Numerical PDE. DYKwak. September 6, 2007

Transcription

1 Numerical PDE DYKwak September 6, 2007

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3 Contents 1 Finite Difference Method nd order linear p.d.e. in two variables Finite difference operator Error of difference operator Elliptic equation Basic finite difference method for elliptic equation Treatment of irregular boundaries(dirichlet boundary conditions Convection -diffusion equation Numerical Difficulties Upwind difference scheme Parabolic p.d.e s Discretization of parabolic p.d.e, General Case Hyperbolic Equation Method of Characteristics Finite Element Methods Model examples Iterative method for SPD system Acceleration of convergence Computation of the iterates Method of steepest descent Conjugate gradient Error analysis Preconditioning Application to Conjugate Gradient Method Variational formulation of boundary value problems Euler- Lagrange equation Weak solution Inhomogeneous boundary condition iii

4 iv CONTENTS Question Ritz-Galerkn Method Inhomogeneous boundary conditions Noncoercive forms Finite Element Method Finite element basis functions Assembly of stiffness matrix Numerical Integration Fractional order interpolation Trace theorem Exact solution near a singular point Mixed Method Raviart Thomas Mixed Method Mixed Formulation Mixed Triangular Element Raviart-Thomas Projection Global estimates Duality argument-brezzi-douglas-marini for RT element Hybrid form of mixed methods Trace estimate

5 CONTENTS 5 Topics to be covered (1) Elliptic and parabolic p.d.e (2) FEM for Model problem, (a) Variational formulation (b) Essential BC (c) Natural BC (d) Matrix Assembly (3) FDM for parabolic p.d.e (a) Explicit scheme (b) Implicit scheme (c) Stability (4) FEM for parabolic p.d.e (5) Iterative Methods(Conjugate gradient, PCG, condition number) (6) Domain decomposition Methods(Overlapping DDM) (a) Multiplicative Schwarz Method (b) Additive Schwarz Method (7) Domain decomposition Methods(Nonoverlapping DDM) (a) Basic Method (b) l 1/2 0 Method (8) Other topics (a) Upwind scheme (b) interpolation theory, fractional order space (c) Trace inequality (d) Poincaré inequality, inverse estimate (e) Finite Elements for Stokes equation (9) Multigrid Methods (10) Mixed FEM

6 6 CONTENTS (11) DG Method References (1) Finite element method, by Axelsson and Barker (2) Clae Johnson (3) Iterative methods for solving linear systems by Anne GreenBaum (4) Bramble, et al. (5) Widlund et al.

7 Chapter 1 Finite Difference Method 1.1 2nd order linear p.d.e. in two variables General 2nd order linear p.d.e. in two variables is given in the following form: L[u] = Au xx + 2Bu xy + Cu yy + Du x + Eu y + F u = G According to the relations between coefficients, the p.d.es are classified into 3 categories, namely, elliptic if AC B 2 > 0 i.e., A, C has the same sign hyperbolic if AC B 2 < 0 parabolic if AC B = 0 Furthermore, if the coefficients A, B and C are constant, it can be written as [ x, [ ] [ A B u ] y ] + Du B C x + Eu y + F u = G. Auxiliary condition B.C. Interface Cond I.C. x u y we say well posed if a solution exists. There are basically two class of method to discretize it, (1) Finite Difference method (2) Finite Element method 7

8 8 CHAPTER 1. FINITE DIFFERENCE METHOD 1.2 Finite difference operator Let u(x) be a function defined on Ω R n. Let U i,j be the function defined over discrete domain {(x i, y j )} such that U i,j = u(x i, y j ). Such functions are called grid functions. We introduce some operators on the grid functions. Shift operator. S + U i = U i+1, S U i = U i 1, S + S + U i = (S + ) 2 U i = U i+2 In case of two variables, S + x U i,j = U i+1,j, S x U i,j = U i 1,j S + x S y U i,j = U i+1,j 1. Averaging operator Difference operator µ + U i = U i+1 + U i, right average 2 µ U i = U i + U i 1, left average 2 µ 0 U i = U i+1 + U i 1, central average 2 δ + U i = U i+1 U i h i+1, forward difference δ U i = U i U i 1 h i, backward difference δ 0 U i = U i+1 U i 1 h i + h i+1, central difference δ 2 U i = 2(δ+ δ ) h i + h i+1, central 2nd difference Difference equation is an equation of type F (δu i, U i ) = 0. Difference scheme is a set of difference equations from which one can determine the unknown grid functions. Example Consider the following second order differential equation : u (x) = f(x), u(a) = c, u(b) = d.

9 1.2. FINITE DIFFERENCE OPERATOR 9 Given a mesh a = x 0 < x 1 < < x N = b, x i = x i+1 x i = h, we have U i 1 2U i + U i+1 h 2 = f i = f(x i ), i = 1, N 1, U 0 = c, U N = d which determines U i uniquely. We obtain an (N 1) (N 1) matrix equations. 2 1 U 1 f 1 c = h U N 1 f N 1 d Above equation can be written as L h U h = F h, called a difference equation for a given differential equation. Exercise Write down a matrix equation for the same problem with second boundary condition changed to the normal derivative condition at b, i.e, u (b) = d. Example (Heat equation). We consider u t = σu xx, for 0 < x < 1, 0 < t < T u(t, 0) = u(t, 1) = 0 u(0, x) = g(x), g(0) = g(1) = 0 Let x i = ih, i = 0,, N, x = 1/N and t n = n t, t = T J. Then we have the following difference scheme U n+1 i Ui n t [ U n = σ i 1 2Ui n + Ui+1 n ] x 2, for i = 1, 2,, N 1 and n = 1, 2,, M 1 where Ui n boundary condition and initial condition we have u(t i, x n ). From the In vector notation U 0 i = g(x i ), U n 0 = 0, U n N = 0. Ui n+1 = Ui n + σ t [ U n x 2 i 1 2Ui n + Ui+1 n ]. U n+1 h = U n h σ t x 2 AU n h where A is the same matrix as in example 1. If n = 0, right hand side is known. Thus Uh n t = (I σ x 2 A)n G, G = (g(x 1 ),, g(x N 1 )) T.

