Journal of Applied Mathematics and Physics, 2014, 2, 903-909 Published Online August 2014 in SciRes. http://www.scirp.org/journal/jamp http://dx.doi.org/10.4236/jamp.2014.29102
Positive Solutions for Singular Boundary Value Problems of Coupled Systems of Nonlinear Differential Equations Ying He School of Mathematics and Statistics, Northeast Petroleum University, Daqing, China Email:
[email protected] Received 13 June 2014; revised 15 July 2014; accepted 26 July 2014 Copyright © 2014 by author and Scientific Research Publishing Inc. This work is licensed under the Creative Commons Attribution International License (CC BY). http://creativecommons.org/licenses/by/4.0/
Abstract We establish the existence of positive solutions for singular boundary value problems of coupled systems
− p t u′ ′ − q t u = f t , v + e t , ) 1( ) 1( ) 1( ( 1( ) ) ′ − ( p2 ( t ) v ′ ) − q2 ( t ) v = f 2 ( t , u ) + e2 ( t ) , α u ( 0 ) − β u′ ( 0 ) = 0, γ u (1) + δ u′ (1) = 0, 1 1 1 1 ′ ′ 0. α 2 v ( 0 ) − β 2 v ( 0 ) = 0, γ 2 v (1) + δ 2 v (1) = The proof relies on Schauder’s fixed point theorem. Some recent results in the literature are generalized and improved.
Keywords Positive Solutions, Second-Order Boundary Value Problems, Coupled Systems, Schauder’s Fixed Point Theorem
1. Introduction In this paper, we consider the existence of positive solutions for coupled singular system of second order ordinary differential equations How to cite this paper: He, Y. (2014) Positive Solutions for Singular Boundary Value Problems of Coupled Systems of Nonlinear Differential Equations. Journal of Applied Mathematics and Physics, 2, 903-909. http://dx.doi.org/10.4236/jamp.2014.29102
Y. He
− p t u ′ ′ − q t u = f t , v + e t , ) 1( ) 1( ) 1( ( 1( ) ) ′ − ( p2 ( t ) v′ ) − q2 ( t ) v = f 2 ( t , u ) + e2 ( t ) , α u ( 0 ) − β u ′ ( 0= ) 0, γ 1u (1) + δ1u ′ (1=) 0, 1 1 ′ − v 0 v 0 α β 2 ( ) 2 ( = ) 0, γ 2 v (1) + δ 2 v′ (1=) 0.
(1.1)
Throughout this paper, we always suppose that
( S1 )
pi ( t ) ∈ C1 ([ 0,1] , R ) , pi ( t ) > 0, qi ( t ) ∈ C ([ 0,1] , R ) , qi ( t ) ≤ 0, ei ( t ) ∈ C ([ 0,1] , R ) α i , βi , γ i , δ i ≥ 0,
and βi γ i + α i γ i + α= 1, 2 ) . f1 , f 2 ∈ C ([ 0,1] × ( 0, +∞ ) , ( 0, +∞ ) ) , and may be singular near the zero. iδ i > 0 ( i In recent years, singular boundary value problems to second ordinary differential equations have been studied extensively (see [1]-[3]). Some classical tools have been used in the literature to study the positive solutions for second order singular boundary value problems of a coupled system of differential equations. These classical methods include some fixed point theorems in cones for completely continuous operators and Schauder fixed point theorem, for example, see [4]-[6] and literatures therein. Motivated by the recent work on coupled systems of second-order differential equations, we consider the existence of singular boundary value problem. By means of the Schauder fixed point theorem, we study the existence of positive solutions of coupled system (1.1).
2. Preliminary We consider the scalar equation − ( p ( t ) u ′ )′ − q ( t ) u = e (t ).
(2.1)
α u ( 0 ) − β u ′ ( 0= ) 0, γ u (1) + δ u ′ (1=) 0,
(2.2)
with boundary conditions
Suppose that u is a positive solution of (2.1) and (2.2). Then
u ( t ) = ∫0G ( t , s ) e ( s ) ds. 1
where G ( t , s ) can be written by
G ( t , s ) := here I = [ 0,1] , Q= I × I and = Q1
1 m ( t ) n ( s ) , ω m ( s ) n ( t ) ,
( t , s ) ∈ Q1 , ( t , s ) ∈ Q2 .
