Legendre Conditions for a Variational Problem with One-sided Phase Constraints

Society of Japan

Vol. 38, No. 4, December 1995

LEGENDRE CONDITIONS FOR A VARIATIONAL PROBLEM WITH ONE-SIDED PHASE CONSTRAINTS

Hidefumi Kawasaki Sayuri Koga Kyushu University

(Received January 28, 1994; Revised November 28, 1994)

Abstract This paper is concerned with a variational problem with inequality phase constraints. We present second-order necessary conditions (Legendre condition) for weak minimal solutions of the variational problem. Our optimality conditions give an information about the Hessian of the integrand at not only inactive points but also active points. Since the present constraints do not include X, our conditions differ from the Legendre-Clebsch condition.

1. Introduction

This paper is concerned with a variational problem with one-sided phase constraints:

mInImIze

10

1

f(t, x, ±)dt

subject to x(O) =

Xo,

x(l) = Xl, X E

Xo,

(Po)

a(t) :::; x(t)

Vt

E

[0,1],

where f(t, X, ±) is.a continuous function defined on RHn+n, X

_o

is the space of all n-dimensional vector-valued absolutely continuous functions X' :

[0,1]

-+ Rn equipped with the norm

Ilxllxo = max Ilx(t)11

+

esssupll±(t)lI,

tE[O,I] tE[O,I]

(1.1)

where esssup denotes the smallest number M such that 11±(t)11 :::; M for almost everywhere

[0,1].

The end-points Xo and Xl are given points in Rn, a :

[0,1]

-+ Rn is a given continuous function and a(t):::; x(t) means the component wise inequalities. We assume that

f(t,x,±)

is twice continuously differentiable w.r.t. X and

:r.

We give second-order necessary conditions (Legendre condition) for weak minimal solutions of (Po). A local solution

x

in the sense of the norm 11 . Ilxo is said to be weak, and

x

in the sense of the norm maxtE[O,I]llx(t)11 is said to be strong. We encounter the one-sided phase constriant, for example, production planning and planning mathematics of the employment, see pp.

234,253

in

[18].

In the literature, we can find many second-order necessary optimality conditions (Weier-strass condition, Legendre-Clebsch condition, Legendre condition, maximum principle) for variational problems or optimal control problems with various types of constraints: equality, inequality or set constraints,

[1] [2] [4] [5] [6] [7] [8] [9] [10] [12] [13] [18] [19] [20] [21] [22]

[23] [25].

All of them except

[23]

dealt with strong minimal solutions or mixed constraints g( t, x, ±) :::; 0, where g!E is of full rank for the active constraints, which is never satisfied by the present constraint a( t) - x( t) :::; 0. We note that Pales and Zeidan

[23]

dealt with optimal control problem where the objective function is a supremum-type function.

484 H.Kawasaki & S.Koga

Our optimality conditions give an information about fxx at not only inactive points

(a(t)

<

x(t»

but also active points

(aj(t)

=

Xj(t)

for some i), where

aj(t)

denotes the i-th

component of

a(t).

2 Preliminary results

Let X, V and W be Banach spaces. Let /( be a closed convex cone of V with non-empty

interior

int /(.

We denote by

V*

the topological dual space of

V.

The polar cone of /( is

defined by /(0 =

{v*

E

V*;

<

v*,v

>::;

0 Vv E /(}.

In this section, we consider the following abstract optimization problem with generalized equality and inequality constraints:

(P)

minimize

F(

x)

subject to

G(x)

E /(,

H(x)

=

O.

where F : X - R, G : X - V and H : X - Ware twice continuously Frechet differentiable

mappings. We denote by

F'(x)

and

F"(x)

the first and second Fnkhet differential of

F(x),

respectively.

Now, let x be a weak minimal solution of (P). A direction y E X is said to be critical at

x

if

F'(x)y

= 0,

G'(x)y

E clcone(I< -

G(x», H'(x)y

= 0,

where clcone A denotes the closure of the conical hull of A.

A feasible solution

x

is said to satisfy the Mangasarian-Fromovitz condition (regular) if

(i)

H'(x):

X - W is onto

( ii)

3z

E X $.

t.

H'(x)z

= 0,

G(x)

+

G'(x)z

E int I<.

The following theorem can be found in e. g. Ben-Tal and Zowe [3], Kawasaki

[Ill].

Theorem 2.1

Let

x

be a regular minimal solution of

(P).

Then, for each critical direction

y

at

x

satisfying

G'(x)y

E

cone(I< - G(x»,

there exist v·

E /(0

and

w· E

W* such that

where

3 Main theorems

L'(x)

= 0,

L"(x)(y,y) :::::

0,

<

v*,G(x)

>=

0,

<

v*,G'(x)y

>=

0,

L(x)

=

F(x)+

<

v*, G(x)

>

+

<

w*, H(x)

> .

(2.1 )

(2.2)

(2.3)

(2.4)

(2.5)

Let

x(t)

be a weak minimal solution satisfying

a(O)

<

Xo and a(l)

<

Xl for

(Po).

