Method of characteristic - what am i doing wrong? Problem:
$$xu_x + u^2u_y=1$$
$$u(x,0)=1$$
My attempt:
Characteristic ODEs are:
$$\frac{dx}{dt} = x \\
\frac{dy}{dt} = z^2 \\
\frac{dz}{dt} = 1$$
with ICs:
$$x(0,s) = s \\
y(0,s) = 0 \\
z(0,s) = 1$$
so the solutions to ODEs come out to be:
$$x(t,s) = se^t \\
z(t,s) = t+1 \\
y(t,s) = \frac{(t+1)^{3}}{3}$$
so,
$$u(x,t) = (3y)^{\frac{1}{3}}$$
this satisfies the equation but does not satisfy $u(x,0) = 1$.
In order to satisfy both the equation and $u(x,0) = 1$, I need the solution to be
$$u(x,t) = (3y+1)^{\frac{1}{3}}$$
so help me figure out where I'm missing this '$1$'.
 A: Your solution to your $y$ equation is wrong.
You have
$$\begin{align}
\frac{dy}{dt} &= u^{2} \\
&= (t + 1)^{2} \\
\implies y(t) &= \frac{(t + 1)^{3}}{3} + C
\end{align}$$
With $y(0) = 0$
$$\begin{align}
\implies y(0) &= \frac{(0 + 1)^{3}}{3} + C \\
&= \frac{1}{3} + C \\
&= 0 \\
\implies C &= \frac{-1}{3} \\
\implies y(t) &= \frac{(t + 1)^{3}}{3} - \frac{1}{3} \\
\end{align}$$
And some notational issues..
If you have reparameterised your curve using $t$, why then do you have initial conditions in $s$? It's not a big deal, but could confuse you in harder problems. Also, where did a $z$ variable come from? Your function is in $u$.. Again, not a big deal if you are aware of what you are doing, but it could confuse you when doing harder problems and can also confuse people reading your work.
A: Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{du}{dt}=1$ , letting $u(0)=0$ , we have $u=t$
$\dfrac{dx}{dt}=x$ , letting $x(0)=x_0$ , we have $x=x_0e^t=x_0e^u$
$\dfrac{dy}{dt}=u^2=t^2$ , letting $y(0)=f(x_0)$ , we have $y=\dfrac{t^3}{3}+f(x_0)=\dfrac{u^3}{3}+f(xe^{-u})$
$u(x,0)=1$ :
$0=\dfrac{1^3}{3}+f(xe^{-1})$
$f\left(\dfrac{x}{e}\right)=-\dfrac{1}{3}$
$f(x)=-\dfrac{1}{3}$
$\therefore y=\dfrac{u^3}{3}-\dfrac{1}{3}$
$u^3=3y+1$
