3-D ADI Poisson Solver

Transcription

1 3-D ADI Poisson Solver A Project Report Submitted in partial fulfilment of the requirements for the Degree of Master of Technology in Computational Science by V.Karthik Supercomputer Education and Research Center Indian Institute of Science BANGALORE June 7, 211

2 Acknowledgements I would like to thank my advisor Dr.Atanu Mohanty for his valuable guidance and support through out the project. I would like to mention that his passion for teaching, interest towards students encourages us to perceive our studies and research with much more interest. I would like to thank all the people of SERC for providing this platform of study to us. I would like to thank my friends Govind, Sufal, Avinash Bhat, Mritunjay, Avinash Dash for all the nice time we had together studying and having fun. Special thanks to my friends Appala Naidu and Brahmanandam. I would like to thank all my family members for being with me all the time and giving continuous support and encouragment. Finally, I would like to thank god for his blessings through out. i

3 Abstract Alternating Direction Implicit(ADI) method is a Finite Difference Method(FDM) to solve partial differential equations. We employ ADI method to solve Poisson s equation for dirichlet boundary conditions.the discretised equation is modified near the boundaries to incorporate arbitrary domain shapes. The ADI Poisson solver developed, works for both two or three dimensions. ADI method performs much better than traditional iterative methods because of usage of efficient tri-diagonal solvers ii

4 Contents Acknowledgements Abstract i ii 1 Introduction Poisson s Equation Applications of Poisson s equation Properties of Poisson s equation Analytical Solution to Poisson s Equation Solution strategies Solution of Laplace equation on square domain: Solution of Poisson equation on square domain: Solution of Poisson equatin on square domain with given boundary conditions Solution of Laplace equation on an annular ring: Solution of Laplace equation on circular domain: Solution of Poisson s equation on circular domain: Solution of Poisson s equation using Finite Difference Methods Finite Difference Methods Mesh/Grid Generation Discretised Poisson s equation Matrix Formulation Iterative Methods Jacobi Method Gauss-Siedel Method Successive-over Relaxation Method Convergence of Iterative Methods Alternating Direction Implicit(ADI) Method Motivation Procedure Matrix Formulation iii

5 CONTENTS iv 4.4 ADI Method Implementation Solution of Laplace equation on Square domain Problem Statement Numerical Solution Error Analysis Choice of relaxation parameter Solution of Poisson Equation on square domain Problem Statement Governing Equation and Numerical Solution Error Analysis ADI Method on various domains ADI Solver Implementation Steps FDM formula for irregular boundary Solution of Laplace equation on a ring domain Solution of Poisson equation on a circular disk domain Solution of Laplace equation on a hyperbola domain ADI Method on 3-D Domains D Poisson Equation D ADI method Solution of Laplace Equation on the Cube Conclusions and Future Work Conclusion Future Work References 57

6 List of Tables 4.1 Comparision of iterations taken by various iterative methods Cyclic ADI parameters performance comparision v

7 List of Figures 2.1 Superposition principle Superposition principle Solution of Laplace Equation on Square domain Analytical Solution of Poisson Equation on Square domain Domain Grid Square Domain Laplace equation Solution using ADI method on Square domain Error plot for ADI method in Square domain Error plot for ADI Method in the interior of Square domain Poisson function for square plate Numerical solution of Poisson equation on square domain Error plot for Poisson equation on square domain Error plot in the interior of the domain for Poisson equation on square domain A curved boundary with mesh Mesh point O with neighbours NEWS Ring Domain Laplace equation Solution using ADI method on ring domain Error plots for solution of Laplace equation on ring domain Poisson equation Solution using ADI method on circular disk Error plots for solution of Poisson equation on circular disk Hyperbola Domain Laplace equation Solution using ADI method on hyperbola domain Error plots for solution of Laplace equation on hyperbola domain A point in 3D domain Solution of Laplace Equation on Cube domain using ADI method Error plot on Cube domain vi

8 Chapter 1 Introduction 1.1 Poisson s Equation Poisson s Equation is an elliptic linear inhomogeneous partial differential equation of the second order. It is given by 2 u = f (1.1) where 2 (also denoted with ) stands for the Laplacian operator, and f is called as load/source function. The solution of the equation u, is the unknown scalar potential function. In Cartesian coordinate coordinate system, the Poisson equation is given by where x, y, z are the independent space dimensions. 2 u x + 2 u 2 y + 2 u = f(x,y,z) (1.2) 2 z2 When the Poisson equation is satisfied by a scalar potential in a given domain Ω. We can find the scalar potential inside the domain by solving the Poisson equation with the help of boundary conditions prescribed at the boundary Ω. 1

9 Chapter 1. Introduction Applications of Poisson s equation Poisson equation arises in a variety of branches mathematical and physical contexts ranging through elasticity and solid mechanics, fluid mechanics, electromagnetism, potential theory, heat conduction, geometry, probability, number theory, and many more [4, 5]. Poisson equation is mainly used to describe the equilibrium physical phenomena under the presence of source particles or load function. It is used to describe the steady state conditions of a given system as it involves no time variable. Electrostatics: In electrostatics, it is often necessary to find the electric field ( E) for a given charge distribution in a region[3]. Directly solving for E requires solving two equations. E = E ρ = 4πǫ As E is a vector, it involves solving for all the vector components which is cumbersome. This process can be simplified with help of Poisson s equation. We can find the electric potential in the region by solving Poisson s equation. 2 V(x,y,z) = 1 4πǫ ρ(x,y,z) where ρ(x,y,z) is a charge distribution in the given volume. Then electric field is given simply by the gradient of the potential. The advantages of second method is that it involves only scalars and requires to solve only one partial differential equation(poisson s equation) involving voltage instead of two involving gradient and curl.

