Use energy method to solve modified heat equation A question in my PDE class is 
Suppose $a(x) = a_{ij}(x)$ takes values in the class of symmetric, positive definite $n \times n$ matrices. Consider the PDE
$$u_t = \sum_{i,j=1}^n \frac{\partial}{\partial x_i}\left( a_{ij}(x) \frac{\partial u}{\partial x_j} \right)$$
in a bounded domain $\Omega$, with intial condition $u=u_0(x)$ at $t=0$ and a Dirichlet boundary condition $u=g$ at $\partial \Omega$. Use the "energy method" to show there can be at most one solution.
In the past, we have used the energy method to show uniqueness by multiplying both sides by $u$ and integrating:
$$\int_{\Omega}uu_t= \frac12 \frac{d}{dt}\int_{\Omega}u^2 = \int_{\Omega} u \sum_{i,j=1}^n \frac{\partial}{\partial x_i}\left( a_{ij}(x) \frac{\partial u}{\partial x_j} \right),$$
and attempting to show somehow that the right hand side must be negative, therefore if there are two solutions $u_1$ and $u_2$, their difference has zero initial state, and $\frac{d}{dt}\int_{\Omega}u^2 dx \leq 0$ shows that it must remain there, so their difference is the zero solution, and therefore the solution is unique.
Here I am trying to use the positive definiteness of $a(x)$ to show that the right hand side must be negative. If we can write it in the form $\int_{\Omega} -\langle a(x)\vec{v}, \vec{v}\rangle$ we will be done. But so far I have not been able to do so. 
If we apply the chain rule we get 
$$\frac12 \frac{d}{dt}\int_{\Omega}u^2 = \int_{\Omega} u \sum_{i,j=1}^n \left( \frac{\partial a_{ij}}{\partial x_i} \frac{\partial u}{\partial x_j} + a_{ij}(x)\frac{\partial^2 u}{\partial x_i \partial x_j}\right),$$
but I'm not getting anywhere with that. Any ideas?
 A: As dls mentioned you're lacking an integral on the RHS. With this notice that
$$
\sum_{i,j=1}^n \partial_i(a_{ij} \partial_j u) =\text{div}(a \nabla u), 
$$
so that, integrating by parts and using the zero boundary conditions, we get that the RHS is equal to 
$$
-\int_{\Omega} \langle \nabla u, a \nabla u \rangle dx \leq 0.
$$
Edit: Notice that you  want to prove that if $u(x,0)=0$ for every $x\in \Omega$ and $u(y,t)=0$ for every $y\in \partial \Omega$ and $t>0$ then $u\equiv 0$. So take such an $u$, then your argument gives
$$
\frac{1}{2}\frac{d}{dt} \int_{\Omega} |u|^2 dx = \int_{\Omega} u \text{div}(a\nabla u) dx.
$$
Call $v=a\nabla u$. The divergence theorem, applied to the vector field $uv$, gives 
$$
\int_{\Omega}u \text{div}(v)dx= -\int_{\Omega}\langle \nabla u, v \rangle dx + \int_{\partial \Omega}\langle uv,\nu \rangle d \sigma.
$$
Since we have that $u(y,t)=0$ for every $y\in \partial \Omega$ the boundary integral is zero for every $t$. Therefore we have
$$
\frac{1}{2}\frac{d}{dt}\int_{\Omega} |u|^2 dx =- \int_{\Omega} \langle \nabla u,a\nabla u\rangle dx \leq 0.
$$
