Brezis' exercise 8.27.6: how to characterize the solution of the ODE $-u'' + u-\lambda_0 u(0)=f$?

Question

Let $I$ be the open interval $(-1, 1)$. Consider the bilinear form on $H^1_0 (I)$ $$ a(u, v) = \int_I [ u'v' + uv + \lambda u(0) v], $$ where $\lambda \in \mathbb R$ is fixed. I am trying to solve a problem in Brezis' Functional Analysis

Exercise 8.27

1. Check that $a(u, v)$ is a continuous bilinear form on $H_0^1(I)$.

2. Prove that if $|\lambda|<\sqrt{2}$, the bilinear form $a$ is coercive.

3. Deduce that if $|\lambda|<\sqrt{2}$, then for every $f \in L^2(I)$ there exists a unique solution $u \in H^2(I) \cap H_0^1(I)$ of the problem $$ (1) \quad \begin{cases} -u'' + u-\lambda u(0)=f \quad \text {on} \quad I, \\ u(-1)=u(1)=0. \end{cases} $$

4. Prove that there exists a unique value $\lambda=\lambda_0 \in \mathbb{R}$, to be determined explicitly, such that the problem $$ (2) \quad \begin{cases} -u'' + u=\lambda u(0) \quad \text {on} \quad I, \\ u(-1)=u(1)=0. \end{cases} $$ admits a solution $u \not \equiv 0$.

5. Prove that if $\lambda \neq \lambda_0$, then for every $f \in L^2(I)$ there exists a unique solution $u \in H^2(I) \cap H_0^1(I)$ of $(1)$.

6. Analyze completely problem (1) when $\lambda=\lambda_0$.

In my below attempt of (6.), a necessary condition for $(1)$ to have a solution (in case $\lambda=\lambda_0$) is $(Sf)(0)=0$. Could you elaborate on how to completely describe the solutions of $(1)$?

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For $f \in L^2 (I)$, there is a unique solution $u \in H^2(I)$ of $$ (4) \quad \begin{cases} -u'' + u=f \quad \text {on} \quad I, \\ u(-1)=u(1)=0. \end{cases} $$ Clearly, the map $S: L^2 (I) \to L^2(I), f \mapsto u$ is linear. The equation $(1)$ is equivalent to $$ u = S(f + \lambda_0 u(0))=Sf + \lambda_0 u(0) S \mathbf{1}. $$

Let $v:=S \mathbf{1}$. It follows from (4.) that $\lambda_0 = \frac{1}{v(0)}$. Then $(1)$ is equivalent to $$ u=Sf+ \frac{u(0)}{v(0)} v. $$

For $(1)$ to have a solution, it has to be the case that $u(0) = (Sf)(0) + \frac{u(0)}{v(0)} v(0)$ or equivalently $(Sf)(0)=0$.

Answer 1

Now we assume $(Sf)(0)=0$. As you can see from your attempt, a solution of $(1)$ (in case $\lambda=\lambda_0$) has to be of the form $w = Sf + \alpha v$ for some $\alpha \in \mathbb R$. Substituting $w$ into $(1)$, we get $$ \begin{align*} -w'' + w &= f+\lambda_0 w(0) \\ \iff -(Sf + \alpha v)'' + (Sf + \alpha v) &= f + \lambda_0 (Sf + \alpha v)(0) \\ \iff -(Sf)'' - \alpha v'' + Sf + \alpha v &= f + \lambda_0 (Sf)(0) + \alpha \lambda_0 v(0). \end{align*} $$

Notice that we have $v:=S \mathbf{1}$ and $(Sf)(0)=0$, so the last equality trivially holds.