Mathematical methods in the physical sciences 3rd edition Mary L. Boas
Chapter 9 Calculus of variations
Lecture 10 Euler equation
1
1. Introduction
- Geodesic: a curve for a shortest distance between two points
along a surface
1) On a plane, a straight line
2) On a sphere, a circle with a center identical to the sphere
3) On an arbitrary surface, ???  In this case, we can use the
calculus of the variation.
cf. Because the geodesic is the shortest value, finding the geodesic
is relevant to finding the max. or min. values.
2
- In the ordinary calculus with f(x), how can you find the max. or min?
(or how can you make the quantity stationary)
First, obtain the first derivative of f(x), and find the stationary points.
: Stationary point to make f’(x)=0. f(x) becomes Max. and Min.
points and more.
cf. But, we do not know if a given stationary point is a Max, Min, or
a point of the inflection with a horizontal tangent.
3
- What is the quantity which we want to make stationary in this chapter?
dy 

 where y'  
dx 

x2
I   F ( x, y, y' )dx
x1
2
2
2
Ex. 1) shortest distance
ds

dx

dy

1

y
'
dx
 

I 
x2
x1
1  y'2 dx;
F ( x, y, y' )  1  y'2
Ex. 2) brachistochrone problem: brachistos=shortest, chronos=time
e.g., In what shape should you bend a wire joining two given
points so that a bead will slide down from one point to the other in
the shortest time? We must minimize “time”.
ds : element of arc length, v=ds/dt : velocity
1
1
dt  ds 
1  y '2 dx ,
v
v
 dt  
1
1
1  y '2 dx; F ( x, y. y ' ) 
1  y '2
v
v
4
Ex. 3) a soap film suspended between two rings
What is the shape of the surface?
The answer is the shape to minimize the surface area.
Other examples
- chain suspended between two points hangs so that its center of
gravity is as low as possible.
- Fermat’s principle in optics. (light traveling between two given
points follows the path requiring the least time.
5
2. Euler equation
1) Geodesic on a plane
x2
x2
x1
x1
y  y( x) to minimize I   ds  
1  y'2 dx
We call this y(x) ‘extremal’.
We define a completely arbitrary function passing two points.
Y ( x )  y ( x )  ε ( x )
y ( x) : desired extremal,  : parameter,
 ( x) : a function which is zero at x1 and x 2
I 
x2
x1
1  Y '2 dx (functionof parameter )
; when   0, Y ( x)  y ( x).
6
Then, our problem is to make I() take its min. when  = 0.
dI
 0,   0.
d
x2 1
dI
1
dY '
 dY ' 

2Y ' 
dx. Y ' ( x)  y ' ( x)   ' ( x),
  ' ( x)

2
x
1
d
2 1 Y'
d
 d 
x 2 y ' ( x ) ' ( x )
 dI 
dx  0 (Y '   0  y )
   x1
2
d

   0
1  y'
7
x 2 y ' ( x) ' ( x)
 dI 
dx  0 (Y '   0  y )
   x1
2
 d   0
1  y'
u  y' / 1  y' ,
2
d 
y'
du 
dx  1  y '2
dv   ' ( x)dx,

dx,


v   ( x),
 cf. udvdx uv  vudx 

 

x2
y'
d 
y'
x2
 dI 

( x) x1    ( x)
  
x1
dx  1  y '2
 d   0
1  y '2

dx  0


- first term is zero because (x1)= (x2)=0.
- (x) is an arbitrary function. In order for the second term to be zero,
d  y'
dx  1  y'2


  0,


y'
1  y'
2
 const. or y'  const.
**From this, y(x) (geodesic on a plane) is a straight line.
8
2) Generalization
x2
x2
x1
x1
I   F ( x, y, y ' )dx   F ( x, Y , Y ' )dx for Y ( x)  y ( x)   ( x).
x 2  F dY
x 2  F
dI
F dY ' 
F

 

dx     ( x) 
 ' ( x) dx

x1
x1
d
Y '
 Y d Y ' d 
 Y

x 2  F

F
 dI 
Using dI / d  0 and Y  y at ε  0,       ( x) 
 ' ( x) dx  0
y'
 d  0 x1  y

x2
For the second term,

x2
x1
x 2 d  F 
F
F

 ( x)dx.
 ' ( x)dx 
 ( x)  
x
1
y '
y '
dx  y ' 
x1
x 2  F
d F 
 dI 
 ( x)dx  0
     

x
1
 d    0
 y dx y ' 
For arbitrary ,
F d F

 0. Euler (or Euler - Lagrange)equation
y dx y '
9
Example
geodesic on a plane
y  y( x) to minimize I  
x2
x1
1  y'dx
F
y'
F
F ( x)  1  y ' .

