POTENTIAL THEORY IN
Chapter
The
O
THREE DIMENSIONS
of the
preceding chapter, when generalized to three or more
considerably complicated. The development of this
the
19th
theory during
century was motivated to a considerable extent by
intuition.
The
physical
study of fields of force and velocity of fluid flows
led to the theorems on integration in severable variables which are in this
chapter. More modern expositions of this material lean heavily on algebraic
developments of the late 19th and early 20th centuries. Although the
mathematics has significantly improved with the introduction of the notions
of differential forms and invariance, the intuition provided by concrete
interpretations has been lost. We shall lean heavily on the interpretation
by fluid flows, thereby sacrificing some mathematical rigor for a little bit of
concreteness.
We certainly should point out that the importance of the
of
forms by far transcends its use in putting the divergence
differential
subject
theorem on firm ground. This theory has had major impact on all branches
of modern research mathematics and physics. We have however selected to
complete our story rather than begin to suggest a new one.
A fluid flow is given by a function {x0 t) defined for x0 in some domain
We require that
D in R3 and t on an interval in R about the origin.
theory
dimensions becomes
,
(i)
(ii)
(iii)
<J> is continuously differentiable
<t>(x0 0) x0 all x0 e D,
in all variables,
=
,
,
for fixed t, the transformation x0
nonsingular differential.
->
<Kx0 0
,
is one-to-one and has
611
a
612
Potential Theory in Three Dimensions
8
The value
which
was
the space position at time t of the particle
We shall refer to x0 as the particle coordinate
(b(x0 r) represents
,
at x0 at time t
=
0.
<|)(x0 /) as the space coordinate. Condition (ii) asserts that the
0.
Condition (iii) asserts that
and space coordinates coincide at t
the relation between particle and space coordinates at any time t is invertible:
we can recapture the initial position of a particle from its position at any
and to
x
=
,
particle
We shall denote the inverse of
time.
x0
=
=
<|> by \|/ :
x
=
<t>(x0 t)
,
if and
only if
\(>(x, t).
The
curve
given by x
=
c/>(x0 t)
,
is the path of motion of the particle x0 The
{dfy/dt)(x0 t). If we fix the time t,
.
of x0 at time / is, of course,
the collection of velocity vectors forms
velocity
spatial coordinates)
velocity of the particle at
course
is the
to
,
a
field, denoted by v(x, t) (referring of
called the velocity field of the flow. v(x,
We have already noted that
x at time t.
t)
d<t>(x0 0
,
v(x, 0
=
dt
xo
=
(8.1)
<l>(x. ')
velocity field is independent of time, we say that the flow is steady.
velocity field of a flow completely determines the flow : the path of
motion x
u(?) of a particle x0 is the solution of the differential equation
If the
The
=
t-
=
dt
o(0)
v(u, f)
(8.2)
=
x0
By (8.1) the solution is given by u(0
=
<j>(x0 t),
,
for
(8.1)
can
be rewritten
as
3<p(x0 t)
,
v(<Kxo,0,0
Thus the
equation
Equation (8.2).
=
-
dt
of flow is
recaptured
from the
velocity
field
by solving
recapitulates what we have already learned about fluid
flows. In the subsequent section we shall develop the mathematics required
We shall see
to study the evolution through time of a given mass of fluid.
that the various laws of conservation of physics (mass, energy) correspond
to mathematical theorems (divergence theorem, Stokes' theorem).
This introduction
8.1
8.1
Divergence
Let
us
we
x
.
.
Kp)
density
,
=
a
hm
A-p
of
Equation of Continuity
613
Continuity
flowing through a domain in R3 according to the
According to reasonable physical assumptions, if
at a
mass
.
Equation
and the
fluid
<b(x0,t).
=
define the
and the
with
begin
equation
Divergence
point
p
as
the limit
A
-
vol A
the domain A shrinks
uniformly down to p, then the mass of any domain
given by integration of the density function p. In our case, that of a fluid
in motion, we shall express the
density of the fluid at the point x at time t
as p(x, t).
Thus, for any domain D, the mass of fluid in D at time t is
as
is
f p(x, 0 dV
We
can also consider the
density at a particle: p((p(x0, t), t) is the density
of the fluid at time t at the particle
(originally at) x0
(More generally, we
always have this option of referring measurable quantities to either the spatial,
or the particle coordinates.
This option is a source of some confusion, as
well as deepening, of our
understanding.)
.
The law of conservation of matter asserts that the mass of a
given object
independent of time. If we fix a domain D, the space occupied at time t
by the fluid originally in D is the domain D,
{t>(x0, /): x0 e D}. The
mass of fluid in
Dt is
is
=
f
JDt
P(x, 0
dV
Since mass must be conserved, this must be
law of conservation of mass can be
expressed
d_ f
dt JD,
p(x, t)dV
=
through
Thus the
0
(8.3)
for any domain D. We would prefer to
functions of points, rather than domains.
how to carry
independent of t.
by this equation :
the differentiation
state this as an
equation involving
In order to do that
implied
in
(8.3).
we
The
must know
problem
with
614
Potential
8
in Three Dimensions
Theory
(8.3) is that we have a variable domain of integration. This can be solved
by replacing that integral by one over D. We shall now briefly interrupt
this discussion with a description of the formula for change of variables in
an integral.
This will allow us to compute (8.3).
now
that we are given a one-to-one transformation y
F(x) of a
Suppose
domain D onto a domain A. We assume that F is continuously differentiable, and its differential is everywhere nonsingular. We shall require also
that dF(x) is orientation-preserving : that is, that it maps the standard basis
Ejl -> E2 -> E3 into a right-handed system. Writing x (x1, x2, x3),
y
(y1) y2, j3), the image of E( under the linear transformation d{x) is
just (d'Fjdxi)(x). Thus we require that
=
=
=
be
a
5F
5F
5F
right-handed system, which is
d(yl, y\ v3)
same as
hypotheses
the
of variable F.
we
have the
If/ is
an
asking that
8F
dF
/5F
With these
change
the
\
following formula for integration
integrable function on A, then
under
//(F(x))det|^^
(8.4)
We shall defer the derivation of this formula to the end of this section.
The
J/(y)^
=
motivating idea is that it is true in the small: if the function /is constant, and
the transformation F is a linear transformation, and D is a rectangle, then
(8.3) just says that the volume of the parallelepiped F(D) is det F vol(D)
(an easily verified fact). The general case follows by locally approximating
by this case and summing over the whole domain.
Examples
1. Find
\Bx2y* dV,
coordinates for this
x
=
r
sin 6
cos
'
'
=
,,
,^
K*'0'
I
\
computation
y
/sin 9
-.
.,,
<fi
cos
sin 9 sin
cos
where B is the unit ball.
9
=
<f>
<p
r
We
use
spherical
:
sin 9 sin
cj>
z
=
r cos
9
r cos
9
cos
<p
r
sin 9 sin
r cos
9 sin
<f>
r
sin 9
cj> \
cos <f> \
0
/
-sin 9
8.1
Divergence and
the
Equation of Continuity
615
so
det
d(x, y,z)
^
r
=
^rr
0
sin
3(r, 9, cj>)
f x2};4 dJ/
f
=
f
[
'o J-ti 'o
'B
[
=
f
r8 dr
J0
J-Jt
1
16
_
r8 sin6 0 cos2 0 sin5 </> Jr d</> d0
In
'
'
l05
JD(*2
2-
^2)
-
16
dx
sin6 0 J0
Jo
In
_
~
~
9
\
cos2 0 sin5 4> d$
dy,
945
where
D
=
{0
< x <
1,
x
-
1 < y
<
x}
be
comes
,1
,
1
1
uv
,
du dv
=
z Jo
2
J0 Jo
under the
3.
change
JB(x2
+
y2
of variable
+
z2)
<^x i/v
u
=
x
y,
dz, where
v
=
x
+ y.
B is the domain
B={x2 +y2< l,0<z<2}
This
can
be
easily computed
in
cylindrical
f (x2 + y2 + z2)dxdydz=\n \ f
(r2
coordinates :
+
z2)r
d9 dr dz
=27t({r3+r)dr
19n
_
We return to
the sake of
our
fluid flow
compution,
(x1, x2, x3)
=
given by
x
=
<Kx0 0,
We
saau
express
it, for
in coordinates :
<b(V, V. V, 0
(8-5)
616
Potential
8
Since
Theory
reduces to the
(8.5)
identity for
d{xl, x2, x3)
6{Xq
,
Xq
,
in Three Dimensions
=
t
=
0,
we
have
1
(8.6)
x0 )
Thus, the determinant
J((x0)
=
d(xl, x2, x3)
det
0{Xq
Xq
,
)
Xq
,
positive for all small t so we can apply
computation of (8.3) for fixed small
servation law expressed by
is
,
the
0=jtL p(x'
dv
=
e
L p((Kx
'
the
change of variable
/.
We
now
' t)Jt dv
=
have the
L it {pJt)
formula to
mass con
dv
(The final equation follows since differentiation under the integral is now
allowable.) Since this must be true for every domain D, the integrand is
identically zero :
dt
We
let
us
{pjt)
(8.7)
0
=
dt
t
=
0, using (8.6).
d{x\ x2, x3)
_d_
J,
dt
O(X0
The determinant is the usual
8x'/dx0J.
one
that derivative for
explicitly compute
can
First,
consider
,
Xq
sum
only
one
Xq
)
(8.8)
r
=
0
of products of the various
partial derivatives
product will have three terms; in each
Each term is
term is differentiated with respect to t.
The derivative of such
of which
,
a
of the form
dr3
5T2
diyds1)
where
!
=
{r1, r2, r3}
0
is
ds2
a
r
=
0
(8.9)
5s3
permutation
of
{x1, x2, x3},
and
{s1, s2, s3}
a
permuta^
8.1
Divergence
{x0\ x02, x03}. According
tion of
dr
=
ds
(
=
to
and the
Equation of Continuity
617
(8.6)
dr
0 if
s
# r0
1 if
=
d~s
0
(
=
s
=
r
0
Thus the only relevant terms (8.9) are those where y2
r02, s3 r03 and,
fortiori, sl r0K Finally, by the equality of mixed partial derivatives,
=
=
a
=
d_ / &c|_\
W07
dt
_
~
,=0
dv'
_d_ /axj\
dx0l [it)
dx0
(=o
(v1, v2, v3) is the velocity field of the flow (recall Equation (8.1)).
Thus, the computation of (8.8) is complete: there are only three relevant
terms, for r1
x1, x2, x3, respectively, and we have
where
v
=
=
dv1
d
dlJ'
dx0
(=o
Definition 1.
Let
domain in R3.
The
div
The
v
name
dv2
+
=
>
&
^^
dv3
+
dx0
dx0
r
=
(8.10)
0
(u1, v2, v3) be a differentiable vector field defined in
divergence of v is the function defined by
v
=
-.
dx1
will appear
discussion in the
a
presently to be justified.
following assertion.
We
now
summarize
our
Proposition 1. (Equation of Continuity) Let v(x, t) be the velocity field of a
fluidflow, andp(x, t) its density. The law of mass conservation takes this form :
dp
-
Proof.
law of
+
Referring
mass
8t
div(pv)
dp
=
3
dp
-+l>^ + pdivv
to the
preceding discussion
conservation asserts that
{p{4>{Xo,t),t)J,{Xo))=0
=
we
0
have
(8.11)
seen
from
(8.7)
that the
618
Potential Theory in Three Dimensions
8
for all t, x0
Evaluating
.
at t
=
0, this becomes
8
0
_
(p(<J>(Xo t), /))|.-o /o(Xo)
+
,
8(
Jt{Xo)\t=o
/s(*(Xo 0, 0) ^ Jr.(Xo)l
=
,
o
(8.12)
=
1
^(xo,0)5-(x0,0)+ J(xo,0)
2
=
8t
Sx'
1
+
ot
p(xo,0)divv(xo,0)
expression follows from our computation above terminating in (8.10),
1 x0
<t>(x0 0). Now, we could have started our clock
/0(x0)
0 except that our formulas
at any time; there is nothing special about the time t
hold
for all (x, 0 since it is
must
there.
are most easily computed
Thus, (8.12)
valid for all (x0 0). Thus (8.1 1) is true. We leave the first equality as an exercise.
The second
and the fact that
=
=
,
,
=
,
Equation (8.11)
be referred to the
can
3
dp
dt
x
=
coordinates of the motion:
dp
dx\
OX
Ot
i=l
(t>(x0,r)
particle
x
=
<t>(x0,t)
dx
+
p(<p(x0 t), t) div
(x0 0
=
,
,
0
which compresses into
-
p(<Kx0 ,0,0
p(Kxo 0, 0 div
+
,
(x0 0
,
=
(8. 1 3)
0
This relates the time rate of change of density at a particle with the rate of
change of its position. A fluid flow is called incompressible if the same mass
occupies the same volume. For an incompressible fluid flow we
always
constant for any initial domain D.
jDt dV is
must therefore have that
Sx,
JD dt'
dt JD
dt JDt
for every domain D. Thus div v
for a flow to be incompressible.
(8.13)) this is the same
dependent of time.
Corollary
if div
v
=
0.
1.
v
as
is the
Thus
(8.14)
JD
0 is the necessary and sufficient condition
By the equation of continuity (in the form
=
asking
that the
density
at a
particle
is also in
velocity offlow of an incompressible fluid if and only
8.1
Corollary
particle
2.
The
Divergence and the Equation of Continuity
fluid
incompressible if and only if
of the fluid.
is
the
density
619
at a
is constant under all flows
Now the
integral \D div v dV is the rate of expansion of the fluid in D,
computation (8.14). (Hence, the name divergence.) We
could also calculate the "infinitesimal expansion" of D by calculating the
amount of fluid which enters during an
infinitesimal amount of time, and
subtracting from it the amount of fluid that leaves. The mathematical
expression of this will be an integral over the boundary of the domain D.
according
to our
"
The fact that this is the
is
its
a
"
divergence theorem, which
\D div v dV is the
same as
We shall return to this theorem and
fundamental fact in calculus.
in Section 8.5.
implications
Examples
Consider the flow
4.
x
=
x0(l
+
If D is the
A
=
{(*o(l
0
+
0>o
y
J'oCl
=
of
original position
+
0
ty0, >'0(1
+
the
given by
-
0
-
equations
+ tx0
a mass
0
+ tx0
z
=
z0e'
of fluid,
,
z0
e') ; (x0
,
y0
,
z0)
e
and the volume of D, is
f
Jot
dv=ttetJKx'y-z\dv
jd
z0)
d(x0
,
=
Since
dt
div
f e'(l
vol(Dt)
v(x, r)
2f2)
-
=
y0
,
dV
=
e'(l
-
22) vol(D)
\d div v dV for every
4 e'(l
~
ot
2'2)
=
W
~4t~
domain D,
2'2)
5. For this flow:
x
=
x0e'
'
y
=
y0e
z
=
z0e'
+
x0(l
-
e')
we
have
D}
Potential
8
Theory
in Three Dimensions
have
we
dx
T~
(-*o
=
e
~~
,
y
e
,
Z0
e
~
x0
e
)
v(x, 0 (x, y,z xe ') and div v
1 vol(A), so vol(Z),)
D, (5/50 vol(A)
at
time t, the equation
function
density
so
=
=
=
find p in terms of its initial values. Let
Then, according to (8.13), if p(x0 t) is the
,
dp
^
p-l
+
p(x0 0)
=
=
,
Thus, for any domain
1.
=
e'
vol(>).
If
of
p(x, 0 is the
allows
us to
continuity
be
p(x0 0) p(x0)
given.
particle density, we have
=
,
0
p(x0)
Thus
p(x0 0
=
,
6.
a
=
P(x0)e-'
Suppose
an
{a1, a2, a3).
Then the
speed
v(x, 0)
0(x)a
where
=
cj>
we
have
div
v
=
is
a
p(x, 0
=
e-'p(xe-', ye',
z
x(e"'
-
-
1)0
incompressible fluid flows steadily in the direction
is, the path lines are parallel to the vector a.
constant along the paths.
For the velocity field is
That
is
scalar function
(the speed), and since
fll
"2
fl3
S
|4
|^
5xx
5xz
5xJ
+
+
=
v
is
divergence free,
0
But then
#(x)(a)
=
<Vtf>(x), a>
for all x,
so
the
of motion.
paths
cj)
=
0
is constant
along
the lines
parallel
to a; but these
are
8.1
Integration Under
jd
g(x,
z)
y,
(u,
w)
F(x, y, z) be an orientation-preserving change
in the domain D in x, y, z space.
