$y u_x-xu_y=0,u=g $ on $ \Omega $ has a unique solution in neighborhood of $\Omega$ for every differentiable function g: $\Omega \rightarrow R$ if
1.$\Omega =\{(x,0):x>0\}$
2.$\Omega =\{(x,y):x^2+y^2=1\}$
3.$\Omega =\{(x,y):x+y=1,x>1\}$
4.$\Omega =\{(x,y):y=x^2,x>0\}$
What i have tried I use Lagrange's method $$\frac{dx}{y}=\frac{dy}{-x}=\frac{du}{0}$$ $\implies u=c_1,x^2+y^2=c_2$
Solution is of the form $u= \phi(x^2+y^2)$ where $\phi$ has to be match with condition given . For option 1 after apply condition $g= \phi (x^2)$ $\implies \phi(x)= g(√x)$
solution becomes $u = g(√x^2+y^2)$ So option 1 looks correct to me
for option 2 After apply condition
$g= \phi(1)$ from this i got no solution option 2 looks wrong to me
I am stuck at 3rd and 4th option please help me for these and also check my explanation for option 1and 2 . or is there other method to solve such problems
In case 2) you can only get solutions if $g$ is constant on the unit circle, and then one has infinitely many solutions. So there is never a unique solution.
In case 3) the only point of tangency to a characteristic circle would be at $x=y=\frac12$ which is not on the ray. Thus you can parametrize a neighborhood of the ray using circle segments or the full circles $x^2+y^2=r^2$, $r>1$.
In case 4), the parabolic segment is likewise perpendicular to the characteristic circles, thus giving a unique solution on $\Bbb R^2\setminus\{(0,0)\}%$