pdepe: why does spatial discretization fail?

2015 4 6
1 个回答
回答已采纳
更新时间：2015 4 7
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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;

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 采纳的回答

Bill Greene
Bill Greene 2015-4-6

1 个投票

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;

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