! o o r z > x = 0 f o r z = 0 i s c h a n g e d i n t o ~ - z 0 f o r z = ~ o ( x , y ) , < 0 f o r z < 0 , 0 f o r

ARKIV FOR MATEMATIK Band 6 nr 22

1.65 12 1 Communicated 13 October 1965 b y L. GXI~DING and L. CARLESO~

M i n i m i z a t i o n p r o b l e m s f o r t h e f u n c t i o n a l

supxF(x,f (x),f'(x)). ^(II)

By Gu~NAR AnoNsso~

Introduction

The present paper is a continuation of the paper [1]. This means t h a t we shall t r y to minimize the functional

/t(/) =sup~/~(z,/(x),/'(x))

over the class :~ of all absolutely continuous functions/(x) which satisfy the boundary conditions/@1)

=Yl

and/(Xz) =Y2-

I n Chapter 1 we consider the question of the existence of a minimizing function, which was left open in [1].

I n Chapter 2 some of the previous results are generalized to a wider class of func- tions

F(x,y,z):

the previous condition

~F

~z

> 0 for z > 0

[!o orz> x

= 0 for z = 0 is changed into ~-z 0 for z = ~o(x, y),

< 0 for z < 0 , 0 for

z<o:,(x,y).

An existence proof for absolutely minimizing functions is given in the general case.

I n Chapter 3 we examine further the properties of a.s. minimals. We also consider uniqueness questions for a.s. minimals and minimizing functions. For reasons of simplicity, this chapter is restricted to the case

o(x,y)=-0.

I n Chapter 4, finally, we give some examples in order to illustrate the previous exposition.

We shall now introduce notations for some conditions on

F(x,y, z).

They are as- sumed to hold for x E J and for all real y and z. Here J is an interval which will be stated explicitly in most cases. If it is not explicitly given, then the choice of J will be clear from the context.

1. F(x,y,z)

is continuous and

~F/~z

^exists.

~F(x, y, z)

is

2A. ~z

> 0 if

= 0 if

< 0 if z > 0 , Z ~ 0 , z < 0 .

409

c. AROrCSSO~,

Minimization problems. I I

2 B . There is a continuous f u n c t i o n

co(x, y)

such t h a t

OF(x, y, z)

Oz

f!

^{O if}

z>o~(x,y),

is 0 if z = o ~ ( x , y ) , 0 if

z<co(x,y).

3. liml~l_,~r

F(x,

y, z ) = + oo if x a n d y are fixed.

T h e conditions 1 a n d 3 will be assumed to hold t h r o u g h o u t this paper, t o g e t h e r with 2 A or 2 B . 1 2 A is a s s u m e d to hold in C h a p t e r 1 a n d Chapter 3; in Chapter 2 we shall use t h e m o r e general condition 2 B. Xt is easy t o see (using t h e H e i n e - B o r e l covering theorem) t h a t

lim ( inf

F(x, y, z))= +

I z l ~ ~x~<~

l y K K

for every c o m p a c t interval [:r c J a n d for every K > 0.

Chapter 1. The question of the existence of a minimizing function T h e assumptions on

F(x,y,z)

in this c h a p t e r are m a i n l y t h e same as in [1]. W e shall assume t h a t t h e following conditions are satisfied t h r o u g h o u t t h e c h a p t e r : 1, 2 A a n d 3. T h e interval J will be chosen as

Xl<~X<~X ~.

1. W e shall n o w give an example which shows t h a t t h e existence of a minimizing function does n o t follow f r o m these properties of

F(x, y, z).

A counterexample.

^Choose

F(x, y, z) =z2/(y 2 +

1) 2 - cosx - x ( 2 ~ - x ) c o s x - ~ ( y ) (2~ - ~ ( y ) ) ,

where~(y)=arctgy+~/2;

x l = 0 a n d

I x2=2~'

Yl = 0 [ Y2 = 0.

L e t /E :~. W e h a v e

H(I)

>I F ( ~ , / ( ~ ) , 0) = 1 + ~ 2 - ~ ( / ( ~ ) ) [ 2 ~ - ~ ( / ( ~ ) ) ] > 1,

since 0 < ~ ( / ( ~ ) ) < z . H e n c e H ( / ) > I for all /E:~. Suppose n o w t h a t 0 < ~ < g / 2 a n d consider t h e function

1 T h e r e is one exception: E x a m p l e 5 in C h a p t e r 4.

ARKIV FOR MATEMATIK. Bd 6 nr 22 tg x

tg (2 :~ - x )

I t is easy to verify t h a t

for 0 ~ < x - ~ - ~, for 2 - - (~ ~ < x ~ + (5, for 3 - ~ + ~ ~<x~<27~.

H(g~) = F ( z , g~(ze), O) = 1 + ~2.

This proves t h a t infr~ ~ H ( / ) = 1 and t h a t there is no minimizing function. Compare the r e m a r k to Theorem 1.1.

2. I t follows from this example that, in order to ensure the existence of a mini- mizing function, we m u s t impose suitable extra conditions on F(x,y,z). We shall describe a few such conditions (they are separately sufficient), but we are not aiming at a complete discussion of the question.

I n each existence proof, we shall need

Lemma 1.1. Let {/~(x)}~ be a sequence o//unctions in :~ such that supl.<~<~ H(/~) <

and suppose that ]~(x) -+/(x) uni/ormly /or x I <<. x <~ x 2 . Then / 6 :~ and H(/) <~ lim~_~ ~ H(/r).l Proo/. Since ]/r(x)] ~<K1 a n d H(/~)<~K 2 for all v, it follows from the properties of F t h a t ]/:(x)] ~<K a. Hence all/~(x) a n d / ( x ) satisfy a uniform Lipschitz condition.

Clearly /6:~, and the rest of the proof follows easily. (Use the fact t h a t if /'(xo) exists, then we have

~(xo, /(xo), z) >~ F(Xo, /(xo), /'(xo))

for all z >~/'(xo) or for all z <<./'(xo), together with a continuity argument.)

L e t us now return to the minimization problem and introduce the notation M 0 =inf~E~ H(/). Suppose t h a t {/~}F is a minimizing sequence, i.e. lim~_~ ~ H ( / , ) = M 0.

I f the f u n c t i o n s / , ( x ) are uni/ormly bounded, then it follows (as above) t h a t t h e y are equicontinuous, and then there is a uniformly convergent subsequence. L e t the limit function b e / ( x ) . I t follows from the l e m m a t h a t H ( / ) = M 0.

Hence, in order to prove t h a t there is a minimizing function, it is sufficient to prove t h a t there is a minimizing sequence of uniformly bounded functions.

I n m a n y cases, it can be seen from the function F(x, y, 0) t h a t there is a minimizing function. This possibility is illustrated b y the following theorem.

Theorem 1.1. Suppose that there is a number Y1 >~max (Yl, Y2) such that m a x F(x, Yl, O) = m a x (F(~, Yl, 0), F(/~, Y1, 0))

/or all :r and fi satis/ying xl <~<fl<~x ~. Suppose also that there is a Y2 ~< rain (Yl, Y2) with the same property. Then there is a minimizing/unction.

number 1 In other words, the functional H(]) is lower semi-continuous.

411

G. AROrr

Minimization problems. I I

Proo/.

Y1 if / ( x ) > Y1, L e t / ( x ) e ~ and form

g(x)

= /(x) if

Y2 <~/(x) <~ Y1,

[Y2 if

/(x)<Y2.

Clearly

H(g)<.H(/)

and the rest of the proof is obvious.

I t is not difficult to find different conditions on

F(x,

y, 0) which lead to the same result. (For instance,

F(x,

y, 0) increasing in y for y >/]71 and all x, decreasing in y for Y ~< Ye and all x, are also sufficient conditions for the existence of a minimizing function.)

3. Suppose that there is a number

M > M o

(M0=infrE~

H(/))

such t h a t

H ( / ) < M

implies t h a t max

I/(x) l<~g,

where K depends on M but not on/(x). Then, clearly, every minimizing sequence is bounded and there is a minimizing function.

I t is evident t h a t every number M > M 0 has this property if limlzl_~

F(x, y, z) = +

uniformly for all x and y. But there m a y exist numbers M with this property even if the limit is not uniform. We can describe the result of this section thus: if the roots zl and z2 of the equation

F(x, y, z ) = M

do not increase too fast in absolute value when [Yl -+c~, then the attainable set

E(M)

is bounded, and then there is a mini- mizing function. For instance, it will follow from Theorem 1.2' t h a t there is a minimizing function if liml~l_~r

F(x,

y, y) = + ~ uniformly in x.

