Chapter 7

Div, grad, and curl

7.1 The operator and the gradient:Recall that the gradient of a differentiable scalar field on an open set D inRn is given by the formula:

=(

x1,

x2, . . . ,

xn

). (7.1)

It is often convenient to define formally the differential operator in vectorform as:

=(

x1,

x2, . . . ,

xn

). (7.2)

Then we may view the gradient of , as the notation suggests, as theresult of multiplying the vector by the scalar field . Note that the orderof multiplication matters, i.e.,

xjis not

xj.

Let us now review a couple of facts about the gradient. For any j n,xj

is identically zero on D iff (x1, x2, . . . , xn) is independent of xj. Conse-quently,

= 0 onD = constant. (7.3)Moreover, for any scalar c, we have:

is normal to the level set Lc(). (7.4)Thus gives the direction of steepest change of .

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7.2 Divergence

Let f : D Rn, D Rn, be a differentiable vector field. (Note that bothspaces are n-dimensional.) Let f1, f2, . . . , fn be the component (scalar) fieldsof f . The divergence of f is defined to be

div(f) = f =nj=1

fjxj

. (7.5)

This can be reexpressed symbolically in terms of the dot product as

f = ( x1

, . . . ,

xn) (f1, . . . , fn). (7.6)

Note that div(f) is a scalar field.Given any n n matrix A = (aij), its trace is defined to be:

tr(A) =ni=1

aii.

Then it is easy to see that, if Df denotes the Jacobian matrix, then

f = tr(Df). (7.7)Let be a twice differentiable scalar field. Then its Laplacian is defined

to be

2 = (). (7.8)It follows from (7.1),(7.5),(7.6) that

2 = 2

x21+2

x22+ +

2

x2n. (7.9)

One says that is harmonic iff 2 = 0. Note that we can formallyconsider the dot product

= ( x1

, . . . ,

xn) (

x1, . . . ,

xn) =

nj=1

2

x2j. (7.10)

Then we have

2

2 = ( ). (7.11)

Examples of harmonic functions:

(i) D = R2; (x, y) = ex cos y.Then

x= ex cos y,

y= ex sin y,

and 2x2

= ex cos y, 2y2

= ex cos y. So, 2 = 0.(ii) D = R2 {0}; (x, y) = log(x2 + y2) = log(r).

Then x

= xx2+y2

, y

= yx2+y2

, 2x2

= (x2+y2)2x(2x)

(x2+y2)2= (x

2y2)(x2+y2)2

, and 2y2

=(x2+y2)2y(2y)

(x2+y2)2= (x

2y2)(x2+y2)2

. So, 2 = 0.These last two examples are special cases of the fact, which we mention

without proof, that for any function f : D C which is differentiable in thecomplex sense, the real and imaginary part, 0} or D = {z = x+ iy| y = 0 x < 0}. Butthe union of D and D is C {0} as in (ii) .(iii) D = Rn {0}; (x1, x2, ..., xn) = (x21 + x22 + + x2n)/2 = r for somefixed R.

Then xi

= r1 xir

= r2xi, and2x2i

= ( 2)r4xi xi + r2 1.Hence 2 = ni=1 (( 2)r4x2i + r2) = ( 2 + n)r2.So is harmonic for = 0 or = 2 n ( = 1 for n = 3).

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7.3 Cross product in R3

The three-dimensional space is very special in that it admits a vector prod-uct, often called the cross product. Let i,j,k denote the standard basis ofR3. Then, for all pairs of vectors v = xi + yj + zk and v = xi + yj + zk,the cross product is defined by

v v = det(

i j kx y zx y z

)= (yz yz)i (xz xz)j + (xy xy)k. (7.12)

Lemma 1 (a) vv = vv (anti-commutativity) (b) ij = k, jk = i,k i = j (c) v (v v) = v (v v) = 0.

Corollary: v v = 0.

Proof of Lemma (a) v v is obtained by interchanging the second andthird rows of the matrix whose determinant gives vv. Thus vv=vv.

(b) i j = det(

i j k1 0 00 1 0

), which is k as asserted. The other two identities

are similar.(c) v (v v) = x(yz yz) y(xz xz) + z(xy xy) = 0. Similarly

for v (v v).Geometrically, vv can, thanks to the Lemma, be interpreted as follows.

Consider the plane P in R3 defined by v,v. Then v v will lie along thenormal line to this plane at the origin, and its orientation is given by the righthand rule: If the fingers of your right hand grab a pole and you view themfrom the top as a circle in the v v-plane that is oriented counterclockwise(i.e. corresponding to the ordering (v, v) of the basis) then the thumb pointsin the direction of v v.

Finally the length ||v v|| is equal to the area of the parallelogramspanned by v and v. Indeed this area is equal to the volume of the paral-lelepiped spanned by v, v and a unit vector u = (ux, uy, uz) orthogonal to vand v. We can take u = v v/||v v|| and the (signed) volume equals

det

ux uy uzx y zx y z

=ux(yz yz) uy(xz xz) + uz(xy xy)=||v v|| (u2x + u2y + u2z) = ||v v||.

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More generally, the same argument shows that the (signed) volume of theparallelepiped spanned by any three vectors u, v, v is u (v v).

