Is solution of $div(k(x,y) \nabla(u(x,y)))=0$ harmonic function, for $k(x,y) > 0$?

Question

I have heat transfer equation $$ \mbox{div}(k(x,y) \nabla(u(x,y))) = 0, $$ with Dirichlet boundary conditions, non constant coefficient $k(x,y) \in L^2(\Omega)$ in some good domain $\Omega$. Is solution function $u(x,y)$ harmonic? How can I prove it?

Answer 1

If you look at the BVP for some $C \in \mathbb{R}$: $$ div(k(x,y) \nabla u(x,y))=f(x,y) \; \; (x,y) \in \Omega \\ u(x,y)=C \; \; (x,y) \in \partial \Omega $$ and you would assume that $u(x,y)$ is harmonic, you can conclude by using the maximum principle that $u=C$. Now taking any $f(x,y) \neq 0$ we get that the $0=f(x,y)$ contradiction by plugging in $u(x,y)$ into the PDE.
If you want to know anything else please let me know, as I feel like you wanted to know something else rather than this rather straight-forward answer. Maybe your equation was homogenous in the sense that $f=0$?