a priori estimates involving Sobolev norms

Question

Let $\sigma,$ $f$ and $g$ be $C^{2}(\overline{\Omega})$ functions, with $0<\frac{1}{M} < \sigma < M.$ We have the Dirichlet problem:

$\text{div}\sigma \nabla u=f, \hspace{3mm} \text{in} \hspace{1mm} \Omega$

$u=g \hspace{3mm} \text{in} \hspace{1mm} \partial \Omega$

I want to derive an a priori estimate of the form:

$\Vert u \Vert _{H^{1}(\Omega)} \leq C \left( \Vert f \Vert _{H^{-1}(\Omega)} +\Vert g \Vert _{H^{1/2}(\partial \Omega)}\right)$

I do not know if this is possible, but it would be very helpful to me.

The Fourier transform approach is messy and does not get me anywhere. I have tried with an strategy based upon the Rellich theorem, very much like in the proof of Poincaré inequality. Is this a good idea? I do not know what else to try.

Thanks for the help.

Answer 1

Take $w$ a solution to the problem $\text{div}\sigma \nabla w = 0$ and $w=g$ on $\partial \Omega$, let also $v$ be a solution of $\text{div} \sigma\nabla v = f$ with zero boundary conditions. Obviously we can write $u=v+ w$, so we obtain the appropriate estimates for the last ones.

1. $v$ is the easiest, since by definition of weak solution we have $$ \int_\Omega \sigma \nabla v \cdot \nabla \varphi = \langle-f, \varphi\rangle, \qquad \forall \varphi \in H^1_0( \Omega). $$ Choosing $v=\varphi$ (by the boundary conditions on $v$) we arrive at $$ M^{-2}\int_\Omega |\nabla v|^2 = \langle -f, v\rangle \leq \| f\|_{H^{-1}} \| v\|_{H^1} $$ and so we get, by Poincare's inequality, $\| v\|_{H^1} \leq C\| f\|_{H^{-1}}$.

2. For $w$ we proceed as follows: Write $w= G- w_1 $, where $G$ is an $H^1 $ extension of $g$ to $\Omega$ and $w_1$ solves $\text{div}\sigma \nabla w_1 =\text{div}\sigma \nabla G$ with zero boundary conditions. By the definition of weak solution we get $$ \int_\Omega \sigma \nabla w_1 \cdot \nabla \varphi = \int_\Omega \sigma \nabla G \cdot \nabla \varphi, \qquad \forall \varphi\in H^1_0(\Omega) $$ so taking $\varphi=w_1$ we get $\| \nabla w_1 \|_2 \leq C \| \nabla G\|_2$. Therefore $\| \nabla w\|_2 \leq C\| \nabla G\|_2$. Since $G$ was any extension of $g$, we conclude that $\|\nabla w\|_2 \leq C\| g\|_{H^{1/2}}$. To control the $L^2$ norms of $w$ simply note that, by Poincare, $\| w_1\|_2 \leq C\| \nabla w_1\|_2 \leq C \| G\|_{H^1}$ and trivially $\| G\|_2 \leq \| G\|_{H^1}$, now just take the infimum over the extensions $G$ to conclude as before.