pdepe: why does spatial discretization fail?

조회 수: 6 (최근 30일)
Wim Bergmans
Wim Bergmans 2015년 4월 6일
댓글: Wim Bergmans 2015년 4월 7일
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;

채택된 답변

Bill Greene
Bill Greene 2015년 4월 6일
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;

추가 답변 (0개)

카테고리

Help CenterFile Exchange에서 PDE Solvers에 대해 자세히 알아보기

태그

제품

Community Treasure Hunt

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

Start Hunting!

Translated by