Open Access

Positive solutions of a fractional thermostat model

Boundary Value Problems20132013:5

DOI: 10.1186/1687-2770-2013-5

Received: 17 October 2012

Accepted: 29 December 2012

Published: 16 January 2013

Abstract

We study the existence of positive solutions of a nonlinear fractional heat equation with nonlocal boundary conditions depending on a positive parameter. Our results extend the second-order thermostat model to the non-integer case. We base our analysis on the known Guo-Krasnosel’skii fixed point theorem on cones.

1 Introduction

Fractional calculus has been studied for centuries mainly as a pure theoretical mathematical discipline, but recently, there has been a lot of interest in its practical applications. In current research, fractional differential equations have arisen in mathematical models of systems and processes in various fields such as aerodynamics, acoustics, mechanics, electromagnetism, signal processing, control theory, robotics, population dynamics, finance, etc. [13]. For some recent results in fractional differential equations, see [412] and the references therein.

Infante and Webb [13] studied the nonlocal boundary value problem
u = f ( t , u ) , t ( 0 , 1 ) , u ( 0 ) = 0 , β u ( 1 ) + u ( η ) = 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_Equa_HTML.gif
which models a thermostat insulated at t = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq1_HTML.gif with the controller at t = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq2_HTML.gif adding or discharging heat depending on the temperature detected by the sensor at t = η https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq3_HTML.gif. Using fixed point index theory and some results on their work on Hammerstein integral equations [14, 15], they obtained results on the existence of positive solutions of the boundary value problem. In particular, they have shown that if β 1 η https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq4_HTML.gif, then positive solutions exist under suitable conditions on f. This type of boundary value problem was earlier investigated by Guidotti and Merino [16] for the linear case with η = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq5_HTML.gif where they have shown a loss of positivity as β decreases. In the present paper, we consider the following fractional analog of the thermostat model:
C D α u ( t ) = f ( t , u ( t ) ) , t [ 0 , 1 ] , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_Equ1_HTML.gif
(1)
where 1 < α 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq6_HTML.gif, D α C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq7_HTML.gif denotes the Caputo fractional derivative of order α and f C ( [ 0 , 1 ] × [ 0 , ) , [ 0 , ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq8_HTML.gif subject to the boundary conditions
u ( 0 ) = 0 , β C D α 1 u ( 1 ) + u ( η ) = 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_Equ2_HTML.gif
(2)

where β > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq9_HTML.gif, 0 η 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq10_HTML.gif are given constants.

We point out that for α = 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq11_HTML.gif, we recover the second-order problem of [13]. We use the properties of the corresponding Green’s function and the Guo-Krasnosel’skii fixed point theorem to show the existence of positive solutions of (1)-(2) under the condition that the nonlinearity f is either sublinear or superlinear.

2 Preliminaries

Here we present some necessary basic knowledge and definitions for fractional calculus theory that can be found in the literature [13].

Definition 2.1 The Riemann-Liouville fractional integral of order α > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq12_HTML.gif of a function g : ( 0 , ) R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq13_HTML.gif is given by
I α g ( t ) = 1 Γ ( α ) 0 t ( t s ) α 1 g ( s ) d s https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_Equb_HTML.gif

provided the integral exists.

Definition 2.2 The Riemann-Liouville fractional derivative of order α > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq12_HTML.gif of a function g : ( 0 , ) R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq13_HTML.gif is given by
D 0 + α g ( t ) = 1 Γ ( n α ) ( d d t ) n 0 t g ( s ) ( t s ) α n + 1 d s ( n 1 < α < n , n = [ α ] + 1 ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_Equc_HTML.gif

where [ α ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq14_HTML.gif denotes the integer part of the real number α.

Definition 2.3 The Caputo derivative of order α > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq12_HTML.gif of a function g A C n 1 [ 0 , ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq15_HTML.gif is given by
D α C g ( t ) = 1 Γ ( n α ) 0 t ( t s ) n α 1 g ( n ) ( s ) d s ( n 1 < α < n , n = [ α ] + 1 ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_Equd_HTML.gif

where [ α ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq14_HTML.gif denotes the integer part of the real number α.

