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An m-point boundary value problem of Neumann type for a p-Laplacian like operator
Nonlinear Analysis 56 (2004) 1071 – 1089
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An m-point boundary value problem of Neumann type for a p-Laplacian like operator M. Garc)*a-Huidobroa , C.P. Guptab;∗ , R. Man)asevichc a Departamento
de Matem aticas, Ponticia Universidad Cat olica de Chile, Casilla 306, Correo 22, Santiago, Chile b Department of Mathematics, 084, University of Nevada, Reno, Reno, NV 89557, USA c Departamento de Ingenier *a Matem atica, Universidad de Chile and Centro de Modelamiento Matem atico, Casilla 170, Correo 3, Santiago, Chile Received 13 August 2003; received in revised form 27 September 2003; accepted 6 November 2003
Abstract Let and be two odd increasing homeomorphism from R onto R with (0) = 0; (0) = 0, and let f : [0; 1] × R × R → R be a function satisfying Caratheodory’s conditions. Let ai ∈ R; i ∈ (0; 1); i = 1; 2; : : : ; m − 2; 0 ¡ 1 ¡ 2 ¡ · · · ¡ m−2 ¡ 1 be given. We are interested in the problem of existence of solutions for the m-point boundary value problem: ((x )) = f(t; x; x ); t ∈ (0; 1); x (0) = 0; (x (1)) =
m−2
ai (x ( i )):
i=1
We note that this non-linear m-point boundary value problem is always at resonance since the associated m-point boundary value problem ((x )) = 0; t ∈ (0; 1); x (0) = 0; (x (1)) =
m−2
ai (x ( i ))
i=1
has non-trivial solutions x(t) = ; ∈ R (an arbitrary constant). Our results are obtained by a suitable homotopy, Leray–Schauder degree properties, and a priori bounds. ? 2003 Elsevier Ltd. All rights reserved. MG-H was supported by Fondecyt grant 1030593, Ch. G. was supported by Fondecyt grant 7030029, and RM was supported by Fondap M.A. and Milenio grant-P01-34. ∗
Corresponding author. E-mail addresses: [email protected] (M. Garc)*a-Huidobro), [email protected] (C.P. Gupta), [email protected] (R. Man)asevich). 0362-546X/$ - see front matter ? 2003 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2003.11.003
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1. Introduction Let and be two odd increasing homeomorphisms from R onto R with (0) = 0; (0) = 0 and let f : [0; 1] × R × R → R be a function satisfying Caratheodory’s conditions. Let ai ∈ R; i ∈ (0; 1); i = 1; 2; : : : ; m − 2; 0 ¡ 1 ¡ 2 ¡ · · · ¡ m−2 ¡ 1 be given. We are interested in the problem of existence of solutions for the m-point boundary value problem: ((x )) = f(t; x; x ); t ∈ (0; 1); m−2
x (0) = 0; (x (1)) =
ai (x ( i )):
(1)
i=1
Notice that problem (1) is always at resonance in the sense that the associated m-point boundary value problem ((x (t))) = 0; 0 ¡ t ¡ 1; m−2
x (0) = 0; (x (1)) =
ai (x ( i ))
(2)
i=1
has the non-trivial solution x(t) = , where ∈ R is an arbitrary constant. Let us consider the problem x = f(t; x; x );
t ∈ (0; 1);
x (0) = 0; x (1) =
m−2
ai x ( i );
(3)
i=1
that corresponds to problem (1) when (u) = (u) ≡ u. The diDerential equation in problem (3) has been studied extensively by many authors with diDerent boundary conditions, namely, m−2 x(0) = 0; x (1) = ai x ( i ) or i=1
x (0) = 0; x(1) =
m−2
ai x( i ) or
i=1
x(0) = 0; x(1) =
m−2
ai x( i );
i=1
for example in [2,3,7–14]. A natural generalization for problem (3) is (p (x )) = f(t; x; x ); t ∈ (0; 1); x (0) = 0; x (1) =
m−2 i=1
ai x ( i ):
(4)
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where p ¿ 1, and (p (x )) = (|x |p−2 x ) denotes the so-called one-dimensional p-Laplace operator. We recall that this operator is generated by the function p : R → R deGned by p (s) = |s|p−2 s, if s = 0 and p (0) = 0. We refer to [1,5,6] as examples of recent works for (4) with diDerent boundary conditions. Another generalization for problem (3) is x = f(t; x; x ); t ∈ (0; 1); x (0) = 0; (x (1)) =
m−2
ai (x ( i )):
(5)
i=1
Problem (5) is in the spirit of the nonlinear boundary value problem studied by Gaines-Mawhin in [4]. Thus the motivation to study problem (1) comes from the fact that problems (3)–(5) are special cases of (1). Additional motivation comes from the challenge to Gnd simple conditions on the function f, for the existence of a solution, as the non-linear homeomorphisms and generating, respectively, the diDerential operator and the boundary condition in (1) are diDerent (in general). We shall denote by C[0; 1] (resp. C 1 [0; 1]) the classical space of continuous (resp. continuously diDerentiable) real-valued functions on the interval [0; 1]. The norm in C[0; 1] is denoted by | · |∞ . Also, we shall denote by L1 (0; 1) the space of real-valued (equivalence classes of) functions whose absolute value is Lebesgue integrable on (0; 1). The Brouwer and Leray–Schauder degree shall be, respectively, denoted by degB and degLS . This paper is organized as follows. In Section 2 we present an abstract formulation of the boundary value problem (1) and establish a general continuation Lemma. In Section 3 we apply this Lemma to obtain several existence theorems for the boundary value problem (1). In all the cases the key point is to obtain a priori bounds. This is done under suitable growth conditions together with coerciveness conditions. We illustrate some of our results, by means of simple examples containing the p-Laplace operator. Example 1. For p ¿ 1 and q ¿ 1 let us consider the problem (p (x )) = (t)p (x) − A|x |p−1 − B; t ∈ (0; 1); x (0) = 0; q (x (1)) =
m−2 i=1
ai q (x ( i ));
06
m−2
ai 6 1;
(6)
i=1
where : [0; 1] → R is a nonnegative continuous function such that (t) ¿ 0 ¿ 0 for all t ∈ [0; 1], and A and B are positive constants with A ¡ 1. Then, if ( 1 =1 − A)p −1 (1 + (A=0 )p −1 ) ¡ 1, the boundary value problem (6) has 1 at least one solution x ∈ C [0; 1].
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Example 2. Let us consider the problem (p (x )) = g(t; x)(p (|x |) + 1); t ∈ (0; 1); x (0) = 0; q (x (1)) =
m−2
ai q (x ( i ));
06
i=1
m−2
ai 6 1;
(7)
i=1
where g is any continuous function such that ug(t; u) ¿ 0, for all |u| ¿ M ¿ 0; where M is a constant, p ¿ 1; and q ¿ 1. Then problem (7) has at least one solution x ∈ C 1 [0; 1]. Remark 3. A model for the function g is the following: g(t; x) = 1 + t 2 + |x|m−2 x; where m ¿ 2. In particular we can take m ¿ p and have lim|x|→∞ g(t; x)=p (x) = ∞, uniformly for t ∈ [0; T ]. Example 1 is a direct consequence of Theorem 5 and Example 2 is a consequence of Theorem 7. We leave to the interested reader to verify the assumptions of the corresponding theorems. 2. Abstract formulation and a deformation lemma We begin this section by formulating a general continuation theorem for the solvability of the boundary value problem (1). Let f∗ : [0; 1] × R × R × [0; 1] → R be a given function satisfying Caratheodory’s conditions, i.e. (i) for all (s; r; ) ∈ R ×R ×[0; 1] the function f∗ (·; s; r; ) is measurable on [0; 1], (ii) for a.e. t ∈ [0; 1] the function f∗ (t; ·; ·; ·) is continuous on R × R × [0; 1], and (iii) for each R ¿ 0 there exists a Lebesgue integrable function R : [0; 1] → R such that |f∗ (t; s; r; )| 6 R (t) for a.e. t ∈ [0; 1] and all (s; r; ) ∈ R × R × [0; 1] with |s| 6 R, and |r| 6 R. We suppose, further, that f∗ (t; s; r; 1) = f(t; s; r) for all (t; s; r) ∈ [0; 1] × R × R: Here the function f is as given in the introduction. To formulate our continuation lemma we introduce an operator B : C 1 [0; 1] × [0; 1] → R by m−2 m−2 B(x; ) = ((x (1)) − ai (x ( i ))) + (1 − )((x (1)) − ai (x ( i ))): i=1
i=1
(8)
For ∈ [0; 1] we consider the family of boundary value problems: ((x )) = f∗ (t; x; x ; );
t ∈ (0; 1);
x (0) = 0; B(x; ) = 0:
(9)
We also deGne a function F : R → R by setting 1 m−2 i F() = f∗ (t; ; 0; 0) dt − ai f∗ (t; ; 0; 0) dt: 0
i=1
0
We have the following continuation lemma.
