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Effect of geometry on the behavior of steady Newtonian fluid in a multiply connected domain Montazer, Mohammad 2014

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Effect of geometry on thebehavior of steady Newtonianfluid in a multiply connecteddomainbyMohammad MontazerB.Sc., Sharif University of Technology, 2010A THESIS SUBMITTED IN PARTIAL FULFILLMENT OFTHE REQUIREMENTS FOR THE DEGREE OFMASTER OF SCIENCEinThe Faculty of Graduate and Postdoctoral Studies(Mathematics)THE UNIVERSITY OF BRITISH COLUMBIA(Vancouver)March 2014c? Mohammad Montazer 2014AbstractWe will start with the prototype problem of flow of a Newtonian fluid in theannular region between two infinitely long circular cylinders, with the givenvelocity and temperature on the boundaries of the domain.Then we will try to find out how does geometry would affect the behaviorof the flow inside of the domain. We will explore two invariant mappingsKT and K?, such that under appropriate conditions on the boundary, themapping KT would preserve solution of temperature field from one domainto another and the mapping k? would preserve solution of velocity field.We will prove that if a mapping is conformal, it would preserve theconvection-diffusion equation in both domains. After that, we?ll find whichsubsets of the conformals would also preserve the velocity field as well. Inorder to answer that question, we will obtain the required condition for themapping , such that it would preserve both velocity and temperature fields, from one domain to another.iiPrefaceThis dissertation is original, unpublished, independent work by the author,M. Montazer and his supervisor G.M. Homsy.iiiTable of ContentsAbstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiPreface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiiTable of Contents . . . . . . . . . . . . . . . . . . . . . . . . . . . . iv1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 Mathematical overview . . . . . . . . . . . . . . . . . . . . . . 43 Orthogonal sets of coordinates . . . . . . . . . . . . . . . . . . 73.1 Jacobian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73.2 Polar decomposition . . . . . . . . . . . . . . . . . . . . . . . 83.3 Gradient, divergence and laplacian operators . . . . . . . . . 114 Change of coordinates and conformal mapping . . . . . . . 134.1 Hermitian part of the jacobian and conformal mapping . . . . 134.2 Combination of a conformal map with general orthogonal changeof coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . 174.3 Modified polar coordinates . . . . . . . . . . . . . . . . . . . 215 Temperature field . . . . . . . . . . . . . . . . . . . . . . . . . . 255.1 Convection-diffusion equation in general orthogonal coordi-nates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 256 Velocity field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 296.1 General solution of the biharmonic . . . . . . . . . . . . . . . 296.2 Intersection of invariant of temperature and velocity field . . 307 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34ivTable of ContentsBibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35vChapter 1IntroductionThere are flows in domains and boundary conditions that exhibit internalstagnation points. The prototype problem was investigated by Ballal andRivlin (1976). They carried out a detailed analysis of the flow of a Newtonianfluid in the annular region between two infinitely long circular cylinders withparallel axes, resulting from the uniform rotation of one, or both, of thecylinders about their axes.They didn?t apply any restrictions on the geometry of the system andobtained the results with the neglect of inertial effects and for the linearizedinertial approximation. They also calculated the resultant of the forces ex-erted by the fluid on the cylinders and the distribution of their normal andtangential components over the cylinders. Finally they fully analyzed thestreamline patterns and obtained a number of conditions, under which stag-nation points, separation points and eddies can exists.Before Ballal and Rivlin, there were some studies on the problem fromdifferent standpoints. Reynolds (1886) investigated the problem from thestandpoint of lubrication theory, making the assumption that the radii ofthe cylinders are nearly equal. This restriction has become known as thelubrication approximation.Apparently, the first attempt to solve the problem without restrictingthe geometry of the system was due to Zhukoswski (1887), who initiatedthe use of a bipolar coordinate system, in discussing the problem. Froma mathematical point of view, that approach determines a streamfunctionwhich satisfies a biharmonic equation and has constant values on two circularcoordinate lines of the bipolar system.One can make another generalization to the problem and ask, if insteadof two eccentric cylinders, we have a multiply connected domain with theknown velocity on the two boundaries of the domain, how the behavior ofthe flow does not depend on the geometry of the domain. We recall thatfrom a mathematical standpoint, rotation of the cylinders would do nothingbut determining the velocity of the boundaries, and therefore the problem1Chapter 1. Introductionfor a general multiply connected domain is conceptually the same with theproblem of two eccentric cylinders.Attempts have also been made to analyze the modification in the results,which follow from taking into account the effects of inertia. Kamal (1966)was the first person who discussed the problem for arbitrary geometry of thesystem (i.e. for arbitrary ratio of the radii of the cylinders and arbitrarydistance between their axes). Sood and Elrod (1970) have used a finitedifference method to solve the full Navier-Stokes equations for the problemnumerically.Its well known that time modulation of such flows results in chaotic ad-vection if the steady flow contains internal stagnation points. Aref explainedthe development of the chaotic advection in his famous paper in 2001. Whena particle moves with the fluid, we speak of advection, sometimes passiveadvection to emphasize that the particle is so light and inert that it can donothing but follow the fluid.The particle velocity is given by the rate of change of its position, andthe fluid velocity is given by