Palack University, Olomouc Department of Mathematical Analysis

Preprint Series No.5/1999

Nonsmooth lower and upper functions and solvability of certain nonlinear second order BVP's Irena Rachunkov


and Milan Tvrd y

Summary. In the paper we present some new existence results for nonlinear second order

generalized periodic boundary value problems of the form (0.1), (0.2). These results are based on the method of lower and upper functions associated with the problem and their relation to the Leray-Schauder topological degree of the corresponding operator. Our main goal consists in a fairly general denition of lower and upper functions as couples of functions from A C ]
BV ] Some conditions ensuring the existence of such functions are indicated, as well. AMS Subject Classi
cation. 34 B 15, 34 C 25 Keywords. Second order nonlinear ordinary dierential equation, periodic solution, generalized periodic boundary value problem, lower and upper solution, lower and upper functions, strict lower and upper functions, Leray-Schauder topological degree. a b

a b :

0. Introduction In this paper we give existence theorems for the generalized periodic boundary value problem (0.1) (0.2)

u = f (t u u ) u(a) = u(b) u (a) = w(u (b)): 0

00

0

0

Using these results (Theorems 4.1 - 4.3) we can get both the existence and multiplicity for solutions of various periodic problems and their generalizations. One of such possible applications is shown in Corollary 4.4 which generalizes some results of 3], for other applications see 8] or 9]. Supported by the grant No. 201/98/0318 of the Grant Agency of the Czech Republic by the grant No. 201/97/0218 of the Grant Agency of the Czech Republic

y Supported

1

The main tool of our arguments is a connection between the existence of lower and upper functions for (0.1), (0.2) (called also lower and upper solutions by some authors) and the Leray-Schauder topological degree of an operator associated with (0.1), (0.2). The notions of lower and upper functions of the second order boundary value problems have a long history starting in 1931 when G. Scorza Dragoni 10] used them for the Dirichlet problem. So far there have been a lot of denitions introduced. Classically, we understand lower and upper functions as C 2 -functions. Dierential equations with Carathodory right hand sides or with singularities involved their generalization, for example as A C 1 -functions, C 1 -functions having left and right second derivatives or W 2 1 -functions. The majority of existence results was gained under the ordering assumption that a lower function is less than or equal to an upper one. During the last two decades the extension to nonordered or reversely ordered lower and upper functions was attained. See 1] and the references mentioned there. Here, we introduce a denition (cf. Denition 1.7) of lower and upper functions of the problem (0.1), (0.2) which generalizes those of 1], 4], 5], 6] or 7] and consider the both cases of their ordering as well as the non-ordered one.

1. Preliminaries Throughout the paper we assume: ;1 < a < b < 1 w : R 7! R is continuous and nondecreasing and f : a b]  R 2 7! R fulls the Carathodory conditions on a b]  R 2 i.e. f has the following properties: (i) for each x 2 R and y 2 R the function f (: x y) is measurable on a b] (ii) for almost every t 2 a b] the function f (t : :) is continuous on R 2  (iii) for each compact set K R 2 the function mK (t) = sup ( ) K jf (t x y)j is Lebesgue integrable on a b]: Furthermore, we keep the following notation: L a b] is the Banach space of Lebesgue integrable functions on a b] equipped with the usual norm denoted by k:kL : Furthermore, for k 2 N f0g, C a b] and A C a b] are the Banach spaces of functions having continuous k -th derivatives on a b] and of functions having absolutely continuous k-th derivatives on a b] respectively. As usual, the corresponding norms are dened by x y 2

k

k

kxkC = k

X k

i

=0

max jx( ) (t)j and kxkAC k = kxkC k + kx(  ] i

t2 a b

k

+1)

kL :

The symbols C a b] or A C a b] are used instead of C 0 a b] or A C 0 a b]: Moreover, B V a b] is the set of functions of bounded variation on a b]: For u 2 B V a b]

using and uac denote its singular and absolutely continuous parts, respectively. Furthermore, if u 2 B V a b] then its one-sided derivatives are denoted by D+ u and D u: Car( a b]  R 2 ) is the set of functions satisfying the Carathodory conditions on a b]  R 2 : Finally, for a given Banach space X and its subset M cl(M ) stands for the closure of M and @M denotes the boundary of M: If  is an open bounded subset in C 1 a b] and the operator T : cl() 7! C 1 a b] is compact, then deg(I ; T ) denotes the Leray-Schauder topological ;

degree of I ; T with respect to  where I stands for the identity operator on a b]: For a denition and properties of the degree see e.g. 2]. By a solution of (0.1),(0.2) we understand a function u 2 A C 1 a b] satisfying (0.1) for a.e. t 2 a b] and having the property (0.2). The following estimate will be helpful later. 1.1. Lemma. Let a function m 2 L a b] and sets U (t)  R t 2 a b] be such that m(t) < 0 on a subset of a b] of a positive measure, C 1

(1.1) and (1.2)

m(t) < f (t x y) for a.e. t 2 a b] and any (x y) 2 U (t)  R

w(y)  y

for all y 2 ;kmkL kmkL ]:

Let u be an arbitrary solution of (0.1), (0.2) such that u(t) 2 U (t) for all t 2 a b]: Then

ku kC < kmkL :

(1.3)

0

If we suppose m(t) > 0 on a subset of a b] of a positive measure and

(1.4) and (1.5)

m(t) > f (t x y) for a.e. t 2 a b] and any (x y) 2 U (t)  R

w(y) y

for all y 2 ;kmkL kmkL ]

instead of (1.1) and (1.2), then the estimate (1.3) remains valid, as well. Proof. We shall restrict ourselves only to the proof of the former assertion. The latter can be proved by a similar argument. Let u be an arbitrary solution of (0.1), (0.2) such that u(t) 2 U (t) for all t 2 a b] and let (1.1) and (1.2) be fullled. Then

