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Hybrid asymptotic-numerical analysis of pattern formation problems Moyles, Iain 2015

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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© 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 Φ0 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 O(1) . . . . . . . . . . . . 923.2.5 Real Eigenvalue Summary . . . . . . . . . . . . . . 943.2.6 Complex Eigenvalues . . . . . . . . . . . . . . . . 943.2.7 Complex Eigenvalues: 0 ≤ m < mb− . . . . . . . . 1003.2.8 Complex Eigenvalues: m > mb− . . . . . . . . . . 1013.2.9 Eigenvalue Summary . . . . . . . . . . . . . . . . . 1023.2.10 Numerical Computation of Eigenvalues . . . . . . 1043.2.11 Computing Eigenvalues, τ 6= 0 . . . . . . . . . . . 1063.2.12 Adding Saturation . . . . . . . . . . . . . . . . . . 1153.3 Eigenvalues Associated with Φ0 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, τ = 0 . . . . . . . . . 1544.2.2 Explicit Stripe Eigenvalues, τ > 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 τ = 0 and b = 0 of numerical and asymptoticcomputations of mb− and mb+ for a variety of exponent sets,, R, and r0 with D = 1 for all. The (n) refers to numericcomputations of (3.16) using eigs in Matlab. mb−(a) is com-puted via Newton’s 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− , 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− , (4.30) for mb+ , and (4.34) for mdom . . . . . . . . . . 1596.1 Numerical-analytic comparison of integrating∫ 10 cos(σ) log |σ|dσusing the product integration method with linear interpola-tion. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2306.2 Numerical-analytic comparison of integrating∫ 10 cos(σ) log |σ|dσusing 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(∆σ2 log ∆σ) 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 σˆ = 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(∆σ2 log ∆σ) 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 chooseLˆ = 20 as a sufficient representation of infinity. Note we cansolve the equation on [0, Lˆ] 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 σ. Here we take D = 1, R = 1 and exponentset (2, 1, 2, 0). . . . . . . . . . . . . . . . . . . . . . . . . . 412.8 Growth of Hˆ versus − log(r0) for R = 1 and when σ = 5. . 422.9 Right-hand side to (2.43) for various saturation values, σ 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, σ 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 σ. The dashed curve represents an asymp-totic approximation for r0  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(θ) = cos(6θ). Here we take exponent set(2, 1, 2, 0), R = 1, D = 1, r0 = 0.5, σ = 10, and ε = 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(θ) = cos(6θ). Here we take exponentset (2, 1, 2, 0), R = 1, D = 1, r0 = 0.5, σ = 10, and ε = 0.01. 673.1 Computation of f(µ) from (3.19b) for various o with m = 0,r0 = 0.5, and  = 0.025. We set m = 0 solely to satisfyµ = λ 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(µ) when m = 0, r0 = 0.5,  =0.025, and o = 2 along with the asymptotic expression (3.28)demonstrating the simple pole at µ = ν0. . . . . . . . . . . 843.3 Numerical computation of the derivative of (3.30) with respectto m. For a given value of m, R, and θλ, we compute theorder derivative of (3.30) over r0 ∈ [0, R] and then take themaximum value over that interval. The figure shows eachmaximal value of the derivative as a function of θλ 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  = 0.025. We set m = 0 solely to satisfydealing with a single variable λI . 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 λ in (3.16) for the case τ = 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 λ in (3.16) using Newton’s 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),  = 0.025,D = 1, R = 1, and r0 = 0.5. . . . . . . . . . . . . . . . . . 1113.7 Numerical computation of eigenvalues near m = mb− for τ >τ ∗mb− and τ < τ∗mb− . In all experiments (2, q, o, s) = (2, 1, 2, 0), = 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’s method on (3.59). The black x markpoints where Re(λ) = 0. In all experiments (2, q, o, s) =(2, 1, 2, 0),  = 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 τ = 0. In all cases(2, q, o, s) = (2, 1, 2, 0),  = 0.025, R = 1, r0 = 0.5, and D = 1 1173.11 Computation of eigenvalues for b = 0.2 and τ 6= 0. In all cases(2, q, o, s) = (2, 1, 2, 0),  = 0.025, R = 1, r0 = 0.5, and D = 1 1184.1 Eigenvalues for τ = 0 computed from (4.27) versus m for = 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 =∞ respectively. . . . . . . . . . . . . . . . . . . . . 1584.2 Plot of λ versus m for q = 1 (solid curve) and q = 2 (dashedcurve). The parameters here are s = 0, l =∞, and τ = 2. . 1704.3 Plot of τHm and λIH from (4.52) for m in 0 < m < mb− andl =∞. The parameter values are s = 0 and  = 0.05 while thesolid and dashed curves are for q = 1 and q = 2 respectively. 1714.4 Plot of eigenvalues λ versus m for l = 0.1 (solid curve) andl = 0.8 (dashed curve). The parameter values are q = 1, s = 0, = 0.05, and τ = 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 λ versus m for q = 2, s = 0,  = 0.05, l = 5, andτ = 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 λ versus m for q = 2, s = 0,  = 0.05, l = 5,and τ = 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 τ 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 0 = 0.05, D0 = 1, τ = 0.1, andd0 = 2. This corresponds to  = 0.05, l = 1/2, τ = 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 0 = 0.05, D0 = 1,τ = 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 0 = 0.05, D0 = 1, τ = 0.1, andd0 = 3. This corresponds to  = 0.05, l = 1/2, τ = 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 0 = 0.05, D0 = 1,τ = 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 0 = 0.05, D0 = 0.1, τ = 0.1,and d0 = 2. This corresponds to  = 0.05√10 ≈ 0.1581,l =√10/2 ≈ 1.58, τ = 0.1, and d = 2√10 ≈ 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 0 = 0.05, D0 = 0.1,τ = 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 0 = 0.01, D0 = 0.04,and τ = 0.1. This corresponds to  = 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 0 = 0.025, D0 = 1, and τ = 0.1.This corresponds to  = 0.025, R = 1, and τ = 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 0 = 0.05, D0 = 1,and τ = 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 0 = 0.025, D0 = 1, τ = 0.1, andσ = 25. This corresponds to  = 0.025, R = 1, and τ = 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 0 = 0.025, D0 = 1,τ = 0.1, and σ = 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 0 = 0.025, D0 = 1, τ = 0.1, andσ = 950. This corresponds to  = 0.025, R = 1, and τ = 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 0 = 0.05, D0 = 1, τ =0.1, and σ = 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),  = 0.05,τ = 0, and R = 1 with r0 = 0.213 and σ = 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 0 = 0.025, D0 = 1, τ = 0.1, andσ = 950. This corresponds to  = 0.025, R = 1, and τ = 0.1.We take as an initial radius r0 = 0.5 + 0.02 cos(6θ). . . . . 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 0 = 0.01, D0 = 0.01, τ = 0.1, andσ = 5910. This corresponds to  = 0.1, R = 10, and τ = 0.1.We take as an initial radius r0 = 0.5 + 0.02 cos(6θ). . . . . 2086.1 This shows the value of integrating (6.38) with ∆σ = 0.02 forall the possible discrete values of σ∗ ∈ [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 (σ∗) as defined in (6.39). The function has anabsolute maximum at σ∗ = 0.5 and and absolute minimum atσ∗ = 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 (−1, 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 (∆t =1 × 10−2) 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 (∆t = 1×10−2) 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 σˆ = 0, R = 1,r0 = 1/2, D = 1, exponent set (2, 1, 2, 0) and time step 1 ×10−3. 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 σˆ = 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 σˆ = 0. There is anunstable equilibrium at r0/R ≈ 0.044 (r0 ≈ 0.176) and a stableequilibrium at r0/R ≈ 0.69 (r0 ≈ 2.76). . . . . . . . . . . . 251xxiiiList of Figures6.9 Circle evolution using the GMS model with R = 4, D = 1,exponent set (2, 2, 2, 0), and σˆ = 0. The boundary curve atR = 4 has been omitted. . . . . . . . . . . . . . . . . . . . 2516.10 Circle evolution under the GMS model with σˆ = 10, R =1, r0 = 1/2, D = 1, exponent set (2, 1, 2, 0) and time step1× 10−3. 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 σˆ = 10,R = 1, r0 = 1/2, D = 1, and exponent set (2, 1, 2, 0). Sinceσˆ 6= 0 a Newton’s 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 × 10−2.The lines represent the numerical solution for different valuesof the saturation parameter σˆ (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(3σ) 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(6θ) using the GMS model with R = 1, D = 1, ex-ponent set (2, 1, 2, 0), and σˆ = 10. The boundary curve atR = 1 has been omitted. . . . . . . . . . . . . . . . . . . . 2596.15 Perturbation of a circle with perturbed radius r = 1/2 +0.3 cos(2θ) using the GMS model with R = 1, D = 1, ex-ponent set (2, 1, 2, 0), and σˆ = 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 σˆ = 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 σˆ = 10. . . . . . . . . . . . . . 2626.18 Non-concentric circle evolution with centre [−1, 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(6θ)using the GMS model with R = 10, D = 1, exponent set(2, 1, 2, 0), and σˆ = 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´e, 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’dlike 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’m 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’s 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´nard convection patterns(cf. [21], [13]). Unlike the difficulty in experimentally confirming theoreti-cal models of patterns in chemical reactions, the Rayleigh-Be´nard 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’s 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’s 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’s 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∆v − v +vpuq(1.1a)τut = Du∆u− u+vous(1.1b)with v an activator and u an inhibitor. The coefficients Di are the respectivediffusivities, τ is the inhibitor time constant, and the exponents (p, q, o, s)satisfy [20],p > 1, q > 0, o > 0, s ≥ 0, p− 1q<os+ 1. (1.2)We refer to the activator as autocatalytic because it encourages its owngrowth, satisfying the local-self enhancement for pattern formation. How-ever the activator also induces production of the inhibitor which ultimatelylimits the destabilizing autocatalytic behaviour, satisfying the long-range in-4Chapter 1. Introductionhibition requirement.Typically the onset of pattern formation is initiated by a Turing instabil-ity of a spatially homogeneous steady-state [74]. For the GM model, thiswas conducted in [24] where it was concluded that these types of solutionsare unstable when Du  Dv. In modern nomenclature, this is known as thesemi-strong regime of the GM model which is the focus of the work in thisthesis. Final patterns that ultimately form from perturbations to spatiallyhomogeneous initial data are strongly localized and qualitatively very differ-ent from the initial structure (cf. [38], [29]). Therefore there is vast interestin understanding these patterns far from the Turing regime. For the semi-strong description, we take Dv = 2  1 and Du = D = O(1) and make thefollowing scalings which are uniquely determined by appropriate order onebehaviour of the system (cf. [33]),v = Nq(p−1)(s+1)−qo vˆ, u = N(p−1)(p−1)(s+1)−qo uˆ; (p− 1)(s+ 1)− qo < 0,where N is the dimension of the localized domain. Substituting these scaledvariables into (1.1) simplifies to (dropping the hats),vt = 2∆v − v + vpuq(1.3a)τut = D∆u− u+1Nvous. (1.3b)We remark that the scaling is chosen so that the localized activator is of orderunity as  tends to zero. We will also consider a weakly saturated variant of51.1. Main Contribution and Summary of Previous Work(1.3) where we replacevpuq→ vpuq(1 + σvp),where the term weak saturation is denoted when σ = O(1), specifically thatσ  Npq(p−1)(s+1)−qo .1.1 Main Contribution and Summary ofPrevious WorkMost of the early mathematical work of the GM model was done in the limitD → ∞, known as the shadow regime (cf. [59], [26]). In this scenario, theinhibitor is spatially constant and (1.3) resolves to a single parabolic PDE forthe activator. The solution to this problem concentrates on a finite numberof points in a way that is related to a ball-packing problem. Moving from theshadow regime and taking D finite, multiple spike solutions were consideredin a one dimensional variant of (1.3) for which τ = 0 (cf. [33], [32]). Itwas concluded that an N spike pattern is stable if D < DN , some criticaldiffusivity related to the properties of Green’s functions. The analysis of thiswork was based upon the time constant τ being zero. As τ increases theinhibitor takes longer to notice changes in the activator concentration and itwas postulated that this should destabilize the pattern. Indeed the extensionfor τ 6= 0 in [79] showed this unstable behaviour for τ large and included theconditions for which a Hopf bifurcation occurs on a quasi-equilibrium spikepattern.61.1. Main Contribution and Summary of Previous WorkOne extension of the study of (1.3) to two dimensions involved analyzinga stripe solution centered at the mid-line of a rectangular domain (cf. [15],[39]). The construction of this solution was such that the cross-section was a1-spike pattern constructed in the one dimensional analysis. The stability ofthis stripe solution was analyzed in [39] in terms of both breakup and zigzaginstabilities which are generated by a class of even and odd eigenfunctionsrespectively. This work also shows that the addition of saturation can sta-bilize a stripe from equilibrium. Unlike the one dimensional case however,studies in two-dimensional domains are not as abundant, particularly fornon-stripe geometries. Steady-state ring solutions to N -D radially symmet-ric domains with N ≥ 2 was investigated in [58] but with no accompanyingstability analysis information. The stability of homoclinic stripes and ringshas been considered for other models with the Grey-Scott model ([52], [40]),the Schnakenberg model [11], and the Swift-Hohenberg model [22], but lesswork is available for the GM model. Many of these models share similarcharacteristics but the presence of a saturation parameter in the GM modelintroduces unique aspects from previously studied models.Through the use of matched asymptotic analysis (cf. [9]), we analyze theexistence and stability of ring solutions to breakup and zigzag instabilitiesfor the saturated Gierer-Meinhardt model. The analysis of breakup instabil-ities has a similar structure to that of a stripe and leads to the same con-clusions but relies on a different Green’s function formulation. Specifically,the ring geometry involves modified Bessel functions which are less analyti-cally tractable than the exponential Green’s functions that are generated on71.1. Main Contribution and Summary of Previous Worka rectangular domain. As such we require numerical computations of keymonotonicity properties to show regions of stability and instability for thebreakup. The analysis of zig-zag eigenvalues associated with the odd eigen-functions for the ring geometry shows a significant difference in derivation ascompared to the stripe in [39] or [33]. The latter analysis relies on differenti-ating the global problem for the activator for which the Laplacian coefficientsfor a stripe are constant. In contrast, the polar form of the Laplace oper-ator has non-constant coefficients arising from the 1/r curvature term andtherefore we present a more natural formulation relying on the inner regionasymptotic analysis of (1.3).Unlike the stripe, there is no equivalent universal mid-line with which toplace a ring equilibrium. Instead, the equilibrium values depend on the pa-rameters of the problem and, when there is no saturation, undergo a saddle-node bifurcation with which equilibrium radii only exist on D > 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(−2) 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) curveΓ 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 Ωring 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(θ).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 τ , 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 λ = 0 and, as weshow, are of order 2. 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’s 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’s 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 = 2∆v − v + g(u, v) (2.1a)τut = D∆u− u+1f(u, v) (2.1b)on a domain Ω ⊂ R2 subject to Neumann boundary conditions on ∂Ω. τis an effective time scale delay between the activator and inhibitor while Dand 2 are the diffusivities of the inhibitor and activator respectively. As wasdiscussed in Chapter 1, we consider D = O(1) while  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 Γ.To do this, we will consider a boundary fitted coordinate system (cf. Figure142.1. Choosing a Coordinate System2.1, [22], [35], [27]) wherex ≡ γ1(s) + ηnˆx, y ≡ γ2(s) + ηnˆy.Here s is the arclength of the curve with 〈γ1(s), γ2(s)〉 being the parametriza-tion of Γ and η is the signed normal distance from the curve with normalnˆ = 〈nˆx, nˆy〉. We consider the normal to be the inward pointing normalrelative to Γ and so η > 0 denotes inside the curve for a single curve.Γ = 〈γ1(s), γ2(s)〉tˆnˆ1Figure 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 satisfytˆ ≡ 〈γ′1, γ′2〉, nˆ ≡ 〈−γ′2, γ′1〉;tˆ′ = κnˆ, nˆ′ = −κtˆwhere prime denotes differentiation with respect to s and κ is the signed cur-vature (positive for convex curves). Using this information we can computethe (s, η) derivatives for (x, y) to get the Jacobian,J =xs xηys yη =(1− κη)γ′1 −γ′2(1− κη)γ′2 γ′1 ; det J = 1− κηby noting that |tˆ| =√γ′21 + γ′22 = 1. Therefore, by the inverse functiontheorem,sx syηx ηy =γ′1(1−κη)γ′2(1−κη)−γ′2 γ′1. (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∆ = ∂xx + ∂yy = ∂ηη −κ(1− κη)∂η +1(1− κη)∂s(∂s(1− κη)). (2.3)It is worth noting that there is a slight issue with this formulation and thatis the singularity (1 − κη) 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 ∈ 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 η away from the centre of the circle x. Therefore xis no longer uniquely defined by a single (s, η) 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 whereη  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 Γ, i.e.for some slow time scale T = a()t, we have that the boundary fitted coordi-nates satisfy (η, s) = (η(T ), s(T )). Quasi-steady solutions are typical whenconsidering front motion problems (cf. [65], [31], [11]). We rescale our timederivative,∂∂t→ aη˙ ∂∂η+ as˙∂∂swhere 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()region near Γ can be defined with the local distance coordinate ηˆ = η . We172.2. Asymptotic Expansion of Steady-Statewill also define the activator and inhibitor variable near the front,u˜(ηˆ, s) = u(ηˆ, s), v˜(ηˆ, s) = v(ηˆ, s).In the local coordinate frame, the system (2.1) becomes,aη˙v˜ηˆ + as˙v˜s =v˜ηˆηˆ −κ(1− κηˆ) v˜ηˆ+2(1− κηˆ)∂∂s(v˜s(1− κηˆ))− v˜ + g(u˜, v˜) (2.4a)aτ η˙u˜ηˆ + 2aτ s˙u˜s =Du˜ηˆηˆ −Dκ(1− κηˆ) u˜ηˆ+2D(1− κηˆ)∂∂s(u˜s(1− κηˆ))− 2u˜+ f(u˜, v˜). (2.4b)We expand the inner solutions as follows:v˜ ∼ v˜0+v˜1+. . . u˜ ∼ u˜0+u˜1+. . . , κ ∼ κ0+κ1+. . . , η˙ ∼ η˙0+η˙1+. . .and note that since we want the curve motion to be on a sub-order one timescale, it is most natural to take a = 2. To first order we get,v˜0ηˆηˆ − v˜0 + g(u˜0, v˜0) = 0 (2.5a)u˜0ηˆηˆ = 0. (2.5b)We can solve (2.5b) to get,u˜0 = Aηˆ +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˜0 = 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˜0ηˆ(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 − w + f(w) = 0, −∞ < y <∞, w → 0 as |y| → ∞,w′(0) = 0, wm = w(0) > 0 (2.6)and assume f(w) is C2 smooth on w > 0 with f(0) = 0 and f ′(0) < −1. Ifwe define Q(w) ≡ f(w)−w then a unique, positive, homoclinic orbit solutionexists when1. Q(0) = 0, Q′(0) < 02. For s > 0 Q(s) = 0, Q′(s) > 0; Q(w) < 0, for 0 < w < s3. Q(w) > 0 for s < w < wm with wm satisfying∫ wm0 Q(w) dw = 0.The requirements on f(w) are needed so that w decays exponentially in thefar-field. When f ′(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 ′(0) is not finite such as f(w) = w logw we can solve (2.6) exactly and showthat w → exp(−y2) 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′(0) = 0. Ifwe take w = v˜0 and f(w) = g(U0, v˜0) 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˜0 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 |ηˆ| → ∞.Continuing the expansion of (2.4) we have at O(),Lv˜1 = κ0v˜0ηˆ − gu(U0, v˜0)u˜1 + v˜0ηˆη˙0 (2.7a)u˜1ηˆηˆ = −1Df(U0, v˜0) (2.7b)whereLv˜1 = v˜1ηˆηˆ − v˜1 + gv(U0, v˜0)v˜1.202.2. Asymptotic Expansion of Steady-StateConsider the problem for the homoclinic orbit (2.5a) and differentiate,(v˜0ηˆ)ηˆηˆ − v˜0ηˆ + gv(U0, v˜0)v˜0ηˆ = Lv˜0ηˆ = 0. (2.8)Here