I'm trying to solve the Cauchy problem:
$u_{x}+xu_{y}=0, x \in\mathbb{R}$
$u(x,-1)=exp(x)$
Parametrization of the surface: $(x,y,z)=(s,-1, e^{s})$
I got to these projected characteristics: $x=\tau+s$ $y=-\tau^{2}/2+s\tau-1$
So now I want to express \tau and s in terms of x and y. Now I'm stuck at the quadratic equation. I understood that the solution should be unique so my first step was to set the discriminant to zero but then I get z=1. If I should choose one of the signs, how to choose it? What does it depend on?
$$u_{x}+xu_{y}=0 $$ Characteristic equations: $\quad \frac{dx}{1}=\frac{dy}{x}=\frac{du}{0} $
First family of characterisics curves, from $\quad \frac{dx}{1}=\frac{dy}{x}\quad\to\quad \frac{x^2}{2}-y=c_1$
Second family of characterisic curves, from $\quad du=0\quad\to\quad u=c_2$
General solution of the PDE : $\quad u=F\left(\frac{x^2}{2}-y\right)\quad$ where $F(X)$ is any differentiable function.
Condition : $\quad u(x,-1)=e^x =F\left(\frac{x^2}{2}-(-1)\right) $
In this condition, $\quad X=\frac{x^2}{2}+1 \quad\to\quad x=\pm\sqrt{2(X-1)}\quad\to\quad F(X)=e^{\pm\sqrt{2(X-1)}}$
Sign $+$ if $x>0$ or sign $-$ if $x<0$.
Puting this function into $\quad u=F\left(\frac{x^2}{2}-y\right)\quad$ with $\quad X=\frac{x^2}{2}-y\quad$ leads to : $$u(x,y)=e^{\pm\sqrt{2\left(\frac{x^2}{2}-y-1\right)}}= e^{\pm\sqrt{x^2-2y-2}} $$