Let us denote b y

El(M)

the set of points (x, y) such that

a) xl<~x<~x~,

b) there is an absolutely continuous function

g(t),

joining the points @1, Yl) a n d

(x,y),

such t h a t

sup

F(t, g(t), g'(t)) <~ M.

t

The set

Ee(M )

is defined analogously, but with

(x2,y~)

instead of

(xl, Yl).

^{We are}

interested in the set

E(M) = EI(M ) N E2(M ).

I t is clear that

H(/) <~ M

implies t h a t max

I/(x) I <~ K

if and only if

E(M)

is bounded, and in t h a t case the constant K is determined by

E(M).

We shall first examine

E(M)

in the case

F = F(y, z).

Later, we shall t r y to carry over some of the results to the general case.

Using some results from [1], we can easily prove the following theorem (concerning

dPM(t), ~pM(t)

and the integrals below, see section 2 of [1]):

Theorem 1.2.

Assume that F is independent o / x and that M > Mo.

A)

E(M) is unbounded above (i.e. it contains points with arbitrarily large ordinates) i/and only i/the integrals

f ~ dt and 12= f~ dt

11

J~, O ~ ( t ) J ~ -

~,(t)

are well-de/ined, convergent and

I i + I ~ < x 2 - x 1.

(1)

A R K I V FOR M A T E M A T I K . B d 6 n r 2 2

B)

E(M) is unbounded below i / a n d only i/the integrals

i3=f _ - V,~(t) ^{dt and i4=fu} _ opt(t) ^dt are well-de/ined, convergent and

I3 + I4 < x 2 - x 1.

(2)

Proo].

We restrict the proof to the first p a r t of the theorem. Assume first t h a t

B(M)

is unbounded above. L e t (x0, Yo) C

E(M)

and Yo > m a x

(Yl,

Y2). I t follows from Theorem 2 in [1] t h a t

f yV~ dt

1 ~ ~< x ~ xl a n d

f y . dt

~<x 2 - x 0.

, - ~ , ~ ( t )

I f we add these inequalities and let Y0 tend to infinity, then we have proved one p a r t of the assertion.

Next, suppose t h a t

I i + I 2 < x 2 - x 1.

I t follows from the discussion of " t h e at- tainable cone" in section 2B of [1] t h a t

(x, y ) E E I ( M )

if x 1 + I l < x < x 2 and

Y>~Yl.

Similarly, it follows t h a t

(x, y)EE2(M )

if

x 1 <x<~x 2 - I 3

and

Y>~Y2.

Clearly, all points (Xl + I1, y), where y ~> m a x (Yl, Y2), belong to

E(M).

This completes the proof of Theorem 1.2, since the reasoning in the case B is analogous.

A consequence of the theorem is t h a t

E(M)

is bounded if a n d only if neither (1) nor (2) is satisfied.

L e t us now consider the general case

F = F ( x , y, z).

I t can be

"reduced"

to the previous case simply b y neglecting the dependence on x. This m a y be considered to be a crude method, but, nevertheless, it works in m a n y cases.

The functions

~gM(y)

and

~fM(Y)

^{w e r e} defined in [1]. Clearly, we can introduce here the corresponding functions

O(x,y,M)

and ~(x, y, M). Now fix y and M and suppose t h a t U =

{x IF(x,

y, 0) ~<M} is non-empty.

De/inition: I ~(y'

M) = max~

~ v O(x, y, M),

[ fl(y,

M) = minx ~ v ~f(x, y, M).

I t is easy to verify t h a t these functions are continuous and t h a t they are monotonic in M, as are (I)~(y) and ~v~(y). Finally, an argumentation, similar to the one used in the special case F =

F(y,z),

leads to the following result:

Theorem 1.2'.

Assume that M > M o.

A)

I / E ( M ) is unbounded above, then the integrals dt

I1= , a(t,M) and

dt 12=

~ - f l ( t , M )

are well.de/ined, convergent and I 1 + 12 <<-x2 -x l .

B)

I / E ( M ) is unbounded below, then the integrals

413

c. ARO~SSOr~, Minimization problems. I I

i3=f dt

_ - f l ( t , M ) and Y~ dt

I ~ = _ a(t, M ) are well-defined, convergent and I a + I a <~ x 2 - x 1,

An application of this theorem is given in Example 1 in Chapter 4. Compare also the counterexample.

Chapter 2. A more general case. P r o o f of the existence of absolutely minimizing functions

]. I n this section we shall generalize some of the previous results to a more general class of functions F(x, y, z).

We are not going to change the condition limN_~cF(x, y, z ) = + c o , which has turned out to be very useful and which is a rather natural condition. However, the assumption t h a t F(Xo, Yo, z) has a minimum at z = 0 for all (x0, Y0), is more restrictive, and we shall assume instead t h a t F(x0, Yo, z) has a minimum at z=eo(x0,yo) , where oo(x,y) is a continuous function. Compare Example 5 in Chapter 4.

Thus, let us assume t h a t the following conditions hold throughout this chapter:

F(x, y, z) E C1; 2B and 3.

The minimization problem is meaningful, since we have H(/) >~ m a x [F(Xl, Yl, O~(Xl, Yl)), F(x2, Y2, c~

for every admissible function/(x).

We shall now carry over some of the results of the paper [1] and of the previous chapter to this more general case.

The analogues of the Theorems 5 and 6 in [1] are Theorem 5'. L e t / ( x ) be an admissible/unction such t h a t : a) /'(x) is continuous on I (x 1 <~x <~x2),

b)

F z ( x , / ( x ) , / ' ( x ) ) > ~ O on I , o r < O on I ,

c) F(x,/(x),/'(x))

= M on I .

T h e n / ( x ) is a m i n i m i z i n g / u n c t i o n 1 (not necessarily unique).

Theorem 6'. Suppose that/(x) E ~ satisfies:

a) /'(x) is continuous on I (x 1 ~x<~x~), b) F z ( x , / ( x ) , / ' ( x ) ) # 0 on I ,

c) F(x,/(x),/'(x))

= M on I .

T h e n / ( x ) is a unique m i n i m i z i n g / u n c t i o n . 1

The proofs are analogous to those of the case ~o(x, y)=~0 and they are omitted.

Lemma 4 in [1] holds in this case without changes. The proof is simple.

1 It also follows that ](x) is an a.s. minimal on I.

ARKIV FOR MATEMATIK, B d 6 n r 22 A c o n s e q u e n c e of t h i s l e m m a is t h a t we can n e g l e c t sets of m e a s u r e zero. This m e a n s t h a t we can j o i n t w o f u n c t i o n s a t a f i n i t e n u m b e r of p o i n t s w h e r e t h e y a r e e q u a l b y t a k i n g one or t h e o t h e r on successive i n t e r v a l s . T h e n t h e s e p o i n t s will n o t affect t h e v a l u e of t h e f u n c t i o n a l for t h e r e s u l t i n g f u n c t i o n . (This s i t u a t i o n will o c c u r in t h e n e x t section.)

:Next, we m a k e a j u m p t o T h e o r e m 9, w h i c h is c a r r i e d o v e r w i t h o u t changes:

T h e o r e m 9'. I / ] ( x ) is an a.s. minimal on the interval 1 (x 1 ~ x <~x2), then:

1) /(x)EC 1 on I,

2) the differential equation

dF(~,/(x),/'(~))

Fz(X, / ( x ) , / ' ( x ) ) = 0 dx

is satisfied on I in the sense that i / F ~ ( x , / ( x ) , / ' ( x ) ) =#0 on a sub-interval I 1 c I, then F ( x , /(x), /'(x)) is constant on 11.

W e shall p r o v e t h e a s s e r t i o n s 1) a n d 2) u n d e r t h e following, w e a k e r a s s u m p t i o n s

on/(x):

I) /(x) satisfies a L i p s c h i t z c o n d i t i o n on [Xl, x2],

I I ) t h e r e is a c o n s t a n t K > 0 such t h a t if [tl, t2] is a n a r b i t r a r y s u b - i n t e r v a l of I , t h e n t h e r e a r e two (not n e c e s s a r i l y different) m i n i m i z i n g f u n c t i o n s u(x) a n d v(x) be- t w e e n ( t i , / ( t i ) ) a n d (t2,/(t2) ) such t h a t [u(x)[ ~ K , Iv(x)] ~ K a n d u(x) <~/(x) <~v(x) for t 1 ~< x ~< t 2.

(These a s s u m p t i o n s a r e s a t i s f i e d if /(x) is a n a.s. m i n i m a l , since we c a n choose K = m a x z ~ ]/(x) l a n d u ( x ) = v ( x ) = / ( x ) in t h a t case.)

T h e r e a s o n w h y we shall p r o v e t h i s m o r e g e n e r a l r e s u l t is t h a t i t will be useful in t h e n e x t section.

