Finding error like unrecognized function or variable ' tridiagonal'

Question

Aman Murkar 2021-10-2

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1465234-finding-error-like-unrecognized-function-or-variable-tridiagonal

编辑： C B 2021-10-3

Program:

% solution of 2D elliptical solution
% using Line Over Relaxation Method(LSOR)
% ep is accepted error%Tridiag: Tridiagonal equation zsolver banded system
clc;
clear all;
eps = 0.001;
omega = input(' enter the omega value: ');
beta = input (' enter the beta value: ');
n= 10000;
nx = 21;
ny = 42;
T(1:nx, 1:ny-1) = 0;
TN(1:nx, 1:ny-1) = 0;
T(1:nx, ny)= 100;
TN(1:nx, ny) = 100;
% its number of iteration
coeff = ( 2*(1+beta^2));
for iterations = 1:n
    for j = 2:ny-1
        a(1:nx-2) = -coeff;
        b(1:nx-3)=  omega;
        c(1:nx-3)= omega;
        for i = 2:nx-1 
            r(i-1) = - coeff*(1-omega)*T(i,j)-omega*beta^2*T(i,j+1)-omega*beta^2*TN(i,j-1);
        end
        r(1)= r(1)-omega*TN(1,j);
        r(nx-2)= r(nx-2)-omega*TN(nx,j);
        y = tridiagonal(c,a,b,r);
        for k = 1:nx-2
            TN(k+1,j)= y(k); 
        end
        
    end
    error = abs(TN-T);
    totalerror = sum(error,'all');
    if totalerror <= eps
        break
    end
    T=TN;
end
iterations;
contour(TN');

RESULTS;

enter the omega value: 1.3

enter the beta value: 1

Unrecognized function or variable 'tridiagonal'.

Error in LSOR (line 28)

y = tridiagonal(c,a,b,r);

4 个评论
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C B 2021-10-2

@Aman Murkar your have defined function as Tridiagonal and calling it using tridiagonal

C B 2021-10-2

在 MATLAB Online 中打开

Plus i have changed a part of code please check if its as per requirement or not

%         b(1:nx-3)=  omega;
%         c(1:nx-3)= omega;
         b(1:nx-2)=  omega;
         c(1:nx-2)= omega;

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Answer 1

C B 2021-10-2

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1465234-finding-error-like-unrecognized-function-or-variable-tridiagonal#answer_800059

编辑：C B 2021-10-2

在 MATLAB Online 中打开

function main
% solution of 2D elliptical solution
% using Line Over Relaxation Method(LSOR)
% ep is accepted error%Tridiag: Tridiagonal equation zsolver banded system
clc;
clear all;
eps = 0.001;
omega = input(' enter the omega value: ');
beta = input (' enter the beta value: ');
n= 10000;
nx = 21;
ny = 42;
T(1:nx, 1:ny-1) = 0;
TN(1:nx, 1:ny-1) = 0;
T(1:nx, ny)= 100;
TN(1:nx, ny) = 100;
% its number of iteration
coeff = ( 2*(1+beta^2));
for iterations = 1:n
    for j = 2:ny-1
        a(1:nx-2) = -coeff;
%         b(1:nx-3)=  omega;
%         c(1:nx-3)= omega;
         b(1:nx-2)=  omega;
         c(1:nx-2)= omega;
        for i = 2:nx-1 
            r(i-1) = - coeff*(1-omega)*T(i,j)-omega*beta^2*T(i,j+1)-omega*beta^2*TN(i,j-1);
        end
        r(1)= r(1)-omega*TN(1,j);
        r(nx-2)= r(nx-2)-omega*TN(nx,j);
        y = Tridiagonal(c,a,b,r);
        for k = 1:nx-2
            TN(k+1,j)= y(k); 
        end
        
    end
    error = abs(TN-T);
    totalerror = sum(error,'all');
    if totalerror <= eps
        break
    end
    T=TN;
end
iterations;
contour(TN');
end
function x = Tridiagonal(e,f,g,r)
% Tridiagonal: Tridiagonal equation solver banded system
% x = Tridiagonal(e,f,g,r): Tridiagonal system solver.
% input:
% e = subdiagonal vector
% f = diagonal vector
% g = superdiagonal vector
% r = right hand side vector
% output:
% x = solution vector
n=length(f);
% forward elimination
for k = 2:n
    factor = e(k)/f(k-1);
    f(k) = f(k) - factor*g(k-1);
    r(k) = r(k) - factor*r(k-1);
end
% back substitution
x(n) = r(n)/f(n);
for k = n-1:-1:1
    x(k) = (r(k)-g(k)*x(k+1))/f(k);
end
end

3 个评论
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Aman Murkar 2021-10-2

thanks I got iterations too by doing some changes thanks for helping me.

Now the main part comes that can you tell me specifically what mistakes do i did as i am just learning matlab and how I can avoid such mistakes? Hope you will guide me.

Thanks again

C B 2021-10-2