Heat Equation Backward Difference-Numerical Analysis-MATLAB Code, Exercises of Mathematical Methods for Numerical Analysis and Optimization

This is solution to one of problems in Numerical Analysis. This is matlab code. Its helpful to students of Computer Science, Electrical and Mechanical Engineering. This code also help to understand algorithm and logic behind the problem. This code includes: Gauss, Siedel, Iterative, Technique, Algorithm, Initial, Approximation, Matrix, Equations, Iterations, Tolerance

Typology: Exercises

2011/2012

Uploaded on 07/31/2012

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% HEAT EQUATION BACKWARD-DIFFERENCE ALGORITHM 12.2
%
% To approximate the solution to the parabolic partial-differential
% equation subject to the boundary conditions
% u(0,t) = u(l,t) = 0, 0 < t < T = max t,
% and the initial conditions
% u(x,0) = F(x), 0 <= x <= l:
%
% INPUT: endpoint l; maximum time T; constant ALPHA; integers m, N.
%
% OUTPUT: approximations W(I,J) to u(x(I),t(J)) for each
% I = 1, ..., m-1 and J = 1, ..., N.
syms('OK', 'FX', 'FT', 'ALPHA', 'M', 'N', 'M1', 'M2', 'N1');
syms('H', 'K', 'VV', 'I', 'W', 'L', 'U', 'J', 'T', 'Z');
syms('I1', 'FLAG', 'NAME', 'OUP', 'X', 's', 'x');
TRUE = 1;
FALSE = 0;
fprintf(1,'This is the Backward-Difference Method for Heat
Equation.\n');
fprintf(1,'Input the function F(X) in terms of x.\n');
fprintf(1,'For example: sin(pi*x)\n');
s = input(' ','s');
F = inline(s,'x');
fprintf(1,'The lefthand endpoint on the X-axis is 0.\n');
OK = FALSE;
while OK == FALSE
fprintf(1,'Input the righthand endpoint on the X-axis.\n');
FX = input(' ');
if FX <= 0
fprintf(1,'Must be positive number.\n');
else
OK = TRUE;
end;
end;
OK = FALSE;
while OK == FALSE
fprintf(1,'Input the maximum value of the time variable T.\n');
FT = input(' ');
if FT <= 0
fprintf(1,'Must be positive number.\n');
else
OK = TRUE;
end;
end;
fprintf(1,'Input the constant alpha.\n');
ALPHA = input(' ');
OK = FALSE;
while OK == FALSE
fprintf(1,'Input integer m = number of intervals on X-axis\n');
fprintf(1,'and N = number of time intervals - on separate lines.\n');
fprintf(1,'Note that m must be 3 or larger.\n');
M = input(' ');
N = input(' ');
if M <= 2 | N <= 0
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% HEAT EQUATION BACKWARD-DIFFERENCE ALGORITHM 12.

% To approximate the solution to the parabolic partial-differential % equation subject to the boundary conditions % u(0,t) = u(l,t) = 0, 0 < t < T = max t, % and the initial conditions % u(x,0) = F(x), 0 <= x <= l: % % INPUT: endpoint l; maximum time T; constant ALPHA; integers m, N. % % OUTPUT: approximations W(I,J) to u(x(I),t(J)) for each % I = 1, ..., m-1 and J = 1, ..., N. syms('OK', 'FX', 'FT', 'ALPHA', 'M', 'N', 'M1', 'M2', 'N1'); syms('H', 'K', 'VV', 'I', 'W', 'L', 'U', 'J', 'T', 'Z'); syms('I1', 'FLAG', 'NAME', 'OUP', 'X', 's', 'x'); TRUE = 1; FALSE = 0; fprintf(1,'This is the Backward-Difference Method for Heat Equation.\n'); fprintf(1,'Input the function F(X) in terms of x.\n'); fprintf(1,'For example: sin(pi*x)\n'); s = input(' ','s'); F = inline(s,'x'); fprintf(1,'The lefthand endpoint on the X-axis is 0.\n'); OK = FALSE; while OK == FALSE fprintf(1,'Input the righthand endpoint on the X-axis.\n'); FX = input(' '); if FX <= 0 fprintf(1,'Must be positive number.\n'); else OK = TRUE; end; end; OK = FALSE; while OK == FALSE fprintf(1,'Input the maximum value of the time variable T.\n'); FT = input(' '); if FT <= 0 fprintf(1,'Must be positive number.\n'); else OK = TRUE; end; end; fprintf(1,'Input the constant alpha.\n'); ALPHA = input(' '); OK = FALSE; while OK == FALSE fprintf(1,'Input integer m = number of intervals on X-axis\n'); fprintf(1,'and N = number of time intervals - on separate lines.\n'); fprintf(1,'Note that m must be 3 or larger.\n'); M = input(' '); N = input(' '); if M <= 2 | N <= 0

fprintf(1,'Numbers are not within correct range.\n'); else OK = TRUE; end; end; if OK == TRUE W = zeros(1,M); L = zeros(1,M); U = zeros(1,M); Z = zeros(1,M); M1 = M-1; M2 = M-2; N1 = N-1; % STEP 1 H = FX/M; K = FT/N; VV = ALPHAALPHAK/(HH); % STEP 2 for I = 1 : M W(I) = F(IH); end; % STEP 3 % STEPS 3 through 11 solve a tridiagonal linear system % using Crout reduction L(1) = 1+2VV; U(1) = -VV/L(1); % STEP 4 for I = 2 : M L(I) = 1+2VV+VVU(I-1); U(I) = -VV/L(I); end; % STEP 5 L(M1) = 1+2VV+VVU(M2); % STEP 6 for J = 1 : N % STEP 7 % current t(j) T = JK; Z(1) = W(1)/L(1); % STEP 8 for I = 2 : M Z(I) = (W(I)+VVZ(I-1))/L(I); end; % STEP 9 W(M1) = Z(M1); % STEP 10 for I1 = 1 : M I = M2-I1+1; W(I) = Z(I)-U(I)W(I+1); end; end; % STEP 11 fprintf(1,'Choice of output method:\n'); fprintf(1,'1. Output to screen\n');