Let $y\in R^N$. Show that $V(x)=\frac{|y^2|-|x|^2}{|y-x|^N}$ is hamonic over $R^N\setminus\{y\}$.
This is an exercise of my first course in PDE.
My doubt
I know that the Laplace's equation is: $$ \Delta u=\sum_{j=1}^N \frac{\partial^2u}{\partial x_i^2} $$
In other hand I have that
$$ V(x)=\frac{|y^2|-|x|^2}{|y-x|^N} $$
When I have to calculate the derivative of $V(x)$ this is becoming a monster! I believe that exists other way to verify using spherical mean but I don't know how to do this.
You want to show that $V(x)$ satisfies laplaces equation on $\mathbb{R}^N\setminus\{y\}$.
That would make it harmonic.
OR, you need to show that the mean integral of $V$ over a sphere boundary of a sphere in $\mathbb{R}^N$, centred at $x_0$ is equivalent to $V(x_0)$.
$\frac{\partial}{\partial x_i}V(x)=\large\frac{\partial}{\partial x_i}(\frac{|y^2|-(x_1^2+...+x_N^2)}{(y_1-x_1)^2+...+(y_N-x_N)^2})$
$=\frac{(-2x_i)((y_1-x_1)^2+...+(y_N-x_N)^2)+2(y_i-x_i)(|y^2|-(x_1^2+...+x_N^2))}{((y_1-x_1)^2+...+(y_N-x_N)^2)^2}$
Can you do the next differentiation step? since it looks like things will cancel nicely when you take the sum of all of the second derivatives.