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) .

568 CHAPTER 29. CALCULUS OF VECTOR FIELDSdefinition in terms of coordinates. Be sure you understand that for f a vector field, div fis a scalar field meaning it is a scalar valued function of three variables. For a scalar fieldf, Vf is a vector field described earlier. For f a vector field having values in R*, curl f isanother vector field.Example 29.1.2 Let f (x) = xyi+(z—y) 7+ (sin (x) +z)k. Find div f and curl f.First the divergence of f isA (xy) | A(z—y) | A(sin(x) +z)Ox + oy + Oz=yt+(-I+l=y.Now curl f is obtained by evaluating4 j koa oo oa _Ox oy Oz ~—xy z—-y sin(x)+z(5 (sin (x) +z) — < -»)) J (5. (sin (x) +z) — < (w)) +k ($6-»-F ow) = —4— cos (x) j — xk.29.1.1 Vector IdentitiesThere are many interesting identities which relate the gradient, divergence and curl.Theorem 29.1.3 Assuming f,g are a C? vector fields whenever necessary, the fol-lowing identities are valid.1.V-(Vx f)=02. VxVo=03. Vx (Vx f) =V(V-f)—VW’f where Vf is a vector field whose it component isV fi.4.V-(f xg)=9:(Vxf)—-f(V xg)5. Vx(fxg)=(V-g) f-(V- f)gt+(a-V) f-(f-VY)gProof: These are all easy to establish if you use the repeated index summation conven-tion and the reduction identities.V-(Vxf) O(V x f); = 0; (€:jn0j fe) = €ijn 9 (Oj fe)E jikOj (Oifk) = —E:jnO; (Ait) = —EijnO: (Oj fx)-V-(Vxf).