568 CHAPTER 29. CALCULUS OF VECTOR FIELDS

definition in terms of coordinates. Be sure you understand that for f a vector field, divfis a scalar field meaning it is a scalar valued function of three variables. For a scalar fieldf , ∇ f is a vector field described earlier. For f a vector field having values in R3,curlf isanother vector field.

Example 29.1.2 Let f (x) = xyi+(z− y)j+(sin(x)+ z)k. Find divf and curlf .

First the divergence of f is

∂ (xy)∂x

+∂ (z− y)

∂y+

∂ (sin(x)+ z)∂ z

= y+(−1)+1 = y.

Now curlf is obtained by evaluating∣∣∣∣∣∣i j k∂

∂x∂

∂y∂

∂ zxy z− y sin(x)+ z

∣∣∣∣∣∣=i

(∂

∂y(sin(x)+ z)− ∂

∂ z(z− y)

)−j

(∂

∂x(sin(x)+ z)− ∂

∂ z(xy)

)+

k

(∂

∂x(z− y)− ∂

∂y(xy)

)=−i− cos(x)j− xk.

29.1.1 Vector Identities

There are many interesting identities which relate the gradient, divergence and curl.

Theorem 29.1.3 Assuming f,g are a C2 vector fields whenever necessary, the fol-lowing identities are valid.

1. ∇ · (∇×f) = 0

2. ∇×∇φ = 0

3. ∇× (∇×f) = ∇(∇ ·f)−∇2f where ∇

2f is a vector field whose ith component is∇

2 fi.

4. ∇ · (f ×g) = g·(∇×f)−f ·(∇×g)

5. ∇× (f ×g) = (∇ ·g)f− (∇ ·f)g+ (g·∇)f− (f ·∇)g

Proof: These are all easy to establish if you use the repeated index summation conven-tion and the reduction identities.

∇ · (∇×f) = ∂i (∇×f)i = ∂i(ε i jk∂ j fk

)= ε i jk∂i (∂ j fk)

= ε jik∂ j (∂i fk) =−ε i jk∂ j (∂i fk) =−ε i jk∂i (∂ j fk)

= −∇ · (∇×f) .