# Calculating $p$ and $q$ from the Hamiltonian $H = p^2q^2$ [closed]

Suppose the Hamiltonian is: $H = p^2q^2$. We want to obtain the solution of equations of motion from the Hamiltonian as $p(t)$ and $q(t)$.

How can I go about solving this problem?

I have obtained the equations of motion as follows

$$\dfrac{\partial H}{\partial q} = -\dot{p} = 2p^2q$$

and

$$\dfrac{\partial H}{\partial p} = \dot{q} = 2pq^2$$

I' don't know how I should proceed further.

Edit: The answer is given as

$$p(t) = Be^{-2At}, \qquad q(t) = \dfrac{A}{B}e^{2At}$$

## closed as off-topic by ACuriousMind♦, Jon Custer, user36790, Floris, WolpertingerOct 12 '16 at 19:06

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• Hint: $pq$ is a constant. Why? There are at least two ways of showing this from what you have written above. – Phoenix87 Oct 10 '16 at 8:52

## 1 Answer

Just solve as you normally would solve coupled differential equations. Say you want to solve for $p$; differentiate the equation $-\frac{\dot{p}}{2p^2} = q$. Plug the value of $\dot{q}$ from the second equation; you will have a $q$ term which you can easily substitute in terms of $p$ and $\dot{p}$ from the first equation itself. This is the lengthier method. You can definitely use the comment by 'phoenix'

• Thank you for the quick reply. I had totally forgotten the concept of coupled differential equations. I'll be able quite easily solve in both the ways. – Marcus Oct 10 '16 at 13:09
• Yes. If you want to know how $pq$ just as a side note, notice that the Hamiltonian itself does not depend on time explicitly, hence $H=p^2q^2$ is a constant of the motion, ie $pq$ is (also) a constant. Or you can see from $\frac{\dot{p}}{\dot{q}}$ = $\frac{dp}{dq}$=$-\frac{p}{q}$ and take it from there – Prasad Mani Oct 10 '16 at 13:45