7
$\begingroup$

Let $\mathcal{P}$ be a "small particle" trapped in a $n$-dimensional potential. We will assume the dynamics of $\mathcal{P}$ are well described by the stochastic differential equation \begin{align*} \mathrm{d} x_t & = v_t \mathrm{d}t \\ m\mathrm{d} v_t& = -\nabla \Psi(x_t) \mathrm{d}t -\gamma v_t \mathrm{d}t + \sigma \mathrm{d} W_t \end{align*} where $x_t$ and $v_t$ are the position and velocity vectors, $\Psi(x)$ is a bell-shaped potential with a single minimum at $\Psi(0)$ and $W_t$ is a vector Wiener process. Now define the energy of the particle as \begin{align}\label{eq:energy} h(x_t, v_t) = \Psi(x_t) + \frac{1}{2}mv_t^2. \end{align} I am interested in calculating the time it takes for $\mathcal{P}$ to reach a certain energy $r$, i.e., the stopping time $$ \tau_h = \inf\{ t \geq 0: h(x_t, v_t) = r \}. $$ Calculating $\mathbb{E}(\tau_h)$ would be enough for now. So far I have tried calculating \begin{align}\label{eq:en_evolution} h(x_t, v_t) = h(x_0, v_0) + \int_0^t \mathrm{d} h(x_s, v_s), \end{align} and then, by taking expected values, I hoped I'd be able to solve by $\mathbb{E}(\tau_h)$. With the Wiener process I got the right and well-known result of the exit time from a ball of radius $r$ (this is, in fact, a naive way of using Dynkin's formula), but with the process I described before I didn't get too far. Assuming the initial conditions are distributed according to the invariant (Gibbs) measure, one gets, of course, that $$ \mathbb{E}[h(x_t, v_t)] = \mathbb{E}[\Psi(x_t)] + \frac{1}{2}k_B T, $$ which is true, and reassuring, but not very useful, since we lost $\mathbb{E}(\tau_h)$.

I'm not especially used to calculations like this, so any suggestions of "tricks" and/or common techniques for calculating hitting times would be appreciated. I would also be grateful for any references where similar exit time calculations are performed (for non trivial processes).

$\endgroup$
2
$\begingroup$

Here is a common method used for calculating exit time/stopping times of stochastic processes.

We need to find a non-random function $u(h)$ which is a function of the stochastic process on which the stopping criteria is applied. We will use Itô's lemma to find a differential equation satisfied by $u$ and the stopping criteria will be the boundary condition for this function.

For the stochastic process specified in the question, we can apply Itô's lemma to get,

$du(h_t) = u_h(h_t)dh_t + \frac{1}{2}u_{hh}(h_t)d[h,h]_t$

we can now write the dynamic of $h_t$ as following,

\begin{equation} h_t = \Psi(x_t) + \frac{1}{2}mv_t^2 \\ dh_t = \nabla \Psi(x_t)dx_t + mv_tdv_t + \frac{1}{2}md[v,v]_t \\ dh_t= v_t\nabla \Psi(x_t)dt + v_t(-\nabla \Psi(x_t)dt - \gamma v_tdt+\sigma dW_t) + \frac{\sigma^2}{2m}dt\\ d[h,h]_t = \sigma^2v_t^2dt \end{equation}

Plugging the above into the differential of $u$, we get

\begin{equation} du = u_h(-\gamma v_t^2+\frac{\sigma^2}{2m})dt + u_hv_t\sigma dW_t+ \frac{1}2 u_{hh}\sigma^2v_t^2dt \end{equation}

Integrating the above between $t=0$ and $t=\tau$, we get,

\begin{equation} u(h(\tau)) = u(h(0)) + \int_0^\tau{\left(u_h(-\gamma v_t^2+\frac{\sigma^2}{2m})+u_{hh}\sigma^2v_t^2\right)dt} + \int_0^\tau{u_hv_t\sigma dW_t} \end{equation}

Now $h(\tau) = r$ and if we can find $u(h)$ which satisfies the differential equation,

$u_h(-\gamma v^2+\frac{\sigma^2}{2m})+u_{hh}\sigma^2v^2=-1$,

and boundary condition, $u(r)=0$. Then we can take expectation of the Stochastic Integral equation and use the fact that expectation of a well-behaved Itô integral is $0$, and we will get

$\mathbb{E}[\tau] = u(h(x_0,v_0))$.

We can use the relation between $h$ and $v$ to eliminate $v$ from the ODE. Alternate formulation of the function $u$ could be as a function of $(x,v)$ which will lead to a PDE based on similar application of Itô's lemma.

Special Case: If we can assume some relation between $\gamma, \sigma$ and $m$, then the established ODE could be simplified to get nice results. For example if $\gamma = 0$, then $u(h) = \frac{2m}{\sigma^2}(r-h)$ satisfies the ODE and boundary condition. $\mathbb{E}[\tau]$ would be then equal to $\frac{2m}{\sigma^2}(r-h(x_0,v_0))$.

$\endgroup$

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.