vector field operations

7/30/2019 Vector field operations

1/24

in other coordinates 1

Chapter 15

in other coordinates

On a number of occasions we have noticed that del is geometrically determined it doesnot depend on a choice of coordinates for Rn. This was shown to be true for f, the gradientof a function from Rn to R (Section 2H). It was also verified for F, the divergence of afunction from Rn to Rn (Section 14B). And in the case n = 3, we saw in Section 13G that itis true for F, the curl of a function from R3 to R3.

These three instances beg the question of how we might express in other coordinatesystems for Rn. A recent example of this is found in Section 13G, where a formula is given for F in terms of an arbitrary right-handed orthonormal frame for R3. We shall accomplishmuch more in this chapter.

A very interesting book about , by Harry Moritz Schey, has the interesting title Div,Grad, Curl, And All That.

A. Biorthogonal systems

We begin with some elementary linear algebra. Consider an arbitrary frame {1, 2, . . . , n}for Rn. We know of course that the Gram matrix is of great interest:

G = (i j).This is a symmetric positive definite matrix, and we shall denote its entries as

gij = i j.Of course, G is the identity matrix we have an orthonormal frame.

We also form the matrix whose columns are the vectors i. Symbolically we write

= (1 2 . . . n).

We know that is an orthogonal matrix we have an orthonormal frame (Problem 420).Now denote by the transpose of the inverse of , and express this new matrix in terms

of its columns as = (1 2 . . . n).

We then have the matrix product

t =

t1...

tn

(1 . . . n)

= (i j).


2/24

2 Chapter 15

But since t equals the inverse of , we conclude that

i j = ij.

DEFINITION. Two frames {1, . . . , n} and {1, . . . , n} are called a biorthogonal systemif

i j = ij.Clearly, {1, . . . , n} and {1, . . . , n} form a biorthogonal system the frame {1, . . . , n}

is an orthonormal one. So the present definition should be viewed as a generalization of theconcept of orthonormal basis.

Also it is clear that the relation between the is and the is given here is completelysymmetric.

PROBLEM 151. Let a frame for R2 be {, a + b}, where of course b = 0. Computethe corresponding {1, 2} which produces a biorthogonal system. Sketch all four vectorson a copy ofR2.

PROBLEM 152. Let {1, 2, 3} be a frame for R3. Prove that the biorthogonalframe is given by

3 =1

2[1, 2, 3] etc.

PROBLEM 153. Suppose {1, . . . , n} is an orthogonal frame. Show that thecorresponding vectors 1, . . . , n are given as

i =i

i2 .

PROBLEM 154. Given a frame {1, . . . , n} denote

J = det .

Prove that det G = J2.


3/24


Finally, a little more notation. It has become customary to denote the inverse of G = (gij)

byG1 = (gij)

(called raising the indices). That is,

nk=1

gikgkj =n

k=1

gikgkj = ij.

Inasmuch as

G = t,

we conclude that

= (t)1

= G1.

In terms of entries of these matrices this equation means that

()ij =n

k=1

()ikgkj .

Thus we find that the columns satisfy

j =

nk=1

kgkj

.

Rewriting this equation gives

i =n

j=1

gijj.

The corresponding inverse equation is of course

i =n

j=1

gijj.

PROBLEM 155. Prove that for all v Rn

v =n

i=1

(i v)i.


4/24

4 Chapter 15

The result of this problem is that when you wish to express v as a linear combination of the

basis vectors i, the corresponding coefficients are given directly in terms of inner productswith the is. You might think of the formal expression

ni=1

(i ( ))i

as representing the identity linear function on Rn. In other words,

PROBLEM 156. Prove that

I =n

i=1

iti .

B. The gradient

We continue with an arbitrary biorthogonal system on Rn, and maintain the notation ofSection A.

Suppose that Rnf R is differentiable at a point x Rn. Our task is to express the

vector f(x) Rn as a linear combination of the vectors i in the given frame. This is quitean easy task. In fact, Problem 155 gives immediately

f(x) =n

i=1

f(x) ii.

The inner products in this formula are of course directional derivatives of f in the directionsi. Thus, in the notation of Section 2C

f(x) =n

i=1

Df(x; i)i.

PROBLEM 157. Show that everything in this formula can be expressed entirely interms of the frame {1, . . . , n} by a double sum

f(x) =n

i,j=1

gijDf(x; j)i.


