If $\nabla f(x,y) \cdot (x,y) = f(x,y)$, then $f$ is first degree homogeneous

Question

Given $f$ differentiable in $\mathbb{R}^2 \setminus \{(0,0)\}$. We're asked to prove that if $\nabla f(x,y) \cdot (x,y) = f(x,y)$ for all $\mathbb{R}^2$, then $f$ is an homogeneous function, i.e. $f(t(x,y))=tf(x,y)$ for $t>0$ and $(x,y) \neq (0,0)$.

My prof. explained that proving the consequent amounts to proving that

$$g(t) = \frac{f(tx_0, ty_0)}{t}, t>0,$$

is a constant function, which in turn amounts to showing that its derivative is 0 everywhere. But how do we justify that $g(t) = f(x_0,y_0)$?

Answer 1

Well, once you show that $g$ is constant, you'll have $$f(x_0,y_0) = g(1) = g(t) = \frac{f(tx_0,ty_0)}{t}$$for all $t >0$, and so $f(tx_0,ty_0) = tf(x_0,y_0)$. Since $(x_0,y_0)$ was arbitrary to begin with, we're done. Now, we use the chain and quotient rules to get $$g'(t) = \frac{f(tx_0,ty_0) - (tx_0,ty_0)\cdot \nabla f(tx_0,ty_0)}{t^2} = 0.$$

Answer 2

We can even solve the differential equation.

In polar coordinates the equation becomes $ r\hat{\mathbf{r}} \cdot \left( \hat{\mathbf{r}} \partial_r f + \hat{\mathbf{\alpha}} \partial_\alpha f \right) = f, $ i.e. $ r \, \partial_r f = f . $

The solutions to this are given by $ f = C(\alpha) \, r = C(\arctan(y/x)) \, \sqrt{x^2+y^2} , $ where $C$ is a differentiable function.

Thus, $$ f(tx, ty) = C(\arctan((ty)/(tx))) \, \sqrt{(tx)^2+(ty)^2} = C(\arctan(y/x)) \, t \, \sqrt{x^2+y^2} = t \, f(x,y). $$