(10)
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Lemma 4. Let
⊂ C 1 [0; 1] be a bounded open set and set
0
=
1075
∩ R. Assume that
(i) for each ∈ (0; 1) problem (9) has no solution x ∈ @ , (ii) the equation F() = 0, has no solution with ∈ @ ∩ R, and (iii) the Brouwer degree degB (F; 0 ; 0) = 0. Then, problem (1) has at least one solution in K . Proof. Let us deGne "∗ : C 1 [0; 1] × [0; 1] → C 1 [0; 1] by setting 1 ∗ ∗ f ($; x($); x ($); ) d$ " (x; )(t) = x(0) + # 0
m−2
−
i ∗ f ($; x($); x ($); ) d$ ai # 0
i=1
+ (1 − )
m−2
−
ai
i=1
+
t
0
0
f∗ ($; x($); x ($); ) d$
i
0
−1
1
f∗ ($; x($); x ($); ) d$
s ∗ f ($; x($); x ($); ) d$ ds;
(11)
0
where # is the increasing homeomorphism from R onto R given by # = ◦ −1 . We note, from our assumption that the function f∗ satisGes Caratheodory’s conditions, so that for (x; ) ∈ C 1 [0; 1] × [0; 1] the function t → f∗ (t; x(t); x (t); ) is in L1 (0; 1). s Accordingly, the function s ∈ [0; 1] → 0 f∗ ($; x($); x ($); ) d$ is absolutely continuous on [0; 1], and hence the operator "∗ is well deGned. Next, let us suppose that x is a solution to problem (9) for some ∈ (0; 1]. By integrating the equation in (9) and using that x (0) = 0; we have that t f∗ ($; x($); x ($); ) d$: (12) (x (t)) = 0
Evaluating x (t) from (12) and substituting (x (t)) and x (t) into B(x; ) = 0, we Gnd m−2 1 i ∗ ∗ f ($; x($); x ($); ) d$ − ai # f ($; x($); x ($); ) d$ # 0
+ (1 − ) = 0:
0
i=1 1
∗
f ($; x($); x ($); ) d$ −
0
m−2 i=1
ai
0
i
∗
f ($; x($); x ($); ) d$ (13)
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It is then immediate to see from (11), (12), and (13) that x satisGes x(t) = "∗ (x; )(t);
t ∈ [0; 1]:
(14)
Conversely, let us suppose that for some ∈ (0; 1]; x ∈ C 1 [0; 1] satisGes (14). We Grst see from Eqs. (11) and (14), by evaluating at t = 0, that (13) is satisGed. Next, by diDerentiating Eq. (14), we obtain that (12) holds and hence evaluating this equation at t = 0 we see that x (0) = 0. Finally, by (14) evaluated at t = 0, and combining with (12), we obtain that x satisGes B(x; ) = 0. Also, since (12) implies that (x ) is absolutely continuous on [0; 1], we Gnally obtain and by diDerentiation that ((x (t))) = f∗ (t; x(t); x (t); ); t ∈ (0; 1): Accordingly, for ∈ (0; 1]; we have proved that x ∈ C 1 [0; 1] is a solution of the boundary value problem (9) if and only if x(t); t ∈ [0; 1], is a solution of Eq. (14). We observe, by using standard arguments, that "∗ : C 1 [0; 1] × [0; 1] → C 1 [0; 1] is a completely continuous operator. Now, if there is a function x ∈ @ which satisGes problem (1) then we are done. Accordingly, let us assume that problem (1) has no solution on @ . Since, now, f∗ (t; s; r; 1) = f(t; s; r) for all (t; s; r) ∈ [0; 1] × R × R we see that assumption (i) of the lemma implies that x = "∗ (x; ) for all u ∈ @
and ∈ (0; 1]:
We, next, assert that x = "∗ (x; 0) for all x ∈ @ . Indeed, let x ∈ @ be such that x = "∗ (x; 0). It, then, follows from (11) that x(t) = , x (t) = 0, and x ∈ @ ∩ R, where 1 m−2 i f∗ ($; x($); x ($); 0) d$ − ai f∗ ($; x($); x ($); 0) d$ = x(0) + 0
i=1
0
= + F();
(15)
which is not possible since it implies that F() = 0, contradicting assumption (ii) of the lemma. We thus get that x = "∗ (x; ) for all x ∈ @
and ∈ [0; 1]:
Finally, we use the homotopy invariance property of Leray–Schauder degree to get degLS (I − "∗ (·; 1); ; 0) = degLS (I − "∗ (·; 0); ; 0) = degB (I − "∗ (·; 0)|R ; = degB (F;
0 ; 0)
0 ; 0)
= 0:
Thus, the mapping " ≡ "∗ (·; 1) : C 1 [0; 1] → C 1 [0; 1] has at least one Gxed-point in and hence problem (1) has at least one solution in . This completes the proof of the lemma.
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3. Existence theorems m−2 Theorem 5. In problem (1), let the constants ai ; i = 1; : : : ; m − 2 satisfy 0 6 i=1 ai 6 1, and let f : [0; 1]×R×R → R be a continuous function satisfying the following conditions: (i) there exist non-negative functions d1 (t); d2 (t), and r(t) in L1 (0; 1) such that |f(t; u; v)| 6 d1 (t)(|u|) + d2 (t)(|v|) + r(t); for a.e. t ∈ [0; 1] and all u; v ∈ R, (ii) there exist constants ¿ 0; A; B ¿ 0, and a u0 ¿ 0 such that for all u with |u| ¿ u0 , all t ∈ [0; 1] and all v ∈ R, one has |f(t; u; v)| ¿ (|u|) − A(|v|) − B; (iii) there exists an R ¿ max{u0 ; B=} such that f(t; ; 0)f(t; −; 0) ¡ 0 for all ¿ R and all t ∈ [0; 1]. (iv) Suppose, further, d2 L1 (0; 1) ¡ 1, and the function ( : [0; ∞) → [0; ∞), dened by
d1 L1 (0; 1)
r L1 (0; 1) ((z) = −1 (z) + 1 − d2 L1 (0; 1) 1 − d2 L1 (0; 1) A d1 L1 (0; 1) A r L1 (0; 1) B −1 + (z) + + (1 − d2 L1 (0; 1) ) (1 − d2 L1 (0; 1) ) satises the condition lim sup z→∞
((z) ¡ 1: z
(16)
Then problem (1) has at least one solution x ∈ C 1 [0; 1]. Remark 6. Notice that condition (16) implies that d1 L1 (0; 1) + d2 L1 (0; 1) ¡ 1. The converse is not necessarily true for general . Proof. We consider the family of boundary value problems (9) with f∗ (t; u; v; ) = f(t; u; v) for all (t; u; v; ) ∈ [0; 1]×R×R×[0; 1], i.e. we consider the family of problems: ((x )) = f(t; x; x ); t ∈ (0; 1); ∈ [0; 1]; x (0) = 0;
B(x; ) = 0:
(17)
We shall show that the family of problems (17) satisfy the conditions of Lemma 4 to conclude that problem (1) has at least one solution in C 1 [0; 1]. Let x ∈ C 1 [0; 1] be a solution to problem (17) for some ∈ (0; 1). Then, by integrating the equation in (17) from 0 to t, t ∈ [0; 1], using x (0) = 0, assumption (i), and the fact that is an odd increasing homeomorphism from R onto R, we Grst Gnd that (|x (t)|) 6 ( x ∞ ) d1 L1 (0; 1) + ( x ∞ ) d2 L1 (0; 1) + r L1 (0; 1)
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for all t ∈ [0; 1], and then that
d1 L1 (0; 1)
r L1 (0; 1)
x ∞ 6 −1 : ( x ∞ ) + 1 − d2 L1 (0; 1) 1 − d2 L1 (0; 1)
(18)
By integrating the equation in (17) on the interval [0; t] and using that x (0) = 0; we see that t (x (t)) = f($; x($); x ($)) d$; 0
from which B(x; ) = 0 becomes m−2 1 i f($; x($); x ($)) d$ − ai # f($; x($); x ($)) d$ # 0
+ (1 − )
0
i=1 1
0
f($; x($); x ($)) d$ −
m−2 i=1
ai
0
i
f($; x($); x ($)) d$
= 0:
(19)
We claim that there must exist a $1 ∈ (0; 1) such that f($1 ; x($1 ); x ($1 )) = 0. Indeed, assume this is not true. Now, let us assume m−2without loss of generality that f($; x($); x ($)) ¿ 0 for all $ ∈ [0; 1]. Next, if T = i=1 ai = 0, then we get ai = 0 for i = 1; 2; : : : ; m − 2 since 0 6 ai 6 1 for i = 1; 2; : : : ; m − 2. Accordingly, we get from (19) that 1 1 0=# f($; x($); x ($)) d$ + (1 − ) f($; x($); x ($)) d$ ¿ 0; (20) 0
0
in view of the assumption f($; x($); x ($)) ¿ 0 for all $ ∈ [0; 1]. This is a contradiction, m−2 so there must exist a $1 ∈ (0; 1) such that f($1 ; x($1 ); x ($1 ))=0, when T = i=1 ai =0. m−2 Now let T = i=1 ai = 0 and let us set m−2 i ai # f($; x($); x ($)) d$ I = 0
i=1
and let + and − in [0; 1] be such that
+ i f($; x($); x ($)) d$ = max f($; x($); x ($)) d$|i = 1; m − 2 0
and
0
Then
0
−
f($; x($); x ($)) d$ = min
#
0
0
i
f($; x($); x ($)) d$|i = 1; m − 2 :
+ I f($; x($); x ($)) d$ 6 6 # f($; x($); x ($)) d$ ; T 0
−
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t ai . Considering the continuous function S(t)=#( 0 f($; x($); x ($)) d$), I we obtain that S( − ) 6 6 S( + ) and thus there must exists a t ∈ (0; 1) such that T I = TS(t ). Similarly there is an +0 ∈ (0; 1) such that +0 m−2 i ai f($; x($); x ($)) d$ = T f($; x($); x ($)) d$: m−2
where T =
i=1
0
i=1
0
In this form (19) can be written as t 1 # f($; x($); x ($)) d$ − T# f($; x($); x ($)) d$ 0
0
+ (1 − )
1
+0
f($; x($); x ($)) d$ + (1 − T )
0
+0
f($; x($); x ($)) d$
= 0:
(21)
Hence by using the fact that T 6 1; # is increasing and the function f is continuous and positive, we obtain from (21) the contradiction 1 1 f($; x($); x ($)) d$ ¡ f($; x($); x ($)) d$; 0
0
proving our claim, that is, the existence of $1 ∈ (0; 1) such that f($1 ; x($1 ); x ($1 )) = 0. Next let us suppose that |x(t)| ¿ u0 for all t ∈ [0; 1]. This along with assumption (ii) then gives B A (|x($1 )|) 6 + ( x ∞ ): (22) Since, now, for t ∈ [0; 1], t x (s) ds; x(t) = x($1 ) + $1
we see from Eqs. (18) and (22) that
x ∞ 6 |x($1 )| + x ∞ B A 6 −1 + ( x ∞ )
d1 L1 (0; 1)
r L1 (0; 1) −1 ( x ∞ ) + + 1 − d2 L1 (0; 1) 1 − d2 L1 (0; 1) A d1 L1 (0; 1) A r L1 (0; 1) B 6 −1 ( x ∞ ) + + (1 − d2 L1 (0; 1) ) (1 − d2 L1 (0; 1) )
d1 L1 (0; 1)
r L1 (0; 1) −1 + ( x ∞ ) + 1 − d2 L1 (0; 1) 1 − d2 L1 (0; 1) = (( x ∞ ):
(23)
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Next, let us suppose that there exists a t0 ∈ [0; 1] such that |x(t0 )| 6 u0 . Then we see from t |x(t)| = x(t0 ) + x (s) ds 6 |x(t0 )| + x ∞ t0
6 u0 + x ∞ and Eq. (18) that
x ∞ 6 u0 + −1
for t ∈ [0; 1]
d1 L1 (0; 1)
r L1 (0; 1) ( x ∞ ) + 1 − d2 L1 (0; 1) 1 − d2 L1 (0; 1)
6 u0 + (( x ∞ ):
(24)
It, now, follows from our assumption lim supz→∞ ((z)=z ¡ 1 and Eqs. (23) and (24) that there exists a constant C, independent of ∈ (0; 1] such that
x ∞ 6 C: Then by using (18) there is a constant, which we again denote by C, independent of ∈ (0; 1] such that
x C 1 [0; 1] 6 C: We have thus shown that if x(t) ∈ C 1 [0; 1] is a solution to problem (17), equivalently for the equation x = "∗ (x; ), for some ∈ (0; 1] then
x C 1 [0; 1] 6 C; where C is a constant independent of ∈ (0; 1]. It follows that there exists an R0 ¿ R, where R is as in assumption (iii), such that for ∈ (0; 1] the family of problems (17), or equivalently the equations x = "∗ (x; ), ˜ ⊂ C 1 [0; 1] for have no solution on the boundary of the bounded open set = B(0; R) ˜ every R ¿ R0 . Accordingly, we see that the family of problems (17) satisfy condition (i) of Lemma 4. Next, we see from assumption (iii) that for all || ¿ R, it holds that |f(t; ; 0)| ¿ 0 for all t ∈ [0; 1], hence writing F() as deGned in (10) in the form 1 m−2 1 F() = (1 − T ) f(t; ; 0) dt + ai f(t; ; 0) dt (25) 0
i=1
i
we see that F() is either strictly positive or strictly negative for all such that || ¿ R. Accordingly, we see that f∗ (t; u; v; ) = f(t; u; v) satisGes condition (ii) of Lemma 4. ˜ and F(−R) ˜ have oppoFinally, we see from assumption (iii) and (25) that F(R) site signs. It follows immediately that the Brouwer degree degB (F; 0 ; 0) = 0. Thus, condition (iii) of Lemma 4 is also satisGed. Thus it follows from Lemma 4 that problem (1) has at least one solution in K . This completes the proof of the theorem. Our next existence theorem gives a diDerent set of conditions on f to which our results apply. In fact, we are able to relax the sublinear assumption (i) in Theorem 5 in order to allow superlinear behavior of f with respect to the variable u.