we see that v˜0ηˆ is a homogeneous solution to (2.7a) and therefore wewill require an orthogonality condition with the source terms. This conditionisκ0∫ ∞−∞v˜20ηˆ dηˆ + η˙0∫ ∞−∞v˜20ηˆ dηˆ −∫ ∞−∞gu(U0, v˜0)v˜0ηˆu˜1 dηˆ︸ ︷︷ ︸I= 0. (2.9)If we defineG ≡∫ v˜00gu(U0, x) dx (2.10)then we can simplify the final integral and use integration by parts to getI = −∫ ∞−∞Gu˜1ηˆ dηˆby noting that since v˜0 is even then G(−∞) = G(∞) = 0. We now define anew function,Gˆ(ηˆ) ≡∫ ηˆ0G(x) dx (2.11)so that integrating I by parts once more, we haveI = − u˜1ηˆGˆ∣∣∣∞−∞+∫ ∞−∞Gˆu˜1ηˆηˆ dηˆ.212.2. Asymptotic Expansion of Steady-StateAgain, since v˜0 is even then G is as well, which ensures Gˆ is an odd function,so thatI = −Gˆ(∞) (u˜1ηˆ(∞) + u˜1ηˆ(−∞))−1D∫ ∞−∞Gˆf(U0, v˜0) dηˆ, (2.12)where we have simplified the last integral by using (2.7b). However, since v˜0is even then so to is f(U0, v˜0) so the final integrand is odd and hence vanishesover the domain. Finally then we can write the solvability condition (2.9) asη˙0 = −κ0 −Gˆ(∞)∫∞−∞ v˜20ηˆ dηˆ(u˜1ηˆ(∞) + u˜1ηˆ(−∞)) (2.13)which prescribes the leading order velocity of the curve. We can also relatethe difference in u˜1ηˆ by integrating (2.7b),(u˜1ηˆ(∞)− u˜1ηˆ(−∞)) = −1D∫ ∞−∞f(U0, v˜0) dηˆ. (2.14)Finally, we will consider the expansion at O(2) as this will be needed whenanalyzing the spectrum of the linearization for a radial geometry,Lv˜2 =κ0v˜1ηˆ + κ1v˜0ηˆ + κ20ηˆv˜0ηˆ + η˙0v˜1ηˆ + η˙1v˜0ηˆ + s˙0v˜0s − v˜0ss− 12guu(U0, v˜0)u˜21 − guv(U0, v˜0)u˜1v˜1 −12gvv(U0, v˜0)v˜21 − gu(U0, v˜0)u˜2,(2.15a)u˜2ηˆηˆ =1DU0 + κ0u˜1ηˆ −1Dfu(U0, v˜0)u˜1 −1Dfv(U0, v˜0)v˜1. (2.15b)Once again, this will have an orthogonality relationship with v˜0ηˆ producinga condition for the velocity correction η˙1 but we do not derive this here as222.2. Asymptotic Expansion of Steady-Statewe will consider a singularity limit where we take  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 η = O(1), we have thatv ≡ 0 (to within exponential order) and so we only need to consider theproblem for the inhibitor u:D∆u− u+ 1f(u, v) = 0 (2.16)where we assume that f(u, 0) = 0 so that the inhibitor solution does notexperience global blowup as → 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)=f(u, v˜(η))=→0∞, η = 00, elseand so (cf. [33])lim→0f(u, v)= Aδ(η)with δ(η) the Dirac mass centered at η = 0. To find A, we integrate over asmall domain including zero and scale to the inner coordinate,lim→0∫ 0+0−f(u(η), v(η))dη = lim→0∫ 0+/0−/f(u˜(ηˆ), v˜(ηˆ)) dηˆ = A.232.2. Asymptotic Expansion of Steady-StateExpanding out f(u˜, v˜),f(u˜, v˜) = f(U0, v˜0) +  (fu(U0, v˜0)u˜1 + fv(U0, v˜0)v˜1) + . . . =→0f(U0, v˜0)so thatA =∫ ∞−∞f(U0, v˜0) dηˆ,and finally (2.16) becomesD∆u− u = −(∫ ∞−∞f(U0, v˜0) dηˆ)δ(η). (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  as (2.17)captures the entire global problem. Upon solving u we will need to match tothe inner region viau(ηˆ, s) ∼ u(0, s) + ηˆ ∂u∂η∣∣∣∣η=0±+ · · · = u˜0(±∞) + u˜1(±∞),where the ± indicates approaching the curve from either side of η = 0. Uponperforming the matching we have u(0, s) = U0 andu˜1ηˆ(±∞) =∂u∂η∣∣∣∣η=0±. (2.18)Using this with (2.14) we have that[∂u∂n]η=0= − 1D∫ ∞−∞f(U0, v˜0) dηˆ,242.2. Asymptotic Expansion of Steady-Statewhere [·]η=a indicates a jump from η = a+ to η = a−. This is the equiva-lent singularity structure to match the Dirac measure in (2.17). Combiningeverything we can write the problem for u asD∆u− u = 0, x ∈ Ω (2.19a)∂u∂n= 0, x ∈ ∂Ω (2.19b)u = U0(s), x ∈ Γ (2.19c)[∂u∂n]Γ= − 1D∫ ∞−∞f(U0, v˜0) dηˆ, x ∈ Γ. (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 = κ0 +H(∂u∂n∣∣∣∣η=0++∂u∂n∣∣∣∣η=0−)(2.19e)where we have definedH ≡ Gˆ(∞)∫∞−∞ v˜20ηˆ dηˆ. (2.20)Note that if V is the normal velocity measured with respect to the originthen η˙ = −V . This is because for a single curve, if η˙ > 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 = −η˙.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 + σv2), f(u, v) =vous(2.21)for some exponent set (2, q, o, s) and saturation parameter σ > 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˜0 = Uq0w and recast (2.5a)for w withwηˆηˆ − w +w21 + bw2= 0, w′(0) = 0, lim|ηˆ|→∞w = 0, (2.22)whereb = U2q0 σ > 0 (2.23)is a modified saturation parameter. Notice if σ = 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(ηˆ) =32sech 2(ηˆ2). (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− 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± =1±√1− 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† = 2. If we classify the rootsof (2.25) by linear theory, we have that Q′(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∗ = 0, we haveQ′(w∗) =2− w∗w∗which is zero when w∗ = w† owing to the degeneracy of the root. Thereforethis root cannot be classified by linear theory but we have that Q′′(w†) =−1/2 and therefore for b = 1/4 since w = 0 and w = w† 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 ≤ b < 1/4,since w+ > w† (w− < w†) then Q′(w+) < 0 (Q′(w−) > 0) and thereforew = w− 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−. To satisfy condition 3, werequire∫ 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† to degenerate into w+ and w−then 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 ≤ 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, Lˆ] (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−20 −15 −10 −5 0 5 10 15 2000.20.40.60.811.21.41.6 Homoclinic orbit for b= 0ρw(a) b = 0−20 −15 −10 −5 0 5 10 15 2000.20.40.60.811.21.41.61.82 Homoclinic orbit for b= 0.12807ρw(b) b = 0.12807−20 −15 −10 −5 0 5 10 15 2000.511.522.5 Homoclinic orbit for b= 0.19211ρw(c) b = 0.19211−20 −15 −10 −5 0 5 10 15 2000.511.522.533.5 Homoclinic orbit for b= 0.21132ρw(d) b = 0.21132Figure 2.2: Solutions to (2.22) for different values of b. Here we choose Lˆ = 20as a sufficient representation of infinity. Note we can solve the equation on[0, Lˆ] and use symmetry because the functions are even.The homoclinic orbit problem holds for any curve Γ 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∫ ∞−∞f(U0, v˜0) dηˆ = Uβ0∫ ∞−∞wo dηˆ,∫ ∞−∞v˜20ηˆ dηˆ = U2q0∫ ∞−∞w2ηˆ dηˆ; (2.28a)β =qo− s. (2.28b)Furthermore, we can actually simplify Gˆ in (2.11) which we first write using292.3. Quasi-Steady State Profiles for the Gierer-Meinhardt Modelthe homoclinic asGˆ(∞) = −qU2q−10∫ ∞0∫ w(ηˆ)0v21 + bv2dv dηˆ = qU2q−10∫ ∞0W(w) dηˆwhere W is defined byW ≡∫ w0v2(1 + bv2)dv. (2.29)To simplify this, consider the homoclinic orbit problem (2.22), multiply bywηˆ, and integrate to getw2ηˆ − w2 + 2W(w) = 0.If we integrate again, we get∫ ∞0W(w) dηˆ = 14(∫ ∞−∞w2 dηˆ −∫ ∞−∞w2ηˆ dηˆ)(2.30)where we have exploited thatW(w) is an even function. Finally then we canuse this expression for Gˆ with (2.28a) in H to write (2.20) asH = − q4U0(∫∞−∞w2 dηˆ∫∞−∞w2ηˆ dηˆ− 1)≡ − q4U0Hˆ. (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˜0 then we explicitly remove the U0 dependence in a tractableway. With this simplification we now have the inhibitor problem isD∆u− u = 0, x ∈ Ω (2.32a)∂u∂n= 0, x ∈ ∂Ω (2.32b)u = U0(s), x ∈ Γ (2.32c)[∂u∂n]Γ= − 1DUβ0A, (2.32d)V0 = κ0 −q4U0Hˆ(∂u∂n∣∣∣∣η=0++∂u∂n∣∣∣∣η=0−), (2.32e)where we have definedA =∫ ∞−∞wo dηˆ. (2.33)2.3.1 Inhibitor Problem on a Circular CurveWe start by considering the curve Γ to be a circle of radius r0 inside a circulardomain 0 ≤ r ≤ R. In this case, the inward normal is nˆ = −rˆ anddudn∣∣∣∣ηˆ=0±= −ur(r∓0 )where we note that ηˆ+ is slightly on the inside of the curve which from theradial coordinate perspective is r−0 and the opposite is true for ηˆ−. 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∂∂r(r∂u∂r)+1Dr2∂2u∂θ2− 1Du = 0, 0 ≤ r ≤ R, r 6= r0, (2.34a)∂u∂r= 0, r = R (2.34b)u = U0(θ), r = r0 (2.34c)[∂u∂r]r0= − 1DUβ0A, (2.34d)dr0dt= − 1r0− q4U0Hˆ(∂u∂r∣∣∣∣r=r+0+∂u∂r∣∣∣∣η=r−0).(2.34e)Here we have noted that the curvature κ0 and normal velocity V are, κ0 = 1r0 ,dr0dt = −V 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 → 0+ asu(r) =AI0(r√D), 0 ≤ r ≤ r0E(α0I0(r√D)+K0(r√D)), r0 ≤ r ≤ R, α0 = −K0(R√D)′I0(R√D)′where 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 ′(u(r)) =dfdu∣∣∣∣u=u(r), f(u(r))′ =df(u(r))dr= f ′(u(r))dudr. (2.35)Enforcing continuity and the jump condition (2.34d) at r0 we haveu(r) =r0DUβ0AG0(r; r0) (2.36)where we have used the Wronskian relationship,W (r) = I0(r√D)K0(r√D)′−K0(r√D)I0(r√D)′= −1r. (2.37)Here G0(r; r0) is the Green’s functionG0(r; r0) =J0,1(r)J0,2(r0), 0 ≤ r ≤ r0J0,1(r0)J0,2(r), r0 ≤ r ≤ R(2.38)whereJ0,1(r) = I0(r√D), J0,2(r) = α0I0(r√D)+K0(r√D). (2.39)We can determine the value of U0 by solving u(r0) = U0 to getU0 =(Dr0AG0(r0; r0))1/(β−1). (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’s Method. Regardless we can use (2.23)to define (cf. [39])G˜(b) ≡ bA2qβ−1 =(Dr0G0(r0; r0)) 2qβ−1σ (2.41)and numerically (Figure 2.3) we see that dAdb > 0. Therefore,dG˜db > 0 and sofor each σ there is a unique b and vice-versa. An elegant proof in appendixB of [82] shows analytically that dG˜db > 0 when o = 2. The monotonicity ofG˜ 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= − 1r0− q4Hˆ(J ′0,1(r0)J0,1(r0)+J ′0,2(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√D= `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 (σ = 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 1−2−1.5−1−0.500.51r0/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 1−2−1.5−1−0.500.511.52r0/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’s method continuation on r0 starting from the smallest root r0s. We canapproximate these roots asymptotically by definingF (r) =(J ′0,1(r)J0,1(r)+J ′0,2(r)J0,2(r))=I1 (r)I0 (r)+α0I1 (r)−K1 (r)α0I0 (r) +K0 (r)(2.44)372.3. Quasi-Steady State Profiles for the Gierer-Meinhardt Modeland then for r  1 we haveF (r) ∼ − 1(α0 + log(2)− log(r)− γ)rwith γ ≈ 0.5772156649 the Euler-Mascheroni constant. Therefore from(2.43),dr0dT∼(qHˆ4(α0 + log(2)− log(r)− γ)− 1)1r0+O(r0). (2.45)and we can approximate r0s by setting the expression in brackets to zero,r0s ≈ exp(α0 + log(2)− γ −qHˆ4). (2.46)As R tends to infinity, α0 tends to zero and so there is a limiting small radius,r0s ≈ 1.1229 exp(−q) (2.47)where we have noted that when σ = 0 then Hˆ = 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 σ = 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 ≈ 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 σ is the staticparameter and not b which varies as U0 varies. Therefore, Hˆ 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 Hˆ) with (2.23). We solve theproblem (2.40) for U0 numerically using Newton’s 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’s method steps). Something we notice immediately from (2.41) isthat the denominator tends to zero as r0 tends to zero and thus for anyσ > 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  σ=0.01σ=0.5σ=2σ=5σ=10σ=25Figure 2.7: Modified saturation parameter b as a function of r0 for varioussaturation values σ. Here we take D = 1, R = 1 and exponent set (2, 1, 2, 0).When σ = 0 we had that as r0 tended to zero then F (r0) from (2.44) tendedto zero as well since Hˆ was fixed. Therefore dr0/dt tended to negative infinityas evidenced by Figures 2.4 and 2.5. However, when σ 6= 0 then since b tendsto the critical value as r0 tends to zero then Hˆ tends to infinity. While itcan be hard to analyze the growth of Hˆ 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  Hˆ− log(r0)Figure 2.8: Growth of Hˆ versus − log(r0) for R = 1 and when σ = 5.We plot (2.43) in Figure 2.9 for saturation values σ = 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 σ = 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 1−5−4−3−2−1012345r0/Rdr 0/dt  `=1`=4`=10`=20(a) σ = 0.50 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−5−4−3−2−1012345r0/Rdr 0/dt  `=1`=4`=10`=20(b) σ = 50 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−5−4−3−2−1012345r0/Rdr 0/dt  `=1`=4`=10`=20(c) σ = 100 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−5−4−3−2−1012345r0/Rdr 0/dt  `=1`=4`=10`=20(d) σ = 25Figure 2.9: Right-hand side to (2.43) for various saturation values, σ 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 1−5−4−3−2−1012345r0/Rdr 0/dt  `=1`=4`=10`=20(a) σ = 0.50 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−5−4−3−2−1012345r0/Rdr 0/dt  `=1`=4`=10`=20(b) σ = 50 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−5−4−3−2−1012345r0/Rdr 0/dt  `=1`=4`=10`=20(c) σ = 10Figure 2.10: Right-hand side to (2.43) for various saturation values, σ 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). Ifσ  1 then the behaviour should be similar to σ = 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 σ = 0.5. Since dr0/dt starts positive, this rootis necessarily stable and exists prior to R = Rc, the saddle-node bifurcationpoint when σ = 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  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 σ = 10and σ = 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 Hˆ = Hˆ(r0).1 2 3 4 5 6 700.511.522.533.5Rr 0  NumericAsymptotic(a) σ = 101 2 3 4 5 6 700.511.522.533.5Rr 0  NumericAsymptotic(b) σ = 8Figure 2.11: Bifurcation diagram to (2.43) for exponent set (2, 1, 2, 0) anddifferent values of σ. The dashed curve represents an asymptotic approxi-mation for r0  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, θ) =∞∑n=−∞Un(r) exp(inθ), U0 =∞∑n=−∞an exp(inθ),Uβ0 =∞∑n=−∞fn exp(inθ)where we treat the Uβ0 term separately just for simplicity. Using this expan-sion in (2.34) we get for each eigenmode n,1rddr(rdUndr)− n2Unr2− UnD= 0, (2.48a)dUndr∣∣∣∣r=R= 0, (2.48b)Un(r0) = an, (2.48c)[dUndr]r=r0= − 1DfnA. (2.48d)This is the n > 0 analogue of the radially symmetric case and so if we definethe functionsJn,1(r) = In(r√D), Jn,2(r) = αnIn(r√D)+Kn(r√D)(2.49a)462.3. Quasi-Steady State Profiles for the Gierer-Meinhardt Modelwhereαn = −Kn(R√D)′In(R√D)′ =Kn+1(R√D)− n√DR Kn(R√D)n√DR In(R√D)+ In+1(R√D) (2.49b)then the solution can be written asUn(r) =r0DfnAG0,n(r; r0)with G0,n(r; r0), the Green’s functionG0,n(r; r0) =Jn,1(r)Jn,2(r0), 0 ≤ r ≤ r0Jn,1(r0)Jn,2(r), r0 ≤ r ≤ 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 = aβ0 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 ≈1NN∑j=1exp(−2piin(j − 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(−2piin(j − 1)N).Using this, we can get the Fourier coefficients we desire by numerically solv-ing,FU0 = diag(Y)FUβ0 , (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 Γ to be a perturbed circle given byr = r0 + εh(θ), ε 1 (2.52)482.3. Quasi-Steady State Profiles for the Gierer-Meinhardt Modelstill inside a global circular domain 0 ≤ r ≤ 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  tends to zeroand so for our machinery to work with this perturbed circle geometry, werequire  ε 1. We can write the inner normal to Γ asnˆ =〈−1, εh′(θ)r0+εh(θ)〉√1 + ε2h′(θ)2(r0+εh(θ))2,where prime here indicates differentiation with respect to θ. We can use thisnormal vector to write∂u∂n=−ur + εh′(θ)(r0+εh(θ))2uθ√1 + ε2h′(θ)2(r0+εh(θ))2.We consider solving (2.32) with a formal expansionu(r, θ) = u0(r) + εu1(r, θ) + ε2u2(r, θ) + . . .where we explicitly note that we want to consider a perturbation from theradially symmetric inhibitor solution hence why u0 is independent of θ. Using492.3. Quasi-Steady State Profiles for the Gierer-Meinhardt Modelthe condition (2.32c) we haveU0 = u(r0 + εh(θ)) ∼ u0(r0) + ε(u0r(r)h(θ) + u1(r))r0+ ε2(12u0rr(r)h(θ)2 + u1r(r)h(θ) + u2(r))r0= U00 + εU01(θ) + ε2U02(θ). (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∣∣∣∣ηˆ=0±∼ − u0r(r∓0 ) + ε(−u0rr(r∓0 )h(θ)− u1r(r∓0 ))+ ε2(−12u0rrr(r∓0 )h(θ)2 − u1rr(r∓0 )h(θ) +u1θ(r∓0 )h′(θ)r20+u0r(r∓0 )h′(θ)22r20− u2r(r∓0 ))(2.54)where we recall that because we are using the inner normal, ηˆ = 0± corre-sponds to r = r∓0 . Finally we use (2.53) to writeUβ0 ∼ Uβ00 + εβUβ−100 U01 + ε2βUβ−2002((β − 1)U201 + 2U00U02). (2.55)In the presence of saturation, both A and Hˆ 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 σ ∼ U2q00σ + ε2qU2q−100 U01σ + ε2qU2q−200((2q − 1)U201 + 2U00U02)σ= b0 + εb1 + ε2b2, (2.56)and then writeA ∼ A0 + εdA0dbb1 + ε2(dA0dbb2 +12d2A0db2b21), (2.57a)Hˆ ∼ Hˆ0 + εdHˆ0dbb1 + ε2(dHˆ0dbb2 +12d2Hˆ0db2b21), (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(ε):1r∂∂r(r∂u1∂r)+1r2∂2u1∂θ2− 1Du1 = 0, r 6= r0, (2.58a)∂u1∂r∣∣∣∣r=R= 0, (2.58b)[u1]r0 = −h(θ)[du0dr]r0, (2.58c)[∂u1∂r]r0= −h(θ)[d2u0dr2]r0− A0DUβ−100 A¯0U01,(2.58d)512.3. Quasi-Steady State Profiles for the Gierer-Meinhardt Modelwhere we have definedA¯0 = β +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¯0r0[d2G0dr2]r0= A¯0r0 (J0,1(r0)J0,2(r0)′′ − J0,1(r0)′′J0,2(r0))=A¯0r0(K0(r0√D)′′I0(r0√D)−K0(r0√D)I0(r0√D)′′)where we defineA¯0 =A0DUβ00, (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 ′(r) = K0(r√D)′′I0(r√D)−K0(r√D)I0(r√D)′′=1r2.Therefore,[d2u0dr2]r0=A¯0r0(2.61)522.3. Quasi-Steady State Profiles for the Gierer-Meinhardt Modeland the derivative jump condition (2.58d) becomes[∂u1∂r]r0= −h(θ)A¯0r0− A¯0U00A¯0U01. (2.62)Since the problem (2.58) is linear in the θ dependence, we can perform aneigenfunction expansion,u1(r, θ) =∞∑n=−∞V1n(r) exp(inθ), U01(θ) =∞∑n=−∞U1n exp(inθ),h(θ) =∞∑n=−∞Hn exp(inθ)for integer eigenmodes, n. Upon this expansion, (2.58) becomes1rddr(rdV1ndr)− n2r2V1n −1DV1n = 0, r 6= r0, (2.63a)dV1ndr∣∣∣∣r=R= 0, (2.63b)[V1n]r0 = HnA¯0, (2.63c)[dV1ndr]r0= −HnA¯0r0− A¯0U00A¯0U1n. (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 αn by (2.49) then we canwriteV1n =AJn,1(r), 0 ≤ r < r0BJn,2(r), r0 < r ≤ 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−Jn,1(r0) Jn,2(r0)−J ′n,1(r0) J ′n,2(r0)︸ ︷︷ ︸=MAB =HnA¯0−HnA¯0r0 −A¯0U00 A¯0U1n =A¯1nB¯1n .Using the Wronskian relationship (2.37), which holds for any n, we have thatdetM = 1/r0 and thereforeAB =A¯1nr0J ′n,2(r0)− B¯1nr0Jn,2(r0)A¯1nr0J ′n,1(r0)− B¯1nr0Jn,1(r0) .If we define G0,n(r; r0) as in (2.50) and G1,n viaG1,n(r; r0) =Jn,1(r)J ′n,2(r0), 0 ≤ r < r0J ′n,1(r0)Jn,2(r), r0 < r ≤ R. (2.64)then we can writeV1n(r) = A¯1nr0G1,n(r; r0)− B¯1nr0G0,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¯1nr0(〈G1,n(r; r0)〉r0 +〈dG0dr〉r0)+(A¯1n +A¯0r0U00A¯0U1n)G0;n(r0, r0),where 〈·〉a indicates the average value across r = a. If we notice thatA¯0r0 =U00G0(r0; r0),then we can solve for U1n and getU1n =HnU00G0(r0; r0)(1− A¯0G0;n(r0; r0)G0(r0; r0))−1(G0,n(r0; r0)r0+〈G1,n(r, r0) +dG0dr〉r0).(2.66)Typically, we consider the perturbation h(θ) to be a finite combination ofsinusoidal modes and if this is the case then the correction at O(ε) willadd components in each of those modes. However, unless an n = 0 mode isexplicitly part of h(θ), the perturbation at this order can not describe verticalshifting as there is no cross-mode influence. Therefore, we will consider theexpansion at O(ε2) 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(ε2) is1r∂∂r(r∂u2∂r)+1r2∂2u2∂θ2− 1Du2 = 0, r 6= r0 (2.67a)∂u2∂r∣∣∣∣r=R= 0, (2.67b)[u2]r0 = −h(θ)[∂u1∂r]r0− h(θ)22[d2u0dr2]r0,(2.67c)[∂u2∂r]r0=12h′(θ)2r20[du0dr]r0+h′(θ)r20[∂u1∂θ]r0− h(θ)[∂2u1∂r2]r0− 12h(θ)2[d3u0dr3]r0− A¯0U200(A¯1U201 + A¯0U00U02), (2.67d)where we defineA¯1 =12β(β − 1) + qb0(2q − 1)A0dA0db+2q2b20A0d2A0db2.As with the O(ε) 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(θ)2 A¯02r0+ h(θ)A¯0A¯0U00U01. (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[∂u1∂θ]r0=∞∑n=−∞in [V1n]r0 exp(inθ) = A¯0∞∑n=−∞inHn exp(inθ) = A¯0h′(θ).Next, we need to determine[∂2u1∂r2]r0=∞∑n=−∞(A¯1nr0[d2G1,ndr2]r0− B¯1nr0[d2G0,ndr2]r0)exp(inθ)[d3u0dr3]r0= A¯0r0[d3G0dr3]r0.Since the Wronskian relationship (2.37) also holds for n 6= 0 we haveW0,n(r) = In(r√D)Kn(r√D)′−Kn(r√D)In(r√D)′= −1r(2.69)which we can differentiate to getW ′0,n(r) =In(r√D)Kn(r√D)′′−Kn(r√D)In(r√D)′′=[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 ′′0,n(r) = In(r√D)Kn(r√D)′′′−Kn(r√D)In(r√D)′′′+ In(r√D)′Kn(r√D)′′−Kn(r√D)′In(r√D)′′=[d3G0,ndr3]r+ In(r√D)′Kn(r√D)′′−Kn(r√D)′In(r√D)′′= − 2r30+W1,n(r0),where we define the new Wronskian,W1,n(r) = In(r√D)′Kn(r√D)′′−Kn(r√D)′In(r√D)′′. (2.70)Furthermore,[d2G1,ndr2]r0= In(r0√D)′Kn(r0√D)′′−Kn(r0√D)′In(r0√D)′′,and therefore[d3G0dr3]r0= − 2r30−W1,0(r0),[d2G1,ndr2]r0= W1,n(r0).To determine the Wronskian (2.70), consider that y = u0 = In(r√D)andy = v0 = Kn(r√D)satisfyd2ydr2+1rdydr−(n2r2+1D)y = 0582.3. Quasi-Steady State Profiles for the Gierer-Meinhardt Modeland differentiate this expression. Therefore u = In(r√D)′and v = Kn(r√D)′satisfyd2udr2+1rdudr−(n2 + 1r2+1D)u+2n2r3u0 = 0,d2vdr2+1rdvdr−(n2 + 1r2+1D)v +2n2r3v0 = 0respectively. We define W1,n(r) = u′v − uv′ and so multiplying the firstexpression by v and the second by u we havedW1,ndr+W1,nr− 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[∂2u1∂r2]r0=∞∑n=−∞(A¯1n(n2r2+1D)− B¯1nr0)exp(inθ),[d3u0dr3]r0= −A¯0(2r20+1D).We can simplify the first expression,[∂2u1∂r2]r0=A¯0r20(h(θ)− h′′(θ)) + A¯0Dh(θ) +A¯0A¯0U00r0U01592.3. Quasi-Steady State Profiles for the Gierer-Meinhardt Modeland finally write[∂u2∂r]r0=A¯0h′(θ)22r20+A¯0r20h(θ)h′′(θ)− A¯02Dh(θ)2− A¯0A¯0U00r0h(θ)U01 −A¯0U200(A¯1U201 + A¯0U00U02). (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(θ) and g(θ) have a fourier se-ries given byf(θ) ∼∞∑n=−∞an exp(inθ), g(θ) ∼∞∑m=−∞bm exp(imθ)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(θ)g(θ) =∞∑m=−∞∞∑n=−∞anbm exp(i(n+m)θ).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(θ) iscomposed of finite sinusoidal modes, the Fourier series will terminate. Thisallows us to use Definition 2.3.2.1 for quadratic products of h(θ) at O(ε2).