Can someone solve the following equation exactly? $$\frac{d^2x}{dt^2}+\omega_0^2 x+\alpha x^3=0$$ , with the initial conditions: $x(t_n)=x_n$ and $\dot{x}(t_n)=v_n$.
I have some references:
The exact solution of the equation of motion for the anharmonic oscillator
Elliptic Functions and Solutions of Certain Nonlinear Equations (Page. 526)
It seems that I should get a Jacobi elliptic function cn, if $\omega_0^2>0$ and $\alpha>0$. But I don't know how to derive to this solution......You can see how I solve it as the following:
Multiply $\dot{x}$ on the whole equation: $$\ddot{x}\dot{x}+\omega_0^2x\dot{x}+\alpha x^3\dot{x}=0$$ Integrate over the time, with the initial conditions: $$\frac{1}{2}(\dot{x})^2+\frac{\omega_0^2}{2}x^2+\frac{\alpha}{4}x^4=\frac{v_n^2}{2}+\frac{\omega_0^2}{2}x_n^2+\frac{\alpha}{4}x_n^4$$ or $$(\dot{x})^2+\omega_0^2 x^2+\frac{\alpha}{2}x^4=C_n^2$$ with $C_n=\sqrt{v_n^2+\omega_0^2 x_n^2+\frac{\alpha}{2}x_n^4}$.
Then, $$\dot{x}=\frac{dx}{dt}=\sqrt{C_n^2-\omega_0^2 x^2-\frac{\alpha}{2}x^4}$$ $\Rightarrow dt=\frac{dx}{\sqrt{C_n^2-\omega_0^2 x^2-\frac{\alpha}{2}x^4}}$ $$t=\int_0^x\frac{du}{\sqrt{C_n^2-\omega_0^2 u^2-\frac{\alpha}{2}u^4}}+D$$ Then, $$C_n t=\int_0^x\frac{du}{\sqrt{1-\left(\frac{\omega_0}{C_n}\right)^2u^2-\frac{\alpha}{2C_n^2}u^4}}+D$$ Note that $$sn^{-1}(x)=\int_0^x\frac{du}{\sqrt{(1-u^2)(1-k^2u^2)}}=\int_0^x\frac{du}{\sqrt{1-(1+k^2)u^2+k^2u^4}}$$ If we compare the above equation ($C_nt=...$) with the inverse $sn(x)$, then the parameter $k$ should satisfy $$k^2=-\frac{\alpha}{2C_n^2},\; 1+k^2=\left(\frac{\omega_0}{C_n}\right)^2$$ But $k$ cannot satisfy these two equations at the same time, so I think there's something wrong......(Can someone find the bug?)
If we have dealt with the above issue, then the solution would be this form: $$x(t)=x_n+a\cdot sn[C_n(t-t_n)]$$ $$p=m\dot{x}=a\cdot mC_n\cdot cn[C_n(t-t_n)]\cdot dn[C_n(t-t_n)]$$ (Note that $sn(0)=0$, and $cn(0)=dn(0)=1$.)
With the initial condition: $\dot{x}(t_n)=v_n$ or $\dot{p}(t_n)=m\dot{x}(t_n)=p_n$, we can find $a=p_n/(mC_n)$.
Finally, I get $$x=x_n+\frac{p_n}{mC_n}sn[C_n(t-t_n)]$$ $$p=p_n\cdot cn[C_n(t-t_n)]\cdot dn[C_n(t-t_n)]$$
There must be something wrong......Can someone correct these?