Lemma 2.1 Let g L 1 ( 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq16_HTML.gif and α , β > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq17_HTML.gif.
  1. (i)

    If α = n N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq18_HTML.gif, then I n g ( t ) = 1 ( n 1 ) ! 0 t ( t s ) n 1 g ( s ) d s https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq19_HTML.gif.

     
  2. (ii)

    If α = n N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq18_HTML.gif, then D n C g ( t ) = g ( n ) ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq20_HTML.gif.

     
  3. (iii)

    D α C I α g ( t ) = g ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq21_HTML.gif.

     
  4. (iv)

    I α I β g ( t ) = I α + β g ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq22_HTML.gif.

     
Remark 2.1 In addition to the above properties, the Caputo derivative of a power function t k https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq23_HTML.gif, k N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq24_HTML.gif, is given by
D α C t k = { Γ ( k + 1 ) Γ ( k α + 1 ) t k α , for  k > n 1 , 0 , for  k n 1 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_Eque_HTML.gif

where n 1 < α < n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq25_HTML.gif, n = [ α ] + 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq26_HTML.gif.

Lemma 2.2 For α > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq12_HTML.gif, the general solution of the fractional differential equation D α C u ( t ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq27_HTML.gif is given by
u ( t ) = c 0 + c 1 t + c 2 t 2 + + c n 1 t n 1 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_Equf_HTML.gif

where c i R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq28_HTML.gif, i = 0 , 1 , 2 , , n 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq29_HTML.gif ( n 1 < α < n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq25_HTML.gif, n = [ α ] + 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq26_HTML.gif).

Lemma 2.3
I α C D α u ( t ) = u ( t ) + c 0 + c 1 t + c 2 t 2 + + c n 1 t n 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_Equ3_HTML.gif
(3)

for some c i R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq28_HTML.gif, i = 0 , 1 , 2 , , n 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq29_HTML.gif ( n 1 < α < n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq30_HTML.gif, n = [ α ] + 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq26_HTML.gif).

We start by solving an auxiliary problem to get an expression for the Green’s function of boundary value problem (1)-(2).

Lemma 2.4 Suppose f C [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq31_HTML.gif. A function u C [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq32_HTML.gif is a solution of the boundary value problem
C D α u ( t ) = f ( t ) , u ( 0 ) = 0 , β C D α 1 u ( 1 ) + u ( η ) = 0 , t [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_Equg_HTML.gif
if and only if it satisfies the integral equation
u ( t ) = 0 1 G ( t , s ) f ( s ) d s , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_Equh_HTML.gif
where G ( t , s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq33_HTML.gif is the Green’s function (depending on α) given by
G ( t , s ) = β + H η ( s ) H t ( s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_Equ4_HTML.gif
(4)

and for r [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq34_HTML.gif, H r : [ 0 , 1 ] R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq35_HTML.gif is defined as H r ( s ) = ( r s ) α 1 Γ ( α ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq36_HTML.gif for s r https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq37_HTML.gif and H r ( s ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq38_HTML.gif for s > r https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq39_HTML.gif.

Proof Using (3) we have, for some constants c 0 , c 1 R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq40_HTML.gif,
u ( t ) = I α f ( t ) + c 0 + c 1 t = 0 t ( t s ) α 1 Γ ( α ) f ( s ) d s + c 0 + c 1 t . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_Equ5_HTML.gif
(5)
In view of Lemma 2.1, we obtain
u ( t ) = 0 t ( t s ) α 2 Γ ( α 1 ) f ( s ) d s + c 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_Equi_HTML.gif

Since u ( 0 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq41_HTML.gif, we find that c 1 = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq42_HTML.gif.

