Fourier Transform $\displaystyle{F}^{-1}({e^{i\xi^3t}})$

Question

The problem is$$u_t+u_{xxx}=0,u(x,0)=f(x),$$ Use Fourier Transform we get$$\overline{u}=e^{i\xi^3t}\overline{f},$$I want to solve out $u$ . Thus I want to know $\displaystyle{F}^{-1}({e^{i\xi^3t}})$,then I can express $u$ in the form of convolution.can anyone help me?

Answer 1

Your procedure have problems. Since if you solve this PDE by Fourier Transform you will miss its most general solution.

So your procedure should be modified as follows:

Let $u(x,t)=X(x)T(t)$ ,

Then $X(x)T'(t)+X'''(x)T(t)=0$

$X(x)T'(t)=-X'''(x)T(t)$

$\dfrac{T'(t)}{T(t)}=-\dfrac{X'''(x)}{X(x)}=i\xi^3$

$\begin{cases}\dfrac{T'(t)}{T(t)}=i\xi^3\\X'''(x)+i\xi^3X(x)=0\end{cases}$

$\begin{cases}T(t)=c_4(\xi)e^{it\xi^3}\\X(x)=\begin{cases}c_1(\xi)e^{ix\xi}+c_2(\xi)e^{-\frac{ix\xi}{2}}\sinh\dfrac{\sqrt{3}x\xi}{2}+c_3(\xi)e^{-\frac{ix\xi}{2}}\cosh\dfrac{\sqrt{3}x\xi}{2}&\text{when}~\xi\neq0\\c_1x^2+c_2x+c_3&\text{when}~\xi=0\end{cases}\end{cases}$

$\therefore u(x,t)=C_1x^2+C_2x+C_3+\int_{-\infty}^{\infty}C_4(\xi)e^{i(t\xi^3+x\xi)}~d\xi+\int_{-\infty}^{\infty}C_5(\xi)e^{\frac{i(2t\xi^3-x\xi)}{2}}\sinh\dfrac{\sqrt{3}x\xi}{2}d\xi+\int_{-\infty}^{\infty}C_6(\xi)e^{\frac{i(2t\xi^3-x\xi)}{2}}\cosh\dfrac{\sqrt{3}x\xi}{2}d\xi$

$u(x,0)=f(x)$ :

$C_1x^2+C_2x+C_3+\int_{-\infty}^{\infty}C_4(\xi)e^{ix\xi}~d\xi+\int_{-\infty}^{\infty}C_5(\xi)e^{-\frac{ix\xi}{2}}\sinh\dfrac{\sqrt{3}x\xi}{2}d\xi+\int_{-\infty}^{\infty}C_6(\xi)e^{-\frac{ix\xi}{2}}\cosh\dfrac{\sqrt{3}x\xi}{2}d\xi=f(x)$

$\int_{-\infty}^{\infty}C_4(\xi)e^{ix\xi}~d\xi=f(x)-C_1x^2-C_2x-C_3-\int_{-\infty}^{\infty}C_5(\xi)e^{-\frac{ix\xi}{2}}\sinh\dfrac{\sqrt{3}x\xi}{2}d\xi-\int_{-\infty}^{\infty}C_6(\xi)e^{-\frac{ix\xi}{2}}\cosh\dfrac{\sqrt{3}x\xi}{2}d\xi$

$C_4(\xi)=2\pi\int_{-\infty}^{\infty}f(x)e^{-i\xi x}~dx+C_1\delta''(\xi)+iC_2\delta'(\xi)-C_3\delta(\xi)-2\pi\int_{-\infty}^{\infty}e^{-i\xi x}\int_{-\infty}^{\infty}C_5(\xi)e^{-\frac{ix\xi}{2}}\sinh\dfrac{\sqrt{3}x\xi}{2}d\xi~dx-2\pi\int_{-\infty}^{\infty}e^{-i\xi x}\int_{-\infty}^{\infty}C_6(\xi)e^{-\frac{ix\xi}{2}}\cosh\dfrac{\sqrt{3}x\xi}{2}d\xi~dx$

$\therefore u(x,t)=C_1x^2+C_2x+C_3+2\pi\int_{-\infty}^{\infty}e^{i(t\xi^3+x\xi)}\int_{-\infty}^{\infty}f(x)e^{-i\xi x}~dx~d\xi+C_1\int_{-\infty}^{\infty}\delta''(\xi)e^{i(t\xi^3+x\xi)}~d\xi+iC_2\int_{-\infty}^{\infty}\delta'(\xi)e^{i(t\xi^3+x\xi)}~d\xi-C_3\int_{-\infty}^{\infty}\delta(\xi)e^{i(t\xi^3+x\xi)}~d\xi-2\pi\int_{-\infty}^{\infty}e^{i(t\xi^3+x\xi)}\int_{-\infty}^{\infty}e^{-i\xi x}\int_{-\infty}^{\infty}C_5(\xi)e^{-\frac{ix\xi}{2}}\sinh\dfrac{\sqrt{3}x\xi}{2}d\xi~dx~d\xi-2\pi\int_{-\infty}^{\infty}e^{i(t\xi^3+x\xi)}\int_{-\infty}^{\infty}e^{-i\xi x}\int_{-\infty}^{\infty}C_6(\xi)e^{-\frac{ix\xi}{2}}\cosh\dfrac{\sqrt{3}x\xi}{2}d\xi~dx~d\xi+\int_{-\infty}^{\infty}C_5(\xi)e^{\frac{i(2t\xi^3-x\xi)}{2}}\sinh\dfrac{\sqrt{3}x\xi}{2}d\xi+\int_{-\infty}^{\infty}C_6(\xi)e^{\frac{i(2t\xi^3-x\xi)}{2}}\cosh\dfrac{\sqrt{3}x\xi}{2}d\xi=2\pi\int_{-\infty}^{\infty}e^{i(t\xi^3+x\xi)}\int_{-\infty}^{\infty}f(x)e^{-i\xi x}~dx~d\xi-2\pi\int_{-\infty}^{\infty}e^{i(t\xi^3+x\xi)}\int_{-\infty}^{\infty}e^{-i\xi x}\int_{-\infty}^{\infty}C_5(\xi)e^{-\frac{ix\xi}{2}}\sinh\dfrac{\sqrt{3}x\xi}{2}d\xi~dx~d\xi-2\pi\int_{-\infty}^{\infty}e^{i(t\xi^3+x\xi)}\int_{-\infty}^{\infty}e^{-i\xi x}\int_{-\infty}^{\infty}C_6(\xi)e^{-\frac{ix\xi}{2}}\cosh\dfrac{\sqrt{3}x\xi}{2}d\xi~dx~d\xi+\int_{-\infty}^{\infty}C_5(\xi)e^{\frac{i(2t\xi^3-x\xi)}{2}}\sinh\dfrac{\sqrt{3}x\xi}{2}d\xi+\int_{-\infty}^{\infty}C_6(\xi)e^{\frac{i(2t\xi^3-x\xi)}{2}}\cosh\dfrac{\sqrt{3}x\xi}{2}d\xi$

Answer 2

You should have a look at the Airy function. It transforms like this:

$$ \mathcal{F} [\textrm{Ai}(x)](k) = e^{\frac{i}{3}(2\pi k)^{3}}$$

and with suitable manipulation it can be used to solve your problem.