Uniqueness of solution $u \in H^2(0, 1)$ to partial differential equation.

Question

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, 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 $k = k_0$. 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 $u$?

Answer 1

The solution is unique if k is nonnegative.

Let $u, v $ be the solutions, consider w = u - v, then w solves $$\begin{cases} -(pw')' + qw = 0 & \text{on }(0, 1), \\ w'(0) = kw(0),\,w(1) = 0,\end{cases}$$ 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.

Answer 2

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 $v_0(0)=1$.

Answer 3

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 $f \in \mathcal{R}(L_{\alpha,\beta})=\mathcal{N}(L_{\alpha,\beta})^{\perp}$. The solution is not unique because $u+u_0$ is also a solution if $u_0 \in \mathcal{N}(L_{\alpha,\beta})$.

In your case, you have designed $k$ so that $\mathcal{N}(L_{\alpha,\beta})$ is one-dimensional, spanned by $v_0$. So there is a solution of $Lu=f$ iff $(f,v_0)=0$, which is to say $\int_{0}^{1}f(t)v_0(t)dt=0$. If there is a solution $u$, it is not unique because $u+\alpha v_0$ is a solution for all constants $\alpha$.