other considerations, which involves the solu-tion of some set of partial differential equations, such as the Euler equations,the Navier-Stokes equations or the Stokes equations. For 2D, incompress-ible flow, the velocity is derived from a streamfunction, ?. Combining thesewith the advection equations turns the problem into something we recog-nize from dynamics, namely Hamilton?s canonical equations for a one degreeof freedom systems. In short, two-dimensional kinematics of advection byan incompressible flow is equivalent to the Hamiltonian dynamics of a one-degree-of-freedom system. Since the advection equations are kinematic innature, this applies regardless of whether the dynamics of the fluid itself isviscous or inviscid.It was immediately clear that the context in which to look for chaoticadvection experimentally was viscous flows, in fact, Stokes flow, in which theentire flow field is prescribed by the motion of the boundaries, seemed ideallysuited. The stokes flow between rotating cylinders, one situated eccentricallywithin the other, was a perfect candidate.Chaotic advection has seen numerous applications in diverse areas of fluidmechanics. Some of the most important may be those addressing the stirringof fluids on geophysical or planetary scales. Numerical simulations of stirringdue to convective motions in Earth?s mantle show every sign of chaotic ad-vection, in the sense of folds and layers and smaller and smaller scales. Onecan mention the most significant advance in the theory in this area, as the2Chapter 1. Introductionuse of ideas from topology to ?build in? chaotic advection in the design ofstirring devices.Another important study in this area is the paper of Kaper and Wiggins.They applied new tools from the field of adiabatic dynamical systems the-ory to make quantitative predictions of various important mixing quantitiesin quasi-steady Stokes flows which possess slowly varying saddle stagnationpoints. Many of those quantities can be obtained before experiments or nu-merical simulations are performed using only knowledge of the streamlinesin steady state flows and externally determined flow parameters.Study of transport of a passive scalar such as temperature or concentra-tion (T,C) in such flows is only just a beginning. There exists the possibilityof significant increase in transport rates. Christov and Homsy considered theproblem of transport of heat or mass from circulating droplets that are bothsettling and subject to an axial electric field. The electric field can be eithersteady or oscillatory in time and drives an electrohydrodynamic flow, calledthe Taylor circulation, which augments the Hadamard circulation caused bysteady translation.They solved the convective diffusion equation numerically, by an effectivefinite-difference scheme that allows a wide range of parameters to be covered.The results were characterized by the asymptotic rate of extraction of heator mass from the droplet. The enhancement factor was defined as the ratioof this rate to that of a stagnant drop. For steady drops, they found thattransport remains diffusion controlled, but the enhancement factor was sig-nificantly higher with the Taylor flow than without. For modulated electricfield the enhancement factor was not a simple function of parameters andexhibited spectral ?resonant peaks? at particular values of w for which theenhancement factor was extremely large.There are two main questions for the problem of flow in a multiply con-nected domain. The first one is if chaotic advection in this class of domainsis a general feature or Balal and Rivlin is an special case of. The secondquestion is how to formulate a general class of flow transport problems incomplex time dependent non-trivial geometries. In the following chapters,we will answer these two questions.3Chapter 2Mathematical overviewWe want to find out effect of geometry on boundary value problems for somedifferential operators. We focus on two operators, LT which describes thetemperature field and L? which describes the streamfunctions and thereforethe velocity field on domain. Both operators are assumed to act on a multiplyconnected domain D ? R2 which has two boundaries ?1,?2. As it is clear,LT and L? are scalar field operators. The velocity at each point of thedomain is obtained easily from the streamfunction. So we can think of L?as an operator describing the velocity field.If f1, f2 are the temperature functions on the boundaries ?1,?2 and g1, g2are the velocity functions on ?1,?2 , then those boundary conditions woulduniquely determine the temperature field T and the streamfunction ? onthe whole domain. Thus we can explore the important of geometry and theboundary conditions on this problem. We are looking for two mappings calledK? and KT , such that if K? maps one domain with its boundary conditionsto the second domain with its boundary conditions, then it would map thecomplete streamfunction from one domain to the other domain. The sameconsideration hold for KT ; if it maps the first domain with its boundary con-ditions to the second domain with its boundary conditions, then it would mapthe whole temperature field from one domain to the other domain. Usuallythere are two different ways for solving these sorts of differential equationswith complicated boundaries, depending on the set of coordinates we employ.One way is to use cartesian coordinates. In this case, the boundaries ?1,?2are non-trivial functions of our coordinates and thus we have to solve theproblem using some numerical techniques. Such methods can hardly help usin our goal. The second way to solve these problems, is to change coordinatesin such a way as to allow analytical treatment. Such a change of coordinateswould usually map our initial domain to a strip; therefore the boundaries inthe new coordinates would be trivial, and it just remains to find the equiva-lent differential operators in the new coordinates. For some technical reasons,representation of differential operators in orthogonal coordinates are much4Chapter 2. Mathematical overvieweasier than non orthogonal ones; thus we will use orthogonal coordinates(?, ?). We will express the different operators in the new coordinates whichdepend on two parameters called H?, H? which are scaling functions of thenew coordinates. We will show that they are equal to the hermitian partof polar decomposition of jacobian. Thus for each operator, we have twodegrees of freedom which come from H? and H?.Lets ?