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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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    I edited my post at any rate not to confuse or to make you believe I've got the solution.. which is not true...2012-08-03

3 Answers 3

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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](http://mathworld.wolfram.com/LemniscateConstant.html).2012-08-04
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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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    @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]`2012-08-04
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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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    @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.2012-08-04