Deriving isotimic lines/surfaces of PDE solution 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?  

 A: 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.