(?, ?) = 0 be the differential equation of the operator L. So ?would count as one degree of freedom. The second degree of freedom comesfrom the geometry. As explained before, lets K be the invariant map whichpreserves the solution of the operator L on two different domains. K will bethe second degree of freedom that we are looking for.For summary, we recall that we have an operator which is a function ofa differential equation and some boundary conditions, say L(?, f1, f2) whereL is the operator, ? is the differential equations and f1, f2 are the boundaryconditions. This problem is a balance between two parameters (H?, H?)which come from our coordinate transformation and two parameters (?,K)where ? = 0 is the differential equation and K is an invariant on the solutionon L and comes from geometry. So we have:{H? , scale factorH? , scale factor}??{? = 0 , differential equationK , geometric invariant on solution of L}(2.1)Now let?s look at our own problem. We have two operators LT , L? withdifferential equations and boundary conditions ?T , f1, f2 and ??, g1, g2 re-spectively.For the streamfunction ?, we don?t have any problem and we can writeL?(??, g1, g2) = 0 and we are looking for a geometric invariant on the solu-tion of L? called K?. That?s a difficult but straightforward problem. Thesituation is a bit more complicated for the temperature operator LT . LetV = V (?, ?) be the velocity field obtained from the streamfunction ?. ThenV is a vector field on the same domain as L?. For the temperature fieldLT , not only does it depend on ?T , f1, f2 but also on V = V (?, ?); in otherwords we have LT (?T , V, f1, f2). This means that the two differential equa-tions ?T , ?? are coupled to each other. So one cannot directly talk about aninvariant KT just in terms of the ?T and the temperature boundary condi-tions f1, f2, because they all depend on the velocity field, and the velocitydepends on the operator L?(??, g1, g2).The way we approach the problem is by breaking the coupled prob-5Chapter 2. Mathematical overviewlem into two independent problems, one for L?(??, g1, g2) and the other forLT (?T , f1, f2). We first solve the problem for LT and then for L?.In the first problem, we treat V (?, ?) as a given vector field, exactly likethe given functions f1, f2. Determining the velocity field, we would have asituation like equation (2.1), and we would try to find a geometric invariantKT , such that if it maps the velocity field V (?, ?) as well as the temperatureboundary conditions f1, f2 from one domain to the other, then it would mapthe whole temperature field from one domain to the other. In the secondproblem, we would look for a geometric invariant K?, such that it wouldpreserve the solution of L? and would map streamlines and velocity field fromone domain to another. We also know that KT would map the temperaturefield from one domain to another, if and only if , KT would also map thestreamlines and velocity field from one domain to the other.We recall defining velocity boundary conditions (g1, g2) would uniquelydefine velocity field in the whole domain and any element of K? would mapvelocity field from one domain to another.Once the velocity field is given, defining temperature boundary conditions(f1, f2) would uniquely define temperature field in the whole domain and anyelement of KT would map the whole temperature field from one domain toanother.Considering these two points, we can conclude once we define velocityand temperature boundary conditions (f1, f2, g1, g2) , any element of K??KTwould map both velocity and temperature fields, from one domain to another.6Chapter 3Orthogonal sets of coordinatesLet (?, ?) be a change of coordinates. In order to find convection-diffusionequation in the new coordinates i.e. ?T = 0, ?? = 0 in the new coordinates,we first have to find the equivalent gradient , divergence and laplacian opera-tors in our new system. We start with the important concept of the jacobianoperator and its polar decomposition for a coordinate transformation .3.1 JacobianAssume{? = ?(x, y)? = ?(x, y)(3.1)is a coordinate transformation, and the unit vectors in the (x, y), (?, ?) coor-dinates are (~ex, ~ey), (~e?, ~e?) respectively. Let the matrix J , be the jacobian ofour coordinate transformation:J =??????(x, y)?x??(x, y)?y??(x, y)?x??(x, y)?x???? =(?x ?y?x ?y)(3.2)Now lets find (~e?, ~e?) in terms of (~ex, ~ey) and the jacobian. If we define:{H? =??2x + ?2yH? =??2x + ?2y(3.3)73.2. Polar decompositionThen we have:~e? =???x~ex +???y~ey?(???x)2+(???y)2=?x~ex + ?y~ey??2x + ?2y=?x~ex + ?y~eyH?(3.4)~e? =???x~ex +???y~ey?(???x)2+(???y)2=?x~ex + ?y~ey??2x + ?2y=?x~ex + ?y~eyH?(3.5)So in matrix notation we have:(~e?~e?)=????xH??yH??xH??yH????(~ex~ey)(3.6)or equivalently:(H?~e?H?~e?)=(?x ?y?x ?y)(~ex~ey)= J.(~ex~ey)(3.7)While orthogonality is an important concept, let?s find out an equivalentidentity for ~e?.~e? = 0. We have:~e?.~e? = 0 ?? ~e?.~e? =(?xH?~ex +?yH?~ey)(?xH?~ex +?yH?~ey)=1H?.H?(?x?x + ?y?y) = 0 ?? ?x?x + ?y?y = 0 (3.8)3.2 Polar decompositionWe recall from linear algebra that the polar decomposition of a square com-plex matrix J , is of the form J = UH, where U is a unitary matrix, and His a positive-semidefinite Hermitian matrix.Intuitively, the polar decomposition separates J , into a component thatstretches the space along a set of orthogonal axes, represented by H, and a83.2. Polar decompositionrotational component represented by U . The decomposition always exists,and so long as J is invertible, it is unique, with H positive definite. Onecan also decompose J in the form J = H ?U where U is the same as beforeand H ? = UHU?1 is again a positive-semidefinite Hermitian matrix. Thefirst and second decomposition are known as the right polar decompositionand left polar decomposition, respectively. Here we work with the left polardecomposition of the jacobian. Again we recall that two matrix (H,U) are(Hermitian,unitary) if and only if:{H = H?T (hij = h?ji) , H? is conjugate of H.UU?T = U?TU = I , U? is conjugate of U.