← 2018 Paper 1

UPSC 2018 Maths Optional Paper 1 Q8a — Step-by-Step Solution

12 marks · Section B

Vector identities (curl of grad, div of curl, product rules) · Vector Analysis · asked 3× in 13 yrs · Read the full method →

Question

Let v⃗=v1i^+v2j^+v3k^\vec v=v_1\hat i+v_2\hat j+v_3\hat k. Show that curl⁡(curl⁡v⃗)=grad⁡(div⁡v⃗)−∇2v⃗\operatorname{curl}(\operatorname{curl}\vec v)=\operatorname{grad}(\operatorname{div}\vec v)-\nabla^2\vec v.

Technique

Componentwise expansion of the double curl; complete div⁡\operatorname{div} and Laplacian by adding/subtracting ∂2v1/∂x2\partial^2 v_1/\partial x^2; invoke cyclic symmetry.

Solution

We prove the identity componentwise (the i^\hat i-component; the others follow by cyclic symmetry x→y→z→xx\to y\to z\to x, v1→v2→v3→v1v_1\to v_2\to v_3\to v_1).

Step 1 — Compute curl⁡v⃗\operatorname{curl}\vec v

curl⁡v⃗=∇×v⃗=(∂v3∂y−∂v2∂z)i^+(∂v1∂z−∂v3∂x)j^+(∂v2∂x−∂v1∂y)k^.\operatorname{curl}\vec v=\nabla\times\vec v= \Big(\tfrac{\partial v_3}{\partial y}-\tfrac{\partial v_2}{\partial z}\Big)\hat i +\Big(\tfrac{\partial v_1}{\partial z}-\tfrac{\partial v_3}{\partial x}\Big)\hat j +\Big(\tfrac{\partial v_2}{\partial x}-\tfrac{\partial v_1}{\partial y}\Big)\hat k.

Step 2 — i^\hat i-component of curl⁡(curl⁡v⃗)\operatorname{curl}(\operatorname{curl}\vec v)

The i^\hat i-component of ∇×w⃗\nabla\times\vec w is ∂w3∂y−∂w2∂z\dfrac{\partial w_3}{\partial y}-\dfrac{\partial w_2}{\partial z}, with w⃗=curl⁡v⃗\vec w=\operatorname{curl}\vec v:

[curl⁡(curl⁡v⃗)]1=∂∂y(∂v2∂x−∂v1∂y)−∂∂z(∂v1∂z−∂v3∂x)\big[\operatorname{curl}(\operatorname{curl}\vec v)\big]_1 =\frac{\partial}{\partial y}\Big(\tfrac{\partial v_2}{\partial x}-\tfrac{\partial v_1}{\partial y}\Big) -\frac{\partial}{\partial z}\Big(\tfrac{\partial v_1}{\partial z}-\tfrac{\partial v_3}{\partial x}\Big) =∂2v2∂y ∂x−∂2v1∂y2−∂2v1∂z2+∂2v3∂z ∂x.=\frac{\partial^2 v_2}{\partial y\,\partial x}-\frac{\partial^2 v_1}{\partial y^2} -\frac{\partial^2 v_1}{\partial z^2}+\frac{\partial^2 v_3}{\partial z\,\partial x}.

Step 3 — Add and subtract ∂2v1∂x2\dfrac{\partial^2 v_1}{\partial x^2}

[curl⁡(curl⁡v⃗)]1=∂∂x(∂v1∂x+∂v2∂y+∂v3∂z)⏟= ∂x(div⁡v⃗)−(∂2v1∂x2+∂2v1∂y2+∂2v1∂z2)⏟= ∇2v1.\big[\operatorname{curl}(\operatorname{curl}\vec v)\big]_1 =\underbrace{\frac{\partial}{\partial x}\Big(\frac{\partial v_1}{\partial x}+\frac{\partial v_2}{\partial y}+\frac{\partial v_3}{\partial z}\Big)}_{=\,\partial_x(\operatorname{div}\vec v)} -\underbrace{\Big(\frac{\partial^2 v_1}{\partial x^2}+\frac{\partial^2 v_1}{\partial y^2}+\frac{\partial^2 v_1}{\partial z^2}\Big)}_{=\,\nabla^2 v_1}.

Indeed adding +∂2v1/∂x2+\partial^2v_1/\partial x^2 inside the first bracket completes div⁡v⃗\operatorname{div}\vec v, and the same term subtracted completes ∇2v1\nabla^2 v_1. Hence

[curl⁡(curl⁡v⃗)]1=[grad⁡(div⁡v⃗)]1−[∇2v⃗]1.\big[\operatorname{curl}(\operatorname{curl}\vec v)\big]_1=\big[\operatorname{grad}(\operatorname{div}\vec v)\big]_1-\big[\nabla^2\vec v\big]_1.

Step 4 — Conclude

By cyclic symmetry the same holds for the j^\hat j- and k^\hat k-components, so

Answer

  curl⁡(curl⁡v⃗)=grad⁡(div⁡v⃗)−∇2v⃗.  \boxed{\;\operatorname{curl}(\operatorname{curl}\vec v)=\operatorname{grad}(\operatorname{div}\vec v)-\nabla^2\vec v.\;}
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