I've given this Hamiltonian with one degree of freedom: $$ H(x,p)=\frac{p^2}{2}+\frac{\omega_0^2x^2}{2}+\lambda\left(\frac{p^2}{2}+\frac{\omega_{0}^2x^2}{2}\right)^2 $$ I need to find the general solution for the motion of this particle $(x(t),p(t))$. It's required to solve the equations of motion, not only find them. I showed that $f = p^2+\omega_{0}^2x^2 $ is time independent by computing: $$\{f,H\}=0$$ But I'm having a hard getting the final solution. I tried doing: $$ 2p\dot{p}+2\omega_{0}^2x\dot{x}=\dot{f}=0 $$ $$ \dot{p}=-\frac{\omega_{0}^2x\dot{x}}{p} $$ And then: $$ \newcommand{\pder}[2][]{\frac{\partial#1}{\partial#2}} \dot{p}=-\pder[H]{x} $$ Same for $\dot{x}$. Yet, I don't obtain nicer differential equations and I don't think it's the right approach. Could someone give me a hint?

  • $\begingroup$ Hi. Do you need to find the equations of motion or solve the equations of motion? Thanks. $\endgroup$ – Constantine Black Nov 22 '16 at 20:34
  • $\begingroup$ I need to solve them $\endgroup$ – user3903647 Nov 22 '16 at 20:34

I want simplification, so I'll take $\omega_0 = 1$. You can rescale $x$ and always get this kind of simplification, if $\omega_0$ is a constant. The Hamiltonian then becomes

$$H(x,p) = \frac{p^2}{2} + \frac{q^2}{2} + \lambda \left( \frac{p^2}{2} + \frac{q^2}{2} \right)^2 \, .$$

Hamilton's equations are given by

\begin{align} \begin{cases} \dot{q} &= p + 2 \lambda p \left( \frac{p^2}{2} + \frac{q^2}{2} \right) \\ \dot{p} &= -q - 2 \lambda q \left( \frac{p^2}{2} + \frac{q^2}{2} \right) \end{cases} \quad \Rightarrow \quad \begin{cases} \dot{q} &= p \left[1 + 2 \lambda \left( \frac{p^2}{2} + \frac{q^2}{2} \right) \right] \\ \dot{p} &= -q \left[1 + 2 \lambda \left( \frac{p^2}{2} + \frac{q^2}{2} \right) \right] \end{cases} \, . \end{align}

Assuming $q, \dot{p} \neq 0$, divide one equation by the other to get

$$ \frac{ \left( \frac{dq}{dt} \right)}{ \left(\frac{dp}{dt} \right)} = \frac{p \left[1 + 2 \lambda \left( \frac{p^2}{2} + \frac{q^2}{2} \right) \right]}{-q \left[1 + 2 \lambda \left( \frac{p^2}{2} + \frac{q^2}{2} \right) \right]} = -\frac{p}{q} \, .$$

The solution will, therefore, be the same as for the SHO, with the exception that the new frequency will increase as the orbit gets farther away from the origin.

As I believe this is homework, I'll give no further details. The problem is basically solved.

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