10 Chapter 1. Introduction 3 Derivation of Poisson s equation in Electrostatics In a given volume, Gauss s law in differential form states that the divergence of the electric field in a region is proportional to charge enclosed in the region. E = 1 4πǫ ρ (1.3) whereρstandsforchargedensity. Intheabsenceofchangingmagneticfield B, Faraday s law of induction gives E = B t = Since curl of electric field is zero, the electric field can be given by the gradient of the electric potential. E = V (1.4) Combining the equations 1.2 and 1.2, we get the Poisson equation for electric potential 2 V = 1 4πǫ ρ (1.5) Stationary Heat Equation Poisson s equation also governs heat flow problems that are steady, that is, time independent. The thermal equilibrium of bodies(eg. flat plates) can be described by the Poisson s equation in the presence of heat sources: α 2 T(x,y) = h(x,y) where T(x,y) represents the temperature, h(x,y) represents the external heat source and α stands for the thermal conductivity.

11 Chapter 1. Introduction 4 Steady state Wave equation The steady state displacement of a membrane(like drum skin) follows Poisson s equation. The inhomogenuous term represents an external forcing over the surface of the membrane. The governing equation is given by 2 u(x,y) = P(x,y) T P(x, y) represents pressure, T represent Tension and u(x, y) represents the displacement of the membrane. Fluid Dynamics The velocity potential of a steady incompressible, irrotational planar flow of fluid follows Laplace equation. The gradient of the velocity potential gives the velocity. Elasticity Poisson s equation arises in the study of torsion of prismatic bars. The stress distribution over a non-circular cross-section of a twisted bar can be determined by finding the stress function Ψ(x, y) which satisfies the two-dimensional Poisson equation. 2 Ψ = 2Gα G is shear modulus. α is the angle of twist per unit length Gravitational Potential To study the forces acting in a gravitational field filled with masses, Poisson equation is also useful.the gravitational potential is given by the equation 2 ψ = 4πGρ

12 Chapter 1. Introduction 5 ψ stands for Gravitational potential, G stands for universal gravitational constant and ρ stands for mass density. Besides above,poisson s equation has applications in image processing, number theory etc. 1.3 Properties of Poisson s equation Poisson s equation and its homogeneous counter-part Laplace equation has several interesting properties which help in finding their solutions or make their study interesting and useful in many areas which are worth mentioning here. 1. Harmonic functions: The Solutions of Laplace equations are called as harmonic functions. They have continuous second-order partial-derivaties. 2. Mean value property: The value of the harmonic function at any interior point is equal to the mean value of the function on any circle in the domain with its center at that point. 3. Maximum modulus property: Solution of Laplace equation cannot have maximum or minimum in the interior of the domain. Consequently, the maximum and the minimum are taken on the boundary of the domain 4. Analytic functions: The real and imaginery part of analytic functions are harmonic. They remain harmonic under conformal mapping so that conformal mapping becomes a powerful tool in solving boundary value problems for Laplace equation by mapping complex domains on to simpler domains like circle. The complex representation of potential has several interpretations. For example, in electrostatics, the equal real parts correspond to equi-potential lines and the equal imaginery parts correspond to lines of electric force.in heat problems, they are isotherms and lines of heat flow. 5. Linearity : Laplace equation is second order linear PDE. So the principle of superposition hold good. So new solutions can be obtained by the summation of two

13 Chapter 1. Introduction 6 existing solutions. This is very useful in obtaining solution of Laplace equation like complex problems can be constructed by summing simple solutions. 6. Uniqueness: We can derive from above properties that Laplace equation and Poisson s equation has unique stable solution for Dirichlet or Neumann boundary conditions for closed surfaces. So that if a solution is found to Laplace equation by any method then any other solution is just different represention of the same solution.

14 Chapter 2 Analytical Solution to Poisson s Equation In this chapter we introduce various traditional analytical methods used to solve partial differential equations. Finding analytical solutions helps in verifying the correctness of solution obtained from the numerical method. 2.1 Solution strategies Some general techniques used to solve Partial Differential Equations analytically are as follows 1. Separation of variables: This is the most common method used to solve partial differential equations on simple domains. The idea is to express the solution as the product of the functions of the variables involved. For example, in two variables, Solution u(x,y) can be written as u(x,y) = F(x)G(y) This separates out the partial differential equation in to two or three ordinary differential equations which are related by a common constant(usually eigen values). 7

15 Chapter 2. Analytical Solution to Poisson s Equation 8 These ordinary differential equations are solved using the boundary conditions that are already transformed in to simpler conditions involving only one of the variable. Thus separation of variables is a very useful technique when domain boundary is symmetric to the variables involved and boundary conditions specified involve those variables only. 2. Conformal mapping: These are useful to reduce a complicated domain in to a simpler domain like disk. Complex analysis is useful in this method. Components of analytic functions satisfy laplace equation and they remain analytic in conformal maps. So that Laplace equation is solved on the simpler domain and inverse mapped to get the solution on the complicated domain. 3. Green s function : Green s function characterise the response of the system for a unit impulse. In simpler terms, it gives the solution of the system for a unit load at a particular point which when integrated over the domain with required scaling gives the solution for the given load. Solution of Laplace equation on circular disk can be derived in this method with the help of Poisson s Integral formula and also inhomogeneous equation like Poisson equation can be solved using this method. But this method involves integrals which may be tough to evaluate. 4. Integral transforms Integral transforms like Fourier transforms are useful when the domain considered is infinite domain. 5. Superposition principle Laplace equation is linear so that combination of solutions is also a solution. We can find solution for complicated boundary condition by splitting it in to smaller problems. Solution of Laplace equation 2 U = on square domain of length L with two boundaries x =,y = L having potential one can be found by U = U 1 +U 2 where 2 U 1 = with U 1 (x,l) = and 2 U 2 = with U 2 (,y) = 1. This is described in figure (2.1). Solution of Poisson equation for given boundary conditions can be found by adding

16 Chapter 2. Analytical Solution to Poisson s Equation 9 Figure 2.1: Superposition principle solution of poisson equation for homogeneous zero boundary condition and particular solution of laplace equation for given boundary conditions.this is described in figure(2.2) Figure 2.2: Superposition principle Mathematically solution of 2 U = f(x,y) for U = g(x,y) on boundary Ω is given by U = U 1 +U 2 where 2 U 1 = f(x,y) for U 1 (x,y) = on boundary Ω and 2 U 2 = with U 2 (x,y) = g(x,y) on Ω.

