MA20219: Analysis 2B – Vector Analysis
Karsten Matthies
Department of Mathematical Sciences
University of Bath
Spring 2014
This course deals with concepts, results and proofs in vector integration and vector differentiation.
It is [fundamental] to most areas of applied Maths and many areas of pure Maths and a [prerequisite]
for a large number of courses in Years 3 and 4.
We will take the point of view of rigorous analysis, but the questions are motivated by applications
and we will also need to calculate many examples.
Please take this course seriously and do not fall behind with the problem sheets. The
syllabus for this course is very dense and only through constant practice will you be able to
grasp the multitude of methods and concepts.
Please revise MA20218 (Analysis 2a) and MA10208 (Vector Algebra).
Main references
(a) Advanced calculus by Gerald B. Folland, Prentice-Hall.
(b) Vector Calculus by J. Marsden, A. Tromba, W.H. Freeman.
(c) Calculus of several variables by Serge Lang, Springer.
Other useful textbooks:
(a) Introduction to calculus and analysis, Vol.2/1 and Vol.2/2 by F. John and R. Courant, Springer.
(b) Vector Analysis and Cartesian Tensors by D.E. Bourne and P.C. Kendall, Stanley Thornes.
(c) Vector Analysis and an Introduction to Tensor Analysis by M.R. Spiegel, Schaum’s outline series,
McGraw-Hill.
See moodle or Course webpage:
http://people.bath.ac.uk/km230/ana.bho/ana.html
for lecture notes, example sheets, diary and more.
Contents
1 Derivatives and Parametrisation
1.1 First derivatives and Gradients . . . . . .
1.2 Divergence and Curl; the ∇–Operator . .
1.3 Second order derivatives . . . . . . . . . .
1.4 Curves in the plane . . . . . . . . . . . . .
1.5 Surfaces and Curves in Space . . . . . . .
1.6 Transformations and Coordinate Systems
1.7 Differentiation in Curvilinear Coordinates
.
.
.
.
.
.
.
1
1
2
3
3
4
6
6
2 Integration in higher dimensions
2.1 Riemann Integration in R2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Multiple Intergrals and Iterated Integrals . . . . . . . . . . . . . . . . . . . . . . . . .
2.3 Change of Variables for Multiple Integrals . . . . . . . . . . . . . . . . . . . . . . . . .
8
8
10
11
3 Line and Surface Integrals; Integral Theorems
3.1 Arc Length and Line Integrals . . . . . . . . . .
3.2 Green’s Theorem . . . . . . . . . . . . . . . . .
3.3 Surface Integrals . . . . . . . . . . . . . . . . .
3.4 Divergence Theorem . . . . . . . . . . . . . . .
3.5 Stokes’ Theorem . . . . . . . . . . . . . . . . .
3.6 Some Applications to Physics . . . . . . . . . .
3.7 Integrating Vector Derivatives . . . . . . . . . .
13
13
14
14
16
16
16
17
2
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
Chapter 1
Derivatives and Parametrisation
In this course we will be dealing with functions in more than one unknown whose function value
might be vector–valued as well.
1.1
First derivatives and Gradients
[Folland 2.2] & [Marsden 2.6]
Definition 1.1. Let Ω ⊂ Rn be open, a function F : Ω → Rn is called a vector field on Ω. A function
f : Ω → R is called a scalar field on Ω. (Usually n will be 2 or 3.)
Example 1.2. (a) Typical examples of vector fields are gravitational field G(x), electrical field
E(x), magnetic field B(x), and velocity field V (x).
(b) Typical examples of scalar fields are speed |V (x)|, kinetic energy
potential ϕE (x)
1
2
m |V (x)|2 , and electrical
Definition 1.3. Let U ⊂ Rn be open, a function F : U → Rm has a partial derivative at x =
(x1 , . . . , xn )T ∈ U with respect to xi , if the limit
1
(F (x1 , . . . , xi−1 , xi + h, xi+1 , . . . , xn ) − F (x1 , . . . , xi−1 , xi , xi+1 , . . . , xn ))
h→0 h
lim
exists, this defines the function ∂∂xFi (x).
Remark 1.4. (a) Remember the definition of Fréchet derivative DF (x) for F = (f1 , . . . , fm )T from
Analysis 2A, §3. It is represented by the Jacobian matrix
 ∂f1

∂f1
. . . ∂x
(x)
∂x1 (x)
n


..
JF (x) =  ...
.
.
∂fm
∂x1 (x)
...
∂fm
∂xn (x)
∂fi
(b) If all ∂x
(x0 ) with i = 1, . . . , m and j = 1, . . . , n exist and are continuous in x0 ∈ U with U
j
open, then F is Fréchet differentiable at x0 .
Definition 1.5. A function F : U → Rm with U ⊂ Rn open is called continuously differentiable
on U (short C 1 ), if the partial derivatives with respect to x1 , . . . , xn exist and the partial derivatives
are continuous functions on U .
Definition 1.6. Let f : Ω → R be a continuously differentiable scalar field on an open set Ω ⊂ Rn .
Then
∂f
∂f
grad f ≡ ∇f :=
e1 + . . . +
en
(1.1)
∂x1
∂xn
is the gradient of f on Ω, which is itself a vector field on Ω.
1
Definition 1.7. Let Ω ⊂ Rn be open and let f : Ω → R be a continuously differentiable scalar field
on Ω ⊂ Rn and let â be a unit vector in Rn . Then
f (x0 + hâ) − f (x0 )
h→0
h
Dâ f (x0 ) := lim
(1.2)
is the directional derivative of f in the direction â at x0 ∈ Ω, i.e. the rate of change of f in the
direction of â. Moreover, for any a ∈ Rn \ {0} we define Da f := Dâ f where â = a/|a|.
Proposition 1.8. Let f : Ω → R be differentiable in x0 ∈ Ω, then directional derivatives exist. They
are given for a ∈ Rn \ {0} by
a
Da f (x0 ) = ∇f (x0 ) ·
.
(1.3)
|a|
Example 1.9. Find Da f (x0 ) for f (x) := 2x2 + 3y 2 + z 2 , a := (1, 0, −2)T , and x0 := (2, 1, 3)T .
Find ∇f , then normalise a. Finally evaluate (1.3) at x0 .
Proposition 1.10. Let f be a differentiable scalar field at x0 , then
max
|Da f | = |∇f |
a ∈ Rn \ {0}
(1.4)
and the maximum is attained in the direction a = ∇f 6= 0 or in any direction if ∇f = 0.
Remark 1.11. (geometric interpretation of grad).
(a) Proposition 1.10 shows that ∇f (x0 ) gives the direction and magnitude of the largest directional derivative of f at x0 , i.e. the largest rate of change of f .
