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(1)

EXTREMA OF FUNCTIONS

Differential calculus <~> Variational calculus

An extremum problem consists of finding the largest or smallest value of a quantity. (Mini-Max)

For functions of one or two variables, the function can be graphed and we immediately see where the function

(2)

attains its extreme values. These may be inside of a domain, or the extrema may be located at points on the boundary.

For functions of more than two variables, graphing is not possible, so we must report to performing a comparison of the values at a point with neighboring values.

That is, we examine the local rate of change of a function.

These ideas carry over directly to finding extrema of integral functionals.

To find the extrema of a function inside an interval, we look for local stationary behavior. At a point x where a

(3)

function attains a local extremum, given an infinitesimal change, the value of the function should remain the same;

otherwise we do not have an extremum.

This is the same as examing the local linearization of the function. At an extremum, the function should be flat.

The rate of change in every possible direction must be zero. Since there are only two directions, this is easy.

Analytically, this means that the differential of the function is equal to zero.

Hence, a necessary condition that the function f x( ) be stationary at a point x0 is that the derivative f x( )0 is

(4)

equal to zero. The location where this happens is called a critical point. This condition is only necessary, since the condition implies that at the critical point the function can have a local maximum, a local minimum, or an inflection point. Further examination is required, namely checking out the local curvature at the critical point.

This involves the second-derivative test. For functions defined on finite intervals, the values at the endpoints must be checked separately, since the derivates are not defined there.

(5)

For functions of two variables, say f x y( , ), let us assume that a point ( ,x y0 0) is a critical point. In order for the function to be stationary at this point, we must examine the variation of the function as we move an infinitesimal

amount in any possible direction.

So let r = x i+y j be any vector that will denote some fixed but arbitrary direction. We can use a small parameter, ε , to test the variation of the function under a infinitesimal displacement:

*r = ε (x i+y j)

(6)

Now the function

0 0

( , ) ( , )

f r ε = f x + ∆ε x y + ∆ε y

is a function of a single variable ε .

This can be thought of as cutting a slice through the surface defined by f x y( , ) along the direction of the vector r. This curve is parameterized by ε . Note that at ε =0, , f r( ,0) = f x y( , )0 0 .

Thus we have reduced the analysis to searching condition that f r( , )ε : ( parameter :

ε

) is stationary at ε =0 is that f ′(r, 0) 0= .

(7)

(Taylor series expansion !)

The rate of change with respect to ε is

df f f

x y

dε x y

= ∆ +

(3.5) Setting ε =0 in Eqn (3.5), we find that

f f 0

x y

x y

∆ + ∆ =

(3.6)

in which the partial derivatives are evaluated at the point

0 0

( , )x y .

(8)

The left-hand side of Eqn (3.6) is equal to zero for an arbitrary direction specified by x and y if and only if the partial derivatives are equal to zero at ( ,x y0 0). That is,

0 0 0 0

( , ) 0, ( , ) 0

f f

x y x y

x y

= =

(3.7)

The condition (3.7) give a necessary and sufficient condition for local stationary behavior of f(x,y) at the point ( , )x y0 0 .

(9)

These conditions extend to higher dimensions for functions of n variables ( x1,….,xn ), as

(x )0 0.. .. 1,..., )

f for i n

x

= =

We now generalize these concepts to establish criteria for stationary values of scalar quantities that depend on entire functions as arguments.

Before formally developing the theory, let us consider a simple motivational example.

(10)

* Remember : , d , δ

Lagrangian eqn in hybrid form : Page.115

~

1

[ ( )] x0 ( , ( ), ( ))

I y x =

x F x y x y x dx (3.2)

Lagrangian of the integral functional ~ Fundamental objective of the “calculus of variations” is to establish conditions under which an integral functional attains an “extreme” value.

(11)

: Maximum or minimum.

Ex : Of all the continuously differentiable curves passing through point P1 and P2 , find the function whose arclength is minimal.

Ex: total time of travel along the curve is obtained

2

1

2

* 1 [ '( )]

2 ( )

x x

t y x dx

gy x

=

+

: F(y, y`) !

Notice that in each of these problems the quantity “to be

(12)

Minimized” was formulated in terms of an integral.

NECESSARY CONDITIONS FOR AN EXTREMUM

Establishing local stationary behavior of a functional is a generalization of locating the critical points of a function.

