Does PDEPE solve interlinked PDEs?

2018 9 21
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更新时间：2018 9 27
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Hi every one. I Have two interlinked PDEs that I want to solve with pdepe. I already saw in mathwork that pdepe is capable of solving two pdes but my case is a little bit different. In my case, not only the fluxs are combinations of u1 and u2, but also in the time-dependent part I have both u1 and u2. In below form:
∂/∂x (f( x,t,u1,u2,∂u1/∂x,∂u2/∂x))+S1=C1 ∂(u1,u2)/∂t
∂/∂x (f( x,t,u1,u2,∂u1/∂x,∂u2/∂x))+S2=C2 ∂(u1,u2)/∂t
I was wondering if pdepe can solve these sort of pdes?

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Shokufe
Shokufe 2018-9-24
编辑：Shokufe 2018-9-24

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Here are the general formulations of my pdes:
∂/∂x ((ρo K kro/μo) (∂po/∂x) ) + qo = ∂(so)/∂t
.
∂/∂x ((ρw K krw/μw) (∂pw/∂x) ) + qw = ∂(sw)/∂t
.
∂/∂x ((ρg K krg/μg) (∂pg/∂x) ) + qg = ∂(1-so-sw)/∂t
.
And the u is defined as: u=[po;so;sw]
Other parameters are functions of u1 or u2. For example:
pw=f(u1,u2)
pg=f(u1,u2)

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Torsten
Torsten 2018-9-24
编辑：Torsten 2018-9-24
∂(1-so-sw)/∂t =
-(∂(so)/∂t+∂(sw)/∂t)=
-(∂/∂x ((ρo K kro/μo) (∂po/∂x) ) + qo + ∂/∂x ((ρw K krw/μw) (∂pw/∂x) ) + qw )
Thus the third equation must be included without time-derivative term (c=0) as
∂/∂x ((ρo K kro/μo) (∂po/∂x) ) + qo + ∂/∂x ((ρw K krw/μw) (∂pw/∂x) ) + qw + ∂/∂x ((ρg K krg/μg) (∂pg/∂x) ) + qg = 0
Shokufe
Shokufe 2018-9-27
Many thanks, Torsten, that is correct! I applied the pdepe form for the pdes but I'm getting warnings and NAN answers which I think it might be due to the boundary conditions. However, if I couldn't solve it I may get back to you again. Thanks so much again!

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