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in that plane has angular velocity -2
=
-
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
=
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)
(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
=
,
=
equation of motion,
Curl and Rotation we
633
have
x0
,
deb
-^(x0,0) and
(0,1,0)
=
solves the differential
{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)
=
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
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
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
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
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
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
.
,
E
=
ds
]7/du1
dv2
du2
dvA
\ ds
ds
ds
ds
ds
orthogonal
at p if
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
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
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
-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
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
(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
=
,
=
,
=
=
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
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
(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
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
=
=
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
=
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,
=
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
J j)
that
assume
=
f 2 '!
JHDi>
since each part of 8Dt which is not with the opposite orientation.
J j)
=
=
=
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
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
0<x
Z is the surface
0
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
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)
=
=
-"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
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
=
=
=
=
=
=
=
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>
=
[
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
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
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
,
=
,
DAt
=
.
,
=
{x:
x
=
/ <
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
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
-
,
-
,
,
E<
by letting the covering get arbitrarily fine, integral: and
Jefl<
-
,
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
-
times
stantaneous flow into
we
at time t
=
/
=
values is the in
0.
0 is
JJ>
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>
[
-
4>(x, 0)], dS>
=
f
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)
=
{(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
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,
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 ,
-
,
=
-
=
(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
=
,
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
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
f
Let pu
.
f
=
.
.
=
,
div
p be
2
f
v
a
dV
partition of unity subordinate
=
2
f
f divvrfK=2 f div(p,v)rfK 2 'd =
^D
i
i
to
Bi,
.
.
.
,
B
.
Then
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
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
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 +
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-\
( =
=
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
-
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.
/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
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
=
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 =
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,
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
=
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
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
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
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
Theorem 8.5.
regular
(Green's Identities)
Letfi
g be two
C2 functions defined on
\ f% dS Jdf U*g +
JdD
ON
Proof.
f fj%dS=\JSD dS JD\ dbf{fVg) dV =
Jeo
But,
as
is
oN
easily computed (see Exercise 10):
div(/V#) =/div Vg +
Theorem 8.5 is proven.
Corollary
a
Then
domain D.
1.
(i) Ifg is harmonic, \dDf{dg/dN) dS E
=
<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
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
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
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
=
the
reasoning
same
as
in Euclidean geometry. u=0on
If g is another function in Mt ,g
0.
.
E2
u> + 2E
-
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
#
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
cos
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
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
->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
'
"
'
[(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
(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
/ on an
interval in R about the
origin.
properties: (i) (ii)
j)(x0 0)
all x0
=
x0 For fixed /, x0 ,
-
6
D.
is one-to-one and has
a
nonsingular
differ
ential. The vector field
d
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
=
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
!
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
f <curl
=
divergence theorem.
of a
regular domain
f
Z> in
dS
f /||dS= oN
JSd
values
E2{u)
=
dF
on
g be two
+
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
=
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
.
=
You need
=
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)
=
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. =
=
=
=
.
ANSWERS TO SELECTED EXERCISES
Chapter SECTION
1.
2.
5.
1
1.1
(a) (b) (c) (d) (e)
0, -7) (4, 3/4) (8,6 ,1) (0,1 ,2,1) (11, 13, -2)
(a) (b) (c)
(-z 0,z) (0, -z,z) (-z, w,z, w)
(a)
/I \o
(b)
(f) (g) (h) (i) (j)
-
{4~2y~ 3z,y, z) (4,3) (5,2) (5-y,y, -1- w, w) no
solutions
-
/I
\o
-6 1
0
-4
^
1
(c)
\
-11 ,
/
/l
2
1
0
1
0
0
1
V
0
0
0
0
0
6
1
0
-4
0
1
7
4
0
0
1
0
694
5\
-10/3
/
0 0
Chapter 6.
(a) (b) (c) (d) (e)
section
11.
12.
(a) (b) (c) (d) (a) (b) (c)
(32/78, -5/78, 35/78) (-3 5x5, 2 + (10/3)x5, -(7/2) 4x5, 1/2, x5) (-3 5x5, 8/3 + (10/3)x5, -4 4x5, 1, x5) -
-
-
-
no
solutions
no
solutions
1.2
(120/52,4/52) (4/5,-8/5) (-51/22,29/22) (-39/5,-43/5) 3x+7y =
x-y y
-
2x
=
-l
8 14
=
(c)
(b)
14.
1
0
0
0
12
15
14
-4
-5
2
3
19
-1
(d)
(c)
(a)
(24)
/
\
4
24
24
12
8
12
12
6
4
2
-6
-6
-3
-2
-1
48
48
24
16
8
42
42
21
14
V
(b)
(d) 15.
(a)
doesn't exist
I
0
0
3
-1
,-20/78
7/78
(c)
0
1/6 1/6 1/2
-1/2 1/12 5/12 1/4
0 0
1/8 0
696
Answers to Selected Exercises
0>)
/
|
16.
0
0
0
\
1/3
0
0
1
1
-1
I
1
0
-1
\
0
0
0
I 1/2/ 0
conditions
(a)
no
(b) (c)
6'=0 2b2 + b3 + b*=0 b1 + 3b2 2b* 0 4b2-b* + b5=0 -4bl 25b2 + 14b3 + 1064 -
(d)
=
-
=
0
17.
Since the index d of A is at most n, if P row reduces A we obtain ,Mx Pb. d rows of iM are zero, b must satisfy the (nonSince (at least) the last m d entries of Pb are zero. vacuous) conditions that the last m
19.
Ifx
=
20.
If
=
=
x
2x'i,r(x)=2x'7,('i) 2 x'E, T{x)
=
0.
2"=! x'-Ei+i + x"Ei,
=
,
so
T is
the conditions.
section
21.
1.4
(a)
4
22.
(a)
3
23.
(a) (b) (c) (d)
(b) 3 (c) 3 (d) 3 (b) 3 (c) 3 {xeRi;xl + x2=x3 + xi} x2 {x e R*; -3X1 + x3 0, 2X1 {xe,R3;x1 + 2x2-x3=0} {xeJ?5;x3=x5,x2=0,x4=0}
24.
(a)
No
25.
(a) (b) (c)
0 6v3 + 5y4) c{4vi -v2 a{v2 vt) + b{2vi + v3 + vt) 0 avi + b{2v2 2v3 + v*)
26.
The
27.
(a) (b)
given vectors form a basis for R5. (-1, 1/2, 0, 0, 1) (0, -1/2, 1, 0, 0) (0,-1,0,1) (1,2,1,0)
SECTION
29.
(a) (b)
=
(b)
Yes
-
(c)
x*
-
=
0}
No
=
=
-
0
=
1.5 #
=
R
=
K
=
r
=
{6x1
=
17x*,x2
=
-2x*, .v3=x4/3}
R3
{x1=0, x2=0} 1 lx1, 2x3 12x* {6x2 =
-
=
4x*
-
39X1}
uniquely determined by
Chapter (c)
(d)
30.
31.
