UPSC 2026 Maths Optional Paper 1 Q1c — Step-by-Step Solution

10 marks · Section A

Partial derivatives · Calculus · asked 9× in 14 yrs · Read the full method →

Question

If $u(x, y, z) = (x^2 + y^2 + z^2)^{-1/2}$ , then find the value of $\dfrac{\partial^2 u}{\partial x^2} + \dfrac{\partial^2 u}{\partial y^2} + \dfrac{\partial^2 u}{\partial z^2}$ .

Technique

Write $u=r^{-1}$ with $r=\sqrt{x^2+y^2+z^2}$ and use the chain rule through $\dfrac{\partial r}{\partial x}=\dfrac{x}{r}$ . Compute $u_{xx}$ once and obtain $u_{yy},u_{zz}$ by the symmetry of $r$ in $x,y,z$ ; add and use $x^2+y^2+z^2=r^2$ . State the exclusion of the origin.

Solution

Step 0 — Notation and domain.

Put

r=\sqrt{x^2+y^2+z^2}\;(>0),\qquad u=r^{-1}=(x^2+y^2+z^2)^{-1/2}.

$u$ is defined and infinitely differentiable only for $(x,y,z)\neq(0,0,0)$ — at the origin $r=0$ and $u$ is not even finite, so the whole computation is carried out on $\mathbb{R}^3\setminus\{(0,0,0)\}$ . This exclusion is part of the answer, not a footnote.

From $r^2=x^2+y^2+z^2$ , differentiating partially with respect to $x$ : $2r\,r_x = 2x$ , so

\frac{\partial r}{\partial x}=\frac{x}{r},\qquad\text{and similarly}\qquad \frac{\partial r}{\partial y}=\frac{y}{r},\quad \frac{\partial r}{\partial z}=\frac{z}{r}.

Step 1 — First partial derivative.

\frac{\partial u}{\partial x}=\frac{d}{dr}\big(r^{-1}\big)\cdot\frac{\partial r}{\partial x} = -r^{-2}\cdot\frac{x}{r} = -\frac{x}{r^{3}}.

(Direct check without the $r$ -substitution: $\dfrac{\partial u}{\partial x} = -\tfrac12 (x^2+y^2+z^2)^{-3/2}\cdot 2x = -x(x^2+y^2+z^2)^{-3/2}$ . ✓)

Step 2 — Second partial derivative.

Differentiate $u_x=-x\,r^{-3}$ by the product rule:

\frac{\partial^2 u}{\partial x^2} = -\left[\,1\cdot r^{-3} + x\cdot\frac{\partial}{\partial x}\big(r^{-3}\big)\right] = -\left[\,r^{-3} + x\left(-3r^{-4}\cdot\frac{x}{r}\right)\right],

\frac{\partial^2 u}{\partial x^2} = -\frac{1}{r^{3}}+\frac{3x^{2}}{r^{5}}.

By the complete symmetry of $r$ in $x,y,z$ (the same computation with $y$ or $z$ in place of $x$ ):

\frac{\partial^2 u}{\partial y^2} = -\frac{1}{r^{3}}+\frac{3y^{2}}{r^{5}},\qquad \frac{\partial^2 u}{\partial z^2} = -\frac{1}{r^{3}}+\frac{3z^{2}}{r^{5}}.

Step 3 — Add the three.

\nabla^{2}u=\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}+\frac{\partial^2 u}{\partial z^2} = -\frac{3}{r^{3}}+\frac{3\left(x^{2}+y^{2}+z^{2}\right)}{r^{5}}.

Since $x^{2}+y^{2}+z^{2}=r^{2}$ ,

\nabla^{2}u = -\frac{3}{r^{3}}+\frac{3r^{2}}{r^{5}} = -\frac{3}{r^{3}}+\frac{3}{r^{3}} = 0.

Hence $u$ satisfies Laplace’s equation everywhere except the origin: $u=1/r$ is a harmonic function on $\mathbb{R}^3\setminus\{\mathbf 0\}$ — the Newtonian (gravitational / Coulomb) potential of a unit point source at the origin.

Step 4 — The general result, as a check and for depth.

For $u=r^{n}$ the identical computation gives $u_x = n\,x\,r^{n-2}$ and

\frac{\partial^2 u}{\partial x^2}=n\,r^{n-2}+n(n-2)\,x^{2}r^{n-4},

so, adding the three and using $x^2+y^2+z^2=r^2$ ,

\nabla^{2}\left(r^{n}\right)=3n\,r^{n-2}+n(n-2)\,r^{n-2}=n(n+1)\,r^{n-2}.

Putting $n=-1$ : $\nabla^{2}\left(r^{-1}\right)=(-1)(0)\,r^{-3}=0$ ✓ — consistent with Step 3. (Note $n=0$ and $n=-1$ are the only exponents making $r^n$ harmonic in $\mathbb{R}^3$ .)

Spot check at a point. Take $(x,y,z)=(1,2,-2)$ , so $r=\sqrt{1+4+4}=3$ , $r^3=27$ , $r^5=243$ :

u_{xx}=-\tfrac{1}{27}+\tfrac{3(1)}{243},\qquad u_{yy}=-\tfrac{1}{27}+\tfrac{3(4)}{243},\qquad u_{zz}=-\tfrac{1}{27}+\tfrac{3(4)}{243}.

\text{Sum}=-\frac{3}{27}+\frac{3(1+4+4)}{243}=-\frac{1}{9}+\frac{27}{243}=-\frac{1}{9}+\frac{1}{9}=0\ \checkmark

Remark (depth, one line). The exclusion is essential: over all of $\mathbb{R}^3$ , in the sense of distributions $\nabla^{2}\!\left(\frac{1}{r}\right)=-4\pi\,\delta(\mathbf r)$ , so the origin carries the whole source. Classically, $\nabla^2 u$ is simply undefined there.

Answer

\boxed{\;\frac{\partial^{2}u}{\partial x^{2}}+\frac{\partial^{2}u}{\partial y^{2}}+\frac{\partial^{2}u}{\partial z^{2}}=0\quad\text{for all }(x,y,z)\neq(0,0,0);\ \ u=r^{-1}\text{ is harmonic off the origin.}\;}