Electronic Journal of Differential Equations, Vol. 2008(2008), No. 106, pp. 1–8.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu (login: ftp)
POSITIVE SOLUTIONS FOR MULTIPOINT BOUNDARY-VALUE
PROBLEM WITH PARAMETERS
JUANJUAN XU, ZHONGLI WEI
Abstract. In this paper, we study a generalized Sturm-Liouville boundaryvalue problems with two positive parameters. By constructing a completely
continuous operator and combining fixed point index theorem and some properties of the eigenvalues of linear operators, we obtain sufficient conditions for
the existence of at least one positive solution.
1. Introduction
Multipoint boundary-value problems for ordinary differential equations arise in
different areas of applied mathematics and physics. For example, the vibrations of
a guy wire of uniform cross-section and composed of N parts of different densities
can be set up as a multipoint boundary-value problem; many problem in the theory of elastic stability can be handled as multipoint boundary-value problems too.
Recently, the existence and multiplicity of positive solutions for nonlinear ordinary
differential equations have received a great deal of attention. To identify a few
cases, we refer the readers to [5, 9, 10, 11] and references therein.
Li [4] studied the following boundary-value problem (BVP for short):
u(4) (t) + βu00 − αu = f (t, u(t)),
00
0 < t < 1,
00
u(0) = u(1) = u (0) = u (1) = 0,
(1.1)
where the function f ∈ C([0, 1] × [0, +∞), [0, +∞)), α, β ∈ R and satisfy β <
2
2π 2 , α ≥ − β4 , πα4 + πβ2 < 1. By applications of the fixed point index theory,
sufficient conditions for existence of at least one positive solution are established.
2000 Mathematics Subject Classification. 34B15, 39A10.
Key words and phrases. Multipoint; positive solution; eigenvalue; parameters.
c
2008
Texas State University - San Marcos.
Submitted May 2, 2008. Published August 7, 2008.
Supported by grants 10771117 from the National Natural Science Foundation of China,
and 306001 from the Foundation of School of Mathematics, Shandong University.
1
2
J. XU, Z. WEI
EJDE-2008/106
Ma [6] studied the existence of positive solution for BVP:
u(4) (t) + αu00 − βu = f (t, u(t)),
u(0) =
m−2
X
αi u(ξi ),
u(1) =
i=1
u00 (0) =
m−2
X
βi u(ξi ),
(1.2)
i=1
αi u00 (ξi ),
u00 (1) =
i=1
2
0 < t < 1,
m−2
X
m−2
X
βi u00 (ξi ),
i=1
2
− α4 ,
αi , βi , ξi > 0 (i = 1, 2, . . . , m − 2) are
where α, β ∈ R and α < 2π , β ≥
constants, and f ∈ C([0, 1] × [0, +∞), [0, +∞)). The main tool is also the fixed
point index theory.
Motivated by the results mentioned above, we are concerned with the existence
of at least one positive solution for the following generalized Sturm-Liouville BVP:
u(4) (t) − βu00 + αu = f (t, u(t)),
au(0) − bu0 (0) =
m−2
X
αi u(ξi ),
cu(1) + du0 (1) =
i=1
au00 (0) − bu000 (0) =
m−2
X
0 < t < 1,
m−2
X
βi u(ξi ),
i=1
αi u00 (ξi ),
cu00 (1) + du000 (1) =
i=1
m−2
X
(1.3)
βi u00 (ξi ),
i=1
where f ∈ C([0, 1]×[0, +∞), [0, +∞)) satisfying f (t, u) 6≡ 0 and α, β ≥ 0, a, b, c, d ∈
[0, +∞) and ρ := ac + bc + ad > 0, ξi ∈ (0, 1), αi , βi ∈ [0, +∞) (i = 1, 2, . . . , m − 2)
are constants.
To study (1.3), we set up an integral equation which is equivalent to (1.3). By
using the classical fixed point index theorem and combining some knowledge about
eigenvalue of linear operator, we obtain a sufficient condition for the existence of
at least one positive solution.
Following theorems are needed.
Theorem 1.1 ([3]). Let E be a Banach space, and let P ⊂ E be a cone. Assume
Ω(P ) is a bounded open set in P . Suppose that A : Ω(P ) → P is a completely
continuous operator. If there exists ψ0 ∈ P \{θ} such that ϕ − Aϕ 6= µψ0 , for all
ϕ ∈ ∂Ω(P ), µ ≥ 0, then the fixed point index satisfies i(A, Ω(P ), P ) = 0.
