calculus of variations
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Calculus of VariationsTRANSCRIPT
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Chapter 9 Calculus of variations
Mathematical methods in the physical sciences 3rd edition Mary L. Boas
Lecture 10 Euler equation
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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.
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- 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.
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- What is the quantity which we want to make stationary in this chapter?
dx
dyydxyyxFI
x
x' where )',,(
2
1
Ex. 1) shortest distance
22
1
2
222
'1)',,( ;'1
'1
yyyxFdxyI
dxydydxds
x
x
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
222 '11
)'.,( ;'11
,'111
yv
yyxFdxyv
dtdxyv
dsv
dt
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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.
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2. Euler equation
1) Geodesic on a plane
2
1
22
1'1 minimize to)(
x
x
x
xdxydsIxyy We call this y(x) ‘extremal’.
We define a completely arbitrary function passing two points.
2 and 1at zero ishich function w a :)(
parameter, : extremal, desired :)(
)(ε)()(
xxx
xy
xxyxY
).()( 0,when ;
)parameter of(function '12
1
2
xyxY
dxYIx
x
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Then, our problem is to make I() take its min. when = 0.
)'( 0'1
)(')('
)(''
),(')(')(' .'
'2'1
1
2
1
.0 ,0
0
2
1 20
2
1 2
yYdxy
xxy
d
dI
xd
dYxxyxYdx
d
dYY
Yd
dI
d
dI
x
x
x
x
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)'( 0'1
)(')('0
2
1 20
yYdxy
xxy
d
dI x
x
- first term is zero because (x1)= (x2)=0.- (x) is an arbitrary function. In order for the second term to be zero,
const.or const.'1
' ,0
'1
'22
y'
y
y
y
y
dx
d
**From this, y(x) (geodesic on a plane) is a straight line.
0'1
')()(
'1
'
cf.
),( ,'1
' ,)(' ,'1/'
2
1 2
2
120
2
2
x
x
x
xdx
y
y
dx
dxx
y
y
d
dI
dxuvuvudvdx
xvdxy
y
dx
ddudxxdvyyu
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2) Generalization
2
1
2
1
2
1
2
1
)(''
)('
'
).()()(for )',,()',,(
x
x
x
x
x
x
x
x
dxxY
Fx
Y
Fdx
d
dY
Y
F
d
dY
Y
F
d
dI
xxyxYdxYYxFdxyyxFI
For arbitrary , equation Lagrange)-Euler(or Euler .0'
y
F
dx
d
y
F
0)(''
)( ,0at and 0/ Using2
10
x
xdxx
y
Fx
y
F
d
dIεyYddI
0)('
.)('
)('
)(''
term,second For the
2
10
2
1
2
1
2
1
x
x
x
x
x
x
x
x
dxxy
F
dx
d
y
F
d
dI
dxxy
F
dx
dx
y
Fdxx
y
F
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Example geodesic on a plane
2
1'1 minimize to)(
x
xdxyIxyy
.0'1
' ,0 ,
'1
'
' .'1)(
22
2
y
y
dx
d
y
F
y
y
y
FyxF
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Chapter 9 Calculus of variations
Mathematical methods in the physical sciences 3rd edition Mary L. Boas
Lecture 11 Application of Euler equation
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3. Using the Euler equation
1) Other variables
.0
/ where)',,(
θ
F
θ'
F
dr
d
dr,dθθ'drrF
.0
where),,(
x
F
x'
F
dt
d
dx/dt,x'dtx'xtF
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.
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.const.1
011'
1
' ,0
,0'
equation Euler theUsing
22
2222
22222
Kr
rdr
d
r
rr
dr
drr
dr
dF
dr
dF
FF
dr
d
Example 1. Find the path followed by a light ray if n (refractive index) is prop. to r-2 (polar coord.).
.0 without1
.1 1set
222
22222222
FrrF
drrrdrdrrdsrcndsdt
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const.arcsin n table,integratio theUsing
1
,)1('or 1
22
22222222
Kr
rK
K
dr
d
KrKrK
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2) First integrals of the Euler equation
const.'
0'
,0 If
y
F
y
F
dx
d
y
F
In this case, we can integrate the Euler equation once.Such a equation (F/ y’) is called a first integral of the Euler equation.
