Solving 2 coupled pde
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I want to solve a system of 2 coupled pde (in MATLAB) of the format:
c1*(df/dt)+c2*(df/dz)+c3*(f)+c4*(g)=0
(dg/dt)=c5*f+c6*g
with Initial conditions as
f(0,t)=1, g(z,0)=0 and f(z,0)=0
0<f,g,z,t<1
I am getting the error:
Warning: Failure at t=5.414867e-02. Unable to meet integration tolerances without reducing
the step size below the smallest value allowed (1.110223e-16) at time t.
> In ode15s at 669
In pdepe at 317
In pdex4 at 13
Warning: Time integration has failed. Solution is available at requested time points up to
t=5.414805e-02.
> In pdepe at 323
In pdex4 at 13
Error using surf (line 75)
Data dimensions must agree.
Error in pdex4 (line 18)
surf(x,t,u1);
My code is
function pdex4
c1=(0.63*150*(10^-6))/(6.93*(10^-5)*3600);
c2=1;
c3=((1-0.63-0.27)*0.0045*150*(10^-6)*5021.8)/(6.93*(10^-5));
c4=-((1-0.63-0.27)*0.0045*150*(10^-6)*4.62*(10^3))/(1.03*6.93*(10^-5));
c5=(0.0045*3600*1.03*5021.8)/(4.62*(10^3));
c6=-0.0045*3600;
m = 0;
x = linspace(0,1,10);
t = linspace(0,1,1000000);
options=odeset('NonNegative',[]);
sol = pdepe(m,@pdex4pde,@pdex4ic,@pdex4bc,x,t,options);
u1 = sol(:,:,1);
u2 = sol(:,:,2);
figure
surf(x,t,u1);
title('u1(x,t)')
xlabel('Distance x')
ylabel('Time t')
figure
surf(x,t,u2);
title('u2(x,t)')
xlabel('Distance x')
ylabel('Time t')
function [c,f,s] = pdex4pde(x,t,u,DuDx)
c1=(0.63*150*(10-6))/(6.93*(10^-5)*3600);
c2=1;
c3=((1-0.63-0.27)*0.0045*150*(10^-6)*5021.8)/(6.93*(10^-5));
c4=-((1-0.63-0.27)*0.0045*150*(10^-6)*4.62*(10^3))/(1.03*6.93*(10^-5));
c5=(0.0045*3600*1.03*5021.8)/(4.62*(10^3));
c6=-0.0045*3600;
c = [c1; 1];
f = [-c2*u(1); 0];
s = [c3*u(1)+c4*u(2); c5*u(1)+c6*u(2)];
function u0 = pdex4ic(x)
u0 = [0; 0];
function [pl,ql,pr,qr] = pdex4bc(xl,ul,xr,ur,t)
pl = [-1; ul(2)-1];
ql = [1; 0];
pr = [-1; ur(2)-1];
qr = [1; 1];
Can you please tell me what changes should I make.
3 件のコメント
Torsten
2015 年 8 月 19 日
I think the structure of your matrices A and B is wrong.
But yes: ode15s can handle sparse ODE systems resulting from the discretization of PDEs.
Best wishes
Torsten.
回答 (1 件)
Walter Roberson
2015 年 8 月 19 日
Those boundary conditions are incompatible.
You have f(0,t)=1, g(z,0)=0 and f(z,0)=0 . So f(z,t)=1 when z = 0. Consider then t = 0, then f(z,0) = 1 by the first boundary condition when z = 0. But in the third boundary condition you said f(z,0) = 0 for all z which has to include the case of z = 0.
With your boundary conditions being incompatible you are not going to be able to meet the integration tolerances.
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