MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 '14 at 22:21
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 '14 at 22:24
Here is the earlier question:… Normally you should wait more than 9 hours before crossposting - some of the math.SE people are sleeping or busy. – S. Carnahan Jun 23 '14 at 23:26
up vote 2 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.

share|cite|improve this answer

In the linear case, $V(x)=x$, the Schrodinger equation can be solved by Lewis-Riesenfeld approach. See as well as and references cited therein.

Some other cases is considered in (Exactly Solvable Time-Dependent Problems: Potentials of Monotonously Decreasing Function of Time, by Tae Jun Park).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.