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Uncertainty Principles state that a function and its Fourier transform cannot be simultaneously sharply localised. A well known result due to G.H.Hardy says that if $f(x)=O(e^{-\alpha^2|x|^2})$, $\hat{f}(\xi)=O(e^{-4\beta^2|\xi|^2})$ and $\alpha\beta>\frac{1}{4}$, then $f\equiv 0$. And this can be rewriten in terms of the solution of free Schrödinger equation in $\mathbb{R}^n\times(0,+\infty)$, $$ i\partial_tu+\Delta u=0\\ u(0)=f $$ in the following way: If $u(x,0)=O(e^{-\alpha^2|x|^2})$,$u(x,1)=O(e^{-\beta^2|x|^2})$ and $\alpha\beta>\frac{1}{4}$, then $u\equiv 0$

This is because that $u(x,1)$ can be expressed through the Fourier transform of the initial value (since the evolution kernel is $t^{-\frac{n}{2}}e^{\frac{i|x-y|^2}{4t}}$)

We can also apply Hardy Uncertainty (and the backward uniqueness arguments for heat equation)to prove the optimal decay result for solutions of free heat equation:

If $f$ and $e^{\delta^2 |x|^2}e^\Delta f$ are in $L^2$ for some $\delta\geq \frac{1}{2}$, then $f\equiv 0$

I'm interested to know that if there are any other types of initial problems sharing such properties,i.e., if we know $u(x,0)$ and $u(x,t)$ are both decreasing very fast(such as exponentially decay), then we can obtain $u\equiv 0$

For example, I'm considering the following problem $$ i\partial_tu=(-\Delta)^2 u\\ u(0)=f $$ Now the main difficulty is that we don't have the explicit expression of the solution compared with the Schrödinger equation. And it may include the following type oscillatory integral $$ u(x,t)\approx \int_{\mathbb{R}^n}|x-y|^{-\frac{n}{3}}e^{it|x-y|^{\frac{4}{3}}}f(y)dy, \quad \text{when x is large,} $$
So, is there some method to estimate this oscillatory integral operator? And I'm curious to know whether or not the uniqueness result is valid in this case.

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1 Answer 1

You should have a look at http://arxiv.org/abs/1110.4873

It appeared also peer reviewed journal but may not be freely accessible. There are more papers of Vega and collaborators on Hardy's uncertainty principle where you might find your answer.

I have not clarified whether this answers your question exactly but it is at least very related and you might be able to work your way from there.

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