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I have to solve this PDE, that doesn't have any boundary condition.

$ u_y + uu_x=1 $

$ u=0$ in $y=x^2$

I already found that

$u(s)=s$

$x(s)= \frac{s^2}{2} + x_0$

$y(s)= s+ x_0^2$

With the conditions $x(0)=x_0,$ $y(0)=x_0^2,$ $u(0)=0$

The problem now is that I can't find a way to transform the function $u(s)$ to a function $u(x,y)$.

Does someone as an idea? Thank you

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    How exactly did you obtain your solutions in a variable s – 2017-02-23
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    @Triatticus I used the methods of characteristics. I considered the curve (x(s),y(s),u(s)) in $\mathbb{R}^3 and I supposed those conditions for s=0. – 2017-02-23
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    Ah I see now I was going to suggest that but wasn't sure of what you did – 2017-02-23
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    According to the given condition, I found this solution on the form of an implicit equation : $$u-y+\left(x-\frac{1}{2}u^2\right)^2=0$$. Is it consistent with your result ? – 2017-02-23

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You do have a boundary condition, it is $u=0$ on the parabola $y=x^2$.

To finish the problem, you need to find for a given point $(x,y)$ in the plane which characteristic curve passes through that point, and the value of $s$ at which it exactly hits $(x,y)$. So you need to solve the system of equations

$$x = \frac{s^2}{2}+x_0$$ $$y = s + x_0^2$$

for $s=s(x,y)$ and $x_0=x_0(x,y)$ as functions of $x$ and $y$. Then your solution is $u(x,y)=s(x,y)$.

EDIT: Let me add that if you sketch your characteristic curves, you will find that they pass through the boundary $y=x^2$ twice, and the value of $u$ is not zero both times. So your solution will only be defined locally, and is not unique.