How to conclude this solution is periodic?

Question

I've met the following problem in finishing my argument. The expression I ended with is $$\frac{\mathrm d}{\mathrm d\xi}U(\xi)=\pm\sqrt{C_1-\frac{U^4(\xi)}{2}},$$ with $U:\mathbb R\to\mathbb R$ and $C_1$ is non zero because otherwise I would have no square root unless $U\equiv 0$, which is not admissible in my situation. Therefore I would say the solution $U$ is periodic, and also I would like to find some bounds on the period.

The first problem I ask is the following: how would you proceed in showing that $U$ is periodic?

The second question is to find some bounds on the period. I mean: separating the variables and choosing the $+$ sign one gets $$\frac{\mathrm d U}{\sqrt{C_1-\frac{U^4}{2}}}=\mathrm d \xi,$$ but then I derived no useful informations about the period. Could you help me?

Thank you in advance.

Answer 1

For simplicity I will denote $U$ by $x$. Suppose $x$ is a non-zero solution of your ODE. Then $x$ solves the Hamiltonian equation $$\tag{HS} \ddot{y}(t)+y^3(t)=0, $$ and in particular $$ \frac{1}{2}\dot{x}^2(t)+\frac{1}{4}x^4(t)=h $$ for some constant $h>0$ not dependending on $t$. Therefore $$ (x,\dot{x})=(\pm h^{1/4}\sqrt{2\sin\theta},\sqrt{2h}\cos\theta), $$ for some function $\theta:\mathbb{R} \to [2k\pi,(2k+1)\pi],\ k \in \mathbb{Z}$. Since $x$ solves (HS), it follows that $\theta$ solves the ODE $$ \ddot{\theta}=4\sqrt{h}\cos\theta. $$ The function $$\tag{1} t \mapsto \phi(t)=\theta(t/2h^{1/4})-\frac{\pi}{2} $$ then solves the (mathematical) pendulum equation $$\tag{P} \ddot{\phi}=-\sin\phi. $$ We recall that any solution $\phi$ of (P) has constant energy, i.e. $$ E(t)=\frac{1}{2}\dot{\phi}^2(t)+1-\cos\phi(t)=E(0) \quad \forall t. $$ It is well known that (P) admits periodic, homoclinic, and heteroclinic solutions. Let $\phi_\tau$ be a periodic solution of (P) with period $\tau>0$. Then $$ x_\tau(t)=\pm h\sqrt{\cos\phi_\tau(2h^{1/4}t)} $$ is a periodic solution of (HS) with period $T_h=\tau/2h^{1/4}$ and energy $h$.

Let $x^T$ be a periodic solution solution of (HS) with period $T=T_h>0$, and energy $h>0$. Setting $$ r=\sqrt{2}h^{1/4}, $$ we have \begin{eqnarray} T_h&=&2\sqrt{2}\int_{-r}^r\frac{dy}{\sqrt{4h-y^4}}=4\sqrt{2}\int_0^r\frac{dy}{\sqrt{4h-y^4}}\cr &=&\frac{4\sqrt{2}}{r}\int_0^1\frac{ds}{\sqrt{1-s^4}}=4h^{-1/4}\int_0^1\frac{ds}{\sqrt{1-s^4}}. \end{eqnarray}

Answer 2

The original equation after modding out constants, etc is the familiar elliptic integral of first kind(for a more detailed introduction see this note. For your application we only need to know if it is periodic, I do not have a definite answer (the addition law for elliptic integrals is not really related). But judged by the plot of this function it seems highly unlikely it would be periodical. I guess if you are certain the integral is right, working with the information from the wikipedia page might be helpful.

Answer 3

Since $U$ and $U'$ satisfy $$ 2\,U'^2+U^4=2C_1\tag{1} $$ $U'$ is dependent on $U$ in such a way that $(U,U')$ follows the path below clockwise:

$\hspace{3.5cm}$

If the solution is $C^2$, then $U$ won't stop at $U^4=2C_1$, and if $U$ can get to that point in a finite amount of time, it will follow this oval and be periodic.

Using the substitution $(2C_1)^{1/4}t^{1/4}=U$, we get the period to be $$ \begin{align} 4\int_0^{\sqrt[4]{2C_1}}\frac{\mathrm d U}{\sqrt{C_1-\frac{U^4}{2}}} &=(2/C_1)^{1/4}\int_0^1\frac{t^{-3/4}\,\mathrm{d}t}{\sqrt{1-t}}\\ &=(2/C_1)^{1/4}\,\mathrm{B}(1/4,1/2)\\ &=(2\pi^2/C_1)^{1/4}\frac{\Gamma(1/4)}{\Gamma(3/4)}\tag{2} \end{align} $$ where $$ \sqrt[4]{2}\,\mathrm{B}(1/4,1/2)=6.236338999021645\tag{3} $$