Brezis' exercise 8.27.5: if $\lambda \neq \lambda_0$ and $f \in L^2(I)$, then there exists a unique solution $u \in H^2(I) \cap H_0^1(I)$

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 $(1)$.

There are possibly subtle mistakes that I could not recognize in my below attempt of (4.) and (5.). Could you please have a check on it?

---

4. If $\lambda u(0)=0$ then $(2)$ clearly has a unique solution $u=0$. Next we consider $\lambda u(0)\neq 0$. Let $v= \frac{u}{\lambda u(0)}$. Then $(2)$ is equivalent to $$ (3) \quad \begin{cases} -v'' + v=1 \quad \text {on} \quad I, \\ u(-1)=u(1)=0. \end{cases} $$ Using Maple or Mathematica, we get $$ v (x) = 1-\frac{e^{x+1} + e^{-x+1}}{1+e^2}, \quad x \in \bar I. $$ Then $v(0)=\frac{(1-e)^2}{1+e^2}$. Clearly, $\lambda_0 = \frac{1}{v(0)}= \frac{1+e^2}{(1-e)^2}$.

5. 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 $$ u = S(f + \lambda u(0))=Sf + \lambda u(0) S \mathbf{1}. $$ We have from (4.) that $S \mathbf{1}=v$. Then $(1)$ is equivalent to $u=Sf+\lambda u(0) v$. In particular, $u(0)=(Sf)(0)+ \lambda u(0) v(0)$ or equiavelently $(1-\lambda v(0)) u(0) = (Sf)(0)$. If $\lambda \neq \lambda_0 = \frac{1}{v(0)}$ then $u(0) = \frac{(Sf)(0)}{1-\lambda v(0)}$ is well-defined. In this case, $u$ is well-defined and is the unique solution of $(1)$.