PDE problem, for every $f \in L^2(0, 1)$, does problem admit a unique solution $u \in H^2(0, 1)$? Assume that $p \in C^1([0, 1])$ with $p(x) \ge \alpha > 0$ for all $x \in [0, 1]$ and $q \in C([0, 1])$ with $q(x) \ge 0$ for all $x \in [0, 1]$. Let $v_0 \in C^2([0, 1])$ be the unique solution of$$\begin{cases} -(pv_0')' + qv_0 = 0 & \text{on }[0, 1], \\ v_0(0) = 1,\,v_0(1) = 0.\end{cases}\tag*{$(*)$}$$Set $k_0 = v_0'(0)$. From some post elsewhere from MSE (actually, right here), we know that $k_0 \le -\alpha/p(0)$. We now investigate the problem$$\begin{cases} -(pu')' + qu = f & \text{on }(0, 1), \\ u'(0) = ku(0),\,u(1) = 0,\end{cases}\tag*{$(**)$}$$where $k \in \mathbb{R}$ is fixed and $f \in L^2(0, 1)$ is given. Assume $k \neq k_0$. For every $f \in L^2(0, 1)$, does problem $(**)$ admit a unique solution $u \in H^2(0, 1)$?
 A: Suppose $p$, $q$ are as given. Let $\alpha$, $\beta$ be real arguments. Define $$L_{\alpha,\beta} : \mathcal{D}(L_{\alpha,\beta})\subset L^2[0,1]\rightarrow  L^2[0,1]\\
 L_{\alpha,\beta}f = -(pf')'+qf
$$
on the domain $\mathcal{D}(L_{\alpha,\beta})$ consisting of all twice absolutely continuous functions $f \in L^2$ for which $L_{\alpha,\beta}f\in L^2$ and
$$
              \cos\alpha f(0)+\sin\alpha f'(0)= 0 \\
              \cos\beta f(1)+\sin\beta f'(1)= 0.
$$
Then $L_{\alpha,\beta}$ is selfadjoint in the strictest Functional Analytic sense. $L_{\alpha,\beta}$ has a closed range and, as is true for selfadjoint operators,
$$
          \mathcal{R}(L_{\alpha,\beta})^{\perp} = \mathcal{N}(L_{\alpha,\beta}),\\ \overline{\mathcal{R}(L_{\alpha,\beta})}=\mathcal{N}(L_{\alpha,\beta}).
$$
The range of $L_{\alpha,\beta}$ is closed because $L_{\alpha,\beta}$ is a Fredholm operator of index $0$. So $L_{\alpha,\beta}u = f$ has a solution $u$ iff $f \in \mathcal{R}(L_{\alpha,\beta})=\mathcal{N}(L_{\alpha,\beta})^{\perp}$.
In your case, you have designed $k$ so that $\mathcal{N}(L_{\alpha,\beta})=\{0\}$. So there is a solution $u\in H^2(0,1)$ of $Lu=f$ for every $f\in L^2$, and the solution is unique.
