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# A result on three solutions theorem and its application to p-Laplacian systems with singular weights

Eun Kyoung Lee1 and Yong-Hoon Lee2*

Author Affiliations

1 Department of Mathematics Education, Pusan National University, Busan, 609-735, Korea

2 Department of Mathematics, Pusan National University, Busan, 609-735, Korea

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Boundary Value Problems 2012, 2012:63  doi:10.1186/1687-2770-2012-63

 Received: 16 February 2012 Accepted: 18 May 2012 Published: 22 June 2012

This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

### Abstract

In this paper, we consider p-Laplacian systems with singular weights. Exploiting Amann type three solutions theorem for a singular system, we prove the existence, nonexistence, and multiplicity of positive solutions when nonlinear terms have a combined sublinear effect at ∞.

MSC: 35J55, 34B18.

##### Keywords:
p-Laplacian system; singular weight; upper solution; lower solution; three solutions theorem

### 1 Introduction

In this paper, we study one-dimensional p-Laplacian system with singular weights of the form where φ p ( u ) = | u | p 2 u , λ is a nonnegative parameter, h i , i = 1 , 2 is a nonnegative measurable function on ( 0 , 1 ) , h i 0 on any open subinterval in ( 0 , 1 ) and f , g C ( R + , R + ) with R + = [ 0 , ) . In particular, h i may be singular at the boundary or may not be in L 1 ( 0 , 1 ) . It is easy to see that if h i L 1 ( 0 , 1 ) , then all solutions of ( P λ ) are in C 1 [ 0 , 1 ] . On the other hand, if h i L 1 ( 0 , 1 ) , then this regularity of solutions is not true in general; for example, even for scalar case, if we take h ( t ) = ( p 1 ) t 1 | 1 + ln t | p 2 , p > 2 and λ = 1 , f 1 , then h L 1 ( 0 , 1 ) , and the solution u for corresponding scalar problem of ( P λ ) is given by u ( t ) = t ln t which is not in C 1 [ 0 , 1 ] .

For more precise description, let us introduce the following two classes of weights;

We note that h given in the above example satisfies h A B but h L 1 ( 0 , 1 ) . The main interest of this paper is to establish Amann type three solutions theorem [4] when h i A B with possibility of h L 1 ( 0 , 1 ) . The theorem generally describes that two pairs of lower and upper solutions with an ordering condition imply the existence of three solutions. Recently, Ben Naoum and De Coster [6] have proved the theorem for scalar one-dimensional p-Laplacian problems with L 1 -Caratheodory condition which corresponds to case h L 1 ( 0 , 1 ) ; Henderson and Thompson [18] as well as Lü, O’Regan, and Agarwal [23] - for scalar second order ODEs and one-dimensional p-Laplacian with the derivative-dependent nonlinearity respectively; and De Coster and Nicaise [11] - for semilinear elliptic problems in nonsmooth domains. For noncooperative elliptic systems ( p = 2 ) with k i 1 and Ω bounded, one may refer to Ali, Shivaji, and Ramaswamy [3]. Specially, for subsuper solutions which are not completely ordered, this type of three solutions result was studied in [26].

The three solutions theorem for our system ( P λ ) or even for corresponding scalar p-Laplacian problems is not obviously extended from previous works mainly by the possibility h L 1 ( 0 , 1 ) . Caused by the delicacy of Leray-Schauder degree computation, the crucial step for the proof is to guarantee C 1 regularity of solutions, but with condition h A B , C 1 regularity is not known yet. Due to the singularity of weights on the boundary, the C 1 regularity heavily depends on the shape of nonlinear terms f and g. Therefore, the first step is to investigate certain conditions on f and g to guarantee C 1 regularity of solutions. Another difficulty is to show that a corresponding integral operator is bounded on the set of functions between upper and lower solutions in C 0 1 [ 0 , 1 ] . To overcome this difficulty, we give some restrictions on upper and lower solutions such that their boundary values are zero. As far as the authors know, our three solutions theorem (Theorem 1.1 in Section 2) is new and first for singular p-Laplacian systems with weights of A B class.

To cover a larger class of differential system, we consider the systems of the form where F , G : ( 0 , 1 ) × R R are continuous. We give more conditions on F and G as follows: ( F 1 ) = For each t ( 0 , 1 ) , F ( t , u ) and G ( t , u ) are nondecreasing in u.; (H) = There exist h 1 , h 2 A B and f , g C ( R , R + ) such that

0 lim s 0 f ( s ) φ p ( | s | ) < , 0 lim s 0 g ( s ) φ p ( | s | ) <

and

| F ( t , u ) | h 1 ( t ) f ( u ) , | G ( t , u ) | h 2 ( t ) g ( u ) ,

for all t ( 0 , 1 ) and u R .; ( F 2 ) = F ( t , u ) u > 0 and G ( t , u ) u > 0 , for all ( t , u ) ( 0 , 1 ) × R .. We now state our first main result related to three solutions theorem as follows. See for more details in Section 2.

Theorem 1.1Assume (H), ( F 1 ) and ( F 2 ). Let ( α 1 , α ¯ 1 ) , ( β 2 , β ¯ 2 ) be a lower solution and an upper solution and ( α 2 , α ¯ 2 ) , ( β 1 , β ¯ 1 ) be a strict lower solution and a strict upper solution of problem (P) respectively. Also, assume that all of them are contained in C 0 1 [ 0 , 1 ] × C 0 1 [ 0 , 1 ] and satisfy ( α 1 , α ¯ 1 ) ( β 1 , β ¯ 1 ) ( β 2 , β ¯ 2 ) , ( α 1 , α ¯ 1 ) ( α 2 , α ¯ 2 ) ( β 2 , β ¯ 2 ) , ( α 2 , α ¯ 2 ) ≰ ( β 1 , β ¯ 1 ) . Then problem (P) has at least three solutions ( u 1 , v 1 ) , ( u 2 , v 2 ) and ( u 3 , v 3 ) such that ( α 1 , α ¯ 1 ) ( u 1 , v 1 ) ( β 1 , β ¯ 1 ) , ( α 2 , α ¯ 2 ) ( u 2 , v 2 ) ( β 2 , β ¯ 2 ) , ( α 1 , α ¯ 1 ) ( u 3 , v 3 ) ( β 2 , β ¯ 2 ) and ( u 3 , v 3 ) ≰ ( β 1 , β ¯ 1 ) , ( u 3 , v 3 ) ≱ ( α 2 , α ¯ 2 ) .