Let A = {F(x, y, z) :
of coordinates valid
(x, y, z) e D}. If g
c
621
Coordinate Change
a
Let
Theorem 8.1.
Divergence and the Equation of Continuity
is
dx
a
v,
=
function
dy
continuous
c
dz
=
ja
g(F~\u,
on
v,
D, then
d(x
w) det
'
v'
z")'
v,
w)
,
d(u,
du dv dw
The
proof consists in a series of reductions terminating in the oneenough to show that for any point peZ), this theorem is true
for some rectangle centered at p. For, once this is shown, we may cover D by
finitely many such rectangles RU...,R. If {pi,
pn} is a partition of unity
subordinate to {Ri,
The theorem is thus
R}, then pt g is zero outside Rt
true for each pi- g.
Summing over j, we obtain the general result.
Thus we may concentrate our attention on a particular point p0 in D, which we
take to be the origin. If the theorem is valid for the coordinate changes u
F(x),
y
G(u), then it is also true for the composed mapping y G(F(x)), simply because
Proof.
variable
case.
It is
.
.
.
,
...,
.
=
=
=
8{x\ x2, x3) 8{u\ u\ u3)
8{x\ x2, x3)
'
=
8{u\ 2, 3) {8y\y2,y3)
8{y\y\y3)
We will
decompose
our
mapping into a composition of four special cases, for each
The general result will follow by composing these
of which the theorem is easy.
mappings.
First of
all, let
W
Then F
=
identity.
(F
T be the linear
8{x,
y,
z)
8{u,
v,
w)
T)
T"1 and F
mapping
(x)
(U,
The theorem is
V.
w) = 0
easily
T has the
seen
only prove it for FT.
Our situation is now this: we
we
property that its Jacobian
to be true for a linear
at 0 is the
mapping (Problem 5),
so
need
G(x,
y,
z) defined
8{u,
'
v,
w)
0(x,
y,
z)
at the
>)=I
It follows that
8{u,
y,
z)
3(x,
y,
z)
8{u,
v,
z)
8{x,
y,
z)
(0)
(0)
=
I
=
I
are
given
origin such that
a
change of coordinates {u,v,w)
=
622
8
Potential
in Three Dimensions
Theory
Thus, by the inverse mapping theorem there is a neighborhood B of 0 in which
(x, y, z), {u, y, z), {u, v, z), (, v, w) are all bona fide orientation-preserving coordinate
systems. If we denote the respective coordinate changes as follow
Fi(x, y, z)
F*{u, y, z)
F3(, v, z)
then F
only
we
F3
=
=
=
=
F2
(,
(,
(,
z)
z)
w)
y,
v,
v,
Each F<
Fi.
changes only
to prove the theorem for each
F(
shall do it only once.
we do our computation.
Now, here
u
h{x,
=
y,
one
coordinate at
a
time, and
Since the proof of each
.
case
we
need
is the same,
Let
z)
v=y
w =z
be
a
coordinate change defined
R
=
If
now
g is
J
{{u,
=
a
v,
w):
Let
u
=
h{x,v, w),a<,x<a, b<v<b, c^w<c}
continuous function
g{x,
y,
Now, according
J
rectangle
{-a^x<a, -b<y<,b, -c<z<c)
centered at the origin.
A
on a
g{x,
-a
z) dx dy
dz
=
on
j J j
to the theorem of
y,
R,
g{x,y,z)dx dydz
change of variable in
z)dx=\JH-a.y.z) g{h~l{x, y, z, v, w))
one
OU
(8.15)
dimension
(,
y,
z) du
Thus (8.15) becomes
r"
rc
J-b J-c
=
J*
[ (<->
8h~l
LJli(-a,i,,w)
9{h~l{u,
g{h '(,
v, w, v,
v, w, v,
w)) det
w))
-.
{u,
OU
.
V,
w)
(h,
v,
w)
du dv dw
du
]
J
dv dw
8.1
The last
Divergence and the Equation of Continuity
623
equation follows from
o{u,
v,
w)
S(x,
y,
z)
Idh
8h
8h\
dx
dy
8z
0
0
so
(8{x,y,z)\
j
J8{u,v,w)\-i
8h'1
(8h\-1
^(*^))-(*^)) =\Fx)
8u
EXERCISES
Compute the
1.
area
of these domains, using either spherical
or
cylin
drical coordinates :
(a)
y2 + z2 :xyz
(b) 1 >x2 + y2-z2>0
(c) x2 + y2 < z < 1
(d) a2x2 + b V + c2z2 < 1
2. Integrate / over the domain Z>:
D={x2 + y2 + z2<l}
(a) f{x)=x2y2z'
0<z<l}
D={x2 + y2<\
(b) f{x)=xyz
0<z<x2 + y2}
D={a2x2 + b2y2<l
(c) f{x)=x2+y2-z2
Z
{0<x(x2 + ^2 + z2)<l}
(d) /(x) / sin2 0 cos2 <
=
=
3. What is the
0 <
z
<
whose
mass
of
a
parabolic section:
a{x2 + y2)
is
density
4. Find the
proportional
mass
to the distance from the xy
of the ball of radius 1, whose density is
plane?
p(x)=(l
+ r)_1.
5. Let
x
=
x0
y=yoe'
+ ty0
tz0
z
=
z0 e
'
+ fx0
equations of a flow in space.
(a) Compute the velocity field v(x, t).
(b) Compute the divergence of the flow.
the
(c) Assuming an initial density function which is constant, find
density function p(x, t).
t
1 ?
(d) What is the mass of the fluid in the unit cube at time
is
flows
fluid
these
Which
of
incompressible?
6.
(a) v(x, /) =( z, x, y)
(*2 x2, z y, z)
(b) v(x, 0
(c) x=x0e' + (l-0.>'o,J'=}'oe-*'2 + (l-Ozo,z e-"2zo
z
azo(l + 0
x0 sin /
y
y0 cos t
(d) x x0cosf + ;y0sin/
sin
cs
r>
ze')
',
*.y
0
C*
(e) v(x,
be the
=
=
-
-
=
=
=
=
=
624
Potential
8
Theory
in Three Dimensions
1. Find the volume at time t
sphere under these flows:
(a) Exercise 6(a).
8. Show that
potential
=
1 of the
mass
of fluid
originally in the unit
(b) Exercise 6(c).
(c) Exercise 6(d).
only if it is the
incompressible flow. {Hint: div V/= A/.)
C2 function in R3 is harmonic if and
a
of the vector field of
an
PROBLEMS
1
A radial field is
.
v(x)
=
field of the form
^(l|x||)x
Find all
incompressible
2. If L is
a
point
at any
a
radial fields.
line in R3, a flow around the axis L is one whose velocity field
is tangent to the cylinder with central line L. Show that the
flow of Exercise 6(d) is a flow around the
which is incompressible.
=
x0
+
axis.
incompressible flow whose path
3. Find the
x
z
u
y
=
y0 + sin
z
u
=
Find another such flow
lines
are
=
e
o--1
=
curves
z0
4. Find the incompressible flow whose path lines
cylindrical coordinates)
z
the
are
the
curves
(in
0O
(see Figure 8.1).
5. Prove Theorem 1 for the coordinate
nonsingular
6. In the
u
=
was
h{x,
y,
proof of
u
=
T(x),
where T is
a
Theorem 1,
a
function
z)
found.
It
was
Express 8h~1j8u in
8.2
change
linear transformation.
tacitly
8hj8x > 0. Why is that so ?
original functions (, v, w)
G(x, y, z).
assumed that
terms of the
=
Curl and Rotation
The
divergence
of the
of the fluid in flow
its rotation around
x
=
velocity field
as we
a
<t>(x0,0
have
seen.
given axis.
of a flow
measures
We shall
now
the rate of expansion
compute
an
indicator of
Suppose
(8.16)
8.2
Figure
Curl and Rotation
625
8.1
is the equation of motion of the flow. Let x0 be any
point, and n a direction
We shall compute the average angular velocity
(unit) vector at the point x0
in the plane orthogonal to n at the point x0 in terms of the
field v.
.
velocity
We take
x
0 for convenience.
=
Since
we are
interested in the motion around
the axis n, relative to the motion of 0, we must work in coordinates relative
to 0.
What is the same, we shall subtract from the above motion a motion
of translation by the image of 0, so that 0 remains fixed.
Since translation
involves
no
Thus
replace (8.16) by
we
x
so
to
Cr
,
moved to
to
a
Cr
a
a
=
our
computation
<Kx0 0
a
=
,
motion the
circle of radius
be
will be valid for the
original
motion.
the flow
r
<K0, 0
origin
is fixed.
centered at 0
lying in the plane IT(n) orthogonal
time t, the particle originally at a has
point
Cr
a onto the line tangent
Let L be the projection of \|/(a, 0
on
.
After
(8.17)
a
\|/(a, 0(see Figure 8.2). Let 9{t) be the angle at 0 in n(n) between a and
Thus 9{t) is the angle in the plane orthogonal to n through which a
at a
+ L.
our new
be
Let
n.
x|/(x0 0
=
that in
Let
rotation,
626
Potential Theory in Three Dimensions
8
Figure
has moved
fl,rt
9{t)
(relative
sin
=
to
-XL
0) during
.
=
sin
8.2
Thus
the time t.
_t<x|/(a,Q-<|/(a),T>
when T is the unit tangent vector to Cr at a. Dividing by
t -> 0, we obtain the angular velocity for the particle a in Il(n)
L
(-.-'?)
=
according
to
(8.17).
=
j
The
sum over
<v(a, 0)
all of
This number, calculated for small
as
fast
as r
-
Cr
n
r
If
of this
angular velocity is called
circ(Cr). Thus
ds
gives
at 0.
Cr
and is denoted
v(0, 0), T>
-
rotation of the flow around
to zero
letting
<v(a,0)-v(0,0),T>
the total circulation of the flow about
circ(Cr)
and
f(a,0),T
(l-fL/r2)1'2
t=0
t
as
(8.18)
us some
we
idea of the instantaneous
suitably normalize ((8.17) tends
0), and take the limit as r -> 0 we will have the same kind
given by a point function, rather than a function
of information, but it will be
of circles.
Definition 2.
each
point
x0 in
Let
v
be the
D, and unit
velocity field
vector
n
of
a
flow in
a
domain D.
define the curl of the flow about
n
For
at x0
8.2
Curl and Rotation
621
to be
i
/
curl
>
v(x0 n)
i=
,
hm
circ(C-)
r-o
where
~
(8.19)
r-
Cr is the circle of radius
r
centered at x0 in the
plane orthogonal
to
n.
Example
7.
x
=
Let
Consider the flow
x0
us
v{x,t)
cos / +
take x0
=
{y,
y0 sin /
=
-x,
(Figure 8.3)
y
(1, 0, 0) and
=
n
y0
=
E3
1)
Figure
cos t
8.3
.
-
x0 sin t
Then,
as we
z
have
=
z0 + t
already
seen
628
Potential
8
If
Cr
we
=
in Three Dimensions
Theory
take
1 +
\x
r
cos-, y
=
sin
r
r
-, z
=
0
r
then
;irc(Cr)
j
<v(x)
=
I
/ I r sin
=
f
=
v(l, 0, 0), T>
-
r
*(-r>i0=
ds
cos-, 0
J, (-sins,
cos
s,
0)\ ds
-27ir2
Thus the xy
sense
constant
plane rotates around (1, 0, 0) in the negative
angular velocity), as t changes. If now we take
n
=
(with
E1; we
have
Cr
=
!x
1,
=
l
y
/Ir
I
=
Thus there is
in the
plane orthogonal
Thus n
a x p,
For
a
time
cos
?",
-
-
,
1, 1
J, 1 0,
r
-
sin
-
,
r cos
-
J
) ds
plane.
=
=
to
=
=
sin-
shall compute the curl explicitly in terms of the velocity field v.
0 and let a =(a1, a2, a3), p
(/J1, jS2, /J3) be two unit vectors
Again
n
r
=
rotation in this
no
take x0
basis.
z
0
=
we
cos-,
r
circ(Cr)
Now,
r
=
(a2/?3
we
-
n so
that
a
-
p
-
n
is
a
right-handed
orthonormal
so
a3/?2, a3^1
a1^3, a1^2
-
-
<x2pl)
shall compute relative to this basis.
(8.20)
Cr has this parametriza-
tion
s
x
=
x(s)
=
r cos
-
r
s
a
+
r
sin
-
r
p
(8.21)
Curl and Rotation
8.2
629
The tangent vector is
s
T(s)
Expanding
=
sin
-
s
a
-
+ cos
r
the
v(x, 0)
r
field in terms of this basis :
velocity
v(0, 0)
-
P
-
=
v"{x)a
t>"(x)p
+
+
t;"(x)n
Then
circ
(Cr)
J"
=
<v(x, 0)
-
v(0, 0), T(x)>
ds
Cr
ft- V{x{s))
=
Now, substitute 9
s/r
=
sin
in the
+
-
v\x{s)) cos
and
integral
-
j
ds
approximate
(8.22)
the v"
(v
=
a,
/?)
by their differentials:
v\x{9))
=
vv{0)
dv\0){x{9))
+
+
e\\\x\\)
where
||x||-V(x)->0
Since
t>v(0)
=
||x||
=
these
we
have
dv\0){a) +rsin9- <fov(0)(P)
9
r cos
(8.23)
->0
0, using (8.21) for x{9),
t>v(x(0))
Substituting
as
expressions
into
(8.22),
we
=
ev(||x||)
obtain
,.2*
circ
(C,)
=
f
Jo
+
[- dv*{0){<*)
f
JQ
+
du"(0)(p)]r2
=
Jo
Jo
(-a(x)
cos
7tr2[-rfi>'(0)(p)
+
r
,
9 sin 9>de
"[-dt)"(0)(p) sin2 9 + dv\0){a.) cos2 9~\r2 d9
r2n
+
cos
\ *[-e"(x)
Jn
+
9 +
e"(x) sin 9)r d9
M0)()]
cos
9 +
e"(x) sin 0]
d9
630
Potential
8
in Three Dimensions
Theory
Dividing by nr2, and letting
(8.23)
and
curl
This
can
r
->
0, the second
disappears because
term
of
obtain
we
v(0, n)
=
dv\H>){<)
<foa(0)(P)
-
(8.24)
be rewritten in terms of the vector
=
<v(x, 0)
v(0, 0), >
-
Let
v
=
(v1, v2, v3)
in terms
Then
of the standard Euclidean coordinates.
v*{x, 0)
n.
[V(x, 0)
=
-
i>;(0, 0)]a;
so
dC(0)(p)=
I^o1
i=l
OXJ
Similarly,
dv\){)=i^
(8.24)
can
be
expanded
3
out
as
3
dv'
dv1
curlv(0,n)=I^aT-I^^
3
i= i
dv1
Referring back
to
derived from
with the
definition and
Definition 3.
R3,
we
a
.
dx3
2
v
.
(8.20)
fl.,3x
we
see
,^3
that this is the inner
given unit
proposition.
If
v
=
(v1, v2, v3)
defined the vector field curl
/dv2
dv3
dv3
vector
is
a
v
by
dv1
n.
gpl
product of
a
vector
We collect these results in
a
vector field defined in a domain in
dv1
dv2\
curlv=l^-^'^-ax^-^j
,
(8>26)
8.2
2.
Proposition
// v
around the direction
Proof.
is the
velocity field of a fluid flow, the
given by curl <v(x0 0, n>.
at time t is
n
A flow with
curl
631
of v
at
x0
,
Equation (8.25) is just <curl
Definition 4.
Curl and Rotation
field
velocity
n>.
v,
v
is called irrotational if curl
v
=
0.
Examples
8. Let
curl
v
=
v(x)
=
{-y,
x,
in
1) (as
Example 6).
(0,0, -2)
Thus for any plane n
{p: <p x, n>
in that plane has angular velocity -2<n,
rotation is about the z axis.
=
=
-
In
x
Then
general, curl v(x) spans the axis of the
and its magnitude is the angular velocity.
"
0} through x, the rotation
E3>. Thus the maximum
infinitesimal
"
rotation about
9. Let
X
=
x0{l
be the
+
t)
+
y0{l
equations
of
-
a
e')
y
flow.
