I have the following PDE:
$$\frac{\partial\psi}{\partial t}=i\frac{\partial^2 \psi}{\partial x^2}$$
Consider $\psi(x,t)$ the solution for a given initial condition $\psi_0(x)$. I am interested in finding the isotimic curves for $|\psi(x,t)|^2$ given $\psi_0$. To be more precise, starting from a point $P_0=(x_0,t=0)$ where the initial condition is $\psi_0(x_0)$ I want to be able to say something about the curve that passes through $P_0$ and for which it is true that $|\psi(x(t), t)| = |\psi_0(x_0)|$.
My attempt so far is the following. Given a curve $\Gamma$ that satisfies my condition, then $\nabla_{x,t}|\psi(\hat x,\hat t)|^2\bot \Gamma$ for $(\hat x, \hat t)\in \Gamma$. From this observation and using the PDE I can compute the direction of the tangent to that curve for $t=0$. I'm not going into details, it's just a manipulation of the PDE to get $\partial_t |\psi(x,t=0)|^2$. This however is not enough as I wish to characterize the curve for more than $t=0$.
Some issues I encountered so far:
- I want to be able to characterize the curve $\Gamma$ without solving the PDE (not even with a numeric scheme), otherwise my problem would be irrelevant.
- The gradient gives only the direction of the normal to $\Gamma$ but it sais nothing of the numeric value for the "curvature" of $\Gamma$. Is there a way to link those?
It took me some time to come with an answer to this question and i'll post it here in case someone else stumbles into something like this.
I consider the notation $F(x,t) = |\psi(x,t)|^2$ for simplicity.
The curves $\Gamma$ defined in the question are isotimic, thus $F$ does not change its value on $\Gamma$. This means that the gradient in the direction tangent to $\Gamma$ in every point on that curve is orthogonal with $\nabla_{x,t}F$, or in other words, the scalar product of the tangent and $\nabla_{x,t}F$ is $0$. I'll write this:
$$ \nabla_{x,t}F\cdot T_{\Gamma}=0 $$
Now, in order to actually do something with this relation I went with the following particular case. Consider the curves defined as $\Gamma(g(t), t)$. Although not all the isotimic curves are defined $\forall t\in \mathbb{R}$, I can take some finite interval for $t$ and for $x$ on which my choice can be valid. Under this form, $T_{\Gamma}=(\partial_tg(t), 1)$. Substitute this in our main formula and we get:
$$ \partial_x F \cdot \partial_t g(t) + \partial_t F=0 $$ $$ \partial_t g(t) = -\frac{\partial_t F}{\partial_x F} $$
Ok, what now? Well, if we assume $g(t)$ is analytic, we can write it as a Taylor series:
$$ g(t)=\sum_{n=0}^{\infty}\frac{t^n}{n^!}\partial_t^n g(t=0) $$
Now we have a formula that allows us to compute the terms $\partial_t^n g(t=0)$ up to what precision we want, and after that we can use them in the Taylor series to compute $g(t)$. Plug $g(t)$ into $\Gamma$ and it should be considered solved, given the particular choices I made along the way.