Open Access

Variational approach to second-order impulsive dynamic equations on time scales

Boundary Value Problems20132013:119

DOI: 10.1186/1687-2770-2013-119

Received: 27 February 2013

Accepted: 23 April 2013

Published: 9 May 2013

Abstract

The aim of this paper is to employ variational techniques and critical point theory to prove some conditions for the existence of solutions to a nonlinear impulsive dynamic equation with homogeneous Dirichlet boundary conditions. Also, we are interested in the solutions of the impulsive nonlinear problem with linear derivative dependence satisfying an impulsive condition.

MSC:34B37, 34N05.

Keywords

impulsive dynamic equations second-order boundary value problem variational techniques critical point theory time scales

1 Introduction

This paper is concerned with the existence of solutions of second-order impulsive dynamic equations on time scales. More precisely, we consider the following boundary value problem:
( P ) { u Δ Δ ( t ) + λ u σ ( t ) = f ( t , u σ ( t ) ) ; Δ -a.e.  t [ 0 , T ] T κ 2 , u ( 0 ) = 0 = u ( T ) , u Δ ( t j + ) u Δ ( t j ) = I j ( u ( t j ) ) , j = 1 , 2 , , p , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equa_HTML.gif

where the impulsive points t j J https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq1_HTML.gif are right-dense points in an arbitrary time scale T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq2_HTML.gif, with t 0 = 0 < t 1 < t 2 < < t p < t p + 1 = T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq3_HTML.gif. Here f : [ 0 , T ] T × R R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq4_HTML.gif and I j : R R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq5_HTML.gif, j = 1 , , p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq6_HTML.gif, are continuous functions.

It is well known that the theory of impulsive dynamic equations provides a natural framework for mathematical modeling of many real world phenomena. The impulsive effects exist widely in many evolution processes in which their states are changed abruptly at certain moments of time.

Applications of impulsive dynamic equations arise in biology (biological phenomena involving thresholds), medicine (bursting rhythm models), pharmacokinetics, mechanics, engineering, chaos theory, etc. As a consequence, there has been a significant development in impulse theory in recent years. We can see some general and recent works on the theory of impulsive differential equations; see [19] and the references therein.

For a second-order dynamic equation, we usually consider impulses in the position and velocity. However, in the motion of spacecraft, one has to consider instantaneous impulses depending on the position that result in jump discontinuities in velocity, but with no change in position. The impulses only on the velocity occur also in impulsive mechanics. An impulsive problem with impulses in the derivative is considered in [10].

Moreover, we are interested in the solutions of the impulsive nonlinear problem in time scale with derivative dependence satisfying an impulsive condition. We can see, for example, recent works on the theory of impulsive differential equations in [1, 3, 6, 8, 11].

There have been several approaches to studying the solutions of impulsive dynamic equations on time scales, such as the method of lower and upper solutions, fixed-point theory [1214]. Sobolev spaces of functions on time scales, which were first introduced in [15], opened a very fruitful new approach in the study of dynamic equations on time scales: the use of variational methods in the context of boundary value problems on time scales (see [16, 17]) or in second-order Hamiltonian systems [18]. Moreover, the study of the existence and multiplicity of solutions for impulsive dynamic equations on time scales has also been done by means of the variational method (see, for example, [19, 20]).

The aim of this paper is to use variational techniques and critical point theory to derive the existence of multiple solutions to (P); we refer the reader to [2124] for a broad introduction to dynamic equations on time scales and to [25, 26] for variational methods and critical point theory.

The paper is organized as follows. In Section 2 we gather together essential properties about Sobolev spaces on time scales proved in [15, 27, 28] which one needs to read this paper.

The goal of Section 3 is to exhibit the variational formulation for the impulsive Dirichlet problem. As we will see, all these problems can be understood and solved in terms of the minimization of a functional, usually related to the energy, in an appropriate space of functions. The results presented in the part where we address the linear problem are basic but crucial to revealing that a problem can be solved by finding the critical points of a functional. Moreover, we prove some sufficient conditions for the existence of at least one positive solution to (P).

To finish, in Section 4, we present an impulsive nonlinear problem with linear derivative dependence. We transform the problem into an equivalent one that has no dependence on the derivative, and then we prove that the problem has at least one solution. Also, with additional conditions in nonlinearities and impulse functions, we can show the existence of at least two solutions by using the mountain pass theorem.

2 Preliminaries

Let T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq2_HTML.gif be an arbitrary time scale. We assume that T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq2_HTML.gif has the topology that it inherits from the standard topology on . Assume that a < b https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq7_HTML.gif are points in T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq2_HTML.gif and define the time scale interval [ a , b ] T = { t T : a t b } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq8_HTML.gif. We denote J 0 = [ a , b ) T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq9_HTML.gif.

Below we set out some results proved in [15, 27] about Sobolev spaces on time scales.

Definition 2.1 Let p R ¯ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq10_HTML.gif be such that p 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq11_HTML.gif and u : J R ¯ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq12_HTML.gif. We say that u belongs to W Δ 1 , p ( J ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq13_HTML.gif if and only if u L Δ p ( J 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq14_HTML.gif, and there exists g : J κ R ¯ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq15_HTML.gif such that g L Δ p ( J 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq16_HTML.gif and
J 0 ( u φ Δ ) ( s ) Δ s = J 0 ( g φ σ ) ( s ) Δ s φ C 0 , rd 1 ( J k ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equb_HTML.gif
with
C 0 , rd 1 ( J k ) = { u : J R : u C rd 1 ( J k ) , u ( a ) = u ( b ) = 0 } , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equc_HTML.gif

and C rd 1 ( J κ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq17_HTML.gif is the set of all continuous functions on J such that they are Δ-differentiable on J κ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq18_HTML.gif and their Δ-derivatives are rd-continuous on J κ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq18_HTML.gif.

