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Writing $int_Omega nabla u^T M nabla v$ in terms of $H^1$ inner product of $u$ with another function

Mathematics Asked by StopUsingFacebook on November 2, 2021

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 $ntimes n$ 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
$$int_Omega nabla u^T M nabla v = int_Omega nabla u^T nabla varphi + int_Omega uphi?$$
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.

One Answer

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 $(Mnabla v)_i = sum_{j=1}^n a^{ij} v_{x_j}$. Therefore, $$ begin{align} int nabla u^T M nabla v & = int nabla u cdot (Mnabla v) \& = int sum_{i=1}^n u_{x_i} sum_{j=1}^n a^{ij} v_{x_j} \ & = int sum_{i,j=1}^n u_{x_i}a^{ij} v_{x_j} \ & = int sum_{i=1}^n u_{x_i}a^{ii} v_{x_i} + int sum_{ineq j} u_{x_i}a^{ij} v_{x_j}\ & = int sum_{i=1}^n u_{x_i}a^{ii} v_{x_i} + int u left(sum_{ineq j} a^{ij} v_{x_ix_j} + a_{x_i}^{ij}v_{x_j}right) end{align} $$ 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 $v$.

Answered by mathdoge on November 2, 2021

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