Theorem 1.2 ([3]). Let E be a Banach space, and let P ⊂ E be a cone. Assume
Ω(P ) is a bounded open set in P with θ ∈ Ω(P ). Suppose that A : Ω(P ) → P is a
completely continuous operator. If Aψ 6= µψ, for all ψ ∈ ∂Ω(P ), µ ≥ 1, then the
fixed point index satisfies i(A, Ω(P ), P ) = 1.
We shall organize this paper as follows. In Section 2, we present some preliminaries and lemmas for use later. Finally, we obtain the main result and state the
proof.
2. Preliminaries
In this section, we state some useful preliminary results and change the BVP (1.3)
into the fixed point problem in a cone. First, we state the following hypothesis to
assumed in this paper.
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POSITIVE SOLUTIONS
3
(H1) α, β ≥ 0 and α ≤ β 2 /4.
Remark 2.1. From (H1), it follows that
α
π4
β
π2
+
> −1.
Lemma 2.2. Under assumption (H1) there exist unique ϕ1 , ϕ2 , ψ1 , ψ2 satisfying
−ϕ00i (t) + λi ϕi = 0,
ϕi (0) = b,
00
−ψi (t) + λi ψi
0 < t < 1,
ϕ0i (0)
= 0,
ψi0 (1)
ψi (1) = d,
= a,
0 < t < 1,
= −c,
for i = 1, 2. Also on [0, 1], ϕ1 , ϕ2 , ψ1 , ψ2 ≥ 0, where λ1 , λ2 are the roots for the
polynomial equation λ2 − βλ + α = 0; i.e.,
p
p
β + β 2 − 4α
β − β 2 − 4α
λ1 =
, λ2 =
.
2
2
Moreover, ϕ1 , ϕ2 are nondecreasing on [0, 1] and ψ1 , ψ2 are nonincreasing on [0, 1].
Proof. From (H1), √
we have λ1 , λ2 ≥
√ 0. By computations we get that: If λi > 0,
then ϕi (t) = b cosh λi t + √aλ sinh λi t,
i
p
p
c
ψi (t) = d cosh λi (1 − t) + √ sinh λi (1 − t), (i = 1, 2);
λi
if λi = 0, then ϕi (t) = b + at, ψi (t) = d + c − ct, (i = 1, 2).
It is obvious that on [0, 1], ϕ1 , ϕ2 , ψ1 , ψ2 ≥ 0 and ϕ1 , ϕ2 are nondecreasing on
[0, 1], ψ1 , ψ2 are nonincreasing on [0, 1].
We denote
ψ1 (0) ϕ1 (0)
ψ2 (0) ϕ2 (0)
,
ρ1 = 0
, ρ2 = 0
ψ1 (0) ϕ01 (0)
ψ2 (0) ϕ02 (0)
P
Pm−2
− m−2 αi ϕ1 (ξi )
ρ1 − i=1 αi ψ1 (ξi )
i=1
P
P
∆1 = ,
m−2
m−2
ρ1 − i=1 βi ϕ1 (ξi )
− i=1 βi ψ1 (ξi ) P
Pm−2
− m−2 αi ϕ2 (ξi )
ρ2 − i=1 αi ψ2 (ξi )
i=1
Pm−2
Pm−2
∆2 = .
ρ −
β ϕ (ξ )
−
β ψ (ξ ) 2
i=1
i
2
i
i=1
i
2
i
Assume that
Pm−2
Pm−2
(H2) ∆1 < 0, ρ1 − i=1 αi ψ1 (ξi ) > 0, ρ1 − i=1 βi ϕ1 (ξi ) > 0;
Pm−2
Pm−2
(H3) ∆2 < 0, ρ2 − i=1 αi ψ2 (ξi ) > 0, ρ2 − i=1 βi ϕ2 (ξi ) > 0,
Similar to [8], we can get the following two lemmas by direct calculations.
