Higher order derivatives; Laplacian

At a Glance

• Frequency: 2 sub-parts across 2 of 13 years (2013, 2021)
• Priority tier: T3
• Marks (count): 10 (2)
• Average solve time: ~5 min
• Difficulty mix: easy 2
• Section: B | Dominant type: computation

Why This Chapter Matters

Both UPSC questions on this topic are compulsory 10-mark items that take under six minutes each. The 2013 question asks for $\nabla^2(r^n)$ — a clean formula derivation worth knowing cold. The 2021 question requires computing $\nabla\cdot(\mathbf{r}/r)$ first and then the Laplacian of the result; the problem statement contains a probable typo, but knowing the two-step calculation makes it straightforward. These are the easiest 10 marks in the Paper 1 vector analysis section.

Minimum Theory

Radial Laplacian formula. For any function $f$ depending only on $r = |\mathbf{r}|$ in $\mathbb{R}^3$:

$\nabla^2 f(r) = f''(r) + \frac{2}{r}f'(r) = \frac{1}{r^2}\frac{d}{dr}\!\left(r^2\frac{df}{dr}\right).$

This formula is derived from spherical coordinates; it applies whenever $f$ is spherically symmetric.

Power law: $\nabla^2(r^n)$. With $f(r) = r^n$, $f'(r) = nr^{n-1}$, $f''(r) = n(n-1)r^{n-2}$:

$\nabla^2(r^n) = n(n-1)r^{n-2} + \frac{2}{r}\cdot nr^{n-1} = n(n+1)r^{n-2}.$

Special cases: $n=0$ and $n=-1$ are both harmonic ($\nabla^2 = 0$), corresponding to the constant function and Newton’s $1/r$ potential.

Product rule for divergence. For a scalar $\phi$ and vector $\mathbf{F}$:

$\nabla\cdot(\phi\mathbf{F}) = \nabla\phi\cdot\mathbf{F} + \phi\,\nabla\cdot\mathbf{F}.$

In 3D: $\nabla\cdot\mathbf{r} = 3$ and $\nabla(r^{-1}) = -\mathbf{r}/r^3$.

Question Archetypes

laplacian-radial (2 question(s); 2013, 2021)

Recognition Cues

• ”$\nabla^2$ of a power of $r$” — apply the formula $\nabla^2(r^n) = n(n+1)r^{n-2}$ directly.
• “Show $\nabla^2[\nabla\cdot(\mathbf{r}/r^k)] = \ldots$” — compute the inner divergence first to reduce to a radial function, then apply $\nabla^2$.
• Both questions appear in Section B as compulsory 10-mark items.

Solution Template

1. If the argument of $\nabla^2$ is already $r^n$: quote $f'(r) = nr^{n-1}$, $f''(r) = n(n-1)r^{n-2}$; apply the radial formula.
2. If the argument involves $\nabla\cdot(\ldots)$ first: compute the inner divergence using the product rule $\nabla\cdot(\phi\mathbf{F}) = \nabla\phi\cdot\mathbf{F}+\phi\nabla\cdot\mathbf{F}$; reduce to a radial function $g(r)$; then compute $\nabla^2 g(r)$.
3. Check special cases: $n=0, -1, 1, 2$ to verify signs.

Worked Example

2013 Paper 1, 2013-P1-Q8a (10 marks)

Calculate $\nabla^2(r^n)$ and find its expression in terms of $r$ and $n$, where $r$ is the distance of any point $(x,y,z)$ from the origin and $n$ is a constant.

Strategy. $r^n$ depends only on $r$; use the 3D radial Laplacian.

Step 1 — Derivatives. $f(r) = r^n$, $f'(r) = nr^{n-1}$, $f''(r) = n(n-1)r^{n-2}$.

Step 2 — Apply.

$\nabla^2(r^n) = f''(r) + \frac{2}{r}f'(r) = n(n-1)r^{n-2} + \frac{2}{r}\cdot nr^{n-1} = n(n-1)r^{n-2} + 2nr^{n-2}.$

$\boxed{\nabla^2(r^n) = n(n+1)\,r^{n-2}.}$

Step 3 — Special cases. $n=0$: $0\cdot 1 \cdot r^{-2}=0$ ✓ (constant is harmonic). $n=-1$: $(-1)(0)r^{-3}=0$ ✓ (Newton potential is harmonic for $r\ne 0$). $n=2$: $2\cdot 3=6$, and direct check $\nabla^2(x^2+y^2+z^2)=2+2+2=6$ ✓.

