Neither Div nor Curl nor Both Constitute the Derivative

Charles S. Barnett

Adjunct Mathematics Instructor

Las Positas College

Presented at the 41st annual fall conferenceCalifornia Mathematics Council, Community Colleges

Monterey, California

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December 13 and 14, 2013

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AbstractThe Div and Curl operators are derivative-like but are only aspects of the derivative of a map from 3-space to 3-space. The derivative of such a given map is a linear map from 3-space to 3-space. By employing the Frechet difference quotient at the outset, we can construct that derivative without use of the heavy machinery of advanced analysis. The process, results, and properties parallel those of the one-dimensional case.

Attribution

1.H. K. Nickerson, D. C. Spencer, and N. E. Steenrod, Advanced Calculus, Dover Publications, 2011. Unabridged republication of the work originally published in 1959 by the D. Van Nostrand Company.

2.Spivak, Michael, Calculus on Manifolds: A Modern Approach to Classical Theorems of Advanced Calculus,” W. A. Benjamin, Inc., New York, Amsterdam, 1965.

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3.H. M. Schey, Div, Grad, Curl, and All That: An Informal Text on Vector Calculus, W.W. Norton & Company, Inc., New York, 1973.

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Outline

Alternate definition of the derivative of a real-valued function of a real variable (the f :R→R case)

The Frechet derivative of a scalar-valued function of a vector (the f :R3→R case)

The Frechet derivative of a vector-valued function of a vector (the F :R3→R3 case)

The divergence and curl of a vector field

An application: the divergence and curl of a steady flow

Generalizations and the chain rule

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Scalar-valued Functions of a Scalar (1)

Situation: R represents the real numbers. f :R→R.

Goal: Define the derivative of f at x , f ' ( x ) .

For ( x , y ) ∈R2 consider the following limit:

f ' ( x , y )= limh⟶0

1h [ f ( x+hy )−f (x ) ]

If this limit exists, then f ' ( x , y ) is called “the derivative of f at x with respect to y .”

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Scalar-valued Functions of a Scalar (2)

Claim: f ' ( x , y ) is linear in the variable y .

Pick and fix a in R and consider f ' ( x ,a y ). Show first that f ' ( x , a y ) = a f ' ( x , y ). Well,

f ' ( x ,a y )=limh→ 0

1h [ f ( x+hay )−f ( x ) ]

¿ limh→0

aha [ f ( x+hay )− f (x ) ]

¿a limt →0

1t [ f ( x+ty )−f (x ) ] (t=ha )

¿a f ' ( x , y ) .

Therefore, f ' ( x ,ay )=a f ' ( x , y )

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Scalar-valued Functions of a Scalar (3)Now let y and z denote two non-zero elements of R and consider f ' ( x , y+z ).

f ' ( x , y+z )=limh→0

1h [ f ( x+hy+hz )−f ( x ) ]

¿ limh→0

1h [ f ( x+hy+hky )−f ( x) ]for some k ∈R

¿ limh→0

1h [f (x+h (1+k ) y )−f ( x ) ]

¿ (1+k ) f ' ( x , y )

¿ f ' ( x , y )+k f ' ( x , y )

¿ f ' ( x , y )+ f ' ( x ,ky )

¿ f ' ( x , y )+ f ' ( x , z ).

So, f ' ( x , y ) is linear in the second variable.Scalar-valued Functions of a Scalar (4)

Definition: 8

The derivative of f at x, denoted by f ' ( x ), is the linear map from R→R defined by

( f ' ( x ) ) ( y )=f ' ( x , y ) .

Examples (old chestnuts in new shells):

1. f is a constant function, f ( x )=k, where k is a constant. Then ( f ' ( x ) ) ( y )=0, i.e. f ' ( x ), sends all y to zero for every x∈ R .

Argument: f ( x+hy )−f ( x )=k−k=0.

2. f is a linear map, i.e., f ( x )=ax, where a∈R is an arbitrary but fixed number.

Then f ' ( x ) is a constant map, i.e.,

( f ' ( x ) ) ( y )=ay for every x∈ R , y ∈R.