0'
FF
dr
d
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Example 2. Find the curve so that the surface area of revolution is minimized.
p1p2curve, y(x)
revolving the curve about x-axis
a surface of revolution
nintegratio of variable:,'1 ,'12
'1 form usual of instead
,'11
2
22
2
22
yxyFdyxyI
dxyds
dyxdyds
ydsI
dydx
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,'
.'1 ,'1
'
.0'1
'0
'
21
2
1
21
2212
2
0
cy
c
dy
dxx
xcxycx
yx
x
yx
dy
d
x
F
x
F
dy
d x
F
‘catenary line ( 현수선 )’
.cosh
,cosh n table,integratio theUsing
1
21
21
11
c
cxcy
cc
ycx
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- Like the above, if F(y,y’) does not have the independent variable x, we had better change to y as integration variable.
dyxdydy
dxdx
xy
dx
dy
dy
dxx ' ,
'
1' ,'
1
Example 3. . no where,'1 2
xdxy
yI
const.1'
'
01'
'
'0
'
.)',(1'
,1'''1'1
2
2
/
2222
xy
x
xy
x
dy
d
x
F
dy
d
x
F
x
F
dy
d
dyxyFy
dyxIdyxdyxydxy
xF
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Example 4. Find the geodesics on the cone z^2=8(x^2+y^2) using the cylindrical coordinate.
,'99
98
,8,8,8
22222
222222222222
22
drrdrdrdsI
drdrdrdrdrdzdrdrds
drdzrzrz
zzx
yyxr
zzryrx
zrzyx
,tan,
,sin,cos
),,(),,( :coordinate lCylindrica
122
,3
9)('
),'9('4 ,const.'9
'
' ,0
'
22
24242
22224
22
2
Krr
KdrdKrKr
rKrKr
rFF
dr
d
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.cosor 3
cos
n)integratio of const.( arccos1
3 table, theFrom
Krr
K
r
K
KK
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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.
‘gravity’
reference
mgy
dt
dsmmv
energy potential
,energy kinetic2
212
21
The sum of two energies is zero initially and therefore zero at any time and position.
.2or 0221 gyvmgymv
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The integral we want to minimize is
3)section 3, (example '1
2
1
2
2
1
2x
xdx
y
y
ggy
ds
v
dsdt
.1'
'
01'
'
'0
'
.)',(1'
,1'''1'1
equation)Euler theof integral(First
2
2
0/
2222
cxy
x
xy
x
dy
d
x
F
dy
d
x
F
x
F
dy
d
dyxyFy
dyxIdyxdyxydxy
xF
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')21arccos(2
1 table, theFrom
,or 1
'
1'
'
2
2
2
ccyc
yc
yx
yc
y
ydydx
cy
cy
dy
dxx
cxy
x
c’=0 for (x1, y1) = (0,0)
“This is the equation of a cycloid.”
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- 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.”
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When a circle rolled a little,
)cos1(cos
),sin(sin
aaaBCACABy
aaaPBOAx
aPAOA
‘parametric equation of a cycloid’
- Parametric equation of a cycloid
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.cos21 )21arccos(let we
),21arccos(2
1 chronebranchisto for theresult previous theFrom 2
cycy
cyc
yc
yx
cos12
1 ,sin
2
1
2
1sin
2
1
sin4
1cos1
4
1cos1cos12
4
1
),cos1(2
1 ,cos21
22
22
2
22
cy
cccx
cccy
c
yc
ycy
“parametric equation of brachistochrone or parametric equation of cycloid.”
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- Cycloids differ from each other only in size, not in shape.
22
2
y
x21
1
y
x23
3
y
x
- 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.
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Chapter 9 Calculus of variations
Mathematical methods in the physical sciences 2nd edition Mary L. Boas
Lecture 12 Lagrange’s equations
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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.
.0'
,0'
z
F
z
F
dx
d
y
F
y
F
dx
d
- 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.
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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.
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Example 1. Equation of motion of a single particle moving (near the earth) under gravity. (three dimensional motion)
.
,222
21
222212
21
mgzzyxmVTL
mgzVzyxmmvT
gz
y
x
mgzmdt
d
ymdt
d
xmdt
d
.
const.,
const.,
or
,0
,0
,0
equation sLagrange' .0 ,0 ,0
z
L
z
L
dt
d
y
L
y
L
dt
d
x
L
x
L
dt
d
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- 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.
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Example 2. Equation of motion in terms of polar coordinate variable r, .
. , 2222
222
22222 rrdt
dr
dt
dr
dt
dsvdrdrds
centripetal v^2/r Coriolis acceleration
).,(
).,( ,
22221
22221
rVrrmVTL
rVVrrmT
.0,0
.0 ,0
22
V
mrdt
d
r
Vmrrm
dt
d
LL
dt
d
r
L
r
L
dt
d
.2,
122 :motion ofequation
:motion ofequation
2
2
2
rrarra
V
rmarrm
Vrrrm
r
Vmarrmr
r
r
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Example 3. m1 moves on the cone. (spherical coord. , , ) m2 is joined to m1 and move vertically up and down. (z-component)
,: ,2/3cos: :
.: ,sin: :
21
2222
22412222222
22
1
lzmzmV
zvmdt
dsvmT
Here, the cone ( =30) is a constraint for motion.