(1.6)

m(t) < u (t) for a.e. t 2 a b]: 00

Certainly, there is t0 2 (a b) such that u (t0 ) = 0: Hence 0

(1.7)

;kmkL ;

and (1.8)

;kmkL ;

Z

t

Z

jm(s)j ds < u (t) for t 2 (t0 b] 0

t0

t0

t

jm(s)j ds < ;u (t) for t 2 a t0 ): 0

In particular, with respect to (0.2),

w(u (b)) = u (a)
;

Z

b

t0

jm(s)j ds ;

Z

t

a

jm(s)j ds  ;kmkL :

This together with (1.7) yields

;kmkL < u (t) for all t 2 a b]:

(1.11)

0

On the other hand, in virtue of (1.9),(1.10) and (1.2) we have for t 2 t0 b]

u (t) < u (b) + 0

0

Z

b

t

jm(s)j ds
0 a.e. on a b] and w fulls(1.2), then ( ) could not be lower functions of (0.1), (0.2). Analogously, f (t (t) (t)) < 0 a.e. on a b] with (1.5) can be true for no upper functions ( ) of (0.1), (0.2). Let us denote (1.19) and (1.20)

L : x 2 A C 1 a b] 7! (x ; x x(a) ; x(b) x (a)) 2 L a b]  R 2 00

0

F : x 2 C 1 a b] 7! Fx 2 L a b]  R 2 where (Fx)(t) = (f (t x(t) x (t)) ; x(t) 0 w(x (b))) a.e. on a b]: 0

0

Then L is a linear bounded operator and the operator F is continuous. After a careful computation we can check that if we put (1.21)

8> ;e2 >< ;0 (t s) = > ; >: ; e2

t;s

a

b;s;t

and (1.22)

;



+e e2 ; e2 2 (e ; e )2 ;e2 + e2  ; 2 e ;e 2 (e ; e )2

a;s;t

s

b

t;s

s

a

a

;1 (t) = ; e

+e + (e ; e )2

2a+b;t b

if t < s

b

b

a

t

+b

a

(e

t;s

b

; e ) if t > s s;t

and ;2 (t) = ; e e ; e+ e on a b] +b;t

a

b

t

a

then max j;0 (t s)j + sup  ]

t s2 a b







 @ ;0(t s)   @t  < 1   ]

t s2 a b

max j;1 (t)j + j;1 (t)j + max j;2 (t)j + j;2 (t)j < 1  ]  ] 0

0

t2 a b

t2 a b

and for any (y r1 r2 ) 2 L a b]  R 2 the unique solution of the linear boundary value problem

x ; x = y(t) x(a) ; x(b) = r1 00

is dened by

x(t) =

Z a

b

x (a) = r2 0

;0 (t s)y(s)ds + ;1 (t)r1 + ;2 (t)r2 on a b]:

Furthermore, the operator L+ dened by L+ : (y r1 r2 ) 2 L a b]  R 2 7! L+ (y r1 r2 ) 2 C 1 a b]

(1.23) where

(L+ (y r1 r2 ))(t) =

Z

b

a

;0 (t s)y(s)ds + ;1 (t)r1 + ;2 (t)r2 on a b]

is linear and bounded and the operator L+ F : C 1 a b] 7! C 1 a b] is compact. The problem (0.1),(0.2) is equivalent to the operator equation

;I ; L+Fx = 0

and if for some open bounded set   C 1 a b] the relation deg(I ; L+ F ) 6= 0

(1.24)

is true, then the problem (0.1), (0.2) possesses at least one solution in :

2. Strict lower and upper functions and topological degree The following denition is motivated by the similar one used in 1] for the periodic problem x = f (t x) x(a) = x(b) x (a) = x (b): 00

0

0

2.1. Denition. Lower functions (1 1) of (0.1), (0.2) such that 1 is not a solution of this problem are called strict lower functions of (0.1), (0.2) if there exists " > 0 such that (2.1)

1 (t)  f (t x y) for a.e. t 2 a b] and all (x y) 2 1 (t) 1 (t) + "]  1 (t) ; " 1 (t) + "]: 0

Analogously, upper functions (2 2 ) of (0.1), (0.2) are said to be strict upper functions of (0.1), (0.2) if 2 is not a solution of this problem and there exists " > 0 such that (2.2)

2 (t) f (t x y) for a.e. t 2 a b] and all (x y) 2 2 (t) ; " 2 (t)]  2 (t) ; " 2 (t) + "]: 0

In this section we want to prove theorems giving sucient conditions for (1.24) in terms of strict lower and upper functions of (0.1),(0.2). We shall need the following two lemmas.

2.2. Lemma. Let (1 1) and (2 2) be respectively strict lower and upper functions of the problem (0.1), (0.2) such that

1 (t) < 2 (t) on a b]:

(2.3)

Then for any solution u of (0.1), (0.2) fullling

1 (t) u(t) 2 (t) on a b]

(2.4)

we have 1 (t) < u(t) < 2 (t) on a b]: Proof. i) Suppose

(2.5)





u(t0 ) ; 2 (t0 ) = max u(t) ; 2 (t) = 0 and t0 2 (a b):  ] t2 a b

In particular, u (t0 ) ; 2 (t0 ;)  0  u (t0 ) ; 2 (t0 +) and thus, with respect to (1.16), 0

(2.6)