5/24


C. The divergence

We want to discuss a vector field f defined on an open subset ofRn. We can thus regardf as a function from Rn to Rn, and as such it has a derivative. At a point x in its domain, thederivative Df(x) is a linear transformation ofRn to Rn, represented in terms of the standardcoordinate basis e1, . . . , en, by the n n Jacobian matrix

fixj

.

It is the trace of this matrix which is f, the divergence of f. This observation is the keyto our representation of f, and we need a simple fact from linear algebra:

THEOREM. Suppose A is a real n n matrix, and regard the is and is as columnvectors. ThentraceA =

ni=1

Ai i.

PROOF. We use the matrices and from Section A, so that

Ai j = ji entry of the matrix tA.Thus

ni=1

Ai i = trace(t

A)

= trace(At) (Problem 42)

= trace(A),

since and t are inverses.QED

As a result we now have

f(x) =n

i=1fixi

=n

i=1

Df(x)i i

=n

i=1

Df(x; i) i.


6/24

6 Chapter 15

D. The curl

Before going into the representation of curl, we summarize what we have obtained to thispoint. In terms of a fixed biorthogonal system we have found

f(x) = f(x) =n

i=1

Df(x; i)i

and

divF(x) = F(x) =n

i=1

DF(x; i) i.

These two formulas can be incorporated as a single expression

=n

i=1

i times D( ; i)

where and the right side are both regarded as operators. They operate on scalar-valuedfunctions to produce the gradient and on vector fields to produce the divergence. In thegradient situation times is scalar multiplication, whereas in the divergence situation it isdot product.

A good way to remember this formula is to replace the directional derivative

Df(x; i)

by the symbolf

i

Then we have

=n

i=1

i times

i.

This leads us to an educated guess for curl, where we just use the same formula. Thus we

suppose that {1, 2, 3} and {1, 2, 3} form a biorthogonal system for R3

. Then we have

THEOREM. For a vector field F onR3,

curlF(x) = F(x) =3

i=1

i DF(x; i).


7/24


PROOF. Every vector field F can be expressed in the given frame in the form

F(x) =3

i=1

fj(x)j.

Thus it suffices to prove the formula for the special case of vector fields of the form fj(x)j,for j = 1, 2, 3. Thus we assume

F(x) = f(x)j.

We have from the product rule of Problem 1314,

F(x) =

f(x)

j (since j is constant)

and our formula for gradient gives

F(x) =3

i=1

Df(x; i)i j

=3

i=1

i Df(x; i)j

=3

i=1

i D(f j)(x; i)

=3

i=1

i DF(x; i).

QED

PROBLEM 158. Are you surprised that the formula for curl does not require theframe {1, 2, 3} to be right-handed? Explain what happens if 3 is replaced with 3.


1 2 = J3,2 3 = J1,3 1 = J2.


8/24


9/24


Since uj is the jth coordinate ofT1(x), the gradient uj is the jth row of the Jacobian matrix

for T1

. Therefore, since the chain rule implies

(T1)(T(u))T(u) = I,

we have the matrix equation

12...

n

(1 2 . . . n) = I.

That is,

i j = ij.

In other words, the frames at x given by

{1(T(x)), . . . , n(T(x))} and {1(x), . . . , n(x)}

are biorthogonal. Heres a representative sketch in the x1x2 plane:


10/24

10 Chapter 15

=

constant

= constant

.

.

.

.

.

.

The`sare

tangentto

thecoordinate

curves

i

The`sare

orthogonalto

thecoordinate

curves

i

=

constant

= constant

Our point of view is going to be that when we work on Rn and use the coordinatesu1, . . . , un, we are interested in calculating everything in terms of the natural vectors i(u)and the natural derivatives /uj .

Of course, each j is a vector field on Rn in fact, a gradient field. And each i T is

also a vector field on Rn, but not necessarily a gradient field.


11/24


We still maintain the notations introduced in Section A. Thus

gij(u) = i(u) j(u),

and the matrix inverse to (gij) has entries

gij(u).

Furthermore, it follows that, with x = T(u),

i(x) =n

j=1gij(u)j(u)

and

i(u) =n

j=1

gij(u)j(x).

F. The gradient

The formula for f goes over with no change from Section B. We read it off from Problem157: with x = T(u),

f(x) =

ni,j=1 g

ij

(u)Df(x; j(u))i(u).

We notice that the chain rule gives

Df(x; j(u)) = f(T(u)) T(u)uj

=

ujf(T(u)).

