Uniqueness of solution $u \in H^2(0, 1)$ to partial differential equation. Assume that $p \in C^1([0, 1])$ with $p(x) \ge \alpha &gt; 0$ 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$$\begin{cases} -(pv_0')' + qv_0 = 0 &amp; \text{on }[0, 1], \\ v_0(0) = 1,\,v_0(1) = 0.\end{cases}\tag*{$(*)$}$$Set $k_0 = v_0'(0)$. From some post elsewhere from MSE, we know that $k_0 \le -\alpha/p(0)$. We now investigate the problem$$\begin{cases} -(pu')' + qu = f &amp; \text{on }(0, 1), \\ u'(0) = ku(0),\,u(1) = 0,\end{cases}\tag*{$(**)$}$$where $k \in \mathbb{R}$ is fixed and $f \in L^2(0, 1)$ is given. Assume $k = k_0$. We have that$$[(**) \text{ has a solution }u \in H^2(0, 1)] \iff \left[\int_0^1 fv_0 = 0\right].$$My question is, is there uniqueness of $u$?
No, $u$ is not unique.  Take any $u(t)$ that satisfies (**).  Then $u(t) + \theta v_0(t)$ is
also a solution for all $\theta \in \mathbb{R}$.  As the equations are linear, the only
thing to check is the condition: 

$$ u’(0) + \theta v_0’(0) = ku(0) + \theta k = k u(0) + \theta k v_0(0) = k[u(0)+\theta v_0(0)] $$

which uses the fact that $k=k_0=v_0'(0)$ and $v_0(0)=1$.
The solution is unique if k is nonnegative. 

Let $u, v $ be the solutions, consider w = u - v, then w solves
$$\begin{cases} -(pw')' + qw = 0 &amp; \text{on }(0, 1), \\ w'(0) = kw(0),\,w(1) = 0,\end{cases}$$
Now
$$\int_0^1(-(pw')' + qw) w dx= 0$$
Integrating by parts
$$ -pw'w|_{x =0}^{x =1} + \int_0^1(pw'w' + qw^2) dx = 0$$
We know $-pw'w|_{x =0}^{x =1} = pw'(0)w(0) = pk(w(0))^2$

Therefore,
$$pk(w(0))^2 + \int_0^1(p(w')^2 + qw^2) dx = 0$$
So w = 0. Hence, the solution is unique.

If k is negative, it is not clear if it has a unique solution.