It also follows that
D α 1 C u ( t ) = I 1 u ( t ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_Equj_HTML.gif
Using the boundary condition β C D α 1 u ( 1 ) + u ( η ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq43_HTML.gif, we get
c 0 = β 0 1 f ( s ) d s + 0 η ( η s ) α 1 Γ ( α ) f ( s ) d s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_Equk_HTML.gif
Finally, substituting the values of c 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq44_HTML.gif and c 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq45_HTML.gif in (5), we have
u ( t ) = β 0 1 f ( s ) d s + 0 η ( η s ) α 1 Γ ( α ) f ( s ) d s 0 t ( t s ) α 1 Γ ( α ) f ( s ) d s = 0 1 G ( t , s ) f ( s ) d s , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_Equl_HTML.gif

where G ( t , s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq33_HTML.gif is given by (4). This completes the proof. □

Remark 2.2 We observe that H r https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq46_HTML.gif is continuous on [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq47_HTML.gif for any r [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq48_HTML.gif. Thus, G ( t , s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq33_HTML.gif given by (4) is continuous on [ 0 , 1 ] × [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq49_HTML.gif.

Remark 2.3 By taking α = 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq11_HTML.gif, we get
u ( t ) = β 0 1 f ( s ) d s + 0 1 ( η s ) f ( s ) d s 0 t ( t s ) f ( s ) d s = 0 1 G ( t , s ) f ( s ) d s https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_Equm_HTML.gif
and G ( t , s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq33_HTML.gif in this case coincides with the one obtained in [13] for the boundary value problem
u ( t ) = f ( t ) , u ( 0 ) = 0 , β u ( 1 ) + u ( η ) = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_Equn_HTML.gif
Remark 2.4 We observe that for each fixed point s [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq50_HTML.gif, G t = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq51_HTML.gif for t s https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq52_HTML.gif and G t < 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq53_HTML.gif for t > s https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq54_HTML.gif and deduce that G ( t , s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq33_HTML.gif is a decreasing function of t. It then follows that
max t [ 0 , 1 ] G ( t , s ) = G ( 0 , s ) = { β , s > η , β Γ ( α ) + ( η s ) α 1 Γ ( α ) , s η , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_Equo_HTML.gif
and
min t [ 0 , 1 ] G ( t , s ) = G ( 1 , s ) = { β Γ ( α ) ( 1 s ) α 1 Γ ( α ) , s > η , β Γ ( α ) + ( η s ) α 1 ( 1 s ) α 1 Γ ( α ) , s η . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_Equp_HTML.gif
Consequently, by looking at the behavior of G ( t , s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq33_HTML.gif with respect to s, we get
min t , s [ 0 , 1 ] G ( t , s ) = β Γ ( α ) ( 1 η ) α 1 Γ ( α ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_Equq_HTML.gif
and
max t , s [ 0 , 1 ] G ( t , s ) = β Γ ( α ) + η α 1 Γ ( α ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_Equr_HTML.gif
To establish the existence of positive solutions of problem (1)-(2), we will show that G ( t , s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq33_HTML.gif satisfies the following property introduced by Lan and Webb in [17]:
  1. (A)
    There exist a measurable function ϕ : [ 0 , 1 ] [ 0 , ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq55_HTML.gif, a subinterval [ a , b ] [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq56_HTML.gif and a constant λ [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq57_HTML.gif such that
    | G ( t , s ) | ϕ ( s ) t , s [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_Equs_HTML.gif
     
and
G ( t , s ) λ ϕ ( s ) t [ a , b ] , s [ 0 , 1 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_Equt_HTML.gif

Lemma 2.5 If β Γ ( α ) > ( 1 η ) α 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq58_HTML.gif, then G ( t , s ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq59_HTML.gif for all t , s [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq60_HTML.gif, and G ( t , s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq33_HTML.gif satisfies property (A).

