GEODESICS IN 3-SPACE

THESIS FUR THE DEGREE OF M. A.
John ‘William Zimmcr

1933

( Yc’aggwoozdfi Er \

MSU

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GEODESICS IN 3-SPACE

A Thesis
Submitted to the Faculty
of
MICHIGAN STATE COLLEGE
of
AGRICULTURE AND APPLIED SCIENCE
In Partial Fulfillment of the
Requirements for the Degree
of
Master of Arts

by
John William Zimmer

1933

ACKNOWLEDGMENT
To Doctor James Ellis Powell

whose suggestions and help are

responsible for this thesis.

i.(}:3€3&§!3

INTRODUCTION

TABLE OF CONTENTS

I. GEODESICS AS AN ORDINARY PROBLEN OF THE CALCULUS
OF VARIATIONS

lo
2.

10.

Statement of the problem

Euler's equation

The Weierstrass-Erdman corner condition
The necessary condition of Weierstrass
The necessary condition of Legendre

The necessary condition of Jacobi
Sufficiency conditions

Geodesics on a sphere

An illustrative example

An illustrative example; the pseudo sphere

IIo GEODESICS; A PROBLEM OF LAGRANGE

1.
2.
3.

The Lagrange problem

The problem of finding geodesics
The equations of the geodesics
The transversality conditions
Normality of the extremals

Analogue of the Weierstrass condition

(OmKIUIIbCfiOJ

12
13
15
16

BO
20
20

25
25
27

7- The Clebsch condition
8- The necessary condition of Mayer
9. Determination of conjugate points
10- Sufficient conditions for an arc to be a
geodesic
11. Application to a sphere
12. Application to a cylinder

III. AN INVERSE PROBLEM
1. The inverse problem of the calculus of
variations

2. The problem.of Beltrami

IV. AN ISOPERIMETRIC PROBLEM
1. Statement of the isoperimetric problem
2. The problem of Minding
3. The necessary and sufficient conditions
4. Application to a plane
5. Application to a sphere

BIBLIOGRAPHY

28
28
29

31
32
34

36

36
37

50
51
54
55
56

58

INTRODUCTION

We shall define geodesics as curves of shortest dis-
tance on a surface. That is, a geodesic is a curve on a sur-
face such that the distance between two points measured along
it is less than the distance measured along a neighboring
curve joining the two points. The problem of finding geodesics
is then a problem of the calculus of variation. We shall show
that the definitions usually given in metric differential ge-
ometry are necessary conditions that a curve be a geodesic
according to our definition. They are not always sufficient.

It is not the purpose of this paper to give all the
properties of geodesics that have been discovered. In the
bibliography will be found a list of papers giving these prOp-
erties. we shall confine ourselves to a discussion of these
properties that can be derived by the methods of the calculus
of variations.

In part I the problem is set up as an ordinary problem
in the calculus of variations. Necessary conditions and suf-
ficient conditions are given for a curve to be a geodesic. The
results obtained are applied to several examples, including the
sphere and pseudo-sphere. In part II the problem is treated as
a Lagrange problem. In part III an interesting inverse problem
connected with geodesics is considered- The problem is to find

1

all surfaces on which the geodesics can be represented lin-
early in u and v. Part IV is a discussion of the isoperi-
metric problem of finding curves of shortest distance join-
ing two points A and B and enclosing with a given curve,

joining A and B, a given area.

I GEODESICS AS AN ORDINARY PROBLEM OF THE
CALCULUS OF VARIATIONS
1. statement gfuthg,problem- The equations of our
surface S are taken in the Gaussian form
X=X'(a, V),
(l) S. y any (a, V),
2' ext“, V),
We shall assume that IMHO, ytum) and It“. v) have as many
derivatives as necessary to carry through the analysis. We
shall also treat only those surfaces for which A”: EG‘FZ
is everywhere different from zero?' The length of a curve
joining two points on S is given by the integral

(2) iufizrv’wv" d“ - H‘

The problem of the calculus of variations, then, is

 

to find among all admissible arcs joining the points 1 and
2, one which minimizes the integral (2). Such curves we
shall call geodesics. We shall discuss the necessary con-

ditions and sufficient conditions that a curve be a geo-

desico

 

a! L-Po Eisenhart, Differential geometry 9; M and, m,

p. 71. (Hereafter referred to as Eisenhart.)

# x. Eisenhart, pp. 220-221.

2. mlgr's gguationo With the function 7‘ in the
form given in (2) the Euler equation

(3) i fv' "fv " fv’u *fv'VV'*fv‘v' V’I-fy =0,

becomes
I ,3
2(EG-F‘)V’ 46%.? +3F§-§ -2F§-§ 433-5”
6 C '3! F 5' 11
(4) 143ng +25}; +2F§-§-G;—7 -E§—§ -4;§-; -6§;—,)V

E”
+0535- rzrgf mg; 45’;- -2;§§)v’

F’ 35' Eli _
-r£!£?§h:‘-/7§;: -£:55;-C9J
which may be written in the form

v" 116355 *Fagff ”22%)V’3
35‘ II
+[(§%;‘-+F%5‘Z‘f{)‘ ”5517 F591"
2h" 2/!"

 

 

(5)

 

+{2(Ei?£ ’F-aaf) "(63%; +F§~§ -zr§;f]y’
2/" 2H‘

 

F a: at _
+[ng2 ‘éjza‘: ‘f-a—pl—O.

 

Expressing (5) in terms of Christoffel symbols of the second
kind we get

v”- {2.2} v” + (12.21 - {24) v"
+0422] _, [1,1])1/ +[Izl] =0,

Geodesics are defined by means of (6) in metric differential

(6)

geometryfg Since we have shown that Euler's equation reduces
to the form of (6) for our problem the ordinary definition
of a geodesic becomes a necessary condition. In other words,
a curve on the surface must satisfy (6) before it can be a
geodesic. This condition, however, is not sufficient. We

shall call solutions of equation (6) extremals.

3. figflgierstrags-Erdmann garner gongitignf’ A

corner is defined to be a point on the geodesic at which V"
is discontinuous but has a right and left hand limit.

We shall define admissible arcs as those arcs that
are continuous and have only a finite number of corners. If

the geodesic has a corner (u,v) the Weierstrass-Erdmann cor-

 

y- Eisenhart,p. 205.
* G-A. Bliss, Calculus g£_Variations. p. 143. (Hereafter

referred to as Bliss Io)

ner condition says that

 

 

 

(7) fV’( a) V, V,<‘¢‘O)) st’(“.a\/; V’(“+0))p
using F: V! +2FV' +6.14 , (7) becomes
FfGVL __ FfGV’

(a) —, -, - ——-L, . .
yE f2/V— *GVL V?f2f¥*f6Wf

If we square both sides of (8) and then multiply both sides
by (5' and. + 6 V93)(E f2FV—’ 4- G v:’) we get

[5/1 + 25%: v; rta’ Vf + 2/3 v1 HGI‘ V; v!

