Open Access

Existence of solutions for nonlocal p-Laplacian thermistor problems on time scales

Boundary Value Problems20132013:1

DOI: 10.1186/1687-2770-2013-1

Received: 1 September 2012

Accepted: 18 December 2012

Published: 4 January 2013

Abstract

In this paper, a nonlocal initial value problem to a p-Laplacian equation on time scales is studied. The existence of solutions for such a problem is obtained by using the topological degree method.

Keywords

existence p-Laplacian time scales topological degree

1 Introduction

In this paper, we are concerned with the existence of solutions of the following nonlocal p-Laplacian dynamic equation on a time scale T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq1_HTML.gif:
( ϕ p ( u ( t ) ) ) = λ a ( t ) f ( u ( t ) ) ( 0 T f ( u ( s ) ) s ) k , t ( 0 , T ) T , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_Equ1_HTML.gif
(1.1)
with integral initial value
u ( 0 ) = 0 T g ( s ) u ( s ) s , u ( 0 ) = A , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_Equ2_HTML.gif
(1.2)

where ϕ p ( ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq2_HTML.gif is the p-Laplace operator defined by ϕ p ( s ) = | s | p 2 s https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq3_HTML.gif, p > 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq4_HTML.gif, ϕ p 1 = ϕ q https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq5_HTML.gif with q the Hölder conjugate of p, i.e., 1 p + 1 q = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq6_HTML.gif, λ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq7_HTML.gif, k > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq8_HTML.gif, f : [ 0 , T ] T R + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq9_HTML.gif is continuous ( R + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq10_HTML.gif denotes positive real numbers), a : [ 0 , T ] T R + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq11_HTML.gif is left dense continuous, g ( s ) L 1 ( [ 0 , T ] T ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq12_HTML.gif and A is a real constant.

This model arises in ohmic heating phenomena, which occur in shear bands of metals which are deformed at high strain rates [1, 2], in the theory of gravitational equilibrium of polytropic stars [3], in the investigation of the fully turbulent behavior of real flows, using invariant measures for the Euler equation [4], in modeling aggregation of cells via interaction with a chemical substance (chemotaxis) [5]. For the one-dimensional case, problems with the nonlocal initial condition appear in the investigation of diffusion phenomena for a small amount of gas in a transparent tube [6, 7]; nonlocal initial value problems in higher dimension are important from the point of view of their practical applications to modeling and investigating of pollution processes in rivers and seas, which are caused by sew-age [8].

The study of dynamic equations on time scales has led to some important applications [911], and an amount of literature has been devoted to the study the existence of solutions of second-order nonlinear boundary value problems (e.g., see [1218]).

Motivated by the above works, in this paper, we study the existence of solutions to Problem (1.1), (1.2). Compared with the works mentioned above, this article has the following new features: firstly, the main technique used in this paper is the topological degree method; secondly, Problem (1.1), (1.2) involves the integral initial condition.

The paper is organized as follows. We introduce some necessary definitions and lemmas in the rest of this section. In Section 2, we provide some necessary preliminaries, and in Section 3, the main results are stated and proved.

Definition 1.1 For t < sup T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq13_HTML.gif and r > inf T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq14_HTML.gif, define the forward jump operator σ and the backward jump operator ρ, respectively,
σ ( t ) = inf { τ T τ > t } T , ρ ( r ) = sup { τ T τ < r } T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_Equa_HTML.gif

for all t , r T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq15_HTML.gif. If σ ( t ) > t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq16_HTML.gif, t is said to be right scattered, and if ρ ( r ) < r https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq17_HTML.gif, r is said to be left scattered. If σ ( t ) = t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq18_HTML.gif, t is said to be right dense, and if ρ ( r ) = r https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq19_HTML.gif, r is said to be left dense. If T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq1_HTML.gif has a right scattered minimum m, define https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq20_HTML.gif ; otherwise, set https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq21_HTML.gif . If T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq1_HTML.gif has a left scattered maximum M, define https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq22_HTML.gif ; otherwise, set https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq23_HTML.gif .

