# Prove the ODE under given boundary conditions has a unique non-trivial solution

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.

• Does it make any sense to speak of an eigenvalue of a $\textbf{non-linear}$ mapping? I mean: is there a theory at hand? – Peter Melech Jun 16 '17 at 8:27
• It is my first time encounter such nonlinear operator, and I don't know whether there is existing theory on the 'eigenvalues' of this particular type of nonlinear operators. I saw the title 'eigenvalues of nonlinear operators' in some literatures, so I just used the terminology here. Maybe there is alternative way to interpret the problem. – user31899 Jun 16 '17 at 8:47

## 1 Answer

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 $$1>\int_{\frac32λ}^0\frac{dy}{\sqrt{-2V(y)}}.$$

• Thank you for providing an energy formulation of the problem! You claim that $V$ decreases for positive $y$, does this mean your argument applies only to the cases $\lambda\leq 0$? Besides, why does the initial velocity $v_0<0$? (May $v_0>0$ and it becomes negative somewhere between $t=0$ and $t=1$?) – user31899 Jun 20 '17 at 21:17