pdepe: why does spatial discretization fail?

6 visualizaciones (últimos 30 días)
Wim Bergmans
Wim Bergmans el 6 de Abr. de 2015
Comentada: Wim Bergmans el 7 de Abr. de 2015
Hi,
when trying to solve a PDE very similar to the heat equation pdepe yields the following:
>> pdetest
Error using pdepe (line 293)
Spatial discretization has failed. Discretization supports only parabolic and elliptic equations, with flux term involving spatial derivative.
Error in pdetest (line 5)
sol = pdepe(0,@pde,@pdeic,@pdebc,xmesh,tspan);
What am I missing here? Please see the used code below:
function pdetest
xmesh = linspace(-4,3,82);
tspan = linspace(0,1,12);
sol = pdepe(0,@pde,@pdeic,@pdebc,xmesh,tspan);
function [c,f,s] = pde(x,t,u,DuDx)
c = 1;
f = 0.0968*DuDx;
s = 0;
function u0 = pdeic(x)
xmesh = linspace(-4,3,82);
[~, index] = min(abs(xmesh-x));
initial_values = [0.3639 0.3720 0.3801 0.3884 0.3968 ...
0.4054 0.4141 0.4229 0.4319 0.4411 0.4504 0.4599 0.4696 ...
0.4794 0.4894 0.4995 0.5098 0.5204 0.5310 0.5419 0.5530 ...
0.5642 0.5756 0.5873 0.5991 0.6111 0.6234 0.6358 0.6485 ...
0.6614 0.6745 0.6878 0.7014 0.7152 0.7292 0.7435 0.7580 ...
0.7727 0.7878 0.8030 0.8186 0.8344 0.8504 0.8668 0.8834 ...
0.9003 0.9175 0.9350 0.9528 0.9709 0.9893 1.0081 1.0271 ...
1.0465 1.0662 1.0863 1.1067 1.1274 1.1485 1.1700 1.1918 ...
1.2140 1.2366 1.2596 1.2829 1.3067 1.3309 1.3555 1.3805 ...
1.4059 1.4318 1.4581 1.4849 1.5121 1.5398 1.5680 1.5966 ...
1.6258 1.6554 1.6856 1.7163 1.7475];
u0 = initial_values(index);
function [pl,ql,pr,qr] = pdebc(xl,ul,xr,ur,t)
pl = 0.3639;
ql = 0;
pr = 1.7475;
qr = 0;

Respuesta aceptada

Bill Greene
Bill Greene el 6 de Abr. de 2015
I think you want the following in your pdebc function:
function [pl,ql,pr,qr] = pdebc(xl,ul,xr,ur,t)
pl = ul-0.3639;
ql = 0;
pr = ur-1.7475;
qr = 0;

Más respuestas (0)

Etiquetas

Productos

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by