The
y0e-'
=
z
=
velocity field
z0(l
+
t)
is
thus
-e'-(2
+
0e2'\
cu,1,(,,,0=(0.0.-''-1^)
so again the rotation at any point is about the z axis.
Notice that the
1
We can consider that as the initial
equations break down at t
1 spinning off the xy
point of the motion : the fluid came, at t
plane with infinite angular velocity.
=
.
=
The form of curl
previous chapter.
v
-
recalls the discussion of closed and exact forms in the
If
we
consider
the
associated to the vector field v, then curl
v
differential
=
1-form
co
=
<v, dx}
0 is the necessary condition for
632
co
Potential
8
to be the differential of
In
sufficient).
by the field is
We
function
a
particular, if
the
(and by Poincare's
field
lemma it is
is conservative, then the
flow
locally
induced
irrotational.
make
can
in Three Dimensions
Theory
physical
d\/dt
acceleration field
a
sense
=
Newton's law this is
of this statement
by referring it to the
velocity field. By
of the flow rather than the
essentially the field of forces which generates the flow.
have seen, if this field is conservative, then the work done by the flow
in moving a mass from one point to another is precisely what is needed; it
As
we
is the
can
be
the
same as
in energy level.
For this to be the case no work
the
mass
; hence the field is irrotational.
rotating
change
expended in wastelessly
In the
of
theory
electromagnetism
the existence of two fields, the electric
E, and the magnetic H, is postulated. Certain relations between these
fields, corroborated by experimental evidence form the basic laws of the
subject.
These
Maxwell's
are
equations.
Two of these
are
5H
curl E +
a
=
dt
{cr
a
div H
0,
=
0
constant), which state that the rate of change of the magnetic
by the rotation of the electric field, and that the mag
suitable
"
field is determined
netic flow" is
incompressible.
important relations
divergence which are easily derived.
Here
several
are
curl
V/=
div curl
div
v
=
between the
gradient, curl,
and
0
(8.27)
0
(8.28)
(8.29)
V/= A/
curl/v =/curl
v
div(/v)=/divv
x v
(8.30)
<V/,v>
(8.31)
+ Vf
+
Example
10.
A
=
Suppose
{-x,0,y)
is the acceleration field of
motion, assuming
divergence
an
a
initial
fluid in motion.
velocity
and curl of the flow.
field of
Find the
equations
of
(0, 1,0), and find the
8.2
If
x
is the
<b(x0 0
=
,
<b(x0 0)
=
equation of motion,
Curl and Rotation
we
633
have
x0
,
deb
-^(x0,0)
and
(0,1,0)
=
solves the differential
<b(x0 /)
,
{x,y,z)"
{-x,0,y)
=
The
general solutions
x
=
A0
y
=
Ai+ B^
z
=
A2
cos t
+
+
B2t
B0
+
sin
=
y
=
z
=
x0
t
+
give
these
as
the
equations
of motion:
cos t
y0 + t
t2
The
are
^-t2 -jt3
The initial conditions
x
equation
*o + y0
velocity
V(x, 0
divVfx, 0
t3
+
2
=
y
field is
(-xtanr, \,ty)
-tan/
=
curlV(x,/)=(-/,0,0)
n/2 the holocaust arrives.
moving generally in the positive
Notice that at
our
fluid is
/
=
(/
<
0)
and back
y
direction, rotating
parallel to the x axis and spinning
again toward it when / > 0.
clockwise around the line
from it
Before that moment,
away
634
8
Potential
Theory
in Three Dimensions
EXERCISES
Compute the curl for these fluid flows:
z
tz0
z0e~' + tx0
(a) x x0 + 0>o
y=y0e'
(b) v{x,y, z)
{-z, x,y)
(c) v(x, y, z) {y, z, x)
(d) The flow described in Exercise 6(b).
(e) The flow of Exercise 6(c).
(f ) The flow of Exercise 6(e).
10. Verify Equations (8.27)-(8.31).
1 1 Find the equations of motion and analyze the flow as in Example 8
given this acceleration field and initial velocity :
V(xo)=0
(a) \ {-y,x, 1)
(b) A
(x, z, x)
V(x0) (0, 0, 1)
12. Compute the rotation at x0 about the E2 axis for the flow of Ex
ample 6.
9.
=
=
=
=
.
=
=
=
PROBLEMS
Suppose we are given a time-independent field of forces F in a medium
1 ).
density (say
By Newton's law the fluid will flow according
the equation F
A.
Let D be a small ball of fluid. The kinetic energy
7.
of constant
to
=
=
of D at time
z
Z-
f
/
is
llv||2rfK
Jd,
where
v
moving
is the velocity field of the flow. Show that the work done by F in
Dk is equal to the change in kinetic energy. {Hint :
D to
0/g,(||v||2)
=
<v,F>.)
Verify these identities :
(a) curl gVf= Vg x V/
(b) curl/V/=0
8.
9. Show that if u,
v are
curl-free vector fields, then
u x v
is
divergence
free.
10. Show that in
is
11
x
a
ball,
a
vector field is
a
gradient if and only if its curl
zero.
=
.
Let M be
a
3
x
3
matrix, and consider the flow
exp(M0x0
(a) Compute the divergence and curl of the velocity field of the flow.
(b) Show that the flow is divergence free if and only if tr M 0
(c) Show that the flow is curl free if and only if M is symmetric.
=
8.3
Surfaces
635
12. Consider the flow
x
exp(M/)x0
=
where M is
symmetric matrix
that the velocity field of the flow is conservative and has
the potential function
a
(a) Show
n(x)=-<Mx,x>
(b) Show that the flow in an eigenspace with eigenvalue a is in a
straight line either toward the origin {a < 0), or away from the origin
(>0).
(c) Diagram the flow lines for such a flow in the plane in case the
eigenvalues (i) are the same; (ii) have the same sign; (iii) have opposite
signs.
8.3
Surfaces
A surface in R3 is
have been
the notion in this
text) a subset
point has some
neighborhood which can be put into one-to-one correspondence with a
domain in the plane. We shall assume that this correspondence is smooth.
It is given by a continuously differentiable mapping with a nonsingularity
(as
we
of R3 which is two dimensional.
condition
on
(i)
(ii)
a
x
map
we mean
that every
its differential.
Definition 5.
under
using
By this
x
=
A surface patch in R3 is the
x{u, v) with these properties :
image
of
a
domain D in R2
is one-to-one.
continuously differentiable.
dx/du, dx/dv are independent at every point, {u, v) are
called the parameters for the surface patch. The curves u
constant, and
v
constant are called the parametric curves.
(iii)
x
is
The vectors
=
=
A surface is
a
set I in
is, every point
patch.
p
on
Notice that if
we
R3 which
has
fix
u
=
a
c,
can
be covered
neighborhood
by surface patches, that
N such that Z
then the function
J>(t?)
=
n
N is
a
surface
x(c, v) parametrizes
a
636
Potential
8
curve
(since <b is
d<b
dx
dv
dv
Theory
in Three Dimensions
also one-to-one and
is
everywhere nonzero). The vector dx/dv is thus the tangent vector to the
parametric curve u constant. Condition (iii) asks that the curves u c,
v
c' at any point have independent tangents. Another way of phrasing
=
=
=
(iii)
is that the 2x3 matrix
du
dx
w
has rank 2.
Examples
11. The
sphere: x2+}>2
write
(0, 0, 1)
(1 x2 j2)1'2. Thus
surrounding (0, 0, 1) :
we can
x
=
x{u, v)
=
+
z as a
we
z2
=
l
(Figure 8.4).
function of
can
use
{u,v,{l-u2- v2)112)
Figure
8.4
Near the
point
and y on the plane: z
to
define a surface patch
x, y
x
=
8.3
Surfaces
637
Figure 8.5
which coordinatizes the upper
disk u2 + v2 < 1
Since
hemisphere
as
u,
v
range
through
the
.
dx
x
=
x
=
(i,o, -(i- 2-2r1/2)
=
du
^
(0,l, -V(l-u2-v2)-"2)
=
dv
independent. Every point on the sphere can be put
patch, by permuting the roles of (x, y, z) above.
the
example,
point ( 1 0, 0) lies in the surface patch given by
these vectors
in such
For
x
=
are
surface
a
,
x{u, v)
Spherical
=
(-(1
u2
-
coordinates
v2)1'2,
-
can
u,
u2
v)
+
v2
<
1
be used to coordinatize the whole
sphere
except for the points (0, 0, 1):
x
=
x{9, d>)
12. The
a2x2
+
is also
z=
(cos 9
=
cos
d>,
cos
9 sin
<f>,
sin 9)
ellipsoid (Figure 8.5)
b2y2
+
c2z2
=
1
easily parametrized by spherical coordinates (again except
+C"1):
(cos
cos u
u cos v
'
a
b
sin
v
sin u\
,_W
for
638
Potential
8
in Three Dimensions
Theory
Figure 8.6
13. The
paraboloid
coordinated by
is
x
=
x(u, v)
Since x
=
14. The
=
{u,
V,
U2
z
xv
cone
(x2
=
+
y2 (Figure 8.6)
is
a
surface
patch: it
V2)
+
(1, 0, 2u),
z
x2
=
=
(0, 1, 2v), they
+
are
independent.
y2)112 (Figure 8.7)
can
be coordinatized,
except for the vertex, by
x
=
x(u, v)
We
might
vertex
of the
=
(,
n,
{u2
+
v2)1'2
u
# 0,
v
# 0
ask if there is any way to coordinatize a neighborhood of the
cone.
It is quite difficult to show that there exists no function
which does so, but there is one important implication of the differentiability
of such a function which is easy to check out. The differentiability implies
good approximability by
existence of
a
linear functions, thus we should anticipate the
(a plane) which comes "nearest" the surface at
linear surface
given point. This is the tangent plane; which we shall now describe by
limiting arguments as in the case of the tangent line to a curve.
Suppose p is a point on a surface and q, r are two nearby points. The
three points p, q, r (in general) determine a plane. As q, r tend to p, this
plane will (in general) attain a limiting position: this is the tangent plane.
We now compute this process with coordinates.
Suppose the function
x
Z
near
coordinatizes
We
x(w* u2) {u1 ,u2)eD
may assume p
x(0, 0)
p.
0.
Let q
r
The
x{u\ u2),
x{vl, v2).
plane Il(q, r) through p, q, r is then
a
=
=
,
,
=
=
=
8.3
the set of all vectors
q
x r
=
perpendicular
x{ul, u2)
x
In order to take the limit
x{u\ u2)
=
Xl(0V
to
x{v\ vz)
we
+
639
Surfaces
(8.32)
approximate
x2(0)m2
+
x
by its differential
(||u||)
"
where /
q
where
*e(0
x r
we
important
-
=
0
as /
(Xl(0)
-?
x
0.
Equation (8.32)
x2{$j){ulv2
have combined all the
hV)
-
error
becomes
+ R
terms in the
(8.33)
expression
R(u, v)
=
||u|M||v||)
+
||v||82(||u||)
+
The
83(||u||)b(||t||)
where the 8f all have the same behavior: /_1(0 - 0 as / -* 0.
Now, so as to treat the remainder R as an insignificant remainder,
be careful with the term ulv2 -u2vl.
case
R.
behavior of R is this:
we
must
It may, for example, be zero, in which
the remainder becomes very significant. Thus we must assume that
Figure 8.7
640
Potential
8
this terms tends to
wV-wV
it suffices to
tend to
q,
in Three Dimensions
zero more
r
slowly than
R
as
q,
r
-
p.
Since
sin((||u||X||v||)
=
that the
assume
zero as
by u^v2
Theory
->
angle
between the coordinate
vectors
does not
Then, under this assumption, we can divide (8.33)
n(q, r) as the plane through p orthogonal to the
p.
wV, obtaining
vector
Xl(0)
where R1
-*
orthogonal
x
0
to
x2(0)
as
q,
r
+
-*
{dx/du1)
R1
x
Let p be
Definition 6.
Thus the
p.
{dx/du2)
a
limiting position of LT(q, r) is
it is the plane spanned by
the
plane
at p:
point on a surface Z coordinatized by x x(m1, u2).
plane spanned by the vectors dx/du1, dx/du2
=
The tangent plane to X at p is the
at p.
Proposition 3. Let j>be a point on the surface Z, and let Il(q, r) be the plane
spanned by two points q, r on Z so that the angle between q p and r p is
nonzero.
If q, r - p so that this angle remains bounded away from zero,
then II(q, r) tends to the plane tangent to Z at p.
Of course the
angle assumption is crucial,
Problem 28 exhibits the
difficulty
obtained without it.
Examples
1 5. There is
z
=
{x2
+
no
tangent plane
to the cone
y2)1'2
(Figure 8.7). For, if we take qx
(/, 0, /), q2
(0, /, /),
plane spanned by qx and q2 is the plane spanned by (1,0, 1),
(0, 1, 1) for all / -> 0. Thus this is a candidate for the tangent plane.
However, if we consider now the points qt
( /, 0, /), %2
(0> U 0
for / > 0, the candidate we obtain is the plane spanned by ( 1, 0, 1),
(0, 1,1). Since these two planes are distinct, there can be no
tangent plane (Figure 8.8).
at its vertex
=
=
the
=
=
8.3
Figure
16. The
cylinder x2
by using cylindrical
+
x{u, v)
(cos u, sin
x
( sin u, cos u, 0)
x
(0, 0, 1)
x
=
=
=
y2
=
u,
641
8.8
1 is
coordinates
Surfaces
surface.
a
It
can
be coordinatized
:
v)
-
=
The tangent plane at x{u,
x x x
(cos u, sin u, 0).
v)
is the
plane orthogonal
to the vector
=
17. If
x
=
by its family
x
=
x{s, t)
=
x{s) is
the
equation of
of tangent lines is
x(s)
+
/T(s)
a
a
curve, the
surface.
It is
"
surface swept out"
parametrized by
642
Potential
8
Theory
in Three Dimensions
We have
x5
T(s)
=
//cN(s)
+
xf
=
T(s)
so long as k # 0, s, t are patch coordinates for all s, t > 0.
This
surface is called the developable defined by the curve. Its tangent
plane at the point {s, t) is the same as the osculating plane to the
Thus,
at
curve
x{s).
surface, and p a point on the surface. We shall denote the
by T(p). If x x{u, v) parametrizes Z in a neighbor
tangent plane
hood of p, with p
x{u0 v0), then the vectors dx/du{u0 v0), dx/dv{u0 v0)
the
span
plane 7Xp). The inner product on R3 induces an inner product
It will be valuable to us to see how to
on this plane just by restriction.
If t
this
inner
in
terms
of the basis xu x
express
product
axu + bxv is
a vector in T{p) its length is given by
Let Z be
a
to Z at p
=
=
,
,
,
||t||2
Suppose
x
Let
by
<t, t>
that C is
=
=
=
fl2<x, x>
a curve on
Z.
+
2ab(xu, x>
Choose
a
+
=
.
2>2<x, x>
parametrization
of C:
(8.34)
0^s<L
g{s)
{u{s), v(s))
x
and
=
,
be the
(w, v) coordinates of g(s).
Then
(8.34) is the
same as
(8.35)
x{u{s), v{s))
the chain rule, the tangent to C is
du
T
=
dv
xu + xvds
ds
and
||T||2
We shall
=
use
<x,
these
x>(^)2
+
2<xu, x>
following notational
f^
+
<x,
x>(^2
(8.36)
conventions relative to coordinates
onZ:
=<xu,x>
F=<x,x>
G
=
<x,x>
(8.37)
8.3
In terms of this notation
puting
the
lengths
4.
Proposition
Let C be
the
of
have this way, intrinsic to the surface, for
curves on
Let I. be
Z
a curve on
length of C
we
a
f
com
Z:
surface patch parametrized by x x(, v).
x
x{u{t), v{t)). a<t<b. Then
=
parametrized by
=
Jdudv\2
ldv\2
1/2
J.Hsr) +2Fta*) +GU)
The
643
is
f\Jdu\2
Proof.
Surfaces
length of
(8.38)
dt
C is
IITIIA
Jb
which is,
by (8.36), given by (8.38).
(borrowed from the differential form
integrand
gives arc length along a curve. This
the length of any curve C is Jc ds. According to (8.38) we
adopt the
We shall
convention
that ds is the
notation)
means just
can
that
which
be assured that
\r,(du\2
,
for any parameter
ds2
=
/
Edu2
Definition 7.
=
C2:u
=
=
v
u2{s)
v
==
X
can
also write this
dv2
=
=
dvt
r
ds
as
(8.39)
vx{s)
v2{s)
are
du!
