How does one compute the Laplacian of
$$\dfrac{(\mathbf \mu \cdot \mathbf r)}{r^3} \;\; \text{where} \;\; r = \Vert \mathbf r \Vert?$$
I am aware that the Laplacian is defined to be $\Delta f=\sum_i \partial_i^2f$ but am a little confused about this computationally in this case.
I have got as far as saying $\partial_i^2[1/r^3(\mu_1r_1+\mu_2r_2+...+\mu_nr_n)]$, but don't know how to compute this.
On
There are basically two approaches to a problem like this; the first, to write out everything in terms of coordinates and simply grind away with the computation of all the requisite derivatives and so forth; the second, to as much as possible utilize standard and well-known identities satisfied by $\nabla$ to reduce the amount of such explicit calculation as much as possible; I generally prefer the second approach, since I feel it is usually cleaner, less work, more likely to yield easily understood results, and less likely to be error prone. Hopefully, these notions are well-illustrated in the following.
Before proceeding, I must hand it to Joonas Ilmavirta for a particularly elegant solution based upon the formula $\nabla^2 (fg) = f\nabla^2 g + 2 \nabla f \cdot \nabla g + g\nabla^2 f$; very nice indeed, +1. The calculations presented in the present answer do not make direct use of this identity; neither to they rely on single-coordinate-at-a-time calculation of $\nabla u$ and $\nabla^2 u$ as done so concisely by our other poster, marco trevi (and a +1 for that effort as well!). Instead, I have tried in the following to use a small number of general identities satisfied by $\nabla$, and expressed in equations (3), (8), and (18); they are all easy to prove by resorting to Cartesian coordinates; also see here for more information. By repeated application of such general formulas, the amount of direct calculation in a coordinate system is is substantially reduced; only (7), (12), and (28) are done in the $x_i$-system, and they are all relatively simple evaluations.
The following derivation might be a little long-winded, but I have tried to show every step in detail, including the derivation of $\nabla \vert \mathbf r \vert^\alpha = \alpha \vert \mathbf r \vert^{\alpha - 2} \mathbf r$, which is taken is given in Joonas Ilmavirta's argument. In practice, once one is used to the method, working as much as possible with higher-level, abstract entities such as $r$ etc. (rather than $\sqrt{\sum x_i^2}$), generally proves to involve less work.
Incidentally, as anyone who reads this post will soon discover, I took the liberty of performing the calculations for the slightly more general case
$\nabla^2 (\dfrac{\mu \cdot \mathbf r}{r^m}), \tag{0}$
where I take $0 \ne m \in \Bbb R$.
Having typed on enough, I present:
Setting
$\phi = \dfrac{\mu \cdot \mathbf r}{r^m}, \tag{1}$
and noting that
$\nabla^2 \phi = \nabla \cdot \nabla \phi, \tag{2}$
we first compute $\nabla \phi$ making use of the product rule for gradients:
$\nabla(fg) = f\nabla g + g\nabla f, \tag{3}$
which we apply to (1), thus:
$\nabla (\dfrac{\mu \cdot \mathbf r}{r^m}) = \dfrac{1}{r^m} \nabla (\mu \cdot \mathbf r) + (\mu \cdot \mathbf r) \nabla(\dfrac{1}{r^m}); \tag{4}$
we evaluate the gradients occurring in (4): first, it is easy to see that
$\nabla (\mu \cdot \mathbf r) = \mu; \tag{5}$
since
$\mu \cdot \mathbf r = \sum_1^n \mu_i x_i, \tag{6}$
we have
$\dfrac{\partial}{\partial x_j}\mu \cdot \mathbf r = \mu_j; \tag{7}$
(5) follows directly from (7). As for the second term in (4), we may use the formula
$\nabla (f(g)) = \dfrac{df}{dg} \nabla g, \tag{8}$
where $g:\Omega \to \Bbb R$ and $f:\Bbb R \to \Bbb R$, $\Omega \subset \Bbb R^n$ being some open set. Like (3), (8) may also be verified one coordinate at at time, using the chain rule:
$\dfrac{\partial}{\partial x_j}f(g) = \dfrac{df}{dg} \dfrac{\partial f}{\partial x_j}, \; \; 1 \le j \le n; \tag{9}$
(8) follows readily from (9). Thus
$\nabla (\dfrac{1}{r^m}) = \dfrac{d(r^{-m})}{dr} \nabla r = -\dfrac{m}{r^{m + 1}} \nabla r. \tag{10}$
It is easy to compute $\nabla r = \nabla \Vert \mathbf r \Vert$, again using (8):
