MATLAB: How i can solve the following PDE using periodic boundry coditions

error

clear all,close all

x=linspace(0,1,50);

t=linspace(0,1,50);

m=0;

eqn=@(x,t,u,dudx)transistorPDE(x,t,u,dudx,C);

ic=@(x)transistorIC(x,C);

sol=pdepe((m,eqn,ic,transistorBC,x,t);

function [c,f,s]=transistorPDE(x,t,u,dudx,C)

a=0.1;

c=1;

f=0;

s=-a*dudx;

end

function u0=transistorIC(x,C)

L=1;

u0=exp(-(x-L/2)^2);

end

function [pl,ql,pr,qr]=transistorBC(xl,ul,xr,ur,t)

pl=ul;

ql=0;

pr=pl;

qr=0;

end

Best Answer

Periodic boundary conditions require that when something leaves one boundary with a certain velocity, it enters the boundary on the other side with the same velocity. In other words:

So, you would need to modify your values for the qL and qR coefficients in transistorBC to incorporate the flux term (which generally includes the above partial derivative). The boundary conditions are written in the standardized form:

The flux term f is defined in the main pdefun function. In your case, you have f = 0 so would not be able to do this without also redefining your flux term (which would also change what equation you're solving).

Related Solutions

MATLAB: Pdepe help! (boundry condition in PDE solver)

Your boundary condition function includes the line

pl = ul-TT(i);

TT(i) is not defined within view of the function, so that is what is causing the error. To make that work you need to pass TT in as an extra parameter. But, still, the index i is undefined. So I would recommend just doing something like:

function [pl,ql,pr,qr] = pdex1bc(xl,ul,xr,ur,t)
Tg = 25; % Or Tg = h(t) if you can express it as a function of t
pl = ul-Tg; 
ql = 0; 
pr = 0; 
qr = 1;

If you really want to pass in TT as a parameter, change the header of the boundary function to

function [pl,ql,pr,qr] = pdex1bc(xl,ul,xr,ur,t,TT)

and then define the variable

bc = @(xl,ul,xr,ur,t) pdex1bc(xl,ul,xr,ur,t,TT);

This allows pdepe to pass in the 4 inputs it expects, and allows you to pass in the extra TT input from the workspace. You'll need to define the index i somewhere within pdex1bc as well. The call to pdepe then becomes

sol = pdepe(m,@pdex1pde,@pdex1ic,bc,x,t);

Also, although these don't cause errors some other possible issues are:

The constant k appears in your flux term f = ..., but not in your original equation.
Your source term should just be s = 0 from the original equation, but you have lines to define y, F, and s.

MATLAB: How to solve non-linear PDE

hi,

try :

 doc pdepe

Generally the PDEs in matlab follow the general formuale :

 c(x,t,u,du/dx).du/dt=(x^-m).d/dx[(x^-m)f(x,t,u,du/dx)]+s(x,t,u,du/dx)

Where the s is the source term and f is the flux term.

given a PDE , you have to make an analogy between your equation and the general form above, so for example in your case we have : s=0; m=0;c=1; and f=u^0.8*Diveregence(u) .

You have missing conditions in your problem : Initial conditions and Boundary conditions , i tried to write for you the sample with default conditions in Math(c) documentation , adapt it based on you I.C:

 % function ( M-file)
function SOL=PDEX1()
r=linspace(0,5,100);
t=linspace(0,30,100);
m=0;
SOL= pdepe(m,@pdex1pde,@pdex1ic,@pdex1bc,r,t);
function [c,f,s] = pdex1pde(r,t,u,M)
% du/dt=div(u^(0.8)du/dr)
c =1;
D=gradient(u);
M =(u.^(0.8)).*D;
f= M;
s =0;
function u0 = pdex1ic(r)
u0 = sin(pi*r);
function [pl,ql,pr,qr] = pdex1bc(xl,ul,xr,ur,t)
pl = ul;
ql = 0;
pr = pi * exp(-t);
qr = 1;

change now the initial/ boundary conditions.

Your equation seems like it has a Nusselt number no? anyway we r waiting for the result interpretation,

 % In the workspace try :
 >>S=PDEX1();
 >>surf(S);

I hope this helps.

Best Answer

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MATLAB: Pdepe help! (boundry condition in PDE solver)

MATLAB: How to solve non-linear PDE

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