,
 0,
2
y'
y
1  y'
2
d  y'
dx  1  y'2


  0.


10
Mathematical methods in the physical sciences 3rd edition Mary L. Boas
Chapter 9 Calculus of variations
Lecture 11 Application of Euler equation
11
3. Using the Euler equation
1) Other variables
 F (r , , ' )dr
where θ'  dθ / dr,
d  F  F
 0.


dr  θ'  θ
 F (t, x, x')dt
where x'  dx/dt,
d  F  F
 0.


dt  x'  x
Note.
1. The first derivative is with respect to the integration variable in the
integral.
2. The partial derivatives are with respect to the other variables and
its derivatives.
12
Example 1. Find the path followed by a light ray
if n (refractive index) is prop. to r-2
(polar coord.).

dt   nds set c  1   r 2 ds   r 2 dr2  r 2 d 2   r 2 1  r 2 2 dr.
F  r 2 1  r 2 2 without  
Using theEuler equation
F
 0.

d  F  F
 0,


dr   '  
F
d  F  d  r  2 1  r 2 2
 0,



dr   '  dr 
 '

1 r 
2
2
 d  r  2 r 2 
 
 dr  1  r 2 2

 d 

  
2 2
 dr  1  r  

  0

 const. K .
13
 2  K 2 1  r 2 2  or  '2 (1  K 2 r 2 )  K 2 ,
 
d
K

dr
1  K 2r 2
Using theintegration table,
  arcsin Kr  const.
14
2) First integrals of the Euler equation
d  F  F
0


dr   '  
If
F
 0,
y
d F
F
0
 const.
dx y '
y '
In this case, we can integrate the Euler equation once.
Such a equation (F/ y’) is called a first integral of the Euler equation.
15
Example 2. Find the curve so that the surface area of revolution is minimiz
I   2yds
ds  1 
  dy 
dx 2
dy
1  x'2 dy,
insteadof usual form ds  1  y '2 dx
I   2y 1  x'2 dy, F  y 1  x'2 , y : variableof integration
curve, y(x) p2
p1
revolving the curve about x-a
a surface of revolution
16
F
0
d F F
d  yx'
x


 0 
dy x' x
dy  1  x'2
yx'
1  x '2
x' 
dx

dy
y 2 x2  c11  x'2 .
 c1 ,
c1
y c
2

  0.

2
1
,
‘catenary line (현수선)’
Using theintegration table, x  c1 cosh1
 y  c1 cosh
y
 c2 ,
c1
x  c2
.
c1
17
- Like the above, if F(y,y’) does not have the independent variable
x, we had better change to y as integration variable.
1
dx  dy 
x' 
  ,
dy  dx 
Example 3.
I 
y' 
1
dx
, dx  dy  x' dy
x'
dy
1  y '2
dx,
y
where no x.
1  y ' dx  1  y ' x' dy  x' 1dy, I  
2
2
2
x'2 1dy
  F ( y, x' )dy.
y

d  F  F
d  F  d 
x'
F / x
0
 0  



dy  x'  x
dy  x'  dy  y x'2 1 
x'
y x' 1
2
 const.
18
Example 4. Find the geodesics on the cone z^2=8(x^2+y^2)
using the cylindrical coordinate.
z 2  8r 2 , z  r 8 , dz  dr 8 ,
ds2  dr2  r 2 d 2  dz2  dr2  r 2 d 2  8dr2  9dr2  r 2 d 2
I   ds  
9dr2  r 2 d 2  
9  r 2 '2 dr,
d  F 
F
r 2 '

