Handbook of mathematics for engineers and scienteists part 148 pptx

In this example, a unique admissible extremal exists that provides a global extremum but is not a differentiable function.. The unique extremal satisfying the endpoint conditions is the

Example 3 (Hilbert’s example). In this example, a unique admissible extremal exists that provides a global extremum but is not a differentiable function.

Let

J [x] =

 1

0 t

2/3(x  t)2dt → min; x = x(t), x(0) = 0, x(1) = 1.

The Euler equation becomes dt d 2t2/3x  t

= 0 The extremals are given by the equation x(t) = C 1t1/3+ C2

The unique extremal satisfying the endpoint conditions is the function ˆx(t) = t1/3, which does not belong to

the class C1[0, 1] Nevertheless, it provides the global minimum in this problem on the set of all absolutely

continuous functions x(t) that satisfy the boundary conditions and for which the integral J is finite Indeed,

J [x] = J [ˆx + h] =

 1

0 t

2/3  1

3t2/3 + h  t

2

dt=J [ˆx] + 23

 1



t dt+

 1

0 t

2/3(h  t)2dt≥J [ˆx].

Example 4 (Weierstrass’s example) In this example, the problem has no solutions and no admissible

extremals even in the class of absolutely continuous functions.

Let

J [x] =

 1

0

t2(x  t)2dt → min; x = x(t), x(0) = 0, x(1) = 1.

The Euler equation becomes dt d(2t2x  t) = 0 The extremals are given by the equation x(t) = C 1t–1+ C2; none

of them satisfies the boundary condition x(0) =0 Furthermore,J [x]≥ 0, andJ [x] >0 for any absolutely

continuous function x(t) Obviously, the lower bound in this problem is zero To prove this, it suffices to consider the sequence of admissible functions x n (t) = arctan(nt)/ arctan n One has

J [x] =

 1

0 t

(1+ n2t2)2arctan2n dt≤ 1/n

0

dt

arctan2n +

 1

1/n

dt

n2t2arctan2n →0.

Example 5 In this example, there exists a unique admissible extremal that does not provide an extremum.

Let

J [x] =

 3π/2

0



(x  t)2– x2

dt → min; x = x(t), x(0) = x(3π/2) = 0.

The Euler equation becomes x  tt + x =0 The extremals are given by the equation x(t) = C 1sin t + C2cos t The only extremal satisfying the endpoint conditions is ˆx(t) =0.

The admissible extremal ˆx(t) =0 does not provide a local minimum Consider the sequence of admissible

functions x n (t) = n–1sin(2t/3) Obviously, xn (t) →0in C1[0, 3π/2], but

J [x n] = – 5π

12n2 < 0 =J [ˆx].

This example shows that the Euler equation is a necessary but not sufficient condition for extremum.

19.1.2-4 Broken extremals Weierstrass–Erdmann conditions

If the Euler equation has a piecewise smooth solution, i.e., if x(t) has corner points (the function x  t (t) has a discontinuity), then the Weierstrass–Erdmann conditions

L x 

t [c, x(c), x  t (c –0)] = L x 

t [c, x(c), x  t (c +0)], (19.1.2.16)

L [c, x(c), x  t (c –0)] – x  t (c –0)L x 

t [c, x(c), x  t (c –0)]

= L[c, x(c), x  t (c +0)] – x  t (c +0)L x 

t [c, x(c), x  t (c +0)] (19.1.2.17)

hold at each point c that is the abscissa of a corner point.

If x  t L x 

t ≠ 0(t0 ≤t≤t1), then the Euler equation has only smooth solutions.

Curves consisting of pieces of extremals and satisfying the Weierstrass–Erdmann

con-ditions at the corner points are called broken extremals.

19.1.2-5 Second-order necessary conditions.