( ) zrcavf v1} +;'c v.11 * “'6”: v:1 +e3w} yd] =

9
[zzrc y.’ + zs‘vc' +1 F’vl 4'1}cm v; v.’ + 2:6' v,’ vi“

I I I3

-F’G v,” + 2rc'v,’ v.’ + c v,‘v.] ,

Equation (9) reduces to

(10) H2(V1*V-')[2F+6(V1+\L’)]=o

Equation (10) can be zero when

(11) V,’ = v.’. ,

01'

(12) 2F +6(v;+v.1)=o.

 

 

From (12) we find that V! nary; ran" = VI+2FVI {-61/1'
and Fiat/1 = —(F*6'V1) . Therefore (8) becomes

 

(13) -(F+GV.’) -' L" 6",
V£*2FV,’ +614“ VE+ZFVH6VF °

From (12) and (13) we have that V.’ = V..’ :. 'Z'é: . .Hence we

conclude that a geodesic can not have corners.

4. The necessagy condition 9; Weierstrass."L Let

Gabe a geodesic. The necessary condition of Weierstrass

is that for every (:4, saw) of 6,3 the function
(14) [Nana/'1") =-' {(“.V.V')-ftu.m/’) -(V'—V')Fv'(u.v.v')

I
must be greater than or equal to zero for every (M,V.V)*(“.WV).

For the present problem (14) becomes

 

 

_ hr:- rams-r9
V: r zrvv 6V”"

 

(l5) {heart/7 = V£+2FV96VR - VfrZFv'fG'v"

which may be written as

(16) [(u’Ky:V') g [EanZQGV‘ ZE+2fw+€V7 —/£’+(V9-W)/*6WVJ
f V'rGV' '

Since the denominator of the right hand side of (16) is the

 

 

 

expression ford.» , and is always taken to be positive, we

need only to show that

 

*31183 I p0 1310

 

 

V’Z‘V'4'5‘V'i IEH’FV’itgw' ’/ E + (VLV’IF' HS‘v'V'.

(17)
V{ I +(V'—V')F+ cwvy“

If we write the right hand side of (17) as
and express the left hand side as a single radical, we may

then compare the two radicands. We get

[5V"*fF‘V'V’ +£6w'7/

(18)
2E6wV‘+ V"; up... 7'. WV“.

Setting h” = [G-F" we may write (18) as

(19) (;1*//ajvl‘+#/3y’ V’*(F‘+l/‘)V"7/

2(F‘fh")wV’. V’V" .zrw-V'. Me,

which reduces to

(20) H‘tw-V’NO-

We have shown that for every V'i’ V’ , E(¢-,V,Vj V970

everywhere on our surface. The Weierstrass condition is,

therefore, not a condition on the minimizing curve at all.

p
5. The necessary condition gkaegendre- The
Legendre condition states that if 611 is a geodesic then

at every (my, v') of 6'” the condition

(21) {WW ( a, V, V')’,0

*BliBB I, p‘ 1310

must be satisfied.

For our problem (21) is

£‘6’—/7‘

(22) fyoyl =
(E f-ZF'V’+ G 51")?!

H2
(5 arw. swat '

Therefore fr'y' > a everywhere on the surface.
Hence Legendre's condition places no restriction on the

curves that can be geodesics.

6. 1h2,necessary condition 9; Jacob ?' If we con-
sider the single infinity of geodesics
through the point 1, and if this family
has an envelope, then the geodesic
can have no contact point 3 with the
envelope between 1 and 2. For, by the
use of the envelope theorem the com-

;*
posits are 6”, +34, f 6-,, = ,1-

 

But.an is not a geodesic and may be replaced by an are

which is shorter than 1),, . This would give the composite

 

u Bliss I, p. 141.

10

are 6,, +D” v- 6,2 a smaller value than the geodesic 6,: .
In every neighborhood of 6,, there is such an are 6),, +D”
+ 65;. giving a shorter distance from 1 to 2 than the geo-
desic. Hence the geodesic cannot be the shortest distance.
This contradicts our definition of a geodesic.

Tb make sure that .D,, is not a geodesic we make
use of a well known property of a second order differential

if
equation. Euler's equation for this geodesic is
(23) {Wet "' fr‘y V’ * fww V” ’fy =0.

Equation (23) can be solved for V" since {radio . If an
equation of the type of (23) can be solved for v”’, then
there is one and only one solution through an arbitrarily
selected initial point and direction (143 J 1/, J (4’) . Hence
if D” were a geodesic it would be necessary for it to coin-
cide with 6&3 . Thus, all the geodesics through the point 1
would, by the same property, be tangent to and coincide with
5i, . Then there would be no one-parameter family of geo-
desics as the theorem supposes.

If the Jacobi condition is to be of much value we
will need a convenient method for determining whether the

geodesic has a contact point 3, called a conjugate point of l,

 

* G.A. Bliss, grinceton @lloguium 1,ectures.

11

with the envelope of the family of geodesics through the
point 1. From the calculus we know that if we have a one-
parameter family of curves V=\/(«,d) the contact point of
any one of these curves with the envelope of the family is

a root of V,‘(u,.q when x, is the value of d defining
that particular member of the family.*

It is not always easy to determine a l-parameter
family of extremals through the point 1 even if we do have
the 2-parameter family of solutions of Euler's equation.

For that reason we need a method of finding the points conju—
gate to 1 from the original 2-parameter family of solutions
V = V( a, 0-, b).

If we were to write the 2-parameter family of ex-
tremals as a l-parameter family it would be necessary to
choose the two functions at“) andbc-r) such that
(24) Vlunx) a w u, aw), 5(a))
satisfies the equation
(25) V. = Wand»).

If we differentiate (24) and (20) with respect to.“ we have

V.. = V. (a, a, b)a.’ + v, (a... a... b) ti

0 : qu(“c a do b)al + Vb(“4 “J b) b’,

 

* Wilson, Advanced Calculus, p. 136.

12

where 4' and 6’ are derivatives of d and b with respect to «X .
When v. vanishes the determinant of the coefficients of d'and
6’ must also vanish. We can now state that the points (3)
conjugate to a point 1 on the geodesic are determined by the

zeros for a: a, of the determinant
4

Va‘“.4o,5-) Vb(“a¢0ob0)

(26) A(“1u1)=
V4 (69,40, 60) V5 (“hang be)

 

 

when V( u.’ a, 6) is a 2-parameter family of extremals contain-
ing the particular extremal for parameter values 4., b. . We
shall find later that Jacobi's condition does not hold for all

extremals.

7. §ufficiency conditions. In this section we shall
give conditions that insure that our minimizing arc is a geo-
desic. Tb accomplish this let us first define a field. A
field is a region é? of (Lgyo-space which has associated with
it a l-parameter family of extremal arcs all of which inter-
sect a fixed curve.z> and which have the further property
that through each point (a, v) of 6‘ there passes one and but
one extremal of the family.*

By applying the fundamental sufficiency theorem of the

calculus of variations we have the following theorem:'*

 

¥Bliss I, p. 132.
*nBliss I, p. 133.