Proof If β Γ ( α ) > ( 1 η ) α 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq58_HTML.gif, then G ( t , s ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq59_HTML.gif for all t , s [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq61_HTML.gif. We choose [ a , b ] = [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq62_HTML.gif, and we have
| G ( t , s ) | = G ( t , s ) β Γ ( α ) + η α 1 Γ ( α ) : = ϕ ( s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_Equu_HTML.gif
and
G ( t , s ) λ ϕ ( s ) s , t [ 0 , 1 ] , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_Equv_HTML.gif
where
λ = β Γ ( α ) ( 1 η ) α 1 β Γ ( α ) + η α 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_Equ6_HTML.gif
(6)

 □

Lemma 2.6 If β Γ ( α ) = ( 1 η ) α 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq63_HTML.gif, then G ( t , s ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq64_HTML.gif for all t , s [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq60_HTML.gif, and G ( t , s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq33_HTML.gif satisfies property (A).

Proof We choose [ a , b ] = [ 0 , b ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq65_HTML.gif with η b < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq66_HTML.gif. Following the arguments in the previous lemma, we have
| G ( t , s ) | β Γ ( α ) + η α 1 Γ ( α ) : = ϕ ( s ) t , s [ 0 , 1 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_Equw_HTML.gif
Also, by taking
λ = β Γ ( α ) ( b η ) α 1 β Γ ( α ) + η α 1 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_Equx_HTML.gif
we obtain
G ( t , s ) λ ϕ ( s ) t [ 0 , b ] , s [ 0 , 1 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_Equy_HTML.gif

 □

Lemma 2.7 If β Γ ( α ) < ( 1 η ) α 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq67_HTML.gif, then G ( t , s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq33_HTML.gif changes sign on [ 0 , 1 ] × [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq68_HTML.gif, and G ( t , s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq33_HTML.gif satisfies property (A).

Proof We choose [ a , b ] = [ 0 , b ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq65_HTML.gif with η b < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq69_HTML.gif such that β Γ ( α ) > ( b η ) α 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq70_HTML.gif. We have
| G ( t , s ) | max { β Γ ( α ) + η α 1 Γ ( α ) , ( 1 η ) α 1 β Γ ( α ) Γ ( α ) } : = ϕ ( s ) t , s [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_Equz_HTML.gif
and
G ( t , s ) λ ϕ ( s ) t [ 0 , b ] , s [ 0 , 1 ] , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_Equaa_HTML.gif
where
λ = min { β Γ ( α ) ( b η ) α 1 β Γ ( α ) + η α 1 , β Γ ( α ) ( b η ) α 1 ( 1 η ) α 1 β Γ ( α ) } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_Equab_HTML.gif

For the main results, we use the known Guo-Krasnosel’skii fixed point theorem [18]. □

Theorem 2.1 Let E be a Banach space and let P E https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq71_HTML.gif be a cone. Assume Ω 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq72_HTML.gif, Ω 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq73_HTML.gif are open bounded subsets of E such that 0 Ω 1 Ω ¯ 1 Ω 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq74_HTML.gif, and let T : P ( Ω ¯ 2 Ω 1 ) P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq75_HTML.gif be a completely continuous operator such that
  1. (i)

    T u u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq76_HTML.gif, u P Ω 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq77_HTML.gif and T u u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq78_HTML.gif, u P Ω 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq79_HTML.gif; or

     
  2. (ii)

    T u u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq78_HTML.gif, u P Ω 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq77_HTML.gif and T u u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq76_HTML.gif, u P Ω 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq79_HTML.gif.

     

Then the operator P has a fixed point in P ( Ω ¯ 2 Ω 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq80_HTML.gif.

3 Main results

We set
f 0 = lim u 0 + min t [ 0 , 1 ] f ( t , u ) u , f 0 = lim u 0 + max t [ 0 , 1 ] f ( t , u ) u , f = lim u max t [ 0 , 1 ] f ( t , u ) u , f = lim u min t [ 0 , 1 ] f ( t , u ) u . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_Equac_HTML.gif

We now state the main result of this paper.

Theorem 3.1 Let f ( s , u ( s ) ) C ( [ 0 , 1 ] × [ 0 , ) , [ 0 , ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq81_HTML.gif. Assume that one of the following conditions is satisfied:
  1. (i)

    (Sublinear case) f 0 = https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq82_HTML.gif and f = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq83_HTML.gif.