(b) A point x0 ∈ Ω where ∇f (x0 ) = 0 is called a stationary point (since Proposition 1.10 shows
that Da f (x0 ) = 0 for all directions a ∈ Rn \ {0}).
1.2
Divergence and Curl; the ∇–Operator
[Folland 5.4] & [Marsden 4.4]
Let us consider ∇f again, but now think as ∇ as an operator acting on a scalar field f . Formally
we could write
∂
∂
∂
∂ T
∇ =
e1 + . . . +
en = (
,...,
) .
(1.5)
∂x1
∂xn
∂x1
∂xn
This is called nabla or the del-operator, and applying it to a scalar field f we have
∇f =
∂f
∂f
e1 + . . . +
en = grad f .
∂x1
∂xn
Definition 1.12. Let F be a C 1 vector field on an open set Ω ⊂ Rn . We define the divergence of F
to be the scalar field
∂F1
∂Fn
div F := ∇ · F =
+ ... +
.
(1.6)
∂x1
∂xn
.
Definition 1.13. Let F be a C 1 vector field on an open set Ω ⊂ R3 . We define curl of F to be the
vector field
i
∂ j∂ k∂ ∂F3 ∂F2
∂F1 ∂F3
∂F2 ∂F1
curl F := ∇×F = ∂x ∂y ∂z =
−
i+
−
j+
−
k . (1.7)
∂y
∂z
∂z
∂x
∂x
∂y
F F F 1
2
3
Example 1.14. Find div and curl of F (x) = (−y, x, z)T .
2
Remark 1.15. (a) In manipulating with the ∇–operator many rules from ordinary vector algebra
apply, but not all in general
∇ · (gF ) 6= g∇ · F 6= F · ∇g.
(b) See problem sheets for rules (don’t use vector algebra to prove them!).
(c) Applications: In many fields of mathematical physics, e.g. fluid flow, electromagnetic field
propagation, etc...
1.3
Second order derivatives[Folland 5.4]
Applying the ∇–operator twice gives five possible second derivatives:
A
div(grad f ) = ∇ · (∇f ) = ∇2 f
B
curl(grad f ) = ∇ × (∇f )
C
div(curl F ) = ∇ · (∇ × F )
D
curl(curl F ) = ∇ × (∇ × F )
E
grad(div F ) = ∇(∇ · F )
(1.8)
Note. Quantities such as grad(curlF ) or curl(divF ) are meaningless.
Proposition 1.16. Let Ω be an open subset of R3 , let f be a twice continuously differentiable (C 2 )
scalar function, let F be a C 2 -vector field on Ω, then
curl(grad f ) = 0,
(1.9)
div(curl F ) = 0,
(1.10)
2
2
2
2
T
grad(div F ) − curl(curl F ) = ∇ F = (∇ F1 , ∇ F2 , ∇ F3 ) .
(1.11)
Remark 1.17. The operator
∇2 = ∇ · ∇ =
∂2
∂2
∂2
+
+
∂x2 ∂y 2 ∂z 2
(1.12)
in (1.8) and (1.11) is called the Laplace–operator (also denoted ∆). It is very important in mathematical physics. Many of the basic partial differential equations (PDEs) of mathematical physics
involve it, e.g. the Laplace Equation
∇2 φ =
1.4
Curves in the plane
∂2φ ∂2φ ∂2φ
+ 2 + 2 = 0.
∂x2
∂y
∂z
(1.13)
[Folland 3.2]
We aim to describe curves and tangents. How can we represent curves in the plane?
(a) As the graph of a function y = f (x) or x = f (y), where f is C 1 .
(b) As the level set (locus) of an equation F (x, y) = 0, where F is C 1 (i.e. {(x, y) | F (x, y) = 0}).
(c) Parametrically, as the range of a C 1 function f : (a, b) → R2 .
Example 1.18. (a) F (x, y) = x2 + y 2 − c defines a smooth curve (circle) if c > 0, a single point for
c = 0, and an empty set for c < 0.
3
(b) G(x, y) = x2 − y 2 − c describes hyperbolas: two disjoint curves if c 6= 0. Two intersecting lines
for c = 0.
(c) H(x, y) = y 3 − x2 or f (t) = (t3 , t2 )T cusp at origin.
(d) g(t) = (sin2 t, cos2 t)T is C 1 , range is a line for t ∈ [0, π/2], but the same line is covered twice
again for t ∈ (π/2, 3π/2].
Common feature: problems can occur when ∇F = 0 on set F = 0 or if g 0 = 0.
Theorem 1.19 (Implicit Function Theorem). Let U ⊂ Rm and W ⊂ Rn be open sets, let F : U ×W →
Rn be C 1 , x0 ∈ U , y 0 ∈ W and suppose F (x0 , y 0 ) = 0. Suppose that D2 F (x0 , y 0 ) is invertible. Then
there exist open sets U0 ⊂ U and W0 ⊂ W of x0 and y 0 respectively and a C 1 map f : U0 → W0 such
that for all x ∈ U0 , y = f (x) is the unique solution F (x, y) = 0 in W0 and moreover, for all x ∈ U0
Df (x) = −D2 F (x, f (x))−1 D1 F (x, f (x)).
Proposition 1.20. (a) Let F be a real-valued C 1 function on an open set in R2 , and let S =
{(x, y)T | F (x, y) = 0}. If x0 ∈ S and ∇F (x0 ) 6= 0, then there is a neighbourhood1 N of x0 in
R2 such that S ∩ N is the graph of a C 1 function f (either y = f (x) or x = f (y)).
(b) Let f : (a, b) → R2 be a C 1 function. If f 0 (t0 ) 6= 0, there is an open interval I containing t0
such that S = {f (t) | t ∈ I} is the graph of a C 1 function f (either y = f (x) or x = f (y)).
Definition 1.21. A set S ⊂ R2 is a smooth curve, if S is connected2 and every x0 ∈ S has a
neighbourhood N such that S ∩ N is the graph of a C 1 function f (either y = f (x) or x = f (y)).
Remark 1.22. (a) Proposition 1.20 gives a connection of the three possible ways to represent curves.
The tangent vector at a point x0 with representation x0 = (x, f (x))T is then given by (1, f 0 (x))T .
In the parametric representation of a C 1 function f : (a, b) → R2 the tangent is given by f 0 (t)
at x0 = f (t). In the level set description by a C 1 function F , for x0 ∈ {(x, y) | F (x, y) = 0},
the normal is given by ∇F (x0 ). See problem sheet 3.
(b) The conditions ∇F (x0 ) 6= 0 and f 0 (t0 ) 6= 0 are sufficient but not necessary: If G(x, y) =
F (x, y)2 = (x2 + y 2 − 1)2 , then ∇G = 2F ∇F = 0 on S. If f : (−1, 1) → R2 has f 0 (t) 6= 0, then
still g(t) = f (t3 ) has g 0 (0) = 0 independent of f .