 const. K , 4r 4 '2  K 2 (9  r 2 '2 ),

  0,
dr   ' 
 '
9  r 2 '2
 '2 (r 4  K 2 r 4 )  9 K 2   d  
3Kdr
r r K
2
2
,
Cylindrical coordinate: ( x, y, z )  (r , , z )
x  r cos , y  r sin  , z  z
 y
r  x 2  y 2 ,  tan1  , z  z
x
19
From the table,     3K 
1
K
arccos
K
r
(  const.of integration)
    K
   
cos
or r cos

  K.
 3  r
  
20
4. Brachistochrone problem: cycloids
A bead slides along a curve from (x1,y1)=(0.0) to (x2,y2). Find the
curve to minimize the during time.
2
 ds 
kineticenergy  12 m v2  12 m  ,
 dt 
potentialenergy  m gy
The sum of two energies is zero initially and therefore zero at any time an
1
2
mv2  mgy 0 or v  2gy.
reference
‘gravity’
21
The integral we want to minimize is
 dt  
ds

v 
ds

2 gy
1
2g

x2
x1
1  y '2
dx (example3, section3)
y
(First integral of theEuler equation)
1  y ' dx  1  y ' x' dy  x' 1dy, I  
2
2
2
x'2 1dy
  F ( y, x' )dy.
y

d  F  F
d  F  d 
x'
F / x  0
0
 0  
 



dy  x'  x
dy  x'  dy  y x'2 1 
x'
y x' 1
2
 c.
22
x'
y x ' 1
2
x' 
 c
dx
cy

dy
1  cy
or dx 
From the table, x  
ydy
,
y
 y2
c
y
1
 y 2  arccos(1  2cy)  c'
c
2c
c’=0 for (x1, y1) = (0,0)
“This is the equation of a cycloid.”
23
- Cycloid A circle (radius a) rolls along x-axis. It start at origin
O. Place a mark on the circle at O. As the circle rolls, the mark
traces out a cycloid.
“trace of a mark on the circle when the circle rolls.”
24
- Parametric equation of a cycloid
When a circle rolled a little,
OA  PA  a
x  OA  PB  a  a sin   a(  sin  ),
y  AB  AC  BC  a  a cos  a(1  cos )
‘parametric equation of a cycloid’
25
From the previous result for the branchistochrone x  
we let arccos(1  2cy)  
1  2cy  cos , y 

 1  2cy  cos .
1
(1  cos ),
2c
y
1
 y 2  arccos(1  2cy),
c
2c

y
1
1
1
2
 y 2  2 21  cos   1  cos   2 1  cos2    2 sin 2 
c
4c
4c
4c
1
1
1
1
x   sin       sin  , y  1  cos 
2c
2c
2c
2c
“parametric equation of brachistochrone or parametric equation
of cycloid.”
26
- Cycloids differ from each other only in size, not in shape.
- Rather surprisingly, when a bead slides from origin to P3 in the
least time, it goes down to P2 and backs up to P3 !!
At P2, x/y = /2.
For P1, x/y < /2.
For P3, x/y > /2.
x1 

y1 2
x2 

y2 2
x3 

y3 2
27
Mathematical methods in the physical sciences 2nd edition Mary L. Boas
Chapter 9 Calculus of variations
Lecture 12 Lagrange’s equations
28
5. Several dependent variables; Lagrange’s
equation
- Necessary condition for a minimum point in ordinary calculus,
for an one-variable function z=z(x), dz/dx=0,
for a two-variable function z=z(x, y), z/x=0 and z/y=0.
- The similar idea is applied to Euler equation.
When for F=F(x, y, z, dy/dx, dz/dx) we find two curves y(x) and z(x)
to minimize I =  F dx,
we need two Euler equations.
d  F  F

 
 0,
dx  y'  y
d  F  F
 0.


dx  z '  z
29
It is a very important application to mechanics
; Lagrangian based on Hamilton’s principle
- Lagrangian: L = T – V where T : kinetic energy, V : potential
energy
- Hamilton’s principle: any particle or system of particles always
moves
in such a way that I =  L dt is stationary.
In this case, Euler equation is called Lagrange’s equation.
30
Example 1. Equation of motion of a single particle moving (near the
earth) under gravity. (three dimensional motion)

mx
 V  m gz
  m gz.
T  12 m v2  12 m x 2  y 2  z 2 ,
L  T V 
1
2
2
 y 2  z 2
d  L  L
d  L  L
d  L  L