: Local maxima and minima are found by setting the derivative of the function w.r.t independent variable equals to 0 ~ slop =0 (Fig.3.4) !

(13)

- Value of the function is stationary.

: At a critical point an infinitesimal variation of the independent variable results in no change in the value of the function. --- Function space !

For integral functionals, the arguments or inputs ( x )are entire functions belonging to a specified admissible set.

~ we must rely on the notion of local stationary behavior.

(14)

Reference

 Hildebrant, Method of Applied Mathematics

Fundamental Lemma of the Calculus of Variations.

G x( ) : Continuous function defined in the interval [ , ]x x1 2 and if

2

1

( ) ( ) 0

x

x

G x η x dx =

(15)

for all smooth functions η( )x satisfying the boundary conditions

1 2

( )x ( )x 0...then G x.. ( ) 0

η =η =

for all points( * Governing eqn ! ~ Lagrange equation)

Focus : Minimizing (Maximizing) the functional

1

[ ( )] x0 ( , ( ), ( ))

I y x =

x F x y x y x dx

with specified BC

~ Apply the local stationary behavior of a functional.

(16)

Let y x( ) : Admissible function that minimizes I y[ ]. Suppose y x( ) : Another admissible ftn as in (Fig. 3.5)

“close” in some sense to y x( )

Admissible ~ y x( ) : continuously differentiable with BCs.

Then

(17)

( ) ( ) ( )

y x = ∗y x +εη x : η( )x Differentiable

: Perturbing function

εη

( )x : Variation of function y x( ) . Since y x*( ) is extremal ~ Local minimum value of

[ ( )]

I y x ,

: I y[ *( )x +εη( )]x I y[ *( )]x for all ε near 0.

Once a variation εη( )x is fixed, but arbitrary, then [ *( ) ( )]

I y x +εη x

(18)

actually becomes a function of a single real variable!

: Reduced the problem to a single real variable for

local stationary behavior with the zero-slope criterion for functions.

[ d ( * )] 0 0

d I y εη ε

ε + = =

~>

(19)

1

[ * ] x0 ( , * , * )

I y +εη =

x F x y +εη y +εη dx (3.8) : Derivative of Eqn (3.8) wrt the ε is

(or

1

( ) x0 ( , , )

I ε =

x F x y y dx ,

1 1

0 0

( ) x [ ] x [ ]

x x

dI F y F y F F

dx dx

d y y y y

ε η η

ε ε ε

∂ ∂ ∂ ∂

= + = +

∂ ∂ ∂ ∂

∫ ∫

(20)

1 0

.. 0,

( )] x [ ]

x

For

dI F F

d y y dx

ε

ε η η

ε

=

= +

1 0

[ * ]

[ ]

x x

dI y F F

d y y dx

εη η η

ε

+ = +

Integrating by parts and (applying the BCs)

1 2

0[ ( ) ] 1

x x

x x

F d F F

y η dx y η dx y η

∂ − ∂ + ∂

′ ′

∂ ∂ ∂

(21)

For an extreme value of the functional,

1

0 [ ( )] ( )

x x

F d F

y dx y η x dx

: η( )x is an arbitrary function that vanishes at the endpoints of the interval, the Fundamental Lemma of the Calculus of Variations allow us to conclude

( ) 0....(3.9)

F d F

y dx y

=

(22)

~ An ordinary differential equation that represents the necessary condition for an admissible y x( ) to be a minimizing function.

: Euler-Lagrange equation Back to the future ! P. 115

, ,

x t

y y q q

− >

′− > !

(23)

Ex : (3.10) : p.169

SPECIAL CASES OF THE EULER-LAGRANGE EQUATION

( , , ) ( , ) ( , ) ( , ) ( )

F F x y y F F x y F F y y F F x y F F y

=

=

=

=

=

Eqn(3.13)! Jacobi energy integral

(24)

δ ∫ ∫

=

δ

?

THE VARIATIONAL OPERATOR(?)

NATURAL BOUNDARY CONDITIONS(?) GENERALIZATIONS(?)

SEVERAL INDEPENDENT VARIABLES

VARIATIONAL PROBLEMS WITH CONSTRAINTS

2-11

(25)

Sol)

2 2 2

1 2

[(2 ) (2 ) ]

12 3

Iz = M a + a = Ma

Total energy : 1

(

2 2) 1 2 2 2

2 c cc 2 3

T = M x + y + Ma θ

Generalized moments :

(26)

2

( ) ˆ ( ) ˆ

2 ˆ

( )

3

c c x

c c y

Mx Mx Q

My My Q

Iθ Ma θ Qθ

= =

= =

= =

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

= = =

(27)

Angular velocity : 2 2

3 ˆ 3 2 3

2 2 2

Q aJ J

Ma Ma Ma

θ = θ = =

Problem 4-10

(28)

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.