(3.9)As we are working with the space of real valued functions, H? = H andU? = U , thus left polar decomposition of J becomes:???J = HUHT = H , H is HermitianUT = U?1 , U is Unitary(3.10)In two dimensional space, a unitary matrix would be defined by a polar angleX(?, ?) or X(x, y) (depending on our coordinate) and would be in one of thefollowing forms:UR =(cos(X) sin(X)sin(X) ? cos(X))(Reflective unitary operator) (3.11)orUN =(cos(X) ? sin(X)sin(X) cos(X))(Non? Reflective unitary operator)(3.12)The notation we used (Reflective/Non-Reflective unitary operator) is notstandard, but a reflection is actually a reflective unitary operator, while ro-tation and transport are non-reflective unitary operators. In fact, if theeigenvalues of matrix J have the same sign, the corresponding unitary ma-trix for J would be non-reflective and if the eigenvalues have opposite signs,the corresponding unitary matrix of J would be reflective. This easily followsfrom the fact that:sign(|J |) = sign(|H|.|U |) = sign(|H|).sign(|U |) = sign(|U |) = |U |?? sign(|J |) = |U | (3.13)93.2. Polar decompositionIn the above argument, we used the fact that sign(|H|) = +1 because His positive-semidefinite and sign(|U |) = |U | because U is unitary. As wementioned in (3.10), the Hermitian part of polar decomposition of jacobianfor a general coordinate transformation, is a 2 ? 2 symmetric matrix of theform:H =(H1 H??H?? H2)(3.14)We want to show that if (?, ?) are orthogonal coordinates, i.e. equation (3.8)holds, using notations defined in (3.3), the Hermitian matrix H would beequal to:H =(H? 00 H?)(3.15)In order to prove (3.21) , first recall:JJT = (HU)(HU)T = (HU)(UTHT ) = H(UUT )HT = H.I.H = H2=(H1 H??H?? H2)(H1 H??H?? H2)=(H21 +H2?? (H1 +H2)H??(H1 +H2)H?? H22 +H2??)(3.16)On the other hand we have:JJT =(?x ?y?x ?y)(?x ?x?y ?y)=(?2x + ?2y ?x?x + ?y?y?x?x + ?y?y ?2x + ?2y)=(H2? ?x?x + ?y?y?x?x + ?y?y H2?)(3.17)Now if (?, ?) are orthogonal, then (3.8), (3.16), (3.17) imply:JJT=(H2? ?x?x + ?y?y?x?x + ?y?y H2?)=(H2? 00 H2?)=(H21 +H2?? (H1 +H2)H??(H1 +H2)H?? H22 +H2??)??(H2? 00 H2?)=(H21 +H2?? (H1 +H2)H??(H1 +H2)H?? H22 +H2??)(3.18)From equation (3.18) we conclude(H1 +H2)H?? = 0 (3.19)103.3. Gradient, divergence and laplacian operatorsSo we have:???H1 = ?H2orH?? = 0(3.20)The first case cannot happen because H is positive semidefinite, thus wehave: H?? = 0 and from (3.18) we have:(H2? 00 H2?)=(H21 00 H22)?? H =(H1 00 H2)=(H? 00 H?)(3.21)This last equality follows from the fact that H is positive semidefinite andalso the way we defined H?, H? in (3.3).We want to highlight our last equality in the following theorem:Theorem 1. If???? = ?(x, y)? = ?(x, y)is an orthogonal set of coordinate, J =(?x ?y?x ?y)is the jacobian of the coordinate transformation and J = HU is the left polardecomposition of J , then for the Hermitian matrix H we have:H =(H? 00 H?)=(??2x + ?2y 00??2x + ?2y)(3.22)3.3 Gradient, divergence and laplacianoperatorsNow we want to determine different operators in the new set of coordinates(?, ?). For this purpose, we define new operators as follows:????? ? gradient in (?, ?) coordinate (3.23)???? ? divergence in (?, ?) coordinate (3.24)?2?? ? laplacian in (?, ?) coordinate (3.25)(F1, F2)?? ? F1~e? + F2~e? (3.26)113.3. Gradient, divergence and laplacian operatorsFor the gradient we have:??? =(??x,??y)xy=(??x??y)1?2(~ex~ey)2?1=(???.???x+???.???x???.???y+???.???y)1?2(~ex~ey)2?1=(??????)1?2(?x ?y?x ?y)2?2(~ex~ey)=(??????)1?2(H?~e?H?~e?)=(H????,H????)????????? =(H????,H????)??(3.27)where we used (3.7) in the last identity.For the divergence, we consider a vector field function~F = Fx~ex + Fy~ey = F?~e? + F?~e? (3.28)we know:~?xy ? ~Fxy =?Fx?x+?Fy?y(3.29)and we want to find the equivalent operator in the (?, ?) coordinate. Therequired computations are complicated and we quote the results of chapter4 of reference [14].??? ? F = H?H?[???(F?H?)+???(F?H?)]. (3.30)Finally for the laplacian we have:?2??=??? ? ~???=??? ?(H????,H????)??=H?H?[???(H?H????)+???(H?H????)]?? ?2??() = H?H?[???(H?H????())+???(H?H????())]. (3.31)12Chapter 4Change of coordinates andconformal mappingIn this chapter we follow two goals. The first is to establish conditions forwhich the matrix H in theorem 1 is a multiple of the identity matrix, or inother words, when do we have H? = H?. We show that this is the case, ifand only if the change of coordinates (x, y)? (?, ?) is a conformal mapping.Our second goal is to see, if one has an orthogonal change of coordinates(?, ?) and we send it by a conformal map to new coordinates (r, s), how doesit affect the Hermitian part of the jacobian in the new coordinates (r, s).4.1 Hermitian part of the jacobian andconformal mappingConsider an orthogonal change of coordinates (?, ?) and assumeH=(H? 00 H?)is its Hermitian matrix, we want to see when does the equality H? = H?holds, or in other words, it holds if and only if (?, ?) comes from a change ofcoordinates which is conformal.Theorem 2. Let f be a one-to-one mapping of an open set U in R2 onto f(U)such that for each x ? U, Jf (x) is non-singular where Jf (x) is the jacobianat point x. Then f is conformal if and only if, for all vectors v, w ? R2,?Jf (x) ? v, Jf (x) ? w? = e2?(x)?v, w? (4.1)for some real-valued function ? on U .Proof. Suppose ?Jf (x) ? v, Jf (x) ? w? = e2??v, w?. If ? is the angle between vand w and ? the angle between Jf (x) ?v and Jf (x) ?w, then, since |Jf (x) ?v| =134.1. Hermitian part of the jacobian and conformal mappinge?|v|,cos ? =?Jf (x) ? v, Jf (x) ? w?|Jf (x) ? v||Jf (x) ? w|=e2??v, w?e?|v|e?|w|=?v, w?|v| ? |w|= cos ? (4.2)so that f is angle preserving. Conversely, suppose that f is angle preserving,that, at each point x ? U,B(v, w) = ?Jf (x) ? v, Jf (x) ? w? is a symmetricbilinear form which vanishes on orthogonal pairs v, w. Thus ?Jf (x)?v, Jf (x)?w?is proportional to ?v, w? at each point. Again since f is angle preserving,?Jf (x) ? v, Jf (x) ? w? and ?v, w? have the same sign. Thus, denoting theproportionality factor by ~e2?(x), we have ?Jf (x) ? v, Jf (x) ? w? = e2?(x)?v, w?as desired.Note 1: The above theorem and our argument for proving it, is valid forany open set U in Rn instead of R2, but we wish to use it in another theoremwhich is only valid in R2, and our original problem is also in R2. Thus wedefined it in R2 instead of Rn.Note 2: In the above theorem, the function e? is called the characteristicfunction of the conformal map f , and later we will show that it is equal toe?(x) = H? = H?. Furthermore, e?