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.

  • $\begingroup$ 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. $\endgroup$ – Cookie Feb 2 '15 at 16:37
  • $\begingroup$ Looks good. Thanks for pointing out the errors. $\endgroup$ – Umberto P. Feb 2 '15 at 17:03
  • $\begingroup$ There's another one (I didn't modify), I think. Shouldn't the trace operator $T$ be mapped from $H^1(U) \to L^2(\color{red}{\partial}U)$? $\endgroup$ – Cookie Feb 2 '15 at 19:38
  • $\begingroup$ Yeah, I just fixed it. $\endgroup$ – Umberto P. Feb 2 '15 at 19:46
  • $\begingroup$ 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. $\endgroup$ – Cookie Feb 3 '15 at 20:12

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