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Theorem 7. Let f : [0; 1] × R × R → R in the boundary value problem ((x )) = f(t; x; x ); t ∈ (0; 1); x (0) = 0; (x (1)) =
m−2
ai (x ( i ));
06
i=1
m−2
ai 6 1
(26)
i=1
be a continuous functions satisfying the following conditions: (i) there exists M ¿ 0 such that for all |u| ¿ M , all v ∈ R, and all t ∈ [0; 1], one has uf(t; u; v) ¿ 0: (ii) For any K ¿ 0, there exist non-negative functions b1 (t); b2 (t) in L1 (0; 1) such that |f(t; u; v)| 6 b1 (t) + b2 (t)(|v|) for a.e. t ∈ [0; 1] and all u ∈ [ − K; K], and v ∈ R. Then problem (26) has at least one solution x ∈ C 1 [0; 1]. Proof. We consider the family of boundary value problems (9) with f∗ (t; u; v; ) = f(t; u; v) for all (t; u; v; ) ∈ [0; 1]×R×R×[0; 1], i.e. we consider the family of problems: ((x )) = f(t; x; x ); t ∈ (0; 1); ∈ [0; 1]; x (0) = 0; B(x; ) = 0:
(27)
We shall show that the family of problems (27) satisfy the conditions of Lemma 4 to conclude that problem (26) has at least one solution in C 1 [0; 1]. Our Grst step is to deGne an open set ⊂ C 1 such that (i) of Lemma 4 holds. To this end let x be a solution to problem (27) for ∈ (0; 1]. We claim that |x(0)| 6 M; where M is as in assumption (i) of the theorem. Indeed, suppose Grst that x(0) ¿ M . It then follows from x (0) = 0 and assumption (i) of the theorem, that x (t) is strictly increasing and positive on [0; .] for some positive ., and accordingly, x(t) ¿ M and strictly increasing on [0; .]. Let us deGne tM := sup{t ∈ [0; 1]|x is strictly increasing and positive on (0; t]}: Clearly, tM ¿ .; x(tM ) ¿ M and x is strictly increasing on [0; tM ]. We claim that tM =1. Indeed, if tM ¡ 1, then, since x(tM ) ¿ M , there will be a / ¿ 0 with 0 ¡ tM −/ ¡ tM + / ¡ 1 such that x(t) ¿ M on [tM −/; tM +/]. Accordingly, we see from the equation and assumption (i) of the theorem that x is strictly increasing and positive on [tM −/; tM +/] and hence on (0; tM + /], contradicting the deGnition of tM . Hence tM = 1 and x is strictly increasing and positive for all t ∈ (0; 1]. Taking now into account the second boundary condition, we have that either (x (1)) −
m−2 i=1
ai (x ( i )) ¡ 0
or
(x (1)) −
m−2 i=1
ai (x ( i )) ¡ 0
(28)
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and thus, using that 0 6 x (1) ¡
max
i=1;:::; m−2
m−2 i=1
ai 6 1, we obtain
x ( i );
a contradiction to the fact that x is strictly increasing for all t ∈ (0; 1]. Since the same argument applies in the case that x(0) ¡ − M , the claim follows. Let now tM := sup{t ∈ [0; 1] | |x(s)| 6 M for all s ∈ [0; t]}: Clearly, x(tM ) = ±M . By integrating the equation over (0; t), t 6 tM , and using assumption (ii), we obtain that t |(x (t))| 6 b2 (s)|(x (s))| ds + b1 L1 [0; 1] 0
and thus, by Gronwall’s Lemma we Gnd that |(x (t))| 6 b1 L1 [0; 1] b2 L1 [0; 1] eb2 L1 [0;1] := C0
for all t ∈ [0; tM ]:
(29)
If tM = 1, it follows that x is bounded in C 1 [0; 1] and we can proceed as in the proof of Theorem 5 in order to satisfy assumption (i) of Lemma 4. Thus we may assume that tM ¡ 1. Let now t ∈ [tM ; 1], and suppose for instance that x(tM ) = M . Assume that x has a local maximum at some point t0 ∈ (tM ; 1). Then x (t0 ) = 0 and since x(t0 ) is a positive maximum with x(t0 ) ¿ M , (t − t0 )x (t) 6 0; for t ∈ (0; 1] close to t0 : This certainly implies that ((x )) (t0 ) 6 0 and thus 0 ¿ x(t0 )((x )) (t0 ) = x(t0 ) f(t0 ; x(t0 ); 0) ¿ 0; which is a contradiction. Hence x(t) ¿ M for all t ∈ (tM ; 1], and then, by arguing as above we again have that x is strictly increasing and positive for all t ∈ (tM ; 1], and by the second boundary condition it must be that (28) holds. This implies that either (x (1)) ¡ ai (x ( i )) or (x (1)) ¡ ai (x ( i )) x ( i )¿0
and since
x ( i )¿0
ai 6 1;
x ( i )¿0
we conclude that x (1) ¡ max x ( i ): x ( i )¿0
We claim that it must be that the i where this maximum is achieved is less than or equal to tM . Indeed, if this is not the case we obtain a contradiction to the fact that x is strictly increasing for all t ∈ (tM ; 1]. Therefore, by (29) we Gnd that x (1) ¡ −1 (C0 ); proving that the derivative x is uniformly bounded in [0; 1]. Finally, by writing t x(t) = x(0) + x (s) ds; t ∈ [0; 1]; 0
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we obtain that |x(t)| 6 M + −1 (C0 )
for all t ∈ [0; 1]:
It follows that there is a R˜ ¿ M + −1 (C0 ), such that for ∈ (0; 1] the family of problems (27), or equivalently the equations x = "∗ (x; ), have no solution on the ˜ ⊂ C 1 [0; 1]. We see that the family of boundary of the bounded open set = B(0; R) problems (27) satisfy condition (i) of Lemma 4. As in the proof of Theorem 5, we see from assumption (i) that for all ; || ¿ M , it holds that |f(t; ; 0)| ¿ 0 for all t ∈ [0; 1], hence by the continuity of f and since (10) in this case becomes 1 m−2 i F() = f(t; ; 0) dt − ai f(t; ; 0) dt 0
i=1
= (1 − T )
1
0
f(t; ; 0) dt +
0
m−2 i=1
ai
1
i
f(t; ; 0) dt;
m−2 where T = i=1 ai , we conclude that F() is either strictly positive or strictly negative for all || ¿ M . Accordingly, we see that f∗ (t; u; v; ) = f(t; u; v) + g(t; u) satisGes condition (ii) of Lemma 4. Also, again by (i), it holds that f(t; ; 0)f(t; −; 0) ¡ 0 for all || ¿ M , implying that condition (iii) of Lemma 4 is fulGlled. Our next existence theorem concerns the multi-point boundary value problem: ((x )) + f(t; x; x ) = q(t); t ∈ (0; 1); x (0) = 0; (x (1)) =
m−2
ai (x ( i ));
06
i=1
m−2
ai 6 1;
(30)
i=0
where f is continuous, q(t) ∈ L1 (0; 1). We deGne
1 1 qK = max q(s) ds|+ ∈ [0; m−2 ] 1−+ + and
q = min
1 1−+
+
1
q(s) ds|+ ∈ [0; m−2 ] :
Theorem 8. Let q(t) ∈ L1 (0; 1) and f : [0; 1] × R × R → R in problem (30) be a continuous function and satisfying the following conditions: (i) there exist non-negative functions d1 (t); d2 (t), and r(t) in L1 (0; 1) such that |f(t; u; v)| 6 d1 (t)(|u|) + d2 (t)(|v|) + r(t) for a.e. t ∈ [0; 1] and all u; v ∈ R,
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(ii) there exists a d ¿ 0 such that f(t; u; v) ¿ qK for u ¿ d; f(t; u; v) ¡ q for u 6 − d for a.e. t ∈ [0; 1] and all v ∈ R. Suppose, further, d2 L1 (0; 1) ¡ 1, and the function 0 : [0; ∞) → [0; ∞), dened for z ∈ R by
d1 L1 (0; 1)
r
˜ L1 (0; 1) ; 0(z) = −1 (z) + 1 − d2 L1 (0; 1) 1 − d2 L1 (0; 1) where r(t) ˜ = r(t) + q(t), satises the condition lim sup z→∞
0(z) ¡ 1: z
(31)
Then problem (30) has at least one solution x ∈ C 1 [0; 1]. Remark 9. As in Remark 6 condition (31) implies that d1 L1 (0; 1) + d2 L1 (0; 1) ¡ 1. Proof. We shall show that for f∗ (t; u; v; ) = q(t) − f(t; u; v) for all (t; u; v; ) ∈ [0; 1] × R × R × [0; 1], the family of problems ((x )) = (q(t) − f(t; x; x )); t ∈ (0; 1); ∈ [0; 1]; x (0) = 0; B(x; ) = 0:
(32)
satisGes conditions (i) – (iii) of Lemma 4, and where B(x; ) is as deGned in (8). Let x(t) be a solution of (32) for some ∈ (0; 1). Then, by integrating the equation in (32) from 0 to t ∈ (0; 1] and using that x (0) = 0, we obtain that t (33) (x (t)) = (q($) − f($; x($); x ($))) d$: 0
m−2 As in the proof Theorem 5, by setting T = i=1 ai , we obtain from here that there are +0 ; + ∈ [0; m−2 ] such that +0 m−2 i ai (q($) − f($; x($); x ($))) d$) = T (q($) − f($; x($); x ($))) d$ i=1
0
0
m−2 ai # i=1
i
0
=T#
0
(q($) − f($; x($); x ($))) d$
+
(q($) − f($; x($); x ($))) d$ ;
(34)
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and thus B(x; ) = 0 is equivalent to 1 (q($) − f($; x($); x ($))) d$ (1 − T )# 0
1 (q($) − f($; x($); x ($))) d$ +T # 0
−#
0
+
+(1 − ) (1 − T ) +T
1
+0
(q($) − f($; x($); x ($))) d$ 0
1
(q($) − f($; x($); x ($))) d$
(q($) − f($; x($); x ($))) d$
= 0:
(35)
Next we claim that there exists a t˜ ∈ [0; 1] such that − d 6 x(t˜) 6 d:
(36)
Indeed, if we assume that x(t) ¿ d for all t ∈ [0; 1] then in view of the Grst part of assumption (ii) of the theorem we would have 1 + (q($) − f($; x($); x ($))) d$ (q($) − f($; x($); x ($))) d$ − 0
=
0
1
+
(q($) − f($; x($); x ($))) d$ ¡ 0;
implying (by the monotonicity of #) that the second line in (35) adds to a negative number, and 1 1 (q($) − f($; x($); x ($))) d$ ¡ 0; (q($) − f($; x($); x ($))) d$ ¡ 0; +0
0
which cannot be since these three inequalities would imply B(x; ) ¡ 0; which is a contradiction. Similarly the assumption x(t) ¡ − d for all t ∈ [0; 1] leads to a contradiction. This proves (36). As an immediate consequence of (36) we obtain that
x ∞ 6 d + x ∞ :
(37)
Next, we see from assumption (i) of the theorem that ˜ |q(t) − f(t; x(t); x (t))| 6 d1 (t)(|x(t)|) + d2 (t)(|x (t)|) + r(t)
(38)
and hence by integrating the equation in (32) from 0 to t ∈ [0; 1] and using that x (0)=0, we obtain that ˜ L1 (0; 1) : ( x ∞ ) 6 ( x ∞ ) d1 L1 (0; 1) + ( x ∞ ) d2 L1 (0; 1) + r
(39)
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Thus, Grst solving for x ∞ , and then combining with (32) we Grst Gnd that
r
˜ L1 (0; 1)
d1 L1 (0; 1)
x ∞ 6 −1 ( x ∞ ) + 1 − d2 L1 (0; 1) 1 − d2 L1 (0; 1)
(40)
and then
x ∞ 6 d + 0( x ∞ ): Accordingly, we get using our assumption lim sup z→∞
0(z) ¡ 1; z
that there must exist a z0 ¿ 0 such that
x ∞ 6 z0 :
(41)
Finally, by (40) and (41) there is a positive constant R0 ¿ d (where d is as in assumption (ii)) such that
x C 1 [0; 1] 6 R0 : = B(0; R) ⊂ C 1 [0; 1], then for each 0 ¡ ¡ 1 the Thus for any Gxed R ¿ R0 if family of problems (32) has no solution on @ . Thus (i) of Lemma 4 is satisGed. Next, we observe that for our case 1 m−2 i (q($) − f($; ; 0)) d$ − ai (q($) − f($; ; 0)) d$ F() = 0
i=1
= (1 − T )
0
1
(q($) − f($; ; 0)) d$ +
0
m−2 i=1
ai
1
i
(q($) − f($; ; 0)) d$:
(42)
Then assumption (ii) imply that for all ∈ R, with || ¿ d, F() ¡ 0:
(43)
Since this holds when = R; we have that (ii) of Lemma 4 is satisGed. That (iii) of Lemma 4 holds is immediate from (43). We have thus proved that the family of problems (32) satisGes all the conditions of Lemma 4 and hence existence of a solution for problem (1) follows from Lemma 4. This completes the proof of the theorem. Remark 10. Theorem 5 continues to hold if we replace assumption (ii) by the following: “there exists a d ¿ 0 such that f(t; u; v) ¡ q for u ¿ d; f(t; u; v) ¿ qK for u 6 − d; for a.e. t ∈ [0; 1] and all v ∈ R.”
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We next consider the interesting case when = in problem (1), that is, we consider the problem ((x )) = f(t; x; x ); t ∈ (0; 1);
x (0) = 0; (x (1)) =
m−2
ai (x ( i )):
(44)
i=1
where f is Caratheodory, i.e., it satisGes: (i) for all (s; r) ∈ R × R the function f(·; s; r) is measurable on [0; 1], (ii) for a.e. t ∈ [0; 1]f(t; ·; ·) is continuous on R×R, and (iii) for each R ¿ 0 there exists a Lebesgue integrable function R : [0; 1] → R such that |f(t; s; r)| 6 R (t) for a.e. t ∈ [0; 1] and all (s; r) ∈ R × R with |s| 6 R, and |r| 6 R. Theorem 11. Suppose in problem (14) that the following conditions are satised: (i) there exist non-negative functions d1 (t); d2 (t), and r(t) in L1 (0; 1) such that |f(t; u; v)| 6 d1 (t)(|u|) + d2 (t)(|v|) + r(t); for a.e. t ∈ [0; 1] and all u; v ∈ R, (ii) there exists a d ¿ 0 such that for all x ∈ C 1 [0; 1] with mint∈[0; 1] |x(t)| ¿ d it holds that 1 m−2 i f($; x($); x ($)) d$ = 0: (45) ai f($; x($); x ($)) d$ − 0
i=1
0
(iii) For every R ¿ 0, there exists || ¿ R such that F()F(−) ¡ 0, where 1 m−2 i F() = f($; ; 0) d$ − ai f($; ; 0) d$: 0
i=1
0
(iv) Suppose in addition d2 L1 (0; 1) ¡ 1, and the function 0 : [0; ∞) → [0; ∞), dened for z ∈ R by
r L1 (0; 1)
d1 L1 (0; 1) 0(z) = −1 ; (z) + 1 − d2 L1 (0; 1) 1 − d2 L1 (0; 1) satises the condition lim sup z→∞
0(z) ¡ 1: z
Then problem (44) has at least one solution x ∈ C 1 [0; 1]. Proof. We will consider the family of boundary value problems (9) with f∗ (t; x; x ; ) = f(t; x; x ) and where B(x; ); see (8), becomes B(x; ) = (x (1)) −
m−2 i=1
ai (x ( i )):
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Thus (9) becomes the family of problems: ((x )) = f(t; x; x ); t ∈ (0; 1); ∈ [0; 1]; x (0) = 0; (x (1)) =
m−2
ai (x ( i )):
(46)
i=1
Let x ∈ C 1 [0; 1] be a solution to problem (46) for some ∈ (0; 1). By integrating the equation in (46) from 0 to t, t ∈ [0; 1], using x (0) = 0, we obtain as before that (33) holds. Next since for this case, # = ◦ −1 = id, the second boundary condition in (46) reads 1 m−2 i f($; x($); x ($)) d$ − ai f($; x($); x ($)) d$ = 0: 0
i=1
0
Hence, from (ii), it must be that mint∈[0; 1] |x(t)| 6 d. Then we see that for some t0 ∈ [0; 1] it holds that for t ∈ [0; 1] t |x(t)| = |x(t0 ) + x (s) ds| 6 |x(t0 )| + x ∞ t0
6 d + x ∞ : Also, by integrating the equation in (46) from 0 to t, assumption (i), and the fact that is an odd increasing homeomorphism from R onto R, we Grst Gnd that (|x (t)|) 6 ( x ∞ ) d1 L1 (0; 1) + ( x ∞ ) d2 L1 (0; 1) + r L1 (0; 1) for all t ∈ [0; 1], and then that
d1 L1 (0; 1)
r L1 (0; 1) : ( x ∞ ) +
x ∞ 6 −1 1 − d2 L1 (0; 1) 1 − d2 L1 (0; 1) Hence,
x ∞ 6 d +
−1
d1 L1 (0; 1)
r L1 (0; 1) ( x ∞ ) + 1 − d2 L1 (0; 1) 1 − d2 L1 (0; 1)
(47)
6 d + 0( x ∞ ):
(48)
As in the proof of Theorem 8, this implies that there exists R0 ¿ 0 such that
x C 1 [0; 1] 6 R0 : Since by assumption (iii) we have now that all the assumptions of Lemma 4 are satisGed, we conclude that problem (44) has at least one solution in C 1 [0; 1]. Remark m−212. Note that this last theorem holds with no restriction on the value of ai ∈ R or i=1 ai .
