Consider the first order PDE
\begin{cases} u_x + u_y = u^2 \\ u(x,0)=v(x) \end{cases}
for $x \in \mathbb{R}$ and $y>0$. For which functions $v \in C^{\infty}(\mathbb{R})$ does exists a solution $u(x,y) \in C^1(\mathbb{R} \times [0,\infty))$ ?
I applied the method of characteristics to the equation above, and found:
$$u(x,y)=\frac{v(x-y)}{1 - v(x-y) y}$$
How can I estabilish some conditions on $v$? I would just say that it's sufficient to have $$v(x-y) y < 1$$ for every $x \in \mathbb{R}$ and $y >0$, but I don't know if it is what expected.
Your condition is sufficient. But remark that if $v(\cdot)$ is bounded, then your condition $v(x-y)y < 1$ is locally valid on some interval $y\in [0,\delta)$. For example if $v(x) \equiv 1$, then $$u(x,y) = \frac{1}{1-y}$$ Which solves $$u_y =u_x+u_y= u^2, \ u(x,0) = 1$$ This is actually an ODE in $y$ (so you can apply uniqueness and existence theorems) and it is easy to see that the solution blows up at $y=1$ and is therefore only valid on $[0,1)$ (if we only consider $y\geq 0$). Just like non-linear ODEs, non-linear PDEs can also have a finite blow up time.