Existence of a solution for a three-point boundary value problem for a second-order differential equation at resonance

Boundary Value Problems20132013:130

DOI: 10.1186/1687-2770-2013-130

Received: 20 February 2013

Accepted: 30 April 2013

Published: 20 May 2013

Abstract

We present a new existence result for a second-order nonlinear ordinary differential equation with a three-point boundary value problem when the linear part is noninvertible.

MSC:34B10, 34B15.

Keywords

nonlinear ordinary differential equation three-point boundary value problem problem at resonance existence of solution

1 Introduction

The study of multi-point boundary value problems for linear second-order ordinary differential equations goes back to the method of separation of variables [1]. Also, some questions in the theory of elastic stability are related to multi-point problems [2]. In 1987, Il’in and Moiseev [3, 4] studied some nonlocal boundary value problems. Then, for example, Gupta [5] considered a three-point nonlinear boundary value problem. For some recent works on nonlocal boundary value problems, we refer, for example, to [615] and references therein.

As indicated in [16], there has been enormous interest in nonlinear perturbations of linear equations at resonance since the seminal paper of Landesman and Lazer [17]; see [18] for further details.

Here we study the following nonlinear ordinary differential equation of second order subject to the three-point boundary condition:
u ( t ) = f ( t , u ( t ) ) , t [ 0 , T ] , u ( 0 ) = 0 , α u ( η ) = u ( T ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_Equ1_HTML.gif
(1)

where T > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq1_HTML.gif, f : [ 0 , T ] × R R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq2_HTML.gif is a continuous function α R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq3_HTML.gif and η ( 0 , T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq4_HTML.gif.

In this paper we consider the resonance case α η = T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq5_HTML.gif to obtain a new existence result. Although this situation has already been considered in the literature [19], we point out that our approach and methodology is different.

2 Linear problem

Consider the linear second-order three-point boundary value problem
u ( t ) = σ ( t ) , t [ 0 , T ] , u ( 0 ) = 0 , α u ( η ) = u ( T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_Equ2_HTML.gif
(2)

for a given function σ C [ 0 , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq6_HTML.gif.

The general solution is
u ( t ) = c 1 + c 2 t 0 t ( t s ) σ ( s ) d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_Equa_HTML.gif

with c 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq7_HTML.gif, c 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq8_HTML.gif arbitrary constants.

From u ( 0 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq9_HTML.gif, we get c 1 = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq10_HTML.gif. From the second boundary condition, we have
( T α η ) c 2 = 0 T ( T s ) σ ( s ) d s α 0 η ( η s ) σ ( s ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_Equ3_HTML.gif
(3)

2.1 Nonresonance case

If α η T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq11_HTML.gif, then
c 2 = 1 T α η [ 0 T ( T s ) σ ( s ) d s α 0 η ( η s ) σ ( s ) d s ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_Equb_HTML.gif
and the linear problem (2) has a unique solution for any σ C [ 0 , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq12_HTML.gif. In this case, we say that (2) is a nonresonant problem since the homogeneous problem has only the trivial solution as a solution, i.e., when σ = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq13_HTML.gif, c 1 = c 2 = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq14_HTML.gif and u = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq15_HTML.gif. Note that the solution is given by
u ( t ) = 0 T g ( t , s ) σ ( s ) d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_Equ4_HTML.gif
(4)
with
g ( t , s ) = { t ( T s ) T α η t α ( η s ) T α η ( t s ) , 0 s < min ( η , t ) , t ( T s ) T α η t α ( η s ) T α η , 0 t < s < η < T , t ( T s ) T α η ( t s ) , 0 η < s < t T , t ( T s ) T α η , max ( η , t ) < s T . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_Equc_HTML.gif

For T = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq16_HTML.gif this is precisely the function given in Lemma 2.3 of [20] or in Remark 12 of [21].

