A R K I V FOR MATEMATIK Band 6 nr 2

A R K I V FOR MATEMATIK Band 6 nr 2

1.64141 Communicated 11 November 1964 by L. GJ~RDING 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 SUpx F(x, f(x),f'(x))

By GU~NAn AaoNsso~

l . Introduction

L e t F(x, y, z) be a g i v e n f u n c t i o n , defined for z 1 ~ x ~ x 2 a n d all values of y a n d z.

L e t f u r t h e r :~ be a class of f u n c t i o n s / ( x ) , all of which are defined o n x 1 ~ x ~ x ~ a n d are sufficiently regular. F o r e v e r y /C :~, we define t h e f u n c t i o n a l

H(/) = sup F(x, /(x), /'(x)).

X l ~ X ~ X z

W e are i n t e r e s t e d i n the p r o b l e m to m i n i m i z e H(/) over :7. F o r example, we will t r y to answer these questions: Does there exist a m i n i m i z i n g f u n c t i o n ? Is it u n i q u e ? H a s it a n y special properties? W h a t is t h e v a l u e of infr~ s H(/)? F o r reasons of b r e v i t y , m a n y of t h e results are not g i v e n i n t h e m o s t general form. W e shall o n l y consider real f u n c t i o n s a n d real variables.

If g(x) is c o n t i n u o u s a n d n o n - n e g a t i v e o n x I ~ x ~x2, t h e n

(? )

m a x g ( x ) = lim (g(x))~dx 1/~.

X l ~ X ~ X ~ Tt"-)'Or X l

This suggests t h a t we should a p p r o x i m a t e t h e f u n c t i o n a l H(/) w i t h the sequence of f u n c t i o n a l s

(? )

H~(I) = [F(x,/(x),/'(x))]ndx ~/~, n = 1, 2, 3 , . . . .

X~

T h e E u l e r e q u a t i o n corresponding to Hn(/) = rain is

)

dx [F(x, y, y')~] - ~yy [F(x, y, y')~] = 0, which c a n be w r i t t e n as

n ( n - 1 ) F n - 2 [ d x Y' dx

L e t us p u t t h e expression i n b r a c k e t s e q u a l to zero a n d t h e n let n t e n d to i n f i n i t y .

c. ARONSSON,

Minimization problems

T h e n we get (formally) a new e q u a t i o n

d ! !

_ _ o

dx (F(x, y, y )) F~.(x, y, y ) = O. (*)

W e w a n t to s t u d y the connection (if there is any) between this differential equation a n d the minimization problem.

I t m i g h t be expected t h a t functions

/(x),

such t h a t

F(x, /(x), /'@))-constant,

should be i m p o r t a n t for the problem. This is true, as we shall see in sections 2 a n d 3.

I n section 4, we shall i n t r o d u c e a class of functions which minimize the functional

"on

every i n t e r v a l " a n d prove t h a t such a function m u s t satisfy the e q u a t i o n (*) in a certain sense.

A similar problem has been studied in [1].

2. T h e special case

F = F ( y , y ')

W e shall start with a s t u d y of the case where F is i n d e p e n d e n t of

x,

since this case is simpler t h a n the general a n d since we are interested in the interaction between y a n d y'.

2 A. The minimization problem

L e m m a 1:

Suppose that:

1)

/(x) is continuous/or a ~ x ~ b , /'(x) is continuous/or a < x < b,

2) / ' ( x ) > 0 / o r

a < x < b ,

3)

g(x) is absolutely continuous on a ~ x ~ b,

4)

g(a) </(a), g(b) ~/(b) a n d / $ g .

Then there exist numbers tl, t 2 on the open interval (a, b) such that

l)

/(tl)=g(t2) ,

I I )

g'(t~) exists and/'(tl)

< g ' ( 4 ) .

Proo/:

Clearly, we m a y assume t h a t

g(a) - / ( a ) , g(b) - / ( b )

a n d / ( a ) <g(x)

</(b)

for a < x < b. For, if this is n o t the case, we define

p - m a x {x [

g(x) </(a) },

q - m i n

{x I x >~p, g(x) >~ ](b)},

and, instead of g(x), we consider

gl(X)=g(p+(q-p)(x-a) [ b - a ]

1). (If

gl=/,

t h e n the result is trivial; if gl ~ / , t h e n the proof below applies to

gl,

a n d t h e n the result for g follows.) N o w

y - / ( x )

has a continuous inverse

x - ~ ( y )

f o r / ( a ) ~ y ~ / ( b ) .

~'(y)

is continuous a n d positive for

/(a)<y</(b).

Hence ~(y) is absolutely continuous on

](a) <~y <~/(b).

F o r m the function

q~(x)-at(g(x)).

I t is absolutely continuous on

a ~ x ~ b

a n d

q/(x)=ot'(g(x)).g'(x)

a.e. B y assumption, there exists a n u m b e r

xo, a<xo<b,

such

ARKIV FOR MATEMATIK. B d 6 n r 2 that/(Xo) ~:g(x0); say/(xo)

<g(xo)

(the other case is t r e a t e d similarly). T h e n we h a v e

~(x0) > x 0 a n d

floq~'(x)dx=q~(Xo)-q~(a)=q~(Xo)-a>xo-a.

So there m u s t exist a n u m b e r ~ such t h a t :

a<~<xo,

g'(~), ~'(~) exist a n d ~ ' ( ~ ) > 1.

T h u s B u t

~'(g(~)).g'(~)>l.

~'(g(~))=/,[~(g(~))], 1 which gives us

g'(~) >/'[~(g(~))].

Obviously, the n u m b e r s t 1 = a(g(~)) a n d t~ = ~ will h a v e the required properties.

Remark:

If we change the conditions 2 a n d 4 t o / ' ( x ) < 0 a n d

g(a) >/(a), g(b) ~/(b),

respectively, a n d the assertion I I to

g'(ta)</'(tl)

^, t h e n we get a n o t h e r f o r m of t h e l e m m a which follows f r o m the preceding b y the substitution t = - x .

N o w let us consider a function F =

F(y, z)

a n d let us impose a few conditions u p o n it:

1) F is defined a n d continuous for all y a n d z.

2) exists for all y a n d z a n d ~ is = 0 if z = 0

< 0 if z < 0 .

Let [xDx2]

be the interval m e n t i o n e d in the i n t r o d u c t i o n a n d let

Yl,Y~

be a n y t w o numbers. F r o m n o w on, the class ~ of admissible funetions is defined as follows:

~ is the class of all absolutely continuous functions o n X l ~ Z ~.x2, which satisfy t h e b o u n d a r y c o n d i t i o n s / ( x l ) - Y l a n d / ( x 2 )

=Y2.

L e t / ( x )

E ~-

a n d let E be the set where

['(x)

exists. I t should be noticed t h a t x 1 a n d x~ belong to E if the one-sided derivatives in question exist.

N o w

H ( / ) =

SUpx~E

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

is well-defined a n d obviously

H(/) >~

m a x

(F(y 1, 0), F(y 2, 0)).

Therefore infr~ ~

H(/)

is finite, a n d the questions m e n t i o n e d in the i n t r o d u c t i o n are meaningful.

W i t h the use of L e m m a 1, we can easily prove t h e following simple theorem:

Theorem h

Suppose that F(y,z) satisfies the conditions 1 and 2 stated above. Suppose /urther that/(x) is an admissible/unction such that

a)

/'(x) is continuous/or x 1 ~ x ~x2,

b)

/'(x):#O /or xl <x<x2,

c)

F(/(x), /'(x)) = M /or x 1 ~ x ~ x 2 (M is any constant).

Then/(x) is a unique minimizing/unction in :~. I.e.: i/g(x) is a di//erent element o/

5, then H(g) > H(/).

G. ARONSSON, Minimization problems

Proo/: Consider t h e case /'(x)>0. A c c o r d i n g to L e m m a 1 t h e r e e x i s t t D t 2 such t h a t / ( t ~ ) =g(t2), 0

<ff(tl)<g'(t2).

