# Brezis (Variational Formulation for Boundary Value Problems)

Brezis considers the inhomogeneous Dirichlet problem, \begin{align} -\Delta u+u=f\quad &\text{in }\Omega,\\ u=g\quad &\text{on }\partial\Omega, \end{align} where $$f$$ is given on $$\Omega$$ and $$g$$ is given on $$\partial\Omega$$. In order to build up to the proposition which proves a unique weak solution exists for this problem Brezis first wishes to construct a closed convex set in $$H^{1}(\Omega)$$. He does this as follows:

Suppose that there exists a function $$\tilde{g}\in H^{1}(\Omega)\cap C(\overline{\Omega})$$ such that $$\tilde{g}=g$$ on $$\partial\Omega$$. and consider the set, \begin{align} K=\{v\in H^{1}(\Omega)\,|\,v-\tilde{g}\in H^{1}_{0}(\Omega)\}. \end{align} It follows from Theorem 9.17 that $$K$$ is independent of the choice of $$\tilde{g}$$ and depends only on $$g$$.

Where I require assistance: Theorem 9.17 states,

Theorem 9.17: Suppose $$\Omega$$ is of class $$C^{1}$$. Let, \begin{align} u\in W^{1,\,p}(\Omega)\cap C(\overline{\Omega})\quad\text{with }1\leq p<\infty. \end{align} Then the following properties are equivalent: \begin{align} \text{(i)}&\,u=0\text{ on }\partial\Omega,\\ \text{(ii)}&\, u\in W^{1,\,p}_{0}(\Omega). \end{align}

The way I see this working is that to show that $$v-\tilde{g}\in H^{1}_{0}(\Omega)$$ we can show that $$v-\tilde{g}=0$$ on $$\partial\Omega$$ and property (ii) from Theorem 9.17 follows (since the proof of (i)$$\implies$$(ii) does not require any assumptions on the smoothness of $$\Omega$$). Then from earlier we have $$v=\tilde{g}=g$$ on $$\partial\Omega$$, and so we drop the dependence on $$\tilde{g}$$. However, we do not know that $$v-\tilde{g}\in H^{1}(\Omega)\cap C(\overline{\Omega})$$ so how are we reaching this conclusion?

Showing $$K$$ is convex:

Suppose $$u,v\in K$$. Consider $$t\in\mathbb{R}$$, then, \begin{align} tv+(1-t)u=t(v-u)+u. \end{align} Take $$\tilde{g}$$ as before, then \begin{align} t(v-u)+u-\tilde{g}=t(v-\tilde{g})-t(u-\tilde{g})+(u-\tilde{g})\in H_{0}^{1}(\Omega), \end{align} since $$H^{1}_{0}(\Omega)$$ is a linear space. Hence $$tv+(1-t)u\in K$$ for $$t\in [0,1]$$.

• Just be clear, you are asking why $K$ is independent of the choice of $\tilde{g}$? – StarBug Aug 21 '19 at 7:41
• That is correct. – Zeta-Squared Aug 21 '19 at 8:00
• OK. I guess the problem in your argument is that you want to apply Thm 9.17 to $v-\tilde{g}$, which is not a continuous function and Thm 9.17 therefore not applicable. But you only need to apply Thm 9.17 to the difference of two continuous extensions (see my answer below, hope it helps). – StarBug Aug 21 '19 at 8:09
• You probably should ask the Extra Question as a new question. It doesn't really have anything to do with the first one. The comment you quote from Brezis is definitely not sufficient to understand why a weak solution is a classical solution when the data is sufficiently smooth. Most textbooks have a whole chapter devoted to this question (regularity). – StarBug Aug 21 '19 at 8:21
• Yes, I have created a new question for this now. – Zeta-Squared Aug 21 '19 at 8:24

To show that $$K$$ is independent of the choice of extension $$\tilde{g}$$, consider another extension $$\hat{g}\in H^1(\Omega)\cap C(\overline{\Omega})$$ with $$\hat{g}=g$$ on $$\partial\Omega$$. Put $$\hat{K}:=\{v \in H^1(\Omega)\ |\ v-\hat{g}\in H^1_0(\Omega) \}.$$
We have to verify that $$K=\hat{K}$$. To this end consider $$v\in K$$. Then $$v-\tilde{g}\in H^1_0(\Omega)$$. From Theorem 9.17 we deduce that also $$\hat{g}-\tilde{g}\in H^1_0(\Omega)$$. Since $$H^1_0(\Omega)$$ is a linear space, we obtain $$v-\hat{g} = v- \tilde{g} + (\tilde{g}-\hat{g}) \in H^1_0(\Omega).$$ Consequently, $$v\in\hat{K}$$. We conclude $$K\subset\hat{K}$$. The opposite inclusion follows by a similar argument, and we thus obtain $$K=\hat{K}$$.
• While what you have here is fine, does it really show the dependence of $K$ on $g$? I believe the dependence on $g$ Brezis talks about is that for every $v\in K$ $v=g$ on $\partial\Omega$, however this does not come across, at least for me, from your argument. – Zeta-Squared Aug 21 '19 at 8:20
• Just to be absolutely clear: $K$ depends on $g$. However, $g$ does not appear explicitly in the definition of $K$, but only implicitly via the extension $\tilde{g}$. Since such an extension is not unique, it is only natural to ask whether a different extension might yield a different set $K$. This is what is typically behind the phrase "$K$ is independent of the choice of $\tilde{g}$". – StarBug Aug 21 '19 at 8:35
• The question "is $v=g$ on $\partial\Omega$ for every $v\in K$" from your comment above is an entirely different question. In order to answer it, you first have to define what the value/meaning of $v$ is on the boundary $\partial\Omega$, which is by no means trivial since you only know $v\in H^1(\Omega)$. For this pupose you need the notion of the trace of $v$. It seems to me, however, that Brezis is trying to avoid introducing this notion (it is complicated). – StarBug Aug 21 '19 at 8:43
• One extra question regarding $K$. Brezis mentions it is nonempty, is this just because $\tilde{g}\in K$ always? (I should mention for this example $\Omega$ is an open bounded set in $\mathbb{R}^{N}$). – Zeta-Squared Aug 21 '19 at 8:47