Properties of the solution of the schrodinger equation

Question

I'm considering the Free Schrodinger equation: $$ i\partial_{t}u+\Delta{u}=0,~\mathbb{R}\times\mathbb{R}$$ with $$ u(0,\cdot)=u_{0} \in C^{\infty}_{0}(\mathbb{R}).$$ The solution is given by: $$ u(t,x)=\mathcal{F}^{-1}[e^{-iy^{2}t}\mathcal{F}(u_0)(y)](x)=\frac{1}{\sqrt{4\pi it}}\int_{\mathbb{R}}e^{i|x-y|^{2}/4t}u_{0}(y)dy $$ where $\mathcal{F}$ denotes the Fourier transform. How can I show that, for $t \neq 0$, $u(t,\cdot)$ is not compactly supported? ( $\notin C^{\infty}_{0} )$

Answer 1

The following proof is from Linares & Ponce's Introduction to Nonlinear Dispersive Equation.

Fix $t\neq 0$. Let us begin by rewriting the solution $u(t,x)$ as follows \begin{align} u(t, x) =&\ \frac{1}{\sqrt{4\pi i t}} \int_\mathbb{R} e^{i|x-y|^2/4t}u_0(y)\ dy = \frac{e^{i|x|^2/4t}}{\sqrt{4\pi i t}} \int_\mathbb{R} e^{-ixy/2t} [e^{i|y|^2/4t}u_0(y)]\ dy \\ =&\ \frac{e^{i|x|^2/4t}}{\sqrt{4\pi i t}} (\widehat{e^{i|\cdot|^2/4t}u_0(\cdot)})\left(\frac{x}{2t}\right). \end{align}

Since $u_0 \in C_0^\infty(\mathbb{R})$, then $e^{i|y|^2/4t}u_0(y) \in C_0^\infty(\mathbb{R})$. Thus, by Paley-Wiener, we know that $(\widehat{e^{i|\cdot|^2/4t}u_0(\cdot)})$ can't have compact support (since it extends to an entire function on $\mathbb{C}$) which means $u(t, x)$ also can't have compact support.