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Given the ODE: $$\ddot x(t)-x(t)=0$$ the solution is: $$x(t)=C_1\exp(-t)+C_2\exp(t)$$ If we square the $x(t)$ we have: $$\ddot x(t)-x(t)^2=0$$ and the solution is given by: $$x(t)=6\wp(t+C_1;0,C_2)$$ and so for $x(t)^3$ which gives: $$x(t)=\operatorname{sn}\left(\left(\frac{1}{2}i\sqrt t +C_1\right)C_2,i\right)$$ More generally, is it possible to find closed form solutions for the equation: $$\ddot x(t)-x(t)^n=0$$ ?

Thanks.

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2  
For WeierstrassP, are you looking for $\wp$? That's \wp. –  Ben Millwood Jul 5 '12 at 11:40
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I would hardly consider any expression involving the Weierstrass elliptic functions or the Jacobi elliptic functions as "closed form"... But you can always write it as some implicit function depending on the hypergeometric function ${}_2F_1$. –  Willie Wong Jul 5 '12 at 11:42
    
@Ben Millwood: Yes –  Riccardo.Alestra Jul 5 '12 at 11:48
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This is the answer given by Mathematica: $y=y(x)$ is implicitly given by $$ \frac{(n+1) y(x)^2 \left(c_1 n+c_1+2 y(x)^{n+1}\right){}^2 \, _2F_1\left(\frac{1}{2},\frac{1}{n+1};1+\frac{1}{n+1};-\frac{2 y(x)^{n+1}}{n c_1+c_1}\right){}^2}{\left(c_1 n+c_1\right) \left(c_1 (n+1)+2 y(x)^{n+1}\right){}^2}=\left(c_2+x\right){}^2. $$ In general, I do not believe that $y$ may be written down in terms of elementary or special functions. Moreover, I guess that a rigorous answer (i.e. a proof that it is impossible to find a closed form) is really hard to build. –  Siminore Jul 5 '12 at 11:58
    
@Siminore: I suppose you mean x(t). If you expand your comment I will accept is as an answer. –  Riccardo.Alestra Jul 5 '12 at 13:15

3 Answers 3

up vote 3 down vote accepted

As an addendum of sorts to Mathlover's answer: the differential equation

$$(y^\prime)^2=2\left(\frac{y^{n+1}}{n+1}+c_1\right)$$

is in fact the very sort of equation that is solved by Abelian (hyperelliptic) functions, in the sense that the integral

$$\int\frac{\mathrm dt}{\sqrt{t^{n+1}+c}}$$

is a hyperelliptic (Abelian) integral (which, as already noted, can be represented in terms of the Gaussian hypergeometric function), and the original differential equation is solved by the inversion of this integral; i.e., with Abelian functions. Since Mathematica and Wolfram Alpha know nothing about Abelian functions, they are unable to proceed further, and they just leave an implicit equation as a result.

As expected, when $n=2$ or $3$, the hyperelliptic integral becomes an elliptic integral, and thus the differential equation is expected to have elliptic function solutions. I'll carry out the inversion explicitly for those two cases.


For $n=2$, we have, after absorption of arbitrary constants, the expression

$$\int\frac{\mathrm dy}{\sqrt{y^3+C_1}}=\sqrt{\frac23}x+C_2$$

To make the integral on the left a bit more recognizable, we multiply by a constant on both sides:

$$\frac12\int\frac{\mathrm dy}{\sqrt{y^3+C_1}}=\frac12\left(\sqrt{\frac23}x+C_2\right)$$

which turns into

$$\int\frac{\mathrm dy}{\sqrt{4y^3+4C_1}}=\frac{x}{\sqrt 6}+\frac{C_2}{2}$$

and we now recognize the Weierstrass elliptic integral corresponding to the cubic $4y^3-g_2 y-g_3$ on the left side, with the invariants $g_2=0$ and $g_3=-4C_1$. Inversion yields

$$y=\wp\left(\frac{x}{\sqrt 6}+\frac{C_2}{2};0,-4C_1\right)$$

or, after absorption of arbitrary constants,

$$y=\wp\left(\frac{x}{\sqrt 6}+C_2;0,C_1\right)=6\wp\left(x+C_2;0,C_1\right)$$

where the homogeneity relation for $\wp$ was used to obtain the final expression.


