Why is the solution to PDE $au_x + bu_y = 0$ $f(bx-ay)$ and not $f(bx+ay)$?

Question

Why is the solution to the following PDE

$$au_x + bu_y = 0,$$ where $a$ and $b$ are both non-zero

$f(bx-ay)$ instead of $f(bx+ay)$?

I just do not understand the logic behind the solution.

Answer 1

The function $u^-(x,y) = f(bx-ay)$ solves $au_x+bu_y=0$, while $u^+(x,y) = f(bx+ay)$ doesn't, for the same reason that $x^+=3$ solves $4x-12=0$, while $x^-=-3$ doesn't - it's just a matter of plugging in.

In the equation $4x-12=0$, the symbol $x$ represents a number. Any solution to the equation is some number with the property that, if you plug that number in, you get a true statement. For example, $x=3$ solves the equation because $4\cdot3-12=12-12=0$. This happens to be the only solution, in this case. In particular, $x=-3$ is not a solution because $4(-3)-12=-12-12=-24\neq0$.

Similarly, in the equation $au_x+bu_y=0$, the symbol $u$ represents a function. A solution to the equation is a function with the property that, if you plug that function in, you get a true statement. For example, if $u(x,y) = f(bx-ay)$ then, by the chain rule, we get $$u_x = f'(bx-ay)b \: \text{ and } \: u_y = f'(bx-ay)(-a).$$ Thus, $$au_x+bu_y = af'(bx-ay)b + bf'(bx-ay)(-a) = abf'(bx-ay)-abf'(bx-ay)=0$$ but, if $u(x,y)=f(bx+ay)$, $$au_x+bu_y = af'(bx+ay)b + bf'(bx+ay)b = abf'(bx-ay)+abf'(bx-ay)\neq0.$$