As an application of Theorem 1.1, we study the existence, nonexistence, and multiplicity of positive radial solutions for the following quasilinear system on an exterior domain: where Ω = { x R N : | x | > r 0 } , r 0 > 0 , 1 < p < N , Δ p z = div ( | z | p 2 z ) , k i C ( [ r 0 , ) , ( 0 , ) ) , i = 1 , 2 and f , g C ( R + , R + ) with R + = [ 0 , ) .

In recent years, the existence of positive solutions for such systems has been widely studied, for example, in [1] and [27] for second order ODE systems, in [3,7,9,10,13,14,16] and [8] for semilinear elliptic systems on a bounded domain and in [5,15,17] and [2] for p-Laplacian systems on a bounded domain.

For a precise description, let us give the list of assumptions that we consider. (k) = k i KA KB , where

; ( f 1 ) = f 0 = lim s 0 + f ( s ) s p 1 = 0 and g 0 = lim s 0 + g ( s ) s p 1 = 0 ,; ( f 2 ) = lim s f ( ρ ( g ( s ) ) 1 p 1 ) s p 1 = 0 for all ρ > 0 ,; ( f 3 ) = f and g are nondecreasing..

Condition ( f 2 ) is sometimes called a combined sublinear effect at ∞ and simple examples satisfying ( f 1 ) ∽ ( f 3 ) can be given as follows:

f ( w ) = { w r , w 1 , w q , w 1 , g ( z ) = { z γ , z 1 , z δ , z 1 ,

where r , γ > p 1 and q δ < ( p 1 ) 2 , and also

{ f ( z ) = arctan ( z r ) , g ( w ) = w q ,

where r , q > p 1 .

Among the reference works mentioned above, Hai and Shivaji [17] and Ali and Shivaji [2] (with more general nonlinearities) considered problem ( P E ) with case k i 1 and Ω bounded. For C 1 monotone functions f and g with lim s f ( s ) = = lim s g ( s ) and satisfying condition ( f 2 ), they proved that there exists λ > 0 such that the problem has at least one positive solution for λ > λ .

We first transform ( P E ) into one-dimensional p-Laplacian systems ( P λ ) with change of variables z ( r ) = z ( | x | ) , w ( r ) = w ( | x | ) , u ( t ) = z ( ( r r 0 ) N + p p 1 ) and v ( t ) = w ( ( r r 0 ) N + p p 1 ) where h i is given by

h i ( t ) = ( p 1 N p ) p r 0 p t p ( N 1 ) N p k i ( r 0 t ( p 1 ) N p ) .

It is not hard to see that if k i in ( P E ) satisfies (k), then h i in ( P λ ) satisfies h i A B , for i = 1 , 2 . Mainly by making use of Theorem 1.1, we prove the following existence result for problem ( P λ )

Theorem 1.2Assume h i A B , i = 1 , 2 , ( f 1 ), ( f 2 ) and ( f 3 ). Then there exists λ > 0 such that ( P λ ) has no positive solution for λ < λ , at least one positive solution at λ = λ and at least two positive solutions for λ > λ .

As a corollary, we obtain our second main result as follows.

Corollary 1.3Assume (k), ( f 1 ), ( f 2 ) and ( f 3 ). Then there exists λ > 0 such that ( P E ) has no positive radial solution for λ < λ , at least one positive radial solution at λ = λ and at least two positive radial solutions for λ > λ .

We finally notice that the first eigenfunctions of make an important role to construct upper solutions in the proofs of Theorem 1.2 and Theorem 1.1. This is possible due to a recent work of Kajikiya, Lee, and Sim [19] which exploits the existence of discrete eigenvalues and the properties of corresponding eigenfunctions for problem (E) with h i A B .

This paper is organized as follows. In Section 2, we state a C 1 -regularity result and a three solutions theorem for singular p-Laplacian systems. In addition, we introduce definitions of (strict) upper and lower solutions, a related theorem and a fixed point theorem for later use. In Section 3, we prove Theorem 1.2.

### 2 Three solutions theorem

In this section, we give definitions of upper and lower solutions and prove three solutions theorem for the following singular system where F , G : ( 0 , 1 ) × R R are continuous.

We call ( u , v ) a solution of (P) if ( u , v ) ( C [ 0 , 1 ] × C [ 0 , 1 ] ) ( C 1 ( 0 , 1 ) × C 1 ( 0 , 1 ) ) , ( φ p ( u ( t ) ) , φ p ( v ( t ) ) ) C 1 ( 0 , 1 ) × C 1 ( 0 , 1 ) and ( u , v ) satisfies (P).

Definition 2.1 We say that ( α , α ¯ ) is a lower solution of problem (P) if ( α , α ¯ ) ( C 1 ( 0 , 1 ) × C 1 ( 0 , 1 ) ) ( C [ 0 , 1 ] × C [ 0 , 1 ] ) , ( φ p ( α ( t ) ) , φ p ( α ¯ ( t ) ) ) C 1 ( 0 , 1 ) × C 1 ( 0 , 1 ) and

{ φ p ( α ( t ) ) + F ( t , α ¯ ( t ) ) 0 , t ( 0 , 1 ) , φ p ( α ¯ ( t ) ) + G ( t , α ( t ) ) 0 , t ( 0 , 1 ) , α ( 0 ) 0 , α ¯ ( 0 ) 0 , α ( 1 ) 0 , α ¯ ( 1 ) 0 .

