Using a multivariate chain rule to solve a partial differential equation

Question

(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

Answer 1

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.)

Answer 2

The chain rules are: $$ \frac{\partial f}{\partial x} = \frac{\partial f}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial f}{\partial v} \frac{\partial v}{\partial x} = b \frac{\partial f}{\partial u} + a \frac{\partial f}{\partial v} \\ \frac{\partial f}{\partial y} = \frac{\partial f}{\partial u} \frac{\partial u}{\partial y} + \frac{\partial f}{\partial v} \frac{\partial v}{\partial y} = -a \frac{\partial f}{\partial u} + b \frac{\partial f}{\partial v} $$ Can you take it from here? No?
Perhaps a reference may help : First-Order Equations: Method of Characteristics .