Theorem 2.1 Assume that p R ¯ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq10_HTML.gif and p 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq11_HTML.gif. The set W Δ 1 , p ( J ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq13_HTML.gif is a Banach space with the norm defined for every x W Δ 1 , p ( J ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq19_HTML.gif as
x W Δ 1 , p : = x L Δ p + x Δ L Δ p . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equ1_HTML.gif
(2.1)
Moreover, the set H Δ 1 ( J ) : = W Δ 1 , 2 ( J ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq20_HTML.gif is a Hilbert space with the inner product given for every ( x , y ) H Δ 1 ( J ) × H Δ 1 ( J ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq21_HTML.gif by
( x , y ) H Δ 1 : = ( x , y ) L Δ 2 + ( x Δ , y Δ ) L Δ 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equ2_HTML.gif
(2.2)
Proposition 2.1 Assume that p R ¯ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq10_HTML.gif with p 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq11_HTML.gif, then there exists a constant K > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq22_HTML.gif, only dependent on b a https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq23_HTML.gif, such that the inequality
x C ( J ) K x W Δ 1 , p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equd_HTML.gif

holds for all x W Δ 1 , p ( J ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq19_HTML.gif, and hence the immersion W Δ 1 , p ( J ) C ( J ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq24_HTML.gif is continuous.

Definition 2.2 Let p R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq25_HTML.gif be such that p 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq11_HTML.gif, define the set W 0 , Δ 1 , p ( J ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq26_HTML.gif as the closure of the set C 0 , rd 1 ( J κ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq27_HTML.gif in W Δ 1 , p ( J ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq13_HTML.gif. We define H 0 , Δ 1 ( J ) : = W 0 , Δ 1 , 2 ( J ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq28_HTML.gif.

The spaces W 0 , Δ 1 , p ( J ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq29_HTML.gif and H 0 , Δ 1 ( J ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq30_HTML.gif are endowed with the norm induced by W Δ 1 , p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq31_HTML.gif, defined in (2.1), and the inner product induced by ( , ) H Δ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq32_HTML.gif, defined in (2.2). These spaces satisfy the following properties.

Proposition 2.2 (Poincare’s inequality)

Let p R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq25_HTML.gif be such that p 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq11_HTML.gif. Then there exists a constant L > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq33_HTML.gif, only dependent on b a https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq23_HTML.gif, such that
u W Δ 1 , p L u Δ L Δ p u W 0 , Δ 1 , p ( J ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Eque_HTML.gif

Proposition 2.3 (Corollary 3.3 in [27])

If u H 0 , Δ 1 ( J ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq34_HTML.gif, then
a b ( u σ ) 2 ( t ) Δ t 1 λ 1 a b ( u Δ ) 2 ( t ) Δ t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equf_HTML.gif

holds, where λ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq35_HTML.gif is the smallest positive eigenvalue of problem u Δ Δ ( t ) = λ u σ ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq36_HTML.gif; t J κ 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq37_HTML.gif and u ( a ) = 0 = u ( b ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq38_HTML.gif.

In the Sobolev space H 0 , Δ 1 ( J ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq30_HTML.gif with a = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq39_HTML.gif and b = T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq40_HTML.gif, consider the inner product
( u , v ) = 0 T u Δ ( t ) v Δ ( t ) Δ t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equg_HTML.gif

inducing the norm https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq41_HTML.gif.

It is the consequence of Poincare’s inequality that
u H 0 , Δ 1 ( J ) u W Δ 1 , 2 ( J ) c u Δ L Δ 2 ( J 0 ) c u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equ3_HTML.gif
(2.3)
and
u u W Δ 1 , 2 ( J ) 2 u H 0 , Δ 1 ( J ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equ4_HTML.gif
(2.4)

3 Variational formulation of (P) and existence results

Firstly, to show the variational structure underlying an impulsive dynamic equation, we consider the lineal problem
( LP ) { u Δ Δ ( t ) + λ u σ ( t ) = h ( t ) , Δ -a.e.  t J κ 2 , u Δ ( t j + ) u Δ ( t j ) = d j , j = 1 , 2 , , p , u ( 0 ) = u ( T ) = 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equh_HTML.gif

where we consider J with a = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq39_HTML.gif and b = T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq40_HTML.gif and d j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq42_HTML.gif, j = 1 , , p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq43_HTML.gif, are fixed constants.

Suppose that u C rd ( J ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq44_HTML.gif is such that u ( 0 ) = 0 = u ( T ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq45_HTML.gif. Moreover, assume that for every j = 0 , 1 , , p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq46_HTML.gif, u j : = u | ( t j , t j + 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq47_HTML.gif is such that u j H Δ 2 ( t j , t j + 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq48_HTML.gif.