Lemma 2.3. Let (H1)-(H2) hold. Then for any g ∈ C[0, 1], the problem
−u00 (t) + λ1 u(t) = g(t),
au(0) − bu0 (0) =
m−2
X
i=1
has a unique solution u(t) =
R1
αi u(ξi ),
0 < t < 1,
cu(1) + du0 (1) =
m−2
X
βi u(ξi ),
(2.1)
i=1
G1 (t, s)g(s) ds + A1 (g)ϕ1 (t) + B1 (g)ψ1 (t), where
(
1 ϕ1 (t)ψ1 (s), 0 ≤ t ≤ s ≤ 1,
G1 (t, s) =
ρ1 ϕ1 (s)ψ1 (t), 0 ≤ s ≤ t ≤ 1,
0
4
J. XU, Z. WEI
EJDE-2008/106
P
Pm−2
m−2 R 1
i=1 αi 0 G1 (ξi , s)g(s) ds ρ1 − i=1 αi ψ1 (ξi )
Pm−2
,
Pm−2 R 1
i=1 βi 0 G1 (ξi , s)g(s) ds
− i=1 βi ψ1 (ξi ) Pm−2 R 1
P
1 − m−2
α
G
(ξ
,
s)g(s)
ds
α
ϕ
(ξ
)
i
1
i
i
1
i
i=1
i=1
0
R
B1 (g) :=
P
P
,
1
m−2
m−2
∆1 ρ1 − i=1
βi ϕ1 (ξi )
β
G
(ξ
,
s)g(s)
ds
i 0
1 i
i=1
1
A1 (g) :=
∆1
and where g ≥ 0, u(t) ≥ 0, t ∈ [0, 1].
The proof of the above lemma follows by routine calculations.
Lemma 2.4. Let (H1), (H3) hold. Then for each g ∈ C[0, 1], the problem
−u00 (t) + λ2 u(t) = g(t),
au(0) − bu0 (0) =
m−2
X
i=1
has a unique solution u(t) =
αi u(ξi ),
0 < t < 1,
cu(1) + du0 (1) =
m−2
X
βi u(ξi ),
(2.2)
i=1
R1
G2 (t, s)g(s) ds + A2 (g)ϕ2 (t) + B2 (g)ψ2 (t), where
(
1 ϕ2 (t)ψ2 (s), 0 ≤ t ≤ s ≤ 1,
G2 (t, s) =
ρ2 ϕ2 (s)ψ2 (t), 0 ≤ s ≤ t ≤ 1,
P
R1
Pm−2
1 m−2
α
G
(ξ
,
s)g(s)
ds
ρ
−
α
ψ
(ξ
)
i 0
2
i 2 i i=1
R1 2 i
A2 (g) :=
Pm−2
Pi=1
,
m−2
∆2 i=1 βi 0 G2 (ξi , s)g(s) ds
− i=1 βi ψ2 (ξi ) P
Pm−2 R 1
1 − m−2
αi ϕ2 (ξi )
αi 0 G2 (ξi , s)g(s) ds
i=1
i=1
B2 (g) :=
Pm−2
Pm−2 R 1
,
∆2 ρ2 − i=1 βi ϕ2 (ξi )
i=1 βi 0 G2 (ξi , s)g(s) ds
0
and g ≥ 0, u(t) ≥ 0, t ∈ [0, 1].
The proof of the above lemma follows by routine calculations.
Remark 2.5. Suppose that (H2) and (H3) hold. It follows that Ai (g), Bi (g) (i =
1, 2) are increasing.
Lemma 2.6. Assume that (H1)–(H3) hold. Then (1.3) has a unique solution
Z 1Z 1
Z 1
u(t) =
G2 (t, τ )G1 (τ, s)f (s, u(s)) ds dτ +
G2 (t, τ )A1 (f )ϕ1 (τ ) dτ
0
0
0
(2.3)
Z 1
+
G2 (t, τ )B1 (f )ψ1 (τ ) dτ + A2 (h)ϕ2 (t) + B2 (h)ψ2 (t),
0
where G1 , G2 , A1 , A2 , B1 , B2 are defined as above,
Z 1
h(t) =
G1 (t, s)f (s, u(s)) ds + A1 (f )ϕ1 (t) + B1 (f )ψ1 (t).
0
Obviously, u(t) ≥ 0 for all t ∈ [0, 1]. Let E = C[0, 1] and P = {u ∈ E, u ≥ 0}.
It is obvious that P is a cone in E. Define T : E → E,
Z 1Z 1
Z 1
T u(t) =
G2 (t, τ )G1 (τ, s)f (s, u(s)) ds dτ +
G2 (t, τ )A1 (f )ϕ1 (τ ) dτ
0
0
0
(2.4)
Z 1
+
G2 (t, τ )B1 (f )ψ1 (τ ) dτ + A2 (h)ϕ2 (t) + B2 (h)ψ2 (t),
0
where h(t) =
R1
0
G1 (t, s)f (s, u(s)) ds + A1 (f )ϕ1 (t) + B1 (f )ψ1 (t).
