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I've been asked to come up with a way to numerically solve the differential equation $$\frac{\partial T}{\partial t} = \frac{\partial}{\partial x} \left( \chi(T) \frac{\partial T}{\partial x} \right) $$ for a function $T(x, t)$ with the boundary condition $\chi(T) \frac{\partial T}{\partial x} = P$ at $x=x_R$, where $P$ is some constant. My thinking at this point was that \begin{align} \frac{\partial T}{\partial t} \biggr|_{x=x_R} & = \frac{\partial}{\partial x} \left( \chi(T) \frac{\partial T}{\partial x}\biggr|_{x=x_R} \right)\\ &=\frac{\partial}{\partial x} \left( P \right)\\ &= 0 \\ \implies T(x_R, t) &=constant \end{align} and in particular that $T(x_R, t) = T(x_R, 0)$ but when I asked my lecturer about it they said that I "couldn't just apply that equation to that point" and I'm not sure what he means. Should I instead be evaluating $$\frac{\partial T}{\partial t} \biggr|_{x=x_R} = \frac{\partial}{\partial x} \left( \chi(T) \frac{\partial T}{\partial x} \right)\biggr|_{x=x_R} $$

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  • $\begingroup$ You first need to define the domain. And then define the boundary condition on that domation as well as the initial condition for $t=0$. $\endgroup$ – MrYouMath Nov 24 '16 at 16:01
  • $\begingroup$ You should also clarify how nice $\chi$ is. $\endgroup$ – Ian Nov 24 '16 at 16:19
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The problem is not applying the heat equation to the boundary point. The problem is a more basic one that has nothing to do with PDE:

Your argument basically is that if $ f (0)=0$ then $ \frac{d}{dx}_{|x=0 }f (x)=\frac{d}{dx} 0=0$. This is obviously nonsense.

EDIT: Woops, just read the final part of your question. Yes, that's what you should be evaluating.

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