(x) =|Jf (x) ? v||v|for all v 6= 0.Note 3: In theorem 2, we used a property of a symmetric bilinear formon Rn which we will prove in the following theorem. Again we?ll expressthe theorem in R2, but like theorem 2, whenever we use v, w ? R2, we canreplace it with v, w ? Rn and the argument is still valid.Theorem 3. Let B be a symmetric bilinear form on R2 which vanishes onall orthogonal pairs of vectors. Then there exists a real number ? such thatB(v, w) = ??v, w?Proof. For fixed v ? R2, v 6= 0, letL(w) = B(v, w)?B(v, w)?v, v??v, w? (4.3)Then L(v) = 0 and L(w) = 0 for every w orthogonal to v; thus L vanisheson a basis and L = 0. Similarly, holding w fixed, we haveB(w, v)?B(w,w)?w,w??w, v? = 0 (4.4)144.1. Hermitian part of the jacobian and conformal mappingfor all v. ThereforeB(v, v)?v, v?=B(w,w)?w,w?(4.5)for all non-zero vectors v, w. Calling this common value ?, we have B(v, w) =??v, w?.In the next step, we want to show another equivalency for the conformalmappings. There we use the well-known ?inverse function theorem? for thejacobian. So we recall it belowTheorem 4. Let f be a C1 , Rn-valued function defined on an open setU ? Rn, and let x0 ? U . Suppose |Jf (x0)| 6= 0. Then there exists an open setV containing x0 and an open set W containing f(x0) such that f : V ?? Wis one-to-one and onto. Moreover, f?1 : W ? V is of class C1.The proof of the ?inverse function theorem? is available in any standardtextbook of multivariable calculus. Now combining it with theorem 2, weprove our main theorem for conformal mapping.Theorem 5. Let f : U ? R2 ?? R2 be of class C1 with non-vanishingjacobian. Then f as a map is conformal if and only if f , as a function ofthe complex variable z, is holomorphic or anti-holomorphic.Proof. By the inverse function theorem, f is locally one-to-one. Accordingto theorem 2, we must show that f satisfies?Jf (z) ? V, Jf (z) ?W ? = e2?(z)?V,W ? (4.6)for some function ? and for all vectors V and W , if and only if the componentfunctions u and v of f satisfy?u?x= ??v?y,?u?y= ??v?x. (4.7)Recall that Jf (z) ? V is given by????u?x?u?y?v?x?v?x???(v1v2)=????u?xv1 +?u?yv2?v?xv1 +?v?yv2??? (4.8)154.1. Hermitian part of the jacobian and conformal mappingand ?Jf (z) ? V, Jf (z) ?W ? = e2?(z)?V,W ? becomes(?u?xv1 +?u?yv2)(?u?xw1 +?u?yw2)+(?v?xv1 +?v?yv2)(?v?xw1 +?v?yw2)= e2?(z) (v1w1 + v2w2) (4.9)If this holds for all V and W in R2, then, choosing V = W =(10)andV =W =(01)we have(?u?x)2+(?v?x)2= e2?(z) ,(?u?y)2+(?v?y)2= e2?(z) (4.10)Similarly, setting V =(10)and W =(01)yields?u?x?u?y+?v?x?v?y= 0 (4.11)Equating the two expressions in (4.10), multiply by(?v?y)2, and using (4.11),we have:(?u?y)2(?v?y)2+(?v?y)4=(?u?x)2?(?v?y)2+(?v?x)2?(?v?y)2=(?u?x)2?(?v?y)2+(?u?x)2?(?u?y)2(4.12)Therefore?u?x= ??v?y. If?u?x=?v?y, (4.11) gives?u?y= ??v?xunless?u?x=?v?y= 0, in which case (4.10) gives?v?x= ??u?y. similarly, if?u?x= ??v?y,then?u?y=?v?xunless?u?x=?v?y= 0, and then?v?x= ??u?y. Consequently,conformality implies holomorphy or anti-holomorphy. Conversely, substitu-tion of (4.7) into the left side of (4.9) yields the right side of (4.9) withe2? =(?u?x)2+(?u?y)2164.2. Combination of a conformal map with general orthogonal change of coordinatesTheorem 5 says a mapping f is conformal if and only if it is either holo-morphic i.e. J =(?x ?y??y ?x)or anti-holomorphic i.e. J =(?x ?y?y ??x).Now our next step is to show, for the Hermitian matrix of the jacobian i.e.H =(H? 00 H?), H? = H? if and only if the jacobian is in one of those forms.But that is also trivial; since (?, ?) is orthogonal, then from (3.8) we have :?y = ??x?x?y, and we can conclude:H2? ?H2? = (?2x + ?2y)? (?2x + ?2y) = (?2x + ?2y)?(?2x +?2x?2x?2y)= (?2x + ?2y)??2x?2y(?2x + ?2y)=(?2x + ?2y)?2y(?2y ? ?2x)(4.13)Thus we have H? = H? if and only if ?2y = ?2x.In addition, ?2x + ?2y = H2? =H2? = ?2x + ?2y we can conclude H? = H? if and only if{?2x = ?2y?2y = ?2x. Addingorthogonality (3.8) to this equation we have H? = H? if and only if J =(?x ?y??y ?x)or J =(?x ?y?y ??x)which is equivalent to holomorphy or anti-holomorphy. Together with theorem (4.4),we can conclude a mapping isconformal if and only if Hermitian part of jacobian is a multiple of identity.We highlight this result in the following theorem.Theorem 6. Let f :{U ? R2 ?? R2(x, y) ?? (?, ?)be of class C1 with non-vanishingjacobian. Then f as a map is conformal if and only if the Hermitian partof the left polar decomposition of jacobian is a multiple of the identity i.e.HJ =(H? H??H?? H?)= H(1 00 1).4.2 Combination of a conformal map withgeneral orthogonal change of coordinatesLet (?, ?) be an orthogonal change of coordinates and f a conformal mapwhich sends (?, ?) to a new set of coordinates (r, s) . While conformals are174.2. Combination of a conformal map with general orthogonal change of coordinatesangle preserving, we can conclude (r, s) is also an orthogonal set of coor-dinates. If H??, Hrs, Hf are the Hermitian parts of the left polar decom-position of the jacobian for the following mappings respectively: (x, y) ??(?, ?) , (x, y) ?? (r, s) , (?, ?) ?? (r, s), then we want to find out a rela-tionship between H?? , Hrs and Hc . We first start with the jacobian of thosetransformations. Assume we have the following transformations:???(x, y) ?? (?, ?) , J?? = H?? ? U??(?, ?) ?? (r, s) , Jf = Hf ? Uf(x, y) ?? (r, s) , Jrs = Hrs ? Urs(4.14)then we have:J?? =(?x ?y?x ?y), Jrs =(rx rysx sy), Jf =(r? r?s? s?)(4.15){rx = r??x + r??xry = r??y + r??y??(rx ry)=(r? r?)(?x ?y?x ?y)(4.16){sx = s??x + s??xsy = s??y + s??y??(sx sy)=(s? s?)(?x ?y?x ?y)(4.17)(4.18)combining these two together we have:(rx rysx sy)=(r? r?s? s?)(?x ?y?x ?y)?? Jrs = Jf ? J?? (4.19)Now we have a relationship between the jacobian of those transformations,but our ultimate goal is to find a relationship between H?? , Hrs and Hf .Therefore, let?s combine (4.19) with (4.14) and recall the fact that (r, s) isan orthogonal set of coordinates. If we define:H?? =(H? 00 H?), Hrs =(Hr 00 Hs), Hf =(Hc 00 Hc)(4.20)then we have:H2rs = Jrs ? JTrs = (Jf ? J??)(Jf ? J??)T = Jf ? (J?? ? JT??) ? JTf(Hf ? Uf )(H2??) (UTf HTf)= H2c ? Uf(H2? 00 H2?)U?1f=(H2r 00 H2s)?? Uf(H2? 00 H2?)U?1f =(H2rH2c00 H2sH2c). (4.21)184.2. Combination of a conformal map with general orthogonal change of coordinatesSince Uf is unitary, it is uniquely determined by a function Xf , if f is holo-morphic let?s set Uf = Ufh and if anti-holomorphic, let?s set Uf = Ufa. Ineach case we have:Ufh =(cos(Xf ) sin(Xf )? sin(Xf ) cos(Xf )), Ufa =(cos(Xf ) sin(Xf )sin(Xf ) ? cos(Xf ))(4.22)Substituting (4.22) into (4.21), for f holomorphic we have:(H2rH2c00 H2sH2c)=(cos(Xf ) sin(Xf )? sin(Xf ) cos(Xf ))(H2? 