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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?
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  • $\begingroup$ Could you explain why you want to find these curves? I ask because the equation you've given is the Schrodinger equation: its $L^\infty$ norm is known to decay as $t^{-1/2}$ for $L^1$ initial data, so your curve cannot exist for all time (unless $\psi_0(x_0) = 0$: and even then, I'm highly skeptical). $\endgroup$ – Strants Jul 13 '18 at 15:42
  • $\begingroup$ I am aware that some curves are defined only for $t<t_max$ if they even are at all (this problems occurs when there are intensity local maxima and minima) but to answer the question: the problem that i'm working on is related to characterize the trajectory of local maxima of anything related to the shape of $|\psi(x,t)|^2$ for this equation, non-linear and 3-dimentional versions of it. The problem comes from physics (laser propagation) and I want a fast method that based on which i can say something about the evolution of maxima position. $\endgroup$ – Victor Palea Jul 13 '18 at 18:05
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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.

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