Fourier transforms in PDEs

Question

Given the following PDE: $x²\frac{\partial u}{\partial x} + \frac{\partial u}{\partial t} = 0 $

with the following initial conditions: u(x,0) = f(x)

Is it possible to apply a Fourier transform with respect to $x$ to solve it? I wonder that because of that $x²$ which seems to make it too hard to integrate in the Fourier Transform. None of the examples I found had a variable factor multiplied by the derivative whose FT I would be supposed to take.

Or maybe I should put it this way: How do I take the Fourier Transform of $x^2\frac{\partial u(x,t)}{\partial x}$ with respect to $x$?

Answer 1

Let ${\cal{F(u)}}=\int_{-\infty}^\infty u(x,t)e^{ix\xi}dx=U(\xi, t),$ then ${\cal{F(u_x)}}=i\xi\, U(\xi, t)$, where $i=\sqrt{-1}$. Thus, the Fourier transform of $x^2u$ becomes ${\cal{F(x^2u_x)}}=\frac{d^2}{d\xi^2}\bigg(i\xi\, U(\xi, t)\bigg)=i\xi U_{\xi\xi}+2i\,U_{\xi}$ provided that $x^2u,$ $xu$ and $u$ vanish as $|x|\to\infty$, where we have used the rule ${\cal{F(x^n g(x))}}=(-i)^n\frac{d^n}{d\xi^n}G(\xi)$.

Answer 2

Transforming the derivative in space we get

$$\int_{-\infty}^{\infty} \frac{\partial u}{\partial x} e^{-2\pi i\xi x} dx = -2\pi i\xi \ \hat u(\xi,t).$$ Now from here, you convolute this with the fourier transform of $x^2$, namely $$\int_{-\infty}^{\infty} x^2 e^{-2\pi i\xi x} dx =\dots$$ which you can compute via integration by parts. Also, you could just look this stuff up on a table (if possible, which in this case it is).