I would appreciate if someone can give me help with the following problem:
Find all solution $u(x, y)$ for $$ \left\{ \begin{array}{ll} u_y=F(u_x) \\ u(x, 0)=h(x) \end{array} \right., $$ where $F$ and $h$ are suitable functions such that problem makes sense.
There is a hint for this: the solution has the form $u=F(p)-pF'(p) +h(x+yF'(p))$, where $p$ is given implicit by $p=h'(x+yF'(p))$.
I have been working in for some days, but couldnt solve. Thanks for any help.
Differentiating the PDE w.r.t. $x$ gives $$ p_y = F(p)_x = F'(p) p_x , $$ where $p=u_x$. Let us follow the method of characteristics for the scalar conservation law $p_y-F'(p) p_x=0$. Thus,
$\frac{\text d y}{\text d t} = 1$, letting $y(0)=0$, we know $y=t$.
$\frac{\text d p}{\text d t} = 0$, letting $p(0)=h'(x_0)$, we know that $p = h'(x_0)$ is constant along characteristics.
$\frac{\text d x}{\text d t} = -F'(p)$, letting $x(0)=x_0$, we know that the characteristics $x = x_0 - y F'(p)$ are straight lines in the $x$-$y$ plane.
The latter rewrites as $x_0 = x + y F'(p)$. Thus, using $p = h'(x_0)$, we get $$ p = h'(x + y F'(p)) \, . $$ Let us compute the evolution of $u(x,y)$ along characteristics $$ \frac{\text d u}{\text d t} = u_y \frac{\text d y}{\text d t} + u_x \frac{\text d x}{\text d t} = F(p) - p F'(p) \, , $$ which is a constant. Integrating the previous identity w.r.t. $t=y$ gives $u = (F(p) - p F'(p)) t + h(x_0)$, i.e. $$ u = (F(p) - p F'(p)) y + h(x + y F'(p)) \, . $$ Note that there must be a mistake in OP, since the proposed solution for $u$ does not satisfy the initial condition.