This gives

H e n c e

H(g) >~ F(g(t2) , g'(t2) ) > F(/(t~),/'(t:)) = M = H ( / ) . H(g) > H(I).

T h e o t h e r case is t r e a t e d a n a l o g o u s l y (see t h e r e m a r k to L e m m a 1). This t h e o r e m s h o u l d b e c o m p a r e d w i t h t h e t h e o r e m s of s e c t i o n 3.

I n o r d e r to give a m o r e s y s t e m a t i c t r e a t m e n t of t h e case F = F(y,z), we i n t r o d u c e a n o t h e r c o n d i t i o n on F :

3) lira F ( y , z ) - + c~ for all y.

As is e a s i l y seen (using t h e c o n d i t i o n s 1 a n d 2 also), t h i s i m p l i e s t h a t t h e l i m i t s l i m F ( y , z ) = + ~ , l i m F ( y , z ) = + c~

Z - - - ) + o r Z - - - ) -

a r e u n i f o r m for b o u n d e d y.

I n t h e r e s t of section 2, we shall a l w a y s a s s u m e t h a t F(y, z) satisfies t h e c o n d i t i o n s 1, 2 a n d 3 g i v e n a b o v e .

W e n o w i n t r o d u c e t w o a u x i l i a r y functions:

De/inition: I / F ( y , O ) < M , then we set OM(y)=the positive number z such that F ( y , z ) - M , tFM(y ) - t h e negative number z such that J T ( y , z ) - M , and i/ F ( y , O ) = M then we set r (If F ( y , O ) > M , t h e n t h e e q u a t i o n F ( y , z ) = M h a s no s o l u t i o n z.)

L e m m a 2: OM(y) and t F u ( y ) are continuous /unctions o/ y and M on the set where they are de/ined.

T h e p r o o f is s i m p l e a n d we o m i t it.

W e will n o w t r y to give a c o m p l e t e s o l u t i o n of t h e m i n i m i z a t i o n p r o b l e m u n d e r t h e a s s m n p t i o n s 1, 2 a n d 3 a b o u t F(y,z). W e shall p a y m o s t a t t e n t i o n t o t h e case Yl <Y2 a n d t h e c o r r e s p o n d i n g r e s u l t s for t h e case Yl > Y2 will be g i v e n l a t e r w i t h o u t proofs, since t h e r e a s o n i n g is v e r y s i m i l a r in b o t h cases. F i n a l l y , we shall consider t h e case Yl = Y~-

A) L e t us n o w s u p p o s e Yl < Y2- (v2 (it

I n t e g r a l s of t h e t y p e .]vl(I)M(t) t u r n o u t to b e v e r y useful. L e t us a g r e e to call t h e i n t e g r a l a b o v e w e l l - d e f i n e d if a n d o n l y if (PM(t)> 0 a.e. on Yl < t ~<Y2" T h e n (])~(t) is n o n - n e g a t i v e , m e a s u r a b l e a n d f i n i t e a.e. L e t us use t h e n o t a t i o n 1

f

^{Y~ dt} y~ d)M(t~ = s

T h u s s > 0 a l w a y s a n d t h e p o s s i b i l i t y s = + ~ is n o t e x c l u d e d .

ARKIV FOR MATEMATIK. B d 6 nr 2

L e m m a 3:

Assume that M~-+M, MI>M2>Ma>... and that, /or all

v, s

is well-de/ined and

s

<~ C (C independent o/ v). Then

s

is also well-de/ined and

l : ( M ) - l i m ~ _ ~ ~(M~).

Proo/:

I f

yl<~y<~y.~

then, clearly,

F(y,O)<M,

for all v a n d

F(y,O)<~M.

This means t h a t

OPM(y )

is defined for

yl<~y-<,y2.

I f

(I)M(y)>O,

then, according to L e m m a 2,

1 1

CPM,(y) OM(y)'

a n d if OM(y)= 0, t h e n ^{- - -1 - - - ~ C x 3 .} (D.~(y)

F o r e v e r y y we have ^{1 1 1}

OMI(Y) OM~(y) r ....

B u t it is also t r u e t h a t

r',~ dy

- - < ~ C

J ~, q%~(y)

I t follows f r o m B e p p o - L e v i s t h e o r e m t h a t lira 1

~-~ (P~.(y) Y~ ~< Y ~< Y2 a n d t h a t

-h(y)

exists finite a.e. on

fY~ h(y) dy

= lim d y , O ~ ( y ) - - lim s

(Y~ dy

Y l v - - > ~ v - - - > o o

F r o m the preceding we see t h a t h ( y ) =

OM(Y)

1 a.e., which completes the proof.

Theorem 2:

I/there exists an admissible/unction/(x) such that

supx

F(/(x), /' (x) ) 4 M, then ~(~I) is well-de/ined and I=(M)4x2-x 1.

Proo/:

^{I f}

M ~ = M + l / v

^for

v = l ,

2, 3 . . . t h e n

H(I)<M~.

Hence, if

y l 4 y 4 y 2 , F(y,O) <M~

a n d

(~Mv(y)

> 0 .

We m a y assume t h a t

yl~/(x)~y2.

T a k e an a r b i t r a r y u. F r o m the inequality

F(/(x),/'(x)) <M,

it follows t h a t / ' ( x )

<O~,(/(x))

(a.e.). Therefore

1'(~)

b,,~(l(x)) < ]" (])

l "y dt

N o w ~ ( y ) = | - - is a c o n t i n u o u s l y differentiable function of y for Yl ~< Y < Y2.

Jy, O~,~(t)

T h u s ~ ( x ) = ~(/(x)) is absolutely c o n t i n u o u s on x 1 ~<x ~ x 2. I t follows f r o m (1) t h a t

~F'(x) 4 1 a.e.

So ~F(x~) -~F(xl) ~x~-Xl.

G. ARO~SSOrr Minimization problems

:But ~ ( x l ) =(~(Yl) = 0 a n d

~F(x2)

=(P(Y2) =

I~(M~). This gives

s < x2 - x r Now the assertion follows from the preceding lemma.

Theorem 3: I / ~(M) is well-de/ined and /inite, then there is a /unction /(x) with these properties:

1) /(x) is de/ined and strictly increasing/or 0 ~ x ~ F.(M), 2) / ( 0 ) - Y D / ( s =Y2,

3) /'(x) is continuous/or 0 ~ x ~ IZ(M),

4) F(/(x), ] ' ( x ) ) - M /or O<~x<F.(M) (i.e./'(x)=(I)M[/(x)]).

Proo/: Form the function el(y) = ~

dt/(~M(t ).

Clearly, it is continuous and strictly increasing for Yl ~< Y ~< Y,. Further, ~(Yl)= 0 and ~(Y2)= E(M). If we define/(x) as the inverse function of ~(y), then it follows at once t h a t / ( x ) satisfies 1 and 2. I n order to s t u d y / ' ( x ) , take an x 0 on the interval 0~<x0~< I~(M) and let {~}~ be a sequence such that ~--~x o and ~ # x o for all v. S e t / ( ~ ) = ~ and/(Xo) =Yo (~-->Yo but ~:4=yo).

Then we have

/ ( ~ ) - / ( X o ) _ ~ - Y o _ 1 _

~ - x o ~V(~]~) - ~o(y o) ~ ( ~ v ) - ~O(Yo)

~v -- YO

l ('~ dt "

~ - YoJy. O~(t) I t follows from the continuity of

(I)M(t)

that if

(~)M(Yo)>0,

then

~(~v) - ~ ( Y o ! _ _ > 1 _ 1

~ v - y o O M ( Y o ) OM(/(Xo))' b u t if

OM(Yo): O,

^{then ~(~) - qO(yo) _+ ~ .}

~Tv - Yo Hence

in both cases, and we have

/(~)-/(Xo)

~r )

~ v - - X 0

l'(xo)

=r B u t then it follows t h a t / ' ( x ) is continuous and

F(/(x), / ' ( x ) ) = M for O < x < E ( M ) . This completes the proof.

