Neither Div nor Curl nor Both Constitute the Derivative
Charles S. Barnett
Adjunct Mathematics Instructor
Las Positas College
Presented at the 41st annual fall conference
California Mathematics Council, Community Colleges
Monterey, California
December 13 and 14, 2013
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Abstract
The 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:
1
[𝑓 (𝑥 + ℎ𝑦) − 𝑓 (𝑥 )]
ℎ⟶0 ℎ
𝒻 ′ (𝑥, 𝑦) = lim
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, 𝒶𝑦). Show first that 𝒻 ′ (𝑥, 𝒶𝑦) = 𝒶𝒻 ′ (𝑥, 𝑦).
Well,
1
)
𝒻 𝑥, 𝒶𝑦 = lim [𝑓 (𝑥 + ℎ𝑎𝑦) − 𝑓 (𝑥 )]
ℎ→0 ℎ
′(
𝑎
[𝑓(𝑥 + ℎ𝑎𝑦) − 𝑓 (𝑥 )]
ℎ→0 ℎ𝑎
= lim
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1
= 𝑎 lim [𝑓(𝑥 + 𝑡𝑦) − 𝑓 (𝑥 )]
𝑡→0 𝑡
(𝑡 = ℎ𝑎)
= 𝑎𝒻 ′ (𝑥, 𝑦) .
Therefore, 𝒻 ′ (𝑥, 𝑎𝑦) = 𝑎𝒻 ′ (𝑥, 𝑦)
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).
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𝒻 ′ (𝑥, 𝑦 + 𝑧) = lim [𝑓(𝑥 + ℎ𝑦 + ℎ𝑧) − 𝑓 (𝑥 )]
ℎ→0 ℎ
1
= lim [𝑓(𝑥 + ℎ𝑦 + ℎ𝑘𝑦) − 𝑓 (𝑥 )]
ℎ→0 ℎ
for some 𝑘 ∈ ℝ
1
= lim [𝑓(𝑥 + ℎ(1 + 𝑘)𝑦) − 𝑓(𝑥 )]
ℎ→0 ℎ
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= (1 + 𝑘)𝒻 ′ (𝑥, 𝑦)
= 𝒻 ′ (𝑥, 𝑦) + 𝑘𝒻 ′ (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:
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):
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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:
3. 𝑓 (𝑥 ) = 𝑥 2 implies that 𝑓 ′ (𝑥 ) = (2𝑥 ), i.e.,
𝑓 ′ (𝑥 )(𝑦) = (2𝑥 )𝑦 all (𝑥, 𝑦) ∈ ℝ2
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Argument: 𝑓(𝑥 + ℎ𝑦) − 𝑓 (𝑥 ) = (𝑥 + ℎ𝑦)2 − 𝑥 2
= 𝑥 2 + 2ℎ𝑥𝑦 + ℎ2 𝑦 2 − 𝑥 2
= 2ℎ𝑥𝑦 + ℎ2 𝑦 2
Now, divide by ℎ and pass to the limit to see that
𝒻 ′ (𝑥, 𝑦) = (2𝑥 )𝑦 ⟹ 𝑓 ′ (𝑥 ) = 2𝑥.
Scalar-valued Functions of a Scalar (6)
Best affine approximation to 𝒇
Pick and fix 𝑥0 ∈ ℝ. Then the mean value theorem says that
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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.
′(
1
[𝑓(𝑋 + ℎ𝑌) − 𝑓(𝑋)]
ℎ⟶0 ℎ
𝒻 𝑋, 𝑌) = lim
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) = 𝑎𝒻 ′ (𝑋, 𝑌) for 𝑎 ∈ ℝ.
Pick and fix 𝑎 in ℝ. If 𝑎 = 0, the difference quotient is 0. If 𝑎 ≠ 0, then
1
𝒻 𝑋, 𝑎𝑌) = lim [𝑓(𝑋 + ℎ𝑎𝑌) − 𝑓 (𝑋)]
ℎ→0 ℎ
𝑎
[𝑓(𝑋 + ℎ𝑎𝑌) − 𝑓 (𝑋)]
= lim
ℎ→0 ℎ𝑎
′(
1
= 𝑎 lim [𝑓(𝑋 + 𝑡𝑌) − 𝑓 (𝑋)]
𝑡→0 𝑡
= 𝑎𝒻 ′ (𝑋, 𝑌)
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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),
where
1
𝒻 ′ (𝑋, 𝑌 + 𝑍) = lim ℎ [𝑓(𝑋 + ℎ𝑌 + ℎ𝑍) − 𝑓(𝑋)].