(a) (b) (c) (d) If
K
=
R
=
#
=
R
=
K:
{x1 0, 3x2 + 2x4 0, 3x3 + 4x4 x2 + x3 x* {x1 0} {8x1 + x4 + 6x5=0, 8x2 + 5x4 + =
=
-
=
I
697
0}
=
-
7x5
=
0, x3 + 2x4 + 2x5
0}
=
R3
(17/6, -2, 1/3, 1),
R: Eu
E2,E3
K:
(0,0, 1,0), (0,0,0, X), R: (1, -11/6, -39/2,0), (0, 2, 4, 1) K:{0, -2/3, -4/3, 1), R: (1,1,0,0), (-1,0, 1,0), (1,0,0, 1) AT: (1, 5, 16, 8, 0), (3, -7, -8,0, 4), R: EUE2, E3 nonzero, its range is all of
/is
R,
so
its rank is 1.
Thus its nullity is
n-\. 32.
i
{Ei-Er.
section
=
2,...,n}
1.6
(d)
/
1/9
1/9 0
-2/9
/0
(c)
1
1
1
0
-1
0
0
0
34.
(a) (b)
35.
By induction
(0,2,1) (-17,5,1)
zero
we can
for all / <j + k
section
37.
-
.
=
1.7
Eigenvalue 2 (a) 3
-1
6/9
>
show that A" has the property that its {i, j) entries 1 Once k > n, these are all the entries.
{I+A)~l= 2( ti-D'A' 0 (
0
1/9 -1/12 -1/9 5/18
(9/8,1/2,-7/8) (-1/2,1/2,1)
(c) (d)
n=l
36.
2/9 -2/9
-1/3 1/6
Eigenvectors (0, 1, -2) (1, -2, -4)
(1,1,-4)
are
698
Answers to Selected Exercises 1
(b)
(1,0,1,-1) (2,0,2,-3) (0,0,1,-1)
-1
0 2
(0, 2, 0, 0) (1,0, 0,0), (0,0, 1,-1) (0,1,0,0)
1
(c)
4
basis of
no
eigenvectors
2
(d)
(1,0,1), (0,1,1) (1, 1, -2)
-2
39.
If G represents the standard use the fact that
basis,
we use
Exercise 38 to find AGF, AaE and
-Ao^Ao*)-1 (c)
0
0
1/2
0
0
0
0
1/2
0
-1
1/2
(b)
(-32
\ =
TF
(AS)-lTEAE'
section
42.
=
cl, when
is
=
a
basis of
(5 + 30/34
cis(-2/3)/4 cis(-7)
(3-0/10 1
if and
43.
zz
45.
(a) (b) (c)
cis(/c7r/5) 1,3,5,7,9 1, i(2l'8)cis(7r/16)
(d)
/,(l +
=
only if
z'1
=
(-l+0/V2 =
iV3)/2
4
eigenvectors,
c{AEF)-'{AEF)
(d) (e)
1 + /
3
0
1.8
(a) (b) (c)
1
-1/2
-5
I
If TE
0
1/2
-1/2 \ 7/16 -1/2 j
-1/2 -5/16 1/2
41.
-1/2
z
=
cl
then for any basis F,
Chapter
46.
(e) (f )
2501'6
(a)
(1,0,0,-0
(5)1/2cis[iarctan(-|)]
(13, 5), (1,1)
(c)
(2, 1, 0), ((1 +
+
(l/3)arctan(l/3)]
i/V3)/2,
k
3
-
-2 +
1.9
v2
v3
Vt
v5
v6
13
24
20
5
2
17
41
40
0
4
34
7
5
0
0
5
-15
"2
67
3 y*
Vi
0
Us
0
21
v6
Table of
u x Vj v2
Vi
^3
(6, -5,-5)
Vi
(5,4, -7) (-5, 13,7)
V2
v3
Vi v2
v3 Vi
Vs
v6
l>7
(9,2, -10) (-1, 3,0) (0, 7, 0) (-2, 6,0) 1, 0) (3,
(11,--72,25) (-41, -123,0) (-25. -222, 35) (-67, -201,0) (63,--21, 50) (-5, -15,0)
-
-
v6
49.
(37, 16, -28)/17
50.
(a) (b) (c) (d) vt:
(9,2, -10) (3, 9, 0) 00,--5, -14)
(-6,2,5) (-12,4,10) (-15,5,21) (-15,5,20)
Vs
51.
//V3)/2,_3 + i/V\ 1)
-
Table of
48.
0, 2, 4
=
//V3, 1U(1 (1, -4, 5, 1), (1, 3, 0, 5), ((-3 +V21)/4, ((-3-V21)/4, -2-\/21/2,0, 1)
(d)
47.
cis[/c7r/3
(b)
SECTION
1
9X1 + 2x2
10x3
-
=
0
12x1-4x2-10x3=0 3X1
x2
-
=
3xx + 9x2 x
=z,
2y
v2 : x
+ 3y
v3 :
y
=
Va.:
x
v5 : z v6 : x
y7:x
=
5x
=
0,
0
=
0
z z
+
4y
=
=
=
0
=
0
=
=
=
0
7z
0, + 3y 0, 2z + 5y 0, 3x y 0 0, y + 3y 0, 7x + 5z =
=
=
,
,
,
V21/2,
0, 1),
700
Answers
52.
7X1
53.
4X1 + x2
54.
(a) (b) (c)
55.
x2
-
to
-
-
Selected Exercises
2x3
=
3x3
=
17
2
<x,,2-i>=0 <x,3-'.> 3x3=0,2x' + 2x2 + 5x3=0 =
x1 + x2 + x2 0, x3 =
The planes
(a)
0
given by the equations l (b) x=y (C) x 0 intersection of (a) and (b) is given by equations (a) and (b), x
The
=
are
+y+
z
=
=
56.
(a)
58.
The area of a parallelogram (either) included angle.
of side
59.
False if
and w, but
60.
Apply are
section
61.
x
(a) (f) (k)
+ y=2z
u
the
always
(b)
=
y
is perpendicular to
equation
v
c>
x
(c)
z
=
j
det b
z
=
0
lengths
I
v
a, b is ab sin
to each pair, and
where 6 is
w.
observe that there
same.
1.11 open open
(b) (g)
neither
(c) closed (h) closed
open
(d) (i)
closed open
open
(29, -3,26)/14
63.
(Ill, -22, lll)/34
64.
(0, 1, D/21'2, (1, -l.D/31'2 (b) (0, 1, 0, l)/2"2, (1, 0, 1, 0)/2"2, (-1, -2, 1, 2)/10"2 (c) (0, 1 0, 0, 0), (0, 0, 0, 1 0), (0, 0, 2, 0, l)/5"2 (d) (1, 2, 3, 4)/30"2, (2, 1, 0, -l)/6"2, (1, -3, 3, -l)/20"2 ,
(a)
0,
is not perpendicular to
62.
65.
etc.
,
K:
(10, -16, 16, ll)/477"2 (0, 0, 1, 0), (1, 0, 0, l)/2"2, (1, 2, 0, -l)/6"2 (b) JC:(1,0. -1, 0)/2"2, (0, 1, 0, D/21'2 R: (-1, -2, 1, 0)/6"2, (3, 12, -3, 2)/156"2 R:
-
(e) (j)
closed open
Chapter
section
2.1
1.
(a) does not exist (g) 1 (h) 3
2.
No, takex
4.
Yes
5.
0
SECTION
9.
701
2
Chapter
8.
