On the existence of non-trivial solution for a linear second order boundary value problem I came across with a multiple choice question:
The ODE$$-y^{\prime \prime}+(1+x) y=\lambda y, x \in(0,1), y(0)=y(1)=0$$ has a non-zero solution
(1) $\forall \lambda \in[0,1]$
(2) $\forall \lambda<0$
(3) for some $\lambda \in[2, \infty)$
(4) for a countable number of $\lambda^{\prime} s$
Option 1 and 2 can be rejected since for any $\lambda<1$, we have $Q(x)=\lambda-(1+x)<0$, which is not possible since $y$ has already two zeros in [0,1]. I have seen several answers on the same question in this platform, but none of them convincing me for the existence of non-trivial solutions for $\lambda \geq 2$.
Can we have some special regularity conditions for a second order linear boundary value problem:
$$y''+Q(x)y=0,y(a)=y_1,y(b)=y_2$$
to have a non-trivial solution?
The same difficulty of existence of non-trivial solutions I have faced with the 2 and 4 th option in another multiple choice question:
Consider the eigenvalue problem $$\left(\left(1+x^{4}\right) y^{\prime}\right)^{\prime}+\lambda y=0, x \in(0,1), y(0)=0, y(1)+2 y^{\prime}(1)=0.$$ Then which of the following statements are true?

*

*all the eigenvalues are negative


*all the eigenvalues are positive


*there exists some negative eigenvalues and some positive eigenvalues


*there are no eigenvalues
Since regular SLP problem cannot have negative eigen values, we can reject 1 and 3. Also for $\lambda=0,$ I got $y=0$ is only solution by some alterations in the boundary conditions to get an IVP:
$$(1+x^4)y''+4x^3y'=0, y(0)=0,y'(0)=-y(1),$$
 A: Any non-trivial solution of the BVP will have $y'(0)\ne 0$ (else the IVP gives the zero solution). Thus the solution can be scaled to $y'(0)=1$.
Denote $y(x;λ)$ the solutions of the IVP $y''+( λ-1-x)y=0$, $y(0)=0$, $y'(0)=1$. Finding a solution to the BVP is equivalent to finding a root of $y(1;λ)$. This is a continuous function in $λ$. Also the roots $z_n(λ)$ of $y(x;λ)$ depend continuously on $λ$. As there can be no double roots (would imply zero solution), there will also be no fold points or other singularities in the paths of the roots.
Set $\omega_n=\pi n$. Then with the Sturm-Picone comparison theorem, and a suitably small $ε,δ>0$,

*

*if $λ-1<ω_n^2$ then $y$ will have at most $n-1$ positive roots in $(0,1]$, $z_n(λ)>1$,

*

*as between any two roots of $y$ there has to be one of the $n-1$ roots of $\sin((n\pi-ε) (x+\delta))$. $δ>0$ is chosen with $(n\pi-ε) (1+\delta)<n\pi$, so that there are roots just outside the interval on both sides.



*if $λ-2>ω_n^2$ then $y$ will have at least $n$ positive roots in $(0,1]$, $z_n(λ)\le 1$,

*

*as between any two of the $(n+1)$ roots of $\sin((n\pi+ε)(x-δ))$ in $(0,1]$ there has to be a root of $y$. Here $δ>0$ is chosen so that $(n\pi+ε)(1-δ)>n\pi$, so that the two roots close to the interval boundaries are just inside the interval.



In the transition from one case to the next the $n$th root $z_n(λ)$ has to transition into the interval, thus $z_n(λ)=1$ at some $λ\in(ω_n^2+1,ω_n^2+2)$. Which in turn means the solution $y$ will have a root at the interval boundary $1$, giving a solution to the eigenvalue problem. As there is one distinct solution for each $n$, case (4) is true, and obviously also case (3).
Illustration for the case $n=2$

The dotted curves are the solutions for blue: $λ=(2\pi)^2$ and red: $λ=(2\pi)^2+3$. The thin lines are suitably shifted sine functions $\sin((2\pi\pm0.1)(x\mp0.01))$ clearly separating the cases of 1 and 2 positive roots.
