bl \u00E2\u0088\u0086\u00CF\u0083, (6.40d)\u00CF\u0086k+1i,m = (n\u00CB\u00860 \u00C2\u00B7 n\u00CB\u00860i )\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ef(Uk+10i,m , v\u00CB\u009C0) d\u00CE\u00B7\u00CB\u0086, (6.40e)where Wi,j = wi,j + wc+i,j + wc\u00E2\u0088\u0092i,j . The superscript 0 on the normal vectoris used because the relative orientation of curves does not change and sothe direction of the normals is set by the initial configuration. Notice thatseparating the boundary component is convenient here because the boundarycurve does not change in time, hence there is no k+ 1 on the zbl term. Also,2356.2. Numerical Formulation of Curve Motion Problemwe leave the integral in (6.40e) undiscretized because it is an integration over\u00CE\u00B7\u00CB\u0086 and not over the curve where the unknown values are specified. Finally,we can write the Neumann boundary and velocity condition as0 =M\u00E2\u0088\u0091j=1Lj2piDN\u00E2\u0088\u0091l=0\u00CE\u00B1lK1(zk+1l,j , zbl)\u00CF\u0086k+1l,j \u00E2\u0088\u0086\u00CF\u0083 +\u00CF\u0086k+1bi2+L\u00E2\u0088\u0082\u00E2\u0084\u00A62piDN\u00E2\u0088\u0091l=0\u00CE\u00B1lK1(zbl , zbi)\u00CF\u0086k+1bl \u00E2\u0088\u0086\u00CF\u0083, (6.40f)V k+1i,m = \u00CE\u00BAk+1i,m + 2H(Uk+10i,m )( M\u00E2\u0088\u0091j=1Lj2piDN\u00E2\u0088\u0091l=0\u00CE\u00B1lK1(zk+1l,j , zk+1i,m )\u00CF\u0086k+1l,j \u00E2\u0088\u0086\u00CF\u0083+L\u00E2\u0088\u0082\u00E2\u0084\u00A62piDN\u00E2\u0088\u0091l=0\u00CE\u00B1lK1(zbl , zk+1i,m )\u00CF\u0086k+1bl \u00E2\u0088\u0086\u00CF\u0083). (6.40g)While we will consider integrals accurate up to O(\u00E2\u0088\u0086\u00CF\u00832), one can easily ex-tend this to higher order methods by using the appropriate weights \u00CE\u00B1 andW . Given an initial curve we can compute the solution at the next time stepusing Newton\u00E2\u0080\u0099s method. When forming the Jacobian for Newton\u00E2\u0080\u0099s method,notice that we indeed have enough equations (the system (6.40)) to solve forthe unknown vector u = [z,V,L,U0, \u00CF\u0086, \u00CF\u0086b], where the components are thediscrete values on the curve. Due to the possibility of non-uniqueness of theU0 problem, we cannot guarantee that solution bifurcations will not occur(i.e. we cannot guarantee the Jacobian to have full rank).To initialize a curve, we need a parametrization of some curve in R2, z0 =2366.2. Numerical Formulation of Curve Motion Problem\u00E3\u0080\u0088x(\u00CE\u00B8), y(\u00CE\u00B8)\u00E3\u0080\u0089 with \u00CE\u00B8 \u00E2\u0088\u0088 [0,\u00CE\u0098]. Write the unit arclength parametrization as\u00CF\u0083 =\u00E2\u0088\u00AB \u00CE\u00B80\u00E2\u0088\u009Adxdv2+ dydv2dv\u00E2\u0088\u00AB \u00CE\u00980\u00E2\u0088\u009Adxd\u00CE\u00B82+ dyd\u00CE\u00B82d\u00CE\u00B8,which we can interpolate for discrete values of \u00CF\u0083 \u00E2\u0088\u0088 [0, 1]. We can pre-computevalues to arbitrary accuracy by refining in \u00CE\u00B8 along with higher order quadra-ture and interpolation. Having an initial unit arclength curve, we can de-termine the initial values of U0 and \u00CF\u0086 for the static curve by using New-ton\u00E2\u0080\u0099s method to solve (6.40d) and (6.40e). Again, due to the possible non-uniqueness of solutions, it can be difficult to find an initial guess which willallow Newton\u00E2\u0080\u0099s method to converge. Assuming this problem converges aninitial vector u0 can be used in the dynamic Newton\u00E2\u0080\u0099s method for the fullproblem.Validating Numerical Formulation with the Mullins SekerkaProblemBefore solving the problem using the GMS model, we will validate our codewith a different curve motion problem, the Mullins-Sekerka (MS) problemformulated in [55]. This problem is a singular perturbation limit of the Cahn-Hilliard problem [4] and it (along with similar problems) has been studiedanalytically (cf. [65], [78], [60], [3]) and numerically (cf. [84], [77], [2]). Thedifferential equation for motion is still (6.15) with the tangential velocitycondition (6.16) and equal arclength (6.33) as this is problem independent.2376.2. Numerical Formulation of Curve Motion ProblemHowever, the Dirichlet (6.5a), and jump conditions (6.5b) get replaced with\u00CE\u00BA =\u00E2\u0088\u00AB\u00CE\u0093\u00CE\u00A6(x, q)V (q) dqs + C(t), (6.41a)[\u00E2\u0088\u0082u\u00E2\u0088\u0082n]= \u00E2\u0088\u0092V, (6.41b)where here the fundamental solution for Laplace\u00E2\u0080\u0099s problem is used becausethe MS problem solves Laplace\u00E2\u0080\u0099s equation away form the interface insteadof the modified Helmholtz equation. Notice in these new boundary condi-tions that the integral density so happens to be the normal velocity exactly,and so there is no secondary velocity expression required. Furthermore, theNeumann boundary conditions do not apply to this problem since the onlyrequirement is that the solution is bounded in the far-field. This bounded-ness requirement is the reason for C(t) appearing in (6.41a) as otherwise thesolution will be logarithmic there (cf. [34],[84]). However, introducing thisunknown, time dependent, constant also requires a closure condition whichis ([34]),\u00E2\u0088\u00AB\u00CE\u0093V (q) dqs = 0.This condition is analogous to the zero net-flux Fredholm condition requiredfor Neumann problems of the Laplace operator on bounded domains. Weomit the details for the numerical formulation as they are nearly identical tothe equations for the reaction diffusion models in section 6.2.2. For compari-son reference we note that if \u00CE\u0093 is a multiconnected domain of two concentriccircles in free-space with radius R1 and R2 (R2 > R1) then we can find a2386.2. Numerical Formulation of Curve Motion Problemradially symmetric exact solution [84],u(r) =\u00EF\u00A3\u00B1\u00EF\u00A3\u00B4\u00EF\u00A3\u00B4\u00EF\u00A3\u00B4\u00EF\u00A3\u00B4\u00EF\u00A3\u00B4\u00EF\u00A3\u00B2\u00EF\u00A3\u00B4\u00EF\u00A3\u00B4\u00EF\u00A3\u00B4\u00EF\u00A3\u00B4\u00EF\u00A3\u00B4\u00EF\u00A3\u00B3\u00E2\u0088\u0092 1R1 , 0 \u00E2\u0089\u00A4 r \u00E2\u0089\u00A4 R1\u00E2\u0088\u0092 1R1 +(1R1 +1R2) log(rR1)log(R2R1) , R1 \u00E2\u0089\u00A4 r \u00E2\u0089\u00A4 R21R2 , r \u00E2\u0089\u00A5 R2, (6.42)with equations for the radii given by,dR1dt= \u00E2\u0088\u0092 1R1(1R1 +1R2)log(R2R1) ,dR2dt= \u00E2\u0088\u0092 1R2(1R1 +1R2)log(R2R1) .Notice thatddt(R21 +R22) = 0,and so,R2 =\u00E2\u0088\u009AA2 +R21,where A2 = R210 + R220 is the sum of the initial values for each radius. Wecan therefore write the problem for R1(t) as,t = \u00E2\u0088\u009212\u00E2\u0088\u00AB R1(t)R10x2\u00E2\u0088\u009Ax2 + A2x+\u00E2\u0088\u009Ax2 + A2log(x2 + A2x2)dx. (6.43)We can solve this numerically and invert to get R1(t) (and hence R2(t))for any time t and thus have the analytic solution. Figure 6.3 shows theanalytic solution (6.42) as well as the numerical solution from the integral2396.2. Numerical Formulation of Curve Motion Problemequation technique for two times t = 0 and t = 0.2. The outer radius istaken to be R2 = 2 while the inner radius is R1 = 1. The lines represent thenumeric solution while the circles represent the analytic solution. Here wechose N = 50 points on each curve with a time step of \u00E2\u0088\u0086t = 1\u00C3\u0097 10\u00E2\u0088\u00923. Thefigure shows an excellent agreement between the numerical formulation andthe analytic solution.