# Writing $\int_\Omega \nabla u^T M \nabla v$ in terms of $H^1$ inner product of $u$ with another function

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.

• 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. – Neal Jul 15 '20 at 2:40

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 $$v$$.