3
$\begingroup$

I am trying to prove the following:

Given that $f \in C^1(E) $ where E is a open simply connected subsets of the plane. Show that the system $\dot x=f(x)$ is a hamiltonian if and only if $\nabla \cdot f=0$ for all $x \in E.$

So $\nabla \cdot f= \partial f/ \partial xe_x +\partial f/ \partial y e_y =0$. I am confused as to where to go from here in order to prove it is a hamiltonian...

$\endgroup$
  • $\begingroup$ can you state your definition of what it means for a system to be Hamiltonian? $\endgroup$ – BaronVT Nov 19 '13 at 18:41
  • $\begingroup$ $\dot x= \partial H/ \partial y$ and $\dot y =- \partial H/ \partial x$ $\endgroup$ – user75514 Nov 19 '13 at 18:48
5
$\begingroup$

Ok, so maybe there's some notational confusion here; in particular, since you write $\nabla \cdot f$, this seems to indicate that $f$ is a vector field, and $x$ is a vector. So maybe if we switch from $x$ to $(x,y)$, and write $f = (f_x(x,y),f_y(x,y))$ things become clearer. (the subscripts are to indicate components, not derivatives)

Then the statement is:

Show that $(\dot x,\dot y) = (f_x(x,y),f_y(x,y))$ is Hamiltonian if and only if $\nabla \cdot f = 0$ for all $(x,y) \in E$.

Then, we want to show that $f_x = \frac{\partial H}{\partial y}$ and $f_y = -\frac{\partial H}{\partial x} $ for some $H$ if and only if $\nabla \cdot f = 0$.

One direction is clear: if the system is Hamiltonian, then $$ \nabla \cdot f = \partial_x f_x + \partial_y f_y = \partial_x \left(\frac{\partial H}{\partial y}\right) + \partial_y\left(-\frac{\partial H}{\partial x}\right) = 0.$$

Ok, to spell out the suggestion I made in a comment below:

Fix some $(x_0,y_0) \in E$ and define $H(x_0,y_0) = 0$ ($H$ is only determined up to a constant, so this is fine). Now, to define $H(x,y)$, let $\gamma : [0,1] \to E$ be a path such that $\gamma(0) = (x_0,y_0)$ and $\gamma(1) = (x,y)$, and define

$$ H(x,y) = \oint_\gamma -f_y\,dx + f_x\,dy $$

this will automatically satisfy the conditions $\frac{\partial H}{\partial y} = f_x$ etc. What remains to be shown is that $H$ is well-defined - i.e. that this definition depends only on the endpoint $(x,y)$ and not the particular choice of path $\gamma$.

So, let $\gamma_1$ and $\gamma_2$ be two such paths, and let $C$ be the path that goes from $(x_0,y_0)$ to $(x,y)$ along $\gamma_1$, and then comes back to $(x_0,y_0)$ along $\gamma_2$ (in the reverse direction). Then $C$ is a closed curve bounding some region $D \subset E$. By Green's theorem, $$ \oint_C -f_y\,dx + f_x\,dy = \int\int_D \frac{\partial (f_x)}{\partial x} -\frac{\partial (- f_y)}{\partial y}\,dxdy = \int\int_D \nabla\cdot f \,dxdy = 0 $$

showing that $$ \oint_{\gamma_1} -f_y\,dx + f_x\,dy = \oint_{\gamma_2} -f_y\,dx + f_x\,dy $$ and we're done.

$\endgroup$
  • $\begingroup$ Yep, sorry I was vague on the clarification. I got that direction as well. It's the reverse direction that is stumping me. I think there is a way to construct the Hamiltonian system from the divergence equation by some sort of integral. $\endgroup$ – user75514 Nov 19 '13 at 19:14
  • $\begingroup$ Ok, I think I figured it out, but I'm on my mobile; I'll write more when I'm back at a computer. Basically, fix some point in E, set H to 0 there, and then use a path integral starting there to define H at other points. You can use Green's theorem and the vanishing divergence to show that definition is independent of the path chosen. $\endgroup$ – BaronVT Nov 19 '13 at 20:13
  • $\begingroup$ How do I use a path integral to define H at other points? $\endgroup$ – user75514 Nov 19 '13 at 20:28
  • $\begingroup$ @user75514 Ok, I've written out the details now. $\endgroup$ – BaronVT Nov 19 '13 at 20:47
  • 1
    $\begingroup$ Thanks! It was very clear! $\endgroup$ – user75514 Nov 19 '13 at 21:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.