\u00E2\u0088\u00922 \u00E2\u0088\u00921.5 \u00E2\u0088\u00921 \u00E2\u0088\u00920.5 0 0.5 1 1.5 2\u00E2\u0088\u00922\u00E2\u0088\u00921.5\u00E2\u0088\u00921\u00E2\u0088\u00920.500.511.52xyConcentric circles t=0t=0.2Figure 6.3: The solution to the Mullins-Sekerka problem for concentric circleswith an outer radius R2 = 2 and inner radius R1 = 1. The solid blue curveis the numeric solution at t = 0 and the red dashed curve is the numericsolution at t = 0.2. The hollow circles are the analytic solution as computedwith (6.42) and (6.43),In a more technical comparison, we consider the global truncation errors forthe same concentric circles with R1 = 1 and R2 = 2 compared to the analytic2406.2. Numerical Formulation of Curve Motion Problemsolution (6.42). Standard backward Euler global truncation analysis wouldpredict an error on the order of \u00E2\u0088\u0086t but since the source term for the equationof motion has a spatial error \u00E2\u0088\u0086\u00CF\u00832 log \u00E2\u0088\u0086\u00CF\u0083 then we actually predict a globalerror of O(\u00E2\u0088\u0086\u00CF\u00832 log \u00E2\u0088\u0086\u00CF\u0083+ \u00E2\u0088\u0086t). In Table 6.3 we take \u00E2\u0088\u0086t = \u00E2\u0088\u0086\u00CF\u00832 for a variety ofN values so that the overall convergence should be \u00E2\u0088\u0086\u00CF\u00832 log \u00E2\u0088\u0086\u00CF\u0083 and indeedthat is observed.N xerr xrat Verr Vrat Lerr Lrat CPU Time8 2.90E-03 \u00E2\u0080\u0094 1.45E-01 \u00E2\u0080\u0094 1.61E-01 \u00E2\u0080\u0094 0.18916 7.96E-04 3.64 3.42E-02 3.93 4.11E-02 3.93 0.52032 2.12E-04 3.75 8.22E-03 3.96 1.04E-02 3.96 2.4364 5.49E-05 3.87 2.01E-03 3.98 2.61E-03 3.98 21.4128 1.40E-05 3.93 4.98E-04 3.99 6.53E-04 3.99 352N xerr xrat Verr Vrat Lerr Lrat CPU Time8 5.38E-04 \u00E2\u0080\u0094 7.00E-02 \u00E2\u0080\u0094 3.16E-01 \u00E2\u0080\u0094 0.18916 1.76E-04 3.05 1.66E-02 4.21 7.97E-02 3.96 0.52032 5.10E-05 3.45 4.01E-03 4.14 2.00E-02 3.99 2.4364 1.36E-05 3.74 9.86E-04 4.07 5.01E-03 3.99 21.4128 3.52E-06 3.88 2.44E-04 4.04 1.25E-03 4.00 352Table 6.3: The global truncation error for solving the MS problem withconcentric circles R1 = 1 (top table), R2 = 2 (bottom table) solving toT = 0.0469. We define xerr as the error in the x-component of the curveposition. The error in the y-component is the same and omitted. Verr andLerr are the errors in the normal velocity and curve length respectively. Therat suffix for each indicates the ratio of successive errors to the previous one.The convergence is O(\u00E2\u0088\u0086\u00CF\u00832 log \u00E2\u0088\u0086\u00CF\u0083) as expected. The CPU timings reflectthe computation of both curves and does not included anything that can beprecomputed such as the singular scalar logarithmic integrals.Next, we demonstrate some of the other qualitative results of the MS problemobtained in [84], such as non-concentric circles favouring growth of the largercircle at the expense of the smaller one (Figure 6.4) and the tendency of2416.2. Numerical Formulation of Curve Motion Problemnon-circular curves to become circular (Figure 6.5).\u00E2\u0088\u00922 0 2 4 6 8 10\u00E2\u0088\u009210123456789xyNon\u00E2\u0088\u0092Concentric circles t=0t=1.5Figure 6.4: Evolution of non concentric circles with MS. The first circle iscentered at (\u00E2\u0088\u00921, 0) with radius R1 = 1 and the second circle is centered at(6, 6) with radius R2 = 2. The initial curve is in a blue solid line while thefinal curve at time t = 1.5 (\u00E2\u0088\u0086t = 1 \u00C3\u0097 10\u00E2\u0088\u00922) is in a red dashed line. Astime evolves, an effect known as Ostwald ripening occurs [62] which favoursgrowth of larger objects at the expense of shrinking small objects.2426.2. Numerical Formulation of Curve Motion Problem\u00E2\u0088\u00923 \u00E2\u0088\u00922 \u00E2\u0088\u00921 0 1 2 3\u00E2\u0088\u00922\u00E2\u0088\u00921.5\u00E2\u0088\u00921\u00E2\u0088\u00920.500.511.52xyEllipse t=0t=2Figure 6.5: Evolution of an ellipse to MS with major axis 3 and minor axis1. The initial curve is in a blue solid line while the final curve at time t = 2(\u00E2\u0088\u0086t = 1 \u00C3\u0097 10\u00E2\u0088\u00922) is in a red dashed line. The curve becomes more circularas time evolves which is a consequence of the area preserving and lengthshrinking property of the MS model [84].We emphasize that our aim in comparing results is not to demonstrate thesuperiority of our technique over the proposed method in [84]. However, weindicate a few significant comparisons between our two methods. Firstly, theformulation of solving (6.15) in [84] is based on looking at the tangent angleand does an implicit-explicit (IMEX) splitting so that only stiff terms aresolved implicitly resulting in a diagonal Jacobian of the discretization of thedifferential equation. However, ultimately they are still forced into solvingintegral equations to update the velocity which results in the same denseJacobian that we have implemented. While not using the same splitting2436.3. Solving the GMS Modeltechnique for (6.15), we will have a sparse Jacobian for the differential equa-tion. Furthermore, by using an IMEX splitting, the scheme in [84] requires alower bound that the time step not be larger than 2.5\u00C3\u0097 10\u00E2\u0088\u00923. However, ourmethod, being fully implicit allows us to take larger time steps due to thegeneral stronger stability of fully implicit schemes (as noted in Figures 6.4and 6.5). Finally, by not using the tangent angle formulation, we do not haveto consider curve reconstruction and the full curve update happens withinthe Newton method solver.6.3 Solving the GMS ModelWe will now solve the saturated Gierer-Meinhardt model with saturationparameter \u00CF\u0083\u00CB\u0086, (2.32) discussed in 2.3, which introduced the saturation criteriafor which homoclinic orbit solutions w existed. We will now discuss how toincorporate the limiting saturation into solving the full model.6.3.1 Including Saturation and ComputingHomoclinic OrbitsHaving chosen the GMS model we can write (6.13) as\u00CF\u0086m(\u00CF\u0083\u00E2\u0088\u0097) = (n\u00CB\u0086 \u00C2\u00B7 n\u00CB\u0086i)U\u00CE\u00B20\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ewo d\u00CE\u00B7\u00CB\u0086, \u00CE\u00B2 = qo\u00E2\u0088\u0092 s.We can also write (2.20) as H = \u00E2\u0088\u0092 q4U0 H\u00CB\u0086 with H\u00CB\u0086 defined in (2.31). Aswe iterate a solution using Newton\u00E2\u0080\u0099s method, the value of b is not beingcontrolled but rather is subject to an update on U0. Therefore, it is possible2446.3. Solving the GMS Modelthat U0 could be computed such that b > bc which is invalid. To remedythe possibility of exceeding the critical b value, we consider a mapping of avariable c \u00E2\u0088\u0088 (\u00E2\u0088\u0092\u00E2\u0088\u009E,\u00E2\u0088\u009E) to b \u00E2\u0088\u0088 [0, bc] via,b =bc2(cos(c) + 1).This form is chosen since there are not any steep gradients in the functionfor Newton\u00E2\u0080\u0099s method to overshoot and diverge like there are for exponentialfunctions. If the critical b value is reached exactly, the tangent is horizontalbut a homoclinic cannot exist precisely at this value so this case can beignored. Similarly there is a horizontal tangent issue at b = 0 but this willnot be a problem for \u00CF\u0083\u00CB\u0086 > 0 as long as we do not initialize near b = 0. If \u00CF\u0083\u00CB\u0086 = 0then b = 0 is a solution for all vectors u and time t. One notorious issue witha trigonometric function in Newton\u00E2\u0080\u0099s method is the convergence to a singleroot, however since any period maps to the set of b values we are interested,we are not concerned with the particular value of c that converges. We willsupplement the addition of a new variable to the system via the equation(2.23),b\u00E2\u0088\u0092 U2q0 \u00CF\u0083\u00CB\u0086 = 0.Another advantage of considering b as a separate variable is that the homo-clinic only depends on b and can therefore be precomputed. Otherwise, wewould have to consider the dependence of the homoclinic on U0 and thusthere would need to be an ODE computation at each iteration significantlyslowing down the formulation. In the current form, the time required for aNewton solve is equivalent whether b = 0 or b 6= 0. While computing deriva-2456.3. Solving the GMS Modeltives for the Newton solve, almost all of the terms appear algebraically exceptfor b (or c) which has a dependence in the numerically computed homoclinicsolution. Computing the c derivatives we getd\u00CF\u0086mdc= (n\u00CB\u0086 \u00C2\u00B7 n\u00CB\u0086i)U\u00CE\u00B20\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Erwo\u00E2\u0088\u00921wb d\u00CE\u00B7\u00CB\u0086(\u00E2\u0088\u0092bc2sin(c)),dHdc= \u00E2\u0088\u0092 q4U0\u00EF\u00A3\u00AB\u00EF\u00A3\u00AC\u00EF\u00A3\u00AD\u00E2\u0088\u00AB\u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009E 2wwb d\u00CE\u00B7\u00CB\u0086\u00E2\u0088\u00AB\u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew2\u00CE\u00B7\u00CB\u0086 d\u00CE\u00B7\u00CB\u0086\u00E2\u0088\u0092\u00E2\u0088\u00AB\u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew2 d\u00CE\u00B7\u00CB\u0086\u00E2\u0088\u00AB\u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009E 2w\u00CE\u00B7\u00CB\u0086wb\u00CE\u00B7\u00CB\u0086 