I'm going through exercise 6 here. I've managed to show, as part (a) suggests, that the second order linear homogeneous ODE $$ u''(t)+p(t) u'(t)+q(t)u(t)=0$$ reduces, via the clockwise polar coordinates substitution $$u(t)=r(t) \sin(\theta(t)),\quad u'(t)=r(t) \cos(\theta(t)), $$ to the first order system $$\theta'(t)=\cos^2(\theta(t))+p(t) \cos(\theta(t)) \sin(\theta(t))+q(t) \sin^2(\theta(t)), \\ r'(t)/r(t)=-p(t) \cos^2(\theta(t))+(1-q(t))\cos(\theta(t)) \sin(\theta(t)). $$ I want to show that if $q(t)>p^2(t)/4$ the solutions are oscillatory. Completing the square gives a positive definite quadratic for $\theta'(t)$: $$\theta'(t)=\left[ \cos(\theta(t))+\frac{p(t)}{2} \sin(\theta(t)) \right]^2+\left[q(t)-\frac{p^2(t)}{4} \right] \sin^2(\theta(t))>0. $$ This means that $\theta(t)$ is strictly increasing. The exercise doesn't mention this explicitly, but suppose $p,q$ are continuously defined over the entire real line. My question is: could it be that there exists $$\Theta:=\sup_{t \in \mathbb{R}} \theta(t)<+\infty $$ such that $\theta(t) \uparrow\Theta$ as $t \to \infty$ (that is, $(u,u')$ stop oscillating from some point)? I suspect this is possible if $q(t)=\frac{p^2(t)}{4}+\epsilon(t)$ where $\epsilon(t) \to 0^+$ sufficiently quickly as $t \to \infty$. However, I couldn't find a particular example where this happens.
Thank you!
The change of variables $$y(t)=\mathrm{e}^{\int p(t)/2 \mathrm{d}t} u(t) $$ results in the ODE $$y''(t)+\left( q(t)-\frac{p^2(t)}{4}-\frac{p'(t)}{2} \right)y(t)=0.$$ Consider the particular case where $$\begin{align} p(t)&=t, \\ q(t)&=\frac{t^2}{4}+\frac{1}{2}. \end{align} $$ We have $q(t)>p(t)^2/4 $ for all $t$, however, $y''=0$ implies that $y$ (and hence $u$) can change its sign at most once. In fact, the solution can be found explicitly in this case: $$\begin{align} u(t)&=\mathrm{e}^{-t^2/4} \left(c_1+c_2 t \right) \\ u'(t)&=\mathrm{e}^{-t^2/4}\left(-\frac{1}{2} c_1 t+\frac{1}{2}c_2(2-t^2) \right) \end{align}$$ This means that $$\theta(t)=\arg \left[-c_1 t+c_2(2-t^2)+2(c_1+c_2t)i \right] $$ is bounded, for any choice of a branch of the complex argument.
Thus it appears that the authors' claim is not precisely right.