We also say that ( β , β ¯ ) is an upper solution of problem (P) if ( β , β ¯ ) ( C 1 ( 0 , 1 ) × C 1 ( 0 , 1 ) ) ( C [ 0 , 1 ] × C [ 0 , 1 ] ) , ( φ p ( β ( t ) ) , φ p ( β ¯ ( t ) ) ) C 1 ( 0 , 1 ) × C 1 ( 0 , 1 ) and it satisfies the reverse of the above inequalities. We say that ( α , α ¯ ) and ( β , β ¯ ) are strict lower solution and strict upper solution of problem (P), respectively, if ( α , α ¯ ) and ( β , β ¯ ) are lower solution and upper solution of problem (P), respectively and satisfying φ p ( α ( t ) ) + F ( t , α ¯ ( t ) ) > 0 , φ p ( α ¯ ( t ) ) + G ( t , α ( t ) ) > 0 , φ p ( β ( t ) ) + F ( t , β ¯ ( t ) ) < 0 , φ p ( β ¯ ( t ) ) + G ( t , β ( t ) ) < 0 for t ( 0 , 1 ) .

We note that the inequality on R 2 can be understood componentwise. Let D α β = { ( t , u , v ) | ( α ( t ) , α ¯ ( t ) ) ( u , v ) ( β ( t ) , β ¯ ( t ) ) , t ( 0 , 1 ) } . Then the fundamental theorem on upper and lower solutions for problem (P) is given as follows. The proof can be done by obvious combination from Lee [20], Lee and Lee [21] and Lü and O’Regan [22].

Theorem 2.2Let ( α , α ¯ ) and ( β , β ¯ ) be a lower solution and an upper solution of problem (P) respectively such that( a 1 ) = ( α ( t ) , α ¯ ( t ) ) ( β ( t ) , β ¯ ( t ) ) , for all t [ 0 , 1 ] ..Assume ( F 1 ). Also assume that there exist h F , h G A B such that( a 2 ) = | F ( t , v ) | h F ( t ) , | G ( t , u ) | h G ( t ) , for all ( t , u , v ) D α β ..Then problem (P) has at least one solution ( u , v ) such that

( α ( t ) , α ¯ ( t ) ) ( u ( t ) , v ( t ) ) ( β ( t ) , β ¯ ( t ) ) , for all t [ 0 , 1 ] .

Remark 2.3 It is not hard to see that condition (H) implies the following condition;

For each M > 0 , there exists C M > 0 such that

| F ( t , u ) | C M h 1 ( t ) φ p ( | u | ) , | G ( t , u ) | C M h 2 ( t ) φ p ( | u | ) ,

for t ( 0 , 1 ) and | u | M .

Lemma 2.4Assume (H) and ( F 2 ). Let ( u , v ) be a nontrivial solution of (P). Then there exists a > 0 such that bothuandvhave no interior zeros in ( 0 , a ] [ 1 a , 1 ) .

Proof Let ( u , v ) be a nontrivial solution of (P). Suppose, on the contrary, that there exist sequences ( t n ) , ( s n ) of interior zeros of u and v respectively with t n , s n 0 . We note that both sequences should exist simultaneously. Indeed, if one of the sequences say, ( t n ) , does not exist, then assuming without loss of generality, u > 0 on ( 0 , a ] for some a > 0 , we get φ p ( v ( s ) ) = G ( t , u ( t ) ) > 0 for t ( 0 , a ] by ( F 2 ). From the monotonicity of φ p , we know that v is concave on the interval. Thus v should have at most one interior zero in ( 0 , a ] , a contradiction. With this concave-convex argument, we know that ( t n , t n 1 ) ( s n , s n 1 ) , u v 0 on ( t n , t n 1 ) ( s n , s n 1 ) and if t n and s n are local extremal points of u and v on ( t n , t n 1 ) and ( s n , s n 1 ) respectively, thus both t n and s n are in ( t n , t n 1 ) ( s n , s n 1 ) . We consider the case that t n s n , t n s n and u , v > 0 in ( t n , t n 1 ) ( s n , s n 1 ) . All other cases can be explained by the same argument. If M = max { u , v } , then by using Remark 2.3, we have

u ( t n ) = t n t n φ p 1 ( s t n F ( r , v ( r ) ) d r ) d s t n t n φ p 1 ( s n t n F ( r , v ( r ) ) d r ) d s C M t n t n φ p 1 ( s n t n h 1 ( r ) v ( r ) p 1 d r ) d s C M ( t n t n φ p 1 ( s n t n h 1 ( r ) d r ) d s ) v ( t n ) (2.1)

and similarly,

v ( t n ) C M ( s n t n φ p 1 ( s s n h 2 ( r ) d r ) d s ) u ( t n ) . (2.2)

Therefore, it follows from plugging (2.2) into (2.1) that

u ( t n ) ( C M ) 2 ( t n t n φ p 1 ( s n t n h 1 ( r ) d r ) d s ) ( s n t n φ p 1 ( s s n h 2 ( r ) d r ) d s ) u ( t n ) . (2.3)

Since h i A , for sufficiently large n, we obtain

( C M ) 2 ( t n t n φ p 1 ( s n t n h 1 ( r ) d r ) d s ) ( s n t n φ p 1 ( s s n h 2 ( r ) d r ) d s ) < 1 / 2 .

This contradicts (2.3) and the proof is done. □

Theorem 2.5Assume (H) and ( F 2 ). If ( u , v ) is a solution of (P), then ( u , v ) C 0 1 [ 0 , 1 ] × C 0 1 [ 0 , 1 ] .

Proof Let ( u , v ) be a nontrivial solution of (P). Then u , v C 0 [ 0 , 1 ] C 1 ( 0 , 1 ) so that it is enough to show

| u ( 0 + ) | < , | u ( 1 ) | < , | v ( 0 + ) | < , | v ( 1 ) | < .