d\u00CE\u00B7\u00CB\u0086(\u00E2\u0088\u00AB\u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ew2\u00CE\u00B7\u00CB\u0086 d\u00CE\u00B7\u00CB\u0086)2\u00EF\u00A3\u00B6\u00EF\u00A3\u00B7\u00EF\u00A3\u00B8(\u00E2\u0088\u0092bc2sin(c)).In this expression we consider the c derivative using chain rule since the bdependence on the homoclinic is more natural. To actually compute thehomoclinic we consider (2.22) along with the same expression differentiatedwith respect to b,wb\u00CE\u00B7\u00CB\u0086\u00CE\u00B7\u00CB\u0086 \u00E2\u0088\u0092 wb +2wwb(1 + bw2)2\u00E2\u0088\u0092 w4(1 + bw2)2= 0.We then write these two second order differential equations as a first or-der system and solve them using a standard boundary value solver such asbvp4c in MATLAB. We supplement this system with the boundary condi-tions w\u00CE\u00B7\u00CB\u0086(0) = 0 and wb\u00CE\u00B7\u00CB\u0086 = 0 and to appropriately capture the exponentialdecay in the far-field, we prescribe a mixed boundary conditionw\u00CE\u00B7\u00CB\u0086(L\u00CB\u0086) = \u00E2\u0088\u0092w(L\u00CB\u0086),where L\u00CB\u0086\u001D 1 is chosen to sufficiently represent infinity. Note that since w iseven (see Lemma 2.2.0.1), we only solve the system on the domain \u00CE\u00B7\u00CB\u0086 \u00E2\u0089\u00A5 0 andrecast the necessary integrals, which are also of even functions, to be on the2466.3. Solving the GMS Modelsame interval. Plots of various homoclinic orbits are in Figure 2.2 in section2.3.6.3.2 GMS ResultsWe are now in a position to solve (6.40) subject to the choice of f and Hin section 6.3.1. We will start by comparing to analytic results for radiallysymmetric solutions on a circular domain of radius R. For this problem, it iseasier to consider a polar coordinate system (r \u00E2\u0088\u0088 [0, R] and \u00CE\u00B8 \u00E2\u0088\u0088 [0, 2pi]) witha localization on some circle r = r0. In this case then, the normal coordinate\u00CE\u00B7 would be \u00CE\u00B7 = r0 \u00E2\u0088\u0092 r. When b = 0 and p = 2 we have that the homoclinicorbit that solves (2.22) isw(\u00CF\u0081) =32sech 2(\u00CF\u00812),where \u00CF\u0081 = r\u00E2\u0088\u0092r0\u000F is the inner region radial coordinate (i.e. \u00CF\u0081 = \u00E2\u0088\u0092\u00CE\u00B7\u00CB\u0086). Usingthis homoclinic orbit we can write the problem (2.19) as\u00E2\u0088\u0086u\u00E2\u0088\u0092 uD= 0, 0 \u00E2\u0089\u00A4 r \u00E2\u0089\u00A4 R \ r0 (6.44a)dudr= 0, r = R (6.44b)u = U0, r = r0 (6.44c)[dudr]= \u00E2\u0088\u0092 1DU\u00CE\u00B20\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009Ewo d\u00CF\u0081, r = r0, (6.44d)where we are seeking radially symmetric solutions and so we take the Laplaceoperator to have only a radial dependence. This is subject to the velocity2476.3. Solving the GMS Modelcondition (2.43,dr0dt= \u00E2\u0088\u0092 1r0\u00E2\u0088\u0092 qU0(dudr\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3r=r+0+dudr\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3r=r\u00E2\u0088\u00920)where we have computed H explicitly since the homoclinic orbit is knownanalytically. The solution to this is given by (2.42),u(r) = U0G0(r; r0)G0(r0; r0),with U0 given by (2.40) and G0 by (2.38). Figure 6.6 shows a comparison withthis analytic solution and the numerical scheme for exponent set (2, q, o, s) =(2, 1, 2, 0) and for parameters R = 1, r0 = 1/2 and D = 1. Figure 6.7 showsthe value of U0 computed numerically and analytically using (2.42) for t = 0and t = 0.119.2486.3. Solving the GMS Model\u00E2\u0088\u00921 \u00E2\u0088\u00920.8 \u00E2\u0088\u00920.6 \u00E2\u0088\u00920.4 \u00E2\u0088\u00920.2 0 0.2 0.4 0.6 0.8 1\u00E2\u0088\u00921\u00E2\u0088\u00920.8\u00E2\u0088\u00920.6\u00E2\u0088\u00920.4\u00E2\u0088\u00920.200.20.40.60.81xy t=0t=0.119Figure 6.6: Circle evolution under the GMS model with \u00CF\u0083\u00CB\u0086 = 0, R = 1,r0 = 1/2, D = 1, exponent set (2, 1, 2, 0) and time step 1 \u00C3\u0097 10\u00E2\u0088\u00923. Thelines represent the numerical solution while the circles represent the analyticsolution computed using (2.42). The outer black line represents the boundarycurve r = R = 1.2496.3. Solving the GMS Model0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10.120.130.140.150.160.170.18\u00CF\u0083U 0 NumericAnalytic(a) t = 0, exact value U0 = 0.15250 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10.20.210.220.230.240.250.26\u00CF\u0083U 0 NumericAnalytic(b) t = 0.119, exact value U0 = 0.2336Figure 6.7: U0 values computed numerically (solid line) and analytically(dashed-line) using (2.40) for the GMS model with \u00CF\u0083\u00CB\u0086 = 0, R = 1, r0 = 1/2,D = 1, and exponent set (2, 1, 2, 0).We can also showcase the existence of equilibrium discussed in 2.3.1 whichwe can verify analytically using the velocity expression (2.43). One such plotof (2.43), for the exponent set (2, 2, 2, 0), R = 4, and D = 1 is in Figure 6.8which shows a stable equilibrium at r0 \u00E2\u0089\u0088 2.76. We demonstrate this is foundqualitatively in Figure 6.9 where an initial circle with radius r0 < 2.76 grows(Figure 6.9a) while one with initial radius r0 > 2.76 shrinks (Figure 6.9b).2506.3. Solving the GMS Model0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1\u00E2\u0088\u00922\u00E2\u0088\u00921.5\u00E2\u0088\u00921\u00E2\u0088\u00920.500.511.52r0/Rdr0/dtFigure 6.8: Slope-field for circle evolution using the GMS model with R = 4,D = 1, exponent set (2, 2, 2, 0) and \u00CF\u0083\u00CB\u0086 = 0. There is an unstable equilibriumat r0/R \u00E2\u0089\u0088 0.044 (r0 \u00E2\u0089\u0088 0.176) and a stable equilibrium at r0/R \u00E2\u0089\u0088 0.69(r0 \u00E2\u0089\u0088 2.76).\u00E2\u0088\u00922.5 \u00E2\u0088\u00922 \u00E2\u0088\u00921.5 \u00E2\u0088\u00921 \u00E2\u0088\u00920.5 0 0.5 1 1.5 2 2.5\u00E2\u0088\u00922.5\u00E2\u0088\u00922\u00E2\u0088\u00921.5\u00E2\u0088\u00921\u00E2\u0088\u00920.500.511.522.5xy t=0t=0.2(a) Circle with radius r0 = 2.\u00E2\u0088\u00924 \u00E2\u0088\u00923 \u00E2\u0088\u00922 \u00E2\u0088\u00921 0 1 2 3 4\u00E2\u0088\u00924\u00E2\u0088\u00923\u00E2\u0088\u00922\u00E2\u0088\u0092101234xy t=0t=0.2(b) Circle with radius r0 = 3.Figure 6.9: Circle evolution using the GMS model with R = 4, D = 1,exponent set (2, 2, 2, 0), and \u00CF\u0083\u00CB\u0086 = 0. The boundary curve at R = 4 has beenomitted.2516.3. Solving the GMS ModelWhen b 6= 0, the equations (2.42) still hold for the radially symmetric solu-tion except now (2.40) is a nonlinear equation since the homoclinic orbit wdepends on b which ultimately depends on U0 via (2.23). We therefore callthe radially symmetric results for the b 6= 0 case pseudo-analytic becausewe require a Newton\u00E2\u0080\u0099s method solver for U0. The results for this case arepresented in Figure 6.10 with the same parameter regime as for Figure 6.6but with the addition of saturation \u00CF\u0083\u00CB\u0086 = 10. Figure 6.11 shows the value of U0computed numerically and using (2.40). Notice the effect of the saturationdrastically alters the curve inhibitor U0 value and that the values withoutsaturation in Figure 6.7 would lead to b > bc with \u00CF\u0083\u00CB\u0086 = 10.2526.3. Solving the GMS Model\u00E2\u0088\u00921 \u00E2\u0088\u00920.8 \u00E2\u0088\u00920.6 \u00E2\u0088\u00920.4 \u00E2\u0088\u00920.2 0 0.2 0.4 0.6 0.8 1\u00E2\u0088\u00921\u00E2\u0088\u00920.8\u00E2\u0088\u00920.6\u00E2\u0088\u00920.4\u00E2\u0088\u00920.200.20.40.60.81xy t=0t=0.119Figure 6.10: Circle evolution under the GMS model with \u00CF\u0083\u00CB\u0086 = 10, R = 1,r0 = 1/2, D = 1, exponent set (2, 1, 2, 0) and time step 1 \u00C3\u0097 10\u00E2\u0088\u00923. Thelines represent the numerical solution while the circles represent the analyticsolution computed using (2.42). The outer black line represents the boundarycurve r = R = 1.2536.3. Solving the GMS Model0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10.060.070.080.090.10.110.12\u00CF\u0083U 0 NumericAnalytic(a) t = 0, exact value U0 = 0.0977.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10.080.090.10.110.120.130.14\u00CF\u0083U 0 NumericAnalytic(b) t = 0.119, exact value U0 = 0.1120Figure 6.11: U0 values computed numerically (solid line) and analytically(dashed-line) using (2.40) for the GMS model with \u00CF\u0083\u00CB\u0086 = 10, R = 1, r0 = 1/2,D = 1, and exponent set (2, 1, 2, 0). Since \u00CF\u0083\u00CB\u0086 6= 0 a Newton\u00E2\u0080\u0099s method wasused to solve the analytic value.To see the effect on the motion and U0 values as \u00CF\u0083\u00CB\u0086 is varied, we plot thecurves in Figure 6.12a and U0 values in Figure 6.12b at t = 0.1 for differentvalues of the saturation parameter. The effect is such that the velocity andU0 decreases for increasing saturation.