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