Proo]. 1 Consider a n a r b i t r a r y p o i n t x o on t h e i n t e r v a l x 1 ~ x < x 2 a n d s u p p o s e t h a t t h e u p p e r a n d lower r i g h t - h a n d d e r i v a t i v e s of /(x), :r a n d /~, a r e d i f f e r e n t a t x 0.

(Clearly, t h e y a r e finite.) Choose a n u m b e r 7 ~:o)(Xo,/(Xo)) such t h a t g > ~ >ft. T h e i n i t i a l - v a l u e p r o b l e m F(x, g(x), g'(x))=constant, g(xo)=/(x0), g'(xo)=~, h a s a solu- t i o n ~(x) in a n e i g h b o u r h o o d of x 0. Since ~ > ~ >fi, t h e r e a r e p o i n t s x > x o, a r b i t r a r i l y close t o x0, w h e r e /(x)=~0(x). B u t since ~, ~=eo(x0,/(xo)), i t follows f r o m t h e condi- t i o n I I a n d f r o m T h e o r e m 6' t h a t / ( x ) =q~(x) on [Xo, x0+~t ] for s o m e ~ > 0 . T h e con- t r a d i c t i o n shows t h a t / ( x ) h a s a r i g h t - h a n d d e r i v a t i v e . S i m i l a r l y , i t follows t h a t / ( x ) h a s a l e f t - h a n d d e r i v a t i v e for x 1 < x ~<x 2.

O u r n e x t s t e p is to p r o v e t h a t t h e o n e - s i d e d d e r i v a t i v e s a r e equal. Consider a p o i n t x0, a n d a s s u m e t h a t o:=/'(xo)+4fi=/'(xo)-. Choose a n u m b e r y b e t w e e n a n d ft. C l e a r l y

F(x0,/(Xo), y) < m a x [F(xo,/(xo) , o:), F ( x o, ](Xo), fl)].

S e l e c t a sequence of i n t e r v a l s [Pn, q~] such t h a t p~ <xo <q~, q~-p~-->O

a n d /(q~) - / ( P ~ )

q~ - p~

1 The reasoning will be similar to the one used in [1], but not identically the same.

2s: 4 415

G. ARONSSO~,

Minimization problems. II

I t is obvious t h a t

lim

M(pn, qn; /(P,), /(q,)) < F(xo,/(xo), ~).

n ---> o v

(The n o t a t i o n M(...) was i n t r o d u c e d in [1].) L e t us assume, for example, t h a t g > y > f l a n d

F(xo,/(xo) ,

~) > F(x0,/(xo)

,

y). L e t the functions

Un(X)

correspond to t h e intervals [p~, q~] according to I I . I t follows t h a t lim~_~r (SUpx

u'~(x))>~ ~z.

Thus

Consequently

lira

H(un) ~ F(xo, /(xo), ~).

n ---> ~ r

lim

H(u~)

> lira

M(pn, qn; /(Pn), /(qn)),

n .--> o r n ---> ~

which is a contradiction.

The reasoning is analogous in the other cases. This proves t h a t / ' ( x ) exists.

Suppose n o w t h a t

Fz(xo,/(xo) ,/'(xo) ) #0.

T h e n we assert t h a t / ( x )

E C 2

in a neigh- b o u r h o o d U of x 0 a n d t h a t

F(x, /(x), /'(x))

is c o n s t a n t in U. This is p r o v e d in t h e same w a y as T h e o r e m 8 in [1]. W e omit the details.

N o w t h e only thing left to prove is t h a t / ' ( x ) is continuous a t those points where

Fz(X, /(x), /'(x))

= 0 . Assume t h e n t h a t

]'(Xo)=o~(x

o,/(x0) ). Assume f u r t h e r t h a t there are n u m b e r s ~n->x0, ~ > x 0 such t h a t

/'(~)>~/'(Xo)+~.

I f the functions

Un(X)cor-

respond to the intervals x 0 ~ x ~ according to I I , t h e n we h a v e sup~

u'~(x) >~/'(xo) § ~.

H e n c e l i m n ~

H(Un) ~ F(Xo,/(Xo) ,/'(Xo)

+~). B u t this contradicts the obvious relation lim

M(xo, ~n; /(xo), /(~n)) <~ F(xo, /(Xo), /'(Xo)).

n - - > o ~

T h e other cases can be t r e a t e d similarly. This completes the proof.

W e h a v e t h u s established some i m p o r t a n t properties of a.s. minimals. B u t we h a v e n o t e x a m i n e d the properties of minimizing functions in general. T h e reason for this is evident f r o m E x a m p l e 2 in Chapter 4.

I n t h e previous c h a p t e r we t r e a t e d the question of the existence of a minimizing function for the case

co(x,y)~O.

I t is easy to verify t h a t L e m m a 1.1 holds in t h e present case. This means t h a t if there is a minimizing sequence of u n i f o r m l y b o u n d e d functions, t h e n there is a minimizing function. I n particular, this is t r u e if

E(M)

is b o u n d e d for some

M > M o . 1

W e shall n o t discuss generalizations of the other criteria in Chapter 1.

2. A problem which we h a v e n o t y e t discussed, is t h a t of t h e existence of absolutely minimizing functions. W e shall give an existence proof for the general case.

Theorem 2.1.

Assume that F(x, y, z) E C 1 and satisfies the conditions 2 B and 3 / o r X I <~x<~X 2 and all y, z. (Thus J=[X1,X2]. )

Assume/urther that, /or every choice o/(~1,~1) and (~2,~2), there is a minimizing/unc- tion between these points, and, i/ there are several such/unctions, that they are uni/ormly bounded. (The bounds will depend on ~1, ~h, ~ and ~2")

1 A sufficient condition for this is that limlz[-~r

F(x,y,

z) = + c~ uniformly for all x and y.

ARKIV FSR MATEMATIK. B d 6 n r 22

Then there is an absolutely m i n i m i z i n g / u n c t i o n / o r every minimization problem in the strip X I <~ x ~ X2, - ~ < y < ~ .

Proo/.

I n order to m a k e the proof more lucid, we shall divide it into several parts.

l) Consider the minimization problem between

(Xl,

Yl) and (x2, Y2). L e t ~ be the class of minimizing functions, which, b y assumption, is not empty.

P u t

[

u(x) ~ inf [(x) a n d

lv (x) = sup/(x).

f e T n

I t is clear t h a t

u(x)

and

v(x)

belong to ~ .

Consequently there are always a smallest minimizing function

u(x)

and a greatest minimizing function

v(x).

2) We shall introduce two notations in order to simplify the expressions later on.

L e t

p(x)

be a function on an interval I . Let [Xl,X2] be a sub-interval of I and consider the minimization problem between (Xl,

p(xl) )

and

(x2, p(x2)).

L e t

u(x)

be the smallest and

v(x)

the greatest minimizing function, as above. We shall agree to say t h a t

p(x)

has the p r o p e r t y A on I if

p(x)>~u(x)

on Ix1, xe]/or

every

sub-interval [xi, x2] of I and t h a t

p(x)

has the p r o p e r t y B on I if

p(x) <~v(x)

on

[xl, xeJ/or every

sub-interval Ix1, x2] of I . (Compare subharmonic and superharmonic functions.)

3) Consider our minimization problem on an interval I with some b o u n d a r y values. I t is obvious t h a t the greatest minimizing function

v(x)

is not less t h a n the

greatest

minimizing function

Vi(X )

between a n y two points (tl,

v(ti) )

and

(t2, v(t2) ).

Similarly, the smallest minimizing function

u(x)

is not greater t h a n the

smallest

minimizing function

u1(x )

between a n y two points

(tl, u(tl) )

and

(t2, u(t2) ).

4) We shall now construct an a.s. minimal.

From now on, we consider a /ixed minimization problem, namely between

(Xl, y~) and (x2, Y2). L e t G be the class of those minimizing functions which have the p r o p e r t y A on

[xl,x2]. G

is not empty, since the greatest minimizing function belongs to G.

We introduce the/unction

h(x)

= inf

g(x)

geG

and we assert that h(x) is an a.s. minimal.

I t is clear t h a t

h(x)

is a minimizing function.

5) Now let us prove t h a t

h(x)

has the properties A and B.

I) Choose an a r b i t r a r y interval it1, t2] a n d let

u(x)

be the smallest minimizing func- tion between

(tl, h(tl) )

a n d

(t2, h(t2) ).

L e t

g(x)CG.

We assert t h a t

g(x)>~u(x)

on t 1 ~<x ~ t 2. I f this were not true, then there would be an interval s 1 ~<x ~<s 2 such t h a t U(81)=g(81) , U(S2)=g(82) and

u(x)>g(x)

for

81<x<82.

