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Working in finite differences, I am using a transformation on the temperature variable

$\Theta = \int_{T0}^T \kappa(T)dT$

to linearize the steady-state heat equation into a Poisson equation

$-\nabla\cdot(\kappa\nabla T)=-S \rightarrow \nabla^2 \Theta= -S$

The Dirichlet and Neumann boundary conditions transform in a way such that they have the same form. However, it seems that a Dirichlet condition for a heat sink pulls out too much heat from the system. So I'm trying a convective boundary condition, but this Robin boundary condition causes a problem

$ -\kappa \nabla T = h(T-T_f) \rightarrow -\nabla \Theta=h[T(\Theta) - T_f] $

where $h$ is a heat transfer coefficient, $T(\Theta)$ is the inverse transform back to the temperature variable, and $T_f$ the ambient temperature outside the boundary. The inverse transform is nonlinear making the boundary condition nonlinear as well.

I'm just trying to find some ideas on how I could possibly implement this sort of boundary condition in this case. Can I Just discretize the derivative in the boundary condition as in a Neumann except that instead of a ghost point I have a point outside the domain where the temperature is known ($T_f$) and then solve iteratively?

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  • $\begingroup$ Not sure if this helps/works, but what happens if you just linearize the boundary condition in $\Theta$? Does that fail? $\endgroup$ – Kirill Jun 18 '14 at 18:29
  • $\begingroup$ You mean to have the B.C. be $h[\Theta - \Theta_f]$ right? This is what I initially thought would happen to the B.C. but ran into problems. Then after looking around I found it apparently doesn't transform this way. I've never ran into a Robin B.C. before though, always gotten around with Dirichlet or Neumann. $\endgroup$ – user41125 Jun 18 '14 at 18:43
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    $\begingroup$ No, I was thinking $-\nabla\Theta=h \frac{dT}{d\Theta}(\Theta_f)(\Theta-\Theta_f)$, the Taylor series for the rhs. $\endgroup$ – Kirill Jun 18 '14 at 19:08
  • $\begingroup$ Hadn't thought of it actually. I'll play around with it and see what happens. Thanks a lot for the idea. $\endgroup$ – user41125 Jun 18 '14 at 19:10

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