2
=
(b)
0
(c)
0
(d)
no
limit
(e)
1
(f) 1
_1
2.2
-1/2 Form the
new sum
in this way: at any stage, if the sum is 1, add the first used, and if the sum is less than 1, add positive terms
negative
term not yet
until the
sum
is 1.
The
resulting series
is
1 1 IIIIIIII1JLJ.J. + + + + + + + + + + 2 4_2 8 4 8 8 8_4 T6 T6 T6+16_4
The terms
come
second bracket
in blocks.
'"
The first bracket encloses the first block, the The nth block consists of 2""1
begins the second block.
copies of 11111 2
10.
'
2"+2
Since the
Yes.
negative
2"+2
2"+2
terms is
sum
2"+2
of the
oo, we
positive
can
terms is +oo, and the
of the
so
less than 10,000 we add positive is not less than 10,000 add negative terms until 10,000 is
SECTION
sum
that at any stage, if the sum is terms until 10,000 is passed, and if the sum
rearrange
passed.
2.3
11.
(b), (d), (f ), (g), (h), (1), (m) converge (a), (c), (e), (i), (j), (k), (n) diverge
14.
(a) (f) (j)
|z|
=
0
U+z|
(b) |z|
(c)
allz
(d)
(h)
|z|
all
(e)
z
(i)
allz
|z|
702
Answers to Selected Exercises 2.7
SECTION
15.
(a)
tt
(b)
2/3
17.
(a) (f)
0
(b)
1/16
18.
(a)
1/6
19.
(a)
3tt/16
SECTION
(b)
1/60
e-{H2e)-3/2
(c)
1/48
(b)
df/8x
1/10 (c)
l
8f/8y
(d)
cos(xx)
y
x
yzx(>'^-1,
Since VA
=
(d)
tt/10
(d)
1/24
xy
In
x
x'y2 In
In y
x
+
x^lnx)2] .
.
.
,
i
dhjdx") for
any function
just
as
for functions of
one
variable.
By Exercise 23
25.
5/9
26.
101/2/(l
+
101'2)
2.9
29.
(a)
30.
(b), (c), (f)
0
(b)
0
converge;
(c)
1
x
=
x
=
(d)
0
(a), (d), (e) diverge
2.11
(a) (b)
h,
df
8g
^{f-f)J-fVf+f^f)
31.
5/6
x2 + 2yx
(3/z/fix1,
The proof is
SECTION
(e)
cos(xy)
x^zy*-1
2xy + y2
x^tx'-'+xlnx
SECTION
1/4
8f/8z
xz
yz
8
24.
4/3
2.8
(a) (b) (c) (d)
23.
(c)
(d)
l/3)j3M[(l+sec26>)3'2-lld0
20.
21.
2/3
(c)
(l)(x_1 + a/x_i) 2x_i/3 + a/3x2
(e)
1
we
need only show that
Chapter 32 32.
U) (a)
^_3^.1_3__:___
(b)
X"
=
(c)
X
=
(d)
x
33.
(a) (b)
all
34.
0
,
1
X_i
.
..
1 ^2-i + 2x-i + 3
x--x
2xZ^i'X>=1 x3_ i
.
X_,
3x2_i
=
-x_i
5
+ 4-
5x*_i
=
2.
(a) (b) (c) (d) (e) (f) (a) (b) (c)
(d) (e) (f)
1
x
8F
F{x, g{x))
-
=
3
=
0
8F
(x, g{x))
+
(x, g{x))g'{x)
-
sin(x + y)/(l -
3
SECTION
1.
4x_i
(xy + tanxy)/x2 + sin(x + y)) -y/x ye">l{xe*y 1)
(a)
(b) (c) (d)
Chapter
3x.i + 2
-
points except on the line i) all except (1, -1) (ii) all points (iii) no points
d
35.
2x2_ i
-
j
3.1 ce"
(sin /, cos /, 1) { a sin t,b cost) (2/,3/2) (l,2/,3/2) (cos /, -sin /, 0) | c | exp [(Rec)/ ], arg c 21/2,7r/2 {a2 sin2 / + b2 cos2 z)1'2 |/|(4 + 9/2)1'2, arc cos(4/ + 18/2)/2 |/|[(4
(1 + 4f2 + 9/4)"2 1 w/2
+
9?2)(1 + 9/2)]1'2
3
703
704 3.
Answers to Selected Exercises The tangent to (a) at the point e" is parallel point {a cos /, b sin /) precisely when
lb
1 T
=
-
i[mc
arctan 1
tan/
-
\a
4.
Never
5.
{ab)l>2
6.
2
7.
(-1, D,(-1,D
8.
min(l/fl, \/b, 1/c)
10.
Eigenvalues -4.516
(b)
14
(c)
3 +
10 3
(d)
1
11.
(a)
-
1 + -
(-1,1) (0.383, 0.924)
4(2)1/2 4(2)1/2 2(10)1/2 2(10)1/2
(0.924, -0.383) (0.585,0.811) (0.811, -0.585)
2 + 21'2
(-0.383, 0, 0.924) (0, 1, 0) (0.924, 0, 0.383) (-0.501, -0.382, -0.777) (0.838, -0.438, -0.325) (-0.216, -0.814, 0.540)
1
2-21'2 7.411
(b)
0.313
-1.724
SECTION
:}.2
12.
x
13.
0.1987
-
J
Eigenvectors (0.685, 0.729) (0.729, -0.685) (1,1)
11.516
(a)
1
x3/3 + x5/5,
-
x
+ x2
14.
1.7320
16.
[-0.0312,0.0322], |x|< 0.1
17.
|x| ^0.125
SECTION
18.
(a) (b) (c) (d)
-
x3 + x*
3.4 y
=
y
=
x
=
z
to the
=
(-l/2)exp(-x2) + 3/2 xcosx
/2/2,y -
ie" +
+ sinx
/3/3 + l,z /4/4 (1 + 02/3/3 + l +
=
=
i
tangent to (b) at the
Chapter 19.
20.
(a) (b) (c) (d)
y
y
=
(e) (f) (g)
y
=
y
=
y
=
(h)
y
=
(c-x)-
=
tan y + sec y
x3 +
y3
=
=
K{t&n
sin(x-(l/3)x3+C) Cexp(x + x3/3)
(i)
[ln(l-x)-x-c]-1 -\n{cex) tany + secy A:exp(-2cosx)
(a)
y
=
(b) (c) (d)
y
=
=
exp(-x2/2) fgexp(z2/2)cos/rf/ (sec
x
cos
x)/2
(e)
exp(-x2/2) Js exp(/2/2) dt y exp(/(x)) Jg exp(-/(0 2it) dt + exp(l/l where /(x) (exp(l /)x)/(l 0 y ln(x+e-l)
(a)
y
=
y
=
(b) 22.
x)
sec
K$exp{t2/2)dt+C
y
=
x
-
=
-
=
21.
+
x
C
(a)
(b)
-
0
-
=
-e*/2 + e-*/6 + e2*/3
e'{cosW2t + sinh\/2//\/2)
KJ
=
ci
exp(4 + 20/
([ ) +c2exp(4- 2i)t(].\
aVo a^O
fe) a
(c)
-
=
+a/o)'(-V)
=Ciexp(1
0
+ Cjexp(1
-Va)'U)
\
/\
/
\y2]
\e'{-c2t + c1)J
c2e-
a^O
^) ft)
=
ci
e'{
1
-
j+
c2
exp(l
+
c3
exp(l
+
-
\/2a)/ j a/V2a V2d)/ j -a/V2a
J
\-fl/V2a/
C3)/e' + Cie']
23.