EJDE-2008/106
POSITIVE SOLUTIONS
5
We can easily obtain that u is a positive solution of (1.3) if and only if u is a
fixed point of T in P .
Define L : E → E,
Z 1
Z 1Z 1
G2 (t, τ )A1 (u)ϕ1 (τ ) dτ
G2 (t, τ )G1 (τ, s)u(s) ds dτ +
Lu(t) =
0
0
0
(2.5)
Z 1
G2 (t, τ )B1 (u)ψ1 (τ ) dτ + A2 (e)ϕ2 (t) + B2 (e)ψ2 (t),
+
0
where e(t) =
R1
G1 (t, s)u(s) ds + A1 (u)ϕ1 (t) + B1 (u)ψ1 (t).
0
Lemma 2.7. Suppose that (H1)–(H3) hold. Then T : P → P is completely continuous. Also L : P → P is completely continuous.
Lemma 2.8. Suppose that (H1)–(H3) hold. Then for the operator L defined by
(2.5), the spectral radius r(L) 6= 0 and L has a positive eigenfunction corresponding
to its first eigenvalue λ∗ = r(L)−1 .
Proof. It is easy to see that there is t1 ∈ (0, 1), such that G1 (t1 , t1 )G2 (t1 , t1 ) >
0. Thus there exists [α, β] ⊂ (0, 1) such that t1 ∈ (α, β) and G1 (t, τ )G2 (τ, s) >
0, t, τ, s ∈ [α, β].
Take u ∈ E such that u(t) ≥ 0 for all t ∈ [0, 1], u(t1 ) > 0 and u(t) = 0 for all
t ∈ [0, 1]\[α, β]. Then for t ∈ [α, β],
Z 1Z 1
Z 1
Lu(t) =
G2 (t, τ )G1 (τ, s)u(s) ds dτ +
G2 (t, τ )A1 (u)ϕ1 (τ ) dτ
0
0
1
Z
+
Z
0
G2 (t, τ )B1 (u)ψ1 (τ ) dτ + A2 (e)ϕ2 (t) + B2 (e)ψ2 (t)
0
βZ β
≥
Z
β
G2 (t, τ )G1 (τ, s)u(s) ds dτ +
α
α
β
Z
+
G2 (t, τ )A1 (u)ϕ1 (τ ) dτ
α
G2 (t, τ )B1 (u)ψ1 (τ ) dτ + A2 (e)ϕ2 (t) + B2 (e)ψ2 (t) > 0.
α
So there exists a constant c > 0 such that for t ∈ [0, 1], c(Lu)(t) ≥ u(t). From
Krein-Rutmann Theorem [3], we know that the spectral radius r(L) 6= 0 and L has
a positive eigenfunction corresponding to its first eigenvalue λ∗ = r(L)−1 .
3. Main Result
Theorem 3.1. Suppose that (H1)–(H3) hold, and f0 > λ∗ , f∞ < λ∗ , where λ∗
is the first eigenvalue of L defined by (2.5). Then (1.3) has at least one positive
solution, where
f (t, u)
f (t, u)
f0 = lim inf
min
, f∞ = lim sup max
.
u
u
u→0+ t∈[0,1]
u→+∞ t∈[0,1]
Proof. From f0 > λ∗ , there exists r1 > 0, such that f (t, u) ≥ λ∗ u for all t ∈ [0, 1],
u ∈ [0, r1 ]. Let u ∈ ∂Br1 ∩ P . Then
Z 1Z 1
Z 1
T u(t) =
G2 (t, τ )G1 (τ, s)f (s, u(s)) ds dτ +
G2 (t, τ )A1 (f )ϕ1 (τ ) dτ
0
Z
+
0
1
0
G2 (t, τ )B1 (f )ψ1 (τ ) dτ + A2 (h)ϕ2 (t) + B2 (h)ψ2 (t)
0
6
J. XU, Z. WEI
1
Z
≥ λ∗ [
1
Z
1
Z
G2 (t, τ )G1 (τ, s)u(s) ds dτ +
+
G2 (t, τ )A1 (u)ϕ1 (τ ) dτ
0
0
1
Z
EJDE-2008/106
0
G2 (t, τ )B1 (u)ψ1 (τ ) dτ + A2 (e)ϕ2 (t) + B2 (e)ψ2 (t)]
0
= λ∗ (Lu)(t).
