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Overdamped Langevin dynamics:

$$ \frac{dx}{dt} = -\frac{1}{\gamma} V'(x) + \sqrt{\frac{2k_B T}{\gamma}} \eta(t) $$

Underdamped Langevin:

$$ m \frac{d^2x}{dt^2} = -\gamma \frac{dx}{dt} - V'(x) + \sqrt{2 \gamma k_B T} \eta(t) $$

where $\eta(t)$ is a random variable.

Why do we use Ito chain rule in the first case and standard chain rule in the second one to compute the time evolution of functions of $x(t)$?

Is it because the last one is a second order equation so the noise term in the solution is of order $dt^{3/2}$ and negligible?

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In the first case you have $$\mathrm{d}x_t = -\frac{1}{m\gamma}V'(x_t)\mathrm{d}t + \sqrt{\frac{2k_BT}{m^2\gamma}}dW_t$$ so indeed for $f(t, x)$ you would have by Ito $$\mathrm{d}f_t = \frac{\partial f}{\partial t}\mathrm{d}t + \frac{\partial f}{\partial x} \mathrm{d}x_t + \frac{1}{2}\frac{\partial^2f}{\partial x^2}(\mathrm{d}x_t)^2$$

As for the latter case, we have $$\mathrm{d}\mathbf{x}_t= A(\mathbf{x}_t)\mathrm{d}t + \sigma\mathrm{d}W_t$$ where $\mathbf{x}_t = (x_t, v_t)^\mathrm{T}$, $A(x_t) = \begin{pmatrix}0 & 1 \\ -\frac{V'(x_t)}{m} & -\frac{\gamma}{m}x_t\end{pmatrix}$, $\sigma = (0, \sqrt{2 \gamma k_BT}/m)^\mathrm{T}$.

Now $f(t, x)$ by Ito is: $$\mathrm{d}f_t = \frac{\partial f}{\partial t}\mathrm{d}t + \frac{\partial f}{\partial x} \mathrm{d}x_t + \frac{1}{2}\frac{\partial^2f}{\partial x^2}(\mathrm{d}x_t)^2$$ but $(\mathrm{d}x_t)^2 = 0$ (the noise is in increments to velocity, i.e. $g(t, v)$ would have the noise term in Ito).

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  • $\begingroup$ can you expand on why (dx-t)^2=0? $\endgroup$ – iacolippo Oct 2 '16 at 13:29
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    $\begingroup$ @iacolippo Because $\mathrm{d}x_t = v_t\mathrm{d}t$. $\endgroup$ – alarge Oct 2 '16 at 13:32

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