(a)
ci=exp[(l-0/2]
(b) (c)
ci
=
yi
=
ya
=
lo)+c2e'|l)+C3e'jO _
_
(1 1/2, c2 [(1 + Oexp(l [(1 + Oexp(l =
-
-
-
-
V2a/a)/4 (- 1 /)exp(l + z)/]/2 Oexp(l + i)t]/2i
V2a/a)/4, c3 i)t + (1 i)t (1 -
-
-
=
-
3
705
706
Answers to Selected Exercises
/5+Vl7\ (d)
4
/y^
9V17-17 :
34
w
exp(l+\/l7)//2 5+Vl7
\
*
/ 4
9V17-17
exp(l-Vl7)?/2 5-V17
34
24.
w
fc)-H )+*-G)
(d)
/yi\
/-
fcr-T
c2e4<{ lj+c3e-2,|-l
+
(e)
\yj
=
Cl
eXP^6 +
c2
(f )
(g)
yt
=
y2
=
i3\/5)t j
exp(6
i{c exp(4
.
*VM 2+/3V5
-
70/ +
(c exp(4
c
70/
2
_
/3-\/5
exp(4 + 7/)/)
c
exp(4
+
70/)
yi=Re(cexp(3-20) y2 Im(ci exp(3 20) y3 Re(c2 exp(l 2i)) y4 Im(c2 exp(l 20) =
=
-
=
(h)
/yA y2
=
e'(c3 /2/2 +
c2 /
+
Cl)F, + e'{c3 1 + c2)E2
+
c3
e'F3
Chapter (>)
26.
27.
Cie'Ei +
=
y
=
e-*/2j'/Vrfz + c, x2/2
(a) (b) (c) (d) (e)
x
c2e* + *
or c2 e*
=
29.
j>
=
30.
y
=
+
+
+
+ c2
c2
e~2x +
c2
(sin
x
-
3
cos
x)/10
cix
4x2/9 + 5/9x + c,x
2x2 In
x/3
c2xj"exp[(l-z)e']/z2
+
e-'{exp[(l
x)e'] +
-
x
j e' exp[(l
-
t)e<] dt} +
x
-
1
4.1
x{9)
2.
x{6)=
3.
6 arc cosf/"1 by taking 0 < 6
=
(cos 0, 0,
sin
=
0<6< 2n
6)
a cos
1) < -n;
r
>
i
cos
(1-/,/) Z cos 1) (1 + Z, sin 1 + (1 Z,1-Z,Z) {2a, at, Z) (Z, 1,0 (1 + 2z, -2z, 1 + Z) -
axis
(a)
x
y axis
(c) (d)
y axis
x=y,
J parametrizes the
in the lower
curve
z
=
0
in the upper half-plane -n < 6 < 0.
half-plane by taking
te"lb
e"lb{b sin t +
(b)
0<6<2tt
(cos 8, sin 6, 1)
-
=
=
7.
ci
c3
+
1.
(a) (b) (c) (d) (e) (f)
c2)z + c3]E3
4
SECTION
6.
-
c1x2 + c2x3 + (lnx)(-x2 + x3)
y
z{t)
e'E2 + e'[{ci
Cie2' + c2e-2'-l/4
28.
5.
c2
3.7
y
Chapter
707
/M
yA= SECTION
4
z)/(cos2
/
+ b2 sin2
z)1'2
708
Answers to Selected Exercises
SECTION
4.2 ds
8"
(a)
2a
(1 +
dd~
6 +
cos
a2)112
{l + acosey
ds
(b)
^=(5 + 4cos0)1/2
(c)
{6a)s
=
{l-e-)l2112 (x4 + cos2(l/x))1/2
ds
(6b)dx (6d) ds {6f)s
10.
x~2
J" (a2 + cos2(0/2))1/2 dd $ {St2 + l)112 dt
=
=
(ja)s 7fl>I
=
(z2 + l)1/2
c" =
'
dt
J_1-7+T
(a)
a=21/2e'
(b)
aT
=
(d)
aw
=
ar
=
aN
=
ar
=
=
flr
(9Z2 + 16T2
4+18Z2
11.
(a) (c)
ar
(T+970I72Sgnr'GN= [(1
sin
-
=
4.5
(a) (b) (c) (d) (e) (f)
y
(a) (b) (c) (d) (e)
y
=
cx
x
+
y2/2 c y2=c
13.
0/2]1'1
|sin Z 1(2/2 + cos 2/)"2 -sin2z/(2 + cos2z)1/2 [(z4 + 5z2 + 8)/(Z2 + 2)]"2 z/(2 + Z2)1'2, a
=
=
4 + 9Z2
-[(l+sinz)/2]"2
SECTION
12.
'"
-xy'
xy' + y 0 1 x exp(-y/xy') x sin y cos sin x(sin y + xy' cos y) + + y ') exp(x y) y'/y(l y l+y'=0 =
=
=
=
=
2x + x
=
c{ y
cexp[-f*{t + smt)-l]dt -
1)
=
csc(tt/4
-
x)
x
Chapter 14.
x2 +
c2 (y c)2 (y2-x2)y' + 2xy -
+
2a2
17.
(a)
18.
(a),(b)x2-y2
19.
(a) (e)
ex
x2-y2=c
=
xy
(x
+
(a) (d) (e) (f)
y
=
y
=
SECTION
4.6
21.
22.
23.
24.
=
=
xy
(x,
2V3xy
+
0
(b)
y2-x2=c
=
(c)
c
x
+
c
-
xy=0
(d)
sinx=0
l e*
l, r 0,r
=
y,
cos0
=
=
c
z)
=
l
(b) x2 + y2=c2 {ae\ b-t, ce1)
(c)
(a)
(-x, 1, -z) ((1 + t)x, (1 + t)y, 2/(x2 + y2))/{\ (0, 1, -ztanz) (-x, -j, -z(l + tan z))
(a) (b) (c)
exp(z2/2)(x0 y0 z0) (x0 cos / y0 sin z, y0 (x0e_t, yoC, z0e')
(a)
=
^ ^\n(x- ~A
=
(c)
xy=0
y
y)y=0
(b) (c) (d)
(a) (e)
(c)
y
(l+'j^V
,
0>)
(c)
xz
=
c, yz
(x, -x2/2)
+
=
d
(d)
(l,y,Vl-z2)
z)2)
,
cos Z
+
x0
sin Z,
Z
+
z0)
5
SECTION
2.
e
(b)
(a)
Chapter 1.
r
(a) (d)
J
22-l
y2
20.
0
=
(y-x-1)2
16.
=
709
=
{x + y)2 "
5
5.1
(b)
|x|
|z|
(f) (b)
|x|
|x|>l
(c)
x<0
< 1
allz
(c)
lmz>0
(d)
-19<x<-17
710 3.
Answers to Selected Exercises
(a)
both
tiate for
4.