The solution of the minimization problem is given in the following Theorem 4:

a) Given M, a necessary and su//ieient condition/or the existence o/ a /unction /E :~

such that H(/) = M , is that ~(M) •x 2 - x l ;

b) there is a number M o such that F.(M) <~x 2 - x 1 i / a n d only i / M >~ Mo;

c) M0-infr~ ~ H(/);

ARKIV FOR MATEMATIK. B d 6 n r 2 d)

there exists a minimizing/unction/o (which can be chosen continuously di//erentiable);

e) the minimizing/unction is unique i] and only i / s 2 - x l ;

f) i] the minimizing/unction is not unique, then

M0=maxy,<y<y ~ F(y,0),

but the con- verse is not true.

Proo/:

a) This is seen from Theorems 2 and 3. I n general, the function in Theorem 3 must be translated in the x-direction and continued as a constant to give the function mentioned above.

b) I t follows from the properties of

F(y,z)

and the definition of

OM(Y)

that

OM(y)

is an increasing function of M and hence s is decreasing. Let E be the set of all numbers M such that s ~ x 2 - x 1. I t is clear that E is bounded from below. Let M 0 = inf E. Lemma 3 implies that M 0 E E and, since s is decreasing, the assertion follows.

c) and d) Consequences of a), b) and Theorem 3.

e) Suppose first t h a t s 2 - x 1. I t is clear t h a t the function in Theorem 3 can be translated in the x-direction and continued as a constant in different ways so as to give us different minimizing functions.

Suppose then t h a t I:(M0)=x,~-x

1.

If

/(x)

is the function from Theorem 3 with

M = M o ,

then we assert that

h ( x ) = / ( x - x 0

is the only minimizing function. Let

g(x) E ~

and assume for example g(~)<h(~) for some ~ between x 1 and x~. According to our choice of

h(x)

we have

f ^{Y~ h(~) OM~ dt} x 2 - ~"

If

H ( g ) = M ,

then, according to Theorem 2,

f ~ dt

I t follows t h a t

Y~ dt ~Y~ dt

h(~) OM(t) < j h(~) O~o(t)"

This implies that M >M0, which completes the proof of assertion e).

f) First we shall prove that if the minimizing function is not unique, i.e. s

x 2 - x l ,

then M0=maxyl<u<y ~

F(y,O).

Assume then that M0>maxyl<~<y ~

F(y,O).

This means that

(YPMo(Y)>0

^:[or

yl<~y<~ya.

Since

F(y,z)

and

(PM(Y)

are continuous functions, there must be a number 5 > 0 such t h a t

1/Oi(y)

is uniformly continuous for Yl ~< Y ~ Y2 and I M - M 0 1 ~<5. But then there also must exist a number M 1 < M 0 such t h a t

s

1. But this contradicts b).

Hence

M 0 = max

F(y,O).

(1)

Yl<~Y~Y~

I n order to show that the converse assertion is not true, we give an example where (1) holds and the minimizing function is unique.

Choose

F ( y , z ) = y 2 + z

e , x l = y l = 0 , x 2 = ~ and y ~ = l .

a . AROr~SSON,

Minimization problems

P e r h a p s t h e easiest w a y to see t h a t y - s i n x is t h e u n i q u e m i n i m i z i n g f u n c t i o n is t o a p p l y T h e o r e m 1. (But, of course, t h e s a m e conclusion can be d r a w n w i t h t h e a i d of e) above.)

C l e a r l y

M 0 = 1 - m a x F(y, 0).

0~<y~<l

This c o m p l e t e s t h e p r o o f of T h e o r e m 4.

B) yl>y2.

T h e i n t e g r a l s is n o w r e p l a c e d b y

s ff I

^at

(=(~'dt~

a n d o u r c o n v e n t i o n s are c o r r e s p o n d i n g to t h o s e of t h e f o r m e r case. Thus, for e x a m p l e , 0 < s + ~ .

T h e e x a c t a n a l o g u e s of L e m m a 3 a n d T h e o r e m s 2, 3 a n d 4 are n o w o b t a i n e d in a v e r y n a t u r a l m a n n e r b y s u b s t i t u t i n g s for C(M) a n d (in T h e o r e m 3)

~FM(y )

for

(~PM(Y)"

I n T h e o r e m 3, we m u s t also c h a n g e t h e w o r d i n c r e a s i n g i n t o decreasing. T h e proofs are p r a c t i c a l l y t h e s a m e in b o t h cases.

C) Yl=Y2.

Clearly, t h e f u n c t i o n ](x)=Yl is a m i n i m i z i n g f u n c t i o n . L e t us w r i t e M o - F ( y 1, 0).

As r e g a r d s t h e q u e s t i o n of u n i q u e n e s s , it follows f r o m T h e o r e m 4 a n d its a n a l o g u e in case B t h a t t h e m i n i m i z i n g f u n c t i o n is u n i q u e if a n d o n l y if none of t h e c o n d i t i o n s

a n d fi b e l o w is satisfied:

~) b o t h i n t e g r a l s a n d dy __

~ y , r ~ , --'FM0(y)

b o t h i n t e g r a l s (Y' d y a n d ( y l d?]

P)

e x i s t finite for s o m e ~ > O;

e x i s t f i n i t e for s o m e (5 > 0.

2 B. Determination of the attainable cone

L e t us now l e a v e t h e m i n i m i z a t i o n p r o b l e m a n d consider a n o t h e r question.

S u p p o s e t h e r e are g i v e n a p o i n t (x 0, Y0) a n d a n u m b e r M >~ F ( y o, 0). D e n o t e b y E t h e set of all p o i n t s (x, y) such t h a t

a) x>~Xo,

b) t h e r e is a n a b s o l u t e l y c o n t i n u o u s f u n c t i o n /(t), j o i n i n g t h e p o i n t s (Xo, Yo) a n d (x,y), such t h a t supt F(/(t), /'(t)) <~M.

Our t a s k is to d e t e r m i n e t h e set E. As a c o n v e n i e n t n a m e for t h e set E we use

" t h e a t t a i n a b l e cone". I f x >~x 0 t h e n c l e a r l y (x, y0)E E. I f x > x 0 a n d y >Y0 t h e n we k n o w a l r e a d y t h a t ( x , y ) E E if a n d o n l y if ,~odt/(~M(t ) is w e l l - d e f i n e d a n d ~<x x 0.

Of course, t h e r e is a s i m i l a r c o n d i t i o n for t h e case y < Y0. I f (x, Yl)E E a n d (x, y2)E E, t h e n (x, y) C E for e v e r y y b e t w e e n Yl a n d Y2.

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

W e define

g(x)

= inf {y:

(x, y) C E}

a n d

h(x)

= sup {y: (x, y) C E}. The f u n c t i o n s

g(x)

a n d

h(x)

m a y t a k e on infinite values. I f

g(x)

is finite, t h e n

(x,g(x))EE

a n d t h e s a m e is t r u e for

h(x).

This follows f r o m t h e c r i t e r i o n a b o v e .

F o r t h e d e t e r m i n a t i o n of E , i t is t h u s sufficient t o d e t e r m i n e

g(x)

a n d

h(x),

a n d we shall confine our discussion to

h(x).

L e t us use t h e n o t a t i o n ~(y) = x 0 + S~0

dt/OgM(t)"

T h e d i v i s i o n i n t o v a r i o u s cases b e l o w a n d t h e f a c t s a b o u t

h(x)

follow e a s i l y f r o m t h e c o n d i t i o n ~(y)~<x a n d t h e p r o p e r t i e s of ~(y).

A) F o r e v e r y

Y>Yo, cp(y)

is infinite or n o t defined. T h e n

h(x)=Y0

for all

x>~x o.

