What initial value do I have to take at the beginning? In my lecture notes there is the following example on which we have applied the method of characteristics: 
$$u_t+2xu_x=x+u, x \in \mathbb{R}, t>0 \\ u(x,0)=1+x^2, x \in \mathbb{R}$$ 
$$$$ 
$$(x(0), t(0))=(x_0,0)$$ 
We will find a curve $(x(s), t(s)), s \in \mathbb{R}$ such that $\sigma '(s)=\frac{d}{ds}(u(x(s), t(s))=u_x(x(s), t(s))x'(s)+u_t(x(s), t(s))t'(s)$ 
$$x'(s)=2x(s), x(0)=x_0 \\ t'(s)=1, t(0)=0 \\ \sigma '(s)=2xu_x+u_t=x(s)+u(s), \sigma(0)=u(x(0), t(0))=1+x_0^2$$ 
$$\dots \dots \dots \dots \dots$$ 
$$t(s)=s \\ x(s)=x_0e^{2s} \\ \sigma(s)=x_0 e^{2s}+(1+x_0^2-x_0)e^s$$ 
If $\overline{s}$ is the value of $s$ such that $(x(\overline{s}), t(\overline{s}))=(x_1, t_1)$ then we have $$x_0e^{2\overline{s}}=x_1 \ , \  \overline{s}=t_1 \\ \Rightarrow x_0=x_1e^{-2t_1}, \overline{s}=t_1$$ 
So for $s=\overline{s}$ we have $$\sigma(\overline{s})=u(x_1, t_1)=x_1 e^{t_1}+x_1^2e^{-3t_1}-x_1e^{t_1}$$ 
$$$$ 
I want to apply this method at the following problem: 
$$u_x(x, y)+(x+y)u_y(x, y)=0, x+y>1 \\ u(x, 1-x)=f(x), x \in \mathbb{R}$$ 
What initial value do I have to take at the beginning? 
$(x(0), y(0))=(x_0, 1-x_0)$ ? 
Or something else?
 A: Yes, you are correct. The characteristic curves must start at the boundary of the domain where the initial conditions are given. In your case that's the line $x + y = 1$, for which $x_0 \mapsto (x_0, 1 - x_0)$ is a valid parametrization. 
A: you have found the characteristic curve $(ae^{2t}, t)$ parametrised by $t.$ you can verify that $$\frac{d}{dt} u(ae^{2t}, t) = u_t + 2ae^{2t}u_x = x + u = ae^{2t} + u  $$ the solution is $$u(ae^{2t}, t) = Ce^{-t} + ae^{2t} $$ using the initial condition $u(a,0) = 1 + a^2,$  gives you $C = 1 -a +a^2.$  therefore $$u(ae^{2t}, t) = (1-a+a^2)e^{-t} + ae^{2t}, x = ae^{2t}\to u(x,t)=\left(1 - xe^{-2t}+x^2e^{-4t}\right)e^{-t}+x $$
A: Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dx}{dt}=1$ , letting $x(0)=0$ , we have $x=t$
$\dfrac{dy}{dt}=x+y=t+y$ , we have $y=y_0e^t-t-1=y_0e^x-x-1$
$\dfrac{du}{dt}=0$ , letting $u(0)=F(y_0)$ , we have $u(x,y)=F(y_0)=F((x+y+1)e^{-x})$
$u(x,1-x)=f(x)$ :
$F(2e^{-x})=f(x)$
$F(x)=f(\ln2-\ln x)$
$\therefore u(x,y)=f(\ln2-\ln((x+y+1)e^{-x}))=f(x-\ln(x+y+1)+\ln2)$