(c) In general case (ii) with ∇F 6= 0 does not guarantee that S is a smooth curve. For F (x, y) =
(x2 + y 2 − 1)(x2 + y 2 − 4) the locus is disconnected.
(d) In case (iii) {f (t) | a < t < b} is connected [exercise], but f may not be injective. Examples:
f (t) = (cos t, sin t)T , g(t) = (t3 − t, t2 )T .
1.5
Surfaces and Curves in Space
[Folland 3.3]
Again we would like to have alternatives to represent surfaces in 3−space.
(a) As the graph of a function z = f (x, y) (or y = f (x, z) or x = f (y, z)), where f is C 1 .
(b) As the level set (locus) of an equation F (x, y, z) = 0, where F is C 1 (i.e. {(x, y, z) | F (x, y, z) =
0}).
(c) Parametrically, as the range of a C 1 function f : R2 → R3 .
1
A set is a neighbourhood of a point, if it is an open set containing this point
A set S is disconnected if S = S1 ∪ S2 such that S1 6= ∅, S2 6= ∅ and S̄1 ∩ S2 = S1 ∩ S̄2 = ∅ for some S1 , S2 . A set is
connected, if it is not disconnected.
2
4
We will need additional conditions for (ii) and (iii) to guarantee smoothness.
let U ⊂ R2 be an open set and let
S = {x ∈ R3 | x = f (u), u ∈ U }.
Consider the linear case first
f (u, v) = ua + vb + c,
a, b, c ∈ R3
We need a, b linearly independent to span a surface. Now for general smooth f : U → R3 with U
open near a point (u0 , v0 ):
∂f
∂f
(u0 , v0 ), b =
(u0 , v0 ), c = f (u0 , v0 ),
∂u
∂v
∂f
∂f
where ∂u (u0 , v0 ) is the tangent vector to the ’coordinate curve’ x = f (u, v0 ) and ∂v (u0 , v0 ) is the
∂f
∂f
tangent vector to the ’coordinate curve’ x = f (u0 , v). We will assume that ∂u (u0 , v0 ) and ∂v (u0 , v0 )
T
are linearly independent for all (u, v) ∈ U , this is equivalent to
∂f
∂f
(u, v) 6= 0 for all (u, v)T ∈ U
×
∂u
∂v
f (u0 + u, v0 + v) ≈ ua + vb + c,
a=
Proposition 1.23. (a) Let F be a real-valued C 1 function on an open set in R3 , and let S =
{(x, y, z)T | F (x, y) = 0}. If x0 ∈ S and ∇F (x0 ) 6= 0, then there is a neighbourhood N of
x0 in R3 such that S ∩ N is the graph of a C 1 function f (either z = f (x, y), y = f (x, z) or
x = f (y, z)).
h
i
∂f
∂f
(b) Let f be a C 1 from an open set U in R2 into R3 . If ∂u × ∂v (u0 , v0 ) 6= 0, there is a neighbourhood of (u0 , v0 )T such that the set {f (u, v) | (u, v)T ∈ N } is the graph of a C 1 function
f.
Definition 1.24. A set S ⊂ R3 is a smooth surface, if S is connected and every x0 ∈ S has a
neighbourhood N such that S ∩ N is the graph of a C 1 function f (either z = f (x, y), y = f (x, z) or
x = f (y, z)).
Remark 1.25.
(a) Proposition 1.23 gives the alternative representations.
(b) The tangent plane of surface at a point a can be given by n · (x − a) = 0, where n is a (nonzero)
normal vector to S. If S is given parametrically by a map f (u, v), then ∂u f and ∂v f are tangent
to S, hence the normal can be obtained as the cross product of these vectors.
Example 1.26. The unit sphere S = {x ∈ R3 | |x| = 1} can be parameterised by speherical
coordinates
f (θ, ϕ) = (cos ϕ sin θ, sin ϕ sin θ, cos θ)T
(1.14)
with ϕ ∈ (−π, π) the longitude and θ ∈ (0, π) the co-latitude.
Definition 1.27. A set S ⊂ R3 is a smooth curve, if S is connected and every x0 ∈ S has a
neighbourhood N such that S ∩ N is the graph of C 1 functions f and g (either y = f (x),z = g(x) or
x = f (y), z = g(y) or x = f (z), y = g(z)).
Proposition 1.28. (a) Let F, G be two real-valued C 1 function on an open set in R3 , and let S =
{(x, y, z)T | F (x, y, z) = G(x, y, z) = 0}. If ∇F (x0 ) and ∇G(x0 ) are linearly independent at
x0 ∈ S, then there is a neighbourhood N of x0 in R3 such that S ∩N is the graph of C 1 functions
f, g y = f (x),z = g(x) ( or with coordinates permuted).
(b) Let f : (a, b) → R3 be a C 1 mapping. If f 0 (t0 ) 6= 0, there is an open interval I containing t0 such
that S = {f (t) | t ∈ I} is the graph of C 1 functions f, g y = f (x),z = g(x) ( or with coordinates
permuted).
Remark 1.29. The general concept of a ’smooth k-dimensional object’ is a manifold, see e.g. [Folland].
5
1.6
Transformations and Coordinate Systems
[Folland 3.4]
We study smooth mappings from Rn to itself. We have two interpretations: transformations and
coordinate systems.
Example 1.30. A transformation moves the points in Rn around
(a)
f : R2 → R2 , f (u, v) =
√ T
1 √
3u − v, u + 3v
2
is a rotation by π/6.
(b)
f (u, v) = (u cos v, u sin v)T
Another interpretation is as a coordinate system. Now points stay the same, but we give them different
labels.
Definition 1.31. We call f : U → V with U, V ⊂ Rn open a coordinate system if f is bijective, f is
C 1 and f −1 : V → U is C 1 .
Theorem 1.32 (Inverse Mapping Theorem). Let U and V be open sets in Rn , a ∈ U , b = G(a).
Suppose that G : U → V is C 1 and DG(a) is invertible. Then there exist neighbourhoods M ⊂ U
and N ⊂ V of a and b respectively such that G is bijective from M to N , and the inverse map
G−1 : N → M is also C 1 .
Moreover, if y = G(x) ∈ N , then D(G−1 )(y) = [DG(x)]−1 .
1.7
Differentiation in Curvilinear Coordinates
[Bourne, pp. 118–136]
Motivation. So far we have only looked at scalar fields and vector fields in Cartesian coordinates
(x, y, z). What if we have curvilinear coordinates (u, v, w) defined by
r(u, v, w) = x(u, v, w) i + y(u, v, w) j + z(u, v, w) k,
(1.15)
e.g. spherical polar coordinates, etc. ?