0
,


0
,
 0. Lagrange's equation
 
 


dt  x  x
dt  y  y
dt  z  z
d
 dt mx   0,

d
or
 my   0,
dt

d
 dt mz   m g  0,

 x  const.,


 y  const.,


 z   g.
31
- In some cases, it would be simpler to use elementary method
from Newton’s equation. However, in some cases with many
variables it would be much simpler to use Lagrange’ equation,
because we treat one scalar function, Lagrangian L = T – V.
32
Example 2. Equation of motion in terms of polar coordinate variable r, .
 ds   dr 
 d 
ds2  dr2  r 2 d 2 , v 2        r 2    r 2  r 2 2 .
 dt   dt 
 dt 
2

2
2

T  12 m r 2  r 2 2 , V  V (r , ).


 L  T  V  12 m r 2  r 2 2  V (r , ).
d L L

 0,
dt r r
d L L

 0.
dt  
d
mr   m r 2  V  0,
dt
r



d
V
m r2 
 0.
dt


V
r equation of motion: m r  r 2  m ar 
r
 equation of motion: mr 2  2rr   
ar  r  r 2 ,
V



1 V
 m r  2r  m a  
r 
a  r  2r.
centripetal v^2/r
Coriolis acceleration
33
Example 3. m1 moves on the cone. (spherical coord. , , )
m2 is joined to m1 and move vertically up and down. (z-componen
Here, the cone ( =30) is a constraint for motion.
sin 30  12 , cos30 
1
2
3 ,  0,
 m1  zm 2  l  const. zm 2   zm 2  l   .
Using the above,
2
ds


T : m1 : v 2      2   2 2   2 sin 2  2   2  14  2 2 , m2 : v 2  z 2   2 .
 dt 
V : m1 : z   cos   3 / 2, m2 : z  l   ,


L  T  V  12 m1  2  14  22  12 m2  2  12 m1g 3  m2 gl   .
34


L  T  V  12 m1  2  14  2 2  12 m2  2  12 m1 g 3  m2 g l   .
d  L  L
d
m1  2 1
  
 0  m1   m2   
 2 m1 g 3  m2 g  0,
dt    
dt
4
d  L  L
d
2 
  

0

m

 / 4  0 or
1
dt    
dt


 2  const.
35
cf. Rolling disk
2  3
1
1 2 1
1
1
1
2
2 2
2 y
T  My  I  My  MR   My  MR  2   My 2
2
2
2
4
2
4
R  4
 y  R 
V   Mgy sin 
L  T V 
3
My 2  Mgy sin 
4
d  L  L
 
0
dt  y  y
y 

d 3

 My   Mg sin   0,
dt  2

2
g sin  .
3
36
cf. Atwood’s machine I
dx
v1 
 x ,
dt
d l  x 
v2 
  x,
dt
v
a
 ,
where v  v1  v2  x.
1
1
1
1
1
1  x 
2
2
T  m1v1  m2 v2  I 2  m1 x 2  m2 x 2  I  
2
2
2
2
2
2 a
2
V  m1 gx  m2 g l  x 
2
1
1
1  x 
L  T  V  m1 x 2  m2 x 2  I    m1 gx  m2 g l  x 
2
2
2 a
37
2
1
1
1  x 
L  T  V  m1 x 2  m2 x 2  I    m1 gx  m2 g l  x 
2
2
2 a
d  L  L
0
 
dt  x  x
L 
I 
  m1  m2  2  x 
x 
a 
d  L  
I 
    m1  m2  2  x
dt  x  
a 
L
 g m1  m2 
x
I 
g m1  m2 

  m1  m2  2  x  g m1  m2   x 
I 
a 


m

m

 1

2
a2 

38
cf. Atwood’s machine II
m1 : T 
1
2
m1 x1 ,
2
m2 : T 
1
2
m2 x2  x1  ,
2
m3 : T 
1
2
m3  x2  x1  ,
2
V  m1 gx1
V  m2 g x2  l1  x1 
V  m2 g l2  x2  l1  x1 
39
cf. Swing atwood’s machine
T