ℎ→0
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 that
𝑓(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 nonzero 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.
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So, 𝒻 ′ (𝑋, 𝑌) is linear in 𝑌 for every 𝑋.
Scalar-valued Functions of a Vector (4)
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).
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Scalar-valued Functions of a Vector (5)
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 at
3
′(
𝒻 𝑋, 𝑌) = ∑ 𝑦𝑖
𝑖=1
𝜕𝑓
.
𝜕𝑥𝑖
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.
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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
and 𝑓 ′ (𝑋): ℝ3 → ℝ.
Best affine approximation to 𝒇
Pick and fix 𝑋0 ∈ ℝ3 . Then
𝑓(𝑋) = 𝑓(𝑋0 ) + (𝑓 ′ (𝑋0 ))(𝑋 − 𝑋0 ) + 𝑒𝑟𝑟𝑜𝑟.
Or, in terms of the gradient,
𝑓(𝑋) = 𝑓(𝑋0 ) + ∇𝑓(𝑋0 ) ∙ (𝑋 − 𝑋0 ) + 𝑒𝑟𝑟𝑜𝑟.
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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):
1
ℱ ′ (𝑋, 𝑌) = lim [𝐹 (𝑋 + ℎ𝑌) − 𝐹 (𝑋)]
ℎ→0 ℎ
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.
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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) = ∑3𝑖=1 𝑓𝑖 (𝑋)𝐴𝑖 ,
where the 𝑓𝑖 represent the component functions. Let 𝒻𝑖′ (𝑋, 𝑌) represent the
derivative with respect to 𝑌 of the component functions. Earlier I sketched an
argument that showed that the 𝒻𝑖′ are linear in the variable 𝑌. It can be shown
that ℱ ′ (𝑋, 𝑌) inherits linearity in 𝑌 from the 𝒻𝑖′ . So,
1
ℱ 𝑋, 𝑌) = lim [𝐹 (𝑋 + ℎ𝑌) − 𝐹 (𝑋)]
ℎ→0 ℎ
′(
is linear in the variable 𝑌.
Definition:
The derivative 𝐹 ′ (𝑋) of 𝐹 at 𝑋 is the linear transformation
𝐹 ′ (𝑋): ℝ3 → ℝ3 defined by
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(𝐹 ′ (𝑋))(𝑌) = ℱ ′ (𝑋, 𝑌) .
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 = ∑3𝑖=1 𝑋𝑖 𝐴𝑖 , 𝑌 = ∑3𝑖=1 𝑦𝑖 𝐴𝑖 , 𝐹 (𝑋) = ∑3𝑖=1 𝑓𝑖 (𝑋)𝐴𝑖 ,
where the scalar-valued 𝑓𝑖 are the component functions of 𝐹.
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Then
𝜕𝑓
(𝐹 ′ (𝑋))(𝑌) = ∑3𝑖=1 ∑3𝑗=1 𝑦𝑗 𝜕𝑥 𝑖 𝐴𝑖 .
(∗)
𝑗
Vector-valued Functions of a Vector (5)
I have here invoked both the fact (shown earlier) that, for functions 𝑓: ℝ3 → ℝ,
𝜕𝑓
′
𝒻 ′ (𝑋, 𝑌) = ∑3𝑖=1 𝑦𝑖
, and the linearity of 𝐹 (𝑋 ) to arrive at Eqn.(∗).