We will show | u ( 0 + ) | < . Other facts can be proved by the same manner. Suppose | u ( 0 + ) | = . By Lemma 2.4 and the concave-convex argument, we may assume without loss of generality that there exists a ( 0 , 1 ) such that u , v , u , v > 0 on ( 0 , a ] . Then for given ε > 0 , by the fact h i B , i = 1 , 2 , there exists δ ( 0 , a ) such that

0 δ t p 1 h i ( t ) d t < ε , i = 1 , 2 .

Let M = max { u , v } . Then integrating (P) over ( s , t ) ( 0 , δ ) and using Remark 2.3, we have

u ( s ) p 1 u ( t ) p 1 + C M s t h 1 ( τ ) ( v ( τ ) τ ) p 1 τ p 1 d τ u ( t ) p 1 + C M ( v ( s ) s ) p 1 s t h 1 ( τ ) τ p 1 d τ u ( t ) p 1 + C M ε ( v ( s ) s ) p 1 , (2.4)

where we use the fact that ( v ( τ ) τ ) p 1 is decreasing since v is concave. From u ( 0 + ) = and (2.4), we know lim s 0 + ( v ( s ) s ) p 1 = . This implies that conditions u ( 0 + ) = and v ( 0 + ) = are equivalent. From (2.4), we have

( s u ( s ) v ( s ) ) p 1 ( s v ( s ) ) p 1 u ( t ) p 1 + C M ε .

Thus we have

lim sup s 0 + ( s u ( s ) v ( s ) ) p 1 C M ε .

Since ε is arbitrary, we have

lim sup s 0 + ( s u ( s ) v ( s ) ) p 1 = 0 . (2.5)

Using the fact v ( 0 + ) = , with same argument, we have

lim sup s 0 + ( s v ( s ) u ( s ) ) p 1 = 0 . (2.6)

On the other hand, we observe the inequality

( α + β ) 1 p 1 C p ( α 1 p 1 + β 1 p 1 ) , for α , β 0 , (2.7)

where

C p = { 1 if p 2 , 2 2 p p 1 if 1 < p < 2 .

Since h i A , we may choose b ( 0 , min { a , 1 2 } ) such that

( C M ) 1 p 1 C p 0 b ( s 1 2 h i ( τ ) d τ ) 1 p 1 d s < 1 2 . (2.8)

Integrating (P) over ( s , t ) with 0 < s < t < b and using Remark 2.3, we get

u ( s ) p 1 u ( t ) p 1 + C M v ( t ) p 1 s t h 1 ( τ ) d τ ,

here we use the fact that v ( t ) is increasing in ( 0 , b ) . Using (2.7), we have

u ( s ) C p u ( t ) + ( C M ) 1 p 1 C p v ( t ) ( s 1 2 h 1 ( τ ) d τ ) 1 p 1 . (2.9)

Integrating (2.9) over ( 0 , t ) with respect to s and using (2.8), we have

u ( t ) C p t u ( t ) + ( C M ) 1 p 1 C p v ( t ) 0 t ( s 1 2 h 1 ( τ ) d τ ) 1 p 1 d s C p t u ( t ) + 1 2 v ( t ) . (2.10)

Similarly, we have

v ( t ) C p t v ( t ) + 1 2 u ( t ) . (2.11)

Adding (2.10) and (2.11), we have

0 < 1 2 C p < t u ( t ) + t v ( t ) u ( t ) + v ( t ) t u ( t ) v ( t ) + t v ( t ) u ( t ) , (2.12)

on ( 0 , b ) . From (2.5) and (2.6), we see that the right-hand side of (2.12) goes to zero as t 0 . This is a contradiction and the proof is complete. □

Now, we consider the three solutions theorem for singular p-Laplacian system (P). For ν L 1 ( 0 , 1 ) , if

ζ ( x ) = 0 1 φ p 1 ( x 0 s ν ( τ ) d τ ) d s ,

then the zero of ζ ( x ) , denoted by ξ ( ν ) is uniquely determined by ν. Define A : L 1 ( 0 , 1 ) C 0 1 [ 0 , 1 ] by taking

A ( ν ) ( t ) = 0 t φ p 1 ( ξ ( ν ) 0 s ν ( τ ) d τ ) d s .

It is known that A is completely continuous [24]. Define X C 0 1 [ 0 , 1 ] × C 0 1 [ 0 , 1 ] with norm ( u , v ) X = u + v . We note that

| u ( t ) | 2 t ( 1 t ) u , for all u C 0 1 [ 0 , 1 ] . (2.13)

If F and G satisfy condition (H), then for ( u , v ) X , from Remark 2.3 and (2.13), we get

0 1 | F ( t , v ( t ) ) | d t 0 1 h 1 ( t ) f ( v ( t ) ) d t 0 1 h 1 ( t ) C 0 | v ( t ) | p 1 d t 2 p 1 C 0 v p 1 0 1 t p 1 ( 1 t ) p 1 h 1 ( t ) d t .

This implies F ( , v ( ) ) L 1 ( 0 , 1 ) and by similar computation, we also get G ( , u ( ) ) L 1 ( 0 , 1 ) . This fact enables us to define the integral operator for problem (P) and the regularity of solutions (Theorem 2.5) is crucial in this argument. Now, define an operator T by

T ( u , v ) = ( A ( F ( t , v ( t ) ) ) , A ( G ( t , u ( t ) ) ) ) ,

then we see that T : X X and completely continuous.

Lemma 2.6Assume (H), ( F 1 ) and ( F 2 ). Let ( α , α ¯ ) and ( β , β ¯ ) be a strict lower solution and a strict upper solution of problem (P) respectively such that ( α , α ¯ ) X , ( β , β ¯ ) X and ( α , α ¯ ) ( β , β ¯ ) . Then problem (P) has at least one solution ( u , v ) X such that

( α , α ¯ ) ( u , v ) ( β , β ¯ ) .

