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Let $\Omega \subset \mathbb{R}^n$ be a smooth domain, $u,v \in H^1_0(\Omega)$ with the usual inner product and let $M=M(x)$ be a $n\times n$ matrix with entries $m_{ij}\colon \Omega \to \mathbb{R}$ which are as smooth as necessary. Furthermore, $M(x)$ is positive-definite and invertible for every $x$ .

Is it true that there exist $\varphi, \phi \in H^1_0(\Omega)$ such that $$\int_\Omega \nabla u^T M \nabla v = \int_\Omega \nabla u^T \nabla \varphi + \int_\Omega u\phi?$$ If so how we can relate $\varphi$ and $\phi$ with $M$ and $v$?

Outside of when $M$ is a multiple of the identity, I don't know.

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  • $\begingroup$ If $\Omega$ is bounded and $M$ is symmetric and continuous then this follows from Riesz with $\varphi = \phi$. I imagine that you should be able to prove the general statement for $\Omega$ bounded by tracing through the proof of Lax-Milgram. $\endgroup$ – Neal Jul 15 '20 at 2:40
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I am not quite sure my approach, but I do think that we need more restrictions on $M$. Since it would be quite long, instead of putting it on the comment, I write down my idea here.

Let $v \in C_c^\infty(\Omega)$ for simplicity. Note that the $i$-th entry of $(M\nabla v)_i = \sum_{j=1}^n a^{ij} v_{x_j}$. Therefore, $$ \begin{align} \int \nabla u^T M \nabla v & = \int \nabla u \cdot (M\nabla v) \\& = \int \sum_{i=1}^n u_{x_i} \sum_{j=1}^n a^{ij} v_{x_j} \\ & = \int \sum_{i,j=1}^n u_{x_i}a^{ij} v_{x_j} \\ & = \int \sum_{i=1}^n u_{x_i}a^{ii} v_{x_i} + \int \sum_{i\neq j} u_{x_i}a^{ij} v_{x_j}\\ & = \int \sum_{i=1}^n u_{x_i}a^{ii} v_{x_i} + \int u \left(\sum_{i\neq j} a^{ij} v_{x_ix_j} + a_{x_i}^{ij}v_{x_j}\right) \end{align} $$ So in order to have the specific form you require, we need $a^{ii} = \text{const}$ for $i= 1, \cdots, n$, and $a_{x_i}^{ij} = 0$ for all $i \neq j$, provided we have $v \in C^\infty_c$. However, if $v \in H_0^1$ only, we may not be able to do integration by parts in the last equality, since we may not have the second derivative of $v$.

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