     
  2. (ii)

    (Superlinear case) f 0 = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq84_HTML.gif and f = https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq85_HTML.gif.

     

If β Γ ( α ) > ( 1 η ) α 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq58_HTML.gif, then problem (1)-(2) admits at least one positive solution.

Theorem 3.2 Let f ( s , u ( s ) ) C ( [ 0 , 1 ] × [ , + ) , [ 0 , ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq86_HTML.gif. Assume that one of the following conditions is satisfied:
  1. (i)

    (Sublinear case) f 0 = https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq82_HTML.gif and f = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq83_HTML.gif.

     
  2. (ii)

    (Superlinear case) f 0 = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq84_HTML.gif and f = https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq85_HTML.gif.

     

If β Γ ( α ) ( 1 η ) α 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq87_HTML.gif, then problem (1)-(2) admits a solution which is positive on an interval [ 0 , b ] [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq88_HTML.gif.

Proof of Theorem 3.1 Let C [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq89_HTML.gif be the Banach space of all continuous real-valued functions on [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq47_HTML.gif endowed with the usual supremum norm https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq90_HTML.gif.

We define the operator T : C [ 0 , 1 ] C [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq91_HTML.gif as
T u ( t ) = 0 1 G ( t , s ) f ( s , u ( s ) ) d s , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_Equad_HTML.gif

where G ( t , s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq33_HTML.gif is defined by (4).

It is clear from Lemma 2.4 that the fixed points of the operator T coincide with the solutions of problem (1)-(2).

We now define the cone
P = { u | u C [ 0 , 1 ] , u ( t ) 0 , min t [ 0 , 1 ] u ( t ) λ u } , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_Equae_HTML.gif

where λ is given by (6).

First, we show that T ( P ) P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq92_HTML.gif.

It follows from the continuity and the non-negativity of the functions G and f on their domains of definition that if u P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq93_HTML.gif, then T u C [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq94_HTML.gif and T u ( t ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq95_HTML.gif for all t [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq96_HTML.gif.

For a fixed u P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq97_HTML.gif and for all t [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq96_HTML.gif, the fact that G ( t , s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq33_HTML.gif satisfies property (A) leads to the following inequalities:
T u ( t ) = 0 1 G ( t , s ) f ( s , u ( s ) ) d s λ 0 1 ϕ ( s ) f ( s , u ( s ) ) d s λ 0 1 max t [ 0 , 1 ] G ( t , s ) f ( s , u ( s ) ) d s λ max t [ 0 , 1 ] 0 1 G ( t , s ) f ( s , u ( s ) ) d s = λ T u . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_Equaf_HTML.gif

Hence, T ( P ) P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq98_HTML.gif.

We now show that T : P P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq99_HTML.gif is completely continuous.

In view of the continuity of the functions G and f, the operator T : P P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq100_HTML.gif is continuous.

Let Ω P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq101_HTML.gif be bounded, that is, there exists a positive constant M > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq102_HTML.gif such that u M https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq103_HTML.gif for all u Ω https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq104_HTML.gif. Define
L = max 0 t 1 , 0 u M | f ( t , u ) | + 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_Equag_HTML.gif
Then for all u Ω https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq105_HTML.gif, we have
| T u ( t ) | 0 1 G ( t , s ) f ( s , u ( s ) ) d s L 0 1 G ( t , s ) d s https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_Equah_HTML.gif

for all t [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq96_HTML.gif. That is, the set T ( Ω ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq106_HTML.gif is bounded.

