Let $M^2=(0,1)^2$. Recall that a chart is a diffeomorphism $\varphi:M^2 \to M^2$. Given a chart $\varphi:(M^2,g_0)\to (M^2,g_0)$ for $g_0$ the Euclidean metric, consider the curves $\varphi^{-1}(u,t)=\alpha_t(u)=e^{\frac{t}{\log u}}$ for $t>0$ and $u\in(0,1)$.
Does there exist a local isometry $F:M^2\to E^2$ for $E^2=(0,1)\times S^1$ s.t. for a sequence of fiber-wise mappings $\psi_{t_0}, \cdot\cdot\cdot,\psi_{t_f}:E^2\to {E^{2}}'$ we have $(\psi_{t_0}\circ\pi_{t_0})\circ \cdot\cdot\cdot\circ (\psi_{t_f} \circ \pi_{t_f})$ with $t_0t_f=1$ generates an evolution of $\alpha_t(u)$ under $t \frac{\partial}{\partial t} \alpha_t(u)=-u\frac{\partial}{\partial u}\alpha_t(u)$ with Cauchy data $\alpha_t(u)|_{t=t_0}$?
I think $(E^2,h_0)$ should have the metric $h_0=dr^2+ f^2(r)dx^2$. Then in order to get the projective requirements, we must determine the right conditions on $f^2(r)$. My best guess is that the radii on $E^2$ must asymptotically pinch at $p'=(1,0,0)$ and $q'=(0,1,1)$ where we have the limits on the base:
$$ \lim_{u \to 0^+} \alpha_t(u)=1 $$
$$ \lim_{u\to 1^-} \alpha_t(u)=0. $$
These limits hold $\forall t$ suggesting we should try to mirror this invariant on the total space and make it compatible with the projections onto the base. In order to answer the above question, we need $(\psi_{t_0}\circ\pi_{t_0})\circ \cdot\cdot\cdot\circ (\psi_{t_f} \circ \pi_{t_f})$ to fix points $p',q'$ and respectively$,p,q,$ $~\forall t$.
Intuitively all of this should work but I am having trouble capturing precisely that the sequence of fiber-wise mappings generates a "shadow" of curves that diffuse on the base space according to the pde. The part that is difficult is that on the base space we'll have a family of curves simultaneously diffusing backwards in time and forwards in time, but when we view this from $E^2$ we see that the fibers are simply moving circularly around the asymptotically pinched cylinder.
