Unique solution of Cauchy problem in a neighbourhood of given set Consider the  Cauchy problem $yu_x-xu_y=0$ where $u=g$ on $S=${$(x,y): x+y=1, x>1$}
Prove that the Cauchy problem has a unique solution in a neighbourhood of $S$ for every differentiable function $g:S\rightarrow \mathbb R$.

I assumed the solution of the form $u(x,y)=X(x).Y(y)$ and on plugging it into $yu_x-xu_y=0$, I got $$u(x,y)=Aexp(\frac{k(x^2+y^2)}{2})$$

Then I wanted to use the given condition  $u=g$ on $S=${$(x,y): x+y=1, x>1$} in my solution, but stuck at this point. Please help how to proceed from here.
 A: First, you have to find the general solution of your first-order semilinear PDE on a suitable open domain which contains $S.$ You'll find that the general solution is
$$
u(x,y)=F(x^2+y^2)
$$
where $F$ is a continuously differentiable function in one variable (I'd recommend my book on PDEs to see how to handle first-order semilinear PDEs.) 
Next, a reasonable way to treat your initial condition is to rewrite it as
$$
u(s,1-s)=g(s), \quad (s > 1),
$$
where $g$ is a continuously differentiable function.
Now let us a take any solution $u(x,y)=F(x^2+y^2)$ and see which function $F$ meets the initial condition. We have that
$$
F(s^2+(1-s)^2)=F(2s^2-2s+1)=g(s)
$$
for all $s > 1.$ The function 
$$
\alpha(s)=2s^2-2s+1
$$
on $(1,+\infty)$ is evidently invertible, its inverse is
$$
\alpha^{-1}(s)=\frac 12 + \frac{\sqrt{2s-1}}2, \quad (s > 1).
$$
Then
$$
F(\alpha(s))=g(s) \Rightarrow F(s)=g(\alpha^{-1}(s)).
$$
for all $s > 1$ (why?) Finally, the function
$$
u(x,y)=g(\alpha^{-1}(x^2+y^2))=g\left(\frac 12 +\frac 12 \sqrt{2x^2+2y^2-1} \right)
$$
is the solution we were looking for.