0

u (t0 ) = lim0 2 (t) = 2 (t0 ): 0

t!t

Hence, if " > 0 is such that (2.2) is true, then there is  2 (0 b ; t0 ] such that the relations 2 (t) ; " u(t) 2 (t) and 2 (t) ; " u (t) 2 (t) + " are satised 0

for all t 2 t0 ;  t0 + ] and consequently, making use of (2.2), (2.6) and the monotonicity of sing we get for any t 2 t0 t0 + ] 2 (2.7) 0

Z ;  Z ;u (t) ;  (s) ds f (s u(s) u (s)) ; 2 (s) ds = 2 t

t

0

0

00

t0

0

t0

sing = u (t) ; 2 (t) ; u (t0 ) + 2 (t0 ) = u (t) ; 2 (t) + sing 2 (t) ; 2 (t0 ) u (t) ; 2 (t): By (1.16), (2.7) and (2.4) we have ac

0

ac

0

0

0

0  u(t) ; 2 (t) =

Z  t



u (s) ; 2 (s) ds  0 on t0 t0 + ]

t0

0

i.e. u(t) = 2 (t) on t0 t0n+ ]: o Let us put t = sup 2 t0 b] : u(t) = 2 (t) on t0 ] : Then t  t0 +  u(t ) = 2 (t ) and u (t ) = 2 (t ;): Let us assume that t < b: Then, by (1.16), we have u (t )  2 (t +): If u (t ) > 2 (t +) were valid, then 0 = u(t0 );2 (t0 ) = u(t ) ; 2 (t ) could not be the maximum value of u(t) ; 2 (t) on a b] and this would contradict the assumption (2.5). Thus, u (t ) = 2 (t +): Repeating the above considerations with t in place of t0 we would obtain further that there is  2 (0 b ; t ] such that u(t) = 2 (t) on t t +  ] a contradiction with the denition of t : It means that t = b and u(t) = 2 (t) on t0 b]: Similarly, we could prove that u(t) = 2 (t) on a t0 ] i.e. u(t) = 2 (t) on a b]: This contradicts our assumption that 2 is not a solution of the problem (0.1), (0.2) on a b] i.e. u(t) < 2 (t) on (a b): ii) Suppose 



0







0









0









0















(2.8)











0 = u(b) ; 2 (b) = u(a) ; 2 (a) = max u(t) ; 2 (t) :  ] t2 a b

This is possible only if u (a) 2 (a+) and u (b)  2 (b;): On the other hand, by (0.2) and (1.18) we have 0  u (a) ; 2 (a+)  w(u (b)) ; w(2 (b;))  0 and hence (2.9) u (a) = 2 (a+): Similarly as in part i) of the proof, we can deduce from the relations (2.8) and (2.9) that u(t) 2 (t) on a b]: This being impossible by Denition 2.1, we conclude that u(t) < 2 (t) on a b]: iii) Similarly we can show that under our assumptions the relation u(t) > 1 (t) is true for all t 2 a b] as well. 0

0

0

0

0

2.3. Lemma. Let (1 1) and (2 2) be respectively strict lower and upper functions of (0.1), (0.2) such that (2.3) is true. Let us put

8 f (t  (t) y) ;  (t) < 1 1 fe(t x y) = : f (t x y) ; x f (t 2 (t) y) ; 2 (t)

(2.10)

if x < 1 (t) if 1 (t) x 2 (t) if 2 (t) < x:

Then fe 2 Car( a b]  R 2 ) and for any solution u of the problem

u ; u = fe(t u u )

(2.11)

00

0

(0:2)

the relations (2.4) are satised. Proof. In view of (2.10), we have fe 2 Car( a b]  R 2 ): Let u be an arbitrary solution of the problem (2.11) and let





u(t0 ) ; 2 (t0 ) = max u(t) ; 2 (t) > 0:  ]

(2.12)

t2 a b

By (0.2) and (1.18) it suces to consider the cases t0 2 (a b) and t0 = a: If t0 2 (a b) then similarly as in the proof of Lemma 2.2 we obtain that lim 0 2 (t) = 2 (t0 ) = u (t0 ): If t0 = a then like in the second part of the proof of Lemma 2.2 we get u (a) = 2 (a+): In particular, in both cases, if " > 0 is such that (2.2) is satised, then there is  2 (0 b ; t0 ] such that u (t) 2 2 (t) ; " 2 (t) + "] and u(t) > 2 (t) on t0 t0 + ]: Hence, owing to (2.10) we have t!t

0

0

0

u (t) ; 2 (t) = f (t 2 (t) u (t)) + u(t) ; 2 (t) ; 2 (t) > f (t 2 (t) u (t)) ; 2 (t)  0 a.e. on t0 t0 + ] 00

0

0

0

0

0

and like in (2.7) for t 2 (t0 t0 + ] we obtain 0< Consequently, 0
c 8 w(;c) if y < ;c < we(y) = : w(y) if jyj c w(c) if y > c:

and

Let fe be given by (2.10), where we put g instead of f and choose c > kmkL such that (1 1 ) and (2 2 ) are strict lower and upper functions of (2.14)

u = g(t u u ) u(a) ; u(b) = 0 u (a) = we(u (b)): 00

0

0

0

Now consider the parameter system of boundary value problems (2.15)

u ; u = fe(t u u ) u(a) ; u(b) = 0 u (a) = we(u (b)) 2 0 1]: 00

0

0

0

Dening for x 2 C 1 a b] and for a.e. t 2 a b]

(Fe x)(t) = (fe(t x(t) x (t)) 0 we(x (b))) 0

0

we get a continuous operator Fe : C 1 a b] 7! L a b]  R 2 and the system (2.15) can be rewritten as the parameter system of operator equations

u ; L+ Fe u = 0 2 0 1]:

For 2 0 1] a function u 2 satises the relation

Z

u(t) =

b

a b] is a solution to (2.15) if and only if it

C 1

;0 (t s)fe(s u(s) u (s))ds + ;2 (t)we(u (b)) 0

a

0

!