Let

ul(x )

be the smallest minimizing function between (Sl,

u(sl) )

and

(se, u(s2) ).

I t follows from the definition of G t h a t

g(x)~>Ul(X

). B u t we also have

u ( x ) ~ u l ( x ).

(See p a r t 3 of the proof.) This gives

g(x)>~ul(x ) ~ u ( x )

and we have a contradiction.

Thus

u(x) <~g(x),

and the inequality

u(x) <~h(x)

follows from the definition of

h(x).

This proves t h a t

h(x)

has the p r o p e r t y A.

I I ) Assume t h a t

h(x)

has not the p r o p e r t y B. This means t h a t there is an interval It1, t2] such t h a t the corresponding greatest minimizing function

v(x)

does not satisfy 417

c . AROr~SSOrr Minimization problems. II

t h e inequality

v(x)>~h(x).

I t follows f r o m p a r t 3 of the proof t h a t we m a y assume t h a t

h(x)>v(x)

for t I < X < t 2. L e t us f o r m the function

h(x) for

x<~q, p(x)=]v(x)

for

tl<x<t2,

[h(x) for

x ~ t 2.

I f we can prove t h a t

p(x)E G,

t h e n this will contradict the definition of

h(x),

since we h a v e

p(x) < h(x)

on (tl, t2). I t is clear t h a t

p(x)

is a minimizing function on [xl, x~], a n d it remains to prove t h a t

p(x)

has t h e p r o p e r t y A. So let us consider a n interval Is1, s2] a n d let the corresponding smallest minimizing function be

ul(x ).

W e h a v e

ul(sl)=p(sl)<h(sO

^{a n d}

ul(s2)=p(%)<~h(% ).

^Since

h(x)

has t h e p r o p e r t y

A,

^this

implies t h a t

h(x)~>Ul(X

). If p ( ~ ) < u l ( ~ ) for some ~E(Sl, S2), t h e n we m u s t h a v e

p(~)=v(~)<ul(~ ).

Consequently there m u s t be an interval [rl, r2] such t h a t

v(rl)

= u l ( r 0 ,

v(r2) -ul(r2)

a n d

v(x) < ul(x )

for r 1 < x < r 2. L e t

u2(x )

a n d

v2(x )

corre- spond to this interval a n d these b o u n d a r y values in the usual manner. T h e n we h a v e

udx)

<u2(x) <v2(x) < v(x).

B u t this contradicts t h e inequality

v(x)<ul(x

). This proves t h a t

h(x)

has the pro- p e r t y B.

6) I t is obvious t h a t

h(x)

satisfies the conditions I a n d I I which, as we h a v e seen, g u a r a n t e e t h a t t h e assertions in T h e o r e m 9' are true. Thus

h(x)E C 1

a n d the differential equation

dE(x, h(x), h'(x))

dx 9 Fz(x, h(x), h'(x)) = 0

is satisfied in t h e sense described there.

7) W e are n o w in a position to prove t h a t

h(x)

is an a.s. minimal. Consider an a r b i t r a r y interval t 1 < x < t 2. L e t M 0 be t h e m i n i m u m value of the functional, as usual, a n d p u t

M = m a x

F(x, h(x), h'(x)).

t x ~ x ~ t ~

W e w a n t to prove t h a t M = M 0.

I f

F(t 1,

h(tl), h'(t 0) = M, t h e n t h e result follows at once, since

h(x)

has the properties A a n d B, a n d t h e same is true for t 2. So let us assume t h a t

F(t 1,

h(tl),

h'(tl))<M

a n d t h a t

F(t~, h(t~), h'(t2))<M.

I t follows f r o m p a r t 6 t h a t there is a n u m b e r

2, tl<~<t2,

such t h a t h'(~)=r h(~)) a n d F(~, h(~), h'(~)) = M . L e t us assume t h a t M > M o

This means t h a t there is no minimizing function co(x) for which 0)(2)=h(~). (I.e.

(2, h(~)) CE(Mo). )

L e t K be the class of those minimizing functions a)(x), for which 0)(2)<h(~). (K is n o t e m p t y , see p a r t 5.) P u t k ( x ) = s u p ~ K co(x). I t is clear t h a t k ( x ) E K a n d t h a t k(~) < h(~).

P u t

/ 81 = sup {x ix < ~,

h(x)

= k(x)}

a n d )

[ s2 = inf {x ] x > ~,

h(x) = k(x)}.

ARKIV FOR MATEMATIK. B d 6 n r 22 W e introduce the function

h(x) for

x < s 1, q(x)= jk(x)

^for

s l < x < s 2,

[h(x) for x ~ s 2.

If we can prove t h a t

q(x)E G,

t h e n this will c o n t r a d i c t the definition of h(x), since

q(x)

<h(x) on (s 1, s2). (Compare p a r t 4.)

So all we have to prove is that q(x) has the property A.

Consider t h e n a n a r b i t r a r y interval [rl, r2] a n d let

u(x)

be t h e smallest minimizing f u n c t i o n between

(rl, q(rl) )

a n d

(r2, q(r2) ).

W e h a v e

u(rl)=q(ri)<~h(rl)

a n d

u(r2)

=q(r2)

<~h(r2).

Since

h(x)

has the p r o p e r t y A, it follows easily t h a t

h(x) >~u(x)

for r~ ~< x ~< r e.

So if we assume t h a t

u(xo)>q(x0)

for some x0, t h e n we m u s t h a v e

s 1 <xo<s 2

a n d

q(xo) =k(x0).

Consequently there m u s t be a n interval (~h, ~2) such t h a t s 1~<~1 <~2 ~<s2, u(~h) = k(~l), u(~2) = k(~2) a n d

u(x) > k(x)

for ~1 < x <V2.

If

H(k; ~ , ~2) <~H(u; ~ , ~2),

t h e n we get a contradiction to the definition of

u(x).

Suppose t h e n t h a t L e t us form the function

H(k; ~1,

~ 2 ) > H ( u ; ~1,

~2)"

[

~(X) for Z ~ l ,

kl(x ) = j u ( x )

for ~ 1 < x ~ 2 ,

[k(x) for x ~ 2 .

Clearly, kl(X )

is a minimizing function between (tl,

h(tl) )

a n d (t~,

h(t2) ).

F u r t h e r , we h a v e kl(~)~<h(~ ) (for, as we m e n t i o n e d above,

u(x)~h(x)

for all x, where

u(x)

is defined). H e n c e kl(x ) EK. :But this contradicts the definition of k(x). Hence we get a contradiction in a n y case.

This completes the proof of T h e o r e m 2.1.

3. The m e t h o d of a p p r o x i m a t i n g a m a x i m u m b y a sequence of integral m e a n values is well known. I n our case it means t h a t we should consider t h e functional

H(/)

as the limit of the sequence of functionals

H n ( l ) = 1 ,

[F(x,/(x),f(x))]ndx] 11~,

n = 1 , 2 , 3 , . . . .

W e have used this a p p r o a c h in [1] to derive t h e differential e q u a t i o n

(dF/dx). F z = O.

I t can also be used to give a different existence proof for a.s. minimals. This is v e r y n a t u r a l since a minimizing function in the calculus of variations a u t o m a t i c a l l y is minimizing on e v e r y sub-interval. However, we shall n o t c a r r y t h r o u g h this proof since it is more complicated t h a n the one already given, a n d since it requires stronger conditions on

F(x, y, z).

A n o t h e r result which can be p r o v e d with this "integral m e t h o d " is the following:

if/o(X) is t h e only a.s. minimal for

H(/)

a n d if/~(x) minimizes

Hn(/)

for n = 1, 2, 3, ..., t h e n limn_~ ~/~(x) =/o(X) uniformly.

419

c. ARONSSON, Minimization problems. II

Chapter 3. Further examination of absolutely minimizing functions.

Uniqueness questions

For reasons of simplicity, this chapter will treat only the case

(o(x,y)=-O.

N o doubt, the results can be modified for the general case.

1. The following theorem gives us further information on the structure of absolutely minimizing functions:

Theorem 3.1.

Assume that F(x, y, z) satis/ies these conditions:

F(x, y, z) is an analytic/unction o/ x, y and z in a complex neighbourhood o/the set o/ values

{

X l ~ X ~ X 2

y, z real;

2 A and 3.

Assume/urther that/(x) is an absolutely minimizing/unction on the compact interval xl ~ x ~ x 2.

