I have a couple of questions in regards to the following passage from Partial Differential Equations by Strauss:
When we say that the solution is constant on each such line, are we restating that $bx-ay$ is equal to the constant as stated in the previous sentence?
Why does it follow that $u(x,y)$ depends on $bx-ay$ only?
Let us solve $$au_{x} + bu_{y} = 0, $$
where $a$ and $b$ are constants not both zero. The quantity $au_{x} + bu_{y}$ is the directional derivative of $u$ in the direction of the vector $V = (a, b) = ai + bj$. It must always be zero. This means that $u(x, y)$ must be constant in the direction of $V$. The vector $(b, −a)$ is orthogonal to $V$. The lines parallel to $V$ (see Figure 1)
have the equations $bx – ay$ = constant. (They are called the characteristic lines.) The solution is constant on each such line. Therefore, $u(x, y)$ depends on $bx – ay$ only. Thus the solution is $u(x, y) = f (bx − ay)$, where $f$ is any function of one variable.
1) For any real number $c$ the following is true: $u(x,y)=u(x',y')$ whenever $(x,y)$ and $x',y')$ satisfy the equation s$bx-ay=c$ and $bx'-ay'=c$. This is what is meant by sayning that the solutio is consatn t on the lines $bx-ay=$ constant.
2). Define $f$ as follows: $f(t)=u(x,y)$ where $(x,y)$ is any pair such that $bx-ay=t$. [One such choice is $y=0,x=\frac t b]$. We have to make sure that this is a well defined function. In other words we have to show that $bx-ay=t$ and $bx'-ay'=t$ imply that $u(x,y)=u(x',y')$. But we already know (by 1))that this is true.
It is obvious from the definition of $f$ that $u(x,y)=f(bx-ay)