Journal of Applied Mathematics & Bioinformatics, vol.5, no.2, 2015, 85-97
ISSN: 1792-6602 (print), 1792-6939 (online)
Scienpress Ltd, 2015
Positive solutions of singular (k,n-k) conjugate
eigenvalue problem
Shujie Tian 1 and Wei Gao 2
Abstract
Positive solution of singular nonlinear (k,n-k) conjugate eigenvalue problem is
studied by employing the positive property of the Green’s function, the fixed point
theorem of concave function and Krasnoselskii fixed point theorem in cone.
Mathematics Subject Classification: 34B18
Keywords: Eigenvalue; Positive solution; Fixed point theorem
1 Introduction
This paper deals with the following singular (k,n-k) conjugate eigenvalue
problem
(=
−1) n − k y ( n ) ( x) λ h( x) f ( y )
1
2
0 < x < 1,
(1.1)
School of Mathematics and Statistics, Northeast Petroleum University, Daqing 163318,
China. Ε-mail: outset2006@126.com
School of Mathematics and Statistics, Northeast Petroleum University, Daqing 163318,
China. Ε-mail: hit_gaowei@163.com
Article Info: Received : March 14, 2015. Revised : April 24, 2015.
Published online : June 20, 2015.
86
Positive solutions of singular (k,n-k) conjugate eigenvalue problem
(i )
( j)
y =
(0) 0, y =
(1) 0
0 ≤ i ≤ k − 1, 0 ≤ j ≤ n − k − 1,
(1.2)
Where 1 ≤ k ≤ n − 1 is a positive number and λ > 0 is a parameter.
If the conjugate eigenvalue problem (1.1), (1.2) has a positive solution y ( x)
for a particular λ , then λ is called an eigenvalue and y ( x) a corresponding
eigenfunction of (1.1), (1.2). Let Λ be the set of eigenvalue of the problem (1.1),
(1.2), i.e.
Λ = { λ > 0 ; (1.1), (1.2) has a positive solution}.
In recent years, the conjugate eigenvalue problem (1.1), (1.2) has been studied
extensively, for special case of λ = 1 , the existence results of positive solution of
the problem (1.1), (1.2) has been established in [1-6], and as for twin positive
solutions, several studies to the problem (1.1), (1.2) can be found in [7-9]. For the
case of λ > 0 , eigenvalue intervals characterizations of the problem (1.1), (1.2)
has been discussed in [10] by using Krasnoselskii fixed point theorem if f ( y ) is
superlinear or sublinear. In this paper, by employing property of Green function,
the fixed point theorem of concave function and Krasnoselskii fixed point theorem
in cone, we give eigenvalue characterizations under different hypothesis condition,
and we may allow that f ( x) is singular at x = 0,1 . By using different method
from [10] we establish not only existence of positive solution but also multiplicity
of positive solutions of the problem (1.1), (1.2).
Our assumptions throughout are:
(H1) h( x) is a nonnegative measurable function defined in (0,1) and do not
vanish identically on any subinterval in (0,1) and
1
1
0
0
0 < ∫ s n − k (1 − s ) k h( s )ds < ∫ s n − k −1 (1 − s ) k −1 h( s )ds < +∞ ;
(H2)
f :[0, +∞) → [0, +∞) is
a
nondecreasing
and f ( y ) > 0 for y > 0 ;
f ( y)
f ( y)
(H
=
lim
==
0 , f ∞ lim
= 0.
3) f 0
y →0
y →∞
y
y
continuous
function
Shujie Tian and Wei Gao
87
By a positive solution y ( x ) of the problem (1.1), (1.2), we means that y ( x)
satisfies
(a) y ( x) ∈ Ck −1[0,1)  Cn − k −1 (0,1]  Cn −1 (0,1) , f ( y ) > 0 in (0,1) and (1.2)
holds;
(b) y ( n −1) ( x) is locally absolutely continuous in (0,1) and
(−1) n − k y ( n ) ( x) =
λ h( x) f ( y ( x)) a.e. in (0,1).
The main results of this paper are as follows.
Theorem 1 Assume that (H1), (H2) and (H3) hold. Then there exist two positive
numbers λ∗ , λ ∗ with 0 < λ∗ ≤ λ ∗ < +∞ such that
(i).
(1.1),(1.2) has no positive solution for λ ∈ (0, λ∗ ) ;
(ii). (1.1),(1.2) has at least one positive solution for λ ∈ (λ∗ , λ ∗ ] ;
(iii). (1.1),(1.2) has at least two positive solutions for λ ∈ (λ ∗ , +∞) ;
(iv). (1.1),(1.2) has nonnegative solution for λ = λ∗ .
