Unique solution of the Cauchy problem in the neighbourhood of the given set $\Gamma$

Question

The Cauchy problem $ \begin{cases}y\frac{\partial u}{\partial x}-x\frac{\partial u}{\partial y}=0 \\ u=g \;on \;\Gamma \end{cases}$

has a unique solution in a neighbourhood of $\Gamma$ for every differentiable function $g:\Gamma \to R\; if \\ 1. \Gamma =\{(x,0):x\gt 0\} \\ 2. \Gamma =\{(x,y):x^2 +y^2=1\} \\3.\Gamma =\{(x,y):x+y=1,x\gt 1\}\\4. \Gamma =\{(x,y):y=x^2,x\gt 0\}$

My Attempt: I solved the pde by taking $\frac{dx}{y}=\frac{dy}{-x}=\frac{du}{0}\\ u(x,y)=F(x^2+y^2)$

I don't know how to check the uniqueness in the neighbourhood of the given set. Thanks

Answer 1

In case 1. , you've derived \begin{eqnarray} g(x,0) = u(x,0) = F(x^2 + 0^2) = F(x^2). \end{eqnarray} Hence $F(w) = g(\sqrt{w})$ ($w>0$). Therefore \begin{eqnarray} u(x,y) = F(x^2 + y^2) = g\left(\sqrt{x^2 + y^2}\right). \end{eqnarray}