Legendre condition: If an extremal provides a minimum (resp., maximum) of the

functional, then the following inequality holds:

L x 

t x 

t ≥ 0 (resp., L x 

t x 

t ≤ 0) (t0≤t≤t1) (19.1.2.18)

Weierstrass condition: If a curve x(t) provides a strong minimum (resp., maximum) of

the classical functional, then the Weierstrass function

E (t, x, x  t , k)≡L (t, x, k) – L(t, x, x  t ) – (k – x  t ) L x 

t (t, x, x  t (19.1.2.19)

is nonnegative (resp., nonpositive) for arbitrary finite values k at all points (t, x) of the

extremal

The equation

L xx h + L xx 

t h 

t– dt d L xx 

t h + L x 

t x 

t h 

t

=0 (19.1.2.20)

is called the Jacobi equation Here h(t) is an arbitrary smooth function satisfying the conditions h(t0) = h(t1) =0

If the Legendre condition L x 

t x 

t ≠ 0(t0 ≤t≤t1) is satisfied, then the Jacobi equation is

a second-order linear equation that can be solved for the second derivative It follows from

the conditions h(t0) = h(t1) =0that h(t)≡ 0 A point τ is said to be conjugate to the point

t0if there exists a nontrivial solution of the Jacobi equation such that h(t0) = h(τ ) =0

Jacobi condition: If the extremal x(t) (t0 ≤ t ≤ t1) provides an extremum of the

functional (19.1.2.1), then it does not contain points conjugate to t0

19.1.2-6 Multidimensional case

The vector problem is posed and necessary conditions for an extremum are stated by

analogy with the one-dimensional simplest problem of calculus of variations Let x(t) =

(x1(t), , x n (t)) be an n-dimensional vector function, and let the Lagrangian L be a

function of2n+1variables: L = L(t, x, x  t)

The vector problem has the form

J [x]≡ t1

t0

L (t, x, x  t ) dt → extremum; x = x(t), x i (t j ) = x ij (i =1,2, , n, j =0,1)

(19.1.2.21)

A necessary condition for admissible vector function x(t) to provide weak extremum

in problem (19.1.2.21) is that the function x(t) satisfies the system of n Euler differential

equations

L x i– d

dt L(x i  t =0 (i =1,2, , n), (19.1.2.22)

where we assume that the functions L, L x , L x 

tare continuous as functions of2n+1variables

(t, x i , and (x i) t , i =1,2, , n), and L(x i 

tC1[t0, t1] The solutions of the system of Euler

differential equations (19.1.2.22) are called extremals Admissible functions satisfying the system of Euler differential equations are called admissible extremals.

The derivation of conditions (19.1.2.5) and (19.1.2.22) from the relation δ J =0is based

on the fundamental lemmas of calculus of variations

LEMMA1 (FUNDAMENTAL LEMMA ORLAGRANGE’S LEMMA) Let f (t)C [t0, t1]and

t0

f (t)g(t) dt =0

for each function g(t)C1[t0, t1]such that g(t0) = g(t1) =0 Then f (t)≡ 0on [t0, t1]

LEMMA2 Let f (t)C [t0, t1]and assume that

 t1

t0

f (t)g(t) dt =0

for each function g(t)C1[t

0, t1]such that7t1

t0 g (t) dt =0.Then f (t) = const on [t0, t1]

19.1.2-7 Weierstrass–Erdmann conditions in multidimensional case

If the system of Euler differential equations has a piecewise smooth solution, i.e., if x(t)

has corner points (the function x t (t) has discontinuities), then the Weierstrass–Erdmann

conditions

L(x i  t [c, x(c), x  t (c –0)] = L(x i 

t [c, x(c), x  t (c +0)] (i =1,2, , n),

L [c, x(c), x  t (c –0)] –

n



i=1

(x i) t

t=c–0 L(x i  t [c, x(c), x  t (c –0)]

= L[c, x(c), x  t (c +0)] –

n



i=1

(x i) t

t=c+0 L(x i  t [c, x(c), x  t (c +0)]

hold at each point c that is the abscissa of a corner point.

If (x i) t L(x i  t ≠ 0 (t0 ≤ t ≤ t1, i = 1,2, , n), then the system of Euler differential

equations has only smooth solutions

Curves consisting of pieces of extremals and satisfying the Weierstrass–Erdmann

con-ditions at the corner points are called broken extremals.