2 Preliminary Notes
In this section, we provide some properties of the Green’s function for the
problem (1.1),(1.2) which are needed later, and state the fixed point theorems
required. As shown in [6], the problem (1.1),(1.2) is equivalent to the integral
equation
1
y ( x) = λ ∫ G ( x, s )h( s ) f ( y ( s ))ds,
0
where
(2.1)
88
Positive solutions of singular (k,n-k) conjugate eigenvalue problem
s (1− x )
1

t n − k −1 (t + x − s ) k −1 dt , 0 ≤ s ≤ x ≤ 1,
 (k − 1)!(n − k − 1)! ∫0

G ( x, s ) = 
x (1− s )
1

t k −1 (t + s − x) n − k −1 dt , 0 ≤ x ≤ s ≤ 1.
∫
 (k − 1)!(n − k − 1)! 0
(2.2)
Moreover, the following results have been offered by Kong and Wang [6].
Lemma 2.1 For any x, s ∈ [0,1] , we have
α ( x) g ( s ) ≤ G ( x, s ) ≤ β ( x) g ( s ),
(2.3)
 ng ( s )
, 0 ≤ s ≤ x ≤ 1,
∂G ( x, s )  1 − s
≤
∂x
 ng ( s ) , 0 ≤ x ≤ s ≤ 1.
 s
(2.4)
where
α ( x) =
x k (1 − x) n − k
x k −1 (1 − x) n − k −1
s n − k (1 − s ) k
, β ( x) =
, g ( x) =
.
n −1
min{k , n − k}
(k − 1)!(n − k − 1)!
Let K be a cone in Banach space E . We say that a mapΨ is a nonnegative
continuous concave function on K , if it satisfied:Ψ : K → [0, +∞) is continuous
and
Ψ (α x + (1 − α ) y ) ≥ αΨ ( x ) + (1 − α )Ψ ( y )
for all x, y ∈ K and 0 ≤ α ≤ 1 .
Theorem 2 [12] Le K be a cone in Banach space E . For given R > 0 , define
K R =∈
{u K ; u < R} . Assume that T : K R → K R is a completely continuous
operator and Ψ is a nonnegative continuous concave function on Κ such
thatΨ ( y ) ≤ y for all y ∈ K R . Suppose that there exist 0 < a < b ≤ R such that
(A)
{ y ∈ K (Ψ, a, b);Ψ ( y ) > a} ≠ φ , andΨ (Ty ) > a for all y ∈ K (Ψ, a, b) where
K (Ψ, a, b=
)
(B) Ty < r for all y ∈ K r ;
{ y ∈ K ;Ψ ( y ) ≥ a;
y ≤ b} ;
Shujie Tian and Wei Gao
89
(C) Ψ (Ty ) > a for all y ∈ K (Ψ, a, R) with Ty > b ,
then T has at least three fixed points y1 , y2 and y3 in K R satisfying y1 ∈ K r ,
y2 ∈ { y ∈ K (Ψ, a, R);Ψ ( y ) > a} and y3 ∈ K R \ ( K (Ψ, a, R) ∪ K r ) .
Theorem 3 [13] Let E be a Banach space, and K ⊆ E a cone in E .
Assume Ω1 , Ω 2 are
open
subset
of E with 0 ∈ Ω1 , Ω1 ⊂ Ω 2 ,
and
let Φ : K ∩ (Ω 2 \ Ω1 ) → K be a completely continuous operator such that either
(I) Φu ≤ u for u ∈ K ∩ ∂Ω1 , and Φu ≥ u for u ∈ K ∩ ∂Ω 2 ; or
(II) Φu ≥ u for u ∈ K ∩ ∂Ω1 , and Φu ≤ u for u ∈ K ∩ ∂Ω 2 ,
then Φ has a fixed point in K ∩ (Ω 2 \ Ω1 ) .
3 Main Results
Let α = min α ( x) , β = max β ( x) and γ =
1 3
x∈[ , ]
4 4
x∈[0,1]
α
. Define the cone in Banach space
β
C[0,1] given as
{ y ( x) C[0,1]; y ( x) ≥ 0} ,
P =∈
K=
{ y ( x) ∈ C[0,1]; min y ( x) ≥ γ y } .