13

Theorem. I; 6,, is an extremal of a fi_el_._d 5‘ t_h_§2_ 5,;
_i_.§_ shorter than any 939; admissible gag C1,, in J3 joining
1h}: points I 229.. _2_. 1119.; 1—3, 6;, 15 a geodesic.

This follows since [(1.9% V,’ V') 7a everywhere on our

surface.

theorem. Li; a curve 6,, ig a solution g_f_ Euler's

 

 

a}; a, then 6!: i_s_ a geodesic.

This theorem is a consequence of the ordinary suf-
ficiency theorem for a strong relative minimum in the calcu-
lus of variations since [(1.9% VIV') >0 and fwydu, V, V920

everywhere on the surface.

8. geodesics 9;; a sphere. If we choose the (UN) co-

ordinate system as indicated in figure 2, th equations of

32
the sphere are

X: 44...} u can”,
(27) gi-‘dwwm' V)

18¢ Gnu.

 

 

The linear element is

 

 

2_ a ‘ ‘-1 .1
(28) 4,, -aduramuaarx

The problem is to find among all admissible arcs
V:V(a.) , Lanna-l4,

one which minimizes the integral

14

“1
fa. V/rmzu. V“ 44-.
a

l

 

Euler's equation is

M, = C,

(29) y/ +M’KLV'2

Solving (29) for V’ we get

v’- C
(so) - WW.)

which gives, by a quadrature

(31) V‘mmL f5,
tau-“-
V—C'
when K = -IC—’.'
1

With the values of sin V and cos V computed from (81),

and with the aid of (27), we can write the equations of the

geodesic as
x: aKConuCocC -—aco.u,e.:...c [luau ~17,

 

 

21:: QKMILMC-i-dwudo-QC sz‘c-Kzu

(58) 2: am».

If in (32) we multiply x by cos c , y by sin C , z
by-K and add we have

(33) X can 6- +JAALIC +(-K)z=a.

Equation (33) is of the form
Ax +3y+€z:0,

15

which is the equation of a plane through the center of the
sphere. The intersection of such a plane with a sphere is
a great circle. Hence, the geodesics on a sphere are great
circles.

Suppose we have two points 1 and 2 on a sphere. The
great circles through 1 have the point at the other extremity
of the diameter through 1 as an envelope. Therefore to ob-
tain the shortest distance we must use a portion of the are
less than a semi-circle.

For a sphere the determinant (26) is

1
I
(I ~42) V/ —c,’,e..:x=u.

 

(34)

 

I
(z-c,2)V/-c,5m;i2a" I

 

 

I

The zeros of (34) are u:- “Ign/pa‘. This checks with the

above statement.

9. Ln_illustrative example. Given the surface of

revolution
x=L¢MV }

y: “JAG—V)

(35) z=é[um_,¢3(u.+m).,

to find the geodesics. The linear element is

(36) da’ = u’da.‘ “1‘00" .

16

The problem is to minimize the integral

‘5
(:57) / “Vl+v"du.

l
Euler's equation is

I

(38) W—% :C .

Solving (38) for V, and then integrating, we have
(39) v = ,2? c, (u . Vai-c‘) .

The determinant (26) is

 

I
(4 163-63 +0.41%“) '
c
(40) ' 23'

 

I
u,Vu;‘-c‘+(u,‘+c‘) , I.

 

 

The zeros of (40) are use“, . Therefore, there are no
conjugate points.

Some of the geodesics are shown in figure 4, page 17.

lo. A§_illustrative exam 1 ; th§_pseudo-sphere.
The equations of our surface are
X= umv.
(41) yeumv,

’lr—zfj - ’.fifi7521
Z " I a ’2? a fi

18
The linear element is
2
(42) 442:3!1du +u‘do-2.

The problem is to find the minimizing arcs for the

integral

a.
(4'2) /)£". * “W7 dd"
9‘:

The Euler equation is

 

 

tc‘v’ duh-:0

(44) ’. + u V”
pa

Solving (44) for V"and integrating we have

V __ l/c.¢'--¢:IL

(45) "' Ca

+6}.

The determinant (26) becomes

_z__ /
6" Va 3-63 ’

‘

(46) I!

62 “$3.61 1 I

 

 

The zero of which is u:=cg . Therefore there are no conjugate
points. That is, a solution of Euler's equation is always a

geodesic.

Figure 5, page 19 shows the surface with some of the

geodesics drawn.

19

 

 

 

II cronrsros; A PROBLEM or LAGRANGE
t
l. Ihe Lagrange problem. The Lagrange problem in

the calculus of variations is that of finding among the arcs
5.11.11) (x, :x:x,,' i=1-" n),

satisfying a set of differential equations

(47) %(XJ jl“"jn, WWW” («=/--'rn4n),

and joining two fixed points in (X, y, —---j..) space, one

which minimizes an integral of the form

‘3
(48) x 'f (x, y, ""jn, y,’----y.,') d4.

2. Ihe problem 9: finding geodesics. Let

(49) S.’ ¢(A;J,z)=a ,

be a surface on which we wish to find the geodesics. We
assume that f has continuous derivatives up to and in-
cluding those of the fifth order and that

g! + £2 +£z #0’
at G.A. Bliss, J‘hq Eroblem 9; Lagrange .11}. 15h; Qalculus pi
Iariations. American Journal of mathematics, Vol. LII
No. 4, October 1930 p. 674 (Hereafter referred to as

Bliss II)
20

21

everywhere on the surface.
The form of the Lagrange problem that we are inter-

ested in is to find among the arcs

x: X($),
(50) y = 3(5), 5, 9. 5:5,
2: 2(5).

satisfying the equations

(51) ?(X’5’I2)‘0 J

X’z+ y’2+2”-/= 04
and having end points satisfying the equations

(52) SI = Xl-“I =J1-p’ 31”” =X1‘“1 ‘Jz‘fiz‘zz'Yz =01

one which minimizes the integral

is
(53) j = .4: @-

This is a problem with variable end points in (s, xd‘qfigyspace?
the parameter 5 measures distance along the are starting at
the point 1. In order to reduce this to the regular Lagrange
problem we replace the first of the equations (51) by the

equivalent equations

(54) fl,x’+ fiy’+¢zz'=o,

 

22

*-
(55) X, ”d! =3: '5: = 21" )7 = 4'2“: =j1-fla =0.

We assume here that 7} as a at the end point 2. This does
not restrict the generality as one of they;" J ¢73 J g is
different from zero at every point. We can now replace

equations (53) with the equations

(56) 5’ :Xl-“I=Jl‘pl =1.'n=xa‘°‘z=y‘-pz=o.

3. The equations p:_f_ the geodesics. As a first
necessary condition that an are be a geodesic, we have the

followimg theorem:

Ihgorem. [or every geodesic 6;, there exists 9, sgt

pi constants c, , 62.9 and 9, function

(57) F: / +A,($)(%X'+%gl+?22’) +_3:_2(5) (X’zfj’:f2'.-/)J
such that

S
6’21;de4 *9,

£57“ *‘2.

