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