For each u Ω https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq104_HTML.gif and t 1 , t 2 [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq107_HTML.gif such that t 1 < t 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq108_HTML.gif, we have
| T u ( t 2 ) T u ( t 1 ) | = | 0 t 2 ( t 2 s ) α 1 Γ ( α ) f ( s , u ( s ) ) d s + 0 t 1 ( t 1 s ) α 1 Γ ( α ) f ( s , u ( s ) ) d s | 1 Γ ( α ) 0 t 1 ( ( t 2 s ) α 1 ( t 1 s ) α 1 ) | f ( s , u ( s ) ) | d s + 1 Γ ( α ) t 1 t 2 ( t 2 s ) α 1 | f ( s , u ( s ) ) | d s L Γ ( α ) ( 0 t 1 ( ( t 2 s ) α 1 ( t 1 s ) α 1 ) d s + t 1 t 2 ( t 2 s ) α 1 d s ) = L α Γ ( α ) ( ( t 2 t 1 ) α + t 2 α t 1 α + ( t 2 t 1 ) α ) = L Γ ( α + 1 ) ( t 2 α t 1 α ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_Equai_HTML.gif

Clearly, the right-hand side of the above inequalities tends to 0 as t 1 t 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq109_HTML.gif and therefore the set T ( Ω ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq106_HTML.gif is equicontinuous. It follows from the Arzela-Ascoli theorem that the operator T : P P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq100_HTML.gif is completely continuous.

We now consider the two cases.
  1. (i)

    Sublinear case ( f 0 = https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq82_HTML.gif and f = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq83_HTML.gif).

     
Since f 0 = https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq82_HTML.gif, there exists ρ 1 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq110_HTML.gif such that f ( t , u ) δ 1 u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq111_HTML.gif for all 0 < u ρ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq112_HTML.gif, where δ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq113_HTML.gif satisfies
δ 1 ( β Γ ( α ) ( 1 η ) α 1 Γ ( α ) ) 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_Equ7_HTML.gif
(7)
We take u P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq93_HTML.gif such that u = ρ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq114_HTML.gif, then we have the following inequalities:
T u = 0 1 G ( t , s ) f ( s , u ( s ) ) d s δ 1 0 1 G ( t , s ) u ( s ) d s δ 1 u ( β Γ ( α ) ( 1 η ) α 1 Γ ( α ) ) u . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_Equaj_HTML.gif

Let Ω 1 = { u C [ 0 , 1 ] | u < ρ 1 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq115_HTML.gif. Hence, we have T u u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq76_HTML.gif, u P Ω 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq116_HTML.gif.

Since f ( t , ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq117_HTML.gif is a continuous function on [ 0 , ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq118_HTML.gif, we can define the function:
f ˜ ( t , u ) = max z [ 0 , u ] { f ( t , z ) } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_Equak_HTML.gif
It is clear that f ˜ ( t , u ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq119_HTML.gif is non-decreasing on ( 0 , ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq120_HTML.gif and since f = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq83_HTML.gif, we have (see [19])
lim u { max t [ 0 , 1 ] f ˜ ( t , u ) u } = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_Equal_HTML.gif
Therefore, there exists ρ 2 > ρ 1 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq121_HTML.gif such that f ˜ ( t , u ) δ 2 u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq122_HTML.gif for all u ρ 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq123_HTML.gif, where δ 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq124_HTML.gif satisfies
δ 2 ( β Γ ( α ) + η α 1 Γ ( α ) ) 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_Equ8_HTML.gif
(8)
Define Ω 2 = { u C [ 0 , 1 ] | u < ρ 2 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq125_HTML.gif and let u P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq97_HTML.gif such that u = ρ 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq126_HTML.gif. Then
T u = 0 1 G ( t , s ) f ( s , u ( s ) ) d s 0 1 G ( t , s ) f ˜ ( s , u ) d s δ 2 u ( β Γ ( α ) + η α 1 Γ ( α ) ) u . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_Equam_HTML.gif

Hence, we have T u u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq78_HTML.gif, u P Ω 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq127_HTML.gif.

Thus, by the first part of the Guo-Krasnosel’skii fixed point theorem, we conclude that (1)-(2) has at least one positive solution.
  1. (ii)

    Superlinear case ( f 0 = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq84_HTML.gif and f = https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq85_HTML.gif).

     

Let δ 2 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq128_HTML.gif be given as in (8).