5.2
[1' 5.4. F}.a +63,

I

H

(58) F3!

 

*Bliss II, p. 703

23

m gahhsfigd y; gm m pi Qt. 1h; gunctions (An A,)
gr; M ppph identically _z_e_;‘_g_, 2.2.8. unigue 211g continuous
except possibly g1; values p_f; s defining corners 9; 6’2.

This theorem is the first necessary condition of
the calculus of variations problem. In order to write the
function F in the form (57) it is necessary that the mini-
mizing arc be normal. In section 4 we shall show that every

minimizing arc is normal.

For the function F of (57) the equations (58) become
I s I
A: $3: “*Azx 57/5: Az(flxx *¢xgg'*¢:xz’)d& 3+CIJ
I ‘ ' l I
(59) A, $5 “(,3 =1; A,(¢.gx +4333 +7312).a +c21
5
1,56. + A, 22-1 A, (new my .59.. 194 +6..
By an integration by parts we can write (59) as
s
11x, =jéfljl’fi’h fc’)
I 5 l -—
(so) 1.31 1'ij 4+ch
, 5 l
)1‘?’ :j "/(Ifld‘cfcg'
5/

From equations (60) we can state the theorem:
Iheorem. h.geodesic can not have corners.
From (60) we have that 111'; Agy', A, I'must be

continuous on S, S, . Later we show that A; s/ and there-

24

fore X', ff ', zlmust be continuous.

If we now take the derivative of both sides of (60)

with respect to 5 we get the Euler-Lagrange multiplier rule.

At every point of 532 the equations
670ml, 2) =0 .
x” a“ fir ,
(6°) , .7” v“ 9’2,
2” =7“ g,

must be satisfied. This proves the theorem:
Theorem. Ah every point 93: .9. geodesic the principle

normal 9; the geodesic must coincide with the surface normal.
The derivatives x”! 3’2 3” and the factor/acnare

continuous on S,’ 5, since the determinant

A2 0 0 fl 2x'

0 1 0 ¢ 23‘
(61) ’ ’

a a A: «p, 22'

fir ‘25 V9 0 0
2X’ 23’ 22’ 0 0

3 ¢(¢x: 4’ fl: *fl") 7‘0.
at 131133 II, p. 684.

 

 

25

1 r
4. The transversality condition. The transversality

condition for our problem is that the determinant

-m *1. , -1, r. -Anx',".‘.7: 4.32 4.54213 run-1,; '1' ’3 “W34. 79 “’3'
l 0' 0| 0 £7 (a ‘0
C? / 0 (0 C7 0 ‘0
0 ‘7 / 01 C7 «0 (7
O ‘7 a I (7 ‘7 ‘0
o c: o 0 6’ / 0
0 c7 '0 ‘7‘ ‘7 ‘9 I

 

equals zero. This determinant reduces to

A2 =F(’:) =7].

5. uprmh;ipy p;_th§ gxtrema 5” Every extremal arc

of our problem is normal. The condition for normality is
that there exists six sets of values of (t’ f}, 2(5), 21(5),), (5))

with f: ’ f1 arbitrary and p, , 7, , % satisfying the equations
c! i
25? (5’17. *2; 2. *9‘373)‘0:

xlz’f glyzl" 217’! = a 1
and making the determinant

(63)

 

yBliSB II, p0 6930

1* Bliss II, p. 693 and p. 704.

 

 

3‘7: be $3 ‘‘‘‘ In.

7” (51) 712 (5.) 7,, (5.) ————— y“ (5.)

71! (SJ 7;; (5.) 7,, (5,) ----- 7,.(5)

731(5) 731 (5.) 7,, (5,) -—— -- 7... cs.)

x,’ f,, *2/54) x; [11* 7,.(54) X,’fl,+z,(s,) ...___._x,‘};.+7,‘(s‘)
3.‘ t. we» :1.’ n. was» 3; )5. +7....) ----- g; h. an...)

 

different from zero. In (62) the arguments are the Ly, z,

X’, y: 2' of the extremal arc. The equation (62) can

written

% 7,’ 437,: 1* fiz’ + (flxx’ *fi,y’+ 75,297,
+(gxx’+¢,,y’+¢,,2')g, +(q>,,x’+ ¢,yg’+%32')73=0’
X’z ’ +57; 4- 2’3’ 5Q

The matrix

9% ,

be

is of rank 2 at every point of the extremal since the

directions 49,. :¢, ; $1 and X ’;g’,z' are orthogonal.

another function Yang/,2, x,’ y’, 2') such that

Take

27

is of rank 3 at every point. Consider the set of equations

¢x7,’ *QP; (4&2, +(%.X'+¢qg’+¢xel')z
+(%M’%.y'%.wz +<a.r’*¢zw'+a.zvz=o.

xlzl+ylzlfztzl=o’

9;. z’+ 997.” 1M: + m, +537. mafia»-

Let 715.). 7,6.) , 7,'( 5,) , 7, (5,) be arbitrary and determine
{(5) such that the above equations are satisfied. Since the
sets[, , 5,, , 29'.) , puts) , 2"ch , 7". (5.) are arbitrary we can
certainly choose them so that the determinant is different
from zero if 3(5.)¢0. If g’(s,)=0 and 1:25,) #0 we can
vary the above proof. If g’¢s.):x’(53)=o , then 225,)” .

- l I I_ ‘

Since 7;, x + gvgy +32 .0 on the surface we would nave

fl, =- a at the point on the extremal determined by 5-5,, .

This contradicts our hypothesis that ¢1¢aat the point 2.

6. hnalogue g; the leierstrass condition. The

function for this problem is

f = / + A. (fii’*@1”+¢329 + firm/“.29
" *1 (QXX‘ +7335 Q, 2') -f (X'3+3"+ 2")
- (X'-x‘)( A. 7:. +x') -(Y'—y')(1'¢g .3.)

‘(2'-2')(x. 4’; 1* 2').

28
This reduces to
E: f [UV—X91 + (gt-3")" +(2’-Z’)‘] .

Hence we have that [(sz', xjy,’ Z; [17: 21))20 for
every (XIJflZ'Jv‘: (0, a, o) everywhere on the surface.

7. The Qlebsch condition. The expression
(63) 57.77,”, ”3441.76 +5475»; rzfly 77.74 + 2574.775 +2544. 77;,
becomes

7Z”+-Z§’i*25'.

Therefore we have the eXpression (63) greater than zero
for all sets (75,771,767 #(Qa,o) everywhere on the surface.
The conditions given in section 6 and in this
section are stronger than the ordinary conditions. The
ones stated here are usually denoted by z: and [Z ’ re-
spectively. They are of importance in proving the suf-

ficiency conditions.

8. Ihe necessary condition p§_§hzer. The necessary
condition of Mayer corresponds to the Jacobi condition of

the ordinary problem.
Theorem. 1;,the hyperameteh family 9; geodesics

 

29

throhgh_the_point ;.has.hh_envelope, thg geodesic joining ;_
and.§rcan havg hp_contact point §_with the envelope between

_]-._ and g.
This follows from the Hayer condition of the Lagrange
problem and the fact that the determinant (61) is everywhere

‘*
different from zero on the surface.