Moreover, for R > 0 large enough,

deg ( I T , Ω , 0 ) = 1 ,

where Ω = { ( u , v ) X | ( α , α ¯ ) ( u , v ) ( β , β ¯ ) , ( u , v ) X < R } .

Proof Define γ : [ 0 , 1 ] × R R given by

and also define

F ( t , v ( t ) ) = F ( t , γ ¯ ( t , v ( t ) ) ) , G ( t , u ( t ) ) = G ( t , γ ( t , u ( t ) ) ) .

Let us consider the following modified problem We first show that there exists a constant R > 0 such that if ( u , v ) is a solution of ( P ¯ ), then ( u , v ) Ω . In fact, every solution ( u , v ) of ( P ¯ ) satisfies ( α , α ¯ ) ( u , v ) ( β , β ¯ ) on [ 0 , 1 ] . From (H), ( F 1 ) and the fact that ( α , α ¯ ) X , ( β , β ¯ ) X , we get

| φ p ( u ( t ) ) | = | t 0 t F ( τ , v ( τ ) ) d τ | t 0 t max { | F ( τ , α ¯ ( τ ) ) | , | F ( τ , β ¯ ( τ ) ) | } d τ 0 1 h 1 ( t ) max t [ 0 , 1 ] { f ( α ¯ ( t ) ) , f ( β ¯ ( t ) ) } d t 0 1 c 1 h 1 ( t ) max t [ 0 , 1 ] { | α ¯ ( t ) | p 1 , | β ¯ ( t ) | p 1 } d t 2 p 1 c 1 max { α ¯ p 1 , β ¯ p 1 } 0 1 t p 1 ( 1 t ) p 1 h 1 ( t ) d t < .

Similarly, we see that v is bounded. Therefore, ( u , v ) X < R , for some R > 0 . Thus it is enough to show that

( α , α ¯ ) ( u , v ) ( β , β ¯ ) .

Assume, on the contrary, that there exists t 0 ( 0 , 1 ) such that

min ( u ( t ) α ( t ) ) = u ( t 0 ) α ( t 0 ) = 0 .

Then choosing t 1 ( t 0 , 1 ) with ( u α ) ( t 1 ) 0 , we get the following contradiction:

0 [ φ p ( u ( t 1 ) ) φ p ( α ( t 1 ) ) ] [ φ p ( u ( t 0 ) ) φ p ( α ( t 0 ) ) ] = t 0 t 1 F ( t , v ( t ) ) φ p ( α ( t ) ) d t t 0 t 1 F ( t , α ¯ ( t ) ) φ p ( α ( t ) ) d t < 0 .

Now, assume u ( 0 ) = α ( 0 ) . Since u ( t ) > α ( t ) on t ( 0 , 1 ) and u ( 0 ) = α ( 0 ) = 0 , there exists t 2 ( 0 , 1 ) such that u ( t 2 ) α ( t 2 ) and we get the same contradiction from the above calculation by using 0 instead of t 0 . For u ( 1 ) = α ( 1 ) case, we also get the same contradiction. Consequently, we get α u . The other cases can be proved by the same manner. Taking Ω = { ( u , v ) X | ( α , α ¯ ) ( u , v ) ( β , β ¯ ) , ( u , v ) X < R } , we see that every solution of ( P ¯ ) is contained in Ω. We now compute deg ( I T , Ω , 0 ) . For this purpose, let us consider the operator T ¯ : X X defined by

T ¯ ( u , v ) ( t ) = ( A ( F ( t , v ( t ) ) ) , A ( G ( t , u ( t ) ) ) ) .

Then it is obvious that T ¯ is completely continuous. We show that there exists R ¯ > 0 such that R ¯ > R and T ¯ ( X ) B ( 0 , R ¯ ) . Indeed, since A ( F ( 0 , v ( 0 ) ) ) = 0 = A ( F ( 1 , v ( 1 ) ) ) , there is t ˜ ( 0 , 1 ) such that d d t A ( F ( t , v ( t ) ) ) | t = t ˜ = 0 . By integrating

d d t φ p ( d d t A ( F ( t , v ( t ) ) ) ) = F ( t , v ( t ) )

from t ˜ to t, we have

| φ p ( d d t A ( F ( t , v ( t ) ) ) ) | = | t ˜ t F ( τ , v ( τ ) ) d τ | 0 1 h 1 ( t ) f ( γ ¯ ( t , v ( t ) ) ) d t 0 1 h 1 ( t ) C 1 | γ ¯ ( t , v ( t ) ) | p 1 d t 0 1 h 1 ( t ) C 1 max { | β ¯ ( t ) | p 1 , | α ¯ ( t ) | p 1 } d t C 2 max { β ¯ p 1 , α ¯ p 1 } 0 1 t p 1 ( 1 t ) p 1 h 1 ( t ) d t .

Similarly, we see that d d t A ( G ( t , u ( t ) ) ) is bounded. Therefore, we get

deg ( I T ¯ , B ( 0 , R ¯ ) , 0 ) = 1 .

Since every solution of ( P ¯ ) is contained in Ω, the excision property implies that

deg ( I T ¯ , Ω , 0 ) = deg ( I T ¯ , B ( 0 , R ¯ ) , 0 ) = 1 .

Since T ¯ = T on Ω, we finally get

deg ( I T , Ω , 0 ) = deg ( I T ¯ , Ω , 0 ) = 1 .

This completes the proof. □

We now prove three solutions theorem for (P).