{( t , s ) ∈ Q 0 ≤ t ≤ s ≤ 1} ,
Q = 2
{( t , s ) ∈ Q 0 ≤ s ≤ t ≤ 1} .
Lemma 2.1. Suppose that ( S1 ) holds, then the Green’s function G ( t , s ) , defined by (2.3) possesses the following properties: 1): m ( t ) ∈ C 2 ( I , R ) is increasing and m ( t ) > 0, x ∈ ( 0,1] . 2): m ( t ) ∈ C 2 ( I , R ) is decreasing and n ( t ) > 0, x ∈ [ 0,1) . 3): ( Lm )( t ) ≡ 0, m ( 0 ) = β , m′ ( 0 ) = α . 4): ( Ln )( t ) ≡ 0, n (1) = −γ . δ , n′ (1) = 5): ω is a positive constant. Moreover, p ( t ) ( m′ ( t ) n ( t ) − m ( t ) n′ ( t ) ) ≡ ω . 6): G ( t , s ) is continuous and symmetrical over Q . 7): G ( t , s ) has continuously partial derivative over Q1 , Q2 . 8): For each fixed s ∈ I , G ( t , s ) satisfies LG ( t , s ) = 0 for s ≠ t , t ∈ I . Moreover, R= R= 0 1 (G ) 2 (G ) for s ∈ ( 0,1) . 9): Gt′ has discontinuous point of the first kind at t = s and 1 − , Gt′ ( s + 0, s ) − Gt′ ( s − 0, s ) = p (s)
904
s ∈ ( 0,1) .
Y. He
We define the function γ i ( t ) : [ 0,1] → R by
= γ i (t )
G ( t , s ) ei ( s ) ds, i ∫= 0 i 1
1, 2,
which is the unique solution of
− p t u ′ t ′ − q t = ei ( t ) , 0 < t 0, for all ( t , s ) ∈ [ 0,1] × [ 0,1] ,
Following from Lemma 2.1 and
i= 1, 2
Let us fix some notation to be used in the following: For a given function h ∈ C [ 0,1] , we denote the essential supremum and infimum by h∗ and h∗ . if they exist. Let, γ i∗ = min γ i ( t ) , γ i* = max γ i ( t ) . t
t
3. Main Results 1) γ 1∗ ≥ 0 , γ 2∗ ≥ 0 . Theorem 3.1. We assume that there exists bi ≥ 0 , bˆi ≥ 0 , and 0 < α i < 1 such that bˆi ( t )
( H1 )
u
αi
≤ fi ( t , u ) ≤
bi ( t ) uαi
, for all u > 0, a.e. t ∈ ( 0,1) , i = 1, 2
If γ 1∗ ≥ 0 , γ 2∗ ≥ 0 , then there exists a positive solution of (1.1). Proof A positive solution of (1.1) is just a fixed point of the completely continuous map = A ( u , v ) ( A1u , A2 v ) : C [ 0,1] × C [ 0,1] → C [ 0,1] × C [ 0,1] defined as
= ( A1u )( t ) :
e1 ( s ) ds ∫ G1 ( t , s ) f1 ( s, v ( s ) ) ds + γ 1 ( t ) ; ∫0G1 ( t , s ) f1 ( s, v ( s ) ) += 0
= ( A2 v )( t ) :
e2 ( s ) ds ∫ G2 ( t , s ) f 2 ( s, u ( s ) ) ds + γ 2 ( t ) ; ∫0G2 ( t , s ) f 2 ( s, u ( s ) ) += 0
1
1
1
1
By a direct application of Schauder’s fixed point theorem, the proof is finished if we prove that A maps the closed convex set defined as
{( u, v ) ∈ C [0,1] × C [0,1] : r1 ≤ u ( t ) ≤ R1 , r2 ≤ v ( t ) ≤ R2 , for all t ∈ [0,1]}
K=
into itself, where R1 > r1 > 0 , R2 > r2 > 0 are positive constants to be fixed properly. For convenience, we introduce the following notations
= βi ( t ) Given
, s ) bi ( s ) ds, βˆi ( t ) ∫= G ( t , s ) bˆi ( s ) ds, i ∫0Gi ( t= 0 i 1
1
( u, v ) ∈ K , by the nonnegative sign of
= ( A1u )( t ) Note for every = ( A1u )( t )
1, 2.
