Show that Hill's Equation $u'' + a(t)u=0$ if $a(t)<0$ for all $t$ then $u\to\infty$ as $t\to\infty$

Question

(a) Consider the Hill's equation $$u'' + a(t)u = 0,$$ where $a(t+T) = a(t)$ for all $t$. Show that, if $a(t)<0$ for all $t$, then the solution satisfying the initial condition $$u(0)=u'(0)=1$$ is unbounded as $t\to\infty$.

Suppose $a(t) < 0$ for all $t$. Then, consider the expression, \begin{align} u'(t) = 1 - \int_0^t a(s)u(s)ds, \tag{1}\end{align} so that $$u''(t) = -a(t)u(t)$$ Now we'll show $u'(t) \geq 1$ for all $t \geq 0$. By contradiction, suppose there exists some $t_0 > 0$ such that $u'(t_0) \leq \frac{1}{2}$, then $u(s) = 1 + \frac{s}{2}$ for all $0 \leq s \leq t_0$, then since $a(s) < 0$ for all $s$, and $u(s) \geq 1$ on the interval $(0,s)$, which is a contradiction since $u(t_0)\leq\frac{1}{2}$. Then, the integral $\int_0^t a(s)u(s) \leq 0$ so that $u'(t) \geq 1$ for all $t\geq 0$ and, hence, \begin{align*} \lim_{t\to\infty} u(t) &= \lim_{t\to\infty} \int u'(t) dt \\ &= \lim_{t\to\infty} \int \left( 1 - \int_0^t a(s)u(s)ds \right) \\ &\to \infty \end{align*} since $a(s)<0$ for all $s$ and $u(s)>0$ for all $s\in(0,t)$ we know that $$-\int_0^t a(s)u(s) \geq 0$$ Then, $$\lim_{t\to\infty} \int \left( 1 - \int_0^t a(s)u(s)ds \right) \geq \lim_{t\to\infty} \int_0^t dt \to \infty$$ so $u$ is unbounded as $t\to\infty$.

(b) Next suppose that $a(t)>0$ for all $t$ and $$\int_0^T a(t)dt < \dfrac{4}{T}$$ It may be shown that all solutions are bounded as $t\to\infty$. Use this result and that of part (a) to estimate the stable and unstable zones in the $\delta-\epsilon$ plane for the Mathieu equation and Missner's equation.

I am not sure how to proceed with this part. Also, I do not feel too confident in my proof of part (a) the proof by contradiction felt a little wonky since I never actually reached a contradiction. Any advise would be appreciated.

Answer 1

Your contradiction is that you supposed there is some $t_0>0$ with $u'(t_0) \leq \frac{1}{2}$ and obtained that $u'(t)\geq 1$ for all $t>0$. I don't see how your limit shows that is unbounded (although if you do show that it will get your result.)

Answer 2

Let us synthesize the counterexample.

Substitution $$u' = uv$$ to the Hill equation creates the Riccati equation of $v$: $$v' + v^2 = a(t).\tag1$$ (the earlier example see there).

The homogeneous equation $(1)$ has common solution $$v_c(t) = \dfrac1{t+const},$$ and that gives reason to try the function $v$ in the form of $$v = \dfrac1{t + w(t)},$$ Then $$a(t) = v' + v^2 = \dfrac{-w'(t)}{(w(t) + t)^2}.\tag2$$ The function $a(t)$ must be bounded at $t\to\infty$ and $T$-periodic with $a(t) < 0$.

This can be achieved if you take in account piecewise continuous functions $a$. For example, substitution of the function $$w(t) = \sin\left(\dfrac{3π}{T}\left(t\ \mathrm{mod}\ \dfrac{T}{3} - \dfrac T6\right)\right) + \dfrac{17}{18} T + \dfrac13((t + \dfrac{2T}{3}\mathrm{mod}\ T - t\ \mathrm{mod}\ T)\tag3$$ to $(2)$ (see also Wolfram Alpha graph for $T= 18$) gives the counterexample $a(t)$.