HW help, PDE solving with Fourier Transforms

For the discussion of math. Duh.

Moderators: gmalivuk, Moderators General, Prelates

User avatar
the tree
Posts: 801
Joined: Mon Apr 02, 2007 6:23 pm UTC
Location: Behind you

HW help, PDE solving with Fourier Transforms

Postby the tree » Fri Oct 15, 2010 8:53 am UTC

Right so I've got the heat equation [imath]u_t = u_xx[/imath], and some initial condition [imath]u(x,0) = f(x)[/imath].

Now according to my notes, my solution should be:
[math]\mathcal{F}^1 \left( F(k)e^{ikx +k^2 t} \right)[/math]
(where [imath]F(k)= \mathcal{F}(f(x))[/imath])

But with the initial condition [imath]f(x) = 1 \, \mbox{if} \, x \in (-1,2) , \, 0 \, \mbox{o/w}[/imath], [imath]F(k)= \tfrac{i}{k}(e^{2ik} - e^{ik})[/imath], I wind up with the godawful integral:

[math]\frac{i}{\pi} \int_{-\infty}^{\infty} \!{\frac {{{\rm e}^{ik \left( x-2 \right) + \left( {k}^{2}-ik
\right) t}}+{{\rm e}^{ik \left( x+1 \right) + \left( {k}^{2}-ik
\right) t}}}{k}}{dk}[/math]

Can anyone tell me where I've gone wrong please?

User avatar
william
Not a Raptor. Honest.
Posts: 2418
Joined: Sat Oct 14, 2006 5:02 pm UTC
Location: Chapel Hill, NC
Contact:

Re: HW help, PDE solving with Fourier Transforms

Postby william » Sat Oct 23, 2010 1:09 am UTC

The Fourier transform of a uniform distribution is going to look like [math]F(k)=\frac{e^{ix_1k}-e^{ix_2k}}{k}[/math] so your integral looks about right.
SecondTalon wrote:A pile of shit can call itself a delicious pie, but that doesn't make it true.


Return to “Mathematics”

Who is online

Users browsing this forum: No registered users and 13 guests