9. Determination p;.conjpgate points. A point 3

where the extremal has contact with the envelopeD is called
a conjugate point of 1. In order to apply the theorem of
this section it is necessary to have a method of determining
the conjugate points. 1

Let
x: X (5, al, 43,43, bl’bgié') J

3:.- 3(5, a,, 43,533. bubabz),
Z: 2 (S, 4,) alias) bl‘b 46..)4

be a 6-parameter family of extremals containing 6,, for digs-.1230

(i: 4 2, 3), Suppose also that the determinant

 

*Bliss II, p. 722.

1., (3. a, u
a, (S, 4.6)

1c, (s, 4.6)

1‘: ”a, “'4.

1‘0 p3.) *Jé’

 

”'4 fia.‘ 11!.

is different from zero at the point 1 on G): .

points 3 conjugate to 1 are determined by roots

X4, (5,4,6)
54, (s, a, 6)
24, (5,a, b)
A“: fraud:
Ad, 799+ 34,

I
A“: $19 ”a

of the function *(64)

’4, (5, 4., b)

k, (‘0 ‘0, 60)

X" (5, 40,60)

3" (seed; h)

 

2" (5,, ‘0; 60)

14, (S, 1.... be)
344$, as. be)
23, (3 do. be)
X4, (sue... 5.)
3a“. 4a. he)

243 (S, 00, be)

x4, (5. e. A.)
#4, ( s, 4,6)
74, (5. d. b)
14, fl., 4' x1,
A“; 79‘: I”;

I
A‘J ?“J + 2‘,

’3 (5. 4o. ‘0)
94, (s, 4.. A.)
20., (5. do. be)
1., ( 5,, do, be)

94, (‘1)‘OJ 5.)

z a, (5“ e9: 6.)

X6, (5.. 4.5)
50, (5,46)
24 (5, a, 6)
A" f". + I;
*5, 71b. ’71.,

A5, fiélfl‘:

x6, ( s, as, b.)
55, (5. do. A.)
2,, (s, 4., 6.)
X6, (5., 40.90)

y“ (5” do)“,

1"(3~40, 60)

30

X5, (5. a...) I" (5, 4,1.)
951(5. a, 6) 3,, (5, a. c)
26; (s, 2,1») 2"“, a, 5)
Abzfibt-f x4 16%.)”;
2‘: fi‘z*‘y‘; 1% ’95:,"
“.7: a +14 *9 and?»
Then the
5, 9* 5,

xbzcs, awn.) X6 ($4.. 6.)
5., (s. 4.. a.) 5., (c a... a.)
lo, (5. lab.) 21., (8. tab.)
16, (she... b.) x” (9.. “a. A.)
y;,($,,l.,b) is, (s , ¢., 6.)
2.. (a, 1,, p.) 2,, (5., c... a.)

 

 

 

*Blise II, p. 729.

31

10. gufficient conditions for 9.2!}. are _i_:p_ h_e_ §._ geodesic.

A Mayer field is a region 5‘ of($,X,y,z).space containing only

interior points and having associated with it a set of functions
,b,- (s,.x,y,z) , 2.. (s,x,g,z),

with the following properties:

(a) they have continuous first partial derivatives
in 3 ;

(b) the sets (5, 1.5, 7, pass”) defined by the points
( 5., :93, z) in 55 are all admissible;

(c) the integral

1* =/[F(S,X,H,3.b.U + (436-17844?) F1), (ax, 3.3.)”. U].

a
formed with these functions is independent of the path.

By applying the sufficiency theorem of the Lagrange
problem we have the two following theorems which correspond

H
to those of part 1.

Theorem. 1; 5;: _i_._s_ Ell. gxtremal 2f_ 9 field 3 then

is shorter than any other admissible arc Cm hp 5‘ bin-

4:
ing the points .1; and _2_._. Ihat _i_s_, 6,1 ;_s_ _a_ geodesic.
This follows from the fact that 6714,9971) 70 at

 

every point of the surface.

 

*Bliss II, p. 730.
*fiBliss II, p. 731.

32

Iheorgm. if; 21.1. are £12 satisfies equations (fig) and
hesrena‘uliconiusate islheimeelehaasreié. 13213.1}. 612

_i__ 2. geodesic.

11. Application t__§_ sphere. The Euler-Lagrange

equations fer a sphere are

(65)

X3r51f3‘-dz=o.

If we differentiate the equation of a sphere twice
with respect to s and make the substitution xx’-/«X‘.yy"wf

and zz”==/u.z’ we have
, ....L
fl “3 '
Putting in this value of)‘n and solving (65) we find
X9 6', 60¢.-2- f 63M i.)

S
(66) y: C’MZ+CVM%J

If we multiply the equations (66) by

’4 =C,Cg “Ceca",

(37) B = CzCJ'C,C‘,

It

respectively we find that
Ax + .8, 1.61:0.

Conversely, given values ofA ,3, and C we can solve (67)
for 63,6}, 6,, 6'9, cf, 6" . Therefore the geodesics on
a sphere are among the arcs of great circles.

For the sphere (64) becomes

Cot-2- 0 0 min-g 0 O
0 at: a a was. 0
O o c.45- 0 0 Mg.

datgl a a 4k“ is o O
0 can" a a ”.25. o
0 o antgL a a ,a;~§%

 

33

 

34

the zeros of which are u. a egg/re" . Thus, as in the prob-
lem in part I, the geodesic connecting two points of a sphere
is that portion of the great circle, less than a semi-circle,

through the points.

12. Application hp.h,cylinder. For a cylinder the

Euler-Lagrange equations are

The solutions of (68) are
S I
X: C’MVI-Q! 2" + CKMVI’Cfg'J
C s
y: C, c... Vl-cfas- + 6:4,...(1—63‘3)

2: C35 +C‘,

If for convenience we set /(= .——-l-";" the determinant

(64) becomes

35

tee/(s 0 iii MKS -£.L&. cuss 4.th a
a 1-: 4W1
c c
0 (ac/(S —-‘—‘- . 5 .. c m: 0 ° Ks
.w-wf‘f" mi? “*
o o 5 o 0'
cars flair/es. - 5L5"- GaKS law: a
4|”-ch 4"]..2" I I
O 8 ‘- Coah". 0
c“UV/ aw7:§?4u~w¥% ‘zmfira? .uamkfi
C) o 5’ <9 <0

 

The zeros of which are 5 = grew-{3% . Hence, from (89), the

geodesics joining the two points will make less than one com-

plete turn and go in the direction of the smaller angle.

 

III AN INVERSE'PROBLEE

We know that on a plane the geodesics are all the
straight lines. Therefore in the vacoordinate system the
geodesics are represented by equations linear in xandj .