1
1
Mr 2  m r 2  r 2 2 ,
2
2
V  grM  grm cos 
40
cf. Double pendulum
x1  l sin  , y1  l cos ,
x2  l sin   l sin  , y2  l cos  l cos ,
T



1
1
2
2
m x1  y12  m x2  y 22
2
2

 




 

2
2
2
2
1
1
m l cos  l sin   m l cos  l cos  l sin   l sin 
2
2

1 2 2 1
m l   m l 2 2  l 2 2  2lcos cos  sin  sin  
2
2

1 2 2 1
m l   m l 2 2  l 2 2  2l cos   
2
2




V  mgy1  mgy2  2mglcos  mglcos
41
cf. Prob. 19
T
1 2 1 2
mx  my
2
2
Using some approximations,
x  l sin   l  x  l & y  l sin   l  y  l

1
 m l  2  2
2

1
2
V  m gl1  cos   m gl1  cos   k  x  y 
2
1 
1  1
2
 m gl  2   m gl  2   kl 2    
2 
2  2
42
6. Isoperimetric problems
Ex. To find a curve to make largest area ( y dx = Max.) with a given
length ( ds = l)
cf. Lagrange multiplier (Max. or min. (stationary point)
problem withf a( xconstraint)
, y, z ),
g ( x, y , z )  c  F  f   g
F
F
F
 0,
 0,
 0.
x
y
z
x2
I   F ( x, y, y' )dx : integralto makestationary
x1
x2
J   G( x, y, y' )dx : given constant value.
x1
By using the Lagrange multiplier method,
 F  G dx
x2
x1
should be stationary.
F  G should satisfy the Euler equation.
- Good news: Sometimes we do not need to fi
43
Example 1. Find the shape of the curve of constant length joining
two points x_1 and x_2 on the x-axis which, with the x axis, encloses
the largest area.
x2
I   ydx,
x1
J 
x2
x1
ds  
x2
x1
1  y '2 dx  l.
F  y , G  1  y ' 2 , F  G  y   1  y ' 2 .

y'

d  y'
F  G  
F  G   1  
and
2
y'
y
dx 1  y'2
1  y'


 1  0


The curve length is fixed.
(l > x2-x1)
x1
x2
44
y '
1  y'
2
2


 x  c  2 y '2  x  c  1  y '2 ,


y '2 2  x  c   x  c  ,
dy 
2
2
( x  c)dx
  x  c 
2
2
,
y  c'   2  x  c  ,
2
 y  c'2  2  x  c2  ( x  c)2  ( y  c' )2  2 .
45
7. Variation notation
-  : differentiation with respect to . just like the symbol d in a
differential except that , not x, is a differential variable.
dI
d ,
d
Y ( x,  )  y ( x)   ( x),
I 
Y ' ( x,  )  y ' ( x)   ' ( x)
 Y 
 d    x d ; just like a differential dY if  is a variable.
   0
d
d
 Y ' 




 x d    ' x d
y '  
d



'
x
d

,

y


dx
dx
   0
y  
F
F
y 
y ' ;
y
y '
just a totaldifferential dF  F /   0 d of F x, Y ( x,  ),Y ' ( x,  ) at   0.
F 
46
x2
x2
x1
x1
I    Fdx   Fdx
 F
F 
  
y 
y ' dx
x1
y ' 
 y
x 2  F

F
    x d 
 ' x d dx.
x1
y '
 y

x2
x 2  F

F
 dI 
cf.       ( x) 
 ' ( x) dx  0
y'
 d  0 x1  y

- The meanings of two statement are the same
(a) I is stationary; that is, dI/d=0 at =0.
(b) The variation of I is zero; that is, I=0
47
H. W (Due 13th of Nov.)
Chap. 9
2-1
3-4,5
5-4 (G1), 5(G2), 9(G3)
6-1, 2(G4)
48
Problem
5-4 Use Lagrange’s equations to find the equation of motion of a
simple pendulum.
5-5. Find the equation of motion of a particle moving along the x
axis if the potential energy is V=(1/2)kx^2.
5-9 A mass m moves without friction on the surface of the corner r
= z under gravity acting in the negative z direction. Find the
Lagrangian and Lagrange’s equation in terms of r, .
6-2 The plane area between the curve and a straight line joining the
points is a maximum.
49