Definition 1.2 For x : T R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq24_HTML.gif and https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq25_HTML.gif , we define the delta derivative of x ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq26_HTML.gif, x ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq27_HTML.gif, to be the number (when it exists) with the property that for any ε > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq28_HTML.gif, there is a neighborhood U of t such that
| [ x ( σ ( t ) ) x ( s ) ] x ( t ) [ σ ( t ) s ] | < ε | σ ( t ) s | https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_Equb_HTML.gif
for all s U https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq29_HTML.gif. For x : T R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq24_HTML.gif and https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq30_HTML.gif , we define the nabla derivative of x ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq26_HTML.gif, x ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq31_HTML.gif, to be the number (when it exists) with the property that for any ε > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq28_HTML.gif, there is a neighborhood V of t such that
| [ x ( ρ ( t ) ) x ( s ) ] x ( t ) [ ρ ( t ) s ] | < ε | ρ ( t ) s | https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_Equc_HTML.gif

for all s V https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq32_HTML.gif.

Definition 1.3 If F ( t ) = f ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq33_HTML.gif, then we define the delta integral by
a t f ( s ) s = F ( t ) F ( a ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_Equd_HTML.gif
If Φ ( t ) = f ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq34_HTML.gif, then we define the nabla integral by
a t f ( s ) s = Φ ( t ) Φ ( a ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_Eque_HTML.gif

Throughout this paper, we assume that T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq1_HTML.gif is a nonempty closed subset of with 0 T k https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq35_HTML.gif, T T k https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq36_HTML.gif.

Lemma 1.1 (Alternative theorem)

Suppose that X is a Banach space and A is a completely continuous operator from X to X. Then for any λ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq37_HTML.gif, only one of the following statements holds:
  1. (i)
    For any y X https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq38_HTML.gif, there exists a unique x X https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq39_HTML.gif, such that
    ( A λ I ) x = y ; https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_Equf_HTML.gif
     
  2. (ii)
    There exists an x X https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq39_HTML.gif, x 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq40_HTML.gif, such that
    ( A λ I ) x = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_Equg_HTML.gif
     

2 Preliminaries

Let E = C l d ( [ 0 , T ] T , R ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq41_HTML.gif be a Banach space equipped with the maximum norm u = max lim [ 0 , T ] T | u ( t ) | https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq42_HTML.gif.

Consider the following problem:
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_Equ3_HTML.gif
(2.1)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_Equ4_HTML.gif
(2.2)

where y C ( [ 0 , T ] T ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq43_HTML.gif, 0 T g ( s ) s 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq44_HTML.gif.

Integrating Eq. (2.1) from 0 to t, one obtains
ϕ p ( x ( t ) ) ϕ p ( x ( 0 ) ) = 0 t y ( s ) s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_Equh_HTML.gif
Using the initial condition (2.2), we have
x ( t ) = ϕ p 1 ( ϕ p ( A ) 0 t y ( s ) s ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_Equi_HTML.gif
Integrating the above equality from 0 to t again, we obtain
x ( t ) 0 T g ( s ) x ( s ) s = 0 t ϕ p 1 ( ϕ p ( A ) 0 τ y ( s ) s ) τ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_Equ5_HTML.gif
(2.3)

Let F ( t ) : = 0 t ϕ p 1 ( ϕ p ( A ) 0 τ y ( s ) s ) τ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq45_HTML.gif.

Define an operator K : C l d ( [ 0 , T ] T ) C l d ( [ 0 , T ] T ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq46_HTML.gif by
( K x ) = 0 T g ( s ) x ( s ) s , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_Equj_HTML.gif
then (2.3) can be rewritten as
( I K ) x ( t ) = F ( t ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_Equ6_HTML.gif
(2.4)

Thus, x ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq26_HTML.gif is a solution to (2.1), (2.2) if and only if it is a solution to (2.4).

Lemma 2.1 I K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq47_HTML.gif is a Fredholm operator.

Proof To prove that I K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq47_HTML.gif is a Fredholm operator, we need only to show that K is completely continuous.

It is easy to see from the definition of K that K is a bounded linear operator from C l d ( [ 0 , T ] T ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq48_HTML.gif to C l d ( [ 0 , T ] T ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq48_HTML.gif. Obviously, dim R ( K ) = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq49_HTML.gif. So, K is a completely continuous operator. This completes the proof. □

Lemma 2.2 Problem (2.1), (2.2) admits a unique solution.

Proof Since Problem (2.1), (2.2) is equivalent to Problem (2.4), we need only to show that Problem (2.4) has a unique solution.

Using Lemma 2.1 and the alternative theorem, it is sufficient to prove that
( I K ) x ( t ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_Equ7_HTML.gif
(2.5)

has a trivial solution x 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq50_HTML.gif only.