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, θ) =∞∑k=−∞V2k(r) exp(ikθ), U02(θ) =∞∑k=−∞U2k exp(ikθ),and the base problem is identical to (2.63) so we can immediately writeV2k =AJk,1(r), 0 ≤ r < r0BJk,2(r), r0 < r ≤ R.The continuity and derivative jump conditions for each mode k become[V2k]r0 =A¯02r0∑n∈NHnHk−n +A¯0A¯0U00∑n∈NHnU1k−n = A¯2k, (2.73a)[dV2kdr]r0=− A¯02r20∑n∈Nn(k − n)HnHk−n −A¯0r20∑n∈N(k − n)2HnHk−n− A¯02D∑n∈NHnHk−n −A¯0A¯0U00r0∑n∈NHnU1k−n− A¯0U200(A¯1∑n∈NU1nU1k−n + A¯0U00U2k)= B¯2k. (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(θ) = cos(6θ) then there are twomodes at O(ε) of n = −6 and n = 6. Various quadratic combinations ofthese modes leads to k = −12, k = 0, and k = 12 as the only possible modesthat can occur at O(ε2). Therefore, if k = −12 then N = {−6} since it isonly through this mode that frequencies exp(−12iθ) can occur but for k = 0then the set N = {−6, 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(θ) = cos(3θ) + cos(6θ) which has admissiblemodes at O(ε2) of k = 0,±3,±6,±9,±12 and here for k = 0, the set isN = {±3,±6} while for k = 9, the set is N = {3, 6}. In similar fashion to(2.63) at O(ε), we can use (2.73) to write,V2k(r) = A¯2kr0G1,k(r; r0)− B¯2kr0G0,k(r; r0). (2.74)As with U1n, we can solve U2k by taking the average value of (2.53) at O(ε2)U2k =A¯0r02d2G0dr2∑n∈NHnHk−n + r0∑n∈NHn(A¯1k−ndG1,k−ndr− B¯1k−ndG0,k−ndr)+ r0A¯2kG1,k − r0B¯2kG0,k.If we defineB˜2k = B¯2k +A¯0A¯0U00U2kthen we can solveU2k =(1− A¯0G0,k(r0; r0)G0(r0; r0))−1(A¯0r02〈d2G0dr2〉r0∑n∈NHnHk−n+r0∑n∈NHn(A¯1k−n〈dG1,k−ndr〉r0− B¯1k−n〈dG0,k−ndr〉r0)+r0A¯2k 〈G1,k〉r0 − r0B˜2kG0,k(r0; r0)). (2.75)Thus we have completely solved the inhibitor problem up to O(ε2). 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 = κ0 −q2U0Hˆ〈dudn〉η=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]),κ0 =∣∣∣∣dtˆds∣∣∣∣ =r2 + 2r′2 − rr′′(r2 + r′2)3/2where tˆ is the unit tangent vector and once again prime is differentiationwith respect to s. Using the near-circle radius (2.52) we have thatκ0 ∼1r0− ε(h′′(θ) + h(θ)r20)+ ε2(4h(θ)h′′(θ) + h′(θ)2 + 2h(θ)22r30). (2.76)We can make a velocity expansion as followsV0 ∼ V00 + εV01 + ε2V02and 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(ε) isV01 =−(h′′(θ) + h(θ)r20)− qHˆ02U200(1− 2qb0Hˆ0dHˆ0db)〈du0dr〉r0U01+qHˆ02U00(h(θ)〈d2u0dr2〉r0+〈∂u1∂r〉r0)(2.77)All of the necessary values to compute V1 have been obtained and can there-fore be directly substituted. Notice that if r0  1 then to leading orderV01 ∼ −(h′′(θ) + h(θ)r20),and for typical h(θ) = cos(mθ) thenV01 ∼(m2 − 1r20)cos(mθ),which is in phase with h(θ) 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  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(ε2) isV02 =(4h(θ)h′′(θ) + h′(θ)2 + 2h(θ)22r30)+q2((U201U200− U02U200)Hˆ0 −U01U200dHˆ0dbb1+1U00(12d2Hˆ0db2b21 +dHˆ0dbb2))〈du0dr〉r0+q2U200(U00dHˆ0dbb1 − U01Hˆ0)(h(θ)〈d2u0dr2〉r0+〈∂u1∂r〉r0)− qHˆ02U00(−12h(θ)2〈d3u0dr3〉r0−h(θ)〈∂2u1∂r2〉r0+h′(θ)r20〈∂u1∂θ〉r0+h′(θ)22r20〈du0dr〉r0−〈∂2u2∂r2〉r0)(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 ε(taken here to be 0.01). In all of our simulations we take σ = 10, R = 1,D = 1, exponent set (2, 1, 2, 0), and r0 = 0.5 and we consider the near circleperturbation h(θ) = cos(6θ). Figure 2.12 shows the corrections at each orderof ε for U0 and b.652.3. Quasi-Steady State Profiles for the Gierer-Meinhardt Model0 1 2 3 4 5 60.09730.09740.09750.09760.09770.09780.09790.0980.09810.0982θU 0  NumericU00U00 + εU01U00 + εU01 + ε2U02(a) U0 corrections0 1 2 3 4 5 60.09460.09480.0950.09520.09540.09560.09580.0960.09620.0964θb  Numericb0b0 + εb1b0 + εb1 + ε2b2(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(θ) = cos(6θ).Here we take exponent set (2, 1, 2, 0), R = 1, D = 1, r0 = 0.5, σ = 10, andε = 0.01.We see that indeed the correction at O(ε) introduces the sinusoidal pertur-bation only whereas the correction at O(ε2) allows for the vertical shift cor-rection. Figure 2.13 shows the corrections at each order of ε 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 5 600.511.522.53θV 0  NumericV00V00 + εV01V00 + εV01 + ε2V02(a) Velocity corrections0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 100.511.522.53θV 0  NumericV00V00 + εV01V00 + εV01 + ε2V02(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(θ) = cos(6θ). Here we take exponent set (2, 1, 2, 0), R = 1, D = 1,r0 = 0.5, σ = 10, and ε = 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 = 2 1r∂∂r(r∂v∂r)+2r2∂2v∂θ2− v + v2uq(1 + σv2)(3.1a)τut = D1r∂∂r(r∂u∂r)+Dr2∂2u∂θ2− u+ 1vous(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 radiusρ =r − r0(T )and 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 θ and so we can perform a Fourier expansion as follows:v ∼ v˜e(r − r0, T)+ φ(r − r0, t)exp(imθ) (3.2a)u ∼ ue(r, T ) +M (r, t) exp(imθ) (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 θdictates that m is an integer on (−∞,∞), 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 = 2t. Substituting our693.1. Linear Stability Formulationexpansion (3.2) into (3.1) we get2φT − φρdr0dT=1r0 + ρ∂∂ρ((r0 + ρ)φρ) (3.3a)− 2m2(r0 + ρ)2φ− φ− qv˜2eu˜q+1e (1 + σv˜2e)M(r0 + ρ)+2v˜eu˜e(1 + σv˜2e)2φ, (3.3b)2τMT =Dr∂∂r(rMr)−Dm2r2M −M + 1ovo−1euseφ(r − r0)− 1svoeus+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 φ and M using the long time scale.We now consider a WKB ansatz (cf. [9]) for each eigenfunction,φ(ρ, T ) = Φ(ρ, T ) exp(ϕ(T )2), M(r, T ) = N(r, T ) exp(ϕ(T )2). (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)2ΦT + ΦϕT − Φρdr0dT=1r0 + ρ∂∂ρ((r0 + ρ)Φρ)− 2m2(r0 + ρ)2Φ− Φ− qv˜2eu˜q+1e (1 + σv˜2e)N(r0 + ρ)+2v˜eu˜e(1 + σv˜2e)2Φ, (3.5a)2τNT + τNϕT =Dr(rNr)r −Dm2r2N −N + 1ovo−1euseΦ(r − r0)− 1svoeus+1eN. (3.5b)Since N is O(1) to leading order, we seek a natural asymptotic expansion ofthe inner activator function and phase,Φ ∼ Φ0 + Φ1 + . . . ϕ ∼ ϕ0 + ϕ1 + . . .to finally get the leading order problemΦ0ϕ0T =Φ0ρρ −2m2r20Φ0 − Φ0 +2w(1 + bw2)2Φ0 −qU q−10 w2(1 + bw2)N(r0), (3.6a)τNϕ0T =Dr(rNr)r −Dm2r2N −N + 1ovo−1euseΦ0(r − r0)− 1svoeus+1eN,(3.6b)where we have used in (3.6a) that to leading order u˜e ∼ U0 and v˜e ∼ 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(2m2)713.1. Linear Stability Formulationbecause if m 1 then this can be an O(1) term. If we define,L0bΦ0 = Φ0ρρ − Φ0 +2w(1 + bw2)2Φ0, (3.7)then we can write (3.6a) asL0bΦ0 − qU q−10w2(1 + bw2)N(r0) =(ϕ0T +2m2r20)Φ0, lim|ρ|→∞Φ = 0, (3.8)and so we can think of ϕ0T = λ as the eigenvalues of (3.8). In (3.8) wedon’t 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 haveφ(ρ, T ) ∼ Φ0 exp(∫ T0λ(s) ds/2), (3.9)where we have assumed that ϕ0(0) = 0 without loss of generality. Noticethat if the eigenvalues were not time-dependent we would get the standardexp(λt) 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,1ovo−1euseΦ(r − r0)=ov˜e( r−r0)o−1u˜e( r−r0)s Φ =→0Aδ(r − 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 = oUβ−q0∫ ∞−∞wo−1Φ0 dρ.In a similar fashion we have that1svoeus+1eN(r) =→0sUβ−10 N(r0)Aδ(r − r0), A =∫ ∞−∞wo dρ,and so we can write (3.6b) as1r(rNr)r −m2r2N − θ2λN=(sUβ−10DN(r0)A−oUβ−q0D∫ ∞−∞wo−1Φ0 dρ)δ(r − r0), (3.10)whereθλ =√1 + τλD. (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(oUβ−q0D∫ ∞−∞wo−1Φ0 dρ−sUβ−10DN(r0)A)G¯0,m(r; r0), (3.12)733.1. Linear Stability FormulationwhereG¯0,m(r; r0) =J¯m,1(θλr)J¯m,2(θλr0), 0 ≤ r ≤ r0J¯m,1(θλr0)J¯m,2(θλr), r0 ≤ r ≤ R,withJ¯m,1(θλr) = Im (θλr) , (3.13a)J¯m,2(θλr) = α¯mIm (θλr) +Km (θλr) , (3.13b)and α¯m given byα¯m = −Km (θλR)′Im (θλR)′ =(Km+1 (θλR)− mθλRKm (θλR)mθλRIm (θλR) + Im+1 (θλR)). (3.13c)If we notice thatAUβ−10 r0D=1G0(r0; r0),with G0 defined by (2.38) then we can write (3.12) asN(r) =(oU1−q0G0(r0; r0)A∫ ∞−∞wo−1Φ0 dρ−sG0(r0; r0)N(r0))G¯0,m(r; r0).(3.14)By evaluating this expression at r = r0, we solve for N(r0) to getN(r0) = oU1−q0(s+J0,1(r0)J0,2(r0)J¯m,1(θλr0)J¯m,2(θλr0))−1 ∫∞−∞wo−1Φ0 dρA (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 Φ0. The structure of (3.8)dictates that Φ0 will admit both an even and an odd solution (cf. [30]).Since w is even then if Φ0 is odd, N(r0) vanishes and the assumption thatit is O(1) to leading order fails. If Φ0 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 − w + f(w) = 0, wy(0) = 0, w → 0 as |y| → ∞, w(0) > 0and assume this has a homoclinic orbit solution via Lemma 2.2.0.1. Theassociated linearized operatorLΦ = Φyy − Φ + f ′(w)Φ = λΦhas a discrete simple positive eigenvalue λ0 > 0 associated to a positive eigen-function Φ0. There is also a discrete eigenvalue λ1 = 0 with the eigenfunctionΦ1 = w′. Furthermore when f ′(0) is finite, a continuous spectrum exists onRe(λ) ≤ −1 + f ′(0) < 0 with Im(λ) = 0.A corollary to Lemma 3.1.0.1 is that if other discrete eigenvalues λj, j > 1exist then −1 + f ′(0) < λj < 0. See [14] for other discretely computedeigenvalues with f(w) = wp. Since the principal eigenvalue is the only onefor which Re(λ) > 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 Φ0 Even3.2 Eigenvalues Associated with Φ0 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 defineχm = qo(s+J0,1(r0)J0,2(r0)J¯m,1(θλr0)J¯m,2(θλr0))−1,so that we can write (3.8) asL0bΦ0 −χmAw2(1 + bw2)∫ ∞−∞wo−1Φ0 dρ =(λ+2m2r20)Φ0, (3.16)subject to far-field decay conditions. We call (3.16) the non-local eigenvalueproblem (NLEP) for λ 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(λ) < 0. If wedefine the quantitiesµ = λ+2m2r20, A(Φ0) =∫ ∞−∞wo−1Φ0 dρ (3.17)763.2. Eigenvalues Associated with Φ0 Eventhen we can write (3.16) asL0bΦ0 −χmAw2(1 + bw2)A(Φ0) = µΦ0.We notice that L0b is a self-adjoint operator and with this in mind, multiplythe eigenvalue problem with a function ψ, satisfying the same boundaryconditions as Φ0, and integrate∫ ∞−∞(L0bΦ− µΦ0)ψ dρ =χmA∫ ∞−∞w2(1 + bw2)A(Φ0)ψ dρ.Since L0b is self-adjoint,∫ ∞−∞(L0bΦ0 − µΦ0)ψ dρ =∫ ∞−∞(L0bψ − µψ)Φ0 dρ,and so∫ ∞−∞(L0bψ − µψ)Φ0 −χmAw2(1 + bw2)A(Φ0)ψ dρ = 0.Define ψ such that(L0b − µ)ψ =w2(1 + bw2), (3.18)so that∫ ∞−∞w2(1 + bw2)(Φ0 −χmA A(Φ0)ψ)dρ = 0.Since everything outside the brackets in the integrand is positive, the integralcan only vanish ifΦ0 =χmA A(Φ0)ψ.773.2. Eigenvalues Associated with Φ0 EvenWe can use this to write (3.16) asχmA A(Φ0)L0bψ −χmA A(Φ0)χmAw2(1 + bw2)∫ ∞−∞wo−1ψ dρ =χmA A(Φ0)µψ.Dividing through by χmA A(Φ) and using (3.18) we get,1χm−∫∞−∞wo−1ψ dρA = 0.We define Cm(λ) and f(µ) by,Cm(λ) ≡1χm=1qo(s+J0,1(r0)J0,2(r0)J¯m,1(θλr0)J¯m,2(θλr0)), (3.19a)f(µ) ≡∫∞−∞wo−1ψ dρA , (3.19b)and so the eigenvalue problem (3.16) becomes a root finding problemgm(λ) ≡ Cm(λ)− f(µ) = 0, (3.20)subject to (3.18) where µ is given by (3.17). Note that we are only interestedin roots satisfying Re(λ) > 0 which correspond to unstable eigenvalues.3.2.1 Removing Saturation: The case b = 0We begin by considering σ = 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 Φ0 Eventerm in (3.16) does not vanish. First when b = 0, we write (3.7) asL0Φ0 = L00Φ0 = Φ0ρρ − Φ0 + 2wΦ0, (3.21)so that the NLEP (3.16) simplifies toL0Φ0 −χmA w2∫ ∞−∞wo−1Φ0 dρ = µΦ0. (3.22)For this limiting case, the root finding problem (3.20) remains unchangedexcept that now the function ψ from (3.19b) satisfies(L0 − µ)ψ = 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 λ is purely real then the root finding problem remains completely un-changed. When b = 0 then we can determine the discrete eigenvaluesL0Ψ = νΨ using Lemma 3.1.0.1 explicitly (cf. [14], [47]). The continuousspectrum exists on Re(ν) < −1 and the discrete eigenvalues and eigenfunc-tions satisfy,ν0 =54, Ψ0 = w3/2; ν1 = 0, Ψ1 = w′; ν2 = −34, Ψ2 =(1− 56w)w1/2.(3.24)793.2. Eigenvalues Associated with Φ0 EvenThe principal eigenfunction Ψ0, is a solution to the homogeneous problem(3.23) when µ = ν0 and since the operator is self-adjoint, there will only bea solution ψ if the compatibility condition,∫ ∞−∞w2Ψ dρ = 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 µ = ν0. Properties off(µ) in (3.19b) have been analyzed in [79] where the asymptotic structure off(µ) for µ 1 is given byf(µ) ∼ 1 +(1− 12o)µ+ κcµ2 + . . . (3.25)withκc =∫∞−∞wo−1ψc dρA , ψc = L−30 (w2). (3.26)In general, ψc 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(µ) are also provided in Proposition 3.5:f ′(µ) > 0, µ ∈ [0, ν0) (3.27a)f ′′(µ) > 0, µ ∈ [0, ν0) (3.27b)f(µ) < 0, µ ∈ (ν0,∞). (3.27c)803.2. Eigenvalues Associated with Φ0 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(µ) for o = 1, 4, and 5.0 0.5 1 1.5 2 2.5 3 3.5 4−10−8−6−4−20246810µf(a) o = 10 0.5 1 1.5 2 2.5 3 3.5 4−10−8−6−4−20246810µf(b) o = 40 0.5 1 1.5 2 2.5 3 3.5 4−10−8−6−4−20246810µf(c) o = 5Figure 3.1: Computation of f(µ) from (3.19b) for various o with m = 0,r0 = 0.5, and  = 0.025. We set m = 0 solely to satisfy µ = λ 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 µ =813.2. Eigenvalues Associated with Φ0 Evenν0 we consider a small parameter δ such thatµ = ν0 + δµ1, ψ = ω(δ)(ψ0 + δψ1),with ω(δ) 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 ν0, we can only achievesolution compatibility if the right hand-side of (3.23) is subdominant to ψ0.If we expand our substitution,L0ψ0 − ν0ψ0 + δ(L0ψ1 − ν0ψ1 − ν1ψ0) +O(δ2) =w2ω,then a reasonable balance is to take ω = 1/δ. The leading order problemthen results in ψ0 = Ψ0, the eigenfunction for ν0. At next order the problemis,L0ψ1 − ν0ψ1 = w2 + µ1ψ0.We notice that ψ1 = Ψ0 is a solution to the homogeneous problem and so, inorder for a solution to exist, we must have that the compatibility condition,∫ ∞−∞w2Ψ0 dρ+ µ1∫ ∞−∞Ψ20 dρ = 0,is satisfied. This yields thatµ1 = −∫∞−∞w2Ψ0 dρ∫∞−∞Ψ20 dρ.823.2. Eigenvalues Associated with Φ0 EvenIf we rewriteδ =µ− ν0µ1,then we have that to leading order,ψ ∼ µ1µ− ν0Ψ0,for µ near ν0. With this we can compute,f(µ) ∼ µ1µ− ν0∫∞−∞wo−1Ψ0 dρ∫∞−∞wo dρ, µ ≈ ν0. (3.28)Therefore we have that there is a simple pole to f(µ) at µ = ν0 satisfyingf → ∞ as µ → ν−0 and f → −∞ as µ → ν+0 . In Figure 3.2 we show f(µ)computed numerically for m = 0, r0 = 0.5,  = 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 Φ0 Even0 0.5 1 1.5 2 2.5 3 3.5 4−10−8−6−4−20246810µf  numericasymptoticFigure 3.2: Numerical computation of f(µ) when m = 0, r0 = 0.5,  = 0.025,and o = 2 along with the asymptotic expression (3.28) demonstrating thesimple pole at µ = ν0.Returning to the root problem g given by (3.20), the structure can dependsignificantly on the magnitude of m and specifically centers around the term2m2/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  1,µ = λ+2m2r20≈ λ.843.2. Eigenvalues Associated with Φ0 EvenFirst note that if λ > ν0 then since Cm(λ) > 0 and f(λ) < 0 from (3.27) thenwe conclude that gm(λ) > 0 when λ > ν0. Therefore, to leading order, alleigenvalues to (3.22) satisfy λ < ν0. We begin by finding the neutral stabilitypoint λ = 0. From the asymptotic expansion of f(µ) given by (3.25), we havethat f(0) = 1 and sogm(0) = Cm(0)− 1.Furthermore, when λ = 0 then θλ = 1/√D and when m = 0 we have thatJ¯m,i(θλr) = 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,∂Cm∂m= − C¯0(J¯m,1(θλr0)J¯m,2(θλr0))2∂∂m(J¯m,1(θλr0)J¯m,2(θλr0)),whereC¯0 =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,∂∂mIm (z)∣∣∣∣m=0= −K0 (z) ,∂∂mKm (z)∣∣∣∣m=0= 0,and so in this case∂Jm∂m∣∣∣∣m=0= −2α¯0I0 (θλr0)K0 (θλr0)−K0 (θλr0)2 + I0 (θλr0)2∂α¯m∂m∣∣∣∣m=0,(3.29)853.2. Eigenvalues Associated with Φ0 EvenwhereJm(r0) = J¯m,1(θλr0)J¯m,2(θλr0). (3.30)To understand this better we require the order derivative of α¯0,dα¯mdm∣∣∣∣m=0=∂Km(θλR)∂m∣∣∣m=1−K0 (θλR)I1 (θλR)−K1 (θλR)(∂Im(θλR)∂m∣∣∣m=1+ I0 (θλR))I1 (θλR)2 .Once again from [1] we can compute that∂Im (z)∂m∣∣∣∣m=1= K1 (z)−I0 (z)z,∂Km (z)∂m∣∣∣∣m=1=K1 (z)z,and using the Wronskian relationship (2.37) we can finally write thatdα¯mdm∣∣∣∣m=0=1(θλR)2I1 (θλR)2 (1− θλR)− α¯20.For θλR > 1, this expression is negative and for θλR < 1, if we use the smallargument asymptotics for the modified Bessel functions [1], we havedα¯mdm∣∣∣∣m=0∼ − 14(θλR)3,and therefore the order derivative of α¯0 is always negative so (3.29) is alwaysnegative and therefore∂Cm∂m∣∣∣∣m=0> 0.863.2. Eigenvalues Associated with Φ0 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 ∈ [0, R] via Figure 3.3 whichplots the maximum of the order derivative of (3.30) for r0 ∈ [0, R] versus θλfor various values of R. This derivative is always negative and tending tozero so therefore we have that ∂Cm∂m > 0 for all m.5 10 15 20 25−0.03−0.02−0.0100.010.020.03θλmax r 0∈[0,R]∂J ∂m  m=5m=10m=20m=50(a) R = 0.15 10 15 20 25−0.03−0.02−0.0100.010.020.03θλmax r 0∈[0,R]∂J ∂m  m=5m=10m=20m=50(b) R = 15 10 15 20 25−6−4−20246 x 10−3θλmax r 0∈[0,R]∂J ∂m  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 θλ, we compute the order derivative of(3.30) over r0 ∈ [0, R] and then take the maximum value over that interval.The figure shows each maximal value of the derivative as a function of θλ forvarious values of R.Since the derivative is always positive, we can have at most one root to Cm.873.2. Eigenvalues Associated with Φ0 EvenIf we consider m large then we can use the small argument asymptotics forIm (x) and Km (x) which are valid for x √m+ 1 [1]. These expansionsare,Im (x) ∼1Γ(m+ 1)(x2)m, Km (x) ∼Γ(m)2(2x)m(3.31)and so thereforeCm(λ) ≈m1C¯0Jm(r0)∼ C¯02m1 +( r0R)2m  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− such thatgmb− (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 λ = 0 is the largest eigenvalue when m = mb− .To investigate this, we need to consider the λ derivative and second deriva-tive for Cm. In a similar fashion to what we did with the order derivative,we can conclude numerically that∂Cm∂λ> 0,∂2Cm∂λ2< 0, (3.34)when τ 6= 0. When τ = 0 then Cm(λ) = Cm(0) since the eigenvalue onlyoccurs in a product with τ and therefore dCmdλ = 0 when τ = 0. Using these883.2. Eigenvalues Associated with Φ0 Evenexpressions along with (3.27) we have that∂2gm∂λ2=∂2Cm∂λ2− ∂2f∂λ2< 0. (3.35)Since this expression holds for all m, it must hold at m = mb− and so gmb−can have at most 2 roots of which we already know λ = 0 is one. As λ tendsto ν0 from the left then gmb− tends to negative infinity. Ifdgmb−dλ < 0 whenλ = 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−which violates the maximal root condition. Conversely, if the derivative ispositive when λ = 0 then there must be a second crossing to achieve thefar-field behaviour and therefore a second, larger, root must exist. Whenτ = 0, we know that dCmdλ = 0 and so by (3.27)dgmb−dλ∣∣∣∣τ=0< 0, (3.36)always. If τ  1 then θλ  1 and we can use the large argument asymptoticexpansions for the Bessel functions [1],Im (x) ∼exp(x)√2pix, Km (x) ∼√pi2xexp(−x)to writeCm(λ) ∼ 2C¯0r0θλ1(1 + exp(−2θλ(R− r0)))≈ 2C¯0r0θλ  1 τ  1.