Argument: f ( x+hy )−f ( x )=a ( x+hy )−ax=hay.Scalar-valued Functions of a Scalar (5)

Examples (old chestnuts in new shells), cont’d:

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3. f ( x )=x2 implies that f ' ( x )=(2 x ), i.e.,

f ' ( x ) ( y )=(2x ) y all ( x , y ) ∈R2

Argument: f ( x+hy )−f ( x )=( x+hy )2−x2

¿ x2+2hxy+h2 y2−x2

¿2hxy+h2 y2

Now, divide by h and pass to the limit to see that

f ' ( x , y )=(2x ) y⟹ f ' ( x )=2 x.

Scalar-valued Functions of a Scalar (6)Best affine approximation to f

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Pick and fix x0∈ R. Then the mean value theorem says that

f ( x )=f (x0 )+f ' (x0 ) (x−x0 )+error.

You will see direct parallels to this equation as we work our way up to functions from R3⟶R3.

Scalar-valued Functions of a Vector (1)Situation: f :R3→R

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Goal: Define the derivative of f at X, f ' ( X ). For ( X ,Y ) ∈R3×R3, consider the following limit that defines function f ' :R3×R3→R.

f ' ( X ,Y )= lim

h⟶0

1h [ f (X+hY )−f ( X ) ]

If this limit exists, then f ' ( X ,Y ) is called “the derivative of f at X with respect to Y”.

Scalar-valued Functions of a Vector (2)

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Claim: f ' ( X ,Y ) is linear in the variable Y .

First, show that f ' ( X ,aY )=a f ' (X ,Y ) for a∈R.

Pick and fix a in R. If a=0, the difference quotient is 0. If a≠0, then

f ' ( X ,aY )=limh→ 0

1h [ f ( X+haY )−f ( X ) ]

¿ limh→0

aha [ f (X+haY )−f ( X ) ]

¿a limt →0

1t [ f (X+tY )−f ( X ) ]

¿a f ' (X ,Y )

Scalar-valued Functions of a Vector (3)Now, consider f ' ( X ,Y +Z ). We must show that f ' ( X ,Y +Z )=f ' ( X ,Y )+ f ' ( X ,Z ),

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where

f ' ( X ,Y +Z )=limh→0

1h [ f (X+hY +hZ )−f (X ) ].

There exists a version of the mean value theorem that we will need:

If f ' ( X+tY ,Y ) exists for 0≤ t ≤1, then there exists θ ,0<θ<1 such thatf ( X+Y )−f (X )=f ' ( X+θY ,Y )(¿)

This version can be established fairly easily from the standard (R→R ) version.

In the proof of linearity for the (R→R ) case I used the fact that for any two non-zero real numbers x , y there exists a third, k, say, such that x+ y=x+kx. The above (R3→R ) version ( ¿ ) of the mean value theorem is used similarly to complete the proof of linearity for the present case. The details, somewhat involved, appear on page 5 of the handout.

So, f ' ( X ,Y ) is linear in Y for every X.

Scalar-valued Functions of a Vector (4)

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Definition:

The derivative f ' ( X ) of f at X is the linear map

f ' ( X ) :R3→R

defined by

( f ' ( X ) ) (Y )=f ' ( X ,Y ) .

Scalar-valued Functions of a Vector (5)

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The Gradient

Let Y ∈R3

Expand Y in the standard basis.

Invoke the fact that f ' (X ,Y ) is linear in Y and expand f ' ( X ,Y ) via Y ’s components y1 , y2 , y3 .

Arrive atf ' ( X ,Y )=∑

i=1

3

y i∂ f∂x i

.

A theorem from linear algebra says that if T :R3→R is linear, then there exists a unique vector B such that T (Y )=B ∙Y .

So f ' ( X ,Y )=B ∙Y , which implies that B=∇ f (X ) .

Hence, f ' ( X ,Y )=∇ f (X ) ∙ Y .

Scalar-valued Functions of a Vector (6)Observe that knowing ∇ f (X ) is equivalent to knowing f ' ( X ), but ∇ f (X ) ∈R3 andf ' ( X ) :R3→R.

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Best affine approximation to f

Pick and fix X 0∈R3 . Then

f ( X )=f ( X0 )+( f ' (X0 )) ( X−X0 )+error.

