solve $uu_{x_1}+u_{x_2} = 1$ using Method of Characteristics I am solving this PDE: $$uu_{x_1}+u_{x_2} = 1; u(x_1,x_1)= \frac {x_1}2$$
For the characteristic equations I have:
$$(x^1)' = z$$
$$(x^2)' = 1$$
$$z' = 1$$
Attempting to solve these ODEs I get: $$x^2(s) = x^1_o + s$$ but I am confused on how to solve the first and third characteristic ODEs. 
Also, how do I know what z(s) is?
Sorry, I am new to PDEs and this method but am trying to learn.
 A: $$
\begin{cases}
\dfrac {dx_1}{ds}=u \\
\dfrac {dx_2}{ds}=1  \\
\dfrac {du}{ds}=1
\end{cases}
$$
$$ \implies 
\begin{cases}
\dfrac {dx_1}{ds}=u \\
x_2=s+x_2(0)  \\
u=s+u(0)
\end{cases}
$$
First equation is easy to integrate
$$\frac {dx_1}{ds}=u $$
$${dx_1}=\int u ds $$
$$\implies x_1=\int s+u(0) ds \\$$
$$ x_1=\frac { s^2}2+u(0)s+x_1(0) \\$$
Apply intitial conditions now
$$
\begin{cases}
x_1(0)=r \\
x_2(0)=r  \\
u(0)=\frac {r}2
\end{cases}
$$
Therefore:
$$
\begin{cases}
 x_1=\dfrac { s^2}2+s\dfrac r2+r \\
x_2=s+r  \\
u=s+\dfrac {r}2
\end{cases}
$$
$$
\begin{cases}
 x_1=\dfrac { s^2}2+s\dfrac {r}2+r \\
x_2=s+r \\
\end{cases}
\implies
 \begin{cases}
 s=2\dfrac { x_1-x_2}{x_2-2} \\
r=x_2- s\\
\end{cases}
$$
From the third equation we have:
$$u(x_1,x_2)=s+\dfrac r2$$
$$u(x_1,x_2)=\frac s2+\dfrac {x_2}2$$
$$u(x_1,x_2)=\dfrac { x_1-x_2}{x_2-2}+\dfrac {x_2}2$$
Finally,
$$\boxed {u(x_1,x_2)=\dfrac { x_1-2}{x_2-2}+\dfrac {x_2}2-1}$$
A: We are given
$$uu_{x_1}+u_{x_2} = 1$$
$$u(x_1,x_1)= \frac{x_1}2$$
The initial data curve is $\Gamma: \langle s,s,s/2 \rangle$. Therefore, the Jacobian evaluated with the boundary data is
$$J\Bigg|_\Gamma=\begin{vmatrix}
{x_1}_s & {x_1}_t \\
{x_2}_s & {x_2}_t
\end{vmatrix}\Bigg|_\Gamma=\begin{vmatrix}
1 & \dfrac{s}{2} \\
1 & 1
\end{vmatrix}=1-\frac{s}{2}$$
So, there is a unique solution in the neighborhood of $\Gamma$ provided $s\neq 2$ . The characteristic equations are 
\begin{cases} 
      \dfrac{dx_1}{dt}=u, &  x_1(s,0)=s \\[1em]
      \dfrac{dx_2}{dt}=1, & x_2(s,0)=s \\[1em] 
      \dfrac{du}{dt}=1, & u(s,0)=\dfrac{s}{2}
   \end{cases}
For the third equation,
$$\frac{du}{dt}=1 \implies du=dt\implies u = t+c_1(s)$$
Plugging in the initial condition of $ u(s,0)=\dfrac{s}{2} \implies  c_1(s)=\dfrac{s}{2}$. Hence,
$$\boxed{u=t+\frac{s}{2}}\tag{1}$$
For the second equation,
$$\frac{dx_2}{dt}=1 \implies dx_2 = dt \implies x_2=t+c_2(s)$$
Plugging in the initial condition of $ x_2(s,0)=0 \implies  c_2(s)=s$. Hence,
$$\boxed{x_2=t+s}\tag{2}$$
For the first equation,
$$\frac{dx_1}{dt}=u \implies dx_1 =u\,dt \implies x_1=\frac{t^2}{2}+\frac{st}{2}+c_3(s)$$
Plugging in the initial condition of $ x_1(s,0)=s \implies  c_3(s)=s$. Hence,
$$\boxed{x_1=\frac{t^2}{2}+\frac{st}{2}+s}\tag{3}$$
Combining $(1),(2),(3),$ we see that
$$s=\frac{x_2^2-2x_1}{x_2-2}$$
$$t=\frac{2(x_1-x_2)}{x_2-2}$$
and upon solving we find
$$\boxed{u=u(x_1,x_2)=t+\frac{s}{2}=\frac{2(x_1-x_2)}{x_2-2}+\frac{x_2^2-2x_1}{2(x_2-2)}=\frac{x_1-2}{x_2-2}+\frac{x_2}{2}-1}\tag{4}$$
in which the solution does not exist for $x_2=2$ as we found from our initial parametrization.
