# Uniqueness theorems related to Hardy Uncertainty Principle

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.