10 10 CHAPTER 1. FINITE DIFFERENCE METHOD This is called forward Euler or explicit scheme. If we change the right hand side to [ U n+1 i 1 U n+1 i Ui n t = σ ] n+1 2Ui + Ui+1 n+1 x 2 Ui n+1 = Ui n + σ t [ U n+1 n+1 x 2 i 1 2Ui + Ui+1 n+1 ]. (I + σ t x 2 A)n U n h = G, G = (g(x 1),, g(x N 1 )) T. This is called backward Euler or implicit scheme Error of difference operator For u C 2, use the Taylor expansion about x i u i+1 = u(x i + h i ) = u(x i ) + h i u (x i ) + h2 i 2 u (ξ), ξ (x i, x i+1 ) u i+1 u i h i u (x i ) = h i 2 u (ξ). Expand about x i+1, u i = u i+1 h i u (x i ) + h i 2 u (θ). These are first order accurate. To derive a second order scheme, expand about x i+1/2, u i+1 = u i+1/2 + h i 2 u (x i+1/2 ) (h i 2 )2 u (x i+1/2 ) (h i 2 )3 u (3) (ξ) Subtracting, u i = u i+1/2 h i 2 u (x i+1/2 ) (h i 2 )2 u (x i+1/2 ) 1 6 (h i 2 )3 u (3) (ξ). u i+1 u i h i = u (x i+1/2 ) + O(h 2 i ). Thus we obtain a second order approximation to u (x i+1/2 ). By translation, we have u i+1 u i 1 2h i u (x i ) = O(h 2 i /6) if h i = h i+1. H.W. Do the same for irregular mesh.(use weighted difference) Assume h i = h i+1 and we substitute the solution of differential equation into the difference equation. Using u = f we obtain ( u i 1 + 2u i u i+1 ) h 2 f(x i )

11 1.2. FINITE DIFFERENCE OPERATOR 11 = 1 h 2 ( u i + hu i h2 2 u i + h3 6 u(3) h4 24 u(4) (θ 1 ) + 2u i ) + 1 h 2 ( u i hu i h2 2 u i h3 6 u(3) h4 24 u(4) (θ 2 )) f(x i ) = u i f(x i ) h2 24 (u(4) (θ 1 ) + u (4) (θ 2 )) truncation error = h2 24 max u(4). We let τ h = L h u F h and call it the truncation error(discretization error). Definition We say a difference scheme is consistent if the truncation error approaches zero as h approaches zero, in other words, if L h u f 0 in some norm. Truncation error measures how well the difference equation approximates the differential equation. But it does not measure the actual error in the solution. Use of different quadrature for f. Instead of f(x i ) we can use 1 12 [f(x i 1) + 10f(x i ) + f(x i+1 )] = 5 6 f(x i) + µ 0 6 f(x i) where µ 0 f(x i ) is the average of f which is f(x i ) + O(h 2 ). H.W Show for uniform grid( use u = f) u i 1 + 2u i u i 1 h 2 = 1 12 [f(x i 1) + 10f(x i ) + f(x i+1 )] + Ch 4 max u (6) (x). Nonuniform grid(irregular mesh) We use central difference scheme to get u (x i+1/2 ) u (x i ) u i+1 u i h i+1 ( u i+1 u i h i+1 and u (x i 1/2 ) u i u i 1 h i. Thus, u i u i 1 h i )/( h i+h i+1 2 ). H.W. Find truncation error for u (x i ) in case of nonuniform grid. L h u f = 2[ h i u i+1 + (h i + h i+1 )u i h i+1 u i 1 ]/h i h i+1 (h i + h i+1 ) f(x i ) = 2[ h i (u i + h i+1 u i + h2 i+1 2 u i + h3 i+1 6 u(3) i + O(h 4 )) + (h i + h i+1 )u i h i+1 (u i h i u i + h2 i 2 u i h3 i 6 u(3) i + O(h 4 ))]/h i h i+1 (h i + h i+1 ) f(x i ) = u (2) i f i (h i+1 h i ) + O(h 2 i + h 2 i+1).

12 12 CHAPTER 1. FINITE DIFFERENCE METHOD H.W Use 1 3 [f(x i) + f(x i+1 ) + f(x i 1 )] for the right hand side. truncation error? What is the Definition L h is said to be stable if U h C L h U h C F h for all h > 0 where U h is the solution of the difference equation, L h U h = F h. Also note that L h is stable if and only if L 1 h is bounded. Definition A finite difference scheme is said to converge if U h u 0 as h 0. U h u is called a discretization error. Theorem (P. Lax). Given a consistent scheme, stability is equivalent to convergence. Proof. Assume stability. From L h u f = τ h, L h U h F h = 0, we have L h (u U h ) = τ h. Thus, u U h C L h (u U h ) = C τ h 0. Hence the scheme converges. Obviously a convergent scheme must be stable. From the theory of p.d.e, we know u C f. Hence U h U h u + u O(τ h ) + C f C f C F h. 1.3 Elliptic equation Basic finite difference method for elliptic equation In this chapter, we only consider finite difference method. First consider the following elliptic problem:(dirichlet problem by Finite Difference Method) u u = f in Ω = g on Ω (1) Approx. D.E. (u xx + u yy ) = f at each interior mesh pt. (2) The unknown function is to be approximated by a grid function u

13 1.3. ELLIPTIC EQUATION 13 (3) Replace the derivative by difference quotient. u(x + h) = u(x) + hu x (x) + h2 2 u xx(x) + h3 6 u xxx(x) + O(h 4 ) u(x h) =... u(x + h) 2u(x) + u(x h) h 2 = u xx (x) + O(h 2 ). u xx (x, y) = [u(x + h, y) 2u(x, y) + u(x h, y)]/h 2. u yy (x, y) = [u(x, y + h) 2u(x, y) + u(x, y h)]/h 2 (x, y + h) (x h, y) (x, y) (x + h, y) (x, y h) Figure 1.1: 5-point Stencil This picture is called, Molecule, Stencil, Star, etc. For each point (interior mesh pt), approx 2 u = u by 5-point stencil. By Girshgorin disc theorem, the matrix is nonsingular. L[u] is called differential operator while L h [u] is called finite difference operator, e.g., L h [u](x, y) = [ 4u(x, y) + u(x + h, y) + u(x h, y) + u(x, y + h) + u(x, y h)]/h 2 or more generally, L[u] = [ x, y ] Diag{a 11, a 22 } [ u x u y ] + cu = (a 11 u x ) x (a 22 u y ) y + cu Uniform meshes u x (x) (u x ) x (x). = u(x+h) u(x h) 2h. = ux(x+ h 2 ) ux(x h 2 ) h Central difference