Definition 3.1 We say that u is a classical solution of (LP) if the limits u Δ ( t j + ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq49_HTML.gif and u Δ ( t j ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq50_HTML.gif exist for every j = 1 , 2 , , p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq51_HTML.gif and it satisfies the equation on (LP) for Δ-almost everywhere (Δ-a.e.) t J κ 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq52_HTML.gif.

Take v H 0 , Δ 1 ( J ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq53_HTML.gif, multiply the equation by v σ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq54_HTML.gif and integrate between 0 and T:
0 T u Δ Δ v σ + λ 0 T u σ v σ = 0 T h v σ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equi_HTML.gif
Taking into account that v ( 0 ) = 0 = v ( T ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq55_HTML.gif and integrating by parts, we get
0 T u Δ Δ v σ = j = 0 p t j t j + 1 u Δ Δ v σ = j = 0 p [ ( u Δ v ) | t j + t j + 1 + t j t j + 1 u Δ v Δ ] = j = 1 p d j v ( t j ) + 0 T u Δ v Δ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equj_HTML.gif
Hence,
0 T u Δ v Δ + λ 0 T u σ v σ = 0 T h v σ j = 1 p d j v ( t j ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equk_HTML.gif
We define the bilinear form a : H 0 , Δ 1 ( J ) × H 0 , Δ 1 ( J ) R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq56_HTML.gif by
a ( u , v ) = 0 T u Δ v Δ + λ 0 T u σ v σ , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equ5_HTML.gif
(3.1)
and the linear operator l : H 0 , Δ 1 ( J ) R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq57_HTML.gif by
l ( v ) = 0 T h v σ j = 1 p d j v ( t j ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equ6_HTML.gif
(3.2)

Thus, the concept of weak solution for the impulsive problem (LP) is a function u H 0 , Δ 1 ( J ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq34_HTML.gif such that a ( u , v ) = l ( v ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq58_HTML.gif is valid for any v H 0 , Δ 1 ( J ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq53_HTML.gif.

We can prove that a defined by (3.1) and l defined by (3.2) are continuous, and, from Proposition 2.3, that a is coercive if λ > λ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq59_HTML.gif.

Consider φ : H 0 , Δ 1 ( J ) R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq60_HTML.gif defined by
φ ( v ) = 1 2 a ( v , v ) l ( v ) = 1 2 0 T ( v Δ ) 2 + λ 2 0 T ( v σ ) 2 0 T h v σ + j = 1 p d j v ( t j ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equ7_HTML.gif
(3.3)

We can deduce the following regularity properties which allow us to assert that the solutions to (LP) are precisely the critical points of φ.

Lemma 3.1 The following statements are valid.
  1. 1.
    φ is differentiable at any u H 0 , Δ 1 ( J ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq34_HTML.gif and
    ( φ ( u ) , v ) = 0 T u Δ v Δ + λ 0 T u σ v σ 0 T h v σ + j = 1 p d j v ( t j ) = a ( u , v ) l ( v ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equl_HTML.gif
     
  2. 2.

    If u H 0 , Δ 1 ( J ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq34_HTML.gif is a critical point of φ defined by (3.3), then u is a weak solution of the impulsive problem (LP).

     

We will use the following result in linear functional analysis, which ensures the existence of a critical point of φ.

Theorem 3.1 (Lax-Milgram theorem)

Let H be a Hilbert space and let a : H × H R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq61_HTML.gif be a bounded bilinear form. If a is coercive, i.e., there exists α > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq62_HTML.gif such that a ( u , u ) α u 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq63_HTML.gif for every u H https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq64_HTML.gif, then for any σ H https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq65_HTML.gif (the conjugate space of H) there exists a unique u H https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq64_HTML.gif such that
a ( u , v ) = φ , v , v H . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equm_HTML.gif
Moreover, if a is also symmetric, then the functional J : H R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq66_HTML.gif defined by
J ( v ) = 1 2 a ( u , v ) φ , v , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equn_HTML.gif

attains its minimum at u.

By the Lax-Milgram theorem, we obtain the following result.

Theorem 3.2 If λ > λ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq59_HTML.gif then the problem (LP) has a weak solution u H 0 , Δ 1 ( J ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq34_HTML.gif for any h L Δ 2 ( J 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq67_HTML.gif. Moreover, u H Δ 2 ( J ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq68_HTML.gif and u is a classical solution and u minimizes the functional (3.3), and hence it is a critical point of φ.

3.1 Impulsive nonlinear problem

We consider the nonlinear Dirichlet problem
( P ) { u Δ Δ ( t ) + λ u σ ( t ) = f ( t , u σ ( t ) ) ; Δ -a.e.  t J κ 2 , u Δ ( t j + ) u Δ ( t j ) = I j ( u ( t j ) ) , j = 1 , 2 , , p , u ( 0 ) = 0 = u ( T ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equo_HTML.gif

We assume that λ > λ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq59_HTML.gif.

A weak solution of (P) is a function u H 0 , Δ 1 ( J ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq69_HTML.gif such that
0 T u Δ v Δ + λ 0 T u σ v σ = j = 1 p I j ( u ( t j ) ) v ( t j ) + 0 T f ( t , u σ ( t ) ) v σ ( t ) Δ t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equp_HTML.gif

for every v H 0 , Δ 1 ( J ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq53_HTML.gif.