𝜕𝑥𝑖
Observe from Eqn. (∗) that, relative to the standard basis, (𝐹 ′ (𝑋))(𝑌) is
represented by
𝜕𝑓1
𝜕𝑥1
𝜕𝑓2
[𝐹 ′ (𝑋)][𝑌] =
𝜕𝑥1
𝜕𝑓3
[𝜕𝑥1
𝜕𝑓1
𝜕𝑥2
𝜕𝑓2
𝜕𝑥2
𝜕𝑓3
𝜕𝑥2
𝜕𝑓1
𝜕𝑥3
𝑦1
𝜕𝑓2
[𝑦2 ] ,
𝜕𝑥3 𝑦
3
𝜕𝑓3
𝜕𝑥3 ]
where here and below I use [∙] to indicate the matrix of ∙ .
Best affine approximation to F
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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)
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). Then
1 ′
′( )
𝐹+ 𝑋 = [𝐹 (𝑋) + 𝐹 ′ (𝑋)∗ ] = 𝑠𝑦𝑚𝑚𝑒𝑡𝑟𝑖𝑐 𝑝𝑎𝑟𝑡
2
and
1
𝐹−′ (𝑋) = 2 [𝐹 ′ (𝑋) − 𝐹 ′ (𝑋 )∗ ] = skew-symmetric part.
Observe that 𝐹 ′ (𝑋) = sum of the two parts.
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Divergence and Curl of a Vector Field (2)
To simplify typography, until further notice let
𝑓𝑖𝑗 ≡
𝜕𝑓𝑖
.
𝜕𝑥𝑗
Then the matrix representation of 𝐹 ′ (𝑋) is
𝑓11
[𝐹 ′ (𝑋)] = [𝑓21
𝑓31
𝑓12
𝑓22
𝑓32
𝑓13
𝑓23 ] ,
𝑓33
and the matrix of 𝐹 ′ (𝑋)∗ is
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𝑓11
[𝑓12
𝑓13
𝑓21
𝑓22
𝑓23
𝑓31
𝑓32 ] .
𝑓33
Divergence and Curl of a Vector Field (3)
Then a bit of matrix algebra leads to the matrix representations of 2𝐹+′ (𝑋) and
2𝐹−′ (𝑋)
They are
2𝑓11
[2𝐹+′ (𝑋)] = [𝑓12 + 𝑓21
𝑓13 + 𝑓31
𝑓12 + 𝑓21
2𝑓22
𝑓23 + 𝑓32
𝑓13 + 𝑓31
𝑓23 + 𝑓32 ]
2𝑓33
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0
[2𝐹−′ (𝑋)] = [𝑓21 − 𝑓12
𝑓31 − 𝑓13
𝑓12 − 𝑓21
0
𝑓32 − 𝑓23
𝑓13 − 𝑓31
𝑓23 − 𝑓32 ] .
0
Divergence and Curl of a Vector Field (4)
Definition:
The Divergence of 𝐹 is a map 𝐷𝑖𝑣 𝐹: ℝ3 → ℝ, defined by
𝐷𝑖𝑣 𝐹 (𝑋) = 𝑇𝑟𝑎𝑐𝑒 𝐹 ′ (𝑋).
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
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the diagonal elements of any matrix representation. Hence, the familiar version
of the Divergence:
𝐷𝑖𝑣 𝐹 (𝑋) = 𝑓11 + 𝑓22 + 𝑓33
=
𝜕𝑓1 𝜕𝑓2 𝜕𝑓3
+
+
.
𝜕𝑥1 𝜕𝑥2 𝜕𝑥3
Observe also that 𝐷𝑖𝑣 𝐹 (𝑋) = 𝑇𝑟𝑎𝑐𝑒 𝐹+′ (𝑋) .
Divergence and Curl of a Vector Field (5)
Definition:
The Curl of 𝐹 at 𝑋 is defined by
2𝐹−′ (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, 2𝐹−′ (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 skewsymmetric 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 2𝐹−′ (𝑋) and you see that
𝐶𝑢𝑟𝑙 𝐹 (𝑋) = (𝑓32 − 𝑓23 )𝑖 + (𝑓13 − 𝑓31 )𝑗 + (𝑓21 − 𝑓12 )𝑘
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or, in the usual notation,
𝜕𝑓3 𝜕𝑓2
𝜕𝑓1 𝜕𝑓3
𝜕𝑓2 𝜕𝑓1
𝐶𝑢𝑟𝑙 𝐹 (𝑋) = (
−
)𝑖 + (
−
)𝑗 + (
−
)𝑘
𝜕𝑥2 𝜕𝑥3
𝜕𝑥3 𝜕𝑥1
𝜕𝑥1 𝜕𝑥2
Divergence and Curl of a Steady Flow: an application (1)
Consider a vector field 𝐹: ℝ3 → ℝ3 and the flow induced by the differential
equation
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𝑑𝑌
= 𝐹 (𝑌) .
