Fourier transform of partial differential equation How would I find the Fourier Transform of the pde
$$
u_t = u_{xx} + \alpha\cdot u_x,\quad\alpha &gt; 0\,?
$$
I know the definition of the Fourier transform, but I'm not sure how to use it in the context of partial differential equations.
If $f : \mathbb R\to\mathbb C$ is a "nice" function (i.e. $f\in H^1$ or $f\in C^1$ with $f(\pm\infty) = 0$) the Fourier transform of its derivative is by partial integration
$$
\widehat{f'}(\omega) = \int f'(x)e^{-2\pi ix\omega}\,dx = 2\pi i\omega\int f(x)e^{-2\pi ix\omega}\,dx = 2\pi i\omega\cdot\hat f(\omega).
$$
So if $U(\omega,t)$ denotes the Fourier transform of $u(x,t)$ with respect to $x$, applying the Fourier transform with respect to $x$ to $u_t = u_{xx} + \alpha u_x$ gives
$$
U_t = -4\pi^2\omega^2 U + 2\pi i\omega\alpha U = (-2\pi + i\alpha\omega)2\pi U.
$$
Just ignore $t$ and apply the transform as usual, you'll get something like

$$ U_t = -\omega^2 U + i\alpha \omega U $$

where $U(\omega,t)$ is the Fourier transform of $u(x,t)$, given by

$$ U(\omega,t) = \frac{1}{\sqrt{2\pi}}\int_{\Bbb R} u(x,t) e^{-i\omega x}dx $$

Some constants may apply, depending on your definition of the Fourier transform.