We now consider the functional
φ ( u ) = 1 2 a ( u , u ) l ( u ) = 1 2 0 T ( u Δ ) 2 + λ 2 0 T ( u σ ) 2 + j = 1 p 0 u ( t j ) I j ( t ) d t 0 T F ( t , u σ ( t ) ) Δ t , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equ8_HTML.gif
(3.4)

where F ( t , u ) : = 0 u f ( t , x ) d x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq70_HTML.gif.

One can deduce, from the properties of H, f and I j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq71_HTML.gif, the following regularity properties of φ.

Proposition 3.1 The functional φ defined by (3.4) is continuous, differentiable, and weakly lower semi-continuous. Moreover, the critical points of φ are weak solutions of (P).

Theorem 3.3 Suppose that f is bounded and that the impulsive functions I j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq71_HTML.gif are bounded. Then there is a critical point of φ, and (P) has at least one solution.

Proof Take M > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq72_HTML.gif and M j > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq73_HTML.gif, j = 1 , 2 , , p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq51_HTML.gif, such that
| f ( t , u ) | M ( t , u ) [ 0 , T ] T × R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equq_HTML.gif
and
| I j ( u ) | M j u R , j = 1 , 2 , , p . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equr_HTML.gif
Using that λ > λ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq59_HTML.gif, there exists α > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq74_HTML.gif such that for any u H 0 , Δ 1 ( J ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq34_HTML.gif
φ ( u ) α 2 u 2 + j = 1 p 0 u ( t j ) I j ( t ) d t 0 T F ( t , u σ ( t ) ) Δ t α 2 u 2 j = 1 p M j | u ( t j ) | M 0 T | u σ ( t ) | Δ t . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equs_HTML.gif
Thus, using Proposition 2.1, (2.3) and m = max j = 1 , , p { M , M j } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq75_HTML.gif, we have
φ ( u ) α 2 u 2 m ( j = 1 p | u ( t j ) | 0 T | u σ ( t ) | Δ t ) α 2 u 2 m ( p u C ( J ) + T u C ( J ) ) α 2 u 2 m ρ u C ( J ) α 2 u 2 m ρ K u H 0 , Δ 1 α 2 u 2 m ρ K c u , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equt_HTML.gif

where ρ = p + T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq76_HTML.gif.

This implies that lim u φ ( u ) = + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq77_HTML.gif, and φ is coercive. Hence (Th. 1.1 of [26]), φ has a minimum, which is a critical point of φ. □

Theorem 3.4 Suppose that f is sublinear and the impulsive functions I j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq71_HTML.gif have sublinear growth. Then there is a critical point of φ and (P) has at least one solution.

Proof Let a , b , a j , b j > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq78_HTML.gif, and γ , γ j [ 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq79_HTML.gif, j = 1 , 2 , , p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq51_HTML.gif, such that
| f ( t , u ) | a + b | u | γ and | I j ( u ) | a j + b j | u | γ j t [ 0 , T ] T , u R . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equu_HTML.gif
Again, using that λ > λ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq59_HTML.gif, Proposition 2.1, (2.3) and m = max j = 1 , , p { a , a j } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq80_HTML.gif, m ˜ = max j = 1 , , p { b , b j } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq81_HTML.gif, we have
φ ( u ) α 2 u 2 β u δ u γ + 1 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equv_HTML.gif

where β = m ρ K c https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq82_HTML.gif and δ = m ˜ ρ K γ + 1 c γ + 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq83_HTML.gif.

Since γ + 1 < 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq84_HTML.gif, then lim u φ ( u ) = + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq85_HTML.gif for every u H 0 , Δ 1 ( J ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq34_HTML.gif. □

4 Impulsive nonlinear problem with linear derivative dependence

Consider the following problem:
( NP ) { u Δ Δ ( t ) + g ( t ) u Δ ( σ ( t ) ) + λ u σ ( t ) = f ( t , u σ ( t ) ) ; Δ -a.e.  t J κ 2 , ( u Δ ( t j + ) u Δ ( t j ) ) = I j ( u ( t j ) ) , j = 1 , 2 , , p , u ( 0 ) = 0 = u ( T ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equw_HTML.gif

where f and I j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq71_HTML.gif, j = 1 , , p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq6_HTML.gif are continuous and g is continuous and regressive.

