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Problem with Finite Difference Scheme

조회 수: 1 (최근 30일)
Danila Zharenkov
Danila Zharenkov 2013년 11월 6일
마감: MATLAB Answer Bot 2021년 8월 20일
Hello, I'm trying to build the finite difference scheme for this task
I build the finite difference scheme like this
The main problem is boundary conditions for theta, because boundary values on its edges don't depend of other counted values. Here is my code, maybe I made a mistake in it?
clc;
clear all;
%constants
r0=0.1;
a=1.01;
b=1;
mu=1;
c=100;
%step-size for r
n=100;
r_max=1;
hr=r_max/n;
r=0:hr:r_max;
nr=max(size(r));
%step-size for theta
m=200;
th_max=2*pi;
hth=th_max/m;
th=0:hth:th_max;
nth=max(size(th));
%step-size for t
l=100;
t_max=5;
ht=t_max/l;
time=0:ht:t_max;
nt=max(size(time));
u=zeros(nr,nth,nt);
v=zeros(nr,nth,nt);
%Initialization
for i=1:nr
for j=1:nth
if ((i*hr<=r0)&&(0<=th(j))&&(th(j)<=pi/6))
u0=0;
elseif ((i*hr<=r0)&&(pi/6<th(j))&&(th(j)<2*pi))
u0=1;
elseif ((i*hr>r0)&&(0<=th(j))&&(th(j)<=2*pi))
u0=0;
end
u(i,j,1)=u0;
end
end
for i=1:nr
for j=1:nth
if ((r(i)<=r0)&&(0<=th(j))&&(th(j)<=pi/6))
v0=1;
elseif ((r(i)<=r0)&&(pi/6<th(j))&&(th(j)<2*pi))
v0=0;
elseif ((r(i)>r0)&&(0<=th(j))&&(th(j)<=2*pi))
v0=0;
end
v(i,j,1)=v0;
end
end
for t=1:nt-1
for i=2:nr-1
for j=2:nth-1
u(i,j,t+1)=u(i,j,t)+ht*((u(i+1,j,t)-2*u(i,j,t)+u(i-1,j,t))/hr^2+(1/(i*hr)))*((u(i+1,j,t)-u(i,j,t))/hr)+(1/(i*hr)^2)*((u(i,j+1,t)-2*u(i,j,t)+u(i,j-1,t)/hth)+u(i,j,t)*(1-u(i,j,t)-c*v(i,j,t)));
v(i,j,t+1)=v(i,j,t)+ht*((v(i+1,j,t)-2*v(i,j,t)+v(i-1,j,t))/hr^2+(1/(i*hr)))*((v(i+1,j,t)-v(i,j,t))/hr)+(1/(i*hr)^2)*((v(i,j+1,t)-2*v(i,j,t)+v(i,j-1,t)/hth)+a*v(i,j,t)*(1-c*u(i,j,t)-b*v(i,j,t)));
%Boundary
u(1,j,t)=u(2,j,t);
u(nr,j,t)=u(nr-1,j,t);
v(2,j,t)=v(1,j,t);
v(nr,j,t)=v(nr-1,j,t);
u(i,nth,t)=u(i,1,t);
v(i,nth,t)=v(i,1,t);
end
end
end

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