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

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 Torsten Posts: 1,717 Registered: 11/8/10
Re: System of 1D PDEs with coupled boundary conditions
Posted: Aug 23, 2013 5:26 AM

"Luminita T." wrote in message <kv789d\$mae\$1@newscl01ah.mathworks.com>...
> "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

flux1 and flux2 are scalar values prescribed by you - so if the flux term for the third equation is
f(3)=a3*du3/dx,
you can set
p(3)=flux1 + flux2
q(3) = 1
in the function where you prescribe the boundary conditions.

Best wishes
Torsten.

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