MATLAB: Undefined function ‘ln’ for input arguments of type ‘double’. Sorry I kindly no idea what was wrong with the coding. Kindly help please ,Thanks a lots.

% NONLINEAR FINITE-DIFFERENCE ALGORITHM 11.4
%



% To approximate the solution to the nonlinear boundary-value problem
%
% Y&#39;&#39; = F(X,Y,Y&#39;), A&lt;=X&lt;=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&#39;
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 &gt; 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.

pi/2

Input Y( 0.0000000000e+00).

Input Y( 1.5707963268e+00).

exp(1)

Input an integer > 1 for the number of

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

Input Tolerance.

10^-4

Input maximum number of iterations.

Choice of output method:

1. Output to screen

2. Output to text File

Please enter 1 or 2.

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);

MATLAB: Undefined function ‘ln’ for input arguments of type ‘double’. Sorry I kindly no idea what was wrong with the coding. Kindly help please ,Thanks a lots.

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