Show that if $f(x,y)$ is a monomial then $Df = Nf$

Question

$D = x∂_x + y∂_y$

A monomial $x^my^n$ with total degree $m + n $

So then they ask show that, if $f(x,y)$ is a monomial then $Df = Nf$

$N = m + n$ is the total degree of $f$

Answer 1

I understand the question as the checking for some class of solutions for a pde.

$x\partial_xf+y\partial_yf=Nf$ for some N. Trying with $f(x,y)=x^my^n$

$\partial_xf=mx^{m-1}y^n$ and $\partial_yf=x^mny^{n-1}$

$x\partial_xf+y\partial_yf=xmx^{m-1}y^n+yx^mny^{n-1}=mx^my^n+nx^my^n=(m+n)f$

It's the same relation if $n+m=N$.

Answer 2

This is a special case of Euler's homogeneous function theorem: If $\Phi:\>{\mathbb R}^n\setminus\{{\bf 0}\}\to{\mathbb R}$ satisfies $$\Phi(t\, {\bf x})=t^\alpha\>\Phi({\bf x})\qquad(t>0, {\bf x}\ne0)$$ for some constant $\alpha\in{\mathbb R}$ then $$\nabla\Phi({\bf x})\cdot {\bf {\bf x}}=\alpha\,\Phi({\bf x})\qquad({\bf x}\ne{\bf 0})\ .$$