" Consider the following equation in the plane
$x\frac{\partial u}{\partial y}-y\frac{\partial u}{\partial x}=f(x^2+y^2)$
where $f(t)$ is a $C^\infty$ function of the real variable $t$ such that $f(t)=0$ if $t$ < 1 or $t$ > 2, and $f(3/2) = 1$.
Show that the equation has no distribution solution in $\mathbb{R^2}$\{0}. "
This is the question 2.1 in the book basic linear partial differential equations by Francois Treves. I'm reading the book this summer. He gives the hint that rewrite $\iint |f(x^2+y^2)|^2 dx dy$ by introducing u, assumed to satisfy the equation.
I'm a beginner of the distribution theory and actually I don't know how to approach this problem even with the hint, because I even don't know whether the equation has a smooth solution. In view of the form of the hint, I cannot help using Plancherel Theorem, but it only works for $L^2$, and I don't know how to do the Fourier Transform on a distribution. All I know about is the Fourier Transform on a tempered distribution. Any help would be appreciated.
Here's a hint to get you started. What you know is that $\iint_D |f(x^2+y^2)|^2\,dx\,dy > 0$ for $D$ a disk centered at the origin of radius $3$, say. Rewrite this integral (assuming $f\ge 0$) by switching to polar coordinates: $$\iint_D |f(x^2+y^2)|^2\,dx\,dy = \iint_D f(x^2+y^2)\left(x\frac{\partial u}{\partial y} - y \frac{\partial u}{\partial x}\right) dx\,dy = \iint_D f(r^2)\frac{\partial u}{\partial\theta}\,r\,dr\,d\theta.$$ Now apply Green's Theorem (integration by parts) in an appropriate way to show that this integral is $0$, thereby arriving at a contradiction.