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References [1] B. Chuan-zhi Bai, F. Jin-xuan, Existence of multiple positive solutions for nonlinear m-point boundary value problems, Appl. Math. Comput. 140 (2003) 297–305. [2] W. Feng, J.R.L. Webb, Solvability of three-point boundary value problems at resonance, Nonlinear Anal. T.M.A. 30 (1997) 3227–3238. [3] W. Feng, J.R.L. Webb, Solvability of m-point boundary value problems with nonlinear growth, J. Math. Anal. Appl. 212 (1997) 467–480. [4] R.E. Gaines, J. Mawhin, Ordinary diDerential equations with nonlinear boundary conditions, J. DiDerential Equations 26 (2) (1977) 200–222. [5] M. Garcia-Huidobro, C. Gupta, R. Manasevich, Solvability for a non-linear three-point boundary value problem with p-Laplacian-like operator at resonance, Abstract Appl. Anal. 6 (2001) 191–213. [6] M. Garcia-Huidobro, R. Manasevich, A three point boundary value problem containing the operator −((u )) , Discrete Continuous Dynam. Systems, submitted for publication. [7] C.P. Gupta, Solvability of a three-point boundary value problem for a second order ordinary diDerential equation, J. Math. Anal. Appl. 168 (1992) 540–551. [8] C.P. Gupta, A note on a second order three-point boundary value problem, J. Math. Anal. Appl. 186 (1994) 277–281. [9] C.P. Gupta, A second order m-point boundary value problem at resonance, Nonlinear Anal. T.M.A. 24 (1995) 1483–1489. [10] C.P. Gupta, Existence theorems for a second order m-point boundary value problem at resonance, Internat. J. Math. Math. Sci. 18 (1995) 705–710. [11] C.P. Gupta, S. Ntouyas, P.Ch. Tsamatos, On an m-point boundary value problem for second order ordinary diDerential equations, Nonlinear Anal. T.M.A. 23 (1994) 1427–1436. [12] C.P. Gupta, S. Ntouyas, P.Ch. Tsamatos, Solvability of an m-point boundary value problem for second order ordinary diDerential equations, J. Math. Anal. Appl. 189 (1995) 575–584. [13] C.P. Gupta, S. TroGmchuk, Solvability of multi point boundary value problem of Neumann type, Abstract Anal. Appl. 4 (1999) 71–81. [14] Ma Rayun, C. Nelson, Existence of solutions of nonlinear m-point boundary value problems, J. Math. Anal. Appl. 256 (2001) 556–567.
www.elsevier.com/locate/na
An m-point boundary value problem of Neumann type for a p-Laplacian like operator M. Garc)*a-Huidobroa , C.P. Guptab;∗ , R. Man)asevichc a Departamento
de Matem aticas, Ponticia Universidad Cat olica de Chile, Casilla 306, Correo 22, Santiago, Chile b Department of Mathematics, 084, University of Nevada, Reno, Reno, NV 89557, USA c Departamento de Ingenier *a Matem atica, Universidad de Chile and Centro de Modelamiento Matem atico, Casilla 170, Correo 3, Santiago, Chile Received 13 August 2003; received in revised form 27 September 2003; accepted 6 November 2003
Abstract Let and be two odd increasing homeomorphism from R onto R with (0) = 0; (0) = 0, and let f : [0; 1] × R × R → R be a function satisfying Caratheodory’s conditions. Let ai ∈ R; i ∈ (0; 1); i = 1; 2; : : : ; m − 2; 0 ¡ 1 ¡ 2 ¡ · · · ¡ m−2 ¡ 1 be given. We are interested in the problem of existence of solutions for the m-point boundary value problem: ((x )) = f(t; x; x ); t ∈ (0; 1); x (0) = 0; (x (1)) =
m−2
ai (x ( i )):
i=1
We note that this non-linear m-point boundary value problem is always at resonance since the associated m-point boundary value problem ((x )) = 0; t ∈ (0; 1); x (0) = 0; (x (1)) =
m−2
ai (x ( i ))
i=1
has non-trivial solutions x(t) = ; ∈ R (an arbitrary constant). Our results are obtained by a suitable homotopy, Leray–Schauder degree properties, and a priori bounds. ? 2003 Elsevier Ltd. All rights reserved. MG-H was supported by Fondecyt grant 1030593, Ch. G. was supported by Fondecyt grant 7030029, and RM was supported by Fondap M.A. and Milenio grant-P01-34. ∗
Corresponding author. E-mail addresses: [email protected] (M. Garc)*a-Huidobro), [email protected] (C.P. Gupta), [email protected] (R. Man)asevich). 0362-546X/$ - see front matter ? 2003 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2003.11.003
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1. Introduction Let and be two odd increasing homeomorphisms from R onto R with (0) = 0; (0) = 0 and let f : [0; 1] × R × R → R be a function satisfying Caratheodory’s conditions. Let ai ∈ R; i ∈ (0; 1); i = 1; 2; : : : ; m − 2; 0 ¡ 1 ¡ 2 ¡ · · · ¡ m−2 ¡ 1 be given. We are interested in the problem of existence of solutions for the m-point boundary value problem: ((x )) = f(t; x; x ); t ∈ (0; 1); m−2
x (0) = 0; (x (1)) =
ai (x ( i )):
(1)
i=1
Notice that problem (1) is always at resonance in the sense that the associated m-point boundary value problem ((x (t))) = 0; 0 ¡ t ¡ 1; m−2
x (0) = 0; (x (1)) =
ai (x ( i ))
(2)
i=1
has the non-trivial solution x(t) = , where ∈ R is an arbitrary constant. Let us consider the problem x = f(t; x; x );
t ∈ (0; 1);
x (0) = 0; x (1) =
m−2
ai x ( i );
(3)
i=1
that corresponds to problem (1) when (u) = (u) ≡ u. The diDerential equation in problem (3) has been studied extensively by many authors with diDerent boundary conditions, namely, m−2 x(0) = 0; x (1) = ai x ( i ) or i=1
x (0) = 0; x(1) =
m−2
ai x( i ) or
i=1
x(0) = 0; x(1) =
m−2
ai x( i );
i=1
for example in [2,3,7–14]. A natural generalization for problem (3) is (p (x )) = f(t; x; x ); t ∈ (0; 1); x (0) = 0; x (1) =
m−2 i=1
ai x ( i ):
(4)
M. Garc *a-Huidobro et al. / Nonlinear Analysis 56 (2004) 1071 – 1089
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where p ¿ 1, and (p (x )) = (|x |p−2 x ) denotes the so-called one-dimensional p-Laplace operator. We recall that this operator is generated by the function p : R → R deGned by p (s) = |s|p−2 s, if s = 0 and p (0) = 0. We refer to [1,5,6] as examples of recent works for (4) with diDerent boundary conditions. Another generalization for problem (3) is x = f(t; x; x ); t ∈ (0; 1); x (0) = 0; (x (1)) =
m−2
ai (x ( i )):
(5)
i=1
Problem (5) is in the spirit of the nonlinear boundary value problem studied by Gaines-Mawhin in [4]. Thus the motivation to study problem (1) comes from the fact that problems (3)–(5) are special cases of (1). Additional motivation comes from the challenge to Gnd simple conditions on the function f, for the existence of a solution, as the non-linear homeomorphisms and generating, respectively, the diDerential operator and the boundary condition in (1) are diDerent (in general). We shall denote by C[0; 1] (resp. C 1 [0; 1]) the classical space of continuous (resp. continuously diDerentiable) real-valued functions on the interval [0; 1]. The norm in C[0; 1] is denoted by | · |∞ . Also, we shall denote by L1 (0; 1) the space of real-valued (equivalence classes of) functions whose absolute value is Lebesgue integrable on (0; 1). The Brouwer and Leray–Schauder degree shall be, respectively, denoted by degB and degLS . This paper is organized as follows. In Section 2 we present an abstract formulation of the boundary value problem (1) and establish a general continuation Lemma. In Section 3 we apply this Lemma to obtain several existence theorems for the boundary value problem (1). In all the cases the key point is to obtain a priori bounds. This is done under suitable growth conditions together with coerciveness conditions. We illustrate some of our results, by means of simple examples containing the p-Laplace operator. Example 1. For p ¿ 1 and q ¿ 1 let us consider the problem (p (x )) = (t)p (x) − A|x |p−1 − B; t ∈ (0; 1); x (0) = 0; q (x (1)) =
m−2 i=1
ai q (x ( i ));
06
m−2
ai 6 1;
(6)
i=1
where : [0; 1] → R is a nonnegative continuous function such that (t) ¿ 0 ¿ 0 for all t ∈ [0; 1], and A and B are positive constants with A ¡ 1. Then, if ( 1 =1 − A)p −1 (1 + (A=0 )p −1 ) ¡ 1, the boundary value problem (6) has 1 at least one solution x ∈ C [0; 1].
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Example 2. Let us consider the problem (p (x )) = g(t; x)(p (|x |) + 1); t ∈ (0; 1); x (0) = 0; q (x (1)) =
m−2
ai q (x ( i ));
06
i=1
m−2
ai 6 1;
(7)
i=1
where g is any continuous function such that ug(t; u) ¿ 0, for all |u| ¿ M ¿ 0; where M is a constant, p ¿ 1; and q ¿ 1. Then problem (7) has at least one solution x ∈ C 1 [0; 1]. Remark 3. A model for the function g is the following: g(t; x) = 1 + t 2 + |x|m−2 x; where m ¿ 2. In particular we can take m ¿ p and have lim|x|→∞ g(t; x)=p (x) = ∞, uniformly for t ∈ [0; T ]. Example 1 is a direct consequence of Theorem 5 and Example 2 is a consequence of Theorem 7. We leave to the interested reader to verify the assumptions of the corresponding theorems. 2. Abstract formulation and a deformation lemma We begin this section by formulating a general continuation theorem for the solvability of the boundary value problem (1). Let f∗ : [0; 1] × R × R × [0; 1] → R be a given function satisfying Caratheodory’s conditions, i.e. (i) for all (s; r; ) ∈ R ×R ×[0; 1] the function f∗ (·; s; r; ) is measurable on [0; 1], (ii) for a.e. t ∈ [0; 1] the function f∗ (t; ·; ·; ·) is continuous on R × R × [0; 1], and (iii) for each R ¿ 0 there exists a Lebesgue integrable function R : [0; 1] → R such that |f∗ (t; s; r; )| 6 R (t) for a.e. t ∈ [0; 1] and all (s; r; ) ∈ R × R × [0; 1] with |s| 6 R, and |r| 6 R. We suppose, further, that f∗ (t; s; r; 1) = f(t; s; r) for all (t; s; r) ∈ [0; 1] × R × R: Here the function f is as given in the introduction. To formulate our continuation lemma we introduce an operator B : C 1 [0; 1] × [0; 1] → R by m−2 m−2 B(x; ) = ((x (1)) − ai (x ( i ))) + (1 − )((x (1)) − ai (x ( i ))): i=1
i=1
(8)
For ∈ [0; 1] we consider the family of boundary value problems: ((x )) = f∗ (t; x; x ; );
t ∈ (0; 1);
x (0) = 0; B(x; ) = 0:
(9)
We also deGne a function F : R → R by setting 1 m−2 i F() = f∗ (t; ; 0; 0) dt − ai f∗ (t; ; 0; 0) dt: 0
i=1
0
We have the following continuation lemma.