2.2 Resonance case

If T = α η http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq17_HTML.gif, then (3) is solvable if and only if
0 T ( T s ) σ ( s ) d s = α 0 η ( η s ) σ ( s ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_Equ5_HTML.gif
(5)
and then (2) has a solution if and only if (5) holds. In such a case, (2) has an infinite number of solutions given by
u ( t ) = c t 0 t ( t s ) σ ( s ) d s , c R . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_Equd_HTML.gif
In particular ct, c R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq18_HTML.gif is a solution of the homogeneous linear equation
u ( t ) = 0 , t [ 0 , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_Eque_HTML.gif
satisfying the boundary conditions
u ( 0 ) = 0 , α u ( η ) = u ( T ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_Equf_HTML.gif
Note that
u ( T ) u ( η ) = c 2 T 0 T ( T s ) σ ( s ) d s c 2 η + 0 η ( η s ) σ ( s ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_Equg_HTML.gif
and then
c 2 = 1 T η [ u ( T ) u ( η ) + 0 T ( T s ) σ ( s ) d s 0 η ( η s ) σ ( s ) d s ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_Equh_HTML.gif
We now use that u ( T ) = T η u ( η ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq19_HTML.gif to get
1 T η [ u ( T ) u ( η ) ] = 1 T u ( T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_Equi_HTML.gif
and
c 2 = 1 T η [ 0 T ( T s ) σ ( s ) d s 0 η ( η s ) σ ( s ) d s ] + 1 T u ( T ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_Equj_HTML.gif
Hence the solution of (2) is given, implicitly, as
u ( t ) = 0 T t ( T s ) T η σ ( s ) d s 0 η t ( η s ) T η σ ( s ) d s 0 t ( t s ) σ ( s ) d s + t T u ( T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_Equk_HTML.gif
or, equivalently,
u ( t ) = 0 T k ( t , s ) σ ( s ) d s + t T u ( T ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_Equ6_HTML.gif
(6)
where
k ( t , s ) = { s , 0 s < min ( η , t ) , t , 0 t < s < η T , t ( T s ) T η ( t s ) , 0 η < s < t T , t ( T s ) T η , max ( η , t ) < s T . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_Equl_HTML.gif

We note that k C ( [ 0 , T ] × [ 0 , T ] , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq20_HTML.gif and k ( t , s ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq21_HTML.gif for every ( t , s ) [ 0 , T ] × [ 0 , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq22_HTML.gif.

3 Nonlinear problem

Defining the operators:
F : C [ 0 , T ] C [ 0 , T ] , [ F u ] ( t ) = f ( t , u ( t ) ) , u C [ 0 , T ] , t [ 0 , T ] , K : C [ 0 , T ] C [ 0 , T ] , [ K σ ] ( t ) = 0 T k ( t , s ) σ ( s ) d s , σ C [ 0 , T ] , t [ 0 , T ] , L : C [ 0 , T ] C [ 0 , T ] , [ L u ] ( t ) = t T u ( T ) , u C [ 0 , T ] , t [ 0 , T ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_Equm_HTML.gif
the nonlinear problem is equivalent to
u = N u , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_Equn_HTML.gif

where N = K F + L http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq23_HTML.gif.

We note that (6) can be written as
u ( t ) t T u ( T ) = 0 T k ( t , s ) σ ( s ) d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_Equo_HTML.gif
and the nonlinear problem (1) as
u ( t ) t T u ( T ) = 0 T k ( t , s ) f ( s , u ( s ) ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_Equp_HTML.gif

This suggests to introduce the new function v ( t ) = u ( t ) t T u ( T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq24_HTML.gif. To find a solution u, we have to find v and u ( T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq25_HTML.gif.

For every constant c R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq18_HTML.gif, we solve
v ( t ) = 0 T k ( t , s ) f ( s , v ( s ) + s T c ) d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_Equ7_HTML.gif
(7)
and let φ ( c ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq26_HTML.gif be the set of solutions of (7). This set may be empty (no solution), a singleton (unique solution) or with more than one element (multiple solutions). For every v c φ ( c ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq27_HTML.gif, we consider
u c ( t ) = v c ( t ) + t T c , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_Equq_HTML.gif
and hence
u c ( t ) = 0 T k ( t , s ) f ( s , u c ( s ) ) d s + t T c . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_Equr_HTML.gif
If c = u c ( T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq28_HTML.gif, then u c http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq29_HTML.gif is a solution of the nonlinear problem (1). We then look for fixed points of the map
c R u c ( T ) R . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_Equs_HTML.gif

For c R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq18_HTML.gif fixed, we try to solve the integral equation (7).

Assume that there exist a , b C [ 0 , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq30_HTML.gif and α [ 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq31_HTML.gif such that
| f ( t , u ) | a ( t ) + b ( t ) | u | α http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_Equ8_HTML.gif
(8)

for every t [ 0 , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq32_HTML.gif, u R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq33_HTML.gif.