NOTATION. In this context we denote the pull-backoff by the coordinate transformation

x = T(u) to be the functionf(u) = f(T(u)).

Thus we have

f(x) =n

i,j=1

gij(u)f

uji(u).


12/24

12 Chapter 15

This is exactly the sort of formula we want. Everything on the right side of the equation is in

terms of the coordinates we are working with.

G. Spherical coordinates

We pause for a significant example. We use our standard spherical coordinates for R3:

x = r sin cos ,

y = r sin sin ,

z = r cos .

We then have the frame

1 = (sin cos , sin sin , cos ),2 = r(cos cos , cos sin , sin ),3 = r sin ( sin , cos , 0).

Thus gij = 0 ifi = j, andg11 = 1,

g22 = r2,

g33 = r2 sin2 .

Furthermore, g

ij

= 0 ifi = j, andg11 = 1,

g22 =1

r2,

g33 =1

r2 sin2 .

Thus

f(x,y,z) = f

r1 +

1

r2f

2 +

1

r2 sin2

f

3.

Frequently this formula is displayed in terms of theorthonormal

frame associated with {1, 2, 3}.Namely, definei =

ii .

Then it becomes

f = f

r1 +

1

r

f

2 +

1

r sin

f

3.


13/24


In fact, a nice alternative for this notation is to say that in general

ui =Tui

Tui ,

so the formula becomes

f = f

rr +

1

r

f

+

1

r sin

f

.

PROBLEM 1511. Show that J = r2 sin .

PROBLEM 1512. Carry out the same analysis for the case of cylindrical coordinatesfor R3:

x = r cos ,

y = r sin ,

z = z.

Conclude that

f = f

rr +

1

r

f

+

f

zz.

H. The divergence

A new situation arises in the curvilinear representation of divergence and curl. Namely weassume from the start that the vector field in question is represented in terms of curvilinearcoordinates and the frame {1, . . . , n} associated with them. Thus we treat the vector fieldF(x) by first expressing it in the form

F(x) =n

i=1

Fi(u)i(u), where x = T(u).

We then want to express the scalar F in terms of derivatives /ui involving the coefficientsFi(u). Since the vector i(u) comes from T by differentiation and is liable to be subject todifferentiation again, we need to assume T is of class C2. We shall make this assumption forthe remainder of the chapter.

It turns out that there is a huge simplification available if we first establish the remarkable


14/24

14 Chapter 15

LEMMA. (J1i) = 0.

PROOF. The vector field we are dealing with in this context is

G(x) = J1(u)i(u), where x = T(u).

We give a proof of this lemma based on the divergence theorem. To set this up, suppose h isany C1 real-valued function on Rn which is zero outside a small ball. We shall prove that

Rnh Gdx = 0. ()

Since G is a continuous function, this will prove the result. For if G(x0) > 0 for somex0, then G(x) > 0 for all x in some neighborhood of x0. We can then choose a suitable hto make the integrand in () always 0 and positive in a neighborhood of x0, so the integralin () will be positive, a contradiction. Likewise if G(x0) < 0.

The function h in this calculation is frequently called a test function. It is unimportant initself, but is used to test the crucial function G.

We now turn to the proof of (). First,

Rn

(hG)dx = 0

follows from the divergence theorem. Actually, just the fundamental theorem of calculus isneeded, as a typical term in the Cartesian formula for divergence is

Rn

xj(hG ej)dx,

and performing the xj integration first gives the result, since h is zero outside a small ball.

Since the product rule gives

(hG) = h G + h G,

the proof of () now reduces to proving thatRn

h Gdx = 0.


15/24


In this integral we change variables with x = T(u), and we obtainRn

h Gdx =Rn

h(T(u)) G(T(u))| det T(u)|du

=

Rn

h(T(u)) J1(u)i(u)|J(u)|du

= Rn

h(T(u)) Tui

du

chain rule=

Rn

ui(h(T(u))du

FTC= 0.

QEDNow we compute the divergence of a typical summand in F(x): the product rule gives

(Fi(u)i(u)) = (J(u)Fi(u) 1J(u)

i(u))

= (J(u)Fi(u)) 1J(u)

i(u),

the other term being zero thanks to the lemma. And now we invoke the chain rule to get

1

J(u)(J(u)Fi(u)) i(u) =1

J(JFi) T

ui

=1

J

ui(JFi).