17 Chapter 2. Analytical Solution to Poisson s Equation 1 6. Eigen function expansions: Solution of inhomogenuous differential equations like Poisson s equation can be found by expressing the source/load function as an expansion of eigen functions of the linear operator and finding the coefficients/eigen values. Fourier series is used in rectangular domain and Fourier-Bessel series is used in circular domains. 2.2 Solution of Laplace equation on square domain: Problem Statement To solve Laplace equation u xx +u yy = (2.1) in the interior of a square domain of length L with the Dirichlet boundary conditions. u(,y) = for < y L (2.2) u(l,y) = for < y L (2.3) u(x,) = for x L (2.4) u(x,l) = V for < x < L (2.5) i.e. three sides are kept at zero boundary conditions and another side(y = L) is maintained at a constant potential V. Separation of Variables By the method of separation of variables, we split the partial differential equation (2.1) in to two ordinary differential equations and solve for the potential u in the interior of the domain. The first step is to assume the solution as the product of functions which depend on only one variable i.e. u(x,y) = F(x)G(y) (2.6)

18 Chapter 2. Analytical Solution to Poisson s Equation 11 Substituting in equation (2.1) we get, G(y)F (x)+f(x)g (y) = Dividing by U(x,y) we get, F (x) F(x) + G (y) G(y) = We write the above equation as F (x) F(x) = G (y) G(y) = λ2 Since the LHS in the above equation involes only x variable and RHS is only a function of y variable, we conclude that this can happen only if λ 2 is constant. This forms the two ordinary differential equations F (x)+λ 2 F(x) = (2.7) G (y) λ 2 G(y) = (2.8) The solution of (2.7) is given by F(x) = Asin(λx)+Bcos(λx) (2.9) Applying boundary conditions (2.2), we get B= in the above equation. Applying the boundary condition (2.3), we get λ = nπ L The solution of (2.8) is given by where n = 1,2... F(x) = Ce λy +De λy = C 1 sinh(λy)+d 1 cosh(λy)

19 Chapter 2. Analytical Solution to Poisson s Equation 12 Using boundary condition (2.4), we get D 1 = in the above equation. Now using equation (2.6), the solution of Laplace equation can be given as follows: u(x,y) = c n sinh n=1 ( ) nπy sin L ( ) nπx L (2.1) Fourier Series In order to determine the solution completely, we need to determine the coefficients c n. It can be determined by the fourth boundary condition given.assume on y = L, the potential is f(x). We can observe that equation (2.1) is in the form of fourier sine series. Then the coefficient c n is given by c n = 2 L f(x)sin Lsinh(nπ) ( nπx For the given problem f(x) = V which gives the solution, L ) dx (2.11) u(x,y) = n=1,n:odd ( ) 4V nπy nπ sinh(nπ) sinh sin L ( ) nπx L (2.12) 1 u(x,y) y x Figure 2.3: Solution of Laplace Equation on Square domain. Boundary Conditions: u(x,y) = on three sides and u(x,1) = 1 No. of terms in fourier series used = 2

20 Chapter 2. Analytical Solution to Poisson s Equation Solution of Poisson equation on square domain: Problem Statement To solve Poisson equation u xx +u yy = 1 4πǫ ρ(x,y) (2.13) in a square domain of length L with homogeneous zero boundary conditions and constant charge density ρ in the interior of the domain. i.e. ρ(x,y) = ρ for < x < L, < y < L (2.14) Eigenfunctions The solution to the Poisson s equation 2 u = q(x,y) for homogeneous dirichlet boundaryconditionscanbegivenintermsofeigenfunctionsofthelaplacianoperator: sin ( nπx L i.e. Let the solution vector u be: Then, u = m,n=1 K m,n sin ( ) nπx sin L ( ) mπy L ) sin ( mπy L ) 2 u = u xx +u yy = (m2 +n 2 )π 2 K L 2 m,n sin = (m2 +n 2 )π 2 u L 2 ( nπx L ) sin ( ) mπy So in order to find the solution of Poisson equation, we express charge density in terms of eigen functions sin ( ) ( ) nπx L sin mπy L with the help of two dimensional Fourier sine series ρ(x,y) = ( ) ( ) nπx mπy Q n,m sin sin L L m,n=1 L (2.15)

21 Chapter 2. Analytical Solution to Poisson s Equation 14 where Q n,m is given by the fourier series coefficient: Q n,m = 4 L 2 L L ( ) nπx q(x,y)sin sin L Then the solution to Poisson s equation is given by, ( mπy L ) dxdy (2.16) u(x,y) = m,n=1 Q n,m L 2 ( ) nπx (m 2 +n 2 )π sin sin 2 L ( ) mπy L (2.17) For the given problem, fourier series coefficients of unit charge density is given by, Q n,m = 4 L 2 L = L ρ 4πǫ sin ( nπx ) sin ( mπy L L 16 ρ n,m: oddotherwisezero nmπ 2 4πǫ ) dxdy Thus from the equation (2.17), the solution for the given problem is u(x,y) = m,n=1, odd ρ 16L 2 ( ) nπx 4πǫ mn(m 2 +n 2 )π sin sin 4 L ( ) mπy L (2.18) 2 x Voltage y 2 2 x 4 6 Figure 2.4: Analytical Solution of Poisson Equation on Square domain with zero boundary value and f(x,y) = 9e9. No. of terms in fourier series used = 1

