Solve for Differential Equations by using regular perturbation method

Question

carry out regular perturbation calculation for $\epsilon$ satisfying $$x''(t)+x(t)=\epsilon x^2(t)$$ correct to the second order in the small parameter $\epsilon$. then use the result to perform a renormalization calculation, straightforwardly assuming that the lowest order correction to the period of the system is of order $\epsilon^2$, not (the erroneous) $\epsilon$.

Answer 1

This system is conservative, multiply with $2 x'$ and integrate to get $$ (x')^2+x^2(1-ϵ\tfrac23x)=R^2=const. $$ Write this as $(x')^2+u(x)^2=R^2$. This circle equation can now be parametrized like a circle, with $x'(t)=R\cos\varphi(t)$, and then consequently $R\sinφ(t)=u(x(t))=x(t)\sqrt{1-ϵ\tfrac23x(t)}$.

---

Now find the inverse function $v$ to $u$ so that $x=v(R\sinφ)$. Then its derivative and the circle equation lead to the identity $$ R\cosφ=x'=v'(R\sinφ)R\cosφ\,φ', \\ 1=v'(R\sinφ)\,φ'. $$ This allows to compute the period of $x$ as $$ T=\int_0^{2\pi}v'(R\sinφ)\,dφ $$

---

To find the first terms of the expansion of $v$, bring $u=x\sqrt{1-ϵ\tfrac23x}$ into fixed-point form and apply the binomial series for the power $-1/2$ to give $$ x=u\left(1-2\fracϵ3x\right)^{-1/2}=u\left(1+\fracϵ3x+\frac{ϵ^2}6x^2 + \frac{5}{54}(ϵx)^3 + \frac{35}{648}(ϵx)^4 +O(ϵ^5)\right) $$ Iterating this relation gives, starting from $x=u+O(ϵ)$, \begin{align} x&=u\left(1+\fracϵ3u+O(ϵ^2)\right) \\&\vdots\\ x=v(u)&=u + \frac13ϵu^2 + \frac{5}{18}ϵ^2u^3 + \frac{8}{27}ϵ^3u^4 + \frac{77}{216}ϵ^4u^5 +O(ϵ^5)\\ v'(u)&=1 + \frac23ϵu + \frac{5}{6}ϵ^2u^2 + \frac{32}{27}ϵ^3u^3 + \frac{385}{216}ϵ^4u^4 +O(ϵ^5) \end{align} so that, using that odd powers of the sine integrate to zero and the constant part of $\sin^{2k}φ$ is $\frac{\binom{2k}{k}}{2^{2k}}$, \begin{align} T&=2\pi+\int_0^{2\pi}\left[\frac23ϵR\sinφ + \frac{5}{6}(ϵR)^2\sin^2φ + \frac{32}{27}(ϵR)^3\sin^3φ + \frac{385}{216}(ϵR)^4\sin^4φ +O(ϵ^5)\right]\,dφ \\ &=2\pi\left(1+\frac{5}{6}(ϵR)^2\frac{\binom{2}{1}}{2^2}+\frac{385}{216}(ϵR)^4\frac{\binom{4}{2}}{2^4}+O(ϵ^6)\right) =2\pi\left(1+\frac{5}{12}(ϵR)^2+\frac{385}{576}(ϵR)^4+O(ϵ^6)\right). \end{align}