We use the following abbreviated notation:

All vectors except gradient vectors are column vectors. For each a E Rn, we denote aT the

transpose of a and by ai denotes the i-th component of a. For each t

E [0, 1],

we define

JR(t) = {i; 3b

>

0 s.t. ;1:;

>

a; on (t, t

+

b)},

h(t) =

{i;

3b> 0 s.t. :I:i

>

ai on (t - b,t)}.

A function

x

is said to be piecewise smooth if

[0, 1]

is divided into a finite number of

subintervals and the function has continuous derivative on each subintervals.

Theorem 3.1 Let x(t) be a piecewise smo.o.th weak minimal so.lutio.n satisfying a(O)

<

x(O) and a(l)

<

x(l) fo.r (Po). Put

A(t) =

lot

Ix(r)T dr -fx(tf. (3.1 ) Then

(i)

Ai(t) is no.ndecreasing and increases only o.n {t ; ai(t) = Xi(t)},

(ii)

~Tlxx(t- O)~ ;::: 0 V~ satisfying ~i = 0 fo.r

if/.

h(t), (iii) ~Tlxx(t

+

O)~ ;::: 0 V~ satisfying ~i

=

0 fo.r i

f/.

JR(t).

We can find (i) in

[18].

This condition can be regarded as an inequality version of the

Euler-Lagrange equation. So we call it the Euler-Lagrange condition for the one-sided phase constraints.

For the sake of better understanding, let us consider the case of n

=

1. Then (ii) and

(iii) amount, respectively, to

lxx(t - 0) ;::: 0 if 3b

>

0, x> a on (t - b, t), fxx(t

+

0) ;::: 0 if 3b

>

0, x> a·on (t, t

+

b).

(3.2)

(3.3)

Fig. 3.1 is a standard picture of x(t). In Fig. 3.1, the above conditions assert that

h:x ;:::

0

on [0, t2] U [t3, 1]. The following theorem gives an information about lxx on [t2' t3].

Theorem 3.2 Let x(t) be a piecewise smo.o.th weak minimal so.lutio.n satisfying a(O)

<

Xo and a(l)

<

Xl fo.r (Po). Let EL(t) deno.te the sd o.f all indices

if/.

h(t) fo.r which the Euler equatio.n w. r. t. X·i

(3.4)

ho.lds a. e. o.n (t - b, t] fo.r so.me b

>

O. Then we may replace, zn (ii) o.f the previo.us theo.rem, ~i

=

0 b~f

~i ;::: 0 for i E EL(t), ~i = 0 for i E h(t) U EL(t).

Similarly, Id ER(t) deno.te the set o.f all indices

i f/.

JR(t) fo.r which the Euler equatio.n w.

r. t. Xi ho.lds a. e. o.n It, t

+

15) fo.r so.me 15

>

0, then we may replace, in (iii) o.f the previo.us theo.rem, ~i = 0 by ~i;::: 0 fo.r i E ER(t),~i = 0 fo.r i E JR(t) U ER(t).

It is clear that we may replace [0, 1] with any bounded closed interval

[e, d]

in the above

486 H.Kawasaki & S.Koga

Example 3.1 Let us conS1der the following problem on

[0, 2j:

minimize

10

2 {4x - j;2}dt Take subject to x(O) = x(2) = 1, x(t) ~ 0 Vt E [0,2J. { _t2 - t

+

1 x(t) = 0 _t2

₊

_{5t - 5} on

[0

/5-1J , 2 ' on[~ 5-

0

J 2 ' 2 ' on

[5-0

2 , 2J , see Fig 3. 2. Then

>.(t)

defined by (3.1) is

{ -2

>.(t)

= 4t 10

on[O~J

_{, 2 '} on[~

5-0J

2 ' 2 ' on

[5-/5

2] 2 ' ,

which is nondecreasing, see Fig 3.3. Hence x(t) satisfies the Euler-Lagrange condition. But it follows from Theorem 3.1 that x(t) is not a weak minimal solution, since

IH(t) ::::::: -2

<

O.

Theorem 3.1 can not exclude the following non-minimal solution.

Example 3.2 Let us consider the following problem on [-2,2J.

minimize

12

(t

2 _-1)±2dt -2

subject to x( -2) = x(2) = 1, a(t) ::; x(t) Vt E [-2,2],

where a(t) is given by Fig.

9.4.

Take x(t) ::::::: 1. Then

>.(t)

defined by (3.1) is identically zero. Thus x(t) satisfies the Euler equation

(3.4).

Moreover, x(t) satisfies all conditions of Theorem 3.1. But

it

follows from Theorem 3.2 that x(t) is not a weak minimal solution, since lxx(t) = 2(t2 - 1)

<

0 on (-1,1).

4 Proofs of the theorems

Define F : Xo - t R, G: Xo - t (C[O, lW, H : Xo - t R 2n and J(

c

(C[O, l])n by

F(x) =

10

1

f(t,x,±)dt, G(x)=a-x,

H(x)

=

(x(O) - xo, x(1) -

Xl),

J( = {v E (C[O,I]t;v(t)::; OVt E [0, I)}.