B) ~(y) is d e f i n e d a n d f i n i t e for

yo<~y< Y < o~

(clearly ~ ( Y ) = oo).

h(x)

= i n v e r s e of q0(y) for all

x>~x o

lira

h(x)= Y.

x--> or

C) ~(g) d e f i n e d a n d f i n i t e for Y0 ~<Y ~< Y < oo

i n v e r s e of ~v(y) for

Xo<~X<~v(Y ), h(x)=

Y for x~>~v(Y).

D) ~(g) d e f i n e d a n d f i n i t e for all

Y>~Yo"

1. l i m ~v(y)=

y--->~

h ( x ) = i n v e r s e of ~v(y) for all

x>~x o

lira

h(x)= c~.

x--->oo

2. lira ~v(y) = X <

y---> ~

i n v e r s e of ~v(y) for x 0 ~<x < X

h ( x ) =

§ o~ for

x>~X.

Clearly,

h(x)

is s t r i c t l y i n c r e a s i n g on e v e r y i n t e r v a l w h e r e i t is d e f i n e d as t h e i n v e r s e of ~v(y).

F u r t h e r , t h e r e l a t i o n

F(h(x), h'(x))=M

holds a t e v e r y p o i n t w h e r e

h(x)

is finite.

3. S o m e s u f f i c i e n t c o n d i t i o n s for a g i v e n f u n c t i o n to m i n i m i z e t h e f u n c t i o n a l i n t h e case F =

F(x, y, z)

L e t us n o w r e t u r n to t h e g e n e r a l case w h e r e F is allowed to d e p e n d on x also.

As we h a v e seen in section 2, m o n o t o n i c f u n c t i o n s / ( x ) , such t h a t

F(x, [(x), ['(x)) -

c o n s t a n t , a r e of g r e a t i m p o r t a n c e . S u c h a f u n c t i o n m i n i m i z e s H([), as is seen f r o m T h e o r e m 5. R o u g h l y s p e a k i n g , T h e o r e m 6 shows t h a t if

](x)

is also s t r i c t l y m o n o t o n i c , t h e n

](x)

is t h e u n i q u e m i n i m i z i n g f u n c t i o n . C o m p a r e also T h e o r e m 1.

Of course, i t can be d e d u c e d f r o m t h e c o n d i t i o n

F(x,/(x), ]'(x))=constant

t h a t

[(x)

has a c e r t a i n degree of r e g u l a r i t y (at least for t h e f u n c t i o n s

F(x,y,z)

t h a t we s t u d y ) . B u t since t h i s is e a s y to do a n d n o t v e r y i m p o r t a n t for t h e t h e o r e m s , we

G. AnOr~SSON,

Minimization problems

shall o m i t a discussion of t h i s a n d i n s t e a d choose t h e c o n d i t i o n s o n / such as to m a k e t h e proofs simple. P e r h a p s we shall r e t u r n to t h i s q u e s t i o n in a l a t e r connection.

N o w l e t us a s s u m e t h a t t h e f u n c t i o n

F(x,y,z)

satisfies t h e c o n d i t i o n s : 1) F is c o n t i n u o u s for

xl<~x<~x ~

a n d all y a n d z,

2)

~F/~z

e x i s t s for

xl~x<~x 2

a n d all y a n d z.

> 0 if z > 0 ,

F u r t h e r , ~F__ is = 0 if z = 0 ,

~z < 0 if z < 0.

W e shall use t h e s a m e n o t a t i o n s as in s e c t i o n 2 a n d t h e s a m e d e f i n i t i o n s of a d m i s - sible f u n c t i o n s a n d t h e f u n c t i o n a l . I t follows t h a t

H(9)

~>max

(F(xl,yl,0), F(x2,y2,0))

for a n y g E 5.

T h e o r e m 5:

Suppose that F satisfies the conditions 1 and 2 above. Let / be an admis- sible/unction such that:

a)

/'(x) exists/or x I ~X~'~X2,

b)

/(x) is monotonic (not necessarily strictly),

c)

F(x, /(x), / ' ( x ) ) - i /or xl <~x<~x 2.

Then/(x) is a minimizing/unction. (/(x)

n e e d

not

be t h e o n l y one.)

Proo/:

L e t us choose t h e ease where

/(x)

is n o n - d e c r e a s i n g . L e t h be a d i f f e r e n t e l e m e n t of :~. T h e n we h a v e , for e x a m p l e , h ( ~ ) < / ( ~ ) for s o m e ~:, x 1 < ~ <x~.

T h e n t h e r e m u s t e x i s t a n x 0 ~ x 2 such t h a t

h(xo)=/(x0)

b u t

h(x)</(x)

for ~ ~ x < x 0.

H e n c e lira

h' (x) >~ /' (Xo) >~ O.

X--~X0

I f / ' ( X o ) - 0 , t h e n i t follows a t once t h a t

l i m

F(x,

h(x),

h'(x)) >~

F ( x 0,/(Xo),/'(x0)) = M . (1)

X'-~Xo

If/'(Xo) > 0 , t h e n we t a k e a ~ > 0 a n d a sequence ~ - - ~ x 0 such t h a t h ' ( ~ ) >/'(Xo) - ~ > 0 . This m e a n s t h a t

F ( ~ , h(~), h ' ( ~ ) ) > F(~,,, h ( ~ ) , / ' ( x o ) - ~ ) . T h e r i g h t m e m b e r t e n d s to

F(xo, /(Xo), /'(Xo)-~).

I f we l e t (5-~0, t h e n (1) follows again. T h u s

H(h) >~ F(x o,/(Xo),/'(xo) ) = i =H(/).

This c o m p l e t e s t h e proof.

T h e o r e m 6:

Suppose that F(x,y, z) satisfies the conditions:

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

_ _ _ _ Ix1 ~< x ~< x 2 ,

A)

F, ~F~y, ~F~z ezist and are continuous/or [y, z arbitrary.

~F [ > 0 if z > 0 , B) ~ z

is l = 0

if z = 0 ,

< 0 if z < 0 .

Suppose/urther that/(x) is an admissible/unction such that:

a ) / ' ( x )

is continuous and ~ 0/or x 1 <~ x <~ x2,

b)

F(x, /(x), /'(x)) = M /or xl <-<x<x2.

Then/(x) is a unique minimizing/unction in 5.

(Compare T h e o r e m 1.)

Proo/:

Following our habit, w e will c a r r y o u t the proof only for the case /' > 0 . Assume n o w t h a t

gE :~, g~-/and H(g)<~M.

We w a n t to derive a contradiction f r o m this.

L e t g(~)>/(2) for some 2, X l < ~ < X 2 . T h e n there m u s t be a n u m b e r x o such t h a t x 1 ~< x o < ~ a n d

g(x) > ](x)

for x o < x ~ ~ b u t

g(xo) =/(xo).

Hence lim

g'(x) ~ / t

(Xo).

X'-~X.+{}

:But since

F ( x , g , g ' ) < M

^{a n d}

/'(xo)>0

it follows w i t h the use of B ) t h a t lim g (x)-~/(Xo).

x.-.).x o

' X ' X

C o n s e q u e n t l y lira g ( ) = / ( o ) .

x--~xo

F o r m ~v(x) =g(x)

-/(x).

If g'(x) exists, t h e n we a p p l y the m e a n value t h e o r e m to the function

9 (t) = F(x,/(x) +t~(x),/'(x) +tqJ'(x))

between t = 0 a n d t = 1. This gives

~(x). F~(x, /(x) +Oq~(x), /'(x) +Oq)'(x)) +~'(x) Fz(...)

40,

i.e.

q)'(x) Fz(... ) <~ -q)(x) Fy(...).

(1)

L e t 5 be a n u m b e r > 0 such t h a t x o + 5 ~< ~, a n d write Ja =

[Xo,

Xo + b]. T h e n ~0(x)~> 0 for

xEJ~.

F o r m the function

C(~) = sup ~'(x).

xeJ6

Clearly O<~(x)<(~C((~) for

xEJ(~.

