Prove that the boundary value problem, \begin{cases} u_{t} -\nabla\cdot(k(x) \nabla u) = r(x,t) && \text{in } \Omega_{T},\\ u(x,0) = g(x) && \text{in } \ \Omega,\\ u(\sigma,t) = 0 && \text{in } \ \partial \Omega \times (0,T], \end{cases} has a unique solution. (Note: $k(x) > 0$.)
Attempt at solution
I must use the energy method in higher dimension, along with Green's identity. Let $w = u-v$ solve \begin{cases} u_{t} -\nabla\cdot(k(x) \nabla u) = 0 && \text{in } \Omega_{T},\\ u(x,0) = 0 && \text{in } \ \Omega,\\ u(\sigma,t) = 0 && \text{in } \ \partial \Omega \times (0,T], \end{cases} Then by the energy method , I want to show that $E'(t) \le 0 \implies E(t) \le 0 $, but by definition of energy $E(t) \ge 0$ which would imply that $w(x,t) \equiv 0$ from inside the integral.
\begin{align} E(t) &:= \int_{\Omega} w^{2}(x,t) \ dx \\ E'(t) &= \int_{\Omega}\frac{d}{dt} w^{2} (x,t) \ dx \\ &=2 \int_{\Omega} w w_{t} \ dx\\ \end{align} Using $w_{t} = \nabla \cdot(k(x) \nabla u)$, \begin{align} E'(t) &= 2\int_{\Omega} w (\nabla \cdot (k(x)\nabla w)) \ dx \\ &= 2\int_{\Omega} w (k(x) \Delta w + \nabla w \nabla k(x)) \ dx \\ &= 2 \int_{\Omega} w k(x) \Delta w \ dx + 2\int_{\Omega} w \nabla w \nabla k(x) \ dx \end{align} Greens Identity states that $$ \int_{\Omega} w \Delta w \ dx = \int_{\partial \Omega} w \partial_{n} w \ d\sigma - \int_{\Omega} \lvert \nabla w \rvert^{2} \ dx$$ therefore, \begin{align} E'(t) &= 2 \int_{\partial \Omega} k(x) w \partial_{n} w \ d\sigma - 2\int_{\Omega} k(x) \lvert \nabla w \rvert^{2} \ dx + 2 \int_{\Omega} w \nabla w \nabla k \ dx \end{align} Since the given BVP has Dirichlet homogenous boundaries , the first term vanishes since $w \partial_{n} w = 0 $ on $\partial w$, therefore $$ E'(t)= - 2\int_{\Omega} k(x) \lvert \nabla w \rvert^{2} \ dx + 2 \int_{\Omega} w \nabla w \nabla k(x) \ dx.$$ Now if I show that the latter expression is less or equal to $0$ then I'm done. The first integral is evidently less than $0$ since $k(x) >0$ ,but I don't know how to show that the second integral is less than the first or less than $0$, and I'm not sure if I went wrong somewhere in the proof.
Edit
I believe I found a solution. \begin{align} E'(t) &= 2\int_{\Omega} w(x) \text{div} (k(x) \nabla w) \\ &= 2\int_{\Omega} w(x) \sum_{i} (k(x) w_{x_{i}})_{x_{i}} \\ &= 2\int_{\partial \Omega} w(x) k(x) \sum_{i} w_{x_{i}} \ d\sigma - 2 \int_{\Omega} k(x)\sum_{i}(w_{x_{i}})^{2} \ dx \\ &= - 2 \int_{\Omega} k(x)\sum_{i}(w_{x_{i}})^{2} \ dx \\ E'(t) &= -2 \int_{\Omega} k(x) \Delta(w)^{2}, \end{align} the rest follows.
I'm adding proof verification tag , if someone could confirm.
You used Green's identity wrong in differentiating $E$. It should be
$$E'(t) = 2\int_\Omega w(\nabla \cdot (k(x)\nabla w)) \, dx = -2\int_\Omega k(x)|\nabla w|^2\, dx$$
from which $E'(t)\leq 0$ is clear.
EDIT: The version of Green's identity I used is
$$\int_\Omega w \nabla F \, dx = \int_{\partial \Omega} w F\cdot \nu \, dS - \int_\Omega \nabla w \cdot F\,dx$$
for a vector field $F$ (here, $F=k\nabla w$), which is just the Divergence Theorem.