How to show that applying Hamiltonian dynamics recovers the original wave equation

Question

Problem

Consider the wave equation:

$$ \frac{\partial^2 u}{\partial t^2} = c^2 \frac{\partial^2 u}{\partial x^2}\tag{1}$$

with $ u = u(t, x)$ over domain $x \in [0, l] = \Omega$. This can be represented as a Hamiltonian system with generalized coordinates $p = \dot{u}$ and $q = u$.

Then the Hamiltonian is defined as:
$$ \mathcal{H}(p, q) = \int_{\Omega}\left[ \frac{1}{2} p^2 + \frac{1}{2}c^2 \left(\frac{\partial q}{\partial x}\right)^2\right] \; dx \tag{2}$$

with dynamics $$\dot{q} = \frac{\delta \mathcal{H}}{\delta p }\quad\text{and}\quad\dot{p} = – \frac{\delta \mathcal{H}}{\delta q }.\tag{3}$$
(See 6.1 in https://arxiv.org/abs/1407.6118)

I am trying to recover the original form (1) using the Hamiltonian formulation above.

Here is what I've tried:

I have a background in mathematics, but less so in physics (other than an intro sequence that did not cover Hamiltonian mechanics). I am currently a graduate student trying to understand this derivation for my research.

I originally started by treating the dynamics for $\dot{q}$ and $\dot{p}$ as standard partial derivatives, but ran into a couple of issues; at which point I tried using the concept of the total variation, and was starting from this definition $$\delta \mathcal{H} = \frac{\partial \mathcal{H}}{\partial p} \delta p + \frac{\partial \mathcal{H}}{\partial q} \delta q. \tag{4}$$

But I seem to be missing something and am getting stuck:

1. the integral over $\Omega$ is still present in my derivation so it seems like the dynamics of $\dot{q}$ require integrating over the entire domain $\Omega$ – this seems wrong.
2. When taking the derivative with respect to $q$, I obtain the term $\frac{\partial}{\partial q}\left( \left(\frac{\partial q}{\partial x}\right)^2\right)$ which I think should yield $ 2\left(\frac{\partial q}{\partial x}\right) \cdot \frac{\partial}{\partial q}\left(\frac{\partial q}{\partial x}\right)$. However I don't know exactly how to evaluate $\frac{\partial}{\partial q}\left(\frac{\partial q}{\partial x}\right)$ and this doesn't seem to lead to the right result.

Assistance with showing the derivation, or even just recommended readings/references would be greatly appreciated.

Answer 1

$$ \frac{\partial^2 u}{\partial t^2} = c^2 \frac{\partial^2 u}{\partial x^2}\tag{1}$$

with $ u = u(t, x)$ over domain $x \in [0, l] = \Omega$. This can be represented as a Hamiltonian system with generalized coordinates $p = \dot{u}$ and $q = u$.

the Hamiltonian is defined as: $$ \mathcal{H}(p, q) = \int_{\Omega}\left[ \frac{1}{2} p^2 + \frac{1}{2}c^2 \left(\frac{\partial q}{\partial x}\right)^2\right] \; dx \tag{2}$$

with dynamics $$\dot{q} = \frac{\delta \mathcal{H}}{\delta p }\quad\text{and}\quad\dot{p} = - \frac{\delta \mathcal{H}}{\delta q }.\tag{3}$$

I am trying to recover the original form (1) using the Hamiltonian formulation above.

To orient yourself, recall the standard Hamiltonian equations of motion for a system with $N$ coordinates labeled by index $i$: $$ \frac{\partial H}{\partial p_i} = \dot q_i $$ $$ \frac{\partial H}{\partial q_i} = -\dot p_i\;. $$

When we start to treat the index $i$ as continuous rather than discrete, we switch to a continuous label $x$ and we switch from a sum over $i$ to an integral over $x$. We also switch to saying that $H$ is a functional of $q(x)$ rather than a function of the $q_i$, etc.

The continuum generalization of Hamilton's equations of motion are: $$ \frac{\partial H}{\partial p(x)} = \dot q(x) $$ and $$ \frac{\partial H}{\partial q(x)} = -\dot p(x)\;, $$ where I am going to keep using the $\partial$ notation rather than the $\delta$ notation, even for functional derivative, but you can use whatever notation you like.

When we take the "functional derivative" we consider what happens to the functional $H$ when the function $q(x)$ is changed to $q(x)+\delta q(x)$ and we define the functional derivative $\frac{\partial H}{\partial q(x)}$ as: $$ \delta H \equiv \int \frac{\partial H}{\partial q(x)}\delta q(x) dx + O(\delta q^2) \tag{5} $$

For example, in your Hamiltonian, let $q \to q+\delta q$. Then $$ H \to H + \int dx c^2\frac{\partial q}{\partial x}\frac{\partial \delta q}{\partial x} + O(\delta q^2) \tag{6} $$ $$ = H - \int dx c^2\frac{\partial^2 q}{\partial x^2}\delta q + O(\delta q^2)\;, \tag{7} $$ which shows that the functional derivative wrt q(x) is: $$ \frac{\partial H}{\partial q(x)} = -c^2\frac{\partial^2 q}{\partial x^2} $$

Similarly: $$ \frac{\partial H}{\partial p(x)} = p(x) \equiv \dot q(x) $$

Then, also using Hamilton's canonical equations, we see that: $$ -c^2\frac{\partial^2 q}{\partial x^2} = \frac{\partial H}{\partial q(x)} = -\dot p(x) = -\ddot q(x) $$

Or, cancelling the minus sign and switching back to $u=q$: $$ c^2 \frac{\partial^2 u}{\partial x^2} = \ddot u $$