We assume that λ > m λ 1 / M https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq86_HTML.gif. Here, m = min t J e g ( t , 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq87_HTML.gif, M = max t J e g ( t , 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq88_HTML.gif, where e g ( t , 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq89_HTML.gif is an exponential function. Note that, as g is regressive, e g ( , 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq90_HTML.gif is the solution of the problem
y Δ = g ( t ) y , y ( 0 ) = 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equx_HTML.gif
We transform the problem (NP) into the following equivalent form:
( NPE ) { ( e g ( t , 0 ) u Δ ( t ) ) Δ + λ e g ( t , 0 ) u σ ( t ) = e g ( t , 0 ) f ( t , u σ ( t ) ) ; Δ -a.e.  t J κ 2 , ( u Δ ( t j + ) u Δ ( t j ) ) = I j ( u ( t j ) ) , j = 1 , 2 , , p , u ( 0 ) = u ( T ) = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equy_HTML.gif
Obviously, the solutions of (NPE) are solutions of (NP). Consider the Hilbert space H 0 , Δ 1 ( J ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq30_HTML.gif with the inner product:
( u , v ) = 0 T e g ( t , 0 ) u Δ ( t ) v Δ ( t ) Δ t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equz_HTML.gif
and the norm induced
u = ( 0 T e g ( t , 0 ) | u Δ ( t ) | 2 Δ t ) 1 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equaa_HTML.gif
A weak solution of (NPE) is a function u H 0 , Δ 1 ( J ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq69_HTML.gif such that
0 T e g ( t , 0 ) u Δ ( t ) v Δ ( t ) Δ t + λ 0 T e g ( t , 0 ) u σ ( t ) v σ ( t ) Δ t = j = 1 p e g ( t j , 0 ) I j ( u ( t j ) ) v ( t j ) + 0 T e g ( t , 0 ) f ( t , u σ ( t ) ) v σ ( t ) Δ t . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equab_HTML.gif
Hence, a weak solution of (NP) is a critical point of the following functional:
ψ ( u ) = 1 2 A ( u , u ) j = 1 p e g ( t j , 0 ) 0 u ( t j ) I j ( t ) d t 0 T e g ( t , 0 ) F ( t , u σ ( t ) ) Δ t , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equ9_HTML.gif
(4.1)
where
F ( t , u ) = 0 u f ( t , ξ ) d ξ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equac_HTML.gif
and
A ( u , u ) = 0 T e g ( t , 0 ) u Δ ( t ) v Δ ( t ) Δ t + λ 0 T e g ( t , 0 ) u σ ( t ) v σ ( t ) Δ t . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equad_HTML.gif

It is evident that A is bilinear, continuous and symmetric.

Lemma 4.1 (Theorem 38.A of [29])

For the functional F : M X R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq91_HTML.gif with M not empty, min u M F ( u ) = a https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq92_HTML.gif has solutions in case the following hold:
  1. (i)

    X is a reflexive Banach space.

     
  2. (ii)

    M is bounded and weak sequentially closed.

     
  3. (iii)

    φ is sequentially lower semi-continuous on M.

     

Lemma 4.2 (Analogous to Lemma 2.2 of [5])

There exist constants β > α > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq93_HTML.gif such that
α u 2 A ( u , u ) β u 2 , u H 0 , Δ 1 ( J ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equae_HTML.gif

Proof In fact, by Poincare’s inequality, if λ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq94_HTML.gif, we can take α = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq95_HTML.gif, β = 1 + λ M λ 1 m https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq96_HTML.gif; if m λ 1 M < λ < 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq97_HTML.gif, then we can take α = 1 + λ M λ 1 m https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq98_HTML.gif and β = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq99_HTML.gif. □

Lemma 4.3 If u H 0 , Δ 1 ( J ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq34_HTML.gif, then there exists a constant δ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq100_HTML.gif such that u 0 δ u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq101_HTML.gif, where
u 0 = max t [ 0 , T ] T | u ( t ) | . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equaf_HTML.gif
Proof The result is followed by the following inequalities:
| u ( t ) | 0 T | u Δ ( s ) | Δ s ( 0 T 1 e g ( t , 0 ) ) 1 2 ( 0 T e g ( t , 0 ) | u Δ ( s ) | 2 Δ s ) 1 2 T m u = δ u . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equag_HTML.gif

 □

Lemma 4.4 The functional ψ defined by (4.1) is continuous, continuously differentiable and weakly lower semi-continuous.

Theorem 4.1 Suppose that λ > m λ 1 M https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq102_HTML.gif, f and I j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq71_HTML.gif are bounded, j = 1 , 2 , , p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq51_HTML.gif, then (NP) has at least one solution.

Proof Take B > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq103_HTML.gif and B j > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq104_HTML.gif, j = 1 , , p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq6_HTML.gif, such that
| f ( t , u ) | B ( t , u ) [ 0 , T ] T × R , | I j ( u ) | B j u R , j = 1 , 2 , , p . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equah_HTML.gif
For any u H 0 , Δ 1 ( J ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq34_HTML.gif, using Lemma 4.3 and Proposition 2.3, we have
ψ ( u ) α 2 u 2 ( M δ j = 1 p B j + B M T m λ 1 ) u . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equai_HTML.gif

This implies that lim u ψ ( u ) = + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq105_HTML.gif, and ψ is coercive. Hence, ψ has a minimum, which is a critical point of ψ. □

We will apply the mountain pass theorem in order to obtain at least two critical points of ψ.

Suppose that X is a Banach space (in particular, a Hilbert space) and ϕ : X R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq106_HTML.gif is differentiable and c R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq107_HTML.gif. We say that ϕ satisfies the Palais-Smale condition if every bounded sequence { u k } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq108_HTML.gif in the space X such that lim k ϕ ( u k ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq109_HTML.gif contains a convergent subsequence.

Theorem 4.2 (Mountain pass theorem)

Let ϕ C 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq110_HTML.gif be such that it satisfies the Palais-Smale condition. Assume that there exist u 0 , u 1 X https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq111_HTML.gif and a bounded neighborhood Ω of u 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq112_HTML.gif such that u 1 Ω https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq113_HTML.gif and
inf { ϕ ( u ) : u Ω } > max { ϕ ( u 0 ) , ϕ ( u 1 ) } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equaj_HTML.gif

Then there exists a critical point u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq114_HTML.gif of ϕ.

