how can i solve this PDE (IVP) by using Fourier transform?

Question

$$ u_t=\alpha^2u_{xx}\\ u(x,0)=e^{-x^2} $$

where t>0
i tried to solve it, but not certain
and i got hint
$$ \int_{-\infty}^{\infty} e^{-\xi^2+i\xi x}d\xi=\sqrt\pi e^{-x^2/4} $$

thanks for your help

Answer 1

Differential equation to be solved: $$\partial_t u(t,x) = \alpha^2 \partial_x^2 u(t,x).$$

Doing a Fourier transform in $x$ gives $$\partial_t \hat{u}(t,\xi) = -\alpha^2 \xi^2 \hat{u}(t,\xi).$$

For fixed $\xi$ this is an ordinary differential equation of order 1 in $t$ with solution: $$\hat{u}(t,\xi) = \hat{u}(0,\xi) e^{-\alpha^2 \xi^2 t}.$$

The initial condition becomes $$\hat{u}(0,\xi) = \mathcal{F}\{ e^{-x^2} \} = \sqrt{\pi}e^{-\xi^2},$$ so we get the solution $$\hat{u}(t,\xi) = \sqrt{\pi} e^{-\xi^2} e^{-\alpha^2 \xi^2 t} = \sqrt{\pi} e^{-(1+\alpha^2t)\xi^2}.$$

Now we take the inverse Fourier transform: $$ u(t,x) = \mathcal{F}^{-1}\{ \sqrt{\pi} e^{-(1+\alpha^2t)\xi^2} \} = \frac{1}{\sqrt{1+\alpha^2 t}} e^{-x^2}. $$

Answer 2

First of all apply the method of separation of variables, then apply all your boundary conditions and after that compute the value of constant term using Fourier transform The equation given in question is 1 dimensional heat equation