SEVERAL INDEPENDENT VARIABLES  Remember ~ t:x,y…; q, u etc Multi-dimensional problems An integral functional may have an input consisting of a function

SEVERAL INDEPENDENT VARIABLES

 Remember ~ t:x,y…; q, u etc Multi-dimensional problems

An integral functional may have an input consisting of a function ^{u x y( , )} that depends on several independent variables. In such a case, the integral is an area integral with the domain of integration defined over a region of the plane. Local stationary behavior at an extremum leads to the necessary condition

^{( ) ( ) 0}

x y

F F F

u x u y u

∂ − ∂ ∂ − ∂ ∂ =

∂ ∂ ∂ ∂ ∂ (3.19)

Integrating by part in this case is replaced by its higher dimensional analogue, known as gradient theorem. Note that the necessary condition for the extremum (3.19) is now a partial differential equation. Subscripts on the function ^{u x y( , )} in (3.19) refer to partial derivatives with respect to the independent variables ^x and ^y

A similar derivation may be done for functions ^{u x y z( , , )} of the independent variables. The Euler-Lagrange equation in this case is

^{( ) ( ) ( ) 0}

x y z

F F F F

u x u y u z u

∂ − ∂ ∂ − ∂ ∂ − ∂ ∂ =

∂ ∂ ∂ ∂ ∂ ∂ ∂

Practice: examine the functional with input function

( , )

u x y given by

^{[ ( , )] [( )2 ( )2} 2 uf(x, y)]dxdy 0

D

u u

I u x y

x y

∂ ∂

= + + =

∂ ∂

∫∫

where the integral is taken over some domain D in the plane. Derivation ~~> !!

VARIATIONAL PROBLEMS WITH CONSTRAINTS

There are two types of extremum problems involving constraints.

Two Types of Constraint! : Subsidiary conditions that must be satisfied during the variational process may be stated as integral constraints or as equation constraints.

Find an extremal ^{y x( )} of the functional :

2

1

( ) ^x ( , , ) I y =

∫

satisfying the boundary conditions ^{y x( )0 = y y x0, ( )1 = y1}

and subject to the constraint that the functional

2

1

( ) ^x ( , , )

J y =

∫

: Isoperimetric problems : To find the simple closed curve of constant length maximizing the enclosed area.

For ^{y x( )} to be an extremal,

2

1

( ) ^x ( , , ) 0 I y x F x y y dx

δ =

∫

Since the integral constraint (3.20) must be maintained at a constant value.

2

1

( , , ) 0

x

x δG x y y dx^′ =

∫

For all admissible functions. Multiplying Eq.(3.22) by a constant ^λ and adding to Eqn.(3.21) result in

2

1

[ ( , , ) ( , , )] 0

x

x δ F x y y^′ +λδG x y y dx^′ =

∫

or equivalently in

₀^{1[ ( , , ) ( , , )] 0}

x

x F x y y G x y y dx

δ

∫

Hence the problem reduces to finding an extremal of the auxiliary functional :

I y^{[ ]}+λJ y^{[ ]}=

∫

 Thus the Euler-Lagrange equation for the auxiliary functional is

( ) ( ( )

F G d F G ) 0

y dx y

λ λ

∂ + − ∂ + =

∂ ∂ ′

Or

^{F,y +λG,y −( ,F y′ +λG, )y′ ,x = 0} (3.23)

Eqn (3.23) is a differential eqn with parameter ^λ .

The value of the constant λ is determined by substituting the solution of (3.23) into the integral constraint (3.20)

A more general isoperimetric problem consists of finding the extremals of the functionals

^{[ ] x}₀¹ ( , ... y , ,... )¹ _{n n} I y^ =

∫

under the conditions that the ^m functionals

1

0 1

[ ] ^x ( , ... y , ,... )

k x n n k

J y^ =

∫

Take on the specified values ^Ck for ^{k =1,...., .m} It is also assumed that each of the input functions are assigned prescribed values at the endpoints. As a direct extension to the above, we construct the auxiliary functionl

1

[ ] [ ] [ ]

m

k k k

I y_λ I y λ J y

=

= +

∑

  

with the augmented Lagrangian ^{( , , ) ( , , )} ₁ ^{( , , )}

m

k k k

F x y y_λ F x y y λ G x y y

=

′ = ′ +

∑

     

Based on the Euler-Lagrange equations for several input functions (3.18), the associated equations for the extremals are

^{( ) 0 : 1,...,}

i i

F d F

i n

y dx y

λ λ

∂ − ∂ = =

∂ ∂ ′

The m Lagrange multipliers ^λk are determined by substituting the solutions into the integral constraints (3.24).