22 Chapter 2. Analytical Solution to Poisson s Equation Solution of Poisson equatin on square domain with given boundary conditions Solution of Laplace equation on an annular ring: To solve Laplace equation 2 u = on a ring domain with inner radius R 1 and outer radius R 2 with boundary conditions. 1 r = R 1 f(r,θ) = (2.19) r = R 2 We express Laplace equation in polar coordinates so that boundary becomes symmetric to the variables involved. Laplace equation in polar coordinates is given by: General solution 1 r { r r ( )} r u r 2 θ 2u = Using separation of variables in polar coordinates, let us say, u(r,θ) = R(r) Θ(θ) gives us two ordinary differential equations, r 2 R +rr µr = (2.2) Θ +µθ = (2.21) Solution of equation(2.21) is given in terms of trigonmetric functions because it needs to be periodic Θ(θ +2π) = Θ(θ). Equation (2.2) is in the form of second order Euler ordinary differential equation. The solutions can be given in terms of eigen functions as follows:

23 Chapter 2. Analytical Solution to Poisson s Equation 16 n =, Θ(θ) = 1, R(r) = 1 and R(r) = lnr n >, Θ(θ) = sin(nθ),cos(nθ) R(r) = r n and R(r) = r n Thus general solution of Laplace Equation on ring is: u(r,θ) = a + n a n r n cos(nθ)+ n a n r n cos(nθ) + b lnr + n b n r n sin(nθ)+ n b n r n sin(nθ) (2.22) Since the given boundary conditions does not depend on the parameter θ we can say solution is in the form u(r,θ) = c 1 +c 2 lnr Solving with the boundary conditions (2.19), we get u(r,θ) = lnr 2 lnr lnr 2 lnr 1 (2.23) Solution of Laplace equation on circular domain: Problem Statement To solve Laplace equation 2 u r + 1 u 2 r r u r 2 θ = (2.24) 2 on a circular domain of radius R with boundary conditions u(r,θ) = f(θ). For a circular disk, the terms containing lnr, r n in equation (2.22) have singularities as r so that solution on disk is given by, u(r,θ) = a + (a n r n cos nθ+b n r n sin nθ) (2.25) n=1 The coefficients can be determined by using the boundary conditions and finding fourier coefficients. Solution on circular disk can also be given in terms of Green s function which is given

24 Chapter 2. Analytical Solution to Poisson s Equation 17 by Poisson s kernel, P(x,s). The influence due to a small element s on the boundary at a point x in the interior of the boundary is given by P(x,s) = R2 x 2 x s 2 where the denominator measures squared distance from x to s and numerator roughly measures distance from x to boundary. In polar coordinates, s can be given by single angular coordinate α [ π,π) If x has coordinates (x,y), then P(x,s) = R 2 (x 2 +y 2 ) (x Rcos(α)) 2 +(y Rsin(α)) 2 Converting to polar coordinates x is represented by (r,θ) and dividing by R 2 we get, P(x,s) = 1 ( ) 2 r R 1 2 ( r R) cos(θ α)+ ( r R ) 2 So the solution on circular disk is given by Poisson s Integral formula: u(r,θ) = 1 π f(α) 2π π 1 ( ) 2 r R 1 2 ( r R) cos(θ α)+ ( r R ) 2 dα (2.26) The solution can be viewed as a sort of convolution of boundary function with Poisson s kernel in circular domain Solution of Poisson s equation on circular domain: Problem Statement To solve Poisson equation 2 u = q(r,θ) i.e. 2 u r + 1 u 2 r r u = q(r,θ) (2.27) r 2 θ2

25 Chapter 2. Analytical Solution to Poisson s Equation 18 on a circular domain of radius R with homogeneous zero boundary conditions and unit source function inside the domain i.e. q(r,θ) = 1 for r < R Bessel Equation derivation Let f(r,θ) be an eigen function of the Laplacian operator then the eigen vector equation is given by f = λ 2 f (2.28) for some eigen value λ 2. Applying separation of variables, f(r) = R(r)Θ(θ) in the polar coordinates form we get, R (r)θ(θ)+ 1 r R (r)θ(θ)+ 1 r 2R(r)Θ (θ) = λ 2 R(r)Θ(θ) Dividing by R(r)Θ(θ) R (r) R(r) + R (r) rr(r) + Θ (θ) r 2 Θ(θ) = λ2 Multiplying with r 2 and rearranging the terms we get, r 2R (r) (r) R(r) +rr R(r) +λ2 r 2 = Θ (θ) Θ(θ) To satisfy the periodic boundary condtions on θ, both LHS and RHS should be equal to a negative constant, so that equations become, Θ (θ) m 2 Θ(θ) = (2.29) r 2 R (r)+rr (r)+(λ 2 r 2 m 2 )R(r) = (2.3) Equation (2.3) is known as Bessel s equation. The solution of (2.3) is given in terms of Bessel s function of first kind J n (r) and Bessel functions of second kind Y n (r). The solution of (2.29) is given by cos(nθ) and sin(nθ). Y n (x) is unbounded at the centre, so that Y n (x) is not a solution for the circular disk. In order to satisfy homogeneous zero boundary conditions we need J n (λr) =. So

