I am reading through Chapter 8 of Brezis. Particularly Section 8.4, which deals with some examples of boundary value problems on an interval. For the most part I understand the steps in the proofs contained in 8.4, however I am lacking the motivation and reasoning behind constructing the proofs as they are. I refer to Proposition 8.17 as an example.

Consider the problem, \begin{align}(21)\qquad\begin{cases} -u''+u=f\quad\text{on }I=(0,1),\\\ u'(0)=u'(1)=0. \end{cases}\end{align}

Proposition 8.17

Given $f\in L^{2}(I)$ there exists a unique function $u\in H^{2}(I)$ satisfying (21). Furthermore, $u$ is obtained by, \begin{align} \min_{v\in H^{1}(I)}\bigg(\frac{1}{2}\int_{I}(v'^{2}+v^{2})-\int_{I}fv\bigg) \end{align} If, in addition, $f\in C(\overline{I})$, then $u\in C^{2}(\overline{I})$.

Proof. If $u$ is a classical solution of (21) we have, \begin{align} (22)\qquad\int_{I}u'v'+\int_{I}uv=\int_{I}fv\quad\forall v\in H^{1}(I). \end{align} We use $H^{1}(I)$ as our function space: there is no point in working in $H^{1}_{0}$ as above since $u(0)$ and $u(1)$ are a priori unkonwn. We apply the Lax_Milgram theorem with the bilinear form $a(u,v)=\int_{I}u'v'+\int_{I}uv$ and the linear function $\varphi:v\mapsto\int_{I}fv$. In this way we obtain a unique function $u\in H^{2}(I)$. Using (22) once more we obtain, \begin{align} (23)\qquad\int_{I}(-u''+u-f)v+u'(1)v(1)-u'(0)v(0)=0\quad\forall v\in H^{1}(I). \end{align} In (23) begin by choosing $v\in H^{1}_{0}$ and obtain $-u''+u=f$ a.e Returning to (23), there remains, \begin{align} u'(1)v(1)-u'(0)v(0)=0\quad\forall v\in H^{1}(I). \end{align} Since $v(0)$ and $v(1)$ are arbitrary, we deduce that $u'(0)=u'(1)=0$. $\blacksquare$

My questions are:

1. Why do we begin by assuming $u$ is a classical solution if that is a result of the proposition?

2. Once we show that there exists a weak solution $u\in H^{2}(I)$ what are we aiming to do in the second part of the proof? i.e. from (23) onwards. By this I mean, we have showed the same proposition for the same PDE with homogeneous Dirichlet conditions. So what stops me from showing this same property in that proposition as opposed to this proposition (if you have the text I am referring to Proposition 8.15).

  • $\begingroup$ Giving this question a bump. If anyone can help that would be greatly appreciated. $\endgroup$ – Zeta-Squared Jul 22 at 8:58

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