How to terminate my code?

조회 수: 26 (최근 30일)
Nathi
Nathi 2023년 5월 2일
편집: Nathi 2023년 5월 6일
clc;
close all;
clear all;
%-----------SPACEGRID---------%
Y=14; nodey=56; DY=Y/nodey; DX=0.05;
%--------TIMEGRID-------------%
TMAX=100; NT=500; Dt=TMAX/NT; t=0:Dt:TMAX;
error(1)=1; tol=1e-5;
MI=100;
u_prev=zeros(nodey,NT); v_prev=zeros(nodey,NT);
t_cur=0; t_prev=t_cur*ones(nodey,NT); T_surf=ones(1,length(t));
P=zeros([1,nodey]); Q=P; R=P; S=P;
T=t_prev;
k=0;
%while k<MI %HOW TO USE THIS iterate loop to converge 1 to 0
for m=1:NT
for n=1:nodey
if m>n
R(n)=(Dt*v_prev(n,m)/4*DY)-(Dt/(2*DY^2));
elseif n>m
P(n)=(-Dt*v_prev(n,m)/4*DY)-(Dt/(2*DY^2));
elseif n==m
Q(n)=1+(Dt*u_prev(n,m)/2*DX)+(Dt/(DY^2));
end
end
end
for m=2:NT
if m==1
for n=1:nodey
if n==1
S(n)=(-Dt*u_prev(n,m)*-t_cur+t_prev(n,m)-t_cur/2*DX)+(Dt/2*DY^2*t_prev(n+1,m)-2*t_prev(n,m)+t_cur)-(Dt*v_prev(n,m)/4*DY*t_cur-t_prev(n+1,m)-t_prev(n,m))-(Dt/4*DY*v_prev(n,m))-(Dt/2*DY^2*t_cur);
elseif n==nodey
S(n)=(-Dt*u_prev(n,m)*-t_cur+t_prev(n,m)-t_cur/2*DX)+(Dt/2*DY^2*T_surf(m)-2*t_prev(n,m)+t_prev(n-1,m))-(Dt*v_prev(n,m)/4*DY*t_prev(n-1,m)-T_surf(m)+t_prev(n,m))-(-Dt/4*DY*v_prev(n,m))-(Dt/2*DY^2*T_surf);
else
S(n)=(-Dt*u_prev(n,m)*-t_cur+t_prev(n,m)-t_cur/2*DX)+(Dt/2*DY^2*t_prev(n+1,m)-2*t_prev(n,m)+t_prev(n-1,m))-(Dt*v_prev(n,m)/4*DY*t_prev(n-1,m)-t_prev(n+1,m)+t_prev(n,m));
end
end
else
for n=1:nodey
if n==1
S(n)=(-Dt*u_prev(n,m)*-T(n,m-1)+t_prev(n,m)-t_prev(n,m-1)/2*DX)+(Dt/2*DY^2*t_prev(n+1,m)-2*t_prev(n,m)+t_cur)-(Dt*v_prev(n,m)/4*DY*t_cur-t_prev(n+1,m)+t_prev(n,m))-(Dt/4*DY*v_prev(n,m))-(Dt/2*DY^2*t_cur);
elseif n==nodey
S(n)=(-Dt*u_prev(n,m)*-T(n,m-1)+t_prev(n,m)-t_prev(n,m-1)/2*DX)+(Dt/2*DY^2*T_surf(m)-2*t_prev(n,m)+t_prev(n-1,m))-(Dt*v_prev(n,m)/4*DY*t_prev(n-1,m)-T_surf(m)+t_prev(n,m))-(-Dt/4*DY*v_prev(n,m))-(Dt/2*DY^2*T_surf(m));
else
S(n)=(-Dt*u_prev(n,m)*-T(n,m-1)+t_prev(n,m)-t_prev(n,m-1)/2*DX)+(Dt/2*DY^2*t_prev(n+1,m)-2*t_prev(n,m)+t_prev(n-1,m))-(Dt*v_prev(n,m)/4*DY*t_prev(n-1,m)-t_prev(n+1,m)+t_prev(n,m));
end
end
end
T(:,m)=crank(P,Q,R,S);
Dt=0.2+Dt;
t_prev=T;
error(m) = max(abs(T(:,m)-T(:,m-1))./max(1,abs(T(:,m-1))));
if error(m)<tol
break
end
end
%k=k+1;
% end
function x = crank(e,f,g,r)
%input
% e = subdiagonal vector
% f = diagonal vector
% g = superdiagonal vector
% r = right hand side vector
% output:
% x = solution vector
n=length(f);
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
x(n) = r(n)/f(n);
for k = n-1:-1:1
