Why solutions of $y''+(w^2+b(t))y=0$ behave like solutions of $y''+w^2y=0$ at infinity

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.

Here is an outline of an approach that almost gets there (with some pieces left to be filled in). I will show that we can find solutions $\phi(t)$ that are arbitrarily close to $\sin wt$ for all sufficiently large $t$ under the assumption that $\lim_{t\to\infty}b(t) = 0$.

Write the solution to the ODE as $\phi(t) = \sin w t + \delta y$ and impose the initial conditions $\delta y(t_*) = \delta y'(t_*) = 0$ for some $t_*$ to be fixed later on. Then

$$\delta y'' + (w^2+b)\delta y = -b\sin w t$$

This system is equivalent to a ball rolling without friction attached to a spring with varying spring-constant and we have some driving force $b(t)\sin wt$.

This physics-analogy motivates us to try to find an 'energy equation' for this system. Multiply by $\delta y'$ to find

$$\frac{d}{dt}(\delta y'^2 + w^2\delta y^2) = -b(t)\left(2\sin w t\frac{d\delta y}{dt} + \frac{d\delta y^2}{dt}\right)$$

Integrating over $[t_*,t]$ yields

$$(\delta y'^2 + w^2\delta y^2) \leq \max_{t'\in [t_*,t]}|b(t')|\left(2|\delta y| + \delta y^2\right)$$

Now since $y$ is bounded (I will just assume this here, see the other answer for the proof. Physically this is because the energy is bounded.) and $b\to 0$ (Edit: $b\to 0$ does not follow from continuity, but I will need to assume this here) we get that for any $\epsilon > 0$ there exist a $t_*$ s.t. for $t>t_*$ we have

$$\delta y'^2 + w^2\delta y^2 < \epsilon$$

and since $w$ is a constant it also follows that we can find a $t_*$ s.t.

$$\delta y'^2 + \delta y^2 < \epsilon$$

for all $t>t_*$. Writing out this equation we get

$$(\phi(t)-\sin wt)^2 + (\phi'(t)-w\cos wt)^2 < \epsilon$$

for $t>t_*$. Thus there exists solutions that are arbitrarely close to $y=\sin w t$ for all sufficiently large $t$, however I have not yet found a way to show that we can find one particular solution where the difference goes to zero at infinity.

The following proof does not originate from me but the hint in Coddington and Levinson's ODE book. I write it out in full here for my own and perhaps others' benefit.

Let $y_0(t)=\sin(\omega t)$, and $$y_{i+1}(t) = \sin(\omega t)+\frac{1}{\omega}\Big(\cos(\omega t)\int_t^\infty \sin( \omega s) b(s) y_i(s) ds- \sin(\omega t) \int_t^\infty \cos(\omega s) b(s) y_i(s) ds\Big).$$

For large $T$ such that $$\int_T^\infty |b|<\frac{1}{2},$$ we have $|y_1(t)|<\infty$ and a uniform Cauchy sequence $y_i(t)$ where \begin{align*} &\big|y_{i+1}(t)-y_i(t)\big| \\ =& \frac{1}{\omega}\Big(\cos(\omega t)\int_t^\infty \sin( \omega s) b(s) (y_i(s)-y_{i-1}(s)) ds- \sin(\omega t)\int_t^\infty \cos(\omega s) b(s)(y_i(s)-y_{i-1}(s))ds\Big) \\ <& \frac{K}{2^i} \end{align*} for some positive constant $K$ and $\forall t \ge T$. So $y_i(t)$ approaches to a function $y_\infty$ uniformly where $|y_\infty|<K$ and $$y_\infty(t) = \sin(\omega t)+\frac{1}{\omega}\Big(\cos(\omega t)\int_t^\infty \sin( \omega s) b(s) y_\infty(s) ds- \sin(\omega t) \int_t^\infty \cos(\omega s) b(s) y_\infty(s) ds\Big).$$

(I will finish the proof later.)