suppose that any solution $u(x,y)$ to $$ au_{x} + b u_{y} = 0,$$ satisfies $u(1,2) = u(3,6)$. What is $b/a$?
$$au_{x} + bu_{y} = \nabla u \cdot \binom{a}{b} = 0, $$ which implies that $u$ is constant along lines of the form $bx- ay =c \ \ \forall \ c \in \mathbb{R}$. Therefore the solution is $$ u(x,t) = g(bx - ay) , \qquad \text{for } g \text{ arbitrary and } \ g: \mathbb{R} \to \mathbb{R}.$$ \begin{align} \because u(1,2) = u(3,6) &\implies g(b-2a) = g(3b-6a), \\ &\implies b-2a = 3b -6a \tag{1}\\ &\therefore \frac{b}{a} = 2\end{align}
I'm quite confused about how to justify $(1)$, I note that $b-2a$ is a linear combination of $3b-6a$ but I also know that for instance $f(x) = x^{2}$ we have $f(2) =f(-2) \nRightarrow 2=-2 $. Any insights ?
$$au_x+bu_y=0$$ The general solution is straightaway found thanks to the method of characteristics : $$u(x,y)=F(bx-ay)$$ $F$ is an arbitrary function.
$$u(1,2)=u(3,6)\quad\implies\quad F(b-2a)=F(3b-6a)$$ Let $X=b-2a$ $$F(X)=F(3X)$$ Since this equality must be satisfied for all above solutions $u(x,y)$ , that is for all functions $F(X)$ thus : $$F(X)=F(3X)\quad\implies\quad X=0$$ $$b-2a=0$$ $$\frac{b}{a}=2$$