Consider the manifold $(\Bbb R^2, \mathrm{can})$.
I derived a PDE on this manifold:
$$r^2 \frac{\partial ^3\Psi(r,s)}{\partial r^3}=s^2 \frac{\partial \Psi(r,s)}{\partial s}$$
and found a particular analytic solution:
$$\Psi(r,s)=2 \sqrt{\frac{r}{s}}K_1(2\sqrt{r s}) $$
for $K_1$ being a modified Bessel function of the second kind. This solution can be verified with Mathematica software.
Letting $r$ act as a parameter that indexes the "leaves" we can view this particular solution as a foliation of a submanifold of the positive quadrant of the Euclidean plane, and then it can be seen that as $r \to 0$ then $\Psi(r,s) \to 1/s.$
So there is a "wall" at this hyperbola $1/s$ that I can't extend my foliation past. And I would like to extend the particular solution to all of the strictly positive quadrant.
I thought about this, and it seems to me that there is enough information to get the extension because the PDE is defined on the entire manifold and the metric is globally defined.
But letting $r$ go negative results in complex numbers.
Taking $\frac{1}{\Psi(r,1/s)}$ seems to solve the problem because it is a foliation and also analytic and essentially extends the foliation to the rest of the strictly positive quadrant. I'm not sure that it solves the same PDE however. My preliminary checks suggested that it does not.
So I suppose my questions are:
Does taking $\frac{1}{\Psi(r,1/s)}$ qualify as an analytic extension of the foliation defined through $\Psi(r,s)?$ I think yes, and I can think of other analytic foliations that get the job done. So is there a way to decide whether $\frac{1}{\Psi(r,1/s)}$ is a unique extension in some respect?
How do you define a vector field whose flow lines are $\Psi(r,s)$ and $\frac{1}{\Psi(r,1/s)}?$