Given $y(t)$ over $t\in [0,1]$ which satifies $$y''+\lambda y-y^2=0$$ subject to boundary conditions $y(0)=0=y'(1)$. In which $\lambda$ is a real-valued parameter.
(a).Prove that this boundary value problem has a unique non-trivial solution for each $\lambda\leq 0.$
(b).Suppose $c$ is the supremum over $\lambda\in \mathbb{R}$ such that the above boundary value problem has a unique non-trivial solution. Show that $c>0$ and compute $c$.
Note: I attempted to interpret $\lambda$ as an eigenvalue of the non-linear operator $F[\cdot]:=-\frac{d^2}{dt^2}(\cdot)+\cdot^2$ over the functional space $C[0,1]$ equiped with the corresponding boundary conditions. The questions are then asking one to show that the spectrum of $F$ contains $(-\infty,c)$ for some $c$ positive. Besides, the eigenspace of each $\lambda<c$ is $1$. But I am not aware of any way to tackle the quadratic nonlinearity.
This second order equation is conservative in that it is the motion without friction under the potential $$ V(y)=\frac12λy^2-\frac13y^3, \ y''=-V'(y) $$ The start at $y=0$ has potential energy $0$ so that all energy is kinetic, $E=\frac12v_0^2$. At $t=0$ the kinetic energy is zero, so that $E=V(y_1)$. As $0$ is a maximum of the potential, i.e., an unstable stationary point, and the potential falls continuously for positive $y$, one needs $y_1<0$, $v_0<0$, and as $V(\frac32λ)=0$ is the second point with potential $0$ and negative potential in between,one needs indeed $y_1<\frac32λ$. Now one has to ensure that the point $y_1$ is actually reachable at time $t=1$, $$ 1=\int_{y_1}^0\frac{dy}{\sqrt{2(V(y_1)-V(y))}} $$ for which it is necessary that $$ 1>\int_{\frac32λ}^0\frac{dy}{\sqrt{-2V(y)}}. $$