TITLE: Why solutions of $y''+(w^2+b(t))y=0$ behave like solutions of $y''+w^2y=0$ at infinity
QUESTION [14 upvotes]: Assume $w>0$ and $b(t)$ be continuous on $[0,+\infty)$ and $\int_0^\infty |b(t)| dt <\infty$
show that $y''+(w^2+b(t))y=0$ has solution $\phi(t)$ such that $$\lim_{t\to\infty} [(\phi(t)-\sin(wt))^2+(\phi'(t)-w\cos(wt))^2]=0$$
thanks

REPLY [2 votes]: Here's another proof (apologies for any similarities, but I think it's different enough). Put $Y = y + i\dot y/w$ and write $\beta(t) = b(t)/w$. Then with initial conditions $Y(0) = Y_0$ the differential equation can be written
\begin{align*}
 \dot Y = -iwY - i\beta y,\qquad Y(0) = Y_0. \tag{1}
\end{align*}
Since
\begin{align*}
 {d\over dt}(Ye^{iwt}) &= (\dot Y + iw Y)e^{iwt} = -i\beta y e^{iwt},
\end{align*}
we can write
\begin{align*}
 Y(t)e^{iwt} - Y(s)e^{iws} &= -i\int_{s}^t \beta(r)y(r)e^{iwr}\,dr.
\end{align*}
Set $B = \int_0^\infty |\beta(r)|\,dr$. In a moment I will prove that
\begin{align*}
 |Y(t)|\leq e^B|Y(s)|, \qquad s,t\geq0. \tag{2}
\end{align*}
Suppose for now that this is proved. Then for any time $t_0$,
\begin{align*}
 |Y(t)e^{iwt} - Y(s)e^{iws}| \leq e^B|Y(t_0)| \int_s^t|\beta(r)|\,dr, \tag{3}
\end{align*}
which since the right side tends to $0$ as $s,t\to\infty$ implies that $Y(t)e^{iwt}$ has a limit $Y(\infty)$ as $t\to\infty$. If $z$ is an arbitrary point in the plane, I claim that we can make $Y(\infty) = z$. (This is enough to answer the question because $Y(t)$ converges to the curve $Y(\infty)e^{-iwt}$.) To prove the claim, define a map $\Lambda\colon\mathbb C\to \mathbb C$ by setting $\Lambda(\zeta) = Y_\zeta(\infty)$, where $Y_\zeta$ is the solution to the ODE $(1)$ with initial condition $Y_\zeta(0) = \zeta$. Observe that $\Lambda$ is a linear map (using $Y_\zeta(t) + cY_\xi(t) = Y_{\zeta + c\xi}(t)$, which is because the left side is a solution to $(1)$ passing through $\zeta + c\xi$ at time zero and because there is only one such solution), and that the inequality $(2)$ implies that $\Lambda$ is invertible: $e^{-B}|\zeta|\leq |\Lambda(\zeta)|$. An invertible linear map on the plane is surjective, so the claim is proved.
It remains to prove $(2)$. For this I'll use Gronwall's inequality twice. Gronwall's inequality says that if $u(t)$ is real-valued and satisfies $\dot u(t)\leq f(t)u(t)$ for $a\leq t\leq b$, then $u(t) \leq e^{\int_a^tf(s)\,ds}u(a)$ for $t\in [a,b]$. The proof involves writing $$u(t) = e^{\int_a^tf(s)\,ds}u(a) + e^{\int_a^tf(s)\,ds}\int_a^t(\dot u(s) - f(s)u(s))e^{-\int_a^sf(r)\,dr}\,ds$$
and observing that the integral is negative for $t\in[a,b]$. 
On now to the proof of $(2)$. Fix $s$. I'll first use Gronwall's inequality to show that $|Y(t)|\leq e^B|Y(s)|$ for $t\geq s$. From what we showed above,
\begin{align*}
 \left|{d\over dt} (Ye^{iwt})\right| = |-i\beta ye^{iwt}| \leq |\beta||Y|.
\end{align*}
Thus
\begin{align*}
 {d\over dt}|Y| &= {d\over dt}|Ye^{iwt}| = \operatorname{Re}\left({\bar{Y}e^{-iwt}\over |Y|}{d\over dt} (Ye^{iwt})\right)\leq \left|{d\over dt} (Ye^{iwt})\right|\leq |\beta||Y|,
\end{align*}
and applying Gronwall's inequality on the half-line $t\geq s$ gives the desired bound. All that's left is to show that $|Y(t)|\leq e^B|Y(s)|$ for $t\leq s$. To do this put $U(t) = Y(s-t)$ for $t\in[0,s]$, so that $U$ satisfies the reverse time equation
\begin{align*}
 \dot U(t) = iwU(t) + i\beta(s-t)\operatorname{Re}U(t)
\end{align*}
Thus differentiating $Ue^{-iwt}$ and running the same proof as before gives ${d\over dt}|U|\leq |\beta(s-t)||U|$, and then Gronwall's inequality gives $|Y(s - t)|\leq e^B |Y(s)|$ for $t\in[0,s]$, which is equivalent to $Y(t)\leq e^BY(s)$ for $0\leq t\leq s$. With that, $(2)$ is proved.