26 Chapter 2. Analytical Solution to Poisson s Equation 19 we consider a sequence of roots of Bessel function λ = k nm which are irrational. Then solution to (2.28) is given by Fourier-Bessel expansion f(r,θ) = n= m=1 ( ) knm r A nm J n cos(nθ) + R where A nm and B nm are real-valued coefficients. n=1m=1 ( ) knm r B nm J n sin(nθ) R To find the solution of Poisson s equation given in equation (2.27), we find Fourier- Bessel expansion of q(r, θ). q(r,θ) = n= m=1 ( ) knm r A nm J n cos(nθ) + R The coefficients are given by the formula A nm = B nm = n=1m=1 ( ) knm r B nm J n sin(nθ) R 2 π ( ) R knm r q(r,θ)j πr 2 Jn+1(k 2 n cos(nθ)rdrdθ nm ) π R 2 π ( ) R knm r q(r,θ)j πr 2 Jn+1(k 2 n sin(nθ)rdrdθ nm ) π R Then the solution of Poisson s equation is given by, u(r,θ) = n= m=1 R 2 ( ) A nm knm r J knm 2 n cos(nθ) R n=1m=1 R 2 ( ) B nm knm r J knm 2 n sin(nθ) R (2.31) Consider a unit radius circular disk with uniform charge density then its Fourier-Bessel expansion in terms of J ( kmr )is given by R q(r,θ) = 1 = n= 2 k m J 1 (k m ) J Then the solution of Poisson s equation is given by ( ) km r R u(r,θ) = n= 2R 2 k 3 mj 1 (k m ) J ( ) km r R (2.32) where radius R = 1

27 Chapter 3 Solution of Poisson s equation using Finite Difference Methods It is not always possible to solve differential equation analytically. For example, it is not possible to get analytical solution to Poisson s Equation for L-shaped domain. Even when analytical solutions exist it may be tough to calculate i.e. it may involve evaluating complicated integrals. In such situations, numerical methods are employed to get approximate solutions to some desired level of accuracy. Various numerical methods exist for solving differential equations like finite element methods, finite difference methods, boundary element method etc. Here we employ finite difference methods to solve Poisson equation. 3.1 Finite Difference Methods The Finite Difference Methods(FDM) involve approximating the equation involving derivates with difference equation. For example, df f(x+h) f(x) is taken as dx h 2

28 Chapter 3. Solution of Poisson s equation using Finite Difference Methods Mesh/Grid Generation To apply the difference equation we first divide the solution region of potential function u(x,y) in to rectangles or meshes. For a 2D mesh, consider the distance between two adjacent mesh points as δx = h x,δy = h y. Then coordinates of any point P with index (i,j) on 2D mesh is given by x = i h x y = j h y xmax xmin ymax ymin where h x = h y = no of divisions+1 no of divisions+1. Here xmax, xmin, ymax, ymin stand for boundary extremes. Figure 3.1: Domain Grid 3.3 Discretised Poisson s equation Now the discretised Poisson s equation at point P(x,y) for the potential u(x,y) is given by centralised difference formula is given by, u i+1,j 2u i,j +u i 1,j h 2 x + u i,j+1 2u i,j +u i,j 1 h 2 y = f(x,y) (3.1)

29 Chapter 3. Solution of Poisson s equation using Finite Difference Methods22 where u i,j = u(x i,y j ) and f(x,y) is the load function. With the above difference formula, the accuracy of the calculated solution depends on the number of divisions of the mesh and the error is O(h 2 ). 3.4 Matrix Formulation A system of equations are obtained when the difference formula is applied at each interior mesh point. For 1 i,j no of divisions we get N 2 number of equations,(n = no of divisions). These equations can be put in matrix form as follows: AU = B, Vector of unknowns : U = [u 11,u 21,...,u N1,u 12,...u N2,...,u NN ] T, Right hand side vector : B = h 2 [f 11,...,f N1,...,f 1N,...,f NN ] T

30 Chapter 3. Solution of Poisson s equation using Finite Difference Methods23 The N 2 N 2 matrix A is given by A = (3.2) The solution of Poisson s equation can thus be found by solving above equations. 3.5 Iterative Methods To solve the above system of equations usual elimination methods cannot be applied. Especially when the number of divisions become more, the amount of storage required to store intermediate matrices produced during the computation by elimination methods become very large. Iterative methods are employed to solve system of equations. This is because they take advantage of the sparsity structure of the A matrix. They require storage of just the same as A matrix (5N 2 in our problem) and can be terminated when we get the

31 Chapter 3. Solution of Poisson s equation using Finite Difference Methods24 solution with sufficient amount of accuracy. In an iterative method, we begin with an initial vector X (), and generate a sequence of vectors X () X (1) X (2) which converge towards the desired solution X. Each iteration requires very less amount of work comparable to that of multiplication of A with a vector which is very less when matrix A is sparse. So we can carry out a large number of iterations Jacobi Method This is the simplest of the iterative methods. A system of n equations where i th equation(1 i n) is given by n a ij x j = b i j=1 can be solved by transforming each equation in to a suitable format. The i th equation is re written as x i = b i n j=1j n a ij x j a ii (3.3) Jacobi method involves finding next iteration vector X (I+1) from current vector X (I) by substituting x (I+1) in LHS and x (I) in RHS of equation 3.3. So each iteration of the Jacobi method is given by n explicit steps. The i th step involves calculating x (I) i the formula: x (I+1) i = b i n j=1 j n a ij x (I) j using a ii (3.4)

32 Chapter 3. Solution of Poisson s equation using Finite Difference Methods Gauss-Siedel Method Gauss Siedel Method uses new values as soon as they are computed. It involves using (i-1) new values of X in RHS in the calculation of x i. x (I+1) i = i 1 b i j=1 a ij x (I+1) j n j=i+1 a ij x (I) j a ii (3.5) Successive-over Relaxation Method In this method, in order to improve convergence, we extrapolate the solution vector obtained from Gauss-Siedel method. Extrapolation is done in form of weighted average of the previous iteration solution and new solution from Gauss-Siedel method with the help of relaxation parameter ω. x (I+1) j = (1 ω)x (I) j +ω x (3.6) where ω x is obtained using gauss-siedel method as in Convergence of Iterative Methods In order to study the convergence of iterative methods, iterative method can be formulated in terms of matrix notation. For each iterative method, we have a new matrix M which expresses iteration step for solution of matrix equation AX = B as MX (I+1) = (M A)X (I) +B X (I+1) = (I M 1 A)X (I) +B