#### Proof of Theorem 1.1

Define

and let us consider Then noting that every solution ( u , v ) of ( P ˜ ) satisfies ( α 1 , α ¯ 1 ) ( u , v ) ( β 2 , β ¯ 2 ) , we may choose K 1 , K 2 > 0 , by (H) such that

Let λ 1 and μ 1 be the first eigenvalues of for i = 1 , 2 respectively and let e 1 and e 2 be corresponding eigenfunctions with e 1 = e 2 = 1 . Since e 1 , e 2 C 0 1 [ 0 , 1 ] are positive and concave [19], we may choose M 1 , M 2 > 0 such that ( M 1 e 1 , M 2 e 2 ) ( β 2 , β ¯ 2 ) ( M 1 e 1 , M 2 e 2 ) ( α 1 , α ¯ 1 ) and for t ( 0 , 1 )

We show that ( M 1 e 1 , M 2 e 2 ) and ( M 1 e 1 , M 2 e 2 ) are a strict upper solution and a strict lower solution of ( P ˜ ) respectively. Indeed,

φ p ( M 1 e 1 ( t ) ) + F ( t , γ ¯ 1 ( t , M 2 e 2 ( t ) ) ) = φ p ( M 1 e 1 ( t ) ) + F ( t , β ¯ 2 ( t ) ) λ 1 h 1 ( t ) φ p ( M 1 e 1 ( t ) ) + h 1 ( t ) f ( β ¯ 2 ( t ) ) λ 1 h 1 ( t ) φ p ( M 1 e 1 ( t ) ) + h 1 ( t ) K 1 φ p ( | β ¯ 2 ( t ) | ) < 0 .

Similarly, we get

φ p ( M 2 e 2 ( t ) ) + G ( t , γ 1 ( t , M 1 e 1 ( t ) ) ) < 0 .

Moreover,

Similarly, we also get

φ p ( M 2 e 2 ( t ) ) + F ( t , γ ¯ 1 ( t , M 2 e 2 ( t ) ) ) > 0 .

For R > 0 , large enough, define

Then by Theorem 2.2, there exist two solutions ( u 1 , v 1 ) and ( u 2 , v 2 ) of (P) satisfying ( α 1 , α ¯ 1 ) ( u 1 , v 1 ) ( β 1 , β ¯ 1 ) and ( α 2 , α ¯ 2 ) ( u 2 , v 2 ) ( β 2 , β ¯ 2 ) . Therefore, by Lemma 2.6, we get

deg ( I T ˜ , Ω 1 , 0 ) = deg ( I T ˜ , Ω 2 , 0 ) = deg ( I T ˜ , Ω 3 , 0 ) = 1 ,

and by the excision property, we have

deg ( I T ˜ , Ω 3 ( Ω 1 Ω 2 ) , 0 ) = 1 .

This completes the proof.

### 3 Application

In this section, we prove the existence, nonexistence, and multiplicity of positive solutions for ( P λ ) by using three solutions theorem in Section 2. Let us define a cone

K = { u C [ 0 , 1 ] | u are concave and u ( 0 ) = 0 = u ( 1 ) } ,

and define A λ , B λ : K C [ 0 , 1 ] by taking

where σ v and σ u are unique zeros of

respectively. And define T λ : K × K C [ 0 , 1 ] × C [ 0 , 1 ] by

T λ ( u , v ) = ( A λ ( v ) , B λ ( u ) ) .

Then it is known that T λ : K × K K × K is completely continuous [25] and ( u , v ) = T λ ( u , v ) in K × K is equivalent to the fact that ( u , v ) is a positive solution of ( P λ ). We know from Theorem 2.5 that under assumptions h i A B i = 1 , 2 and ( f 1 ), any solution ( u , v ) of problem ( P λ ) is in C 0 1 [ 0 , 1 ] × C 0 1 [ 0 , 1 ] .

Remark 3.1 If ( u , v ) is a solution of ( P λ ), then u = A λ ( B λ ( u ) ) and v = B λ ( A λ ( v ) ) .

For later use, we introduce the following well-known result. See [12] for proof and details.

Proposition 3.2LetXbe a Banach space, K an order cone inX. Assume that Ω 1 and Ω 2 are bounded open subsets inXwith 0 Ω 1 and Ω ¯ 1 Ω 2 . Let A : K ( Ω ¯ 2 Ω 1 ) K be a completely continuous operator such that either

(i) A u u , u K Ω 1 , and A u u , u K Ω 2

or

(ii) A u u , u K Ω 1 , and A u u , u K Ω 2 .

ThenAhas a fixed point in K ( Ω ¯ 2 Ω 1 ) .

Lemma 3.3Assume h i A B , i = 1 , 2 , ( f 2 ) and ( f 3 ). Let R be a compact subset of ( 0 , ) . Then there exists a constant b R > 0 such that for all λ R and all possible positive solutions ( u , v ) of ( P λ ), one has ( u , v ) b R .

Proof If it is not true, then there exist { λ n } R and solutions { ( u n , v n ) } of ( P λ n ) such that ( u n , v n ) . We note that

where Λ = max { λ 1 p 1 | λ R } and

Q i = 0 1 2 φ p 1 ( s 1 2 h i ( τ ) d τ ) d s + 1 2 1 φ p 1 ( 1 2 s h i ( τ ) d τ ) d s , i = 1 , 2 .

This implies both u n and v n . Moreover, by the above estimation,

u n Λ Q 1 φ p 1 ( f ( Λ Q 2 φ p 1 ( g ( u n ) ) ) ) .

Thus we get

1 φ p ( Λ Q 1 ) f ( Λ Q 2 φ p 1 ( g ( u n ) ) ) φ p ( u n ) 0

as u n and this contradiction completes the proof. □

Lemma 3.4Assume h i A B , i = 1 , 2 , ( f 2 ) and ( f 3 ). If ( P λ ) has a lower solution ( α , α ¯ ) C 0 1 [ 0 , 1 ] × C 0 1 [ 0 , 1 ] for some λ > 0 , then ( P λ ) has a solution ( u , v ) such that ( α , α ¯ ) ( u , v ) .

