Let $T(\mathbf{x},t)$ denote the temperature at location $\bf x$ and time $t$ in a closed and bounded region $R$ with thermal conductivity $k(\mathbf{x})>0$, density $\rho$ and specific heat capacity $c$, where $\rho$ and $c$ are positive constants. The boundary of R is $\partial R$.
Suppose that we have derived the heat equation $$\rho c{\partial T\over \partial t}=\nabla\cdot(k(\mathbf{x}) \nabla T), \quad \mathbf{x} \in R$$, assume that the heat energy source $S(\mathbf{x})>0$, the heat equation at steady state is given by $$0=\nabla \cdot(k(\mathbf{x}) \nabla T)+\rho S(\mathbf{x}), \quad \mathbf{x} \in R$$
(1)We want to show that if there exists a solution of the steady state with a same boundary condition, then the solution is unique.
(2)We want to show that if the boundary condition is $\mathbf{n} \cdot(k(\mathbf{x}) \nabla T)=0, \quad \mathbf{x} \in \partial R$ (i.e. no heat flux into $R$) then there is no solution.
So I tried to use $\phi=\phi_1-\phi_2$ where the latter two are supposed to be solutions, trying to show that $\phi:=0$.
It used to work in some basic heat equations, because in that case $\phi$ would also be a solution to the heat equation, then we can adapt Divergence Theorem and consider $\phi^2$, finding LHS being $0$ and RHS being $(\nabla\phi)^2$ and thus $\nabla\phi:=0\Rightarrow \phi \text{ is a constant}$, then combining the boundary condition to have $\phi:=0$.
But now here $\phi$ is not a solution, so I cannot summarise the RHS being $0$. So I am not sure how to proceed. I tried to consider $\nabla^2( \phi^2) ,\,\,\left(\nabla\cdot(k\mathbf{x})\nabla\phi^2\right)$ but neither of them work.
Any hint would be appreciated.
Now I solved it... Still the same procedure...
For (1), using the same notation: $\phi$ being the difference of two assumed solutions, then $\phi=0$ on $\partial R$ and $\nabla \cdot(k(\mathbf{x}) \nabla \phi)=0$ in $R$. If we consider $\phi^2$ then firstly $\nabla \cdot(k(\mathbf{x}) \nabla (\phi^2))=0$, then simplifying gives $(\nabla \phi)^2=0$ so $\phi$ is a constant, combining the boundary condition $\phi:=0$ hence the uniqueness.
For (2), similarly applying the Divergence theorem for the given boundary condition gives $\iiint_R \nabla\cdot k(\nabla T)\,\mathrm{d}V=0$ and $\iiint_R \nabla\cdot k(\nabla (T^2))\,\mathrm{d}V=0$. Then substituting using the same idea gives the integral of $\nabla (T^2)$ in $R$ is $0$. So $T$ is constant, but then it never satisfies the equation, as $S$ is defined to be strictly greater than $0$.
Thanks so much for @Jose27 's hint.