I am currently trying to solve this problem but am having difficulties with it. The question reads:
Let $\textbf x=(x_1, x_2, ..., x_n)$ denote a point in $\mathbb R^n$. Consider the function: $$u(\textbf x) = \frac{1}{|\textbf x|}$$ Suppose $n=3$. Show that for all $\textbf x\neq0$, we have that $\Delta u=0.$ Show that the analogous statement in $\mathbb R^2$ is false.
Now as I believe, $\Delta u$ is just the divergence of the gradient of $u$. However, am I supposed to use the chain rule to find the gradient of $u$? I am definitely confused by the absolute value sign as well. My thought is that the gradient should look like: $$\left\langle u'(x_1)\cdot dx_1, u'(x_2)\cdot dx^2, u'(x_3)\cdot dx^3\right \rangle$$ But the above doesnt make any sense to me. If anyone could lead me in the right direction it would be appreciated! Thank you
Let $u(x)=u(|x|)$, set $|x|=r=\sqrt{(x_{1}^{2} +....+x_{n}^{2})}$. Then, use chain rule:
$$\Delta u(|x|)= \frac{d^{2} u(|x|)}{dr^{2}} +\frac{n-1}{r}\frac{du(|x|)}{dr}.$$
If you have some trouble check ''Partial Differential Equations'' by L.C. Evans.
Now, we know that $\displaystyle u(|x|)=\frac{1}{|x|}=\frac{1}{r}.$
$$\frac{d^{2} u(|x|)}{dr^{2}}=\frac{2}{r^{3}}$$
$$\frac{d u(|x|)}{dr}=-\frac{1}{r^{2}}.$$
Now set n=3 as you want, and we got $$\Delta u(x)=0. $$
Now, let's set n=2. Then,
$$\Delta u(x)=\frac{1}{r^{3}}.$$