Note. In this section we will always use the notation xi + yj + zk for vectors in Cartesian
coordinates rather than (x, y, z)T .
Definition 1.33. A triple (u, v, w) ∈ D with D open together with a Cartesian map r : D → Ω ⊂ R3
as defined in (1.15) is called a set of orthogonal curvilinear coordinates (OCCs) on Ω, if
(a) r is a continuously differentiable bijection with continuously differentiable inverse r −1 .
(b) The vectors r u , r v , r w are mutually orthogonal, i.e.
r u · r v = r u · r w = r v · r w = 0.
Example 1.34. Spherical polar coordinates (ρ, θ, φ):
x(ρ, θ, φ) = ρ sin θ cos φ,
y(ρ, θ, φ) = ρ sin θ sin φ,
6
z(ρ, θ, φ) = ρ cos θ .
Remark 1.35.
(a) It is useful to introduce unit vectors
eu :=
ru
,
|r u |
ev :=
rv
,
|r v |
ew :=
rw
.
|r w |
(1.16)
Let F : Ω → R3 be a vector field on Ω. Since eu , ev , ew are orthonormal, we can use them as a
basis for representing F , i.e.
F = Fu eu + Fv ev + Fw ew
(1.17)
where Fu , Fv , Fw are the components of F along the coordinate lines Cu , Cv and Cw .
(b) This is why we do not use the notation F = (F1 , F2 , F3 )T here. It does not carry any information
on the coordinate system we work in.
(c) How do we find Fu , Fv and Fw ? Note that since eu , ev and ew are orthonormal,
F · eu = Fu eu · eu + Fv ev · eu + Fw ew · eu = Fu .
(1.18)
Similarly, F · ev = Fv and F · ew = Fw .
Example 1.36. Express the vector field F = z i in spherical polar coordinate form.
Proposition 1.37. Let r be an OCC with open range Ω ⊂ R3 . Let f be a C 1 -scalar field on Ω and
let F be a C 1 vector field on Ω, then
1 ∂f
1 ∂f
1 ∂f
eu +
ev +
ew
(1.19)
grad f =
|r
|
∂u
|r
|
∂v
|r
u
v
w | ∂w
(a)
1
∂
∂ ∂ div F =
|r v ||r w |Fu +
|r u ||r w |Fv +
|r u ||r v |Fw
(1.20)
|r u ||r v ||r w | ∂u
∂v
∂w
(b)
(c)
(d)
|r u |eu |r v |ev |r w |ew 1
curl F =
∂/∂u ∂/∂v ∂/∂w |r u ||r v ||r w | |r |F |r |F |r |F u u
v v
w w
∂ |r v ||r w | ∂f
∂ |r u ||r w | ∂f
∂ |r u ||r v | ∂f
1
2
+
+
∇ f =
|r u ||r v ||r w | ∂u
|r u | ∂u
∂v
|r v | ∂v
∂w
|r w | ∂w
(1.21)
(1.22)
(a) For a scalar field f : Ω → R find ∇f and ∇2 f in spherical polar coordinates.
Example 1.38.
(b) Use spherical polar coordinates to show that f (x) = |x|−1 satisfies the Laplace Equation ∇2 f = 0
for x ∈ R3 \ {0}.
(c) Let
F
:=
ρ2 cos θ
eρ +
1
ρ
eθ +
Find curlF .
7
1
ρ sin θ
eφ
in
spherical
polar
coordinates.
Chapter 2
Integration in higher dimensions
Remember Riemann-integration of bounded functions on compact intervals, see Analysis 2A.
2.1
Riemann Integration in R2
[Folland 4.2]
We will develop the theory of multiple integrals. For notational convenience we develop the twodimensional case and then sketch the extension to higher dimensions.
Definition 2.1. (i) A set R = [a, b]×[c, d] = {(x, y) ∈ R2 | x ∈ [a, b], y ∈ [c, d]} is called a rectangle.
A partition P = (x0 , . . . , xJ ; y0 , . . . , yK ) with
a = x0 < . . . < xJ = b; c = y0 < . . . < yK = d
of R is a subdivision into sub-rectangles Rjk = [xj−1 , xj ] × [yk−1 , yk ] with area ∆Ajk := (xj −
xj−1 )(yk − yk−1 ).
(ii) Given a partition P , we set for a bounded function f : R → R
mjk = inf{f (x, y) | (x, y) ∈ Rjk };
We define the lower Riemann sum LR (P, f ) =
P
P
UR (P, f ) = Jj=1 K
k=1 Mjk ∆Ajk
Mjk = sup{f (x, y) | (x, y) ∈ Rjk }
PJ
j=1
PK
k=1 mjk ∆Ajk
and the upper Riemann sum
(iii) The lower Riemann integral of f over R is LR (f ) = supP LR (P, f ) and the upper Riemann integral
UR (f ) = inf P UR (P, f ) taking the suppremum/infimum over all partitions.
(iv) A bounded function f : R → R is called (Riemann)-integrable on R if UR (f ) = LR (f ), then
denote this value by
ZZ
ZZ
f dA or
f dxdy.
R
R
Definition 2.2. (i) Let S be a subset of R2 (or Rn ), then the characteristic function or the indicator function
of S is the function χS given by
1 x∈S
χ(x) =
0 otherwise.
(ii) Suppose S ⊂ R2 bounded, f a bounded function on R2 . Let R be a rectangle that contains S.
We say that f is integrable on S if χS f is integrable on R, then
ZZ
ZZ
f dA =
χS f dA
S
R
8
Remark 2.3. The definition of
S as χs f = 0 outside S.
RR
S
f dA does not depend on the choice of R or the values of f outside
Theorem 2.4. (a) If f1 , f2 are integrable on the bounded set S and c1 , c2 ∈ R then c1 f1 + c2 f2 is
integrable on S and
ZZ
ZZ
ZZ
f2 dA.
f1 dA + c2
[c1 f1 + c2 f2 ]dA = c1
S
S
S
(b) Let S1 , S2 be bounded with S1 ∩ S2 = ∅. Let f be a bounded function. If f is integrable on S1
and on S2 then f is integrable on S1 ∪ S2 , in which case
ZZ
ZZ
ZZ
f dA.
f dA +
f dA =
S1 ∪S2
S2
S1
(c) If f, g are integrable on S and f (x) ≤ g(x) for x ∈ S, then
ZZ
ZZ
f dA ≤
gdA
S
(d) If f is integrable on S, then so is |f |, and |
S
RR
S
f dA| ≤
RR
S
|f |dA.
(e) If f is continuous on a rectangle R, then f is integrable on R.
