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  Non-averaged pertubative solution to Hamilton-Jacobi equation?

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When perturbing a fully integrable system whose action-angle variables you known, you can find the response to that perturbation by averaging the influence of the perturbation over the angle variables and finding new, deformed action coordinates that will incorporate the influence of the perturbation up to the case of resonances (occurs when the perturbation is periodic over the angle coordinate at some point). This is called Poincaré-Lindstedt method, or in some other cases a Lie-series or Linstedt-Deprit expansion.

However, I am interested in a problem where it is possible to perturbatively solve the Hamilton-Jacobi equation without averaging.


For example, let's say I have the following Hamiltonian modeling the motion of a relativistic particle in $1+1$ dimension $t,x$, that had an additional degree of freedom $y$ activated

$$H = -p_t^2 + p_x^2 + \epsilon (p_t + p_x + \sin(x) \sin(k y)(-p_t^2 + p_x^2 +1))p_y  + \mathcal{O}(\epsilon^2)$$

with $\epsilon$ and $k$ some constants. Now I attempt to solve the respective Hamilton-Jacobi equation iteratively in ascending order in $\epsilon$. Furthermore, the equation is solved on the hypersurface $H = -1$ (proper-time parametrization). The zeroth order reads

$$-S_{(0),t}^2 + S_{(0),x}^2 = -1$$

And it is fully separable as $S_{(0)} = x p_x - t\sqrt{1 + p_x^2} + C$ (choosing the physical root of particles travelling forward in time). Now let us pass to the leading-order perturbation. We assume that the new action is $S = S_{(0)} + S_{(1)}$ where $S_{(1)}$ is $\mathcal{O}(\epsilon)$. Then we obtain

$$-2S_{(0),t}S_{(1),t} +2 S_{(0),x}S_{(1),x} + \epsilon (S_{(0),t} + S_{(0),x})S_{(1),y} = 0$$

This can be easily solved  and after the dust settles, we have the new $\mathcal{O}(\epsilon)$-valid action

$$S = xp_x - t\sqrt{1 + \epsilon(\sqrt{1 + p_x^2} + p_x)p_y + p_x^2} + y p_y  + C \;\;\;\; (*)$$

Since the perturbative Hamilton-Jacobi problem is fully separable, we might be lead to the impression that there will be no issue with resonances and this simply presents us with the full solution for orbits at $O(\varepsilon)$ order.

However, if you take a look at the corresponding equations of motion, you will obtain

$$\dot{x} = p_x + \epsilon(1 + \sin(x) \sin(k y) p_x) p_y$$

$$\dot{t} = -p_t + \epsilon(1 - \sin(x) \sin(k y) p_t) p_y$$

$$\dot{y} = \epsilon(p_t + p_x - \sin(x) \sin(k y) (-p_t^2 + p_x^2)) = \epsilon (p_t + p_x) + \mathcal{O}(\epsilon^2)$$

Now we see that even though the perturbative solution to the Hamilton-Jacobi equation was fully separable, the trajectories themselves are not!

On the other hand, we can find a zeroth order solution for $y$, plug it into the equations, average over $y \in (0,2\pi/k)$ and we will get the average equations

$$\dot{\bar{x}} = p_x + \epsilon p_y$$

$$\dot{\bar{t}} = -p_t + \epsilon p_y$$

However, no such averaging will converge if $k$ is a rational number because the $\sin(x)\sin(k y)$ term will not average out to zero over any period.


Now comes my question - is the perturbative (non-averaged) solution of the Hamiltoni-Jacobi equation $(*)$ valid even around resonances? What is happening in my example? Do you know any similar examples in the literature?

asked Oct 29 in Theoretical Physics by Void (1,570 points) [ revision history ]
edited Oct 29 by Void

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