(a) Let $a$ and $b$ be some parameters for which $a^2 + b^2$ > 0 and use the
substitution $u = bx−ay$ and $v = ax+by$, to rewrite the partial differential equation
$af_{x} + bf_{y} = 0$, into a p.d.e. in the variables $u$ and $v$ instead.
(b) Solve the p.d.e. in part (a).
What i tried
Based on the multivariate chain rule since $u=(x,y)$ and $v=(x,y)$ I drew a tree diagram to releate $f_{x}$ and $f_{y}$ into $u$ and $v$.
What i got was $f_{x}=u_{x}+v_{x}$ and $f_{y}=u_{y}+v_{y}$
From here i just work out the value of $u_{x}+v_{x}$ to get $f_{x}$ and the same for $f_{y}$.
But im unsure whether this is correct and how to continue from here. Could someone please explain and guide me to the right answer. Thanks
We have to express the function $(x,y)\mapsto f(x,y)$ in terms of the new variables $u$, $v$. To this end we have to invert the defining equations $$u=bx-ay,\quad v=ax+by$$ and obtain $$x(u,v)={bu+av\over a^2+b^2},\quad y(u,v)={-au+bv\over a^2+b^2}\ .$$ Our function $f$ now appears as $$\tilde f(u,v):=f\bigl(x(u,v),y(u,v)\bigr)\ .$$ Using the chain rule we find that $$\tilde f_v=f_x\>x_v+f_y\>y_v={1\over a^2+b^2}(f_x a+f_y b)=0\ .$$ Therefore the given PDE appears in the new variables as $$\tilde f_v=0\ .\tag{1}$$ The solutions to $(1)$ are the functions depending solely on $u$: $$\tilde f(u,v)=g(u)\ ,\tag{2}$$ where $g:\>{\mathbb R}\to{\mathbb R}$ is an arbitrary function of one real variable. In order to get the solutions to the original PDE we rewrite $(2)$ in terms of $x$, $y$, and obtain $$f(x,y)=\tilde f\bigl(u(x,y),v(x,y)\bigr)=g\bigl(u(x,y)\bigr)=g(bx-ay)\ .$$ (Of course a geometric argument starting from $(a,b)\cdot\nabla f(x,y)\equiv0$ would lead to the same result.)