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I would like to see if this idea has any applications:

So CR equations are given by:

$$ \frac{\partial u}{\partial x} =\frac{\partial v}{\partial y} ; \ \frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$$

And Hamilton equations are given by:

$$ \dot{p}=-\frac{\partial H}{\partial q} ; \ \dot{q}= \frac{\partial H}{\partial p}$$

Now these conjugate equations ask for an analogy between them, so I tried the next thing:

I am trying to find $u(p,q),v(p,q)$ s.t:

$$\dot{p}=-\frac{\partial H}{\partial q}=\frac{\partial u}{\partial p}=-\frac{\partial v}{\partial q}$$ $$\dot{q}= \frac{\partial H}{\partial p} = \frac{\partial u}{\partial q} =\frac{\partial v}{\partial p}$$

So we can see that we can find a complex-analytic representation of the hamiltonian, $H$:

$$f(p,q)=u(p,q) + \imath (H(p,q)+c)$$

The only restriction is for $u(p,q)$ which by integration we can find it.

Does this represnetation has any applications in mathematical physics or complex analysis? I guess it's already known.

Thanks in advance, Alan.

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Have a look at Section 11.1.1 of my lectures where I explain a relationship between the CR equations and the Hamilton equations which I found useful in many concrete investigations. – Liviu Nicolaescu Nov 28 '12 at 14:20
up vote 2 down vote accepted

Doesn't this only work if $H(p,q)$ is a harmonic function in the plane?

If that's the case, we have

$\{u,H\}=\frac{\partial H}{\partial p}\frac{\partial u}{\partial q}-\frac{\partial H}{\partial q}\frac{\partial u}{\partial p}=|\nabla H|^2=-|\nabla u|^2$

where $\{\cdot,\cdot\}$ is the Poisson bracket. Since $\frac{du}{dt}=\{u,H\}$ this tells us how fast $u$ is changing.

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