I recently asked why $u(x, y) = A(x - y)$ is the general solution to this problem:
Solve the PDE $u_x + u_y = 0$ in the domain $y > \phi(x)$, $x \in \mathbb{R}$ given that $u = g(x)$ on the curve $y = \phi(x)$, where $\phi(x) = \frac{x}{1 + |x|}$.
From the generous responses of Batominovski and Mattos, I tried to develop my own understanding of this. I am seeking advice from others as to whether my reasoning is correct. In addition to my reasoning please also critique my language. I am not sure if the way I have thought about everything is correct so this might be reflected in my use of mathematical language (I might have described certain things incorrectly, confused concepts, and so on).
Starting with the easy part, we use separation of variables on two of the characteristic equations:
$\dfrac{dx}{dt} = 1$
$x(t) = t + C_1(\gamma)$
$\dfrac{dy}{dt} = 1$
$y(t) = t + C_2(\gamma)$
$C_1(\gamma)$ and $C_2(\gamma)$ are constants along any specific characteristic curve but this constant is different for different characteristic curves. That is why I set $C_1$ and $C_2$ as being dependent on $\gamma$. I will say that when $t = 0$, then we are on the initial curve and $\gamma$ will change along the initial curve as it passes through different characteristic curves.
If I am correct in my thinking, the $y$ characteristic must inherit its parameterisation from the curve $y = \phi(x) = \frac{x}{1 + |x|}$. However we can parameterise $x$ in a way that is convenient. I will set $x = \gamma$ when $t = 0$ and $y = \phi(x) = \frac{x}{1 + |x|}$ when $t = 0$.
So we have $x(0) = C_1(\gamma) = \gamma$ and $y(0) = C_2(\gamma) = \frac{x}{1 + |x|} = \frac{\gamma}{1 + |\gamma|}$ (since $x = \gamma$ when $t = 0$). This means that our characteristic curves are $x(t) = t + \gamma$ and $y(t) = \frac{\gamma}{1 + |\gamma|}$.
If I am not mistaken, these are our change of variables, which map from $(s, t) \to (x, y)$.
We now try to find the general solution by using separation of variables on the third characteristic equation.
We have $\dfrac{du}{dt} = 0$.
Using separation of variables gives $u(x, y) = A(C_3)$.
Now we come to the part that was of main confusion in the recently asked question: Why is $A$ dependent on $C_3$? My reasoning is this:
The values of $C_1(\gamma)$ and $C_2(\gamma)$ in the characteristic equations change from one characteristic to another but are constant on any single characteristic. In other words, the values of $C_1(s)$ and $C_2(s)$ distinguish the different characteristics. $A$ is also a constant along any single characteristic but will be different for different characteristics. Therefore, $A$ will change depending on $C_1(\gamma)$ and $C_2(\gamma)$ in the characteristic equations. Therefore, let $C_3 = C_1 - C_2$. We then have $x - y = C_1 - C_2 = C_3$. Therefore, we have our result that $u = A(C_3) = A(x - y)$.
Thank you for kindly reviewing.