1 3
x∈ , 
4 4
We define the operator T : P → P by
1
(Ty )( x) := λ ∫ G ( x, s )h( s ) f ( y ( s ))ds .
0
(3.1)
Lemma 3.1 Suppose that (H1)-(H3) hold. Then T : P → P is a completely
continuous mapping and T(K) ⊂ K . Moreover, for y ∈ K we have
(Ty )( x) ∈ C k -1[0,1)  C n-k -1 (0,1]  C n-1 (0,1) ,
90
Positive solutions of singular (k,n-k) conjugate eigenvalue problem
(n)
(−1) n − k (Ty )=
( x) λ h( x) f ( y ( x)) a.e. x ∈ (0,1) ,
(Ty )(i ) (0)= 0, (Ty )( j ) (1)= 0, 0 ≤ i ≤ k − 1, 0 ≤ j ≤ n − k − 1 .
Proof We only prove T ( K ) ⊂ K . The proof of the remainder of Lemma 3.1 can
be found in [6].
For y ∈ K , by employing (2.3) we have
1
min (Ty )( x) ≥ λ min α ( s ) ∫ g ( s )h( s ) f ( y ( s ))ds
1 3
x∈[ , ]
4 4
1 3
x∈[ , ]
4 4
≥
0
1
λα
max ∫ G ( x, s )h( s ) f ( y ( s ))ds
β x∈[0,1] 0
= γ Ty ,
□
this implies T ( K ) ⊂ K .The proof is complete.
It follows from the lemma 3.1 that we know that T ( K ) ⊂ K and fixed point in
K of T is a solution of the problem (1.1),(1.2) and vice versa.
Lemma
3.2
Suppose
that
(H1),
(H2)
and
(H3) hold.
Then
there
exists 0 < λ ∗ < +∞ such that the problem (1.1), (1.2) has at least two positive
solutions for λ ∈ (λ ∗ , +∞) .
Proof It follows from lim
y →+∞
all y ≥ R1 ,
f ( y)
= 0 that there exists R1 > 0 such that f ( y ) ≤ ε y for
y
1
where ε satisfies ελβ ∫ g ( s )h( s )ds < 1 .
then f ( y ) ≤ M + ε y
0
for all y ≥ 0 . By (H3) we get lim
y →0
thus, there exists 0 < b < +∞
exists 0 < λ < +∞ such that
∗
have
such that
Let M = max f ( y ) ,
0 ≤ y ≤ R1
y
y
= lim
= +∞ ,
f ( y ) y →+∞ f ( y )
b
y
, and hence there
= min
f (b) y ≥0 f ( y )
ab 34
−1
λ =[
1 g ( s ) h( s )ds ] . Clearly, for all y ≥ 0 we
∫
f (b) 4
∗
Shujie Tian and Wei Gao
91
f ( y) ≥
b
y.
f (b)
(3.2)
We shall now show that the conditions of Theorem 2 are satisfied. Choose
1
1
0
0
R=
max{b + 1, 2 M λβ (1 − ελβ ∫ g ( s )h( s )ds ) −1 ∫ g ( s )h( s )ds} . For y ∈ K R we have
1
(Ty )( x) ≤ λ max ∫ G ( x, s )h( s ) f ( y ( s ))ds
x∈[0,1] 0
1
≤ λ max β ( x) ∫ g ( s )h( s )( M + ε y ( s ))ds
x∈[0,1]
0
1
≤ λβ ( M + ε y ) ∫ g ( s )h( s )ds
0
1
=
λβ ( M + ε R) ∫ g ( s )h( s )ds < R ,
0
this shows T ( K R ) ⊂ K R ⊂ K R .
LetΨ : K → [ 0, +∞ ) be defined by
Ψ ( y ) = min
y ( x) ,
1 3
x∈[ , ]
4 4
clearly,Ψ is a nonnegative continuous concave function on K such thatΨ ( y ) ≤ y
for all y ∈ K .
It is noted that
y ( x=
)
1
(γ b + b) ∈ { y ∈ K (Ψ , γ b, b);Ψ ( y ) ≥ γ b} ≠ φ ,
2
let y ∈ K (Ψ , γ b, b) , then min =
y ( x) Ψ ( y ) ≥ γ b and y ≤ y ≤ b . Using this together
1 3
x∈[ , ]
4 4
with (3.2), for λ > λ ∗ we get
Ψ (Ty )( x) = min
(Ty )( x)
1 3
x∈[ , ]
4 4
3
≥ λ min α ( x) ∫14 g ( s )h( s ) f ( y ( s ))ds
1 3
x∈[ , ]
4 4
≥
λα b
3
4
1
4
f (b) ∫
4
g ( s )h( s ) y ( s )ds
92
Positive solutions of singular (k,n-k) conjugate eigenvalue problem
≥
=
λαγ b 2
f (b)
3
4
1
4
∫
g ( s )h( s )ds
λγ b
> γb.
