Q:

How to show that a given PDE doesnot have continously differentiable solution

If $iu_x-u_y=0$ is a given PDE with $u(0,s)=g(s)$ as boundary condition and $g(s)$ is not analytic then I have to show that  the given pde has no $C^1$ solution. 

I know the given curve is not characteristic and the solution can be found using the method of characterstics as $u(x,y)=g(x+iy)$.
I know that $g$ is not analytic but how to claim that $u$ is not even $C^1$ ?
Any help will be appreciated. 

A:

Setting $u = U + i V$, $U,V\in\mathbb R$, the PDE is $$ iU_x-U_y - V_x -iV_y = 0,$$
i.e. $U_x = V_y, V_x = -U_y$.
So the PDE is another way of writing the Cauchy-Riemann equations, and the solution is analytic at any point where this equation holds. Hence, if
$$ iu_x - u_y=0$$
held for each $x>0$, and the derivatives were continuous at $x=0$, this would imply that $$ i u_x - u_y=0$$ held at $x=0$ as well, i.e. $g$ would be analytic, contradicting the assumption.