Gi and fi , i = 1, 2 we have
∫ G1 ( t , s ) f1 ( s, v ( s ) ) ds + γ 1 ( t ) ≥ ∫ G1 ( t , s ) 1
1
0
0
bˆ1 ( s ) α1
v
ds ≥ ∫ G1 ( t , s ) 1
(s)
bˆ1 ( s )
0
α1
R2
1 ds ≥ βˆ1∗ α R2 1
( u, v ) ∈ K b1 ( s )
∫0G1 ( t , s ) f1 ( s, v ( s ) ) ds + γ 1 ( t ) ≤ ∫0G1 ( t , s ) vα ( s ) ds + γ 1 ≤ ∫0G1 ( t , s ) 1
1
∗
1
1
b1 ( s ) r2α1
ds + γ 1∗ ≤ β1∗
1 + γ 1∗ r2α1
Similarly, by the same strategy, we have
= ( A2 v )( t )
bˆ2 ( s )
∫0G2 ( t , s ) f 2 ( s, u ( s ) ) ds + γ 2 ( t ) ≥ ∫0G2 ( t , s ) uα ( s ) ds ≥ ∫0G2 ( t , s ) 1
1
2
905
1
bˆ2 ( s ) R1α 2
ds ≥ βˆ2∗
1 R1α 2
Y. He
= ( A2 v )( t )
Thus
b2 ( s )
∫0G2 ( t , s ) f 2 ( s, u ( s ) ) ds + γ 2 ( t ) ≤ ∫0G2 ( t , s ) uα ( s ) ds + γ 2 ≤ ∫0G2 ( t , s ) 1
1
∗
1
2
( A1u, A2 v ) ∈ K
b2 ( s ) r1α 2
ds + γ 2∗ ≤ β 2∗
1 + γ 2∗ r1α 2
if r1 , r2 , R1 , R2 are chosen so that 1 1 βˆ1∗ ⋅ α1 ≥ γ 1 , β1∗ ⋅ α1 + γ 1∗ ≤ R1 R2 r2
βˆ2∗ ⋅
1 1 ≥ γ 2 , β 2∗ ⋅ α + γ 2∗ ≤ R1 . α2 R1 r1 2
1 Note that βˆi∗ > 0 , βi∗ > 0 and taking R = R= R2 , r= r1= r2 , r = , it is sufficient to find R > 1 such 1 R that βˆ1∗ ⋅ R1−α1 ≥ 1, β1∗ ⋅ Rα1 + γ 1∗ ≤ R
βˆ2∗ ⋅ R1−α 2 ≥ 1, β 2∗ ⋅ Rα 2 + γ 2∗ ≤ R. and these inequalities hold for R big enough because α i < 1 . 2) γ 1∗ ≤ 0 , γ 2∗ ≤ 0 . The aim of this section is to show that the presence of a weak singular nonlinearity makes it possible to find positive solutions if γ 1∗ ≤ 0 , γ 2∗ ≤ 0 . Theorem 3.2. We assume that there exists bi ≥ 0 , bˆi ≥ 0 , and 0 < α i < 1 such that ( H1 ) is satisfied. If γ 1∗ ≤ 0 , γ 2∗ ≤ 0 and 1
1−α α βˆ1∗ 1 2 1 r1∗ ≥ α1α 2 ⋅ 1 − , ∗ α1 α1α 2 β 2
( )
βˆ2∗ r2∗ ≥ α1α 2 ⋅ α2 β1∗
( )
1 1−α1α 2
1 1 − . α1α 2
then there exists a positive solution of (1.1). Proof In this case, to prove that A : K → K it is sufficient to find 0 < r1 < R1 , 0 < r2 < R2 such that βˆ1∗ β1∗ r , + ≥ ≤ R1 γ ∗ 1 1 R2α1 r2α1
βˆ2∗
R1α 2
If we fix R1 =
β1∗
r2α1
, R2 =
β 2∗
r1α 2
β 2∗
+ γ 2∗ ≥ r2 ,
r1α 2
≤ R2
, then the first inequality of (3.3) holds if r2 satisfies
βˆ2∗ ( β1∗ )
−α 2
r2α1α 2 + γ 2∗ ≥ r2
or equivalently
r2 − γ 2∗ ≥ g ( r2 ) :=
βˆ2∗
(β ) ∗ 1
α2
r2α1α 2
The function g ( r2 ) possesses a minimum at
βˆ2∗ r20 : α1α 2 ⋅ = α2 β1∗
( )
Taking r2 = r20 , then (3.3) holds if