on a b]

where ;0 and ;2 are dened by (1.21) and (1.22). Therefore there is r 2 (0 1) such that

n

1  K (r) = x 2 C 1 a b] : kxkC 1 < r

o

and for any 2 0 1] any solution u to (2.15) belongs to K (r): Thus, the operator I ; L+ Fe is a homotopy on K (r)  0 1] and

;

deg I ; L+ Fe

K (r) = deg ;I K (r) = 1:

Now, let = 1 and let u be an arbitrary solution of the corresponding problem (2.15). We can apply Lemma 2.3 and get (2.4). Hence u is a solution of (2.14). Since g(t x y) > m(t) for a.e. t 2 a b] and all (x y) 2 1 (t) 2 (t)]  R 1 (t) u(t) 2 (t) on a b] and we(y) = w(y) for y 2 ;kmkL kmkL ] we can use Lemma 1.1 and get ku kC < kmkL < c: It follows that u is a solution of (0.1), (0.2). Consequently, we can make use of Lemma 2.2 to show that 1 (t) < u(t) < 2 (t) on a b]: To summarize, for = 1 and for any solution u of (2.15) we have u 2 1 : Since Fe = F on cl(1 ) this means that 0

;



deg I ; L+ F 1 ;  ; = deg I ; L+ Fe 1 = deg I ; L+ Fe

K (r) = 1:

The case that (1.4) and (1.5) are satised instead of (1.1) and (1.2) could be treated in a similar way. Now, we prove an analogous theorem provided 1 2 are ordered in the opposite way, i.e. (2.16)

2 (t) < 1 (t) for all t 2 a b]:

2.5. Theorem. Let (1 1) and (2 2) be respectively strict lower and upper functions of (0.1),(0.2) satisfying (2.16). Further, let us assume that either (1.1)

and (1.2) or (1.4) and (1.5) are satised with m 2 L a b] and U (t) R : Let A 2 R be such that k1 kC + k2 kC + (b ; a)kmkL A and let

n

2 = x 2 C 1 a b] : kxkC < A kx k < kmkL 0

o

C

and there exists t 2 a b] such that 2 (t ) < x(t ) < 1 (t ) : x

Then (2.17)

x

x

x

deg(I ; L+ F 2 ) = ;1:

Proof. Put Ae = A + (b ; a): Assume (1.1) and (1.2) and consider an auxiliary equation (2.18) u = g (t u u ) where 8 > f (t x y) + jm(t)j if x  Ae + 1 >< f (t x y) + (x ; Ae)jm(t)j if Ae < x < Ae + 1 g (t x y ) = > f ( t x y ) if ; Ae x Ae >> f (t x y) + (Ae + x) f (t x y) + jm(t)j] if ; Ae ; 1 < x < ;Ae : ;jm(t)j if x ;Ae ; 1: 00

0

We have g 2 Car( a b]  R 2 ) and (2.19) g(t x y) > ;(jm(t)j + 1) for a.e. t 2 a b] and all (x y) 2 ;(Ae + 2) (Ae + 2)]  R : The couples of functions (1 1 ) and (2 2 ) are respectively strict lower and upper functions to the problem (2.18), (0.2). Furthermore, in virtue of the assumption (1.1), also (3 3 ) = (;(Ae + 2) 0) and (4 4 ) = (Ae + 2 0) are respectively strict lower and upper functions to the problem (2.18), (0.2) which are "well-ordered", i.e. 3 (t) < 4 (t) on a b]: Let us dene sets

n n

o

n

o

o

 = x 2 C 1 a b] : kxkC < Ae + 2 kx kC < kmkL + 1

and

1 = x 2  : 1 (t) < x(t) on a b]

2 = x 2  : x(t) < 2 (t) on a b] and an operator

0

G : x 2 C 1 a b] 7! Gx 2 L a b]  R 2

where

(G x)(t) = (g(t x(t) x (t)) 0 w(x (b))) a.e. on a b]: 0

0

Owing to Theorem 2.4 we have deg(I ; L+ G ) = deg(I ; L+ G 1 ) = deg(I ; L+ G 2 ) = 1: Let us denote  =  n cl(1  2 ): Then

n

 = x 2  : there is t 2 a b] such that 2 (t ) < x(t ) < 1 (t ) x

x

x

o

x

and by the additivity of the degree we have deg(I ; L+ G ) = deg(I ; L+ G ) ; deg(I ; L+ G 1 ) ; deg(I ; L+ G 2 ) = ;1: Let u be a solution to (2.18), (0.2) and let u 2 : Then there is t 2 (a b) such that 2 (t ) u(t ) 1 (t ): Consequently, for any t 2 a b] we have u

u

(2.20)

u

u



ju(t)j = u(t ) +

Z

t

u

tu



u (s)ds k1 kC + k2 kC + (b ; a)ku kC 0

0

wherefrom by (2.19) and Lemma 1.1 the relation kukC < Ae follows. Therefore u is a solution of (0.1), (0.2) and using Lemma 1.1 and (2.20) once more we get ku kC < kmkL and kukC < A i.e. u 2 2 : Consequently, the excision property 0

of the degree yields

deg(I ; L+ G 2 ) = ;1 wherefrom, since G = F on cl(2 ) we obtain (2.17). In the case that (1.4) and (1.5) are satised instead of (1.1) and (1.2) we can argue similarly. The case (2.21) there are r and s 2 a b] such that 1 (r) < 2 (r) and 2 (s) < 1 (s) is treated by the following theorem.

