Period $T$ of oscillation with cubic force function How would I find the period of an oscillator with the following force equation?
$$F(x)=-cx^3$$
I've already found the potential energy equation by integrating over distance:
$$U(x)={cx^4 \over 4}.$$
Now I have to find a function for the period (in terms of $A$, the amplitude, $m$, and $c$), but I'm stuck on how to approach the problem. I can set up a differential equation:
$$m{d^2x(t) \over dt^2}=-cx^3,$$
$$d^2x(t)=-{cx^3 \over m}dt^2.$$
But I am not sure how to solve this. Wolfram Alpha gives a particularly nasty solution involving the hypergeometric function, so I don't think the solution involves differential equations. But I don't have any other leads.
How would I find the period $T$ of this oscillator?
 A: Starting from
  $$ \frac{1}{2} \left( v(x)^2 - v_0^2 \right)= - \frac{c}{m} x^4 $$
with initial velocity $v_0$ when $x=0$, the time relationship is
$$ t = \int_0^x \frac{1}{v(x)}\,{\rm d} x $$
I use in intermediate variable $\xi$ for distance $x = \sqrt[4]{\frac{2 m v_0^2}{c}}\, \xi $ .
I integrate the energy relationship to get
$$ t = \int_0^x \frac{1}{\sqrt{v_0^2 - \frac{c x^4}{2 m} }} \,{\rm d} x=\sqrt[4]{ \frac{2 m}{c v_0^2} } \int_0^\xi \frac{1}{\sqrt{1-\xi^4}}\,{\rm d} \xi $$
$$ t = \sqrt[4]{ \frac{2 m}{c v_0^2} }\; {\rm EllipticF}( \sin^{-1}\xi, -1) $$
Note that the elliptic integral has a taylor expansion of
$${\rm EllipticF}(x,m) \approx x + \frac{m}{6} x^3 - \frac{m}{30} x^5 + \ldots $$
which makes the above solution approximately (for small displacements)
$$ t = \frac{ \tau}{2\pi} \; \sin^{-1}\xi $$
with period $\tau = 2 \pi \sqrt[4]{ \frac{2 m}{c v_0^2} }$ and final solution
$$ x =  \sqrt[4]{\frac{2 m v_0^2}{c}}\,\sin \left( \frac{2 \pi t}{\tau} \right) $$
A: Since
$$\frac1 2mv^2+U(x)=U(A)$$
We have
$$dt=\frac{dx}v=\frac{dx}{\sqrt{2(U(A)-U(x))/m}}=\frac{dx}{\sqrt{c(A^4-x^4)/(2m)}}$$
Then
$$\frac T4=\int_0^{\frac T4}dt=\int_0^A\frac{dx}{\sqrt{\frac{c}{2m}(A^4-x^4)}}$$
Thus
$$T=4\int_0^A\frac{dx}{\sqrt{\frac{c}{2m}(A^4-x^4)}}$$
