Homework 3 Problem 6

#### Problem

Consider the mechanical system with configuration manifold $Q = R$ and Hamiltonian $H:T^{\star}Q \to R : (q, p) : (q, p) \mapsto p^2$. Verify explicitly that flow along the corresponding Hamiltonian vector field is area-preserving.

#### Solution

To show the Hamiltonian flow $\Phi_t$ is volume preserving we begin by finding the Hamiltonian vector field $X_H$.

(1)
\begin{align} &dH = 2p \mathbf{d}p \\ &i_{X_H} \Omega = \mathbf{d}q \wedge \mathbf{d}p (X_H, \cdot) = \mathbf{d}q(X_H) \cdot \mathbf{d}p(\cdot) - \mathbf{d}p(X_H) \cdot \mathbf{d}q(\cdot)\\ &\implies X_H = 2p \frac{\partial}{\partial q} \end{align}

This vector field can be interpreted as the following equations of motion on $T^{\star}Q$,

(2)
\begin{align} &\dot{q} = 2p\\ &\dot{p} = 0. \end{align}

Solving this linear differential equation yields

(3)
\begin{align} &q(t) = 2p_0t + q_0\\ &p(t) = p_0. \end{align}

Hence, the flow operator $\Phi_t$ is

(4)
\begin{align} \Phi_t(q,p) : T^{\star}Q \to T^{\star}Q : (q,p) \mapsto (2pt + q, p). \end{align}

Finally, we are ready to solve the original problem. Let $A$ be a region in $T^{\star}Q$. The area of of $A$ is

(5)
\begin{align} \int_A \mathbf{d}q \mathbf{d}p \end{align}

Instead of finding the area of $\Phi_t(A)$ by integrating over $\Phi_t(A)$ (i.e. $\int_{\Phi_t(A)} \mathbf{d}q \mathbf{d}p$), we can compute it by integrating over $A$ with the assistance of $\Phi_t$ and integration_by_substitution.

(6)
\begin{align} \int_{\Phi_t(A)} \mathbf{d}q \mathbf{d}p = \int_A \mathtt{det}(D\Phi_t) \mathbf{d}q \mathbf{d}p \end{align}

First we compute $\mathtt{det}(D\Phi_t)$

(7)
\begin{align} &D\Phi_t = \left[\begin{matrix} 1 & 2t \\ 0 & 1\end{matrix}\right]\\ &\implies \mathtt{det}(D\Phi_t) = 1 \end{align}

Thus

(8)
\begin{align} \int_{\Phi_t(A)} \mathbf{d}q \mathbf{d}p = \int_A \mathtt{det}(D\Phi_t)\mathbf{d}q \mathbf{d}p = \int_A \mathbf{d}q \mathbf{d}p. \end{align}

Volume is preserved, and Hyrule has been saved.

[kcrane] Here's another solution that does a couple things differently:

1. it walks you through the mechanics of finding the Hamiltonian vector field a lot more explicitly
2. it does a much simpler check that this vector field is area preserving by simply checking that its divergence is zero everywhere. I.e., if a smooth vector field u is divergence-free then flows through it preserve volume. (You can find a proof of this fact in many places, including Chorin and Marsden, A Mathematical Introduction to Fluid Mechanics, pp. 8-11.)

Given a Hamiltonian vector field $H$, the Hamiltonian vector field is defined as

(9)
\begin{align} i_{X_H} \Omega = \mathrm{d}H \end{align}

i.e., the Hamiltonian vector field is the vector field which, when fed to the 2-form $\Omega$, becomes a thing that behaves exactly like the exterior derivative of H. (What is $\Omega?$ it's the canonical symplectic 2-form $\Omega = dq^dp$.) So to figure out what $X_H$ is, we're going to

1. compute $i_{X_H}\Omega$
2. compute $dH$
3. figure out what $X_H$ must be so that the left hand side looks like the right hand side!

Here we go…

We were given a Hamiltonian $H=p^2$. Since $H$ is a 0-form, its exterior derivative is just the differential of $p^2$

(10)
\begin{align} \mathrm{d}H = \mathrm{d}(p^2) = \frac{\partial}{\partial q}p^2 \mathrm{d}q + \frac{\partial}{\partial p}p^2 \mathrm{d}p = 2p \mathrm{d}p \end{align}

We can compute the contraction of $\Omega$ with $X_H$ by imagining we have some additional vector field $v$, doing the usual evaluation of a 2-form, and then figuring out what 1-form would produce the same results if it were fed $v$. First, let's give names to the components of $X_H$ and $v$:

(11)
\begin{align} X_H = x_q \frac{\partial}{\partial q} + x_p \frac{\partial}{\partial p} \end{align}
(12)
\begin{align} v = v_q \frac{\partial}{\partial q} + v_p \frac{\partial}{\partial p} \end{align}

Now evaluate $\Omega$:

(13)
\begin{align} \Omega(X_H,v) = \mathrm{d}q(X_H) \mathrm{d}p(v) - \mathrm{d}q(v) \mathrm{d}p(X_H) = x_q v_p - v_q x_p. \end{align}

So $i_{X_H}$ must be

(14)
\begin{align} x_q \mathrm{d}p - x_p \mathrm{d}q. \end{align}

(Why? Think about what happens when you feed $v$ to this form.)

All we've done at this point is written out the equation that defines our Hamiltonian vector field explicitly in terms of its components $x_q$ and $x_p$. I.e., this equation

(15)
\begin{align} i_{X_H} \Omega = \mathrm{d}H \end{align}

became this one

(16)
\begin{align} x_q \mathrm{d}p - x_p \mathrm{d}q = 2p \mathrm{d}p. \end{align}

It should be clear that the only way this equality can be satisfied is if $x_q=2p$ and $x_p=0$. Since we've figured out what the components are, we've figured out what the vector field looks like!

(17)
\begin{align} X_H = 2p \frac{\partial}{\partial q}. \end{align}

How do we check that this vector field is volume-preserving? Easy! As noted at the beginning, we can just check that it's divergence-free, i.e., that

(18)
\begin{align} \nabla \cdot X_H = 0, \end{align}

which turns out to be true:

(19)
\begin{align} \left( \frac{\partial}{\partial q}, \frac{\partial}{\partial p} \right) \cdot \left( 2p, 0 \right) = \frac{\partial}{\partial q} 2p + \frac{\partial}{\partial p} 0 = 0. \end{align}
page revision: 19, last edited: 09 Apr 2007 16:15