Or, in terms of the gradient,

f ( X )=f ( X0 )+∇ f (X 0 )∙ (X−X0 )+error .

Vector-valued Functions of a Vector (1)Situation: F :R3→R3

For (X ,Y ) ∈R3×R3, consider the following limit that defines F ' ( X ,Y ):

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F ' ( X ,Y )= limh→0

1h [ F (X+hY )−F ( X ) ]

if the limit exists.

Examples:

1.F :R3→R3 is a constant function.

F (X )=B∈ R3 for every X. B is a fixed vector.

Then F ( X+hY )−F ( X )=B−B=0∈R3.

So, F ' ( X ,Y )=0∈R3.

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Vector-valued Functions of a Vector (2)2. F :R3→R3 is a linear map.

F (X )=T ( X ), where T is linear.

Then F ( X+hY )−F ( X )=T ( X+hY )−T (X ).¿T ( X )+hT (Y )−T (X )=hT (Y ).

Now, divide by h and pass to the limit to see that

F ' ( X ,Y )=T (Y ) for all (X ,Y ) ∈R3×R3

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Vector-valued Functions of a Vector (3)Claim: F ' ( X ,Y ) is linear in the variable Y .

Let { A1 , A2 , A3 } represent the usual basis in R3. Then

F (X )=∑i=1

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f i ( X ) A i ,

where the f i represent the component functions. Let f i' ( X ,Y ) represent the

derivative with respect to Y of the component functions. Earlier I sketched an argument that showed that the f i

' are linear in the variable Y . It can be shown that F ' ( X ,Y ) inherits linearity in Y from the f i

'. So,

F ' ( X ,Y )= limh→0

1h [ F (X+hY )−F ( X ) ]

is linear in the variable Y .

Definition:

The derivative F ' ( X ) of F at X is the linear transformation

F ' ( X ) :R3→R3defined by(F ' (X ) ) (Y )=F' (X ,Y ) .

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Vector-valued Functions of a Vector (4)Examples (restatement in terms of F 'rather than F '):

1.The derivative of a constant function is the zero vector. So F ' ( X ) transforms all of R3 into the zero vector.

2.The derivative of a linear function is a constant.F (X )=T ( X ) , T linear, implies that

F ' ( X )=T for every X.

Now we can get back to more familiar, classical ground. F ' ( X ) is a linear map. Given a basis, a linear map can be represented by a matrix relative to that basis. So, as above, let { A1 , A2 , A3 } represent the standard basis. Then

X=∑i=1

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X i A i ,Y =∑i=1

3

y i A i ,F ( X )=∑i=1

3

f i ( X ) A i ,

where the scalar-valued f i are the component functions of F.

Then

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(F ' (X ) ) (Y )=∑i=1

3

∑j=1

3

y j

∂ f i

∂ x jAi . (¿ )

Vector-valued Functions of a Vector (5)I have here invoked both the fact (shown earlier) that, for functions f :R3→R , f ' ( X ,Y )=∑

i=1

3

y i∂ f∂x i

, and the linearity of F ' ( X ) to arrive at Eqn.( ¿ ).

Observe from Eqn. ( ¿ ) that, relative to the standard basis, (F ' (X ) ) (Y ) is represented by

[ F ' ( X ) ] [Y ]=[∂ f 1∂ x1

∂ f 1∂ x2

∂ f 1∂ x3

∂ f 2∂ x1

∂ f 2∂ x2

∂ f 2∂ x3

∂ f 3∂ x1

∂ f 3∂ x2

∂ f 3∂ x3

][ y1y2y3] , where here and below I use [ ∙ ] to indicate the matrix of ∙ .

Best affine approximation to F

As in the previous (R→R ) and(R3→R ) cases, we have a best approximation via Taylor’s formula:

F ( X )=F ( X0 )+(F ' ( X0 )) (X−X0 )+error .

Divergence and Curl of a Vector Field (1)22

Given F :R3→R3, then F ' ( X ) is a linear map from R3→R3. Linear maps may be broken into the sum of their symmetric and skew-symmetric parts.

