Periodic responses of $\ddot{z}=\mathrm{Re}(w{e}^{-i\omega t})$ and the affine classification of all solutions

Write $w_j={\alpha }_j+i{\beta }_j$ with ${\alpha }_j=\mathrm{Re}{w}_j$, ${\beta }_j=\mathrm{Im}{w}_j$ real and fix the component index $j$; by [F1] $e^{-i\omega t}=\cos \omega t-i\sin \omega t$, so the product rule of [F2] with $(a,b)=({\alpha }_j,{\beta }_j)$ and $(c,s)=(\cos \omega t,-\sin \omega t)$ gives $z_{p,j}(t)=\mathrm{Re}(-w_j{\omega }^{-2}{e}^{-i\omega t})=-{\omega }^{-2}({\alpha }_j\cos \omega t+{\beta }_j\sin \omega t)$ for every real $t$. The maps $t\mapsto \omega t$, $\sin$ and $\cos$ are smooth, and the sum, scalar-multiple and product rules of [F3] give the derivatives of finite combinations, so each $z_{p,j}$ is $C^{\mathrm{\infty }}$ and hence $z_p$ is $C^{\mathrm{\infty }}$ by [F7].

Differentiating the display of step 1.1 componentwise with the rules of [F3] gives $z_{p,j}^{\mathrm{\prime }}(t)={\omega }^{-1}({\alpha }_j\sin \omega t-{\beta }_j\cos \omega t)$ and $z_{p,j}^{\mathrm{\prime }\mathrm{\prime }}(t)={\alpha }_j\cos \omega t+{\beta }_j\sin \omega t=\mathrm{Re}(({\alpha }_j+i{\beta }_j)(\cos \omega t-i\sin \omega t))=\mathrm{Re}(w_j{e}^{-i\omega t})$ for every $j$ and every real $t$, the middle equality by the product rule of [F2] and the last by [F1]; hence $z_p^{\mathrm{\prime }\mathrm{\prime }}(t)=\mathrm{Re}(w{e}^{-i\omega t})$ componentwise.

Since $\omega (t+T)=\omega t+2\pi$, the display of step 1.1 and the period formulas $\cos (\omega t+2\pi )=\cos \omega t$, $\sin (\omega t+2\pi )=\sin \omega t$ of [F4] give $z_{p,j}(t+T)=z_{p,j}(t)$ for every $j$ and every $t\in \mathbb{R}$, so $z_p$ is $T$-periodic; with steps 1.1 and 2.1 this is claim (a).

Let $z:\mathbb{R}\to {\mathbb{R}}^n$ be $C^2$ with $z^{\mathrm{\prime }\mathrm{\prime }}(t)=\mathrm{Re}(w{e}^{-i\omega t})$ on $\mathbb{R}$ and set $h:=z-z_p$; by step 2.1, $h^{\mathrm{\prime }\mathrm{\prime }}=0$ on $\mathbb{R}$. Fix $j$; the component $h_j$ is $C^2$, so $h_j^{\mathrm{\prime }}$ is continuous on $\mathbb{R}$ by [F6] and differentiable at every point of $\mathbb{R}$, which is the interior of $\mathbb{R}$ by [F5], with $(h_j^{\mathrm{\prime }}{)}^{\mathrm{\prime }}=h_j^{\mathrm{\prime }\mathrm{\prime }}=0$; [F5] applied on the order-convex interval $\mathbb{R}$ therefore makes $h_j^{\mathrm{\prime }}$ a constant $b_j\in \mathbb{R}$. The map $h_j-b_j\mathrm{i}\mathrm{d}$ is continuous on $\mathbb{R}$ by [F6] and differentiable at every point with derivative $h_j^{\mathrm{\prime }}-b_j=0$, the derivative of $b_j\mathrm{i}\mathrm{d}$ being $b_j$ by [F3]; [F5] makes $h_j-b_j\mathrm{i}\mathrm{d}$ a constant $a_j$, so $h_j=a_j+b_{j}t$ and $z=z_p+a+bt$ with $a=(a_1,\dots ,a_n)$, $b=(b_1,\dots ,b_n)$. If also $z=z_p+a^{\mathrm{\prime }}+b^{\mathrm{\prime }}t$ on $\mathbb{R}$, evaluating at $t=0$ gives $a=a^{\mathrm{\prime }}$ and then differentiating at $t=0$ gives $b=b^{\mathrm{\prime }}$, so the pair $(a,b)$ is unique; this is claim (b).

Let $z=z_p+a+bt$ be a solution as in step 3.1. If $b=0$, then $z=z_p+a$ is $T$-periodic by step 2.2. Conversely, if $z$ is $T$-periodic, then for every $t\in \mathbb{R}$ the equality $z(t+T)=z(t)$ together with $z_p(t+T)=z_p(t)$ of step 2.2 gives $b(t+T)=bt$, hence $bT=0$ and $b=0$ because $T>0$. The $T$-periodic solutions are therefore exactly the maps $z_p+a$ with $a\in {\mathbb{R}}^n$; their oscillating part is $z_p$ itself, whose defining formula $z_p(t)=\mathrm{Re}(-w{\omega }^{-2}{e}^{-i\omega t})$ exhibits the complex amplitude vector $-w\mathrm{/}{\omega }^2$; this is claim (c).