(29)

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 + λG = J : ( :λ Lagrange multiplier.. )

x ' is absent -> J yyJ = C

(30)

~~ 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

(31)

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

1

1

cosh( )

C gx

y g C g

µ λ

µ µ

=

~~~

Practice

(32)

- 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

(33)

( , , ) 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

= + = + + +

(34)

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)

(35)

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)

(36)

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

= = = ±

= ± = ±

(37)

~

2-11

Sol)

2 2 2

1 2

[(2 ) (2 ) ]

12 3

Iz = M a + a = Ma

Total energy : 1

(

2 2) 1 2 2 2

2 c cc 2 3

T = M x + y + Ma θ

Generalized moments :

(38)

2

( ) ˆ ( ) ˆ

2 ˆ

( )

3

c c x

c c y

Mx Mx Q

My My Q

Iθ Ma θ Qθ

= =

= =

= =

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

= = =

(39)

Angular velocity : 2 2

3 ˆ 3 2 3

2 2 2

Q aJ J

Ma Ma Ma

θ = θ = =

Problem 4-10

(40)

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.

(41)

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 + λG = J : ( :λ Lagrange multiplier.. )

x ' is absent -> J yyJ = C

(42)

~~ 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

(43)

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

1

1

cosh( )

C gx

y g C g

µ λ

µ µ

=

~~~

Practice

(44)

- 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

(45)

( , , ) 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

= + = + + +

(46)

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)

(47)

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)

(48)

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

= = = ±

= ± = ±

(49)

~

HAMILTON’S PRINCIPLE

Remember : , d , δ

Lagrangian eqn in hybrid form : Page.115

1

[ ( )] x0 ( , ( ), ( ))

I y x =

x F x y x y x dx (3.2)

(50)

Lagrangian of the integral functional ~ Fundamental objective of the “calculus of variations” is to establish conditions under which an integral functional attains an “extreme” value.

: Maximum or minimum.

Ex : Of all the continuously differentiable curves passing through point P1 and P2 , find the function whose arclength is minimal.

Thus the total time of travel along the curve is obtained

(51)

* xx12 1 [ '( )]2 ( ) 2

t y x dx

gy x

= +

: F(y, y`) !

Notice that in each of these problems the quantity “to be Minimized” was formulated in terms of an integral.

EXTREMA OF FUNCTIONS

NECESSARY CONDITIONS FOR AN EXTREMUM

Establishing local stationary behavior of a functional is a generalization of locating the critical points of a function.

: Local maxima and minima are found by setting the

(52)

derivative of the function w.r.t independent variable equals to 0 ~ slop =0 (Fig.3.4) !

- Value of the function is stationary.

: At a critical point an infinitesimal variation of the independent variable results in no change in the value of the function. --- Function space !

For integral functionals, the arguments or inputs are entire functions belonging to a specified admissible set.

(53)

~ we must rely on the notion of local stationary behavior.

Fundamental Lemma : Calculus of Variations.

( )

G x : Continuous function in [ , ]x x1 2

2

1

( ) ( ) 0

x

x

G x η x dx =

for all smooth functions η( )x with η( )x1 =η( )x2 = 0...then G x.. ( ) 0 for all points( Govering eqn ! for example G = T-V )

(54)

Focus : Minimizing (Maximizing) the functional

1

[ ( )] x0 ( , ( ), ( ))

I y x =

x F x y x y x dx with BC

~ Apply the local stationary behavior of a functional.

Let y( )x : Admissible function that minimizes I y[ ]. Suppose y x( ) : Another admissible ftn as in (Fig. 3.5)

“close” in some sense to y( )x

Admissible ~ y x( ) : continuously differentiable with BCs.

Then

(55)

( ) ( ) ( )

y x = ∗y x +εη x : η( )x Differentiable

: Perturbing function εη( )x : Variation of function y x( ). Since y* ( )x is extremal ~ Local minimum value of

[ ( )]

I y x ,

: I y[ *( )x +εη( )]x I y[ *( )]x for all ε near 0.