(10)
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Lemma 4. Let
⊂ C 1 [0; 1] be a bounded open set and set
0
=
1075
∩ R. Assume that
(i) for each ∈ (0; 1) problem (9) has no solution x ∈ @ , (ii) the equation F() = 0, has no solution with ∈ @ ∩ R, and (iii) the Brouwer degree degB (F; 0 ; 0) = 0. Then, problem (1) has at least one solution in K . Proof. Let us deGne "∗ : C 1 [0; 1] × [0; 1] → C 1 [0; 1] by setting 1 ∗ ∗ f ($; x($); x ($); ) d$ " (x; )(t) = x(0) + # 0
m−2
−
i ∗ f ($; x($); x ($); ) d$ ai # 0
i=1
+ (1 − )
m−2
−
ai
i=1
+
t
0
0
f∗ ($; x($); x ($); ) d$
i
0
−1
1
f∗ ($; x($); x ($); ) d$
s ∗ f ($; x($); x ($); ) d$ ds;
(11)
0
where # is the increasing homeomorphism from R onto R given by # = ◦ −1 . We note, from our assumption that the function f∗ satisGes Caratheodory’s conditions, so that for (x; ) ∈ C 1 [0; 1] × [0; 1] the function t → f∗ (t; x(t); x (t); ) is in L1 (0; 1). s Accordingly, the function s ∈ [0; 1] → 0 f∗ ($; x($); x ($); ) d$ is absolutely continuous on [0; 1], and hence the operator "∗ is well deGned. Next, let us suppose that x is a solution to problem (9) for some ∈ (0; 1]. By integrating the equation in (9) and using that x (0) = 0; we have that t f∗ ($; x($); x ($); ) d$: (12) (x (t)) = 0
Evaluating x (t) from (12) and substituting (x (t)) and x (t) into B(x; ) = 0, we Gnd m−2 1 i ∗ ∗ f ($; x($); x ($); ) d$ − ai # f ($; x($); x ($); ) d$ # 0
+ (1 − ) = 0:
0
i=1 1
∗
f ($; x($); x ($); ) d$ −
0
m−2 i=1
ai
0
i
∗
f ($; x($); x ($); ) d$ (13)
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It is then immediate to see from (11), (12), and (13) that x satisGes x(t) = "∗ (x; )(t);
t ∈ [0; 1]:
(14)
Conversely, let us suppose that for some ∈ (0; 1]; x ∈ C 1 [0; 1] satisGes (14). We Grst see from Eqs. (11) and (14), by evaluating at t = 0, that (13) is satisGed. Next, by diDerentiating Eq. (14), we obtain that (12) holds and hence evaluating this equation at t = 0 we see that x (0) = 0. Finally, by (14) evaluated at t = 0, and combining with (12), we obtain that x satisGes B(x; ) = 0. Also, since (12) implies that (x ) is absolutely continuous on [0; 1], we Gnally obtain and by diDerentiation that ((x (t))) = f∗ (t; x(t); x (t); ); t ∈ (0; 1): Accordingly, for ∈ (0; 1]; we have proved that x ∈ C 1 [0; 1] is a solution of the boundary value problem (9) if and only if x(t); t ∈ [0; 1], is a solution of Eq. (14). We observe, by using standard arguments, that "∗ : C 1 [0; 1] × [0; 1] → C 1 [0; 1] is a completely continuous operator. Now, if there is a function x ∈ @ which satisGes problem (1) then we are done. Accordingly, let us assume that problem (1) has no solution on @ . Since, now, f∗ (t; s; r; 1) = f(t; s; r) for all (t; s; r) ∈ [0; 1] × R × R we see that assumption (i) of the lemma implies that x = "∗ (x; ) for all u ∈ @
and ∈ (0; 1]:
We, next, assert that x = "∗ (x; 0) for all x ∈ @ . Indeed, let x ∈ @ be such that x = "∗ (x; 0). It, then, follows from (11) that x(t) = , x (t) = 0, and x ∈ @ ∩ R, where 1 m−2 i f∗ ($; x($); x ($); 0) d$ − ai f∗ ($; x($); x ($); 0) d$ = x(0) + 0
i=1
0
= + F();
(15)
which is not possible since it implies that F() = 0, contradicting assumption (ii) of the lemma. We thus get that x = "∗ (x; ) for all x ∈ @
and ∈ [0; 1]:
Finally, we use the homotopy invariance property of Leray–Schauder degree to get degLS (I − "∗ (·; 1); ; 0) = degLS (I − "∗ (·; 0); ; 0) = degB (I − "∗ (·; 0)|R ; = degB (F;
0 ; 0)
0 ; 0)
= 0:
Thus, the mapping " ≡ "∗ (·; 1) : C 1 [0; 1] → C 1 [0; 1] has at least one Gxed-point in and hence problem (1) has at least one solution in . This completes the proof of the lemma.
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3. Existence theorems m−2 Theorem 5. In problem (1), let the constants ai ; i = 1; : : : ; m − 2 satisfy 0 6 i=1 ai 6 1, and let f : [0; 1]×R×R → R be a continuous function satisfying the following conditions: (i) there exist non-negative functions d1 (t); d2 (t), and r(t) in L1 (0; 1) such that |f(t; u; v)| 6 d1 (t)(|u|) + d2 (t)(|v|) + r(t); for a.e. t ∈ [0; 1] and all u; v ∈ R, (ii) there exist constants ¿ 0; A; B ¿ 0, and a u0 ¿ 0 such that for all u with |u| ¿ u0 , all t ∈ [0; 1] and all v ∈ R, one has |f(t; u; v)| ¿ (|u|) − A(|v|) − B; (iii) there exists an R ¿ max{u0 ; B=} such that f(t; ; 0)f(t; −; 0) ¡ 0 for all ¿ R and all t ∈ [0; 1]. (iv) Suppose, further, d2 L1 (0; 1) ¡ 1, and the function ( : [0; ∞) → [0; ∞), dened by
d1 L1 (0; 1)
r L1 (0; 1) ((z) = −1 (z) + 1 − d2 L1 (0; 1) 1 − d2 L1 (0; 1) A d1 L1 (0; 1) A r L1 (0; 1) B −1 + (z) + + (1 − d2 L1 (0; 1) ) (1 − d2 L1 (0; 1) ) satises the condition lim sup z→∞
((z) ¡ 1: z
(16)
Then problem (1) has at least one solution x ∈ C 1 [0; 1]. Remark 6. Notice that condition (16) implies that d1 L1 (0; 1) + d2 L1 (0; 1) ¡ 1. The converse is not necessarily true for general . Proof. We consider the family of boundary value problems (9) with f∗ (t; u; v; ) = f(t; u; v) for all (t; u; v; ) ∈ [0; 1]×R×R×[0; 1], i.e. we consider the family of problems: ((x )) = f(t; x; x ); t ∈ (0; 1); ∈ [0; 1]; x (0) = 0;
B(x; ) = 0:
(17)
We shall show that the family of problems (17) satisfy the conditions of Lemma 4 to conclude that problem (1) has at least one solution in C 1 [0; 1]. Let x ∈ C 1 [0; 1] be a solution to problem (17) for some ∈ (0; 1). Then, by integrating the equation in (17) from 0 to t, t ∈ [0; 1], using x (0) = 0, assumption (i), and the fact that is an odd increasing homeomorphism from R onto R, we Grst Gnd that (|x (t)|) 6 ( x ∞ ) d1 L1 (0; 1) + ( x ∞ ) d2 L1 (0; 1) + r L1 (0; 1)
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for all t ∈ [0; 1], and then that
d1 L1 (0; 1)
r L1 (0; 1)
x ∞ 6 −1 : ( x ∞ ) + 1 − d2 L1 (0; 1) 1 − d2 L1 (0; 1)
(18)
By integrating the equation in (17) on the interval [0; t] and using that x (0) = 0; we see that t (x (t)) = f($; x($); x ($)) d$; 0
from which B(x; ) = 0 becomes m−2 1 i f($; x($); x ($)) d$ − ai # f($; x($); x ($)) d$ # 0
+ (1 − )
0
i=1 1
0
f($; x($); x ($)) d$ −
m−2 i=1
ai
0
i
f($; x($); x ($)) d$
= 0:
(19)
We claim that there must exist a $1 ∈ (0; 1) such that f($1 ; x($1 ); x ($1 )) = 0. Indeed, assume this is not true. Now, let us assume m−2without loss of generality that f($; x($); x ($)) ¿ 0 for all $ ∈ [0; 1]. Next, if T = i=1 ai = 0, then we get ai = 0 for i = 1; 2; : : : ; m − 2 since 0 6 ai 6 1 for i = 1; 2; : : : ; m − 2. Accordingly, we get from (19) that 1 1 0=# f($; x($); x ($)) d$ + (1 − ) f($; x($); x ($)) d$ ¿ 0; (20) 0
0
in view of the assumption f($; x($); x ($)) ¿ 0 for all $ ∈ [0; 1]. This is a contradiction, m−2 so there must exist a $1 ∈ (0; 1) such that f($1 ; x($1 ); x ($1 ))=0, when T = i=1 ai =0. m−2 Now let T = i=1 ai = 0 and let us set m−2 i ai # f($; x($); x ($)) d$ I = 0
i=1
and let + and − in [0; 1] be such that
+ i f($; x($); x ($)) d$ = max f($; x($); x ($)) d$|i = 1; m − 2 0
and
0
Then
0
−
f($; x($); x ($)) d$ = min
#
0
0
i
f($; x($); x ($)) d$|i = 1; m − 2 :
+ I f($; x($); x ($)) d$ 6 6 # f($; x($); x ($)) d$ ; T 0
−
M. Garc *a-Huidobro et al. / Nonlinear Analysis 56 (2004) 1071 – 1089
1079
t ai . Considering the continuous function S(t)=#( 0 f($; x($); x ($)) d$), I we obtain that S( − ) 6 6 S( + ) and thus there must exists a t ∈ (0; 1) such that T I = TS(t ). Similarly there is an +0 ∈ (0; 1) such that +0 m−2 i ai f($; x($); x ($)) d$ = T f($; x($); x ($)) d$: m−2
where T =
i=1
0
i=1
0
In this form (19) can be written as t 1 # f($; x($); x ($)) d$ − T# f($; x($); x ($)) d$ 0
0
+ (1 − )
1
+0
f($; x($); x ($)) d$ + (1 − T )
0
+0
f($; x($); x ($)) d$
= 0:
(21)
Hence by using the fact that T 6 1; # is increasing and the function f is continuous and positive, we obtain from (21) the contradiction 1 1 f($; x($); x ($)) d$ ¡ f($; x($); x ($)) d$; 0
0
proving our claim, that is, the existence of $1 ∈ (0; 1) such that f($1 ; x($1 ); x ($1 )) = 0. Next let us suppose that |x(t)| ¿ u0 for all t ∈ [0; 1]. This along with assumption (ii) then gives B A (|x($1 )|) 6 + ( x ∞ ): (22) Since, now, for t ∈ [0; 1], t x (s) ds; x(t) = x($1 ) + $1
we see from Eqs. (18) and (22) that
x ∞ 6 |x($1 )| + x ∞ B A 6 −1 + ( x ∞ )
d1 L1 (0; 1)
r L1 (0; 1) −1 ( x ∞ ) + + 1 − d2 L1 (0; 1) 1 − d2 L1 (0; 1) A d1 L1 (0; 1) A r L1 (0; 1) B 6 −1 ( x ∞ ) + + (1 − d2 L1 (0; 1) ) (1 − d2 L1 (0; 1) )
d1 L1 (0; 1)
r L1 (0; 1) −1 + ( x ∞ ) + 1 − d2 L1 (0; 1) 1 − d2 L1 (0; 1) = (( x ∞ ):
(23)
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M. Garc *a-Huidobro et al. / Nonlinear Analysis 56 (2004) 1071 – 1089
Next, let us suppose that there exists a t0 ∈ [0; 1] such that |x(t0 )| 6 u0 . Then we see from t |x(t)| = x(t0 ) + x (s) ds 6 |x(t0 )| + x ∞ t0
6 u0 + x ∞ and Eq. (18) that
x ∞ 6 u0 + −1
for t ∈ [0; 1]
d1 L1 (0; 1)
r L1 (0; 1) ( x ∞ ) + 1 − d2 L1 (0; 1) 1 − d2 L1 (0; 1)
6 u0 + (( x ∞ ):
(24)
It, now, follows from our assumption lim supz→∞ ((z)=z ¡ 1 and Eqs. (23) and (24) that there exists a constant C, independent of ∈ (0; 1] such that
x ∞ 6 C: Then by using (18) there is a constant, which we again denote by C, independent of ∈ (0; 1] such that
x C 1 [0; 1] 6 C: We have thus shown that if x(t) ∈ C 1 [0; 1] is a solution to problem (17), equivalently for the equation x = "∗ (x; ), for some ∈ (0; 1] then
x C 1 [0; 1] 6 C; where C is a constant independent of ∈ (0; 1]. It follows that there exists an R0 ¿ R, where R is as in assumption (iii), such that for ∈ (0; 1] the family of problems (17), or equivalently the equations x = "∗ (x; ), ˜ ⊂ C 1 [0; 1] for have no solution on the boundary of the bounded open set = B(0; R) ˜ every R ¿ R0 . Accordingly, we see that the family of problems (17) satisfy condition (i) of Lemma 4. Next, we see from assumption (iii) that for all || ¿ R, it holds that |f(t; ; 0)| ¿ 0 for all t ∈ [0; 1], hence writing F() as deGned in (10) in the form 1 m−2 1 F() = (1 − T ) f(t; ; 0) dt + ai f(t; ; 0) dt (25) 0
i=1
i
we see that F() is either strictly positive or strictly negative for all such that || ¿ R. Accordingly, we see that f∗ (t; u; v; ) = f(t; u; v) satisGes condition (ii) of Lemma 4. ˜ and F(−R) ˜ have oppoFinally, we see from assumption (iii) and (25) that F(R) site signs. It follows immediately that the Brouwer degree degB (F; 0 ; 0) = 0. Thus, condition (iii) of Lemma 4 is also satisGed. Thus it follows from Lemma 4 that problem (1) has at least one solution in K . This completes the proof of the theorem. Our next existence theorem gives a diDerent set of conditions on f to which our results apply. In fact, we are able to relax the sublinear assumption (i) in Theorem 5 in order to allow superlinear behavior of f with respect to the variable u.