2.6. Theorem. Let (1 1) and (2 2) be respectively strict lower and upper functions of (0.1), (0.2) satisfying (2.21). Further, let us assume that either (1.1) and (1.2) or (1.4) and (1.5) are satised with m 2 L a b] and U (t) R : Let A 2 R be such that k1 kC + k2 kC + (b ; a)kmkL A and let

n

3 = x 2 C 1 a b] : kxkC < A kx k < kmkL and there exist 0

o

C

r s 2 a b] such that 1 (r ) > x(r ) and 2 (s ) < x(s ) : x

x

x

x

x

x

Then

deg(I ; L+ F 3 ) = ;1:

Proof. Let g G Ae 1 2 and  have the same meaning as in the proof of Theorem 2.5. Taking into account that in the case (2.21),  n cl(1  2 ) is the set of all x 2  for which there exist r and s 2 a b] such that 1 (r ) > x(r ) and 2 (s ) < x(s ) it is easy to see that the proof of this theorem can be completed by an argument analogous to that used in the proof of Theorem 2.5. x

x

x

x

x

x

3. Lower and upper functions and topological degree In this section we give proper modications of the results described in the previous section to the case of lower and upper functions which need not be strict. 3.1. Lemma. Let the assumptions of Theorem 2.4 be fullled but with (1 1) and (2 2 ) not necessarily strict. For a.e. t 2 a b] and any  2 0 1] let us put (3.1) !1 (t  ) = sup R 1 ( ) jf (t 1 (t) 1 (t)) ; f (t 1 (t) z )j (3.2) !2 (t  ) = sup R 2 ( ) jf (t 2 (t) 2 (t)) ; f (t 2 (t) z )j: Furthermore, let us dene (3.3) 8 >> f (t 1(t) y) ;1(t) ; !1(t 1 (t) ; x ) if x < 1(t) 1 (t) ; x + 1 < ;x if x 2 1 (t) 2 (t)] h(t x y) = > f (t x y) >: f (t 2(t) y) ;2(t) + !2(t x ; 2 (t) ) if x > 2(t) x ; 2 (t) + 1 and z2

j

t ;z j

z2

j

t ;z j

8 w(;kmk ) + y + kmk >< L L we(y) = > w(y) : w(kmkL ) + y ; kmkL

for y < ;kmkL for jyj kmkL for y > kmkL :

Then h 2 Car( a b]  R 2 ) and for any solution u of the problem

u(a) = u(b) u (a) = we(u (b))

u ; u = h(t u u )

(3.4)

00

0

0

0

the relations (2.4) and (1.3) are true. Proof. The functions ! : a b]  0 1] 7! R + (i = 1 2) given by (3.1) and (3.2) are nondecreasing in the second variable and belong to the class Car( a b] 0 1]): Hence h 2 Car( a b]  R 2 ) as well. Let u be an arbitrary solution of (3.4) and suppose i





u(t0 ) ; 2 (t0 ) = max u(t) ; 2 (t) > 0:  ] t2 a b

In virtue of (0.2) and (1.18) it suces to consider the cases a < t0 < b and

t0 = a: As in the proof of Lemma 2.3 we have u (t0 ) = lim0 2 (t) = 2 (t0 ) 0

t!t

in the former case and u (a) = 2 (a+) in the latter. Making use of the continuity of 2 u and u we conclude that in both cases there are  > 0 and 2 (0 1) such that for all t 2 t0 t0 + ] we have j2 (t) ; u (t)j < < u(ut)(t;) ; (t2)(t+) 1 < u(t) ; 2 (t) 2 and, with respect to (3.2), 0

0

0

jf (t 2 (t) 2 (t)) ; f (t 2 (t) u (t))j !2 (t j2 (t) ; u (t)j) !2 (t u(ut)(t;) ; (t2)(t+) 1 ): 0

0

2

Consequently, by means of (1.15), for any t 2 t0 t0 + ] we get

u (t) ; 2 (t) = u(t) + f (t 2 (t) u (t)) + !2 (t u(ut)(t;) ; (t2)(t+) 1 ) ; 2 (t) ; 2 (t) 2 > + f (t 2 (t) 2 (t)) ; 2 (t) > 0: 00

0

0

0

0

Like in the proof of Lemma 2.3 this yields a contradiction with the assumption that u(t0 ) ; 2 (t0 ) is the maximal value of u(t) ; 2 (t) on a b]: Thus, the relation u(t) 2 (t) is true on a b]: Similarly we can show that 1 (t) u(t) on a b] as well, i.e. u satises (2.4). Therefore u is a solution of (0.1) on a b]: Moreover, we satises (1.2) or (1.5) for all y 2 R : Hence by Lemma 1.1 we get (1.3). 3.2. Lemma. Let the assumptions of Lemma 3.1 be fullled. Then for any  > 0 the couples (1 ;  1 ) and (2 +  2 ) are respectively strict lower and upper functions to the problem (3.4). Proof. Let (1 1 ) and (2 2 ) be respectively lower and upper functions to the problem (0.1), (0.2) such that (2.3) is true. Let an arbitrary  > 0 be given and let us dene e2 (t) = 2 (t) +  on a b]: Obviously, the couple (e2 2 ) satises the boundary conditions (1.18). Further, making use of (1.15) and (3.2), we get for a.e. t 2 a b]

 ) e2 (t) + h(t e2 (t) 2 (t)) =  + f (t 2 (t) 2 (t)) + !2 (t  + 1 > f (t 2 (t) 2 (t))  2 (t): This means that (e2 2 ) are upper functions to (3.4) and e2 is not a solution of 0

(3.4). Now, let us put " = 2+1 . Since " < 2 for any t 2 a b] and any couple 2 (x y) 2 R 2 such that (3.5) jx ; e2 (t)j < " and jy ; 2 (t)j < " x ; 2 (t) and hence also we obtain x ; 2 (t) > 2 and jy ; 2 (t)j < x ; 2 (t) + 1 !2 (t jy ; 2 (t)j) !2 (t x ;x ; (t2)(t+) 1 ): 2 Consequently, for a.e. t 2 a b] and all (x y) 2 R 2 fullling (3.5) we can compute x + h(t x y)  x ; 2 (t) + !2 (t x ;x ; (t2)(t+) 1 ) ; !2 (t jy ; 2 (t)j) + f (t 2 (t) 2 (t)) 2 > f (t 2 (t) 2 (t))  2 (t) 







0

i.e. the functions (e2 2 ) are strict upper functions to the problem (3.4). Analogously we could show that for any  > 0 the functions (1 ;  1 ) are strict lower functions of (3.4).