It is the purpose of this part to determine whether
we can so choose acsv-parameter system on a surface such that
all the geodesics will be represented by equations linear in
Leandv'. Beltrami was the first to propose and solve this
problemf* He asked if it were possible to get a one to one
correspondence between a sphere and a plane such that the
geodesics on a sphere will correspond to straight lines of
the plane. Darboux solved the problem as an inverse problem
of the calculus of variationef* The method of this paper is
essentially that of Darboux.

l. Ihg,inverse problem p§.php_calculus p;.variations.
The inverse problem of the calculus of variations is, having

given a second order differential equation

 

x‘E. Beltrami, Risoluzione del Problema: Riportare i punti di
una superficie sopra un piano in modo che 1e linee gecdetiche
vengano rappresentate da linee rette. (Annali di Matematica
pura ed applicata pubblicati da B. Tortolini, t.VII,p.185;1866)
aH‘Darboux, Theorie Des Surfaces, Vol. 3, pp. 53-63.

36

s7
(69) v”:— ch.v,V').

defining a 2-parameter family of extremals, to find a
function f(u,vv7such that (69) is the Euler equation for

the problem of minimizing the integral

“1.
Ram, V')du.

“I
2. The problem p_f_’_ heltram . The necessary and suf-
ficient condition that the geodesics be represented linearly

in u and V is that the equation of the geodesics be
(70) V”: 0.

We shall first find a function {(uficv‘) such that
the Euler equationé fw - {van a is v”=a . The re-

quirement that Euler's equation reduces to (70) is

 

I.
‘3; V, + 31f "a :0.

(71) pva au‘av’ 3V

If we take the partial derivative of (71) with respect to V'

3‘!
and set M: 57, we get

 

3M an
—_ V’ + :0.
(72) a V’ 3 H-

To solve (72) we first solve the equations *

 

#Wilson, Advanced galculus, pp. 267-268.

38

(73) fig 0‘" r—éfi’=-¢¢'-,
I v' a a

From (73) we have
0’: 0,
dm =0,
llhac==db7

which gives M=c; ,V'scz and VLuV’I-CJ . The solution

of (72) is, therefore,
(74) M = Wv’; v-av'),

where f defines an arbitrary function of v’ and V—uV’.
Thus, in order that v’S-o be the Euler equation, we must

have

31'}: _, ’ .. .
(75) 3V"— 47(Vjvuv)'

 

From.part I we know that the geodesics are extremals for

minimizing the integral

 

“t
j V! +2FV‘+6'V" at“ .
0‘

Equation (75) then requires that

33*
3 V"8 1

 

£6 4"

(76) (5+ ZFV’ +GV'2)%

39

01‘

a
Ezrv'f-GV" , (6‘11) 3
(£6 -r8)’/: 3 V' ‘

 

We shall not consider surfaces for which EG-F’:o. Since
ts-F’ is an invariant of the surface it can not be made
zero by a transformation of parameters.
3" " . . ,
Since (57.) is an arbitrary function of V,
V-uv’ we can write (77) as

[fl/V’IGV" ._ c ._ a.
(78) (EG‘F‘)? "' 7(V, V “V

Since the left hand side of (78) is a quadratic in V’ the
right hand side must be also. This is possible if and only
if 7 is a quadratic in V ' and V—u V’ . V will then assume

the following form
A v" +3V’+ C(v-uw)‘ +0 (v—u V’) f K J

from which we have

inn/06V" .
(56 v")?!

 

(79) (la mam/1) 4- (.3 —Du-zcuV) v’ +(A +cu’) V" =

Equating the coefficients of powers of V'in (79) we get

40

E = 1
(£6..th Cy {'DV-I'KJ
(80
) __’r__ —- 19 Cu .2
([G’Fz);§-.§ V’qu
6 :2 A + Caz .
(gs-r945
[6 -F‘
Com ti _—
pu ng (EC—F‘) “from (80) we get
I
(81 _____,___ =Acv1+ AD
mm ( wwv mew ran ”—4-:-

E g cv’fbv-HI ,

 

 

R!

(82). R2
6 = A *6“: J

1?!

H’ = [JG-F“:- 35-...

41

We can now prove the following theorem:

:heorem. gh§.gg;1,surfaoes ghggg geodesics g§g_bg_
represefigg pl linear eguations as Lh£s_e of constant tota_l_
curvature.

. The total curvature,/(t , expressed in terms of E, F

and G, is given by
9925 , __ .1 1.5a a; 5 a: *
(83) t:- 2_’-// En L& H 3344] +DI/[// Du. // 3V ‘7} 34]]¢

Using (80) we get

 

._£: 19"- I 6‘ _ C ’ . f ac 1
5,, :5; "W awn“?! K 73%“! (2Acx+£2__+AD)v

~2Acw3 +(CDK --f3)u2 -3Acp v1 4%...“ -2.3C/()u- T _——3—’-’,+Am/j

’—

25. :6 gist [( 9521, gas/()u’ + 2Ac‘u v2 +2ACD¢4V

-2 Age y + (AN—2AM 42.1.3344 wan],

’74-? =- M {(g, D" ~3CD/0a; +(ZCDz-44'3K) «av-2.34%
-23¢D a V +( «36K 3.331)“ + 3A CD v‘+(24cx my! 1332:) v

+2.4:IV3 +ADK {- £3’Dj’

 

*Iisenhart, p. 155.

42

4L.3£’ 2 ._ 3 - 1 3.. 3 .;g z
Hfi=kfiafl2cfi :52)u=v 24c 1/ Many +(CDK 4)“

+ (“z/ICK - £2? —- ANN +(é‘92- 236K)“ ‘91-? -—AM’]
2 2 4‘ "

h

fi ‘3 ..._!_ 2 a , a :
El/S— _x‘sIzBCuv +(yck CD)“ V+23c.p.uv

+ (zapx 232.3% -29’c v + (aw—2.9m)“ -g.z}
We find that

(84‘) 42.25-425-11 a; =0

 

fi/ 3“ II 2V AEA’ 51; J
and
/f 3&7 _szji I ‘4 3 1L 1
.- _. - .-.-._ Bcty + ABCDV
5” a ” au- (cV‘+Dv+/()R3§ [ 2

+ (38C’K-%.BCD')L¢’V + (3c: - A}? + 2463K)al/1
.'.- 3
9-5) + (“52" - CD‘A’+2€‘/(‘) «3 4- (C311) ~45? +2ACDK)uV
2
+(213c2k --”-B.D~')uz *(___3:c 7‘ L230 *AECK)V

I I"! 3
*(5-3’5’4551 + 4-323- +2ACK’)« -.%_4+4%9.!}

=;" JR'}QuL
(avian/0m:

Using (83), (84) and (85) we get

43

_LD k. RR“

Kt :- 2” an (4: v1 +DV 1.10723;

3 I -__L 1
W (3.... R 2 m)

— -§_is.. ’
_ACA’ “ Af—

='<Ln¢a1=uu4€

We next prove the converse theorem:

Iheorem. I; 13);; tgt_a_l_ curvature of a surface E.
constant than the geodesics 93 that surface m be expressed
linearly ig,u QEQ,V .