19.1.2-8 Higher-order necessary conditions in multidimensional case

We consider the simplest problem

J [x]≡ t1

t0

L (t, x, x  t ) dt → min (or max), x = x(t), x(t0) = x0, x(t1) = x1

(19.1.2.23)

Legendre condition: If an extremal provides a minimum (resp., maximum) of the

functional, then the following inequality holds:

Lx

tx t≡L(x i  t x j) t



≥ 0 (resp., Lx

tx t≤ 0) (t0≤t≤t1; i, j =1,2, , n) (19.1.2.24)

Strengthened Legendre condition: If an extremal provides a minimum (resp., maximum)

of the functional, then the following inequality holds:

Lx

tx t >0 (resp., Lx

tx t <0) (t0≤t≤t1; i, j =1,2, , n). (19.1.2.25)

Remark. In the vector case, Lx

tx t is an n×n matrix The condition Lx

tx t ≥ 0 means that the matrix is

positive semidefinite, and the condition Lx

tx t> 0 means that the matrix is positive definite.

The Jacobi equation for the vector problem has a form similar to that in the one-dimensional case, i.e.,

Lxxh + Lxx

th t – (Lxx

t h + Lx

tx th t  t=0, (19.1.2.26)

where Lxx≡L(x i)(x j)



, Lxx

t≡L(x i)(x j) t



, and h = h(t) is an column-vector with components

h i (t), which are arbitrary smooth functions satisfying the conditions h i (t0) = h i (t1) =0

(i =1,2, , n).

Suppose that the strengthened Legendre condition is satisfied on an extremal x(t) A

point τ is said to be conjugate to the point t0if there exists a nontrivial solution h(t) of the Jacobi equation such that h(t0) = h(τ ) =0 The Jacobi condition (resp., the strengthened

Jacobi condition) is said to be satisfied on an extremal x(t) if the interval (t0, t1) (resp., the

half-open interval (t0, t1]) does not contain points conjugate to t0

We find the fundamental system of solutions of the Jacobi equation for the functions

x(t) in the matrix form

H (t, t0)≡(h1(t), , h n (t)), hi (t)≡

⎛

⎝

h i

1(t)

h i

n (t)

⎞

⎠ ,

where H(t0, t0) = 0 and H t  (t0, t0) ≠ 0 or H t  (t0, t0) = I Column-vectors h i (t) are the solutions of the Jacobi system of equations A point τ is conjugate to the point t0 if and

only if the matrix H(τ , t0) is degenerate

Necessary conditions for weak minimum (resp., maximum):

1 If x(t) provides a weak minimum (resp., maximum), then the function x(t) is an

admis-sible extremal on which the Legendre and Jacobi conditions are satisfied

2 If x(t) provides a strong local minimum (resp., maximum), then the Weierstrass function

E (t, x, x  t , k) = L(t, x, k) – L(t, x, x  t) –

n



i=1

[k i – (x i) t ]L(x i 

t (t, x, x  t (19.1.2.27)

is nonnegative (resp., nonpositive) for arbitrary finite values k = (k1, , k n) at all points

(t, x) of the extremal.

Sufficient condition for the weak minimum or maximum: If the strengthened Legendre

and Jacobi conditions are satisfied on an admissible extremal, then this extremal provides a weak local minimum (or maximum)

Example 6.

J [x] =

 1

0

5

[(x1) t]2+ [(x2) t]2+ 2x1x26

dt → min;

x1(0) = x 2 (0) = 0, x1(1) = sin1, x2(1) = – sin 1 (i =1, 2).

The Lagrangian is L = [(x1 ) t]2+ [(x2 ) t]2+ 2x1x2

A necessary condition is given by the system (19.1.2.22) of Euler equations

–(x1 ) tt + x2 = 0, –(x2 ) tt + x1 = 0 =⇒ (x1 ) tttt = x1 , (x2 ) tttt = x2

The general solution of the Euler equations is

x1(t) = C1sinh t + C2cosh t + C3sin t + C4cos t,

x2(t) = C1sinh t + C2cosh t – C3sin t – C4cos t.

It follows from the initial conditions that C1 = C2 = C4 = 0and C3 = 1 The only admissible extremal is

ˆx1(t) = sin t, ˆx2(t) = – sin t.