Definition 2.5. A set Z ⊂ R2 is said to be of zero content, if for all > 0, there exists a finite
collection of rectangles R1 , . . . , RM with areas A1 , . . . , AM such that
Z⊂
M
[
M
X
Rm ,
m=1
Am < .
m=1
Theorem 2.6. Suppose f is bounded on the rectangle R. If the set of points in R at which f is
discontinuous has zero content, then f is integrable on R.
Proposition 2.7.
(a) If Z ⊂ R2 has zero content and U ⊂ Z, then U has zero content.
(b) If Z1 , . . . , ZK have zero contents, then ∪K
j=1 Zj has zero content.
(c) If f : (a0 , b0 ) → R2 is of class C 1 , then f ([a, b]) has zero content whenever a0 < a < b < b0 .
Lemma 2.8. The function χS is discontinuous at x if and only if x is in the boundary of S.
Definition 2.9. A set S ⊂ R2 is (Jordan)-measurable, if it is bounded and its boundary has zero
content.
Theorem 2.10. Let S be a measurable subset of R2 . Suppose f is bounded and the set of points in S
at which f is discontinuous has zero content. Then f is integrable on S.
RR
Remark 2.11. We can introduce the area of a bounded set S ⊂ R2 , as area(S) = A(S) = S 1dA =
RR
R χS dA for S ⊂ R, if it exists. More generally, we use the inner area A(S) = LR (χS ) and the outer
area A(S) = UR (χS ). On problem sheet 5, we show A(S) = A(S) if and only if ∂S has zero content,
i.e. S is Jordan measurable. While this is good enough for all uses in this course, the notion of Jordan
measurable is not as wide as we would wish, e.g. it does not contain all open bounded sets.
Remark 2.12. The generalisation from double to n-dimensional integrals is very obvious , only the
notation is more involved. Consider rectangular boxes in Rn given as
R = [a1 , b1 ] × . . . × [an , bn ] = {x ∈ Rn | a1 ≤ x1 ≤ b1 , . . . , an ≤ xn ≤ bn }
9
Q
with n-dimensional volume vol(R) = nj=1 (bj − aj ).
A set ZS⊂ Rn is defined
Pk to have zero content, if for all > 0 there are rectangular boxes R1 , . . . , Rk
k
with Z ⊂ j=1 Rj and j=1 vol(Rj ) < .
Any lower dimensional set S = f (K) given by a C 1 -parametrisation f : U → Rn with U ⊂ Rk
open, k < n and K a compact subset of U has zero content (cf. prop 2.7).
The notation for integrals over bounded sets S ⊂ Rn are
Z
Z
Z
Z
Z
Z
. . . f dV n = . . . f (x)dn x = . . . f dx1 . . . dxn .
S
2.2
S
S
Multiple Intergrals and Iterated Integrals[Folland 4.3]
RR
Motivation: How do we evaluate R f dA?
Let R = [a, b] × [c, d], partition P = (x0 , . . . , xJ ; y0 , . . . , yK ) and x̃ = (x̃1 , . . . , x̃K ), ỹ = (ỹ1 , . . . , ỹJ )
with x̃j ∈ [xj−1 , xj ] and ỹk ∈ [xk−1 , xk ]. Form the sum with ∆xj = xj − xj−1 and ∆yk = yk − yk−1 :
ZZ
f (x, y)dA ≈
R
K X
J
X
f (x̃j , ỹk )∆xj ∆yk
k=1 j=1
≈
K Z
X
k=1
b
Z
d Z b
f (x, ỹk )dx∆yk ≈
c
a
f (x, y)dx dy.
a
Possible problems, e.g. f integrable on R does not imply f (x, yj ) integrable on [a, b] for fixed yj .
Theorem 2.13 (Fubini). Let R = [a, b] × [c, d], and f be integrable on R. Suppose that, for each y ∈
Rb
[c, d] the functions defined by fy (x) = f (x, y) is integrable on [a, b] and the function g(y) = a f (x, y)dx
is integrable on [c, d]. Then
ZZ
Z d Z b
f dA =
f (x, y)dx dy.
(2.1)
R
c
a
If f x (y) = f (x, y) is integrable on [c, d] for each x ∈ [a, b] and h(x) =
[a, b] then
ZZ
Z b Z d
f dA =
f (x, y)dy dx.
R
a
Rd
c
f (x, y)dy is integrable on
(2.2)
c
For the proof we need the generalisation of Riemann’s integrability criterion:
Proposition 2.14. Let f be an integrable function on the rectangle R = [a, b] × [c, d]. Then for all
> 0 there exists δ > 0 such that if partition P = (x0 , . . . , xJ ; y0 , . . . , yK ) is any partition of R with
x̃ = (x̃1 , . . . , x̃K ), ỹ = (ỹ1 , . . . , ỹJ ) with x̃j ∈ [xj−1 , xj ] and ỹk ∈ [yk−1 , yk ] and
max( max (xj − xj−1 ), max (yk − yk−1 )) < δ,
j=1...,J
k=1...,K
then any Riemann sum for f associated to R defined by
S((P, x̃, ỹ), f ) =
J X
K
X
f (x̃j , ỹk )∆xj ∆yk
j=1 k=1
differs from
RR
R f dA
by at most .
Remark 2.15. The integrals on the right hand sides of (2.1) and (2.2) are called iterated integrals. We
evaluate them from the inside out. If the region S of integration is given by two graphs
S = {(x, y) | a ≤ x ≤ b, φ(x) ≤ y ≤ ψ(x)},
10
i
RR
R b hR ψ(x)
then S f dA = a φ(x) f (x, y)dy dx.
If the region S is given by graphs to the left and right, we have
then
RR
S
f dA =
R d hR ν(y)
c
µ(y)
S = {(x, y) | c ≤ y ≤ d, µ(y) ≤ x ≤ ν(y)},
i
f (x, y)dx dy.
Example 2.16. Let T be the triangle with vertices (0, 0)T , (1, 0)T and (1, 2)T . Evaluate
y 2 )dA.
RR
T (xy
+
Remark 2.17. Fubini’s theorem also holds in the n-dimensional case, if R = [a1 , b1 ] × . . . × [an , bn ] and
f is integrable then
Z b1
Z bn
Z
Z
n
f dx1 . . . dxn ,
...
. . . fd V =
S
a1
an
provided all integrals exist.
RRR
Example 2.18. The mass of an object in a set S with density ρ(x) is givenRRR
by m = S ρ(x)d3 x. Its
centre of gravity x̄ can defined as the point with the coordinates x̄j = m−1 S xj ρ(x)d3 x.
Find the mass of the tetrahedron T formed by the three coordinate planes and the plane x+y+2z =
2 if the mass density is given by ρ(x, y, z) = e−z .
