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Re: Re: 2 domain PDE, NDSolve

• To: mathgroup at smc.vnet.net
• Subject: [mg88649] Re: [mg88623] Re: 2 domain PDE, NDSolve
• From: "W_Craig Carter" <ccarter at mit.edu>
• Date: Sun, 11 May 2008 15:14:53 -0400 (EDT)
• References: <g00v09$gkr$1@smc.vnet.net> <g01bt2$l3u$1@smc.vnet.net>

Hello Jeremy,
I started looking at this, but I believe that you may have a typo in
your Dirichlet condition for T2. Assuming that you are solving a 1d
diffusion problem in two connected domains.

> You're right, the (0,1) was meant to mean Derivative[0,1]... There was also
> a typo in BC2.  And so the expressions look like this:
>
> Derivative[0, 1][T1][z, t] == k1/rho1/Cp1*Derivative[2, 0][T1][z, t]
> Derivative[0, 1][T2][z, t] == k2/rho2/Cp2*Derivative[2, 0][T2][z, t]
> T1[z, 0] == T0
> T2[z, 0] == T0
> Derivative[1, 0][T1][0, t] == 0
> -k2*Derivative[1, 0][T2][0, t] == qs
> k1*Derivative[1, 0][T1][L1, t] == k2*Derivative[1, 0][T2][L1, t]

I scaled to remove redundant parameters (you can get rid of T0, make
L2=1, L1=rat, etc)

(*left domain*)
t1de = Derivative[0, 1][T1][z, t] == diff1 Derivative[2, 0][T1][z, t]

(*right domain*)
t2de = Derivative[0, 1][T2][z, t] == diff2 Derivative[2, 0][T2][z, t]

(*ic left*)
icleft = T1[z, 0] == 0

(*ic right*)
icright = T2[z, 0] == 0

(*bc-left left)
dirchleft = Derivative[1, 0][T1][0, t] == 0

(*what ? T2 is defined on (L1,L2)*)
dirichwhat = diff2*Derivative[1, 0][T2][0, t] == qsscaled



In any case, for this problem I suggest that you use mathematica to do
this problem the old-fashioned way:
the solution in each domain will be a fourier series... All you need
to do is match coefficients of the derivative of each series at the
internal interface.... closed form solution.

Sorry if I my assumption was incorrect and you were thinking of a
different problem....

-- 
W. Craig Carter