(a)
not an
=
(b)
2f>x-1
(c)
2
v2( + l
i.to(2/+l)Z! co
n
(e)
9-
=
4+l
0
5.3
-e2x + xe2' + ex
(3/4)e-* e2x
-
3ex
-
2i)el* + (- 1
(7/40)x
-
(l/2)[exp(21'2x)
X++-- +
x'
t
(l/4)e3* {5/2)x2e2x
xex
(6/5)e*
12
11 11.
+ 2xe-* +
2xe2x +
Re[(l
-
(f) (g)
fO,
2 (-!)"
(d)
(a) (b) (c) (d)
both
0
co
5.
(c)
integer
22(n + l)(-x)2" 11
SECTION
both
(b)
x/2tt
504
+
-
-
i)xe'x]
(l/40)xe"3*
exp(-21'2x)]
-
e~*
40360
x
\
(a)
2a"+1 a+ n
+ 2
a"
{n+ 1)( + 2)
3K
ln|<-
,
2a"+1 (D)
tf + 2
n
+ 2
4K
\a\
....
!
a"-*
_
;
{n+\){n + 2)
-|
n\
(d)
integrated for
all x, differen
(c)
( + 1) ( + /c) a:
lnl
(d)
<
!
&2o *"
( + 2)(+l)
k,i
fln-l
(e)
fln + l
.
rt
, + 1
/s:
kl^[n/2)! SECTION
5.5
co
14.
(a)
(b)
2 to
2
z2ll n\
d + /)"-(i-0" ^-j 2in\
11=0
[n/2]
(c)
(d)
2
(f)
2
16.
n
j-i
1
y i 2 jioKn-f)in -
nl{2nV. co
(g)
( l)fc
*o (2A:)!
=o(2 + l)! 00
(e)
*"
,2n + l
2
=o(2 + l)!
sin z sin w cos z cos w cos(z + w) cosh(z + w) cosh z cosh w + sinh z sinh w sin(z + w) sin z cos w + cos z sin w sinh(z + v)= sinh z cosh w + cosh z sinh w =
=
=
712
Answers to Selected Exercises
SECTION
11
5.7
42
!_n, 1)'
+
2" ao
2
W+ D
(2;)(2; + l)]x2 +
-
[*(*
+
D
"
(a)
aa
(b) (c)
y
(d)
y
=
=
2
1+jc)
JJcy+l-A)
(2)
l,ai =2,
x2/4
or
y
a+2
21.
a<
_
(+!)( + 2) ,+%/!
0
=
solution
no
=
(2x2
-
c)-1
=
-(1/ci) 2 (2/ci)"x2* n
20.
2)]x2"+1 +
x2"+l
1 19.
+
"-'
2" ,
(2/ + 0(2;
=
(2n)!y=o
+
1
j=o
ai{ 2i(2+l)! 7^-r-TT7 it fJb n
18.
IT
ln=i n {2n+
=
0
10
(a) (b) (c)
460
(a)
k odd- integer,
3
a0
=
0
or
k
even
integer,
ai
=
0
Chapter 6 SECTION
6.1
1.
/(0)=772/3
(a)
/()
=
(-!)"
(b) /(5)=/(-5)
4 =
l/32
/(3)=/(-3) 5/32 /(l) =/(-l) 10/32 /() 0 all other n =
=
=
(c)
/()
(-l)"e'""=
-^3
27T(
e"""1
(-l)"sin(fi7r) =
,
/it
n
7r
p.
n
p. not
an
integer
x]
Chapter (d) /(0)=W8
/()
=0 =
=
l/rm2 2/tt2
=
if
n
is odd
if
n
=
=
(/) /(D
4/c + 2
nodd
/()=0
(e)
4k
ifn
2)
2/tt(1
n even
(1-0/2 /(-1) (1 + 02 =
=
/() (g) /(0) /() f{n)
=
=
all other
0
=0 =
n
=
-2/mn
(-!)'
2tt(1
2+
2.
/():
(a)
(Re z)3 + (Im z)3
(c)
W
(e)
(f) SECTION
3.
(a)
(b)
4k, k = 0
)
1+n2
r2e-2'9)"
sin fA7r
i1
8
=
e -
Kt)'^"-
(b)
or
2en(-l)"
(i)
2
odd
n=4k + 2
e*
(h) /()
n
l/2
(-D"-rrijr
2
+
-Re 77
..?. (27TT?
e
'
H-
(1 + 4 6.2
-i +
^lmln(^)
-^(^
+
z"2)
+2,i. (2* + D2
6
713
714
Answers to Selected Exercises 27J-2
(c)
(d)
4.
r\*\el*0 +
3
22i-iyrn n*0
1
2(-D"s-^W
7T
jLl
n=-co
(e)
(1+z)2
(a)
r
sin 0 + r2
~2n + l
1
{2n + l)2
2
a,
5.
Im
-
tt-
2 f0
6.3
(a)
(35/2 +
(b)
2/
,
28
(e)
20 + 14
1 2
2
+
2
-
77
=
~
2
tt
-
cos
sin(2n .
sin(4/c
A
+
koAA
+
{-\y
5
cos
30 +10
cosine series
cos
sine series
4 77
(c)
cos(277x)
cos
even
0)/l 6
"
sin(777ix)
1
-
k
\kl2-)
2)0
7.
(1
80)/64
2/c+ 1
See Problem 27, Section 6.5
(b)
cos
.
6.
(a)
60 + (1/2)
2 + 1
0
(cos 50 +
cos
1)0
+ .
40 + 4
2j)6
{-\ycos{k-2])6
12"
( )
cos
'My-**sin{k
m (c)
2(-l)"-2\
(-D-1
\n2
1^ I
TTn^O
SECTION
20
H"1 /tt2
i
(b)
(c)
cos
4ttx)/2
11
=
n
1
sin(27rx) 4 n
{2n + l)sin(2n + 1)ttx
A
(2n + 3)(2n + 1)
Chapter 12+ 2
(-l)"cos(2
~
W
9 l
77
=
n
+
2
l)7rx
(2tf+l)
77
n
=
(sin(4
-
1
+ 2
(e)
(1
cos
1
8
(1+2
sin
cos(4 + 2)t7x
^
772 tb
4
(g)
2t7x)/2
cos 77x
sin 77X
+
/(0)
9.
2
=
+
l)77x
(2 + l)2
^
{2n
r-^7;
3)(2n
+
rr
1)
-
sin(2/7. + l)77x 7
6.4
/sin(77rt 1) /sin(ir-l)
-S
,,
2
(a)
n=l
sin(77+l)\
:
\
TT~ 1 77W+ /
1
77H
sin 77Z sin
-
H 77
TT3
2 n
=
r7T3
(2+ 1)J
0
-
sin
(a)
77X
+
77x
=
i
nn
tt cos
,
m//l
(42
2t7Z
sin
2t7x
1)
9)(4n2
cos(2n + Vnt sin(-2n + V
-
77
r
.
2i =
T
in
cos
;
(42
/-772n2z\
-
Sln
-
9)
/sinew?
-
L)
^^(-^rrvrjvi^i /-77"m2z)\/
2lrnx
sm 77Z sm 77nx
z
l)(4n2
(imx\
27Tnt
1
42
n{n2-2)
^
> V( l)"'1 ; 772 ti
8
sin
77Z
77
128 (e) W
2 ,,
8
1
(d)
sin
1
8
(c)
77X
77
cos 77Z
n2
64
1
(b)
n.
sin(2
W2 cos(2)0 + B2n+i sin(2 + 1)0]
,
10.