(3.37)893.2. Eigenvalues Associated with Φ0 EvenConsider λ small such that f ∼ 1 via (3.25) but large enough (i.e. λ >O(1/τ)) so that for τ  1gm ∼ Cm  1.Specifically then, when m = mb− , and if τ is large, we havedgmb−dλ∣∣∣∣τ1> 0for some range of λ and therefore this must be true when λ = 0 otherwise thetwo-root condition will be violated. Therefore, there must exist some criticalvalue τ = τ ∗mb− which satisfiesdgmb−dλ(0)∣∣∣∣τ=τ∗mb−= 0 (3.38)such that a new root λτ∗mb−> 0 is created and persists for τ > τ ∗mb− . We canfind this point by using a numerical root-finding algorithm to solve (3.33) and(3.38). For example using Newton’s method with r0 = 0.5, R = 1, D = 1,and exponent set (2, 1, 2, 0), we get,mb− ≈ 0.4003 τ ∗mb− ≈ 1.8376. (3.39)While we solve this problem numerically, it is still an asymptotic approxima-tion as we are ignoring the terms 2m2/r20.Now that we have investigated the neutral stability point, we will turn ourattention to the intervals that m = mb− creates. Firstly, we will look at903.2. Eigenvalues Associated with Φ0 Evenm ∈ [0,mb−). When m = mb− we have that Cmb− = 1 and sincedCmdm > 0 forall m then for 0 ≤ m < mb− we must have Cm(0) < 1. Since f(0) = 1 thengm(0) < 0. When τ = 0 then recall Cm(λ) = Cm(0) and therefore via (3.36),there are no positive λ for m ∈ [0,mb−) when τ = 0. When τ is really largethen (3.37) holds for Cm and by a similar logic as when m = mb− there is arange of λ for which gm(λ) > 0. Furthermore, we still have that gm → −∞as λ → ν−0 , and so combining this information there are at least two rootsto gm for τ  1 for m ∈ [0,mb−). Using (3.35) we can restrict this furtherand conclude that there are exactly two roots when τ  1. Of course, thismeans there must exist some τ = τ ∗m such that there is exactly one positiveroot λτ∗m which will occur whengm(λτ∗m) =∂gm∂λ∣∣∣∣λτ∗m= 0.Note that since the only neutral stability point occurs at m = mb− , thesepositive roots cannot transition through λ = 0 and must become real via thecomplex plane.Finally, we consider m ∈ (mb− ,∞), bearing in mind that we are only asymp-totically considering m = O(1). On this region, Cm(0) > Cmb− (0) = 1 andso gm(0) > 1. When τ = 0, then once again (3.36) holds and by the far-fieldbehaviour of gm, there must be exactly one root to gm. When τ 6= 0 then(3.35) states that there can be at most one critical point to gm. Regardlessof the sign of dgmdλ (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 Φ0 Even3.2.4 Real Eigenvalues: m O(1)We now turn our attention to the case when m is large enough that µ ≈ λ isno longer valid as an approximation to (3.17). This approximation ceases tobe valid when m = O(−1) and so we definem˜ = m.Since m  1, we can use (3.32) as an approximation to Cm which meansthat χm = O(−1) 1, and so to leading order (3.22) becomesL0Φ0 =(λ+m˜2r20)Φ0. (3.40)We once again start by looking for a neutral stability point and so settingλ = 0, we know from Lemma 3.1.0.1 that ν0 in (3.24) is the only eigenvalue,eigenfunction pair with Re(λ) > 0 to L0, and so we have a neutral stabilitypoint m˜ = m˜b+ which satisfiesm˜b+ = r0√ν0. (3.41)We can find a correction to this by expandingm˜2b+ ∼ r20ν0 + m˜1, Φ0 ∼ Ψ0 + Φ01,and after substituting into (3.22), Φ01 and m˜1 satisfyL0Φ01 − ν0Φ01 =w22C¯0r0√ν0A∫ ∞−∞wo−1Ψ0 dρ+m˜1r20Ψ0. (3.42)923.2. Eigenvalues Associated with Φ0 EvenHere we have used (3.32) for large m asymptotics to Cm (and hence χm) andhave taken that since r0 < R,( r0R)2mtends to zero as m→∞. The solutionΦ01 = Ψ0 is a homogeneous solution to (3.42) and so we require a solvabilitycondition for the right-hand side. From this we get m˜1 satisfiesm˜1 = −r02C¯0√ν0B,whereB =∫∞−∞w2Ψ0 dρ∫∞−∞wo−1Ψ0 dρA∫∞−∞Ψ20 dρ.Therefore, overall we have,mb+ ∼1√r20ν0 + m˜1 =r0√ν0− B4C¯0ν0. (3.43)Now, unlike the case when m = mb− , λ = 0 is the only eigenvalue whenm = mb+ . This is because if λ > 0 then µ > ν0 and there is no solution to(3.40). In fact, this conclusion holds on the interval m > mb+ because for allλ ≥ 0, µ > ν0 always. Now we consider m˜ < m˜b+ and first note that by thesame logic just discussed, if λ > ν0− m˜ then there are no solutions to (3.40).We therefore restrict our attention to λ < ν0 − m˜. From (3.27) we havef(0) = 1, and from (3.32) we have Cm(0)  1 so gm(0)  1. Once againvia (3.35), we have that gm can have at most one critical point and sincegm → −∞ as λ→ ν−0 then there is exactly one root to gm on m˜ ∈ [0, m˜b+).Asymptotically, this root is given by,λ = ν0 −m˜2r20. (3.44)933.2. Eigenvalues Associated with Φ0 Even3.2.5 Real Eigenvalue SummaryWe have now analyzed all the possible cases for real eigenvalues and wesummarize them here. For 0 ≤ m < mb− with mb− given asymptoticallyby the numerical solution of (3.33) and (3.38), we have that there are noreal, positive eigenvalues when τ is sufficiently small and two positive realeigenvalues when τ is sufficiently large. Furthermore, there exists some τ =τ ∗m where there is exactly one real eigenvalue on this interval. When m = mb−then λ = 0 is the largest eigenvalue for 0 ≤ τ < τ ∗mb− with τ∗mb− satisfying(3.38). When τ > τ ∗mb− , there exists a non-zero positive real eigenvalue as thelargest eigenvalue. When mb− < m < mb+ with mb+ given asymptoticallyby (3.43) then there is exactly one real eigenvalue for all values of τ . Whenm = mb+ , λ = 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 λ is complex. Note that the possibilityof complex eigenvalues is due to the full operator in (3.22) being non self-adjoint. We start by lettingλ = λR + iλI , ψ = ψR + iψI (3.45)943.2. Eigenvalues Associated with Φ0 Evenwhich we substitute into (3.23) to getL0ψR =(λR +2m2r20)ψR − λIψI + w2, (3.46a)L0ψI =(λR +2m2r20)ψI + λIψR. (3.46b)We then definefR =∫∞−∞wo−1ψR dρA , fI =∫∞−∞wo−1ψI dρA , (3.47)and gm = gmR + igmI withgmR = Re(Cm(λ))− fR, gmI = Im(Cm(λ))− fI , (3.48)and Cm(λ) 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, Γ, via[arg f(x)]Γ = 2pi(N0(f)−Np(f)). (3.49)We take as a contour Γ = ΓI ∪ ΓK withΓK :{λ = K exp(it)|t ∈[−pi2,pi2]}, ΓI : −Ki ≤ λ ≤ Ki,traversed counter-clockwise and we consider what happens as K tends toinfinity while assuming that τ is chosen so that gm has no roots on the953.2. Eigenvalues Associated with Φ0 Evenimaginary axis. As K tends to infinity, so does µ, and from (3.23) we havethat ψ tends to zero. Therefore fR and fI tend to zero as well as µ → ∞.Also since |λ|  1 then (3.37) holds andCm(λ) ∼ 2C¯0r0√KτDexp(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 ΓK in the counter clockwise direction. Whent = −pi/2 then from (3.50),arg gm = argCm = −pi4,and conversely when t = pi/2,arg gm = argCm =pi4,and so[arg gm]ΓR =pi2.For the change in argument along ΓI , since Cm(λ) is holomorphic and realvalued when λ is real then Cm(λ¯) = Cm(λ). This similarly holds for f andso we can write[arg gm]ΓI = 2[arg gm]Γ+I963.2. Eigenvalues Associated with Φ0 Evenwhere Γ+I is the positive imaginary axis traversed from infinity to zero. Wetherefore have from (3.49) thatN0(gm) =14+Np(gm) +1pi[arg gm]Γ+I .We are already aware from (3.28) that f(µ) (and hence gm) has a simple polewhen µ = ν0. However, we also determined that if m > mb+ then µ > ν0,and so the pole only exists inside the contour on m ∈ [0,mb+). Therefore,we have the number of roots to gm is given byN0(gm) =54 +1pi [arg gm]Γ+I , m < mb+14 +1pi [arg gm]Γ+I , m > mb+. (3.51)At the start of Γ+I , coming in from infinity, where K  1, we can once againuse (3.37) and writeCm(λ) ∼ 2C¯0r0√iKτD,whereRe(Cm) = Im(Cm) ∼√2C¯0r0√KτD. (3.52)As we already discussed, ψ → 0 as λ→∞ and so for K  1,arg gm = argCm = arctan(Im(Cm)Re(Cm))=pi4.Near the end of Γ+I , traversing towards the origin, we have K  1 and sowe can use (3.32) which, even though derived for m  1, was based off of973.2. Eigenvalues Associated with Φ0 Evensmall argument asymptotics and so will be valid for all m with K  1. Inthis case we haveRe(Cm) =2C¯0m1 +( r0R)m , Im(Cm) = 0.From Proposition 3.1 and 3.2 of [79], we have that for λI  1 and m = O(1),fR(λI) ∼ 1− κcλ2I +O(λ4I), (3.53a)fI(λI) ∼(1− 12o)λI +O(λ3I) (3.53b)where κc is defined by (3.26). If m O(1) then for λI  1 (3.46b) simplifiestoL0ψI =2m2r20ψIwhich 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 λI  1 and gmI = 0. For gmR, recall that we alreadydetermined gm(0) < 0 on 0 ≤ m < mb− and gm(0) > 0 on m > mb− sotherefore,arg gm = arctan(gmIgmR)=pi, m ∈ [0,mb−)0, m ∈ (mb− ,∞) \mb+,near λI = 0 of Γ+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 Φ0 Evenset of properties for fR and fI . Particularly we require that for λI > 0 andm = O(1),∂fR∂λI< 0, fI(λI) > 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 1000.20.40.60.811.21.4λIf  fRfI(a) o = 10 1 2 3 4 5 6 7 8 9 1000.20.40.60.811.21.4λIf  fRfI(b) o = 40 1 2 3 4 5 6 7 8 9 1000.20.40.60.811.21.4λIf  fRfI(c) o = 5Figure 3.4: Computation of fR and fI from (3.47) for various o with m = 0,r0 = 0.5, and  = 0.025. We set m = 0 solely to satisfy dealing with a singlevariable λI . 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 Φ0 Even3.2.7 Complex Eigenvalues: 0 ≤ m < mb−On 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 havedgmRdλI=ddλIRe(Cm)−dfRdλI> 0 (3.55)for all λI 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 τ = 0 then Cm(λ) = Cm(0) and sofor τ = 0,Re(Cm) = Cm(0), Im(Cm) = 0,and so from (3.54),gmI = Im(Cm)− fI = −fI < 0,and so the negative imaginary axis is crossed. Therefore,[arg gm]Γ+I = −5pi4, τ  1.1003.2. Eigenvalues Associated with Φ0 EvenWhen τ  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 ≈ Im(Cm) 1and so for τ  1, the positive imaginary axis is crossed and[arg gm]Γ+I =3pi4, τ  1.Therefore on 0 ≤ m ≤ mb− we have from (3.51)N0(gm) =0, τ  12, τ  1.Therefore, for τ small, there are no complex eigenvalues with positive realpart which eventually transition to two complex eigenvalues with positivereal part. These eigenvalues cannot transition through λ = 0 since, if λ = 0for one value of τ , it is an eigenvalue for all values of τ and therefore, theeigenvalues must cross through the imaginary axis at some value τ = τHmwhere a Hopf bifurcation occurs. We know from analyzing the real eigenval-ues that, eventually for τ large enough, these two complex eigenvalues mustbecome purely real and therefore τHm < τ∗m.3.2.8 Complex Eigenvalues: m > mb−On 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 Φ0 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− with m = O(1), we must have zero crossings. This tells us that[arg gm]Γ+I = −pi4. (3.56)For m = O(−1), we have that (3.32) holds for Cm  1 and since, unlike thecase for purely real eigenvalues, fR does not have a vertical asymptote form 6= mb+ thengmR ≈ Re(Cm) ∼ O(m) 1.Therefore gmR > 0 always and cannot cross the imaginary axis so (3.56)holds as well. Therefore, via (3.51), we haveN0(gm) =1, mb− < m < mb+0, m > mb+.Since we have already determined that a real eigenvalue exists on m ∈(mb− ,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 Φ0 even and σ = 0 via the following principal result:Principal Result 3.2.9.1 Eigenvalue Classification:On 0 < m < mb−:There are no eigenvalues with positive real part when τ is sufficiently small1023.2. Eigenvalues Associated with Φ0 Evenand two complex eigenvalues with positive real part when τ is sufficientlylarge. These eigenvalues undergo a Hopf bifurcation when τ = τHm and as τincreases further, these eigenvalues coincide as a single real eigenvalue whenτ = τ ∗m before splitting on the real axis for τ > τ∗m.If m = mb−:The eigenvalue λ = 0 persists for all τ and when τ < τ ∗mb− then λ = 0 is thelargest eigenvalue. If τ > τ ∗mb− then there exists some λτ > 0 that is purelyreal and positive.On mb− < m < mb+ :There is exactly one real positive eigenvalue on this interval for all τ .On m > mb+:There are no eigenvalues with positive real part on this interval for all τ .The condition that there is always at least one unstable eigenvalue for all r0and all values of τ allows us to classify these eigenvalues as being an instabilityon an O(1) time scale. This is because if mb− < m < mb+ then λ(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 Φ0 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 τ 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 ofΦ0, we first truncate to a finite domain∫ ∞−∞wo−1Φ dρ ≈∫ L−Lwo−1Φ dρ.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 = −L+ kh, k = 0 . . . N, h =2LN.With this in mind we write the integral as∫ L−Lwo−1Φ dρ =N−1∑k=0∫ −L+(k+1)h−L+khwo−1Φ dρand discretize each integral with the trapezoid method [6]. Finally, we sup-plement far-field conditions that Φ0 = 0 at x = ±L.We will begin computations by setting σ = b = 0 since this is what wehave previously analyzed analytically. Furthermore, when τ = 0, χm is in-1043.2. Eigenvalues Associated with Φ0 Evendependent of λ 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(λ)} as a function of mwhen τ = 0 for various parameters computed in Matlab.0 5 10 15 20 25−1−0.500.511.5mmaxRe(λ)(a) (2, q, o, s) = (2, 1, 2, 0),  = 0.025,R = 1, r0 = 0.50 5 10 15 20 25 30−1−0.500.511.5mmaxRe(λ)(b) (2, q, o, s) = (2, 2, 3, 0),  = 0.025,R = 1, r0 = 0.50 5 10 15 20 25 30−1−0.500.511.5mmaxRe(λ)(c) (2, q, o, s) = (2, 4, 3, 3),  = 0.025,R = 1, r0 = 0.50 50 100 150−1−0.500.511.5mmaxRe(λ)(d) (2, q, o, s) = (2, 1, 2, 0),  = 0.025,R =√10, r0 = 9/√10Figure 3.5: Numerical computation of the largest real part of the eigenvalueλ in (3.16) for the case τ = 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 Φ0 Evenobtain mb− and mb+ and compare them to the asymptotic approximations(3.33) for mb− 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)  R = ` r0 mb−(n) mb−(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/√10 1/2√10 0.0485 0.0485 7.0482 7.0711 7.0469(2,1,2,0) 0.025√10√10/2 2.3362 2.3373 69.3076 70.7107 69.3350(2,1,2,0) 0.025√10 2/√10 0.9756 0.9745 27.6761 28.2843 27.6900(2,1,2,0) 0.025√10 9/√10 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 τ = 0 and b = 0 of numerical and asymptoticcomputations of mb− and mb+ for a variety of exponent sets, , R, and r0with D = 1 for all. The (n) refers to numeric computations of (3.16) usingeigs in Matlab. mb−(a) is computed via Newton’s method on (3.33), mb+(a1)is computed via (3.41) while mb+(a2) is computed via (3.43).3.2.11 Computing Eigenvalues, τ 6= 0When τ 6= 0 then χm in (3.16) is a function of λ and the eigenvalue problemis more intricate as χm depends nonlinearly on the product τλ. For this case,we must solveT (λ)Φ = 0,where Φ is the discrete eigenvector and T (λ) are the discretized coefficientsfor (3.16). In order to enforce uniqueness of eigenvector solutions, we require1063.2. Eigenvalues Associated with Φ0 Evensome normalizing constraint,vTΦ = 1.We can write the problem as a block systemF(λ,Φ) =T (λ)ΦvTΦ− 1 = 0and hence our eigenvalues and eigenvectors are roots to F which we can solvewith Newton’s Method. We can write the Jacobian to the problem as,J(λ,Φ) =Tλ(λ)Φ T (λ)0 vTand so the Newton problem is,J(λk,Φk)λk+1 − λkΦk+1 −Φk = −F (λk,Φk), (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(λk+1 − λk)Tλ(λk)Φk + T (λk)Φk+1 − T (λk)Φk = −T (λk)Φk,1073.2. Eigenvalues Associated with Φ0 Evenwhich can simplify toΦk+1 = −uk(λk+1 − λk), (3.58)where we defineuk = T (λk)−1Tλ(λk)Φk.If we left-multiply by vTk thenλk+1 = λk −vTk Φk+1vTk uk,which still appears to depend on the solution Φk+1. However, if we expandout the second equation in (3.57) then we getvTk Φk+1 = 1and soλk+1 = λk −1vTk uk. (3.59)This determines λk+1 and we could use (3.58) to determine Φk+1 but sinceit is just uk up to a constant we instead renormalize via,Φk+1 =ukvTk uk,1083.2. Eigenvalues Associated with Φ0 Evenwhich completes the problem. For simplicity we take,vk =uk−1|uk−1|2to normalize. There are several practical issues to deal with when implement-ing Newton’s method for this problem because there are several eigenvaluesthat exist (including the continuous spectrum that exists below λ = 1). Weare only interested in capturing the eigenvalues at each m for a given τ thathave the largest positive real part. To accomplish this we take as initial datathe τ = 0 case where the problem is linear and the largest eigenvalues are eas-ily computed. We choose an m = m∗ value in the interval mb− < 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∗, we computethe Newton solve by slowly varying τ and finding eigenvalues at m = m∗until a desired final τ value is reached. At this point, we begin anotherNewton solve, fixing τ , for m < m∗ and m > m∗ 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∗, there is very little trouble because once m > mb+we know that there are no eigenvalues with positive real part for all τ and sowe can terminate the Newton iteration quickly after this point. However, onm < m∗, if τ satisfies τHm < τ < τ∗m, the Newton iteration has some stabilityissues. This is due to the existence of complex eigenvalues with positive realpart on this interval of τ . On this region, there are still real eigenvalues for1093.2. Eigenvalues Associated with Φ0 Evenm ≥ mb− 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− and m = mb+). The problem isthat λ = 0 is always an eigenvalue of (3.16) with eigenfunction Φ = w′ (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 ΦTm1Φm0 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’s method as outlined abovefor various values of τ with (p, q, o, s) = (2, 1, 2, 0),  = 0.025, D = 1, R = 1,and r0 = 0.5.1103.2. Eigenvalues Associated with Φ0 Even0 5 10 15 20 25−1−0.500.511.5mmaxRe(λ)(a) τ = 0.40 5 10 15 20 25−1−0.500.511.5mmaxRe(λ)(b) τ = 10 5 10 15 20 25−1−0.500.511.5mmaxRe(λ)(c) τ = 20 5 10 15 20 25−1−0.500.511.5mmaxRe(λ)(d) τ = 30Figure 3.6: Numerical computation for b = 0 of the largest real part of theeigenvalue λ in (3.16) using Newton’s 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), = 0.025, D = 1, R = 1, and r0 = 0.5.When τ  1 as in Figure 3.6a then the plot is almost indistinguishable fromthe linear case of τ = 0 (Figure 3.5a) as should be expected. This helps verifythat the Newton algorithm is working properly. In Figure 3.6c when τ = 2 wesee that at m = mb− the zero eigenvalue is no longer the largest eigenvalue.In (3.39) we estimated that τ ∗mb− = 1.8376 and in this case τ > τ∗mb− . Figure3.7 verifies our predicted τ ∗mb− value as in Figure 3.7a-3.7b τ = 1.82 < τ∗mb−1113.2. Eigenvalues Associated with Φ0 Evenand we indeed still see at m = mb− that λ = 0 as the largest eigenvalue. InFigure 3.7c-3.7d, τ = 1.84 > τ ∗mb− and at m = mb− there is an eigenvaluelarger than 0.0 5 10 15 20 25−1−0.500.511.5mmaxRe(λ)(a) τ = 1.820.3 0.35 0.4 0.45 0.5−0.01−0.008−0.006−0.004−0.00200.0020.0040.0060.0080.01mmaxRe(λ)(b) τ = 1.82 (zoom)0 5 10 15 20 25−1−0.500.511.5mmaxRe(λ)(c) τ = 1.840.3 0.35 0.4 0.45 0.5−0.01−0.008−0.006−0.004−0.00200.0020.0040.0060.0080.01mmaxRe(λ)(d) τ = 1.84 (zoom)Figure 3.7: Numerical computation of eigenvalues near m = mb− for τ > τ ∗mb−and τ < τ ∗mb− . In all experiments (2, q, o, s) = (2, 1, 2, 0),  = 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− as there is certainly another eigenvalue where Re(λ) = 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 Φ0 Evenof the eigenvalues for a range of τ in Figure 3.8. For τ = 1.83, we seethat indeed the second eigenvalue with Re(λ) = 0 has a non-zero imaginarypart and thus there is a Hopf bifurcation at this m value for this particularvalue of τ . We see that when τ = 1.84 > τ ∗mb− and λ = 0 is no longerthe biggest eigenvalue that the eigenvalues on m < mb− are complex andtherefore τ ∗m > 1.84. However, when τ = 30 we that all of the eigenvalue arepurely real so τ ∗m < 30.1133.2. Eigenvalues Associated with Φ0 Even0 0.2 0.4 0.6 0.8 1−1−0.500.511.5mIm(maxRe(λ))(a) τ = 0.40 0.2 0.4 0.6 0.8 1−1−0.500.511.5mIm(maxRe(λ))(b) τ = 10 0.2 0.4 0.6 0.8 1−1−0.500.511.5mIm(maxRe(λ))(c) τ = 1.820 0.2 0.4 0.6 0.8 1−1−0.500.511.5mIm(maxRe(λ))(d) τ = 1.840 0.2 0.4 0.6 0.8 1−1−0.500.511.5mIm(maxRe(λ))(e) τ = 30Figure 3.8: Imaginary part of the eigenvalues with largest real part computedusing Newton’s method on (3.59). The black x mark points where Re(λ) = 0.In all experiments (2, q, o, s) = (2, 1, 2, 0),  = 0.025, D = 1, R = 1, andr0 = 0.5.1143.2. Eigenvalues Associated with Φ0 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  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 ν0(b) be thislargest discrete eigenvalue to b 6= 0 then the upper bound neutral stabilitywave mode is given by,mb+ ∼r0√ν0(b).In section 2.3 we have that the homoclinic orbit solution w(ηˆ) 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′ is even andhas no nodal lines. Therefore when b = bc, we must have that the largesteigenvalue is the zero eigenvalue and thereforelimb→b−cmb+ = 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 Φ0 Even0 0.05 0.1 0.15 0.2 0.2500.20.40.60.811.21.4bmaxλFigure 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),  = 0.025, R = 1, r0 =0.5, and D = 1. In this figure, we have plotted the numerically computedeigenvalues to (3.16) for τ = 0 and various values of b. It can be seen thatby the time b = 0.2 all of the eigenvalues satisfy Re(λ) < 0.1163.2. Eigenvalues Associated with Φ0 Even0 5 10 15 20 25−1−0.500.511.5mmaxRe(λ)  b =0b =0.01b =0.05b =0.1b =0.15b =0.2Figure 3.10: Computation of eigenvalues for b 6= 0 and τ = 0. In all cases(2, q, o, s) = (2, 1, 2, 0),  = 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),  = 0.025, R = 1, r0 = 0.5, and D = 1 for various values of τ .1173.2. Eigenvalues Associated with Φ0 Even0 5 10 15 20 25−1−0.8−0.6−0.4−0.200.2mmaxRe(λ)  τ =0.4τ =2τ =5τ =10τ =30τ =300τ =3000Figure 3.11: Computation of eigenvalues for b = 0.2 and τ 6= 0. In all cases(2, q, o, s) = (2, 1, 2, 0),  = 0.025, R = 1, r0 = 0.5, and D = 1Unsurprisingly, since we saw gm  1 for τ  1, we have that there is athreshold of τ where eigenvalues begin to enter Re(λ) > 0. However, unliker0, and b, τ is a static parameter and so for a fixed τ , there will exist a rangeof b for which all of the eigenvalues satisfy Re(λ) < 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 Φ0 Oddfind a saturation value σ∗ such that for σ > σ∗, the spectrum of the NLEP(3.16) satisfies Re(λ) < 0 for most values of r0.3.3 Eigenvalues Associated with Φ0 OddWe will now consider the second class of eigenfunctions for which Φ0 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 Φ0 is oddthen from (3.15), we have N(r0) = 0 to leading order, and so for Φ0 theleading order problem from (3.8) isL0bΦ0 =(λ+2m2r20)Φ0 = µΦ0.For λ ≥ 0 then µ ≥ 0 and from [14] and section 3.2.12, we have that the onlypossibility for Φ0 to be odd is if µ = 0 and Φ0 = w′ where w is the homoclinicorbit solution (2.22) and prime indicates differentiation with respect to ρ. Wecould once again analyze for m = O(1) and m O(1) but for the latter casenote that µ = 0 impliesλ = −2m2r20< 0and so there are no unstable roots for m 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  and1193.3. Eigenvalues Associated with Φ0 Oddtherefore, to leading order,Φ0 ∼ w′, λ = ω()λˆ, N(r) = Nˆwith ω  1 to be determined. We let N˜(ρ) = Nˆ(r0 + ρ) and make thefollowing expansions,Φ ∼ w′ + Φ1 + 2Φ2 + . . . , N˜(ρ) ∼ N˜0 + N˜1 + . . . .We also consider the asymptotic expansion of the equilibrium solution fromsections 2.2 and 2.3,v˜e ∼ U q0w + v˜1 + 2v˜2 + . . . , u˜e ∼ U0 + u˜1 + 2u˜2 + . . . .We have already satisfied the leading order problem and so we consider theproblem at O() for Φ1 (by expanding (3.5a)) which becomes,L0bΦ1 = −Φ0ρr0− Φ0ρdr0dT− a1Φ0 +qU q−10(1 + bw2)w2N˜0 + λˆΦ0, (3.60)where we have defineda1 ≡2U0(1 + bw2)2((1− 3bw2)U q−10 (1 + bw2)v˜1 − qwu˜1), (3.61)1203.3. Eigenvalues Associated with Φ0 Oddand, at least temporarily, consider ω() = . Next, consider the problem forv˜1 and u˜1 given by (2.7) which, in radial coordinates, isL0bv˜1 = −U q0r0w′ + qU q−10w2(1 + bw2)u˜1 − U q0dr0dTw′, (3.62a)u˜′′1 = −1DUβ0 wo. (3.62b)Upon differentiating (3.62a), we obtainL0bv˜′1 + 2(1− 3bw2)(1 + bw2)3w′v˜1 =−U q0r0w′′ + qU q−102w(1 + bw2)2w′u˜1+ qU q−10w2(1 + bw2)u˜′1 − U q0dr0dTw′′, (3.63)where we note that(L0by)′ = L0by′ + 2(1− 3bw2)(1 + bw2)3w′y. (3.64)We can rearrange (3.63) to geta1w′ = − 1U q0L0bv˜′1 −w′′r0+qU0w2(1 + bw2)u˜′1 −dr0dTw′′, (3.65)so that (3.60) becomesL0bΦ1 =1U q0L0bv˜′1 +qU0w2(1 + bw2)(U q0 N˜0 − u˜′1) + λˆw′, (3.66)1213.3. Eigenvalues Associated with Φ0 Oddwhere we have used that Φ0 = w′. We know Φ1 = w′ is a homogeneous solu-tion to (3.66) and so we must satisfy the following compatibility condition:1U q0∫ ∞−∞w′L0bv˜′1 dρ+qU0∫ ∞−∞w2(1 + bw2)(U q0 N˜0 − u˜′1)w′ dρ+ λˆ∫ ∞−∞w′2 dρ = 0. (3.67)Now since L0b is self-adjoint and w′ a homogeneous solution then,∫ ∞−∞w′L0bv˜′1 dρ =∫ ∞−∞v˜1L0bw′ dρ = 0. (3.68)In this way, (3.67) reduces to− qU0∫ ∞−∞w2(1 + bw2)(U q0 N˜0 − u˜′1)w′ dρ = λˆ∫ ∞−∞w′2 dρ. (3.69)We now wish to understand the behaviour of the function F˜ defined by,F˜0 ≡ U q0 N˜0 − u˜′1,which is easier to obtain if we differentiate,F˜ ′0 = U q0 N˜ ′0 − u˜′′1.First from (3.62b) we have thatu˜′′1 = −Uβ0Dwo,1223.3. Eigenvalues Associated with Φ0 Oddand for N˜0 we turn to the expansion (3.5b) in inner coordinates:4τN˜T + 2τN˜ω()λˆ =D(r0 + ρ)((r0 + ρ)N˜′)′ − 2 Dm2(r0 + ρ)2N˜ − 2N˜+ov˜o−1eu˜esΦ−  sv˜oeu˜s+1eN˜ , (3.70)which to leading order satisfies,N˜ ′′0 = −oUβ−q0Dwo−1w′.If we multiply this by U q0 and integrate we haveU q0 N˜′0 = −Uβ0Dwo,and therefore F˜ ′0 = 0 so F˜0 is a constant. We already know that the derivativeof ue in the global region is discontinuous and therefore F˜0 being constantimplies that Nˆ must be discontinuous as well but in such a way that theappropriate sum is continuous and F˜0 is defined. Having F˜0 be constantmakes the integrand on the left-hand side of (3.69) an odd function and sothe integral vanishes. Therefore we conclude that ω() . Next we considerω() = 2 and by expanding (3.5a) to the proper order we get the problemfor Φ2:L0bΦ2 =qUq−10w2(1 + bw2)N˜1 + a¯2Uq0 N˜0 − a1Φ1 − a2w′ −Φ′1r0+ρw′′r20− dr01dTw′′ − dr00dTΦ′1 +(λˆ+m2r20)w′, (3.71)1233.3. Eigenvalues Associated with Φ0 Oddwhere we define a2 and a¯2 by,a2 ≡2U0(1 + bw2)2((1− 3bw2)U q−10 (1 + bw2)v˜2 − qwu˜2)+q(q + 1)U20w(1 + bw2)2u˜21− 2qU q+10(1− 3bw2)(1 + bw2)3u˜1v˜1 −12bwU2q0(1− bw2)(1 + bw2)4v˜21 (3.72a)a¯2 ≡2qU q+10w(1 + bw2)2v˜1 −q(q + 1)U20w2(1 + bw2)u˜1. (3.72b)In (3.71) we replace r0 = r00 + r01 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(2). 