Then:

a)

/(x) EC 1 on

[xl, x2]

(already proved);

b)

the set {x I F~(x, /(x), /'(x)) # 0 } consists o / a f i n i t e number o/intervals (to be proved now);

c)

the di//erential equation

dF(x, /(x), /'(x)) Fz(x, /(x), /'(x)) = 0 dx

is satis/ied in the sense that F(x, /(x), /'(x)) is constant on each o/these intervals (al- ready proved).

Proo/.

I n order to facilitate the further references, we divide the proof into several parts.

1) Assume t h a t

] ' ( x ) # 0

for

p < x < q

and t h a t / ' ( p )

=/'(q)=0.

Fx(p,/(p), o) < o,

Then ^/

[ Fx(q,/(q), O) >10.

To see this, we recall t h a t / ( x )

EC 2

on

(p,q)

and t h a t

F(x, /(x), /'(x))

is constant there (Theorem 8 in [1]). So we have

Fx(X, /(x), /'(x)) + F~(...) /'(x) + Fz(...)/"(x) = 0

for

p < x < q .

Clearly, there m u s t be points arbitrarily close to p, where

/'(x)

a n d

/"(x)

have the same sign. At such a point

Fz(...)/"(x)>0,

which gives

Fx(x, l(x), l'(x)) + ~'y(...) /'(x)

< 0 . 420

ARKIV FOR MATEMATIK. B d 6 n r 2 2 Since l i m x - ~ v / ' ( x ) = O , t h i s p r o v e s t h e first i n e q u a l i t y , a n d t h e s e c o n d is p r o v e d s i m i l a r l y (there a r e p o i n t s a r b i t r a r i l y close t o q w h e r e

Fz(...)/"(x ) <0,

etc.).

N o w s u p p o s e t h a t t h e r e is a n u m b e r

t>q

such t h a t / ' ( t )

4=0.

T h e n i t follows f r o m t h e a b o v e r e s u l t t h a t we can define

r=inf{xlx>~q,

F~(z,/(z), 0)=0},

a n d i t follows t h a t

q<r<t.

I f

q<r,

t h e n we o b v i o u s l y h a v e

/'(x)=O

for

q~x<~r

( a n d if q = r , t h e n t h i s is t r i v i a l l y true).

I f t h e r e is a n u m b e r

u < p

such t h a t

]'(u)4=0,

t h e n we define s = s u p {x; x < p , F ~ ( x , / ( x ) , 0) = 0 } . W e h a v e / ' ( x ) = 0 for

s ~ x ~ p .

F i n a l l y , we o b s e r v e t h a t

{ Fx(x,/(x),O)~O

for

s<~x~p, Fx(x,/(x),

0)~>0 for

q<~x<~r.

2) W e shall give a n i n d i r e c t p r o o f of t h e t h e o r e m . So we a s s u m e t h a t t h e set {x[ x 1 < x < x2,

/'(x)

4 0 } is t h e u n i o n of a n i n f i n i t e n u m b e r of d i s j o i n t o p e n i n t e r - vals. These i n t e r v a l s m u s t h a v e a l i m i t - p o i n t a n d we m a y a s s u m e t h a t i t is

x=O.

W e m a y also a s s u m e t h a t / ( 0 ) = 0 a n d t h a t t h e r e a r e i n f i n i t e l y m a n y such i n t e r v a l s i n e v e r y i n t e r v a l 0 < x < e. C l e a r l y / ' ( 0 ) = 0 a n d Fx(0, 0, 0) = 0.

T h e f u n c t i o n

Fx(x,

y, 0) will b e i m p o r t a n t for t h e r e s t of t h e proof. I t m a y be i d e n t i c a l l y zero or not. T h i s gives us t w o cases a n d we shall s t a r t w i t h t h e s i m p l e s t of t h e m .

3) A s s u m e t h a t Fx(x, y, 0) ~ 0. T h i s m e a n s t h a t we h a v e

F(x, y, O) = F(O, y, O) =of(y),

w h e r e ~c(y) is a n a l y t i c in a (complex) n e i g h b o u r h o o d of y = 0 .

A) ~c(y)--- c o n s t a n t , i.e.

F(x,

y, 0) --- c o n s t a n t . Consider a n i n t e r v a l

(p, q)

as in 1). P u t

t=89 +q).

T h e n we h a v e

F(t, /(t), /'(t)) > F(t,/(t), 0) =

F(p, tip),

0),

w h i c h gives a c o n t r a d i c t i o n .

B) ~c(y) ~ c o n s t a n t .

I t follows f r o m T h e o r e m 10 in [1] 1 t h a t / ( x ) is m o n o t o n i c , for i n s t a n c e n o n - d e c r e a s - ing.

B u t ~ c ' ( y ) # 0 for 0 < y < ~. H e n c e ~c(y) is s t r i c t l y m o n o t o n i c for 0 ~ y ~ . Consider a n i n t e r v a l

(p,q)

such t h a t

O</(p)</(q)<~.

T h e n we m u s t h a v e

cf(/(p))=cf(/(q))

w h i c h gives a c o n t r a d i c t i o n .

1 I t is c l e a r t h a t t h e c o n d i t i o n o n F z i n T h e o r e m 10 n e e d o n l y b e a s s u m e d t o h o l d for z = 0.

I n f a c t , t h e c o n d i t i o n of T h e o r e m 1.1 i n t h i s p a p e r is s u f f i c i e n t if i t h o l d s f o r a l l y.

4 2 1

G. AROr~SSOr~, Minimization problems. II

4) ~qow let us assume t h a t

F~(x,

y, 0) ~ 0. Clearly

F~(x, y,

0) is a n analytic function a n d we can a p p l y Weierstrass' p r e p a r a t i o n t h e o r e m (see [2], p. 89). I n our case it says t h a t

F~(x, y, O) =x~o~(x, y)W'(x, y)

in a n e i g h b o u r h o o d U of the origin. Here # is an integer ~>0;

~o(x,y)

is analytic in U a n d # 0 ; ~F(x, y) is

either =- 1 or

of t h e f o r m

vt~(x ' y) =ym + A~(x)y,~-i + A2(x)ym-2 + ... + A,~(x),

where t h e functions

Ak(x)

are analytic in a (sufficiently small) n e i g h b o u r h o o d of

x=O

a n d A k ( 0 ) = 0 for all k.

W e k n o w f r o m 2) t h a t

F:~(x,

y, 0) - 0 at t h e points (r,/(r)) a n d

(s, ](s)).

^Therefore we can exclude t h e case ~F ~= 1 and, instead, we inquire a b o u t the zeros of

uz.(x,y ) =ym + A~(x)yr~-I + ... +Am(x).

L e t us, for the present, replace x a n d y b y the complex variables ~ a n d ~]. W e shall t r y to determine ~ as a function of $ f r o m t h e relation ~F(~,~) = 0 . W e are only in- terested in the solutions of this e q u a t i o n in a n e i g h b o u r h o o d of ~ = 9 = 0 .

Consider a complex n e i g h b o u r h o o d V of $ = 0 a n d c u t it along t h e negative real axis. T h e n every root of ~ can be defined as a regular function in V, a n d the roots

~k of ~F(~, ~]) = 0 m a y be written

= ~ C i~l/n~

for k = l , 2 , . . , m . ,~,o, j , k ~ ,

Concerning this expansion, see [3], p. 50; [4], pp. 98-103 a n d [5], Chapter X I I I . (The roots can be divided into cyclic systems, a n d the n u m b e r n can be chosen as t h e p r o d u c t of the n u m b e r of roots in each system.)

W e m a y assume t h a t ~l/n is real for ~ > 0, which gives the expressions

~ ~ - ~ J . k [ x / ^1/n~i ) ~ k = 1, 2, . . . , m.

j = l

Clearly, we need only consider those series in which Cj.g are real for all ?'. (If there are no such series, t h e n there is n o t h i n g left to prove.) Therefore, let us suppose t h a t

~]k(x),

k = 1,2 ...

M,

are real for 0 < x < 0 , a n d t h a t ~k(x) are complex (not real) for k > M a n d 0 < x < 0 . W e k n o w f r o m t h e first p a r t of t h e proof t h a t

Fx(r,/(r),

0)= Fx(8,/(s), 0)=0.

H e n c e t h e points (r,/(r)) a n d

(s,/(s))

m u s t lie on the curves

y =~k(x)

if r a n d s are sufficiently close to x = 0.

ARKIV FOll MATEMATIK. B d 6 n r 22 5) W e shall n o w s t u d y in some d e t a i l t h e f u n c t i o n s

~k(x) -= ~ ~].k

( x l / n ) ~ = g k ( X l / n ) 9

j = l

H e r e t h e f u n c t i o n s

g~(z)

a r e a n a l y t i c for ]z[ < 8. F u r t h e r , e v e r y

g~(z)

is r e a l if z is real.

