PDE: Find all Functions $u \in C^1(\mathbb{R}^2)$ that satisfy $x_2 u_{x_1} - x_1 u_{x_2} = 0$ Can anyone help me with this problem? It's the start of the PDE lecture so there should be some kind of basic way of figuring out all solutions to the PDE. I tried re-writing it as $u_{x_1} = \frac{x_1}{x_2} u_{x_2}$ and tried to integrate it but I didn't really come up with anything I can work with. 
 A: The problem can be made easier if you make the correct change of variables. In this case polar coordinates will help a lot:
$x_1=r\cos\theta, x_2=r\sin\theta$ and using the chain rule:
$\frac{\partial u }{\partial x_1}=\frac{\partial u }{\partial r}\cos\theta-\frac{\sin\theta}{r}\frac{\partial u }{\partial \theta}$
$\frac{\partial u }{\partial x_2}=\frac{\partial u }{\partial r}\sin\theta+\frac{\cos\theta}{r}\frac{\partial u }{\partial \theta}$
which upon plugging in yields:
$x_2u_{x_1}-x_1u_{x_2}=-\frac{u_\theta}{r}=0$ which in turn implies that:
$$u_{\theta}=0\Rightarrow u=f(r)$$
where f is an arbitrary function, so in the original coordinates the most general solution is:
$$u(x_1,x_2)=f(\sqrt{x_1^2+x_2^2})$$
A: The hypothesis say that at any given point $(x,y)$, the gradient $\nabla\! f(x_1,x_2)$ is orthogonal to $(x_2,-x_1)$. This vector is orthogonal to $(x_1,x_2)$, so what you know is that at any given point $\vec x$, the gradient is a multiple of $\vec x$. If you draw a picture, this should point to the conclusion $f$ is radial.
To prove this, consider $u(t) = f(e^{it} \vec x)$ where $\vec x$ is fixed and non-zero. Then $u'(t) = \nabla\! f(e^{it} \vec x)\cdot ie^{it}\vec x$. Since we know $\nabla\!f(e^{it} \vec x)$ points in the direction of $e^{it}\vec x$, which is orthogonal to $ie^{it}\vec x$, we get what we wanted: $u'(t) = 0$ and $f$ is radial.