00 H2?)(cos(Xf ) ? sin(Xf )sin(Xf ) cos(Xf ))=(cos(Xf ) sin(Xf )? sin(Xf ) cos(Xf ))(H2? cos(Xf ) ?H2? sin(Xf )H2? sin(Xf ) H2? cos(Xf ))=(H2? cos2(Xf ) +H2? sin2(Xf )(H2? ?H2?)cos(Xf ) sin(Xf )(H2? ?H2?)cos(Xf ) sin(Xf ) H2? sin2(Xf ) +H2? cos2(Xf ))(4.23)and for f anti-holomorphic we have:(H2rH2c00 H2sH2c)=(cos(Xf ) sin(Xf )sin(Xf ) ? cos(Xf ))(H2? 00 H2?)(cos(Xf ) sin(Xf )sin(Xf ) ? cos(Xf ))=(cos(Xf ) sin(Xf )sin(Xf ) ? cos(Xf ))(H2? cos(Xf ) H2? sin(Xf )H2? sin(Xf ) ?H2? cos(Xf ))=(H2? cos2(Xf ) +H2? sin2(Xf )(H2? ?H2?)cos(Xf ) sin(Xf )(H2? ?H2?)cos(Xf ) sin(Xf ) H2? sin2(Xf ) +H2? cos2(Xf ))(4.24)In order to satisfy (4.23) or (4.24), we have three possibilities as follows:???sin(Xf ) = 0cos(Xf ) = 0H? = H?(4.25)194.2. Combination of a conformal map with general orthogonal change of coordinatesif sin(Xf ) = 0 then(H2r 00 H2s)= H2c(H2? 00 H2?)(4.26)if cos(Xf ) = 0 then(H2r 00 H2s)= H2c(H2? 00 H2?)(4.27)if H? = H? then H2r = H2s = H2cH2? = H2cH2?(4.28)Now let us consider each of the cases. The most trivial one is the third one,it says that if (x, y) ?? (?, ?) and (?, ?) ?? (r, s) are both conformal maps,then the combination of them (x, y) ?? (r, s) is also conformal, with theidentity Hr = H?Hf = Hs for the scale functions.If sinXf = 0 then cosXf = ?1 and for the jacobian we have:Jf =(r? r?s? s?)= Hc(?1 00 ?1)=(?Hc 00 ?Hc), this means r? = s? = 0. Inother words r= r(?), s = s(?), and it implies that the conformal map sendslevel curves{? = const.? = level curves{r = const.s = const.or in other wordsthe mapping f : (?, ?) ?? (r, s) preserves the order of coordinates.If cos(Xf ) = 0 then sin(Xf ) = ?1 and for the jacobian we have: Jf =(r? r?s? s?)= Hc(0 ?1?1 0)=(0 ?Hc?Hc 0), this means r? = s? = 0, inother words r = r(?), s = s(?), and it means the conformal map sends levelcurves{? = const.? = level curves{s = const.r = const.or in other words themapping f : (?, ?) ?? (s, r) reverse the order of coordinates.We can summarize our results in the following theorem:Theorem 7. Lets{(x, y) ?? (?, ?)(x, y) ?? (r, s)be two orthogonal changes of coor-dinates in a way that there is a conformal map (?, ?) ?? (r, s) such thatfrs = fof??. Assume the scaling matrix (i.e. Hermitian part of left polardecomposition of the jacobian) for each transformation is as follows:H?? =(H? 00 H?), Hrs =(Hr 00 Hs), Hf =(Hc 00 Hc)204.3. Modified polar coordinatesIf f preserves the order of coordinates (i.e. it maps{? = const.? ={r = const.s = const.respectively), then we have Hrs = Hc(H? 00 H?)otherwise if freverses the order of coordinates (i.e. it maps{? = const.? ={s = const.r = const.respectively), then we have Hrs = Hc(H? 00 H?).4.3 Modified polar coordinatesHere we want to modify polar coordinates in a way that the mapping (x, y) ??(? = ?(r), ? = ?(?)) is conformal. Assume{f(x, y) = x2 + y2 = ?(r, ?) = ?(r)g(x, y) =yx= ?(r, ?) = ?(?)(4.29)This means we have a mapping (x, y) ?? (?, ?) such that the first coordinates?, is just a function of r and the second coordinates ? , is just a function of?. While (r, ?) are an orthogonal set of coordinates, we can easily conclude(?, ?) are orthogonal too. We will show this fact, using the jacobian of thecoordinate transformation (?, ?). We have:?f?x=?f??????x+?f??????x= f? ? ?x + 0 ? ?x = f? ? ?x = fx (4.30)?f?y=?f??????y+?f??????y= f? ? ?y + 0 ? ?y = f? ? ?y = fy (4.31)?g?x=?g??????x+?g??????x= 0 ? ?x + g? ? ?x = g? ? ?x = gx (4.32)?g?y=?g??????y+?g??????y= 0 ? ?y + g? ? ?y = g? ? ?y = gy (4.33)Putting (4.29) into above equations we have:?x =fxf?=2xf?, ?y =fyf?=2yf?(4.34)?x =gxg?=?yx?2g?, ?y =gyg?=x?1g?(4.35)214.3. Modified polar coordinatesSubstituting (4.34), (4.35) into the jacobian we have:J =(?x ?y?x ?y)=(fxf?fyf?gxg?gyg?)=(1f?00 1g?)(fx fygx gy)(4.36)Since J ? JT = H2, we have:H2 = J ? JT =???1f?001g????(fx fygx gy)(fx fygx gy)T???1f?001g????=???1f?001g????(fx fygx gy)(fx gxfy gy)???1f?001g????=???1f?001g????(f 2x + f2y fxgx + fygyfxgx + fygy g2x + g2y)???1f?001g????=???1f?001g????(f 2x + f2y 00 g2x + g2y)???1f?001g????=?????f 2x + f2yf 2?00g2x + g2yg2??????=(H2? 00 H2?)= H2 (4.37)In the above, we used the equality fxgx + fygy = 0 by replacing the corre-sponding terms from (4.34), (4.35), this means (?, ?) are an orthogonal setof coordinates as desired. Now the mapping (x, y) ?? (?, ?) is conformal, ifand only if in (4.37), H? = H?. Thus the mapping is conformal if and onlyif the following identities hold:1 =H2?H2?=f 2x + f2yf 2?g2x + g2yg2?=f 2x + f2yg2x + g2y?g2?f 2???f 2x + f2yg2x + g2y=f 2?g2?= D2(?, ?).(4.38)224.3. Modified polar coordinatesEquation (4.38) means we have to start with D2(?, ?) =D21(f)D22(g), then wewould have{D21(f) = f2?D22(g) = g2?and we would find (?, ?) by a simple integrationin terms of (?, ?) as follows:???????D21(f) =(dfd?)2D22(g) =(dgd?)2 (4.39)After obtaining (?, ?) from (4.39), we can find the scale function using thefollowing identity:H2 =f 2x + f2yf 2?=g2x + g2yg2?. (4.40)In our specific case we have:D2(f, g) =f 2x + f2yg2x + g2y=4(x2 + y2)x?4(x2 + y2)=4x?4= 4x4 ?? D2(f, g) = 4x4(4.41)while{f = x2 + y2g =yxso we have f = (1 + g2)x2, substituting it into (4.41)we obtain:D2(f, g) =4x4 = 4(x2)2 = 4(f1 + g2)2=(2f)2(1 + g2)2=f 2?g2????????dfd?= (2f)dgd?= (1 + g2)???????d? =df2fd? =dg1 + g2???????? =? df2f=Lnf2=Ln(x2 + y2)2=ln(r2)2= ln(r)? =? dg1 + g2= arctan(g) = arctan(yx) = ?(4.42)234.3. Modified polar coordinatesso the desired conformal map is (x, y) ?? (? = ln r, ? = ?). In order to findthe scale functions, while e2? = x2 + y2 and tan(?) =yx, differentiating thesetwo equations with respect to (x, y) we have:{2?xe2? = 2x2?ye2? = 2y?? H2? = ?2x + ?2y =(e?2?x)2+(e?2?y)2= e?4?(x2 + y2)= e?4?e2? = e?2? =? H? = e?? (4.43)????????x (1 + tan2(?)) = ?yx2= ?e? sin ?e2? cos2(?)=?xcos2(?)?y (1 + tan2(?)) =1x=1e? cos(?)=?ycos2(?)??{?x=?e??sin ??y=e??cos ??? H2? = ?2x + ?2y = e?2?(sin2(?) + cos2(?))= e?2? ?? H? = e??(4.44)Since H? = H? we can see the mapping (x, y) ?? (ln r, ?) is conformal andthe scale function isH = e?? =1r. (4.45)24Chapter 5Temperature field5.1 Convection-diffusion equation in generalorthogonal coordinatesHere we want to express the convection-diffusion equation in a general set ofcoordinates (?, ?). We know the equation in cartesian coordinates isPe ~v.