2021 Paper 1, 2021-P1-Q5e (10 marks)

Show $\nabla^2\!\left[\nabla\cdot\!\left(\dfrac{\mathbf{r}}{r}\right)\right]=\dfrac{2}{r^4}$, where $\mathbf{r}=x\hat\imath+y\hat\jmath+z\hat k$.

Note on the question. The UPSC question as printed states the inner expression as $\mathbf{r}/r$, but computing $\nabla\cdot(\mathbf{r}/r)=2/r$ and then $\nabla^2(2/r)=0$ (for $r\ne 0$) does not give $2/r^4$. The inner expression that makes the question consistent is $\nabla\cdot(\mathbf{r}/r^2)=1/r^2$, since $\nabla^2(1/r^2)=2/r^4$. The intended computation is demonstrated below.

Step 1 — Compute $\nabla\cdot(\mathbf{r}/r^2)$. Use the product rule with $\phi = 1/r^2$ and $\mathbf{F} = \mathbf{r}$:

$\nabla\cdot\!\left(\frac{\mathbf{r}}{r^2}\right) = \nabla(r^{-2})\cdot\mathbf{r} + r^{-2}\nabla\cdot\mathbf{r}.$

$\nabla(r^{-2}) = -2r^{-3}\hat r = -2\mathbf{r}/r^4$; $\quad\nabla\cdot\mathbf{r}=3$; $\quad\nabla(r^{-2})\cdot\mathbf{r} = -2r^2/r^4 = -2/r^2$.

$\nabla\cdot\!\left(\frac{\mathbf{r}}{r^2}\right) = -\frac{2}{r^2} + \frac{3}{r^2} = \frac{1}{r^2}.$

Step 2 — Apply $\nabla^2$ to $1/r^2$. Use $\nabla^2(r^n) = n(n+1)r^{n-2}$ with $n=-2$:

$\nabla^2\!\left(\frac{1}{r^2}\right) = (-2)(-1)r^{-4} = \frac{2}{r^4}.$

$\boxed{\nabla^2\!\left[\nabla\cdot\!\left(\frac{\mathbf{r}}{r^2}\right)\right] = \frac{2}{r^4}.}$

Common Traps

• The 3D radial Laplacian has coefficient $2/r$ in front of $f'(r)$, not $1/r$ (which would be the 2D version). Getting this wrong changes the formula to $n^2 r^{n-2}$ (incorrect).
• The formula $\nabla^2(r^n) = n(n+1)r^{n-2}$ is symmetric: $r^n$ and $r^{-n-1}$ have the same Laplacian. So $1/r$ ($n=-1$) and $r^0$ ($n=0$) are the two harmonic radial functions.
• For the 2021 question, the inner expression is $\mathbf{r}/r^2$ (not $\mathbf{r}/r$); the question as printed has a typo. Recognise this inconsistency and compute the version that gives $2/r^4$.
• Always verify with $n=2$: $\nabla^2(r^2) = 2\cdot 3 = 6$, and Cartesian check gives $6$ immediately.

Marks-Aware Writing

A 10-mark answer must: write the radial Laplacian formula; differentiate $f(r) = r^n$ twice; combine to get $n(n+1)r^{n-2}$; verify one special case (usually $n=-1$ to show it is the harmonic Newton potential). For the 2021 question, additionally compute the inner divergence via the product rule — this step is worth 4–5 marks.

Practice Set

• 2018-P1-Q8a (12 m) — Prove $\nabla\times(\nabla\times\mathbf{v}) = \nabla(\nabla\cdot\mathbf{v}) - \nabla^2\mathbf{v}$; component expansion. ()
• 2025-P1-Q6c-ii (10 m) — Derive wave equations for $\mathbf{E}$ and $\mathbf{H}$ from Maxwell’s equations using curl-curl identity. ()