Since f 0 = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq84_HTML.gif, there exists a constant r 1 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq129_HTML.gif such that f ( t , u ) δ 2 u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq130_HTML.gif for 0 u r 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq131_HTML.gif. Take u P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq93_HTML.gif such that u = r 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq132_HTML.gif. Then we have
T u = 0 1 G ( t , s ) f ( s , u ( s ) ) d s δ 2 0 1 G ( t , s ) u ( s ) d s δ 2 u ( β Γ ( α ) + η α 1 Γ ( α ) ) u . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_Equan_HTML.gif

If we let Ω 1 = { u C [ 0 , 1 ] | u < r 1 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq133_HTML.gif, we see that T u u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq134_HTML.gif for u P Ω 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq135_HTML.gif.

Now, since f = https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq85_HTML.gif, there exists r > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq136_HTML.gif such that f ( t , u ) δ 1 u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq111_HTML.gif for all u r https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq137_HTML.gif, where δ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq113_HTML.gif is as in (7).

Define Ω 2 = { u C [ 0 , 1 ] | u < r 2 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq138_HTML.gif, where r 2 = max ( 2 r 1 , r λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq139_HTML.gif. Then u P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq97_HTML.gif and u = r 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq140_HTML.gif imply that
min u ( t ) λ u = λ r 2 r , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_Equao_HTML.gif
and so we obtain
T u = 0 1 G ( t , s ) f ( s , u ( s ) ) d s δ 1 0 1 G ( t , s ) u ( s ) d s δ 1 u ( β Γ ( α ) ( 1 η ) α 1 Γ ( α ) ) u . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_Equap_HTML.gif

This shows that T u u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq76_HTML.gif for u P Ω 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq141_HTML.gif. We conclude by the second part of the Guo-Krasnosel’skii fixed point theorem that (1)-(2) has at least one positive solution u P ( Ω ¯ 2 Ω 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq142_HTML.gif. □

Remark 3.1

To prove Theorem 3.2, we use the cone
P = { u | u C [ 0 , 1 ] , min t [ 0 , b ] u ( t ) λ u } , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_Equaq_HTML.gif

where b and λ are defined in Lemma 2.6 for the case where β Γ ( α ) = ( 1 η ) α 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq143_HTML.gif, and in Lemma 2.7 for the case where β Γ ( α ) < ( 1 η ) α 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq144_HTML.gif. We skip the rest of the proof as it is similar to the proof of Theorem 3.1.

Example 3.1

Consider the fractional boundary value problem:
{ C D 3 2 u ( t ) = t 2 e u ( t ) + u ( t ) , t [ 0 , 1 ] , u ( 0 ) = 0 , 4 5 C D 1 2 u ( 1 ) + u ( 3 4 ) = 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_Equ9_HTML.gif
(9)

which is problem (1)-(2) with α = 3 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq145_HTML.gif, β = 4 5 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq146_HTML.gif, η = 3 4 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq147_HTML.gif and f ( t , u ( t ) ) = t 2 e u ( t ) + u ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq148_HTML.gif.

First, we note that u = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq149_HTML.gif is not a solution of (9).

Clearly, f 0 = https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq82_HTML.gif and f = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq83_HTML.gif, and we also have β Γ ( α ) ( 1 η ) α 1 = 2 π 5 1 2 0.20898 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq150_HTML.gif.

We take
λ = β Γ ( α ) ( 1 η ) α 1 β Γ ( α ) + η α 1 = 2 π 5 1 2 2 π 5 + 3 2 = 4 π 5 4 π + 5 3 0.13269 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_Equar_HTML.gif

and consider the cone P = { u | u C [ 0 , 1 ] , u ( t ) 0 , min t [ 0 , 1 ] u ( t ) λ u } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-5/MediaObjects/13661_2012_Article_260_IEq151_HTML.gif.

By the first part of Theorem 3.1, we conclude that the boundary value problem (9) has a positive solution in the cone P.

Declarations

Acknowledgements

Dedicated to Professor Jean Mawhin for his 70th anniversary.

The research has been partially supported by Ministerio de Economía y Competitividad, and FEDER, project MTM2010-15314.

Authors’ Affiliations

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

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© Nieto and Pimentel; licensee Springer. 2013

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