resolution of PDE

Question

mouna 2011-5-7

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/7000-resolution-of-pde

I have a system of equations:

(?C/?t)+(u/?)?C/?z+((1-?)/?)*(?s/?f)*Ki*a*(q-KéqC)=0             
?q/?t=-Ki a(q-KéqC)

the initial condition is:

t=0, q=q0 and C=C0,z>0

the boundary conditions are:

z=0;C=0 and q=q0
z=L; ?C/?z=0 ,?q/?z=0

this is my program: i don't find the error so pleaaase help me. thanks in advance.

function modelediffusion
m=0;
z=linspace(0,1,26);
t=linspace(0,2,200);
sol = pdepe(m,@pdexpde,@pdexic,@pdexbc,z,t);
C = sol(:,:,1);
q = sol(:,:,2);
figure
plot(t,C(:,15))
function[g,f,s]= pdexpde(z,t,C,DCDz)
rhos=0.55;
rhof=0.385;
dp=0.03;
a=6/dp;
epsilon=0.45;
Ki=1.4E-7;
u=0.098;
Keq=16.86;
A=Ki*a*(C(2)-(Keq.*C(1)));
B=((1-epsilon)/epsilon)*(rhos/rhof)*A;
g=[1; 1];
f=[0; 0];
s=[((-u)/epsilon).*DCDz-B; A];
function C0 = pdexic(z)
c0=1.5E-3;
q0=2.53E-2;
C0 = [c0; q0];                                 
% -------------------------------------------------------------------------
function [pl,ql,pr,qr] = pdexbc(zl,Cl,zr,Cr,t)
q0=2.53E-2;
pl = [C1(1);q0-Cl(2)];
ql = [1; 0];
pr = [0; 0];
qr = [0; 1];

5 个评论
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Walter Roberson 2011-5-8

The question marks indicate derivative. Except the ones that appear isolate right after a "/" -- I don't know what those are.

Is KéqC a complete variable, or is it Kéq * C ?

Walter Roberson 2011-5-8

((1-?)/?) is a puzzler alright.

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Answer 1

Andrew Newell 2011-5-8

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/7000-resolution-of-pde#answer_9595

在 MATLAB Online 中打开

Here is one error: in this line,

pl = [C1(1);q0-Cl(2)];

the first C1 is C followed by the numeral 1, but it should be C followed by the letter l.

Debugging note: It is a good idea to have just one command per line. That way, when you get an error message, you know which command triggered it (if not which command actually caused it).