(2)

P u t (5~ =

1/v

a n d select, for every

v >~Vo, x~ EJ~

such t h a t

~'(x~) > (1 - ~ ) C(5~) > 0 .

g'(x~) >/'(x~)

^{implies lim}

g'(x~) >~/'(Xo).

v--> r162

G. AnoNssor%

Minimization problems

B u t lira

g'(x) =/'(Xo).

x.-.~xo

H e n c e lira g' (x~) = / ' (x0).

v---> o r

P u t x = x , in (1).

/'(x~)</'(x~)+O~'(x,)<g(x~).

~.~e see t h a t

/'(x,.) + 0 ~ (x,)

^{P _ _ ~}

/ (x0) >0.

^P

F r o m t h e c o n d i t i o n s on F i t follows t h a t

F,(x~, /(x~) +O~(x~), /'(x~) +O~'(x~)) ~ o: > 0

and

IF~(...)]<K for v~v~.

H e r e ~ a n d K are c o n s t a n t s i n d e p e n d e n t of v.

(1) can n o w be w r i t t e n (for r>~ra)

This i m p l i e s t h a t

for v ~ r 1

L e f t m e m b e r R i g h t m e m b e r

H e n c e a n d we h a v e

B u t t h i s gives a c o n t r a d i c t i o n if v - - > ~ .

T h e case g(~) </(~) can be t r e a t e d a n a l o g o u s l y , a n d so t h e p r o o f is complete.

0 <~'(x~) Fz(...) < ~(x~) F,(...).

[~'(x,)[ [F~(...)[ <]~(x,)l fF,(...)l,

>(1 c~,) C(d~) ~.

< ~ C ( ~ ) K (see (2)).

(1 -,~) C((~)~ <~,C(~) K

4. Examination o f functions which minimize the functional on every interval I n t h i s section we shall e x a m i n e m o r e closely t h e c o n n e c t i o n b e t w e e n t h e mini- m i z a t i o n p r o b l e m a n d t h e d i f f e r e n t i a l e q u a t i o n

dF(x,/(x),/'(x)) F~(x,/(x), fl(x)) = 0 dx

d e r i v e d in t h e i n t r o d u c t i o n . T h e m a i n r e s u l t s on this s u b j e c t are t h e T h e o r e m s 8 a n d 9.( 1 )

L e t us first s t a t e t h e c o n d i t i o n s on

F(x, y, z).

W e shall a s s u m e t h a t t h e c o n d i t i o n s 1 a n d 2, g i v e n in section 3 (before T h e o r e m 5) h o l d t h r o u g h o u t section 4, b u t we m u s t r e p l a c e [Xl,Xe] b y t h e i n t e r v a l c o n s i d e r e d in each case. L a t e r on, we shall i m p o s e f u r t h e r c o n d i t i o n s on F .

S u p p o s e t h a t t h e f u n c t i o n / ( x ) is d e f i n e d on t h e i n t e r v a l I a n d let

~ x < ~ f i

be a c o m p a c t s u b i n t e r v a l of I .

(1) See also the Theorems 5 and 6 of section 3.

ARKIV FOR MATEMATIK. B d 6 n r 2

Definition: I f i t is true, for e v e r y such i n t e r v a l [a,fl] (inclusive of I if I is c o m p a c t ) t h a t /(x) is a m i n i m i z i n g f u n c t i o n for o~<~x<~fl a n d a s s i g n e d b o u n d a r y v a l u e s /(a) a n d /(fi), t h e n /(x) is s a i d to m i n i m i z e t h e f u n c t i o n a l in the absolute sense on t h e i n t e r v a l I . T h e f u n c t i o n / ( x ) is s a i d t o be a minimal in the absolute sense on I . This will be a b b r e v i a t e d a.s. minimal in t h e sequel.

L e m m a 4: Let I be a compact interval and suppose that/(x) is absolutely continuous on I. Denote by E the subset of I where/'(x) exists, including endpoints o / 1 i / t h e ap- propriate one-sided derivatives exist. Then

s u p F ( x , /(z), /'(x)) = e s s s u p r ( x , / ( x ) , / ' ( x ) )

x~E xEE

in the sense that i / o n e member is finite, then so is the other and they are equal.

Proof: L e t x 0 be a p o i n t of 1 such t h a t / ' ( x o ) e x i s t s a n d let I s be a s u b i n t e r v a l of I c o n t a i n i n g x 0. T h e n i t is true, for e v e r y (~>0, t h a t / ' ( x ) > / ' ( x 0 ) - c ~ on a s u b s e t of 11 of p o s i t i v e m e a s u r e . S i m i l a r l y , /'(x)<f'(Xo)+(5 h o l d s on a s u b s e t of 11 of p o s i t i v e m e a s u r e .

F r o m this, a n d f r o m t h e c o n d i t i o n s on F ( x , y , z), i t follows e a s i l y t h a t e s s s u p

F(x,/(x),/'(x)) >~ F(x o,/(x o)/'(Xo)),

XE[ l fl E

a n d t h e r e s t of t h e p r o o f is obvious.

Remark: I t follows f r o m t h i s l e m m a t h a t our p r e v i o u s d e f i n i t i o n of H(f) is equi- v a l e n t to t h e d e f i n i t i o n

H(/)

= e s s s u p F ( x , / ( x ) ,

l'(x)).

xEE

I t also follows t h a t , in t h e d e f i n i t i o n of H(/), we can e x c l u d e t h e e n d p 0 i n t s of I f r o m E w i t h o u t c h a n g i n g t h e f u n c t i o n a l in a n y w a y . Therefore, t h e c o n d i t i o n s a) a n d c) in T h e o r e m 1 m a y be w e a k e n e d i n t o

a ' ) /'(x) is c o n t i n u o u s for x 1 < x < x 2 a n d

C')

F ( f ( x ) , / ' ( x ) ) - - M f o r x i < x < x 2.

L e t us n o w consider t h e m i n i m i z a t i o n p r o b l e m on t h e i n t e r v a l x 1 ~ x ~ x 2 w i t h t h e b o u n d a r y v a l u e s Yl a n d Y2, r e s p e c t i v e l y . L e t us use t h e n o t a t i o n

M (xl, x2; Yl, Y2) = inf H (/).

W e shall n e e d t h e following e s t i m a t e s :

L e m m a 5: Let L be the straight line between (xl, Yl) and (x 2, Y2).

P u t t = y ~ - y l

Z 2 - - X 1

Then r a i n F(x, y, t) <~ M ( x 1, x~; Yl, Y2) <~ m a x F(x, y, t).

(x,y)eL (X,y)eL

a . A R O ~ S S O N , Minimization problems

Proo/: The right inequality is obvious. L e t l(x) be the admissible linear function a n d let re(x) be a different admissible function. N o w the left inequality is p r o v e d in the same w a y as T h e o r e m 5. W e only have to substitute l(x) f o r / ( x ) a n d re(x) for h(x). This completes the proof.

L e m m a 6: Suppose that x~-->Xo, ~=-->Xo, x ~ < ~ /or all n, yn-~yo, ~n-->yo and (~7~-Yn)/(~ x~)->z o. (0/ course, F(x,y,z) must be de/ined and satis/y its conditions on an interval containing all the points xo, {x~}~ and { ~ } ~ . ) Then

lim M(x,, ~,; y~, ~n) = F(x0, Y0' %)"

n - - ~

Proo/: This is an i m m e d i a t e consequence of the preceding l e m m a a n d the c o n t i n u i t y of F(x,y,z).

L e m m a 7: Suppose that /(x) is a minimizing /unction on xl<~x~x z. (Boundary values Yl and Y2. ) Suppose/urther that x I ~ ~ < fi ~ x 2. Then

M(zc,fl; /(~), /(fl)) <~M(xl,x2; Yl,Y2).

Proo/: M(g, fl; /(~), /(~)) ~H(/; ~,fl) ~H(/; xi,x2) =M(xa,x2; Yl,Y2), w h e r e ' t h e mean- ing of the notations is obvious.