1
We
(8.39), where E, F, G are given by (8.37) relative to a
x(u, v) on Z is called the first fundamental form of Z.
curves given parametrically by
uy{s)
then their tangents
1
C.
+ 2F dudv + G
parametrization
If Cu C2 are two
u
along
,
The form
x
C\ :
(dv\2V-12
nJdudv\
X
ds
T2
du2
=
X""ds"
+
dv2
X""ds"
644
At
8
a
Potential
point
in Three Dimensions
Theory
of intersection p the vectors
product is
^(p), T2(p)
lie in the tangent
plane
at p and their inner
,,
The
du*l dui2
.
,
<T1; T2>
curves are
E
=
ds
]7/du1
dv2
du2
dvA
\ ds
ds
ds
ds
ds
orthogonal
at p if
<T1; T2>
ds
ds
0
=
Proposition 5. The parametric curves u
surface patch are orthogonal if and only if F
constant,
0.
=
=
Proof.
dvx dv2
j
v
constant
=
on
The tangent line to u
c is spanned by x; the tangent line to v
by x. These lines are orthogonal if and only if <xu, x> F= 0.
=
is spanned
a
=
c
=
Examples
18. The
have ds2
joining
plane
dx2
=
a to
\\f\t)2
we
we
=
0.
dy2.
2-i
rectangular coordinates we
0<s<L, is any curve
g{t),
=/(/),
(as in Chapter 5) the length of L is
curve
by
x we
obtain the
as
dy/dx
=
0; that is, when the
curve
is
a
straight
This conforms with known facts.
line.
19. The
x(w, v)
Here x
=
cylinder
=
(
-
(cos
u, sin u,
sin u,
cos
u,
v)
0),
x
=
(0, 0, 1).
Thus E
so
ds2
length
dx
This is minimized when
=
=
y
1/2
Ch
x
x
dt
parametrize this
.
In the standard
If
have
g'{t)2T12
+
Jn
'o
If
b
z
+
=
du2
Again,
+
the
dv2
length
of
a curve
1/2
fl"*
du
given
as v
=
v{u)
is
=
1
=
G, F
=
0,
8.3
the
Surfaces
645
of minimal
length (called geodesies) on the cylinder are
represented by straight lines in the u, v coordinates. Thus the
typical geodesic on the cylinder is the helix
so
curves
those
x
/, sin /,
(cos
=
20. For the
x
x{u, v)
=
=
have E
we
ds2
=
Once
du2
=
sphere
(cos
1,
F
cos2
+
again,
\yds.
at)
u cos
0, G
=
u
v, cos
=
sin v, sin
u
cos2
u.
u)
Thus
dv2
discover the
we
Let a, b be two
geodesies by minimizing the integral
the sphere; by rotating the sphere
on the longitude v
0.
If y is any curve
points
on
may suppose that a, b lie
joining a to b, the length of y is
we
J
ds
The
=
J
{du2
=
^a
Now
v
cos2
of the
length
f {du2)112
+
f
u
=
dv2)1'2
longitude {u
=
(8.40)
0)
is
du
(8.41)
^a
(8.40)
0 along y; that is,
is always larger than (8.41) unless dv
Thus it is the longitude which is the curve of the
=
is constant.
shortest distance between
clude that the
planes:
geodesies
a
on
By rotating back again we con
sphere are the sections by diametric
and b.
the
the great circles.
Geodesies
The
problem
of
finding
the
geodesies
on
any surface is
more
difficult,
because the general form
Edu2 +2Fdudv
is harder to
analyze.
+
Gdv2
One way to
proceed
is to try to find coordinates so
examples: it has the
that the first fundamental form looks like the above
646
8
Potential
Theory
in Three Dimensions
form
ds2
=
du2
When this is the
+ G
dv2
(8.42)
that the
constant are geodesies
(Problem 17). However, in order to find such coordinates, we must know
what we are looking for; that is, we must know how to find geodesies in the
first place. Thus, this line of reasoning has to be supplemented by the
discovery of a characteristic property of geodesies. We seek such a charac
behavior of a
teristic property by trying to understand the
infinitesimal
geodesic: this (we hope) leads to a differential equation which is solvable.
Then we can carry out our original plan: solving the differential equations
will provide a convenient coordinate system in which we can discover the
curves of minimal length.
We shall, however, not carry through the entire
shall
only derive the basic property.
program here; we
If y is a geodesic, a curve of minimal length, on the surface Z, then,
relative to Z it is a straight line. That is, it would have to be as close to a
straight line as it could be : it should bend only as much as it must in order
to remain on Z.
Thus the rate of change of the tangent, relative to Z, should
be zero. Infinitesimally this says that the normal to the curve has no com
ponent on the tangent plane to Z. We shall now show that a geodesic has
case we can
verify
curves v
=
"
"
this property.
Let y be
Theorem 8.2.
Z.
Then,
o/Z.
at
any
point
p
a
on
geodesic {curve of minimal length) on the surface
orthogonal to the tangent plane
y, the normal to y is
near p so that p
Let pey and let u, v be coordinates for
((0), t>(0)).
0 and so that the
We may choose these coordinates so that y is the curve v
coordinates are everywhere orthogonal (see Problems 9 and 10). Now let a be
Proof.
=
=
(a, 0) to {a, 0) in the uv plane lies on the
=/() defines a curve lying in D and joining
{a, 0) to {a, 0), then x x{u,f{u)), a <Lu<a gives another curve on S, joining
The length of T is no more than that of y, since y
two points of y (Figure 8.9).
is a geodesic.
small
enough
so
that the interval from
domain D of the coordinates.
If r :
v
=
Figure 8.9
8.3
647
Surfaces
We have not yet done enough to investigate the local behavior of y; we must
a whole family of curves including y rather than just one other.
But that
consider
is easy to do
T,:
let T, be the
:
=
x
x(, //())
a
minimum at
F(/)=[
J
is
certainly
at /
-
ot
=
0,
(if it is
so
Let
F(0
be the
differentiable) F'{0)
=
length of T,.
0.
We
now
Then
F{t)
compute this :
\\x + x.tf'{u)\\du
-\\xu + xvtf{u)\\du
t
-a
0, the integrand is
<x +
=
t
differentiable function of /, and
a
=
-a<u<a
-a
FV)=\1
Now,
parametrized by
y is T0 and T is IV
for -1 < / <1.
has
curve
x
//'(),
_
+
2<W(")
^ t-n
2 ||x||
=
x
x
+
f/'()>"2 1.=0
x
/'(), x>
/()
<*^>
llxJI
(8-43)
The last
equation follows from
<x0,x>
=0.
the
assumption
are orthogonal:
secondly, the expression (8.43)
that the coordinates
First, the second term drops out,
derives from
8
0
=
8u
<x x>
=
,
Therefore, from F'(0)
=
<x x> + <x xu>
,
,
0,
we
obtain
f{u)d=o
f'<E^p>
IIX.II
J-a
This
equation must hold for all differentiable
We conclude then that
<x,xM(>=0
functions /such that /(-a) =/()
=
0.
648
along
Potential Theory in Three Dimensions
8
y
(see Miscellaneous Problem 41 of Chapter 2). Now, the normal N to y
plane spanned by xu and xm. Since these are both orthogonal to x,
Further, N is orthogonal to the tangent line of y which is spanned by x
is orthogonal to both x and x so is orthogonal to the tangent plane of S.
is in the
N _L x
.
Thus N
.
,
Examples
21. Find the
Z:j
=
geodesies
parametrize Z by x
parametrize a geodesic T
=
the surface
x2
We
x
on
=
x{u, v)
Zt.
on
=
{u, u2, v).
Let
u
=
u{s),
v
=
v{s)
Then T has the form
(u{s), u*{s), v{s))
and
xs
xss
=
=
(', 2mm', v')
N
=
(, 2{u')2
For T to be
xu
=
+
2uu", v")
geodesic,
a
(1,2m,0)
x
=
this must be
orthogonal
to both
(0,0,1)
Thus, the functions u{s), v{s) parametrizing the geodesic T satisfy these
differential
equations
u" +
2u[_2{u')2
v"
0
=
+
2mm"]
=
0
Notice that from Picard's theorem the
equations
m=-4m(')2
1+4m2
v"
=
have
0
unique
there exists
point.
solutions
a
curve
given
the initial values of u, v, u', v'. Thus,
length in every direction, at every
of minimal
8.3
22. Find the geodesies
Z:z2
=
on
the
Surfaces
649
cone
x2+j;2
Notice that any
plane z + x cos a + y sin a b intersects Z at right
angles (Figure 8.10). Thus the normal to the curve of intersection is
orthogonal to the surface, and such a plane always intersects Z in a
geodesic. More generally, we can compute the equations for
any
geodesic using Theorem 8.2
Z:
x(, v)
x
=
xu
=
{
x
=
(cos
=
=
{v
vsinu,v
m, sin m,
cos m, v
cos u,
sin u,
v)
0)
1)
Figure
8.10
650
Potential
8
If
m
=
=
xs
xss
=
u(s),
{v'
v
Theory
v{s) parametrizes
=
vu' sin u, v' sin
cos u
{v"
2v'u' sin
cos m
v" sin
in Three Dimensions
+ 2dV
u
u
a
geodesic T, then
+ vu'
vu" sin
u
cos m
+ vu"
cos
u
cos m
u,
on
T
v')
v{u')2
cos
v{u')2
u,
sin m,
j;")
The differential
equations are readily computed (and hardly
explicitly) by expressing <xss, x> 0, <xss, x> 0.
=
Surface
solved
=
Area
We would like
to define the area of a surface in a way analogous to
length of a curve. We select a collection of points
xu
xk on Z and replace Z by the polygonal surface Z' whose vertices are
If the points xt,
X[
xt.
xk are very numerous and close to each
other, then the sum of the areas of the faces of Z' is a good approximation
now
the definition of the
.
.
.
,
.
to the area of Z.
such
sums as
everywhere
We
can
the set of
.
.
,
then try to define the area of Z to be the limit of
xt,
xk becomes infinitely numerous and
points
.
.
.
,
dense.
Now this definition
surface
unfortunately does
not
work, there
are
ways of
so
obtain any desired area (for a fuller account see
Rather than give it all up as a hopeless task because
partitioning
Spivak, pp. 128-130).
of this phenomenon, we try a different approach.
First, we study the
of
area
in
the
to
a
approximation
small.hoping generate plausible formula for
surface area (by plausible I mean that approximations to our formula are
also approximations to our notion of area).
If the formula turns out to be
that
of
intrinsic,
is, independent
parametrizations, then it will define a relevant
which
we
call
shall
surface
area.
measure,
Returning to the above approxia
so as to
"
Figure 8.11
8.3
651
Surfaces
mation," let F be one of the faces of Z', and x0 one of its vertices. Let F0
be the projection of F onto Z (see Figure 8.11) and Ft the projection onto
the tangent plane T{x0). If the surface is very smooth, then for small F
these three surfaces have
essentially the same area, and we can confuse the
We may suppose that F0 lies in a patch parametrized by x
x(m, v)
with x0
x(m0 v0). Let D be such that
three.
=
=
,
F=
Confusing
{x{u,v);{u,v)e D)
the surface with
x{u, v)
Now,
x0+
=
F,
we
xu(m0 v0)u
area
(F()
=
+
,
know how to compute
we
\\xu{u0,v0)
x
Chapter
1
Thus,
.
at
be the linear map
,
the
xv{u0,v0)\\
least
to
x
xv{u0 v0)v
area on
This is true because it is true for
28 of
may take
image
area
rectangles,
on
of
a
linear map:
D
as we
have
this coordinate
seen
patch,
in
the
Proposition
area
of Z' is
very close to
Z
||x(m;, i>;)
where the
x
xv{ut, vt)\\
{>,} partition
limit of such
f || x
sums
x
JD
area
{Dt)
the coordinate domain D and
Let Z be
Definition 8.
Din R2.
>D
If Z is
a
surface
||x
x
e
>,.
The
xj du dv
We take this to be the definition of surface
through
(u,, vt)
is
The
a
area
surface
patch
area.
with coordinates u, v,
ranging
of Z is
xj du dv
is
surface, partition Z into pieces Du..., Dk such that each Dt
patch.
area
(Z)
Define
=
area
{Dt)
a
652
Potential
8
in Three Dimensions
Theory
We must show that this definition is
Proposition
The above
6.
independent of the particular partition.
is
definition
independent of
the partition
of
Z
chosen.
Proof.
Suppose
S
is still
a
(fl,n Ei)
=
third
(E()
area
{Dj)
partition
{Di
u
partition.
area
since in each
also
we
n
=
vj
{Dk
u
n
:
H
Ei
u
u
E
Then
.
E)
Clearly,
2 area {Dj
=
E2)
S another way
n
2 area {Dj n
,)
(8.44)
,)
(8.45)
computing relative
case we are
to the
same
coordinates.
area
is the
summing (8.44)
are the lefts, as
over
(see Problem 1 8) to verify that the computation of the
whether it is done in the Dj or Et coordinates. Then,
We leave it
to the reader
same
i, and (8.45) over/, the right-hand sides
the same; and
are
so
of Dj
n
Et
desired.
In accordance with
length,
convention to denote ds
our
shall let dS denote the
we
integrand
as
for surface
the
integrand for
area.
of any coordinate system u, v we have dS
H du dv, where H
||x
It follows from Lagrange's identity (Chapter 1) that also H
{EG
=
=
=
Examples
23. Find the
{x2
+
We
use
x
{R
=
xu
=
x
=
so
y2
z2
(
R
=
u cos
cos u
[EG
-
v, R
-it
cos u
sin v, R sin
v, R sin
sin v, R
F2]112
u
cos u cos
=
'-11/2
du dv
v,
R2 |cos u\.
=
m)
sin v, R
-It/ 2
cos u
J
sphere
R2}
=
cos u cos
R sin
H
of the
spherical coordinates:
(
.71
R2
+
area
4nR2
cos
u)
0)
The
area
is
arc
Thus, in terms
-
x
x||.
F2)112.
8.3
24. The
{z
=
area
of the
x2+y2,
piece
of the
paraboloid
Surfaces
653
is
0<z<l}
The parametrization is
x
=
xr
(/
=
=
In
sin u,
cos u, r
u, sin u,
(cos
2, F
=
0, G
1)
=
=
x
r2,
H
=
(-/ sin
2r.
The
u,
r cos
area
u,
0)
is
.1
2rdrd9
'o
r)
=
2n
'o
EXERCISES
13. Let
/be a C function defined in a domain Z> in R2.
(a) Show that S: {z =/(x, y)} is a surface patch with coordinate x, y.
(b) Compute the first fundamental form and the area element for /.
(c) Show that the element area is given by sec y dx dy, where y is
the angle between the normal to 2 and the z axis.
14. Find the tangent plane, first fundamental form and area element for
these surfaces :
(a) The paraboloid x y2 + z2x2 + y2.
(b) The cone zz
(c) The hyperboloid z x2 y2
: x(, v)
(d)
{u + v2, v + u2, uv)
15. Find the length of the intersection of these surfaces:
1
(a) x2 + y2 + z2
1
*x2 + 2y2 + 2z2
(b) z2=2x2+>-2
z
x2 + 2^2
16. Find the angle between the parametric curves at a general point for
the surface given in Exercise 14(d).
17. Find the area cut off the tip of the paraboloid x2 =y2 + z2 by the
1
plane x + z
=
=
=
=
=
=
=
=
.
18. Find the
area
of these surfaces:
0<z<a.
x2 + y2
tt<u<tt
0<v<tt,
(b) S: x
(, cos , v)
(c) The part of the hyperboloid z x2 y2 inside the unit ball.
(d) The ellipsoid x2 + y2 + 4z2 4.
(a)
The
cone
z2
=
=
=
=
654
Potential
8
in Three Dimensions
Theory
PROBLEMS
13. Recall that
those
curves :
a
differential form M du + N dv determines a family of
0. If ds2
E du2
along which M du+ N dv
curves
=
=
+ IF du dv + G dv2 is the first fundamental form of
that the family of curves
a surface patch show
orthogonal to the family defined by M du + N dv 0
=
is determined by
{EN
du +
FM)
-
14. Let p0 be
patch
{FN
-
point
a
GM) dv=0
on
the surface S.