D i f f e r e n t i a t i o n gives

. _ _ - 1 - - 1

,(X)=gk(xl/n ) 1 "X n

for

O < x < 8 n.

n

P u t t =

x ~/n,

w h i c h g i v e s

~k (x) = g; (0 [ ~ = ~ ( t ) .

T h e f u n c t i o n ~ ( z ) is a n a l y t i c in 0 < I z] < 8 a n d t h e s i n g u l a r i t y a t z = 0 is e i t h e r re- m o v a b l e or a pole. W e h a v e

l i m ~ (x) = l i m F~(t),

x --.-~ + 0 t ' - - > + O

a n d t h i s l i m i t m u s t be real. I t can b e + o% - 0% f i n i t e b u t ~ 0 a n d 0. Since ]'(0) = 0 , we can e x c l u d e f r o m c o n s i d e r a t i o n t h o s e v a l u e s of/c for which we do

not

h a v e ~ ( 0 ) = 0 . T h e r e f o r e we c a n a s s u m e t h a t ~ ( z ) is a n a l y t i c for l zl < ~ .

A c o n s e q u e n c e of t h i s is t h a t t h e r e is a 8 1 > 0 such t h a t ~vk(z ) 4=0 for 0 < ]z] <81, unless ~vk(z)~0. F r o m t h e r e l a t i o n ~ ( x ) =~k(xl/~), we can infer a c o r r e s p o n d i n g r e s u l t for ~ ( x ) .

N o w let us c o m p a r e t h e f u n c t i o n s ~ ( x ) =gk(t) a n d ~ l ( x ) - g l ( t ) . T h e f u n c t i o n s

gk(z)

a n d

gz(z)

a r e a n a l y t i c a n d n o t i d e n t i c a l . H e n c e t h e r e m u s t e x i s t a 82 > 0 such t h a t

9k(z)#gl(z)

for 0 < l z ] <82.

T h i s gives a c o r r e s p o n d i n g r e s u l t for ~k(x) a n d ~z(x).

W e h a v e f o u n d t h a t ~/k(x)=gk(x

1/'~)

a n d ~ ( x ) = ( v k ( x l / ~ ) , w h e r e

gk(z)

a n d ~k(z) a r e a n a l y t i c for Iz[ < 8 . L e t us w r i t e

xl/n=t,

as a b o v e . This gives

~'(x, ~(x), o)= ~'(t ~, g~(t),

o ) = u ( t ) a n d

F(x, ~k(x), ~'k(x) )= F(t ~, gk(t), qJk(t) )=v(t).

Clearly, t h e f u n c t i o n s

u(z)

a n d

v(z)

are a n a l y t i c in a n e i g h b o u r h o o d of z = 0 . H e n c e t h e r e is a 8~ > 0 such t h a t e a c h of t h e f u n c t i o n s

F(x,

~k(x), 0) a n d

F(x, ~(x), ~'k(x))

is e i t h e r increasing, c o n s t a n t or d e c r e a s i n g on t h e whole i n t e r v a l 0 ~< x ~< 83.

W e h a v e e x c l u d e d t h o s e f u n c t i o n s

~k(x),

for w h i c h we d i d n o t h a v e ~]~(0) = 0 . A f t e r a r e n u m b e r i n g , we h a v e ~k(x), k = 1, 2 . . . P , left.

I f we collect o u r results, t h e n we see t h a t t h e r e e x i s t n u m b e r s ~ > 0 a n d % such t h a t : a)

~k(X):4:~l(X)

if 0 < X ~ : r a n d

lc:4:l.

This m e a n s t h a t one of t h e f u n c t i o n s ~k(x) is s m a l l e s t a n d one is g r e a t e s t on t h e whole i n t e r v a l 0 < x ~< ~.

b) ~]k(x) C C 1 on 0 ~<x ~ cr a n d ~k(0) =~/~(0) = 0 . I f ~k(x) ~: 0, t h e n ~ ( x ) 4=0 for 0 < x ~< ~.

c) I f we h a v e p , q, r a n d s as in 1) a n d 0 < s < r ~ < c ~ , t h e n

](r)=~k(r )

for s o m e

k<~p

a n d

](s)=~(s)

for s o m e k 1 ~<P.

423

c. xRo~ssorr

Minimization problems. I I

d) E v e r y function

F(x, ~k(x),

0) is either c o n s t a n t or strictly m o n o t o n i c on 0~<x~<~. The same is true for

F(x,

~k(x), ~ ( x ) ) (1

<~k<~P).

6) W e are interested only in those functions ~k(x) for which one of t h e relations u n d e r c) above really occurs on t h e interval 0 < x ~< ~. I f we exclude t h e others, t h e n we get (after a renumbering) t h e functions ~k(x), k = 1, 2 . . . . , N, left. This does n o t affect t h e validity of t h e s u m m a r y a)-d).

I f N = 1 a n d ~ l ( x ) ~ 0 , t h e n we get a contradiction at once a n d there is n o t h i n g left t o prove.

I f this is n o t t h e case, t h e n t h e greatest of t h e functions ~ ( x ) is > 0 (for 0 < x ~< r162 o r t h e smallest is < 0.

L e t us choose t h e first case a n d let ~N(X) be the function in question. T h e n

~N(X) > 0

a n d ~]N(X) > 0 on 0<x~<r162

I t follows from our choice of

~N(X)

t h a t there is an interval (P0, %) such t h a t / ( % )

=

~N(ro).

W e have t h u s

/(qo)--/(ro)=~N(ro);

/'(qo) = 0 a n d ~N(X) > 0 .

H e n c e

/ ( x ) > ~ ( x )

for q0 - ~ < x < % . B u t we also h a v e / ( P o ) =/(So)<~N(So) (owing t o our choice of ~ ( x ) ) a n d

/'(po)-O.

H e n c e

/ ( x ) < ~ ( x )

for

p o < x < p o §

Conse- q u e n t l y there m u s t be a ~, p 0 < ~ < q 0 , such t h a t / ( ~ ) = ~ ( ~ ) a n d / ' ( ~ ) > ~ N ( ~ ) > 0 .

N o w assume first t h a t

F(x,

~?N(X), 0) is c o n s t a n t or decreasing. T h e n we have

F(~, ~N(~), O) >~ F(r o, ~]N(rO), O) >~ F(q o, ](qo), 0).

(See t h e first p a r t of the proof.) This gives

F(8, /(8), /'(8))

> F(q0,/(qo),

0),

which is a contradiction.

Assume t h e n t h a t

F(x, ~N(X),

0) is increasing. A n obvious consequence of this is t h a t

F(x, ~N(X), ~N(X))

is increasing. L e t us n o w consider t h e minimization problem between t h e points x l = y l = 0 a n d x2=~, Y2=/(~)=~N(~). Since

/(x)

is an absolute minimal, we h a v e

M o = H ( / ) >~F(~,/(~), 1'(~)).

:But

F(x, ~TN, ~7"N)

is increasing, which gives

M o <~ H(~tN ) = F(~, ~IN(~), ~'N(~)) <~ H(/) = M o.

T h u s

H(~tN ) = M o

a n d

~N(X)

is also a minimizing function.

Choose a function r C 1 on 0 ~<x ~<~ such t h a t ~ ( 0 ) = r a n d r 0. F o r m t h e function

g(x)-~N(x)+),~b(x),

where A > 0 is a parameter. W e h a v e

.F(x, g(x), g'(x) ) = F(x, ~N(x), V]'N(X) ) § ~.(a(x)r d-b(x)r ) § R(x, 2.).

H e r e

a(x) = Fy(x, ~N(x), ~7'N(X))

a n d

b(x) = F~(x,

~N(x), ~v(x)) a n d [

R(x, 4) 1 <~ C~ 2

where C is i n d e p e n d e n t of x a n d 2, if 0 < ~ < 1.

Now, recalling the fact t h a t

F(x, ~TN, ~'N)

takes its m a x i m u m o n l y at x = ~ a n d t h a t F~(~, ~N(~), ~TN(~)) > 0 , it is easy to verify t h a t

H(g)

< H ( ~ ) , if A is small enough.

This gives

H(g) < M o

which is a contradiction. This completes the proof.

ARKIV FOR MATEMATIK. B d 6 n r 2 2 2. This section will t r e a t b r i e f l y t h e q u e s t i o n of u n i q u e n e s s for a.s. m i n i m a l s . I t will be s h o w n b y e x a m p l e s t h a t t h e r e m a y be s e v e r a l a.s. m i n i m a l s for a g i v e n mini- m i z a t i o n p r o b l e m . A few sufficient c o n d i t i o n s for u n i q u e n e s s will also be given.