Let us choose our curvilinear coordinates in such a
manner that V': constant are geodesics andcx== constant are
the orthogonal trajectories. When such a curvilinear co-
'ordinate system is used the linear element may be put in the

form
(86) 44‘ = m.’ 4-- Gaw‘.

1
From (83) and (86) thetotal curvature is 7;!- g—f .

By the hypothesis of the theorem we have

(87) ”iii-e =K.

3“‘

We shall consider the three cases K=0 , K 70 , K40.

For /(=o we have

_Léié-eo J
6

244‘

01'

2
(88) 36 :0

got

The solution of (88) is 6' :2 Va «r t; where If and IQ
any two arbitrary functions of V.
For R70 we have

/ 3’6-.£
2-57‘2'42 J

or

é’f + _Q. :0 .
(89) a u’ a‘
The solution of (89) is

(90) 6=KCod§'*V4¢l«-%.

where V and I; are any two arbitrary functions of V.

For K <0 we have

23.5 : -.L
" EL 3 u" 42 J
01‘
31 _. _6. :0 a
(91) 5-5: at

are

44

45

The solution of (91) is

(92) 6=Veé+vz¢'a‘-‘.
If we taKe V"-.;.€+¥ J V’s-7V+¥ we can change
(92) into
-5 -
(93) vaef-€4*Z6£+eg.

 

’ T- 2

If in (90) and (93) we replace a by duwe have the

three forms of the linear element
4.5“ M1+(VR*K)I¢V’ J
(94) d4“ = a[d«‘+(¥,¢.;lu +15 muffle-“j J
‘ =a[d«=+(vdg—e'—‘%+W—%‘4-)’M%

If V': constant are to be geodesics it will be n
necessary for 6 to be zero when u is zero. If G = 0 when
use we have from (88), (90), and (93) K30 . Choosing
the coordinate V in a suitable manner we may reduce‘M'to

unity and write the three forms of (94) as
44’: du.’ + u’dv",

(95) at" = d’[da’+44‘u’udvz],

 

do‘ = a‘fi‘n‘ 49:; 3-326],

46

We next compute the geodesics for the three forms of

(95). For

11¢” = dflL’f'xx‘afir3.,
Euler's equation is

ALL :1 C,
(96) W

Solving for b/'and integrating we get
(97) Aumv *3KMV-I'C-100
For
44‘ = a’[d«’ +4.;n‘udw4 ,

Euler's equation is

#2“, V, -_-, C,

(98) W. . ”w.

Solving (98) for v’ and integrating we get

(99) Vsmmvf—tyflu-ui-C.

If we put K c (99) reduces to

__£L_.
VI " 6/2

47

(100) AMVW“ +3MVWK+C=0o

For
44‘: magnum maze—.1 ,

Euler's equation is

 

 

 

 

)aéa42zta v”' c:
M + M" ’V” J
01'
c
(101) V’=

44441 VM’u —-C‘- .

Solving (101) as we did (98) we get

(103) AMVMKf-flmvwfca'a.
The geodesics for the three forms are
Aumv+3aw VfC-saJ
(103) AMV‘W-w +3oéxyuau¢¢=c¢

A €10.wa +341“un +4 =0)

48

respectively.

The transformations which we must make on the u andv
in order that the equations of the geodesics will be linear

in a andv are evidently

Ei=uca¢v , 7::qu J
(104) E a emu/t...“ , Var-'MVfau-u. J

E=mvuul V=Mku,
respectively.

After making the transformation (104) the linear.

elements (94) take the form

 

44‘ : da‘i'da" J

at“; .. a'[du.’+dw' +cva4u-uac-rf]
.. (Ii-“.3 M”)? J

at.” __, a_'[gh’+dw’—-(vx..~u¢a)’]

 

(l—u“—V’)’
respectively.

Theorem. m only surfaces 9_n_ 32122.13. ho_th_ thg
equations _oi thg geodesics Eh; h_e_ linear it}. (4 9.1.19. V , hi;
the parametric system orthogonal §£g_developable surfaces.

The necessary and sufficient condition that two

fl .. V - .1.) K
curves be orthogonal is that {so . From (82) F: 2 C“ 2

R2

49

SinceF' is to be zero for all «’5‘ and Visit will be necessary
for 3 --= c =D so. since K. =Acx - €14 - grit follows that
“t is zero when F30 . The necessary and sufficient condition
that a surface be deve10pable is K¢=0 . This proves the

theorem.

If we take the sphere with the same uv— coordinates

as in part I and make the transformation
(2:01ku , V‘MVW“J
we find that the equations of the sphere become

a“

X -.- V] +u‘f-V’

 

(105) _ a V J
t( ‘ V9 rcxiwrvr'

 

*4
= Vl'rul'I-Vz

.-

 

Z

The linear element will be

as L" wad...)- ant/ac... at.- .0...» div-fl

‘dfl
4‘ - (Ifu’er’y'

If the equation of the geodesics is computed from (105) it

will take the form V2:- 0.

IV- AN ISOPERIMETRIG PROBLEM

In this part we consider the problem of finding
on a surface the curve of shortest length joining the
points 1 and 2, and enclosing with a given curve Cb, join-
ing 1 and 2, a fixed area. It was in the solution of this
problem that Minding discovered the function to which Bon-

net gave the name geodesic curvature.

l. gtatement g; the isoherimetric problem. The

isoperimetric problem is, given two integrals

“1

a
z= name , Je/‘yama,
“I

u.
to find among all admissible arcs, joining (1) and (8), the
one which gives the integral J'a constant value and minimizes
the integralj'?*

If AD; is a minimizing arc for the problem there

exists constants (Aal)¢(o.a) such that

(106) 0%: FVHszo

between corners of £72, where

 

*‘Minding, grelle, Vol. V (1830), p. 297.
xx G.A. Bliss, Lectures.

50

51
(107) F: Aof’lj.

If the are 5;, is taken to be normal there exists an unique
set of multipliers (1., J) = (I, .U . Hereafter we shall con—
sider only arcs normal on every sub-interval. Abnormal arcs
are of no importance in our problem since an abnormal arc
would also minimize the integral .7.

The other necessary and sufficient conditions are
like those of the ordinary problem of part I with {NIH/AW)
replaced by Fat, tel/1}!) .

2. 1h; problem 91 Minding. Given an arc Co join-

ing two points 1 and 2 on a surface, to find the curve of
shortest length joining l and 2, and enclosing with Co a
given area.

The area will be given by [fl/44060- , where the
integration is to be taken over the (AV-region corresponding
to the area} It is possible, however, to find two functions

flH(a\I ' 3A/..29 J**
.. ) and AN“. V) such that H- 5; V . By an application

of Green's theorem we have***

 

K
fiat If Ham/)is continuous in u we can take ,sz; may) and Ate/Wu),
I

x-H-F-S. Woods, hdvanced galculus, p. 181.

52
[/H‘m‘a” = f/(gg’gT/M)d“°b=/Mw r/Vdv'.

