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?

  • $\begingroup$ This is like the advection equation except that different Fourier modes move at different speeds. In particular the mode with wavenumber $\kappa$ moves with velocity $-\kappa^2$, as you can see by plugging in the ansatz $u(x,t)=a\cos(\kappa(x-vt)+\phi)$. $\endgroup$ – Rahul Jun 28 '17 at 19:19
  • $\begingroup$ Why not simply write it as $u(x,t) = \frac{1}{\sqrt{2\pi}}\int \hat{g}(\xi) e^{-i\xi^3 t + i\xi x}{\rm d}\xi$? Since $g$ is a general function you cannot go much further than this and this form avoid having to explicitly invert $h$. $\endgroup$ – Winther Jun 28 '17 at 21:31

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 $nx$.

  • $\begingroup$ A non-periodic function on $\mathbb{R}$ does not have a Fourier series. $\endgroup$ – Winther Jul 4 '17 at 18:05
  • $\begingroup$ On a given range, a non-periodic function can be expressed as a Fourier series. If you solve numerically the problem, you always solve it on a limited range, not on all reals. $\endgroup$ – JJacquelin Jul 4 '17 at 18:41
  • $\begingroup$ But the domain here is not finite, it's $\mathbb{R}$. If you solve it numerically then you would not write down this expression for the solution in the first place so I don't see your point. $\endgroup$ – Winther Jul 4 '17 at 18:45

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