M. Garc *a-Huidobro et al. / Nonlinear Analysis 56 (2004) 1071 – 1089
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Theorem 7. Let f : [0; 1] × R × R → R in the boundary value problem ((x )) = f(t; x; x ); t ∈ (0; 1); x (0) = 0; (x (1)) =
m−2
ai (x ( i ));
06
i=1
m−2
ai 6 1
(26)
i=1
be a continuous functions satisfying the following conditions: (i) there exists M ¿ 0 such that for all |u| ¿ M , all v ∈ R, and all t ∈ [0; 1], one has uf(t; u; v) ¿ 0: (ii) For any K ¿ 0, there exist non-negative functions b1 (t); b2 (t) in L1 (0; 1) such that |f(t; u; v)| 6 b1 (t) + b2 (t)(|v|) for a.e. t ∈ [0; 1] and all u ∈ [ − K; K], and v ∈ R. Then problem (26) has at least one solution x ∈ C 1 [0; 1]. Proof. We consider the family of boundary value problems (9) with f∗ (t; u; v; ) = f(t; u; v) for all (t; u; v; ) ∈ [0; 1]×R×R×[0; 1], i.e. we consider the family of problems: ((x )) = f(t; x; x ); t ∈ (0; 1); ∈ [0; 1]; x (0) = 0; B(x; ) = 0:
(27)
We shall show that the family of problems (27) satisfy the conditions of Lemma 4 to conclude that problem (26) has at least one solution in C 1 [0; 1]. Our Grst step is to deGne an open set ⊂ C 1 such that (i) of Lemma 4 holds. To this end let x be a solution to problem (27) for ∈ (0; 1]. We claim that |x(0)| 6 M; where M is as in assumption (i) of the theorem. Indeed, suppose Grst that x(0) ¿ M . It then follows from x (0) = 0 and assumption (i) of the theorem, that x (t) is strictly increasing and positive on [0; .] for some positive ., and accordingly, x(t) ¿ M and strictly increasing on [0; .]. Let us deGne tM := sup{t ∈ [0; 1]|x is strictly increasing and positive on (0; t]}: Clearly, tM ¿ .; x(tM ) ¿ M and x is strictly increasing on [0; tM ]. We claim that tM =1. Indeed, if tM ¡ 1, then, since x(tM ) ¿ M , there will be a / ¿ 0 with 0 ¡ tM −/ ¡ tM + / ¡ 1 such that x(t) ¿ M on [tM −/; tM +/]. Accordingly, we see from the equation and assumption (i) of the theorem that x is strictly increasing and positive on [tM −/; tM +/] and hence on (0; tM + /], contradicting the deGnition of tM . Hence tM = 1 and x is strictly increasing and positive for all t ∈ (0; 1]. Taking now into account the second boundary condition, we have that either (x (1)) −
m−2 i=1
ai (x ( i )) ¡ 0
or
(x (1)) −
m−2 i=1
ai (x ( i )) ¡ 0
(28)
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M. Garc *a-Huidobro et al. / Nonlinear Analysis 56 (2004) 1071 – 1089
and thus, using that 0 6 x (1) ¡
max
i=1;:::; m−2
m−2 i=1
ai 6 1, we obtain
x ( i );
a contradiction to the fact that x is strictly increasing for all t ∈ (0; 1]. Since the same argument applies in the case that x(0) ¡ − M , the claim follows. Let now tM := sup{t ∈ [0; 1] | |x(s)| 6 M for all s ∈ [0; t]}: Clearly, x(tM ) = ±M . By integrating the equation over (0; t), t 6 tM , and using assumption (ii), we obtain that t |(x (t))| 6 b2 (s)|(x (s))| ds + b1 L1 [0; 1] 0
and thus, by Gronwall’s Lemma we Gnd that |(x (t))| 6 b1 L1 [0; 1] b2 L1 [0; 1] eb2 L1 [0;1] := C0
for all t ∈ [0; tM ]:
(29)
If tM = 1, it follows that x is bounded in C 1 [0; 1] and we can proceed as in the proof of Theorem 5 in order to satisfy assumption (i) of Lemma 4. Thus we may assume that tM ¡ 1. Let now t ∈ [tM ; 1], and suppose for instance that x(tM ) = M . Assume that x has a local maximum at some point t0 ∈ (tM ; 1). Then x (t0 ) = 0 and since x(t0 ) is a positive maximum with x(t0 ) ¿ M , (t − t0 )x (t) 6 0; for t ∈ (0; 1] close to t0 : This certainly implies that ((x )) (t0 ) 6 0 and thus 0 ¿ x(t0 )((x )) (t0 ) = x(t0 ) f(t0 ; x(t0 ); 0) ¿ 0; which is a contradiction. Hence x(t) ¿ M for all t ∈ (tM ; 1], and then, by arguing as above we again have that x is strictly increasing and positive for all t ∈ (tM ; 1], and by the second boundary condition it must be that (28) holds. This implies that either (x (1)) ¡ ai (x ( i )) or (x (1)) ¡ ai (x ( i )) x ( i )¿0
and since
x ( i )¿0
ai 6 1;
x ( i )¿0
we conclude that x (1) ¡ max x ( i ): x ( i )¿0
We claim that it must be that the i where this maximum is achieved is less than or equal to tM . Indeed, if this is not the case we obtain a contradiction to the fact that x is strictly increasing for all t ∈ (tM ; 1]. Therefore, by (29) we Gnd that x (1) ¡ −1 (C0 ); proving that the derivative x is uniformly bounded in [0; 1]. Finally, by writing t x(t) = x(0) + x (s) ds; t ∈ [0; 1]; 0
M. Garc *a-Huidobro et al. / Nonlinear Analysis 56 (2004) 1071 – 1089
1083
we obtain that |x(t)| 6 M + −1 (C0 )
for all t ∈ [0; 1]:
It follows that there is a R˜ ¿ M + −1 (C0 ), such that for ∈ (0; 1] the family of problems (27), or equivalently the equations x = "∗ (x; ), have no solution on the ˜ ⊂ C 1 [0; 1]. We see that the family of boundary of the bounded open set = B(0; R) problems (27) satisfy condition (i) of Lemma 4. As in the proof of Theorem 5, we see from assumption (i) that for all ; || ¿ M , it holds that |f(t; ; 0)| ¿ 0 for all t ∈ [0; 1], hence by the continuity of f and since (10) in this case becomes 1 m−2 i F() = f(t; ; 0) dt − ai f(t; ; 0) dt 0
i=1
= (1 − T )
1
0
f(t; ; 0) dt +
0
m−2 i=1
ai
1
i
f(t; ; 0) dt;
m−2 where T = i=1 ai , we conclude that F() is either strictly positive or strictly negative for all || ¿ M . Accordingly, we see that f∗ (t; u; v; ) = f(t; u; v) + g(t; u) satisGes condition (ii) of Lemma 4. Also, again by (i), it holds that f(t; ; 0)f(t; −; 0) ¡ 0 for all || ¿ M , implying that condition (iii) of Lemma 4 is fulGlled. Our next existence theorem concerns the multi-point boundary value problem: ((x )) + f(t; x; x ) = q(t); t ∈ (0; 1); x (0) = 0; (x (1)) =
m−2
ai (x ( i ));
06
i=1
m−2
ai 6 1;
(30)
i=0
where f is continuous, q(t) ∈ L1 (0; 1). We deGne
1 1 qK = max q(s) ds|+ ∈ [0; m−2 ] 1−+ + and
q = min
1 1−+
+
1
q(s) ds|+ ∈ [0; m−2 ] :
Theorem 8. Let q(t) ∈ L1 (0; 1) and f : [0; 1] × R × R → R in problem (30) be a continuous function and satisfying the following conditions: (i) there exist non-negative functions d1 (t); d2 (t), and r(t) in L1 (0; 1) such that |f(t; u; v)| 6 d1 (t)(|u|) + d2 (t)(|v|) + r(t) for a.e. t ∈ [0; 1] and all u; v ∈ R,
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(ii) there exists a d ¿ 0 such that f(t; u; v) ¿ qK for u ¿ d; f(t; u; v) ¡ q for u 6 − d for a.e. t ∈ [0; 1] and all v ∈ R. Suppose, further, d2 L1 (0; 1) ¡ 1, and the function 0 : [0; ∞) → [0; ∞), dened for z ∈ R by
d1 L1 (0; 1)
r
˜ L1 (0; 1) ; 0(z) = −1 (z) + 1 − d2 L1 (0; 1) 1 − d2 L1 (0; 1) where r(t) ˜ = r(t) + q(t), satises the condition lim sup z→∞
0(z) ¡ 1: z
(31)
Then problem (30) has at least one solution x ∈ C 1 [0; 1]. Remark 9. As in Remark 6 condition (31) implies that d1 L1 (0; 1) + d2 L1 (0; 1) ¡ 1. Proof. We shall show that for f∗ (t; u; v; ) = q(t) − f(t; u; v) for all (t; u; v; ) ∈ [0; 1] × R × R × [0; 1], the family of problems ((x )) = (q(t) − f(t; x; x )); t ∈ (0; 1); ∈ [0; 1]; x (0) = 0; B(x; ) = 0:
(32)
satisGes conditions (i) – (iii) of Lemma 4, and where B(x; ) is as deGned in (8). Let x(t) be a solution of (32) for some ∈ (0; 1). Then, by integrating the equation in (32) from 0 to t ∈ (0; 1] and using that x (0) = 0, we obtain that t (33) (x (t)) = (q($) − f($; x($); x ($))) d$: 0
m−2 As in the proof Theorem 5, by setting T = i=1 ai , we obtain from here that there are +0 ; + ∈ [0; m−2 ] such that +0 m−2 i ai (q($) − f($; x($); x ($))) d$) = T (q($) − f($; x($); x ($))) d$ i=1
0
0
m−2 ai # i=1
i
0
=T#
0
(q($) − f($; x($); x ($))) d$
+
(q($) − f($; x($); x ($))) d$ ;
(34)
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and thus B(x; ) = 0 is equivalent to 1 (q($) − f($; x($); x ($))) d$ (1 − T )# 0
1 (q($) − f($; x($); x ($))) d$ +T # 0
−#
0
+
+(1 − ) (1 − T ) +T
1
+0
(q($) − f($; x($); x ($))) d$ 0
1
(q($) − f($; x($); x ($))) d$
(q($) − f($; x($); x ($))) d$
= 0:
(35)
Next we claim that there exists a t˜ ∈ [0; 1] such that − d 6 x(t˜) 6 d:
(36)
Indeed, if we assume that x(t) ¿ d for all t ∈ [0; 1] then in view of the Grst part of assumption (ii) of the theorem we would have 1 + (q($) − f($; x($); x ($))) d$ (q($) − f($; x($); x ($))) d$ − 0
=
0
1
+
(q($) − f($; x($); x ($))) d$ ¡ 0;
implying (by the monotonicity of #) that the second line in (35) adds to a negative number, and 1 1 (q($) − f($; x($); x ($))) d$ ¡ 0; (q($) − f($; x($); x ($))) d$ ¡ 0; +0
0
which cannot be since these three inequalities would imply B(x; ) ¡ 0; which is a contradiction. Similarly the assumption x(t) ¡ − d for all t ∈ [0; 1] leads to a contradiction. This proves (36). As an immediate consequence of (36) we obtain that
x ∞ 6 d + x ∞ :
(37)
Next, we see from assumption (i) of the theorem that ˜ |q(t) − f(t; x(t); x (t))| 6 d1 (t)(|x(t)|) + d2 (t)(|x (t)|) + r(t)
(38)
and hence by integrating the equation in (32) from 0 to t ∈ [0; 1] and using that x (0)=0, we obtain that ˜ L1 (0; 1) : ( x ∞ ) 6 ( x ∞ ) d1 L1 (0; 1) + ( x ∞ ) d2 L1 (0; 1) + r
(39)
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Thus, Grst solving for x ∞ , and then combining with (32) we Grst Gnd that
r
˜ L1 (0; 1)
d1 L1 (0; 1)
x ∞ 6 −1 ( x ∞ ) + 1 − d2 L1 (0; 1) 1 − d2 L1 (0; 1)
(40)
and then
x ∞ 6 d + 0( x ∞ ): Accordingly, we get using our assumption lim sup z→∞
0(z) ¡ 1; z
that there must exist a z0 ¿ 0 such that
x ∞ 6 z0 :
(41)
Finally, by (40) and (41) there is a positive constant R0 ¿ d (where d is as in assumption (ii)) such that
x C 1 [0; 1] 6 R0 : = B(0; R) ⊂ C 1 [0; 1], then for each 0 ¡ ¡ 1 the Thus for any Gxed R ¿ R0 if family of problems (32) has no solution on @ . Thus (i) of Lemma 4 is satisGed. Next, we observe that for our case 1 m−2 i (q($) − f($; ; 0)) d$ − ai (q($) − f($; ; 0)) d$ F() = 0
i=1
= (1 − T )
0
1
(q($) − f($; ; 0)) d$ +
0
m−2 i=1
ai
1
i
(q($) − f($; ; 0)) d$:
(42)
Then assumption (ii) imply that for all ∈ R, with || ¿ d, F() ¡ 0:
(43)
Since this holds when = R; we have that (ii) of Lemma 4 is satisGed. That (iii) of Lemma 4 holds is immediate from (43). We have thus proved that the family of problems (32) satisGes all the conditions of Lemma 4 and hence existence of a solution for problem (1) follows from Lemma 4. This completes the proof of the theorem. Remark 10. Theorem 5 continues to hold if we replace assumption (ii) by the following: “there exists a d ¿ 0 such that f(t; u; v) ¡ q for u ¿ d; f(t; u; v) ¿ qK for u 6 − d; for a.e. t ∈ [0; 1] and all v ∈ R.”
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We next consider the interesting case when = in problem (1), that is, we consider the problem ((x )) = f(t; x; x ); t ∈ (0; 1);
x (0) = 0; (x (1)) =
m−2
ai (x ( i )):
(44)
i=1
where f is Caratheodory, i.e., it satisGes: (i) for all (s; r) ∈ R × R the function f(·; s; r) is measurable on [0; 1], (ii) for a.e. t ∈ [0; 1]f(t; ·; ·) is continuous on R×R, and (iii) for each R ¿ 0 there exists a Lebesgue integrable function R : [0; 1] → R such that |f(t; s; r)| 6 R (t) for a.e. t ∈ [0; 1] and all (s; r) ∈ R × R with |s| 6 R, and |r| 6 R. Theorem 11. Suppose in problem (14) that the following conditions are satised: (i) there exist non-negative functions d1 (t); d2 (t), and r(t) in L1 (0; 1) such that |f(t; u; v)| 6 d1 (t)(|u|) + d2 (t)(|v|) + r(t); for a.e. t ∈ [0; 1] and all u; v ∈ R, (ii) there exists a d ¿ 0 such that for all x ∈ C 1 [0; 1] with mint∈[0; 1] |x(t)| ¿ d it holds that 1 m−2 i f($; x($); x ($)) d$ = 0: (45) ai f($; x($); x ($)) d$ − 0
i=1
0
(iii) For every R ¿ 0, there exists || ¿ R such that F()F(−) ¡ 0, where 1 m−2 i F() = f($; ; 0) d$ − ai f($; ; 0) d$: 0
i=1
0
(iv) Suppose in addition d2 L1 (0; 1) ¡ 1, and the function 0 : [0; ∞) → [0; ∞), dened for z ∈ R by
r L1 (0; 1)
d1 L1 (0; 1) 0(z) = −1 ; (z) + 1 − d2 L1 (0; 1) 1 − d2 L1 (0; 1) satises the condition lim sup z→∞
0(z) ¡ 1: z
Then problem (44) has at least one solution x ∈ C 1 [0; 1]. Proof. We will consider the family of boundary value problems (9) with f∗ (t; x; x ; ) = f(t; x; x ) and where B(x; ); see (8), becomes B(x; ) = (x (1)) −
m−2 i=1
ai (x ( i )):
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Thus (9) becomes the family of problems: ((x )) = f(t; x; x ); t ∈ (0; 1); ∈ [0; 1]; x (0) = 0; (x (1)) =
m−2
ai (x ( i )):
(46)
i=1
Let x ∈ C 1 [0; 1] be a solution to problem (46) for some ∈ (0; 1). By integrating the equation in (46) from 0 to t, t ∈ [0; 1], using x (0) = 0, we obtain as before that (33) holds. Next since for this case, # = ◦ −1 = id, the second boundary condition in (46) reads 1 m−2 i f($; x($); x ($)) d$ − ai f($; x($); x ($)) d$ = 0: 0
i=1
0
Hence, from (ii), it must be that mint∈[0; 1] |x(t)| 6 d. Then we see that for some t0 ∈ [0; 1] it holds that for t ∈ [0; 1] t |x(t)| = |x(t0 ) + x (s) ds| 6 |x(t0 )| + x ∞ t0
6 d + x ∞ : Also, by integrating the equation in (46) from 0 to t, assumption (i), and the fact that is an odd increasing homeomorphism from R onto R, we Grst Gnd that (|x (t)|) 6 ( x ∞ ) d1 L1 (0; 1) + ( x ∞ ) d2 L1 (0; 1) + r L1 (0; 1) for all t ∈ [0; 1], and then that
d1 L1 (0; 1)
r L1 (0; 1) : ( x ∞ ) +
x ∞ 6 −1 1 − d2 L1 (0; 1) 1 − d2 L1 (0; 1) Hence,
x ∞ 6 d +
−1
d1 L1 (0; 1)
r L1 (0; 1) ( x ∞ ) + 1 − d2 L1 (0; 1) 1 − d2 L1 (0; 1)
(47)
6 d + 0( x ∞ ):
(48)
As in the proof of Theorem 8, this implies that there exists R0 ¿ 0 such that
x C 1 [0; 1] 6 R0 : Since by assumption (iii) we have now that all the assumptions of Lemma 4 are satisGed, we conclude that problem (44) has at least one solution in C 1 [0; 1]. Remark m−212. Note that this last theorem holds with no restriction on the value of ai ∈ R or i=1 ai .
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