.const.
,0,330cos,30sin
2221
21
21
lzzlz mmmm
Using the above,
.3 21212
22122
412
121 lgmgmmmVTL
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const.or 04/0
,034
0
.3
221
2121
21
21
21212
22122
412
121
mdt
dLL
dt
d
gmgmm
mmdt
dLL
dt
d
lgmgmmmVTL
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cf. Rolling disk
sin
4
3
4
1
2
1
4
1
2
1
2
1
2
1 22
222222
MgyV
RyyMR
yMRyMMRyMIyMT
.sin3
2
,0sin2
30
sin4
3 2
gy
MgyMdt
d
y
L
y
L
dt
d
MgyyMVTL
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cf. Atwood’s machine I
.,,, 2121 xvvvwhere
a
vx
dt
xldvx
dt
dxv
xlgmgxmV
a
xIxmxmIvmvmT
21
22
22
122
222
11 2
1
2
1
2
1
2
1
2
1
2
1
xlgmgxma
xIxmxmVTL
21
22
22
1 2
1
2
1
2
1
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221
2121221
21
221221
21
22
22
1
0
2
1
2
1
2
1
a
Imm
mmgxmmgx
a
Imm
mmgx
L
xa
Imm
x
L
dt
dx
a
Imm
x
L
x
L
x
L
dt
d
xlgmgxma
xIxmxmVTL
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cf. Atwood’s machine II
112222
1233
11222
1222
112
111
,2
1:
,2
1:
,2
1:
xlxlgmVxxmTm
xlxgmVxxmTm
gxmVxmTm
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cos,2
1
2
1 2222 grmgrMVrrmrMT
cf. Swing atwood’s machine
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cf. Double pendulum
,coscos,sinsin
,cos,sin
22
11
llyllx
lylx
cos22
1
2
1
sinsincoscos22
1
2
1
sinsincoscos2
1sincos
2
1
2
1
2
1
222222
222222
2222
22
22
21
21
lllmml
lllmml
llllmllm
yxmyxmT
coscos221 mglmglmgymgyV
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2222
2
22
22
2
1
2
1
2
1
2
1cos1cos1
2
1
sin & sin
ions,approximat some Using
2
1
2
1
klmglmgl
yxkmglmglV
ml
lyllylxllx
ymxmT
cf. Prob. 19
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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 with a constraint)
.0 ,0 ,0
),,( ),,,(
z
F
y
F
x
F
gfFczyxgzyxf
alue.constant vgiven :)',,(
stationary make tointegral :)',,(
2
1
2
1
x
x
x
x
dxyyxGJ
dxyyxFI
By using the Lagrange multiplier method,
equation.Euler esatisfy th should
.stationary be should 2
1
GF
dxGFx
x
- Good news: Sometimes we do not need to find .
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44
.'1 ,'1 ,
.'1 ,
22
2
1
22
1
2
1
yyGFyGyF
ldxydsJydxIx
x
x
x
x
x
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.
x1 x2
The curve length is fixed.(l > x2-x1)
01'1
'1 and
'1
'
' 22
y
y
dx
dGF
yy
yGF
y
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45
,
)(
,'
,'1''1
'
22
2222
2222
2
cx
dxcxdy
cxcxy
ycxycxy
y
.)'()('
,'
222222
22
cycxcxcy
cxcy
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46
7. Variation notation
- : differentiation with respect to . just like the symbol d in a differential except that , not x, is a differential variable.
0.at ),('),,(, of / aldifferenti totalajust
;''
' ,''
'
variable.a is if aldifferenti a likejust ;
)(')('),(' ),()(),(
,
0
0
0
xYxYxFdFdF
yy
Fy
y
FF
dxdxdx
dy
dx
ddxd
Yy
dYdxdY
y
xxyxYxxyxY
dd
dII
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.''
''
2
1
2
1
2
1
2
1
dxdxy
Fdx
y
F
dxyy
Fy
y
F
FdxFdxI
x
x
x
x
x
x
x
x
0)(''
)( cf. 2
10
x
xdxx
y
Fx
y
F
d
dI
- 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
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48
H. W (Due 13th of Nov.)Chap. 9
2-13-4,55-4 (G1), 5(G2), 9(G3)6-1, 2(G4)
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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.