14 14 CHAPTER 1. FINITE DIFFERENCE METHOD u x (x + h 2 ) u x (x h 2 ) = u(x+h) u(x) h = u(x) u(x h) h Nonuniform meshes (a 11 u x ) x = [(a 11 u x )(x + h 2 ) (a 11u x )(x h 2 )]/h u(x + h 2 ) = u(x) + h 2 u x (x) + h2 2 2 u xx + h3 2 3! u(3) + h 1 u(x h 1 ) = u(x) h 1 u x (x) + h2 1 2 u xx h3 2 3! u(3) h 2 h 1 u(x + h 2 ) h 2 u(x h 1 ) = (h 1 h 2 )u(x) + 2h 1 h 2 u x (x) + h 1h 2 2 (h 2 h 1 )u xx + u x (x) = h 1u(x + h 2 ) h 2 u(x h 1 ) (h 1 h 2 )u(x) + O(h) 2h 1 h 2 This is only first order accurate. To consider the second derivatives, we shall lose O(h) accuracy. To get a second order method multiply two equations respectively by h 2 1, h2 2 and subtract to get h 2 1u(x + h 2 ) (h h 2 2)u(x) h 2 2u(x h 1 ) ( h = (h 2 h h 1 h 2 3 2)u xx (x) + 2 h 2 ) 1 + h3 1 h2 2 max u. 6 6 Hence u xx is seconder order accurate. Assume the differential operator is of the form(with γ > 0) L[u] [u xx + u yy ] + γu = f whose discretized form L h [U] = a 0 U(x, y) a 1 U(x + h, y) + = F (x, y) satisfies 1 h γh γh γh rγh 2 (1) L h [u] = L[u] + O(h 2 ) as h 0. u is true solution. (2) AU = F + Bdy, Au = [ u γu + O(h 2 )] + Brdy U 1 U 2 U 3 U 4 = F

15 1.3. ELLIPTIC EQUATION 15 A(U u) = O(h 2 ) = ε Then the discretization error U u = A has the form 1 ε(depends on h) and satisfies U u A 1 ε A 1 O(h 2 ) If we put D = diaga = {a 11,..., a nn }, then D 1 A(U u) = D 1 ε. Write D 1 A = I + B, where B is off diagonal. Then we know B = 4 4+γh 2 < 1 if γ > 0. Thus (D 1 A) 1 = (I + B) 1 exists and Hence (D 1 A) 1 = (I + B) B 4 + γh2 γh 2. U u (D 1 A) 1 D 1 ε 4 + γh2 h 2 γh γh 2 O(h2 ) = O(h 2 ) 0 Thus, we have proved the following result. Theorem Let (1) u C 4 (Ω) (2) r > 0 (3) uniform mesh Then U u = O(h 2 ) as h 0. Generally, when A, B, C are not constant, we can still put the problem into a conservative form as follows: L[u] = Au xx ( + 2Bu xy ) + Cu yy + Du x + Eu y + F u + G = 0 A B = T u (A B C x + B y D)u x (B x + C y E)u y + F u + G, where = it is self-adjoint. ( ) x, y, so that u = ( ux u y ). If A x + B y D = 0 and B x + C y E,

16 16 CHAPTER 1. FINITE DIFFERENCE METHOD Treating the cross term like u xy Assume u xy = u yx, we approximate u y x by δ0 yδxu 0 h where δxu 0 h (P ) = U h (E) U h (W ) 2 x is the central difference. Then from δ 0 yu h (P ) = U h (N) U h (S) 2 y and forward -backward difference formula we get δyδ 0 xu 0 h = 1 [ U h (NE) U h (NW ) U h (SE) U h ] (SW ) 2 y 2 x 2 x Change of variable method to eliminate the cross term One can transform the variable so that the resulting equation in new variable does not have cross term. Lemma Let s = s(x, y), t = t(x, y) be a coordinate transform which is locally one-to-one onto. Denote its derivative by (s,t) (x,y) = P T, Jacobian matrix. Then we have [ ] [ ] [ ] ux us s (x,y) u = = x + u t t x us = P = P u y u s s y + u t t (s,t) u y u t In other words, (x,y) = ( ) / x = / y ( s x s + t x t s y s + t y t ) = ( s x s y t x t y ) ( ) / s = P / t (s,t) Hence T (x,y) = T (s,t) P T and we see that T (x,y) A (x,y)u = T (s,t) P T AP (s,t) u. If A is symmetric, there exists a P such that P T AP = diagonal = {d 1, d 2 }. If we choose s(x, y), t(x, y) so that (s,t) (x,y) = P T, then T (x,y) A (x,y)u = s ( ) u d 1 + ( ) u d 2. s t t Example Transform the problem u xx + 4u xy + u yy = 0 so that it does not have cross term.

17 1.3. ELLIPTIC EQUATION 17 ( ) 1 2 Since A =, its eigenvalues are 3, 1 with corresponding eigenvectors 2 1 ( ) 1 1 (1, 1) and (1, 1), we see that with P =, we have 1 1 ( ) P T 3 0 AP =. 0 1 Hence the transformed equation is P T = ( ) 1 1 = 1 1 s (3 u s ) t ( u t ) = 0. ( ) s/ x s/ y t/ x t/ y. s = x + y, t = x y. ( s t ) = P T ( x y). If s = constant, ds = s x dx + s y dy = 0, so the s = constant line is described as dy dx = sx s y = P 11 P 12. Likewise, if t = constant, dt = t x dx + t y dy = 0 dy dx [ ] = tx t y = P 21 cos λ sin λ P 22. If P =, s = x cos λ y sin λ, t = x sin λ + y cos λ, sin λ cos λ [ ] b cot 2λ = c a a b 2 when A is. b c Treatment of irregular boundaries(dirichlet boundary conditions Figure 1.2: Ω h, regular, irregular Let Ω be a domain with grid. Let Ω h be the set of all grid points in Ω. Definition Two points P, Q on the grid are said to be properly adjacent if they are adjacent and the line segment connecting P, Q belongs to Ω.

