III.D Curl and Divergence Given a Scalar Function F(X, Y, Z)
III.d Curl and Divergence ∂f Given a scalar function f(x, y, z) we have computed its gradient ∇f = i + ∂x ∂f ∂f j + k and discussed some of its practical significance. We now introduce two ∂y ∂z operations on vector fields F = M i + N i + P k.
The first is the divergence of F, denoted by div (F) or ∇ · F and defined by
∂ ∂ ∂ ∂M ∂N ∂P div (F) = i + j + k · (M i + N j + P k) = + + . ∂x ∂y ∂z ∂x ∂y ∂z
The second is the curl of F, denoted by curl (F) or ∇ × F and defined by:
i j k ∂ ∂ ∂ ∇ × F = det ∂x ∂y ∂z M N P ∂P ∂N ∂M ∂P ∂N ∂M = i − + j − + k − . ∂y ∂z ∂z ∂x ∂x ∂y
Observe that for any F, div (F) is scalar while curl (F) is a vector.
We shall not discuss in any detail the physical significance of div and curl . But there are some properties we wish to point out.
Properties:
(1) ∇×F ≡ 0 iff F is conservative. Indeed the components of ∇×F are the three
quatities that are zero precisely when a potential exists.
(2) As a consequence of (1), we have ∇ × (∇f) = 0, i.e., the curl of a gradient is
the zero vector.
(3) By direct calculation, ∇ · (∇ × F) = 0, i.e., the div of a curl is zero.
(4) We can write Green’s Theorem in vector form: Let F = M i + N j + O k, and C is simple closed path in the (x, y) plane. 197 D C
Then
∂N ∂M F · dr = − dA = (curl F) · k dA. Z ZZD ∂x ∂y ZZD
Terminology: If curl F = 0 at a point P in (x, y, z) space, then we say F is irrota- tional at P . If div F = 0 at some point P in (x, y, z) space, then F is incompressible at P . These names come from the physical significance of curl , div , respectively.
Example 1. Find the div and curl of F = x2 i + xy j + xz2 k.
Answer.
∂ ∂ ∂ div F = (x2) + (xy) + (xz2) = 2x + x + 2xz. ∂x ∂y ∂z
Example 2. Is there a vector field G whose curl is given by curl G = x i+y j+z k?
Answer. At first sight, it may seem wise to approach this problem as follows: Try to find G1, G2, G3 such that G = G1 i + G2 j + G3 k has curl G = x i + y j + z k. In reality, it is much better to note that div (curl G) = 0 for any G, while in this case we would have: div (curl G) = 1 + 1 + 1 = 3. So there is no G.
Example 3. Prove the identity div (fF) = fdiv (F) + F · ∇f.
Answer. This cannot be done by choosing examples for f, F (e.g. f(x) = xyz, F = x i + y j + z k, say) and then proving the identity for the chosen f, F! Instead, 198 we write F = M i + N j + P k. Then
div (fF) = div (fM i + fN j + fP k) ∂ ∂ ∂ = (fM) + (fN) + (fP ) ∂x ∂y ∂z ∂f ∂M ∂f ∂N ∂f ∂P = · M + f + · N + f + · P + f ∂x ∂x ∂y ∂y ∂z ∂z ∂M ∂N ∂P ∂f ∂f ∂f = f + + + · M + · N + · P ∂x ∂y ∂z ∂x ∂y ∂z = f div (F) + F · ∇f.
Example 4. Verify the identity: div (F × G) = G · curl F − F · curl G.
Answer. Set
F = M i + N j + P k
Now i j k F × G = det M N P Q R S = i[NS − P R] + j[P Q − MS] + k[MR − NQ].
So
∂N ∂S ∂P ∂R div [F × G] = S + N − R − P ∂x ∂x ∂x ∂x
∂P ∂Q ∂M ∂S + Q + P − S − M ∂y ∂y ∂y ∂y
∂M ∂R ∂N ∂Q + R + M − Q − N ∂z ∂z ∂z ∂z 199 ∂P ∂N ∂M ∂P ∂N ∂M = Q − + R − + S − ∂y ∂z ∂z ∂x ∂x ∂y
∂R ∂S ∂S ∂Q ∂Q ∂R − M − + + N − + + P − + ∂z ∂y ∂x ∂z ∂y ∂x
= G · curl F − F · curl G.
200 Further Exercises:
Find div (F), curl (F) for 1)–3).
2 1) F = (sin x)i + (x + y)j + (sec z)k.
2) F = (sec(xy))i + (csc(yz))j + (tan z sec z)k.
x y xyz 3) F = e y i + e x j + e k.
4) Verify the identity: div (f curl F) = (gradf) · (curl F).
5) Verify the identity: ∇ × (fF) = f∇ × F + (∇f) × F.
6) Verify the identity: curl (fF) = fcurl (F) + (∇f) × F.
7) Find all constant(s) k, if any, for which f = ekz sin(x)sin(2y) is harmonic (i.e., satisfies Laplace’s equation ∇ · (∇f) = 0).
8) Find all constant(s) k, if any, such that the field F = xyi+(y2/2)j+[tan(sin(x2+
y2)) + kezxy]k is incompressible.
9) Find all constant(s) k, if any, such that the field F = x2yi + [(x3/3) + zy]j + [ez +kzy2]k is irrotational. For these k, find the potential f such that ∇f = F.
201