Third-order Linear Parabolic PDE What's the best method to solve analytically an equation of the form
$$f_t=f_x+af_{xx}+bf_{xxx}$$
with $a,b\in\mathbb{R}$ ?
 A: The partial differential equation specified is given by,
$$\frac{\partial f(x,t)}{\partial t}=\frac{\partial f(x,t)}{\partial x} + a \frac{\partial^2 f(x,t)}{\partial x^2}+b\frac{\partial^3 f(x,t)}{\partial x^3}$$
We approach the problem with the Fourier transform, i.e.
$$F(k,t)=\int_{-\infty}^{\infty} \mathrm{d}x \, e^{-ikx} \, f(x,t)$$
The new differential equation in terms of the function in Fourier space is given by,
$$\frac{\partial F(k,t)}{\partial t}=F(k,t)\left(ik-ak^2-ibk^3\right)$$
where we have employed the standard formula for the Fourier transform of a derivative, derived by integration by parts, c.f. Fourier Transform. Can you proceed from here? Notice as the equation does not contain any $k$ derivatives, $F=F(t)$ from the perspective of the equation.

Additional Information
It is clear a particular solution to the equation in Fourier space is simply an exponential given by,
$$F(k,t)=\exp \left[ \left(ik -ak^2-ibk^3 \right)t\right]$$
To convert back to physical space is a daunting task,$^{\dagger}$ the inverse Fourier integral  required:
$$f(x,t)=\int_{-\infty}^{\infty} \frac{\mathrm{d}k }{2\pi}\, \exp \left[ \left(ik -ak^2-ibk^3 \right)t + ikx\right]$$

$\dagger$ As MIT Professor Arthur Mattuck stated, jokingly, "that's conservation of mathematical difficulty!"
A: The way I would do it would be to write the equation in reciprocal space. The equation becomes algebraic (i.e., there are no differential operators). You will get an expression $\omega = \omega(\vec{k})$. Then given an initial $f(x,t_0)$, you can decompose it as $f(x,t_0) = \int f(\vec{k},t_0) e^{i\vec{k}\cdot \vec{x}} d\vec{k}$. Then the solution for $t\ne t_0$ will be $f(x,t) = \int f(k,t_0) e^{-i\omega(\vec{k})t} e^{i\vec{k}\cdot \vec{x}} d\vec{k}$
A: It seems that the best method is the Fourier substitution
$$f(x,t) = \chi(x)\tau(t),$$
then
$${\tau'(t)\over\tau} = {b\chi'''(x)+a\chi''(x)+\chi(x)\over\chi}=-\lambda = const,$$
$$\begin{cases}
\tau'(t)+\lambda\tau = 0\\
b\chi'''(x) + a\chi''(x) + \chi'(x) + \lambda\chi(x) = 0.
\end{cases}$$
That leads to generalize solution in the form of
$$f(x,t) = \int\limits_{\Omega(\lambda)}\left(C_1e^{r_1(\lambda)x} + C_2e^{r_2(\lambda)x} + C_3e^{r_3(\lambda)x} + C_4e^{r_4(\lambda)x}\right)e^{-\lambda t}dt,$$
where $r_i(\lambda)$ are the roots of the algebraic equation
$$br^3+ar^2+r+\lambda = 0.$$
