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I asked this question already on stackexchange, but I did not get any resonance at all, so maybe anybody here can give me a few hints about my problem.

My goal is to solve this PDE for $f:[-1,1] \times \mathbb{R}_{\ge 0}\rightarrow \mathbb{C}$ $$ i\partial_t f(x,t) = -\partial_x^2 f(x,t) + g(t)V(x)f(x,t).$$

I would consider this PDE to be solved if I get two ODEs just depending on either $x$ or $t$. $$f(x,0)$$ is specified a priori and $\int_{[-1,1]} f^*(x,t)f(x,t) dx=1$ for all $t \ge 0$.

Separation of variables seems to fail here and also integral transforms appear to be useless. Despite, I don't want to use perturbative techniques.

I want to have that $g$ is a $C^{\infty}$ function with compact support and $V \in C^{\infty}$.

A simpler setting where an integral transform could maybe work is this one:

If we take $g(t):=\delta(t-t_0)$, where $t_0>0$.

$$ \partial_t f(x,t) = -\partial_x^2 f(x,t) + \delta(t-t_0)V(x)f(x,t).$$

On the other hand, the integral transform(especially the Fourier transform) seems to fail, as the function does not need to be square integrable with respect to time. So, I don't really see if we can do anything about it.

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You will definitely have to impose boundary conditions at $x=\pm 1$ to obtain a well defined problem and have $\int|f|^2$ preserved. The second version could of course be reduced to $i \dot{f}=Hf$ with a time-independent $H$, plus a condition at $t=t_0$ (so this boils down to understanding the spectral theory of $H$), but I have a feeling that's as far as you can go in general. –  Christian Remling Jun 23 at 22:21
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Put differently, I don't think the product structure $g(t)V(x)$ helps a lot, as there are no obvious relations between the spectra/eigenfunctions of $-D^2+cV$ for different $c$'s. –  Christian Remling Jun 23 at 22:24
    
sure, I would consider the standard boundary conditions at $x= \pm 1$, but they shall be time-independent: One possible choice could be $f(1,t) = f(-1,t)$ and also for the derivatives $\partial_1 f(1,t)= \partial_1 f'(-1,t)$. ( though this might restrict the potential to special even functions), this is what I have in mind. But since it is time-independent I thought that this may be unnecessary for the separation process itself. –  Tobias Hurth Jun 23 at 22:24
    
@ChristianRemling sorry, I did not quite understand your comment on the example. Did you mean that it is easy to solve or impossible in general? By the way: Thank you for the disillusionment about the equation ;-) –  Tobias Hurth Jun 23 at 22:31
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Here is the earlier question: math.stackexchange.com/questions/844601/… Normally you should wait more than 9 hours before crossposting - some of the math.SE people are sleeping or busy. –  S. Carnahan Jun 23 at 23:26

2 Answers 2

up vote 4 down vote accepted

I agree with @Christian Remling that the product structure of your potential $V(x,t)=g(t)V(x)$ is not helpful in general, but it would help if $g(t)$ is a monotonically decreasing function of time, see this paper by T.J. Park (2002).

Only a few time-dependent systems have been reported to be analytically solved whose potentials are constant, linear, and quadratic functions of coordinate with arbitrary time-dependences. Here we show that time-dependent potentials of any coordinate-dependence can be solved exactly if the time-dependence is monotonously decreasing. We do this by a unitary transformation of the wavefunction and variable transformations to change the Schrödinger equation to be time-independent in new variables. These variables are then determined by solving a set of simple differential equations.

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In the linear case, $V(x)=x$, the Schrodinger equation can be solved by Lewis-Riesenfeld approach. See http://link.springer.com/article/10.1007%2Fs12648-013-0322-4 as well as http://arxiv.org/abs/quant-ph/0309174 and references cited therein.

Some other cases is considered in http://journal.kcsnet.or.kr/main/j_search/j_abstract_view.htm?code=B021211&qpage=j_search&spage=b_bkcs&dpage=ar (Exactly Solvable Time-Dependent Problems: Potentials of Monotonously Decreasing Function of Time, by Tae Jun Park).

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