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I'm having trouble with solving $u_t+u_{xxx}=0$ for $u(x,t)$ with $u(x,0)=f(x)$. I'm asked to used Fourier Transform method, with the Fourier pair defined by: $F(w)=\frac{1}{2\pi}\int_{-\infty}^{\infty}f(x)e^{iwx}dx$ and $f(x)=\int_{-\infty}^{\infty}F(w)e^{-iwx}dw$.

The answer is they want is in the form: $u(x,t)=\frac{1}{(3t)^{\frac{1}{3}}}\int_{-\infty}^{\infty}f(r)A_i(\frac{x-r}{(3t)^{\frac{1}{3}}})dr$, where $A_i$ is the Airy function.

I got as my last line of working: $u(x,t) = \int_{-\infty}^{\infty} f(r)(\frac{1}{2\pi}\int_{-\infty}^{\infty} e^{i(w^3t+w(x-r))}dw)dr$ using the convolution theorem. But now I have no idea how to make my answer into the required form.

Any help is appreciated!

satokun
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    The Fourier representation of the Airy function $$\mathrm{Ai}(x) = \frac{1}{2\pi}\int e^{\frac{i}{3}\omega^3 + i\omega x}{\rm d}\omega$$ might be useful. Try changing variables to put your last expression on this form. – Winther Nov 05 '15 at 05:21
  • thank you! I have not considered change of variables at all! – satokun Nov 05 '15 at 05:34

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