On the contrary, suppose (2.5) has a nontrivial solution μ, then μ is a constant, and we have
I μ = K μ = μ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_Equk_HTML.gif
The definition of K and the above equality yield
[ 1 0 T g ( s ) s ] μ = 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_Equl_HTML.gif

which is a contradiction to the assumptions 0 T g ( s ) s 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq51_HTML.gif and μ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq52_HTML.gif.

Thus, we complete the proof. □

3 Main results

Throughout this section, we assume that the following conditions hold.

(H1) 0 T | g ( s ) | s = M < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq53_HTML.gif;

(H2) f : [ 0 , T ] T R + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq9_HTML.gif is continuous;

(H3) a : [ 0 , T ] T R + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq11_HTML.gif is left dense continuous and max t [ 0 , T ] T a ( t ) M 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq54_HTML.gif;

(H4) f ( y ) [ c 1 ϕ p ( | y | ) + c 2 ] 1 1 k https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq55_HTML.gif, c 1 , c 2 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq56_HTML.gif and c 1 < ϕ p ( 1 M 2 q 1 T ) λ M 1 T 1 k https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq57_HTML.gif, when k < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq58_HTML.gif;

(H5) f ( y ) [ c 3 ϕ p ( | y | ) ] 1 1 k https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq59_HTML.gif, c 3 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq60_HTML.gif and c 3 < ϕ p ( 1 M 2 q 1 T ) λ M 1 T 1 k https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq61_HTML.gif, when k > 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq62_HTML.gif.

From Lemma 2.2 we know that u ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq63_HTML.gif is a solution to Problem (1.1), (1.2) if and only if it is a solution to the following integral equation:
( I K ) u ( t ) = 0 t ϕ p 1 ( ϕ p ( A ) 0 τ λ a ( s ) f ( u ( s ) ) ( 0 T f ( u ( s ) ) s ) k s ) τ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_Equ8_HTML.gif
(3.1)
Define an operator F : C l d ( [ 0 , T ] T ) C l d ( [ 0 , T ] T ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq64_HTML.gif by
( F u ) ( t ) = 0 t ϕ p 1 ( ϕ p ( A ) 0 τ λ a ( s ) f ( u ( s ) ) ( 0 T f ( u ( s ) ) s ) k s ) τ , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_Equm_HTML.gif
then (3.1) can be rewritten as
( I K ) u ( t ) = ( F u ) ( t ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_Equn_HTML.gif

In order to prove the existence of solutions to (3.1), we need the following lemmas.

Lemma 3.1 F is completely continuous.

Proof Let R 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq65_HTML.gif be an arbitrary positive real number and denote B 1 = { u C l d ( [ 0 , T ] T ) ; u R 1 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq66_HTML.gif. Then we have for any u B 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq67_HTML.gif,
| ( F u ) ( t ) | 0 t | ϕ p 1 ( ϕ p ( A ) 0 τ λ a ( s ) f ( u ( s ) ) ( 0 T f ( u ( s ) ) s ) k s ) | τ 0 T ϕ p 1 ( | ϕ p ( A ) | + | 0 T λ a ( s ) sup u B 1 f ( T inf u B 1 f ) k s | ) τ ϕ p 1 ( | ϕ p ( A ) | + M 1 T λ sup u B 1 f ( T inf u B 1 f ) k ) T . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_Equo_HTML.gif

This shows that F ( B 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq68_HTML.gif is uniformly bounded.

Moreover, for any t [ 0 , T ] T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq69_HTML.gif, we have
| ( F u ) ( t ) | = | ϕ p 1 ( ϕ p ( A ) 0 t λ a ( s ) f ( u ( s ) ) ( 0 T f ( u ( s ) ) s ) k s ) | ϕ p 1 ( | ϕ p ( A ) | + M 1 T λ sup u B 1 f ( T inf u B 1 f ) k ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_Equp_HTML.gif

Thus, it is easy to prove that F ( B 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq68_HTML.gif is equicontinuous. This together with the Ascoli-Arzelà theorem guarantees that F ( B 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq68_HTML.gif is relatively compact in C l d ( [ 0 , T ] T ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq48_HTML.gif.

Therefore, F is completely continuous. The proof of Lemma 3.1 is completed. □

Theorem 3.1 Assume that conditions (H1)-(H5) hold. Then Problem (1.1), (1.2) has at least one solution.