18 18 CHAPTER 1. FINITE DIFFERENCE METHOD Let Ω h be the set of all regular points, i.e. the set of all points P Ω h such that its four adjacent points belong to Ω h and they are properly adjacent to P. Let Ω h = Ω h Ω h be the set of all irregular points. In the following, we let E be the east neighbor point of P in Ω h and let W be the west neighborhood point of P in Ω h, etc. Method 1 We form the difference equation L h U h = f h at all regular points only while we let U h (P ) = g(q) if P is an irregular point. Here Q = P if P Ω. Otherwise, Q is a point of Ω closest to P. Here U h (P ) is not an unknown. Method 2(Collatz-linear interpolation) h 1 h 2 W W P E Figure 1.3: Near irregular boundary We form L h U h = f h at all points of Ω h as follows: First we form L h U h = f h at all regular points of Ω h. If P Ω h is an irregular point lying near west part of Ω, take the point of intersection W of the line segment EP with Ω. Then we let U h (P ) = h 1 U h (E) + h 2 u(w ). h 1 + h 2 h 1 + h 2 Now append this equation to the difference equation. If E happens to belong to Ω also, then U h (P ) is completely determined, hence we do not need to append it to the difference equation. Remark The last equation has nothing to do with the differential equation itself, thus it may break certain properties of matrix.

19 1.3. ELLIPTIC EQUATION 19 Method 3(Shortley-Weller) For an irregular point P, we set 2 u x 2 (P ). = 2 ( U h (E) U h (P ) h 2 U h (P ) u(w ) h 1 ) /(h 1 + h 2 ). Advantage: This equation comes from the differential equation, thus preserving(hopefully) certain properties of the matrix(like positive definiteness, banded structure, diagonal dominance). But usually symmetry breaks down. Method 4 Let P be an irregular point whose west neighbor point W lies outside of Ω h. Assuming u is defined at W, we use extrapolation to get to get U h (W ) = αu h (W ) + βu h (P ), where W is the point of intersection of the line segment from P with Ω. (α = h 2 h 1, β = h 2 h 1 h 1. h 2 = h and h 1 is the distance from P to the boundary.) We substitute U h (W ) in the difference equation. (It is called fictitious point method) Neumann or Robin boundary condition(regular point) P E Figure 1.4: regular boundary point Consider the boundary condition of type u n + γu = g on Ω. Let P be a regular boundary point(boundary point lying on the grids). If the boundary is vertical line, then use one sided difference to get U h (P ) U h (E) h and append it to the difference equation. + γ(p )U h (P ) = g(p )

20 20 CHAPTER 1. FINITE DIFFERENCE METHOD If P is an irregular boundary point(figure 1.3), use U h (E) U h (W ) 2h +γ(p )U h (P ) = g(p ). Now solve it for U h (W ) and substitute it into the difference equation at P to get a new equation. 1 h 2 ( U h (S) U h (W ) + 4U h (P ) U h (N) U h (E)) = f(p ) If P is near corner do the same for north and south derivative. Neumann or Robin boundary condition(irregular point) C C B A A Figure 1.5: irregular boundary point We let C be a grid point not in Ω h. Draw a normal line to Ω and let C be the point of intersection with Ω. Now treat C as a grid point. Extend CC to the closest grid line consisting of AB, letting A denote the intersection of extension and the segment AB. Use u. n (C ) = u(c) U h (A ) CA U h (C. ) = (1 σ)u h (A ) + σu h (C) U h (C) U h (A ) + σ(c )U h (C ) = g(c ) CA where U h (A ) is obtained by interpolation. U h (A ) = (1 α)u h (A) + αu h (B) This is an equation involving unknowns U h (A), U h (B) and U h (C).

21 1.3. ELLIPTIC EQUATION 21 Example u + (x 2 + y 2 )u = 40(2 x y + 4xy)e xy, 0 x, y 1 where u(x, y) = [10 20{(x 1 2 )2 +(y 1 2 )2 }]e xy on the boundary. Compare with error is O(h 2 ). h = mesh U u = = = = = Assuming the error is of the form U u = Mh α, we see U h u u = U h 2 Mhα M( h 2 )α = 2α. These are computable with u replaced by U hmin and [ ] / U h u α = log log 2 u U h 2 Theorem (Maximum Principle). Assume A is positive definite symmetric, c 0. Let u be the solution of elliptic p.d.e. given by L[u] = [ ] u a ij + cu = 0 = T A u + cu. x i i x j Then for (x, y) int Ω j u(x, y) max u(x, y) (x,y) Ω Proof. Assume c > 0. There exists orthogonal matrix P such that P T AP = diag{d 1, d 2 }. If u has a positive maximum at some interior point Q = (x, y ) of Ω, then define ( ) ( ) s x = P T (x, y ) t y so that L[u] = (s,t) P T AP (s,t) u + cu = 0. At Q, u s (Q) = u t (Q) = 0, u ss (Q) 0 and u tt (Q) 0. Hence L[u] = (d 1 u s ) s (Q) (d 2 u t ) t (Q) + c(q)u(q) = 0

22 22 CHAPTER 1. FINITE DIFFERENCE METHOD Remembering, d 1 > 0, d 2 > 0, cu > 0 this is a contradiction. Similarly, u cannot have negative minimum. Now if c 0 not c > 0 we consider a perturbation. Choose α so large that L[e αx ] = (d 1 α 2 + d 2 α 2 c)e αx < 0 and let v = u + Ee αx. L[v] = L[u] + EL[e αx ] < 0 for all E > 0. Suppose v has a pos. max. at Q, an interior point of Ω. Then L[v] = d 1 v ss (Q) d 2 v tt (Q) + c(q)v(q) 0, a contradiction. Hence u(x, y) v(x, y) < max Ω {u + Ee αx }. Let E 0. Then u(x, y) max Ω u. Applying maximum principle to u and u, we obtain the following result. Corollary If then u 0. L[u] = 0 in Ω u = 0 on Ω, As a consequence we have uniqueness of solution. Corollary If u 1, u 2 satisfy then u 1 = u 2. L[u i ] = 0 in Ω u i = g on Ω, A symmetric, positive definite system satisfying L[u] 0 has a unique solution. Theorem (Discrete max. principle). Let the grid function U satisfy the finite difference equation L h [U] Q A(P, Q)U(Q) = 0, i.e, A(P, P )U(P )+A(P, E)U(E)+A(P, S)U(S)+A(P, N)U(N)+A(P, W )U(W ) = 0 for each mesh point P, where coefficient A(P, Q) are generated by finite difference method(e.g, p.955) and A is pos. def. weakly diagonally dominant. Then min P Ω h U(P ) U(P ) max P Ω h U(P ), for all P int Ω h.

