Harmonic function on a ball

Example

Let $n \ge 2$, let $B = \set{x \in \R^n : \abs{x} < 1}$ be the unit open ball, and fix a point $\zeta$ with $\abs{\zeta} = 1$. Consider the function $$u(x) = \frac{1 - \abs{x}^2}{\abs{x - \zeta}^n}, \qquad x \in B,$$ which is the Poisson kernel of the ball with pole $\zeta$, up to a constant factor. It is positive on $B$, because $\abs{x} < 1$ and $x \ne \zeta$. To check that it is harmonic, write $f(x) = 1 - \abs{x}^2$, $g(x) = \abs{x - \zeta}^{-n}$ and $r = \abs{x - \zeta}$. Then $\Delta f = -2n$, $\nabla f = -2x$, $\nabla g = -n r^{-n-2} (x - \zeta)$ and $\Delta g = 2n r^{-n-2}$, where $\Delta$ is the Laplacian. The product rule $\Delta(fg) = f \Delta g + 2 \nabla f \cdot \nabla g + g \Delta f$ gives $$\Delta u = 2n r^{-n-2} \left( 1 - \abs{x}^2 + 2 x \cdot (x - \zeta) - r^2 \right).$$ Since $r^2 = \abs{x}^2 - 2 x \cdot \zeta + 1$, the bracket equals $0$, so $\Delta u = 0$.