$(\exists \lambda - eigenvalue, \lambda=0) \Leftrightarrow ? (\alpha = -1)$

Question

$\left\{ \begin{aligned} -u''=\lambda u\\ u(0)=u'(1)+\alpha u(1) =0\\ \end{aligned} \right.$, on $[0,1]$ ;

$(\exists \lambda \text{ eigenvalue}, \lambda=0) \Leftrightarrow \alpha = -1$
$\lambda$ should be eigenvalue of differential operator u↦−u′′ with domain being the space of twice differentiable functions on [0,1] satisfying the given boundary conditions for some fixed α

$u(x) = A \cos(\sqrt\lambda x) + B \sin(\lambda x)$

$u(0) = 0 \Rightarrow A=0.$

$\left\{ \begin{aligned} u'(x) = B\sqrt\lambda \cos(\sqrt\lambda x) \\ u'(1)+\alpha u(1) =0\\ \end{aligned} \right.$

$B\sqrt\lambda \cos(\sqrt\lambda)-\alpha B \sin(\sqrt\lambda) =0$

Let $\lambda\neq 0 \Rightarrow \operatorname{ctg}(\sqrt \lambda)=-\alpha$ solution $\forall \alpha\neq0.$

Let $\lambda= 0 \Rightarrow u''=0, u(x) = D + Cx, u(0)=0, \Rightarrow D=0, u'(1)+\alpha u(1)=0, \Rightarrow c(1+\alpha)=0,$ in this way if $\alpha\neq-1,$ then $c=0$, what gives a trivial solution $u(x)$. If $\alpha = 1, c$ any, then there exists a non-trivial solution. $\square$

Answer 1

Let's analyze what's being asked. $\lambda = 0$ being eigenvalue is equivalent to saying that there is non-trivial solution $u$ to differential equation \begin{align} u'' &= 0\\ u(0) &= u'(1)+\alpha u(1) = 0 \tag{1} \end{align}

So, you want to prove that there exists non-trivial solution $u$ of $(1)$ if and only if $\alpha = -1$. Therefore, the first part of your solution is completely redundant (and incomprehensible to me). You shouldn't care about general solutions to $-u'' = \lambda u$ for $\lambda \neq 0$.

Thus, the only relevant part is after you write "Let $\lambda = 0$." Then you conclude that $u(x) = D + Cx$. Here you skipped some things that should be written in interest of clear communication, I would write something like this:

Assume that $\lambda = 0$ is an eigenvalue. That means that there exists non-trivial solution $u$ of $(1)$. Since $u'' = 0$, we conclude that $u$ is of the form $u(x) = D + Cx$ for some constants $C,D$, such that not both $C$ and $D$ are $0$.

What you then do is write the boundary conditions of $(1)$ in terms of $C$ and $D$, a correct step: $D = (\alpha +1)C = 0$. You then argue that if $\alpha \neq -1$, this would imply that $C = 0$ and therefore $u = 0$, which is contradiction with the assumption that $u$ was non-trivial. This is fine, but I'd write it directly:

Since $D = 0$ from the boundary conditions and $u$ is non-trivial, we have $C \neq 0$. Therefore, $(\alpha + 1)C = 0$ implies that $\alpha + 1 = 0$, i.e. $\alpha _= -1$.

This finishes one implication and we need to prove the converse. You write "If $\alpha=1,c$ any, then there exists a non-trivial solution." I'll assume that $\alpha = 1$ is a typo, it should say $\alpha = -1$. However, "$c$ any" is a genuine mistake, it should be $C\neq 0$. Then you say it gives a non-trivial solution, but you don't write that solution or verify that it is solution. So, you need to write something like this:

Assume that $\alpha = -1$ and define $u(x) = Cx$ for $C\neq 0$. Then $u'' = 0$ and $u(0) = 0$, $u'(1) + \alpha u(1) = C - C = 0$, so $u$ is a non-trivial solution to $(1)$, and therefore $\lambda = 0$ is an eigenvalue.