Once a variation εη( )x is fixed, but arbitrary, then

[ * ( ) ( )]

I y x +εη x

actually becomes a function of a single real variable!

(56)

: Reduced the problem to a single real variable for

local stationary behavior with the zero-slope criterion for functions.

[ d ( * )] 0 0 d I y εη ε

ε + = =

~>

1

[ * ] x0 ( , * , * )

I y +εη =

x F x y +εη y +εη dx (3.8)

(57)

: Derivative of Eqn (3.8) wrt the ε is

(or I( )ε =xx01F x y y dx( , , ) ,

1 1

0 0

( ) x [ ] x [ ]

x x

dI F y F y F F

dx dx

d y y y y

ε η η

ε ε ε

∂ ∂ ∂ ∂

= + = +

∂ ∂ ∂ ∂

∫ ∫

1 0

.. 0,

( )] x [ ]

x

For

dI F F

d y y dx

ε

ε η η

ε

=

= +

=

(58)

1 0

[ * ]

[ ]

x x

dI y F F

d y y dx

εη η η

ε

+ = +

Integrating by parts and (applying the BCs)

1 2

0[ ( ) ] 1

x x

x x

F d F F

y η dx y η dx y η

+

For an extreme value of the functional,

1

0[ ( )] ( )

x x

F d F

y dx y η x dx

(59)

: η( )x is an arbitrary function that vanishes at the endpoints of the interval, the Fundamental Lemma of the Calculus of Variations allow us to conclude

( ) 0....(3.9)

F d F

y dx y

=

~ An ordinary differential equation that represents the necessary condition for an admissible y x( ) to be a minimizing function.

: Euler-Lagrange equation Back to the future ! P. 115

(60)

, ,

x t

y y q q

− >

′− > !

Ex : (3.10) : p.169

SPECIAL CASES OF THE EULER-LAGRANGE EQUATION

(61)

( , , ) ( , ) ( , ) ( , ) ( )

F F x y y F F x y F F y y F F x y F F y

=

=

=

=

=

Eqn(3.13)! Jacobi energy integral

δ ∫ ∫

=

δ

?

THE VARIATIONAL OPERATOR(?)

NATURAL BOUNDARY CONDITIONS(?)

(62)

GENERALIZATIONS(?)

SEVERAL INDEPENDENT VARIABLES

VARIATIONAL PROBLEMS WITH CONSTRAINTS HAMILTON’S PRINCIPLE

THE VARIATIONAL OPERATOR(?) NATURAL BOUNDARY CONDITIONS

…. + Fyδ yxx12 = 0...(3.15)

(63)

as in Fig.3.6

Then, at x2 for δI = 0, Euler-Lagrange Eqn. is satisfied with Fy =0 : Natural BC or Force BC.(ex: ICBM…) (δ y =0 : Essential BC or Geometric BC)

GENERALIZATIONS

( , , ) : . ! F = F x y y most simple

~> For more generality with several functions ! As in Eqn.(3.16)

(64)

y =( y1, ..., yn ) : independent Ftns Ex : !

If F = F x y y y( , , ,′ ′′)..? : δ I = ? ~ page 172.

Ordinary Differential Eqn ! Partial Differential Eqn (x,t) :Beam bending problem:

4 2

4 2

( , ) ( , )

d y x t d y x t

EI m

dx = dt

~

2

2 2

2

1 ( , ) 1 ( , )

( ) , ( )

2 2

y x t y x t

T m V EI

t x

= =

(65)

SEVERAL INDEPENDENT VARIABLES

VARIATIONAL PROBLEMS WITH CONSTRAINTS

Two Types of Constraint!

(66)

~ Integral constraint Eqn(3.20)

Isoperimetric problem

(F + λG),y [(F + λG), ],y' x = 0...(3.23)

- More general isoperimetric problem Eqn(3.24)

(67)

~ Eqns of constraint

(Non~) holonomic !

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

. ,

, 0

n

yi i

G x y y

or

Taking variation G δ y

=

=

(68)

HAMILTON’S PRINCIPLE

THE VARIATIONAL OPERATOR(?) NATURAL BOUNDARY CONDITIONS GENERALIZATIONS

(69)

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

= =

SEVERAL INDEPENDENT VARIABLES

VARIATIONAL PROBLEMS WITH CONSTRAINTS

~ Integral constraint :

More general isoperimetric problem

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

(70)

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

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