Remark 2.19. Fubini’s theorem tells us that under suitable hypotheses on the integrand f the order
of integration in an iterated integral can be changed, e.g.
Z bZ d
Z dZ b
f (x, y)dydx =
f (x, y)dxdy.
a
Example 2.20. Evaluate
2.3
c
c
a
R2R1
0
−x3 dxdy.
y/2 ye
Change of Variables for Multiple Integrals[Folland 4.4]
We look for a multidimensional version of
Z b
Z
0
f (g(u))g (u)du =
a
g(b)
f (x)dx,
g(a)
where f is continuous on [a, b], g is bijective and C 1 . We reformulate it first to
Z
Z
f (x)dx =
f (g(u))|g 0 (u)|du
I
g −1 (I)
So the question is, what is the right expression for
Z
Z
Z
Z
n
. . . f (x)d x = . . .
f (G(u))[??]dn u
S
G−1 (S)
a b
Example 2.21. Let G be the linear planar map defined the matrix A =
with det A 6= 0.
c d
Example 2.22. Let Gu = Au, where A is an invertible 3 × 3 matrix.
Theorem 2.23. Let A be an invertible n × n matrix and let Gu = Au be the corresponding linear
transformation of Rn . Suppose S is a measurable region in Rn and f is an integrable function on S.
Then G−1 (S) = {A−1 x | x ∈ S} is measurable and f ◦ G is integrable on G−1 (S) and
Z
Z
Z
Z
. . . f (x)dn x = | det A| . . .
f (A(u))dn u.
(2.3)
−1
S
G (S)
11
This immediately leads to a formula for affine functions. If Gu = Au + b then also
Z
Z
Z
Z
n
. . . f (x)d x = | det A| . . .
f (A(u))dn u.
−1
S
G (S)
Considering small sets and using a Taylor expansion G(u + ∆u) = G(u) + DG(u)∆u + o(∆u) this
motivates the following theorem.
Theorem 2.24 (Change of Variables). Given open sets U, V ⊂ Rn , let G : U → V be a bijective
transformation of class C 1 whose derivative DG(u) is bounded on U and is invertible for all u ∈ U .
Suppose that T ⊂ U , S ⊂ V are measurable sets such that G(T ) = S. If f is an integrable function
on S, then f ◦ G is integrable on T , and
Z
Z
Z
Z
. . . f (x)dn x = . . .
f (G(u))| det DG(u)|dn u.
(2.4)
−1
S
G (S)
Remark 2.25. The expression dV = dn V = dn x = | det DG(u)|dn u is called the volume element.
For spherical polar coordinates (cf. example 1.34) Gsph (ρ, θ, φ) = ρ sin θ cos φi+ρ sin θ sin φj +ρ cos θk
this is given by
dV = | det DGsph (u)|dρdθdφ = ρ2 sin θdρdθdφ.
For cylindrical polar coordinates Gcyl (R, θ, z) = R cos θ i + R sin θ j + z k, we obtain
dV = | det DGcyl (u)|dRdθdz = RdRdθdz.
For concrete calculations we need particular care when determining G−1 (S).
Example 2.26. Find the volume of the region S above the surface z = x2 + y 2 and below the plane
z = 4.
We use cylindrical polar coordinates then the conditions mean z ≥ R2 and z ≤ 4.
ZZZ
ZZZ
V =
dV =
RdRdθdz
S
G−1
cyl (S)
Z 4 Z 2π 2 √z
Z 4 Z 2π Z √z
R
=
RdRdθdz =
dθdz
2 0
0
0
0
0
0
2 4
Z 4
z
z
=
2π dz = π
= 8π
2
2 0
0
12
Chapter 3
Line and Surface Integrals; Integral
Theorems
3.1
Arc Length and Line Integrals[Folland 5.1]
Motivation: Determine the arclength of a circle.
Definition 3.1. Let C be a smooth curve with parameter representation g : [a, b] → Rn with g of
class C 1 , then we define its arclength
Z b Z
dg L(C) :=
ds,
dt dt =
a
C
dg where ds = dt dt is the line element.
Remark 3.2. (a) The definition of arclength is independent of the parametrisation, see e.g. problem
sheet 8, homework question 1.
(b) The notion of arclength can be extended to piecewise smooth curves, i.e. for curves with
parametrisation g : [a, b] → Rn if (i) g is continuous and (ii) its derivative exists and is continuous
except finitely many points tj , at which the one-sided limits limt→tj ± g 0 (t) exist.
(c) In the preceding discussion we implicitly assumed g(t) to be injective. If we think of g(t) as the
Rb
path of a particle then a |g 0 (t)|dt is the total distance traveled from time a to b.
Definition 3.3. Let f be a continuous scalar field on a set containing a piece-wise smooth curve
C ⊂ Rn . Then we define the line integral of a scalar field when C is parametrised by g(t), t ∈ [a, b]
by
Z
Z b
dg f ds =
f (g(t)) dt.
dt
C
a
Example 3.4. Integrate f (x, y) = 2xy around the first quadrant of a circle with radius a.
Definition 3.5. Let F be a continuous vector field on a set containing a piece-wise smooth curve
C ⊂ Rn . Then we define the work integral of a vector field when C is parametrised by r(t), t ∈ [a, b]
by
Z
Z b
dr
F · dr =
F (r(t)) ·
dt,
(3.1)
dt
C
a
where dr is the vector line element.
Proposition 3.6. If T̂ is the unit tangent vector to C in (3.1) that points in the direction in which t
is increasing, then
Z
Z
F · dr = (F · T̂ )ds.
C
C
13
Remark 3.7. (a) It follows directly from the previous proposition that a reversal of the orientation
of C (the direction in which t increases) changes the sign of the work integral (in contrast to
the line integral of a scalar field). Otherwise it is independent of parametrisation.
R
(b) If F is a force field then C F · dr is the work done by moving a particle from r(t0 ) to r(te )
along the curve C.
(c) A vector field F : Ω → Rn is called conservative, if
Z
Z
F · dr =
F · dr
(3.2)
C2
C1
for any two smooth curves C1 and C2 within Ω connecting two points x0 and xe in Rn
3.2
Green’s Theorem[Folland 5.2]
Definition 3.8. (a) A simple closed curve in Rn is a curve, that can be parametrised by a continuous map x = g(t), a ≤ t ≤ b, such that g(a) = g(b) but g(s) 6= g(t) unless {s, t} = {a, b}.
(b) A set S is a regular region, if it is compact and if it is the closure of its interior S = intS.
(c) If a simple closed curve is part of the boundary ∂S of a region S ⊂ R2 , it is called positively oriented,
if the orientation of the curve is such that S is on the left with respect to the positive direction
of the curve.