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 + ρ ≈ r00 + ρ. As we did with the O() problem, weconsider the problem for v˜2 and u˜2 given by (2.15a) and (2.15b) respectivelywhich, in radial coordinates, isL0bv˜2 =−v˜′1r00+ρU q0r200w′ − dr00dTv˜′1 −dr01dTU q0w′ − Uq−20 q(q + 1)2w2(1 + bw2)u˜21+2qU0w(1 + bw2)2u˜1v˜1 −1U q0(1− 3bw2)(1 + bw2)3v˜21 + Uq−10 qw2(1 + bw2)u˜2,(3.73a)u˜′′2 =U0D− u˜′1r00+1DUβ−10 swou˜1 −1DUβ−q0 owo−1v˜1. (3.73b)1243.3. Eigenvalues Associated with Φ0 OddAs before, we differentiate (3.73a) and use (3.64) to geta2w′ = − 1U q0L0bv˜′2 −v˜′′1U q0r00+w′r200+ρw′′r200− 1U q0dr00dTv˜′′1− dr01dTw′′ + a¯2u˜′1 −a1U q0v˜′1 +qU0w2(1 + bw2)u˜′2. (3.74)Upon substituting (3.74) into (3.71) and simplifying we getL0bΦ2 =qU0w2(1 + bw2)F˜1 + a¯2F˜0 − a1Φ1 +1U q0L0bv˜′2 +v˜′′1U q0r00− w′r200+1U q0dr00dTv˜′′1 +a1U q0v˜′1 −Φ′1r00− dr00dTΦ′1 +(λˆ+m2r200)w′, (3.75)where we define,F˜1 = U q0 N˜1 − u˜′2. (3.76)To simplify further we return to the problem for Φ1 given by (3.66), whichwe can simplify asL0b(Φ1 −1U q0v˜′1)=qF˜0U0w2(1 + bw2). (3.77)Now, considering the even eigenfunction problem discussed in section 3.2,when µ = 0 there is a function ψ in (3.18) which satisfiesL0bψ =w2(1 + bw2), (3.78)1253.3. Eigenvalues Associated with Φ0 Oddand so we can write (3.77) asL0b(Φ1 −1U q0v˜′1 −qF˜0U0ψ)= 0.Therefore, the function being operated on must be some multiple of the nulleigenfunction w′ which we can take to be zero without loss of generality, andwe have that,Φ1 =1U q0v˜′1 +qF˜0U0ψ. (3.79)We can actually determine ψ analytically by first noticing that when b = 0,we can directly verify that ψ = w. With this in mind, we consider how theL0b operator acts on w even with saturation:L0bw = w′′ − w + 2 w2(1 + bw2)2=w2(1 + bw2)2− 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′′b − wb + 2wwb(1 + bw2)2− w4(1 + bw2)2= 0which impliesL0bwb =w4(1 + bw2)21263.3. Eigenvalues Associated with Φ0 Oddand this is precisely the last term in (3.80). With this in mind we attempt asolution for ψ of the form,ψ = 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− b) w4(1 + bw2)2=w2(1 + bw2)(1 + (c− b)w2)(1 + bw2)which will hold if c = 2b and thus we haveψ = w + 2bwb. (3.81)Substituting (3.79) into (3.75) we getL0bΦ2 =qU0w2(1 + bw2)F˜1 +(a¯2 −qU0(a1ψ +1r00ψ′ +dr00dTψ′))F˜0+1U q0L0bv˜′2 −w′r200+(λˆ+m2r200)w′.1273.3. Eigenvalues Associated with Φ0 OddOnce again, Φ2 = w′ is a homogeneous solution to this and so we need tosatisfy the solvability condition,qU0∫ ∞−∞w2(1 + bw2)w′F˜1 dρ︸ ︷︷ ︸I1+ F˜0∫ ∞−∞a¯2w′ − qU0(a1ψw′ +1r00ψ′w′ +dr00dTψ′w′)dρ︸ ︷︷ ︸I2+(λˆ+m2 − 1r200)∫ ∞−∞w′2 dρ = 0, (3.82)where we have once again used (3.68) to remove the L0bv˜′2 term. First considerthe integral I2 in (3.82). We can use (3.65) to write,∫ ∞−∞a1ψw′ dρ =∫ ∞−∞− 1U q0ψL0bv˜′1 −w′′ψr00+qU0ψu˜′1w2(1 + bw2)− dr00dTw′′ψ dρ=∫ ∞−∞− 1U q0ψL0bv˜′1 +w′ψ′r00+qU0ψu˜′1w2(1 + bw2)+dr00dTw′ψ′ dρ,(3.83)where we have integrated by parts on the second and last term. We also have∫ ∞−∞a¯2w′ dρ =2qU q+10∫ ∞−∞ww′(1 + bw2)2v˜1 dρ−q(q + 1)U20∫ ∞−∞w2w′(1 + bw2)u˜1 dρ=qU q+10∫ ∞−∞v˜1(L0bψ)′ dρ− q(q + 1)U20∫ ∞−∞W ′u˜1 dρ,1283.3. Eigenvalues Associated with Φ0 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∫ ∞−∞a¯2w′ dρ = − qU q+10∫ ∞−∞v˜′1L0bψ dρ+q(q + 1)2U20Hˆ〈duedr〉r00∫ ∞−∞w′2 dρwhere the last term was simplified using (2.12), (2.18), and that∫∞0 W dρ∫∞−∞w′2 dρ=Hˆ4. (3.84)Using the self-adjoint property of L0b we can finalize once more to obtain∫ ∞−∞a¯2w′ dρ = − qU q+10∫ ∞−∞ψL0bv˜′1 dρ+q(q + 1)2U20Hˆ〈duedr〉r00∫ ∞−∞w′2 dρ.(3.85)Combining (3.83) and (3.85) we can simplify I2 in (3.82) asI2 =q(q + 1)2U20Hˆ〈duedr〉r00∫ ∞−∞w′2 dρ− q2U20∫ ∞−∞ψu˜′1w2(1 + bw2)dρ− 2qU0(1r00+dr00dT)∫ ∞−∞w′ψ′ dρ.Using (2.34e) we can re-write the last term and we haveI2 =q(q + 1)2U20Hˆ〈duedr〉r00∫ ∞−∞w′2 dρ− q2U20∫ ∞−∞ψu˜′1w2(1 + bw2)dρ+q2U20Hˆ〈duedr〉r00∫ ∞−∞w′ψ′ dρ.1293.3. Eigenvalues Associated with Φ0 OddFinally we substitute ψ from (3.81) to getI2 =q2U20Hˆ〈duedr〉r00∫ ∞−∞w′2 dρ+3q22U20Hˆ〈duedr〉r00∫ ∞−∞w′2 dρ+q2U202bHˆ〈duedr〉∫ ∞−∞w′w′b dρ−∫ ∞−∞w3(1 + bw2)u˜′1 dρ︸ ︷︷ ︸I3−2b∫ ∞−∞w2wb(1 + bw2)u˜′1 dρ︸ ︷︷ ︸I4.Integrate I4 by parts to getI4 = 2∫ ∞0w2wb(1 + bw2)dρ〈duedr〉r00−∫ ∞−∞(∫ ρ0w2wb(1 + bw2)dx)u˜′′1 dρ,but the last integral vanishes since u˜′′1 is an even function. For the integralI3, we also integrate by parts to split the integral,I3 = −3∫ ∞−∞w2w′(1 + bw2)u˜1 dρ+ 2b∫ ∞−∞w4w′(1 + bw2)2u˜1 dρ,but then integrate by parts on each separate integral again, twice, transfer-ring the derivative back to the u˜1 term. After doing this, and once againremoving integrals involving u˜′′1, we getI3 =(6∫ ∞0W dρ− 4b∫ ∞0∫ w0v4(1 + bv2)2dv dρ)〈duedr〉r00=(32Hˆ∫ ∞−∞w′2 dρ− 4b∫ ∞0∫ w0v4(1 + bv2)2dv dρ)〈duedr〉r00,1303.3. Eigenvalues Associated with Φ0 Oddwhere we have used (3.84) to simplify. Transferring everything together backinto I2 we haveI2 =q2U20Hˆ〈duedr〉r00∫ ∞−∞w′2 dρ− bq2U20〈duedr〉r00(4∫ ∞0w2wb(1 + bw2)dρ+4∫ ∞0∫ w0v4(1 + bv2)2dv dρ− 2Hˆ∫ ∞−∞w′w′b dρ).We can actually simplify this if we differentiate Hˆ in (3.84) with respect to band notice that this is exactly the last term in the expression. Therefore wehave,I2 =q2U20Hˆ〈duedr〉r00∫ ∞−∞w′2 dρ(1− 2bqHˆdHˆdb). (3.86)Finally, we consider I1 in (3.82) which we integrate by parts to getI1 =−Hˆ4∫ ∞−∞w′2 dρ(F˜ ′1(∞) + F˜ ′1(−∞))+∫ ∞−∞∫ ρ0(W(x) dx) F˜ ′′1 dρ.(3.87)To simplify this we need to understand the behaviour of F˜1 given by (3.76)and specifically, its second derivative,F˜ ′′1 = U q0 N˜ ′′1 − u˜′′′2 .1313.3. Eigenvalues Associated with Φ0 OddTherefore, we need to consider the problem for N˜1 which we obtain by ex-panding (3.70) to O(2),N˜ ′′1 =−N˜ ′0r00+sUβ−10DwoN˜0 −oUβ−q0Dwo−1(v˜′1U q0+qF˜0U0ψ)+osUβ−q−10Dwo−1w′u˜1 −o(o− 1)Uβ−2q0Dwo−2w′v˜1. (3.88)If we multiply this by U q0 and subtract u˜′′′2 by differentiating (3.73b) we getF˜ ′′1 =Uβ−10Dwo−1F˜0 (sw − oψ) (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˜ ′1 to simplify I1 we will instead consider the global F problem and relatethem viaF(r00 + ρ) ∼ F(r00) + ρdFdr∣∣∣∣r00+ · · · = F˜0 + F˜1, (3.90)where we note that sided limits apply when functions are not continuous.Using (3.90), we can write (3.87) asI1 =−Hˆ2∫ ∞−∞w′2 dρ(U q0〈dNˆdr〉r00−〈d2uedr2〉r00). (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 ω() = 2 and the eigenvalues, to leading1323.3. Eigenvalues Associated with Φ0 Oddorder, are given byλˆ =1−m2r200− q2U20Hˆ(1− 2bqHˆdHˆdb)〈duedr〉r00F(r00)+qHˆ2U0(U q0〈dNˆdr〉r00−〈d2uedr2〉r00). (3.92)The fact that the eigenvalues exist to leading order at O(2) 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(−2). Since Nˆ is not continuous, but F(r00) = F˜0is, 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〈Nˆ〉r00−〈duedr〉r00.We now need to consider the global problem for Nˆ .3.3.1 Global Inhibitor Eigenvalue ProblemSubstitute N = Nˆ into (3.6b) and simplify to get the base problem for Nˆ is1r(rNˆr)r −m2r2Nˆ − 1DNˆ +12oDvo−1euseΦ(r − r0)− 1sDvoeus+1eNˆ︸ ︷︷ ︸S1= 0, (3.93)where we do not use θλ here as we did for even eigenfunctions because λ =O(2). Focusing solely on the singular terms S1 and expanding to all singular1333.3. Eigenvalues Associated with Φ0 Oddpowers of , we have,S1 =oUβ−q0D2wo−1(·)w′(·) + oUβ−q0Dwo−1(·)Φ1(·)−osUβ−q−10Dwo−1(·)u˜1(·)w′(·)+o(o− 1)Uβ−2q0Dwo−2(·)v˜1(·)w′(·)−sUβ−10Dwo(·)N˜0(·), (3.94)where the dot indicates evaluation at (r − r0)−1, 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−r0)=→0Aδ(r − r0),where δ is the Dirac measure and A is again given by (2.33). Differentiatingthis expression we geto2wo−1(r − r0)w′(r − r0)= Aδ′(r − r0),and the first term of S1 in (3.94) is precisely a multiple of this. If we associateinverse powers of  as leading to Dirac singularities then we expect thisbehaviour for the remaining terms in (3.94) and soS1 =Uβ−q0DAδ′(r − r0) + Aδ(r − r0),1343.3. Eigenvalues Associated with Φ0 Oddwhere we can determine A by integrating all but the first term in (3.94) overthe entire inner domain. This yields thatA =∫ ∞−∞oUβ−2q0Dwo−1v˜1 +oUβ−q−10DqF(r0)wo−1ψ −osUβ−q−10Dwo−1u˜1w′+o(o− 1)Uβ−2q0Dwo−2v˜1w′ − sUβ−10DwoN˜0 dρ,where we have substituted (3.79) for Φ1. We can integrate the third andfourth term by parts and upon simplifying we haveA =Uβ−q−10DF(r0)∫ ∞−∞(qowo−1ψ − swo)dρ.Substituting ψ using (3.81) and recognizing thatdAdb=∫ ∞−∞owo−1wb dρ,we finally have thatA =Uβ−q−10DF(r0)β(A+ 2qbdAdb),where β is given by (2.28b). Putting everything together we have,S1 =Uβ−q0DAδ′(r − r0) +Uβ−q−10DF(r0)A(β +2qbAdAdb)δ(r − r0), (3.95)1353.3. Eigenvalues Associated with Φ0 Oddand therefore the global problem for Nˆ is (by once again taking r0 ≈ r00),1r(rNˆr)r −m2r2Nˆ − 1DNˆ =− Uβ−q0DAδ′(r − r00)− Uβ−q−10DF(r00)A(β +2qbAdAdb)δ(r − r00),which we can rewrite using constants A¯0 (given by (2.60)) and A¯0 (given by(2.59)) used in section 2.3.2 to get:1r(rNˆr)r −m2r2Nˆ − 1DNˆ = − A¯0U q0δ′(r − r00)−A¯0A¯0U q+10F(r00)δ(r − 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 Nˆ 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[dNˆdr]r00=A¯0U q0r00− A¯0A¯0U q+10F(r00) =A¯0U q0r00− A¯0A¯0U q+10U q0〈Nˆ〉r00+A¯0A¯0U q+10〈duedr〉r00. (3.97a)1363.3. Eigenvalues Associated with Φ0 OddTo get the jump in the function itself, we first compute the indefinite integralof (3.96), ignoring constants of integration,rNˆr −m2∫1rNˆ dr − 1D∫Nˆ dr =− A¯0U q0r00δ(r − r00) +A¯0U q0H(r − r00)− A¯0A¯0U q+10r00F(r00)H(r − r00)with H the Heaviside function. We then divide by r, and integrate thisexpression over a small domain centered around r00 to get[Nˆ ]r00 = −A¯0U 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 Nˆsubject 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(ε), 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 Nˆ + V1n,1373.3. Eigenvalues Associated with Φ0 Oddthen we have the following problem for Z,1r(rZr)r −n2r2Z − 1DZ = 0, r 6= r00dZdr∣∣∣∣r=r00= 0,[Z]r00 = 0,[dZdr]r00= −A¯0A¯0U0Z(r00).Solving this problem we haveZ(r) = r00A¯0A¯0U0Z(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,Nˆ = − 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) asλˆ =1−m2r200+q2HmU20Hˆ(1− 2bqHˆdHˆdb)〈duedr〉r00(〈V1m〉r00 +Hm〈duedr〉r00)− qHˆ2HmU0(〈dV1mdr〉r00+Hm〈d2uedr2〉r00). (3.99)1383.3. Eigenvalues Associated with Φ0 OddIt is rather unsurprising that there is such an intimate relationship betweenthe problem for Nˆ 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 Nˆ function. Recall that we definethe concentrated activator curve at r = r0 + ρ∗(θ) where ρ∗ is defined as thelocation of the maximum of the activator and hencev˜(ρ∗)′ = v˜′e(ρ∗) + Φ′(ρ∗) exp(imθ +∫ T0λˆ(s) ds)= 0. (3.100)The radial velocity of the curve will be given bydrdT=dr0dT+ dρ∗dTand differentiating (3.100) with respect to T , we have0 =(v˜′′e (ρ∗) + Φ′′(ρ∗) exp(imθ +∫ T0λˆ(s) ds))dρ∗dT+ Φ′(ρ∗) exp(imθ +∫ T0λˆ(s) ds)λˆ(T ).Using (3.100) we can simplify and writedρ∗dT=v˜′e(ρ∗)v˜′′(ρ∗)λˆ(T ) (3.101)and thereforedrdT=dr0dT+ v˜′e(ρ∗)v˜′′(ρ∗)λˆ(T ).1393.3. Eigenvalues Associated with Φ0 OddIf we think of the coefficient on λˆ playing to role of ε in the near circleperturbation then −λˆ should be the first velocity correction we obtained.Indeed if we compare the negative of (3.99) with (2.77) then if we takeh(θ) = exp(imθ)we see perfect agreement. Notice this required choice of h(θ) to balancecomes from the eigenfunction expansion of the linear stability problem (3.2).The coefficients Hm in the Fourier expansion of h(θ) were arbitrary and, ifdesired, could be extracted from the coefficient in front of λˆ.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 = 2vxx − v + a(u)v2r−3,τut = uxx + (ub − u) +1b(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 formL0Φ− χ(λ)h(w)∫ ∞−∞g(w)Φ dy = λΦ, −∞ < y <∞;Φ→ 0 as |y| → ∞, (4.1)where ,L0Φ = Φyy − Φ + f ′(w)Φ, (4.2)with prime indicating differentiation with respect to w. χ(λ) is a transcenden-tal function of the eigenvalue parameter λ and w(y) is the unique homoclinicofwyy − w + f(w) = 0, −∞ < y <∞;w → 0 as |y| → ∞, 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 → 0+;h(0) = 0, h(w) is C1 as w → 0+. (4.4)The conditions on g(w) and h(w) are such that the integral in (4.1) vanishesas Φ→ 0 and that Φ has exponentially decaying solutions when y is asymp-totically large and Re(λ) > 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 L0ψ = νψ. Specifically, it admits a simplediscrete eigenvalue ν0 with eigenfunction ψ0 of one sign and an eigenvalueν1 = 0 with ψ1 = w′. Consider a choice of g(w) in (4.1) which satisfies (4.4),and is such thatL0g(w) = σg(w), (4.5)for some σ > 0. Since there can only be one positive eigenvalue to L0, wemust have that σ = ν0 and g(w) = ψ0. If we multiply (4.1) by g(w) andintegrate over the entire domain then∫ ∞−∞g(w)L0Φ dy − χ(λ)∫ ∞−∞h(w)g(w) dy∫ ∞−∞g(w)Φ dy = λ∫ ∞−∞g(w)Φ 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,∫ ∞−∞g(w)Φ(σ − χ(λ)∫ ∞−∞h(w)g(w) dy − λ)dy = 0,where we have used (4.5) to simplify. Assuming that∫∞−∞ g(w)Φ dy 6= 0 thenwe have an explicit relationship for λ,λ = σ − χ(λ)∫ ∞−∞h(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∫∞−∞ g(w)Φ dy = 0, where (4.1)reduces toL0Φ = λΦ,i.e. Φ is an eigenfunction of the operator L0. The only positive solution tothis is for λ = ν0 but this would require Φ = g(w) and we would not be ableto satisfy∫∞−∞ g(w)Φ 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′′(w)w2y + g′(w)wyy1444.1. Explicit Non-Local Eigenvalue Formulationthen we can write (4.5) as,g′′(w)w2y + g′(w)wyy + f′(w)g(w) = (1 + σ)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 − w2 + 2W = 0, W =∫ w0f(s) ds.Substituting into (4.8) we get,g′′(w)w2 − 2g′′(w)W + g′(w)w − g′(w)f(w) + f ′(w)g(w) = (1 + σ)g(w).If we integrate by parts and use that g(0) = f(0) = 0 we can simplify to get(w2 − 2W)g′(w) = Σ(w)− f(w)g(w); (4.9a)Σ(w) ≡∫ w0ξ(s) ds, ξ(s) ≡ sg′(s) + (σ + 1)g(s). (4.9b)If we differentiate (4.9a) with respect to w then2(w − f(w))g′(w) + (w2 − 2W)g′′(w) = Σ′(w)− (f(w)g(w))′.Using (4.9a) we can simplify to2wg′(w)− 2f(w)g′(w) + Σ(w)− f(w)g(w)g′(w)g′′(w) = Σ′(w)− (f(w)g(w))′.1454.1. Explicit Non-Local Eigenvalue FormulationDividing by g′(w) and rearranging this expression we havef ′(w)g(w)g′(w)− f(w)g′(w)g′(w)− f(w)g′′(w)g′(w)2=Σ′(w)g′(w)− Σ(w)g′′(w)g′(w)2− 2w. (4.10)We can recognize the expression on the left of (4.10) as,f ′(w)g(w)g′(w)− f(w)g′(w)g′(w)− f(w)g′′(w)g′(w)2= g2(fgg′)′,and the first two expressions on the right of (4.10) as,Σ′(w)g′(w)− Σ(w)g′′(w)g′(w)2=(Σ(w)g′(w))′,so that finally, for a given g(w), we can compute f(w) as the solution to(f(w)g(w)g′(w))′=1g(w)2((Σ(w)g′(w))′− 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) = wα for α > 1 for which this theory can easily be usedto determine f(w).First consider g(w) = w so that from (4.9b),ξ(w) = (σ + 2)w, Σ =(σ + 2)w22.1464.1. Explicit Non-Local Eigenvalue FormulationFrom (4.11) we have(f(w)w)′=1w2(((σ + 2)w22)− 2w)=σw,and therefore,f(w) = σw logw + Aw,where we can set A = 0 without loss of generality. Next consider g(w) = wαfor α > 1. In this case,ξ(w) = (σ + 1 + α)wα, Σ =(σ + 1 + α)wα+1α + 1,and hence from (4.11),(f(w)αw2α−1)′=1w2α(((σ + 1 + α)w2α(α + 1))′− 2w)=2(σ + 1 + α)α(α + 1)w1−2α.Integrating this yields thatf(w) =(σ + 1− α2)(1− α2) w + Aw2α−1,for arbitrary A which we set to be one without loss of generality. Since thisholds for any σ, we can avoid the term that is linear in w by choosing,σ = α2 − 1, (4.12)so that f(w) = w2α−1. 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 = 2∆v − v + vpuq, (4.13a)τut = ∆u− u+vous, (4.13b)and for the time being consider a rectangular domain with homogeneousNeumann conditions on the boundary. The rectangular domain Ω is definedbyΩ ≡ {(x1, x2)| − 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) = wα, α > 1, and use the simplified condition on σgiven 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) = wα is intimately tied to theexponent set o via α = o − 1, and by choosing p = 3, this specifies that1484.2. Explicit Stability Formulation for the Gierer-Meinhardt Model on a Stripeα = 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 = 2∆v − v + v3uq, (4.14a)τut = ∆u− u+v3us, (4.14b)with q > 0 and s ≥ 0. We consider the activator v localized on an O()region around the midpoint x1 = 0 for all x2 (hence the classification of astripe) and define an inner coordinate variable y = x/. 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) ∼ Uγ0w(x1); ue(x1) ∼ U0Gl(x1)Gl(0), (4.15)where w(y) =√2 sech y is the unique homoclinic orbit solution that satisfiesw′′ − w + w3 = 0, −∞ < y <∞;w → 0 as |y| → ∞ , w(0) > 0, w′(0) = 0. (4.16)We define the constants U0 and γ byU ζ0 ≡1b˜Gl(0); b˜ ≡∫ ∞−∞w3 dy =√2pi; ζ ≡ 3q2− (s+ 1) > 0; γ ≡ q2.(4.17)1494.2. Explicit Stability Formulation for the Gierer-Meinhardt Model on a StripeThe Green’s function Gl(x1) in (4.15) satisfiesGlx1x1 −Gl = −δ(x1), |x1| ≤ l; Glx1(±l) = 0,with δ(·) the Dirac measure. This has solutionGl(x1) =cosh(l − |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 + φ(x1) exp(imx2 + λt), u = ue + η(x1) exp(imx2 + λt); 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 problem2φx1x1 − φ+3v2euqeφ− qv3euq+1eη = (λ+ 2m2)φ, |x1| ≤ l; φx1(±l) = 0,(4.20a)ηx1x1 − (1 + τλ+m2)η = −3ve2useφ+sv3eus+1eη, |x1| ≤ l; ηx1(±l) = 0.(4.20b)Once again, as in 3.2, we do not ignore the terms 2m2 in the event thathigh frequency modes lead to instability behaviour. Like the base state,ve, the activator perturbation φ will be entirely localized and so we takeφ(x1) ∼ Φ(−1x1). The leading order problem for Φ from (4.20a) isΦyy − Φ + 3w2Φ− qU q/2−10 w3η(0) = (λ+ 2m2)Φ, −∞ < y <∞,Φ→ 0 as |y| → ∞. (4.21)Here η(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(−1) (see 2.2.1) and therefore,3v2euseφ =3U2γ−s0∫ ∞−∞w2Φ dyδ(x1) =3U1−γ0b˜Gl(0)∫ ∞−∞w2Φ dyδ(x1),sv3eus+1eη =sU ζ0 η(0)∫ ∞−∞w3 dyδ(x1) =sη(0)Gl(0)δ(x1),1514.2. Explicit Stability Formulation for the Gierer-Meinhardt Model on a Stripewhere we have simplified using (4.17). Therefore, η(x1) satisfiesηx1x1 − θ2λη =(sGl(0)η(0)− 3U1−γ0b˜Gl(0)∫ ∞−∞w2Φ dy)δ(x1),|x1| ≤ l; ηx1(±l) = 0, (4.22)where θλ ≡√1 +m2 + τλ is the principal value of the square root. Bythis we mean the branch cut taken along the negative real axis such thatRe(λ) ≤ −1−m2τ which implies η(0) is analytic in Re(λ) ≥ 0. If we considerthe Green’s function problemGλx1x1 − θ2λGλ = −δ(x1), |x1| ≤ l, Gλx1(±l) = 0,which has solution,Gλ(x1) =cosh(θλ(l − |x1|))2θλ sinh(θλl)(4.23)then we can writeη(x1) =(3U1−γ0b˜Gl(0)∫ ∞−∞w2Φ dy − sGl(0)η(0))Gλ(x1).Using (4.18) and (4.23) we can solve for η(0) to yield thatη(0) =3U1−γ0b˜∫ ∞−∞w2Φ dy[s+θλ tanh(θλl)tanh(l)]−1. (4.24)This expression is non-zero as long as∫∞−∞w2Φ dy 6= 0 which as per thediscussion in section 4.1 holds for any unstable eigenvalue Re(λ) > 0. This1524.2. Explicit Stability Formulation for the Gierer-Meinhardt Model on a Stripemeans that Φ 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 asL0Φ− χw3∫ ∞−∞w2Φ dy = (λ+ 2m2)Φ, −∞ < y <∞;Φ→ 0 as |y| → ∞, (4.25a)χ ≡ 3qb˜[s+θλ tanh(θλl)tanh(l)]−1, (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 α = 2 (g(w) = w2)and σ = 3 so thatL0w2 = 3w2.By comparing (4.25a) to (4.7) we takeλ = λ+ 2m2, h(w) = w3to get that (4.25a) becomesλ = 3− 2m2 − 9q2[s+θλ tanh(θλl)tanh(l)]−1, (4.26)where we have simplified that∫∞−∞w5 dyb˜=32.We will now analyze (4.26) for τ = 0 and τ > 0.1534.2. Explicit Stability Formulation for the Gierer-Meinhardt Model on a Stripe4.2.1 Explicit Stripe Eigenvalues, τ = 0When τ = 0, we have that (4.26) becomesλ = 3− 2m2 − 9q2[s+√1 +m2 tanh(√1 +m2l)tanh(l)]−1. (4.27)We begin by finding neutral stability points λ = 0. If m is O(1) then toleading order the neutral stability point (4.27) becomes0 = 3− 9q2[3q2+ κ(z)]−1,wherez =√m2 + 1, z ≥ 1; κ(z) = z tanh(zl)tanh(l)− (ζ + 1),with ζ from (4.17). This expression simplifies so that the neutral stabilitypoint is a root of κ(z). Now κ(1) = −ζ < 0 and κ(z) → ∞ as z → ∞ sothere is at least one root to κ(z). Furthermore,κ′(z) =tanh(zl)tanh(l)+ zlsech 2(zl)tanh(l)> 0and so there is a unique root z− > 1 to κ(z) and hence a unique neutralstability modemb− =√z2− − 1. (4.28)1544.2. Explicit Stability Formulation for the Gierer-Meinhardt Model on a StripeIf z < z− then λ < 0 and so there are no unstable eigenvalues on 0 < m <mb− , and conversely, λ > 0 is z > z−. If we now consider m = O() anddefine m = −1m˜ then (4.27) becomesλ = 3− m˜2 − 9q2[s+√m˜2 + 2 tanh(√m˜2 + 2l−1)tanh(l)]−1. (4.29)If we search for the neutral stability point m˜+b and expand m˜+2b ∼ m˜20 + m˜21then we get m˜0 =√3 and for m˜1 we havem˜21 = −9q2tanh(l)m˜0,and so the upper neutral stability point ismb+ ∼1√m˜20 + m˜21 =√3− 3q4tanh(l). (4.30)If m > mb+ (m < mb+) then λ < 0 (λ > 0) and therefore unstable eigenvaluesexist on mb− < m < mb+ . Next we seek to estimate the dominant wave modemdom where λ achieves its maximum. When m is an O(1) or O(−1) numberwe have that to leading order λ is monotonic and therefore the dominantmode must occur at some intermediate scaling in . Therefore we let,m = −amˆ, 0 < a < 1,1554.2. Explicit Stability Formulation for the Gierer-Meinhardt Model on a Stripeand substitute into (4.27),λ = 3− 2−2amˆ2 − a9q2[as+√mˆ2 + 2a tanh(√mˆ2 + 2a−al)tanh(l)]−1.