(Principle of Least Action)

Hamilton’s Principle

Hamilton’s Principle

The path that a dynamical system taken in moving from point 1 at time t1

to another point 2 at time t2 which satisfies the Newton’s law minimizes

the functional: 2

1

t

tI L dt

L T V

In which the Lagrangian L is:

Where T is the “Kinetic energy” and V is “Potential energy”.

Example (vibrating system): For the system,

n2

i ii 1

1T m x

2

n2

i ii 1

1V k x

2

2 2

1 2

1 1T Mx Mx

2 2

2 2 2

1 1 2 1 2 1 2

1 1 1V k x k ( x x ) k x

2 2 2

L T V

2

1

t

tI (T V )dt

I 0 1 1

2 2

F d F0

x dt x

F d F0

x dt x

1 1 2 2 1 1

1 2 2 2 1 2

dk x k ( x x ) Mx 0

dt

dk x k ( x x ) Mx 0

dt

2

1

t 2 2 2 2 2

1 2 1 1 2 2 1 1 2t

I

1 1M ( x x ) k x k ( x x ) k x dt

2 2

1 1 2 1 2 2

2 2 1 1 2 2

Mx ( k k )x k x

Mx k x ( k k )x

1 1 2 2 1

2 2 1 2 2

x k k k xM 0

x k k k x0 M

1 1 1

2 2 2

x Asin( t )

x B sin( t )

x sin , y cos y

x

l

Datum Δh 2 21

T m( x y )2

x cos , y sin

2 2 2 2 2 21 1T m (cos sin ) m

2 2

h cos ( 1 cos )

V mg h V mg ( 1 cos )

2 21L T V m mg ( 1 cos )

2

L d LE .L . ( ) 0

dt

2d

mg sin ( m ) 0dt

g

sin

Example: Motion of a Pendulum Find the angular acceleration of a pendulum of length .

2

0

1I EIy q( x ) y ( x ) dx

2

Example (Deflection of a simply supported beam):

I 0

l

0EIy y q( x ) y dx 0

l l

0 0

2l

20

EIy y ( EIy ) y

dEIy q( x ) ydx 0

dx

EIy ( x ) q( x ) (Euler equation of the functional above)

EIy ( 0 ) EIy ( ) 0

y ( 0 ) y ( ) 0

Natural boundary conditions

q(x)

y(x)

Eq. EIy ( x ) q( x )

0EIy y ? B.C.

0EIy y ?

,

but

but

but

but

q(x)

x 0y ( 0 ) 0 y 0

x 0EIy 0

xEIy 0

xy ( ) is not specified y 0

x 0EIy 0

x 0y ( 0 ) 0 y 0

xEIy 0

xy ( ) is not specified y 0

0EIy y 0

0, EIy y 0

0( EIy y q( x ) y )dx 0

Integration by parts:

0

0

( EIy y q( x ) y )dx

EIy y 0

by parts again:

0

0 0

( EIy y q( x ) y )dx

EIy y EIy y 0

EIy ( x ) q( x ) 0

How to find the functional from corresponding Euler equation?

dv u

udv

21EIy y EI y

2

q y qy

2

0

0 0

1( EIy q( x ) y )dx

2

EIy y EIy y 0

STRONG FORM 0 0EIy y EIy y 0

2

0

1I EIy q( x ) y ( x ) dx

2

EIy ( x ) q( x ) 0

2

0

1I EIy q( x ) y ( x ) dx

2

Subject to B.C so that. :

WEAK FORM

Now assume an external moment M1 at the end of the cantilever beam:

xEIy y

Now we solve:

l 2

0

1( EIy q( x ) y )dx My 0

2

1xEIy M

q(x)

M1

0EIy y 0

1 x( M y )

1 x( M y )

What does it mean according to Hamilton’s principle?

)(02

py

dx

dyF

dx

d

Another example: small deflections of a rotating string:

y(x): displacement F(x): tension in the string ρ(x): linear mass density w: angular velocity of rotation p(x): intensity of a distributed laod

Multiply by δy and integrate: 0)( 0

)(0

21

2

022

l

py

l

y

l

dxypdxyyydxdx

dyF

dx

d

Integrate the first term by parts:

0)()(2

100

22

0

)(2

1

0

2

lll

dx

dyF

l

dxpydxypwdxydx

dyFy

dx

dyF

If we impose:

0

0

1

0

00

lxlx

xx

dx

dyForyy

dx

dyForyy

0)(2

1

2

1

00

222

ll

ydx

dyFdx

dx

dyFpyyw

The boundary terms vanishes and we obtain:

)(0)(2

1

2

1

0

222

dxdx

dyFpyyw

l

are the so-called “natural boundary conditions” The end conditions 00

l

ydx

dyF

2)(2

1wySpeed of an element of the string= wy

(Kinetic energy of string per unit length)

pyThe work done by p (Potential energy per unit length due to P)

2)(2

1

dx

dyF Potential energy per unit length due to tension F)

00

)(

x

x

Kydx

dyF

In case of yielding support at the end x=0

Since this term would not vanish, we write:

If the slope of the string at the end x=0 were prescribed as y’(0)=α

0)2

1()(

2

1

2

1

0

2

0

222

x

l

Kydxdx

dyFpyyw

000

)()()(

xx

x

FyyFydx

dyF Then

and we write:

0)0()0()(2

1

2

1

0

222

yFdxdx

dyFpyyw

l

ty

Example: motion of an elastic string Use Hamilton’s principls to derive the wave equation for small transverse ascillations of a taut string. Solution: Let ρ : linear density T : tension

Transverse speed of any part of the string :

l

dxt

yT

0

2)(

l

dxdsTV0

)(

But: dxdx

dx

dydxdydxdxds

21

221

22 )(1)()(

Since: ...)(

2

11)(1 2

21

2

dx

dy

dx

dy

ds

dx y

x x

y

Since it is small oscillation we will take the first two terms:

dxdx

dydxds 2)(

2

1

l

dxdx

dyTV

0

2)(2

1

dxdx

dyT

dt

dyVTL

l

0

22 )(2

1)(

2

1

dxdtdx

dyT

dt

dyI

t

t

l

2

1 0

22 )(2

1)(

2

1

Euler-Ostrogradsky:

22

2

1

2

1yTyL

0)()(

y

L

xy

L

ty

L

0)()(

yT

xy

t

0)()(

x

y

xT

t

y

t

2

2

2

2

t

y

Tx

y

Wave equation

Example: A bead of mass m slides freely on a frictionless circular wire of

radius r that rotates in a horizontal plane about a point on the circular

wire with a constant angular velocity w. show that the bead oscillates as a

pendulum of length l=g/w2 about the line joining the center of the circle and the point of rotation.

x r cos t r cos( t )

y r sin t r sin( t )

Since the motion is taken place in a horizontal

plane, the potential energy can be taken as zero, v=0 .

x r sin t r ( ) sin( t )

y r cos t r ( )cos( t )

2 21T m( x y )

2

r

C r

θ Φ=wt

Φ=wt

bead

O

2 2 21

T mr ( ) 2 ( )cos ( t ) t2

2 2 21T mr ( ) 2 ( )cos

2

L d L( ) 0

dt

2mr ( ) sin sin 0

2sin

2 2 2 2 2 2

2

2 2 2 2 2 2

2

r sin t r ( ) sin ( t )

2 r ( ) sin t sin( t )1T m

2 r cos t r ( ) cos ( t )

2 r ( )cos t cos( t )

gPendulum Motion : sin

2

g