Plotting of error vs iteration number matlab
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Nima Vali
el 29 de Oct. de 2020
Comentada: KSSV
el 29 de Oct. de 2020
I am going to plot the error as a function of iteration number in Jacoby calculation of second order diffrential equation. However, I am not getting any plot. Can you help. also, how I can get the plot at iteration number of 0.
clear all; clc ; close all;
%Setting up the parameters
dx=0.1 ; dy = dx ;
Lx=1 ; Ly =1;
x = (0:dx:Lx); y = (0:dy:Ly);
Nx=length(x); Ny=length(y) ;
if (Nx==Ny)
N=Nx;
end
[X Y] = meshgrid(x,y);
X=X' ; Y=Y' ;
%% calculating g(i,j)
g=-2*X.*(1-X)-2*Y.*(1-Y);
%% Analytical solution for tfinal
iterlimit=800;
f_exact=X.*(1-X).*Y.*(1-Y);
%%jacobi iterative method
f=zeros(Nx,Ny);
f_jacobi=zeros(Nx,Ny);
for n=1:iterlimit
for j=2:Ny-1
for i=2:Nx-1
f_jacobi(i,j)=(0.25)*(f(i,j+1)+f(i,j-1)+f(i+1,j)+f(i-1,j))-(0.25)*g(i,j)*dx^2 ;
end
end
f=f_jacobi;
B = reshape(f_jacobi-f_exact,[N*N,1]);
L1= norm(B,1)/numel(B);
plot(n,L1)
title ('L1 error vs iteration')
xlim([0,800])
end
1 comentario
KSSV
el 29 de Oct. de 2020
You code has for loops which can be vectorised......think of your logic and code it again.
Respuesta aceptada
KSSV
el 29 de Oct. de 2020
clear all; clc ; close all;
%Setting up the parameters
dx=0.1 ; dy = dx ;
Lx=1 ; Ly =1;
x = (0:dx:Lx); y = (0:dy:Ly);
Nx=length(x); Ny=length(y) ;
if (Nx==Ny)
N=Nx;
end
[X Y] = meshgrid(x,y);
X=X' ; Y=Y' ;
%% calculating g(i,j)
g=-2*X.*(1-X)-2*Y.*(1-Y);
%% Analytical solution for tfinal
iterlimit=800;
f_exact=X.*(1-X).*Y.*(1-Y);
%%jacobi iterative method
f=zeros(Nx,Ny);
f_jacobi=zeros(Nx,Ny);
L1 = zeros(iterlimit) ;
for n=1:iterlimit
for j=2:Ny-1
for i=2:Nx-1
f_jacobi(i,j)=(0.25)*(f(i,j+1)+f(i,j-1)+f(i+1,j)+f(i-1,j))-(0.25)*g(i,j)*dx^2 ;
end
end
f=f_jacobi;
B = reshape(f_jacobi-f_exact,[N*N,1]);
L1(n)= norm(B,1)/numel(B);
end
plot(1:iterlimit,L1,'*r')
title ('L1 error vs iteration')
xlim([0,800])
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