~?T = ?2T (5.1)In two dimensions, velocity comes from the streamfunction. Thus we have~v =???y~ex ????x~ey (5.2)and substituting (5.2) into equation (5.1) we obtain:Pe(???y~ex ????x~ey)(?T?x~ex ??T?y~ey)= ?2T (5.3)Now we want to find the equivalent equation of (5.3) in a general set of coor-dinates. While in (3.31) we obtained the laplacian in a general orthogonal setof coordinates, it remains to find the left hand side in the (?, ?) coordinates.We have:~v =???y~ex ????x~ey =(???y????x)(~ex~ey)=(???????y+???????y????????x????????x)(~ex~ey)=(?????????)(?y ??x??y ?x)(~ex~ey)=(?????????)(?y ??x??y ?x)(?x ?y?x ?y)?1(H?~e?H?~e?)=(?????????)(?y ??x??y ?x)?1?x?y ? ?y?x?(?y ??y??x ?x)(H?~e?H?~e?)=1?x?y ? ?y?x(?????????)(?2y + ?2x ??y?y ? ?x?x??y?y ? ?x?x ?2x + ?2y)(H?~e?H?~e?)=1?x?y ? ?y?x(?????????)(H2? ?(H1 +H2)H???(H1 +H2)H?? H2?)(H?~e?H?~e?)=sign(|J |)(H2?H2? ? (H1 +H2)2H2??) 12(?????????)(H2? ?(H1 +H2)H???(H1 +H2)H?? H2?)(H?~e?H?~e?)(5.4)255.1. Convection-diffusion equation in general orthogonal coordinatesNote in the last expression we used the following equality:H2?H2? ? (H1 +H2)2H2?? =(?2x + ?2y) (?2x + ?2y)? (?x?x + ?y?y)2= ?2x?2x + ?2x?2y + ?2y?2x + ?2y?2y ? ?2x?2x ? 2?x?x?y?y ? ?2y?2y= ?2x?2y + ?2y?2x ? 2?x?x?y?y= (?x?y ? ?y?x)2 (5.5)Orthogonality of (?,?) implies H?? = 0 and we have:{H2?H2? = (?x?y ? ?y?x)2?x?y ? ?y?x = |J |(5.6)Based on our definition, H?, H? are always positive , so from (5.6) we canconclude?x?y ? ?y?x = sign(|J |)H? ?H? (5.7)Substituting (5.7) in (5.4) and using H?? = 0 we obtain:~v=sign(|J |)(H2?H2? ? (H1 +H2)2H2??) 12(?????????)(H2? ?(H1 +H2)H???(H1 +H2)H?? H2?)(H?~e?H?~e?)=sign(|J |)H?H?(?????????)(H2? 00 H2?)(H?~e?H?~e?). (5.8)265.1. Convection-diffusion equation in general orthogonal coordinatesFrom (3.27) we know that ~???T =(H?~e? H?~e?)????T???T?????; using this equalitytogether with (5.8) and (5.1) we have:~v.~?T =sign(|J |)H?H?(?????????)(H2? 00 H2?)(H?~e?H?~e?)(H?~e? H?~e?)????T???T?????=sign(|J |)H?H?(?????????)(H2? 00 H2?)(H2? 00 H2?)????T???T?????=sign(|J |)H?H?(?????????)(H2?H2? 00 H2?H2?)????T???T?????=H2?H2? sign(|J |)H?H?(?????????)????T???T?????= H?H?sign(|J |)(?????????)????T???T????? (5.9)Substituting (3.31) and (5.9) into (5.1) we have:Pe .H?H? ? sign(|J |)(?????????)????T???T????? = H?H?[???(H?H??T??)+???(H?H??T??)](5.10)Since J = HU , H is positive and U is unitary, we can easily conclude thatsign(|J |) = |U |, replacing this identity in (5.10) we would have:Pe(?????????)????T???T????? = |U |[???(H?H??T??)+???(H?H??T??)](5.11)275.1. Convection-diffusion equation in general orthogonal coordinatesThus we have determined the convection-diffusion equation in a general or-thogonal set of coordinates. We recall in the above, ? is the streamfunctiondefining the velocity field. We already proved that a mapping is conformal ifand only if H? = H?. Therefore the convection-diffusion equation takes theformPe(?????????)????T???T????? = |U |(?2T??2+?2T??2)(5.12)if and only if a mapping is conformal.We know that a conformal map is eitherholomorphic i.e. J =(?x ?y??y ?x)or anti-holomorphic i.e. J =(?x ?y?y ??x).While |U | = 1 for a holomorphic mapping, from (5.12) we can see thata holomorphic mapping preserves the convection-diffusion equation. Letsrecall that convection equation in cartesian coordinates is :Pe(???y????x)????T?x?T?y??? =(?2T?x2+?2T?y2)(5.13)On the other hand, if a mapping (x, y) ?? (?, ?) is anti-holomorphic, thenthe mapping (x, y) ?? (?, ?) would be holomorphic. Thus we can concludethat an anti-holomorphic mapping will preserve the convection-diffusion equa-tion in reversed coordinates. Combining it with the previous result, we canpresent a new definition for a conformal mapping.Theorem 8. Let f :{U ? R2 ?? R2(x, y) ?? (?, ?)be of class C1 with non-vanishingjacobian. Then f will preserve the convection-diffusion equation, in its director reversed coordinate, if and only if f as a map is conformal.Note 1: We know when Pe = 0 , the convection-diffusion equation i.e.Pe ~v.~?T = ?2T would become the heat equation i.e. ?2T = 0. In that case,our theorem would be the well-known definition of conformal mapping i.e. amapping is conformal if and only if it preserve the heat equation ?2T = 0Note 2: The theorem indicates that the desired invariant KT is the con-formal mapping. In the next section, we will look for the KT ?K?.28Chapter 6Velocity field6.1 General solution of the biharmonicIn this section, we focus our attention on stokes flows, i.e. flows for whichinertia may be neglected. We want to find the general solution of biharmonicequationE4? = 0 (6.1)Note for a function of (x, y) variables like ?(x, y) , we define E2? =?xx + ?yy or its equivalent operator in another system of coordinates (likepolar coordinates) and for a function of (?, ?) variables like ?(?, ?) , we defineE2? = ??? + ???.An analytic solution for such a problem can thus be sought as a super-position of separate solutions of this equation in any orthogonal curvilinearcoordinate system. Rather than using the polar angle ? as an independentvariable, it is more convenient to introduce? ? cos ? (6.2)so that the coordinate variables are (r, ?, ?) with ?1 6 ? 6 +1, and the E2operator takes the simplified formE2 ??2?r2+(1? ?2)r2?2??2(6.3)We seek a solution of (6.1) via the method of separation of variables. Forthis purpose, it is convenient to note that E4? = 0 can be split into twosecond-order equationsE2w = 0 (6.4)andE2? = w (6.5)296.2. Intersection of invariant of temperature and velocity fieldObviously, substituting (6.5) into (6.4), we recover (6.1). The most generalsolution of w by solving (6.4) via the method of separation of variables, canbe written as:w =??n=1wnwn =(A?nrn+1 + C?nr?n)Qn(?) (6.6)In equation (6.6), A?n, C?n are constant coefficients (which require impositionof boundary conditions). The polynomial functions Qn(?) for n > 1 are theGegenbauer polynomials which satisfy the equationQn(?) =? ??1pn(?)d? (6.7)In (6.7), pn(?) are the Legendre polynomials (of the first kind) of degree n.The first several Gegenbauer and Legendre polynomials are:p0(?) = 1 Q0(?) = ? + 1p1(?) = ? Q1(?) =12(?2 ? 1)p2(?) =12(3?2 ? 1)Q2(?) =12(?3 ? ?)(6.8)The most general solution of ? by solving (6.5) via the method of separationof variables, can be written as:? =??n=?