Proof It suffices to show the existence of an upper solution ( β , β ¯ ) of ( P λ ) satisfying ( α , α ¯ ) ( β , β ¯ ) . Let ϕ i and i = 1 , 2 be positive solutions of

{ φ p ( u ( t ) ) + λ h i ( t ) = 0 , t ( 0 , 1 ) , u ( 0 ) = 0 , u ( 1 ) = 0 .

(Case I) Both f and g are bounded.

Since ϕ i ( i = 1 , 2 ) are positive concave functions and ( α , α ¯ ) C 0 1 [ 0 , 1 ] × C 0 1 [ 0 , 1 ] , we may choose M > 0 such that M > max { f 1 p 1 , g 1 p 1 } and ( M ϕ 1 , M ϕ 2 ) ( α , α ¯ ) . We now show that ( β , β ¯ ) = ( M ϕ 1 , M ϕ 2 ) is an upper solution of ( P λ ). In fact,

φ p ( β ( t ) ) + λ h 1 ( t ) f ( β ¯ ( t ) ) = M p 1 φ p ( ϕ 1 ( t ) ) + λ h 1 ( t ) f ( M ϕ 2 ( t ) ) = λ h 1 ( t ) [ f ( M ϕ 2 ( t ) ) M p 1 ] λ h 1 ( t ) [ f M p 1 ] 0 .

Similarly,

φ p ( β ¯ ( t ) ) + λ h 2 ( t ) g ( β ( t ) ) λ h 2 ( t ) [ g M p 1 ] 0 .

(Case II) g ( u ) as u .

Using ( f 2 ), choose M > 0 such that ( g ( M ϕ 1 ) ) 1 p 1 ϕ 2 α ¯ , M ϕ 1 α and

f ( ϕ 2 ( g ( M ϕ 1 ) ) 1 p 1 ) ( M ϕ 1 ) p 1 1 ϕ 1 p 1 .

Let ( β , β ¯ ) = ( M ϕ 1 , ( g ( M ϕ 1 ) ) 1 p 1 ϕ 2 ) . Then

φ p ( β ( t ) ) + λ h 1 ( t ) f ( β ¯ ( t ) ) = M p 1 φ p ( ϕ 1 ( t ) ) + λ h 1 ( t ) f ( ( g ( M ϕ 1 ) ) 1 p 1 ϕ 2 ( t ) ) = λ h 1 ( t ) [ f ( ( g ( M ϕ 1 ) ) 1 p 1 ϕ 2 ( t ) ) M p 1 ] λ h 1 ( t ) [ f ( ϕ 2 ( g ( M ϕ 1 ) ) 1 p 1 ) M p 1 ] 0 .

And

φ p ( β ¯ ( t ) ) + λ h 2 ( t ) g ( β ( t ) ) = g ( M ϕ 1 ) φ p ( ϕ 2 ( t ) ) + λ h 2 ( t ) g ( M ϕ 1 ( t ) ) = λ h 2 ( t ) [ g ( M ϕ 1 ( t ) ) g ( M ϕ 1 ) ] λ h 2 ( t ) [ g ( M ϕ 1 ) g ( M ϕ 1 ) ] = 0 .

Thus ( β , β ¯ ) is an upper solution of ( P λ ).

(Case III) g is bounded and f ( u ) as u .

Choose M > 0 such that ( f ( M ϕ 2 ) ) 1 p 1 ϕ 1 α , M > g 1 p 1 and M ϕ 2 α ¯ and let

( β , β ¯ ) = ( ( f ( M ϕ 2 ) ) 1 p 1 ϕ 1 , M ϕ 2 ) .

Then

φ p ( β ( t ) ) + λ h 1 ( t ) f ( β ¯ ( t ) ) = f ( M ϕ 2 ) φ p ( ϕ 1 ( t ) ) + λ h 1 ( t ) f ( M ϕ 2 ( t ) ) = λ h 1 ( t ) [ f ( M ϕ 2 ( t ) ) f ( M ϕ 2 ) ] λ h 1 ( t ) [ f ( M ϕ 2 ) f ( M ϕ 2 ) ] = 0 .

And

φ p ( β ¯ ( t ) ) + λ h 2 ( t ) g ( β ( t ) ) = M p 1 φ p ( ϕ 2 ( t ) ) + λ h 2 ( t ) g ( ( f ( M ϕ 2 ) ) 1 p 1 ϕ 1 ( t ) ) = λ h 2 ( t ) [ g ( ( f ( M ϕ 2 ) ) 1 p 1 ϕ 1 ( t ) ) M p 1 ] λ h 2 ( t ) [ g M p 1 ] 0 .

Consequently, by Theorem 2.2, ( P λ ) has a solution satisfying

( α , α ¯ ) ( u , v ) ( β , β ¯ ) .

□

Lemma 3.5Assume h i A B , i = 1 , 2 , ( f 1 ), ( f 2 ) and ( f 3 ). Then there exists λ ¯ > 0 such that if ( P λ ) has a positive solution ( u , v ) , then λ λ ¯ .

Proof Let ( u , v ) be a positive solution of ( P λ ). Without loss of generality, we may assume λ < 1 . From ( f 1 ), we know that

lim x 0 + f ( ρ φ p 1 ( g ( u ) ) ) φ p ( u ) = 0 , for all ρ > 0 . (3.1)

From (3.1) and ( f 2 ), we can choose M f > 0 such that

f ( Q 2 φ p 1 ( g ( u ) ) ) M f φ p ( u ) , for all u > 0 , (3.2)

where Q i = 0 1 2 φ p 1 ( s 1 2 h i ( τ ) d τ ) d s + 1 2 1 φ p 1 ( 1 2 s h i ( τ ) d τ ) d s , i = 1 , 2 . Using (3.2) and ( f 3 ), we have

u Q 1 φ p 1 ( λ f ( B λ ( u ) ) ) Q 1 φ p 1 ( λ f ( Q 2 φ p 1 ( λ g ( u ) ) ) ) Q 1 φ p 1 ( λ f ( Q 2 φ p 1 ( g ( u ) ) ) ) Q 1 φ p 1 ( λ M f φ p ( u ) ) Q 1 φ p 1 ( λ M f ) u .

