Problem 4

[Rouhollah Ramezani]

Hint: the l.h.e. is $-\partial_\theta u$. Prove it using chain rule  $u_\theta = u_x x_\theta + u_y y_\theta$ and calculate $x_\theta$, $y_\theta$.

$$x=r\cos{\theta} \rightarrow x_{\theta}=-r\sin{\theta}\\
y=r\sin{\theta} \rightarrow y_{\theta}=r\cos{\theta}\\
\Rightarrow yu_x-xu_y=r\sin{\theta}u_x-r\cos{\theta}u_y \\
= -u_x x_\theta - u_y y_\theta \\
=-u_\theta
$$

Note that $\theta$ is defined modulo $2\pi \mathbb{Z}$ and all functions must be $2\pi$-periodic with respect to $\theta$ (assuming that we consider domains where one can travel around origin)

Do you mean:
$$-u_\theta=r^2 \\
\Rightarrow u=-r^2\theta
$$ and since it's not periodic we are done?

[Victor Ivrii]

Do you mean:
$$-u_\theta=r^2 \\
\Rightarrow u=-r^2\theta
$$ and since it's not periodic we are done?

Basically yes, except certain things should be clarified: since our trajectories are closed each is parametrized by some parameter ($\theta$) running from $0$ to $T$ (in our case $T=2\pi$ but it may depend on trajectory) and equation looks like $\partial_\theta u= g(\theta,r)$. So we are looking for periodic solution
$u(\theta,r)=\int g(\theta,r)\,d\theta$.

Note that primitive of periodic function $g(\theta)$  is periodic if and only if average of $g$ over period is $0$:
$\int_0^T g(\theta)\,d\theta=0$. Otherwise this primitive is the sum of a periodic function and a linear function.

Finally, in (a) $g(\theta)=r^2 \sin(\theta)\cos(\theta)$ and integral over period is $0$; in (b) $g(\theta)=r^2$ and  integral over period is not $0$.

So, the source of trouble is: periodic trajectories.

[Di Wang]

is there any clear solution for part c of question four. thank you?

[Victor Ivrii]

is there any clear solution for part c of question four. thank you?

See above