33 Chapter 3. Solution of Poisson s equation using Finite Difference Methods26 Theconvergence ofiterativemethoddependsonthespectralradius(ρ)ofthe(i M 1 A) matrix. For an iterative method to converge, the spectral radius (ρ) should be less than one. The smaller the spectral radius, faster the convergence of iterative method is. Given below are the results from [8], For consistently ordered matrices, Spectral radius of Jacobi method ρ(j) and Gauss-Siedel method ρ(h) are related by ρ(h) = ρ(j) 2 For SOR method, the optimal parameter ω b is given by, ω b = ρ(j) 2 ρ(j) ρ(h(ω b ) = 1+ 1 ρ(j) 2 Therefore Gauss-Siedel method converges twice as fast as Jacobi method and SOR method converges N times as fast as Jacobi method. ADI method, an iterative method to solve Poisson equation is introduced in next chapter and is shown to be superior than above classical iterative methods. 2

34 Chapter 4 Alternating Direction Implicit(ADI) Method Alternating Direction Implicit(ADI) method is an iterative method to solve system of equations arising from the finite difference discretizations of parabolic and elliptic partial differential equations. It was first introduced in mid-195s by Peaceman and Rachford. 4.1 Motivation When dealing with multi-dimensional domains, inversion of coefficient matrix A (3.2) requires lot of effort. But if we do it in steps by solving for each dimension once we get a tridiagonal system to solve which can be carried out in O(n) time. So, as the iteration steps varies in direction every time, the name is given as alternation direction method. Also since the system of equations needs to solved implicitly, this is an implicit method. 4.2 Procedure The idea of the ADI method is to execute each iteration in two or three steps( equal to the number of dimensions). The finite difference formula (6.2) is written in such a way that derivates involving one spatial direction are in left hand side of the equation and 27

35 Chapter 4. Alternating Direction Implicit(ADI) Method 28 remaining terms are carried to the right hand side. u i+1,j 2u i,j +u i 1,j h 2 x u i,j 1 2u i,j +u i,j+1 h 2 y = f(x,y) u i,j 1 2u i,j +u i,j+1 h 2 y = f(x,y) u i 1,j 2u i,j +u i+1,j h 2 y (4.1) (4.2) Using above equations, each iteration is described in two steps. U (I+1 2 ) is obtained by using 4.1 and then U (I+1) is obtained from 4.2 u (I+1 2 ) i+1,j 2u (I+1 2 ) i,j +u (I+1 2 ) i 1,j h 2 x u (I+1) i,j 1 2u (I+1) i,j +u (I+1) i,j+1 h 2 y = f(x,y) u(i) i,j 1 2u (I) i,j +u (I) i,j+1 h 2 y 1 = f(x,y) u(i+ 2 ) i 1,j 2u (I+1 2 ) i,j +u (I+1 2 ) i+1,j h 2 y (4.3) (4.4) So in each iteration, the discrete equations are solved first in one spatial direction and then in the other direction. Also the system of equations should be solved implicitly leading to the terminology, alternating direction implicit method. 4.3 Matrix Formulation The difference equations given in 4.4 when applied to each point on the domain gives rise to system of equations which can be given in matrix equations. Hu = B Vu (4.5) Vu = B Hu (4.6) where H: Horizontal dimension matrix,contains coefficients for unknowns in x direction V : Vertical dimension matrix, contains for coefficients for unknowns in y-direction. B : RHS vector In mathematical terms, w = 2u i,j u i 1,j u i+1,j h 2 x if w = Hu

36 Chapter 4. Alternating Direction Implicit(ADI) Method 29 w = 2u i,j u i,j 1 u i,j+1 h 2 y if w = Vu In simple words, H and V arise as the central difference approximations of the corresponding terms in the Poisson s equation. The H matrix for the the domain described in figure (3.1) is given by: H = h 2 x (4.7) The diagonal entries of H matrix correspond to the current point(x i,y j ) for which the equation is being generated. H matrix row contains a sub diagonal entry for the preceding point(x = x i 1 ) and a super diagonal entry for the next point in x-direction. If the adjacent point is a boundary point, then correspondingly sub-diagonal or super-diagonal entry may be zero and the rhs vector is updated with the value at the boundary.

37 Chapter 4. Alternating Direction Implicit(ADI) Method 3 The V matrix for the the domain described in figure (3.1) is given by: V = h 2 y (4.8) Each row of V matrix corresponds to an equation for a point(x i,y j ) in the interior of the domain in vertical direction. One entry in lower triangular part is non-zero corresponding to the preceding point (y = y j 1 ) in the vertical dimension and an entry in upper triangular part is non-zero corresponding to the next point (y = y j+1 )in the vertical dimension. This matrix formulation can also be viewed as the coefficient matrix A described in (3.2) being split in to parts in the ADI method depending on the number of dimensions of the domain. For two dimensions, A = H +V

38 Chapter 4. Alternating Direction Implicit(ADI) Method 31 The B vector is obtained by evaluation the Poisson load/source function and accumulating the boundary values if the adjacent points to the given point are boundary points. 4.4 ADI Method Implementation Relaxation parameter r is introduced in the equations 4.6 similar to Gauss-Siedel method to enhance the convergence. Now the steps in an iteration of the ADI method can be summarised as follows: (H +ri)u (n+1 2 ) = (ri V)u (n) +B (4.9) (V +ri)u (n+1) = (ri H)u (n+1 2 ) +B (4.1) Each iteration step involves solving a matrix equation which is costly if we employ Gaussian elimination (O(n 3 )wheren = N 2 ). But we can observer that H matrix is tri-diagonal i.e. it contains non-zero entries only in diagonal, sub-diagonal and superdiagonal elements. So to solve system of equations (4.9), we can employ tri-diagonal solve which involves only O(n) work. Algorithm Tri-diagonal-solve can be slightly modified to solve for equations (4.1) where elimination of non-zero lower-triangular entry below this diagonal element is done and corresponding row is modified instead of next row. We can observe that this algorithm also involves only O(n) work. So the implementation of an iteration requires only O(n) to solve the matrix equations given in the steps of ADI method. But it is still costlier than classical elimination methods. This may seem to be a disadvantage. But the advantage of ADI method springs from the fact that it converges at much faster rate by clever choice of relaxation parameter r.