(Anrn+3 +Bnrn+1 + Cnr2?n +Dnr?n)Qn(?) (6.9)where An, Bn, Cn, Dn are constant coefficients. One can find required calcu-lations in order to prove (6.6) and (6.9) in chapter 4 of reference [15].6.2 Intersection of invariant of temperatureand velocity fieldIn the previous sections we proved that the invariant KT is the conformalmapping. In this section, we want to see which subset of conformals will306.2. Intersection of invariant of temperature and velocity fieldpreserve the biharmonic equation (velocity field). Assume that we have aconformal map f and a streamfunction ?, such that the image of the stream-function under the function f would be ?, in other words(x, y) ?? (?, ?)?(x, y) ?? ?(?, ?) (6.10)We want to explore a property for the function f , such that image of thestreamfunction would also satisfy biharmonic i.e. E4? = 0. Again we recallfor a function of (x, y) variables like ?(x, y) , we define E2? = ?xx + ?yyand for a function of (?, ?) variables like ?(?, ?) , we define E2? = ??? + ???.As we discussed before, each conformal mapping would be uniquely definedby its characteristic function ?, such that e? = H and H is the non zeroelement of Hermitian part of the left polar decomposition of the Jacobian ofthe function f .Since ? is the streamfunction and ? is its image, we know H satisfies theidentity(?2??x2+?2??y2)= H2(?2???2+?2???2)(6.11)while streamfunction ? and its image ?, both satisfy the biharmonic equationi.e. E4? = 0, we would have a property on the characteristic function H.This property will explain which elements of KT also belong to K?, inother words, it will define all elements of KT ?K?. In order to obtain thisproperty, we start with equation (6.11), and then we impose the conditionE4? = E4? = 0 (6.12)We have:E2? = (?)xx+yy = ?xx + ?yy = H2 (??? + ???) = H2(?)??+?? ??E4? =E2 ((?)xx+yy)= E2(H2(?)??+??)=(H2(?)??+??)xx+yy= H2(H2(?)??+??)??+??= H2(H2(?)??+??)??+H2(H2(?)??+??)??(6.13)We recall(AB)?? = ((AB)?)? = (A?B + AB?)? = A??B + 2A?B? + AB?? (6.14)316.2. Intersection of invariant of temperature and velocity fieldsubstituting (6.14) in (6.13) we will obtain:E4? = H2[(H2(?)??+??)??]+H2[(H2(?)??+??)??]= H2[(H2)??(?)??+?? + 2(H2)?((?)??+??)? +H2 ((?)??+??)??]+H2[(H2)??(?)??+?? + 2(H2)?((?)??+??)? +H2 ((?)??+??)??]= H2 ? (?)??+?? ?(H2)??+??+H4 ((?)??+??)??+??+ 2H2(H2)?((?)??+??)? + 2H2(H2)?((?)??+??)?= H4E4?+H2L ((?)??+??) (6.15)In equation (6.15), L is an operator which acts on the function (?)??+?? asfollowing:L(T ) =(H2)??+??T + 2(H2)?T? + 2(H2)?T? (6.16)we can summarize equation (6.15) as:E4? = H4E4?+H2L ((?)??+??) (6.17)Equation (6.17) together with equation (6.12) implies:L ((?)??+??) = 0 (6.18)we obtained general solution of streamfunction ? and its laplacian at equa-tions (6.9) and (6.6). On the other hand, equation (6.11) explains the re-lationship between streamfunction ? and its image ?. Substituting (6.6) in(6.11) we will have:(?)??+?? =1H2??n=1wn (6.19)where wn is defined in (6.6). Substituting (6.19) in (6.18) we will obtainL(1H2??n=1wn)= 0 (6.20)326.2. Intersection of invariant of temperature and velocity fieldSince L is a linear operator, and (6.20) should hold for any choice of constantsin (6.6), we can conclude (6.20) is equivalent with:L(rn+1Qn (cos(?))H2)= 0 , L(r?nQn (cos(?))H2)= 0 n = 1, 2, ? ? ?(6.21)equation (6.21) explains, if f : (x, y) ?? (?, ?) is a change of coordinatewhich is canformal and H is it?s scaling function i.e. H2 = ?2x+?2y = ?2r+?2?r?2and similarly H2 = ?2x + ?2y = ?2r + ?2?r?2 , then the conformal map f willsatisfy(6.21), if and only if f ? KT ?K?.33Chapter 7SummaryWe started with the prototype problem of flow of a Newtonian fluid in theannular region between two infinitely long circular cylinders, with the givenvelocity and temperature on the boundaries of the domain.Then we tried to find out how geometry will affect the behavior of the flowinside of the domain. We explored two invariant mappings KT and K?, suchthat under appropriate condition on the boundary the mapping KT wouldpreserve solution of temperature field from one domain to another and themapping k? would preserve solution of velocity field in the domain.We proved that if a mapping is conformal, it preserves the convection-diffusion equation in both domains. After that, we find which subsets of theconformals would also preserve the velocity field as well. In order to answerthat question, we obtained the required condition for the mapping , such thatit would preserve both velocity and temperature fields , from one domain toanother.34Bibliography1] G. I. Taylor, Studies in electrohydrodynamics. I. The circulationproduced in a drop by electric field, Proc. R. Soc. London. Ser. A 291,159 (1966)[2]L. Chang, T. E. Carleson, and J. C. Berg, Heat and mass transfer to atranslating drop in an electric field, Int. J. Heat Mass Transfer 25(1982).[3]T. Ward and G. M. Homsy, Electrodynamically driven chaotic mixingin a translating drop, Phys. Fluids 13, 3521 (2001).[4]R. Kronig and J. C. Brink, On the theory of extraction from fallingdroplets, Appl. Sci. Res. A 2,142 (1951)[5]Y.-F. Pan and A. Acrivos, Heat transfer at high Peclet number inregions of closed streamlines, Int. J. Heat Mass Transfer 11, 439 (1968).[6]C.I. Christov and G. M. Homsy, Enhancement of transport from dropsby steady and modulated electric fields, Phys. Fluids 21,(2009).[7]B. Y. Ballal and R.S. Rivlin, Flow of a Newtonian fluid between eccen-tric rotating cylinders: inertial effects, Archive for Rational Mechanicsand Analysis 62, 237-294, (1966)[8]T. J. Kaper and S. Wiggins, An analytical study of transport in Stokesflows exhibiting large-scale chaos in the eccentric journal bearing, J.Fluid Mech. 253, 211 (1993).[9]H. Aref, The development of chaotic advection, Phys. Fluids 14,(2002).[10]O. Reynolds, Phil. Trans. R. Soc. Lond. 177(I), 157-234 (1886).[11]N. E. Zhukowski, Comm. Math. Soc. Kharkov 17, 31-46 (1887).[12]D. R. Sood and H.G. Elroo, Report 17, Lubr. Res. Lab., ColumbiaU., N.Y., 1-44 (1970).[13]M. M. Kamal, J. Basic Engng., Trans. A.S.M.E. 88, 717-724 (1966).[14]L. P. Lebedev, M. J. Cloud and V. A. Eremeyev, Tensor analysis withapplications in mechanics, World scientific (2010).[15]L. G. Leal, Laminar flow and convective transport processes,Butterworth-Heinemann (1992)35


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