SPECIAL CASES OF THE EULER-LAGRANGE EQUATION

Remember

: , ,

~ ' ...

x t y y q q

Euler Lagrange s Eqns

− > ′− >

> −



!

SPECIAL CASES OF THE EULER-LAGRANGE EQUATION

~ ( , , ) : Lagrange .. Eqn .

( , ), ( ) : tan

( , ) : Jacobi ..integral ( , ), ( ) : ?

F F x y y

F F x y F F y F cons t

y F F y y

F F x y F x

= ′

′ ′ ∂

= = =

∂ ′

= ′

=

Eqn(3.13)! Jacobi energy integral

δ ∫ ∫

δ

THE VARIATIONAL OPERATOR

Referring to Fig.3.5, the small perturbation of the extremal

function y x*( ) ,that is,εη( )x is known as a variation of y x*( ) _.

It is an infinitesimal changes in the values of y x*( ) and denoted by

*:

δ y

( ) *( ) ( )

*( ) *( ) y x y x x

y x y x εη δ

= +

= +

For an infinitesimal change in the value of a function ^{y x( )} , or variation of ^{δ y,} the new value of the functional is given by

I y[ +δ y] = I y[ ]+δ I

That is, a variation in the function y x( ) induces a corresponding infinitesimal change or variation in the value of the functional ^{I y( )}.

Furthermore, a maximum or minimum value of I y[ ]corresponds to local stationary behavior. In other words,

I y[ +δ y]− I y[ ] ≡ δ I = 0.

is a necessary condition for an extremum of I y[ ]_.

Thus local stationary behavior of a functional is implied by setting the variation of the functional equal to zero.

Utilizing the notion that the variational operator behaves like a differential operator, we proceed to establish the necessary condition for an extremum.

The variation of the functional ^{I y[ ]} is computed as

2 2

1 1

( , , ) ( , , )

x x

x x

I F x y y dx F x y y dx

δ =δ

∫

∫

since the interval of integration is held fixed. Using the properties of the variation operator ^δ ,

^{δ F x y y( , , )′ = ∂}_∂^F_y ^{δ y+ ∂}_∂^F_y′^{δ y′}

Note : δ x = 0.

Therfore the variation of ^{I y[ ]} is given by

2

1

( )

x x

F F

I y y dx

y y

δ = ^∂ δ + ^∂ δ ′

∂ ∂ ′

∫

The variational operator also computes with the usual derivative operator, that is,

δ y^′ = εη^′ = (εη)^′ = (δ y)^′

So we have

1²

[ ( ) ]

x x

F F

I y y dx

y y

δ = ^∂ δ + ^∂ δ ^′

∂ ∂ ′

∫

Integrating by parts to eliminate the derivative on the variation

δ y , we have

1^{2 12}

[ ( )] ( )

x x

x x

F d F F

I ydx y

y dx y y

δ = ^∂ − ^∂ δ + ^∂ δ

′ ′

∂ ∂ ∂

∫

Now the variations vanish at the end points. Thus the boundary terms in Equation (3.14) also vanish.

Since ^{δI = 0} for arbitrary variations ^{δ y} , the Fundamental Lemma of the Calculus of Variation requires that :

^∂_∂^F_y ⁻ _dx^{d (∂}_∂^F_y′^{) = 0}

: Euler-Lagrange equations for an extremum of ^{I y[ ]}.

Utilizing the variational operator is much more efficient than using the ε -argument discussed previously.

Necessary conditions can now be derived for integral functionals with more elaborate Lagrangians and also for functionals involving several input functions.

NATURAL BOUNDARY CONDITIONS

In the previous sections, part of the conditions for admissibility consisted of specified boundary conditions. That is , the candidate functions all had to satisfy certain value constraints at the boundary of the domain of integration. Consequently, all the allowable variations had to vanish at these same endpoints. The derivation of the associated necessary condition for an extremum functional ^{I y[ ]} was based on these specifications. These requirement will now be relaxed.

Suppose that in the above variational problem only on the value at one of the endpoints is specified. That is, only one endpoint is designated as part of admissibility.

Without loss of generality, suppose the value of each candidate function is specified at x1, say y(x1) = y1.

The extremal y*(x) can still be infinitesimally varied, but now there is no restrictions on the variation ^{δ y} at the end point x2 (Fig3.6). The minimization of the functional ^{I y[ ]} still requires local stationary behavior.

The difference now is that the collection of admissible

functions is larger. From Eqn(3.14), the variation of the functional is given by

^x1² [ ( )] ( )2 x

F d F F

I ydx y x

y dx y y

δ = ^∂ − ^∂ δ + ^∂ δ

′ ′

∂ ∂ ∂

∫

The extra term in Eqn(3.15) results from no restriction on the variation of y(x) at the endpoint x2 .

It is important to note that the variation of any function at x2 is a local process. That is, the variation of the function inside the domain and varation of the function at the boundary point can be done independently. Since there is no restriction at x2, no variation there at all is

also allowable. The variation in Eqn(3.15) can therefore be considered separately.