23 1.3. ELLIPTIC EQUATION 23 Proof. Solving for U(P ), we have U(P ) = 1 A(P,P ) A(P, Q)U(Q) Q P Thus U(P ) A(P, Q) A(P, P ) max U(Q) max U(Q), Q P Q P Q P since A(P, P ) Q P A(P, X). Corollary implies U = 0. L h (U) = 0 in Ω U = 0 on Ω Theorem Discrete max. principle also hold if L h [U] 0. Proof. Assume U(p 0 ) > U(p) for all p Ω h, p 0 Ω h. Then h U(p 0 ) = 1 h 2 [U(p 1 ) + U(p 2 )] + 1 x h 2 [U(p 3 ) + U(p 4 )] 2( 1 y h x h 2 )U(p 0 ). y But h U = L h [U] 0. Hence ( 1 1 U(p 0 ) 2( ) h h 2 x h 2 [U(p 1 ) + U(p 2 )] + 1 ) 2 x h 2 [U(p 3 ) + U(p 4 )] y y U(p 0 ) which means U(p 0 ) = U(p ν ), ν = 1, 2, 3, 4. Repeat the argument for each p ν instead of p 0 until we arrive at the boundary point of Ω. Then we get U(p) U(p 0 ), p Ω h Ω h, which is a contradiction to the assumption. Note. Minimum principle is obtained when L h [U] 0. Theorem Let u be the solution of L[u] = T A u + cu = 0 in Ω and u = g on Ω and let U be the finite sequence of grid functions satisfying L h (U) = 0, where L h (u) = O(h α ), α > 0(truncation error). If A is constant, diagonally dominant, positive definite, then U u = O(h α ) as h 0.

24 24 CHAPTER 1. FINITE DIFFERENCE METHOD Proof. Let w = U u, then L h [w] = L h [U] L h [u] = L h [u] and w = 0 on Ω. Let s(x, y) r 2 (x x 0 ) 2 (y y 0 ) 2 with (x 0, y 0 ) int Ω, r chosen so large that the circle s = 0 contains Ω. Then L h [s] = L[s] because s is quadratic. (compute it) L[s] = T A s + cs = a 11 s xx (a 12 + a 21 )s xy a 22 s yy + cs = 2(a 11 + a 22 ) + cs 2(a 11 + a 22 ). There exist M > 0 such that L h [u] Mh α by the hypothesis L h [u] = O(h α ). We see that [ Mh α ] s(x, y) L h Mh α L h [u]. 2(a 11 + a 22 ) Also [ ] L h ±w Mhα s(x, y) = ±L h [w] L h [ ] L h [u] Mh α 0 2(a 11 + a 22 ) (Recall w = U u and L h [w] = L h [u] and w = 0 on Ω) Since the discrete maximum principle also holds if L h (U) = 0 is replaced by L h [U] 0, max P Ω [ ±w Thus w Mh α ] [ s max ±w 2(a 11 + a 22 ) P Ω Mhα s 2(a 11 +a 22 ) and u U = w Mh α ] s 2(a 11 + a 22 ) Mh α 2(a 11 + a 22 ) max Ω s Mhα r 2 2(a 11 + a 22 ). Mh α = 2(a 11 + a 22 ) min s 0. Ω Convection -diffusion equation Consider another type of differential equation, namely a special case of convection diffusion equation. ɛu xx + au x = f u(0) = u 0, u(1) = u 1 or with Neumann condition u (1) = 0 if a(1) > 0. If we use the central difference scheme for the first order derivative, we get ( ) U h ɛ i 1 + 2Ui h Ui+1 h h 2 + a U i+1 h U i 1 h = fi h 2h

25 1.3. ELLIPTIC EQUATION 25 ( ɛ h 2 + a ) Ui 1 h + 2ɛ 2h h 2 U i h ( ɛ h 2 a ) Ui+1 h = fi h 2h Thus the sum of off diagonal elements is a ij = ɛ h 2 + a + ɛ 2h h 2 a. 2h j i If a, h is fixed and ɛ 0, it becomes a/h, while a ii = 2ɛ/h 2 0. Thus the resulting matrix is not diagonally dominant and it causes a lot of problems. For example, the resulting linear system is not positive definite and hence it may be more difficult to solve. But, most importantly, the resulting discretization does not yield an accurate approximation to the problem. One way to fix this situation is to keep the Peclet number : ah ɛ < 2 so that the sum of off diagonal elements is less than 2ɛ/h 2 = a ii. The disadvantage of this scheme is that small h enlarges the size of discrete equation Numerical Difficulties (1) The iterative method may fail to converge (2) The solution may exhibit oscillation which are physically unrealistic (3) Taking small mesh size means large problem size which take more time to solve Upwind difference scheme An alternative way to avoid this difficulty is to use backward difference for u x for a > 0 and forward difference for u x for a < 0. This method of choosing difference scheme is called upwind difference scheme. For a > 0 ɛ ( U h i 1 + 2U h i U h i+1 ( ɛ h 2 + a ) Ui 1 h + h h 2 ( 2ɛ h 2 + a h ) ) U h i + a U i h Ui 1 h = fi h h ( ɛ ) h 2 Ui+1 h = fi h The resulting system is irreducibly diagonally dominant, thus it is an M-matrix and hence Jacobi method works. The following examples are taken from (K.W. Morton-Numerical Solution of convection-diffusion problems, Chapman Hall, 1996)

26 26 CHAPTER 1. FINITE DIFFERENCE METHOD Example ɛ u + b u = 0 on (0, 1) (0, 1) with b = b(cos θ, sin θ) for 0 θ < π 2 and discontinuous inflow boundary condition and u y u(0, y) = { 0, y [0, 1 2 ) 1, y ( 1 2 ] = 0 on the rest of the boundary. This leads to an internal layer along y = x tan θ and a boundary layer at x = 1 for y > tan θ when tan θ < 1 2. Example (Heat equation). u + b u t = ɛ u on Ω (0, T ) u(x, y, 0) = u 0 (x, y) where u 0 (x, y) is a circular cone type centered at (1, 0) with b = b(wy, wx) with exact inflow boundary are needed. 1.4 Parabolic p.d.e s Consider a heat equation on a bar. u t = u xx, 0 x 1, 0 < t T. Theorem (Maximum principle). If u satisfies the above condition for t T, then min{f, g, h} = m min u max u M = max{f, g, h} 0 x 1,0 t T 0 x 1,0 t T