Let $I$ be a nonempty open interval, $t_{\ast }\in I$, and let $z:I\to {\mathbb{R}}^n$ be $C^2$ with $z^{\mathrm{\prime }\mathrm{\prime }}=\mathrm{Re}(w{e}^{-i\omega t})$ on $I$; put $h:=z-z_p$ on $I$. Fixing $j$ and repeating the argument of step 3.1 on the order-convex interval $I$, every interior point of which lies in $I$ where the assumed derivatives exist, gives a constant $b_j$ on $I$ equal to $h_j^{\mathrm{\prime }}$, then a constant $a_j$ with $h_j=a_j+b_{j}t$ on $I$; hence $z=z_p+a+bt$ on $I$. Evaluating at $t_{\ast }$ gives $b=z^{\mathrm{\prime }}(t_{\ast })-z_p^{\mathrm{\prime }}(t_{\ast })$ and $a=z(t_{\ast })-z_p(t_{\ast })-b{t}_{\ast }$, and conversely those values reproduce $z$ on $I$, so the pair $(a,b)$ is unique. Define $Z(t):=z_p(t)+a+bt$ for all real $t$ with these $(a,b)$; by steps 2.1 and 2.2, $Z$ is a $C^{\mathrm{\infty }}$ solution on $\mathbb{R}$ extending $z$. If $\tilde{Z}$ is any global $C^2$ solution on $\mathbb{R}$ agreeing with $z$ on $I$, then step 3.1 writes $\tilde{Z}=z_p+\tilde{a}+\tilde{b}t$ with a unique pair $(\tilde{a},\tilde{b})$; restricting to $I$ and using the uniqueness just proved gives $\tilde{a}=a$ and $\tilde{b}=b$, so $\tilde{Z}=Z$ and the extension is unique.

For $Z$ as in step 4.2, step 4.1 gives that $Z$ is $T$-periodic if and only if $b=0$, which by step 4.2 is equivalent to $z^{\mathrm{\prime }}(t_{\ast })=z_p^{\mathrm{\prime }}(t_{\ast })$. If $t$ and $t+T$ both lie in $I$, then $Z(t+T)-Z(t)=bT$ while $Z$ agrees with $z$ at those two arguments, so equality $z(t+T)=z(t)$ forces $bT=0$ and $b=0$. If $I$ contains no such pair, then the condition "$z(t+T)=z(t)$ whenever both $t$ and $t+T$ lie in $I$" has empty scope and is satisfied vacuously; the extension with any fixed nonzero $b$ obeys it and is not $T$-periodic, so the condition determines nothing about periodicity of the global extension.

Fix $t\in \mathbb{R}$ and a component index $j$ and define $G_j(s):=-{\omega }^{-3}({\alpha }_j\sin \omega s-{\beta }_j\cos \omega s)$ for $s\in [t,t+T]$; by the rules of [F3], $G_j$ is differentiable at every point of $[t,t+T]$ with $G_j^{\mathrm{\prime }}(s)=-{\omega }^{-2}({\alpha }_j\cos \omega s+{\beta }_j\sin \omega s)=z_{p,j}(s)$. The map $z_{p,j}$ is differentiable on $\mathbb{R}$ by steps 1.1 and 2.1, hence continuous on $[t,t+T]$ by [F6], hence Riemann integrable on $[t,t+T]$ by [F6]; the fundamental theorem [F6] therefore gives ${\int }_t^{t+T}{z}_{p,j}={\omega }^{-3}[{\beta }_j(\cos (\omega (t+T))-\cos \omega t)-{\alpha }_j(\sin (\omega (t+T))-\sin \omega t)]=0$, the vanishing because $\cos (\omega t+2\pi )=\cos \omega t$ and $\sin (\omega t+2\pi )=\sin \omega t$ by [F4] and $\omega T=2\pi$. Summing over $j$ and dividing by $T$ gives mean zero for $z_p$ over $[t,t+T]$. For the affine part put $P(s):=as+b{s}^2\mathrm{/}2$, differentiable with $P^{\mathrm{\prime }}(s)=a+bs$ by [F3]; the same fundamental theorem, the integrand $s\mapsto a+bs$ being continuous, gives $\frac{1}{T}{\int }_t^{t+T}(a+bs)ds=\frac{P(t+T)-P(t)}{T}=a+\frac{b((t+T{)}^2-t^2)}{2T}=a+b(t+\frac{T}{2})$. Adding the two computations, the mean of $Z$ over $[t,t+T]$ is $a+b(t+T\mathrm{/}2)$ for every real $t$, and it equals $a$ for every $t$ exactly when $b=0$, which by step 5.1 is exactly the periodicity condition; then the mean over every complete period is $a$. This completes the proof of (d) and of the lemma.