Consider the PDE and initial condition
\begin{align}
u_x + u^2 u_y &= 1 \\
u(x,0) &= 1,
\end{align}
where $u = u(x,y)$ and $(x,y) \in \mathbb{R}^2$. I've solved this with the method of characteristics, and the solution is
\begin{equation}
u(x,y) = \sqrt[3]{3y + 1}.
\end{equation}
Also the non-characteristic boundary condition required for (local) existence of the solution is
\begin{equation}
(1,u^2) \cdot (0,1) = u^2 \neq 0.
\end{equation}
Now my question is, isn't this condition automatically true since the boundary condition for the PDE is $u=1$ at the boundary? Or does this condition somehow constrain the result obtained with the method of characteristics? If $y=-\frac{1}{3}$, the solution is zero, so do we have to define the solution only at points where $y \neq -\frac{1}{3}$?
What I'm confused about is if it's enough for the non-characteristic boundary condition to hold on the boundary, or if it has to hold at every point in the domain for local solution to exist.
Best Answer
$$u_x+u^2u_y=1$$
I don't know exactly what you have done. So, I cannot comment it. It is certainly correct since your result is consistent with mine. I will only comment the manner to express and understand the result on my viewpoint.
For me, the general solution of the PDE is (on the form of implicit equation) : $$\frac{1}{3}u^3-y=F(u-x)$$ where $F$ is any differentiable function.
$u(x,0)=1 \quad\implies\quad \frac{1}{3}1^3-0=F(1-x)=\frac{1}{3}$
This implies that the function $F$ is constant : $F(X)=\frac{1}{3}$ any $X$. So, the arbitrary function $F$ is now well determined according to the condition. Putting it into the general solution leads to : $$\frac{1}{3}u^3-y=\frac13$$ $$u^3=3y+1$$
I cannot see any ambiguity up to now :
$\begin{cases} 3u^2u_y=3\quad\implies\quad u^2u_y=1\\ u_x=0\end{cases}\quad\implies\quad u_x+u^2u_y=0+1=1$
Thus, the solution $u^3=3y+1$ satisfies the PDE and the condition (even at $y=-\frac13$ ).
The apparent difficulty comes when you express the solution on the form : $$u=\sqrt[3]{3y+1}$$ due to the derivative :$\quad u_y=\frac{1}{(3y+1)^{2/3}}\quad$ which is not finite at $y=-\frac13$.
But, in the PDE, the term $u^2u_y$ continue to be finite $=(\sqrt[3]{3y+1})^2\frac{1}{(3y+1)^{2/3}}=1$ at limit $y\to -\frac13$.