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

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 2K views
StewartHolmes
Messages
2
Reaction score
0

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.
 
Physics news on Phys.org
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.