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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?

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    $\begingroup$ I may be mistaken but I think $\frac{\partial^{2} \omega}{\partial x_{i}^{2}}$ is wrong. I find : $$ \frac{\partial^{2} \omega}{\partial x_{i}^{2}} = (2x_{i})^{2}k(k-1)\left( \sum_{i=1}^{n} x_{i}^{2} \right)^{k-2} + 2k\left( \sum_{i=1}^{n} x_{i}^{2} \right)^{k-1} $$ $\endgroup$ – jibounet Jul 24 '13 at 7:07
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    $\begingroup$ Just a typo @jibounet $\endgroup$ – Anthony Peter Jul 24 '13 at 7:13
  • $\begingroup$ Yes I saw that just after sending the comment ! Your solution looks good ! I don't see any mistake. $\endgroup$ – jibounet Jul 24 '13 at 7:15
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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.

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