Problem checking that a fuction verifies the Laplace equation. I'm having trouble solving an excersive from a past final that goes like this:
Prove that if $u(x,y) \in \mathbb{C}^2$ satisfies $u_{xx} + u_{yy} = 0$ in $\mathbb{R}² - \{(0,0)\} $ then $v(x,y) = u( \frac{x}{x² + y²} ; \frac{y}{x²+y²})$ also satisfies it.
Here's what I did:
Let $ \alpha (x,y) = ( \frac{x}{x² + y²} , \frac{y}{x²+y²}) $, so
$ \frac{\partial v}{\partial x} = \frac{\partial u}{\partial x} ({\alpha(x,y)}) \frac{\partial \alpha_1}{\partial x} + \frac{\partial u}{\partial y} ({\alpha(x,y)}) \frac{\partial \alpha_2}{\partial x}$
$ \frac{\partial² v}{\partial x²} = \frac{\partial^2 u}{\partial x²} ({\alpha(x,y)}) \frac{\partial \alpha_1}{\partial x} + \frac{\partial u}{\partial x} ({\alpha(x,y)}) \frac{\partial² \alpha_1}{\partial x²} + \frac{\partial^2 u}{\partial xy} ({\alpha(x,y)}) \frac{\partial \alpha_2}{\partial x} + \frac{\partial u}{\partial y} ({\alpha(x,y)}) \frac{\partial² \alpha_2}{\partial x²}$
Likewise for $y$. I did all the operations and went for $ \frac{\partial² v}{\partial x²} + \frac{\partial² v}{\partial y²} $ but couldn't actually make it equal to $0$, so I'm wondering if this is actually the right way to go.
Thanks in advance for any help.
 A: There is something wrong with your second partial derivative. If you write $v(x,y) = u(\alpha(x,y)) =u( \alpha_1(x,y), \alpha_2(x,y))$, then
$$ \frac{\partial v}{\partial x} = \frac{\partial u}{\partial x} ({\alpha(x,y)}) \frac{\partial \alpha_1}{\partial x} + \frac{\partial u}{\partial y} ({\alpha(x,y)}) \frac{\partial \alpha_2}{\partial x}$$
$$ \frac{\partial^2 v}{\partial x^2} = \frac{\partial^2 u}{\partial x^2} ({\alpha(x,y)}) \left(\frac{\partial \alpha_1}{\partial x}\right)^2 + \frac{\partial u}{\partial x} ({\alpha(x,y)}) \frac{\partial^2 \alpha_1}{\partial x^2} + 2 \frac{\partial^2 u}{\partial xy} ({\alpha(x,y)}) \frac{\partial \alpha_2}{\partial x}\frac{\partial \alpha_1}{\partial x} + \frac{\partial u}{\partial y} ({\alpha(x,y)}) \frac{\partial^2 \alpha_2}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} ({\alpha(x,y)}) \left(\frac{\partial \alpha_2}{\partial x}\right)^2$$
A: This approach is doable though it's quite tiresome and requires some attention. I'll try to outline it:
Note that $\Delta = div \nabla$. 
If we write it in Einsteinian notation with $e_j$ - base vectors, $x_j$ - coordinates ($j=1,2$), $r=\sqrt {x_1^2+x_2^2}$ and $f_{,j}:=\partial_{x_j}f$
$$\nabla u(\alpha) = u_{,i}\left(\frac{x_i}{r^2}\right)_{,j}e_j=u_{,i}\left(\frac{\delta_{ij}}{r^2}-2\frac{x_ix_j}{r^4}\right) e_j.$$
Now we apply divergence:
$$\left(u_{,i}\left(\frac{\delta_{ij}}{r^2}-2\frac{x_ix_j}{r^4}\right)\right)_{,j}$$
$$=u_{,ik}\left(\frac{x_k}{r^2}\right)_{,j}\left(\frac{\delta_{ij}}{r^2}-2\frac{x_ix_j}{r^4}\right)+u_{,i}\left(\frac{\delta_{ij}}{r^2}-2\frac{x_ix_j}{r^4}\right)_{,j}.$$
$$
=u_{,ik}\left(\frac{\delta_{kj}}{r^2}-2\frac{x_kx_j}{r^4}\right)\left(\frac{\delta_{ij}}{r^2}-2\frac{x_ix_j}{r^4}\right)+u_{,i}\left(\frac{\delta_{ij}}{r^2}-2\frac{x_ix_j}{r^4}\right)_{,j}$$
Term by term:
$$u_{,ik}\frac{\delta_{ij}}{r^2}\frac{\delta_{kj}}{r^2}=\frac{\Delta u}{r^4}=0,$$
$$-2u_{,ik}\frac{\delta_{kj}}{r^2}\frac{x_ix_j}{r^4}=-2u_{,ik}\frac{x_ix_k}{r^6},$$
$$-2u_{,ik}\frac{x_kx_j}{r^4}\frac{\delta_{ij}}{r^2}=-2u_{,ik}\frac{x_ix_k}{r^6},$$
$$4u_{,ik}\frac{x_kx_j}{r^4}\frac{x_ix_j}{r^4} =4u_{,ik}\frac{x_kx_i}{r^6},$$
which allows to conclude that $$u_{,ik}\left(\frac{\delta_{kj}}{r^2}-2\frac{x_kx_j}{r^4}\right)\left(\frac{\delta_{ij}}{r^2}-2\frac{x_ix_j}{r^4}\right)=0.$$
Now to $$
u_{,i}\left(\frac{\delta_{ij}}{r^2}-2\frac{x_ix_j}{r^4}\right)_{,j}$$
$$=u_{,i}\left(\frac{-2\delta_{ij}x_j}{r^4}-2\frac{x_i }{r^4}-2\frac{x_j\delta_{ij} }{r^4}+8\frac{x_i x_j^2}{r^6}\right)=0.$$
Note that $$\sum_{i=1}^2\sum_{j=1}^2 u_{,i}x_i = 2\sum_{i=1}^2  u_{,i}x_i$$
and $$\sum_{i=1}^2\sum_{j=1}^2 u_{,i}x_i x_j^2=r^2 \sum_{i=1}^2  u_{,i}x_i.$$