The other class of constraints involve finite equations or differential equations. Finite equations of constraint are also called holonomic, whereas differential constraints are called nonholonomic

A holonomic equation of constraint takes the form G x y( , ... y ) 0¹ _n = (3.25)

The constraint (3.25) is required to be satisfied at every point in the domain,including the prescribed boundary conditions. Taking the variation of Equation(3.25), we have

₁ ⁰

n

i

i i

G y

y δ

=

∂ =

∑

Multiplying this constraint equation by an unknown function ^{λ( )x} and adding it to the variation (3.17) does not change the value:

1

0 1 1

{ [ ( )] ( ) } 0

n n

x

i i

x i i i i i

F d F G

y x y dx

y dx y δ λ y δ

= =

∂ − ∂ + ∂ =

∂ ∂ ′ ∂

∑ ∑

∫

Rearranging the terms results in

1

0 1

{ [ ( ) ( ) ] } 0

x n

x i

i i i i

F d F G

x y dx

y dx y λ y δ

=

∂ − ∂ + ∂ =

∂ ∂ ′ ∂

∫ ∑

Now based on the constraint (3.25), only n-1 of the variation ^{δ yi}

are independent. This means that not all of the variations may be arbitrarily varied. However, making the appropriate choice of ^λ , namely, choosing ^{λ( )x} such that

( ) ( ) 0

n n n

F d F G

y dx y λ x y

∂ ∂ ∂

− + =

∂ ∂ ′ ∂

results in

1

0

1

1

{ [ ( ) ( ) ] } 0

x n

x i

i i i i

F d F G

x y dx

y dx y λ y δ

−

=

∂ ∂ ∂

− + =

∂ ∂ ′ ∂

∫ ∑

Now only n-1 variations appear in the integral, so they can be arbitrarily varied. Hence we can obtain the necessary conditions

_i ⁽ _i ^{) ( )} _i ^{0 : 1,...}

F d F G

x i n

y dx y λ y

∂ − ∂ + ∂ = =

∂ ∂ ′ ∂ (3.26)

Furthermore, since the constraint(3.25) does not contain derivatives, Eqn(3.26) are also the Euler-Lagrange Eqns for the arbitrary functional

^{[ ] x}₀^{1 ( , , )} (x) G(x, y)]

I y_λ ^ =

∫

Eq.(3.26) together with the constraint (3.25) represent n+1

Eqns for the extremals ^y1,....yn and the unknown multiplier function ^{λ( ).x} In the case of multiple finite- constraint equations, the modified functional becomes

1

0 1

[ ] ( , , ) (x) G(x, y)

x m x k

k

I y_λ F x y y λ dx

=

 ′ 

=

∫

∑

   

Differential or nonholonomic constraints have the form

G x y( , ... , ... ) 0¹ y y_n ¹′ ′ =y_n

which relate not only the values of the admissible functions but also their derivatives. The conditions for

local stationary behavior in the case of nonholonomic subsidiary conditions become identical with those for holonomic constraints. That is, we introduce multiplier functions ^{λk( )x} and construct the augmented functional

1

0 1

[ ] ( , , ) (x) G( , , )

x m x k

k

I y_λ F x y y λ x y y dx

=

 ′ ′ 

=

∫

∑

    

Since the constraint now involve derivatives, the Euler-Lagrange eqns are

1 1

( ) ( ( ) ) 0 : 1,...

m m

k k

k k

k k

i i i i

G G

F d F

x x i n

y λ y dx y λ y

= =

∂ ∂

∂ ∂

+ − + = =

∂

∑

∑

These eqns are solved simultaneously, together with the m nonholonomic eqns of constraints, for the extremals

1( ),..., _n( ).

y x y x

HAMILTON’S PRINCIPLE

We now make a connection between the Calculus of Variations and Lagrangian formulation

For each particle, ^{Fi + Ri = m xii}

~> Total virtual work (3.28)

: ( _{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): ^{δT +δW = (m xi  i δ xi),t}

Integrating over the time domain and applying BC in time

^{δ x ti( )0 = δ x ti( )1 = 0}

For the actual motion:

( )

1

0

0

x

x

T W dt

δ +δ =

∫ (3.29)

: Hamilton Principle ~ interpretation ! For a conservative system :

^{( )}

1

0

0

x

x

T V dt

δ ∫ − =

For L=T-V :

~~~~~~~

‘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 ? Disadvantages ?

Concept of Energy ; T : ¹₂

∫∫∫

∫∫∫

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 = =

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

= = = ±

= ± = ±

* Practice

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