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).
1) How do I show that $a(u,v)$ is a bounded bilinear form?
2) 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).
My main problem for boundedness is I don't know how to deal with the matrix terms. I can't just say that, eg. $|vMM^Tv| \leq |vM|^2 \leq |v|^2|M|^2$ and use the fact that $\Phi$ is bounded, for example, since I can't split the vector and matrix (or can I?).
For the coercivity condition, how do I deal with the last term in $a$, which has a minus sign? Also I don't know how to deal with the matrix terms in that last term. The second term is fine since it becomes $|\nabla u M|^2 > 0$ since $M$ represents derivative of the diffeomorphism $\Phi$ and has full rank.
Best Answer
We have \begin{align} |a(u,v)|&\leqslant \sup_{\Omega}|f|\lVert u\rVert_{L^2}\lVert v\rVert_{L^2}+\max_{1\leqslant k,j\leqslant n}\sup_{x\in \Omega}\lVert \partial_k\Phi_j(x)\rVert^2\lVert\nabla u\rVert_{L^2}\lVert \nabla v\rVert_{L^2}\\ &+\max_{1\leqslant k,j\leqslant n}\sup_{x\in \Omega}\lVert \partial_k\Phi_j(x)\rVert^2\lVert\nabla u\rVert_{L^2}\lVert v\rVert_{L^2}\sup_{x\in\Omega}\left|\frac{\nabla J(x)}{J(x)}\right|\\ &\leqslant C \lVert\nabla u\rVert_{H^1}\nabla v\rVert_{H^1} ,\end{align} where $$C:=\sup_{\Omega}|f|+\max_{1\leqslant k,j\leqslant n}\sup_{x\in \Omega}\lVert \partial_k\Phi_j(x)\rVert^2+\max_{1\leqslant k,j\leqslant n}\sup_{x\in \Omega}\lVert \partial_k\Phi_j(x)\rVert^2\sup_{x\in\Omega}\left|\frac{\nabla J(x)}{J(x)}\right|.$$
I've not checked all the details, but the following idea can work. We have $$ a(u,u)\geqslant \min_{\Omega}f\int_{\Omega}u^2+\alpha\int_\Omega|\nabla u|^2-2A\lVert u\rVert_{L^2}\lVert \nabla u\rVert_{L^2}, $$ where $\alpha:=\min_{t\in \Omega}\min_{\lVert x\rVert=1}\lVert M(t)x\rVert^2$ and $K$ a constant depending on $J$. Let $C:=M^2+\min_{\Omega}f$, where $M$ will be specified. Then $$a(u,u)+2C\lVert u\rVert_{L^2}\geqslant C\lVert u\rVert_{L^2}+\left(M\lVert u\rVert_{L^2}^2-\frac AM\lVert \nabla u\rVert_{L^2}\right)^2-\frac{A^2}{M^2}\lVert \nabla u\rVert_{L^2}^2+\alpha\lVert \nabla u\rVert_{L^2}^2.$$ So take $M$ such that $\alpha-\frac{A^2}{M^2}>0$, and take $K:=\min\{C, \alpha-\frac{A^2}{M^2}\}$.