N o w we m u s t introduce a new condition on F(x, y, z), a n d this condition is assumed to hold in the rest of section 4:

3) limR~l_~r F(x,y,z)= + ~ for every fixed x a n d y.

As is easily seen, using the conditions 1 a n d 2 also, this means t h a t lim ( inf F(x, y, z))= +

lYI<~K

for every c o m p a c t interval [u,fl] (where F is defined) a n d for every K > 0 .

Theorem 7: I / / ( x ) is an a.s. minimal on an interval containing x o in its interior, then/'(Xo) exists.

Proo/: L e t I be a c o m p a c t interval with x 0 in its interior such t h a t / ( x ) is an a.s.

minimal on I. T h e n / ( x ) a n d F(x, /(x), /'(x)) are b o u n d e d on I . B u t t h e n it follows t h a t / ' ( x ) is also b o u n d e d on I . L e t zr be the greatest a n d fi the smallest of the four derivates o f / ( x ) at x = x 0. T h e n zr a n d fi are finite. Clearly, there exist sequences {p,}

a n d {q~} such t h a t

p~ < x o < q~, q n - p ~ - ~ 0

and /(q.) -/(P-)_+~.

q, - Pn

Of course, corresponding sequences {%}, {s,} exist for ft.

Assume n o w t h a t ~ > f l a n d let ? be a n y n u m b e r such t h a t ~>7>fl. As is easily seen, there m u s t exist sequences {tn}, {u,} such t h a t t~<xo<U,, u~-tn--+O a n d [ / ( u n ) - / ( t , ) ] / ( u , - Q ) = ? . According to L e m m a 6 we h a v e

lira M (pn, qn; /(P.), /(q.) ) = F ( x o , / ( X o ) , ~),

A R K I u FOR MATEMATIK. Bd 6 n r 2 lim M (rn, s~; /(rn), /(sn) ) = F(xo, /(Xo), fl),

n--->oo

a n d lim M (t,, u,; /(t,), /(u~) ) = F ( x o,/(Xo), ~).

n - - > ~

N o w there exist arbitrarily great n u m b e r s nl, n2, n a such t h a t (t~,,unl) c (rn~,s~,) c (P~0, qn~).

Application of L e m m a 7 n o w gives F(xo, /(Xo) , ~) ~ F(xo, /(Xo), fl) ~ F ( x o,/(xo), a).

B u t the inclusion relations can also be chosen in the opposite way, which gives

F(Xo, /(Xo), ~) >~ F(Xo, /(Xo), fl) >~ F(xo, /(xo), ~).

Hence these three n u m b e r s are equal. B u t this contradicts our a s s u m p t i o n t h a t Consequently ~ =fi which means t h a t / ' ( x o ) exists.

Remark: I f / ( x ) is an a.s. minimal on the interval x I ~X~'~Z2, t h e n it follows (with a few modifications in the proof) t h a t the one-sided derivatives in question exist at x 1 a n d x 2.

L e m m a 8: I / / ( x ) is an a.s. m i n i m a l on an open interval containing x o and/'(xo) = 0 , t h e n / ' ( x ) is continuous at x o.

Proo]: L e m m a 6 gives

l i m M ( x o - l , x o § O)

and, since / is an a.s. minimal, we get

lira H I / ; x o - 1 , x o + 1 ) = F(xo '/(Xo), O) n--+oo \ n n /

f r o m which t h e assertion follows.

Remark: This result is obviously true also for an e n d : p o i n t of an interval.

Theorem 8: To our previous conditions on F ( x , y , z ) we add the /ollowing: Fx, Fy and F z exist and are continuous/or all x under consideration and all y, z.

Suppose now t h a t / ( x ) is an a.s. m i n i m a l on an interval which contains x o in its in- terior, and suppose/'(xo) ~:0.

Then

1) /(x) E C 2 on an open interval I containing x o 2) F(x, /(x), /'(x)) =constant on I.

( H e n c e dF(x'/(x)'/'(x))-Odx on I . )

Proo]: Our m e t h o d of proof will be the following: W e c o n s t r u c t two solutions of the differential e q u a t i o n F ( x , y , y ' ) = c o n s t a n t , the first of which is equal to /(x) at x o a n d at some x' > x 0 a n d the second is equal t o / ( x ) at x o a n d at some x" < x 0. Then,

r AnOr~SSON, Minimization problems

using T h e o r e m 6, we prove t h a t / ( x ) a n d these two solutions are identical. We shall confine t h e discussion to the case /'(Xo)>0, since the other case is analogous.

L e t us introduce the n o t a t i o n s Y0 =/(Xo), zo =/'(xo), Mo = F(xo, Yo, %) a n d

~F(x,y,z,M) ~ F(x,y,z) M. T h e n q~(xo,Yo, zo, Mo) = 0 a n d W'~(xo,Yo, Zo, Mo)=F~(xo, Yo, Zo)>O.

Hence the equation 1F(x, y, z, M ) = 0 can be used to define z as a function z-OP(x,y,M) on the set R in x yM-space, defined b y the inequalities Ix x 0 ~<5,

l y - y o l <~5 a n d I M - M o I ~ < 5 , for some 5 > 0 .

W e m a y assume t h a t we have, for .(x,y,M)ER a n d for some 51 > 0 , 0 < % - 5 1 <~(P(x,y,M) <~z o +51.

W e shall also assume t h a t /(x) is an a.s. minimal on [X-Xol <5. The function (I)(x, y, M) is c o n t i n u o u s l y differentiable on R, a n d we have

H e n c e

~x F~' By Fz a n d ~M - F~"

~x ~ 1 a n d ~<C 2 on R.

L e t us consider the differential equation y' =dP(x,y,M) together with t h e initial value y(xo)-Yo- Here the p a r a m e t e r M is a s s u m e d to satisfy I M - M o l ~ .

I t follows f r o m P i c a r d ' s t h e o r e m t h a t there exists a unique solution for I x - x 0 ] ~ 5~, a n d 52 is independent of M. L e t us denote the solution b y y(x, M). I t is also true t h a t for every Xl, such t h a t Ix1 x0] <5~, the solution y(Xl,M ) depends c o n t i n u o u s l y on M. This is p r o v e d b y a s t a n d a r d a r g u m e n t . (Cf. [2], pp. 46, 65, 70.) If ] x - x 0 l ~ 3 ~ 5~, then, clearly, ]y(x,M) Y0] ~K3, where K=zo+51. If f l > 0 is small enough, t h e n there is a 7 > 0 (but 3<52) such t h a t the inequalities I x - x 0 1 < 7 , lY-Y0] ~<KT, i m p l y t h a t

I CP(x,y, Mo+5) ~Zo+t~

a n d [ (I)(x~y,M o - 5 ) <~z o-ft.

(For d) is continuous a n d ~ / ~ M > 0 on R.) This gives the inequalities y(x, M o + 8) ~ Yo + (% +~) (x - xo)

a n d y(x, M o - 5 ) ~ Yo + ( % - fl) (x -xo),

valid for x 0 ~ x ~< x o + 3.

Fix n u m b e r s fl a n d 3 h a v i n g the above properties and, in addition, satisfying the condition

Yo + (Zo fi) 3 </(x o + 3) < Yo + (% + fi) 3.

T h e n we have

y(x o +'v, M o - (~) </(x o + 7) < y(x o + % Mo + (~).

Since y(x o + T, M) depends c o n t i n u o u s l y on M, there is a value M* such t h a t y(x o + T,M*) = / ( x 0 +3).

A R K I V FOR MATEMATIK. B d 6 n r 2 B u t / ( x ) is a n a.s. m i n i m a l a n d n o w it follows f r o m T h e o r e m 6 t h a t / ( x ) = y ( x , M * ) for Xo <~X<~Xo+T.

I n a similar w a y one c a n f i n d n u m b e r s T' a n d M** a n d p r o v e t h a t / ( x ) = y ( x , M * * ) for x o - T ' ~<x ~ x o.

