How to calculate the averaged equations for the weakly nonlinear oscillator $\ddot x+x+\varepsilon (x\dot x^2)=0$?

Question

This is Strogatz exercise $7.6.5:$

For the system $\ddot x+x+\varepsilon h(x,\dot x)=0$, where $h(x,\dot x)=x\dot x^2$ with $0 < ε << 1$, calculate the averaged equations and if possible, solve the averaged equations explicitly for $x(t,ε)$, given the initial conditions $x(0)=a$, $\dot x(0)=0$.

Answer 1

$0)$

$$r=\sqrt{x^2+y^2},$$

$$\phi=\arctan\biggr(\frac{y}{x}\biggr)-t\ (\theta=t+\phi\rightarrow\phi=\theta-t).$$

$1)$

$$h(x,\dot x)=x\dot x^2=r\cos(t+\phi)r^2\sin^2(t+\phi)=r^3\cos(t+\phi)\sin^2(t+\phi).$$

$2)$

$$\dot{\bar r}=\Bigr<\varepsilon h\sin(t+\phi)\Bigr>_t,$$

$$\dot{\bar \phi}=\Bigr<\frac{\varepsilon h}{r}\cos(t+\phi)\Bigr>_t.$$

\begin{align} \rightarrow \dot r & = \Bigr<\varepsilon r^3\cos(t+\phi)\sin^2(t+\phi)\Bigr[\sin(t+\phi)\Bigr]\Bigr>_t \\ & = \varepsilon r^3\Bigr<\cos(t+\phi)\sin^3(t+\phi)\Bigr>_t \\ & = \varepsilon r^3\frac{1}{2\pi}\displaystyle \int_{t-\pi}^{t+\pi}\cos(t+\phi)\sin^3(t+\phi)dt=0 \\ & \rightarrow \dot r=0\rightarrow \frac{dr}{dt}=0\rightarrow dr=0\rightarrow r=r_0. \end{align}

\begin{align} \rightarrow \dot \phi & = \Bigr<\frac{\varepsilon}{r} r^3\cos(t+\phi)\sin^2(t+\phi)\Bigr[\cos(t+\phi)\Bigr]\Bigr>_t \\ & = \varepsilon r^2\Bigr<\cos^2(t+\phi)\sin^2(t+\phi)\Bigr>_t \\ & = \varepsilon r^2\frac{1}{2\pi}\displaystyle \int_{t-\pi}^{t+\pi}\cos^2(t+\phi)\sin^2(t+\phi)dt \\ & = \varepsilon r^2\frac{1}{2\pi}\frac{\pi}{4}=\frac{\varepsilon r^2}{8} \\ & \rightarrow \dot \phi= \frac{\varepsilon r^2}{8}\rightarrow \frac{d\phi}{dt}=\frac{\varepsilon r^2}{8} \\ & \rightarrow d\phi=\frac{\varepsilon r^2}{8}dt \rightarrow \phi=\frac{\varepsilon r^2}{8}t+\phi_0. \end{align}

The amplitude of the closed orbit can be anything and the closed orbit is approximately circular.

$3)$

$$x(0)=a,$$

$$\dot x(0)=0.$$

$$\rightarrow r_0=\sqrt{x^2+y^2}=\sqrt{0+a^2}=a\rightarrow r=a.$$

$$\rightarrow \phi_0=\arctan\Bigr(\frac{0}{a}\Bigr)-0=0\rightarrow \phi=\frac{\varepsilon r^2}{8}t.$$

$4)$

$$x(t)=r(t)\cos(t+\phi)\rightarrow x(t)=a\cos\Bigr(t+\frac{\varepsilon r^2}{8}t\Bigr).$$

Answer 2

This equation, that is the perturbation term, has a curious structure that allows to find a first integral via \begin{align} \ddot x + x(1+ε\dot x^2)&=0\\[1em] \implies \frac{2\dot x\ddot x}{1+ε\dot x^2}+2x\dot x=0\\[1em] \implies \frac1ε\ln|1+ε\dot x^2|+x^2 = C \end{align} For $(x(0),\dot x(0))=(a,0)$ this gives $C=a^2$. Thus it is a consequence that the solutions are periodic, follow the level curves of the first integral, the average radius change over a period is zero in all orders of perturbation.

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To compute the period, first isolate the derivative $$ \dot x=\pm\sqrt{\frac{\exp(ε(a^2-x^2)-1}{ε}}=\pm\sqrt{a^2-x^2}\sqrt{1+\fracε2(a^2-x^2)+\frac{ε^2}6(a^2-x^2)^2+...} $$ A quarter period then computes as \begin{align} \frac{T}4&=\int_0^a\frac{\sqrtε\,dx}{\sqrt{\exp(ε(a^2-x^2)-1}}\\ &=\int_0^{\frac\pi2}\frac{ds}{\sqrt{1+\fracε2a^2\cos^2s+\frac{ε^2}6a^4\cos^4s+...}}\\ &=\frac\pi2-\fracε4a^2\int_0^{\frac\pi2}\cos^2s\,ds+ \frac{ε^2}{96}a^4\int_0^{\frac\pi2}\cos^4s\,ds + \frac{ε^3}{384}a^6\int_0^{\frac\pi2}\cos^6s\,ds - \frac{ε^3}{10240}a^8\int_0^{\frac\pi2}\cos^8s\,ds \mp\dots\\ &=\frac\pi2\left(1-\fracε8a^2+ \frac{ε^2}{256}a^4 + \frac{5ε^3}{6144}a^6 - \frac{7ε^4}{262144}a^8\mp\dots\right) \end{align} The perturbed frequency is then $$ \frac{2\pi}{T}=\frac{\pi/2}{T/4}=1 + \frac{ε}8a^2 + \frac{3ε^2}{256}a^4 + \frac{ε^3}{6144}a^6 - \frac{79ε^4}{786432}a^8+\dots $$

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Series expansion using the CAS Magma (online calculator), with $z=εa^2$,

PS<z>:=PowerSeriesRing(Rationals());
q:=PS!(((Exp(z)-1)/z)^(-1/2));"integrand:",q;
iq := PS![ Coefficient(q,k)*Binomial(2*k,k)/4^k : k in [0..19] ]; "period factor:",iq;
"frequency factor:",1/iq;