}"
0
SECTION ,n
(
772 to
[/(0) +/(-0)]/2 + [/(0) -/(-0)]/2
=
2 n
77X
2n + 1
2,
-
77 =0
8.
1)ttx
2t7x)/2
cos
4 2
+
sin(4 + 2)t7x + sin(4n + 3)t7x)
^4_ ^
(4n + 2)2
\
.
715
1
-
2
~
n,is
0
6
/277X\
7ae*p(-Hl^rml~ j
sm{-rm + L)\
^nrj
716
Answers to Selected Exercises
,
/-772Z
-8L2
,
/-^2'\
(d)
.
12.
,
s
iym-
/-25t72z\
-"X
+
2
,
(c)
l+(
n
L
Sttx
(a)
-
4e
2 exp(-772(2+l)2Z) n
772Z\ exp
4Z,2
=
0
w(2-e(-l)")
=
2L
(b)
sin
77
2l exp( 7722Z)
L
15.
(2/J + 1)77X
+
exp^lzrjsu.r 3exp^-iIj-Jsi:
+
14.
1
\
,
^((2*+ I)2)
(C)
+ 772
1 +
sin(2n+l77x)'
\2 + 1 ,
sin77x
77X
,
ln2n2t\l
\
1
277/tX
-T.?1P(-y-J(S(i?^j
The
sm
general solution is of the form
2 {An sin(n2 + 1)1/2Z + B cos{n2 + l)1'2) sin nx n
=
l
where the A
16.
,
B
On the interval
are
determined
by
the sine series of the initial data.
[77, 77]
2 (e2"-l)W. + Ae"") n=
00
where the A
section
17.
(a)
B
are
determined
2777/64
-
77
n2
-
2 n=i
71
{n2
sin
2|U77 ^
=
p.2)2
2/u. \
1+
277r
2ju.77
^
1 +
(2 lrl")=llTnz (d)
by the Fourier series of the initial data.
6.5
4
(b)
,
r
r"
Chapter
2
277
(e)
n
=
-
l
n
2775
1
2" ^-+1677 9 rr
(f)
n
20.
=
l
6.6
section
19.
(a) (b) (c)
span of
(a)
(sin50
span of
exp(3Z0), exp(-5/0) exp( 3i8), exp( iff)
span of ew
+
cos50)/576 + ^sin0 + cos0
2772
(b)
+
-
27
If
(c)
277 J
22 to
i)V (( 1)e
n2(9
n2 + 6m)
-
2
exp(cos0) n
_
=
-co
(a) (b)
!5 + l
T
1377/4 0
k
2^k~-n){-T Chapter
eXP['?l^)]^
6.7
section
21.
**
7
SECTION
7.1
(a) (b) (c) (d) (e) (f)
[ y sin x + z cos(zx)] dx + cos xdy + x cos(2x) dz [ex+y sin(e*+0 + ey sin(xeO] dx [ex+r sin(x + y) + e<x-aKdx,ay (dx, e<x'a>y + <x, e<Jt'0>>
(g)
2
2.
(a) (b)
2/1000 l/1000e
3.
||/>||2/1000
1
.
7
1
00
-
-
n*( \dxj
if
(c) (d)
2/1000 2/5000
(e) (f)
2/1000e 1/lOOOe
||p||<2, IH/500if \\p\\^2.
xe>
sin(xe")] dy
718
Answers to Selected Exercises 7.2
section
/ 4.
(a)
e"
xe*\
\yex
exj
/
(b)
1
jx2 \
+ x3
/
(d)
1
x'+x2
x'x3
x1x2
2x
2y
2z \
1/x
0
\-z/x2 /
x^O
1/x/
0
1
0
0
-x2/(x')2 -x3/{x1)2 -x-'/tx1)2 (f)
1/x1
0
0
1/x1
Khu...,K) X"-V -A+ d{x1, dx")
-
U1
6.
1
(a)
2 =
M
j0
7.
9.
10.
du
wy
1 + w2
+
:
1 + v2 + w2
t
2
c/2
section
hi,
.
.
.
,
h coordinates.
du1
du/u
1
(C)
y
"' du1 + 2
x" ^ 0 and the
;
hl
2
wv
(b)
x^O
0
0
...,
5.
x3 all different
x1, x2,
-2y\
-y/x2
(e)
1
x' + x3
x2x3
/2x
(c)
xy^l
l+v + r../i
i
-2
,
=
fixed
V2 ~
(1 + w2 + y2)2
w ...2Ml/2
[(1 + W2 + H>2)] c
-
tt~,
du +
1 + v2 ww
dv +
;
, vu
dw
ull2w{w v) ull2v{vw) 77777 dw + 77 77777 dw (1 + u2 + w2)3'2 (1 + y2 + w2)3'2
7.3 are
closed
are
not closed
The real part is exact, the
w2
(1 + w2 + v2)2
edge
(a), (b), (d), (g), (i) (c), (e), (f ), (h), (j)
-
r-:
imaginary part is
not.
Chapter 11.
(a) (b)
13.
15.
(d)
(a) (b) (c)
0
(d)
-z-z2/2
(e) (f)
-a2/2 21/2
-1/y2 1/x
(e) (f)
exp(-x-y) 1/cosx
1 -e cos
1
{e2 sin l)/4
-
-
e2/4 + 5/4
22as/5
-
(b), (c), (f ) are conservative (a), (d), (e), (g) are not conservative
section
14.
(c)
e'/y
719
7.4
section
12.
e
7
(a) (a)
7.5 0
1
(b)
0
(d)
e6(cos2 2)(sin 2)
(e)
77
1-1
(b) (c) 16.
(a) (b) (c) (d)
2ao
b2
a2
18
l+ln21/2-2-"2 Letsin
0o=(51/2-l)/2.
The
area
is (3 sin
20o
+
0O
-
sin3
0o)/6(2)1 '2
16 If
n
is
even
there is
inside.
no
If
n
is
odd, the
area
is
MI section
17.
(a) (b)
7.6 8o*
2
section
7.7
19.
(e4*-l)/4 4/3
(2t7 + 5A/3)/3 (b) 1-2Z-3Z2
(c) (a)
18.
(d) (e)
oo
21/277
(b) (f)
3t7/23'2
(c)
2x + 2y
(c)
(a) (e)
2t7/3
(h) (k)
T7[2 sin(77/10) + 2 sin(377/10) + l]/5
e'[5/-3]/6 (g) 77e-(l
277/(l-a2)1'2_ 77[cos(\/2/2) + sin(-v/2/2)]/-v/2 exp(-V2/2)
0
(d)
+
(d) 77/cos(l/2)/2 a)/2a3
_
(i)
77/3
(j)
n/e
720
Answers to Selected Exercises
Chapter
8
section
8.1
1.
(a)
oo
2.
(a)
2W495
(b)
tt{AB)1I2[-5{A3
+
3.
kna2r6/6
4.
477(ln2-l/2)
5.
(a) (b) (c) (d)
>
{ye-z(l
(a)
11.
+
are
3AB{A
+
=a~2,
B) +
Z2) +
B
=
b~2.