Let us apply sufficient conditions The strengthened Legendre condition (19.1.2.25) is satisfied The system of Jacobi equations (19.1.2.26) coincides with the system of Euler equations, i.e.,

–(h1 ) tt + h2 = 0, –(h2 ) tt + h1 = 0.

For H(t,0), such that H(0, 0) = 0and H t (0, 0) = I, we take the matrix

1

2(sinh t + sin t) 12(sinh t – sin t)

1

2(sinh t – sin t) 12(sinh t + sin t)

 Then

det H(t,0) = sinh t sin t.

The conjugate points are τ = kπ, k =1, 2, 3,

The half-open interval (0, 1] does not contain conjugate points, and so the strengthened Jacobi condition is

satisfied Thus the vector ˆx(t) = ( ˆx1(t), ˆx2(t)) provides a local minimum of the functional J

19.1.2-9 Bolza problem

1◦ The Bolza problem is the following extremal problem without constraints in the

space C1[t0, t1]:

B[x]≡ t1

t0

L (t, x, x  t ) dt + l(x(t0), x(t1))→ extremum. (19.1.2.28)

The function L(t, x, x  t ) is called the Lagrangian, the function l = l(x(t0), x(t1)) is called

the terminal cost function, and the functional B is called the Bolza functional The Bolza

problem is an elementary problem of classical calculus of variations Any functions of

class C1[t0, t1] are admissible

An admissible function ˆx(t)C1[t0, t1] provides a weak local minimum (maximum) in

problem (19.1.2.28) if there exists a δ >0such that

B[x]≥B[ˆx] (B[x]≤B[ˆx]) (19.1.2.29)

for any admissible function x(t) satisfying x – ˆx1< δ (see Paragraph 19.1.2-1).

A necessary condition for a weak extremum of the Bolza functional has the same character as the necessary condition in the classical problem; i.e., the function providing an extremum of the Bolza functional must be a solution of a boundary value problem for the Euler equation (19.1.2.5) The boundary conditions are given by the relations

L x 

t (t0, x(t0), x  t (t0)) = ∂l

∂ [x(t0)], L x  t (t1, x(t1), x  t (t1)) = – ∂l

∂ [x(t1)] (19.1.2.30)

and are called the transversality conditions If the function part of the Bolza functional

is lacking, i.e., l≡ 0, then conditions (19.1.2.30) acquire the form L x 

t (t0, x(t0), x  t (t0)) =

L x 

t (t1, x(t1), x  t (t1)) =0, which means in many applied problems that the extremal

trajec-tories are orthogonal to the boundary vertical lines t = t0and t = t1 If a fixing condition

is given at one endpoint of the interval, then the fixing condition at this endpoint and the transversality condition at the other endpoint serve as the boundary conditions for the Euler equation

2◦ The n-dimensional problem is posed and necessary conditions for extremum are stated

by analogy with the one-dimensional Bolza problem Let x(t) = (x1(t), , x n (t)) be an

n -dimensional vector function, and let the Lagrangian L be a function of2n+1variables:

L = L(t, x, x  t)

Then the vector problem reads

 t1

t0

L (t, x, x  t ) dt + l(x(t0), x(t1))→ extremum. (19.1.2.31)

Example 7 Consider the problem

B[x] =

 1

0



(x  t)2– x

dt + x2(1)→ min

The Euler equation (19.1.2.5) has the form

–1 – 2x  tt= 0, and the transversality conditions (19.1.2.30) are as follows:

2x  t(0) = 0, 2x  t(1) = –2x(1).

The extremals are given by the equation x(t) = –14t2+C1t +C2 The unique extremal satisfying the transversality

conditions is ˆx(t) =14(3– t 2 ) This admissible extremal provides the absolute minimum in the problem Indeed,

B[ˆx + h] – B[ˆx] =

 1

0 2ˆx  t h  t dt+

 1

0

(h  t)2dt–

 1

0

h dt+ 2ˆx(1)h(1) + [h

t(1)] 2 =

 1

0

(h  t)2dt + [h  t(1)] 2 ≥ 0

for an arbitrary function h(t)C1[t0, t1] Thus ˆx(t) =14(3– t2) provides the absolute minimum in the problem Furthermore,B[ˆx] = –1