3.3. Theorem. Let the assumptions of Theorem 2.4 be fullled, but with (1 1)

and (2 2 ) not necessarily strict. Then either the problem (0.1), (0.2) has a solution which belongs to @ 1 or

deg(I ; L+ F 1 ) = 1:

(3.6)

Proof. Let (1 1 ) and (2 2 ) be respectively lower and upper functions to the problem (0.1), (0.2) fullling the relation (2.3). Let us choose an arbitrary  > 0: By Lemma 3.2 the couples (1 ;  1 ) and (2 +  2 ) are respectively strict lower and upper functions to the modied problem (3.4). It means that by Theorem 2.4

deg(I ; L+ H  ) = 1 

where H : x 2 C 1 a b] 7! Hx 2 L a b]  R 2 (Hx)(t) = (h(t x(t) x (t)) 0 we(x (b))) a.e. in a b] o  = x 2 C 1 a b] : 1 (t) ;  < x(t) < 2 (t) +  on a b] and kx kC < km e kL

n

0

0

0



and either m e (t) = m(t) ;  ; !1(t 1) or me (t) = m(t) +  + !2(t 1) (according to whether we assume (1.1), (1.2) or (1.4), (1.5)). On the other hand, by Lemma 3.1 the problem (3.4) does not possess any solution in  n cl(1 ): Moreover, H = F on cl(1 ) and so if the problem (0.1), (0.2) has no solution belonging to @ 1 the modied problem (3.4) has no solution belonging to @ 1 either. Therefore, by the excision property of the degree we have (3.6). 

In the case that 1 and 2 full the relation (2.16) or (2.21), making use of Theorem 3.3 we can modify the proofs of Theorems 2.5 and 2.6 in such a way that we get the following assertions.

3.4. Theorem. Let the assumptions of Theorem 2.5 be fullled, but with (1 1)

and (2 2 ) not necessarily strict. Then either the problem (0.1), (0.2) has a solution which belongs to @ 2 or

deg(I ; L+ F 2 ) = ;1:

Proof. Let (1 1 ) and (2 2 ) be respectively lower and upper functions to the problem (0.1), (0.2) and let m A Ae g G (3 3 ), (4 4 ),  1 2 and  have the same meaning as in the proof of Theorem 2.5. The couples (1 1 ) and (2 2 ) are respectively lower and upper functions to the problem (2.18), (0.2) which need not be strict now. By Theorem 2.4 we have again deg(I ; L+ G ) = 1: Let u be a solution of (0.1), (0.2) such that u 2 @ 2: Then u is also a solution to (2.18), (0.2). Moreover, as in the proof of Theorem 2.5, making use of (2.20) and of Lemma 1.1 we can show that (3.7) kukC < A and ku kC < kmkL : Thus, there exist i 2 f1 2g and t 2 a b] such that (3.8) u(t ) =  (t ) i.e. u 2 @  . On the other hand, let u be a solution of (2.18), (0.2) such that u 2 @ 1 @ 2 : By Lemma 2.2 we have ;(Ae +2) < u(t) < Ae +2 on a b]: Furthermore, by (2.19) and Lemma 1.1 we get ku kC < kmkL + 1: As in the proof of Theorem 2.5 this implies by (2.20) that kukC < Ae i.e. u is a solution of (0.1), (0.2). Now, using (2.20) and Lemma 1.1 once more we obtain again (3.7) and (3.8), i.e. u 2 @ 2 : To summarize, (0.1), (0.2) possesses a solution belonging to @ 2 if and only if (2.18), (0.2) possesses a solution belonging to @ 1  @ 2 : Consequently, if the problem (0.1), (0.2) possesses no solution u such that u 2 @ 2 then making use of Theorem 3.3 we get deg(I ; L+ G 1 ) = 1 and deg(I ; L+ G 2 ) = 1: Finally, by the same argument as in the proof of Theorem 2.5 we can show that any solution u 2  of the problem (2.18), (0.2) belongs to 2 . Therefore deg(I ; L+ G 2 ) = deg(I ; L+ G ) ; deg(I ; L+ G 1 ) ; deg(I ; L+ G 2 ) = ;1 and taking into account that F = G on cl(2 ) we complete the proof. 3.5. Theorem. Let the assumptions of Theorem 2.6 be fullled, but with (1 1) and (2 2 ) not necessarily strict. Then either the problem (0.1), (0.2) has a solution which belongs to @ 3 or deg(I ; L+ F 3 ) = ;1: Proof follows from Theorem 3.3 by a modication of the proof of Theorem 2.6 similar to that used in the proof of Theorem 3.4. 0

u

u

i

0

i

u

4. Existence theorems Theorems 3.3 - 3.5 give directly existence results for our problem (0.1), (0.2). Similarly as in 7] (cf. Theorem 6) it is possible to show the existence of a solution to this problem even in the cases that the strict inequalities (2.3) and (2.16) are replaced by the non-strict ones.