7.4 Curl of vector fields in R3

Let f : D R3, D R3 be a differentiable vector field. Denote by P ,Q,Rits coordinate scalar fields, so that f = P i +Qj +Rk. Then the curl of f isdefined to be:

curl(f) = f = det(

i j kx

y

z

P Q R

). (7.13)

Note that it makes sense to denote it f , as it is formally the crossproduct of with f .

If the vector field f represents the flow of a fluid, then the curl measureshow the flow rotates the vectors, whence its name.

Proposition 1 Let h (resp. f) be a C2 scalar (resp. vector) field. Then(a) (h) = 0.(b) ( f) = 0.

Proof: (a) By definition of gradient and curl,

(h) = det(

i j kx

y

z

hx

hy

hz

)

=

(2h

yz

2h

zy

)i +

(2h

zx

2h

xz

)j +

(2h

xy

2h

yx

)k.

Since h is C2, its second mixed partial derivatives are independent of theorder in which the partial derivatives are computed. Thus, (f) = 0.

(b) By the definition of divergence and curl,

( f) =(

x,

y,

z

)(R

y Qz

,Rx

+P

z,Q

x Py

)

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=(2R

xy

2Q

xz

)+

(

2R

yx+

2P

yz

)+

(2Q

zx

2P

zy

).

Again, since f is C2, 2Rxy

= 2R

yx, etc., and we get the assertion.

Done.

Warning: There exist twice differentiable scalar (resp. vector) fieldsh (resp. f), which are not C2, for which (a) (resp. (b)) does not hold.

When the vector field f represents fluid flow, it is often called irrota-tional when its curl is 0. If this flow describes the movement of water in astream, for example, to be irrotational means that a small boat being pulledby the flow will not rotate about its axis. We will see later in this chapterthe condition f = 0 occurs naturally in a purely mathematical settingas well.

Examples: (i) Let D = R3 {0} and f(x, y, z) = y(x2+y2)

i x(x2+y2)

j. Showthat f is irrotational. Indeed, by the definition of curl,

f = det(

i j kx

y

z

y

(x2+y2)

x(x2+y2)

0

)

=

z

(x

x2 + y2

)i +

z

(y

x2 + y2

)j +

(

x

( xx2 + y2

) y

(y

x2 + y2

))k

=

[(x2 + y2) + 2x2(x2 + y2)2

(x2 + y2) 2y2(x2 + y2)2

]k = 0.

(ii) Let m be any integer 6= 3, D = R3 {0}, andf(x, y, z) = 1

rm(xi + yj + zk), where r =

x2 + y2 + z2. Show that f is not

the curl of another vector field. Indeed, suppose f = g. Then, since fis C1, g will be C2, and by the Proposition proved above, f = ( g) would be zero. But,

f =(

x,

y,

z

)( xrm,y

rm,z

rm

)

=rm 2x2(m

2)rm2

r2m+rm 2y2(m

2)rm2

r2m+rm 2z2(m

2)rm2

r2m

6

=1

r2m(3rm m(x2 + y2 + z2)rm2) = 1

rm(3m).

This is non-zero as m 6= 3. So f is not a curl.

Warning: It may be true that the divergence of f is zero, but f is stillnot a curl. In fact this happens in example (ii) above if we allow m = 3. Wecannot treat this case, however, without establishing Stokes theorem.

7.5 An interpretation of Greens theorem via

the curl

Recall that Greens theorem for a plane region with boundary a piecewiseC1 Jordan curve C says that, given any C1 vector field g = (P,Q) on an openset D containing , we have:

(Q

x Py

)dx dy =

C

P dx+Qdy. (7.14)

We will now interpret the term Qx P

y. To do that, we think of the

plane as sitting in R3 as {z = 0}, and define a C1 vector field f on D :={(x, y, z) R3|(x, y) D} by setting f(x, y, z) = g(x, y) = P i + Qj. Then f = det

(i j kx

y

z

P Q 0

)=(Qy P

x

)k, because P

z= Q

z= 0. Thus we

get:

( f) k = Qy Px

. (7.15)

And Greens theorem becomes:

Theorem 1

( f) k dx dy =

CP dx+Qdy

7.6 A criterion for being conservative via the

curl

Here we just reformulate the remark after Ch. 6, Cor. 1 (which we didntcompletely prove but just made plausible) using the curl.

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Proposition 1 Let g : D R2, D R2 open and simply connected,g = (P,Q), be a C1 vector field. Set f(x, y, z) = g(x, y), for all (x, y, z) R3with (x, y) D. Suppose f = 0. Then g is conservative on D.

Proof: Since f = 0, Theorem 1 implies that CP dx+Qdy = 0 for all

Jordan curves C contained in D. In fact, f = 0 also implies thatCP dx+Qdy = 0 for all closed curves but we wont prove this. Hence f is

conservative. Done.

Example: D = R2 {(x, 0) R2 | x 0}, g(x, y) = yx2+y2

i xx2+y2

j.Determine if g is conservative on D:

Again, define f(x, y, z) to be g(x, y) for all (x, y, z) in R3 such that (x, y) D. Since g is evidently C1, f will be C1 as well. By the Proposition above, itwill suffice to check if f is irrotational, i.e., f = 0, on DR. This wasalready shown in Example (i) of section 4 of this chapter. So g is conservative.

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