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