\u00E2\u0088\u00920.4 \u00E2\u0088\u00920.3 \u00E2\u0088\u00920.2 \u00E2\u0088\u00920.1 0 0.1 0.2 0.3 0.4\u00E2\u0088\u00920.4\u00E2\u0088\u00920.3\u00E2\u0088\u00920.2\u00E2\u0088\u00920.100.10.20.30.4xy \u00CF\u0083\u00CB\u0086 =0\u00CF\u0083\u00CB\u0086 =10\u00CF\u0083\u00CB\u0086 =30\u00CF\u0083\u00CB\u0086 =50(a) Circle curves.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10.040.060.080.10.120.140.160.180.20.22\u00CF\u0083U 0 \u00CF\u0083\u00CB\u0086 =0\u00CF\u0083\u00CB\u0086 =10\u00CF\u0083\u00CB\u0086 =30\u00CF\u0083\u00CB\u0086 =50(b) U0 values.Figure 6.12: Circle evolution under the GMS model with R = 1, r0 = 1/2,D = 1, exponent set (2, 1, 2, 0) and time step 1 \u00C3\u0097 10\u00E2\u0088\u00922. The lines representthe numerical solution for different values of the saturation parameter \u00CF\u0083\u00CB\u0086(0,10,30,50) at t = 0.1. The boundary curve R = 1 has been omitted tomore clearly show the separate curves.2546.3. Solving the GMS ModelWe now showcase the correct convergence behaviour as in section 6.2.2. Table6.4 takes \u00E2\u0088\u0086t = \u00E2\u0088\u0086\u00CF\u00832 for a variety of N values with the GMS model for acircle of radius r0 = 0.5 with parameters R = 1, r0 = 1/2, D = 1, exponentset (2, 1, 2, 0) and saturation \u00CF\u0083\u00CB\u0086 = 10. We see once again that the predictedO(\u00E2\u0088\u0086\u00CF\u00832 log \u00E2\u0088\u0086\u00CF\u0083) error appears. While omitted, the errors for U0, and \u00CF\u0086 alsoconverge to the appropriate order.N xerr xrat Verr Vrat Lerr Lrat CPU Time8 3.66E-03 \u00E2\u0080\u0094 2.40E-01 \u00E2\u0080\u0094 4.53E-02 \u00E2\u0080\u0094 0.816 1.04E-03 3.51 6.45E-02 3.73 1.05E-02 4.31 3.132 2.58E-04 4.04 1.63E-02 3.97 2.64E-03 3.98 14.164 6.30E-05 4.10 4.06E-03 4.00 6.71E-04 3.94 125.5128 1.45E-05 4.34 1.01E-03 4.01 1.75E-04 3.82 1572Table 6.4: The global truncation error for solving the GMS problem on acircle of radius r0 = 0.5 with R = 1, r0 = 1/2, D = 1, exponent set (2, 1, 2, 0)and saturation \u00CF\u0083\u00CB\u0086 = 10 solving to T = 0.0469. We define xerr as the errorin the x-component of the curve position. The error in the y-component isthe same and omitted. Verr and Lerr are the errors in the normal velocityand curve length respectively. The rat suffix for each indicates the ratio ofsuccessive errors to the previous one. The convergence is O(\u00E2\u0088\u0086\u00CF\u00832 log \u00E2\u0088\u0086\u00CF\u0083) asexpected. The CPU timings reflect the computation of both curves and doesnot included anything that can be precomputed such as the singular scalarlogarithmic integrals.Non-radially Symmetric Solutions and Non-UniquenessIt was mentioned in section 6.3.2 that the standard uniqueness theorems forthe modified Helmholtz equation with Dirichlet or Neumann conditions failfor this problem because the boundary conditions are unknowns in the prob-lem. Furthermore in section 2.3.1, we constructed non-radially symmetricsolutions (2.51) using Fourier techniques. We also noted that computing so-2556.3. Solving the GMS Modellutions to (2.51) for arbitrary initial data converged quite consistently to theradially symmetric case. We will instead use these solutions as a verificationthat non-radially symmetric solutions found from the numerical curve motionproblem indeed satisfy the analytic problem. For example, if we attempt tocompute U0 using the numerical curve formulation for an initial U0 configura-tion U00(\u00CF\u0083) = cos(3\u00CF\u0083) then we converge to a non radially symmetric solution(solid curve in Figure 6.13). If we use this computed U0 solution as an initialguess for the non-radially symmetric Newton\u00E2\u0080\u0099s method on (2.51) then it isalso a solution to this problem (dashed curve in Figure 6.13). The agreementshows that indeed the non-radially symmetric solution found is a true oneand not an artifact of the system such as by numerical discretization.2566.3. Solving the GMS Model0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10.20.250.30.350.40.45\u00CF\u0083U 0 Numeric SolutionEigenfunction ExpansionFigure 6.13: Initial circle U0 formulation under the GMS model with R = 4,r0 = 2, D = 1, and exponent set (2, 1, 2, 0). The blue solid curve is thecomputed U0 solution from the numerical interface problem with an initialguess of cos(3\u00CF\u0083) while the red dashed curve is the convergent solution to(2.51) by using the computed solution as an initial guess.Having demonstrated the existence of multiple solutions to this geometry, weare clearly able to violate the uniqueness which can not only lead to differentinitial U0 and velocity configurations but also to solution bifurcations in thedynamic problem. This is currently beyond the scope of this work but issomething of interest to pursue in further analysis.Other ExamplesWe will now showcase examples beyond circular solutions concentric with theorigin of which analytic work is limited or has not been considered. Figure2576.3. Solving the GMS Model6.14 shows a angular dependent radius in a \u00E2\u0080\u009Cflower-pedal\u00E2\u0080\u009D pattern. Veryquickly it can be seen that the dynamic tendency is to circularize and thenshrink. Since we start with an initial unperturbed radius that is quite small,we predicted this circularization in section 2.3.2. However, this behaviourgenerally is in stark contrast to what has been observed in weak diffusionlimits of a stripe [39] and in energy minimizing space filling curves [22].Figure 6.15 also has an angular perturbation in the radius forming a lobestructure, but the dynamic effect is to become an ellipse before shrinking.As the curve shrinks, the ratio of major and minor axes of the ellipse tendto 1 but it is not clear if the curve ever fully circularizes. It is interestingto note that typically this perturbation leads to splitting into two distinctstructures [39] as opposed to the behaviour here.2586.3. Solving the GMS Model\u00E2\u0088\u00920.8 \u00E2\u0088\u00920.6 \u00E2\u0088\u00920.4 \u00E2\u0088\u00920.2 0 0.2 0.4 0.6 0.8\u00E2\u0088\u00920.8\u00E2\u0088\u00920.6\u00E2\u0088\u00920.4\u00E2\u0088\u00920.200.20.40.60.8xy t=0t=0.02t=0.04t=0.07Figure 6.14: Perturbation of a circle with perturbed radius r = 1/2 +0.1 cos(6\u00CE\u00B8) using the GMS model with R = 1, D = 1, exponent set (2, 1, 2, 0),and \u00CF\u0083\u00CB\u0086 = 10. The boundary curve at R = 1 has been omitted.2596.3. Solving the GMS Model\u00E2\u0088\u00920.8 \u00E2\u0088\u00920.6 \u00E2\u0088\u00920.4 \u00E2\u0088\u00920.2 0 0.2 0.4 0.6 0.8\u00E2\u0088\u00920.4\u00E2\u0088\u00920.3\u00E2\u0088\u00920.2\u00E2\u0088\u00920.100.10.20.30.4xy t=0t=0.04t=0.09t=0.14Figure 6.15: Perturbation of a circle with perturbed radius r = 1/2 +0.3 cos(2\u00CE\u00B8) using the GMS model with R = 1, D = 1, exponent set (2, 1, 2, 0),and \u00CF\u0083\u00CB\u0086 = 10. The boundary curve at R = 1 has been omitted.Figure 6.16 shows the evolution under the GMS model starting from anelliptical configuration. As with the perturbation in Figure 6.15 it is notimmediately clear if the curve completely circularizes as it shrinks to theorigin or if the ratio of the major and minor axes just tends to 1.2606.3. Solving the GMS Model\u00E2\u0088\u00920.5 \u00E2\u0088\u00920.4 \u00E2\u0088\u00920.3 \u00E2\u0088\u00920.2 \u00E2\u0088\u00920.1 0 0.1 0.2 0.3 0.4 0.5\u00E2\u0088\u00920.5\u00E2\u0088\u00920.4\u00E2\u0088\u00920.3\u00E2\u0088\u00920.2\u00E2\u0088\u00920.100.10.20.30.40.5xy t=0t=0.02t=0.04t=0.07Figure 6.16: Ellipse with major axis a = 1/2 and minor axis b = 1/4 usingthe GMS model with R = 1, D = 1, exponent set (2, 1, 2, 0), and \u00CF\u0083\u00CB\u0086 = 10.The boundary curve at R = 1 has been omitted.Next, we consider non-concentric circles such as in Figure 6.17. Figure 6.17ashows a non-concentric initial circle placed on the x-axis at [1, 0] with radiusr0 = 1/2 while Figure 6.17b shows the same initial circle but placed at theorigin. The dynamics of the two models look very similar.2616.3. Solving the GMS Model\u00E2\u0088\u00924 \u00E2\u0088\u00923 \u00E2\u0088\u00922 \u00E2\u0088\u00921 0 1 2 3 4\u00E2\u0088\u00924\u00E2\u0088\u00923\u00E2\u0088\u00922\u00E2\u0088\u0092101234xy t=0t=0.2(a) Circle with centre (\u00E2\u0088\u00921, 0) and ra-dius r0 = 1/2.