Let $I$ be the open interval $(0, 1)$. I'm trying to solve a problem in Brezis' Functional Analysis, i.e.,
Exercise 8.21 Assume that $p \in C^1(\bar I)$ and $q \in$ $C(\bar I)$ such that $p(x) \geq \alpha>0$ and $q(x) \geq 0$ for all $x \in \bar I$. Let $v \in C^2(\bar I)$ be the unique solution of $$ (1) \quad \begin{cases} -(p v')' + q v=0 \quad \text {on} \quad \bar I,\\ v(0)=1, v(1)=0. \end{cases} $$
Set $k_0=v'(0)$.
- Check that $k_0 \leq-\alpha / p(0)$.
We now investigate the problem $$ (2) \quad \begin{cases} -(p u')' + q u=f \quad \text {on} \quad I,\\ u'(0)=ku, u(1)=0, \end{cases} $$
where $k \in \mathbb R$ and $f \in L^2 (I)$ are given.
- Assume $k=k_0$. Show that $(2)$ has a solution $u \in H^2(I)$ IFF $\int_I fv=0$.
- Assume now that $k \neq k_0$. Prove that for every $f \in L^2(I)$, problem $(2)$ admits a unique solution $u \in H^2(I)$.
There are possibly subtle mistakes that I could not recognize in my below attempt of (3.). Could you please have a check on it?
Let $K := \{w \in H^1(I) : w(1)=0\}$. Then $K$ is a closed subspace of $H^1(I)$. If $u$ is a classical solution to $(2)$, then $$ \int_I [-(p u')'w + q uw] = \int_I fw, \quad \forall w \in K, $$ which (by integration by parts) is equivalent to $$ (3) \quad kp(0)(uw)(0) + \int_I [p u'w' + q uw] = \int_I fw, \quad \forall w \in K. $$
We define a continuous bilinear form $a$ on $K$ by $$ a(u, w) := kp(0)(uw)(0) + \int_I [p u'w' + (\beta +q) uw], $$
where $\beta >0$ to be chosen. We need an auxilary result (in the same book), i.e.,
Exercise 8.5.1 Let $p \in (1, \infty)$. For each $\varepsilon >0$ there is $C=C(\varepsilon, p)$ such that $$ \| u \|_{L^\infty (I)} \le \varepsilon \| u' \|_{L^p (I)} + C \| u \|_{L^1 (I)}, \quad \forall u \in W^{1, p} (I). $$
By Exercise 8.5.1, we can pick $\beta >0$ large enough such that $a$ is coercive on $K$. By Lax-Milgram theorem, for each $z \in L^2(I)$ there is a unique $Tz \in K$ such that $$ a(Tz, w) = \int_I zw, \quad \forall w \in K. $$
We can further show that $Tz \in H^2(I)$. We claim that the linear map $T : L^2(I) \to L^2(I)$ is compact. By coercivity of $a$, there is $C>0$ such that $\int_I z(Tz) =a(Tz, Tz) \ge C \|Tz\|_{H^1}^2$ for all $z \in L^2(I)$. By C-S inequality, $\|z\|_{L^2} \ge C \|Tz\|_{H^1}$ for all $z \in L^2(I)$. Because $I$ bounded, the injection $H^1(I) \subset L^2(I)$ is compact. Then $T$ is compact.
We have $(3)$ is equivalent to $u=T(f+\beta u)$ and thus to $(I-\beta T)u= Tf$. It suffices to prove that $I-\beta T$ is bijective. By Fredholm alternative, this is equivalent to proving $I-\beta T$ is injective.
Let $z \in L^2(I)$ such that $z = \beta Tz$. Then $z \in H^2(I)$ with $z(1)=0$ and $$ k(pzw)(0)+\int_I [p z'w' + q zw]=0, \quad \forall w \in K, $$
which (by intergration by parts) is equivalent to $$ p (0) w(0)[kz(0)-z'(0)] + \int_I [-(p z')' + qz]w=0, \quad \forall w \in K, $$
which implies $z'(0)= kz(0)$ and $-(p z')' + qz=0$. It remains to prove $z \equiv 0$. Notice that $w \equiv 0$ is the unique solution of the equation $$ (4) \quad \begin{cases} -(p w')' + q w=0 \quad \text {on} \quad I,\\ w(0)=0, w(1)=0, \end{cases} $$
so it suffices to prove $z(0)=0$.
We have $-(p v')' + q v=0$ implies $-\int_I (p v')'z + \int_I qvz=0$. By integration by parts and the fact that $z(1)=0$, we get $$ (5) \quad (pv'z)(0) + \int_I [pv'z'+qvz]=0. $$
We have $-(p z')' + q z=0$ implies $-\int_I (p z')'v + \int_I qzv=0$. By integration by parts and the fact that $v(0)=1,v(1)=0$, we get $$ (6) \quad (pz')(0) + \int_I [pv'z'+qvz]=0. $$
It follows from $(5, 6)$ and the fact that $p(0) >0$ that $z'(0)=v'(0) z(0)$. Recall that $z'(0)= kz(0)$ and $k \neq v'(0)$. This implies $z(0)=0$. This completes the proof.
