Suppose $f: \mathbb{R}^2 \setminus \{ (0,0) \} \rightarrow \mathbb{R}$ is a smooth function, such that $$ x \dfrac { \partial f } { \partial y } - y \dfrac { \partial f } { \partial x } = f$$ Show that $f=0$.

I haven't made any progress on the question, mostly just taking partial derivatives to see if anything noticeable happens; I haven't noticed anything. It was given as a question which could be solved by someone who hasn't done a course in partial differential equations, so there's likely a fairly simple trick that will work. If it helps, $e^{xy} $ satisfies the left hand side set to zero, but I doubt that's useful.


2 Answers 2


HINT: Note that the integral curves of the vector field $(-y,x)$ are circles centered at the origin. The left-hand side is the directional derivative of $f$ along such circles. Consider $g(t) = f(a\cos t,a\sin t)$ for any $a>0$. What is $g'(t)$?

  • $\begingroup$ Thanks for the hint. So I've gotten that $g'(t) = g(t)$ which means that $g(t) = Ce^t$ for some real $C$. But since $g(t+2\pi) = g(t)$, we must have that $C=0$, and so we must have $f = 0$ since $a$ was arbitrary. Is that correct? $\endgroup$
    – Cataline
    Jul 26, 2018 at 23:29
  • 3
    $\begingroup$ Perfect !!! There you go. Often with elementary linear PDE it's useful to think about directional derivatives :) $\endgroup$ Jul 26, 2018 at 23:30

This is not an answer to the question, but a comment but too long to be posted in the comments section.

$$ x \dfrac { \partial f } { \partial y } - y \dfrac { \partial f } { \partial x } = f \tag 1$$ Thanks to the method of characteristics, one can express the general solution on the form : $$f(x,y)=\exp\left(\tan^{-1}\left(\frac{x}{y}\right)\right)F\left(x^2+y^2\right)\tag 2$$ where $F$ is an arbitrary fonction.

The function $F$ has to be determined according to some conditions generally specified in the wording of the problem.

We can directly prove that the function $(2)$ is solution of equation $(1)$ :

$$\frac{\partial f}{\partial x}=-\frac{y}{x^2+y^2}\exp\left(\tan^{-1}\left(\frac{x}{y}\right)\right)F+2xF' \tag 3$$

$$\frac{\partial f}{\partial y}=\frac{x}{x^2+y^2}\exp\left(\tan^{-1}\left(\frac{x}{y}\right)\right)F+2yF'\tag 4$$

Thus $$ x \dfrac { \partial f } { \partial y } - y \dfrac { \partial f } { \partial x } =\\ =x\left(\frac{x}{x^2+y^2}\exp\left(\tan^{-1}\left(\frac{x}{y}\right)\right)F+2yF' \right)-y\left(-\frac{y}{x^2+y^2}\exp\left(\tan^{-1}\left(\frac{x}{y}\right)\right)F+2xF' \right)$$ After simplification : $$x \dfrac { \partial f } { \partial y } - y \dfrac { \partial f } { \partial x } =\exp\left(\tan^{-1}\left(\frac{x}{y}\right)\right)F=f$$ The equation $(1)$ is satisfied. So the general solution of the PDE is : $$f(x,y)=\exp\left(\tan^{-1}\left(\frac{x}{y}\right)\right)F\left(x^2+y^2\right)\qquad\text{with arbitrary function }F.$$ I let you think about the consequence of the wording of the question : "Suppose $f: \mathbb{R}^2 \setminus \{ (0,0) \} \rightarrow \mathbb{R}$ is a smooth function."


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