λ∗
Hence, condition (A) of Theorem 2 is satisfied.
By lim
y →0
f ( y)
= 0 , there exists 0 < r < γ b such that f ( y ) < ε y for all y ≤ r , where
y
1
ε satisfies ελβ ∫ g ( s )h( s )ds < 1 . For y ∈ K r we have
0
1
Ty ≤ λβ ∫ g ( s )h( s ) f ( y ( s ))ds ≤ ελβ y
0
∫
1
0
g ( s )h( s )ds < r ,
this implies that condition (B) of Theorem 2 is satisfied.
Finally, for y ∈ K (Ψ , γ b, b) with Ty > b , we obtain
Ψ (Ty )( x) = min
(Ty )( x)
1 3
x∈[ , ]
4 4
1
≥ λ min α ( x) ∫ g ( s )h( s ) f ( y ( s ))ds
1 3
x∈[ , ]
4 4
≥
0
1
αλ
max ∫ G ( x, s )h( s ) f ( y ( s ))ds
β x∈[0,1] 0
= γ Ty > γ b ．
Therefore, the condition (C) of Theorem 2 is also satisfied. Consequently, an
application of Theorem 2 shows that the problem (1.1), (1.2) has at least three
solutions y1 , y2 , y3 ∈ K R .
Further,
y1 ∈ K r , y2 ∈ { y ∈ K (Ψ,γ b, R );Ψ ( y ) > γ b} and y3 ∈ K R \ ( K (Ψ,γ b, R) ∪ K r ) . This
shows that y2 ( x) and y3 ( x) are two positive solution of the problem (1.1), (1.2), and
y1 ( x) is a nonnegative solution of (1.1), (1.2).
□
Lemma 3.3 Suppose that (H1), (H2) and (H3) hold. If λ is sufficiently small,
Shujie Tian and Wei Gao
93
then λ ∉ Λ .
Proof If λ ∈ Λ , then the problem (1.1),(1.2) has a positive solution yλ ( x) ∈ K and
it satisfies (2.1). We note that (H3) implies the existence of a constant η > 0 such
that f ( y ) ≤ η y for all y ≥ 0 . By employing (2.1) we have
1
yλ ≤ λβ ∫ g ( s )h( s ) f ( yλ ( s ))ds
0
∫
≤ λβη yλ
1
0
g ( s )h( s )ds ,
1
this means λβη ∫ g ( s )h( s )ds ≥ 1 , which contradicts with λ sufficiently small. □
0
Lemma 3.4 Suppose that (H1), (H2) and (H3) hold. Then there exists a λ0 > 0 such
that [λ0 , +∞) ⊂ Λ .
Proof
3
Let K1 =∈
{ y K , y < 1} ,
choose λ0 = (α f (γ ))( ∫14 g ( s )h( s )ds )-1 ,
4
for y ∈ K ∩ ∂K1 and λ ∈ [ λ0 , +∞ ) , then min y ( x) ≥ γ y =
γ , by using (H1) and
1 3
x∈ , 
4 4
(H2) we have
3
Ty ≥ λ0 min α ( x) ∫14 g ( s )h( s ) f ( y ( s ))ds
1 3
x∈ , 
4 4
4
3
≥ λ0α f (γ ) ∫14 g ( s )h( s )ds
4
= 1=
y ,
i.e. Ty ≥ y for y ∈ K ∩ ∂K1 .
It follows from (H3) that there exist R0 > 1 such that f ( y ) ≤ ε y for all y ≥ R0 ,
1
where ε satisfies ελ ∗ β ∫ g ( s )h( s )ds < 1 . Let
0
1
= R0 + λ ∗ β max f ( y ) ∫ g ( s )h( s )ds ,
R
0 ≤ y ≤ R0
0
94
Positive solutions of singular (k,n-k) conjugate eigenvalue problem
and K R =∈
{ y K , y < R} ,then for y ∈ K ∩ ∂K R , we have
1
Ty ≤ λ ∗ β ∫ g ( s )h( s ) f ( y ( s ))ds
0
∫
≤ λ β[
∗
∫
g ( s )h( s ) f ( y ( s ))ds +
0 ≤ y ( s ) ≤ R0
g ( s )h( s ) f ( y ( s ))ds ]
R0 ≤ y ( s ) ≤ R
1
≤ λ ∗ β ( max f ( y ) + ε R0 ) ∫ g ( s )h( s )ds
0 ≤ y ≤ R0
0
1
< λ ∗ β max f ( y ) ∫ g ( s )h( s )ds +R0
0 ≤ y ≤ R0
= R=
0
y ,
i.e. Ty ≤ y for y ∈ K ∩ ∂K R .