27 1.4. PARABOLIC P.D.E S 27 T g(t) Ω h(t) f(x) 1 Figure 1.6: Domain Proof. Put v = u + Ex 2, E > 0 v t 2 v x 2 = 2E < 0 If v attains a maximum at Q int Ω, then v t (Q) = 0, v xx (Q) 0. Thus (v t v xx )(Q) 0, a contradiction. Hence v has maximum at a boundary point of Ω u(x, t) v(x, t) max v(x, t) M + E. Since E was arbitrary, the proof is complete. For minimum, use E instead of E. More general parabolic p.d.e. u t = Au xx + Du x + F u + G { Explicit write down the values of grid function F.D.M Implicit variables implicitly representing the value Let the grid be given by 0 = x 0 < x 1 < x 2 < < x N+1 = 1, x i = ih, uniform grid 0 = t 0 < t 1 <, t j = jk

28 28 CHAPTER 1. FINITE DIFFERENCE METHOD Explicit method { Ui,j+1 U i,j. k = u t U i+1,j 2U i,j +U i 1,j h 2. = uxx U i,j+1 = λu i 1,j + (1 2λ)U i,j + λu i+1,j where λ = k/h 2. Stability: Error doesn t accumulate. In this case solution remains bounded as time goes on. Theorem If u is sufficiently smooth, then u u(x + h, t) 2u(x, t) + u(x h, t) xx h 2 = O(h2 ) as h 0 and u t u(x, t + k) u(x, t) k = O(k) as k 0 Theorem Suppose u is sufficiently smooth, and satisfies u t = u xx 0 < x < 1, t > 0 u(x, 0) = f(x) u(0, t) = g(t) u(1, t) = h(t). If U i,j is the solution of the explicit finite difference scheme, then for 0 < λ 1 2, max u i,j U i,j =. O(h 2 + k) as h, k 0, i, j i.e, finite difference solution converges to the true solution. u = g 1 (y) u = g 2 (y) h 2h 3h 4h u = f(x) Figure 1.7:

29 1.4. PARABOLIC P.D.E S 29 Proof. Put u ij u(x i, t j ). Then from (1) u i,j+1 u i,j k = u t + O(k) (2) u i+1,j 2u i,j + u i,j h 2 = u xx + O(h 2 ) we get Hence u i,j+1 = u i,j + k h 2 (u i+1,j 2u i,j + u i 1,j ) + k(o(k) + O(h 2 )). u i,j+1 = λu i 1,j + (1 2λ)u i,j + λu i+1,j + Ck(k + h 2 ). Let the discretization error be w i,j = u ij U ij so that w i,j+1 = λw i 1,j + (1 2λ)w i,j + λw i+1,j + O(k 2 + kh 2 ). Since 0 < λ 1 2, 0 1 2λ < 1, three coefficient are positive and their sum is 1. (convex combination) We see w i,j+1 λ w i 1,j + (1 2λ) w i,j + λ w i+1,j + M(k 2 + kh 2 ) for some M > 0. If we define w j = max 1 i N w i,j, then w j+1 w j + M(k 2 + kh 2 ) w j 1 + 2M(k 2 + kh 2 ) w 0 + (j + 1)M(k 2 + kh 2 ). Since w 0 = 0, w j+1 (j + 1)kM(k + h 2 ) T M(k + h 2 ), (j + 1)k T. In fact, M = max (1 0 x 1, 0 t T 2 u tt + k2 12 u xxxx ). Remark If λ > 1 2, the solution may not converge. Exercise Prove the formula is unstable for λ > 1 2. Let u(x, 0) = { ε, x = 1 2 0, x 1 2 with g = h = 0

30 30 CHAPTER 1. FINITE DIFFERENCE METHOD Hence U i,j+1 = λu i+1,j + (1 2λ)U i,j + λu i 1,j, λ = k/h 2 U i,j+1 = λ U i+1,j + (2λ 1) U i,j + λ U i 1, j, 1 i N 1. N 1 i=1 N 2 U i,j+1 = λ i=1 N 1 U i+1,j + (2λ 1) i=1 U i,j + λ since U(x i, t) = 0, i = 1, N. Let S(t j ) = N i=1 U(i, j). Then N U i 1,j, S(t j+1 ) = (4λ 1)S(t j ) = (4λ 1) 2 S(t j 1 ) = = (4λ 1) j+1 S(0) = (4λ 1) j+1 ε. Since the number of nonzero U i,j for each j is 2j + 1, there is a point (x p, t j ) such that U(x p, t j ) 1 2j + 1 S(t j) = 1 2j + 1 (4λ 1)j ε which diverges as j since 4λ 1 > 1. Considering the alternating sign, one can see the solution alternates: j = 0 U i,1 = (1 2λ)ɛ, U i 1,1 = λɛ, U i+1,1 = λɛ U i,2 = 2λ 2 ɛ + (1 2λ) 2 ɛ, U i 1,2 = (1 2λ)ɛ + (1 2λ)ɛ = 3λɛ(1 2λ) < 0. Stability of linear system U 1,j+1. U N 1,j+1 = 1 2λ λ... 0 λ 1 2λ λ λ 1 2λ U 1,j. U N 1,j i=2 g(t j ) h(t j ) In vector form, U j+1 = AU j + G j. Assume G j = 0, j = 1, 2,.... Let µ be an eigenvalue of A. Then by G-disk theorem, 1 2λ µ 2λ 2λ 1 2λ µ 2λ 2λ 1 + 2λ + µ 2λ 1 4λ µ 1 If 0 < λ 1 2, then 1 µ 1, hence stable. If λ > 1 2, then µ > 1 is possible. So the scheme may be unstable. The following example show it is actually unstable.

