(a) Show that the Fourier transform of $f(x)=e^{-a^{2} x^{2}}$, for $a>0$, is

$$\tilde{f}(k)=\frac{\sqrt{\pi}}{a} e^{-\frac{k^{2}}{4 a^{2}}},$$

stating clearly any properties of the Fourier transform that you use.

[Hint: You may assume that $\int_{0}^{\infty} e^{-t^{2}} d t=\sqrt{\pi} / 2$.]

(b) Consider now the Cauchy problem for the diffusion equation in one space dimension, i.e. solving for $\theta(x, t)$ satisfying:

$$\frac{\partial \theta}{\partial t}=D \frac{\partial^{2} \theta}{\partial x^{2}} \quad \text { with } \theta(x, 0)=g(x)$$

where $D$ is a positive constant and $g(x)$ is specified. Consider the following property of a solution:

Property P: If the initial data $g(x)$ is positive and it is non-zero only within a bounded region (i.e. there is a constant $\alpha$ such that $\theta(x, 0)=0$ for all $|x|>\alpha)$, then for any $\epsilon>0$ (however small) and $\beta$ (however large) the solution $\theta(\beta, \epsilon)$ can be non-zero, i.e. the solution can become non-zero arbitrarily far away after an arbitrarily short time.

Does Property P hold for solutions of the diffusion equation? Justify your answer (deriving any expression for the solution $\theta(x, t)$ that you use).

(c) Consider now the wave equation in one space dimension:

$$\frac{\partial^{2} u}{\partial t^{2}}=c^{2} \frac{\partial^{2} u}{\partial x^{2}}$$

with given initial data $u(x, 0)=\phi(x)$ and $\frac{\partial u}{\partial t}(x, 0)=0$ (and $c$ is a constant).

Does Property $\mathrm{P}$ (with $g(x)$ and $\theta(\beta, \epsilon)$ now replaced by $\phi(x)$ and $u(\beta, \epsilon)$ respectively) hold for solutions of the wave equation? Justify your answer again as above.