Let F ' ( X )¿ represent the adjoint of F ' ( X ). ThenF

+¿' ( X )=12 [F ' ( X )+F ' ( X )¿ ]=symmetric part ¿

and

F−¿' (X )= 1

2 [F ' (X )−F' (X )¿ ]=¿¿ skew-symmetric part.

Observe that F ' ( X )=¿ sum of the two parts.

Divergence and Curl of a Vector Field (2)

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To simplify typography, until further notice let

f ij≡∂ f i

∂ x j.

Then the matrix representation of F ' ( X ) is

[F ' (X ) ]=[ f 11 f 12 f 13f 21 f 22 f 23f 31 f 32 f 33] ,

and the matrix of F ' ( X )¿ is

[ f 11 f 21 f 31f 12 f 22 f 32f 13 f 23 f 33 ] .

Divergence and Curl of a Vector Field (3)

Then a bit of matrix algebra leads to the matrix representations of 2 F+¿ ' ( X )¿ and 2 F−¿ ' ( X )¿

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They are

¿

¿

Divergence and Curl of a Vector Field (4)

Definition:

The Divergence of F is a map ¿ F :R3→R, defined by

¿ F (X )=Trace F ' ( X ) .

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The Trace of F ' ( X ) is equal to the sum of the roots of the characteristic polynomial of F ' ( X ), a property of the linear map that is invariant to matrix representation. The trace of any linear map from R3→R3 is equal to the sum of the diagonal elements of any matrix representation. Hence, the familiar version of the Divergence:

¿ F (X )=f 11+f 22+ f 33

¿∂ f 1∂x1

+∂ f 2∂ x2

+∂ f 3∂ x3

.

Observe also that ¿F ( X )=TraceF+¿' (X ) .¿

Divergence and Curl of a Vector Field (5)

Definition:

The Curl of F at X is defined by

2 F−¿ ' ( X ) (Y )=[Curl F ( X ) ]×Y (¿ ) ¿

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How do we know that a vector that makes the right side of Eqn. ( ¿ ) equal to the left-hand side exists? Well, 2 F−¿ ' ( X )¿ is a skew-symmetric transformation and item A5 of the Appendix of the handout asserts that such a vector exists for such maps; we name it Curl F ( X ). Also, according to A5, if the matrix of a skew-symmetric map looks like

[ 0 a b−a 0 c−b −c 0 ] ,¿

then the vector at issue (Curl F ( X ) here) is

−ci+bj−ak .(¿∗¿)

Divergence and Curl of a Vector Field (6)Compare the matrix representation of 2 F−¿ ' ( X )¿ and you see that

Curl F (X )=( f 32−f 23 )i+( f 13−f 31) j+( f 21−f 12 )k

or, in the usual notation,27

Curl F (X )=( ∂ f 3∂x2

−∂ f 2∂x3 ) i+( ∂ f 1

∂ x3−

∂ f 3∂ x1 ) j+( ∂ f 2

∂ x1−

∂ f 1∂ x2 )k

Divergence and Curl of a Steady Flow: an application (1)

Consider a vector field F :R3→R3 and the flow induced by the differential equationdYdt

=F (Y ) .(⋆)

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Note that time does not appear on the right hand side of Eqn. (⋆), hence “steady flow.”

Pick and fix X 0∈R3. Expand F around X 0 via Taylor’s formula; substitute the result into Eqn. (⋆) to yieldddt (Y−X0 )=F ( X0 )+F ' (X 0 ) (Y−X 0 )+error .(⋆⋆)

Split F ' (X 0) into the sum of its symmetric and skew-symmetric componentsF ' (X 0)=F+¿' ( X0 )+F−¿' ( X0)(⋆⋆⋆)¿ ¿

and substitute into Eqn. (⋆⋆ ) to obtainddt (Y−X0 )=F ( X0 )+F+¿' (X 0) (Y−X 0)+F−¿' ( X 0) (Y−X0 )+error(⋆ ⋆⋆⋆)¿ ¿

Divergence and Curl of a Steady Flow: an application (2)Now, consider the motion of the points within a small sphere centered at X 0 at time zero, say. How has that spherical test ball changed in a small time t? The terms on the right-hand side yield an approximate answer.