39 Chapter 4. Alternating Direction Implicit(ADI) Method Solution of Laplace equation on Square domain Problem Statement To find the steady state temperature distribution on a unit square plate(length L = 1) with boundary conditions with boundary conditions unit temperature on top edge(y = L, < x < L) and zero on the remaining edges. Figure 4.1: Square Domain Numerical Solution The problem involves finding the solution of Laplace equation 2 u = on the square domain. Solving the problem using ADI method involves generating H and V matrices as given in equation (4.7) and (4.8) and RHS vector B contains non-zero entries only for index j = N rows of the domain given in (3.1). Iterations steps described in (4.9) and (4.1) are applied Error Analysis The relative error plot given in figure(4.3) is obtained by taking ratio of difference between analytical solution (fig. 2.3) and numerical solution (fig. 4.2) with analytical solution.

40 Chapter 4. Alternating Direction Implicit(ADI) Method 33 1 Temperature y x Figure 4.2: Temperature distribution on unit square plate. Solution of Laplace Equation using ADI Method. Side Length =1 No. of Divisions = 99 The relative error plot indicates two peaks at the corners of the square which are due to discontinuity in the function value on the boundary( sudden jump from to 1). Analytically, error can be explained with the help of Gibb s Phenomenon [1] ( overshoot of Fourier series approximation near the discontinuity)..3 Relative Error y x Figure 4.3: Relative error plot between ADI solution and analytical solution for Laplace equation on Square domain. No. of Divisions = 99 Generally two different methods [11] are used to to reduce or localise the error arising due to the discontinuity in the boundary shape/ boundary function

41 Chapter 4. Alternating Direction Implicit(ADI) Method Highly refined mesh near the boundary 2. Higher order approximations(considering more terms in Taylor series). Relative error plot inside the domain is given in figure (4.4) which shows that error when compared with analytical solution is very minimal in the interior of the solution. 4 x 1 4 Relative Error y x Figure 4.4: Error plot in the interior of the domain for Laplace equation on Square domain. x,y [.1,.9] No. of Divisions = Choice of relaxation parameter ADI Method gives faster convergence when we choose relaxation parameters optimally. In this section we compare the performance of ADI method with the classical iterative methods like Jacobi, Gauss Siedel and SOR methods and show that ADI method is superior to them. Number of iterations taken by different methods to generate solution for Laplace Equation in a square domain is shown in the table (4.1). We can observe that Gauss- Siedel method converges twice as fast as Jacobi method and SOR method converges much faster than both the methods(n times faster).

42 Chapter 4. Alternating Direction Implicit(ADI) Method 35 Table 4.1: Comparision of iterations taken by various iterative methods No. of Iterations Divisions Jacobi Gauss-Siedel SOR ADI Method Method Method Method Cyclic ADI parameters ADI method when given cyclic parameters as prescribed in [8] takes very less number of iterations. Relaxation parameter in ADI method is chosen differently for each step upto m = 2 k steps and cyclically repeated. Relaxation parameters are recursively calculated by the formula: r (n) i = r (2n) i + α k 1 jβ k 1 j r 2n i i = 1,2,3...2 j j =,1,...k 1 Lower bound on Eigen values of H and V matrices: α = 4sin 2 π 2(N +1) Upper bound on Eigen values of H and V matrices: β = 4cos 2 π 2(N +1) Calculation of bounds: α = α β = β α j+1 = α j β j β j+1 = α j +β j j =,1,...k 1 2 When relaxation parameters are selected cyclically, the values of relaxation parameters obtained by above algorithm with period m = 8 steps :

43 Chapter 4. Alternating Direction Implicit(ADI) Method 36 Table 4.2: Cyclic ADI parameters performance comparision Number of r=1 m=4 m=8 Divisions Iterations taken Iterations taken Big ADI steps Iterations taken Big ADI steps r[] : r[1] :.1197 r[2] : r[3] : r[4] : r[5] :.317 r[6] : r[7] :.1371 The comparision of iterations taken with constant and varying relaxation parameters is given in table Solution of Poisson Equation on square domain Problem Statement To find the electro static potential inside a square domain of length(l = 5) with homogeneous zero boundary conditions with unit charge density distributed inside the domain Governing Equation and Numerical Solution The electro static potential(voltage) in a closed region with charge density ρ contained inside satisfies Poisson equation 2 V = 1 4πǫ ρ (4.11) where V stands for Voltage or electro static potential. The value of the constant, permitivity of free space is 1 4πǫ is For the given problem, charge density ρ = 1Coulomb. The plot of solution obtained by ADI method is given in 4.6. The potential is in the

44 Chapter 4. Alternating Direction Implicit(ADI) Method 37 Figure 4.5: Poisson function for square plate shape of a dome as the edges of the plate are maintained at ground potential. 2 x Voltage y 2 2 x 4 6 Figure 4.6: Numerical Solution of Poisson equation on square domain of length 5m. No. of Divisions = Error Analysis Figure (4.7) shows relative error plot containing four peaks due to the discontinuity in the shape of the boundary at the corners of the square domain. The reduced error plot in the interior of the region is shown in (4.8).

