# Poisson's equation with Robin boundary conditions

Explain how to define $u \in H^1(U)$ to be a weak solution of Poisson's equation with Robin boundary conditions: \begin{align} \begin{cases} \, \, \, \, -\Delta u = f & \text{in }U \\ u+\frac{\partial u}{\partial v}=0 & \text{on } \partial U. \end{cases} \end{align} Discuss the existence and uniqueness of a weak solution for a given $f \in L^2(U)$.

This is Exercise 5 in Chapter 6 of PDE Evans, 2nd edition.

I would like to define the bilinear form $B[u,v]$, for all $u,v \in H_0^1(U)$. But they did not really give that to the reader, unlike in Exercises 3 and 4 in the textbook.

Should I still define $B[u,v]$? If so, then I can try to satisfy the hypotheses of the Lax-Milgram Theorem, which would allow me to assert the existence of a weak solution to this problem.

It is not appropriate to work in the space $$H^1_0(U)$$ since nonzero boundary conditions are being considered. You will have to assume some regularity of the boundary of $$U$$.

One version of Green's theorem (see e.g. the appendix in Evans) is that $$- \int_U (\Delta u) v \, dx = \int_U Du \cdot Dv \, dx -\int_{\partial U} \frac{\partial u}{\partial \nu} v \, dS.$$

A weak solution to the problem at hand can be proposed by setting $$-\Delta u = f$$ in $$U$$ and $$\dfrac{\partial u}{\partial \nu} = -u$$ on $$\partial U$$ so that $$\int_U fv = \int_U Du \cdot Dv + \int_{\partial U} uv \, dS \quad \forall v \in H^1(U)$$ or a bit more precisely $$\int_U fv = \int_U Du \cdot Dv + \int_{\partial U} \newcommand{\tr}{\mathrm{Tr}\ \! }( \tr u )(\tr v) \, dS \quad \forall v \in H^1(U)$$ where $$\tr : H^1(U) \to L^2(\partial U)$$ is the trace operator.

An appropriate bilinear form is thus given by $$B[u,v] = \int_U Du \cdot Dv + \int_{\partial U} \newcommand{\tr}{\mathrm{Tr}\ \! }( \tr u )(\tr v) \, dS, \quad u,v \in H^1(U).$$ $$B$$ is clearly bounded. As far as coercivity goes, it may be helpful to use the Rellich-Kondrachov theorem. I can follow up with a hint if you like.

It remains to show that there is a constant $$\alpha > 0$$ with the property that $$\|u\|_{H^1}^2 \le \alpha B[u,u]$$ for all $$u \in H^1(U)$$. This can be proven by contradition. Otherwise, for every $$n \ge \mathbb N$$ there would exist $$u_n \in H^1(U)$$ with the property that $$\|u_n\|^2_{H^1} > n B[u_n,u_n]$$. For each $$n$$ define $$v_n = \dfrac{u_n}{\|u_n\|_{H^1}}$$. Then $$v_n \in H^1(U)$$, $$\|v_n\|_{H^1} = 1$$, and $$B[v_n,v_n] < \dfrac 1n$$ and all $$n$$.

Here we can invoke Rellich-Kondrachov. Since the family $$\{v_n\}$$ is bounded in the $$H^1$$ norm, there is a subsequence $$\{v_{n_k}\}$$ that converges to a limit $$v \in L^2(U)$$. However, since $$\|Dv_{n_k}\|_{L^2}^2 < \dfrac{1}{n_k}$$ it is also true that $$Dv_{n_k} \to 0$$ in $$L^2$$. Thus for any $$\phi \in C_0^\infty(U)$$ you have $$\int_U v D \phi \, dx = \lim_{k \to \infty} \int_U v_{n_k} D \phi \, dx = - \lim_{k \to \infty} \int_U D v_{n_k} \phi \, dx = 0.$$ This means $$v \in H^1(U)$$ and $$D v = 0$$, from which you can conclude $$v_{n_k} \to v$$ in $$H^1(U)$$. Since $$\|v_{n_k}\|_{H^1} = 1$$ for all $$k$$ it follows that $$\|v\|_{H^1} = 1$$ as well.

Next, since $$\|\tr v_{n_k}\|_{L^2(\partial U)}^2 < \dfrac{1}{n_k}$$ and the trace operator is bounded there is a constant $$C$$ for which $$\|\tr v\|_{L^2(\partial \Omega)} \le \|\tr v - \tr v_{n_k}\|_{L^2(\partial \Omega)} + \|\tr v_{n_k}\|_{L^2(\partial \Omega)} < \frac{1}{n_k} + C \|v - v_{n_k}\|_{H^1(U)}.$$ Let $$k \to \infty$$ to find that $$\tr v = 0$$ in $$L^2(\partial U)$$.

Can you prove that if $$v \in H^1(U)$$, $$Dv = 0$$, and $$\tr v = 0$$, then $$v = 0$$? Once you have established that fact you arrive at a contradiction, since $$v$$ also satisfies $$\|v\|_{H^1} = 1$$. It follows that $$B$$ is in fact coercive.

• I tried to edit your response to fix (what I thought) were typos, and tried to clarify minor things, to avoid ambiguity and make it easier for me to follow. Lastly, I changed the $\Omega$ to $U$, just to stay consistent with Evans' notation. Are all modifications changes correct? Because, for example, we never had $\Delta u = f$ on $\partial U$, when we were given $-\Delta u = f$ in $U$ by the problem. – Cookie Feb 2 '15 at 16:37
• Looks good. Thanks for pointing out the errors. – Umberto P. Feb 2 '15 at 17:03
• There's another one (I didn't modify), I think. Shouldn't the trace operator $T$ be mapped from $H^1(U) \to L^2(\color{red}{\partial}U)$? – Cookie Feb 2 '15 at 19:38
• Yeah, I just fixed it. – Umberto P. Feb 2 '15 at 19:46
• Fundamental question, I know, but does the $\frac{\partial u}{\partial \nu}$ in the "$u+\frac{\partial u}{\partial \nu}=0$ on $\partial U$" suggest that $\partial U$ is $C^1$? I'm asking since the problem doesn't explicitly state this, and the Trace Theorem requires this in its hypothesis. I want to be absolutely sure that we can conclude $Tu=u\vert_{\partial U}$ from the Trace theorem, as you are doing so in the line after writing "a bit more precisely" in your answer. – Cookie Feb 3 '15 at 20:12