# Evans' PDE Problem 6 Chapter 6 - Existence and uniqueness of weak solutions of Poisson's equation with mixed Dirichlet-Neumann boundary conditions

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 $$T$$?

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. $$C\lVert u\rVert_{H^1}^2\le B[u,u] \quad \forall u\in H^1_{\Gamma_1}(U)$$
• Many thanks for the detailed proof. I don't understand why does the norm $\lVert Tu_{k_j}-Tu\rVert_{L^2(\partial \Gamma_1)}$ is taken on $\partial \Gamma_1$ and not on $\partial U$? Is it due to the boundary conditions? Jun 20, 2021 at 17:13
• Yes, we only know $$\lVert Tv_{k_j}\rVert_{L^2(\partial \Gamma_1)}=0$$.@soundwave Jun 21, 2021 at 1:24
Is it ok to take $$v$$ only in $$H^1_{\Gamma_1}(U)$$ or $$v$$ must be taken in $$H^1(U)$$?
The second integral in your above calculation would not vanish if you take $$v$$ from $$H^1$$; that's why for a proper definition you have no other choice than to take the test function $$v$$ only from $$H^1_{\Gamma_1}$$.
To prove the coercitivity you need a Poincaré inequality in the space $$H^1_{\Gamma_1}$$.