Exit time of a stochastic process defined by a SDE

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).

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,

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

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

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

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

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

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 $$\frac{2m}{\sigma^2}(r-h(x_0,v_0))$$.