Thus the requirement of local stationary behavior ^{δI = 0} results in two necessary conditions.

One follows from the Fundamental Lemma of the Calculus of Variations and is given by the standard Euler-Lagrange Eqn:

^∂_∂^F_y ⁻ _dx^{d (∂}_∂^F_y′^{) = 0}

The other condition follows from the fact that the variation at the right endpoint need not to be zero.

Thus ^{δI = 0} for all admissible variations also dictates that ^{x x2 0}

F y ⁼

∂ =

∂ ′

This additional boundary condition is thus an induced boundary condition and is referred to as a Natural (or Force) boundary condition ( NBC ). This additional condition is inherent to the variational procedure and provides the second condition

necessary for the minimization of a functional.

Boundary conditions specified as part of the variational problem are known as Essential (or Geometric) boundary conditions (EBCs).

Essential boundary conditions are specified in advance and thus require that ^{δ y = 0} at the given boundary.

Ex: string problem ^{+T x( ) dy}_dx^{δ y x l= = 0}

GENERALIZATIONS

Up to now, only functional with simple form of Lagrangian

( , , ) F = F x y y′

~> For more generality with several functions !

(ex: plate, shell ~ geometrically linear, non-linear problems…, elastic ,plastic material properties )

As in

1

0

[ ] ^x ( , , )

I y^ =

∫

( 1, ..., _n)

y = y y : independent Ftns Ex :

1 1

0 0

2

1 1

[ ] ( , , ) ( , , )

[ ( )]

x x

x x

x n x i

i i i

I y F x y y dx F x y y dx

F d F

y dx y y dx

δ δ δ

δ

=

′ ′

= =

∂ ∂

− ′

∂ ∂

∫ ∫

∫ ∑

    

(3.17) Each yi can be varied independently

For each function :

( ) 0... 1,...

i i

F d F

i n

y dx y

∂ − ∂ = =

∂ ∂ ′ (3.18)

Ex: Two input functions

 Inverse operation Ex:

( , , , )..?

F = F x y y y′ ′′ : ^{δ I = ?}

Ordinary Differential Eqn !

Partial Differential Eqn (x,t)

Euler Beam bending, vibration problem:

4 2

4 2

( , ) ( , )

d y x t , d y x t

EI q m

dx = dt

~

2

2 2

2

1 ( , ) 1 ( , )

( ) , ( )

2 2

y x t y x t

T m V EI

t x

∂ ∂

= =

∂ ∂

SEVERAL INDEPENDENT VARIABLES Dimensional problems

VARIATIONAL PROBLEMS WITH CONSTRAINTS

Two Types of Constraint!

~ Integral constraint Eqn(3.20)

Isoperimetric problem

(F + λG),_y −[(F + λG), ],_y' _x = 0...(3.23)

- More general isoperimetric problem Eqn(3.24)

~ Eqns of constraint

(Non~) holonomic !

( , 1..., ) 0...(3.25)

. ,

, 0

n

yi i

G x y y or

Taking variation G δ y

=

=

HAMILTON’S PRINCIPLE

Beam bending problem:

4 2

4 2

( , ) ( , )

y x t y x t

EI m

x t

∂ ∂

∂ = ∂

~

2

2 2

2

1 ( , ) 1 ( , )

( ) , ( )

2 2

y x t y x t

T m V EI

t x

∂ ∂

= =

∂ ∂

~ Integral constraint :

More general isoperimetric problem

~ Eqns of constraint : (Non~) holonomic constraint ! : Boat :

HAMILTON’S PRINCIPLE

Aim:

? Lagrangian formulation ~ Calculus of Variation ?

For a Particle (p.108) :

i i i i

F + R = m x

Energy concept : T, W

Total virtual ( time : fixed) work :

: ( _{i i}) _{i i i i}..( 1,,,3 ) W F R x m x x i N δ + δ = δ =

Remember !

1 2

( ) [ ]..( 1,,3 )

i i i 2

i i i i i i i i

d m x x m x x m x x W m x i N

dt ^ δ = δ + ^ δ ^ =δ +δ =

Eqn(3.28):

( i i i),t

T W m x x

δ +δ = ^ δ

Integrating over the time domain and

applying BC in time

^{δ x ti( )0 = δ x ti( )1 = 0} Finally,

(δT +δW dt) = 0...(3.29) :

∫

‘Hamilton’s Principle’

Advantage of variational point of view:

Hamilton’s principle may be extended to continuous systems with infinite number of DOF !