We may suppose that T has no fixed point on ∂Br1 ∩ P (otherwise, the proof is
complete). Now we show that u − T u 6= µu∗ for all u ∈ ∂Br1 ∩ P , µ ≥ 0.
Otherwise, there exists u1 ∈ ∂Br1 ∩ P , τ0 ≥ 0, such that u1 − T u1 = τ0 u∗ , that
is
u 1 = T u 1 + τ0 u ∗ .
Let τ ∗ = sup{τ : u1 ≥ τ u∗ }, then τ ∗ ≥ τ0 > 0, and u1 ≥ τ ∗ u∗ . Since L(P ) ⊂ P ,
λ∗ Lu1 ≥ τ ∗ λ∗ Lu∗ = τ ∗ u∗ , we have
u1 = T u1 + τ0 u∗ ≥ λ∗ Lu1 + τ0 u∗ ≥ (τ ∗ + τ0 )u∗ .
which contradicts the definition of τ ∗ , so i(T, Br1 ∩ P, P ) = 0.
From f∞ < λ∗ , there exits 0 < σ < 1, r2 > r1 , such that f (t, u) ≤ σλ∗ u for all
t ∈ [0, 1], u ∈ [r2 , +∞). Let L1 u = σλ∗ Lu, u ∈ E, then L1 : E → E is a bounded
linear operator and L1 (P ) ⊂ P . Let
Z 1Z 1
M∗ =
max
G2 (t, τ )G1 (τ, s)f (s, u(s)) ds dτ
u∈B r2 ∩P,t∈[0,1]
Z
+
0
0
1
Z
G2 (t, τ )A1 (f )ϕ1 (τ ) dτ +
0
1
G2 (t, τ )B1 (f )ψ1 (τ ) dτ
0
+ A2 (h)ϕ2 (t) + B2 (h)ψ2 (t),
obviously, 0 < M ∗ < +∞. Let W = {u ∈ P : u = µT u, 0 ≤ µ ≤ 1}, for all u ∈ W ,
d = min{u(t), r2 }, s(u) = {t ∈ [0, 1], u(t) > r2 }, fd
d Then
denote u(t)
(t) = f (t, u(t)).
u(t) = µT u(t) ≤ T u(t)
Z 1Z
Z
=
G2 (t, τ )G1 (τ, s)f (s, u(s)) ds dτ +
0
Z
s(u)
1
+
1
G2 (t, τ )A1s(u) (f )ϕ1 (τ ) dτ
0
G2 (t, τ )B1s(u) (f )ψ1 (τ ) dτ + A2 (hs(u) )ϕ2 (t) + B2 (hs(u) )ψ2 (t)
0
Z
1
Z
+
G2 (t, τ )G1 (τ, s)f (s, u(s)) ds dτ
0
Z
[0,1]/s(u)
1
+
G2 (t, τ )A1[0,1]/s(u) (f )ϕ1 (τ ) dτ
0
Z
1
G2 (t, τ )B1[0,1]/s(u) (f )ψ1 (τ ) dτ
+
0
+ A2 (h[0,1]/s(u) )ϕ2 (t) + B2 (h[0,1]/s(u) )ψ2 (t)
Z 1Z 1
Z 1
≤ σλ∗ [
G2 (t, τ )G1 (τ, s)u(s) ds dτ +
G2 (t, τ )A1 (u)ϕ1 (τ ) dτ
0
Z
+
0
0
1
G2 (t, τ )B1 (u)ψ1 (τ ) dτ + A2 (e)ϕ2 (t) + B2 (e)ψ2 (t)]
0
EJDE-2008/106
Z
POSITIVE SOLUTIONS
1
Z
1
Z
d ds dτ +
G2 (t, τ )G1 (τ, s)f (s, u(s))
+
0
0
Z
+
7
1
G2 (t, τ )A1 (fb)ϕ1 (τ ) dτ
0
1
G2 (t, τ )B1 (fb)ψ1 (τ ) dτ + A2 (fb)ϕ2 (t) + B2 (fb)ψ2 (t)
0
≤ (L1 u)(t) + M ∗ ,
t ∈ [0, 1].
where
1
A1s(u) (f ) :=
∆1
Pm−2 R
i=1 αi R s(u) G1 (ξi , s)g(s) ds
Pm−2
i=1 βi s(u) G1 (ξi , s)g(s) ds
Pm−2
ρ1 − i=1 αi ψ1 (ξi )
Pm−2
,
− i=1 βi ψ1 (ξi ) B1s(u) , A2[0,1]/s(u) , B2[0,1]/s(u) have the similar meaning and
Z
hs(u) (t) =
G1 (t, s)f (s, u(s)) ds + A1s(u) (f )ϕ1 (t) + B1s(u) (f )ϕ2 (t).
s(u)
Thus
(I − L1 )u ≤ M ∗ , t ∈ [0, 1].