Show that
we can
find
surface
a
that the
parametric curves are orthogonal. {Hint: Let
u, v be coordinates near p0 and explicitly find the family of curves u
u{t, c),
v
v{t, c) orthogonal to the curves dv 0 such that (0, c) 0, v{0, c) v0.
Show that v, c are orthogonal coordinates.)
15. Let y be a curve on the surface S.
Find orthogonal coordinates
u, v at a point p0 on y so that (i) y is the curve v
0, (ii) u is arc length
along y.
16. Show that a cube is not a surface along its edges.
near
so
p0
=
=
=
=
=
=
17. Is ds
a
differential 1-form?
18. Find the differential
equations
for the
geodesies
on
the torus
(Figure
8.12):
x
=
y
=
z
=
(1
cos
(1
cos
sin
()sin 8
<^)cos 0
</>
19. Find those
planes which intersect
the
ellipse x2 + a2y2 + b2z2
=
1 in
a
geodesic.
20. Let
{(, v)
mental form ds2
6
R2
=
: u
>
0,
v
>
0} parametrize a surface with first funda
Find the equation of the family of
v2 du2 + u2 dv2
Figure 8.12
8.3
curves
orthogonal
to the curves
form in terms of these
21. Find
ds2
Surfaces
655
uv
constant, and express the fundamental
coordinates.
new
the geodesies
=
the surface with first fundamental form
on
du2 + /() dv2
=
22. Show that the
form Edu2 + G dv2
23. Let
S
: x
=
2
: x
=
be
a
curves v
constant
on a
surface with first fundamental
=
x{u, v)
(, v)e D
x{r s)
{r, s)eA
,
=
geodesies if and only if 8E/8v
0.
surface path with two different coordinates:
are
Show that
Sx
Sx
8u
dudv
8v
=
\8x
8x
8r
8s
J*
{Hint: Define u
u{s, t),v= v{s, t) by this property :
x
x{u, v) with u u{r, s), v v{r, s). Show that
=
if
=
=
3x
8x
t8x
8x\ 8{u, v)\
8r
8s
\8u
8vJ
=
x{r, s) if and only
8{r, s) J
The following problems use the normal to
N orthogonal to the tangent plane.
24. Let y be
x
=
a curve on
the surface 2.
a
surface
:
this is
a
unit vector
Let N represent the normal to
2, and T the tangent to y. The unit surface normal to y is the vector
N x T.
Ny
0.
(a) Show that y is a geodesic on Z if and only if <NV dT/ds}
<NV dT/ds) is called the
(b) In general, the inner produce Kg
geodesic curvature of y on S. Suppose (, v) are orthogonal coordinates
on S and Kg1, k92 are the geodesic curvatures of the lines v
constant,
constant, respectively. Verify Liouville's formula : the geodesic
curvature of the curve y is given by
=
=
,
=
,
=
=
jo
Kg
=
ds
+
Kg1
COS
where 0 is the
{Hint:
Write
T
cos
=
Ti
6 + Kg2
SU1
9
angle between the tangent
to y and the direction x.
0 + T2 sin 0
where Ti, T2
are
the tangents to the
curves v
=
constant,
u
=
constant.
Potential
8
Theory
in Three Dimensions
Then
dl
dTi
ds
ds
0 + -7- sin 0 +
ds
Substitute these
dd
dT2
n
cos
=
(-Ti
sin 0 + T2
cos
0)
ds
expressions into
*9=(g,NxT)
and evaluate at 0
25. If y is
=
0, 0
a curve on
=
n/2.)
the surface 2
we can
decompose dl/ds into its
components tangent to and orthogonal to the surface:
dT
K9Ny + /cN
=
as
where
kn
is called the normal curvature to T.
(a) Show
(b) Show
that the curvature of y is {k,2 + kn2)U2.
that the normal curvature of a curve y depends
tangent to y and is the same as the curvature of the
of 2 with the plane through T and N.
curve
only
on
the
of intersection
(c) Show that the curvature of the curve y is given by kn{T) sec 0,
angle between dT/ds and N.
Using Liouville's formula find the geodesic curvature of a general
on the surface obtained by revolving the curve z
exp( x2) around
where 6 is the
26.
curve
the
z
=
axis.
27. Let 2 be
zero
a
surface such that at every point every curve
Show that 2 is a piece of a plane.
on
2 has
normal curvature.
28. Let p be a point
to select q,
surface 2 and let q, r be two nearby points. It
to zero so that the plane determined by
p, q, r does not converge to the tangent plane (unless 2 is itself a plane).
For example, if y is the curve intersection of some plane II with 2 and if r
is
possible
on a
r
tending
follows q along y then the plane determined by p, q, r is always n, which
need not be the tangent plane to 2. Furthermore, if we move q slightly
off y we can be sure of the same behavior with the requirement that the angle
between q and
to
zero).
r
metrized
plane.
q
=
parametrization) is not zero (however, it must tend
explicit example. 2 is the surface z x2 para
(, v, u2). The tangent plane at p, the origin, is the xy
some
an
by x(k, v)
However, if
(2/,0,4f2),
=
r
=
=
(M2,/2)
plane determined by
(0, 1, 1).
then the
to
(in
Here is
p, q,
r
tends (as
/
-*
0)
to the
plane orthogonal
8.4
8.4
Surface
Integrals
Suppose that /
and Z is
is
Proposition
continuous function defined in
a
6 that the
We
can
following
coordinate choices involved.
Definition 9.
verify by
of/ over
\fdS=
657
a
domain D in
R3,
argument identical to that in
definition makes sense independently of the
Partition Z into subsets
Define the integral
and Stokes' Theorem
and Stokes' Theorem
surface in D.
a
Surface Integrals
an
Z1;
.
.
.
,
Z
of surface
patches
on
Z.
Z to be
j fHdudv
where H du dv is the surface
area
element in the patch
containing Z;
.
Examples
25.
I=\xx2y2z
y2
+
x
+
z2
Example 23,
1,
=
we
1 r"
-
26.
in
I
J-*
/
the
the
hemisphere
same
rn/2
v
sin2
cos5
v
sin
u
Z:
{(x,
parametrization
y,
z):
as
in
du dv
n
u
sin
u
du
=
J0
=
fj.(x
r2" r1
2
*n
cos2
m
sin2 2v dv
Example
=
is
Z
Using
have
cos5
4
0}.
n/2
j*'*.
nil
=
dS, where
z >
*n
+
24
24
y2) dS,
where Z is the
piece
of the
paraboloid given
24.
[r2
cos u
+
r3 sin2 m] dr du
n
=
-
2
Normal and Orientation
Let Z be a surface in /?3. The tangent plane to Z at a point x0 is a twodimensional plane, thus its orthogonal complement is a line, called the
a choice of unit vector lying
continuously with the point. Such a choice is
always possible locally, but is not always possible over the whole surface.
normal line to Z at x0
on
.
The normal vector N is
this line which varies
658
8
Potential
Theory
in Three Dimensions
band
moebius
Figure 8.13
Consider the surface
vertical sides
so
that
continuously
opposite
way to
in the
point
Notice that the
in
Figure 8.13 (called the Moebius band).
rectangle (Figure 8.14) by gluing together the
vertices with corresponding labels abut. There is no
depicted
This is obtained from
a
select
a
normal vector to this surface which does not
direction when traced around the circle in
Figure
8.13.
phenomenon is put in evidence by Figure 8.14:
a right-handed basis gets transformed into a left-handed basis when we cross
the vertical line. We express this by saying that the Moebius band is not
same
kind of
orientable.
Thus in two dimensions
dimension.
under
then.
a
We
change
But
we
at every
we
into the
find
same
of variable in the
a problem which does not exist in one
problem in the discussion of integration
plane, and we successfully sidestepped it
cannot avoid it now.
surface Z in R3
plane
ran
as
a
point.
choice of
We shall refer to
sense
of
an
orientation
on a
rotation in the tangent
This choice is assumed to vary continuously: that is,
a
positive
if vl5 v2 are nowhere collinear continuous vector fields defined on the surface
and the rotation vt -> v2 is positive at x0 it must be so in a neighborhood of
x0
.
A choice of orientation is
equivalent to a choice of normal vector.
positive rotation in the tangent plane
positive if Vj-^v^N is a right-handed system.
For, if a normal N is chosen
as
we
defined
follows: Vj-+v2 is
if an orientation is chosen
Vj x v2 where
If Z is oriented
v1( v2 are unit vectors and the rotation vx - v2 is positive.
and (, v) are coordinates on a patch in Z, we shall say that {u, v) is a positively
Conversely,
we can
define N
=
,
8.4
Surface Integrals
and Stokes' Theorem
659
oriented coordinate system if the rotation
x - x is positive. Here is a fact
relating positively oriented coordinate systems which completes the dis
cussion.
Proposition 7. If {u, v) and (', v') are two positively
systems defined on the oriented surface Z, then
d{u, v)
d{u', v')
oriented coordinate
>0
Examples
27.
If/ is a C1 function defined
graph
on a
domain D in the xy
plane,
then the
T{f):z=f{x,y)
is
surface
patch. We consider it oriented so that the rotation from
(1, 0, df/dx) to xy (0, 1, df/dy) is positive. Then the normal
vector N always points upward out of the surface {N3 >
0) :
x*
a
=
=
-h(ir+f)TH-g'
28. More
generally, we can always orient a surface patch Z: x
x(m, v), {u, v) e D by transferring the orientation from the u, v plane.
That is, we take x
Then
x as the positive sense of orientation.
=
->
the normal to Z is
N=
||x
x
xjr1(x
x
x)
Figure
8.14
660
Potential
8
Theory
in Three Dimensions
"
Just as, in the case of curves, we introduced the
vector length element"
dx
Tds along the curve, we introduce the vector area element dS
Nds on
=
a
=
Notice that, in terms of coordinates
surface.
dS
||x
=
du dv
xj
x
In this way
curves).
x dv
x
we can
If Z is
Definition 10.
xudu
product of the length elements along the parametric
integrate vector fields along oriented surfaces :
it is the vector
(i.e.,
=
oriented surface and
an
around Z, define the flux of
v across
Z
v
is
a
vector field defined
by
J<v, dS} J<v, N>dS
=
The
of the word flux will become apparent in the next section.
significance
Example
Compute the flux of v(x,
29.
/(x, y)
=
x2
+
We take x, y
x2
2/
as
coordinates
f<v,N>rfS=f
h
(y
Jx2+y2S1\
f
=
det
f
'x2+y2sl
t-y*
An
=
4
y2
+
[x3
+
2y2x
-
=
{xy,
zx)
yz,
the
circ
of
dx\
dx
)dxdy
x
dx
dy/
+ 2
y2)x\
0
2x
1
4y
J
/
+ 2
2x2y
-
v2)
4x2y2
(x2
-
8y3]
dx
dx
dy
dy
A
=
Jn
-"O
In Section 8.2
curve).
use
graph
Then
Z.
Suppose that v is the velocity field of a flow and C is
could
the
f \rs cos2 6 sin2 9 dr d9 \6
Jn
'O
closed
across
1
<
on
v(x2
xy
lX}
1
z)
y,
same
(C)
=
a
closed
path (oriented
defined the circulation around
a
circle;
we
definition to define the circulation around C:
f <v, T>
Jr
we
ds
(8.46)
8.4
Surface Integrals
and Stokes' Theorem
661
arc length along C, and T is the tangent vector to C).
In Section 8.2
this idea to define curl v, the "infinitesimal circulation" about a
used
we
point; now we ask if we can recapture the total circulation from the in
finitesimal. A clue is obtained by recognizing the integrand of (8.46) as the
(s
=
differential form associated
to v.
If v
{v1, v2, v3), then, on the curve
Zi/
dx1
ds
<v, dx}. What we are then asking for is the analog
<v, T>
for surfaces of Green's theorem.
Since the curl plays the same role in three
=
=
=
variables that dco plays in two, it is
no
accident that such
a
theorem exists.
Stokes' Theorem
Suppose
that Z is
now
vector field v, and D is
an
oriented surface
lying
in the domain of the
subset of Z bounded by a curve T. For the purposes
must choose an orientation of T.
It will be the natural one
a
of integration we
corresponding to the given orientation of Z : T winds counterclockwise around
Let
D. To be more precise, we shall define the positively directed tangent.
peT and consider a small path y, with tangent vector t at p which crosses T
Then the tangent vector we wish to choose
T such that the rotation T - t is positive (see Figure 8.15). This
and is directed
is that
one
corresponds
to
so
that it
enters D.
the counterclockwise
sense
of rotation about the normal to
so oriented it is a path, de
the tangent plane.
When the boundary
we
in
mind
have
noted dD. Now the theorem
(Stokes' theorem) asserts
of D is
that the circulation around dD is
f <curl v, N>
given by
dS
Figure 8.15
662
8
Potential
in Three Dimensions
Theory
In order to derive this theorem from Green's theorem
conditions of Green's theorem will be met.
of
a
must ensure that the
we
Hence the
following notion
domain.
regular
Definition 11.
Let Z be
domain if it
a
surface in R3.
A subset D of Z will be called
a
finitely many subsets of surface
regular
partitioned
in the plane in the particular
which
to
domains
patches
correspond
regular
coordinate
representation.
Theorem 8.3.
suppose Z is
an
field defined in a domain U in R3, and
surface lying in U with normal N in U. Let D be a
whose boundary dD is a curve. Then
Let
f
J6D
be
v
a vector
oriented
domain in Z
regular
into
be
can
<v, T> ds
=
f <curl v, N>
dS
(8.47)
JD
Since D is regular, there are coordinate patches Si,
2 and a partition
u D of D such that D, <= 2, and Dt corresponds to a regular
Di v
domain in the 2, coordinates. Now let Bi,...,Bm be balls in R3 such that
Proof.
.
.
.
,
D
Dcftu-'uB, and each Bj lies completely inside one of the coordinate patches
Let pi,
2,
Then, since
p be a partition of unity subordinate to this cover.
.
2Pj
.
=
1
on
f
.
.
,
D,
<v, T> ds
=
JSD
2J f iPj v,
JdD
T> ds
f
<curl
v,
N> dS
Thus,
we
This is
x
Then
f
8D appears
2J f <curl(/>,v),
now our
=
T> ds
N)dS
as
=
part of 8Dj for some/ = i and
2
i,j
right-hand sides
f
<curl(pJv),
N> dS
JDt
are
equal termwise:
we may
coordinate patch.
situation. Let S be a surface patch coordinatized by
we are
in
a
{x\u, v), x2{u, v), x3{u, v)), (, v)eN<=
R2
regular domain in TV and D is the subdomain of S corresponding
{x(, v), {u, v) e N}. Let v {vl, v2, v3) be a vector field defined on S.
must verify
a
=
we
on
to show that the
only need
and suppose A is
to A: D
<PjV,
J j)
that
assume
=
f
2
'!
JHDi>
since each part of 8Dt which is not
with the opposite orientation.
J j)
=
=
<v, T> ds
=
f
<curl
v,
N> dS
8.4
This is
just
the
Surface Integrals and Stokes' Theorem
computation that <curl v, N>
we study the left integral :
dS
=
{dwv) du dv
under the
663
change of
First,
variables.
f
Jsd
<v, T> ds
f
Zv' dx'
=
Jtr>
8x'
f
+ Zv<
=\JfliHv'du
ou
8x'
dv
8v
By Green's theorem this is
8x'
r
[du V ~8v) ~Sv\
h
dudv
~8u
(8.48)
Now
8
I
I v'
8xl\
I
~8~u \
8
~8v
The
I
8vJ 8xJ 8x'
_
2
=
8v ]
dv' dx' dx'
8x'\
.
82x
1" v'
;
8xJ 8u 8v
j
d2xl
t
V' ~dv) =y8xJ~8v~8u+V
integrand
in
8u 8
8v
~8v~8i
8u
(8.48) is thus
8xJ 8x'\
8v' I8x] 8xl
~
~8v
} JxJ \8u ~8v
=
~
i< j
du)
8x>
8v]\/8xJ8x>
/0y'
8x>\
~dxl) \8u ~8v~~8v ~8u]
\8~x~1
Hence, after Green's theorem the left integral becomes
<curl v,
But the
xu x
x> du dv
right integral is
f <curl v, N> ||x x
since
x x x
=
||x
x x
x|| dudv
|| N.
The
=
J
proof
<curl
v, x x
x> du dv
is concluded.