x(k) = (r(k)-g(k)*x(k+1))/f(k);
end
for i = 1:length(x)
fprintf('\n%d = %f\n', i, x(i));
end
  댓글 수: 8
Nathi
Nathi 2023년 5월 3일
@Walter RobersonThank you for assit me. Actually, there are two types of convergence here. one is convergence at each time steps and another is convergence at each space steps. In my code, converged at time discretize. That's why i'm using error/tol condition. @per isakson Now I want convergence at each space step i.e., begins with 1 and increase gradually, then diminish gradually to 0 in each 'm'th row. here, my boundary condition are
if m==1
for n=1:nodey
if n==1 %(top boundary condtion i.e (1,1) 1st row and 1st column)
S(n)=(-Dt*u_prev(n,m)*-t_cur+t_prev(n,m)-t_cur/2*DX)+(Dt/2*DY^2*t_prev(n+1,m)-2*t_prev(n,m)+t_cur)-(Dt*v_prev(n,m)/4*DY*t_cur-t_prev(n+1,m)-t_prev(n,m))-(Dt/4*DY*v_prev(n,m))-(Dt/2*DY^2*t_cur);
elseif n==nodey %(bottom boundary condition i.e(nodey,1) last row and 1st col)
S(n)=(-Dt*u_prev(n,m)*-t_cur+t_prev(n,m)-t_cur/2*DX)+(Dt/2*DY^2*T_surf(m)-2*t_prev(n,m)+t_prev(n-1,m))-(Dt*v_prev(n,m)/4*DY*t_prev(n-1,m)-T_surf(m)+t_prev(n,m))-(-Dt/4*DY*v_prev(n,m))-(Dt/2*DY^2*T_surf);
else
S(n)=(-Dt*u_prev(n,m)*-t_cur+t_prev(n,m)-t_cur/2*DX)+(Dt/2*DY^2*t_prev(n+1,m)-2*t_prev(n,m)+t_prev(n-1,m))-(Dt*v_prev(n,m)/4*DY*t_prev(n-1,m)-t_prev(n+1,m)+t_prev(n,m));
end
end
else
for n=1:nodey
if n==1 %(at top boundary condition i.e (1,2:NT)
S(n)=(-Dt*u_prev(n,m)*-T(n,m-1)+t_prev(n,m)-t_prev(n,m-1)/2*DX)+(Dt/2*DY^2*t_prev(n+1,m)-2*t_prev(n,m)+t_cur)-(Dt*v_prev(n,m)/4*DY*t_cur-t_prev(n+1,m)+t_prev(n,m))-(Dt/4*DY*v_prev(n,m))-(Dt/2*DY^2*t_cur);
elseif n==nodey %(at bottom boundary condtion i.e (nodey,2:NT)
S(n)=(-Dt*u_prev(n,m)*-T(n,m-1)+t_prev(n,m)-t_prev(n,m-1)/2*DX)+(Dt/2*DY^2*T_surf(m)-2*t_prev(n,m)+t_prev(n-1,m))-(Dt*v_prev(n,m)/4*DY*t_prev(n-1,m)-T_surf(m)+t_prev(n,m))-(-Dt/4*DY*v_prev(n,m))-(Dt/2*DY^2*T_surf(m));
else
S(n)=(-Dt*u_prev(n,m)*-T(n,m-1)+t_prev(n,m)-t_prev(n,m-1)/2*DX)+(Dt/2*DY^2*t_prev(n+1,m)-2*t_prev(n,m)+t_prev(n-1,m))-(Dt*v_prev(n,m)/4*DY*t_prev(n-1,m)-t_prev(n+1,m)+t_prev(n,m));
end
end
end
If this S get gradually values, then it will store in T. and then if I plot at each column wise in T, it must comes like this "begins with 1 and increase gradually, then diminish gradually to 0 ". For the gradual values and convergence, I'm using "ITERATION".
Nathi
Nathi 2023년 5월 6일
편집: Nathi 2023년 5월 6일
@Walter Roberson and @KSSV Please try and include the below condition . It might work.