24A2 + 6AB +
y{\ -t2) + {t- \){ze' (1 + Z)(x tye'))
-
The
integral
is
2452]/3(2)6
txe<),
-
incompressible.
4t7/3
(b)
477/3
(c)
8^/3
8.2
(e'(l-z) + e-'z(z+l), -1, Ze'(l-Z)-e-')/l-Z3 (1,-1,1) -(1,1,1) (-l,2z,0) {e"2{l + 0/2, e\l /)(e'(l /) + 1), e"2(2 /)) -
-
-
(0,0,ysinz)
If M
=
{a'j),
div
=
=
ai1 + a22 + a33
(a32
a23, ai3
a3\ a2l
at2)
0
section
13.
-
47r/3abc
+ tz- t2x,
curl 12.
Let A
(c)
B3)
(d)
oo
7.
(a) (b) (c) (d) (e) (f)
0
77/2
-3/2
(a), (c)
9.
(c)
fcd-Z3)"1
6.
section
2t7
(b)
(b)
14.
8.3
ds2
=
dS
=
{l+fx2)dx2 + 2fxfydxdy + {l+fy2)dy2 {l+fx2+fy2y2dxdy
Tangent plane
(a)
=0
(1 + 4y2) dy2 + 8yz dy dz + (1 + 4z2) dz2
{2x2+y2)dx2 + 2xydxdy
+
{x2
+
2y2)dy2
Chapter (c)
(p, (-2x, -2y, 1)> =0
(a) (b) (c)
{\ + 4y2 + 4z2)112 dy dz 2ll2dxdy (1 + 4x2 + 4y2)1/2 dx dy
15.
(a)
477/31'2
16.
cos
18.
(a) (b)
(1 + 4x2) dx2
-
=
{2v +
SECTION
19. 20.
(a) (a)
SECTION
25. 28.
(a) (a)
y)/(l + 42 + y2)"2(l + u2 + 4v2)112
277[(l+a2)3'2-l]/3 77 J"_ (1 + sin2 u)1'2 du T
2u +
/\/4 + A/3\"
1
244+Viln(v53T7|). 8.4
21/277/4
(b)
0
e-2
(b)
-4t7/3
(c)
0
8.5
(b) 9TT/2{aby<2 (b) -4W3
0
^{ab)1'2
(c)
721
8xy dx dy + (1 + 4y2) dy2
Area element
0
8
1/18
,1'(;
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726
Index
Energy
Fixed
conservation of, 612 integral, 674
kinetic,
(Theorem 2.15), flows, 385 circulation, 626,
of curves, 376
Laplace's, 467 of continuity, 617 wave, 482
of motion, 311 of a particle, 335
Euclidean inner
R3,
213
Flux of
114
product,
660
curl about n, 626
divergence of, 582, 619 equations of motion, 310, incompressible, 618 irrotation, 631 steady, 612 velocity field, 612, 385
heat, 467, 489
inRn,
(Section 2.11),
Fluid
potential, 557 Envelope of a family Equation
in
theorem
211
501
Equations Equations
point
a
611
field, 660, 667
Force field, 255, 306 Fourier coefficient, 454
111
97
real, 476
Euclidean vector space, 113
Fourier series
Expansion
real, 476 sine, 476, 481 cosine, 476, 481 transform, 523 Frenet-Serret formula, 360 Frenet-Serret frame, in R" (Prob lem 66), 398 Fubini's theorem, 177, 184, 189 Function, 17 analytic, 400, 441, 534 continuous, 160, 201 differentiable, 527 dissipative, 596, 682 domain of, 17 infinitely differentiable, 443 flat, 443 invertible, 17 harmonic, 468 linear, 1
Fourier, 454 Laurent
(Problem 38), Taylor, 246 Exponential, 262 of
601
matrix, 279
a
(Problem 93), 226 Exponential function, definition (Proposition 25), 210 Exponential order, 521 Family
of
of curves, 365
differential
equation of,
370
envelope of, 376 explicit form, 366
implicit form, 366 orthogonal family, tangent
to
a
vector
374
field, 383
253
Fetah, Field, 306 conservative, 557 First fundamental form
Lipschitz, 269 meromorphic, on a
sur
face, 643 First-order linear
Fixed
point,
209
(Chapter 5),
odd,
478
one-to-one, 17
equations,
264
onto, 17
periodic,
452
616
452
Index range, 17 Schwartz test, 523 Fundamental existence and ness
i,85
unique
theorems, 271, 273
Fundamental theorem of
algebra,
61,86
Implicit function theorem, 216 Implicitly defined curves, 317 Incompressible flow, 618 Independence, 43 Infinitely differentiable, 166,
Fundamental theorem of calculus, 172
(Problem 48),
(Proposition 11),
681 655
(Problem 24),
Geometric series, 137
common
Green's identities
divisor,
a
for
a
111 97
improper (Section 2.9),
(Theorem 8.5),
(Definition 13),
of
a
on a
half space, 682 ball (Problems 40 and
195
indefinite, iterated, 177
multiple,
function
179
169
410
681
for
R3,
energy, 674
676 Green's
space,
definite, 170 Dirichlet's, 519
193
Gram-Schmidt process, 117 Great circles, 645 Greatest
in
Integrable function, 174, absolutely, 197 Integral
Geodesies, 645 curvature
vector
114
inR",
Gauss' theorem
abstract
an
608
Garib, 253
443
flat, 443 Inner product in
Gamma function
Gradient,
10
Index, 408
(Theorem 5.2),
111
173
function, 259 surface, 657
vector
test, 199
41),
683, 684 Green's theorem, 567, 571
Integrating factor, 549 Integration, 170 formula for change
of variable,
614, 579, 621 Harmonic functions, 468, 676, 677 Harnack's
principle (Problem 36),
518 Heat
in in
Inverse function theorem
482
7.2),
671
Helix, 315 Homogeneous
of constant coefficient
equation, 262 (Problem 17), Hyperbolic cosine, 424 Hooke's law
Hyperbolic
sine, 424
173
of differential form, 563 Intermediate value theorem, 162
Intersection, 16
equation (Section 6.4)
R2, R3,
multiple (Section 2.7),
256
a
541
function, 17, 168
Invertible function, 168 Invertible matrix, 61 Irrotational flow, 631 Isolated Iterated
singularity, 592 integral, 177
(Theorem
728
Index
Jacobian, 537
eigenvector,
Jordan canonical
kernel, nullity,
form, 81 Jump discontinuity (Problem 30), 517
399
Kernel of a linear transformation, 53 Poisson, 458
Kinetic energy, 501
l\ (Problem 89), 225 (Problem 90), 225
Lagrange multipliers, 234 Laplace transform, 521 Laplace's equation, 467, 672 Laurent expansion (Problem 38), 601
Legendre's equation, 450 polynomials (Problem 60), Length, 203 a
450
196
Linear differential equations
3.6),
(Sec
275
first order, 264 second order (Section
3.7),
289
systems, 278 Linear differential operator, 276 Linear function, 1 Linear span, 40 Linear subspace, 40
basis, 47 dimension,
Lines of force, 309 of
a
fluid flow, 386
Lipschitz function, 269 Local, 112 Logarithm (Example 17), 247 Mass, conservation of, 612 Matrix, 8 change of basis, 82
diagonal (Problem 22), 39 diagonalization of, 77 eigenvalue of, 77 eigenvector of, 77 index (Definition 1), 10 invertible, 61 Jordan canonical form of, 81 multiplication, 31
orthogonal (Problem 29), index; 8 symmetric, 123
289
row
transpose, 123 40
Maximum
Linear systems of differential equa tions, 278 Linear
125
(Problem 65), 693
column index, 8
Liebniz's theorem, 141
tion
Lines of curvature
characteristic polynomial, 76
curve, 331
Limit, 165,