Thus we have our final formula: in case

F(x) =n

i=1

Fi(u)i(u), x = T(u),

then

F =1

J(u)

ni=1

(J(u)Fi)

ui .

Notice how closely this resembles the Cartesian case!For our spherical coordinate example, write

F(x,y,z) = F1(r,,)r + F2 + F3,


16/24

16 Chapter 15

so that the corresponding functions in our formula are actually

F1, F2/r, and F3/r sin , respectively.

Thus

F = 1r2 sin

r(r2 sin F1) +

(r sin F2) +

(rF3)

=1

r2

r(r2F1) +

1

r sin

(sin F2) +

1

r sin

F3

.

PROBLEM 1513. For the case of cylindrical coordinates for R3 let

F(x,y,z) = F1r + F2 + F3z,

and prove that

F = 1r

(rF1)

r+

1

r

F2

+F3z

.

PROBLEM 1514. Suppose that {1, . . . , n} is a frame field for Rn, so that it mayvary from point to point. Suppose that the corresponding frame field {1, . . . , n} yieldsa biorthogonal system. Prove that the basic formula of Section C is still valid:

F =n

i=1

i Fi

.

I. The curl

We continue the notation of the preceding section, so that we are concerned with a vectorfield on R3, but instead of expanding F in terms of the frame {1, 2, 3}, it seems better towork with F(x) itself. We then follow the notation of Section F and define the pull-back

F(u) = F(T(u)).

We shall employ the formula for curl as given in Section D, but with the is and is inter-changed. So we find

F(x) =3

i=1

i DF(x; i).


17/24


Since i = T/ui, the chain rule gives with x = T(u),

F(x) =3

i=1

i(x) F

ui(u).

Next we insert the algebra of Problem 158:

1 = J12 3 etc.,

obtaining

F = J1(

2

3)

F

u1+

.

The formula for vector triple product of Problem 74 gives

J F = 2 F

u13 3 F

u12 + .

The product rule gives

J F = u1

(2 F)3 2u1

F3

u1

(3 F)2 + 3u1

F2 + .

The other eight terms are obtained from these by cycling through the indices 1231.Look at the terms in which F appears undifferentiated: the coefficient of 3 in these terms is

2u1

F + 1u2

F.

But1

u2=

u2

T

u1=

u1

T

u2=

2

u1,

so that the coefficient of 3 just obtained is zero. Notice here that we have the assumptionthat T is of class C2 so that the mixed partial derivatives are equal. Likewise for 1 and 2.Thus we have

J F = u1

(2 F)3 u1

(3 F)2 +


18/24

18 Chapter 15

(four more terms). This expression is precisely the formal determinant

det

1 2 3

u1

u2

u3

1 F 2 F 3 F

.

This gives our final formula for curl:

F = 1J

det

1 2 3

u1

u2

u3

1 F 2 F 3 F

.

Of course, the pattern is just that of the original definition of curl in terms of Cartesiancoordinates.

Now lets examine the spherical coordinate case:

F(x,y,z) = F1(r,,)r + F2 + F3.

Then

1 F = r F = F1,2 F = r F = rF2,3 F = r sin F = r sin F3,

so that

F = 1r2 sin

det

r r r sin

r

F1 rF2 r sin F3

.


19/24


PROBLEM 1515. For the case of cylindrical coordinates in the notation of Problem

1512, show that

F = 1r

det

r r z

r

z

F1 rF2 F3

.

In particular, if the vector field is planar,

F = F1(r, )r + F2(r, ),

conclude that

F =

F2r

+1

rF2 1

r

F1

k.

It is especially interesting to see the formula for the operator in terms of these curvilinearcoordinates. The directional derivative in the direction i(u), as we have noticed, is just /ui.Thus we may rewrite the general formula from Section D in the form

=n

i=1

i times

ui,

or if we replace by its definition,

=n

i=1

ui times ui

.

As always, times means

scalar multiplication if we are calculating gradient, dot product if we are calculating divergence, cross product if we are calculating curl.This version of the expression for is rather easy to remember. If we think of loosely

as /x, then the formula is similar to

x=

ni=1

uix

times

ui,


20/24

20 Chapter 15

or, even more briefly,

x

= ux

times u

.

J. Orthogonal curvilinear coordinates

Many coordinate systems that arise in specific applications have the feature that the curvesin Rn determined by them are orthogonal. This means in our notation that for i = j

i j = Tui

Tuj

= 0.