The problem then is to find among all the admissible arcs

which join 1 and 2 the one which minimizes the integral

a a
[W 4.4 and keeps the integral] (M fNVQda
a, “‘

equal to a constant.

The F-funotion (107) is

 

F: V: +2FV’+6 v" + 1((Me/VV’),

and Euler's equation is

d ( fitEV’ )... _Z-aéfZV'gvf‘fV’zéa-Vc AH
108 — I [Z 3: o
( ) flu VE'fZFV'fGV 2V?+2Fv’f6l/”

 

we can now state and prove the theorem of Minding:

Theorem. Lg order that _a_. curve c joining two points
shall he. the shortest which, together with 2 given curve

 

thrphgh these points, incloses a_ portion o_f_ the surface with
a given area, )5 _ig necessary that the geodesic curvature

 

if; C pg constant.

From a formula derived by Bonnet the geodesic

curvature of 9V“. V)‘ constant is a!

 

+Eisenhart, p. 136.

53

 

.L = _L a ( Fg‘ié “63'; )
9.7 H #5“ ((3.??- 2/35‘55 +6é9’

'3

9
(10) + 2( F3 ~6§§ )}
av , ‘
War-Mgae +6 at”)

 

If we take Pin the form
V - ¢(u)=0,

then
‘2! a". a = —v{
(110) 'av / a ,3“

Substituting (110) in (109), (109) becomes

Frcv’ ) _D_( eri )]
I - .L 2.. - I I ,
(111) (a; " H [2m (anrwrcvw 3V V2? 1” *‘V’

 

 

 

 

Equation (108) can be written as

E+£v’ 2- FfCV’ V
.1. 3-- Vg—Tr-p. *7 (W
A 3 H '3“ ’2 V f V V

I

(112 19 ,‘3F ,aQG
) Q5 4- 2v 2" 4V 7‘,

2 V2,. 2FV’f-6V’z

fl

 

54

Hence, comparing (109) and (112) we have the geodesic cur-

vature constant and equal to X .

Due to the fact that the geodesic curvature is
constant,the curves representing solutions of (109) are

called geodesic circles.

3. 1h§,necessary and sufficient conditions. The
Equnction is

Ewmvjwu = W: ”(MM/w)

 

4’5 sz' + 6v" -A (M +Nv')

rrGV’
"(Vi-v0 (quvw-sv" +1” ’

 

 

which reduces to

Elu,\/,V’V,’A)= VEfZFV'i-GV” ~VE‘f1FV’f6'v"

 

(113) . __ (V’- v') ( F *GV')
VE+2Fw+6 v"

 

 

The {vowof the Legendre condition is

”I.
(E +2Fv’rs v") ’2

 

(114) flowwiwvlz') =-

55

Both (113) and (114) are the exact expressions for
the corresponding functions in the ordinary problem of part I.
Hence, we have the same necessary and sufficient conditions

for the isoperimetric problem as for the ordinary problem.

4. Application t_g_§_ plane. Letting uzx and V83

equation (108) becomes

.4... -—L’ :4.
4‘, Wf;:§71

Integrating we get

I
(115) ”7%, =11“.

 

The solution of (115) is
(116) [1‘ + 113+2€,AX +2¢aAg +q‘+c,1-I =0.

Hence the curve of shortest length that joins two
points in the plane and makes the area with another fixed
curve, joining the points, constant is the arc of a circle
through the two points.

To show that the curves (116) are minimizing curves
we need to show that the sufficient conditions of part I

are satisfied. Since the only zeros of the determinant

56

[1+3
(Plum-6.)“

 

J I

Ha
I.

N, 1* CL
Vii-(1 t,+c,)‘

 

 

 

are .K==x; it follows that there are no points conjugate to

1. Therefore the curves (116) are minimizing curves.

.5. hpplication tp_§_sphere. For a sphere

 

d41= a VI+M‘1uv'i J

and (108) becomes

4 41.13“ v’
(117) M ( V/L——

+4.3“ 144V"

) = A afield-u.

If we integrate both sides of (117) and set ‘344.=I( we have

44‘sz

 

V/«f ‘1“V13 =KMLL+C.

Solving for V"we get

I Hull-f6]

118 V = '
( ) Mu. V2.4... -£K¢«-- *4?“

 

The solution of (118) is

When-4‘ 4.4.. (v-tz) - 6,6... a =I\’.

57

BIBLIOGRAPHY

A. GENERAL

1. Hancoldt, Geodatische Linien auf positive
gekrummten Flachen, Journal fur Mathematik, X01, (1881).

2. Braummuhl, Uber Envellopen geodatischer Linien
hathematische Annalen, XIV.

3; B.F. Kimball, Geodesics on a Toroid, American
Journal of mathematics, LII, pp. 29-53, (1930).

4. o.A. Bliss, The Geodesic lines on the Anchor
Ring, Annals of Mathematics, Ser. 2, IV, (1903).

5. L.P. Eisenhart, Differential Geometry of Curves
and Surfaces.

6. Bolza, variationsrechnung.

7. Laguerre, Sur un genre particulier de surfaces
dont on peut determiner les lignes geodesiques, Bulletin
de la societe mathematique,t, I, p. 81, (1873).

8. Bonnet, Sur quelques proprietes des lignes
geodesiques, Oomptes Rendus, t,XL, p. 1311, (1850).

9. Jordan, Sur la deformation des surfaces, Journal
de Liouville, 3 serie, t. XI, p. 105, (1866).

10. Lie, Untersuchungen uber geodatische Ourven,
Matematische Annalen, t. xx, p. 421.
11. Halphen, Traits des fonctions elliptiques et

58

59

de leurs applications.
B. THE ORDINARY PROBLEM

12. G.A. Bliss, Calculus of Variations.
13. Darboux, Theorie des Surfaces, III.

C. THE LAGRANGE PROBLEM

14. Gvo Bliss, The Problem of Lagrange in the
Calculus of variations, American Journal of Mathematics,

LII, no. 4,(0ct., 1930).
D. THE INVERSE PROBLEM

15. E. Beltrami, Risoluzione del Problems: Riportare
i punti di una superficie sopra un piano in modo che 1e linee
geodetiche vengano rappresentate da linee rette, Annali di
Matematica pura ed applicata pubblicati da B. Tortoline, t.
VII, p.185, (1866).

16. Barboux, Theorie des Surfaces, III.

17. Dini, SOpra un problems che si presenta nella
teoria generalle delle rappresentazioni geografiohe di una
superficie su di un' altra, Annali di Matematica, t. III,

p. 269, (1869).

18. m.J. Miles, Some inverse problems in the Calculus
of Variations, American Mathematical Monthly, XX, pp. 117-123,
(1913).

60

19. Thomas H. Rawles, The invariant integral and the
inverse problem of the Calculus of Variations, American

Mathematical Society Transactions, XXX, pp. 765-784, (1928).
E. THE ISOPERIMETRIC PROBLEM

20. Minding, Crelle, V, (1830).
21. Barboux, Theorie des Surfaces, III.

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