Lecture 20 Heat Equation and Parabolic Problems

Transcription

1 .539 Lecture 2 Heat Equation and Parabolic Problems Prof. Dean Wang. Classification of Differential Equations In most texts the classification is given for a linear second-order differential equation in two indepent variables of the form au + bu " + cu + du + eu + fu = g () The classification deps on the sign of the discriminant, b 4ac < elliptic = parabolic > hyperbolic Example : Poisson equation: u + u = g elliptic problem Heat equation: u = ku parabolic problem Wave equation: u = c u hyperbolic problem Transport equation: u + au = hyperbolic problem (2) 2. Numerical Methods for Heat Equations A simple D Heat equation: u = αu x u x, = η x u, t = g t u, t = g t This is the classic example of a parabolic equation, and many of the general properties seen here carry over to the design of numerical method for other parabolic equations. (3) In practice we solve (3) on a discrete grid with grid points x, t, where x = jh (4a) t = nk (4b) Here h is the mesh spacing on the x-axis and k = t is the time step. Let U u x, t represents the numerical approximation at grid point x, t. One simple numerical discretization of (3) is = αd U = U 2U + U (5) This employs a standard centered difference in space and a forward difference in time. This is an explicit method since we can compute each U explicitly in terms of the previous values:

2 U = U + " U 2U + U (6) Figure shows the stencil of this method. Fig.. An explicit scheme. Another one-step method, which is much more useful in practice, as we will see below, is the Crank-Nicolson method (see Figure 2), = " " = + (7) which can be rewritten as U = U + " U 2U + U + U 2U + U or ru + + 2r U ru = ru + 2r U + ru where r = " () Figure 2 shows the stencil for this method. This is an implicit method and gives a tridiagonal system of equations to solve for all the values U simultaneously. (8) (9) Fig. 2. An implicit method. 2

3 In matrix form this is + 2r r r + 2r r r + 2r r r + 2r r r + 2r r g t + g t + 2r U + ru ru + 2r U + ru = ru + 2r U + ru ru + 2r U + ru + 2r U + r g t + g t ru U U U U U () Since a tridiagonal system of m equations can be solved with O m work, this method is essentially as efficient per time step as an explicit method. Example 2. Solve the following heat equation. u = u x u x, = 2x x 2 2x x u, t = u, t = (2) Matlab program: %% Solve the heat equation % ut = uxx for <= x <=, u() = u() = %% Create the mesh % Create the time-step size t = ; k =.2; % Create the spatial grid x = ; x_ = ; % Initialize the grid spacing h =.5; % Total number of mesh points m = (x_-x)/h + ; x = linspace(x,x_,m); %% Solve the time integration using the explicit method % Initialize U U = zeros(m:); for i = :m if (x(i) <= /2) U(i) = 2*x(i); 3

4 U(i) = 2-2*x(i); % Initial value U() = ; U(m) = ; U_n = zeros(m:); U_n() = ; U_n(m) = ; figure subplot(3,2,); for i = :5 for j = 2:m- U_n(j) = U(j) + k/h^2*(u(j-) - 2*U(j) + U(j+)); % Swap the old and new values U = U_n; if mod(i, ) == plot(x,u); axis([ ]); title({['explicit method (time step size: pause(.); %% Solve the heat equation using the Crank-Nicolson method % Set up the linear system of m-2 equations: AU = F % Create the coefficient matrix A r = k/2/h^2; A = zeros(m-2,m-2); F = zeros(m-2,); U = zeros(m,); for i = :m if (x(i) <= /2) U(i) = 2*x(i); U(i) = 2-2*x(i); U_n = zeros(m,); A(,) = + 2*r; A(,2) = -r; A(m-2,m-2) = + 2*r; A(j,j-) = -r; A(j,j) = + 2*r; A(j,j+) = -r; subplot(3,2,2); % Time integration for i = :5 F() = ( - 2*r)*U(2) + r*u(3); F(m-2) = r*u(m-2) + ( - 2*r)*U(m-); F(j) = r*u(j) + ( - 2*r)*U(j+) + r*u(j+2); 4

