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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).

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.

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What are the condition on $u$? In $H^1$ or $H^1_0$? –  Davide Giraudo Nov 15 '12 at 23:20
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@DavideGiraudo The space is $H^1(\Omega)$. $\Omega$ is technically a compact surface, so there is no boundary. –  soup Nov 15 '12 at 23:23
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For coercivity, we can bound the first term by $\min f\lVert u\rVert^2$, and $\inf_{||x||}||M(x)||>0$ so we can control the other terms. –  Davide Giraudo Nov 15 '12 at 23:26
    
@DavideGiraudo Thank you, I will try it tomorrow (almost midnight here..). In your answer, I believe you want the supremum of $(D\Phi(x))_{ij}$ or something like that and not just $\Phi(x)$ (recall $M$ is the derivative of $\Phi$.) –  soup Nov 15 '12 at 23:30
    
@DavideGiraudo I tried your hint for coercivity but I can't get it. I can't get a positive grad term out.. can you please help? –  soup Nov 16 '12 at 14:48

1 Answer 1

up vote 1 down vote accepted
  1. 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 $$C:=\sup_{\Omega}|f|+\max_{1\leqslant k,j\leqslant n}\sup_{x\in \Omega}\lVert \partial_k\Phi_j(x)\rVert^2+\max_{1\leqslant k,j\leqslant n}\sup_{x\in \Omega}\lVert \partial_k\Phi_j(x)\rVert^2\sup_{x\in\Omega}\left|\frac{\nabla J(x)}{J(x)}\right|.$$

  2. 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 $K:=\min\{C, \alpha-\frac{A^2}{M^2}\}$.

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Thanks Davide, I appreciate it, but I just don't see how you can bound the second integral with that $\alpha.$ –  soup Nov 17 '12 at 17:57
    
I write the integrand as the square of the euclidian norm of $\lVert M\nabla u$. –  Davide Giraudo Nov 17 '12 at 22:29
    
Where is $K$ in the first displayed formula of 2.? –  timur Nov 18 '12 at 2:09
    
I defined it in the last line. –  Davide Giraudo Nov 18 '12 at 13:46
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I has to be true for some $k$, if I red well the OP. –  Davide Giraudo Nov 18 '12 at 19:05

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