For $n=3$, we have (changing the form of one of the arbitrary constants for convenience)

$$\int\frac{\mathrm dy}{\sqrt{y^4+C_1^4}}=\frac{x}{\sqrt 2}+C_2$$

We can try a Weierstrass substitution $y=C_1 \cot\frac{t}{2}$ here:

$$\frac1{2C_1}\int\frac{\mathrm dt}{\sqrt{1-\frac12\sin^2 t}}=\frac{x}{\sqrt 2}+C_2$$

and we recognize the incomplete elliptic integral of the first kind at this point (and absorbing arbitrary constants while we're at it):

$$\frac1{C_1}F\left(2\mathrm{arccot}\frac{y}{C_1}\mid\frac12\right)=\frac{x}{\sqrt 2}+C_2$$

We solve for $y$ in stages:

$$\begin{align*} F\left(2\mathrm{arccot}\frac{y}{C_1}\mid\frac12\right)&=C_1 x+C_2\\ 2\mathrm{arccot}\frac{y}{C_1}&=\mathrm{am}\left(C_1 x+C_2\mid\frac12\right)\\ y&=C_1\cot\left(\frac12\mathrm{am}\left(C_1 x+C_2\mid\frac12\right)\right)\\ y&=C_1\frac{\sin\left(\mathrm{am}\left(C_1 x+C_2\mid\frac12\right)\right)}{1-\cos\left(\mathrm{am}\left(C_1 x+C_2\mid\frac12\right)\right)}\\ y&=C_1\frac{\mathrm{sn}\left(C_1 x+C_2\mid\frac12\right)}{1-\mathrm{cn}\left(C_1 x+C_2\mid\frac12\right)} \end{align*}$$

and that's how Jacobian elliptic functions turn up in the solution.


As it turns out, the hyperelliptic functions for $n$ an odd integer can theoretically be expressed in terms of elliptic functions, but the expressions look rather unwieldy. See Byrd and Friedman for details (p. 252 onwards).

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$$\begin{align*} y''-y^n&=0\\ y'y''&=y'y^n\\ \int y'y'' dx &=\int y'y^n dx\\ \frac{(y')^2}{2} &=\frac{y^{n+1}}{n+1} +c_1\\ y' &=\sqrt {\frac{2y^{n+1}}{n+1} +2c_1 }\\ \int \frac{dy}{\sqrt {\frac{2y^{n+1}}{n+1} +2c_1 }} &=\int dx =x+a\\ \frac{1}{(2c_1)^{1/2}}\int \frac{dy}{ \sqrt{1+\frac{y^{n+1}}{c_1(n+1)}}} &=x+a\\ \int \frac{dy}{ \sqrt{1+\frac{y^{n+1}}{c_1(n+1)}}} &=(2c_1)^{1/2}x+a(2c_1)^{1/2}=(2c_1)^{1/2}x+c_2 \end{align*}$$ After here you can change variable and use the binomial expansion to evaluate integral.

$u=\frac{y^{n+1}}{c_1(n+1)}$

$${(1+u)}^{\alpha}= \sum_{n=0}^\infty \binom{\alpha}{n} u^n=1+ \alpha \frac{u}{1!}+\alpha (\alpha-1) \frac{u^2}{2!}+\dots$$

I avoided doing many calculations, and I preferred to use the quick way: ask Wolfram Alpha. Here is a link to the solution.

$k=\frac{1}{c_1(n+1)}$

$\int \frac{dy}{ \sqrt{1+ky^{n+1}}} =y. _2F_1 (\frac{1}{2},\frac{1}{n+1};\frac{n+2}{n+1}; -ky^{n+1}) = y. _2F_1 (\frac{1}{2},\frac{1}{n+1};\frac{n+2}{n+1}; -\frac{y^{n+1}}{c_1(n+1)})=(2c_1)^{1/2}x+c_2 $

Solution in closed form as Hypergeometric function $$_2F_1 (\frac{1}{2},\frac{1}{n+1};\frac{n+2}{n+1}; -\frac{y^{n+1}}{c_1(n+1)})=\frac{1}{y}(\sqrt{2c_1}x+c_2 )$$

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Yes, $E=\frac{1}{2}|y'|^2 - \frac{1}{n+1}y^{n+1}$ is constant along solutions, since the equation is a conservative system. Very nice remark. –  Siminore Jul 5 '12 at 13:32

This is the answer given by Mathematica: $x=x(t)$ is implicitly given by $$ \frac{(n+1) x(t)^2 \left(c_1 n+c_1+2 x(t)^{n+1}\right){}^2 \, _2F_1\left(\frac{1}{2},\frac{1}{n+1};1+\frac{1}{n+1};-\frac{2 x(t)^{n+1}}{n c_1+c_1}\right){}^2}{\left(c_1 n+c_1\right) \left(c_1 (n+1)+2 x(t)^{n+1}\right){}^2}=\left(c_2+t\right){}^2. $$ In general, I do not believe that $y$ may be written down in terms of elementary or special functions.

Remark: this was already suggested by Willie Wong in a comment. I wrote it just to improve readability. Moreover, I guess that a rigorous answer (i.e. a proof that it is impossible to find a closed form) is really hard to build.

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