Thus we have

λ ¯ 1 φ p ( Q 1 ) M f λ .

□

Lemma 3.6Assume h i A B , i = 1 , 2 , ( f 1 ), ( f 2 ) and ( f 3 ). Then for each R > 0 , there exists λ R > 0 such that for λ > λ R , ( P λ ) has a positive solution ( u , v ) with u > R and v > R .

Proof We know that if ( u , v ) satisfies u = A λ ( B λ ( u ) ) and v = B λ ( u ) , then ( u , v ) is a solution of ( P λ ). Since A λ , B λ : K K are completely continuous, A λ B λ : K K is also completely continuous. Given R > 0 , choose

λ R = max { φ p ( 2 R Γ 2 ) 1 g ( R 4 ) , φ p ( 2 R Γ 1 ) 1 f ( R 4 ) } ,

where Γ i = min t [ 1 4 , 3 4 ] { 1 4 t φ p 1 ( s t h i ( τ ) d τ ) d s + t 3 4 φ p 1 ( t s h i ( τ ) d τ ) d s } . Let Ω 1 = { u C [ 0 , 1 ] | u < R } . If u Ω 1 K , then for t [ 1 4 , 3 4 ] , u ( t ) 1 4 u 1 4 R . From the definition of B λ R ( u ) , we know that B λ R ( u ) ( σ u ) is the maximum value of B λ R ( u ) on [ 0 , 1 ] . If σ u [ 1 4 , 3 4 ] , then from the choice of λ R , we have

If σ u > 3 4 , then we have

B λ R ( u ) 1 4 3 4 φ p 1 ( s 3 4 λ R h 2 ( τ ) g ( R 4 ) d τ ) d s 1 2 Γ 2 φ p 1 ( λ R g ( R 4 ) ) R .

If σ u < 1 4 , then

B λ R ( u ) 1 4 3 4 φ p 1 ( 1 4 s λ R h 2 ( τ ) g ( R 4 ) d τ ) d s 1 2 Γ 2 φ p 1 ( λ R g ( R 4 ) ) R .

By the concavity of B λ R ( u ) , we get for t [ 1 4 , 3 4 ] ,

B λ R ( u ) ( t ) 1 4 B λ R ( u ) 1 4 R . (3.3)

By similar argument as the above, with (3.3), we may show that

A λ R ( B λ R ( u ) ) 1 2 Γ 1 φ p 1 ( λ R f ( R 4 ) ) R = u .

Let Q i = 0 1 2 φ p 1 ( s 1 2 h i ( τ ) d τ ) d s + 1 2 1 φ p 1 ( 1 2 s h i ( τ ) d τ ) d s , i = 1 , 2 . For ε < 1 φ p ( Q 1 ) λ R , from ( f 2 ), we may choose R ˜ > 0 such that R ˜ > R and

f ( Q 2 φ p 1 ( λ R ) φ p 1 ( g ( R ˜ ) ) ) ε φ p ( R ˜ ) .

Let Ω 2 = { u C [ 0 , 1 ] | u < R ˜ } , then Ω 1 ¯ Ω 2 and for u Ω 2 K ,

A λ R ( B λ R ( u ) ) Q 1 φ p 1 ( λ R f ( Q 2 φ p 1 ( λ R ) φ p 1 ( g ( R ˜ ) ) ) ) Q 1 φ p 1 ( λ R ε φ p ( R ˜ ) ) Q 1 φ p 1 ( λ R ε ) R ˜ R ˜ = u .

By Proposition 3.2, ( P λ R ) has a positive solution ( u R , v R ) such that u R > R and v R > R . We know that ( u R , v R ) is a lower solution of ( P λ ) for λ > λ R and by Lemma 3.4, the proof is complete. □

We now prove one of the main results for this paper.

#### Proof of Theorem 1.2

From Lemma 3.6 and Lemma 3.5, we know that the set S = { λ > 0 | ( P λ ) has a positive solution } is not empty and λ = inf S > 0 . By Lemma 3.3 and complete continuity of T, there exist sequences { λ n } and { ( u n , v n ) } such that λ n λ and ( u n , v n ) ( u , v ) in K × K with ( u , v ) a solution of ( P λ ). We claim that ( u , v ) is a nontrivial solution of ( P λ ). Suppose that it is not true, then there exists a sequence of solutions ( u n , v n ) for ( P λ n ) such that ( u n , v n ) ( 0 , 0 ) and λ n λ . As in the proof of Lemma 3.3, we get

1 φ p ( Λ Q 1 ) f ( Λ Q 2 φ p 1 ( g ( u n ) ) ) φ p ( u n ) .

But from ( f 1 ), we have a contradiction to the fact that the right side of the above inequality converges to zero as u n 0 . Thus ( u , v ) is a nontrivial solution of ( P λ ). According to Lemma 3.4 and the definition of λ , we know that ( P λ ) has at least one positive solution at λ λ and no positive solution for λ < λ . To prove the existence of the second positive solution of ( P λ ) for λ > λ , we will use Theorem 1.1. Let λ > λ . Then we have ( α 1 , α ¯ 1 ) = ( 0 , 0 ) a lower solution of ( P λ ) and ( α 2 , α ¯ 2 ) = ( u , v ) a strict lower solution of ( P λ ) in C 0 1 [ 0 , 1 ] × C 0 1 [ 0 , 1 ] satisfying ( α 2 , α ¯ 2 ) ( α 1 , α ¯ 1 ) . For upper solutions, let λ 1 and μ 1 be the first eigenvalues of for i = 1 , 2 respectively and let e 1 and e 2 be corresponding eigenfunctions with e 1 = e 2 = 1