906
(3.1)
1
1−α1α 2
(3.2)
(3.3)
Y. He
γ 2∗
1
βˆ2∗ ≥ g ( r20 ) =α1α 2 ⋅ α2 β1∗
1−α1α 2
( )
1 1 − α 1α 2
Similarly,
r1 − γ 1∗ ≥ h ( r1 ) :=
βˆ1∗
r α1α 2 α1 1
(β ) ∗ 2
h ( r1 ) possesses a minimum at 1
1−α α βˆ ∗ 1 2 = r10 : α1α 2 ⋅ 1 α 1 β 2∗
( )
1
1−α α βˆ1∗ 1 2 1 γ 1∗ ≥ α1α 2 ⋅ 1 − α ∗ 1 α1α 2 β 2
( )
Taking r1 = r10 , r2 = r20 , then the first inequalities in (3.2) and (3.3) hold if γ 1∗ ≥ h ( r1 ) and γ 2∗ ≥ g ( r2 ) , which are just condition (3.1). The second inequalities hold directly from the choice of R1 and R2 , so it
β1∗
remains to prove that = R1
R1 =
β
r20α1
β
∗ 1
∗ 1
= r20α1 βˆ2∗ α1α 2 ⋅ ∗ α 2 β1
( ) 1
1−α1α 2 β1∗ = = α1 α α ⋅ βˆ 1 2 2∗
(
)
> r10 , = R2
= α1 1 α α − 1 2 1
β 2∗
r10α 2
> r20 This is easily verified through elementary computations:
β
∗ 1
=
βˆ2∗ α α ⋅ 1 2 β ∗ α2 1
( )
α1
1−α1α 2
(β ) ∗ 1
(
1+
α1α 2
1−α1α 2
α1α 2 ⋅ βˆ2∗
1
)
1 ∗ 1−α α 1 2 1
(β )
=
α1
1−α1α 2
(
α α ⋅ βˆ 1 2 2∗
)
α1
1
1−α1α 2
1
1−α α 1−α1α 2 βˆ1∗ 1 2 β1∗ 1 r10 , > α1α 2 ⋅ = α α α α1 ⋅ ˆ α1 ∗ 1 ( ) β β 1 2 ∗ 2 2
( )
( )
since βˆi∗ ≤ βi∗ , i = 1, 2 Similarly, we have R2 > r20 .
3) γ 1∗ ≥ 0, γ 2∗ ≤ 0
(γ
∗ 1
≤ 0, γ 2∗ ≥ 0
Theorem 3.3. Assume that
( H1 )
)
is satisfied. If γ 1∗ ≥ 0 , γ 2∗ ≤ 0 and
γ 2∗ ≥ r21 − βˆ2∗ ⋅
(β
∗ 1
r21α1α 2
)
(3.4)
= α1α 2 β1∗ βˆ2∗
(3.5)
+ γ 1∗ r21α1
α2
where 0 < r21 < +∞ is a unique positive solution of equation
(
r21−α1α 2 β1∗ + γ 1∗ ⋅ r2α1
)
1+α 2
then there exists a positive solution of (1.1). Proof We follow the same strategy and notation as in the proof of ahead theorem. In this case, to prove that A : K → K , it is sufficient to find r1 < R1 , r2 < R2 such that βˆ1∗ β 2∗ ≥ r1 , ≤ R2 (3.6) α1 R2 r1α 2
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Y. He
βˆ2∗
R1α 2
β 2∗
If we fix R2 =
β1∗
+ γ 2∗ ≥ r2 ,
r2α1
+ γ 1∗ ≤ R1
(3.7)
, then the first inequality of (3.6) holds if r1 satisfies
r1α 2
βˆ1∗
(β ) ∗ 2
α1
⋅ r1α1α 2 ≥ r1 ,
(3.8)
or equivalently 1
βˆ 1−α1α 2 ∗ 0 < r1 ≤ 1 α β∗ 1 2
(3.9)
( )
If we chose r1 > 0 small enough, then (3.9) holds, and R2 is big enough. If we fix = R1
β1∗
+ γ 1∗ then the first inequality of (3.7) holds if r2 satisfies
r2α1
γ 2∗ ≥ r2 −
βˆ2∗
α2
R1