(d) A simple closed curve C ⊂ R2 is convex, if every straight line intersects C at 2 points at most.A
simple closed curve C ⊂ R2 is semi-convex, if every straight line parallel to the coordinate axes
intersects C at 2 points at most.
Theorem 3.9 (Green). Suppose S is a regular region in R2 with boundary ∂S consisting of finitely
many positively oriented simple closed piecewise smooth curves. Suppose that F is a vector field of
class C 1 on a neighbourhood S. Then
Z
ZZ ∂F2 ∂F1
F · dr =
−
dA
∂x1
∂x2
∂S
S
Example 3.10. Let F (x1 , x2 ) = (−x2 , x1 )T , then with a positively oriented boundary ∂S with
parametrisation r(t) = (r2 (t), r1 (t)) we have for the area of S:
Z
1
A(S) =
(−r2 , r1 )T · dr.
2 ∂S
Compare ‘planimeter’.
3.3
Surface Integrals[Folland 5.3]
Remember the parametrisation of smooth surfaces from section 1.5. We consider G : W → R3 , where
W ⊂ R2 is open connected and the surface is given by x = (x, y, z)T = G(u, v). Let P0 be a point
~ 0 = G(u, v), let P1 be a neighbouring point with OP
~ 1 = G(u + ∆u, v).
on the surface, such that OP
~ 2 = G(u, v + ∆v) and OP
~ 3 = G(u + ∆u, v + ∆v). For
Similarly, let P2 and P3 be the points with OP
∆u and ∆v sufficiently small the area of the surface ∆S can be approximated by
∂G
∂G
∂u (u, v) × ∂v (u, v) .
We hence introduce the surface element
∂G
∂G
dA = (u, v) ×
(u, v) dudv.
∂u
∂v
14
(3.3)
Definition 3.11. Let R be a measurable subset of W in the uv-plane and S = G(R), then
ZZ
ZZ ∂G
∂G
dudv.
Area(G(R)) =
dA =
(u,
v)
×
(u,
v)
∂v
G(R)
R ∂u
(a) With G(u, v) = (x, y, z)T we have


i
j
k
∂(z, x) ∂(x, y) ∂(y, z) ∂G
∂G


∂
x
∂
y
∂
z
=
(u, v) ×
(u, v) = det
u
u
u
∂(u, v) i + ∂(u, v) j + ∂(u, v) k,
∂u
∂v
∂v x ∂v y ∂v z
rh
i
i
i
h
h
∂(z,x) 2
∂(x,y) 2
∂(y,z) 2
+
+
i.e. dA =
∂(u,v) ∂(u,v) dudv.
∂(u,v) Remark 3.12.
(b) The expression in Definition 3.11 is independent of parametrisation. If (u, v) = Φ(s, t), where Φ
is bijective, C 1 and Φ−1 is also C 1 , then Theorem 2.24 gives
∂(u, v) dsdt,
dudv = ∂(s, t) p
∂(y,z) ∂(u,v) ∂(y,z) such that dA = α2 + β 2 + γ 2 dsdt with e.g. α = ∂(u,v)
=
∂(s,t)
∂(s,t) by the chain rule as
required.
Example 3.13. Find the surface area of a cylindrical shell of height b and radius a using G(u, v) =
(a cos u, a sin u, v)T with W = [0, 2π] × [0, b].
Definition 3.14. The surface integral of a scalar field for a continuous function f on a smooth surface
S (parametrised by G : W → S) is given by
ZZ
ZZ
∂G
∂G
f dA =
f (G(u, v)) (u, v) ×
(u, v) dudv.
∂u
∂v
S
W
Remark 3.15. Remember from remark 1.25 that the unit normal vector to a surface is given by
∂G
∂G
1
(u, v) ×
(u, v),
n = G (u, v) × ∂ G (u, v) ∂u
∂v
∂∂u
∂v
the direction is defined up to a factor ±1. We will usually require n to point outwards. A surface
S ⊂ R3 is orientable if a unique normal can be assigned continuously at each point x ∈ S.
Definition 3.16. The surface integral of a vector field for a continuous function F : S → R3 on a
smooth surface S (parametrised by G : W → S) is given by
ZZ
ZZ
∂G
∂G
(u, v) ×
(u, v) dudv,
F · dA =
F (G(u, v)) ·
∂u
∂v
S
W
where dA is the vector surface element.
RR
RR
Remark 3.17. (a) It follows that s F · dA = s (F · n̂)dA, where n̂ is the unit normal vector of S
at a point x ∈ S. Hence the integral is well-defined up to orientation.
(b) The notion of surface integrals can be extended to surface integrals on piecewise smooth surfaces:
If S := S1 ∪ S2 ∪ . . . ∪ Sn and Si ∩ Sj , i 6= j consists of a finite number of smooth curves where
Si is smooth for each i = 1, . . . , n, then we define
ZZ
ZZ
ZZ
ZZ
:=
+
+ ... +
.
S
S1
S2
Sn
ZZ
x · dA , where S is the sphere with radius a, i.e.
Example 3.18. Find
S
S := {x ∈ R3 : x2 + y 2 + z 2 = a2 }.
15
3.4
Divergence Theorem[Folland 5.5]
Theorem 3.19 (Gauss’ Divergence Theorem). Suppose R is a regular region in R3 with piecewise
smooth boundary (surface) ∂R oriented so that the positive normal points out of R. Suppose that F
is a C 1 vector field in a neighbourhood of R. Then
ZZZ
ZZ
ZZ
divF dV
(3.4)
F · ndA =
F · dA =
R
∂R
∂R
Remark 3.20. If we understand F as the flow of some substance, then the left-hand side of (3.4) is the
flux across ∂R from inside R to the outside (the complement of R). Choosing R to be a small ball,
the divergence can be understood as the net outflow at point a, by using
ZZ
ZZ
1
3
divF (a) = lim
F · ndA = lim
F · ndA.
r→0 4πr 3
r→0 V ol(Br )
|x−a|=r
|x−a|=r
3.5
Stokes’ Theorem[Folland 5.7]
Definition 3.21. Let S be an orientable surface with simple boundary (curve) C. Let n̂ be the unit
normal on S. Then S and C are said to be correspondingly oriented, if the surface is to the left
of the curve, when the normal vector points upwards.
Theorem 3.22 (Stokes’ Theorem). Let S ⊂ R3 be an orientable, piecewise smooth surface with
boundary C ⊂ R3 consisting of finitely many correspondingly oriented, simple, piecewise smooth curves.
Suppose that the vector field F is continuously differentiable in a neighbourhood of S. Then
ZZ
Z
curlF · dA =
F · dr .