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Consider first the motion induced byddt (Y−X0 )=F

−¿' (X 0 )(Y−X 0)¿

¿ 12Curl F ( X0 )× (Y −X0 )(⋇)

Equation (⋇) says that the test ball rotates about an axis through X 0 parallel to CurlF (X0 ) with angular velocity 12|CurlF (X0 )|. That motion is volume-preserving.

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Divergence and Curl of a Steady Flow: an application (3)Consider next the motion described byddt (Y−X0 )=F+¿' (X 0) (Y−X 0) .(⋇⋇)¿

F+¿' ( X0 ) ¿ is a symmetric map centered at X 0 . Let λ1 , λ2 , λ3 represent its eigenvalues and B1 ,B2 ,B3 denote the corresponding orthonormal set of eigenvectors. Then the corresponding eigenvalues of the flow described by Eqn. (⋇⋇) are e λ1 t , eλ2t , eλ3 t. That flow distorts the spherical ball slightly into an ellipsoid. The volume of the ball changes by a factor that is the determinant of the transformation. That determinant is exp¿ and its time rate of change when t=0 is trace F+¿' ( X0 )=traceF ' (X 0 )=¿F (X 0)¿.

Finally, consider the motion induced byddt (Y−X0 )=F ( X0 ) .(⋇⋇⋇)

Equation (⋇⋇⋇) says that the ellipsoid experiences a small translation with velocity vector F ( Xo ).

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The General F :Rn→Rm Case and the Chain Rule (1)The chain rule is so important in differential calculus that its multidimensional version deserves a brief discussion. Before doing so, I point out that our restriction to the 1≤n≤3 case above was not necessary; I invoked the restriction because of the emphasis on the undergraduate calculus sequence.

Let n and m represent natural numbers and consider F :Rn→Rm. If F is differentiable at X, then its derivative can be defined via the Frechet difference quotient just as we did in the earlier cases. The process is identical to the one used for the 1≤m ,n≤3 case. The derivative is a linear transformation as before, and all goes through except the discussion of the Curl, which is unique to the three-dimensional case. With this generalization all cases are subsumed, again except for the Curl, by this general case.

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The General F :Rn→Rm Case and the Chain Rule (2)Consider now the chain rule in the general case. Let m ,n , and p represent natural numbers. Situation: F :Rn→Rm is differentiable at X, and G :Rm→Rp is differentiable at F ( X ), then G ∘F is differentiable at X and(G ∘F )' (X )=G' (F ( X ) )∘ F' (X ) , (¿ )

where ∘ represents function composition. Observe that in Eqn. ( ¿ ) G' (F ( X ) ) is a linear map and F ' ( X ) is a linear map; hence the composition of the two maps is a linear map, and all is well. Use brackets to denote matrix representation, and the matrix version of Eqn. ( ¿ ) becomes[ (G ∘F )' ( X ) ]=[G' (F ( X ) ) ] [F' ( X ) ] . ¿

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The General F :Rn→Rm Case and the Chain Rule (3)

Recall the one-variable chain rule.f :R→R ,g :R→R

(g∘ f )' ( x )=g' (f ( x ) ) f ' ( x )(¿∗¿)

where the juxtaposition on the right-hand side of Eqn ( ¿∗¿ ) denotes multiplication, but could just as well have been written

( g∘ f ) ' (x )=g ' ( f ( x ) ) ∘ f ' (x ) (¿∗¿' )

because, in the one-dimensional case, composition of linear functions is multiplication. Compare Eqn. ( ¿ ) and Eqn. (¿∗¿' ) and you see that Eqn. (¿∗¿' ) is just a special case of Eqn. ( ¿ ).

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Nerd’s Bumper StickerAnd God said

∇× E+ ∂B∂t

=0

∇× H−∂ D∂ t

=J

∇ ∙ B=0

∇ ∙ D=ρ

and there was light.

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Closure The derivative is a linear map.

The Divergence and Curl are aspects of that map that account for about 4/9 of its properties, an especially important fraction in applications.

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“Some people can stay longer in an hour than others can in a week.”

William Dean Howells, American novelist, 1837-1920

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