In
view
of
the
□
Theorem
3,
we
know
that
T
has
a
fixed
point y ( x) in K ∩ ( K R \ K1 ) . That is to say, the integral equation (2.1) has at least
one positive solution y ( x) , and hence y ( x) is a positive solution of (k,n-k)
conjugate eigenvalue problem (1.1),(1.2).
Lemma 3.5 Suppose that (H1), (H2) and (H3) hold. Then the problem (1.1),(1.2)
has a nonnegative solution for λ = λ∗ .
Proof Without loss of generality, let {λn }∞n =1 be a monotone decreasing
sequence, lim λn = λ∗ , and { yλn ( x)}∞n =1 be corresponding positive solution sequence
n →∞
where λn ∈ Λ . We claim that { yλn ( x)}∞n =1 is uniformly bounded. If it is not true,
then lim yλn = +∞ . It follows from lim
n →∞
y →+∞
f ( y)
= 0 ; that there exists M > 0 such
y
1
that f ( y ) ≤ M + ε y for all y ≥ 0 , where ε satisfies ελβ ∫ g ( s )h( s )ds < 1 . So we get
0
1
yλn ≤ λn β ∫ g ( s )h( s ) f ( yλn ( s ))ds
0
Shujie Tian and Wei Gao
95
1
≤ λn β ( M + ε yλn ) ∫ g ( s )h( s )ds.
(3.3)
0
Let n → ∞ in (3.3) to yield
1
lim yλn ≤
n →∞
λβ ∫ g ( s )h( s )ds
0
1
1 − ελβ ∫ g ( s )h( s )ds
< +∞ ,
0
which is a contradiction. Thus, there exists a number L with 0 < L < +∞ such
that yλn ≤ L for all n .
It follows from (H1) and (2.4) that we have
|| yλ′ n ||≤ λn ∫
1
0
≤
≤
∂G ( x, s )
h( s ) f ( yλn ( s ))ds
∂x
λ ∗nf ( L)
(k − 1)!(n − k − 1)!
x
1
0
x
( ∫ s n − k (1 − s ) k −1 h( s )ds + ∫ s n − k −1 (1 − s ) k h( s )ds )
1
2nλ ∗ f ( L)
s n − k −1 (1 − s ) k −1 h( s )ds :=
Q,
∫
0
(k − 1)!(n − k − 1)!
this shown that { yλn ( x)}∞n =1 is equicontinuous. Ascoli-Arzela theorem claims
that { yλn ( x)}∞n =1 has a uniformly convergent subsequence, denoted again
by { yλn ( x)}∞n =1 ,
and { yλn ( x)}∞n =1 converges
to y ∗ (x) uniformly
on
[0,1].
Inserting yλn ( x) into (2.1) and letting n → ∞ , using the Lebesgue dominated
convergence theorem, we obtain
1
=
y ∗ ( x) λ ∗ ∫ G ( x, s )h( s ) f ( y ∗ ( s ))ds ≥ 0 ,
0
thus, y ∗ ( x) is a nonnegative solution of (1.1),(1.2).
□
Remark It is possible that y ∗ ( x) = 0 .
Let λ∗ = inf Λ ,
then
from
Lemma
3.3
and
lemma
3.4
we
know λ∗ > 0 and 0 < λ∗ ≤ λ ∗ < +∞ . So (k,n-k) conjugate eigenvalue problem
96
Positive solutions of singular (k,n-k) conjugate eigenvalue problem
(1.1),(1.2) has no positive solution for λ ∈ (0, λ∗ ) , has at least one positive solution
for λ ∈ (λ∗ , λ ∗ ] , has at least two positive solutions for λ ∈ (λ ∗ , +∞) , and has a
nonnegative solution for λ = λ∗ .
Acknowledgements This work is supported by science and technology research
projects of Heilongjiang Provincial Department of Education (12541076).
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