In other words, gij = 0 ifi = j. Then the numbers of importance are

gii(u), 1 i n,and

gij(u) =

0 if i = j,

1gii(u)

if i = j.

Moreover,J2 = g11g22 . . . gnn.

Notably, spherical coordinates and also cylindrical coordinates both fit this situation.Our formulas for all simplify under the assumption of orthogonality. Thus the gradient

is given by

f(x) =n

i=1

1gii(u)

f

uii(u).

The divergence is still the same, no simplification occuring at all.The formula for curl is much simpler. For in the notation

F(x) =3

i=1

Fi(u)i(u)

we now have simplyi F = giiFi.

Thus

F = 1g11g22g33

det

1 2 3

u1

u2

u3

g11F1 g22F2 g33F3

.


21/24


PROBLEM 1516. If you use instead the associated orthonormal frame so that

F(x) =3

i=1

Fi(u)i(u),

show that

F = 1g11g22g33

det

g111

g222

g333

u1

u2

u3

g

11F

1 g

22F

2 g

33F

3

.

K. The Laplacian

Without a doubt the most important second order partial differential operator on Rn isthe Laplace operator, or the Laplacian. It operates on a real-valued function by first formingits gradient and then forming the divergence of the resulting vector field. Thus the notationis 2:

2f =

(

f).

Another common notation is = 2.Thus we have in Cartesian coordinates

2f(x) =n

i=1

2f

x2i.

Of course, we well recognize that

2 has an intrinsic geometric meaning apart from any

coordinate system, since the same is true off and of the divergence of a vector field.Before giving the general formula in curvilinear coordinates we mention the linear case in

which x = T(u) is given by matrix multiplication

x = Au,


22/24

22 Chapter 15

where A = (aij) is a real nonsingular n n matrix. Then our notation gives the following:

j =x

uj=

a1ja2j

...anj

= jth column of A;

i = ui = (A1x)ith entry

=

ai1

ai2

...

ain

= transpose ofith row of A1;

J = det A;

gij =n

k=1

akiakj ;

gij =n

k=1

aikajk .


2f(x) =n

i,j=1

gij2f(u)

uiuj.

PROBLEM 1518. Prove that 2 is invariantwith respect to this change of variables,that is,

2f(x) =n

i=1

2f

u2i,

if and only if A O(n).Now we turn to the general case of curvilinear coordinates. We have from Section F

f(x) =n

i,j=1

gij(u)f

uji(u),


23/24


so the gradient vector field f is represented as in Section H with the corresponding coefficientsdenoted there as Fi(u) given by

Fi(u) =n

j=1

gij(u)f

uj(u).

Therefore the result of Section H gives immediately

2f(x) = 1J

ni,j=1

ui

Jgij

f

uj

.

If we have the special case of orthogonal curvilinear coordinates, then of course

J2 = g11g22 . . . gnn,

gii =1

gii,

so we have

2f = 1g11 . . . gnn

ni=1

ui

g11 . . . gnn

gii

f

ui

.

For spherical coordinates on R3 this gives

2f =

1

r2 sin

r r2 sin f

r +

r2 sin

r2f

+

r2 sin

r

2

sin

2

f

=

1

r2

r

r2

f

r

+

1

r2 sin

sin

f

+

1

r2 sin2

2f

2.

PROBLEM 1519. For the case of cylindrical coordinates in R3, show that

2f = 1r

r

r

f

r

+

1

r22f

2+

2f

z2.

L. Parabolic coordinates

Heres an example that we have not yet mentioned. Its a coordinate system for R2 basedon squaring complex numbers. Namely, in complex notation

x + iy =1

2(u + iv)2.


24/24

24 Chapter 15

In real notation, x = 12 (u2 v2),y = uv.

We assume that u > 0, < v < . Then we produce all points ofR2 except the negativex-axis (, 0] {0}.

PROBLEM 1520. Write out explicit formulas for u, v as functions of x, y. (Becareful to have u > 0.) (See Problem 293.)

PROBLEM 1521. We call u, v parabolic coordinates because the curves in the x

y

plane on which u is constant and also those along which v is constant are parabolas. Provethis, and also prove that the origin is the focus of each of these parabolas.

Our notation from Section E leads us to the frame vectors

1(u, v) = (u, v),

2(u, v) = (v, u).

These vectors are orthogonal, so we have here orthogonal curvilinear coordinates.

PROBLEM 1522. Show that

2f(x, y) = 1u2 + v2

2f

u2+

2f

v 2

.

vector field operations

Documents