I’m attempting to understand a geometric solution of first order PDEs with constant coefficients:
Suppose we have the PDE $$au_x + bu_y=0$$ for $a, b$ real. Geometrically, this can be rewritten in terms of the directional derivative in the direction $(a,b)$: $$(a,b)^T• \nabla u=0$$ Thus we find that our function $u$ is constant along the lines $$c=bx-ay$$ Now then, in the solution presented in my text, we can then state that $u$ depends only our constants $c$, i.e. $$u=f(c)=f(bx-ay)$$ for an arbitrary function $f$. Now I’m not entirely certain why $u$ must depend on $c$. Is it because $u$ is constant along the aforementioned lines and those lines are entirely determined by $c$?
Or alternatively, and perhaps equivalently if I’m not mistaken, is it because $\mathbb R^2$ is spanned by lines in the direction $(a, b)$, and since $u$ is constant on those lines and they are determined by $c$, then $u$ is determined by $c$ also?
Since the function is constant along those lines, you need only consider what its value is on each line. So it's not really a function of $x$ and $y$ anymore, it's a function of which line you are on.
For your alternative explanation, basically yes. You can form a basis of $\mathbb{R}^2$ using the direction the function is constant on and something orthogonal to it. Then you are constant in one of the components so your function becomes 1-dimensional.