(4.31)The only hope for non-monotonic λ is if there is balance in the second andthird term and so we take a = 2/3. Differentiating (4.31) with respect to mˆand setting to zero we getdλdmˆ=− 2mˆ+ 9q2 tanh(l)[2/3s+√mˆ2 + 4/3 tanh(√mˆ2 + 4/3−2/3l)tanh(l)]−2(mˆ tanh(√mˆ2 + 4/3−2/3l)√mˆ2 + 4/3+ −2/3lmˆ sech 2(√mˆ2 + 4/3−2/3l))= 0. (4.32)Substituting mˆ = mˆ0 + 2/3mˆ1 where the mˆ1 correction comes from s intro-ducing terms of O(2/3), we get that to leading order−2mˆ0 +9q tanh(l)2mˆ20= 0,and somˆ0 =(9q tanh(l)4)1/3. (4.33)1564.2. Explicit Stability Formulation for the Gierer-Meinhardt Model on a StripeFor the mˆ1 problem, both tanh(−2/3) and sech (−2/3) terms will remaintheir saturated values up to exponential order in  and therefore−2mˆ1 −9q tanh(l)mˆ1mˆ0− 9q tanh(l)2smˆ30≈ 0,and using (4.33),mˆ1 ≈ −23s tanh(l).Combining everything, we have,mdom ∼ −2/3(9 tanh(l)4q)1/3− 2s tanh(l)3. (4.34)If one were to consider the effect of the walls negligible (taking l→∞) then(4.27) becomesλ = 3− 2m2 − 9q2(s+√1 +m2),and we have thatmb− =√ζ2 + 2ζ, (4.35a)mb+ =√3/− 3q/4, (4.35b)mdom =−2/3(9q/4)1/3 − 2s/3. (4.35c)This limit will allows us to obtain more tractable analytical results when weconsider τ > 0.We now plot (4.27) in Figure 4.1 with  = 0.05, and s = 0 for q = 1 andq = 2. In each case we plot l = 0.5, l = 1, and l = ∞. In Table 4.1 we list1574.2. Explicit Stability Formulation for the Gierer-Meinhardt Model on a Stripethe numeric evaluation (denoted n) of mb− , mb+ and mdom along with theasymptotic approximations (denoted a) given by (4.28) for mb− , (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 35−0.500.511.522.53mλ  l = 0.5l = 1l =∞(a) q = 10 5 10 15 20 25 30 35−0.500.511.522.53mλ  l = 0.5l = 1l =∞(b) q = 2Figure 4.1: Eigenvalues for τ = 0 computed from (4.27) versus m for  = 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 =∞ respectively.1584.2. Explicit Stability Formulation for the Gierer-Meinhardt Model on a Stripeq = 1l mb−(n) mb−(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∞ 1.12 1.12 33.87 33.89 9.60 9.65q = 2l mb−(n) mb−(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∞ 2.85 2.83 33.03 33.14 12.13 12.16Table 4.1: Asymptotic and numerical comparison of the neutral stabilitypoints mb− , 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− , (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 =⌊mdomd2pi⌋where b·c is the floor function, rounding down to the nearest integer and dis the rectangle width.4.2.2 Explicit Stripe Eigenvalues, τ > 0We now consider (4.26) for τ > 0, which becomes a transcendental equationfor λ because of the presence of θλ. For simplicity, we consider the scenariodiscussed in section 4.2.1 where the sidewalls have no effect (l =∞). In this1594.2. Explicit Stability Formulation for the Gierer-Meinhardt Model on a Stripecase (4.26) becomesλ = 3− 2m2 − 9q2(s+√1 +m2√1 + τˆλ), τˆ ≡ τ1 +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 λ separately from(4.36) and write,λ− 3 + 2m2 = − 9q√1 +m2[2s√1 +m2+ 2√1 + τˆλ]−1.If we defined0 = −2s√1 +m2≤ 0, (4.37a)d1 = −9q√1 +m2< 0, (4.37b)β ≡ 3− 2m2, (4.37c)then we can write (4.36) as a root finding problemF(λ) = 2√1 + τˆλ− G(λ), G(λ) ≡ d0 −d1β − λ. (4.38)We will look for the roots to (4.38) that satisfy Re(λ) > 0 by using theargument principle around the same closed contour Γ = ΓI∪ΓK from section1604.2. Explicit Stability Formulation for the Gierer-Meinhardt Model on a Stripe3.2.6 withΓK :{λ = K exp(it)|t ∈[−pi2,pi2]}ΓI : −Ki ≤ λ ≤ Kiwhich we traverse counter-clockwise and take the limit as K tends to infinity.When τ  1, F(λ) ∼ 2√τˆλ and so on ΓK when t = −pi/2, argF = −pi/4and when t = pi/2, argF = pi/4. Therefore, the change in argument of F(λ)over ΓK is pi/2. When β < 0 (m >√3/) then for all Re(λ) > 0, G(λ) isanalytic but when β > 0 (0 < m <√3/) then there is a simple pole atλ = β. Finally, since F is analytic (aside from λ = β) and real valued whenλ is real-valued then F(λ) = F(λ¯). 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(λ) are given byJ =14+ h(β) +1pi[argF ]Γ+I , where h(β) =1, β > 0,0, β < 0, (4.39)and Γ+I is the positive imaginary axis iλI with 0 < λI <∞ traversed down-wards and [argF ] indicates the change in argument of F over the contour.When λ = iλI , λI > 0 we haveG(iλI) = d0 −βd1β2 + λ2I− i λId1β2 + λ2Iand2√1 + iτˆλI = 2(1 + τˆ2λ2I)1/4 exp(iarctan(τˆλI)2).1614.2. Explicit Stability Formulation for the Gierer-Meinhardt Model on a StripeUsing the trig identitiescos2 θ =1 + cos θ2, sin2 θ =1− cos θ2,we can write2√1 + iτˆλI = K+(τˆλI) + iK−(τˆλI), K±(z) =√2(√1 + z2 ± 1)1/2,and therefore we can decompose F(iλI) = FR(λI) + iFI(λI) withFR(λI) ≡ K+(τˆλI)− d0 +βd1β2 + λ2I, FI(λI) ≡ K−(τˆλI) +λId1β2 + λ2I.(4.40)For λI  1 then since τ > 0, FR ∼ FI ∼√2τˆλI and therefore arg(F(iλI))→pi/4 as λI → ∞. At λI = 0, we have that FR(0) = 2− G(0) and FI(0) = 0,so when λ = 0 is a root of F(λ) = 0 which occurs at mb− and mb+ definedby (4.35a) and (4.35b) respectively then G(0) = 2. When m is O(1),G(0) = ζ + 12√1 +m2and so dG(0)/dm < 0. When m is O(−1) thendG(0)dm∣∣∣∣m=mb+= − 2mb+√1 +mb+2+18q2mb+√1 +mb+2(3− 2mb+2)∼ 83q> 0.1624.2. Explicit Stability Formulation for the Gierer-Meinhardt Model on a StripeFinally, if m >√3/, G(0) < 0. Combining everything together we havethatG(0) < 2, mb− < m < mb+ ∪m >√3(4.41a)G(0) > 2, 0 < m < mb− ∪mb+ < m <√3. (4.41b)We are now in a position to classify the roots J to F(λ). If m >√3/ thenβ < 0 and since d0 ≤ 0 and d1 < 0 then by (4.40), FR(λI) > 0 for all λI .Furthermore FR(0) > 0 and so the argument along Γ+I changes from pi/4 to0. Since FR(λI) > 0 always then it cannot enter the negative real plane sowe must have [arg(F)]Γ+I = −pi/4. From (4.39) this means that J = 0 on thisregion. Next consider mb+ < m <√3/. On this interval β > 0 and from(4.41), FR(0) < 0 so the argument changes from pi/4 to pi. We can calculatethatdFRdλI= τˆK′(τˆλI)−2λIβd1(β2 + λ2I)2 , (4.42)which for mb+ < m <√3/ is positive and therefore since FR(λI) → ∞as λI → ∞ then there is a unique root λ∗I to FR. We can find this rootasymptotically by letting m =√3/+m1 with−3q/4 < m1 < 0 and thereforeβ = 3 − 2m2 ∼ −2√3m1. Substituting into (4.40) we get that to leadingorder FR(λ∗I) = K+(τˆλ∗I) = 0 for which there is no solution. Therefore welet λ∗I = λ∗I0 and repeat the asymptotic expansion in (4.40) to get that toleading order,2 +18qm112m21 + λ∗2I0= 0,1634.2. Explicit Stability Formulation for the Gierer-Meinhardt Model on a Stripeand so we get λ∗I0 =√−3m1(4m1 + 3q) which exists for all m1 since m1 >−3q/4. Using this root in FI from (4.40) we have that,FI(λ∗I0) = K−(τˆλ∗I0) +d1λ∗I0β2 + 2λ∗2I0∼ − 9qλ∗I0√3(12m21 + λ2I0)< 0.Since FI(λ∗I) < 0 then [argF ]Γ+I = −5pi/4 and from (4.39), J = 0. Next weconsider mb− < m < mb+ where from (4.41), FR(0) > 0 and since β > 0and d1 < 0 then by (4.42), dFR/dλI > 0. Therefore [argF ]Γ+I = −pi/4 andfrom (4.39) J = 1. This root is actually on the positive real axis and satisfies0 < λ < β. To see this we consider (4.38) with 0 < F(0) < 2 and F → −∞as λ→ β and therefore F has at least one root on 0 < λ < β. If λ > β thenG(λ) < 0 and F > 0 for all λ. We have that√1 + τλ is an increasing concavefunction as well as G ′(λ) > 0 and G ′′(λ) > 0 so that F ′′(λ) < 0. ThereforeF(λ) can have at most one critical point and regardless of the sign of F ′(0),this can only be satisfied if F has exactly one root. Therefore there is exactlyone real root on mb− < m < mb+ . Finally we consider 0 < m < mb− , whereFR(0) < 0 and so the argument goes from pi/4 to pi. Once again dFR/dλI > 0and FR →∞ as λI →∞ so there is exactly one root to FR on this interval.Therefore either [argF ]Γ+I = −5pi/4 and J = 0 or [argF ]Γ+I = 3pi/4 andJ = 2. If τˆ = 0 then from (4.38), F(0) < 0 and F ′(λ) = −G ′(λ) < 0 sothere are no roots with positive real part. This should continue to be truefor τˆ  1 and therefore J = 0 for τˆ  1 on 0 < m < mb− . If τˆ  1 thenF(0) < 0 and F → −∞ as λ → β− but for λ  O(τˆ−1), F ≈ 2√τˆλ  1and since F ′′(λ) < 0 there are exactly two roots with real positive part forτˆ  1. Notice that eigenvalues cannot enter the real plane through λ = 01644.2. Explicit Stability Formulation for the Gierer-Meinhardt Model on a Stripesince λ = 0 is an eigenvalue for all τˆ and therefore this shows that there mustexist a Hopf bifurcation at some τˆ = τˆHm when 0 < m < mb− . We summarizethe eigenvalue conclusions as follows:Principal Result 4.2.2.1 Let J denote the number of roots in Re(λ) > 0of F(λ) = 0 in (4.38). Then, for any τˆ > 0,J =0, m > mb+J =1, mb− < m < mb+J =0 or J = 2, 0 < m < mb− .On mb− < m < mb+ the root is on the positive real axis in the interval0 < λ < β and for 0 < m < mb−, there are no roots with positive realpart for τˆ  1 and two roots with positive real part when τ  1. This lastcondition proves the existence of a Hopf bifurcation at some τˆ = τˆHm .Explicit Representation for Hopf BifurcationWe now turn our attention to evaluating τˆ = τˆHm and showing it is unique.If a Hopf bifurcation occurs then iλI is a root to (4.38) and so from (4.40)we need to set FR = FI = 0. If we do this we get√2(√a+ 1)1/2 = d0 −d1βθ,√2(√a− 1)1/2 = −d1λIθ;a ≡ 1 + τˆ 2λ2I , θ ≡ β2 + λ2I . (4.43)1654.2. Explicit Stability Formulation for the Gierer-Meinhardt Model on a StripeBy dividing these expressions we get(√a+ 1)1/2(√a− 1)1/2 = −d0θd1λI+βλI,which we can simplify to get√a+ 1√a− 1 =1λIA, A = β − d0θd1. (4.44)Using√a− 1 = τˆλI and the first expression in (4.43) we have,τˆ =d212θ2A. (4.45)Next we determine an equation for θ. To do this we square and subtract thefirst two expressions of (4.43),2√a+ 2− 2√a+ 2 = 4 =(d0 −d1βθ2)2− d21λ2Iθ2. (4.46)We can use that λ2I = θ−β2 in (4.46) to get that θ is the root to the quadraticequation M(θ) given byM(θ) = (d20 − 4)θ2 − (2βd0d1 + d21)θ + 2d21β2 = 0. (4.47)In order for a solution λI =√θ − β2 to exist we require θ > β2. A Hopfbifurcation only occurs on 0 < m < mb− and we will only consider thisinterval for M(θ). First we notice that,M(β2) = β4(d20 −2d0d1β+d21β2− 4)= β4(G(0)2 − 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 = −2 then the roots degen-erate and we getθ1 =2β2d1d1 − 4β(4.48)as the unique root to (4.47). At this root we have from (4.44) thatA = β +4β2d1 − 4β=βd1d1 − 4β> 0since d1 < 0 and therefore in this case there is a unique τˆ = τˆHm > 0 andunique Hopf-bifurcation eigenvalue√θ1 − β2. We next consider the tworemaining cases −2 < d0 < 0 and d0 < −2. If −2 < d0 < 0 then M(θ) from(4.47) tends to negative infinity as θ → ±∞ and therefore since M(β2) > 0,by the intermediate value theorem, there are two roots to (4.47), one rootθ2− in β2 < θ2− < ∞ and another root θ2+ in −∞ < θ2+ < β2. Only θ2−given byθ2− =2βd0d1 + d212(d20 − 4)+√(2βd0d1 + d21)2 − 8d21β2(d20 − 4)2(4− d20), (4.49)is a valid solution to M(θ) that satisfies θ > β2 and so we discount the secondroot. We now need to make sure that for this root, A > 0 so that τˆHm = τˆ > 0.First, define θc = d1β/d0 as the unique root to (4.44) when A = 0. Sinced0 < 0 and d1 < 0 then dA/dθ < 0 and so all we need to show to have A > 0when θ = θ2− is that θ2− < θc. If we evaluate M(θc) we getM(θc) =d21β2d0(d0 −d1β)− 4d21β2d20=d21β2d0(G(0)− 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> β(1− 2d0)> βand therefore that θc > β2. This means that θc ∈ (β2,∞) and since θ2− isthe unique root to M(θ) in this interval then M(θc) < 0 implies θ2− < θc andtherefore A > 0 and τˆHm > 0 when θ = θ2−. Finally we consider the intervald0 < −2. On this interval we still have M(β2) > 0 but now M(θ) → +∞as θ → ±∞. We still have from (4.50) that M(θc) < 0 and so for d0 < −2both roots to M(θ), θ3± exist on θ ∈ (β2,∞). In order to satisfy A > 0 wetake the smaller root θ3− < θc also given by (4.49) which leads to the uniqueHopf bifurcation τˆ = τˆHm > 0.We now briefly consider the case when s = 0 and d0 = 0 identically. Inthis scenario, the roots to (4.47) simplify toθ0 = c0 +√c1; c0 = −d218, c1 =d4164+d21β22.We also have that when d0 = 0 that A = β and so we can writeτˆ =d21β2(1c0 +√c1)2=d21β2c20 − 2c0√c1 + c1(c20 − c1)2.We can compute that,(c20 − c1)2 =d41β44,1684.2. Explicit Stability Formulation for the Gierer-Meinhardt Model on a Stripeand soτˆ =d2116β3+12β3√d4164+d21β22+1β=1β(1 +c22+ c√1 +c24), c =d1√8β= − 9q2√2(1 +m2)(3− 2m2).(4.51)We now summarize the results for the Hopf bifurcation.Principal Result 4.2.2.2 For 0 < m < mb−, there exists a unique valueτHm = (1 + m2)τˆHm > 0 for which λ = λI is a root of F(iλI) given by (4.38).The Hopf bifurcation point τˆHm and λIH are given byτˆHm =d212θ2A, λIH =√θ − β2, A ≡ β − d0θd1. (4.52)Here, θ > β2 is the smallest root of (4.47) given by (4.48) if d0 = −2 and(4.49) if d0 6= −2. When s = 0 we can explicitly compute τˆHm via (4.51).As with τ = 0, we plot the eigenvalues of largest real part for τ 6= 0 in Figure4.2 for l = ∞,  = 0.05, τ = 2, and s = 0 for q = 1 and q = 2. ComparingFigure 4.2 to Figure 4.1, we have that for m  1 the curves have similarbehaviour when τ = 0 and τ 6= 0. This is because θλ =√1 +m2 + τλ ≈√1 +m2 for m  1 and τ = O(1) like in the case τ = 2 chosen here.Therefore changing τ should only have a noticeable affect near the lowerthreshold m = mb− given by (4.35a).1694.2. Explicit Stability Formulation for the Gierer-Meinhardt Model on a Stripe0 5 10 15 20 25 30 35−0.500.511.522.53mλ  q = 1q = 2Figure 4.2: Plot of λ versus m for q = 1 (solid curve) and q = 2 (dashedcurve). The parameters here are s = 0, l =∞, and τ = 2.For q = 2 in Figure 4.2 we have that λ → 0 as m → mb− but this is not sofor q = 1. We can understand this behaviour if we plot τHm and λIH 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 301234567mτH m  q = 1q = 2(a)0 0.5 1 1.5 2 2.5 300.511.522.5mλ IH  q = 1q = 2(b)Figure 4.3: Plot of τHm and λIH from (4.52) for m in 0 < m < mb− andl = ∞. The parameter values are s = 0 and  = 0.05 while the solid anddashed curves are for q = 1 and q = 2 respectively.For q = 2, τHm > 2 = τ in Figure 4.3a and so no Hopf bifurcation occurs for0 < m < mb− . Furthermore, since the eigenvalues enter from the Re(λ) < 0complex plane then all of the eigenvalues are stable as evidenced in Figure4.2. For q = 1 however, τ > τHm for all m in 0 < m < mb− 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(λ) > 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(λ) and Gl(λ) defined byFl ≡ 2√1 + τˆλ(tanh(l√1 +m2√1 + τˆλ)tanh(l√1 +m2))− Gl(λ), Gl(λ) ≡ d0l −d1lβ − λ,(4.53)with d0l ≤ 0, d1l < 0 defined byd0l = −2s√1 +m2tanh ltanh(l√1 +m2), d10 = −9q√1 +m2tanh ltanh(l√1 +m2),and β still by (4.37c). We have that m = mb− defined by (4.28) and m = mb+defined by (4.30) are the two values of m for which λ = 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− and mb+ . To findthe roots we can still use (4.39) since Fl still has a pole at λ = β and stillchanges argument by pi/2 over ΓK . We define FRl(λI) ≡ Re(Fl(iλi)) byFRl = CRl(λI)− d0l +d1lββ2 + λ2I,CRl(λI) ≡ Re(2√1 + iτˆλItanh(l√1 +m2√1 + iτˆλI)tanh(l√1 +m2)),1724.2. Explicit Stability Formulation for the Gierer-Meinhardt Model on a Stripeand similarly,FIl = CIl(λI) +d1lλIβ2 + λ2I,CIl(λI) ≡ Im(2√1 + iτˆλItanh(l√1 +m2√1 + iτˆλI)tanh(l√1 +m2)).It was shown in section 3 of [79] that CRl(λI) is a monotonic increasingfunction of λ and therefore, since d1l < 0, dFRl(λI)/dλI > 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 = ∞. Because of thiswe have that Principal Result 4.2.2.1 holds for any l in 0 < l < ∞. Wecan compute roots to (4.26) for any l using Newton’s method. We plot anexample of this for for q = 1 and τ = 2 in Figure 4.4.1734.3. Explicit Stability Formulation for the Gierer-Meinhardt Model on a Ring0 5 10 15 20 25 30 35−0.500.511.522.53mλ  l = 0.1l = 0.8Figure 4.4: Plot of eigenvalues λ versus m for l = 0.1 (solid curve) and l = 0.8(dashed curve). The parameter values are q = 1, s = 0,  = 0.05, and τ = 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− 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 radialdomainΩ ≡ {(r, θ)|0 ≤ r ≤ l, 0 ≤ θ < 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 = Uγ0w(r − r0), u = U0Gl(r; r0)Gl(r0; r0), γ ≡ q/2,where w is still given by (4.16). Here the Green’s function Gl(r; r0) is givenby (2.38) where R is replaced by l and the constant U0 satisfiesU ζ0 =(1r0√2piJ0,1(r0)J0,2(r0)), ζ ≡ 32q − (s+ 1),with J0,i defined by (2.39). We have the dynamic condition for r0 given by(2.43) whereHˆ =∫∞−∞w2 dy∫∞−∞w2y dy− 1 = 2, y = −1(r − r0),and thereforedr0dT= − 1r0− q2(J ′0,1(r0)J0,1(r0)+J ′0,2(r0)J0,2(r0)), T ≡ −2t. (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 1−2−1.5−1−0.500.51r0/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+Φ(r − r0)exp(imθ+λt), u = ue+η(r) exp(imθ+λt), m ∈ 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 Φ even for which Re(λ) > 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 RingbecomesL0Φ− χw3∫∞−∞w2Φ dy∫∞−∞w3 dy=(λ+2m2r20)Φ;χ ≡ 3q(s+J0,1(r0)J0,2(r0)J¯m,1(θλr0)J¯m,2(θλr0))−1, θλ =√1 + τλ, (4.55)where J0,i is given by (2.39) and J¯m,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) becomesλ = 3− 2m2r20− 9q2(s+J0,1(r0)J0,2(r0)J¯m,1(θλr0)J¯m,2(θλr0))−1, (4.56)where we have used that∫∞−∞w5 dy = 3/2∫∞−∞w3 dy. Once again whenτ = 0, we have that (4.56) becomes an explicit expression for λ. We start byfinding neutral stability points λ = 0. When m O(1) then for λ = 0 (4.56)becomesM = ζ + 1− J0,1(r0)J0,2(r0)J¯m,1(r0)J¯m,2(r0)= 0. (4.57)M(0) = ζ > 0 and when m  1 we can use (3.31) for larger order asymp-totics to the modified Bessel functions to get thatJ¯1,m(r0)J¯2,m(r0) ∼12m(1 + (r0/l)2m) ≈ 12m, r0 > 1 (4.58)and soM→−∞ asm→∞. Furthermore, from section 3.2, J¯m,1(r0)J¯m,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− to (4.57) such that λ = 0 when τ = 0. If τ = 0 then we canrewrite (4.56) asλ = −3M(s+J0,1(r0)J0,2(r0)J¯m,1(r0)J¯m,2(r0))−1and if m < mb− ,M > 0 and hence λ < 0 for m in 0 < m < mb− . Conversely,λ > 0 if m = O(1) in m > mb− . When m = −1m˜ with m˜ = O(1) then using(4.58) we have that (4.56) becomes0 ≈ 3− m˜2r20− 9q2(s+2m˜J0,1(r0)J0,2(r0))−1∼ 3− m˜2r20−  9q4m˜J0,1(r0)J0,2(r0)(4.59)and so if we expand m˜ = m˜0 + m˜1 then m˜0 =√3r0 and2m˜0m˜1r20= − 9q4m˜0J0,1(r0)J0,2(r0)and som˜1 = −3q8J0,1(r0)J0,2(r0).Therefore there exists m = mb+ given bymb+ ∼1√3r0 −3q8J0,1(r0)J0,2(r0),such that λ = 0 in (4.56). When τ = 0, if m > mb+ then λ < 0 and ifm = O(−1) m < mb+ then λ > 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− < m < mb+ , Re(λ) > 0. We plot the eigenvaluesin Figure 4.7 for τ = 0 using (4.56) for q = 2 and l = 5. We plot theeigenvalues for the two equilibrium radii r0 ≈ 1.08 and r0 ≈ 2.56 from Figure4.5.0 10 20 30 40 50 60 70 80 90−0.500.511.522.53mλ  r0 = 1.08r0 = 2.56Figure 4.7: Eigenvalues λ versus m for q = 2, s = 0,  = 0.05, l = 5, andτ = 0 using (4.56). The solid curve is for r0 = 1.08 while the dashed curveis for r0 = 2.56.When τ 6= 0 then similar to section 4.2.2 we can defineR(λ) ≡ Cm(λ)− f(µ), µ = λ+2m2r20,Cm(λ) =13q(s+J0,1(r0)J0,2(r0)J¯m,1(θλr0)J¯m,2(θλr0)), f(µ) = − 32(µ− 3) , (4.60)and then eigenvalues of (4.56) become roots to (4.60). We make the split inthis way because then Cm(λ) 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(µ) an explicitfunction in terms of λ. From (4.60) we can show directly thatf ′(µ) > 0, µ ∈ [0, 3)f ′′(µ) > 0, µ ∈ [0, 3)f(µ) < 0, µ ∈ (3,∞)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− < m < mb+ and that on 0 < m < mb− ,there exists a Hopf bifurcation as τ is increased