Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. **There will be no changes to other Yahoo properties or services, or your Yahoo account.** You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.

## Trending News

# Quick fourier transform question?

α and c are constants and u(x, t) → 0 as x → ±∞. The solution found to a previous eqn is

u(x, t) =integral from -infinite to infinite of (Fourier transform of f(x))*e^(−ik[(x−ct)+α(k^2)*t)

Please help me with the following. For the special case f(x) = δ(x), show that the solution can be expressed in terms of the

Airy function Ai(z), deﬁned by (1/pi)*(integral from 0 to infinite of cos(z*k+(k^3)/3)

I think its to do with the real component of the exponential. ALso the fourier transform of delta(x) is 1/(2pi)

THanks! Will best answer

u(x, t) =integral from -infinite to infinite of (Fourier transform of f(x))*e^(−ik[(x−ct)+α(k^2)*t]) sorry missed the square bracket

### 1 Answer

- Anonymous10 years agoFavorite Answer
By Fourier Transform, we can show that by f(x) = δ(x), the given expression can be written as u(x,t) = ∫(-∞, ∞) (Fourier transform of f(x))*e^(−ik[(x−ct)+α(k^2)*t)

See more... http://en.wikipedia.org/wiki/Fourier_transform

Good luck!

Source(s): ☻