\u00E2\u0088\u00924 \u00E2\u0088\u00923 \u00E2\u0088\u00922 \u00E2\u0088\u00921 0 1 2 3 4\u00E2\u0088\u00924\u00E2\u0088\u00923\u00E2\u0088\u00922\u00E2\u0088\u0092101234xy t=0t=0.2(b) Circle with centre (0, 0) and ra-dius r0 = 1/2.Figure 6.17: Circle evolution using the GMS model with R = 4, D = 1,exponent set (2, 1, 2, 0), and \u00CF\u0083\u00CB\u0086 = 10.However, in Figure 6.18a, we place a circle of radius r0 = 1/2 off the axesentirely and centre it at [\u00E2\u0088\u00921, 2] with saturation \u00CF\u0083\u00CB\u0086 = 10. The interesting phe-nomena here is that, as the curve grows, the circle becomes elliptical and themajor axis rotates in the counter-clockwise direction. When the saturationis set to zero (Figure 6.18b), the curve remains a circle and furthermore ac-tually shrinks instead of grows. This demonstrates the effect the saturationcan have on the qualitative curve structure.2626.3. Solving the GMS Model\u00E2\u0088\u00921.4 \u00E2\u0088\u00921.3 \u00E2\u0088\u00921.2 \u00E2\u0088\u00921.1 \u00E2\u0088\u00921 \u00E2\u0088\u00920.9 \u00E2\u0088\u00920.8 \u00E2\u0088\u00920.7 \u00E2\u0088\u00920.61.61.71.81.922.12.22.32.42.5xy t=0t=0.06t=0.13t=0.2(a) Circle with saturation \u00CF\u0083\u00CB\u0086 = 10.\u00E2\u0088\u00921.3 \u00E2\u0088\u00921.2 \u00E2\u0088\u00921.1 \u00E2\u0088\u00921 \u00E2\u0088\u00920.9 \u00E2\u0088\u00920.81.751.81.851.91.9522.052.12.152.22.25xy t=0t=0.043t=0.052t=0.0546(b) Circle with saturation \u00CF\u0083\u00CB\u0086 = 0.Figure 6.18: Non-concentric circle evolution with centre [\u00E2\u0088\u00921, 2] and radiusr0 = 1/2 using the GMS model with R = 4, D = 1, and exponent set(2, 1, 2, 0). The boundary curve at R = 4 has been omitted.Finally, we consider the case of curve buckling. In Figure 5.16 of section5.2.2 we saw a scenario where a perturbed circle elongated into a bucklingtype pattern. Furthermore, we discussed in section 2.3.2 that if this type ofphenomenon were to occur it must be such that r0 is not small as otherwisethe curvature was a stabilizing mechanism. Therefore, we consider a casewith a perturbed circle r = 5+0.2 cos(6\u00CE\u00B8) in a larger circular domain R = 10with exponent set (2, 1, 2, 0) and saturation \u00CF\u0083\u00CB\u0086 = 5. As can be seen in Figure6.19, indeed a buckling pattern forms in this instance. However, it is worthnoting that it is unclear if this curve pattern is stable to breakup. In Figure5.16, the parameter regime was approaching the weak interaction regimewhere curve buckling and curve splitting is known to occur for stripes. Assuch, it is an open problem to investigate stable curve buckling in the semi-strong regime.2636.3. Solving the GMS Model\u00E2\u0088\u009210 \u00E2\u0088\u00925 0 5 10\u00E2\u0088\u009210\u00E2\u0088\u00928\u00E2\u0088\u00926\u00E2\u0088\u00924\u00E2\u0088\u009220246810 t =0t =3t =8t =12Figure 6.19: Perturbation of a circle with perturbed radius r = 5+0.2 cos(6\u00CE\u00B8)using the GMS model with R = 10, D = 1, exponent set (2, 1, 2, 0), and\u00CF\u0083\u00CB\u0086 = 5. The boundary curve at R = 10 has been omitted.Not only have we demonstrated agreement with analytically obtained resultsin the GMS, we have shown that our numerical method can easily extendto curves for which analytical work is limited or unavailable. Furthermore,the generality of the method can easily be extended to other models withdifferent reaction terms f(u, v) and g(u, v). As was stated in Chapter 5, thecomputations using this method have already been scaled to the O(\u000F\u00E2\u0088\u00922) time-scale and therefore avoids the long computation time required for full modelsimulations. Since there is a limiting value of b for the GMS model, onecould expect that curve rupturing may occur. However, in our computationsit seems that U0 always compensates with the saturation parameter \u00CF\u0083\u00CB\u0086 in2646.3. Solving the GMS Modelsuch a way to prevent this rupturing. We therefore conjecture that, in thesemi-strong regime, as long as U0(s) is defined at t = 0 it will continue to bedefined for all points along the curve and therefore curve rupturing cannothappen.265Chapter 7ConclusionsWe have presented a variety of techniques and results to better understandgeneral reaction diffusion equations of the formvt = \u000F2\u00E2\u0088\u0086v \u00E2\u0088\u0092 v + g(u, v)\u00CF\u0084ut = D\u00E2\u0088\u0086u\u00E2\u0088\u0092 u+1\u000Ff(u, v)localized on closed curves in R2. Particularly we have focused our attentionon the saturated Gierer-Meinhardt model,g(u, v) =v2uq(1 + \u00CF\u0083v2), f(u, v) =vous.Firstly, in Chapter 2, we derived a singular limit problem for the inhibitoru given by (2.19) for arbitrary curves using a boundary fitted coordinatesystem. This involved asymptotically translating information from the lo-calized activator v as jump and normal velocity conditions. Taking thisformulation, we derived results in 2.3.1 for a ring where the saturation pa-rameter \u00CF\u0083 was zero and concluded that a saddle-node bifurcation occurred interms of the outer domain radius R for which two equilibrium ring solutionsexisted. These results were similar to those derived in [45] but when we in-cluded saturation, there was not an equivalent analysis in the same work. As266Chapter 7. Conclusionssoon as \u00CF\u0083 > 0, at least one equilibrium solution existed for all R and insteada hysteresis bifurcation occurred. This is demonstrated best in Figure 2.11where we see that as R increases from zero there are regions where thereare one, two, and three equilibrium radii r0. This demonstrates hysteresisbecause if one were to start on the lower branch of solutions where r0 hasthe smallest equilibrium value and increase R then eventually the steady-state solution transitions to the branch where r0 is the largest. If R is thendecreased, the equilibrium will transition back to the lower r0 branch butat a different value of R than the first upper branch transition. Hysteresishas a potentially significant impact on the biological aspect of pattern for-mation. Recall that with scaling, R can change by physically altering thecircular domain radius or by altering the diffusion coefficient. Both of thesecould have biological applications where patterns on an organism change asa result of maturity (domain growth) of the organism or in the presence ofchemicals which could affect the diffusion coefficient. Specifically in [67], itis noted that patterns on a freshwater snail, Theodoxus fluviatilis, changebecause of concentrations of salt in the water they are exposed to. Whilethe specific pattern transitions for this snail are more complex than growingor shrinking ring structures, a diffusion dependent steady-state bifurcationcould help generally explain environmental impact on pattern formation. In[66], it was noted that the size and shape of patterns on a variety of fishcan change in a matter of seconds and spots can have drastically differentdiameters. This qualitative change over a fast time scale can be describedwith a hysteresis like feedback loop. The saturation parameter appears toplay an important role in biological self-regulation, something that previous267Chapter 7. Conclusionsmodels which excluded the effect could not sufficiently explain.Next we showed that since the boundary data for the inhibitor is not pre-scribed a priori, but rather is an unknown of the curve, standard uniquenessresults do not apply. As such, we showed the existence of ring solutions wherethe inhibitor is not radially symmetric. These solutions were verified to existnumerically in Chapter 6. While the ring solutions provided interesting re-sults, they did not rely on the general boundary coordinate model derivationas they could easily be constructed using polar coordinates. Therefore, wenext considered a problem of a near circle r = r0 +\u00CE\u00B5h(\u00CE\u00B8) where \u00CE\u00B5\u001C 1. Whilethis could also be studied using a polar coordinate formulation, it would re-quire carefully tracking the asymptotic consistency of both \u000F and \u00CE\u00B5, where asthe singular boundary coordinate