H e n c e /'(Xo) = r Y0, Mo) = (I)(x0, Y0, M*) = O(x 0, Yo, M**),

which gives us M o = M* = M**. C o n s e q u e n t l y / ' ( x ) = 9 (x,/(x), Mo) for x o - 3' ~< x ~< x o + 3.

This m e a n s t h a t F ( x , / ( x ) , ] ' ( x ) ) = M o for t h e same v a l u e s of x, a n d t h e rest of t h e proof is obvious.

Remark: As is easily seen, t h e t h e o r e m c o n t i n u e s to hold ( b u t w i t h o b v i o u s modi- fications) i n t h e case w h e n x 0 is a n e n d - p o i n t of a n i n t e r v a l , w h e r e / ( x ) is a n a.s.

m i n i m a l .

T h e o r e m 9: Let F ( x , y , z ) satis/y the same conditions as in the previous theorem. I / /(x) is an a.s. m i n i m a l on the interval I , then:

1) /(x) eC

l on I

2) the di//erential equation

d F ( x , / ( x ) , / ' ( x ) ) F z ( x , / ( x ) , / ' ( x ) ) = 0 dx

is satis/ied on I in the/ollowing sense: The second/actor is well-de/ined on I and i/ it is di//erent /rom zero at xo, then the/irst/actor exists and is zero in a neighbourhood o/ x o.

Proo/: T h e t h e o r e m is a consequence of T h e o r e m 7, L e m m a 8 a n d T h e o r e m 8.

Remark: As is s h o w n b y E x a m p l e 6 i n section 5, t h e d e r i v a t i v e s /"(x) a n d d F ( x , / ( x ) , / ' ( x ) ) / d x n e e d n o t exist a t p o i n t s where ] ' ( x ) = 0 .

Corollary: Under the present conditions on F ( x , y , z ) , suppose t h a t / ( x ) is a unique m i n i m i z i n g / u n c t i o n on x 1 <~ x <~ x 2. Then

/ ( x ) E C 1 on [x D x2] and F(x, /(x), /'(x)) =constant on [x 1, x2].

Proo/: O b v i o u s l y , / ( x ) is a n a.s. m i n i m a l o n [Xl, x2]. Hence, b y t h e t h e o r e m , / ( x ) E C 1.

I f it were t r u e t h a t F(xo,/(x0) ,/'(xo) ) < H ( / ) for some x 0, t h e n we could alter /(x) slightly i n a n e i g h b o u r h o o d of x 0 w i t h o u t increasing t h e v a l u e of H(/). B u t this con- t r a d i c t s t h e u n i q u e n e s s , a n d hence we h a v e F(x, /(x), / ' @ ) ) = c o n s t a n t . (Compare T h e o r e m 1 a n d T h e o r e m 6.)

I n t h e t h e o r e m s 5 a n d 6, t h e c o n d i t i o n t h a t / ( x ) is m o n o t o n i c p l a y s a n i m p o r t a n t role. If we impose a s u i t a b l e e x t r a c o n d i t i o n o n F ( x , y , z ) , t h e n e v e r y a.s. m i n i m a l m u s t be m o n o t o n i c :

T h e o r e m 10: Suppose that F(x, y, z) satis/ies all the conditions in Theorem 8 together with the extra condition that ~ F / ~ x does not change s i g n / o r x E I and any y,z. I / n o w /(x) is an a.s. m i n i m a l on the interval I , t h e n / ( x ) is monotonic on I. ( B u t / ( x ) need not be strictly monotonic; compare E x a m p l e 3 in section 5.)

Proo/: According to T h e o r e m 9 we h a v e

/(x) ECI(I).

Assume, for example, t h a t

r AROr~SSO~,

Minimization problems

t h e r e e x i s t x 1 a n d x 2 on I such t h a t / ' ( X l ) > 0 a n d / ' ( x 2 ) < 0 . A s s u m e also

Xl<X 2.

O b v i o u s l y , t h e r e e x i s t x a a n d x 4 such t h a t x 1 ~< x 3 < x 4 ~< x2,/(x3) = / ( x 4 ) , / ' ( x 3 ) > 0 a n d

/'(x4)

< 0 . Consider t h e m i n i m i z a t i o n p r o b l e m o n x s ~<x ~<x 4. I f

g(x) .~/(xs),

t h e n

g(x)

is a d m i s s i b l e a n d , since F x does n o t c h a n g e sign, we h a v e

H(g)

= m a x (F(xs,/(x3), 0),

F(x 4,/(x4),

0)). A s s u m e

H(g) = F(xs,/(Xs)

, 0). B u t since

]'(x3)

> 0 we h a v e

H(/) > H(g)

w h i c h is i m p o s s i b l e , s i n c e / ( x ) is a n a.s. m i n i m a l . H e n c e / ' ( x ) does n o t c h a n g e sign, w h i c h c o m p l e t e s t h e proof.

5. E x a m p l e s

I n t h i s section, we w a n t t o s h o w a few a p p l i c a t i o n s of s o m e of t h e t h e o r e m s a l r e a d y given. W e also m o t i v a t e b y m e a n s of e x a m p l e s t h e i n t r o d u c t i o n of c o n d i t i o n 3 a n d t h e f o r m u l a t i o n of T h e o r e m 9.

Example 1A:

L e t us use T h e o r e m 4 to solve t h e m i n i m i z a t i o n p r o b l e m if

F ( y , y ) - y , _ 2 + ~ , ~ x y , a n d 1

W e h a v e

~PM(y)=l/M----y 2,

a n d

f l o ' V ' d t

s = VM - t 2

is w e l l - d e f i n e d for M / > 89 F o r such v a l u e s of M , we h a v e s = a r c s i n ( 1 / l / ~ ) . To f i n d M0, we m u s t d e t e r m i n e t h e s m a l l e s t M>~ 89 such t h a t

1 ~ g a r c s i n ~ - ~ ~ .

This gives us M . = I . Since s 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 . I n o r d e r t o f i n d it, we use T h e o r e m 3 a n d f o r m

g~(y) = ~ = a r c s i n y.

Since X l = 0 , t h e m i n i m i z i n g f u n c t i o n is t h e i n v e r s e of

rp(y),

n a m e l y

/o(x)=sin

x.

Example 1 B:

T h i s will i l l u s t r a t e t h e ease in T h e o r e m 4 w h e r e t h e r e is no

unique

m i n i m i z i n g f u n c t i o n .

Choose

F(y,

y')

= y ,4 _ 16y2,

X l = - 2 , y x = - l , x 2 = 2 a n d y 2 = l . W r i t e

y'4--16y2

= M w h i c h gives

4

= Vi y 2 + M . F o r t h e e x i s t e n c e of

s 4

dt

9 V1-6 7 M i t is c l e a r l y n e c e s s a r y a n d sufficient t h a t M~> 0.

ARKIV FOR MATEMATIK, B d 6 n r 2

W e n e e d n o t e v a l u a t e t h e i n t e g r a l

F~(M),

since

(1 dt

s = J - 1 2 ~ t ] - < x ~ - xl.

W e see t h a t M 0 = 0 a n d t h a t t h e r e is no u n i q u e m i n i m i z i n g f u n c t i o n . L e t u s d e t e r m i n e a m i n i m i z i n g f u n c t i o n ! F o r m t h e f u n c t i o n

= ( Y dt

g(Y) J

_{o 2v l}

{Compare T h e o r e m 3.) W e g e t

for - - l ~ y ~ < l .

{1/y for y ~> 0,

g(Y)=

- V ~ - I for y~<O.

T h e i n v e r s e f u n c t i o n is

](x)

= x Ix]. I f we d e f i n e / ( x ) as + 1 for x > 1 a n d - 1 for x < - 1, t h e n we g e t a m i n i m i z i n g f u n c t i o n . T h e f u n c t i o n

h(x)= 88 Ix[

(for - 2 ~<x ~< 2) is also a m i n i m i z i n g f u n c t i o n , b u t in c o n t r a s t t o t h e f o r m e r one, i t is c o n t i n u o u s l y dif- f e r e n t i a b l e .