$\nabla r = \nabla \Vert \mathbf r \Vert = \nabla \langle \mathbf r, \mathbf r \rangle^{1/2} = \dfrac{1}{2}\langle \mathbf r, \mathbf r \rangle^{-1/2} \nabla \langle \mathbf r, \mathbf r \rangle; \tag{11}$
but
$(\nabla \langle \mathbf r, \mathbf r \rangle)_j = \dfrac{\partial}{\partial x_j}\sum_1^n x_i^2 = 2x_j, \tag{12}$
which shows that
$\nabla \langle \mathbf r, \mathbf r \rangle = 2\mathbf r; \tag{13}$
thus (11) becomes
$\nabla r = \dfrac{1}{2}\langle \mathbf r, \mathbf r \rangle^{-1/2} (2 \mathbf r) = \dfrac{\mathbf r}{r}, \tag{14}$
which is the unit vector field in the radial direction, in the direction of $\mathbf r$. In any event, (10) and (14) combined yield
$\nabla (\dfrac{1}{r^m}) = -\dfrac{m}{r^{m + 1}} \dfrac{\mathbf r}{r} = -\dfrac{m}{r^{m + 2}} \mathbf r, \tag{15}$
and bringing together (5) and (15) into (4) we find
$\nabla (\dfrac{\mu \cdot \mathbf r}{r^m}) = \dfrac{\mu}{r^m} - \dfrac{(m\mu \cdot \mathbf r)}{r^{m + 2}} \mathbf r. \tag{16}$
Having $\nabla \phi$, we proceed to calculate $\nabla^2 \phi$ as follows:
$\nabla^2 \phi = \nabla \cdot \nabla \phi = \nabla \cdot (\dfrac{\mu}{r^m} - \dfrac{m\mu \cdot \mathbf r}{r^{m + 2}} \mathbf r) = \nabla \cdot (\dfrac{\mu}{r^m}) - \nabla \cdot (\dfrac{m\mu \cdot \mathbf r}{r^{m + 2}}\mathbf r). \tag{17}$
Each of the terms on the right of (17) is of the form $\nabla \cdot (fX)$, where $f$ is a function and $X$ is a vector field on $\Bbb R^n$; for such expressions, we have in general
$\nabla \cdot (fX) = X \cdot \nabla f + f \nabla \cdot X. \tag{18}$
(18) may be verified, as may (3) and (8), by writing out each side in a Cartesian coordinate system. Applying (18) to (17), we have
$\nabla \cdot (\dfrac{\mu}{r^m}) = \mu \cdot \nabla(\dfrac{1}{r^m}) + \dfrac{1}{r^m} \nabla \cdot \mu; \tag{19}$
now
$\nabla \cdot \mu = 0, \tag{20}$
since $\mu$ is constant; from (15),
$\mu \cdot \nabla(\dfrac{1}{r^m}) = -\dfrac{m \mu \cdot \mathbf r}{r^{m + 2}}, \tag{21}$
so that
$\nabla \cdot (\dfrac{\mu}{r^m}) = -\dfrac{m \mu \cdot\mathbf r}{r^{m + 2}}. \tag{22}$
Next,
$\nabla \cdot (\dfrac{m\mu \cdot \mathbf r}{r^{m + 2}}\mathbf r) = \mathbf r \cdot \nabla (\dfrac{m\mu \cdot \mathbf r}{r^{m + 2}}) + (\dfrac{m\mu \cdot \mathbf r}{r^{m + 2}}) \nabla \cdot \mathbf r; \tag{23}$
we compute, in a manner similar to the above:
$\nabla (\dfrac{\mu \cdot \mathbf r}{r^{m + 2}}) = \dfrac{1}{r^{m + 2}} \nabla (\mu \cdot \mathbf r) + (\mu \cdot \mathbf r) \nabla (\dfrac{1}{r^{m + 2}}), \tag{24}$
by (3); again using (8) and (14), we see in a manner similar to (15) that
$\nabla (\dfrac{1}{r^{m + 2}}) = -\dfrac{m + 2}{r^{m + 3}} \nabla \mathbf r = -\dfrac{m + 2}{r^{m + 3}} \dfrac{\mathbf r}{r} = -\dfrac{m + 2}{r^{m + 4}} \mathbf r. \tag{25}$
Using (5) and (25), (24) becomes
$\nabla (\dfrac{\mu \cdot \mathbf r}{r^{m + 2}}) = \dfrac{\mu}{r^{m + 2}} - \dfrac{(m + 2)\mu \cdot \mathbf r}{r^{m + 4}} \mathbf r, \tag{26}$
whence
$m\mathbf r \cdot \nabla (\dfrac{\mu \cdot \mathbf r}{r^{m + 2}}) = \dfrac{m\mu \cdot \mathbf r}{r^{m + 2}} - \dfrac{m(m + 2)\mu \cdot \mathbf r}{r^{m + 4}} \mathbf r \cdot \mathbf r = \dfrac{m\mu \cdot \mathbf r}{r^{m + 2}} - \dfrac{m(m + 2)\mu \cdot \mathbf r}{r^{m + 4}}r^2$ $= \dfrac{(m - m^2 - 2m)\mu \cdot \mathbf r}{r^{m + 2}} = -\dfrac{m(m + 1)\mu \cdot \mathbf r}{r^{m + 2}}. \tag{27}$
We also have
$\nabla \cdot \mathbf r = \sum_1^n \dfrac{\partial x_j}{\partial x_j} = \sum_1^n 1 = n, \tag{28}$
whence
$m(\dfrac{\mu \cdot \mathbf r}{r^{m + 2}}) \nabla \cdot \mathbf r = \dfrac{mn \mu \cdot \mathbf r}{r^{m + 2}}. \tag{29}$
And lo! by virtue of (27) and (29), (23) has turned into
$\nabla \cdot (\dfrac{m\mu \cdot \mathbf r}{r^{m + 2}}\mathbf r) = -\dfrac{m(m + 1)\mu \cdot \mathbf r}{r^{m + 2}} + \dfrac{mn \mu \cdot \mathbf r}{r^{m + 2}} = \dfrac{(mn - m(m + 1))\mu \cdot \mathbf r}{r^{m + 2}}. \tag{30}$
We finally assemble the components we have built so far into the framework of (17); from (22) and (30) we have
$\nabla^2 \phi = \nabla \cdot (\dfrac{\mu}{r^m}) - \nabla \cdot (\dfrac{m\mu \cdot \mathbf r}{r^{m + 2}}\mathbf r)$ $= -\dfrac{m \mu \cdot\mathbf r}{r^{m + 2}} - \dfrac{(mn - m(m + 1)\mu \cdot \mathbf r}{r^{m + 2}} = \dfrac{(m^2 - mn)\mu \cdot \mathbf r}{r^{m + 2}}. \tag{31}$
If we take $m = 3$, (31) agrees with the results of marco trevi and Joonas Ilmavirta.
