Let $I$ be the open interval $(0, 1)$. Assume that $p \in C^1(\bar I)$ and $q \in$ $C(\bar I)$ such that $p(x) \geq \alpha>0$ and $q(x) \geq \alpha>0$ for all $x \in \bar I$. Here $\alpha$ is a constant. Let $f \in L^2 (I)$. I would like to verify a claim in Brezis' exercise 8.26 that the system $$ (1) \quad \begin{cases} -(p u')' + q u=f \quad \text{on} \quad I,\\ u'(0)= u'(1)=0, \end{cases} $$ has a unique (weak) solution in $H^2 (I)$.
There are possibly subtle mistakes that I could not recognize in my below attempt. Could you please have a check on it?
If $u$ is a classical solution to $(1)$, then $$ (2) \quad \int_I [-(p u')'w + q uw] = \int_I fw, \quad \forall w \in H^1 (I), $$ which (by integration by parts) implies $$ (3) \quad \int_I [p u'w' + q uw] = \int_I fw, \quad \forall w \in H^1 (I). $$
We define a continuous symmetric bilinear form $a$ on $H^1 (I)$ by $$ a(u, w) := \int_I [p u'w' + q uw]. $$
We have $$ a(u, u) = \int_I [p |u'| + q |u|^2] \ge \alpha \|u\|^2_{H^1}, $$ which implies $a$ is coercive. By Lax-Milgram theorem, $(3)$ has a unique solution $u \in H^1 (I)$. From $(3)$, we also have $$ \int_I (p u')w' = -\int_I (qu-f)w, \quad \forall w \in H^1 (I), $$ which implies $(p u')'=qu-f$ and thus $pu' \in H^1 (I)$. Because $p \ge \alpha >0$ and $p' \in C (\bar I)$, we get $\frac{1}{p} \in H^1 (I)$.
Corollary 8.10 Let $I$ be a (possibly unbounded) open interval of $\mathbb R$. Let $u, v \in W^{1, p}(I)$ with $1 \leq p \leq$ $\infty$. Then $u v \in W^{1, p}(I)$ and $(u v)^{\prime}=u^{\prime} v+u v^{\prime}$. Furthermore, the formula for integration by parts holds: $$ \int_y^x u^{\prime} v=u(x) v(x)-u(y) v(y)-\int_y^x u v^{\prime} \quad \forall x, y \in \bar{I} . $$
By above corollary (in the same book), we get $u' \in H^1 (I)$ and thus $u \in H^2 (I)$. It follows from $(3)$ and $(p u')'=qu-f$ and integration by parts that $$ (pu'w)(1)-(pu'w)(0)=0, \quad \forall w \in H^1 (I), $$ which implies $$ (pu')(1) = (pu')(0)=0, $$ which implies $$ u'(1) = u'(0)=0. $$