self-adjoint, 125 spectral theorem,
Lioville's formula, 655 Lioville's theorem (Proposition 7), 587
L
of
53
range, 53 rank, 53
-times
differentiable, 240 Kepler's laws (Problem 68),
77
55
transformation, adjoint, 123 eigenspace, 85 eigenvalue, 77 complex, 89
28
for
principle, 449, 586 analytic functions, 512
for harmonic functions, 472 equations, 632
Maxwell's
Mean square convergence, 496 Mean square distance, 496 Mean value property (Proposition
2), 471
Index Mean value
theorem, 166 Meromorphic function, 610 Mixed partial derivatives (Theorem 2.13), 189
Origin, 18 Orthogonal, 97, 1 1 1 Orthogonal curves
Modulus, 88, 111
Orthogonal family
theorem
Morera's 609
31
111
Osculating circle, Osculating plane,
Parseval's
335
(Problems 25
and
curve, 350
closed, 555 integral, 562
Normal line to
a
surface, 657
of motion
692
53
612
function
608
Periodic function, 452 Permutation, 68
507
odd, 69 interchange, 68 even,
Order, 187 521
Picard's theorem
Orientation in
R2, R3,
of
a
curve, 321
of
a
surface, 658
(theorem 3.3), 271 (theorem 3.4), 273 global version (Proposition 9),
579
614
boundary in R2, 567 boundary in R3, 666 of the boundary of a curve surface, 661
286 Plane
of the
in
of the
path,
(of a flow),
(Problem 46),
Open set, 1 1 1 Operator, integral,
Oriented
498
oriented, 555 Period of a harmonic
One-to-one, 17
in
350
Path
a
exponential,
357
equality,
Normal line to
Nullity,
113
Particle motion, 254, 335 Partition of unity, 444, 451
surface, 655 Normal acceleration, 335 to a
62), 656,
(Problem 29),
Partial derivative, 187 Partial differentiation, 187
Normal
Normal curvature
matrix
Pain, 435 Parametrization, 313, 321
679
R2,
of
Orthonormal functions, 491 Orthonormal set of vectors, 1 14
Newton's law, 255, 340 Newton's method, 213 in
family
a
Orthogonal projection,
problem, 473, 675 Newton's gravitational potential,
curve
to
289
Neuman's
to a
surface,
curves, 374
Orthogonal
Neighborhood,
a
644
(Problem 55),
Moving trihedron, 359 Multiplication, matrix, Multiplicity, 409
on
729
555
on a
R3,
97
with chosen
point,
20
Planetary motion, 390 Plane
geometry, 18
730
Index
Poincare's lemma, 564 Point on a surface (Problem 67) elliptic, 693 693
Residue theorem, 592, 593 Riemann integrable, 170
Rn, 16 addition in, 28
hyperbolic, parabolic, 693
linear
subspace of, 40 multiplication, 28 Rodrigues's formula (Problem 65),
Poisson kernel, 458 Poisson transform, 458
scalar
Polar
693
coordinates, 535 Polynomial functions, 406
Population, 252 integers, P, 13 Positively oriented coordinates,
Root, 409 Root test, 151
Positive
Roots of 658
Potential energy, 555 Potential Power
function, series, 149
unity, 406 operation, 9 transformation corresponding to,
Row
555
29
Row-reduced
matrix, reduction, 7
addition and
Row
and
Scalar
multiplication (Pro position 3), 426 Taylor expansions, 246
radius of convergence, 150, 421
Principal
curvatures
(Problem 62),
692
Principal Principle
normal vector, 335 of Mathematical Induc
tion, 13 Product, matrix, 31 Projection, 113 Properties of analytic functions (Theorem 7.10), 588 Radius of convergence, 150, 421
Range,
17
in R", 28 Schwartz test functions, 523 Schwarz's inequality, 120, 223 Schwarz's
lemma, (Problem 60),
610
Second fundamental form
63),
Rearrangement of series, 143 16
closed, 173 volume of, 173 Rectifying plane, 359 Regrouping of series, 143 Regular domain, 568, 571, 662, 668
(Problem
692
Second-order linear
equations, 289 Self-adjoint transformation, 125 Separation of variables, 260, 290 Sequence, 129 of, 131
subsequence of,
Ratio test, 151 Rational numbers, Q, 15 Real numbers, R, 15
Rectangle,
multiplication plane, 20
in the
convergence
Rank, 53
9
130
Series, 137 absolutely convergent, 140
comparison test, 147 conditionally convergent, convergence, 137 Fourier cosine, 481
Fourier sine, 481 137
geometric, of
functions, 401
power, 149
140
Index
ratio test, 151 root
with
Tangent plane
test, 151
Taylor, 246,
to
509
positive 4), 139
terms
(Proposition
equations,
homogeneous system, Singular solution, 373
2
Tests
Skew-symmetric matrix (Problem 32), 302 Solid angle (Problem 34), 673 Spectral theorem for self-adjoi operators, 125
comparison, 147, 198
integral,
199
ratio, 151 root, 151
Topological terms, Topology, 575
Spherical coordinates, 536 Steady flow, 385 Stokes' theorem, 662, 666 Straight line, 24 Subtraction, 22
111
Torsion, 360 Transform
Fourier, 523 Convolution, 520 266
Laplace,
521-22
Poisson, 458, 524 Transpose of a matrix, 123
Surface, 635 arc length, 643 area, definition
curve, 350
a
242
11
approximations,
a
surface, 640 Tangential acceleration, 335 Taylor expansion, 246 Taylor's formula (Theorem 3.1), to
Simultaneous linear
Successive
731
of, 651
definition, 635
developable, 642 elliptic point, 693
Triangle inequality, 120 Trigonometric functions, Taylor ex pansion, 246
Trigonometric polynomial,
454
first fundamental form, 643
geodesies, 645 hyperbolic point,
693
of, 406 444, 451 of, partition nth roots
orthogonal curves, 644 parabolic point, 693 patch, 635 second fundamental form, 692
tangent plane, 640 Symmetric bilinear form, 123 Symmetric matrix, 123 System of coordinates, 534
to a curve,
198,
Union, 6 Unity
normal, 655 orientable, 658 orientation, 658
Tangent
Uniform convergence, 205
322, 324
Variation of parameters, 295
Vector, 20, 28 addition in R2, 21 addition in R", 28 field, 306, 381
independent set, 43 in the plane, 20 product, 100 subtraction in R2, 22
204,
732
Index
Vector
field, 306, 381 curl, 630 flux across a surface, 660 radial, 560, 624
divergence,
58 1
,
670
Vector space, 107, 108 Velocity, 254 335
of
a
particle,
of
a
fluid flow, 385
field of a flow, 612 Volume, 173
Wave
equation, 482 approximation theorem (Problem 6), 467
Weierstrass
Work, 554 Wronskian, 293
