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Re: 2 domain PDE, NDSolve

• To: mathgroup at smc.vnet.net
• Subject: [mg88606] Re: 2 domain PDE, NDSolve
• From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
• Date: Fri, 9 May 2008 07:16:56 -0400 (EDT)
• Organization: The Open University, Milton Keynes, UK
• References: <g00v09$gkr$1@smc.vnet.net>

jeramyk wrote:

> I'm working on solving the basic 1D, transient heat transfer equation in a 
> two layer slab configuration.  One side is insulated, the other has a 
> constant heat flux, and there is not contact resistance between the two 
> layers.
> 
> My formulation looks like this:
> equation1:  T1(0,1)[z,t] == k1/rho1/Cp1*T1(2,0)[z,t]
> equation2:  T2(0,1)[z,t] == k2/rho2/Cp2*T2(2,0)[z,t]
> IC1:  T1[z,0] == T0
> IC2:  T2[z,0] == T0
> BC1:  T1(1,0)[0,t] == 0
> BC2:  -k1*T2(1,0)[L2+L1,t] == qs
> Match1:  k1*T1(1,0)[L1,t] == k2*T2(1,0)[L1,t]
> Match2:  T1(1,0)[L1,t] == T2(1,0)[L1,t]
> 
> T1[z,t] is solve from 0 to L1, T2[z,t] from L1 to L1+L2.
> 
> I've tried NDSolve but it doesn't like the matching conditions (it 
> interprets them as BC's and since they're not on the boundary it has issue 
> with it).

<snip>

Please, could you post genuine Mathematica expressions as well as what 
you tried with NDSolve? (I believe that by writing T1(0,1)[z,t] you mean 
actually Derivative[0, 1][T1][z, t], but I am too lazy today to correct 
all the expressions :-)

Regards,
-- Jean-Marc