Does this second-order boundary-value problem have a unique solution when $k=1$?

Question

Let $I$ be the open interval $(0, 1)$. Let $k \in \mathbb R$. We consider the space $$ V := \{v \in H^1 (I) : v(0) = kv(1)\}, $$ and the symmetric bilinear form $a$ defined on $V$ by $$ a(u, v) = \int_I [ u'v' + uv ] - \left ( \int_I u \right) \left ( \int_I v \right). $$

Clearly, $a$ is continuous. We fix $f \in L^2 (I)$ and consider the problem $$ (1) \quad u \in V \quad \text{and} \quad a(u, v)=\int_I f v \quad \forall v \in V, $$ which (by integration by parts) is equivalent to $$ (2) \quad \begin{cases} -u'' + u-\int_I u = f \quad \text {on} \quad I, \\ u(0)=k u(1) \text { and } u'(1)=k u'(0). \end{cases} $$

In case $k \neq 1$, I have proved that $a$ is coercive and thus $(1)$ has a unique solution by Lax-Milgram theorem.

Now we assume $k=1$. Does $(1)$ have a solution? If yes, is it unique?

Thank you so much for your elaboration!

Answer 1

It is for sure that if $f\equiv 0$, any constants are solutions to (2) for $k=1$ and hence in this case, (2) does not have unique solution. If $f$ is a non-zero constant, it is also easy to see that (2) does not have a solution. For nonconstant $f$, for example, $f(t)=t$, then $$ (*)\quad\begin{cases} -u'' + u-\beta = t \quad \text {on} \quad I, \\ u(0)=u(1) \text { and } u'(1)=u'(0). \end{cases} $$ has the solution $$ u(t)=-\frac{2 (e-1) (\beta +t)+e^{1-t}-e^t}{2-2 e}. $$ Applying $\int_Iu=\beta$ gives $$ \frac12+\beta=\beta $$ which is absurd. Therefore (*) does not have solution for $f(t)=t$.

Update: In order for (2) to have solution, one must impose the condition $\int_If=0$. Even under this, the solution of (2) is not unique. For example, let $f(t)=t-\frac12$ which satisfies the condition. It is not hard to see that $$ (**)\quad\begin{cases} -u'' + u-\beta = t-\frac12 \quad \text {on} \quad I, \\ u(0)=u(1) \text { and } u'(1)=u'(0). \end{cases} $$ has the solution $$ u(t)=-\frac{2 (e-1) (\beta-\frac12 +t)+e^{1-t}-e^t}{2-2 e} $$ which satisfies $\int_Iu=\beta$ for any constant $\beta$. Therefore (**) has infinite many solutions.