TITLE: Correct Solution?
QUESTION [6 upvotes]: Suppose $$\omega = \left(\sum_{i=1}^{n} x_i^2 \right)^k \mid n >2$$ Find $k$ so that 
$$ \sum_{i=0}^{n} \frac{\partial^2 \omega}{\partial x_i^2} = 0 \, \, \, \text{for all} \, \, \, x_i$$
Proposed Solution:
$$ \frac{\partial \omega}{\partial x_i} =2x_i k\left(\sum_{i=1}^{n} x_i^2 \right)^{k-1} $$
$$ \frac{\partial^2 \omega}{\partial x_i^2} = (2x_i)^2 k(k-1)\left(\sum_{i=1}^{n} x_i^2 \right)^{k-2} + 2 k\left(\sum_{i=1}^{n} x_i^2 \right)^{k-1} = 2k\left(\sum_{i=1}^{n} x_i^2 \right)^{k-2}\left(2(k-1)x_i^2 +  \left(\sum_{i=1}^{n} x_i^2 \right)\right)$$
Summing over $i$ yields 
$$ 2k\left(\sum_{i=1}^{n} x_i^2 \right)^{k-2} \sum_{i=0}^{n}\left(2(k-1)x_i^2 + \left(\sum_{i=1}^{n} x_i^2 \right)\right)$$
so the sum is only 0 iff
$$ \sum_{i=0}^{n}\left(2(k-1)x_i^2 +  \left(\sum_{i=1}^{n} x_i^2 \right)\right) = 0$$
$$\sum_{i=0}^{n}(2(k-1)x_i^2 +  \left(\sum_{i=1}^{n} x_i^2 \right)) = 2(k-1) \sum_{i=0}^{n} x_i^2 + \sum_{i=0}^{n} \left(\sum_{i=1}^{n} x_i^2 \right) =  2(k-1) \sum_{i=0}^{n} x_i^2 + n \left(\sum_{i=1}^{n} x_i^2 \right)$$
$$( 2(k-1) + n )\left(\sum_{i=1}^{n} x_i^2 \right) = 0 \to k = 1- \frac{n}{2} $$ 
Is this a correct solution?

REPLY [2 votes]: Your solution is perfect.
Just a side note, what you found is the Green function (fundamental solution) of Laplace equation, so that $-\Delta \omega = 0$ in $\mathbb{R}^n\backslash\{0\}$.
Let $\mathbf{x} = (x_1,\dots,x_n)$, $r = |\mathbf{x}| = \sqrt{x_1^2 + \dots + x^2_n}$. For a radially symmetric function $\omega = u(r)$, we have
$$
u'' + \frac{n-1}{r}u' = 0,
$$ 
this implies when $n\geq 3$
$$
u = c_1 + c_2 \frac{r^{2-n}}{2-n},
$$
and a natural log when $n=2$, a linear function when $n=1$. Hence
$$
\omega =  C\left(\sum^n_{i=1} x_i^2\right)^{1-n/2}.
$$
If we set $c_1=0$, $C$ is often chosen such that, in the distributional sense, $-\Delta \omega = \delta_0(\mathbf{x})$ which is the Dirac delta at the origin.