How to solve $uu_{x_1}+u_{x_2}=1$ with characteristic method? $$uu_{x_1}+u_{x_2}=1\text{ with initial condition }u(x_1,x_1)=\frac{1}{2}x_1$$
I have problem to use following characteristic method to solve it.
Let $$x(s) = (x_1(s), x_2(s))$$$$z(s) = u(x(s))$$$$p(s) = (u_{x_1}(x(s)), u_{x_2}(x(s))$$$$F(x,z,p) = zp_1 + p_2 - 1 = 0$$
Assume, $$\frac{dx}{ds} = F_p = (z,1) $$$$\frac{dz}{ds} = F_p \cdot p = 1$$$$\frac{dp}{ds} = -F_x - F_z p = 0 - p_1(p_1,p_2) = - (p_1^2,p_1p_2)$$
Now, we can have, $$z(s) = s + c_0$$$$\dot x_1 = s + c_0 \Rightarrow x_1 = \frac{1}{2}s^2 + sc_0 + c_1$$$$\dot x_2 = 1 \Rightarrow x_2 = s + c_2$$
I'm stucked here. I don't know what to do next.

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dx_2}{dt}=1$ , letting $x_2(0)=0$ , we have $x_2=t$
$\dfrac{du}{dt}=1$ , letting $u(0)=u_0$ , we have $u=t+u_0=x_2+u_0$
$\dfrac{dx_1}{dt}=u=t+u_0$ , letting $x_1(0)=f(u_0)$ , we have $x_1=\dfrac{t^2}{2}+u_0t+f(u_0)=\dfrac{x_2^2}{2}+(u-x_2)x_2+f(u-x_2)=x_2u-\dfrac{x_2^2}{2}+f(u-x_2)$ , i.e. $u-x_2=F\left(x_1-x_2u+\dfrac{x_2^2}{2}\right)$
$u(x_1,x_2=x_1)=\dfrac{x_1}{2}$ :
$F(x_1)=-\dfrac{x_1}{2}$
$\therefore u-x_2=-\dfrac{x_1}{2}+\dfrac{x_2u}{2}-\dfrac{x_2^2}{4}$
$4u-4x_2=-2x_1+2x_2u-x_2^2$
$(2x_2-4)u=2x_1+x_2^2-4x_2$
$u=\dfrac{2x_1+x_2^2-4x_2}{2x_2-4}$