Examples
30. Calculate
Z:z
=
x2
f
<v, dx), where
0<x<l
Z is the surface
0<y<l
664
Potential
8
and
v(x,
curlv=
xx
=
x,
=
dS
z)
y,
in Three Dimensions
Theory
=
-{y,
We make the
x).
z,
-(1,1,1)
(l,0,2x)
(0,1,0)
=
{xx
x,,)
x
f <v, dx}
dx
dy
{-2x, 0, 1) dx dy
=
f <curl v, dS>
+
x(m, v)
Let N
=
=
(m cos
v,
as
y:
Jy
v
=
(y,
z,
u
=
=
x(u)
<v, dx>
=
=
-"o
dy
=
sin v,
u cos
0
6v)
0
< u <
2n
Then
f <curl v, dS}
curve m
v, sin u, cos
(-sin2
0
1
< u <
Thus, the sought-for integral
x).
(cos
dx
1)
patch
f <v, dx}, where y is the
x
Jo
be the normal to Z.
{N1, N2, N3)
f {N1 + JV2 + TV3) dS
where
f f (2x +
0
31. Let Z be the surface
=
=
's
Jgz
x
computations:
v
=
1
can
be
computed
:
6v)
+ cos
v cos
6
6t>
cos t>
sin
6v)
dv
=
-n
EXERCISES
19. Calculate
\zfdS, where
0<z<l
S:z2=x2+y2
(a) /(x,y,z)=x2 + 2y
x2
S:z
O^z^l
zx
+
y2
(b) /(x, y, z) xy + yz +
S
cos m, w sin u, i> sin )
:
(
x(,
)
(c) /(x, y, z) =xyz
0^w^2t7
0<,u<,2n
20. Calculate J <v, c/S>, where
e*'
S:z
O^x^l
(a) v(x, y, z) (xy, yz, zx)
O^y^l
l
S:x2+y2 + z2
(b) v(x,y,z)=(l,-y,x)
S:z
x2+y2<:l
x2-y2
(c) v(x,y,z) (l,0,y)
=
=
=
=
=
=
=
=
8.4
21
Suppose
.
is
v
a
Surface Integrals and Stokes' Theorem
vector field defined in
a
665
neighborhood of the domain
D.
Show that
f
<curl
22.
(a) Suppose
v,
J 3D
dS}
=
0
that D is
a
regular domain
on a
surface S.
Verify that
for any vector a,
f
-
Jqd
J
(axx,rfx>
=
[ <a, dS>
Jd
(b) Just as we integrated vector functions on the interval,
integrate vector functions on lines and surfaces (and in space).
that, for a regular domain D these vectors are the same :
=
\
-
Jd
xx
we can
Show
dx
Jd
j
{Hint: This follows from part (a).)
23. Show that if u,
f
(uVv, dx>
=
Jsd
f
v are
<V
C1 functions
on
the regular domain D that
Vy, dS>
x
Jd
PROBLEMS
29. If
en
is
a
closed form defined in
R3, show that there is
a
a
neighborhood of the unit sphere
w=dfon the sphere.
in
function /such that
30. Consider the torus T:
x
=
x
=
x(, 0=2 cos u + cos v
x{u, v) (2 + cos Ocos k, (2 + cos Osin u, sin v)
(a) Show that the differentials du, dv are well-defined differential
=
forms
on
T.
(b) If
Jr
co
is
a
closed form defined
T, show that the integrals
Jy
are
constant
all circles
(c)
If
u
to
as
=
is
T ranges
=
ci
du +
c2
over
all circles
v
=
constant, and y ranges
over
constant.
a
closed form there
function /such that
co
on
dv +
df
are
constants ci, c2 and
a
differentiable
666
Potential
8
{Hint: Take
in Three Dimensions
Theory
=
Ci
JY col2n,
defined in part (b).)
31. State and prove
by
cylinder.
32. Verify this
c2
Jy co/2n, where
=
the
fact like that in Problem
a
integrals
are
taken
as
30(c) when Tis replaced
a
restatement of Stokes' theorem: Let
vector field defined in
v
=
(F, G, H) be
a
domain U in R3 and suppose that 2 is an oriented
in U with normal N
(cos a, cos /3, cos y). If D is a regular
a
surface lying
domain in 2, then
f
Fdx+Gdy + Hdz
J 3D
r
\(8H
33. Let D be
is
f
a
<v, Ny> dS
Let
v
8H\
(8G
n
8F\
on
\ <curl(v
=
on
the oriented surface 2.
dS
Show that if
2
N), N> rf5
x
JD
where
The
regular domain
vector field defined
a
'SD
8.5
18F
8G\
m^y--fejC0Sa+\fe-i7JC0^+lix-^jC0Sy
=
v
=
Nv is the unit surface normal
Divergence
be
associated
a
to 8D
Theorem
vector field defined in a domain U
flow.
Let D be
we
choose
can
exterior to the domain D.
this is the chosen normal.
c:
R3,
and
x
=
(b(x0 /)
,
the
domain whose closure is contained in [/
steady
sufficiently differentiable
a
such that dD is
table, since
(see Problem 24).
surface.
Notice that dD is orien-
normal vector the unit vector N which is
as
We shall
For
assume
throughout
this section that
small interval of time A/, let us attempt to
calculate the amount of fluid that passes through dD. For x0 6 D, the
a
particle at the point <b(x0
/) at time 0 for 0 < t < At
since <b(x0 -/ + /)
Thus, the volume
<b(x0 0) x0
through dD at time At is the volume of the domain
-
,
=
,
DAt
=
.
,
=
{x:
x
=
<b(x0, -/):0<
/ <
At, x0
e
passes through x0 ,
of the fluid passing
dD}
We shall
approximate this volume by linearizing locally. That is, we cover
by
neighborhoods Ut and replace Ut n dD by the piece Tt of the
We assume
tangent plane to dD with the same area at some point in Ut
dD
small
,
.
8.5
also that <b is
a
is
a
through Tt
pure translation through
parallelepiped of volume
The
Tl
Divergence Theorem
667
Then the volume which passes
.
^(xj.-AO-^x^O.N)^
where x; is some
out that this is a
Let us point
point in Ut n dD, and AAt is the area of Tt
signed volume ; the sign being positive if the flow is into D
(since N is the exterior normal, and if <b(x;
At) is on the same side of dD
as N, <<b(Xj
At) <b(Xi 0), N> is positive). This is in fact what we want,
for we want to discover the flow into D rather that the flow through dD.
It follows that an approximation to the volume of >A( is
.
-
,
-
,
,
E<<p(x-A/)-<p(x0),N>A^
i
by letting the covering get arbitrarily fine,
integral:
and
Jefl<<Kx
The limit of
-
,
1/A/
A/)
Proposition
replace this by
D,
(8.49)
or
as
A/-0
through positive
the flux into D at time
of D
an
(8.49)
,
,
The flux out
8.
may
<b(x 0) N> ds
-
times
stantaneous flow into
we
at time t
=
/
=
values is the in
0.
0 is
JJ><v,rfS>
The flux out of D is
Proof.
-
f
lim
4r->0
=
A? JSD
lim
-
f
< lim
J3D
f
&t JeD
At-*o
=
<<J>(x, -At)
41-0
A?
4>(x, 0), N> dS
-
<<J>(x, -AO-4>(x,0),rfS>
[<b(x, At)
-
4>(x, 0)], dS>
=
f
<v, dS>
Jsd
Now the flux out of D is the instantaneous rate of flow of fluid out of D.
this should be identical to the instantaneous rate of
physical
On
grounds
expansion
we
of the fluid in D, which is
(as
in Section
8.1)
Jfl div v dV.
Thus,
should expect
jUv,dS> jjjclWvdV
=
(8.50)
668
8
Potential
in Three Dimensions
Theory
and in fact this is the
Equation (8.50)
case.
For suitable domains it is
theorem.
mental theorem of calculus.
call such
domains,
domains in R3
theorem, for
or
an
As in the
is known
the
divergence
easy consequence of the funda
case
of Green's
theorem,
we
finite unions of such domains, regular domains.
regular, but by no means are
arbitrary domain, is not easy to
are
an
as
all
regular.
prove and
The
we
shall
Many
general
shall here
avoid the issue.
A domain D in R3 is
Definition 12.
regular if it
can
be
expressed
in each
of these ways :
D
=
{{x, y, z) : (x, y)
e
Dt
f{x, y)
{(x,
y,
z) : (x, z)
e
D2
r{x, z)<y< s{x, z)}
=
{(x,
y,
z) : {y, z)
e
D3
u{y, z)
Lemma.
regular
f
Jan
If
v
is
are
a
continuously
< x <
v{y, z)}
differentiable.
differentiable vector field defined in
a
neighborhood
of
domain D, then
<v, dS}
Let
Proof.
g{x, y)}
=
where all functions
the
< z <
v
=
=
f
div
v
dV
J n
y'E, + v2E2 + v3E3
Sv1
8v2
8v3
Sx1
8x2
8x3
.
We shall show that for each i,
<v'EdS>
JC
Jd
=
eo
8v'
dV
dx'
Then the lemma will follow by summing over i. To prove the ith case, we use the
appropriate representation of the domain. Since all cases are then the same, we
verify one case, say the third.
Now, using the expression
shall only
D
=
{(x,
y,
z) : (x, y)
e
Di /(x, y)
,
<
z
^ g{x,
y)}
8.5
the
boundary of
D consists of the part
2i:z=/(x,y)
22:z
7(x,y)
The
20 lying
Divergence Theorem
over
669
8D1 and the two surfaces
(x,y)eDi
{x,y)eDi
=
Since E3 is tangent to the surface lying
integral over 20 vanishes.
over
8Dt
at every
point, the left-hand
Now 2i has the parametrization
x
=
(x,y,/(x,y))
(x,y)eZ>,
Since the domain lies above this surface, the exterior normal points
is determined by
xx x x, (see Figure 8.16).
Now
x
so we
=
(l,0,/,)
have dS
f
J*i
=
x,
{f fx
,
<d3E3 dS}
,
-
,
=
-
=
(0,l,/,)
1) dx dy.
f
JDl
v3{x,
Then
y,
f{x, y)) dx dy
Figure
8.16
downward,
so
670
Potential
8
A similar
Theory
computation produces
f
<v3E3 dS>
=
,
JZ2
Now,
we
in Three Dimensions
v3{x,
y,
g{x, y)) dx dy
JD {8v3/dz) dV by Fubini's theorem.
compute
r
f
JDi
8v3
f c"-'1"'' dv3
f
TTdV=\JDl \_J
Jd oz
{x,y,z)dz dxdy
[v3{x,
=
oz
f(x,y)
y,
g{x, y)
-
v3{x,
/( x, y)] dx dy
y,
by the fundamental theorem of calculus. But this is, according to
calculations the same as fD <u3E3 dSy. Thus the lemma is verified.
our
previous
,
(Divergence Theorem) Let v be a continuously differentiable
in a domain D in R3.
Suppose D can be covered by
defined
field
that
each
D n Bf is a regular domain.
balls
such
many
Bu
Bn
finitely
Theorem 8.4.
vector
...,
Then
f <v, dS}
Proof
f
Let pu
<v, ds>
.
f
=
.
.
=
,
div
p be
2
f
v
a
dV
partition of unity subordinate
<pi v, ds>
=
2
f
f divvrfK=2 f div(p,v)rfK 2
'd
=
^D
i
i
to
Bi,
.
.
.
,
B
.
Then
<p v, dsy
f
^DnBt
div(p,v)rfF
1 and p, =0 outside 5(. By the lemma, the
for the customary reasons: Sp,
right-hand sides are the same termwise, so the left-hand sides are the same. We
shall henceforth describe domains of the type referred to in Theorem 8.4 as regular.
=
Examples
all, the result of Exercise 22 follows easily from the
divergence theorem, since div curl v 0. For then
32. First of
=
JdD <curl v, dS}
=
Jj, div curl v dV
=
0
8.5
The
Divergence Theorem
x2
y2
671
33. Let
D
=
{x2 +y2
z2
+
<
l},/(x,y, z)
=
+
+
z2
Then
f
JdD
<V/,rfS>
f divVfdV
=
=
6
f
dV
=
Sn
Jd
Jd
34. Let D be the domain {1 > z
{xy, yz, x). Then div v y + z and
>
x2
+
y2},
and let
v(x, y, z)
=
f
div
dV
v
f
f
J0 LJxi+yiz
=
JaD
f
=
JD
<v, dS}
n
\ z2
Jo
f
=
-"zsl
In
z) dx
-
-
f
^Z=X2+J,2
<(xy, y(x2
+
y2), x),
2f
{y2x2
+
y*)dxdy
=
l
i
Equation
Chapter
dimensional
6 in
case.
at time /
a
thermodynamics,
proportional
q +
<v, dS>
2x, 2y, 1)> dx dy
our
discussion of the heat
derivation in dimensions greater than one.
theorem; with that we can carry through
R3 has
dy] dz
3
Jx2 + y2<.l
The Heat
+
xdxdy-\
(
=
=
<v, dS}
f
=
dz
(y
c
we
postponed its
divergence
our argument just as in the onehomogeneous metallic object Uin
temperature distribution u{x, t). According to the laws of
Thus,
we
suppose
a
the vector field q associated to the flow of heat energy is
gradient of the temperature, but for sign:
to the
Vw
=
0
is this: The increase in temperature of a unit
More specifically, the
to the increase in heat energy.
Another basic
proportional
equation
We had to await the
principle
(8.51)
mass
is
change
672
Potential
8
in energy in any
kp
f
Theory
in Three Dimensions
domain D in
given
a
time interval
/
is
given by
Am dV
JD
Am(x, At) is the change in temperature at x over the period At, p is the
density, and k is the proportionality constant (the specific heat). Thus, the
where
rate of increase of heat energy in D is
Now,
compute (using the law of conservation of energy) the
we can
increase of energy in D; it is the flux into D across the
obtain this basic equation for every domain D:
rate of
Thus
we
for every domain D. Thus the two functions must be the same, and
obtain the heat equation:
we
f <q, dsy
-
kP\d-dv
=
jd ot
JdD
By the divergence theorem and (8.51)
f
div Vu dV
f -^
=
Jd
c
divV
boundary.
we
have
dV
Jodt
(^
\c J
=
dt
As
we
saw
in
Chapter 6, the steady
Laplace's equation:
state
(or equilibrium) temperature
distribution solves
div Vm
=
0
d2u
d2u
d2u
dx2
dy1
dz2
EXERCISES
24. If S is
defined
near
an
S,
oriented surface with normal N and
we
denote
L^dS=i*fdv
for any
regular domain
D.
<V/, N> by S//SN.
/is
Show that
a
C1 function
8.5
25. If v is
vol(fi)
a
f
=
J6D
In
v
=
Divergence Theorem
673
1, then for any regular domain D
<v, dsy
particular,
(x, 0, 0)
vector field such that div
The
we
(0,
y,
may make any
0)
one
of these choices for
v:
(0, 0, z)
Find the volume of these domains,
using the divergence theorem.
(a) The cap z > ax1 + by2
0 < z ^ 3.
(b) The cone z2 ^ ax2 + by2
0<z^l.
(c) The tetrahedron bounded by the planes
x=2y,y 0.
26. Verify this formula for any regular domain:
z
=
0
x
+y+
z
=
l
=
f
4
JD
\
\\x\\dV=
JSD
||x||<x,rfS>
27. Here is another way of expressing the divergence
theorem, which is
free of vector notation. Express N in terms of its direction cosines :
N
=
(cos
a, cos
/J,
cos
y)
Then for any three functions F, G, H,
(*
t*
JdD
{Fcosa. + GcosP +
/ Pi T7
+
Hcosy)dS=\ (
\8x
Jd
P/~*
dy
Pk Tf\
+
dz
) dV
J
28.
Compute
(a) Ji <(x2, y2, z2), dsy where S is the (oriented) surface
with side edge 2, and center at the origin.
(b) J (x cos a y cos y8 z cos y)dS over the sphere
+ (z
l)2
1, where (cos a, cos /?, cos y) is the normal.
of the cube
5: x2 +
y2
=
PROBLEMS
34. Let 2 be
one
a
a
surface which intersects each ray from the
set S.
origin
in at most
The set of rays which intersect S will pierce the unit sphere in
The area of S is the solid angle subtended by S. Show that the
point.
solid angle is given by
f <x,dsy
674
8
Potential
Theory
in Three Dimensions
35. Vector-valued functions
can easily be integrated over any domain,
Verify these formulas for a regular domain D:
coordinate by coordinate.