=r2 − βˆ2∗ ⋅
1 α2
β1∗ ∗ α1 + γ 1 r2
=r2 − βˆ2∗ ⋅
1
=r2 − βˆ2∗ ⋅
α2
β1∗ + γ 1∗ ⋅ r2α1 r2α1
r2α1α 2
( β1∗ + γ 1∗ ⋅ r2α1 )
α2
,
or equivalently
γ 2∗ ≥ f ( r2 ) :=r2 − βˆ2∗ ⋅
r2α1α 2
(3.10)
( β1* + γ 1* ⋅ r2α1 )
α2
According to
f ′ ( r2 ) =1 − βˆ2∗ ⋅ =1 −
1
(β
∗ 1
α1
+ γ ⋅ r2 ∗ 1
)
2α 2
(
α α r α1α 2 −1 β ∗ + γ ∗ ⋅ r α1 1 1 2 1 2 2
)
(
α2
− r2α1α 2 α 2 β1∗ + γ 1∗ ⋅ r2α1
βˆ2∗ α1α 2 r2α1α 2 −1
(β
∗ 1
+ γ 1∗ ⋅ r2α1
)
α2
r2α1 γ 1∗ − =1 − α1α 2 β1∗ βˆ2∗ r2α1α 2 −1 β1∗ + γ 1∗ ⋅ r2α1 1 α1 ∗ ∗ + ⋅ β γ r 1 1 2
(
)
)
α 2 −1
−1−α 2
α1γ 1∗ r2α1 −1
,
we have f ′ ( 0 ) = −∞ , f ′ ( +∞ ) = 1 , then there exists r21 such that f ′ ( r21 ) = 0 , and
(
f ′′ ( r2 ) = − α1α 2 β1∗ βˆ2∗ (α1α 2 − 1) r2α1α 2 − 2 β1∗ + γ 1∗ ⋅ r2α1
(
+α1α 2 β1∗ βˆ2∗ r2α1α 2 −1 ( −1 − α 2 ) β1∗ + γ 1∗ ⋅ r2α1
)
)
−1−α 2
−2 −α 2
γ 1∗α1r2α1 −1 > 0.
Then the function f ( r2 ) possesses a minimum at r21 , i.e., f ( r21 ) = min r2 ∈( 0, +∞ ) f ( r2 ) . Note f ′ ( r21 ) = 0 then we have
(
1 − α1α 2 β1∗ βˆ2∗ r21α1α 2 −1 β1∗ + γ 1∗ ⋅ r21α1
)
−1−α 2
= 0
or equivalently
(
r211−α1α 2 β1∗ + γ 1∗ ⋅ r21α1
)
1+α 2
α1α 2 β1∗ βˆ2∗ =
Taking r2 = r21 , then the first inequality in (3.7) holds if γ 2∗ ≥ f ( r21 ) , which is just condition (3.4). The second inequalities hold directly by the choice of R1 , and it would remain to prove that r21 < R2 and r10 < R1 . These inequalities hold for R2 big enough and r1 small enough.
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Y. He
Remark 1. In theorem 3.3 the right-hand side of condition (3.4) always negative, this is equivalent to proof that f ( r21 ) < 0 . This is obviously established through the proof of Theorem 3.3. Similarly, we have the following theorem. Theorem 3.4. Assume ( H1 ) is satisfied. If γ 1∗ ≤ 0 , γ 2∗ ≥ 0 and
γ 1∗ ≥ r11 − βˆ1∗ ⋅
(β
r11α1α 2 ∗ 2
+ γ 2∗ r11α 2
)
α1
,
where 0 < r11 < +∞ is a unique positive solution of the equation
(
r11−α1α 2 β 2∗ + γ 2∗ r1α 2
)
1+α1
= α1α 2 β 2∗ βˆ1∗ ,
then there exists a positive solution of (1.1).
Funding Project supported by Heilongjiang province education department natural science research item, China (12541076).
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