Finite Difference Algorithm, Boudary Value Problem-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: Finite, Difference, Algorithm, Nonlinear, Approximations, Boundary, Value, Problem, Iterations, Tolerance, Partial

Typology: Exercises

2011/2012

Uploaded on 07/31/2012

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% NONLINEAR FINITE-DIFFERENCE ALGORITHM 11.4
%
% To approximate the solution to the nonlinear boundary-value problem
%
% Y'' = F(X,Y,Y'), A<=X<=B, Y(A) = ALPHA, Y(B) = BETA:
%
% INPUT: Endpoints A,B; boundary conditions ALPHA, BETA;
% integer N; tolerance TOL; maximum number of iterations M.
%
% OUTPUT: Approximations W(I) TO Y(X(I)) for each I=0,1,...,N+1
% or a message that the maximum number of iterations was
% exceeded.
syms('OK', 'AA', 'BB', 'ALPHA', 'BETA', 'N', 'TOL', 'NN');
syms('FLAG', 'NAME', 'OUP', 'N1', 'H', 'I', 'W', 'K', 'X');
syms('T', 'A', 'B', 'D', 'C', 'L', 'U', 'Z', 'V');
syms('VMAX', 'J', 'x', 'y', 'z', 's');
TRUE = 1;
FALSE = 0;
fprintf(1,'This is the Nonlinear Finite-Difference Method.\n');
fprintf(1,'Input the function F(X,Y,Z) in terms of x, y, z\n');
fprintf(1,'followed by the partial of F with respect to y on \n');
fprintf(1,'the next line and the partial of F with respect \n');
fprintf(1,'to z = y-prime on the third line. \n');
fprintf(1,'For example: (32+2*x^3-y*z)/8 \n');
fprintf(1,' -z/8 \n');
fprintf(1,' -y/8 \n');
s = input(' ','s');
F = inline(s,'x','y','z');
s = input(' ','s');
FY = inline(s,'x','y','z');
s = input(' ','s');
FYP = inline(s,'x','y','z');
OK = FALSE;
while OK == FALSE
fprintf(1,'Input left and right endpoints on separate lines.\n');
AA = input(' ');
BB = input(' ');
if AA >= BB
fprintf(1,'Left endpoint must be less than right endpoint.\n');
else
OK = TRUE;
end;
end;
fprintf(1,'Input Y( %.10e).\n', AA);
ALPHA = input(' ');
fprintf(1,'Input Y( %.10e).\n', BB);
BETA = input(' ');
OK = FALSE;
while OK == FALSE
fprintf(1,'Input an integer > 1 for the number of\n');
fprintf(1,'subintervals. Note that h = (b-a)/(n+1)\n');
N = input(' ');
if N <= 1
fprintf(1,'Number must exceed 1.\n');
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% NONLINEAR FINITE-DIFFERENCE ALGORITHM 11.

% To approximate the solution to the nonlinear boundary-value problem % % Y'' = F(X,Y,Y'), A<=X<=B, Y(A) = ALPHA, Y(B) = BETA: % % INPUT: Endpoints A,B; boundary conditions ALPHA, BETA; % integer N; tolerance TOL; maximum number of iterations M. % % OUTPUT: Approximations W(I) TO Y(X(I)) for each I=0,1,...,N+ % or a message that the maximum number of iterations was % exceeded. syms('OK', 'AA', 'BB', 'ALPHA', 'BETA', 'N', 'TOL', 'NN'); syms('FLAG', 'NAME', 'OUP', 'N1', 'H', 'I', 'W', 'K', 'X'); syms('T', 'A', 'B', 'D', 'C', 'L', 'U', 'Z', 'V'); syms('VMAX', 'J', 'x', 'y', 'z', 's'); TRUE = 1; FALSE = 0; fprintf(1,'This is the Nonlinear Finite-Difference Method.\n'); fprintf(1,'Input the function F(X,Y,Z) in terms of x, y, z\n'); fprintf(1,'followed by the partial of F with respect to y on \n'); fprintf(1,'the next line and the partial of F with respect \n'); fprintf(1,'to z = y-prime on the third line. \n'); fprintf(1,'For example: (32+2x^3-yz)/8 \n'); fprintf(1,' -z/8 \n'); fprintf(1,' -y/8 \n'); s = input(' ','s'); F = inline(s,'x','y','z'); s = input(' ','s'); FY = inline(s,'x','y','z'); s = input(' ','s'); FYP = inline(s,'x','y','z'); OK = FALSE; while OK == FALSE fprintf(1,'Input left and right endpoints on separate lines.\n'); AA = input(' '); BB = input(' '); if AA >= BB fprintf(1,'Left endpoint must be less than right endpoint.\n'); else OK = TRUE; end; end; fprintf(1,'Input Y( %.10e).\n', AA); ALPHA = input(' '); fprintf(1,'Input Y( %.10e).\n', BB); BETA = input(' '); OK = FALSE; while OK == FALSE fprintf(1,'Input an integer > 1 for the number of\n'); fprintf(1,'subintervals. Note that h = (b-a)/(n+1)\n'); N = input(' '); if N <= 1 fprintf(1,'Number must exceed 1.\n');

else OK = TRUE; end; end; OK = FALSE; while OK == FALSE fprintf(1,'Input Tolerance.\n'); TOL = input(' '); if TOL <= 0 fprintf(1,'Tolerance must be positive.\n'); else OK = TRUE; end; end; OK = FALSE; while OK == FALSE fprintf(1,'Input maximum number of iterations.\n'); NN = input(' '); if NN <= 0 fprintf(1,'Must be positive integer.\n'); else OK = TRUE; end; end; if OK == TRUE fprintf(1,'Choice of output method:\n'); fprintf(1,'1. Output to screen\n'); fprintf(1,'2. Output to text File\n'); fprintf(1,'Please enter 1 or 2.\n'); FLAG = input(' '); if FLAG == 2 fprintf(1,'Input the file name in the form - drive:\name.ext\n'); fprintf(1,'for example A:\OUTPUT.DTA\n'); NAME = input(' ','s'); OUP = fopen(NAME,'wt'); else OUP = 1; end; fprintf(OUP, 'NONLINEAR FINITE-DIFFERENCE METHOD\n\n'); fprintf(OUP, ' I X(I) W(I)\n'); % STEP 1 A = zeros(1,N); B = zeros(1,N); C = zeros(1,N); D = zeros(1,N); W = zeros(1,N); V = zeros(1,N); Z = zeros(1,N); U = zeros(1,N); L = zeros(1,N); N1 = N-1; H = (BB-AA)/(N+1); % STEP 2 for I = 1 : N

end; % STEP 13 % test for accuracy if VMAX <= TOL I = 0; fprintf(OUP, '%3d %13.8f %13.8f\n', I, AA, ALPHA); for I = 1 : N X = AA+I*H; fprintf(OUP, '%3d %13.8f %13.8f\n', I, X, W(I)); end; I = N+1; fprintf(OUP, '%3d %13.8f %13.8f\n', I, BB, BETA); OK = FALSE; else % STEP 18 K = K+1; end; end; % STEP 19 if K > NN fprintf(OUP, 'No convergence in %d iterations\n', NN); end; end; if OUP ~= 1 fclose(OUP); fprintf(1,'Output file %s created successfully \n',NAME); end;