\
v x
f
fdS=\JD VfdV
dS
=
JdD
I.
curl
I
v
dV
^d
JdD
NrfS=0
36. Let
v
be
divergence-free
a
vector field defined in
a
domain U.
Show
that if y is a closed curve defined in U, then for any regular domain Dona
surface S such that 8D
y, the integral
=
<v, dsy
Jd
always has the
same
value.
37. Show that the function /is harmonic in the domain D if and
for every ball B
Jqb
<=
only if,
D,
<V/rfS>=0
Suppose there is given a flow in R3 with these properties:
(a) The flow has constant velocity outside of some large bounded set.
(b) The flow on the {z 0} plane remains along that plane (no fluid
passes from the upper half space to the lower half space ),Show that
38.
=
f
divv</K
=
0
Jh
where H is the half space
8.6
Dirichlet's
{z > 0}.
Principle
R3, and suppose v is the velocity field of a flow through
The total kinetic energy of the flow
D which is steady (time independent).
is given by the integral
Let D be
a
domain in
2-1'p||v||2 dV
(8-52)
8.6
Dirichlet's
Principle
675
where p is the
density of the fluid (we shall here take p to be constant). An
important physical problem is this: find the flow which minimizes the energy
(8.52) subject to certain conditions being fixed on dD. For example, we may
assume that the normal component of the flow
<v, N> through the boundary
is fixed.
potential
Or
assume that the flow is
conservative, that is, v has a
and
the values of the potential are fixed on the
function,
boundary.
we
may
These
problems are analogous to Neumann's and Dirichlet's problems
respectively (see Chapter 6). Dirichlet's principle is that the flow which
minimizes the energy is the gradient of a harmonic function
(solution of
Laplace's equation). In this section we shall derive Dirichlet's principle and
indicate how the techniques involved can be used to discover the solution
to the problems.
In order to do this, let us make these problems
precise.
Let D be a domain in R3, and/a function defined on D.
I.
(Dirichlet's Problem) Among all C2 functions u defined on D
boundary values/, find the one which minimizes the integral
which
have the
\d \m\\2 dV
II.
that
(8.53)
(Neumann's Problem) Among all C2 functions u defined on D such
<Vm, N> =/on dD, find the one which minimizes the integral (8.53).
In order to study these problems we need (i) to relate boundary data to the
integral (8.53), (ii) to discover an interpretation of (8.53) which will suggest a
technique for minimizing that integral. The first need is filled by the di
vergence theorem, which will take the form of Green's identities (given below).
The interpretation requested in (ii) is that of Euclidean vector spaces and the
technique will be orthogonal projection. Let us describe this idea more
fully.
Let
C2{D)
tinuously
represent the collection of functions which
differentiable
dean vector space by
<m, v)
=
f
JD
on
D.
defining
<Vm, Vi>>
on
are
twice
make this vector space into
it the inner product
We
can
a
con
Eucli
(8.54)
dV
length of Vw in terms of this inner product.
(8.53) by E2(u}. Our problem is to minimize this length
among all functions with the given boundary value/. Let Mf be the space of
functions in C2{D) with boundary value /. Then Mf is a translate of the
{u + g : g e M0}.
space M0 : if u is a function with boundary value/, then Mf
Now it is a simple principle of Euclidean vector spaces that the vector in Mf
which is closest to 0 is orthogonal to Mf, hence also orthogonal to M0.
Then
(8.53)
is the square of the
We shall denote
=
676
Potential
8
Theory
in Three Dimensions
The solution to
our problem will then be that function in
Mf n M0l.
identify M0L as the space of harmonic functions.
There is one fault with our reasoning. The "simple principle" above is
one about finite-dimensional Euclidean vector spaces (recall
Chapter 1),
and it is not necessarily true in the infinite-dimensional case (of which ours
is a prime example). The problem is that there need not be any point in
Mf n M0L; and our argument will be complete once this problem of existence
Finally,
we
can
The mid- 19th century mathematicians such as Dirichlet and
little troubled by such problems; it was during the late 19th
is resolved.
Riemann
were
century that mathematicians began
to think of existence questions as crucial
(with good reason). And it was not until the last decade of that century that
the existence problem was effectively solved. (The reader is referred to the
history by Kellogg (pp. 277-286) for a fuller account.)
The link between the geometry described above and the subject of harmonic
functions comes out of certain computations involving the divergence theorem
(Green's identities). These will now be exposed. We shall adopt one more
notational convention before proceeding (already foreseen in the problems):
if m is defined on the oriented surface Z, then <Vm, N> is the directional
derivative of u in the direction normal to Z. We shall denote it by du/dN.
Theorem 8.5.
regular
(Green's Identities)
Letfi
g be two
C2 functions defined on
\ f% dS Jdf U*g + <v'> v^>j dv
=
JdD
ON
Proof.
f fj%dS=\JSD </%, N> dS JD\ dbf{fVg) dV
=
Jeo
But,
as
is
oN
easily computed (see Exercise 10):
div(/V#) =/div Vg + <V/, V<?>
so
Theorem 8.5 is proven.
Corollary
a
Then
domain D.
1.
(i) Ifg is harmonic, \dDf{dg/dN) dS E </, g}.
0.
(ii) Iffe M0 and g is harmonic, E </, g)
=
=
<8-55)
8.6
(iii) Iff and g
Principle
dN
(8.56)
v
'
dN
jSd
(iv) Iff is orthogonal
is
to every
harmonic, then Ag
=
function
0,
so
f f%rdS=\JD <yfVgydV
in
,
=
the roles of /and g in
,
we
have
Eif,gy
(ii) Now, if fe M0 /has boundary values 0,
also vanishes, and thus E(f gy
(iii) If g is harmonic, we have
M0 f is harmonic.
by (8.55)
=
8N
J 3D
677
harmonic,
are
f%dS=\ g%dS
f
Jsd
Proof.
(i) If g
Dirichlet's
(8.57)
so
the
integral
on
the left of
(8.57)
0.
(8.57).
(8.57) obtaining
If /is also harmonic
we may
interchange
L^ds=E<9'f>
E(f gy.
(8.56) results since E(g,fy
(iv) If ge M0, then by (8.55) (interchanging/ and g),
Thus
=
f
Jd
we
have
gtifdV+E<f,gy=0
0 for every g with boundary value
For
zero.
suppose A/(p) >0 for some p
in D. Let B be a ball in D centered at p in which A/> 0, and let p be a C2 function
0 off B. Then p e M0 , so
1 and p
such that p(p)
orthogonal to M0 then J gAfdV
This implies that A/=0 everywhere.
Now if / is
=
=
\
Jd
Since
pAfdV=
pA/> 0
=
,
\
'b
PAfdV=0
in B, it must be
zero.
Thus
A/(p)
=
p(p)A/(p)
=
0,
a
contradiction.
with the
Corollary 2. The orthogonal complement of M0 in C2{D)
harmonic
functions.
product E </, gy is the space H of
inner
domain in R3 and
(Dirichlet's Principle) Let D be a regular
class
the
be
Let
dD.
offunctions in
on
Mf
suppose f is a continuous function
C2{D) with boundary value f.
Theorem 8.6.
678
Potential
8
Theory
in Three Dimensions
(i) If there is a harmonic function in Mf, it minimizes the energy integral.
(ii) If there is a function in C2{D) which minimizes the energy integral,
must
Proof. These facts follow from
(i) Let u e Mf be such that A
=
dD,
it
be harmonic.
so
M0
u e
g
E2<gy
=
the
reasoning
same
as
in Euclidean geometry.
u=0on
If g is another function in Mt ,g
0.
.
E2<g - + >= E2<g
=
u> + 2E<g -,> + 2<>
-
2<<7-> + 2<>
since uM0. Thus, 2<#> > 2<> for every geMf.
(ii) If e C2(Z>) minimizes the energy integral in Mt it must be
orthogonal to
Forif^e M0, then ugaxe both in Mf, and thus E2(u + gy >,2<>. But
,
M0.
2<"
so
0 >
0>
=
2<, gy + 2<<?>
2<">
2<k, gy + E2igy for all
<t,{t)=2E<u,tgy
#
6
M0
.
Consider for
the function
+ E2<\tgy
0(0 < 0 for all / (positive or negative), and <(0) 0, we must have <^'(0)
<^'(0) 2E(u, gy. Thus u_L M0 so, by Corollary 1, u is harmonic.
Since
But
/ e R
=
=
0.
=
,
In order to solve Dirichlet's
there exists
function in
problem by his principle
C2{D)
it remains to show that
integral. The
carrying
through
finally accomplished by Hermann
technique
and
his
methods
have
had
far
Weyl (1926)
reaching effect in a wide class of
value
for
differential
boundary
problems
partial
equations.
a
for
this
which minimizes the energy
was
Harmonic Functions
Green's identities and Dirichlet's principle in order to derive
properties of harmonic functions (analogous to those in two
dimensions given in Chapter 6). Out of this will come a hint for solving the
Dirichlet problem.
We
can use
the basic
Proposition
domain D.
9.
Let
There is
f be
a
C2 function defined
at most one
on
the
harmonic function with
boundary of a regular
boundary value f.
8.6
Proof.
the
same
If u, v are both harmonic and have the
time harmonic and in Mo
Thus <
Dirichlet's Principle
boundary values/,
-
v,
.
u
-
y>
=
0.
then
u
679
-
v
is at
But
f \\V{u-v\\2dV
E(u-v, u-vy=
Jd
so we
must have
is identical to
V(
0
-
=
0 in D.
Thus
u
-
v
is constant.
Since
u
=
v on
dD,
u
v.
The
gravitational field of a particle
according to Newton, given as
of unit
mass
situated at the point p is,
<8-58)
||x-p||2||x-p||
This field is easily seen to be conservative and divergence free, thus it is the
gradient of a harmonic function, called Newton's gravitational potential.
Writing (8.58) out in coordinates, we have
(x1 p1, x2 p2, x3 p3)
p1)2 + (x2 p2)2 + (x3 p3)2]3'2
-
[(x1
-
=
-
-
and it is not hard to
Up{x)
-
\\x
that this is the
see
pir1
-
-
=
[(x1 -p1)2
gradient
+
of
(x2 -p2)2
+
(x3 -p3)2]-1'2
This
particular function stands at the beginning of a sequence of ideas which
a technique due to Green, for solving Dirichlet's problem.
These
steps were motivated by an inquiry into the nature of gravitational fields (due
to masses more general than that of a particle), the point being to show that
every harmonic function arises as the potential of a gravitational field.
Green's first result is an easy consequence (reminiscent of the Cauchy integral
lead to
formula)
of his identities.
Proposition 10.
LetpeD. Then
Proof.
Once
tained in D.
Let Dbea
again
we
first
Since both h,
flp
regular domain, and h a function harmonic on
remove a
are
small ball 5(p, 0 centered at p and
harmonic in D
-
B(p, e), Corollary
1
D.
con
(iii) applies
680
Potential
8
in that domain.
Thus,
T 8UP
r
This
in Three Dimensions
Theory
8hl
implies that
c
r
on.
L[*
8h~\
Now the second
integral
r
r
n>m\
-
w
dS
can
-
be
^1
en
Li* w-uw\
ds
computed using spherical
^
coordinates centered
at p:
x
Then
=
np(x)
x
=
p +
=
{r cos
r'1.
p +
0
cos
The
</>,
r
sin 0
cos
</>,
r
sin
0)
sphere B(p, e) is given by
sin 6
e(cos 0 cos <f>,
cos
<f>,
sin
0)
and its exterior normal is the radial vector, so djdN
8/8r. The element of area
B{p, e) is dS e2 cos2 </> dd d<f>. Thus the right-hand side of (8.60) is
=
on
=
[
Wo
or
T
Cn
t"
|3A/3r|
-
<
8r\
1
h{x)
J-n '-n/2
Since
r
8K\
.n/2
r
=
\r)
.*
J-.J-K/2
||VA||, the second integrand is bounded
e ->
-h(p)
0
r
\
J-n
which is what
was
our
gA
cos2
or
as e
As for the first term x->p
integral tends to
term will vanish for ->0.
Thus, letting
.a/*
h{x)cos2<f>d6d<f>-e \
</>
->0.
as
d8
d</>
Thus the second
e-*0,
so
A(x)-*-/z(p).
,*/2
cos2
>
</>
dd
d</>
=
-47Th(p)
-n/2
desired.
Now, if D is the ball of radius R centered at p, then np(x)
||x p||-1,
R~x and dUJdN
-R~2. Equation (8.59) becomes
D, np
=
so on
=
*(P)
=
=
hdS
f
tAts
-^-\
|*- dS
4nR2)l{x-U=R
47r^J||x-pli=J?aiV
+
8.6
Dirichlet's Principle
Since h is harmonic in D the second
integral vanishes (Problem
mean value
property for harmonic functions in three
obtain the
Proposition 11. (Gauss' Theorem) If h is harmonic
-B(P, R), then h satisfies the mean value property:
*<*>
f
=
in
a
47)
681
and
we
variables.
neighborhood of
h ds
TTi
4nR Jii__m
=
d
Green's Function
Now, by Corollary l(iii), if A: is any function harmonic
can
be modified
by
-1
Kv)
Thus, if
value
f
=
on
D, then (8.59)
k:
h
8(U>
~
k)
-
dN
47t JdD
m
l
"
-k)8JL
k)dN
dS
(8.61)
k is chosen
np,
so as to solve Dirichlet's
problem with the boundary
the second term will vanish and we obtain an integral formula
for h in terms only of its boundary values. Finally, we could use that
formula to solve Dirichlet's problem with any boundary values. Thus
(8.59) allows us to reduce the general problem to
{np} of specific functions, and for many regular
easily found.
Definition 13.
with the
Let D be
boundary
values
a
domain in R3.
np
,
we
that for
a
certain
family
domains that solution is
If kp solves Dirichlet's
shall call the function
Gp
=
kp
problem
Up
the
Green's function with singularity at p.
Theorem 8.7. Suppose D is
function for every point p in D.
terms of its boundary values:
Kp)
Proof.
=
r-
f
h
4nJBD
*
a
regular domain such that there is a Green's
ifh is harmonic on D, h can be found in
Then
dS
dN
By (8.61),
but the second
integral
vanishes since
kp
U
=
0
on
dD.
682
Potential
8
Theory
in Three Dimensions
Example
35. Let
Then dD
us
=
take D to be the upper half space D
{(x, y,z):z>
{{x, y, 0): (x, y)eR2}. Since the domain is infinite
=
have to restrict attention to functions for which the
If Ht is
sense.
H,
=
then
+
y2
+
z2
<
/,
z >
0}
holds for functions h harmonic
\h8Up
zl[
Jmo,,)[
4n
we
make
large hemisphere :
a
{(x, y, z): x2
(8.61)
integrals
0}.
n"
dN
on
D:
8h\
dN]
zao
+
JLf
4nJ
L^_npSds
L
"dNj
(8.62)'
v
dN
(r=0)
We shall call the function h
dissipative
if the first
integral tends
to 0
/->oo, and the second integral converges. For example, if
||x||2/!(x) and ||x||2V/i(x) are bounded functions on D, h is dissipative
as
(Problem 48). This is true for np, p not the on xy plane.
Now if h is dissipative we can let / -*oo in (8.62) and obtain
(p)
yyj
lf
dN]
47tJ{z=0}L\h8Jh-up8JL]ds
=
Nowifp
np(x)
=
and its
.
is
=
a
"
dN
(*o
=
[(x
(x0, yQ,z0),
x0)2
-
+
{y- yQ)2
boundary values (z
,
Jo
Since
zo)-
Green's function.
=
n9
+
0)
(z
Zo)]"1'2
-
are
the
same as
Thus, the Green's function
is
gp(x)
=
iyx)
-
np(x)
l
[(x-x0)2
+
(y-y0)2
+
(z
+
y0)2
+
(z
z0)2]1/2
1
[(x
-
x0)2
+
(y
-
those for
JTS
where
dissipative, there
for p
(x0 y0 z0)
is harmonic in D and
-
^^2T/2
z0)2]
=
,
,
8.6
Dirichlet's
Principle
683
Now the exterior normal to the
d/dN
=
a*"
plane is the downward vertical,
computation gives
A final
d/dz.
v
<fe
'
"
'
[(x
-
x0)2
+
{y- y0)2
Thus, if h is harmonic and dissipative in the
for any z0
>
n{x0 y0 z0)
,
+
so
z2]3'2
upper half space,
we
have
0
=
-j
2
-T2
+ {y-y0) +
,
2ttJ0 Jo [(x-x0)
j 3^
z2] 3/2
dx
dy
(8.63)
Finally, we remark that (8.59) can be used to solve Neumann's
problem in the same sense. If there is a harmonic function kp for each
p in D such that
dkp
-
dN
=
dIlP
-
an
dD
on
dN
then for any function h harmonic
on
D
we
have
dh
h^=LL^-k^dNds
4V
"
"
Thus h is determined
dN
by
its normal derivative
on
the
boundary.