On
WARNING: GRAPHIC CONTENT! It's not going to be pretty as Joonas' answer (btw thank you for the correction!)
Let's do it in cartesian coordinates. The function is \begin{equation} u(x_1,\dots,x_n)=\frac{\sum_i\mu_ix_i}{\left(\sum_j x_j^2\right)^{3/2}}=\frac{\mu\cdot\mathbf{r}}{r^3} \end{equation} We have that \begin{equation} \frac{\partial\ r^s}{\partial x_k}=sx_k\ r^{s-2} \end{equation} and also \begin{equation} \frac{\partial\ (\mu\cdot\mathbf{r})}{\partial x_k}=\mu_k \end{equation} so, \begin{equation} \frac{\partial u}{\partial x_k}=\frac{\mu_kr^3-3x_kr(\mu\cdot\mathbf{r})}{r^6}=\frac{1}{r^5}\left(\mu_kr^2-3(\mu\cdot\mathbf{r})x_k\right) \end{equation} and \begin{equation} \frac{\partial^2 u}{\partial x_k^2}= \frac{1}{r^{10}}\left[(2\mu_kx_k-3\mu_kx_k-3(\mu\cdot\mathbf{r}))r^5-5x_kr^3(\mu_kr^2-3(\mu\cdot\mathbf{r})x_k)\right]=\\ =\frac{1}{r^7}\left[-\mu_kx_kr^2-3(\mu\cdot\mathbf{r})r^2-5\mu_kx_kr^2+15(\mu\cdot\mathbf{r})x_k^2\right]=\\ =\frac{3}{r^7}\left[(\mu\cdot\mathbf{r})(5x_k^2-r^2)-2\mu_kx_kr^2\right] \end{equation} Summing everything over $k$ we get \begin{equation} \Delta u = \sum_k\frac{\partial^2u}{\partial x_k^2}=\frac{3}{r^7}\sum_k\left[(\mu\cdot\mathbf{r})(5x_k^2-r^2)-2\mu_kx_kr^2\right]=\\ =\frac{3}{r^7}\left[(\mu\cdot\mathbf{r})\sum_k(5x_k^2-r^2)-2r^2\sum_k\mu_kx_k\right]=\\ =\frac{3}{r^7}\left[(\mu\cdot\mathbf{r})(5r^2-nr^2)-2r^2(\mu\cdot\mathbf{r})\right]=\\ =3(3-n)\frac{(\mu\cdot\mathbf{r})}{r^5} \end{equation}
If you have two functions $f$ and $g$, then $\Delta(fg)=f\Delta g+g\Delta f+2\nabla f\cdot\nabla g$. Now we obviously want to choose $f(x)=|x|^{-3}$ and $g(x)=\mu\cdot x$. Recall that $\nabla |x|^\alpha=\alpha|x|^{\alpha-2}x$; to remember this rule, remember that the gradient is a vector and the formula should hold true in dimension one. These choices give \begin{eqnarray} \nabla g(x)&=&\mu\\ \Delta g(x)&=&0\\ \nabla f(x)&=&-3|x|^{-5}x\\ \Delta f(x)&=&-3(\nabla|x|^{-5})\cdot x-3|x|^{-5}\text{div(x)} \\&&\quad=15|x|^{-5}-3n|x|^{-5} \\&&\quad=(15-3n)|x|^{-5}, \end{eqnarray} whence \begin{eqnarray} \Delta(fg) &=& f\Delta g+g\Delta f+2\nabla f\cdot\nabla g \\&=& |x|^{-3}0+(\mu\cdot x)(15-3n)|x|^{-5}-6|x|^{-5}x\cdot\mu \\&=& (9-3n)|x|^{-5}\mu\cdot x. \end{eqnarray}