Since u∗ = λ∗ (Lu∗ ) and 0 < σ < 1, we have r(L1 )−1 > 1; i.e., (I − L1 )−1 exists
and
(I − L1 )−1 = I + L1 + L21 + · · · + Ln1 + . . . .
It follows from L1 (P ) ⊂ P that (I−L1 )−1 (P ) ⊂ P . Therefore, u(t) ≤ (I−L1 )−1 M ∗ ,
t ∈ [0, 1], and W is bounded. We denote by sup W the bound of W .
Select r3 > max{r2 , sup W }, then for all u ∈ ∂Br3 ∩ P , u 6= µT u, 0 ≤ µ ≤ 1;
that is,
1
1
T u 6= u,
≥ 1, ∀u ∈ ∂Br3 ∩ P,
µ
µ
so from Theorem 1.2, we have i(T, Br3 ∩ P, P ) = 1. Therefore,
i(T, (Br3 ∩ P )\(B r1 ∩ P ), P ) = i(T, Br3 ∩ P, P ) − i(T, Br1 ∩ P, P ) = 1.
By the solution properties of the fixed point index, T has at least one fixed point on
(Br3 ∩ P )\(B r1 ∩ P ), which means that the generalized Sturm-Liouville boundaryvalue problem (1.3) has at least one positive solution.
Acknowledgements. The authors would like to thank the anonymous referees for
their kind suggestions and comments on this paper.
References
[1] Chai, Guoqing; Existence of positive solutions for second-order boundary-value problem with
one parameter, J. Math. Anal. Appl., 330 (2007): 541-549.
[2] Guo, Dajun; Sun, Jingxian; Nonlinear Integral Equations, Shandong Science and Technology
Press, Jinan, 1987. (in Chinese)
[3] Guo, D., Lakshmikantham,v.; Nonlinear Problems in Abstract Cones, Academic Press Inc.,
New York, 1988.
[4] Li, Yongxiang; Positive solutions of fourth-order boundary-value problems with two parameters, J. Math. Anal. Appl., 281 (2003): 477-484.
[5] Ma, Huili; Symmetric positive solutions for nonlacal boundary-value problem of fourth order,
Nonlinear Anal., 68(2008): 645-651.
[6] Ma, Huili; Positive solution for m-point boundary-value problem of fourth-order, J. Math.
Anal. Appl., 321(2006): 37-49.
[7] Ma, Ruyun; Nonlocal problems in nonlinear ordinary differential problems, Science press,
Beijing, 2004. (in Chinese)
[8] Ma, Ruyun, Thompson, Bevan; Positive solutions for nonlinear m-point eigenvalue problems,
J. Math. Anal. Appl., 297(2004): 24-37.
8
J. XU, Z. WEI
EJDE-2008/106
[9] Zhang, Xuemei; Feng, Meiqiang; Ge, Weigao; Multiple positive solutions for a class of mpoint boundary-value problems, Appl. Math. Lett. (2007), doi:10.1016/ j.aml.2007.10.019.
[10] Zhang, Xinguang; Liu, Lishan; Eigenvalue of fourth-order m-point boundary-value problems
with two deviatives, Comp. Math. Appl. (2008) doi:10.1016/ j.camwa.2007.08.048.
[11] Zhang, Mingchuan; Wei, Zhongli; Existence of positive solutions for fourth-order m-point
boundary-value problems with variable parameters, Appl. Math. Comp., 190 (2007): 14171431.
Juanjuan Xu
School of Mathematics, Shandong University, Jinan, Shandong 250100, China
E-mail address: jnxujuanjuan@163.com Tel: 86-531-88369649
Zhongli Wei
School of Mathematics, Shandong University, Jinan, Shandong 250100, China
E-mail address: jnwzl@yahoo.com.cn