31 1.4. PARABOLIC P.D.E S 31 Example (Issacson, Keller). Try v(x, t) = Re(e iαx wt ) = cos αx e wt.. v(x, t + t) v(x, t) v(x + x, t) 2v(x, t) + v(x x, t) v t v xx = ( t x 2 e w t ) 1 cos(αx + α x) 2 cos αx + cos (αx α x) = v(x, t) t x 2 ( e w t ) 1 2 cos α x 2 = v(x, t) t x 2 = v(x, t) 1 t {e w t [(1 2λ) + 2λ cos α x]} = v(x, t) 1 [ ( e w t 1 4λ sin 2 α x )] t 2 Thus v is a solution of the difference equation provided w and α satify e w t = 1 4λ sin 2 α x 2. With I.C. v(x, 0) = cos αx, solution becomes ( v(x, t) = cos αxe wt = cos αx 1 4λ sin 2 α x ) t t 2 Clearly, for all λ 1 2, v(x, t) 1. However, if λ > 1 2, then for some x, we have 1 4λ sin 2 α x 2 > 1. So v(x, t) becomes arbitrarily large for sufficiently large t/ t. Since every even function has a cosine series, we may give any even function f(x) of the form f(x) = ν β ν cos(2 ν πx) to get an unstable problem. Implicit Finite Difference Method. Given a heat equation u t = u xx u(0, t) = g(t), t > 0 u(1, t) = h(t) u(x, 0) = f(x), 0 x 1. We discretize it by implicit difference method. U i,j+1 U i,j t = U i+1,j+1 2U i,j+1 + U i 1,j+1 x 2 i = 1,..., N 1. Multiply by t, then with λ = t/ x 2, we have U i,j+1 U i,j = λu i+1,j+1 2λU i,j+1 + λu i 1,j+1 U i,j = λu i+1,j+1 (1 + 2λ)U i,j+1 + λu i 1,j+1 j = 1,..., N 1. e wt

32 32 CHAPTER 1. FINITE DIFFERENCE METHOD This yields a system of N 1 unknowns in {U i,j+1 } i=1 N 1. (1 + 2λ) λ 0 λu 0,j+1 0 U 1,j λ (1 + 2λ) λ +. =. U N 1,j λu N,j+1 U 1,j+1. U N 1,j λ 0 λ (1 + 2λ) Theorem The implicit finite difference scheme is stable for all λ = t/ x 2. (solution remains bounded). Proof. For each j, let U k(j),j be chosen so that U k(j),j U i,j, i = 1,..., N 1. We choose i 0 = k(j + 1) in the following relation. Then U i,j+1 = U i,j + λ{u i+1,j+1 2U i,j+1 + U i 1,j+1 }. (1 + 2λ)U i0,j+1 = U i0,j + λ{u i0 +1,j+1 + U i0 1,j+1} for 1 i N 1. Taking absolute values, (1 + 2λ) U i0,j+1 U i0,j + λ( U i0 +1,j+1 + U i0 1,j+1 ) U i0,j + 2λ U i0,j+1. Thus U i0,j+1 U i0,j U k(j),j and hence U i,j+1 U i0,j+1 U k(j),j for 1 i N 1, and U i,j+1 M = max{f, g, h}, for i = 0 or N, by boundary condition. Repeat the same procedure until we hit the boundary. U i,j+1 U k(j),j U k(0),0 M = max(f, g, h) Using the matrix formulation: We check the eigenvalues of the system AU j+1 = U j + G j Eigenvalue of A satisfies µ + (1 + 2λ) 2λ by G-disk theorem. From this, we see µ 1 and hence the eigenvalues of A 1 is less than one in absolute value. Thus U j+1 A 1 (U j + G j ) = = A j 1 U 0 + A j 1 G 0 + A j 2 G A 1 G j.

33 1.4. PARABOLIC P.D.E S 33 U j+1 A j 1 U 0 + A A 1 max G j remain bounded. Note. A does not have 1 as eigenvalues and all the eigenvalues are positive real. Theorem For sufficiently smooth u, we have u ij U ij = O(h 2 + k) as h and k 0 (for all λ) Proof. Let u ij = u(x i, t j ) be the true solution. Then we have u i,j+1 u i,j k = 1 h 2 {u i+1,j+1 2u i,j+1 + u i 1,j+1 } + O(h 2 + k) Let w i,j = u i,j U i,j be the discretization error. Then w i,j+1 = w i,j + λ{w i+1,j+1 2w i,j+1 + w i 1,j+1 } + O(kh 2 + k 2 ) (1 + 2λ)w i,j+1 = w i,j + λw i+1,j+1 + λw i 1,j+1 + O(kh 2 + k 2 ) Let w j = max i w i,j. Then (1 + 2λ) w i,j+1 w j + 2λ w j+1 + O(kh 2 + k 2 ) and so (1 + 2λ) w j+1 w j + 2λ w j+1 + O(kh 2 + k 2 ). Thus w j+1 w j + C(kh 2 + k 2 ) w 0 + C(j + 1)k(k + h 2 ) w 0 + CT (k + h 2 ) = CT (k + h 2 ) for t = (j + 1)k T Discretization of parabolic p.d.e, General Case Consider c u = t x (p u x ) + y (p u ) γu + f y in Ω [0, T ] I.C. u(x, y, 0) = h(x, y) in Ω B.C. u(x, y, t) = g(x, y, t) for (x, y) Ω.

34 34 CHAPTER 1. FINITE DIFFERENCE METHOD u i,j+1 t = j + 1 t = j u i 1,j u i,j u i+1,j u i 1,j+1 u i,j+1 u i+1,j+1 u i,j Figure 1.8: forward backward Euler method Stencil where c, p, γ, f are functions of x, y and t. Assume 0 < p 0 p(x, y, t) p 1 0 γ(x, y, t) γ 0 < c 0 c(x, y, t) c 1. With Ui,j n = U(x i, y j, t n ) we have ) M h Ui,j n 1 = (p n x 2 i+1/2,j U i+1,j n + pn i 1/2,j U i 1,j n (pn i+1/2,j + pn i 1/2,j )U i,j n + 1 y 2 (p n i,j+1/2 U n i,j+1 + pn i,j 1/2 U n i,j 1 (pn i,j+1/2 + pn i,j 1/2 )U n i,j For 0 θ 1, we let [θc n+1 ) γ n i,j U n i,j ij + (1 θ)c n ij] U i,j n+1 Ui,j n = θm h Ui,j n+1 + (1 θ)m h Ui,j n + θfij n+1 + (1 θ)fij n t For θ = 0, we have For θ = 1, we have c n Ui,j n+1 Ui,j n ij t c n+1 Ui,j n+1 Ui,j n ij t For θ = 1 2, we have Crank-Nicolson. Matrix formulations = M h U n i,j + f n ij forward Euler = M h Ui,j n+1 + fij n+1 For θ = 1, (( ) ) c n+1 ij I M h U n+1 = t backward Euler ( ) c n+1 ij U n + F t n

