Langevin equation in external potential

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The discussion centers on solving the Langevin equation in an external potential, specifically the equation m du/dt = -γu - dV/dx + A(t). The user seeks to derive ensemble averages <Δu> and <Δu^2>, but struggles with transforming the equation back to the time domain due to undefined potential functions. An analytical solution for momentum in terms of the potentials is deemed unattainable, necessitating numerical methods or simulations instead. The user is advised that while they can average the expression for Δu, squaring it for the second moment is not straightforward. Ultimately, numerical approaches are recommended for further analysis.
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I am trying to solve the following equation.

m\frac{du}{dt}=-\gamma u -\frac{dV}{dx} + A(t)
Where u is the momentum, x is position, V is the external potential dependent on position, and A is the random stochastic force dependent on time. There is no initial condition in this problem, but probably limiting conditions to get the constants of integration.

I am basically after the following quantities.

&lt;\bigtriangleup u&gt; and &lt;\bigtriangleup u^2&gt;
Where the brackets denote the ensemble average.

I tried solving the equation for u(x,t), by Fourier transforming both sides into frequency domain, but didn't know how to get it back to the time domain since the potential functions are not defined. Separation of variables would not work due to the potentials as well.

So my question is, can I get an analytical solution for the momentum in terms of the potentials? If not, will the following work?

Attempt sort of... for &lt;\bigtriangleup u&gt;

the equation at the top can be written as

\lim_{\bigtriangleup t\to 0 } m\frac{\bigtriangleup u}{\bigtriangleup t} = -\gamma u(t +\bigtriangleup t) -\frac{\partial V}{\partial x} + A(t + \bigtriangleup t)

expanding the right side around t and multiplying both sides by delta t we get

\lim_{\bigtriangleup t\to 0 } m\bigtriangleup u = -\gamma u(t)\bigtriangleup t -\frac{\partial V}{\partial x}\bigtriangleup t + A(t)\bigtriangleup t + O (\bigtriangleup t^2)

I can then take the average of this, where the average of A(t) is assumed to be zero.

For the second moment, I can't just take the expression for delta t and square it can I? Thanks.

Homework Equations





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Yes, you can square the expression for $\Delta t$ to find the second moment. However, it is not possible to get an analytical solution for the momentum in terms of the potentials. In order to determine the momentum given the potentials, one must solve the equation numerically or by simulation.
 

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