framework is valid on any curve. One of thekey requirements in the near-circle formulation was that when saturation wasconsidered, the effective saturation parameter b given by (2.23) is perturbedby the changing geometry and was a function of the arclength of the curve.After resolving the continuity, jump, and normal velocity conditions we solvethe problem up to O(\u00CE\u00B52). It was necessary to consider the expansion up tothis order because as was evidenced in section 2.3.2, the first order correctiononly accounted for introducing sinusoidal perturbations from the Fourier se-ries of h(\u00CE\u00B8). However, polynomial representations h(\u00CE\u00B8)\u00CE\u00B1 for \u00CE\u00B1 > 1 occurredat O(\u000F2) and introduced n = 0 Fourier modes which overall caused a shift ineach of the boundary inhibitor value, U0, modified saturation parameter, b,and normal velocity V respectively. This was evidenced in Figures 2.12 and2.13. By analyzing the normal velocity correction we saw that when r0 was268Chapter 7. Conclusionssmall enough, the velocity was dominated entirely by the curvature whichcircularized the curve. However, for r0 large (or equivalently, D small), thiscould not be guaranteed to be satisfied and indeed we saw growing curves inChapter 5.Following the formulation of various solutions, we returned to the radiallysymmetric ring solution in Chapter 3 and performed a linear stability anal-ysis. When formulating the stability problem, we could not use standardexp(\u00CE\u00BBt) eigenfunctions since the steady-state was slowly evolving with time-scale T = \u000F\u00E2\u0088\u00922t. Because of this, we needed to use a WKB formulation (3.4)which resulted in the eigenfunction being the evolution of the eigenvalue overtime given by (3.9). The form of the operator used with the activator eigen-functions admitted even and odd solutions which we analyzed separately. Forthe even solutions, in Lemma 3.1.0.1, we saw that only solutions which wereof a single sign could lead to instabilities and in section 3.2, using the globalinhibitor problem we derived a non-local eigenvalue problem for the activa-tor eigenfunctions (3.16). Splitting this eigenvalue problem using functionsCm(\u00CE\u00BB) and f(\u00C2\u00B5) defined by (3.19), we were able to transform this into a rootfinding problem (3.20). We then studied the roots of this problem on thereal axis and in the complex plane after setting saturation b = 0. This leadto Principal Result 3.2.9.1 which has two very key conclusions. Firstly, thereexist neutral stability Fourier modes m = mb\u00E2\u0088\u0092 and m = mb+ such that a realunstable eigenvalue always exists on mb\u00E2\u0088\u0092 < m < mb+ . This is an importantresult because it states that ring solutions will always go unstable to breakupinstabilities for some range of modes when b = 0. A second interesting result269Chapter 7. Conclusionswas that on 0 < m < mb\u00E2\u0088\u0092 there were no unstable eigenvalues for \u00CF\u0084 smallenough which, as \u00CF\u0084 increased, transitioned via a Hopf bifurcation to becomeunstable.In order to verify the conclusions from Principal Result 3.2.9.1, we derivedan algorithm for computing eigenvalues in 3.2.10. In Figures 3.5 and 3.6, weconfirmed the range of unstable eigenvalues existed. In Table 3.1 we per-formed a series of comparisons to the asymptotic and numerically computedneutral stability points which showed excellent agreement with one another.Next we added saturation and showed via Figure 3.9 that the largest eigen-value tended to zero as b tended to its critical value. This had the effect ofstabilizing the ring solution as was seen in Figures 3.10 and 3.11. Full numer-ical simulations in Chapter 5 confirmed the stabilizing effect of saturation tobreakup instabilities.In section 3.3, we showed the odd eigenfunctions produced eigenvalues ofO(\u000F2), called small eigenvalues, which means they would only become rele-vant on the long time-scale T = O(\u000F\u00E2\u0088\u00922). Unlike previous work, such as [39],we were not able to classify these eigenvalues using the outer region awayfrom the curve since the derivative did not commute through the operator,having extra terms due to r\u00E2\u0088\u00921. Therefore, we approached the small eigenval-ues via the inner region only. We concluded that the small eigenvalues givenby (3.99) turned out to precisely be the normal velocity condition for a nearcircle from section 2.3.2. We showed via (3.101) how this conclusion couldbe predicted a priori. This is significant because it means that assuming ring270Chapter 7. Conclusionssolutions are stable to breakup then the full dynamics are captured by thenear circular problems discussed in section 2.3.2.In Chapter 4 we considered a general framework for which the non-localeigenvalue analysis of section 3.2 could be made explicit. Indeed we showedthat under a special class of eigenfunctions g(w) to the linearized operatorof the homoclinic orbit defined in Lemma 3.1.0.1, we could formulate theNLEP explicitly via (4.7). Assuming a form for g(w), we derived a condition(4.11) relating g(w) to f(w) in the problem wyy\u00E2\u0088\u0092w+ f(w) = 0. If this f(w)was such that a homoclinic orbit existed then we could obtain the explicitformulation (4.7). This allowed us to gain analytic insight into the stabilityof stripe solutions (section 4.2) and circle solutions (section 4.3). The sig-nificance of the explicit formulation for these geometries was that we wereable to derive Principal Result 4.2.2.1 which allowed us to extend stabilityresults for parameter regimes not previously explored. Furthermore, in aninfinite stripe domain, we were able to analytically obtain the unique value of\u00CF\u0084 = \u00CF\u0084Hm for which a Hopf bifurcation occurred and presented this in PrincipalResult 4.2.2.2. Such an analytic treatment of the Hopf bifurcation had neverpreviously been considered. We validated the Hopf bifurcation analysis inFigures 4.2 and 4.3 where we showed that for a fixed value of \u00CF\u0084 , and twoexponent values q that complex eigenvalues with positive real part did ordid not exist based on the comparison with the analytically determined Hopfbifurcation value \u00CF\u0084Hm . The results generalized for finite and infinite domainsbut for finite domains we could not classify the unique Hopf bifurcation value.271Chapter 7. ConclusionsFull numerical simulations in Chapter 5 verified breakup instability patternsfor both the stripe and ring in the explicit and implicit formulations. Thiswas done by randomly adding data to steady-state solutions and looking atthe discrete Fourier transform during the curve evolution. In all instancesthe predicted instability modes or bands persisted until secondary insta-bilities of spot dynamics occurred. We define secondary instabilities to beanything which that occurs after the breakup pattern has emerged. Sometypes of observed secondary instabilities were spot collocation motion wherespots continued to move as a ring and spot annihilation where spots weredestroyed until a single spot remained. We tested stability to zig-zag modesby giving a perturbed circle as initial data and allowing it to evolve. Wedemonstrated parameter regimes for which the curve was both stable andunstable to zig-zag modes.Finally in Chapter 6 we derived a general numerical framework for solv-ing quasi-steady solutions in R2. We used a layer potential formulation forsolving any problem of the form (2.19) subject to general conditions,[dudn]\u00CE\u0093= F (U0, V ),\u00E2\u008C\u00A9dudn\u00E2\u008C\u00AA\u00CE\u0093= G(U0, V )where U0 was the inhibitor value on the curve and V , the normal velocity. Weshowed the universality of this formulation by verifying our numerical methodwith the Mullins-Sekerka problem in 6.2.2. One of the intricate details abouta layer potential formulation is that it involves the evaluation of singular2727.1. Future Work and Open Problemsintegrals with logarithmic strength. We extract the logarithmic integralsand evaluate them analytically using a Lagrange polynomial interpolationfor any density functions. Since we are dealing with closed curves, in order toavoid the effects of near-singularities arising from periodicity we also removelogarithmic singularities that are within one period from