# Solve in $H^1(B_2\backslash B_1)$ $-\nabla \cdot (A\nabla u)=f$ for $f\in L^2(B_2\backslash B_1)$

Let $d\geq 2$ and $f\in L^2(B_2\backslash B_1)$ and $A\in \mathbb R^{d\times d}$ a constant matrix positive definite. I recall that $B_r$ is the ball centered at $0$ of radius $r$.

1) Prove that there is a unique solution $u\in H^{1}(B_2\backslash B_1)$ to the system $$\begin{cases}-\nabla \cdot (A\nabla u)=f&\text{in }B_2\backslash B_1\\ u=0&\text{on }\partial (B_2\backslash B_1)\end{cases}$$

I set $\Omega =B_2\backslash B_1$. The weak formulation is $$\int_\Omega (-\nabla \cdot (A\nabla u)\varphi=\int_\Omega f\varphi\iff -\int_{\partial \Omega }\varphi A\nabla u\cdot \nu+\int_\Omega A\nabla u\cdot \nabla \varphi=\int_\Omega f\varphi$$ for all $\varphi\in H^1(B_2\backslash B_1)$.

Q1) Do we have that $-\int_{\partial \Omega }\varphi A\nabla u\cdot \nu=0$ ? If yes, how can I show it rigorously ?

if $-\int_{\partial \Omega }\varphi A\nabla u\cdot \nu=0$, then I set $$a(u,\varphi)=\int_\Omega A\nabla u\cdot \nabla \varphi\quad \text{and}\quad L(u)=\int_\Omega fu.$$

$\bullet$ Using Holder, we have that $$|L(u)|\leq \|f\|_{L^2}\|u\|_{L^2}\leq \|f\|_{L^2}\|u\|_{H^1},$$ and thus $L$ is continuous.

$\bullet$ $$|a(u,\varphi)|=\left|\int_\Omega A\nabla u\cdot \nabla \varphi\right|\leq\int_\Omega |A\nabla u\cdot \nabla \varphi|\leq \underbrace{\|A\nabla u\|_{L^2}\|}_{\leq \|A\|\|\nabla u\|_{L^2}}\nabla \varphi\|_{L^2} \le \|A\|\|\nabla u\|_{L^2}\|\nabla v\|_{L^2}$$ $$\leq \|A\|\|\nabla u\|_{H^1}\|\nabla v\|_{H^1},$$ and thus we have continuity of $a$.

$\bullet$ $$|a(u,u)|=\left|\int_\Omega A\nabla u\cdot \nabla u \right|$$

Q2) Are the two first point correct ? I'm never sure with all my norm estimation. If yes, how conclude the coercivity, i.e. that there is $\nu>0$ s.t. $|a(u,u)|\geq \nu\|u\|_{H^1}^2$ ?

2) Prove that there is a unique solution $u\in H^1(B_2\backslash B_1)$ to the system $$\begin{cases}-\nabla \cdot (A\nabla u)=f&\text{in }B_2\backslash B_1\\ u=0&\text{on }\partial B_2\\ A\nabla u\cdot \frac{x}{|x|}=0&\text{on }\partial B_1\end{cases}.$$ Hint: Prove first that there is $C>0$ s.t. $\|v\|_{L^2(H)}\leq C\|\nabla v\|_{L^2(H)}$ for all $v\in H=\{v\in H^1(B_2\backslash B_1)\mid v|_{\partial B_2}=0\}$.

I proved the Hint, but I don't know how to apply Lax-Milgram's theorem here.

1) Yes because $\varphi$ must be $0$ on the boundary (i.e. in your weak formulation, you must take $\varphi\in H_0^1(B_2\backslash \overline{B_1})$).
2) Your proofs are corrects. For the coercivity, $A$ is definite positive what mean all proper value are in $\mathbb R^+_*$. Therefore if $\lambda _i$ are the eigenvalue, then $|a(v,v)|>\min_{i=1,...,n}\lambda _i\|v\|^2$.
3) The weak for is as in 1) but $\int_{\partial \Omega }\varphi A\nabla u\cdot \nu=\int_{\partial B_2}\varphi A\nabla u\cdot \nu+\int_{\partial B_1}\varphi A\nabla u\cdot \nu=0.$ So you conclude as in 1).