## The problem

The PDE I am working with is given by $\left(\partial_a^b \leftrightarrow\frac{\partial^b}{\partial a^b}\right)$

$$\partial_t \psi = i \partial_x^2 \psi$$ $$\psi(x,t=0) = \psi_0(x)$$ $$\psi:\mathbb{R}^2\rightarrow \mathbb{C}$$

Let $F:\mathbb{R}^2 \rightarrow \mathbb{R}_{\ge 0}$ be a function such that for a solution $\psi$ of the above mentioned PDE, then $F=|\psi|^2$. The solutions $\psi$ that interest me have to be square-integrable.

At this point I am interested in computing $\psi$ for a known $F$, which from my point of view, should be equivalent to replacing the initial condition $\psi_0$ with the now known function $F$.

I have not been able to prove this, but I consider this approach to not hold on for a general case, in the sense that there might not be solutions of the PDE that are compatible with the choice of $F$. This is why I have changed the condition. However this is only an assumption, so if I can be proven wrong it'll be great.

The second attempt is as follows. For a solution $\psi$ of the PDE consider the curves given by $|\psi|^2 = a$, where $a \in [0,\max(|\psi|^2)]$ is a constant. I label this curves as $\Gamma(s)=\left(g(s), h(s)\right)$ where $g,h:\mathbb{R} \rightarrow \mathbb{R}$. Since $|\psi|^2$ is constant on a curve $\Gamma$, then it is true that

$$\nabla_{x,t}|\psi|^2 \perp T_{\Gamma}$$

where $\nabla_{x,t}|\psi|^2=\left( \partial_x |\psi|^2, \partial_t |\psi|^2\right)$ and $T_{\Gamma}(s)=\left(\partial_s g, \partial_s h\right)$ is the tangent of the curve $\Gamma$.

This condition becomes

$$\nabla_{x,t}|\psi|^2 \cdot T_{\Gamma}=0$$

or explicit

$$\partial_s g \cdot \partial_x |\psi|^2 + \partial_s h \cdot \partial_t |\psi|^2 = 0.$$

This is a transport equation which defines, by the choice on $g$ and $h$ the curves on which $|\psi|^2$ is constant, independent of the actual value of $|\psi|^2$ on that curve. Unlike the first attempt where I enforce $F$ and then look for the solution (if possible), here I enforce the curves on which the solution has the same amplitude. This should not be as restrictive as the first approach, but it might also be too little to allow for computing $\psi$.

## Context

For the eventual questions about why do I want to formulate the problem in such a manner, or what is the purpose of it, I will try to give some pre-emptive answers.

There is a solution for this PDE, which although not square-integrable, satisfies the condition $F(x,t) = F(x-t^2)$. More about this here. Following this result, square-integrable versions of this solution have been developed that qualitatively satisfy the same condition, but only for $t<t_{max}\in \mathbb{R}$.

These results do not give an algorithm for generating solutions that have a particular behavior, it is more of a guessing game, where one pluggs an initial condition in the PDE and then hopes to get an oddly interesting solution. This is the reason why I want to replace the initial condition with something similar to what I have presented above.

## My attempts

Following the article I have mentioned in the **Context** section, the choice I made for $g$ and $h$ has been

$$g(s) = x_0 + \frac{s^n}{n!}$$ $$h(s) = s$$

where $n\in \mathbb{N}$. For $n = 2$ a rescaled solution similar to the the one in the article should be obtained. Because of the choice on $g$ and $h$, I can replace $s$ with $t$. The condition becomes

$$t \partial_x F + \partial_t F = 0$$

At this point I considered writing $g$ as

$$g(t) = \sum_{n=0}^{\infty} \frac{t^n}{n!} \partial_t^n g(t=0)$$

I know the values of $\partial_t^n g(t=0), \forall n \in \mathbb{N}$ since I chose $g(t) = x_0 + t^2/2$, so I can derivate the condition with respect to $t$ in order to get to each $\partial_t^n g(t=0)$ term after I evaluate the expression at $t=0$.

By evaluating the condition as it is, it gives

$$\partial_t|\psi(x,t =0)|^2 = 0$$

Derivating once and evaluating it gives

$$\partial_x|\psi|^2(x, t=0) + \partial_t^2 |\psi(x,t=0)|=0$$

and

$$\partial_t^n |\psi|^2(x,t=0) = 0, \forall n>2.$$

These conditions, I thought, could be used in order to compute an initial condition. Since there are an infinite number of such conditions, I thought that in the worst case scenario, if this approach works, I should be able to use a finite number of them in order to get an approximate initial condition that satisfies qualitatively the condition I emposed by choosing $g$ and $h$. As it turns out, the computation is not as simple as would have hoped, since the evaluation of the temporal derivatives can only make use of

$$\partial_t |\psi|^2 = i\left(\partial_x^2\psi \cdot \psi^* - \psi \cdot \partial_x^2 \psi^*\right)$$.

Although I wasn't able to actually compute an initial conditions from conditions for $n \le 2$, I did manage to validate the result by plugging in the initial condition from the article. I did not try for $n>2$ since even using Wolfram Alpha it still was difficult to follow the expressions.

## Conclusion and questions

I am interested in validating this approach by answering to these questions:

- Can I solve the PDE (numerical methods will do) without imposing other conditions, and if not, why is that? I looked for methods of proving that a problem is well-posed, but my background is in engineering and physics, not mathematics so I did not really know where to look for.
- Does this approach already exist and I just didn't look where I had to in order to find it?

## Update

I have considered the following scenarios, first:

$$g(t) = \frac{t^2}{2}$$

with the conditions being

$$\partial_t |\psi_0|^2 = 0$$ and $$\partial_x |\psi_0|^2 + \partial_t^2 |\psi_0|^2 = 0$$

From the first condition $\partial_t |\psi_0|^2 = 0$ I took only the case where $\psi_0 = \psi_0^*$. Adding this to the second condition I got

$$\psi_0 \partial_x^4 \psi_0 + \psi_0 \partial_x \psi_0 - (\partial_x^2 \psi_0)^2 = 0$$

I used a discrete version of this ODE and have computed the initial condition $\psi_0$ using

$$\psi_0^{m+4} = 4\psi_0^{m+3} - 6 \psi_0^{m+2} + 4 \psi_0^{m+1} - \psi_0^{m} + \frac{(\psi_0^{m+3}-2\psi_0^{m+2}+\psi_0^{m+1})^2}{\psi_0^{m+2}} - \Delta x^3(\psi_0^{m+3} - \psi_0^{m+1})$$

and some initial values for $\psi_0^{m=0..3}$ that I chose myself. After a few attempts with the values for $\psi_0^{m=0..3}$, I got to some values that generated the initial condition for the solution that is plotted in the figure.

I have also tested for $g(t) = t$ and managed to change the angle between the "trajectory" a Gaussian beam propagates compared to the propagation axis.

This means that the method works for those cases and probably works for superior orders, but I have not tested them yet.

I consider this question closed.

equivalentto the initial condition? How can that be, the initial condition contains much more information on $\psi$ than just the norm. $\endgroup$ – Carlo Beenakker Jan 23 at 17:39