35 1.5. HYPERBOLIC EQUATION 35 t = j + 1 u i 1,j+1 u i,j+1 t = j u i 1,j u i,j u i+1,j+1 u i+1,j Figure 1.9: Crank-Nicolson method Stencil For θ = 1 2, (( ) ) c n+1 ij I 1 t 2 M h where c ij = 1 2 (cn+1 ij + c n ij ). ( U n+1 cij = t + 1 ) 2 M h U n ( F n + F n+1 ) Exercise (1) Consider a heat equation with k = 1 u t = ku xx 0 < x < 1, t > 0 u(x, 0) = f(x) u(0, t) = g(t) u(1, t) = h(t). where k > 0 is constant. If f(x) = cos πx, g(t) = e kt π 2, h(t) = e kt π 2 it has solution u = e kt π 2 cos πx. Use the following method to compute numerical solution up to T = 1.0. Check u U 2 or u U for each time step j t. (a) Explicit FDM with h = 1 10, 1 20, for k = λh2 where λ = 0.2, 0.4 and 0.6. (b) Implicit FDM with h = 1 10, 1 20, for k = λh2 where λ = 0.2, 0.4, 0.6, 0.8 and 1.6. You can use either Gauss-Seidel type of iteration method or LU-decomposition to solve the system of equations arising in the implicit method. 1.5 Hyperbolic Equation { u t + a 11 u x + a 12 v x = b 1 v t + a 21 u x + a 22 v x = b 2 (1.1)

36 36 CHAPTER 1. FINITE DIFFERENCE METHOD where a i,j and u, v and function of (x, t). u(x, 0) = f(x) t 0 v(x, 0) = g(x) < x < Let w = [u, v] T, b = [b 1, b 2 ] T, A = {a ij }. Then the D.E. is of the form W t + AW x = b. Equation (2.94) is called hyperbolic, if there exist a P such that P 1 AP = diag{λ 1, λ 2 } where λ i (x, t) are real and distinct. Let Z = [z 1, z 2 ] T be related to W by W = P Z, then P t Z + P Z t + A(P x Z + P Z x ) = b P Z t + AP Z x = b (P t + AP x )Z Z t + P 1 AP Z x = P 1 {b (P t + AP x )Z} = β(x, t, Z). Componentwise, (Z i ) t + λ i (Z i ) x = β i, i = 1, 2. Let x i (t) be the solution of the o.d.e. dx i dt = λ i(x i, t) such that x i (t ) = x = λ i (x i, t) (called charac- Let Z i (t) Z i (x i (t), t) be defined along the curve dx i dt teristics). Then dz i dt = Z i x dx i dt + Z i t = λ i(x i (t), t) Z i x + Z i t = β i(x i, t, Z) Thus Z i (x i (t), t) solve the o.d.e (p.d.e on the characterstics) with Z i (0) = Z i (x i (0), 0) = (P 1 W ) i (x i (0), 0). The shaded part is called the Domain of dependence of (x, t ) and its base is called the interval of dependence. A necessary condition for convergence: The numerical domain of dependence must contain the analytic domain of dependence. If the grid point is only in the inner region of the domain of dependence, then changing f by f + δ, g by d + δ near the boundary yields the same solution (numerical).

37 1.5. HYPERBOLIC EQUATION 37 Example Plucked string of wave { u t = cv x v t = cu x u(x, 0) v(x, 0) = f(x) G(σ) dσ = g(x) = 1 c x 0 u tt c 2 u xx = 0 u(x, 0) = f(x) u t (x, 0) = cv x (x, 0) = G(x) [ ] [ ] [ ] [ ] u 0 c u 0 + =. v c 0 v 0 t Eigenvalues of A are ±c. Corresponding to the eigenvector ( ( 1 1), 1 1) Therefore, ( ) 1 1 P =. 1 1 If W = P Z, then P 1 AP = ( ) c 0 0 c x and P 1 = 1 2 ( ) ( ) c 0 Z t + Z 0 c x = 0 z 1 t + c z 1 x = 0 z 2 t c z 2 x = 0 x(t ) = x Along dx 1 dt = c, dx 2 dt = c, dz i dt = 0. Thus x 1 (t) = ct + x ct Z 1 (ct + x ct, t) = Z 1 (x ct, 0) x 2 (t) = ct + x + ct Z 2 ( ct + x + ct, t) = Z 2 (x + ct, 0) Z = P 1 W = 1 2 ( ) ( u v ) Z 1 (x, t) = 1 2 [u(x, t) v(x, t)] = 1 2 [f(x ct ) g(x ct )] Z 2 (x, t) = 1 2 [u(x, t) + v(x, t)] = 1 2 [(f(x + ct ) + g(x + ct )]

38 38 CHAPTER 1. FINITE DIFFERENCE METHOD Hence [ ] u v (x,t ) [ ] [ f(x = P Z = ct ) g(x ct ] ) f(x + ct ) + g(x + ct ) [ = 1 f(x ct ) g(x ct ) + f(x + ct ) + g(x + ct ] ) 2 f(x ct ) + g(x ct ) + f(x + ct ) + g(x + ct ) These are called D Alembert solutions Method of Characteristics Numerical procedure See R.S. Varga or Y. Gregory Ch16 dx i dt = λ i (x i, t), i = 1, 2, dz i dt = β i (x i, t, Z 1, Z 2 ) Assume Z i (t, x) is known at t-th level (say by interpolation). 1st step: Find P 1 ( x 1, t), P 2 ( x 2, t) by 2nd step: x x i t = λ i (x, t ), i = 1, 2 (Backward) Z i (P ) Z i (P i ) t = β i (P i, Z 1 (P i ), Z 2 (P i )) (Forward) solve for Z i (P ), i = 1, 2. t t P (x, t ) t x 1 x 2 x Figure 1.10: Find x 1, x 2