45 Chapter 4. Alternating Direction Implicit(ADI) Method 38 8 x 1 3 Relative Error y 2 2 x 4 6 Figure 4.7: Error plot for Poisson equation on Square domain. No. of Divisions = 1 4 x 1 4 Relative Error y 2 2 x 4 6 Figure 4.8: Error plot in the interior of the domain for Poisson equation on Square domain. No. of Divisions = 1

46 Chapter 5 ADI Method on various domains ADI method was introduced in the previous chapter and was shown to be able to solve Laplace and Poisson equation with good accuracy on square domain. In this chapter, we extend ADI method to various domains in two dimensions. 5.1 ADI Solver Implementation Steps The ADI solver takes the following conditions as inputs. Find Boundary Extremes: This is useful for the generation of mesh. The bounding box of the domain is determined. The 5.2 FDM formula for irregular boundary Finite Difference approximation formula given in equation(6.2) holds when the boundary is aligned with the grid points. If the boundary is curved and doesn t intersect at mesh point as given in the figure (5.1) then the discretisation should take in to account that adjacent point is not at a distance h but a fraction of it, say nh. A generalised discretisation formula taking into account the fractional distances as 39

47 Chapter 5. ADI Method on various domains 4 Figure 5.1: A curved boundary with mesh Figure 5.2: Mesh point O with neighbours NEWS shown in figure (5.2) is given in [4].It is as follows: 2 u o = 2 h 2 [ u R r(r+l) + u N n(n+s) + u ] L l(l+r) + u S lr +ns s(s+n) lrns u o (5.1) 5.3 Solution of Laplace equation on a ring domain Problem statement To find the electrostatic potential in an annular ring domain(figure 5.3) whose outer radius(r 2 = 1) boundary is maintained at ground voltage and inner radius (R 1 =.25)

48 Chapter 5. ADI Method on various domains 41 Figure 5.3: Ring Domain boundary at voltage 1V u(r,θ) = 1, r = R 1 =, r = R 2 Numerical Solution The solution involves solving laplace equation 2 u = in the interior of ring domain. In cartesian coordinates, ring has curved boundary so that discretisation as described in equation (5.1) is applied at points near to the boundary. The solution obtained is given in the figure5.4). The voltage increases from interior boundary to exterior boundary in a logarithmic fashion. Error Analysis The analytical solution for the given problem is obtained from equation(2.23). The error plot and relative error plot for the ring domain is given in figure (5.5).

49 Chapter 5. ADI Method on various domains 42 1 Voltage y 1 1 x Figure 5.4: Voltage inside ring domain. Solution of Laplace Equation using ADI Method. Inner Radius =.25 Outer Radius =1 No. of Divisions = Solution of Poisson equation on a circular disk domain Problem statement To find the solution of Poisson s equation 2 u = 1 on a circular disk of radius R = 1 with homogeneous zero boundary conditions. Numerical Solution The problem can be viewed as uniform charge density inside circular disk for electro static potential or uniform pressure on a membrane. The potential will be maximum at the centre of the disk. The solution plot is given in figure (5.6) Error Analysis The analytical solution for the given problem can be obtained by equation (2.32). Then the error plot is obtained as given in figure(5.7)

50 Chapter 5. ADI Method on various domains 43 Error x Relative Error x y 1 1 (a) Actual Error x 1 1 y 1 1 x (b) Relative Error 1 Figure 5.5: Error plots for solution of Laplace equation on ring domain 5.5 Solution of Laplace equation on a hyperbola domain Problem statement To solve Laplace equation u x x+u y y = in the interior of region bounded by a pair of hyperbolas x 2 y 2 = a 2 and y 2 x 2 = b 2 with boundary conditions U(x,y) = V when x 2 y 2 = a 2 = when y 2 x 2 = b 2 Analytical Solution We can see that the solution is symmetric along the x, y axes so that the solution may not contain any x, y terms. So we can expect a polynomial solution as: U(x,y) = c +c 1 x 2 +c 2 y 2

51 Chapter 5. ADI Method on various domains u(x,y) y x Figure 5.6: Solution of Poisson equation on circular disk. Radius = 1 Source function f(x,y) = 1 No. of Divisions = 1 To satisfy Laplace equation, the necessary condition is c 1 = c 2 So the general solution is: U(x,y) = c +c 1 (x 2 y 2 ) By applying boundary conditions, we get c +c 1 (a 2 ) = V c +c 1 ( b 2 ) = Solving we get, c = V ( b 2 a 2 +b 2 ) c1 = V ( 1 a 2 +b 2 ) Therefore solution is: U(x,y) = V [( ) b 2 ( ) 1 (x + 2 y 2 ) )] a 2 +b 2 a 2 +b 2 Numerical Solution The domain is not closed so it cannot be solved directly with ADI solver. We consider x=-3a, 3a and y= -3b, 3b as boundary extremes and consider them as boundaries. Very few points fall on this line so that the solution is accurate enough. The solution in the interior gives a saddle shape surface i.e. it can not have maximum or minimum in the

52 Chapter 5. ADI Method on various domains 45 Error x Relative Error x y 1 1 (a) Actual Error x y 1 1 x (b) Relative Error in interior Figure 5.7: Error plots for solution of Poisson equation on circular disk Figure 5.8: Hyperbola Domain

53 Chapter 5. ADI Method on various domains 46 interior of the surface. Iso-lines are hyperbolas increasing in magnitude in x direction and decreasing in y direction. The solution plot is given in the figure(5.9). 1 Voltage y 1 1 x Figure 5.9: Solution of Laplace equation on hyperbola domain. a = 2 b = 2 V = 1 No. of Divisions = 1 x [ 3a,3a] = [ 6,6] y [ 3b,3b] = [ 6,6] Error Analysis The error plot indicates four peaks at the corners where the boundary is discontinuous(in the modified version). However the error is very minimum in the interior region(figure:5.1).