## Showing a coercivity condition for this bilinear form

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Suppose $\Omega \subset \mathbb{R}^n$ is a compact domain. Let $f$ and $J$ (and also $\frac 1J$) be $C^1$ functions on $\Omega$. Consider the bilinear form $a:H^1(\Omega) \times H^1(\Omega) \to \mathbb{R}$ $$a(u,v) = \int_\Omega uvf + \int_\Omega \nabla u MM^T\nabla v - \int_\Omega \nabla u MM^T\nabla J \frac{v}{J}$$ where $M = D\Phi$ is the matrix representation of the derivative of a diffeomorphism $\Phi$ between two compact hypersurfaces in $\mathbb{R}^n$ (so $\Phi$ and its derivatives are bounded).

How do I show that there exists a $C$ such that $$a(u,u) + C\lVert u \rVert^2_{L^2(\Omega)} \geq K\lVert u \rVert^2_{H^1(\Omega)}$$ for some $K$. (i.e. that $a$ satisfies a coercivity condition).

I don't know how to show this. How do I deal with the last term in $a$, which has a minus sign? The second term is fine since it becomes $|\nabla u M|^2 > 0$ since $M$ represents derivative of the diffeomorphism $\Phi$ and therefore has full rank. But basically I can't get a positive constant in front of $\lVert \nabla u \rVert_{L^2}$ term.

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1
Cross-posted at math.stackexchange: math.stackexchange.com/questions/237955/…. – Davide Giraudo Nov 17 at 17:28

## 1 Answer

I think the first thing to iron out is that $\Omega$ should be open, since compact would be closed and then even defining $H^1(\Omega)$ is not trivial. Maybe assume $\Omega \subset \mathbb{R}^N$ is open and $J,f$ being $C^1(\overline{\Omega})$.

However, this is not the issue you are interested in. You want to know the relationship between

$\int_\Omega \nabla u MM^t \nabla u\;dx$

and

$||\nabla u||_{L^2(\Omega) }$.

Given $M$, can you show that the eigenvalues of $MM^t$ are strictly positive? The ellipticity constant is just a statement on the eigenvalues of the matrix associated with the coefficients. So what you are looking to do is to prove that this matrix is strictly positive definite. I hope you can compute this successfully.

EDIT: The result you are looking for in this case is to show that $\xi MM^t\xi \geq c$ for all $\xi \in S^{N-1}$, which would imply the coercivity you are looking for. A priori, $MM^t$ positive definite means you can this inequality with $c=0$, but you need a little better if this is your approach. For example, if the eigenvalues of $MM^t$ are ${\lambda_i}_{i=1..N}$, with $\min_i \lambda_i = \tilde{\lambda}>0$, and these eigenvalues correspond to eigenvectors $x_i$ which form an orthogonal basis for $\mathbb{R}^N$, then this is true, as follows.

Any $x\in S^{N-1}$ can be represented as $x=\sum_{i=1}^N c_i x_i$, where $\sum_i c_i^2=1$, and we compute $MM^t x = \sum_i c_i MM^t x_i = \sum_i c_i \lambda_i x_i$, therefore $xMM^tx = \sum_i c_i^2 \lambda_i \geq \tilde{\lambda}$, where we have used $\sum_i c_i^2=1$.

This is only an example - you should verify that the eigenvalues correspond to orthonormal eigenvectors to do this, or adapt the proof to something which is close and true in your case.

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@Daniel Thanks for the answer.  MM^T is symmetric and positive definite (by assumption), so a theorem says all the eigenvalues are positive. This is enough, right?  – soup89 Nov 17 at 22:22
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