Help needed with the following differentiation problem

Question

If $H$ be a homogeneous function of degree $n$, and $u = (x^2+y^2)^{-\frac{1}{2}n}$ , then prove that $\frac{\partial}{\partial x}(H\frac{\partial u}{\partial x}) + \frac{\partial}{\partial y}(H\frac{\partial u}{\partial y}) = 0$

Now, Since $H$ is a homogeneous function of degree $n$ and $u$ is of degree $-n$ , Because $(x^2+y^2)^{-\frac{1}{2}n} = x^{-n}(1 + \frac{y^2}{x^2})^{-\frac{1}{2}n}$ Therefore $Hu$ would be of degree $0$

Hence, $x\frac{\partial (Hu)}{\partial x} + y\frac{\partial (Hu)}{\partial y} = 0$

I don't know how to proceed from here to the form required. Please help.

Answer 1

For a homogeneous function $H:\mathbb{R}^2\rightarrow\mathbb{R}$ we have $$ H =\frac{1}{n}\left( x\frac{\partial H}{\partial x} + y\frac{\partial H}{\partial y}\right) $$

So this PDE becomes

$$ \frac{\partial H}{\partial x}\frac{\partial u}{\partial x} + \frac{1}{n}\left( x\frac{\partial H}{\partial x} + y\frac{\partial H}{\partial y}\right)\frac{\partial^2 u}{\partial x^2} + \frac{\partial H}{\partial y}\frac{\partial u}{\partial y} + \frac{1}{n}\left( x\frac{\partial H}{\partial x} + y\frac{\partial H}{\partial y}\right)\frac{\partial^2 u}{\partial y^2} $$

Answer 2

Writing $u_n = (x^2+y^2)^{-n/2}$ we have $\nabla u_n = -n (x,y) u_{n+2}$ so we get: $$ -\frac{1}{n}\nabla \cdot ( H \ \nabla u) = \nabla \cdot ( (x,y)H \ u_{n+2}) = 2 H u_{n+2} + (x \partial_x + y \partial_y) (H u_{n+2})= 2 H u_{n+2} - 2 H u_{n+2} = 0$$ since $H u_{n+2}$ is homogeneous of degree -2.