Answers keep on iterating

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LightFury Yeji
LightFury Yeji 2021-1-30
回答： Walter Roberson 2021-1-30
This program is made to approximate the solution to the Poisson equation. I get the correct answer but it keeps on repeating the same answers (I think 12 times). I only need one set of these answers (as highlighted). Please help, thank you so much in advance.
%Initialization
a=0;
b=2;
c=0;
d=1;
m=5;
n=6;
N= 100;
TOL = 10^-10;
syms ga(y) gb(y) gc(x) gd(x) f(x,y);
f(x,y)=x*exp(y);
ga(y)=0; %g(a,y)
gb(y)=2*exp(y); %g(b,y)
gc(x)=x; %g(x,c)
gd(x) =x*exp(1); %g(x,d)
%Step 1
h=(b-a)/n;
k=(d-c)/m;
%Step 2;
for i = 1:n-1
xi(i)=a+i*h; %x sub i (k) = xi(k)
end
%Step 3
for j = 1:m-1
yj(j)=c+j*k;
end
%Step 4
for i=1:n-1
for j=1:m-1
wij(i,j)=0;
end
end
%Step 5
lambda=h^2/k^2;
mu=2*(1+lambda);
l=1;
%Step 6
while l <= N
%Step 7
z=(-(h^2)*(f(xi(1),yj(m-1)))+ga(yj(m-1))+lambda*gd(xi(1))+lambda*wij(1,m-2)+wij(2,m-1))/mu;
NORM=abs(z-wij(1,m-1));
wij(1,m-1)=z;
%Step 8
for i=2:n-2
z = (-h^2*f(xi(i),yj(m-1))+lambda*gd(xi(i))+wij(i-1,m-1)+wij(i+1,m-1)+lambda*wij(i,m-2))/mu;
if abs(wij(i,m-1)-z)>NORM
NORM = abs(wij(i,m-1)-z);
end
wij(i,m-1) = z;
%Step 9
z=(-h^2*f(xi(n-1),yj(m-1))+gb(yj(m-1))+lambda*gd(xi(n-1))+wij(n-2,m-1)+lambda*wij(n-1,m-2))/mu;
if abs (wij(n-1,m-1)-z) > NORM
NORM = abs (wij(n-1,m-1)-z);
end
wij(n-1,m-1)=z;
%Step 10
for j=m-2:-1:2
%Step 11
z=(-(h^2)*f(xi(1),yj(j))+ga(yj(j))+lambda*wij(1,j+1)+lambda*wij(1,j-1)+wij(2,j))/mu;
if abs(wij(1,j)-z)>NORM
NORM = abs(wij(1,j)-z);
end
wij(1,j)=z;
%Step 12
for i=2:n-2
z=(-(h^2)*f(xi(i),yj(j))+wij(i-1,j)+lambda*wij(i,j+1)+wij(i+1,j)+lambda*wij(i,j-1))/mu;
if abs (wij(i,j)-z)>NORM
NORM=abs(wij(i,j)-z);
end
wij(i,j)=z;
end
%STEP 13
z=(-h^2*f(xi(n-1),yj(j))+gb(yj(j))+wij(n-2,j)+lambda*wij(n-1,j+1)+lambda*wij(n-1,j-1))/mu;
if abs(wij(n-1,j)-z)>NORM
NORM=abs(wij(n-1,j)-z);
end
wij(n-1,j)=z;
end
%Step 14
z = (-h^2*f(xi(1),yj(1))+ga(yj(1))+lambda*gc(xi(1))+lambda*wij(1,2)+wij(2,1))/mu;
if abs(wij(1,1)-z) > NORM
NORM = abs(wij(1,1)-z);
end
wij(1,1) = z;
%Step 15
for i = 2:n-2
z = (-h^2*f(xi(i),yj(1))+lambda*gc(xi(i))+wij(i-1,1)+lambda*wij(i,2)+wij(i+1,1))/mu;
if abs(wij(i,1)-z)>NORM
NORM = abs(wij(i,1)-z);
end
wij(i,1) = z;
end
%Step 16
z=(-h^2*f(xi(n-1),yj(1))+gb(yj(1))+lambda*gc(xi(n-1))+wij(n-2,1)+lambda*wij(n-1,2))/mu;
if abs (wij(n-1,1)-z) > NORM
NORM= abs(wij(n-1,1)-z) ;
end
wij(n-1,1)=z;
%Step 17
if NORM<=TOL
%Step 18
for i=1:n-1
for j=1:m-1
disp(" "+double(xi(i))+" "+double(yj(j))+" "+double(wij(i,j)))
end
end
%Step 19
end
%Step 20
l=l+1;
end
end

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采纳的回答

Walter Roberson
Walter Roberson 2021-1-30
Yes, that is what you asked it to do. You display output whenever NORM <= TOL for any given iteration of l, and you iterate l up to N = 100, so you can have up to 100 displays of output.
Presumably something inside your loop is making changes so that the next iteration of l will be working with different inputs, so that the output would be different the next time... otherwise it would be pointless to continue iterating l once the tolerance had been met.

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