Then the problem

(Po)

is expressed as (P). Furthermore,

F,

G and

H

are twice continuously

Frechet differentiable and their first and second Frechet differentials are given by

F"(x)(y, y)

=

10

1

{yTlxxY

+

2yTlxxY

+

yTlxxy}dt,

G'(x)y

=

-y, G"(x)(y, .y)

=

0,

H'(x)y

=

(y(O),y(I)),

H"(x)(y,y)

=

(~~),

(4.2) (4.3) ( 4.4) see e. g. Girsanov [11] and Ioffe and Tihomirov [13]. It is easily seen that the Mangasarian-Fromovitz condition is satisfied at

x

if

a(O)

<

Xo and

a(1)

<

Xl. Hence we may apply

Theorem 2.1 to

(Po).

By Riesz's representation theorem, the Lagrange function

L(x)

is represented as

I I I

L(x)

= [

f

dt

+ [

d> .

.T(a --

X)

+

L

Jlf(x(k) -

Xk),

10

10

k=O

(4.5) where Jlo, III E Rn and ,\ : [0,1] - t Rn is a component wise nondecreasing function and

increases only on

{t; ai(t)

=

Xi(t)},

see e. g. Rudin [24], Luenberger

[18].

It follows from (2.2), (4.1), (4.3) and (4.4) that

1 I 1

[ UxY

+

lxy}dt - [ d)\T Y

+

L

Jlfy(k)

= 0

lo

lo

k=O

for all y EX. By :integration by parts, we get

10

1

Ux - l l x dr

+

l d,\T}iJdt

+

{lol

lxdr -

10

1

d,\T

+

Jldy(1)

= 0 for all y with y(O) =

0;

see

[18],

[13]. Hence we have

lx(t) -llx(r)dr

+

,\T(t)

=

constant

=

eT a.e.

(4.6) Now, let r be an arbitrary point of (0,1]. For any s

<

r

sufficiently close to r and sufficiently small c>

>

0, put

{

2J(T-\t-S+C»

on [s-c>,s-~]

hn(t)=

_2y'(71(t-s)

on[s-~,s]

o

otherwise,

(4.7)

and

YlT(t)

= h<T.(t)~, (4.8)

where ~ E Rn is an arbitrary vector which satisfies

~i=O

VitJh(r).

(4.9)

Here we note that

her)

C

h(s).

( 4.10)

Then we see that

G'(x)YlT

=

-yn E

cone (l( -

G(x)).

(4.11)

Indeed, (4.11) is equivalent to

488 H.Kawa~akj & S.Koga

For any i

f/.

h(r),

since

Yui(t)

==

0, the i-th inequality of (4.12) holds. For any i E

h(r),

since 1

Yui(t)

I::;

~ 1

ei

1 and

ai(t)

<

Xi(t)

on

[8 -

a,s],

the i-th inequality of (4.12) holds.

Thus we get (4.11).

Moreover, since

Yu(O)

=

Yu(l)

= 0, we have

where the last equality follows from the complementary condition (2.4). Therefore

Yu

is a

critical direction. Hence, by Theorem 2.1, we have

o ::;

L" (x

)(Yu, Yu)

l

s

T -

2

T -

.

T -

.

2

=

s-u

(e

fxxehu

+

2e

fxxehuhu

+

e

fxxehu)dT.

(4.13)

Putting

M

= max{1

eT]xx(t)e

I;

t

E

[s - a, s]),

we have 1

fLu

er]xxeh~dtl

::; fLu Madt

=

0(a2_{). Similarly the second term of (4.13) is O(a). By mean-value theorem, the third term}

is equal to

-11

S

T-

T-4a

s-u

e

fHedt

=

4e

ixx(tu)e

for some

tu

E (s - a,s). Hence, from (4.13),

Since

tu

- t S as

a

- t 0, we get

T-e

fxx(s)C~

o.

Taking s - t r, we have

T-e

ixx(r - O)e

2

o.

The assertion (iii) is similarly obtained. This completes the proof of Theorem 3.l.

Next, we prove Theorem 3.2. We define

Yu

by (4.8), where

e

E Rn is an arbitrary vector

which satisfies

Vi E EL(r)

Vi E h(r) U EL(r). (4.14)

Then we see that

-Yu

E cone (K - G(x)). ( 4.15)

Indeed, for i E EL(r), the i-th component of the left-hand sided of (4.12) is

aYui(t) - ai(t)

+

Xj(t)

2

Xi(t) - ai(t)

20 Vt.

So the i-th inequality holds. [t follows from the Euler equation (3.4) that

Thus

Yu

is critical. The rest is as in the previous proof. This completes the proof of Theorem

3.2.

The authors would like to thank the referees for their many valuable comments and helpful advices.

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Hidefumi Kawasaki and Sayuri Koga Department of Mathematics

Kyushu University Fukuoka Japan

x(t) t Figure 3.1 x(t) 1

---+---~---~---~---,__..

t

/5-1

2 Figure 3.2 2

492 H.Kawasaki & S.Koga A(t) 10 ... . ....---. -21-_ _ _ _ -2

.J5

-1 2 -1 Figure 3.3 Figure 3.4

5-.[5

2 a(t) 2 t x(t) 2