Diffusion Equation PDE: Solving for u(x, t) with Initial Condition e^(-x^2)

[StewartHolmes]

Homework Statement

Solve
[tex]u_{tt} - 4u_{xx} = 0[/tex], [tex] x \in \mathbb{R}, t > 0 [/tex]
[tex]u(x, 0) = e^{-x^2} [/tex], [tex] x \in \mathbb{R} [/tex]

Homework Equations

General solution to the diffusion equation:
[tex]u(x, t) = \frac{1}{\sqrt{4\pi kt}} \int\limits_{-\infty}^{\infty} e^\frac{{-(x - y)^2}}{4kt} \varphi(y) \, dy[/tex]

The Attempt at a Solution

[tex] u(x, t) = \frac{1}{\sqrt{4\pi kt}} \int\limits_{-\infty}^{\infty} e^\frac{{-(x - y)^2}}{4kt} e^{-y^2} [/tex]

That's about as good as I've got. Integration by parts gets me no further. I've tried to combine the exponents in the integrand, but that leaves me with
[tex] - \frac{x^2 + y^2 - 2xy + 4kty^2}{4kt} [/tex]
I have an example in a textbook where they do similar, then complete the square so that they can substitute [tex]p[/tex], then integrate [tex]\int\limits_{-\infty}^{\infty}e^{-p^2} \, dp[/tex] as [tex]\sqrt{\pi}[/tex]... but I can't complete the square in this case.

 

[hunt_mat]

Start from scratch, it'll be easier. Take Fourier transforms w.r.t x of the PDE and the initial condition, then look for the inverse transform.