Prove that a classical solution of $-\langle\nabla,A\nabla u\rangle=f$ is also a weak one Let


*

*$\Omega\subseteq\mathbb{R}^n$ a domain

*$f\in L^2(\Omega)$

*$A:\Omega\to\mathbb{R}^{n\times n}$ be Borel-measurable and $A(x)$ be symmetric, for all $x\in\Omega$

*$u\in C^2(\Omega)$ with $A\nabla u\in L^1_\text{loc}(\Omega)$ and $$-\langle\nabla,A\nabla u\rangle=f\;\;\;\text{in }\Omega\tag{1}$$


How can we show, that $$-\int_\Omega\langle A\nabla u,\nabla\varphi\rangle\;d\lambda^n=\int_\Omega f\varphi\;d\lambda^n\;\;\;\text{for all }\varphi\in C_0^1(\Omega)\tag{2}\;?$$ Clearly, from $(1)$ we get $$-\int_\Omega\langle\nabla,A\nabla u\rangle\varphi\;d\lambda^n=\int_\Omega f\varphi\;d\lambda^n\;,$$ but I've no idea how we can show $$-\int_\Omega\langle\nabla,A\nabla u\rangle\varphi\;d\lambda^n=-\int_\Omega\langle A\nabla u,\nabla\varphi\rangle\;d\lambda^n$$ I assume we need to apply one of Green's identities, but how?
 A: Recall the vector form of integration by parts in higher dimensions:
$$\int_{\Omega}\nabla\phi\cdot\boldsymbol{v}\,d\Omega= \int_{\partial\Omega}\phi(\boldsymbol{v}\cdot\boldsymbol{n})\,d\Gamma-\int_{\Omega}\phi\nabla\cdot\boldsymbol{v}\,d\Omega.$$
In your case, $\boldsymbol{v} = A\nabla u$ and equality (2) follows immediately, assuming the notation $\phi \in C_{0}^{1}(\Omega)$ means that $\phi$ has compact support, i.e. vanishes on the boundary. 
A: I'm not very familiar with your notation, but maybe this can help you. Recall Green's identity for scalar fields $u$, $\phi$: 
$$\int_\Omega (\nabla \cdot A \nabla u) \, \phi \, \mathrm{d} \Omega =  \int_{\partial \Omega} ( \mathbf{n} \cdot A \nabla u )\, \phi \, d\Gamma - \int_\Omega A \nabla u \cdot \nabla \phi \, \mathrm{d}\Omega,$$ where $\partial \Omega$ is the border of $\Omega$. I think your identity follows now if $\mathbf{n} \cdot A \nabla u = 0$ holds.
I have assumed that:
$$\int_\Omega (\nabla \cdot A \nabla u) \, \phi \, \mathrm{d} \Omega =  \int_\Omega \langle \nabla, A \nabla u\rangle \phi \, \mathrm{d} \Omega $$
Let me know if this was helpful.
Cheers!