Theorem 4.3 Suppose that λ > m λ 1 M https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq102_HTML.gif, then the problem (NP) has at least two solutions if the following conditions hold:

( H 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq115_HTML.gif) There exist constants η > 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq116_HTML.gif and γ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq117_HTML.gif such that for all ( t , u ) [ 0 , T ] T × R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq118_HTML.gif, | u | γ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq119_HTML.gif
0 < η F ( t , u ) u f ( t , u ) , 0 < η 0 u I j ( ξ ) d ξ u I j ( u ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equak_HTML.gif

where j = 1 , 2 , , p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq51_HTML.gif.

( H 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq120_HTML.gif) There exists a positive s η https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq121_HTML.gif such that f ( t , u ) = o ( | u | s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq122_HTML.gif and I j ( u ) = o ( | u | s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq123_HTML.gif uniformly for t [ 0 , T ] T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq124_HTML.gif as | u | https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq125_HTML.gif, j = 1 , 2 , , p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq51_HTML.gif.

( H 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq126_HTML.gif) f ( t , u ) = o ( | u | ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq127_HTML.gif and I j ( u ) = o ( | u | ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq128_HTML.gif uniformly for t [ 0 , T ] T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq129_HTML.gif as | u | 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq130_HTML.gif, j = 1 , 2 , , p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq51_HTML.gif.

Proof From ( H 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq120_HTML.gif), ( H 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq126_HTML.gif) and the continuities of f and I j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq71_HTML.gif, it is easy to see that for any ε > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq131_HTML.gif and ( t , u ) [ 0 , T ] T × R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq132_HTML.gif, there exist C 1 ( ε ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq133_HTML.gif and C 1 j > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq134_HTML.gif such that
| f ( t , u ) | ε | u | + C 1 ( ε ) | u | s , | I j ( u ) | ε | u | + C 1 j ( ε ) | u | s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equal_HTML.gif
Hence, for any ε > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq131_HTML.gif and ( t , u ) [ 0 , T ] T × R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq135_HTML.gif, we have
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equ10_HTML.gif
(4.2)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equ11_HTML.gif
(4.3)

where C 2 ( ε ) = C 1 ( ε ) s + 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq136_HTML.gif and C 2 j ( ε ) = C 1 j ( ε ) s + 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq137_HTML.gif.

From the condition ( H 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq115_HTML.gif), the following hold:
η u f ( t , u ) F ( t , u ) u γ , η u f ( t , u ) F ( t , u ) u γ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equam_HTML.gif
Integrating the above two inequalities with respect to u on [ γ , u ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq138_HTML.gif and [ u , γ ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq139_HTML.gif, respectively (in this case, these are integrals on ), we have
η ln u γ ln F ( t , u ) F ( t , γ ) u γ , η ln γ u ln F ( t , γ ) F ( t , u ) u γ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equan_HTML.gif
That is,
F ( t , u ) F ( t , γ ) ( u γ ) η u γ , F ( t , u ) F ( t , γ ) ( u γ ) η u γ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equao_HTML.gif

Thus there exists a constant a 1 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq140_HTML.gif such that F ( t , u ) a 1 | u | η https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq141_HTML.gif for all | u | γ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq142_HTML.gif.

From the continuity of F ( t , u ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq143_HTML.gif, there exists a constant k > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq144_HTML.gif such that
F ( t , u ) k a 1 | u | η a 1 γ η k | u | γ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equap_HTML.gif
Hence, we have
F ( t , u ) a 1 | u | η a 2 ( t , u ) [ 0 , T ] T × R , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equ12_HTML.gif
(4.4)

where a 2 = a 1 γ η + k https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq145_HTML.gif.

Similarly, there exist a 1 j , a 2 j > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq146_HTML.gif such that
0 u I j ( ξ ) d ξ a 1 j | u | η a 2 j u R . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equ13_HTML.gif
(4.5)

Firstly, we apply Lemma 4.1 to show that there exists ρ such that ψ has a local minimum u 0 B ρ = { u H 0 , Δ 1 ( J ) : u < ρ } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq147_HTML.gif.

Since H 0 , Δ 1 ( J ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq30_HTML.gif is a Hilbert space, it is easy to deduce that B ¯ ρ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq148_HTML.gif is bounded and weak sequentially closed. Lemma 4.4 has shown that ψ is weak lower semi-continuous on B ¯ ρ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq148_HTML.gif and, besides, H 0 , Δ 1 ( J ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq30_HTML.gif is a reflexive Banach space. So, by Lemma 4.1 we can have this u 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq112_HTML.gif such that ψ ( u 0 ) = min { ψ ( u ) : u B ¯ ρ } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq149_HTML.gif.

Now we will show that ψ ( u 0 ) = min { ψ ( u ) : u B ρ } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq150_HTML.gif for some ρ = ρ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq151_HTML.gif.

