Weak solution to $-\Delta u+u=f$

Question

I'd like to show that there is a unique solution $u \in H^1_0$ solving \begin{align} -\Delta u+u=f ~~ \text{in} ~~\Omega \\ u=0 ~~~ \text{on} ~~\partial \Omega \end{align} where $\Omega$ is a bounded domain in $\mathbb{R}^n$ and $f \in L^2(\Omega)$.

Multiplying with a test function $v \in H^1_0(\Omega)$ and integrating yields \begin{equation} \sum_{j=1}^n \int_\Omega (-\frac{\partial^2 u}{\partial x_j^2})v(x)+\int_\Omega uv = \int_\Omega f(x)v(x). \end{equation} Using partial integration we get that the first term can be written as \begin{equation} \sum_{j=1}^n - \int_{\partial \Omega} \frac{\partial}{\partial x_j}u(x)v(x)n(x)dx+\sum_{j=1}^n\int_{\Omega} \frac{\partial}{\partial x_j}u(x) \frac{\partial}{\partial x_j}v(x)dx. \end{equation} Due to the boundary conditions the boundary integral equals $0$. Hence, we can define the bilinear form \begin{equation} a(u,v)=\int_\Omega \sum_{j=1}^n \frac{\partial}{\partial x_j}u(x) \frac{\partial}{\partial x_j}v(x)dx+\int_\Omega u(x)v(x)dx. \end{equation} Is that part correct? I'm just not too sure about the partial integration.

I'm having difficulties to show continuity and coercivity of $a$. For continuity I have \begin{equation} |a(u,v)\leq \int_\Omega \sum_{j=1}^n|\frac{\partial u}{\partial x_j} ||\frac{\partial v}{\partial x_j}| + \int_\Omega|u||v| \leq \sum_{j=1}^n ||\frac{\partial u}{\partial x_j}||_{L^2(\Omega)}||\frac{\partial v}{\partial x_j}||_{L^2(\Omega)}+||u||_{L^2(\Omega)}||v||_{L^2(\Omega)} \end{equation} I don't know how to proceed from there as I'm not sure what to do with the sum. Also, I tried to show the coercivity of $a$ but I didn't succeed.

Answer 1

For economical reasons, I will denote by $u_{x_i} \equiv \frac{\partial u}{\partial x_i}$ the weak partial derivative of $u$ with respect to $x_i$ and by $Du \equiv [u_{x_1},\ldots,u_{x_n}]^T$ its weak gradient.

Your integration by parts is correct, you need to prove that the bilinear form : $$ a:(u,v)\in H_0^1 \times H_0^1 \mapsto \int_{\Omega} Du\cdot Dv + \int_\Omega uv $$ is bounded and coercive, which by Lax-Milgram will imply existence of a unique solution.

For the boundedness part, you're almost there ! Just remember that $\|u\|_{H^1}^2 := \|u\|_{L^2}^2 + \|Du\|_{L^2}^2 $, hence $\|u\|_{L^2}\le \|u\|_{H^1}$ and $\|Du\|_{L^2}\le \|u\|_{H^1}$. Continuing from where you left, we get that $$\|u\|_{L^2}\|v\|_{L^2}\le \|u\|_{H^1}\|v\|_{H^1} $$ and $$\begin{align*}\sum_{i=1}^n \|u_{x_i}\|_{L^2}\|v_{x_i}\|_{L^2}&= \big[\|u_{x_1}\|_{L^2},\ldots,\|u_{x_n}\|_{L^2}\big]^T\cdot\big[\|v_{x_1}\|_{L^2},\ldots,\|v_{x_n}\|_{L^2}\big]^T\\ &\le \left(\sqrt{\sum_{i=1}^n\|u_{x_i}\|_{L^2}^2} \right)\cdot\left(\sqrt{\sum_{i=1}^n\|v_{x_i}\|_{L^2}^2} \right)\\ &= \|Du\|_{L^2}\|Dv\|_{L^2} \le \|u\|_{H^1}\|v\|_{H^1} \end{align*} $$ Where we have used Cauchy-Schwarz inequality (note that we could have used the $L^2$ version of C-S inequality to give a one line proof of the same estimate). Putting the two inequalities together, we conclude that $$|a(u,v)|\le 2\|u\|_{H^1}\|v\|_{H^1} $$ That is, $a$ is a bounded bilinear form. Now for the coercivity part, you need to prove that $|a(u,u)|\ge \alpha \|u\|_{H^1}^2$ for some constant $\alpha >0$, but observe that $$a(u,u) = \int_{\Omega}|Du|^2 + \int_\Omega u^2 = \|Du\|^2_{L^2} + \|u\|^2_{L^2} =: \|u\|^2_{H^1} $$ Can you see why $a$ is coercive now ? If so then congratulations, you can apply Lax-Milgram theorem and conclude that there exists a unique weak solution to your PDE.

(Note : Actually, to fulfill all the requirements of Lax-Milgram theorem, you need to check that the map $v\in H_{0}^1\mapsto \int_{\Omega}fv$ is bounded. I let you check that this is indeed the case here)