the true singularity.In 6.2.1 we show that this introduces an error that has the same order asthe overall numerical quadrature error. We treat the curve evolution entirelyimplicitly which allows us to take relatively large time steps. This is incontrast to work such as [84] where an implicit-explicit splitting technique isused. In 6.3 we use the numerical method for solving the saturated Gierer-Meinhardt model and verify many of the analytically determined conclusionssuch as the r0 equilibrium radii and the circularization of near circular curveswith r0 \u001C 1. However, we then extend results beyond what is currentlyunderstood analytically in 6.3.2. We use an elliptical geometry and showthat it also tends to a circular curve. We also consider the evolution ofnon-concentric circles and interesting behaviour such as that which occurs inFigure 6.18 happens whereby a curve that is an initial circle evolves into arotating ellipse.7.1 Future Work and Open ProblemsThe completion of this work has stimulated some new key results in thepattern formation community, specifically attributed to the introduction ofthe saturation parameter in the Gierer-Meinhardt model. As such, there areseveral open problems that have arisen which warrant some future investi-2737.1. Future Work and Open Problemsgation. Firstly, the general curve tracking framework developed in Chapter6 will hopefully stimulate analytic investigation into other geometries thathave not been previously considered. Of specific interest is the non-concentriccircle problem which dynamically transitioned to an ellipse. Since this dy-namic event can occur with a single non-concentric circle, then perhaps thiscould be recast, using a conformal mapping argument, to a concentric geom-etry which is more available to analysis. In terms of the numerical trackingmethod itself, the dynamic transition of non-concentric circles may demon-strate the existence of solution bifurcations and we are interested in usingtechniques such as pseudo-arclength continuation to search for these bifurca-tion diagrams as a function of the saturation parameter. Another interestingcase study would be the further investigation of buckling states such as thoseevidence in Figure 6.19 and whether the saturation can be chosen such thatboth a buckling pattern forms and that it is stable to breakup.One of the limitations of the curve tracking method implementation is thatif disjoint curves approach each other, there are convergence issues due tonear singular integration of neighbouring logarithmic functions. As such, weare interested in adapting the method to handle near singular integration. Amethod outlined in [68] presents some promising ideas for how we could im-plement this feature. We are also interested in finding a level set formulationfor this problem to compare and contrast the advantages of each method. Insection 2.3.1 we commented that non-radially symmetric inhibitor solutionsexist on a ring but that they are difficult to compute using a standard New-ton algorithm because of the persistence of the constant solution. We are2747.1. Future Work and Open Problemstherefore interested in using a regularized optimization approach where wepenalize the constant solution to find non-radially symmetric solutions.For the explicit eigenvalue formulation in Chapter 4, we discussed that Prin-cipal Result 4.2.2.2 proves the existence of a unique Hopf bifurcation value\u00CF\u0084 = \u00CF\u0084Hm for stripe solutions in an infinite domain. However, no such resultexists for stripe solutions in a finite domain and as such, this remains an openproblem. For the implicit formulation in Chapters 2 and 3, we showed thatincluding saturation not only changes the bifurcation curve from saddle-nodeto one producing hysteresis but also that saturation can stabilize breakup in-stability modes. It is likely that the bifurcation diagram transformation andstability analysis are related and it is an open problem to determine the effectof introducing a second set of stable solutions in Figure 2.11 to the overalllinear stability to breakup modes. In terms of saturation, we have consideredthe exponent set p = 2, but there would be an equivalent saturated solutionfor other values of p and it is an open problem to investigate these under thehomoclinic existence criteria of Lemma 2.2.0.1.On the topic of stability, throughout this thesis we have only consideredpattern formation problems in the semi-strong regime where Dv = \u000F2 \u001C 1and Du = D = O(1). In contrast to this, there is also a weak interactionregime where Du = O(\u000F2) as well. In this regime we expect that stabilityresults for the Gierer-Meinhardt model without saturation as computed forstripes in [39] generalize to arbitrary curves since the underlying differentialequations are identical. 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An efficient boundary integral methodfor the mullins\u00E2\u0080\u0093sekerka problem. Journal of Computational Physics,127(2):246\u00E2\u0080\u0093267, 1996.287Appendix ADerivation of BoundaryProperties for Single LayeredPotentialsThe properties of layered potentials for Laplace\u00E2\u0080\u0099s equation have appearedin a variety of manners and texts (cf. [18], [34], [36]). In this appendix,we will derive the Dirichlet and Neumann jump conditions for the singlelayer potential of Laplace\u00E2\u0080\u0099s problem with the fundamental solution (6.3) ina more systematic way. The connection to the fundamental solution of theHelmholtz problem (6.1) is described in section 6.1. Consider solving thefollowing Dirichlet Laplace problem\u00E2\u0088\u0086u = 0, x \u00E2\u0088\u0088 R2 \ \u00CE\u0093u = f(x), x \u00E2\u0088\u0088 \u00CE\u0093with \u00CE\u0093 a closed curve in R2. The single layer potential for some continuousdensity \u00CF\u0086 is thenu(x) =\u00E2\u0088\u00AB\u00CE\u0093\u00CE\u00A6(x, q)\u00CF\u0086(q) dq,288Appendix A. Derivation of Boundary Properties for Single Layered Potentialsfor all x. To evaluate the Dirichlet condition, we need to understand whathappens as x approaches x0 along its normal direction. Consider some pointx0 on \u00CE\u0093 and define the portion of the curve \u00CE\u0093\u000F,\u00CE\u0093\u000F : {q \u00E2\u0088\u0088 \u00CE\u0093||x0 \u00E2\u0088\u0092 q| < \u000F}, \u000F\u001C 1.Let, x = x0 + \u00CE\u00B1n\u00CB\u0086i, where the subscript indicates that \u00CE\u00B1 > 0 traverses theinward pointing normal. As we near \u00CE\u0093,lim\u00CE\u00B1\u00E2\u0086\u00920+u(x0 + \u00CE\u00B1n\u00CB\u0086i)= lim\u00CE\u00B1\u00E2\u0086\u00920(\u00E2\u0088\u00AB\u00CE\u0093\\u00CE\u0093\u000F\u00CE\u00A6(x0 + \u00CE\u00B1n\u00CB\u0086i, q)\u00CF\u0086(q) dq +\u00E2\u0088\u00AB\u00CE\u0093\u000F\u00CE\u00A6(x0 + \u00CE\u00B1n\u00CB\u0086i, q)\u00CF\u0086(q) dq)= f(x+0 ),where we have used the Dirichlet condition. Now, since the fundamentalsolution is not singular on \u00CE\u0093 \ \u00CE\u0093\u000F then we have thatf(x+0 ) =\u00E2\u0088\u00AB\u00CE\u0093\\u00CE\u0093\u000F\u00CE\u00A6(x0, q)\u00CF\u0086(q) dq +\u00E2\u0088\u00AB\u00CE\u0093\u000Flim\u00CE\u00B1\u00E2\u0086\u00920+\u00CE\u00A6(x0 + \u00CE\u00B1n\u00CB\u0086i, q)\u00CF\u0086(q) dq, (A.1)where we have carried the limit inside the integral since we are not directlyevaluating at the singularity. On \u00CE\u0093\u000F, we have that q = x0 + \u00CE\u00B2t\u00CB\u0086 with \u00E2\u0088\u0092\u000F <\u00CE\u00B2 < \u000F and so on this portion of the curve,\u00CE\u00A6(x0 + \u00CE\u00B1n\u00CB\u0086i, q) = \u00E2\u0088\u009212pilog\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u009A\u00CE\u00B12 + \u00CE\u00B22\u00E2\u0088\u00A3\u00E2\u0088\u00A3\u00E2\u0088\u00A3 \u00E2\u0088\u00BC\u00CE\u00B1\u001C1\u00E2\u0088\u0092 12pilog |\u00CE\u00B2|+O(\u00CE\u00B12),and\u00CF\u0086(q) \u00E2\u0089\u0088 \u00CF\u0086(x0),289Appendix A. Derivation of Boundary Properties for Single Layered Potentialssince it is continuous. The second integral in (A.1) then becomes,\u00E2\u0088\u00AB\u00CE\u0093\u000Flim\u00CE\u00B1\u00E2\u0086\u00920+\u00CE\u00A6(x0 + \u00CE\u00B1n\u00CB\u0086i, q)\u00CF\u0086(q) dq = \u00E2\u0088\u0092\u00CF\u0086(x0)2pi\u00E2\u0088\u00AB \u000F\u00E2\u0088\u0092\u000Flim\u00CE\u00B1\u00E2\u0086\u00920+(log |\u00CE\u00B2|+O(\u00CE\u00B12))d\u00CE\u00B2= \u00E2\u0088\u0092\u00CF\u0086(x0)pi(\u000F