In fact, from (4.2) and (4.3), we obtain
ψ ( u ) α 2 u 2 j = 1 p e g ( t , 0 ) 0 u ( t j ) I j ( t ) d t 0 T e g ( t , 0 ) F ( t , u σ ( t ) ) Δ t α 2 u 2 j = 1 p e g ( t , 0 ) ( ε 2 | u | 2 + C 2 j ( ε ) | u | s + 1 ) 0 T e g ( t , 0 ) ( ε 2 | u | 2 + C 2 ( ε ) | u σ | s + 1 ) α 2 u 2 M ε 2 p δ 2 u 2 δ s + 1 M j = 1 p C 2 j ( ε ) u s + 1 M ε 2 0 T | u σ ( t ) | 2 Δ t M C 2 ( ε ) 0 T | u σ ( t ) | s + 1 Δ t α 2 u 2 M ε 2 p δ 2 u 2 δ s + 1 M j = 1 p C 2 j ( ε ) u s + 1 M ε 2 1 m λ 1 u 2 M C 2 ( ε ) T k s + 1 c s + 1 ( m ) s + 1 u s + 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equaq_HTML.gif
Hence,
ψ ( u ) α M ε ( 1 m λ 1 + δ 2 p ) 2 u 2 M ( δ s + 1 j = 1 p C 2 j ( ε ) + C 2 ( ε ) T k s + 1 c s + 1 ( m ) s + 1 ) u s + 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equar_HTML.gif
We can choose
ε = α 2 M ( 1 m λ 1 + δ 2 p ) , ρ 0 = ( α 8 M ( T k s + 1 c s + 1 ( m ) s + 1 C 2 ( ε ) + δ s + 1 j = 1 p C 2 j ( ε ) ) ) 1 s 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equas_HTML.gif

For any u B ρ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq152_HTML.gif, u = ρ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq153_HTML.gif, we have ψ ( u ) α 8 ρ 0 2 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq154_HTML.gif. Besides, ψ ( u 0 ) ψ ( 0 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq155_HTML.gif. Then ψ ( u ) > α 8 ρ 0 2 > ψ ( 0 ) ψ ( u 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq156_HTML.gif for any u B ρ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq152_HTML.gif. So, ψ ( u 0 ) < inf { ψ ( u ) : u B ρ 0 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq157_HTML.gif. Hence, ψ has a local minimum u 0 B ρ 0 = { u H 0 , Δ 1 ( J ) : u < ρ 0 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq158_HTML.gif.

Next, we will show that there exists u 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq159_HTML.gif with u 1 > ρ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq160_HTML.gif such that ψ ( u 1 ) < inf { ψ ( u ) : u B ρ 0 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq161_HTML.gif.

From (4.4) and (4.5), we have
ψ ( u ) β 2 u 2 j = 1 p e g ( t , 0 ) 0 u ( t j ) I j ( t ) d t 0 T e g ( t , 0 ) F ( t , u σ ( t ) ) Δ t β 2 u 2 j = 1 p e g ( t , 0 ) ( a 1 j | u ( t j ) | η a 2 j ) 0 T e g ( t , 0 ) ( a 1 | u σ ( t ) | η a 2 ) β 2 u 2 m j = 1 p a 1 j | u ( t j ) | η + M j = 1 p a 2 j m a 1 0 T | u σ ( t ) | η Δ t + M a 2 T . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equat_HTML.gif
Thus,
ψ ( u ) β 2 u 2 m j = 1 p a 1 j | u ( t j ) | η + M j = 1 p a 2 j m a 1 u σ L Δ η η + M a 2 T . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equau_HTML.gif
For any u H 0 , Δ 1 ( J ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq34_HTML.gif with u = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq162_HTML.gif, we have
ψ ( N u ) β 2 N 2 m j = 1 p a 1 j N η | u ( t j ) | η + M j = 1 p a 2 j m a 1 N η u σ L Δ η + M a 2 T . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equav_HTML.gif

So, lim N ψ ( N u ) = https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq163_HTML.gif since η > 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq116_HTML.gif. Then there exists N 0 > ρ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq164_HTML.gif such that ψ ( N 0 u ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq165_HTML.gif.

Hence, for the above ρ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq166_HTML.gif, there exists u 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq159_HTML.gif such that u 1 = N 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq167_HTML.gif and ψ ( u 1 ) < 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq168_HTML.gif.

Then, we have max { ψ ( u 0 ) , ψ ( u 1 ) } < inf { ψ ( u ) : u B T 0 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq169_HTML.gif.

The next step is to show that ψ satisfies the Palais-Smale condition.

Let { ψ ( u k ) } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq170_HTML.gif be a bounded sequence such that lim k ψ ( u k ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq171_HTML.gif. Now we show that u k https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq172_HTML.gif is bounded. By (4.1) we have
( ψ ( u k ) , u k ) = A ( u k , u k ) 0 T e g ( t , 0 ) f ( t , u k σ ( t ) ) u k σ ( t ) Δ t j = 1 p e g ( t , 0 ) I j ( u k ( t j ) ) u k ( t j ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equ14_HTML.gif
(4.6)
Thus,
ψ ( u k ) 1 η ( ψ ( u k ) , u k ) = ( 1 2 1 η ) A ( u k , u k ) + Γ 1 + Γ 2 ( 1 2 1 η ) α u k 2 + Γ 1 + Γ 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equaw_HTML.gif
where
Γ 1 = 1 η 0 T e g ( t , 0 ) f ( t , u k σ ( t ) ) u k σ ( t ) Δ t 0 T e g ( t , 0 ) F ( t , u k σ ( t ) ) Δ t , Γ 2 = 1 η j = 1 p e g ( t , 0 ) I j ( u k ( t j ) ) u k ( t j ) j = 1 p e g ( t , 0 ) 0 u k ( t j ) I j ( t ) d t . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equax_HTML.gif
Note that J = ϒ 1 k ϒ 2 k https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq173_HTML.gif, where ϒ 1 k = { t J : | u k ( t ) | < γ } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq174_HTML.gif, ϒ 2 k = { t J : | u k ( t ) | γ } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq175_HTML.gif, and that there exists a constant c such that
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equ15_HTML.gif
(4.7)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equ16_HTML.gif
(4.8)
So, by ( H 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq115_HTML.gif) and (4.7), we have
Γ 1 = 1 η ϒ 1 k e g ( t , 0 ) f ( t , u k σ ( t ) ) u k σ ( t ) Δ t + 1 η ϒ 2 k e g ( t , 0 ) f ( t , u k σ ( t ) ) u k σ ( t ) Δ t ϒ 1 k e g ( t , 0 ) F ( t , u k σ ( t ) ) Δ t ϒ 2 k e g ( t , 0 ) F ( t , u k σ ( t ) ) Δ t 1 η ϒ 1 k e g ( t , 0 ) | f ( t , u k σ ( t ) ) u k σ ( t ) | Δ t ϒ 1 k e g ( t , 0 ) | F ( t , u k σ ( t ) ) | Δ t + 1 η ϒ 2 k e g ( t , 0 ) f ( t , u k σ ( t ) ) u k σ ( t ) Δ t ϒ 2 k e g ( t , 0 ) F ( t , u k σ ( t ) ) Δ t c + c = c 1 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equay_HTML.gif

