Plotting of error vs iteration number matlab

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Nima Vali
Nima Vali 2020-10-29
评论： KSSV 2020-10-29
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 个评论
KSSV
KSSV 2020-10-29
You code has for loops which can be vectorised......think of your logic and code it again.

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采纳的回答

KSSV
KSSV 2020-10-29
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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