nonlinear fourth-order differential equation type Does anybody know how to reduce such nonlinear fourth-order differential equation to well-known type (like Ricatti, Painlevé, sine-Gordon etc)? 
$\frac{{{d^4}\phi }}{{d{x^4}}} + q\frac{{{d^2}\phi }}{{d{x^2}}} - {\omega ^2}\sin\phi  = 0$
 A: $$\frac{{{d^4}\phi }}{{d{x^4}}} + q\frac{{{d^2}\phi }}{{d{x^2}}} - {\omega ^2}\sin\phi  = 0$$
Even in simplified case of $q=0$, the equation $\quad\frac{{{d^4}\phi }}{{d{x^4}}} - {\omega ^2}\sin\phi  = 0\quad$ as no closed form solution, as far as I know. 
The solutions of the second order ODE : $\quad\frac{{{d^2}\phi }}{{d{x^2}}} - {\omega ^2}\sin\phi  = 0\quad$ involve the $am$ Jacobi elliptic function : $\quad \phi=-2\,am\bigg(\frac12(x+c_2)\sqrt{c_1-2\omega^2}\:\bigg|\:\frac{4\omega^2}{2\omega^2+c_1} \bigg)$
As far as I know, no special function was defined for solving similar ODE of higher order such as $\quad\frac{{{d^3}\phi }}{{d{x^3}}} - {\omega ^2}\sin\phi  = 0\quad$ and even more $\quad\frac{{{d^4}\phi }}{{d{x^4}}} - {\omega ^2}\sin\phi  = 0\quad$ and a-fortiori for your equation $\quad\frac{{{d^4}\phi }}{{d{x^4}}} + q\frac{{{d^2}\phi }}{{d{x^2}}} - {\omega ^2}\sin\phi  = 0$.
As usualy done, one can consider the approximate case of small $\phi\quad\to\quad \sin(\phi)\simeq \phi$ so that :
$$\frac{{{d^4}\phi }}{{d{x^4}}} + q\frac{{{d^2}\phi }}{{d{x^2}}} - {\omega ^2}\phi  \simeq 0$$
$$\phi(x)\simeq c_1e^{r_1x}+c_2e^{r_2x}+c_3e^{r_3x}+c_4e^{r_4x}\qquad r_{1,2,3,4}=\pm\sqrt{\frac12\left(-q\pm\sqrt{q^2+4\omega^2}\right)}$$
Of course, some of the $r$ are complex which implies sinusoidal terms is the solution.
Better approximate could be obtained with more or less limited series, but it should be arduous calculus. In practice, for accurate numerical result, it is simpler to directly solve the ODE thanks to numerical method.
A: Hint:
$\dfrac{d^4\phi}{dx^4}+q\dfrac{d^2\phi}{dx^2}-\omega^2\sin\phi=0$
$\dfrac{d\phi}{dx}\dfrac{d^4\phi}{dx^4}+q\dfrac{d\phi}{dx}\dfrac{d^2\phi}{dx^2}-\omega^2(\sin\phi)\dfrac{d\phi}{dx}=0$
$\int\dfrac{d\phi}{dx}\dfrac{d^4\phi}{dx^4}~dx+q\int\dfrac{d\phi}{dx}\dfrac{d^2\phi}{dx^2}~dx-\omega^2\int(\sin\phi)\dfrac{d\phi}{dx}~dx=0$
$\int\dfrac{d\phi}{dx}~d\left(\dfrac{d^3\phi}{dx^3}\right)+q\int\dfrac{d\phi}{dx}~d\left(\dfrac{d\phi}{dx}\right)-\omega^2\int\sin\phi~d\phi=0$
$\dfrac{d\phi}{dx}\dfrac{d^3\phi}{dx^3}-\int\dfrac{d^3\phi}{dx^3}~d\left(\dfrac{d\phi}{dx}\right)+q\int\dfrac{d\phi}{dx}~d\left(\dfrac{d\phi}{dx}\right)-\omega^2\int\sin\phi~d\phi=0$
$\dfrac{d\phi}{dx}\dfrac{d^3\phi}{dx^3}-\int\dfrac{d^2\phi}{dx^2}\dfrac{d^3\phi}{dx^3}~dx+q\int\dfrac{d\phi}{dx}~d\left(\dfrac{d\phi}{dx}\right)-\omega^2\int\sin\phi~d\phi=0$
$\dfrac{d\phi}{dx}\dfrac{d^3\phi}{dx^3}-\int\dfrac{d^2\phi}{dx^2}~d\left(\dfrac{d^2\phi}{dx^2}\right)+q\int\dfrac{d\phi}{dx}~d\left(\dfrac{d\phi}{dx}\right)-\omega^2\int\sin\phi~d\phi=0$
$\dfrac{d\phi}{dx}\dfrac{d^3\phi}{dx^3}-\dfrac{1}{2}\left(\dfrac{d^2\phi}{dx^2}\right)^2+\dfrac{q}{2}\left(\dfrac{d\phi}{dx}\right)^2+\omega^2\cos\phi=c$
$2\dfrac{d\phi}{dx}\dfrac{d^3\phi}{dx^3}-\left(\dfrac{d^2\phi}{dx^2}\right)^2+q\left(\dfrac{d\phi}{dx}\right)^2+\omega^2\cos\phi=C_1$
