I am solving a pde using pdepe in matlab. But I am getting warnings and some peculiar results.

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Rusty 2016-2-2

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https://ww2.mathworks.cn/matlabcentral/answers/266163-i-am-solving-a-pde-using-pdepe-in-matlab-but-i-am-getting-warnings-and-some-peculiar-results

编辑： Rusty 2016-2-2

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Here is my differential equation. A and B are constant. u1 is the lambda and u2 is the Here is my code for the problem

function diff
m = 0;
y = 0:0.01:01;
t = 0:0.01:29;
length(t)
options= odeset('RelTol',1e-13,'AbsTol',1e-8);
options = odeset('NonNegative', 2);
sol = pdepe(m,@diffpde,@diffic,@diffbc,y,t,options);
Lambda = sol(:,:,1);
Velocity = sol(:,:,2);
size(Lambda)
size(t)
size(y)
figure
plot(y, Velocity(end,:))
title('velocity at the last time');
figure
plot(y, Lambda(end,:))
title('Lambda at the last time');
% xlswrite('Velocity', Velocity);
xlswrite('Lambda', Lambda);
end
% -----------------------------------------------------------------------------
function [c,f,s] = diffpde(y,t,u,DuDy)
De= 0.04;
GammaDotNaughtTilda=0.9;
c = [De; 1]; 
f= [0;1].*flip(exp(u)).*DuDy;
s = [1;0]+[-u(1)*GammaDotNaughtTilda;0].*flip(DuDy);
end
% --------------------------------------------------------------
function u0 = diffic(y);
u0 = [0; 0]; 
end
% --------------------------------------------------------------
function [pl,ql,pr,qr] = diffbc(yl,ul,yr,ur,t)
pl = [0; ul(2)]; 
ql = [1; 0]; 
pr = [ur(1); ur(2)-1]; 
qr = [1; 0]; 
end

Here in my code De is the constant A and GammaDotNaughtTilda is B. I am using A as 0.04 and B as 0.9. I have to solve the problem to find the steady state. The problem with my code is that I m not able to find the steady state. After some time I get a warning "Warning: Time integration has failed. Solution is available at requested time points up to t=2.819000e+01. " and the last value of u1 i.e. Lambda I get is 700 always even if I change the values of A and B. Thanks