2

I want to prove that

$x^{2}\frac{\partial^{2}u}{\partial x^{2}}+2xy\frac{\partial^{2}u}{\partial x\partial y}+y^{2}\frac{\partial^{2}u}{\partial y^{2}}=0$

where $u(x,y)=\frac{xy}{x+y}$. I know that it could be done by calculating each partial derivative but I realized that

$x^{2}\frac{\partial^{2}u}{\partial x^{2}}+2xy\frac{\partial^{2}u}{\partial x\partial y}+y^{2}\frac{\partial^{2}u}{\partial y^{2}}=[(x,y)\cdot \nabla]^{2}u$. So my question is if there exists some result that I can use to prove in an easy way that the equality holds.

Iesus Dave Sanz
  • 323

1 Answers1

1

I get $$x^{2}\frac{\partial^{2}u}{\partial x^{2}}+2xy\frac{\partial^{2}u}{\partial x\partial y}+y^{2}\frac{\partial^{2}u}{\partial y^{2}}=[(x,y)\cdot \nabla]^{2}u-[(x,y)\cdot \nabla]u,$$

$$[(x,y)\cdot \nabla]\left(\frac{xy}{x+y}\right)=\frac{xy}{x+y}$$ Then $$x^{2}\frac{\partial^{2}u}{\partial x^{2}}+2xy\frac{\partial^{2}u}{\partial x\partial y}+y^{2}\frac{\partial^{2}u}{\partial y^{2}}=\frac{xy}{x+y}-\frac{xy}{x+y}=0$$

Aleksas Domarkas
  • 2,401