from zero. To compute theeigenvalues for τ > 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,  = 0.05 andr0 = 1.08 and r0 = 2.56 in Figure 4.8 for τ = 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(λ)Figure 4.8: Eigenvalues λ versus m for q = 2, s = 0,  = 0.05, l = 5, andτ = 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 τ 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 = 20∆v − v +vpuqg(v), (5.1a)τut = D0∆u− u+vous(5.1b)on some domain Ω, where g(v) = 1 if saturation is not considered and g(v) =1+σv2 with σ 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 isΩstripe ≡ {(x, y)|0 ≤ x ≤ 1, 0 ≤ y ≤ d0}subject to Neumann conditions on ∂Ω. If we perform computations on a ringthen the domain isΩring ≡ {(r, θ)|0 ≤ r ≤ 1, 0 ≤ θ ≤ 2pi}183Chapter 5. Full Numerical Simulations of the Gierer-Meinhardt Modelsubject to periodicity in θ, Neumann conditions at r = 1 and a compatibilitycondition ∂v/∂r = ∂u/∂r = 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,  =0√D0, d =d0√D0, l =12√D0, (5.2a)where these quantities are used in section 4.2. Similarly for the ring we takeD = 1,  =0√D0, R = l =1√D0, (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 400×400 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, θ) grid we conformal map to the circle via Matlab’spol2cart function.In order to stimulate breakup instabilities we will consider random perturba-1845.1. Stripe Numerical Experimentstions sampled uniformly from [−δ, δ] with δ = 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− andm = mb+ such that there is a range of real unstable eigenvalues (in the ab-sence of saturation) for which Re(λ) > 0 on mb− < 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 ∈ 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), 0 = 0.05, D0 = 1, τ = 0.1, and d0 = 2 which correspond to = 0.05 l = 1/2 in (4.27). We note that this is meant to represent theτ = 0 case but we take τ 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 0 = 0.05, D0 = 1, τ = 0.1, and d0 = 2. This corresponds to = 0.05, l = 1/2, τ = 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 40−4−2024 t: 0kPhase0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2−1−0.500.51 x 10−4y(a)0 20 4000.511.522.5 kdom=4 8kAmplitude0 20 40−4−2024 t: 2.64kPhase0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2−0.01−0.00500.0050.01y(b)0 20 400246810 kdom=4 8kAmplitude0 20 40−4−2024 t: 3.16kPhase0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2−0.04−0.0200.020.04y(c)0 20 40020406080 kdom=2 4kAmplitude0 20 40−4−2024 t: 5kPhase0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2−0.2−0.100.10.2y(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 0 = 0.05, D0 = 1, τ = 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 0 = 0.05, D0 = 1, τ = 0.1, and d0 = 3. This corresponds to = 0.05, l = 1/2, τ = 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 40−4−2024 t: 0kPhase0 0.5 1 1.5 2 2.5 3−1−0.500.51 x 10−4y(a)0 20 4000.511.52 kdom=8kAmplitude0 20 40−4−2024 t: 2.42kPhase0 0.5 1 1.5 2 2.5 3−505 x 10−3y(b)0 10 20 30 4001234 kdom=8kAmplitude0 10 20 30 40−4−2024 t: 2.64kPhase0 0.5 1 1.5 2 2.5 3−0.01−0.00500.0050.01y(c)0 20 400204060 kdom=4kAmplitude0 20 40−4−2024 t: 5kPhase0 0.5 1 1.5 2 2.5 3−0.2−0.100.10.2y(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 0 = 0.05, D0 = 1, τ = 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  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 ≈ 4.24 and therefore from (5.3) with d = 2√10, 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 0 = 0.05, D0 = 0.1, τ = 0.1, and d0 = 2. This corresponds to = 0.05√10 ≈ 0.1581, l =√10/2 ≈ 1.58, τ = 0.1, and d = 2√10 ≈ 6.32 in(4.27) of Chapter 4.1925.2. Ring Numerical Experiments0 20 4000.0050.010.0150.020.025 kdom=34kAmplitude0 20 40−4−2024 t: 0kPhase0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2−1−0.500.51 x 10−4y(a)0 20 4000.10.20.30.4 kdom=8kAmplitude0 20 40−4−2024 t: 2.64kPhase0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2−1−0.500.51 x 10−3y(b)0 10 20 30 4002468 kdom=8kAmplitude0 10 20 30 40−4−2024 t: 4.34kPhase0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2−0.02−0.0100.010.02y(c)0 20 4005101520 kdom=8kAmplitude0 20 40−4−2024 t: 5kPhase0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2−0.0500.05y(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 0 = 0.05, D0 = 0.1, τ = 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(imθ) for m ∈ 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,  = 0.05, and τ = 0.1.This corresponds to taking D0 = 0.04 and 0 = 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 ∈ [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 20−4−2024 t: 5mPhase0 1 2 3 4 5 6 7−0.1−0.0500.050.1$θ$r0= 0.21571(b) r0 = 0.216(c) r0 = 0.5120 50 100012345 mdom=28mAmplitude0 50 100−4−2024 t: 3mPhase0 1 2 3 4 5 6 7−0.02−0.0100.010.02$θ$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 0 = 0.01, D0 = 0.04, and τ = 0.1. This corresponds to  = 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 θ = 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 0 = 0.025, which corresponds to  = 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(−2) 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 belowλ = 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 0 = 0.025, D0 = 1, and τ = 0.1. This corresponds to  = 0.025,R = 1, and τ = 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 20−4−2024 t: 0mPhase0 1 2 3 4 5 6 7−4−2024 x 10−5θr0= 0.5(a)0 10 2000.511.52 mdom=4 6mAmplitude0 10 20−4−2024 t: 5.26mPhase0 1 2 3 4 5 6 7−505 x 10−3θr0= 0.49501(b)0 5 10 15 200246 mdom=4 6mAmplitude0 5 10 15 20−4−2024 t: 6.32mPhase0 1 2 3 4 5 6 7−0.02−0.0100.010.02θr0= 0.49252(c)0 10 201020304050 mdom=1 2 3 4 5 6mAmplitude0 10 20−4−2024 t: 10mPhase0 1 2 3 4 5 6 7−0.2−0.100.10.2θr0= 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 0 = 0.05, D0 = 1, and τ = 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, 0 =  = 0.025,R = 1 and r0 = 0.5. We begin by taking the saturation σ = 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(λ) > 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 σ = 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 0 = 0.025, D0 = 1, τ = 0.1, and σ = 25. This corresponds to = 0.025, R = 1, and τ = 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 20−4−2024 t: 10mPhase0 1 2 3 4 5 6 7−4−2024 x 10−4θr0= 0.47756(a)0 10 2005101520 mdom=4mAmplitude0 10 20−4−2024 t: 26.2mPhase0 1 2 3 4 5 6 7−0.0500.05θr0= 0.43766(b)0 10 200510152025 mdom=4mAmplitude0 10 20−4−2024 t: 37.3mPhase0 1 2 3 4 5 6 7−0.1−0.0500.050.1θr0= 0.35786(c)0 10 200510152025 mdom=4mAmplitude0 10 20−4−2024 t: 50mPhase0 1 2 3 4 5 6 7−0.1−0.0500.050.1θr0= 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 0 = 0.025, D0 = 1, τ = 0.1, and σ = 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 σ = 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 σ, 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 σ = 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 0 = 0.025, D0 = 1, τ = 0.1, and σ = 950. This corresponds to = 0.025, R = 1, and τ = 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 20−4−2024 t: 0mPhase0 1 2 3 4 5 6 7−4−2024 x 10−5θr0= 0.5(a)0 5 10 15 2001234 x 10−4 mdom=4mAmplitude0 5 10 15 20−4−2024 t: 3.48mPhase0 1 2 3 4 5 6 7−1−0.500.51 x 10−6θr0= 0.47756(b)0 5 10 15 2002468 x 10−4 mdom=1mAmplitude0 5 10 15 20−4−2024 t: 12.2mPhase0 1 2 3 4 5 6 7−2−1012 x 10−6θr0= 0.36534(c)0 5 10 15 2001234 x 10−3 mdom=1mAmplitude0 5 10 15 20−4−2024 t: 20mPhase0 1 2 3 4 5 6 7−1−0.500.51 x 10−5θr0= 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 0 = 0.05, D0 = 1, τ = 0.1, and σ = 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 6−0.5−0.4−0.3−0.2−0.100.10.20.30.40.5mmax(Re(λ))Figure 5.14: Eigenvalues of (3.16) for exponent set (2, 1, 2, 0),  = 0.05, τ = 0,and R = 1 with r0 = 0.213 and σ = 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(6θ). 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 0 = 0.025, D0 = 1, τ = 0.1, and σ = 950. This correspondsto  = 0.025, R = 1, and τ = 0.1. We take as an initial radius r0 =0.5 + 0.02 cos(6θ).Conversely, if r0  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 take0 = 0.01 ( = 0.1) to avoid the weak interaction regime where D = O(2)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 σ = 5910. As in Figure 5.15, we initializethe curve with a perturbed radius r0 = 0.5 + 0.02 cos(6θ). 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 0 = 0.01, D0 = 0.01, τ = 0.1, and σ = 5910. This correspondsto  = 0.1, R = 10, and τ = 0.1. We take as an initial radius r0 = 0.5 +0.02 cos(6θ).The delicacy of manipulating 0 in correspondence with D0 to avoid theweak interaction regime and the affect of σ 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(−2) 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∆u− u = 0, x ∈ Ω∂u∂n= 0, x ∈ ∂Ωu = U0(s), x ∈ Γ[∂u∂n]Γ= − 1D∫ ∞−∞f(U0, v˜0) dηˆ, x ∈ Γsubject toV0 = κ0 +H(∂u∂n∣∣∣∣η=0++∂u∂n∣∣∣∣η=0−),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’s 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’s function formu-lation. Typically, when solving these problems on non-standard geometries,the method of images is used to modify the Green’s function and use thecurve data as a source. The method of layer potentials does the opposite ofthis by using the original free space Green’s function (or its derivative) and,instead, considers the source density to be unknown. In order to proceedthen, we need the free space Green’s function or fundamental solution to themodified Helmholtz equation:D∆qΨ−Ψ = −δ(p− q), p, q ∈ R2where the subscript indicates differentiation with respect to q variables. Im-posing that the solution decay at infinity allow us to write the fundamentalsolution asΨ(p, q) =12piDK0( |p− q|√D), (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 φ on Γsuch that at any point x in space,u(x) =∫ΓΨ(x, q)φ(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 Ψ [34]). The subscript s indicates that the integral iswith respect to the arclength of Γ 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’s equation,∆qΦ = −δ(p− q)isΦ = − 12pilog |p− q|. (6.3)The following property holds for the Laplace single layer potential with sourcedensity φ for some curve Γ with Dirichlet boundary conditions u = f(x), x ∈Γ:f(x) =∫ΓΦ(x, q)φ(q) dqs. x ∈ Γ.2126.1. Layer Potential FormulationFurthermore, the limiting values of the normal derivative with the positivenormal on Γ taken as the interior normal of a single curve arelimα→0+∂u∂nx(x+ αnˆxi) =∫Γ∂Φ∂nx(x, q)φ(q) dqs −φ(x)2, (6.4a)limα→0+∂u∂nx(x− αnˆxi) =∫Γ∂Φ∂nx(x, q)φ(q) dqs +φ(x)2, (6.4b)where we define normal derivatives with respect to a coordinate system a as,∂u∂na= ∇au · nˆa.Subtracting (6.4b) from (6.4a) we get[∂u∂n]= − (nˆ · nˆi)φ(x), x ∈ Γ.The subscript i on the normal in (6.4) indicates that we are explicitly con-sidering α > 0 to be traversing the inner normal. It is important to makethis distinction if Γ has multiple curves since nˆ = ±nˆi, 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→ q in Φ(p, q). If we look at the samelimit for the fundamental solution Ψ(p, q) we have [1],Ψ(p, q) ∼p→q1DΦ(p, q) +R(|p− 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 =∫ΓΨ(x, q)φ(q) dqs, x ∈ Γ (6.5a)[∂u∂n]∣∣∣∣Γ= −(nˆ · nˆi)φ(x)D, x ∈ Γ (6.5b)where the extra factor of D comes from the singularity relationship betweenΦ and Ψ.6.1.1 Incorporating Neumann Boundary ConditionsWe have thus far considered how to solve the problem (2.19) for a closed curveΓ 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 Γdoes not have to be a single closed curve, and as such, we can define ∂Ω to beone of these curves. We can therefore write the new layer potential problemasu(x) =∫ΓΨ(x, q)φ(q) dqs +∫∂ΩΨ(x, q)φb(q) dqs. (6.6)We introduce a new density φb here for the integral on the domain boundaryand split the integral in this way so that φb belongs to the, always static,curve ∂Ω and any dynamic curves belong to the set Γ. By replacing (6.2)2146.1. Layer Potential Formulationwith (6.6), we can define the new jump in normal derivative for x ∈ Γ:limα→0+∂u∂nx(x+ αnˆxi) =∫Γ∂Ψ∂nx(x, q)φ(q) dqs −φ(x)2D+∫∂Ω∂Ψ∂nx(x, q)φb(q) dqs (6.7a)limα→0+∂u∂nx(x− αnˆxi) =∫Γ∂Ψ∂nx(x, q)φ(q) dqs +φ(x)2D+∫∂Ω∂Ψ∂nx(x, q)φb(q) dqs (6.7b)where we note that substitution is applicable in the last integral for x ∈ Γand q ∈ ∂Ω since then, Ψ(p, q) is no where singular. It is now straightforwardto see that (6.5) becomesU0 =∫ΓΨ(x, q)φ(q) dqs +∫∂ΩΨ(x, q)φb(q) dqs, x ∈ Γ (6.8a)[∂u∂n]∣∣∣∣Γ= −(nˆ · nˆi)φ(x)D, x ∈ Γ. (6.8b)To incorporate the condition ∂u∂n = 0 on ∂Ω, we once again look at thenormal jump formula. This boundary condition states that as we approachthe boundary from exterior to Ω, the flux must vanish and thereforelimα→0+∂u∂n(x+ αnˆΩe) = 0, x ∈ ∂Ω,2156.1. Layer Potential Formulationwhere nˆΩe is the normal from ∂Ω pointing to the exterior of Ω. Since we areconsidering the exterior normal and nˆΩe = −nˆΩi then we use (6.4b) and get0 =∫Γ∂Ψ∂nΩ(x, q)φ(q) dqs +φb(x)2D+∫∂Ω∂Ψ∂nΩ(x, q)φb(q) dqs, x ∈ ∂Ω.(6.9)6.1.2 Scaled Arclength parametrizationIn order to compute the integrals, we will use a scaled arclength parametriza-tion. First consider that Γ is composed of M distinct closed curves,Γ =M⋃j=1Γj,and write (6.6) asu(x) =M∑j=1∫ΓjΨ(x, qj)φ(qj) dqjs +∫∂ΩΨ(x, q)φb(q) dqs, x ∈ Ω ∪ Γ.(6.10)Define the scaled arclength coordinate as σ = sj/Lj, where Lj is the lengthof curve j. In order to not confuse this with the saturation parameter σ fromthe previous chapters, we will now reserve σˆ for the saturation parameter.Let the parametrization of Γj be written aszj(σ) = 〈z1j(σ), z2j(σ)〉,2166.1. Layer Potential Formulationand the parametrization of ∂Ω aszb(σ) = 〈z1b(σ), z2b(σ)〉.With this in mind, we can write (6.10) for x ∈ Ω ∪ Γ asu(x) =M∑j=1Lj∫ 10Ψ(x, zj(σ))φj(σ) dσ + L∂Ω∫ 10Ψ(x, zb(σ))φb(σ) dσ, (6.11)where φj = φ(zj(σ)). We can write the Dirichlet condition (6.8a) asU0(σ∗) =Lm∫ 10Ψ(zm(σ∗), zm(σ))φm(σ) dσ+M∑j=1j 6=mLj∫ 10Ψ(zm(σ∗), zj(σ))φj(σ) dσ+ L∂Ω∫ 10Ψ(zm(σ∗), zb(σ))φb(σ) dσ, (6.12)where we have isolated the integral with the singularity separately from thesum. We can write the jump condition (6.8b) asφm(σ∗) = (nˆ · nˆi)∫ ∞−∞f(U0(σ∗), v˜0) dηˆ (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∑j=1Lj∫ 10∂Ψ∂nΩ(zb(σ∗), zj(σ))φj(σ) dσ +φb(σ∗)2D+ L∂Ω∫ 10∂Ψ∂nΩ(zb(σ∗), zb(σ))φb(σ) dσ = 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 σ∗ ∈ [0, 1), we have that the motion on curvem is given bydzmdt= V nˆ + Vttˆ, (6.15)2186.1. Layer Potential Formulationand that∫ 10Vt dσ =∫ 10dzmdt· tˆ dσ = 0. (6.16)By prescribing the normal velocity and enforcing equal arclength (|zmσ | = 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),limα→0+∂u∂nx(x+ αnˆxi) + limα→0+∂u∂nx(x− αnˆxi)= 2∫Γ∂Ψ∂nx(x, q)φ(q) dqs + 2∫∂Ω∂Ψ∂nx(x, q)φb(q) dqs.In (2.19e) we need to add the derivative contributions of x+αnˆ with nˆ = ±nˆi.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 (σ∗) = κ+ 2H(U0)(Lm∫ 10∂Ψ∂nz∗m(zm(σ∗), zm(σ))φm(σ) dσ +M∑j=1j 6=mLj∫ 10∂Ψ∂nz∗m(zm(σ∗), zj(σ))φj(σ) dσ+L∂Ω∫ 10∂Ψ∂nz∗m(zm(σ∗), zb(σ))φb(σ) dσ), (6.17)where we have defined, z∗m = zm(σ∗).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∫ 10Ψ(z(σ∗), z(σ))φ(σ) dσ =12piD∫ 10K0(∣∣∣∣z(σ)− z(σ∗)√D∣∣∣∣)φ(σ) dσ, (6.18)which is singular when σ = σ∗. To determine the nature of the singularity,we perform an asymptotic expansion as σ ≈ σ∗,z(σ) ∼ z(σ∗) + L(σ − σ∗)tˆz∗ +L2κ2(σ − σ∗)2nˆz∗ +O(σ − σ∗)3 (6.19)2206.1. Layer Potential Formulationwith tˆz∗ and nˆz∗ , the tangent and normal vectors at z(σ∗) respectively. Herewe have used that∣∣∣∣dzdσ∣∣∣∣ = L,∣∣∣∣∣d2zdσ2∣∣∣∣∣= L2κ,with κ the curvature at z(σ∗) and L the length. We can compute that theasymptotic norm is|z(σ)− z(σ∗)| = L|σ − σ∗|+O(σ − σ∗)3, (6.20)and therefore have that the Bessel function for σ − σ∗  1 has the form [1],K0(∣∣∣∣z(σ)− z(σ∗)√D∣∣∣∣)∼ log(2√DL)− γ − log |σ − σ∗|+O(σ − σ∗)2,where γ is the Euler-Mascheroni constant. To alleviate the logarithmic sin-gularity, we will add and subtract the log term to (6.18), to get∫ 10Ψ(z(σ∗), z(σ))φ(σ) dσ=12piD(∫ 10K0(∣∣∣∣z(σ)− z(σ∗)√D∣∣∣∣)+ log |σ − σ∗|)φ(σ) dσ− 12piD∫ 10log |σ − σ∗|φ(σ) dσ, (6.21)where the integral with the Bessel function is now non-singular ∀σ ∈ [0, 1].Due to the periodicity of the curve, there is a slight issue with this formulationand that is that σ∗ = 0 also induces a singularity at σ = 1. Furthermore, asσ∗ gets close to zero, it begins to notice the effects of a singularity at σ = −12216.1. Layer Potential Formulationas well, though we classify this as a weak singularity since it is outside of thedomain of σ. Since σ∗ ∈ [0, 1), there is no effect from the weak singularityat σ = 1. To remedy the effects of the periodicity inducing singularities, wewill remove the singularities that are a full period from σ∗ on either side andwrite (6.21) as∫ 10Ψ(z(σ∗), z(σ))φ(σ) dσ =12piD(∫ 10K0(∣∣∣∣z(σ)− z(σ∗)√D∣∣∣∣)+ L(σ, σ∗))φ(σ) dσ− 12piD∫ 10log |σ − σ∗|φ(σ) dσ − 12piD∫ 10log |σ − (σ∗ + 1)|φ(σ) dσ− 12piD∫ 10log |σ − (σ∗ − 1)|φ(σ) dσ, (6.22)whereL(σ, σ∗) = log |σ − σ∗|+ log |σ − (σ∗ + 1)|+ log |σ − (σ∗ − 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(σ), z(σ∗)) =K0(∣∣∣z(σ)−z(σ∗)√D∣∣∣)+ L(σ, σ∗), σ 6= σ∗log∣∣∣2√D(σ−(σ∗+1))(σ−(σ∗−1))L∣∣∣− γ, σ = σ∗ 6= 0log∣∣∣2√D(σ+1)L∣∣∣− γ, σ∗ = 0, σ = 0, 1.(6.24)2226.1. Layer Potential FormulationNotice that using the asymptotic form is valid at σ = σ∗ because the errorterms vanish exactly. Finally, also defineS =∫ 10log |σ − σ∗|φ(σ) dσ +∫ 10log |σ − (σ∗ + 1)|φ(σ) dσ+∫ 10log |σ − (σ∗ − 1)|φ(σ) dσ, (6.25)so that (6.22) becomes∫ 10Ψ(z(σ∗), z(σ))φ(σ) dσ =12piD(∫ 10K0(z(σ), z(σ∗))φ(σ) dσ − S).(6.26)Next consider the integral,∫ 10∂Ψ∂nz∗(z(σ∗), z(σ))φ(σ) dσ =12piD3/2∫ 10K1(∣∣∣∣z(σ)− z(σ∗)√D∣∣∣∣)(z(σ)− z(σ∗)) · nˆz∗|z(σ)− z(σ∗)| φ(σ) dσ, (6.27)where K1 is the second order modified Bessel function of the second kind.Using the expansion for z(σ) near σ∗ from (6.19) we get,K1(∣∣∣∣z(σ)− z(σ∗)√D∣∣∣∣)∼√DL1|σ − σ∗| +O(σ − σ∗)(z(σ)− z(σ∗)) · nˆz∗ ∼L2κ2(σ − σ∗)2 +O(σ − σ∗)4,and along with the expansion of the norm (6.20) we haveK1(∣∣∣∣z(σ)− z(σ∗)√D∣∣∣∣)(z(σ)− z(σ∗)) · nˆz∗|z(σ)− z(σ∗)| ∼√Dκ2+O(σ − σ∗).2236.2. Numerical Formulation of Curve Motion ProblemTherefore, we see that unlike the integral involving the zeroth order Besselfunction, K1 is not singular at σ = σ∗ and so we do not have to alter theintegral by removing any singularities. However, to avoid the numericaldifficulties of evaluating directly at σ = σ∗, we define the following functionK1(z(σ), z(σ∗)) =1√DK1(∣∣∣z(σ)−z(σ∗)√D∣∣∣)(z(σ)−z(σ∗))·nˆz∗|z(σ)−z(σ∗)| , σ 6= σ∗κ2 , σ = σ∗, (6.28)where we note that if σ∗ = 0 then due to the periodicity we evaluate thesecond branch if σ = 0 or σ = 1. With (6.28) we have that (6.27) becomes∫ 10∂Ψ∂nz∗(z(σ∗), z(σ))φ(σ) dσ =12piD∫ 10K1(z(σ), z(σ∗))φ(σ) dσ. (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 σ, σi = i∆σ with ∆σ = 1N where N arethe chosen number of grid points. Define zij = zj(σi) as the discretized curvepositions. Using a standard centered difference discretization we havedzjdσ∣∣∣∣σ=σi=zi+1,j − zi−1,j2∆σ+O(∆σ2),d2zjdσ2∣∣∣∣∣σ=σi=zi+1,j − 2zi,j + zi−1,j∆σ2+O(∆σ2),2246.2. Numerical Formulation of Curve Motion Problemwhich we can use to define the unit tangent, normal vectors, and curvatureup to O(∆σ2)tˆi,j =zjσ(σi)Lj=zi+1,j − zi−1,j2Lj∆σ, (6.30)nˆi,j = 〈−tˆi,jy , tˆi,jx〉, (6.31)κi,j =zjσσ(σi) · nˆi,jL2j=12∆σ3L3j(zi+1,j − 2zi,j + zi−1,j) · 〈−tˆi,jy , tˆi,jx〉. (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 ∆t, we willconsider the implicit Backward Euler method so that for each curve m, (6.15)becomes(zk+1m − zkm) · nˆk+1 = ∆tV k+1,for the normal component where the subscript k indicates t = k∆t and∫ 10(zk+1m − zkm) · tˆk+1 dσ = 0,for the tangential component where we have used the zero mean condition(6.16). This is closed by prescribing the equal arclength parametrization,∣∣∣∣dzk+1mdσ∣∣∣∣ = Lm, (6.33)which we can discretize using (6.20) to get|zk+1i+1,m − zk+1i,m | = Lm∆σ. (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,∫ 10f(σ) dσ =N∑i=0αifi∆σ,where fi = f(σi) and αi are the quadrature weights. Unless otherwise spec-ified, since our scheme is accurate to O(∆σ2) from the finite difference dis-cretization, we will consider the quadrature weights to be those that comefrom the trapezoid rule [6],αi =12 , 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 σ 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,∫ 10f(σ) log |σ − (σ∗ + a)| dσ, (6.35)2266.2. Numerical Formulation of Curve Motion Problemwhere σ∗ ∈ [0, 1) is one of the discretized gridpoints (σ∗ = j∆σ) and a =−1, 0, 1 is the shifted singularity being removed. The goal is to develop adiscretization so that we can write (6.35) as∫ 10f(σ) log |σ − (σ∗ + a)| dσ =N∑i=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 σ 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(σ) using Lagrange interpolation andanalytically perform the polynomial-logarithmic integration that results. Westart by writing,∫ 10f(σ) log |σ − σ∗| dσ =N∑k=1∫ k∆σ(k−1)∆σf(σ) log |σ − j∆σ| dσ,where we are explicitly considering the case a = 0 which we will generalizelater. In what follows we consider the case of f(σ) 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(σ) on [(k − 1)∆σ, k∆σ] isf ∼ (σ − (k − 1)∆σ)fk − (σ − k∆σ)fk−1∆σ,and so,∫ 10f(σ) log |σ − σ∗| dσ =N∑k=1fk∆σ∫ k∆σ(k−1)∆σ(σ − (k − 1)∆σ) log |σ − j∆σ| dσ+fk−1∆σ∫ k∆σ(k−1)∆σ(k∆σ − σ) log |σ − j∆σ| dσ. (6.37)Consider evaluating the first integral in this expression by letting u = (σ −(k − 1)∆σ)/∆σ,1∆σ∫ k∆σ(k−1)∆σ(σ − (k − 1)∆σ) log |σ − j∆σ| dσ= ∆σ(log ∆σ∫ 10u du+∫ 10u log |u+ (k − 1)− j| du)=∆σ log ∆σ2+ ∆σψ1,j(k − 1),whereψ1,j(y) =∫ 10u log |u+ y − j| du,is computed analytically for any y and j. Using the same transformation onthe second integral in (6.37) we get1∆σ∫ k∆σ(k−1)∆σ(k∆σ − σ) log |σ − j∆σ| dσ = ∆σ log ∆σ2+ ∆σψ2,j(k − 1),2286.2. Numerical Formulation of Curve Motion Problemwhereψ2,j(y) =∫ 10(1− u) log |u+ y − j| du.Using these functions we can define the following weights:w0,j =∆σ log ∆σ2+ ∆σψ2,j(0),wN,j =∆σ log ∆σ2+ ∆σψ1,j(N − 1),wi,j = ∆σ log ∆σ + ∆σ(ψ1,j(i− 1) + ψ2,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(∆σ2) in line with the Trapezoid rule for regularintegrals. If instead of a = 0, we have a = 1 (a = −1) in (6.35) we define thepositive (negative) complementary functions:ψc±1,j(y) =∫ 10u log |u+ y − j ∓N | du,ψc±2,j(y) =∫ 10(1− u) log |u+ y − j ∓N | du,2296.2. Numerical Formulation of Curve Motion Problemand the positive (negative) complementary weights:wc±0,j =∆σ log ∆σ2+ ∆σψc±2,j(0),wc±N,j =∆σ log ∆σ2+ ∆σψc±1,j(N − 1),wc±i,j = ∆σ log ∆σ + ∆σ(ψc±1,j(k − 1) + ψc±2,j(k)), i 6= 0, N.To demonstrate the error accuracy of the method we will consider the case∫ 10 cos(σ) log |σ| dσ. Analytically we can integrate by parts to get that∫ 10cos(σ) log |σ| dσ = −∫ 10sin(σ)σdσ = −Si(1).From [1] we can write this as a series,−∞∑k=0(−1)k−1(2k − 1)(2k − 1)! ≈ −0.94608307036718301494where 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 ∆σ along with the error ratio demonstrating the∆σ2 convergence.