5 % Solve the linear system U_n(2:m-) = A\F; U = U_n; % Visualize the results if mod(i, ) == plot(x,u); axis([ ]); title({['crank-nicolson method (time step size: pause(.); k =.5; for i = :m if (x(i) <= /2) U(i) = 2*x(i); U(i) = 2-2*x(i); subplot(3,2,3); for i = :4 for j = 2:m- U_n(j) = U(j) + k/h^2*(u(j-) - 2*U(j) + U(j+)); % Swap the old and new values U = U_n; if mod(i, ) == plot(x,u); axis([ ]); title({['explicit method (time step size: pause(.); %% Solve the heat equation using the Crank-Nicolson method % Set up the linear system of m-2 equations: AU = F % Create the coefficient matrix A r = k/2/h^2; A = zeros(m-2,m-2); F = zeros(m-2,); U = zeros(m,); for i = :m if (x(i) <= /2) U(i) = 2*x(i); U(i) = 2-2*x(i); A(,) = + 2*r; A(,2) = -r; A(m-2,m-2) = + 2*r; 5

6 A(j,j-) = -r; A(j,j) = + 2*r; A(j,j+) = -r; U_n = zeros(m,); A(,) = + 2*r; A(,2) = -r; A(m-2,m-2) = + 2*r; subplot(3,2,4); % Time integration for i = :4 F() = ( - 2*r)*U(2) + r*u(3); F(m-2) = r*u(m-2) + ( - 2*r)*U(m-); F(j) = r*u(j) + ( - 2*r)*U(j+) + r*u(j+2); % Solve the linear system U_n(2:m-) = A\F; U = U_n; % Visualize the results if mod(i, ) == plot(x,u); axis([ ]); title({['crank-nicolson method (time step size: pause(.); k =.; for i = :m if (x(i) <= /2) U(i) = 2*x(i); U(i) = 2-2*x(i); subplot(3,2,5); for i = :6 for j = 2:m- U_n(j) = U(j) + k/h^2*(u(j-) - 2*U(j) + U(j+)); % Swap the old and new values U = U_n; if mod(i, ) == plot(x,u); axis([ ]); title({['explicit method (time step size: pause(.); %% Solve the heat equation using the Crank-Nicolson method 6

7 % Set up the linear system of m-2 equations: AU = F % Create the coefficient matrix A %k =.5; r = k/2/h^2; A = zeros(m-2,m-2); F = zeros(m-2,); U = zeros(m,); for i = :m if (x(i) <= /2) U(i) = 2*x(i); U(i) = 2-2*x(i); A(,) = + 2*r; A(,2) = -r; A(m-2,m-2) = + 2*r; A(j,j-) = -r; A(j,j) = + 2*r; A(j,j+) = -r; U_n = zeros(m,); A(,) = + 2*r; A(,2) = -r; A(m-2,m-2) = + 2*r; subplot(3,2,6); % Time integration for i = :6 F() = ( - 2*r)*U(2) + r*u(3); F(m-2) = r*u(m-2) + ( - 2*r)*U(m-); F(j) = r*u(j) + ( - 2*r)*U(j+) + r*u(j+2); % Solve the linear system U_n(2:m-) = A\F; U = U_n; % Visualize the results if mod(i, ) == plot(x,u); axis([ ]); title({['crank-nicolson method (time step size: pause(.); 7

8 Explicit method (time step size:.2) time steps: 5 Crank-Nicolson method (time step size:.2) time steps: Explicit method (time step size:.5) time steps: Crank-Nicolson method (time step size:.5) time steps: Explicit method (time step size:.) time steps: Crank-Nicolson method (time step size:.) time steps: Stability Analysis of Numerical Methods From the above example, we can see that the explicit method becomes unstable when the time-step size is larger than the certain value, while the implicit Crank- Nicolson method is stable no matter what time-step size is used. We can understand why the explicit methods are more restrictive in the time-step size than the implicit methods by performing Fourier analysis. We define U = λ e " (3) So we have U = λ e " = λ e " e " = e " U (4) = λ e " = λ e " e " = e " U (5) U Substituting (3), (4), and (5) into the explicit method (6) gives λu = U + μ e " U 2U + e " U (6) where μ = " (7) Dividing (6) by U results in λ = + μ e " 2 + e " 8

9 = 2μ cos ξh = 4μsin (8) λ ξ is called the amplification factor for the mode ξ (wave number). Numerical stability requites that λ (9) So we must have μ = " (2) Then it requires that the time-step size k In Example 2, h =.5 and α =, so for numerical stability we should have (2) k." =.25 (22) So it can be seen that for k =.5 and., the solutions blowup. Now we perform stability analysis for the implicit Crank-Nicolson method. Substituting (3), (4), and (5) into the explicit method (9) gives rλe " U + + 2r λu rλe " U = re " U + 2r U + re " U (23) Dividing (23) by U gives rλe " + + 2r λ rλe " = re " + 2r + re " (24) So we have λ = "" " " "# = " " = "# "# "# (25) Therefore the Crank-Nicolson is stable no matter what time step is used. 9