𝑑𝑡
(⋆)
Note that time does not appear on the right hand side of Eqn. (⋆), hence “steady
flow.”
Pick and fix 𝑋0 ∈ ℝ3 . Expand 𝐹 around 𝑋0 via Taylor’s formula; substitute the
result into Eqn. (⋆) to yield
𝑑
(𝑌 − 𝑋0 ) = 𝐹 (𝑋0 ) + 𝐹 ′ (𝑋0 )(𝑌 − 𝑋0 ) + 𝑒𝑟𝑟𝑜𝑟 .
𝑑𝑡
(⋆⋆)
Split 𝐹 ′ (𝑋0 ) into the sum of its symmetric and skew-symmetric components
𝐹 ′ (𝑋0 ) = 𝐹+′ (𝑋0 ) + 𝐹−′ (𝑋0 )
(⋆⋆⋆)
and substitute into Eqn. (⋆⋆) to obtain
𝑑
(𝑌 − 𝑋0 ) = 𝐹 (𝑋0 ) + 𝐹+′ (𝑋0 )(𝑌 − 𝑋0 ) + 𝐹−′ (𝑋0 )(𝑌 − 𝑋0 ) + 𝑒𝑟𝑟𝑜𝑟
𝑑𝑡
(⋆⋆⋆⋆)
Divergence and Curl of a Steady Flow: an application (2)
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Now, consider the motion of the points within a small sphere centered at 𝑋0 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.
Consider first the motion induced by
𝑑
(𝑌 − 𝑋0 ) = 𝐹−′ (𝑋0 )(𝑌 − 𝑋0 )
𝑑𝑡
1
= 𝐶𝑢𝑟𝑙 𝐹 (𝑋0 ) × (𝑌 − 𝑋0 )
2
(⋇)
Equation (⋇) says that the test ball rotates about an axis through 𝑋0 parallel to
1
𝐶𝑢𝑟𝑙𝐹 (𝑋0 ) with angular velocity 2 |𝐶𝑢𝑟𝑙𝐹 (𝑋0 )|. That motion is volumepreserving.
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Divergence and Curl of a Steady Flow: an application (3)
Consider next the motion described by
𝑑
(𝑌 − 𝑋0 ) = 𝐹+′ (𝑋0 )(𝑌 − 𝑋0 ) .
𝑑𝑡
(⋇⋇)
𝐹+′ (𝑋0 ) is a symmetric map centered at 𝑋0 . 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 , 𝑒 𝜆2 𝑡 , 𝑒 𝜆3 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(𝑡 𝑡𝑟𝑎𝑐𝑒 𝐹+′ 𝑋0 ) and its time rate of change
when 𝑡 = 0 is 𝑡𝑟𝑎𝑐𝑒 𝐹+′ (𝑋0 ) = 𝑡𝑟𝑎𝑐𝑒 𝐹 ′ (𝑋0 ) = 𝐷𝑖𝑣 𝐹 (𝑋0 ).
Finally, consider the motion induced by
𝑑
(𝑌 − 𝑋0 ) = 𝐹 (𝑋0 ) .
𝑑𝑡
(⋇⋇⋇)
Equation (⋇⋇⋇) says that the ellipsoid experiences a small translation with
velocity vector 𝐹 (𝑋𝑜 ).
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35
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 Sticker
And God said
∇×E+
𝜕𝐵
=0
𝜕𝑡
∇×H−
𝜕𝐷
=𝐽
𝜕𝑡
∇∙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.
________________________________________________________________________
“Some people can stay longer in an hour than others can in a week.”
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William Dean Howells, American novelist, 1837-1920
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