∆σ Error Ratio0.1 7.73× 10−4 —0.05 1.95× 10−4 3.960.025 4.90× 10−5 3.980.0125 1.22× 10−5 3.990.00625 3.08× 10−6 4.00Table 6.1: Numerical-analytic comparison of integrating∫ 10 cos(σ) log |σ|dσusing 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(∆σ4) reduction as would be expected with a standardSimpsons integration rule [6].∆σ Error Ratio0.1 1.70× 10−6 —0.05 1.02× 10−7 16.570.025 6.27× 10−9 16.300.0125 3.89× 10−10 16.150.00625 2.42× 10−11 16.04Table 6.2: Numerical-analytic comparison of integrating∫ 10 cos(σ) log |σ|dσusing 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 = 〈cos(2piσ), sin(2piσ)〉,and integrate∫ 10K0 (|z(σ)− z(σ∗)|) dσ (6.38)for a range of σ∗ ∈ [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 σ∗ = k∆σ, k ∈ [0, N − 1], ∆σ = 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 10.5330.53320.53340.53360.53380.5340.53420.53440.53460.53480.535σ*integral valueFigure 6.1: This shows the value of integrating (6.38) with ∆σ = 0.02 forall the possible discrete values of σ∗ ∈ [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∆σ(σ∗) =∫ 10K0(z(σ), z(σ∗)) dσ − T∆σ(σ∗) ∼∆σ212F (σ∗) +O(∆σ4),where K0(z(σ), z(σ∗)) is given by (6.24), T∆σ is the trapezoid approximationto the integral, andF (σ∗) = K0σ(z(0), z(σ∗))−K0σ(z(1), z(σ∗)). (6.39)F (σ∗) has an absolute maximum at σ∗ = 12 and absolute minimum at σ∗ = 0by recalling that 0 ≤ σ∗ < 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 1−1.5−1−0.50σ*F(σ*)Figure 6.2: The plot of F (σ∗) as defined in (6.39). The function has anabsolute maximum at σ∗ = 0.5 and and absolute minimum at σ∗ = 0.If we consider the difference in F (σ∗) at its maximum and minimum valueswe haveFmax − Fmin =16,and therefore,∆maxE∆σ =∆σ212(Fmax − Fmin) =∆σ272 ∆σ2.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 · nˆk+1i = zki,m · nˆk+1i + ∆tV k+1i,m , (6.40a)N∑i=0αi(zk+1i,m − zki,m) · tˆk+1i ∆σ = 0, (6.40b)|zk+1i+1,m − zk+1i,m | = Lm∆σ. (6.40c)The equations that prescribe the density φ and the value U0 become,Uk+10i,m =Lm2piDN∑l=0(αlK0(zk+1l,m , zk+1i,m )φk+1l,m ∆σ −Wl,iφk+1l,m)+M∑j=1j 6=mLj2piDN∑l=0αlK0(∣∣∣∣∣zk+1l,j − zk+1i,m√D∣∣∣∣∣)φk+1l,j ∆σ+L∂Ω2piDN∑l=0αlK0(∣∣∣∣∣zbl − zk+1i,m√D∣∣∣∣∣)φk+1bl ∆σ, (6.40d)φk+1i,m = (nˆ0 · nˆ0i )∫ ∞−∞f(Uk+10i,m , v˜0) dηˆ, (6.40e)where Wi,j = wi,j + wc+i,j + wc−i,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 overηˆ and not over the curve where the unknown values are specified. Finally,we can write the Neumann boundary and velocity condition as0 =M∑j=1Lj2piDN∑l=0αlK1(zk+1l,j , zbl)φk+1l,j ∆σ +φk+1bi2+L∂Ω2piDN∑l=0αlK1(zbl , zbi)φk+1bl ∆σ, (6.40f)V k+1i,m = κk+1i,m + 2H(Uk+10i,m )( M∑j=1Lj2piDN∑l=0αlK1(zk+1l,j , zk+1i,m )φk+1l,j ∆σ+L∂Ω2piDN∑l=0αlK1(zbl , zk+1i,m )φk+1bl ∆σ). (6.40g)While we will consider integrals accurate up to O(∆σ2), one can easily ex-tend this to higher order methods by using the appropriate weights α andW . Given an initial curve we can compute the solution at the next time stepusing Newton’s method. When forming the Jacobian for Newton’s method,notice that we indeed have enough equations (the system (6.40)) to solve forthe unknown vector u = [z,V,L,U0, φ, φb], 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〈x(θ), y(θ)〉 with θ ∈ [0,Θ]. Write the unit arclength parametrization asσ =∫ θ0√dxdv2+ dydv2dv∫ Θ0√dxdθ2+ dydθ2dθ,which we can interpolate for discrete values of σ ∈ [0, 1]. We can pre-computevalues to arbitrary accuracy by refining in θ along with higher order quadra-ture and interpolation. Having an initial unit arclength curve, we can de-termine the initial values of U0 and φ for the static curve by using New-ton’s 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’s method to converge. Assuming this problem converges aninitial vector u0 can be used in the dynamic Newton’s 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 withκ =∫ΓΦ(x, q)V (q) dqs + C(t), (6.41a)[∂u∂n]= −V, (6.41b)where here the fundamental solution for Laplace’s problem is used becausethe MS problem solves Laplace’s 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]),∫ΓV (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 Γ 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) =− 1R1 , 0 ≤ r ≤ R1− 1R1 +(1R1 +1R2) log(rR1)log(R2R1) , R1 ≤ r ≤ R21R2 , r ≥ R2, (6.42)with equations for the radii given by,dR1dt= − 1R1(1R1 +1R2)log(R2R1) ,dR2dt= − 1R2(1R1 +1R2)log(R2R1) .Notice thatddt(R21 +R22) = 0,and so,R2 =√A2 +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 = −12∫ R1(t)R10x2√x2 + A2x+√x2 + 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 ∆t = 1× 10−3. Thefigure shows an excellent agreement between the numerical formulation andthe analytic solution.−2 −1.5 −1 −0.5 0 0.5 1 1.5 2−2−1.5−1−0.500.511.52xyConcentric 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 ∆t but since the source term for the equationof motion has a spatial error ∆σ2 log ∆σ then we actually predict a globalerror of O(∆σ2 log ∆σ+ ∆t). In Table 6.3 we take ∆t = ∆σ2 for a variety ofN values so that the overall convergence should be ∆σ2 log ∆σ and indeedthat is observed.N xerr xrat Verr Vrat Lerr Lrat CPU Time8 2.90E-03 — 1.45E-01 — 1.61E-01 — 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 — 7.00E-02 — 3.16E-01 — 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(∆σ2 log ∆σ) 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).−2 0 2 4 6 8 10−10123456789xyNon−Concentric circles  t=0t=1.5Figure 6.4: Evolution of non concentric circles with MS. The first circle iscentered at (−1, 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 (∆t = 1 × 10−2) 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−3 −2 −1 0 1 2 3−2−1.5−1−0.500.511.52xyEllipse  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(∆t = 1 × 10−2) 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× 10−3. 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 σˆ, (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) asφm(σ∗) = (nˆ · nˆi)Uβ0∫ ∞−∞wo dηˆ, β = qo− s.We can also write (2.20) as H = − q4U0 Hˆ with Hˆ defined in (2.31). Aswe iterate a solution using Newton’s 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 ∈ (−∞,∞) to b ∈ [0, bc] via,b =bc2(cos(c) + 1).This form is chosen since there are not any steep gradients in the functionfor Newton’s 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 σˆ > 0 as long as we do not initialize near b = 0. If σˆ = 0then b = 0 is a solution for all vectors u and time t. One notorious issue witha trigonometric function in Newton’s 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− U2q0 σˆ = 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 getdφmdc= (nˆ · nˆi)Uβ0∫ ∞−∞rwo−1wb dηˆ(−bc2sin(c)),dHdc= − q4U0∫∞−∞ 2wwb dηˆ∫∞−∞w2ηˆ dηˆ−∫∞−∞w2 dηˆ∫∞−∞ 2wηˆwbηˆ dηˆ(∫∞−∞w2ηˆ dηˆ)2(−bc2sin(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,wbηˆηˆ − wb +2wwb(1 + bw2)2− 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 wηˆ(0) = 0 and wbηˆ = 0 and to appropriately capture the exponentialdecay in the far-field, we prescribe a mixed boundary conditionwηˆ(Lˆ) = −w(Lˆ),where Lˆ 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 ηˆ ≥ 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 ∈ [0, R] and θ ∈ [0, 2pi]) witha localization on some circle r = r0. In this case then, the normal coordinateη would be η = r0 − r. When b = 0 and p = 2 we have that the homoclinicorbit that solves (2.22) isw(ρ) =32sech 2(ρ2),where ρ = r−r0 is the inner region radial coordinate (i.e. ρ = −ηˆ). Usingthis homoclinic orbit we can write the problem (2.19) as∆u− uD= 0, 0 ≤ r ≤ R \ r0 (6.44a)dudr= 0, r = R (6.44b)u = U0, r = r0 (6.44c)[dudr]= − 1DUβ0∫ ∞−∞wo dρ, 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= − 1r0− qU0(dudr∣∣∣∣r=r+0+dudr∣∣∣∣r=r−0)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−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−1−0.8−0.6−0.4−0.200.20.40.60.81xy  t=0t=0.119Figure 6.6: Circle evolution under the GMS model with σˆ = 0, R = 1,r0 = 1/2, D = 1, exponent set (2, 1, 2, 0) and time step 1 × 10−3. 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 10.120.130.140.150.160.170.18σU 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 10.20.210.220.230.240.250.26σU 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 σˆ = 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 ≈ 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 1−2−1.5−1−0.500.511.52r0/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 σˆ = 0. There is an unstable equilibriumat r0/R ≈ 0.044 (r0 ≈ 0.176) and a stable equilibrium at r0/R ≈ 0.69(r0 ≈ 2.76).−2.5 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.5−2.5−2−1.5−1−0.500.511.522.5xy  t=0t=0.2(a) Circle with radius r0 = 2.−4 −3 −2 −1 0 1 2 3 4−4−3−2−101234xy  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 σˆ = 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’s 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 σˆ = 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 σˆ = 10.2526.3. Solving the GMS Model−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−1−0.8−0.6−0.4−0.200.20.40.60.81xy  t=0t=0.119Figure 6.10: Circle evolution under the GMS model with σˆ = 10, R = 1,r0 = 1/2, D = 1, exponent set (2, 1, 2, 0) and time step 1 × 10−3. 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 10.060.070.080.090.10.110.12σU 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 10.080.090.10.110.120.130.14σU 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 σˆ = 10, R = 1, r0 = 1/2,D = 1, and exponent set (2, 1, 2, 0). Since σˆ 6= 0 a Newton’s method wasused to solve the analytic value.To see the effect on the motion and U0 values as σˆ 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.−0.4 −0.3 −0.2 −0.1 0 0.1 0.2 0.3 0.4−0.4−0.3−0.2−0.100.10.20.30.4xy  σˆ =0σˆ =10σˆ =30σˆ =50(a) Circle curves.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10.040.060.080.10.120.140.160.180.20.22σU 0  σˆ =0σˆ =10σˆ =30σˆ =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 × 10−2. The lines representthe numerical solution for different values of the saturation parameter σˆ(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 ∆t = ∆σ2 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 σˆ = 10. We see once again that the predictedO(∆σ2 log ∆σ) error appears. While omitted, the errors for U0, and φ alsoconverge to the appropriate order.N xerr xrat Verr Vrat Lerr Lrat CPU Time8 3.66E-03 — 2.40E-01 — 4.53E-02 — 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 σˆ = 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(∆σ2 log ∆σ) 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(σ) = cos(3σ) 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’s 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 10.20.250.30.350.40.45σU 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(3σ) 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 “flower-pedal” 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−0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8−0.8−0.6−0.4−0.200.20.40.60.8xy  t=0t=0.02t=0.04t=0.07Figure 6.14: Perturbation of a circle with perturbed radius r = 1/2 +0.1 cos(6θ) using the GMS model with R = 1, D = 1, exponent set (2, 1, 2, 0),and σˆ = 10. The boundary curve at R = 1 has been omitted.2596.3. Solving the GMS Model−0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8−0.4−0.3−0.2−0.100.10.20.30.4xy  t=0t=0.04t=0.09t=0.14Figure 6.15: Perturbation of a circle with perturbed radius r = 1/2 +0.3 cos(2θ) using the GMS model with R = 1, D = 1, exponent set (2, 1, 2, 0),and σˆ = 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−0.5 −0.4 −0.3 −0.2 −0.1 0 0.1 0.2 0.3 0.4 0.5−0.5−0.4−0.3−0.2−0.100.10.20.30.40.5xy  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 σˆ = 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−4 −3 −2 −1 0 1 2 3 4−4−3−2−101234xy  t=0t=0.2(a) Circle with centre (−1, 0) and ra-dius r0 = 1/2.−4 −3 −2 −1 0 1 2 3 4−4−3−2−101234xy  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 σˆ = 10.However, in Figure 6.18a, we place a circle of radius r0 = 1/2 off the axesentirely and centre it at [−1, 2] with saturation σˆ = 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−1.4 −1.3 −1.2 −1.1 −1 −0.9 −0.8 −0.7 −0.61.61.71.81.922.12.22.32.42.5xy  t=0t=0.06t=0.13t=0.2(a) Circle with saturation σˆ = 10.−1.3 −1.2 −1.1 −1 −0.9 −0.81.751.81.851.91.9522.052.12.152.22.25xy  t=0t=0.043t=0.052t=0.0546(b) Circle with saturation σˆ = 0.Figure 6.18: Non-concentric circle evolution with centre [−1, 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(6θ) in a larger circular domain R = 10with exponent set (2, 1, 2, 0) and saturation σˆ = 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−10 −5 0 5 10−10−8−6−4−20246810  t =0t =3t =8t =12Figure 6.19: Perturbation of a circle with perturbed radius r = 5+0.2 cos(6θ)using the GMS model with R = 10, D = 1, exponent set (2, 1, 2, 0), andσˆ = 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(−2) 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 σˆ 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 = 2∆v − v + g(u, v)τut = D∆u− u+1f(u, v)localized on closed curves in R2. Particularly we have focused our attentionon the saturated Gierer-Meinhardt model,g(u, v) =v2uq(1 + σv2), 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 σ 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 σ > 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 +εh(θ) where ε 1. Whilethis could also be studied using a polar coordinate formulation, it would re-quire carefully tracking the asymptotic consistency of both  and ε, 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(ε2). 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(θ). However, polynomial representations h(θ)α for α > 1 occurredat O(2) 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(λt) eigenfunctions since the steady-state was slowly evolving with time-scale T = −2t. 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(λ) and f(µ) 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− and m = mb+ such that a realunstable eigenvalue always exists on mb− < 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− there were no unstable eigenvalues for τ smallenough which, as τ 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(2), called small eigenvalues, which means they would only become rele-vant on the long time-scale T = O(−2). 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−1. 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−w+ 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 ofτ = τHm 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 τ , 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 τHm . 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]Γ= F (U0, V ),〈dudn〉Γ= 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  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 valueτ = τHm 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 = 2  1and Du = D = O(1). In contrast to this, there is also a weak interactionregime where Du = O(2) 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. One of the insights of [39] is that there exists a fold2757.1. Future Work and Open Problemspoint of a saddle-node bifurcation Du = Dc such that solutions do not existfor Du < Dc. It has been shown (cf. 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Journal of Computational Physics,127(2):246–267, 1996.287Appendix ADerivation of BoundaryProperties for Single LayeredPotentialsThe properties of layered potentials for Laplace’s 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’s 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∆u = 0, x ∈ R2 \ Γu = f(x), x ∈ Γwith Γ a closed curve in R2. The single layer potential for some continuousdensity φ is thenu(x) =∫ΓΦ(x, q)φ(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 Γ and define the portion of the curve Γ,Γ : {q ∈ Γ||x0 − q| < },  1.Let, x = x0 + αnˆi, where the subscript indicates that α > 0 traverses theinward pointing normal. As we near Γ,limα→0+u(x0 + αnˆi)= limα→0(∫Γ\ΓΦ(x0 + αnˆi, q)φ(q) dq +∫ΓΦ(x0 + αnˆi, q)φ(q) dq)= f(x+0 ),where we have used the Dirichlet condition. Now, since the fundamentalsolution is not singular on Γ \ Γ then we have thatf(x+0 ) =∫Γ\ΓΦ(x0, q)φ(q) dq +∫Γlimα→0+Φ(x0 + αnˆi, q)φ(q) dq, (A.1)where we have carried the limit inside the integral since we are not directlyevaluating at the singularity. On Γ, we have that q = x0 + βtˆ with − <β <  and so on this portion of the curve,Φ(x0 + αnˆi, q) = −12pilog∣∣∣√α2 + β2∣∣∣ ∼α1− 12pilog |β|+O(α2),andφ(q) ≈ φ(x0),289Appendix A. Derivation of Boundary Properties for Single Layered Potentialssince it is continuous. The second integral in (A.1) then becomes,∫Γlimα→0+Φ(x0 + αnˆi, q)φ(q) dq = −φ(x0)2pi∫ −limα→0+(log |β|+O(α2))dβ= −φ(x0)pi( log − )=→00,and therefore the singularity contributes nothing to the integral and as → 0,f(x+0 ) =∫ΓΦ(x+0 , q)φ(q) dq.Similarly, if we approach from below the curve, we havef(x−0 ) =∫ΓΦ(x−0 , q)φ(q) dqand so we have that [u] = 0 and the Dirichlet condition isf(x) =∫ΓΦ(x, q)φ(q) dq, x ∈ Γ. (A.2)Now consider what happens to the normal derivative of u as we approachthe curve from above,limα→0+∂u∂nx(x0 + αnˆi) =∫Γ\Γ∂Φ∂nx(x0, q)φ(q) dq+∫Γlimα→0+∂Φ∂nx(x0 + αnˆi, q)φ(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,∂Φ∂nx(x, q) = − 12pix− q|x− q|2 · nˆx, (A.4)where nˆx is the positively oriented normal at x0. Note that nˆx = ±nˆi de-pending on the orientation and we will proceed with nˆx = nˆi noting thatthere is a minus sign difference if nˆx is the external norm. Substituting xand q on Γ into (A.4),∂Φ∂nx(x0 + αnˆi, x0 + βtˆ) = −12piαα2 + β2.Now if β 6= 0 then for α 1,αα2 + β2∼ αβ2→ 0.However, if β = 0 thenαα2 + β2=1α→α1sgn(α)∞.Therefore,− 12piαα2 + β2= Aδ(β).To find A we integrate around β = 0,∫ 0+0−αα2 + β2dβ = −2piA.291Appendix A. Derivation of Boundary Properties for Single Layered PotentialsTo track the singularity at β = 0, take β = αb and let α → 0. Since α > 0,the integral becomes∫ ∞−∞11 + b2db = pi = −2piA. (A.5)Therefore we have that∂Φ∂nx(x0 + αnˆi, x0 + βtˆ) = −12δ(β).Substituting this into (A.3) we getlimα→0+∂u∂nx(x0 + αnˆi) =∫Γ\Γ∂Φ∂nx(x0, q)φ(q) dq − φ(x0)∫ −12δ(β) dβ=→0∫Γ∂Φ∂nx(x0, q)φ(q) dq −φ(x0)2.Now if we instead approach the curve from below then the only difference isthat the scaling in (A.5) satisfies α < 0 and so the integral becomes∫ −∞∞11 + b2db = −pi = −2piA,and instead we getlimα→0−∂u∂nx(x0 + αnˆi) =∫Γ∂Φ∂nx(x0, q)φ(q) dq +φ(x0)2.292Appendix A. Derivation of Boundary Properties for Single Layered PotentialsTherefore we have the jump conditionlimα→0+∂u∂nx(x+ αnˆi) =∫Γ∂Φ∂nx(x, q)φ(q) dq − φ(x)2x ∈ Γ, (A.6a)limα→0−∂u∂nx(x+ αnˆi) =∫Γ∂Φ∂nx(x, q)φ(q) dq +φ(x)2x ∈ Γ. (A.6b)Notice that indeed as was stated in section 6.1,[∂u∂n]Γ= limα→0+∂u∂nx(x0 + αnˆx)− limα→0−∂u∂nx(x0 + αnˆx),=limα→0+∂u∂nx(x0 + αnˆi)− limα→0−∂u∂nx(x0 + αnˆi) = −φ(x), nˆx = nˆilimα→0+∂u∂nx(x0 − αnˆi)− limα→0−∂u∂nx(x0 − αnˆi) = φ(x), nˆx = −nˆi,= −(nˆx · nˆi)φ(x).We will now finish off by looking at the tangential derivative for completeness,limα→0+∂u∂tx(x0 + αnˆi) =∫Γ\Γ∂Φ∂tx(x0, q)φ(q) dq+∫Γlimα→0+∂Φ∂tx(x0 + αnˆi, q)φ(q) dq. (A.7)Looking at the derivative on Γ,∂Φ∂tx(x0 + αnˆi, x0 + βtˆ) = −12piβα2 + β2∼α1− 12piβ.Using the Cauchy Principal Value for the integration,∫ −1βdβ = 0293Appendix A. Derivation of Boundary Properties for Single Layered Potentialsand there is no contribution on Γ. Therefore,limα→0+∂u∂tx(x+ αnˆi) =∫Γ∂Φ∂nx(x, q)φ(q) dq x ∈ Γ, (A.8a)limα→0−∂u∂nx(x+ αnˆi) =∫Γ∂Φ∂nx(x, q)φ(q) dq x ∈ Γ, (A.8b)and we get [∂u∂tx]Γ= 0,showing that the tangential derivative is continuous.294

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