where c https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq176_HTML.gif, c https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq177_HTML.gif and c 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq178_HTML.gif are constants (independent of k).

Analogously, there exists a constant c 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq179_HTML.gif (independent of k) such that ϒ 2 k c 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq180_HTML.gif.

Hence,
ψ ( u k ) ( 1 2 1 η ) α u k 2 + 1 η ( ψ ( u k ) , u k ) + Γ 1 + Γ 2 ( 1 2 1 η ) α u k 2 1 η ψ ( u k ) u k + c 1 + c 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equaz_HTML.gif

Since ψ ( u k ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq181_HTML.gif is bounded, we have { u k } k = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq182_HTML.gif is a bounded sequence.

Hence, there exists a subsequence { u k } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq183_HTML.gif (for simplicity denoted again by { u k } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq183_HTML.gif) such that { u k } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq183_HTML.gif weakly converges to some u in H 0 , Δ 1 ( J ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq30_HTML.gif. Then the sequence { u k } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq183_HTML.gif converges uniformly to u in C ( J ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq184_HTML.gif.

By (4.6), we have
u k 2 = 0 T e g ( t , 0 ) ( f ( t , u k σ ) u k σ λ u k 2 ) Δ t + ( ψ ( u k ) , u k ) + j = 1 p e g ( t , 0 ) I j ( u k ( t j ) ) u k ( t j ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equba_HTML.gif
So, we have
lim k u k 2 = 0 T e g ( t , 0 ) ( f ( t , u σ ) u σ λ u 2 ) Δ t + j = 1 p e g ( t , 0 ) I j ( u ( t j ) ) u ( t j ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equbb_HTML.gif

Then u k https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq185_HTML.gif converges in H 0 , Δ 1 ( J ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq30_HTML.gif. Since H 0 , Δ 1 ( J ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq30_HTML.gif is a Hilbert space, and the sequence { u k } H 0 , Δ 1 ( J ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq186_HTML.gif satisfies u k u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq187_HTML.gif, then { u k } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq183_HTML.gif converges to u, i.e., u k u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq188_HTML.gif. ψ satisfies the Palais-Smale condition.

Now, by Theorem 4.2, there exists a critical point u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq114_HTML.gif. Therefore, u 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq112_HTML.gif and u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq114_HTML.gif are two critical points of ψ, and they are classical solutions of (NPE). Hence, u 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq112_HTML.gif and u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq114_HTML.gif are classical solutions of (NP). □

Example 4.1 Let T = h Z https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq189_HTML.gif for 0 < h < T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq190_HTML.gif, T h Z https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq191_HTML.gif and t 1 ( 0 , T ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq192_HTML.gif. Thus, J = [ 0 , T ] h Z https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq193_HTML.gif and J κ = [ 0 , T h ] h Z https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq194_HTML.gif. Consider the following boundary value problem:
{ u Δ Δ ( t ) + α u Δ ( t + h ) + λ u ( t + h ) = t u 5 ( t + h ) ; Δ -a.e.  t J κ 2 , ( u Δ ( t 1 + ) u Δ ( t 1 ) ) = u 5 ( t 1 ) , u ( 0 ) = 0 = u ( T ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_Equ17_HTML.gif
(4.9)

where α > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq74_HTML.gif is a constant.

We can see that g ( t ) = α https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq195_HTML.gif is regressive and continuous. If we take η = s = 6 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq196_HTML.gif and λ > λ 1 ( 1 + α h ) T h https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-119/MediaObjects/13661_2013_Article_365_IEq197_HTML.gif, by Theorem 4.3, Eq. (4.9) has at least two solutions.

Declarations

Acknowledgements

The authors are grateful to the referees for their valuable suggestions that led to the improvement of the original manuscript. The research of V Otero-Espinar has been partially supported by Ministerio de Educación y Ciencia (Spain) and FEDER, Project MTM2010-15314.

Authors’ Affiliations

(1)
Departamento de Análise Matemática, Universidade de Santiago de Compostela
(2)
Instituto de Ciencias Matemáticas (CSIC, UAM, UC3M, UCM)

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© Otero-Espinar and Pernas-Castaño; licensee Springer. 2013

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