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# 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

Update:

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

Relevance
• Anonymous
10 years ago
Favorite 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)

Good luck!

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