Suppose $U \subset \mathbb R^n$ is an open, bounded and connected set, with smooth boundary $\partial U$ consisting of two disjoint, closed sets $\Gamma 1$ and $\Gamma 2$. Define what it means for $u$ to be a weak solution of Poisson's equation with mixed Dirichlet-Neumann boundary conditions:
\begin{cases}
-\Delta u = f \ \ \ \text{in $U$} \\
u = 0 \ \ \ \text{on $\Gamma_1$} \\
\frac{\partial u}{\partial \nu} = 0 \ \ \text{on $\Gamma_2$}.
\end{cases}
Discuss the existence and uniqueness of weak solutions.
[Source: Evans' PDE 2nd edition, page 366, problem 6]
My attempt: Let $u \in C^\infty(U)$ be a solution of the problem and let $H^1_{\Gamma_1}(U) = \{v \in H^1(U)\colon v = 0 \text{ on } \Gamma_1 \}$, which is a Hilbert space with respect to the inner product in $H^1$. To define a weak solution of the Poisson's equation multiply it by $v \in H^1_{\Gamma_1}(U)$ and integrate by parts
\begin{align*}
\int_U fv \,dx
&= -\int_U \Delta u \cdot v \,dx \\
&= \int_U Du \cdot Dv \,dx – \int_{\partial U} \frac{\partial u}{\partial \nu} \cdot v \, dS \\
&= \int_U Du \cdot Dv \,dx – \int_{\Gamma_1} \frac{\partial u}{\partial \nu} \cdot v \, dS – \int_{\Gamma_2} \frac{\partial u}{\partial \nu} \cdot v \, dS \\
&= \int_U Du \cdot Dv \,dx
\end{align*}
where the second integral on RHS is zero since $v \in H^1_{\Gamma_1}(U)$ and the third integral is zero since $\frac{\partial u}{\partial \nu} = 0$ on $\Gamma_2$. Is it ok to take $v$ only in $H^1_{\Gamma_1}(U)$ or $v$ must be taken in $H^1(U)$?
Define the bilinear map $B:H^1(U)\times H^1_{\Gamma_1}(U) \to \mathbb R$ by $B[u,v] = \int_U Du \cdot Dv \, dx$. To prove the existence and uniqueness of weak solutions we have to prove that $B$ satisfies the hypotheses of Lax-Milgram, i.e. that $B$ is continuous (linear + bounded) and coercive.
For the boundness we can use Cauchy-Schwarz and the definition of Sobolev norm $||u||_{H^1} = ||u||_{L^2}+||Du||_{L^2}$, which implies $||u||_{H^1} \ge ||Du||_{L^2}$. So
$$
|B[u,v]|
\le \int_U |Du \cdot Dv| \, dx
\le ||Du||_{L^2(U)}||Dv||_{L^2(U)}
\le ||u||_{H^1(U)}||v||_{H^1(U)}.
$$
Is all good until here? What about the proof that $B$ is coercive? I read something about the trace operator $T$, maybe is better to define $B$ through $T$?
Best Answer
We will prove that $B$ is coercive in $H^1_{\Gamma_1}\times H^1_{\Gamma_1}$. We argue by contradiction. If $B$ is not coercive, then there would exist for each integer $k=2,\cdots $ a function $u_k\in H^1_{\Gamma_1}(U)$ staisfying $$ \frac{1}{k}\lVert u_k\rVert_{H^1}^2 \ge B[u_k,u_k]=\int_U |Du_k|^2dx $$ We renormalize by defining $$ v_k:=\frac{u_k}{\lVert u_k\rVert_{L^2}} $$ Then $$ \lVert v_k\rVert_{L^2}=1,\lVert Dv_k\rVert_{L^2}\le \frac{1}{k-1} $$ In particular the functions $\{v_k\}_{k=2}^{\infty}$ are bounded in $H^1(U)$. In view of the Rellich-Kondrachov theorem, there exists a subsequence $\{v_{k_j}\}_{j=1}^{\infty}\subset \{v_k\}_{k=2}^{\infty}$ and a function $v$ in $H^1(U)$ such that $$ v_{k_j}\rightarrow v \quad \text{in } L^2,\quad v_{k_j}\rightharpoonup v \quad \text{in } H^1 $$ It follows that $$ \lVert v\rVert_{L^2(U)}=1,\lVert Dv_k\rVert_{L^2(U)}=0 $$ In fact, $$ \int_U vD_i\phi dx=\lim_{j\to \infty}\int_U v_{k_j}D\phi dx=-\lim_{j\to \infty}\int_{U} Dv_{k_j}\phi dx=0\quad \forall \phi\in C_c^{\infty}(U) $$ Hence, $Dv=0$ a.e. and $v_{k_j}\to v$ in $H^1$. Thus $v$ is constant, since $U$ is connected. According to trace theorem, we have $Tv=v|_{\partial U}=C$, since $v=C$ is $C^1$. Due to $$ \lim_{j\to \infty}\lVert Tv_{k_j}-Tv\rVert_{L^2(\partial \Gamma_1)}\le C\lim_{j\to \infty}\lVert v_{k_j}-v\rVert_{H^1}=0, $$ we conclude that $\lVert v\rVert_{L^2(\Gamma_1)}=0$ and $C=0$, in which case $\lVert v\rVert_{L^2((U)}=0$. This contradiction establishs the estimate. $$ C\lVert u\rVert_{H^1}^2\le B[u,u] \quad \forall u\in H^1_{\Gamma_1}(U) $$