Proof Lemma 2.1 and Lemma 3.1 imply that the operator K + F https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq70_HTML.gif is completely continuous. It suffices for us to prove that the equation
( I ( K + F ) ) u = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_Equ9_HTML.gif
(3.2)

has at least one solution.

Define H : [ 0 , 1 ] × C l d ( [ 0 , T ] T ) C l d ( [ 0 , T ] T ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq71_HTML.gif as
H ( σ , u ) = ( K + σ F ) u , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_Equq_HTML.gif

and it is clear that H is completely continuous.

Set h σ ( u ) = u H ( σ , u ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq72_HTML.gif, then we have
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_Equr_HTML.gif

To apply the Leray-Schauder degree to h σ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq73_HTML.gif, we need only to show that there exists a ball B R ( θ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq74_HTML.gif in C l d ( [ 0 , T ] T ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq75_HTML.gif, whose radius R will be fixed later, such that θ h σ ( B R ( θ ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq76_HTML.gif.

If k < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq58_HTML.gif, choosing R > 2 q 1 ϕ p 1 ( | ϕ p ( A ) | + λ M 1 T 1 k c 2 ) T 1 M 2 q 1 ϕ p 1 ( λ M 1 T 1 k c 1 ) T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq77_HTML.gif, then for any fixed u B R ( θ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq78_HTML.gif, there exists a t 0 [ 0 , T ] T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq79_HTML.gif such that | u ( t 0 ) | = R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq80_HTML.gif. By direct calculation, we have
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_Equ10_HTML.gif
(3.3)
From (H4), we have
| ( h σ u ) ( t 0 ) | ( 1 M ) R 0 T ϕ p 1 ( | ϕ p ( A ) | + λ M 1 ( 0 T f ( u ( s ) ) s ) 1 k ) τ ( 1 M ) R 0 T ϕ p 1 [ | ϕ p ( A ) | + λ M 1 T 1 k ( c 1 ϕ p ( u ) + c 2 ) ] τ > 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_Equ11_HTML.gif
(3.4)
If k > 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq62_HTML.gif, choosing R > 2 q 1 | A | T 1 M 2 q 1 ϕ p 1 ( λ M 1 T 1 k c 3 ) T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq81_HTML.gif, then for any fixed u B R ( θ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq78_HTML.gif, there exists a t 0 [ 0 , T ] T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq79_HTML.gif such that | u ( t 0 ) | = R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq80_HTML.gif. From (H5), we have
| ( h σ u ) ( t 0 ) | ( 1 M ) R 0 T ϕ p 1 ( | ϕ p ( A ) | + λ M 1 ( 0 T f ( u ( s ) ) s ) k 1 ) τ ( 1 M ) R 0 T ϕ p 1 [ | ϕ p ( A ) | + λ M 1 T 1 k c 3 ϕ p ( u ) ] τ > 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_Equ12_HTML.gif
(3.5)
If k = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq82_HTML.gif, choosing R > ϕ p 1 ( | ϕ p ( A ) | + λ M 1 ) T 1 M https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq83_HTML.gif, then for any fixed u B R ( θ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq78_HTML.gif, there exists a t 0 [ 0 , T ] T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq79_HTML.gif such that | u ( t 0 ) | = R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq80_HTML.gif. By direct calculation, we have
| ( h σ u ) ( t 0 ) | ( 1 M ) R 0 T ϕ p 1 ( | ϕ p ( A ) | + λ M 1 ) τ > 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_Equ13_HTML.gif
(3.6)

This implies h σ u θ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq84_HTML.gif and hence we obtain θ h σ ( B R ( θ ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq76_HTML.gif.

Since deg ( h 1 , B R ( θ ) , θ ) = deg ( h 0 , B R ( θ ) , θ ) = ± 1 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq85_HTML.gif, we know that (3.2) admits a solution u B R ( θ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq86_HTML.gif, which implies that (1.1), (1.2) also admits a solution in B R ( θ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-1/MediaObjects/13661_2012_Article_255_IEq74_HTML.gif. □

Declarations

Acknowledgements

This work was supported by NSFC (11271154) and by Key Lab of Symbolic Computation and Knowledge Engineering of Ministry of Education and by the 985 program of Jilin University, and the first author is also supported by the Youth Studies Program of Jilin University of Finance and Economics (XJ2012006).

Authors’ Affiliations

(1)
Institute of Applied Mathematics, Jilin University of Finance and Economics
(2)
Institute of Mathematics, Jilin University

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© Song and Gao; licensee Springer. 2013

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