For v C [ 0 , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq34_HTML.gif, define F c v C [ 0 , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq35_HTML.gif as
[ F c v ] ( t ) = f ( t , v ( t ) + t T c ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_Equt_HTML.gif

Thus, a solution of (7) is precisely a fixed point of K F c = K c http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq36_HTML.gif. Note that K c http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq37_HTML.gif is a compact operator. For v C [ 0 , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq34_HTML.gif, let v = sup t [ 0 , T ] | v ( t ) | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq38_HTML.gif.

For λ ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq39_HTML.gif, if v = λ K c ( v ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq40_HTML.gif we have
v ( t ) = λ 0 T k ( t , s ) f ( s , v ( s ) + s T c ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_Equu_HTML.gif
and
| v ( t ) | k 0 T f ( s , v ( s ) + s T c ) d s k T [ a + b ( v + c ) α ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_Equv_HTML.gif
Hence there exist constants a 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq41_HTML.gif, b 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq42_HTML.gif such that
v a 0 + b 0 ( v + c ) α http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_Equ9_HTML.gif
(9)

for any v C [ 0 , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq43_HTML.gif and λ ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq39_HTML.gif solution of v = λ K c ( v ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq40_HTML.gif. This implies that v is bounded independently of λ ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq44_HTML.gif, and hence by Schaefer’s fixed point theorem (Theorem 4.3.2 of [22]), K c http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq37_HTML.gif has at least a fixed point, i.e., for given c, equation (7) is solvable.

Now suppose f is Lipschitz continuous.

Then there exists l > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq45_HTML.gif such that
| f ( t , x ) f ( t , y ) | l | x y | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_Equ10_HTML.gif
(10)

for every t [ 0 , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq32_HTML.gif and x , y R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq46_HTML.gif.

Then, for v , w C [ 0 , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq47_HTML.gif, we have
| [ K c v ] ( t ) [ K c w ] ( t ) | 0 T k ( t , s ) l | v ( s ) w ( s ) | d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_Equw_HTML.gif
and
K c v K c w k l T v w . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_Equx_HTML.gif

Thus, for l > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq45_HTML.gif small, equation (7) has a unique solution in view of the classical Banach contraction fixed point theorem.

Now, under conditions (8) and (10), set
c R v c C [ 0 , T ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_Equy_HTML.gif

where v c http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq48_HTML.gif is the unique solution of (7), and as a consequence of the contraction principle, this map is continuous.

Define the map
φ : R R , φ ( c ) = v c ( T ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_Equz_HTML.gif
If there exists c R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq18_HTML.gif such that φ ( c ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq49_HTML.gif, then for that c we have v c ( T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq50_HTML.gif, and the function
u c ( t ) = v c ( t ) + t T c http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_Equaa_HTML.gif

is such that u c ( T ) = c http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq51_HTML.gif, and therefore u c http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq29_HTML.gif is a solution of the original nonlinear problem (1).

Now, assume that
lim u ± f ( t , u ) = ± http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_Equ11_HTML.gif
(11)

uniformly on t [ 0 , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq32_HTML.gif.

Then the growth of v http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq52_HTML.gif is sublinear in view of estimate (9). However, c growths linearly. Hence the norm of the function
v c ( s ) + s T c http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_Equab_HTML.gif

growths asymptotically as c.

This implies that lim c ± φ ( c ) = ± http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq53_HTML.gif, and there exists c R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq18_HTML.gif with φ ( c ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq49_HTML.gif.

We have the following result.

Theorem 3.1 Suppose that f satisfies the growth conditions (8) and (10). If (11) holds, then (1) is solvable for l sufficiently small.

Note that condition (11) is crucial since for f ( t , u ) = σ ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-130/MediaObjects/13661_2013_Article_377_IEq54_HTML.gif and, in view of (5), the problem (1) may have no solution.

Declarations

Acknowledgements

This research has been partially supported by Ministerio de Economía y Competitividad (Spain), project MTM2010-15314, and co-financed by the European Community fund FEDER. The author is thankful to the referees for their useful suggestions.

Authors’ Affiliations

(1)
Departamento de Análisis Matemático, Facultad de Matemáticas, Universidad de Santiago de Compostela
(2)
Department of Mathematics, Faculty of Science, King Abdulaziz University

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