3 views (last 30 days)

% 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(' ');

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%Input F which represents the second order ODE

F = inline('y*log(y))','x','y','z');

%s = input(' ');

%Input FY which represents the partial derivative of F

FY = inline('log(y)+y','x','y','z');

%s = input(' ');

%Input FYP which represents the partial derivative of F'

FYP = inline('1','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

W(I) = ALPHA+I*H*(BETA-ALPHA)/(BB-AA);

end;

% STEP 3

K = 1;

% STEP 4

while K <= NN && OK == TRUE

% STEP 5

X = AA+H;

T = (W(2)-ALPHA)/(2*H);

A(1) = 2+H*H*FY(X,W(1),T);

B(1) = -1+H*FYP(X,W(1),T)/2;

D(1) = -(2*W(1)-W(2)-ALPHA+H*H*F(X,W(1),T));

% STEP 6

for I = 2 : N1

X = AA+I*H;

T = (W(I+1)-W(I-1))/(2*H);

A(I) = 2+H*H*FY(X,W(I),T);

B(I) = -1+H*FYP(X,W(I),T)/2;

C(I) = -1-H*FYP(X,W(I),T)/2;

D(I) = -(2*W(I)-W(I+1)-W(I-1)+H*H*F(X,W(I),T));

end;

% STEP 7

X = BB - H;

T = (BETA-W(N-1))/(2*H);

A(N) = 2+H*H*FY(X,W(N),T);

C(N) = -1-H*FYP(X,W(N),T)/2;

D(N) = -(2*W(N)-W(N-1)-BETA+H*H*F(X,W(N),T));

% STEP 8

% STEPS 8 through 12 solve a tridiagonal linear system using

% Crout reduction

L(1) = A(1);

U(1) = B(1)/A(1);

Z(1) = D(1)/L(1);

% STEP 9

for I = 2 : N1

L(I) = A(I)-C(I)*U(I-1);

U(I) = B(I)/L(I);

Z(I) = (D(I)-C(I)*Z(I-1))/L(I);

end;

% STEP 10

L(N) = A(N)-C(N)*U(N-1);

Z(N) = (D(N)-C(N)*Z(N-1))/L(N);

% STEP 11

V(N) = Z(N);

VMAX = abs(V(N));

W(N) = W(N)+V(N);

% STEP 12

for J = 1 : N1

I = N-J;

V(I) = Z(I)-U(I)*V(I+1);

W(I) = W(I)+V(I);

if abs(V(I)) > VMAX

VMAX = abs(V(I));

end;

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:

command window

This is the Nonlinear Finite-Difference Method.

Input left and right endpoints on separate lines.

0

pi/2

Input Y( 0.0000000000e+00).

1

Input Y( 1.5707963268e+00).

exp(1)

Input an integer > 1 for the number of

subintervals. Note that h = (b-a)/(n+1)

9

Input Tolerance.

10^-4

Input maximum number of iterations.

4

Choice of output method:

1. Output to screen

2. Output to text File

Please enter 1 or 2.

1

NONLINEAR FINITE-DIFFERENCE METHOD

I X(I) W(I)

Error using inlineeval (line 14)

Error in inline expression ==> y*log(y))

Error: Unbalanced or unexpected parenthesis or bracket.

Error in inline/subsref (line 23)

INLINE_OUT_ = inlineeval(INLINE_INPUTS_, INLINE_OBJ_.inputExpr, INLINE_OBJ_.expr);

Opportunities for recent engineering grads.

Apply Today
## 1 Comment

## Direct link to this comment

https://se.mathworks.com/matlabcentral/answers/536295-undefined-function-ln-for-input-arguments-of-type-double-sorry-i-kindly-no-idea-what-was-wrong#comment_870247

⋮## Direct link to this comment

https://se.mathworks.com/matlabcentral/answers/536295-undefined-function-ln-for-input-arguments-of-type-double-sorry-i-kindly-no-idea-what-was-wrong#comment_870247

Sign in to comment.