Prove that $u(x,y) = \frac{x}{x^2+y^2}$ is harmonic in $\mathbb{R}^2\setminus\{(0,0)\}$

Question

I want to check if I did this right. I reached the conclusion that $u$ is not harmonic. We know that a function is harmonic if $$\displaystyle\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0$$ For $u(x,y)$, I obtained that $$ \displaystyle\frac{\partial^2 u}{\partial x^2} = -\frac{2x(3y^2 - x^2)}{(x^2+y^2)^3}$$ and $$ \displaystyle \frac{\partial^2 u}{\partial y^2} = -\frac{4x(x^2 - 4y^2)}{(x^2+y^2)^3} $$ Then, $$ \displaystyle\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = \frac{-6xy^2+2x^3-4x^3+16xy^2}{(x^2+y^2)^3} = \frac{10xy^2-2x^3}{(x^2+y^2)^3} = \frac{-2x(x^2 - 5y^2)}{(x^2+y^2)^3} $$

Therefore, $u$ is not harmonic.

Answer 1

Your mistake comes from the computation of $\frac{\partial^2u}{\partial y^2}$.

Indeed, one has :

$$\frac{\partial u}{\partial y}=\frac{0\times(x^2+y^2)-x\times2y}{(x^2+y^2)^2}=\frac{-2xy}{(x^2+y^2)^2}$$

Therefore :

$$ \frac{\partial^2 u}{\partial y^2}=\frac{-2x\times(x^2+y^2)^2-(-2xy)\times(2\times 2y \times (x^2+y^2))}{(x^2+y^2)^4}\\ =\frac{-2x(x^2+y^2)+2xy\times4y}{(x^2+y^2)^3}=\frac{6xy^2-2x^3}{(x^2+y^2)^3}=\frac{2x(3y^2-x^2)}{(x^2+y^2)^3} $$

Alternate solution :

Put $z=x+iy$. Then :

$u(x,y)=\frac{x}{x^2+y^2}= Re(\frac{x}{x^2+y^2}-i\frac{y}{x^2+y^2})=Re(\frac{\bar{z}}{|z|^2})=Re(\frac{1}{z})$

So $u$ is harmonic (it's the real part of a holomorphic function).

Answer 2

Something went wrong with your computation. We can check this more conveniently with the polar coordinates Laplacian:

$$u(r,\theta) = \frac{\cos\theta}{r}$$

$$\Delta u = \frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial u}{\partial r}\right) + \frac{1}{r^2}\frac{\partial^2 u}{\partial \theta^2} = \frac{\cos\theta}{r^3} - \frac{\cos\theta}{r^3}=0$$