Choose F(x, y, z) = (1 + x ~ + y2)2 § z2, 2

x 1 = - 2 , yl=O, x 2 = 2 a n d y ~ = 0 .

I t follows f r o m T h e o r e m 2.1 t h a t t h e r e is a n a.s. m i n i m a l / ( x ) . Clearly, g(x) = - / ( - x ) is also a n a.s. m i n i m a l . Now, if t h e r e is a u n i q u e a.s. m i n i m a l , t h e n we g e t - ] ( - x ) ](x), i.e. /(0) = 0. C o n s e q u e n t l y M o >~ F ( 0 , 0, 0) = 2.

P u t h ( x ) = 1 2 - x for x~>0,

[2+

x for x ~ 0 . C l e a r l y H(h) = 2 / 9 § 1 < 2.

T h e c o n t r a d i c t i o n shows t h a t t h e r e is no u n i q u e a.s. m i n i m a l . (This s i t u a t i o n c a n b e d e s c r i b e d thus: e v e r y m i n i m i z i n g f u n c t i o n m u s t e v a d e t h e m a x i m u m a t x = y = 0 , w h i c h , owing t o t h e s y m m e t r y , l e a d s t o n o n - u n i q u e n e s s . )

P u t F(x, y, z ) = y 2 + z 2 a n d let /o(X) be d e f i n e d as in E x a m p l e 3 in [1]. W e k n o w t h a t all f u n c t i o n s p/o(x+q)(p, q a r e c o n s t a n t s ) a r e a.s. m i n i m a l s . I t is e v i d e n t t h a t t h e r e a r e i n f i n i t e l y m a n y a.s. m i n i m a l s if Y2 = - Y l 44=0 a n d x 2 - x 1 >~r.

I t follows f r o m t h i s e x a m p l e t h a t t h e r e m a y b e s e v e r a l a.s. m i n i m a l s e v e n if 2'(x, y, z) is c o n v e x in y a n d z. I n t h e calculus of v a r i a t i o n s , t h i s c o n d i t i o n l e a d s t o u n i q u e n e s s (for all b o u n d a r y values). I t follows f r o m t h i s e x a m p l e t h a t t h e c o n d i t i o n (Yl - C) (Y2 - C) ~> 0 in T h e o r e m 3.3 c a n n o t b e o m i t t e d .

These e x a m p l e s s h o w t h a t e x t r a c o n d i t i o n s on F(x, y, z) o r on t h e b o u n d a r y v a l u e s m u s t be a d d e d to o u r u s u a l ones in o r d e r to secure uniqueness. Two w a y s to do t h i s a r e s h o w n b y t h e following t h e o r e m s :

T h e o r e m 3.2. Assume that F(x,y,z) satisfies the conditions o/ Theorem 2.1, with o)(x, y ) ~ 0 . Then, as we know ]rom that theorem, there is at least one a.s. m i n i m a l / o r

every choice o / ( x D Yl) and (x2, y~).

Assume now that

~Y(x, y, O)

0 for all x and y.

~x

Then there is a V•IQVE a.s. minimal/or every choice o/(xl, Yl) and (x2, Y2).

Proo/. W e a s s u m e t h a t Fx(x,y,O ) > 0 . W e m a y also a s s u m e t h a t Yl <Y2 (the case yl>y2 is a n a l o g o u s a n d t h e case Yl=Y2 is t r i v i a l ) . L e t /(x) b e a n a.s. m i n i m a l . I f ]'(x) > 0 for x 1 ~ x <~x2, t h e n t h e a s s e r t i o n follows f r o m t h e T h e o r e m s 6 a n d 9 in [1].

I f / ' ( x ) =0 for some x, t h e n i t follows f r o m p a r t 1 of t h e p r o o f of T h e o r e m 3.1 t h a t t h e r e is a n u m b e r ~, x 1 < ~ ~x2, such t h a t

> 0 for xl<~x<~, /'(x) is = 0 for ~ < x ~ < x ~ .

425

a. ARO~SSO~, Minimization problems. I I

I f g(x) is a different a.s. minimal, then the same reasoning holds for g(x). Let ~' correspond to g(x). Assume t h a t ~' <~. This gives g'(xl) >]'(xl) and

B u t we also have

M1: F(Xl,

Yl, /' (xl) ) < F(xl, Yl, g' (x) ) = M e.

M 1 = F ( ~ , Y2, 0) > F(~', Ye, 0) = M e.

The contradiction proves the theorem.

Theorem 3.3. Assume that F(x,y,z) satis/ies these conditions:

F(x,y,z) is analytic, as in Theorem 3.1;

2 A and 3;

the condition concerning existence and boundedness o/minimizing/unctions, which was given in Theorem 2.1;

t > : i / y > C '

~F(x,y,Z) is = i/ y = C ,

@ [ < 0 i/ y < C , (C is a constant);

F(x, ~Yl + ( 1 - 2 ) y~, 4z 1 + (1-4)z2) ~<~F(x, y~, Zl) + ( 1 - 4 ) F(x, y~, z~)

/or all x, Yl, Y2, zl and z e and/or all 4 such that 0 < 4 < 1 , equality holds i/ and only i/

Yl =Y~ and z I =z2.

We consider the minimization problem between (x 1, Yl) and (x 2, Ye). We assume that the boundary values satis/y the inequality

(Yl - C) (Ye - C) >~ O.

Then there is a VNIQV]~ a.s. minimal between (x 1, Yl) and (xe, Ye).

The proof is omitted, since it is rather laborious and since the result is not used in this paper.

3. I n the previous section, we discussed the uniqueness question for a.s. minimals.

We shall now briefly consider the same question for minimizing functions. The following theorem shows t h a t those functions F(x,y,z), for which every minimization problem has a unique solution, constitute a v e r y "small" class.

Theorem 3.4. Let F(x, y, z) satis/y the/ollowing conditions/or X 1 <~x <~X2 and all y,z:

F(x, y, z) E Ce;

2 A and 3;

the condition concerning existence and boundedness o/ minimizing /unctions which was given in Theorem 2.1.

We consider the minimization problem between (x~, y)and (xe, Y2) (X1 <~xl <x2 <<-X2) 9 Then there is a VNTQV~ minimizing /unction /or every choice o/ xl, Yl, x2 and Y2 i/ and only i/

426

ARKIV FOR MATEMATIK. B d 6 n r 2 2

_F(x, y, O) =-constant.

Proof.

1) S u p p o s e t h a t

F(x,

y, 0) - c o n s t a n t . Consider t h e m i n i m i z a t i o n p r o b l e m b e t w e e n (xl, Yl) a n d (x2, Y2). W e k n o w t h a t t h e r e is a n a.s. m i n i m a l / ( x ) a n d we k n o w t h a t / ( x )

E C 1.

If ] ' ( x ) 4 0

for x 1 ~<x ~<x~ t h e n i t follows f r o m T h e o r e m 8 a n d T h e o r e m 6 in [1] t h a t

](x)

is a u n i q u e m i n i m i z i n g f u n c t i o n .

I f t h e r e is a n ~ such t h a t / ' ( r 1 6 2 = 0 , t h e n

] ' ( x ) - 0 .

I n o r d e r to see t h a t , s u p p o s e t h a t

/'(fl) # 0

for s o m e fi > ~. :Put

= sup {x Ix < ~ , / ' ( x ) = o}.

T h e n we g e t

F(/~, / @ , / ' @ ) > P ( ~ , 1 @ , 0) = F ( ~ , / ( r ) , 0).

B u t this c o n t r a d i c t s t h e r e l a t i o n

F ( x , / , / ' )

= c o n s t a n t w h i c h h o l d s o n e v e r y i n t e r v a l w h e r e

]'(x) #0.

So we h a v e

](x)-yl(=y2).

I f

g(x)

is a n a d m i s s i b l e f u n c t i o n a n d g ' ( ~ ) # 0 for s o m e

~, t h e n we g e t

H(g)

>~ F ( $ , g(~), g'($)) > F(~, g(~), 0) = H ( / ) , a n d we h a v e p r o v e d one h a l f of t h e t h e o r e m .

2) S u p p o s e n o w t h a t t h e r e is a u n i q u e m i n i m i z i n g f u n c t i o n for e v e r y choice of xl, x2, Yl a n d Y2.

I f F=(Xl, Yl, 0) =4=0, t h e n we choose Y2 =Yl a n d x 2 such t h a t Fz(X , Yl, 0) # 0 for

xl<~x<<.x ~.

Clearly,

/ ( x ) - Y l

is a m i n i m i z i n g f u n c t i o n , b u t n o t t h e o n l y one. T h i s p r o v e s t h a t

Fx(x , y, O) - O.

