2 variable function only depending on the ratio $x/y$ iff $ \frac{\partial f}{\partial x} x + \frac{\partial f}{\partial y} y = 0$

Question

If a differentiable function of two real variables $f(x,y)$ only depends on the ratio $(x/y)$, i.e. $f(x,y)=g(x/y)$ it is easy to see that

$$ \frac{\partial f}{\partial x} x + \frac{\partial f}{\partial y} y = 0$$

since $\partial f / \partial x = g'(x/y) (1/y)$ and $\partial f / \partial y = g'(x/y) (-x/y^2)$ and:

$$ \frac{\partial f}{\partial x} x + \frac{\partial f}{\partial y} y = g'( x/y ) (x/y - x/y) = 0$$

I suspect the converse is also true, i.e. if $ \frac{\partial f}{\partial x} x + \frac{\partial f}{\partial y} y = 0$ there exists a $g$ such that $f(x,y)=g(x/y)$. Can anyone give me a proof? I don't know how to start. (or a counterexample if I am wrong)

Related questions: What is the geometric interpretation of the solution to PDE $xu_x+yu_y=0$, General solution to the PDE $xU_x+yU_y=0$ — Winther, Dec 12 '16 at 01:57

evilcman · Answer 1 · 2016-12-11T22:23:13.733

I think I have a solution. Let $y=cx$, and consider the function: $G(x,c) = f(x,y) = f(x,cx)$. Now with the notation $f_1(x,y)=\partial f(x,y) / \partial x$ and $f_2(x,y)=\partial f(x,y) / \partial y$ we have by assumption: $$ f_1(x,y)x+f_2(x,y)y=0 \implies f_1(x,cx)x+f_2(x,cx)cx=0 \implies f_1(x,cx)+f_2(x,cx)c=0$$ this in turn implies $\partial G/ \partial x = f_1(x,cx)1 + f_2(x,cx)c=0$ that is the function $f(x,cx)$ does not depend explicitly on x, so it can only depend on $c=y/x$. This is what we wanted to prove.

score 1 · Answer 2 · answered Dec 11 '16 at 19:34

1

Idea: if true, obviously $g(x) = f(x,1)$. Consider now $$f(x,y) - g(x/y) = f(x,y) - f(x/y,1).$$

answered Dec 11 '16 at 19:34

Martín-Blas Pérez Pinilla

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I think I figured out a solution. On the surface it does not seem to involve your hint. Is your idea something similar? – evilcman Dec 11 '16 at 22:01
@evilcman, my idea is to prove that the partial derivatives of $f(x,y)−f(x/y,1)$ are $= 0$. – Martín-Blas Pérez Pinilla Dec 11 '16 at 22:08
@evilcman, but your idea is easier. – Martín-Blas Pérez Pinilla Dec 11 '16 at 22:18

score 0 · Answer 3 · answered Dec 18 '16 at 14:17

This is more a comment that an answer:

The general solution of the PDE $\quad \frac{\partial f}{\partial x} x + \frac{\partial f}{\partial y} y = 0 \quad$ thanks to the method of characteristics is :

Equation of characteristic curves : $\quad\frac{dx}{x}=\frac{dy}{y}=\frac{df}{0} \quad\to\quad \begin{cases}\frac{x}{y}=c_1\\f=c_2\end{cases}\quad$ with independent $c_1,c_2$ on the characteristic curves. Outside, $c_1$ and $c_2$ are related, so that $\quad c_2=F(c_1)=G\big(\frac{1}{c_1}\big)\quad$ where $F$ and $G$ are any differentiable functions.

Thus the general solution is either $\quad f(x,y)=F\big(\frac{x}{y}\big)\quad$ or $\quad f(x,y)=G\big(\frac{y}{x}\big)$ .

Hence, if $\quad \frac{\partial f}{\partial x} x + \frac{\partial f}{\partial y} y = 0 \quad$ they exist an infinity of functions on the form $f(x,y)=F\big(\frac{x}{y}\big)\quad$ or on the form $\quad f(x,y)=G\big(\frac{y}{x}\big)$

2 variable function only depending on the ratio $x/y$ iff $ \frac{\partial f}{\partial x} x + \frac{\partial f}{\partial y} y = 0$

3 Answers3