# IVP Perturbation With Small Non-Linear Term

EDIT: Sorry to bump this without having anything extra to add, but I still cannot reconcile my solution with what was asked (in (2)). Could someone with expertise in this subject take a look? I would definitely appreciate it. In particular, either my approximation derived in the majority of my solution is incorrect, or, in the last paragraph or so, my methodology for determining the center of oscillation is flawed. I've triple checked everything though, and I'm not sure what the issue is. Perhaps it was a typo in the original problem statement from the text?

(BACKGROUND)

I'm asked to do two things with the IVP $\ddot{x}+x+\epsilon x^{2}=0$ where $0<\epsilon\ll 1$ and $x(0)=a>0$ and $\dot{x}(0)=0$.

(1) Find an approximate solution of accuracy $O(\epsilon)$, valid for all time.

(2) Show the center of oscillation is approximately $\epsilon\frac{a^2}{2}$.

I first observe that if a simple $O(\epsilon)$ approximation is to be valid for all time, then the only real possibility for this to happen is if the solution to be approximated is periodic. To that end, I use the Poincaré-Leindstedt perturbation method, since if the system has very strict requirements on initial conditions for periodic solutions, the two parameters given by this method (as opposed to one given by regular perturbations) will allow me to obtain them (i.e. when it comes time to eliminate resonance terms).

EDIT: It turns out that there is no restriction on $a$ for periodic solutions, but this is only discoverable after proceeding with the method.

(ACTUAL QUESTION)

I am pretty confident that my approximation is correct (i.e. that I correctly applied the method). But I have doubts, primarily because in (2) I get a different (though similar) answer for the center of oscillation. So either my methodology for determining it is wrong, or my approximation is wrong. So I included both stages here.

(SOLUTION ATTEMPT)

Our equation is $\ddot{x}+x+\epsilon x^{2}=0$ subject to initial conditions $x(0)=a$ and $\dot{x}(0)=0$. Note that because the system is autonomous, the initial condition on $x'(t)$ is completely general assuming the solution oscillates indefinitely (which are the solution(s) we obtain by letting $a$ be yet undertermined), for in this case $\dot{x}(t)=0$ for infinitely many $t$, and being autonomous, the solution is unchanged by the translation $t\mapsto t-t_{0}$. Introduce a yet to be determined time dilation (or equivalently, frequency) $$\tau=\omega t.$$ Expanding $x$ and $\omega$ asymptotically gives \begin{eqnarray*} x(\tau)=x_{0}+\epsilon x_{1}+O(\epsilon^{2})\\ \omega=\omega_{0}+\epsilon\omega_{1}+O(\epsilon^{2}). \end{eqnarray*} Substituting the asymptotic expansion for $x$ into the equation and immediately discarding $O(\epsilon^{2})$ and higher order terms yields $$\omega^{2}x_{0}''+\epsilon\omega^{2}x_{1}''+x_{0}+\epsilon x_{1}+\epsilon x_{0}^{2}=0,$$ where $'$ denotes differentiation with respect to $\tau$. Substituting the asymptotic expansion for $\omega$ into this expression and ignoring $O(\epsilon^{2})$ and smaller terms then gives $$\omega_{0}^{2}x_{0}''+2\omega_{0}\omega_{1}\epsilon x_{0}''+\epsilon\omega_{0}^{2}x_{1}''+x_{0}+\epsilon x_{1}+\epsilon x_{0}^{2}=0.$$

Collecting like-order terms forms the system of equations \begin{eqnarray*} (1)\;\;\omega_{0}^{2}x_{0}''+x_{0}=0,\\ (2)\;\;\omega_{0}^{2}x_{1}''+2\omega_{0}\omega_{1}x_{0}''+x_{0}^{2}+x_{1}=0. \end{eqnarray*} (1) implies $\omega_{0}=1$ and $x_{0}=A\cos\tau+B\sin\tau$. The initial conditions on $x$ imply $x_{0}(0)=a$ and $x_{k}(0)=x_{k}'(0)=0$. Hence, $$x_{0}=a\cos\tau.$$ Substituting $\omega_{0}$ and $x_{0}$ into (2) and rearranging gives $$x_{1}''+x_{1}=2\omega_{1}a\cos\tau-a^{2}\cos^{2}\tau=2\omega_{1}a\cos\tau-\frac{a^{2}}{2}-\frac{a^{2}}{2}\cos2\tau=f(\tau).$$ To eliminate resonance terms and obtain a periodic solution, we only require $$2\omega_{1}a=0,$$ and this implies $\omega_{1}=0$. This means that $a$ may be regarded as a free parameter for the system, and so there are infinitely many initial conditions which lead to periodic solutions (note that we have only computed up to $O(\epsilon)$ order terms; presumably, eliminating resonance terms from $x_{k}$ ($k>1$) would require some dependence on $a$ for $w_{k}$). At any rate, our equation for $x_{1}$ then becomes $$x_{1}''+x_{1}=-a^{2}\cos2\tau.$$ Solving this equation (using variation of parameters or some other method to obtain a particular solution) with homogeneous initial conditions gives $$x_{1}(\tau)=-\frac{2}{3}a^{2}\sin^{2}\left(\frac{\tau}{2}\right)\left(2+\cos\tau\right).$$ We then form our first order approximation to $x(t)$ by substituting these solutions into our asymptotic expansion for $x$: \begin{eqnarray*} x(t)&=&a\cos\omega t-\epsilon\frac{2}{3}a^{2}\sin^{2}\left(\frac{\omega t}{2}\right)\left(2+\cos\omega t\right)+O(\epsilon^{2})\\ &=&a\cos\omega t +\epsilon\frac{a^{2}}{6}(3-2\cos\omega t-\cos2\omega t)+O(\epsilon^{2}), \end{eqnarray*} where $\omega=1+O(\epsilon^{2})$ and the simplifying identity $\sin^{2}(\frac{\tau}{2})=\frac{1}{2}(1-\cos\tau)$ was used. As for the center of oscillation, we expect it to occur at approximately $\frac{T}{4}$, where $T$ is the period of oscillation, since the oscillation begins at maximal amplitude for $t=0$ (see initial conditions). The period of oscillation is $$T=\frac{2\pi}{\omega}$$ so that evaluating our asymptotic expansion at $\frac{T}{4}=\frac{\pi}{2\omega}$ gives \begin{eqnarray*} x\left(\frac{\pi}{2\omega}\right) &=&a\cos\frac{\pi}{2}+\epsilon\frac{a^{2}}{6}\left(3-2\cos\frac{\pi}{2}-\cos\pi\right)\\ &=& 0+\epsilon\frac{a^2}{6}\left(3-0-(-1)\right)\\ &=& \epsilon\frac{2a^{2}}{3} \end{eqnarray*} which is accurate up to $O(\epsilon^{2})$.

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