I f

Fy(Xl,

Yx, 0) # 0 , t h e n we choose

Yz=Yl

a n d x 2 such t h a t

Fy(x,

Yl, 0) # 0 for x 1 ~<x ~<x z. W e shall p r o v e t h a t / ( x )

=Yl

is n o t t h e o n l y m i n i m i z i n g f u n c t i o n .

I t is no r e s t r i c t i o n to a s s u m e t h a t x 1 =Yl = 0 a n d Fy(0, 0, 0 ) < 0. Consider t h e func- t i o n y = ax2 w h e r e ~ > 0 is a c o n s t a n t . T a y l o r s f o r m u l a gives

F ( x , ~x 2, 2~x) =

F(x, O, O) + ax~F~(x, O, O) + l(~2xaF~(x, O. ax2, O.

2ax) + 2. ax~ .2~XF~z(...) + 4~x~F~z(...)).

I f we consider a s u i t a b l e n e i g h b o u r h o o d of t h e o r i g i n a n d a s s u m e t h a t 0 < ~ < 1, t h e n F ~ ( x , 0 , 0 ) < - K < 0 , a n d t h e m o d u l u s of t h e e x p r e s s i o n in b r a c k e t s is n o t g r e a t e r t h a n K 1 9 ~2. x 2, w h e r e K a n d K 1 a r e i n d e p e n d e n t of x a n d ~.

This m e a n s t h a t

F(x, ax e,

2ax) ~<

F(x,

0, 0) - K . ~x 2 + K 1 ~2x2 =

F(x,

0, 0) -

ax2(K -

~K1).

I f we choose ~ so s m a l l t h a t K - ~ K 1 > 0, t h e n we h a v e

F(x, ax2,

2~r ~<

F(x,

0, 0) for 0 ~ x < ~ , for s o m e 8 > 0 .

A s i m i l a r c o n s t r u c t i o n c a n be c a r r i e d t h r o u g h for @2, Y2), a n d t h e r e s t of t h e p r o o f is obvious.

Therefore, if F ( x , y, 0) is n o t c o n s t a n t , t h e n we m u s t i m p o s e c o n d i t i o n s on t h e b o u n d a r y v a l u e s in o r d e r t o secure u n i q u e n e s s . T h e following t h e o r e m i l l u s t r a t e s t h i s p o s s i b i l i t y .

427

G. ARO~SSOrr Minimization problems. I I

T h e o r e m 3.5. Assume that F(x, y, z) satisfies these conditions:

F(x, y, z) E C1;

2 A and 3;

the condition concerning existence and boundedness o/ minimizing /unctions which was given in Theorem 2.1;

either ~F(x, y, O) >~ 0 / o r all (x, y) or < 0 / o r all (x, y).

~x

Let us denote the admissible linear/unction by l(x). Finally, we assume that m i n F(x, l(x), l'(x)) > m a x F(x, y, 0),

X I ~ X ~ X ~ ( X , y ) e K

where K is the set I xl ~ x <~ x 2

[

Yl <~ Y <~ Y2 (Yl >t Y >1 Y~)"

Then there is a unique minimizing ]unction/(x). Moreover, ]'(x) # 0 /or x 1 ~x<~x2.

Proo/. W e k n o w t h a t t h e r e is a n a.s. m i n i m a l / ( x ) EC 1. I f / ' ( x ) = 0 for some x, t h e n i t follows f r o m t h e T h e o r e m s 9 a n d 10 (with a t r i v i a l change) in [1] t h a t

M 0 ~ m a x F(x, y, 0).

(X,y) E K

B u t i t was p r o v e d in L e m m a 5 in [1] t h a t M0>~minxl<x<x ~ F(x, l(x), l'(x)). So t h e as- s u m p t i o n t h a t ]'(x)~0 for s o m e x l e a d s t o a c o n t r a d i c t i o n . F i n a l l y , i t follows f r o m T h e o r e m 6 in [1] t h a t ](x) is a u n i q u e m i n i m i z i n g f u n c t i o n .

Remark. I t is e a s y t o f i n d n e w r e s u l t s of t h e s a m e t y p e b y using n e w e s t i m a t e s . I t s h o u l d be m e n t i o n e d h e r e t h a t t h e i n e q u a l i t y

M 0~> inf F(x,g(x),g'(x))

x1~x<~x2

h o l d s for e v e r y a d m i s s i b l e m o n o t o n i c f u n c t i o n g(x)EC 1 ( c o m p a r e L e m m a 5 in [1]).

This gives

M 0 >~ sup [ inf F(x, g(x), g'(x))],

g(x) xl<~x~x~

w h e r e t h e s u p r e m u m is t a k e n o v e r all such f u n c t i o n s .

I t is also clear t h a t t h e c o n d i t i o n on F x in t h e t h e o r e m c a n b e r e p l a c e d b y t h ~ c o n d i t i o n in T h e o r e m 1.1 (in t h i s case a s s u m e d t o h o l d for all y).

Chapter 4. Comments and examples

I n t h i s c h a p t e r , we shall i l l u s t r a t e some of t h e t h e o r e m s b y m e a n s of e x a m p l e s . Example 1. S u p p o s e t h a t

F(x, y, z) = G(x, y) + A y 2 + By + z 2,

w h e r e G(x,y) is c o n t i n u o u s a n d b o u n d e d f r o m b e l o w a n d A , B are c o n s t a n t s .

ARKIV FOR MATEMATIK, B d 6 n r 2 2 Is there a minimizing function? W e shall a p p l y T h e o r e m 1.2'. F o r m a l calculation gives

l O(x, y , M ) = § (M - G ( x , y) - A y e - B y ) ~/2, 9 (x, y, M) = - ( M - G(x, y) - A y 2 - By) 1/2.

I f :r is defined for y

~Yl,

t h e n we see f r o m the expression for O(x,y, M) t h a t a(y,M) =O(y), when y--> + ~ . H e n c e t h e integral 11 in T h e o r e m 1.2' c a n n o t exist finite. I t follows in the same w a y t h a t none of t h e integrals Ie, /3 a n d 14 can exist finite.

So we can conclude t h a t E ( M ) is b o u n d e d for all M. Consequently there is a mini- mizing f u n c t i o n for e v e r y choice of xl, x2, Yl a n d Y2-

Example 2. W e h a v e n o t studied t h e properties of minimizing functions in general, we h a v e o n l y studied the special cases of a.s. minimals a n d unique minimizing func- tions. The reason for this is clear f r o m t h e following example:

Choose F(x, y, y') = x + y,2, xi = Yl = 0, x 2 = 1 a n d Y2 = 0. This gives M o >~ F(x~, y~, O) = 1, a n d if g(x)~-0, t h e n H(g)= 1.

Thus we h a v e M 0 = 1. I t follows t h a t i f / ( x ) is an admissible function, t h e n it is a minimizing f u n c t i o n if a n d only if I/'(x) l ~ V i - x a.e.

Clearly, the fact t h a t / ( x ) is a minimizing function implies v e r y little a b o u t / ( x ) . This m o t i v a t e s the i n t r o d u c t i o n of a.s. minimals.

On the other hand, suppose t h a t we h a v e a minimizing function g(x) which belongs to C 1. T h e n the local v a r i a t i o n m e t h o d , which was used in the end of t h e proof of T h e o r e m 3.1, can be applied to g(x). This m e t h o d works also in t h e general case of variable ~o(x,y). F o r instance, it can be used to derive t h e following result: if

Fz(X, g(x), g'(x)).o

for all x, t h e n F(x, g(x), g'(x)) = M o for all x. (This is m e n t i o n e d in [6] with a s k e t c h of a proof.) This result leads us once again to the differential equation (dF/dx)" Fz = 0 for a.s. minimals.

Example 3. W e h a v e p r o v e d in Chapter 2 t h a t if F(x, y, z) satisfies certain condi- tions, t h e n there is an a.s. minimal for e v e r y choice of the b o u n d a r y values, a n d we h a v e also p r o v e d t h a t every a.s. minimal belongs to C 1.

Consequently, there is a minimizing/unction in C 1.

However, there need n o t exist a minimizing function in C 2, as can be seen f r o m the following example:

P u t F(x, y, y') =y,2_cos2x, xl =Yl =0, x 2 =n7~ a n d Y2 = 2 n . W e assert t h a t

/(x)= [cos t l dt

is the only minimizing function. Clearly,/'(x) >~0 a n d F(x, /(x), /'(x)) - 0 . If H(g) <~0, t h e n it follows t h a t g'(x)~/'(x) a.c. This implies t h a t g(x)~/(x), which proves o u r assertion.

Finally, it is easy to verify t h a t / " ( x ) has a j u m p at x =ze/2 § k.~, k = 0,1,2 ... n - 1.

4 2 9