TT=crank(P,Q,R,S);
T(:,m)=[initial value,TT,endpoint];
t_prev is initial point and t_surf is end point
If get any idea about iteration with boundary condition using while loop., please suggest me.

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답변 (1개)

KSSV
KSSV 2023년 5월 3일
You can use for loop instead of while....You know the number of counts right.
clc;
close all;
clear all;
%-----------SPACEGRID---------%
Y=14; nodey=56; DY=Y/nodey; DX=0.05;
%--------TIMEGRID-------------%
TMAX=100; NT=500; Dt=TMAX/NT; t=0:Dt:TMAX;
error(1)=1; tol=1e-5;
MI=100;
u_prev=zeros(nodey,NT); v_prev=zeros(nodey,NT);
t_cur=0; t_prev=t_cur*ones(nodey,NT); T_surf=ones(1,length(t));
P=zeros([1,nodey]); Q=P; R=P; S=P;
T=t_prev;
E = cell(MI,1) ;
for k = 1:MI %HOW TO USE THIS iterate loop to converge 1 to 0
for m=1:NT
for n=1:nodey
if m>n
R(n)=(Dt*v_prev(n,m)/4*DY)-(Dt/(2*DY^2));
elseif n>m
P(n)=(-Dt*v_prev(n,m)/4*DY)-(Dt/(2*DY^2));
elseif n==m
Q(n)=1+(Dt*u_prev(n,m)/2*DX)+(Dt/(DY^2));
end
end
end
for m=2:NT
if m==1
for n=1:nodey
if n==1
S(n)=(-Dt*u_prev(n,m)*-t_cur+t_prev(n,m)-t_cur/2*DX)+(Dt/2*DY^2*t_prev(n+1,m)-2*t_prev(n,m)+t_cur)-(Dt*v_prev(n,m)/4*DY*t_cur-t_prev(n+1,m)-t_prev(n,m))-(Dt/4*DY*v_prev(n,m))-(Dt/2*DY^2*t_cur);
elseif n==nodey
S(n)=(-Dt*u_prev(n,m)*-t_cur+t_prev(n,m)-t_cur/2*DX)+(Dt/2*DY^2*T_surf(m)-2*t_prev(n,m)+t_prev(n-1,m))-(Dt*v_prev(n,m)/4*DY*t_prev(n-1,m)-T_surf(m)+t_prev(n,m))-(-Dt/4*DY*v_prev(n,m))-(Dt/2*DY^2*T_surf);
else
S(n)=(-Dt*u_prev(n,m)*-t_cur+t_prev(n,m)-t_cur/2*DX)+(Dt/2*DY^2*t_prev(n+1,m)-2*t_prev(n,m)+t_prev(n-1,m))-(Dt*v_prev(n,m)/4*DY*t_prev(n-1,m)-t_prev(n+1,m)+t_prev(n,m));
end
end
else
for n=1:nodey
if n==1
S(n)=(-Dt*u_prev(n,m)*-T(n,m-1)+t_prev(n,m)-t_prev(n,m-1)/2*DX)+(Dt/2*DY^2*t_prev(n+1,m)-2*t_prev(n,m)+t_cur)-(Dt*v_prev(n,m)/4*DY*t_cur-t_prev(n+1,m)+t_prev(n,m))-(Dt/4*DY*v_prev(n,m))-(Dt/2*DY^2*t_cur);
elseif n==nodey
S(n)=(-Dt*u_prev(n,m)*-T(n,m-1)+t_prev(n,m)-t_prev(n,m-1)/2*DX)+(Dt/2*DY^2*T_surf(m)-2*t_prev(n,m)+t_prev(n-1,m))-(Dt*v_prev(n,m)/4*DY*t_prev(n-1,m)-T_surf(m)+t_prev(n,m))-(-Dt/4*DY*v_prev(n,m))-(Dt/2*DY^2*T_surf(m));
else
S(n)=(-Dt*u_prev(n,m)*-T(n,m-1)+t_prev(n,m)-t_prev(n,m-1)/2*DX)+(Dt/2*DY^2*t_prev(n+1,m)-2*t_prev(n,m)+t_prev(n-1,m))-(Dt*v_prev(n,m)/4*DY*t_prev(n-1,m)-t_prev(n+1,m)+t_prev(n,m));
end
end
end
T(:,m)=crank(P,Q,R,S);
Dt=0.2+Dt;
t_prev=T;
error(m) = max(abs(T(:,m)-T(:,m-1))./max(1,abs(T(:,m-1))));
if error(m)<tol
break
end
E{k} = error ;
end
%k=k+1;
end
function x = crank(e,f,g,r)
%input
% e = subdiagonal vector
% f = diagonal vector
% g = superdiagonal vector
% r = right hand side vector
% output:
% x = solution vector
n=length(f);
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
x(n) = r(n)/f(n);
for k = n-1:-1:1
x(k) = (r(k)-g(k)*x(k+1))/f(k);
end
for i = 1:length(x)
fprintf('\n%d = %f\n', i, x(i));
end
end
  댓글 수: 1
Nathi
Nathi 2023년 5월 3일
@KSSV I tried your idea, it's better but not reached with my boundary condition. I'm using iteration for convergence at each space steps and gradual values with my boundary condition.