Wave eqn : ^{ρu,tt = µu,xx + f x t( , )}

Euler beam vibration : ^{ρu,tt = EIu,xxxx + f x t( , )}

Merits and Disadvantages

Concept of Energy ; T : ¹₂

∫∫∫

1

2

∫∫∫

Geometrically and Materially Nonlinear problem -> Geometrically Nonlinear problem :

von Karman theory of Plate

2-11

Sol)

2 2 2

1 2

[(2 ) (2 ) ]

12 3

Iz = M a + a = Ma

Total energy : ¹

(

2 ^{c cc} 2 3

T = M x + y +  Ma θ

Generalized moments :

²

( ) ˆ ( ) ˆ

2 ˆ

( )

3

c c x

c c y

Mx Mx Q

My My Q

Iθ Ma θ Q_θ

∆ = =

∆ = =

∆ = =

 

 

 

Virtual work of applied impulsive force J

ˆ ˆ ˆ

ˆ( 2 )

ˆ 0, ˆ ˆ, ˆ 2 ˆ

c x c y c

x y

W J y a Q x Q y Q

Q Q J Q aJ

θ θ

δ = δ + δθ ≡ δ + δ + δθ

= = =

Angular velocity : ^{2 2}

3 ˆ 3 2 3

2 2 2

Q aJ J

Ma Ma Ma

θ = θ = =

Problem 4-10

A cable of fixed length _ suspended between points (-a,b)

and (a,b). The cable is of uniform mass/length µ _.

Determine y x( )

For the equilibrium state which has the minimum potential energy.

sol)

2 2 2 2 2 2

1 (dy) 1 ( )

d dx dy dx y dx

dx ′

= + = + = +



2 2

2 2

1 ( ) : ..

.. int ... 1 ( )

a

a

a

a

F gy y dx Stationary value

with constra G y dx

µ

−

−

= + ′

= + ′ =

∫

∫

Stationary function of ^F + λ^G = ^{J : ( :}λ Lagrange multiplier^{.. )}

x ' is absent -> ^{J − y′}_∂^∂_y^J_′ ^{= C}

~~ Let C = C1

Integrate one more time,

1 2

1

1 2

cosh( )

.. , , :

a

a

gy C gx C

C where C C

µ λ µ

λ

−

+ = ∫ +

Coefficients are determined using Constraint and BCS

y(-a)=y(a)=b

1

1

cosh( )

C gx

y g C g

µ λ

µ µ

= −

~~~

Practice

- Find the largest volume of rectangular solid containing the

ellipsoid given by

2 2 2

2 2 2

( , , ) x y z 1 0 g x y z

a b c

= + + − =

The volume of a rectangular solid with sides (2x,2y,2z) is

( , , ) 23 8

f x y z = xyz = xyz

Then

2 2 2

2 2 2

( , , ) ( , , ) ( , , ) 8 x y z 1

f x y z f x y z g x y z xyz

a b c

λ λ^{ }

= + = +  + + − 

 

2

2 2

2

2 2

2

2 2

( , , ) 2 2

0 : 8 0.. ..8 0....(1)

( , , ) 2 2

0 : 8 0.. ..8 0....(2)

( , , ) 2 2

0 : 8 0.. ..8 0....(3)

f x y z x x

yz or xyz

x a a

f x y z y y

xz or xyz

y b b

f x y z z z

xy or xyz

z c c

λ λ

λ λ

λ λ

∂ = + = + =

∂

∂ = + = + =

∂

∂ = + = + =

∂

From (1)~(3)

2 2 2

2 2 2

4 ...(4)

x y z xyz

a ⁼ b ⁼ c ^{= −} λ

Substitute it in

g x y z( , , ) 0=

Then,

4 4 4

1

12

xyz xyz xyz

xyz

λ λ λ

λ

− − − =

− > = − (5)

Insert (5) into (4)

and

2 2

4 1

... ...

12 3 3

,....

3 3

x xyz a

a xyz or x

b c

y z

= = = ±

= ± = ±

* Problem

Consider the non-holonomic system in Text page ?

Find the equations of motion for the system and solve for x(t),y(t) with

initial conditions

(0) (0) 0

(0) , (0) 0

(0) 0, (0)

o

x y

x v y

ϕ ϕ ω

= =

= =

= =

 

 Sol:

2 2 2

:

cos : sin

cos , sin ds dx dy

dx ds dy ds

x s y s

ϕ ϕ

ϕ ϕ

= +

= =

− > =   =  Eliminate

s 

Then we get, nonholonomic constraint :

sin cos 0

.. sin cos 0

x y

or dx dy

ϕ ϕ

ϕ ϕ

− =

− =

 

hence

1_x

sin ,

cos ,

0 a = ϕ a = ϕ a

=

Also,

2 2

1

1

( ) .. ..

2 2

0

c c

T m x y I with I ml

V

ϕ

= + + =

=

  

Lagrange Equations of Motion :

2

: 2 sin ... : ??

: 1 0 0

2

x mx y

ml t

λ ϕ

ϕ ϕ ϕ ϕ ω ϕ ω

=

= − > = − > = − > =



  

Then

cos

: ??

x 2 t C y

m

λ ω

= ω + =

 

Using Initial Conditions

1 0 2

1 0

... 0 2

& 2 : . t .

C v C

m

C v m const cons

λ

ω ω λ

= + =

=> = => =

Also, T+V=Const = T0 since V = 0