Defining boundary condition for pde for pdepe function
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I have got a pde as show in the function [c,f,s]. Unable to get the solution for the equation and not able figure out the mistake in boundary condition
L = 200;
s1 = 0.5; %equal to k at x=0
s2 = 0;
T = 4;
qr = 0.218;
f = 0.52;
a = 0.0115;
n = 2.03;
ks = 31.6;
x = linspace(0,L,100);
t = linspace(0,T,25);
options=odeset('RelTol',1e-4,'AbsTol',1e-4,'NormControl','off','InitialStep',1e-7)
u = pdepe(0,@unsatpde,@unsatic,@unsatbc,x,t,options,s1,s2,qr,f,a,n,ks);
I'M STRUCK AT THIS POINT. Following gives the editor .m files
% -------------------------------------------------------------------------
function [c,f,s] = trial1(x,t,u,DuDx)
global k n qr a ks p
c=1;
f = k*((DuDx)+1);
s = 0;
m =0.51;
q=qr+(p-qr)*(1+(-a*u)).^-m;
k=ks*((q-qr)/(p-qr))^0.5*(1-(1-((q-qr)/(p-qr))^(1/m))^m)^2;
% -------------------------------------------------------------------------
function u0 = unsatic(x,s1,s2,qr,f,a,n,ks)
u0 = 200+x;
% -------------------------------------------------------------------------
function [pl,ql,pr,qr] = unsatbc(xl,ul,xr,ur,t,s1,s2,qr,f,a,n,ks)
pl = 0;
ql = 1;
pr = ur(1);
qr =0;
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채택된 답변
Torsten
2016년 7월 26일
1. "trial1" is not part of the list of functions you call pdepe with.
2. You will have to include s1,s2,qr,f,a,n,ks in the parameter list for "trial1".
3. in "trial1", you use k before you calculate it.
4. In unsatbc, you set as boundary conditions
u=0 at x=L
and
du/dx = -1/k at x=0
I don't know if this is what you want to set.
Best wishes
Torsten.
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Torsten
2016년 7월 26일
q will become complex because 1+(-a*u) becomes negative and is raised to the power of -m.
Best wishes
Torsten.
추가 답변 (2개)
Zana Taher
2019년 4월 13일
Hi Torsten,
How would you make:
du/dx = 0 at x=0
instead of
du/dx = -1/k at x=0
?
jose luis huayanay villar
2020년 6월 22일
uld you help me embed in the simulink? to carry out a control?
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