Solve $\partial_t u= -\partial_x^3 u$ I want to solve the PDE 
$$\partial_t  u= -\partial_x^3 u\quad \text{s.t.}\quad u(0,x) = g(x).$$
where $u:\mathbb{R}\times\mathbb{R}\rightarrow \mathbb{R}$. I used the Fourier-Transformation $\mathcal{F}_x$ w.r.t $x$ to solve this:
$$\partial_t(\mathcal{F}_xu)(t,\xi) = -i\xi^3(\mathcal{F}_xu)(t,\xi)\quad\text{s.t.}\quad (\mathcal{F}_xu)(0,\xi) = (\mathcal{F}_xg)(\xi)$$
$$\Leftrightarrow\quad(\mathcal{F}_xu)(t,\xi) = (\mathcal{F}_xg)(\xi)\cdot\exp(-i\xi^3t)$$
$$\Leftrightarrow\quad u(x,t) = \frac{1}{\sqrt{2\pi}}(g*\mathcal{F}^{-1}_xh)(x),$$
where $h(t,x) = \exp(-ix^3t)$. Unfortunately I'm not able to calculate $\mathcal{F}^{-1}_xh$ and even WolframAlpha doesn't give me a nice closed form solution. Can this problem be further simplified? Does this equation have a specific name?
 A: HINT : Solution in term of Fourier series.
$$\partial_t  u= -\partial_x^3 u $$
With the method of separation of variables, it is easy to find an infinity of particular solutions on the form :
$$u_{\lambda}(t,x)=A_\lambda\sin(\lambda x+\lambda^3 t)+B_\lambda\cos(\lambda x+\lambda^3 t)$$
where $\lambda$ , $A_\lambda$ , $B_\lambda$ are any constants.
More general solution :
$$u(t,x)=\sum_{\forall \:\lambda}\left(A_\lambda\sin(\lambda x+\lambda^3 t)+B_\lambda\cos(\lambda x+\lambda^3 t) \right)$$
Boundary condition : $ \quad u(0,x) = g(x)$
Expressing $g(x)$ on the form of Fourier series, for example :
$$g(x)=\sum_{n=0}^\infty\left(A_n\sin(n x)+B_n\cos(n x) \right)$$ 
gives $\quad \lambda=n\quad;\quad A_\lambda=A_n \quad;\quad B_\lambda=B_n\quad$ and the solution :
$$u(t,x)=\sum_{n=0}^{\infty}\left(A_n\sin(n x+n^3 t)+B_n\cos(n x+n^3 t) \right)$$
Of course, depending on the range of $x$ considered, the Fourier series can be computed with other coefficients and different variable, for example $n\pi x$ , or $2n\pi x$ , ..., instead of $nx$.
