Does PDEPE solve interlinked PDEs?

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?

[Torsten]

How exactly does d(u1,u2)/dt look like in your case ?

Answer 1

[Shokufe]

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)

[Torsten]

∂(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]

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!