W e h a v e M o = m a x y , < ~ < v ~

F(y,O)

in a c c o r d a n c e w i t h T h e o r e m 4.

Example 2:

L e t us use t h e rules g i v e n in s e c t i o n 2 B t o d e t e r m i n e t h e a t t a i n a b l e cone if ^{F J} ( y , y ) = y _ y4, Xo = 0, Yo = 1 a n d M = 0. ^{1~ t 2}

W e h a v e O0(y ) =y2 a n d hence

q(Y) = x ~ J ~t/~,~ - .'~t t 2 1 - Y-"

S i n c e l i m ~ _ ~ ~(y) = 1 we see t h a t t h e p r e s e n t case is D 2 a n d t h a t X = 1. T h e i n v e r s e of x =~0(y) = 1 -

1/y

is y = 1/(1 - x ) . Therefore,

1 for 0 ~ x < l ,

h(x) = 1 - x

+ ~ f o r x >/1.

I n o r d e r t o d e t e r m i n e g(x), we f o r m t h e f u n c t i o n

~ ( Y ) = X ~ -~F~(t) j~t2 y 1.

Clearly, t h i s c o r r e s p o n d s t o t h e case B. T h e i n v e r s e of x =us =

1/y

- 1 is y = 1/(1 + x ) . H e n c e

g(x)

= 1/(1 + x ) for all x ~>0.

F ' = y~ + y'2

Example 3:

P u t

(y,y)

a n d define t h e f u n c t i o n / o ( x ) as

G. ARO~SSO~, Minimization problems

I

¹

/o(X) = ~ s i n x

2Z

for x ~ 2 ' 7g< x for --

_<~_

" ~ 2 ~ 2~

- 1 for x < - - 2"

C l e a r l y , / o ( x ) E C 1 and/o(X) is m o n o t o n i c o n ~ < x < ~ . F u r t h e r , F(/o(X), / 0 ( x ) ) = l for all x. Now it follows from T h e o r e m 5 that/o(X) is a n a.s. m i n i m a l o n - c~ < x < c~.

Observe t h a t [o'(x) does n o t exist a t x = + ~ / 2 . Compare E x a m p l e 5.

Example 4: This e x a m p l e shows t h a t t h e c o n d i t i o n 3 c a n n o t be o m i t t e d i n T h e o r e m 7. I t will also give a n idea of t h e case where F is i n d e p e n d e n t of y. L e t

F(x, y, Y') = x~ + y,4 y,4 + 1"

T h e c o n d i t i o n s 1 a n d 2 are satisfied, b u t n o t c o n d i t i o n 3.

P u t F(x, y, y') = 1.

4

T h i s gives y, = + r 1 - x ~

-- X 2

F o r m t h e p r i m i t i v e f u n c t i o n

4

VItl

I t is easy to see 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 b e t w e e n ( - 1 , / ( - 1 ) ) a n d (1,/(1)). H e n c e , / ( x ) is a n a.s. m i n i m a l . B u t ['(0) is n o t finite.

Example 8: This is a c o n t i n u a t i o n of E x a m p l e 3. W e n o w w a n t to d e t e r m i n e all a.s. m i n i m a l s for F(y,y')=g2 +y,2.

A s s u m e t h a t /(x) is a n a.s. m i n i m a l o n a n o p e n i n t e r v a l I . I t follows from t h e t h e o r e m s 9 a n d 10 t h a t ](x)ECI(I) a n d t h a t / ( x ) is m o n o t o n i c .

Clearly / ( x ) - - c o n s t a n t is possible. B u t a s s u m e n o w t h a t we h a v e ['(xl)=t=0 for some x 1 E I .

L e t 11 be t h e largest open i n t e r v a l c o n t a i n i n g x 1 where ['(x)=t=0. T h e n [(x)E Ce(I1) a n d [(x) 2 + ['(x) ~ = c o n s t a n t o n 11. D i f f e r e n t i a t i o n gives /"(x) +/(x) = 0. H e n c e there exist A a n d B such t h a t / ( x ) = A sin (x + B) o n 11. S i n c e / ( x ) is m o n o t o n i c , i t follows t h a t 11 is b o u n d e d , s a y I i = ( ~ , f l ) . N o w we assert t h a t ['(x) = 0 o n I - I 1 (if it is n o t e m p t y ) . If this were n o t so, t h e n we would h a v e a different open i n t e r v a l 12 = (y,d) w i t h t h e same properties as 11. L e t us assume t h a t fi ~<~, a n d t h a t / ' ( x ) > 0 o n 11 U 12.

W e h a v e ]'(fl)=['(),)-0. Since [(x) is a s i n e - f u n c t i o n o n I1, it follows t h a t / ( f l ) > 0 . F r o m t h e m o n o t o n i e i t y we conclude t h a t /(y)>~[(~). H e n c e / ( x ) = C sin ( x + D ) h a s t h e properties [()p) > 0 , ['(~,) = 0 a n d ]'(x) > 0 for y < x < d . B u t this is clearly impossible.

C o n s e q u e n t l y / ' ( x ) = 0 o n I - 1 1 . This leads to t h e following result: T h e r e exist n u m -

ARKIV FOR MATEMATIK. B d 6 n r 2 bets p a n d q such t h a t / ( x )

=p/o(x § q)

for all

x E I,

where/o(X) is t h e f u n c t i o n i n t r o - d u c e d i n E x a m p l e 3.

H e n c e we h a v e f o u n d t h a t

the class o/a.s, minimals /or F(y,y') =y2 +y,2 is the class o//unctions o/the/orm p/o(X § q) where p and q are arbitrary real numbers, and constant /unctions

(which c a n n o t be w r i t t e n as

p/o(x + q)

if I is t h e e n t i r e real axis).

Example 6:

This e x a m p l e shows t h a t t h e d e r i v a t i v e s

dF(x,/(x),/'(x))/dx

a n d / " ( x ) i n T h e o r e m 9 n e e d n o t exist for all x.

Choose

F ( x , y , y ' ) = y , 2 _ x

a n d consider t h e f u n c t i o n

0 for x~<0,

](x)= w ~j2

for x>~0.

T h e n

F(x,/(x),/'(x)) = I [x[

for x ~<0,

[

0 for x>~0.

W e assert t h a t / ( x ) is a n a.s. m i n i m a l o n - ~ < x < ~ . To p r o v e t h a t , consider t h e m i n i m i z a t i o n p r o b l e m o n x I ~< x ~< x 2. I f Xl ~> 0 t h e n i t follows from T h e o r e m 5 t h a t

/(x)

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

If x 1 < 0, t h e n we h a v e

H(/)= Ix1/

a n d since e v e r y admissible f u n c t i o n m u s t pass t h r o u g h t h e p o i n t (Xl,0), it follows t h a t inf~

H(g)>~F(Xl,0,O ) = I x~l.

H e n c e

](x)

is a n a.s. m i n i m a l . B u t t h e d e r i v a t i v e s

dF(...)/dx

a n d / " ( x ) do n o t exist a t

x=O.

Therefore, i n T h e o r e m 9, we c a n

not

assert t h a t t h e differentiaI e q u a t i o n

dF(...)/dx. F~(...)=0

is satisfied in t h e classical sense.

R E F E R E N C E S

1. CA~TER, D. S., A m i n i m u m - m a x i m u m problem for differential expressions. Canadian J o u r n a l of Mathematic s 9, 132-140 (1957).

9. PETROVSKI, I, G., Vorlesungen tiber die Theorie der gewhhnlichen Differentialgleichungen.

Teubner, Leipzig 1954.

I n a c o m i n g p a p e r we shall discuss some q u e s t i o n s t h a t h a v e b e e n left o p e n here.

Tryckt den 21 april 1965

Uppsala 1965. Almqvist & ~Viksells Boktryckeri AB