4.1. Theorem. Let the assumptions of Theorem 2.4 be satised but with (1 1) and (2 2 ) not necessarily strict and instead of (2.3) let us assume

1 (t) 2 (t) on a b]:

(4.1)

Then the problem (0.1), (0.2) possesses a solution u such that u 2 cl(1 ) (with 1 given by (2.13). Proof. Consider an auxiliary problem

u = fe(t u u ) (0:2)

(4.2)

00

0

where fe is for a.e. t 2 a b] and any y 2 R given by

 ef (t x y) = ff ((tt x y(t)) y) 2

if x 2 (t) if x > 2 (t):

Clearly, (1 1 ) are lower functions to (4.2). Now, let an arbitrary k 2 N be given. The functions (2 + 1 2 ) are then upper functions to (4.2) and by Theorem 3.3 the problem (4.2) possesses a solution x such that k

k

x (t) 2 1 (t) 2 (t) + 1 ] on a b] and kx kC kmkL : 0

k

k

k

Using the Arzel-Ascoli theorem and the Lebesgue Dominated Convergence Theorem for the sequence fx g we get a solution x 2 cl(1 ) of (0.1), (0.2) as a C 1 limit of a proper subsequence of fx g: k

k

4.2. Theorem. Let the assumptions of Theorem 2.5 be satised but with (1 1) and (2 2 ) not necessarily strict and instead of (2.16) let us assume

(4.3)

2 (t) 1 (t) on a b]:

Then the problem (0.1), (0.2) possesses a solution u such that u 2 cl(2 ) (with 2 given in Theorem 2.5).

Proof. For any k 2 N a.e. t 2 a b] and any x y 2 R put

;

;

g (t x y) = k f (t 2 (t) y) ; f (t x y) x ; (2 (t) ; 1 ) k

and

8 f (t x y) >< fe (t x y) = > ff ((tt 2 ((tt)) yy)) + g (t x y) : f (t x2y) k

k

The couples (1 1 ) and (2 ; functions to

1

k



k

if if if if

x < 2 (t) ; 2 x 2 2 ; 2 2 (t) ; 1 ) x 2 2 ; 1 2 (t)) x  2 (t): k

k

k

k

2 ) are then respectively lower and upper

u = fe (t u u ) (0:2)

(4.4)

00

k

0

and satisfy (2.16). It is easy to verify that for any k 2 N the function fe satises the assumptions for f of Theorem 2.5 with the same m 2 L a b]: Thus by Theorem 3.4 for any k 2 N there are a solution x to the problem (4.4) and a point s 2 a b] such that k

k

k

kx kC A + 1 k

k

kx kC kmkL and 2 (s ) ; 1 x (s ) 1 (s ) 0

k

k

k

k

k

k

where A has the same meaning as in Theorem 2.5. Using the compactness of the interval a b] and the Arzel-Ascoli theorem we get the existence of a subsequence fx ` g in fx g s 2 a b] and x 2 C 1 a b] such that k

k



lim kx ` ; xkC 1 = 0 and

`!1

k

lim s ` = s :

`!1

k



Obviously, x 2 cl(2 ) and by virtue of the Lebesgue Dominated Convergence Theorem, x is a solution of (0.1), (0.2).

4.3. Theorem. Let the assumptions of Theorem 2.6 be satised but with (1 1) and (2 2 ) not necessarily strict. Then the problem (0.1), (0.2) possesses a solution u such that u 2 cl(3 ) (with 3 given in Theorem 2.6). Proof. If 1 and 2 satisfy neither (4.1) nor (4.2), they full (2.21) and hence by Theorem 3.5 we have a solution u 2 cl(3 ) to (0.1), (0.2).

4.4. Corollary. Let z1 z2 2 C a b] (4.5)

m1 = max z1 (t) < m2 = min z2 (t)  ]  ] t2 a b

t2 a b

and let for a.e. t 2 a b] and all x y 2 R

f (t x y) < 0 if x 2 (z1 (t) z2 (t))

(4.6) and (4.7)

f (t x y) > 0 if x < z1 (t) or x > z2 (t): Further, let m 2 L a b] be such that (1.1) is satised with U (t) = z1 (t) z2 (t)] t 2 a b]: Then (i) there are at least two dierent solutions u and v to the periodic boundary value problem

u = f (t u u )

(4.8)

0

00

u(a) = u(b)

u (a) = u (b) 0

0

such that

v(t ) m1 for some t 2 a b]

(4.9)

v

v

and

maxfm2 v(t)g u(t) on a b]

(4.10)

(ii) if we suppose in addition that for any compact K  m2 1)  R there is a nonnegative function h 2 L a b] such that k

(4.11) f (t x1 y1) ; f (t x2 y2 ) > ;h (t)jy1 ; y2 j for a.e. t 2 a b] and all (x1 y1 ) (x2 y2 ) 2 K such that x1 > x2 k

then u is the only solution of (4.8) bounded below by m2 : Proof. (i) Without any loss of generality we may assume that m(t) 0 a.e. on

a b] i.e. we have

f (t x y)  m(t) for a.e. t 2 a b] and all (x y) 2 R 2 : Furthermore, by (4.5) there are r1 r2 such that r1 < min z1 (t) m1 < m2 max z2 (t) < r2 :  ]  ] t2 a b

t2 a b

According to (4.6) the couples (m1 0) and (m2 0) are lower functions of (4.8) and by (4.7) the couples (r1 0) and (r2 0) are upper functions of (4.8). Hence, by Theorems 4.1 and 4.2 there are solutions v and v1 of (4.8) such that

r1 < v(t ) < m1 for some t 2 (a b) and m2 v1 (t) r2 for all t 2 a b]: v

v

Suppose that v and v1 are not ordered on a b] i.e. there is s such that v1 (s ) < v(s ) and set v

v

v

1 (t) = maxfv(t) v1 (t)g for t 2 a b]:

(4.12)

Then 1 2 A C a b] 1 2 B V a b] 1 is not a solution of (4.8) but the functions (1 1 ) are lower functions of (4.8). According to (4.7) we can nd a number r > k1 kC such that (r 0) are upper functions of (4.8). This implies the existence of a solution u of (4.8) satisfying 1 (t) u(t) r on a b]: Provided v and v1 are ordered, we set u = v1 : (ii) Suppose (4.11) and let u1 6= u be a solution of (4.8) such that m2 u1(t) on a b]: Set z (t) = u1 (t) ; u(t) and choose a compact K such that (u(t) u (t)) 2 K and (u1 (t) u1 (t)) 2 K for all t 2 a b]: We can assume that max  ] z (t) = z (t0 ) > 0 and z (t0 ) = 0 for some t0 2 a b): Then there exists t > t0 such that z (t ) 0 and z (t) > 0 on t0 t ]: Now, (4.11) implies 0

0







0

0

0

t2 a b



0

;







z (t) > ;h (t)jz (t)j = ; h (t) sgn(z (t)) z (t) for a.e. t 2 t0 t ]: 00

0

k

Thus,

0

k

0



Z ;   z (t) exp h (s) sgn(z (s)) ds > 0 on t0 t ]



t

0

0

0

k



t0

and

z (t ) exp 0



Z ;   h (s) sgn(z (s)) ds > z (t0 ) = 0 t

0

k

0

t0

a contradiction.

4.5. Remark. Provided z is a constant function for some i 2 f1 2g it is a solution of (4.8). In this case we can set v(t) = z (t): If z1 is not constant, then there exists s 2 a b] such that v(s ) > z1(s ): Similarly, if z2 is not constant, we get u(t ) < z2 (t ) for some t 2 a b]: These observations follow from the i

i

v

u

v

u

v

u

fact that any solution of (4.8) cannot have all its values outside (z1 (t) z2 (t)):

4.6. Remark. In the case that f (t x y) g(t x) the assertion (i) of Corollary 4.4 is fullled under the assumptions (4.5), (4.6) and (4.7). Thus our Corollary 4.4 generalizes Theorem 4.7 from 3]. Further, the assertion (ii) of Corollary 4.4 is true provided g is increasing in x on m2 1) for a.e. t 2 a b]:

4.7. Remark. The lower and upper functions method which is described in this section (cf. Theorems 4.1 - 4.3 and Corollary 4.4) can be used for singular boundary value problems, as well. For multiplicity results for periodic boundary value problems which were obtained by this method, see 8]. 4.8. Remark. Conditions ensuring the existence of constant lower and upper functions of the problem (0.1), (0.2) were mentioned in Remark 1.8. In the proof of Corollary 4.4 we constructed nonconstant lower functions whose rst component was the maximum of two solutions of the problem (4.8) (cf. (4.12)). In general, it is not easy to nd conditions which guarantee the existence of nonconstant lower and upper functions. One of the possibilities is shown in 9] where they are constructed as solutions of linear boundary value problems for generalized linear dierential equations.

References

1] C. De Coster, P. Habets: Lower and upper solutions in the theory of ODE boundary value problems: Classical and recent results, in: Nonlinear Analysis and Boundary Value Problems for Ordinary Di
erential Equations (CISM Courses and Lectures vol 371, SpringerVerlag, Wien 1996), pp. 1-78. 2] J. Cronin: Fixed Points and Topological Degree in Nonlinear Analysis, AMS, 1964. 3] S. Gaete, R. F. Mansevich: Existence of a Pair of Periodic Solutions of an O.D.E. Generalizing a Problem in Nonlinear Elasticity, via Variational Methods, J. Math. Anal. Appl. 134 (1988), 257-271. 4] Ch. Fabry and P. Habets: Lower and Upper Solutions for Second-Order Boundary Value Problems with Nonlinear Boundary Conditions, Nonlinear Analysis, TMA 10 (1986), 985-1007. 5] I. Kiguradze: On Some Singular Boundary Value Problems for Nonlinear Ordinary Dierential Equations of the Second Order (in Russian), Di
erencial'nye Uravnenija 4 (1968), 1753-1773. 6] I. Rachunkov: Lower and Upper Solutions and Topological Degree, J. Math. Anal. Appl., 234 (1999), 311-327. 7] I. Rachunkov: Multiplicity Results for Periodic Boundary Value Problems in the Carathodory Case, Faculty of Science, Palack Univ. Olomouc, Preprint 22 (1998), 1-23. 8] I. Rachunkov: Existence of two positive solutions of a singular nonlinear periodic boundary value problems, J. Comput. Appl. Math., to appear. 9] I. Rachunkov and M. Tvrd: On the existence of nonsmooth lower and upper functions to second order nonlinear boundary value problems, in preparation. 10] G. Scorza Dragoni: Il problema dei valori ai limiti studiato in grande per gli integrali di una equazione dierenziale del secondo ordine, Giorn. di Mat. di Battaglini 69 (1931), 77-112. 11] I. Vrko: Comparison of two denitions of lower and upper functions of nonlinear second order dierential equations, J. of Inequal. & Appl., to appear

Irena Rachunkov, Department of Mathematics, Palack University, 779 00 OLOMOUC, Tomkova 40, Czech Republic (e-mail: [email protected]) Milan Tvrd, Mathematical Institute, Academy of Sciences of the Czech Republic, 115 67 PRAHA 1, itn 25, Czech Republic (e-mail: [email protected])