PROBLEMS
39. Prove
Corollary
Green's Function for
a
2 of Theorem 8.5.
Ball
40. Using a little bit of plane geometry it is possible to discover the
be
Green's function for the unit ball. If P is a point inside the ball, let Q
the point inverse to P in the sphere
P
684
8
Potential
Now let X be
Theory
point on the sphere. Verify that the triangles (see Figure
OXQ are similar (since the angles POX and QOX are the
a
8.17)
OPX and
same
and
1
IQOI
~iPO|
in Three Dimensions
ox
QO
or
=
ox
PO
Conclude that
QX
OQ
PX
OX
41. From the above
n*(x)
=
problem
we
deduce that
7^n*(x)
where q is the point inverse to p in the unit sphere.
in the unit ball B, the Green's function for B is
GP{x)
na(x)
=
^f-U{x)
llp-x||p||
llp-xll
Figure
8.17
Since
IT,(x) is harmonic
8.6
Dirichlet's
Calculate the
the
precise form of Theorem 8.7 (known
ball) if h is harmonic on the unit ball
if
v
.,
l
,
-
Up II2
as
Principle
685
Poisson's formula for
Jn
^^L^IIPlKIIP-xll)^
'llxll
42. Solve Dirichlet's problem for the ball.
43. Solve Neumann's problem for the ball.
44. Find the
steady
state
temperature distribution in the ball if the
surface temperature on the sphere is maintained at
(a) cos <f>, </> is the angle between the point and the north pole.
(b) A{x + 2y), A a constant.
x2 + y2-2z2.
(d) cos 40 sin 2</>, 0, </> spherical coordinates.
45. Suppose D is a domain for which there exists
for all p e D. Show that if p # p'
(c)
GP{p')
=
a
Green's function Gp
GP.{p)
{Hint: Show, by Green's identity that the integral
0
8GP
dGp.
=
is the
Gp
Gp'
8N
dS
8N
same as
8G.1
8GP
[
\g ^
L-- L 8N
g^1n
dS
where B, B' are balls of radius
the fact that
e
centered at p, p',
respectively.
Now, using
1
Go
=
-
+ harmonic
r
compute the limits as p -> 0.)
and
46. Suppose D, D' are domains with Green's functions GD, GV
e
=>
D'
D'. Show that for p
D
allxinZ)'
Gd,p{x)>lGd.p{x)
at p,
47. Show that for h harmonic in the ball of radius R centered
p
8h
on
J||x-,ii=ji
oN
dS
=
0
686
Potential
8
in Three Dimensions
Theory
48. Show that the function h defined
is
dissipative
||x||2//(x)
are
on
the upper half space D
=
{z
>
0}
if
\\x\\2Vh{x)
bounded.
49. Show that if h is harmonic and
dissipative in the upper half space
plane, then h is identically zero.
50. Suppose that h{x, y) is dissipative on the plane. Prove that there
exists a unique dissipative function u continuous on the upper half space
{z > 0} and harmonic for {z > 0} which attains the boundary values h. u is
given by
and
the
zero on
zo
z
0
=
f00
r
u^y'^=2)0 I
h{x,y)
[(x-Xo)2 + (y-yo)2 +
Zo2P'2^
steady state dissipative temperature distribution on the
plane if the temperature on the plane z 0 is maintained at
51. Find the
upper half
exp(x2
8.7
+
y2)-1.
Summary
A fluidflow is
some
=
given by
domain D in
a
R3 and
C1 i?3-valued function <b(x0 /) defined for x0 in
,
/ on an
interval in R about the
origin.
<b has these
properties:
(i)
(ii)
j)(x0 0)
all x0
=
x0
For fixed /, x0
,
-
6
<b(x0 /)
,
D.
is one-to-one and has
a
nonsingular
differ
ential.
The vector field
d<b(x0 0
,
v(x, /)
=
dt
X0
=
(t>'i(X,t)
velocity field of the flow. The flow is steady if v is independent of /.
If y
{vlt v2 v3) is a differentiable vector field, its divergence is the function
is the
=
,
dt>i
dv2
dv3
dx1
dx2
dx3
8.7
is its
p(x, /)
dp
-
dp
divO>v)
v(x, /) is
the
the law of conservation of
density,
+
A flow is
If
of continuity.
equation
=
dp
_
+
-
incompressible
S>i-i +
if the
pdivv
same mass
=
687
Summary
velocity field
mass implies
of
a
flow and
0
always occupies
the
same
volume.
The necessary and sufficient condition for this is div v
0, where v is the
velocity field of the flow. The fluid is incompressible if and only if the
=
density
at a
is constant under all flows of the fluid.
particle
INTEGRATION UNDER A COORDINATE CHANGE.
a
change
of coordinates
continuous
on
I* g{x,
a
(m,
w)
V,
=
F(x, y, z) be
domain D onto the domain A.
If g is
D,
z)
y,
Jn
If v is the
taking
Let
dx
velocity
dy
dz
|
=
JA
field of
a
g{ x(,
v,
w)
det(w)
{u,
v,
du dv dw
w)
flow, the circulation around a
curve
C is defined
as
circ(C)
If
we
at Xq
Jr.
fix the
perpendicular
f
=
<v, T> ds
point
to
n
x0 and vector
of radius
...
v(x0 n)
=
,
lim
r-o
v
n
at x0 let
centered
at x0
.
Cr be the circle in the plane
The curl of the flow about n
is
curl
If
r
=
{v1, v2, v3)
circ(Cr)
r
2
define
/dv2
'dv2
dv3 dv3
dv1
dv1
dv2^
_5t/\
Cmly=\oV3~dx2~'olc1~dT3'dx2 dx1)
<curl v(x0), n>. A
v(x0 n)
A surface patch in R3 is the image of
x(w, v) with these properties:
Then curl
x
=
,
=
(i)
x
is one-to-one
0.
flow is irrotational if curl v
a
C1
under
in
R2
D
map
a domain
=
688
Potential
8
Theory
in Three Dimensions
the vectors x
dx/du, xv dx/dv are independent, (m, v) are called
or
for the surface patch. A surface is a set E in R3
coordinates
parameters
which can be covered by surface patches. The tangent plane to E is the plane
(ii)
=
=
spanned by the vectors x x (this is independent of the particular co
ordinates). The normal N to a surface is a unit vector defined for each point
and orthogonal to the tangent plane there.
,
The form
ds2
defined
Edu2
=
+ 2Fdu dv + G
dv2
surface E with coordinates
on a
=<x,xu>
{u, v) by
G
F=<x,x>
=
<x,x>
is the
first fundamental form of the surface. The parametric curves are
orthogonal if F 0. The length of a curve on E given by u u{t), v v{t) is
=
r,
/*
=
r\Jdu\2
/[*(*)
=
^^dudv
^(dv\2yi2
+2Fd7Jt + G[Tt)\
A geodesic is a curve of minimal length.
point p on y the normal to y is orthogonal
The
area
|
of
dS
a
=
"d
The
domain D
I ||x
integral
of
f /dS
a
JD
These definitions
flux of
an
surface
are
x|| dM dt)
X
E is
!<v,N>dS
on
D is
Xj dM dv
independent
oriented surface and
v across
,
dt
If y is a geodesic on E, then at any
to the tangent plane of E.
is defined by
continuous function /defined
f /||XU
=
JD
If E is
x
'd
on a
=
of the parameters chosen.
v is a vector field defined around E, the
8.7
689
Summary
stokes'
an
theorem.
If v is a C1 vector field defined in a domain
U, and E is
oriented surface in U and D is a regular domain on
E, then
f
'dD
<v, T> ds
f <curl
=
divergence theorem.
of a
regular domain
f <v, N>
Z> in
dS
f /||dS=
oN
JSd
values
E2{u)
=
dF
on
g be two
+
<V/, V<?>]
Let D be
D.
C2 functions defined
Let
a
My
on
a
regular
dV
regular domain in R3 and suppose /is
be the class of C2 functions on D with
given by/.
a
harmonic function in
Ms
,
it minimizes the energy
integral
j \\Vu\\2 dV
If there is
(ii)
v
/,
f UAg
JD
principle.
If there is
(i)
div
Let
continuous function
boundary
dS
JD
green's identities.
domain D. Then
dirichlet's
N>
If v is a C1 vector field defined in a neighborhood
R3 then, with the exterior normal orientation on dD,
f
=
JdD
a
v,
JD
a
C2 function which minimizes the energy integral, it
must
be harmonic.
FURTHER READING
In order to continue the
topics
one
study of the divergence
theorem and further related
must turn to the notations and ideas of differential forms. The
small book
M.
gives
Spivak, Calculus
a
on
Manifolds,
H. K. Nickerson, D. C.
Benjamin, Inc., New York, 1965,
subject. The book
W. A.
clear and direct account of this
Spencer, and
N.
Steenrod, Advanced Calculus,
D. Van Nostrand Company, New York, 1957, was the first to give a complete
For a more recent
account of this subject on an advanced calculus level.
account, with a chapter on potential theory in R", see
L. Loomis and S. Sternberg, Advanced Calculus, Addison-Wesley, Reading,
Mass., 1968.
690
Potential
8
Other references
Theory
in Three Dimensions
are
Munroe, Modern Multidimensional Calculus, Addison-Wesley,
Reading, Mass., 1963.
E. Butkov, Mathematical Physics, Addison-Wesley, Reading, Mass., 1968.
For further study of differential geometry we recommend
M.
S.
E.
Lectures
Classical
Differential Geometry, Addison-Wesley,
Reading,
H. Guggenheimer, Differential Geometry, McGraw-Hill, New York, N.Y.,
Struik,
on
Mass., 1950.
1963.
MISCELLANEOUS PROBLEMS
C1 function defined in a neighborhood of p0 in
dF{p0) # 0. Show that the set
{p : F(p) 0}
is a surface
some neighborhood of p0
{Hint : Choose coordinates
Then the
x, y, z so that the forms dF{p0), dx{p0), dy{p0) are independent.
transformation F(p)
(x(p), y(p), F(p)) is invertible. If G is the inverse
Suppose that
52.
R3 such that
F(p0)
patch in
=
F is
a
0 and
=
=
.
=
to
F, the function
</>{u, v)
=
G{u,
v,
0)
parametrizes .)
53. A family of surfaces in
equation
F{p)
=
a
domain D in R3 is
given implicitly by the
(8.64)
c
where F is C1 in D and
dF{p) =^= 0. For each c, the set (8.64) determines a
Show that the vector field VF is the velocity field of a flow whose
surface.
path lines intersect each surface orthogonally.
54. Find the family of curves which are orthogonal
to these families of
surfaces :
c.
(a) x2 + 2y2 + z2
(c) x2 + y2 c{z + c).
c2
(d) z c cos y.
(b) z2x2
55. Given a family F of curves in space, there may not exist a family of
surfaces orthogonal to F. If say, v is a vector field tangent to the family F
and {F{p)
c] is the family of orthogonal surfaces, show that VFmust be
=
=
=
=
=
collinear with
v.
The condition that
must be satisfied in order for the
v
must be collinear with a
path lines associated
to
v
gradient
to have an
orthogonal family of surfaces. Show that this condition may be written
0.
<curl v, v>
56. Show that the family of path lines of the helical flow
=
(x,
y,
z)
=
(x0
cos t
does not admit
an
+ y0 sin /,
x0
sin
t
+ y0
cos
orthogonal family of surfaces.
/,
z0
+
/)
8.7
57. Show that if the vector field
{II(p)
v
is conservative the
c], where II is a potential for v is orthogonal
58. Show that, although the vector field
v(x,
=
z)
y,
is not
=
(yx,
y,
691
Summary
family of surfaces
path lines.
to the
0)
conservative, the path lines
of its associated flow does admit
a
family
of orthogonal surfaces.
59. Suppose that D is
a star-shaped domain in R3 centered at the
origin.
That is, if p e D, then so is the line segment joining 0 to p in D. Suppose
that v is a C1 vector field defined on D such that div v =0. Define the
vector field u by
f
[v(/p)
x
Show that curl
u
u(p)
=
Jo
rp] dt
=
v.
{Hint: Recall Poincare's lemma (see Theorem 7.5);
this is just a generalization. Differentiate under the integral sign,
condition div v
0 and then integrate by parts.)
use
the
=
60.
Suppose
sphere
f
<curl
that
u
is
C1 vector field defined in
a
a
neighborhood
of
a
Show that
S.
u,
N> dS
=
0
Js
(Use Stokes' theorem one hemisphere at a time.)
61 Every curl-free vector field defined on R3
{0} is a gradient ; however
there is a divergence-free vector field defined there which is not a curl. For
example, take
.
Vo(P)
=
i
Then div
f
v
=
0, but if S is
<Vo,N>dS
=
a
sphere centered
at the
origin
477
Js
so
by Problem 60,
v0
is not
free field defined in R3
-
a
curl.
It
{0}, there is
can
a
that
v
=
curl
u
+ cv0
Can you suggest how to define
c
be shown that if v is any divergence
u and a constant c such
vector field
and u?
692
Potential
8
Theory
in Three Dimensions
Normal Curvature
be
62. Let
a
surface
N be the normal to
be viewed
so
patch
-*
x
differentiable function of u,
as a
jR'-valued linear map of R2.
an
in R3 coordinatized
chosen that xu
->
v.
by x x(h, v). Let
right handed. N can
For p on , dN{p) is thus
=
N is
By defining
8N
dN(/>)(xu)
(/>)
=
0N
dN(/>)(x)
go
=
as a mapping of the tangent space 7*() into J?3.
(a) Show that the range of dN(p) is orthogonal to N(p). {Hint: N
is a unit vector.)
(b) Because of (a) dN(p) can be considered as a linear transformation
of r() to T{Z)P
Show that dN(p) is symmetric :
we
may consider dN
.
<dN(p)v, w>
{Hint:
=
<v, rfN(p)w>
You need
<dN(p)(xn), x>
=
only show that
<x, dN(p)(x)
(c) Show that rfN(p)(v) is kn{\) when rN(v) is the normal curvature
(see Problem 25) of the curve of intersection of the plane through N and
v
with .
symmetric on T()p it has two real eigenvalues and the
corresponding eigenspaces are orthogonal. The eigenvalues are called
the principal curvatures of
at p, and the eigendirections are the principal
Since dN(p) is
,
directions.
63. The second fundamental form
II(v)
=
<dN(p)v, v>
Show that II
II
=
can
v 6
_/SN
8x\
=
\8u~'~8u)
surface is the form
T()(p)
be expressed
L du2 + 2M du dv + N dv2
where
L
for
on a
as
(8.65)
8.7
\
'
dv/
693
dx\
/N Sx\/dN
8u
Summary
'
\ dv du/
/eN sx\
64.
tions
Compute the second fundamental form
on
and find the
principal
direc
these surfaces :
x2 +
1
2y2 + z2
2y=x2
x2
y2 z2
x2-y2 + z2=0
65. (Rodriques' Formula) Show that a curve r on a surface
to a principal direction at every point if and only if dN + kn dx
(Such curves are called lines of curvature.)
(a)
(b)
(c)
(d)
=
=
66. Find the lines of curvature
on
the surface
=
is tangent
along r.
0
:
is the cylinder given by x{u, v)
(a)
(cos u, sin u, v).
is the torus x{u, v)
(b)
(2 + cos )cos v, (2 + cos u (2 + cos u)
cos v, (2 + cos )sin v, sin ).
is the sphere x2 + y2 + z2
1
(c)
67. A point p on a surface
is called an elliptic point if the principal
curvatures have the same sign, a hyperbolic point if the principal curvatures
have different signs and a parabolic point if one principal curvature is zero.
Find examples of all three kinds of points on a torus. Show that p is
elliptic, hyperbolic, parabolic as LN M2 > 0, < 0, =0.
68. Show that at a hyperbolic point in a surface intersects its tangent
plane in two curves with zero normal curvature.
=
=
=
=
.