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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.

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When you say "...to me periodicity is quite obvious from the expression", do you mean that exclusively from the expression you know this or because of this and other info that you did not display you know this? I ask because I see the expression and I've not the slightest idea how only from it it is obvious the function is periodic and even less what its period would be... –  DonAntonio Aug 3 '12 at 13:38
    
maybe you're right.. call it intuition..but I asked exactly because most of the times intuition is wrong.. –  uforoboa Aug 3 '12 at 13:55
    
I edited my post at any rate not to confuse or to make you believe I've got the solution.. which is not true... –  uforoboa Aug 3 '12 at 13:56

3 Answers 3

up vote 4 down vote accepted

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}

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...and that last integral is in fact related to the so-called lemniscatic constant. –  J. M. Aug 4 '12 at 2:06

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:

$\hspace{3.5cm}$enter image description here

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 $$ \sqrt[4]{2}\,\mathrm{B}(1/4,1/2)=6.236338999021645\tag{3} $$

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To produce rob's plot: ParametricPlot[{JacobiCN[u, 1/2], -JacobiSN[u, 1/2] JacobiDN[u, 1/2]}, {u, 0, 4 EllipticK[1/2]}]. –  J. M. Aug 4 '12 at 7:42
    
@J.M.: Thanks for that; it is much simpler than what I actually used, but adding the options I used, it gives the exact same output: ParametricPlot[{{Sqrt[Cos[t]], Sin[t]/Sqrt[2]}, {-Sqrt[Cos[t]], Sin[t]/Sqrt[2]}, {Sqrt[Cos[t]], -Sin[t]/Sqrt[2]}, {-Sqrt[Cos[t]], -Sin[t]/Sqrt[2]}}, {t, 0, Pi/2}, PlotStyle -> RGBColor[.75, 0, 0], AxesLabel -> {"U", "U'"}, ImageSize -> 400] –  robjohn Aug 4 '12 at 10:03

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.

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Are you sure that modding out constants preserves the original nature of the problem? I tell you my point.. $U''+U^3=0$ was the original equation i studied.. then i got, on wolfram alpha, this wolframalpha.com/input/?i=U%27%27%2BU%5E3%3D0.. and the solution looks periodic to me.. Of course if $U$ is not identically zero.. –  uforoboa Aug 3 '12 at 16:12
    
well, modding out constants does not change the problem. I think your step from $U''+U^{3}$ to your formula is not justified. –  Bombyx mori Aug 3 '12 at 19:04
    
@uforoboa: In fact, one particular solution of the DE $y^{\prime\prime}+y^3=0$ is the famed sine lemniscate function $\mathrm{sl}(u)$, of which much has been written about. In terms of the usual Jacobian elliptic functions: $$\mathrm{sl}(u)=\frac1{\sqrt 2}\mathrm{sd}\left(u\mid\frac12\right)$$ The other solution is the cosine lemniscate function, $$\mathrm{cl}(u)=\mathrm{cn}\left(u\mid\frac12\right)$$ Both are, as expected of elliptic functions, doubly periodic. –  J. M. Aug 4 '12 at 2:57

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