(3.5)
S
C
√
R
Example 3.23. Calculate C F · dr for F (x, y, z) = x2 + 1i + xj + 2yk where C is the intersection
of the surfaces z = xy and x2 + y 2 = 1, where C is oriented counterclockwise when seen from above.
Remark 3.24. (a) A curve on the right-hand-side in Stokes’ Theorem can be the boundary of infinitely many surfaces S, e.g. the unit circle in the xy plane for many different paraboloids
z = h(1 − x2 − y 2 ).
(b) Stokes’ Theorem gives a coordinate free description of curl, let D be a disc of radius around
a with normal u and boundary C , then
ZZ
Z
1
1
(curlF (a)) · u = lim 2
(curlF ) · udA = lim 2
F · dr.
→0 π
→0 π
D
C
R
If F is a force field, C F · dr is the work done by F on a particle that moves around C , i.e.
(curlF ) · u represents the tendency of force F to push the particle around C .
3.6
Some Applications to Physics[Folland 5.6]
Example 3.25 (Flow RR
of material). With density ρ(x, t), velocity field v(x, t), the flow of substance
through a surface S is S J · ndA, where J (x, t) = ρ(x, t)v(x, t) is momentum density.
Now we consider regular region R with smooth boundary ∂R
ZZZ
ZZZ
∂
d
ρ (x, t) dV =
ρ (x, t) dV
dt
R ∂t
ZZ R
= −
J · n dA changes of mass only through boundary (mass conservation)
Z Z∂R
Z
= −
div (J ) dV by Divergence Theorem
R
16
Now let R = Br (x) (ball of radius r around x, r → 0). Then if
∂
∂t
(ρ) and div (J ) continuous then
∂ρ
+ divJ = 0
∂t
We call this a conservation law. If this equation holds then
ZZZ
∂
ρ dV = 0
∂t
R
as long as there is no flow out of R. For incompressible fluids, this simplifies to divJ = div (ρv) =
ρdiv (v) = 0.
Example 3.26 (Heat equation). Heat transfer through diffusion, where u(x, t) is temperature, σ is
the specific heat. The flux of heat across a surface S with normal n according to Fourier’s law is
ZZ
−
K∇u · ndA.
S
We use to calculate the rate of change in regular region R ⊂ R3 with smooth boundary ∂R:
ZZZ
ZZ
ZZZ
d
3
σu(x, t)d x
=
K∇u · ndA +
F (x, t)d3 x
dt
R
S
ZZZ
ZZZ
Div. Thm
3
=
K∇ · ∇u(x, t)d x +
F (x, t)d3 x,
R
where F (x, t) is the heat creation. Choosing R to be a small ball around a point x, and assuming
that u, F and ∇2 u are continuous, this means that the following equation must hold pointwise
ρ
3.7
∂ρ
(x, t) = K∇2 u(x, t) + F (x, t).
∂t
Integrating Vector Derivatives[Folland 5.8]
Definition 3.27. A vector field F : Ω → R3 with Ω ⊂ R3 open is called irrotational, if
curlF = 0.
A vector field F : Ω → Rn with Ω ⊂ Rn open is called solenoidal (or divergence-free) if
divF = 0.
Proposition 3.28. A continuous vector field G in an open set R ⊂ R3 is conservative if and only if
G is the gradient of a C 1 -function φ on R. The function φ is called a scalar potential of G.
Proof. See sheet 8, TQ3 and sheet 9, TQ1.
Theorem 3.29. Suppose R is a convex open set in R3 and G is a C 1 vector field on R. Then
curlG = 0,
if and only if G is the gradient of a C 2 function on R.
Proof. If G = ∇φ, then curlG = curl∇φ = 0 by Proposition 1.16.
To show the existence of φ with ∇φ = G for curlG = 0, we pick a point a ∈ R.
Z
φ (x) =
G · dr
L(a,x)
17
where L (a, x) is the line from a to x (using convexity of R). Show G (x) = ∇φ (x). Let h = (h, 0, 0)> ,
small enough such that x + h ∈ R. Consider triangular curve C ⊂ R.
Z
ZZ
G · dr =
curlG · dA = 0
C
Hence φ (x + h) − φ (x) +
R
L(x+h,x) G
φ (x + h) − φ (x)
1
=
h
h
S
· dr = 0. Thus,
Z
1
G · dr =
h
L(x,x+h)
Z
h
G1 (x1 + h, x2 , x3 ) dt
0
with h → 0.
∂xφ (x) = G1 (x)
The other components are analogous.
Theorem 3.30. Suppose R is a convex open set in R3 . A vector field F : R → R3 is divergence free
if and only if there exists a vector field Ψ such that F = curlΨ. Here Ψ is called a vector potential
of F .
Proof. [R restricted] If F = curlψ, then divF = divcurlψ = 0 by Proposition 1.16.
Suppose R = (a1 , b1 ) × (a2 , b2 ) × (a3 , b3 ), divF = 0. Choose ψ = ψ1 i + ψ2 j
curlψ = −∂z ψ2 i + ∂z ψ1 j + (∂x ψ2 − ∂y ψ1 ) k = F1 i + F2 j + F3 k
Rz
Rz
ψ2 (x, y, z) = − c F1 (x, y, t) dt + f (x, y) with c ∈ (a3 , b3 ). ψ1 (x, y, z) = c F2 (x, y, t) dt + g (x, y).
Then
Z z
∂x ψ2 − ∂y ψ1
=
(∂x F1 (x, y, t) + ∂y F2 (x, y, t)) dt + ∂x f (x, y) − ∂y g (x, y)
Zc z
divF =0
∂z F3 (x, y, t) dt + ∂x f (x, y) − ∂y g (x, y)
=
c
FTC
=
F3 (x, y, z) − F3 (x, y, c) + ∂x f (x, y) − ∂y g (x, y)
so choose g ≡ 0
Z
x
F3 (t, y, c) dt with a ∈ (a1 , b1 ) ,
f (x, y) =
a
such that ∂x f (x, y) − ∂y g (x, y) = F3 (x, y, c).
Remark 3.31 (Helmholtz decomposition). Let R be a convex open set in R3 . For any C 2 vector field
F there exist a C 2 function φ and a C 2 vector field Ψ such that F = gradφ + curlΨ. (without proof).
R
1
Remark 3.32. Considering G(x, y, z) = x2 +y
2 (−yi + xj), we see curlG = 0 but C G · dr = 2π, where
C is the unit circle in the xy-plane:



 

Z
Z 2π
Z 2π
cos θ
− sin θ
− sin θ






sin θ
cos θ
cos θ
r (θ) =
.
G · dr =
·
dθ =
1 dθ = 2π
C
0
0
0
0
0
Here R is the complement of the z-axis in R3 , which is not convex.
18