if m==1
for n=1:nodey
if n==1 %(top boundary condtion i.e (1,1) 1st row and 1st column)
S(n)=(-Dt*u_prev(n,m)*-t_cur+t_prev(n,m)-t_cur/2*DX)+(Dt/2*DY^2*t_prev(n+1,m)-2*t_prev(n,m)+t_cur)-(Dt*v_prev(n,m)/4*DY*t_cur-t_prev(n+1,m)-t_prev(n,m))-(Dt/4*DY*v_prev(n,m))-(Dt/2*DY^2*t_cur);
elseif n==nodey %(bottom boundary condition i.e(nodey,1) last row and 1st col)
S(n)=(-Dt*u_prev(n,m)*-t_cur+t_prev(n,m)-t_cur/2*DX)+(Dt/2*DY^2*T_surf(m)-2*t_prev(n,m)+t_prev(n-1,m))-(Dt*v_prev(n,m)/4*DY*t_prev(n-1,m)-T_surf(m)+t_prev(n,m))-(-Dt/4*DY*v_prev(n,m))-(Dt/2*DY^2*T_surf);
else
S(n)=(-Dt*u_prev(n,m)*-t_cur+t_prev(n,m)-t_cur/2*DX)+(Dt/2*DY^2*t_prev(n+1,m)-2*t_prev(n,m)+t_prev(n-1,m))-(Dt*v_prev(n,m)/4*DY*t_prev(n-1,m)-t_prev(n+1,m)+t_prev(n,m));
end
end
else
for n=1:nodey
if n==1 %(at top boundary condition i.e (1,2:NT)
S(n)=(-Dt*u_prev(n,m)*-T(n,m-1)+t_prev(n,m)-t_prev(n,m-1)/2*DX)+(Dt/2*DY^2*t_prev(n+1,m)-2*t_prev(n,m)+t_cur)-(Dt*v_prev(n,m)/4*DY*t_cur-t_prev(n+1,m)+t_prev(n,m))-(Dt/4*DY*v_prev(n,m))-(Dt/2*DY^2*t_cur);
elseif n==nodey %(at bottom boundary condtion i.e (nodey,2:NT)
S(n)=(-Dt*u_prev(n,m)*-T(n,m-1)+t_prev(n,m)-t_prev(n,m-1)/2*DX)+(Dt/2*DY^2*T_surf(m)-2*t_prev(n,m)+t_prev(n-1,m))-(Dt*v_prev(n,m)/4*DY*t_prev(n-1,m)-T_surf(m)+t_prev(n,m))-(-Dt/4*DY*v_prev(n,m))-(Dt/2*DY^2*T_surf(m));
else
S(n)=(-Dt*u_prev(n,m)*-T(n,m-1)+t_prev(n,m)-t_prev(n,m-1)/2*DX)+(Dt/2*DY^2*t_prev(n+1,m)-2*t_prev(n,m)+t_prev(n-1,m))-(Dt*v_prev(n,m)/4*DY*t_prev(n-1,m)-t_prev(n+1,m)+t_prev(n,m));
end
end
end
this boundary condtion must satisfy in T. I need the output like If this S get gradually values, then it will store in T. and then if I plot at each column wise in T, it must comes like this "begins with 1 and increase gradually, then diminish gradually to 0 ".

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