General Problem
I am interested in studying whether solutions to the Fokker-Planck equation:
\begin{equation} \tag{1}\label{fp} \frac{\partial p(x, t)}{\partial t} = \textrm{div}(-p(x, t)\nabla\log q(x)) + \frac{1}{2}\Delta p, \quad p(x, 0) = p_0 \end{equation}
(where $p_0$ and $q$ are probability densities), can intersect at some time $t^\star$, when varying the choice of $q$. More formally, I want to know if there exists a triplet $(q_1, q_2, t^\star)$ such that $p_1(x, t^\star) = p_2(x, t^\star)$, for all $x$, and where $p_1$ (resp. $p_2$) is the solution of \eqref{fp} with $q=q_1$ (resp. $q=q_2$), and $t^\star>0$.
I welcome any pointer to works/fields studying the intersection points of solutions of differential equations (a) belonging to some parametric family and (b) with the same initial conditions.
Simpler Setting
Looking first at simpler ordinary differential equations (ODEs), note that the following ODE: $$\tag{2}\label{ode1} \frac{\textrm{d}x}{\textrm{d}t} + ax = 0, \quad x(0) = 0, \, a \neq 0 $$ is such that no triplet $(a_1, a_2, t^\star)$ (with $a_1, a_2 \ne 0$, and $t^\star > 0$) verifies $x_1(t^\star)=x_2(t^\star)$, where $x_1$ (resp. $x_2$) is the solution to \eqref{ode1} with $a=a_1$ (resp. $a=a_2$).
On the other hand, for the ODE:
$$\tag{3}\label{ode2} \frac{\textrm{d}^2x}{\textrm{d}t^2} + k^2x = 0, \quad x(0) = 0, \, \frac{\textrm{d}x}{\textrm{d}t}\bigg \rvert_{t=0} = 1, \, k \in \mathbb N \setminus \{0\} $$ the triplet $(1, 2, 2\pi)$ verifies $x_1(2\pi) = x_2(2\pi)$, where $x_1$ (resp. $x_2$) is the solution to \eqref{ode2} with $k=1$ (resp. $k=2$).
As this simpler setting is related to my original problem, I would also welcome pointers to any work in this latter simpler setting.
PS: I added the tag Gradient Flows since solutions to \eqref{fp} are also solutions to the Gradient Flow of $KL(\cdot||q)$ with initial condition $p_0$.
(Partial answer, too long for a comment)
Note that your Fokker-Planck equation can be implicitly integrated with respect to time, i.e. $$ p(x,t) = p_0 + \int_0^t\left(\frac{1}{2}\Delta p(x,\tau)-\mathrm{div}(p(x,\tau)\nabla\log q(x))\right)\mathrm{d}\tau, $$ hence for two solutions $p_1,p_2$ : $$ \begin{array}{rcl} 0 &=& p_1(x,t^*)-p_2(x,t^*) \\ &=& \displaystyle \int_0^{t^*}\left(\frac{1}{2}\Delta(p_1(x,\tau)-p_2(x,\tau))-\mathrm{div}\left(p_1(x,\tau)\nabla\log q_1(x)-p_2(x,\tau)\nabla\log q_2(x)\right)\right)\mathrm{d}\tau \end{array} $$ which translates locally as $$ \frac{1}{2}\nabla(p_1-p_2) - p_1\nabla\log q_1-p_2\nabla\log q_2 = A, $$ where $A(x)$ is an incompressible vector field (i.e. $\mathrm{div}A=0$), since $\Delta = \mathrm{div}\,\mathrm{grad}$. It gives you a condition/criterion to check whether the two solutions can coincide or not; I have a feeling that this condition shouldn't be hard to satisfy because of the "gauge freedom" due to $A$.
Final note : if you relax your problem by admitting different initial/boundary conditions, then it becomes a lot more easier for the solutions to intersect; however, that is perhaps a too trivial result for what you need.