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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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    For WeierstrassP, are you looking for $\wp$? That's `\wp`.2012-07-05
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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 ${}_2F_1$](http://www.wolframalpha.com/input/?i=y%27%27%28x%29+%3D+y%28x%29^n).2012-07-05
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    @Ben Millwood: Yes2012-07-05
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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.2012-07-05
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    @Siminore: I suppose you mean x(t). If you expand your comment I will accept is as an answer.2012-07-05

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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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$$\begin{align*} y''-y^n&=0\\ y'y''&=y'y^n\\ \int y'y'' dx &=\int y'y^n dx\\ \frac{(y')^2}{2} &=\frac{y^{n+1}}{n+1} +c_1\\ y' &=\sqrt {\frac{2y^{n+1}}{n+1} +2c_1 }\\ \int \frac{dy}{\sqrt {\frac{2y^{n+1}}{n+1} +2c_1 }} &=\int dx =x+a\\ \frac{1}{(2c_1)^{1/2}}\int \frac{dy}{ \sqrt{1+\frac{y^{n+1}}{c_1(n+1)}}} &=x+a\\ \int \frac{dy}{ \sqrt{1+\frac{y^{n+1}}{c_1(n+1)}}} &=(2c_1)^{1/2}x+a(2c_1)^{1/2}=(2c_1)^{1/2}x+c_2 \end{align*}$$ 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.2012-07-05
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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).