← 2021 Paper 1

UPSC 2021 Maths Optional Paper 1 Q5e — Step-by-Step Solution

10 marks · Section B

Higher order derivatives; Laplacian · Vector Analysis · asked 2× in 14 yrs · Read the full method →

Question

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

Technique

Spherical Laplacian $\nabla^2 f(r)=(1/r^2)(r^2 f')'$ for radial functions.

Solution

Step 1 — Compute $\nabla\cdot(\vec r/r)$

$\vec r/r=\hat r$ (unit radial vector).

$\nabla\cdot\!\left(\dfrac{\vec r}{r}\right)=\nabla\cdot(\vec r r^{-1})=r^{-1}(\nabla\cdot\vec r)+\vec r\cdot\nabla(r^{-1})$.

$\nabla\cdot\vec r=3$ (in 3D).

$\nabla(r^{-1})=-r^{-2}\hat r=-r^{-2}\vec r/r=-\vec r/r^3$.

So $\vec r\cdot\nabla(r^{-1})=\vec r\cdot(-\vec r/r^3)=-|\vec r|^2/r^3=-r^2/r^3=-1/r$.

Therefore $\nabla\cdot(\vec r/r)=3/r-1/r=2/r$.

Step 2 — Compute $\nabla^2(2/r)$

$\nabla^2(1/r)=0$ for $r\ne 0$? Actually $\nabla^2(1/r)=-4\pi\delta(\vec r)$ in the distributional sense; for $r\ne 0$, $\nabla^2(1/r)=0$.

Hmm but we want $\nabla^2(2/r)$, which should be $0$ for $r\ne 0$ if that formula holds. But the question says the answer is $2/r^4$, not 0.

Re-read the question: the Laplacian is of $\nabla\cdot(\vec r/r)=2/r$, not of $1/r$.

Wait — both are $\propto 1/r$. The classical result is $\nabla^2(1/r)=0$ for $r\ne 0$. So $\nabla^2(2/r)=0$, not $2/r^4$.

Let me re-examine. The vector $\vec r/r=\hat r$ has $\nabla\cdot\hat r=2/r$ (standard result in 3D).

$\nabla^2(2/r)$: classical Laplacian.

In spherical: $\nabla^2 f(r)=\dfrac{1}{r^2}\dfrac{d}{dr}\!\left(r^2\dfrac{df}{dr}\right)$.

$f=2/r$, $df/dr=-2/r^2$, $r^2 df/dr=-2$, $d/dr(-2)=0$. So $\nabla^2(2/r)=0$ for $r\ne 0$.

This contradicts the question. Let me re-read.

Re-read the question once more: "$\nabla^2[\nabla\cdot(\vec r/r)]=2/r^4$".

If $\nabla\cdot(\vec r/r)=2/r$, then $\nabla^2(2/r)=0$, not $2/r^4$.

There may be a typo: perhaps the question means $\nabla\cdot[\nabla(\vec r/r)]$ or $\nabla\cdot[\nabla(1/r)]$ or similar. Or the inner expression is $\nabla\cdot(\vec r/r^3)$?

Check $\nabla\cdot(\vec r/r^3)$: this is the divergence of the gravity field. Standard result: $\nabla\cdot(\vec r/r^3)=0$ for $r\ne 0$. Then $\nabla^2(0)=0$. Doesn’t match.

Check the answer $2/r^4$: dimensions of $\nabla^2(1/r)$ is $[1/L^3]$, not $[1/L^4]$. So if the inner is $1/r$, outer Laplacian is $0$ or $\delta$-function.

If the inner is $1/r^2$, then $\nabla^2(1/r^2)=$ ? $f=r^{-2}$, $df/dr=-2r^{-3}$, $r^2 df/dr=-2r^{-1}$, $d/dr(-2r^{-1})=2r^{-2}$, divide by $r^2$: $2/r^4$.

So $\nabla^2(1/r^2)=2/r^4$.

Aha — so the inner expression should be $1/r^2$. Let me recompute $\nabla\cdot(\vec r/r)$:

$\vec r/r$: components $x/r,y/r,z/r$.

$\partial_x(x/r)=1/r+x\cdot\partial_x(1/r)=1/r+x(-x/r^3)=1/r-x^2/r^3$.

Similarly $\partial_y(y/r)=1/r-y^2/r^3$ and $\partial_z(z/r)=1/r-z^2/r^3$.

Sum: $3/r-(x^2+y^2+z^2)/r^3=3/r-r^2/r^3=3/r-1/r=2/r$ ✓.

So $\nabla\cdot(\vec r/r)=2/r$, confirmed.

And $\nabla^2(2/r)=0$ for $r\ne 0$. So the question’s claimed answer $2/r^4$ doesn’t match.

Possible interpretation: Perhaps the intended question is $\nabla\cdot[\nabla(\vec r/r)]$ where $\nabla(\vec r/r)$ is a tensor (gradient of a vector). Or the question means $\nabla^2$ acting on $\vec r/r$ (as a vector Laplacian).

Vector Laplacian: $\nabla^2(\vec r/r)=\hat\imath\nabla^2(x/r)+\hat\jmath\nabla^2(y/r)+\hat k\nabla^2(z/r)$. Each component…

Actually let me check the original Hindi formulation if the question makes more sense. From the question text, it says "$\nabla^2[\nabla\cdot(\vec r/r)]=2/r^4$".

Hmm — given the inconsistency, let me assume the question is correctly stated and find the right interpretation. Perhaps the inner is supposed to be $1/r^2$ rather than $\nabla\cdot(\vec r/r)$.

Going with the problem as stated, with the caveat: the result $\nabla^2(2/r)=0$ for $r\ne 0$ is correct. The claimed answer $2/r^4$ matches $\nabla^2(1/r^2)$, not $\nabla^2(2/r)$.

Possible typo: the inner expression should be $1/r^2$, perhaps written as $\nabla\cdot(\vec r/r^3)$ which would be… let me check: $\nabla\cdot(\vec r/r^3)=$ ?

$\partial_x(x/r^3)=1/r^3+x\partial_x(r^{-3})=1/r^3+x(-3r^{-4}\partial_x r)=1/r^3+x(-3x/r^5)=1/r^3-3x^2/r^5$.

Sum: $3/r^3-3(x^2+y^2+z^2)/r^5=3/r^3-3/r^3=0$.

So $\nabla\cdot(\vec r/r^3)=0$. Not it.

How about $\nabla\cdot(\vec r/r^2)$? $\partial_x(x/r^2)=1/r^2+x\partial_x(r^{-2})=1/r^2+x(-2x/r^4)=1/r^2-2x^2/r^4$.

Sum: $3/r^2-2(x^2+y^2+z^2)/r^4=3/r^2-2/r^2=1/r^2$.

So $\nabla\cdot(\vec r/r^2)=1/r^2$. Then $\nabla^2(1/r^2)=2/r^4$ ✓.

So the original question likely intends $\nabla^2[\nabla\cdot(\vec r/r^2)]=\nabla^2(1/r^2)=2/r^4$. There may be a typo in the PDF (subscript missing).

Either way, demonstrating $\nabla^2(1/r^2)=2/r^4$ is the relevant computation:

Computing $\nabla^2(1/r^2)$

In spherical: $f=1/r^2$, $df/dr=-2/r^3$, $r^2 df/dr=-2/r$.

$\dfrac{d}{dr}(-2/r)=2/r^2$.

$\nabla^2 f=\dfrac{1}{r^2}\cdot 2/r^2=2/r^4$. ✓

Answer