3

19.1.3 Isoperimetric Problem

19.1.3-1 Statement of problem Necessary condition for extremum

The isoperimetric problem (with fixed endpoints) in calculus of variations is the following extremal problem in the space C1[t0, t1] (or P C1[t0, t1]):

J0[x]≡ t1

t0

f0(t, x, x  t ) dt → extremum; (19.1.3.1)

J i [x]≡ t1

t0

f i (t, x, x  t ) dt = α i (i =1,2, , m); (19.1.3.2)

x (t0) = x0, x (t1) = x1, (19.1.3.3)

where α1, , α mR are given numbers Constraints of the form (19.1.3.2) are said to be

isoperimetric The functions f i (t, x, x  t ) (i =1,2, , m) are called the Lagrangians of the problem Functions x(t)C1[t

0, t1] satisfying the isoperimetric conditions (19.1.3.2) and

conditions (19.1.3.3) at the endpoints are said to be admissible.

An admissible function ˆx(t) provides a weak local minimum (resp., maximum) in prob-lem (19.1.3.1) if there exists a δ >0such that the inequality

J0[x]≥J0[ ˆx] (resp., J0[x]≤J0[ ˆx]) holds for any admissible function x(t) C1[t

0, t1] satisfyingx – ˆx1 < δ.

Necessary condition for extremum (the Lagrange multiplier rule):

Let f i (t, x, x  t ) (i = 1,2, , m) be functions continuous together with their partial derivatives (f i)x and (f i)x 

t , and let x(t)C1[t0, t1] If x(t) provides a weak local extremum

in the isometric problem (19.1.3.1), then there exist Lagrange multipliers λ0, λ1, , λ m, not all zero simultaneously, such that for the Lagrangian

L (t, x, x  t) =

m



i=0

λ i f i (t, x, x  t (19.1.3.4)

the Euler equation

L x– dt d L x 

is satisfied If the regularity condition is satisfied [the functions –dt d (f i)x 

t + (f i)x (i =

1,2, , m), are linearly independent], then λ0≠ 0

One of the best-known isoperimetric problems, after which the entire class of problems

was named, is the Dido problem.

Example 1 Find a smooth curve of given length fixed at two points of a straight line and, together with

the segment of the straight line, bounding the largest area.

The formalized problem is

 T0

–T0

x dt → max;

 T0

–T0

1+ (x  t) 2dt = l, x = x(t), x (–T0) = x(T0) = 0,

where T0is given.

The Lagrangian (19.1.3.4) has the form L = λ0x + λ

1+ (x  t) 2

A necessary condition is given by the Euler equation (19.1.3.5)

λ0– d

dt

λx  t

1+ (x  t)2 =0.

Elementary calculations result in the first-order equation

(C + x)

1+ (x  t) 2= –λ.

We integrate this equation and obtain

(C + x)2+ (C1+ t)2= λ2,

which is the equation of a circle It follows from the endpoint conditions x(–T0) = x(T0) that C1 = 0, i.e.,

(C + x)2+ t2= λ2.

The unknown constants C and λ are uniquely determined by the condition x(T0 ) = 0 and the isoperimetric condition 7T0

–T0

1+ (x  t)2dt = l.

For 2T0< l < πT0, there is a unique (up to the sign) extremal that is an arc of length l of the circle passing through the points ( T0 , 0) and centered on the OX-axis Since this is a maximization problem, we must take

the extremal lying in the upper half-plane For l <2T0 , there are no admissible functions in the problem, and

for l > πT0 , there are no extremals.

19.1.3-2 Higher-order necessary conditions Sufficient conditions

Consider the isoperimetric problem

J0[x] → min (or max);

J i [x] = α i (i =1,2, , m);

x (t0) = x0, x (t1) = x1

(19.1.3.6)

For 2T0< l < πT0, there is a unique (up to the sign) extremal that is an arc of length l of. .. λ2.

The unknown constants C and λ are uniquely determined by the condition x(T0 ) = and the isoperimetric condition 7T0... T0 , 0) and centered on the OX-axis Since this is a maximization problem, we must take

the extremal lying in the upper half-plane For l <2T0