log \u000F\u00E2\u0088\u0092 \u000F)=\u000F\u00E2\u0086\u009200,and therefore the singularity contributes nothing to the integral and as \u000F\u00E2\u0086\u0092 0,f(x+0 ) =\u00E2\u0088\u00AB\u00CE\u0093\u00CE\u00A6(x+0 , q)\u00CF\u0086(q) dq.Similarly, if we approach from below the curve, we havef(x\u00E2\u0088\u00920 ) =\u00E2\u0088\u00AB\u00CE\u0093\u00CE\u00A6(x\u00E2\u0088\u00920 , q)\u00CF\u0086(q) dqand so we have that [u] = 0 and the Dirichlet condition isf(x) =\u00E2\u0088\u00AB\u00CE\u0093\u00CE\u00A6(x, q)\u00CF\u0086(q) dq, x \u00E2\u0088\u0088 \u00CE\u0093. (A.2)Now consider what happens to the normal derivative of u as we approachthe curve from above,lim\u00CE\u00B1\u00E2\u0086\u00920+\u00E2\u0088\u0082u\u00E2\u0088\u0082nx(x0 + \u00CE\u00B1n\u00CB\u0086i) =\u00E2\u0088\u00AB\u00CE\u0093\\u00CE\u0093\u000F\u00E2\u0088\u0082\u00CE\u00A6\u00E2\u0088\u0082nx(x0, q)\u00CF\u0086(q) dq+\u00E2\u0088\u00AB\u00CE\u0093\u000Flim\u00CE\u00B1\u00E2\u0086\u00920+\u00E2\u0088\u0082\u00CE\u00A6\u00E2\u0088\u0082nx(x0 + \u00CE\u00B1n\u00CB\u0086i, q)\u00CF\u0086(q) dq, (A.3)290Appendix A. Derivation of Boundary Properties for Single Layered Potentialswhere once again we have interchanged derivatives, limits, and integrals byavoiding evaluating the singularity directly. Now,\u00E2\u0088\u0082\u00CE\u00A6\u00E2\u0088\u0082nx(x, q) = \u00E2\u0088\u0092 12pix\u00E2\u0088\u0092 q|x\u00E2\u0088\u0092 q|2 \u00C2\u00B7 n\u00CB\u0086x, (A.4)where n\u00CB\u0086x is the positively oriented normal at x0. Note that n\u00CB\u0086x = \u00C2\u00B1n\u00CB\u0086i de-pending on the orientation and we will proceed with n\u00CB\u0086x = n\u00CB\u0086i noting thatthere is a minus sign difference if n\u00CB\u0086x is the external norm. Substituting xand q on \u00CE\u0093\u000F into (A.4),\u00E2\u0088\u0082\u00CE\u00A6\u00E2\u0088\u0082nx(x0 + \u00CE\u00B1n\u00CB\u0086i, x0 + \u00CE\u00B2t\u00CB\u0086) = \u00E2\u0088\u009212pi\u00CE\u00B1\u00CE\u00B12 + \u00CE\u00B22.Now if \u00CE\u00B2 6= 0 then for \u00CE\u00B1\u001C 1,\u00CE\u00B1\u00CE\u00B12 + \u00CE\u00B22\u00E2\u0088\u00BC \u00CE\u00B1\u00CE\u00B22\u00E2\u0086\u0092 0.However, if \u00CE\u00B2 = 0 then\u00CE\u00B1\u00CE\u00B12 + \u00CE\u00B22=1\u00CE\u00B1\u00E2\u0086\u0092\u00CE\u00B1\u001C1sgn(\u00CE\u00B1)\u00E2\u0088\u009E.Therefore,\u00E2\u0088\u0092 12pi\u00CE\u00B1\u00CE\u00B12 + \u00CE\u00B22= A\u00CE\u00B4(\u00CE\u00B2).To find A we integrate around \u00CE\u00B2 = 0,\u00E2\u0088\u00AB 0+0\u00E2\u0088\u0092\u00CE\u00B1\u00CE\u00B12 + \u00CE\u00B22d\u00CE\u00B2 = \u00E2\u0088\u00922piA.291Appendix A. Derivation of Boundary Properties for Single Layered PotentialsTo track the singularity at \u00CE\u00B2 = 0, take \u00CE\u00B2 = \u00CE\u00B1b and let \u00CE\u00B1 \u00E2\u0086\u0092 0. Since \u00CE\u00B1 > 0,the integral becomes\u00E2\u0088\u00AB \u00E2\u0088\u009E\u00E2\u0088\u0092\u00E2\u0088\u009E11 + b2db = pi = \u00E2\u0088\u00922piA. (A.5)Therefore we have that\u00E2\u0088\u0082\u00CE\u00A6\u00E2\u0088\u0082nx(x0 + \u00CE\u00B1n\u00CB\u0086i, x0 + \u00CE\u00B2t\u00CB\u0086) = \u00E2\u0088\u009212\u00CE\u00B4(\u00CE\u00B2).Substituting this into (A.3) we getlim\u00CE\u00B1\u00E2\u0086\u00920+\u00E2\u0088\u0082u\u00E2\u0088\u0082nx(x0 + \u00CE\u00B1n\u00CB\u0086i) =\u00E2\u0088\u00AB\u00CE\u0093\\u00CE\u0093\u000F\u00E2\u0088\u0082\u00CE\u00A6\u00E2\u0088\u0082nx(x0, q)\u00CF\u0086(q) dq \u00E2\u0088\u0092 \u00CF\u0086(x0)\u00E2\u0088\u00AB \u000F\u00E2\u0088\u0092\u000F12\u00CE\u00B4(\u00CE\u00B2) d\u00CE\u00B2=\u000F\u00E2\u0086\u00920\u00E2\u0088\u00AB\u00CE\u0093\u00E2\u0088\u0082\u00CE\u00A6\u00E2\u0088\u0082nx(x0, q)\u00CF\u0086(q) dq \u00E2\u0088\u0092\u00CF\u0086(x0)2.Now if we instead approach the curve from below then the only difference isthat the scaling in (A.5) satisfies \u00CE\u00B1 < 0 and so the integral becomes\u00E2\u0088\u00AB \u00E2\u0088\u0092\u00E2\u0088\u009E\u00E2\u0088\u009E11 + b2db = \u00E2\u0088\u0092pi = \u00E2\u0088\u00922piA,and instead we getlim\u00CE\u00B1\u00E2\u0086\u00920\u00E2\u0088\u0092\u00E2\u0088\u0082u\u00E2\u0088\u0082nx(x0 + \u00CE\u00B1n\u00CB\u0086i) =\u00E2\u0088\u00AB\u00CE\u0093\u00E2\u0088\u0082\u00CE\u00A6\u00E2\u0088\u0082nx(x0, q)\u00CF\u0086(q) dq +\u00CF\u0086(x0)2.292Appendix A. Derivation of Boundary Properties for Single Layered PotentialsTherefore we have the jump conditionlim\u00CE\u00B1\u00E2\u0086\u00920+\u00E2\u0088\u0082u\u00E2\u0088\u0082nx(x+ \u00CE\u00B1n\u00CB\u0086i) =\u00E2\u0088\u00AB\u00CE\u0093\u00E2\u0088\u0082\u00CE\u00A6\u00E2\u0088\u0082nx(x, q)\u00CF\u0086(q) dq \u00E2\u0088\u0092 \u00CF\u0086(x)2x \u00E2\u0088\u0088 \u00CE\u0093, (A.6a)lim\u00CE\u00B1\u00E2\u0086\u00920\u00E2\u0088\u0092\u00E2\u0088\u0082u\u00E2\u0088\u0082nx(x+ \u00CE\u00B1n\u00CB\u0086i) =\u00E2\u0088\u00AB\u00CE\u0093\u00E2\u0088\u0082\u00CE\u00A6\u00E2\u0088\u0082nx(x, q)\u00CF\u0086(q) dq +\u00CF\u0086(x)2x \u00E2\u0088\u0088 \u00CE\u0093. (A.6b)Notice that indeed as was stated in section 6.1,[\u00E2\u0088\u0082u\u00E2\u0088\u0082n]\u00CE\u0093= lim\u00CE\u00B1\u00E2\u0086\u00920+\u00E2\u0088\u0082u\u00E2\u0088\u0082nx(x0 + \u00CE\u00B1n\u00CB\u0086x)\u00E2\u0088\u0092 lim\u00CE\u00B1\u00E2\u0086\u00920\u00E2\u0088\u0092\u00E2\u0088\u0082u\u00E2\u0088\u0082nx(x0 + \u00CE\u00B1n\u00CB\u0086x),=\u00EF\u00A3\u00B1\u00EF\u00A3\u00B4\u00EF\u00A3\u00B2\u00EF\u00A3\u00B4\u00EF\u00A3\u00B3lim\u00CE\u00B1\u00E2\u0086\u00920+\u00E2\u0088\u0082u\u00E2\u0088\u0082nx(x0 + \u00CE\u00B1n\u00CB\u0086i)\u00E2\u0088\u0092 lim\u00CE\u00B1\u00E2\u0086\u00920\u00E2\u0088\u0092\u00E2\u0088\u0082u\u00E2\u0088\u0082nx(x0 + \u00CE\u00B1n\u00CB\u0086i) = \u00E2\u0088\u0092\u00CF\u0086(x), n\u00CB\u0086x = n\u00CB\u0086ilim\u00CE\u00B1\u00E2\u0086\u00920+\u00E2\u0088\u0082u\u00E2\u0088\u0082nx(x0 \u00E2\u0088\u0092 \u00CE\u00B1n\u00CB\u0086i)\u00E2\u0088\u0092 lim\u00CE\u00B1\u00E2\u0086\u00920\u00E2\u0088\u0092\u00E2\u0088\u0082u\u00E2\u0088\u0082nx(x0 \u00E2\u0088\u0092 \u00CE\u00B1n\u00CB\u0086i) = \u00CF\u0086(x), n\u00CB\u0086x = \u00E2\u0088\u0092n\u00CB\u0086i,= \u00E2\u0088\u0092(n\u00CB\u0086x \u00C2\u00B7 n\u00CB\u0086i)\u00CF\u0086(x).We will now finish off by looking at the tangential derivative for completeness,lim\u00CE\u00B1\u00E2\u0086\u00920+\u00E2\u0088\u0082u\u00E2\u0088\u0082tx(x0 + \u00CE\u00B1n\u00CB\u0086i) =\u00E2\u0088\u00AB\u00CE\u0093\\u00CE\u0093\u000F\u00E2\u0088\u0082\u00CE\u00A6\u00E2\u0088\u0082tx(x0, q)\u00CF\u0086(q) dq+\u00E2\u0088\u00AB\u00CE\u0093\u000Flim\u00CE\u00B1\u00E2\u0086\u00920+\u00E2\u0088\u0082\u00CE\u00A6\u00E2\u0088\u0082tx(x0 + \u00CE\u00B1n\u00CB\u0086i, q)\u00CF\u0086(q) dq. (A.7)Looking at the derivative on \u00CE\u0093\u000F,\u00E2\u0088\u0082\u00CE\u00A6\u00E2\u0088\u0082tx(x0 + \u00CE\u00B1n\u00CB\u0086i, x0 + \u00CE\u00B2t\u00CB\u0086) = \u00E2\u0088\u009212pi\u00CE\u00B2\u00CE\u00B12 + \u00CE\u00B22\u00E2\u0088\u00BC\u00CE\u00B1\u001C1\u00E2\u0088\u0092 12pi\u00CE\u00B2.Using the Cauchy Principal Value for the integration,\u00E2\u0088\u00AB \u000F\u00E2\u0088\u0092\u000F1\u00CE\u00B2d\u00CE\u00B2 = 0293Appendix A. Derivation of Boundary Properties for Single Layered Potentialsand there is no contribution on \u00CE\u0093\u000F. Therefore,lim\u00CE\u00B1\u00E2\u0086\u00920+\u00E2\u0088\u0082u\u00E2\u0088\u0082tx(x+ \u00CE\u00B1n\u00CB\u0086i) =\u00E2\u0088\u00AB\u00CE\u0093\u00E2\u0088\u0082\u00CE\u00A6\u00E2\u0088\u0082nx(x, q)\u00CF\u0086(q) dq x \u00E2\u0088\u0088 \u00CE\u0093, (A.8a)lim\u00CE\u00B1\u00E2\u0086\u00920\u00E2\u0088\u0092\u00E2\u0088\u0082u\u00E2\u0088\u0082nx(x+ \u00CE\u00B1n\u00CB\u0086i) =\u00E2\u0088\u00AB\u00CE\u0093\u00E2\u0088\u0082\u00CE\u00A6\u00E2\u0088\u0082nx(x, q)\u00CF\u0086(q) dq x \u00E2\u0088\u0088 \u00CE\u0093, (A.8b)and we get [\u00E2\u0088\u0082u\u00E2\u0088\u0082tx]\u00CE\u0093= 0,showing that the tangential derivative is continuous.294"@en . "Thesis/Dissertation"@en . "2015-09"@en . "10.14288/1.0167281"@en . "eng"@en . "Mathematics"@en . "Vancouver : University of British Columbia Library"@en . "University of British Columbia"@en . "Attribution-NonCommercial-NoDerivs 2.5 Canada"@en . "http://creativecommons.org/licenses/by-nc-nd/2.5/ca/"@en . "Graduate"@en . "Hybrid asymptotic-numerical analysis of pattern formation problems"@en . "Text"@en . "http://hdl.handle.net/2429/53715"@en .