MATLAB: How to solve non-linear PDE

differential equationsnonlinearpdeplot

I want to solve the following PDE and plot the U versus r for following equation du/dt=-div(u^0.8 du/dr). was wondering if anyone can help me out? 0=<r<=5, 0=<t<=30 Thanks

Best Answer

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.
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