Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: System of 1D PDEs with coupled boundary conditions
Replies: 5   Last Post: Aug 23, 2013 5:26 AM

 Messages: [ Previous | Next ]
 Luminita T. Posts: 3 Registered: 8/22/13
Re: System of 1D PDEs with coupled boundary conditions
Posted: Aug 23, 2013 4:59 AM

"Torsten" wrote in message <kv70bl\$1pq\$1@newscl01ah.mathworks.com>...
> "Luminita T." wrote in message <kv5999\$7e0\$1@newscl01ah.mathworks.com>...
> > "Torsten" wrote in message <kv589c\$j2g\$1@newscl01ah.mathworks.com>...
> > > "Luminita T." wrote in message <kv56d9\$k7n\$1@newscl01ah.mathworks.com>...
> > > > I've have a system of 4 parabolic PDE with boundary conditions that I would like to integrate in time starting from initial conditions.
> > > >
> > > > I've been trying to put the system into the pdepe() format, but it seems that I can't give coupled boundary conditions, in particular something of the form: a*DuDx(1) + b*DuDx(2) + c*DuDx(3) = 0.
> > > >
> > > > As far as I could see from the help, q(x,t) needs to be specified as a vector-valued function that is element-wise multiplied with f(x,t,u,DuDx), which in my case is just DuDx. This does not allow to express linear combinations of elements of f() for a boundary condition.
> > > >
> > > > If q(x,t) could be a matrix-valued function, then it could be done.
> > > > Is it possible to somehow specify coupled boundary conditions in pdepe() ?
> > > >
> > > > Alternatively, I've been looking at the pdetool() for 2D spatial PDE. It seems a bit more complicated to handle and I'll have to put a bit of time into it. At this point, I can't say directly from the UserGuide (2. Setting up your PDE >>Types of Boundary Conditions) if it can handle the coupled boundary conditions. Any help figuring this out is much appreciated.
> > > >

> > >
> > > Usually, the boundary condition for equation i refers directly to the flux defined in equation.
> > > Thus if f for equation i is given as
> > > f(i)=a*du1/dx + b*du2/dx + c*du3/dx,
> > > you simply have to put
> > > p(i)=0 and q(i)=1
> > > to define the boundary condition
> > > a*du1/dx + b*du2/dx + c*du3/dx = 0.
> > >
> > > Best wishes
> > > Torsten.

> >
> > Hmm, my normal flux functions are f(1) = a1*du1/dx; f(2) = a2*du2/dx, etc. Separate terms. But then maybe I can introduce a "fake" or "helper" equation to have this new flux to use in the boundary condition.

>
> Or if you have flux boundary conditions for all three variables at a common endpoint,
> you can solve for the separate boundary conditions, e.g.
> a1*du1/dx = flux1
> a2*du2/dx = flux2
> a1*du1/dx + a2*du2/dx + a3*du3/dx = 0
> gives
> a3*du3/dx = -flux1 - flux2
>
> Best wishes
> Torsten.

I'm very happy to try again with pdepe() and not having to dig into the pdetool(). Hopefully it will all work by next week.

Yes - the flux boundary conditions for the three variables are at a common endpoint. (Actually the u() functions have originally different domains, but this can be solved by a change of variable, and luckily the transformation allows also to the match the boundary endpoints)

I haven't really understood the last suggestion though. The general boundary equation form is

p(x,t,u) + q(x,t).*f(x,t,u,dx/du) = 0

where the q() vector function is only element wise multiplied with the f(x,t,u,dx/du) flux vector. Thus I don't see how I can write

a3*du3/dx = -flux1 - flux2

as a boundary condition.

best regards
Luminita

Date Subject Author
8/22/13 Luminita T.
8/22/13 Torsten
8/22/13 Luminita T.
8/23/13 Torsten
8/23/13 Luminita T.
8/23/13 Torsten