Electronic Journal of Differential Equations, Vol. 2009(2009), No. 163, pp. 1–9.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
THREE POSITIVE SOLUTIONS FOR A SYSTEM OF SINGULAR
GENERALIZED LIDSTONE PROBLEMS
JIAFA XU, ZHILIN YANG
Abstract. In this article, we show the existence of at least three positive
solutions for the system of singular generalized Lidstone boundary value problems
(−1)m x(2m) = a(t)f1 (t, x, −x00 , . . . , (−1)m−1 x(2m−2) , y, −y 00 ,
. . . , (−1)n−1 y (2n−2) ),
n (2n)
(−1) y
= b(t)f2 (t, x, −x00 , . . . , (−1)m−1 x(2m−2) , y, −y 00 ,
. . . , (−1)n−1 y (2n−2) ),
a1 x(2i) (0) − b1 x(2i+1) (0) = c1 x(2i) (1) + d1 x(2i+1) (1) = 0,
a2 y (2j) (0) − b2 y (2j+1) (0) = c2 y (2j) (1) + d2 y (2j+1) (1) = 0.
The proofs of our main results are based on the Leggett-Williams fixed point
theorem. Also, we give an example to illustrate our results.
1. Introduction
In this article, we study the existence positive solutions for the following system
of singular generalized Lidstone boundary value problems
(−1)m x(2m) = a(t)f1 (t, x, −x00 , . . . , (−1)m−1 x(2m−2) , y, −y 00 , . . . , (−1)n−1 y (2n−2) ),
(−1)n y (2n) = b(t)f2 (t, x, −x00 , . . . , (−1)m−1 x(2m−2) , y, −y 00 , . . . , (−1)n−1 y (2n−2) ),
a1 x(2i) (0) − b1 x(2i+1) (0) = c1 x(2i) (1) + d1 x(2i+1) (1) = 0
(i = 0, 1, . . . , m − 1),
a2 y (2j) (0) − b2 y (2j+1) (0) = c2 y (2j) (1) + d2 y (2j+1) (1) = 0
(j = 0, 1, . . . , n − 1).
(1.1)
where m, n ≥ 1, a(t), b(t) ∈ C((0, 1), [0, +∞)), a(t) and b(t) are allowed to be
singular at t = 0 and/or t = 1; fi ∈ C([0, 1] × Rm+n
, R+ )(R+ := [0, +∞));
+
ai , bi , ci , di ∈ R+ with ρi := ai ci + ai di + bi ci > 0, i = 1, 2.
Singular boundary value problems for ordinary differential equation describe
many phenomena in applied mathematics and physical sciences, which can be found
2000 Mathematics Subject Classification. 34A34, 34B18, 45G15, 47H10.
Key words and phrases. Singular generalized Lidstone problem; positive solution; cone;
concave functional.
c
2009
Texas State University - San Marcos.
Submitted October 7, 2009. Published December 21, 2009.
Supported by grants 10871116 and 10971179 from the NNSF of China.
1
2
J. XU, Z. YANG
EJDE-2009/163
in the theory of nonlinear diffusion generated by nonlinear sources and in the thermal ignition of gases, see [1, 4]. Very recently, increasing attention is paid to
question of positive solutions for systems of second-order or higher order singular
differential equations, see for example [2, 5, 6, 7] and references therein.
In [2], by using Leggett-Williams fixed point theorem [3], Kang et al obtained
at least three positive solutions to the following singular nonlocal boundary value
problems for systems of nonlinear second-order ordinary differential equations.
u00 (t) + a1 (t)f1 (t, u(t), v(t)) = 0,
0 < t < 1,
00
v (t) + a2 (t)f2 (t, u(t), v(t)) = 0, 0 < t < 1,
Z 1
Z
0
0
α1 u(0) − β1 u (0) = 0, γ1 u(1) + δ1 u (1) = g1 (
u(s)dφ1 (s),
α2 v(0) − β2 v 0 (0) = 0,
0
Z 1
γ2 v(1) + δ2 v 0 (1) = g2 (
1
v(s)dφ1 (s)),
0
Z 1
u(s)dφ2 (s),
0
v(s)dφ2 (s)),
0
(1.2)
where ai ∈ C((0, 1), R+ ) is allowed to be singular at t = 0 or t = 1; fi ∈ C([0, 1] ×
R+ × R+ , R+ ) and gi ∈ C([0, 1] × R+ × R+ , R+ ) are such that a1 (t)f1 (t, 0, 0) or
a2 (t)f2 (t, 0, 0) does not vanish identically on any subinterval of (0, 1); αi ≥ 0,
R1
βi ≥ 0, γi ≥ 0, δi ≥ 0, and ρi = αi γi + αi δi + βi γi > 0; 0 u(s)dφ1 (s) and
R1
v(s)dφ1 (s) denote the Riemann-Stieltjes integrals, i = 1, 2.
0
Motivated by [2], we deal with the system of singular generalized Lidstone problems (1.1). To overcome the difficulties of (1.1) resulting from the derivatives of
even orders, as in [8], we use the method of order reduction to transform (1.1)
into an equivalent system of integro-integral equations, then prove the existence
of positive solutions for the resulting system, thereby establishing that of positive
solutions for (1.1)(see the main result in Section 3). The features of this paper
mainly include the following aspects. Firstly, our study is on systems of singular
generalized Lidstone problems. Secondly, a and b are allowed to be singular at t = 0
and/or t = 1. Finally, the system contains two equations, which can be of different
orders. Thus the results presented here are different from those in [2, 5, 6, 7].
The remaining of this paper is organized as follows: Section 2 gives some preliminary results. The main result is stated and proved in Section 3, then followed
by an example to illustrate the validity of our main result.
2. Preliminaries
Given a cone K in a real Banach space E, a map α is said to be a nonnegative
continuous concave functional on K provided that α : K → [0, +∞) is continuous
and
α(tx + (1 − t)y) ≥ tα(x) + (1 − t)α(y)
for all x, y ∈ K and 0 ≤ t ≤ 1.
Let 0 < a < b be given and let α be a nonnegative continuous concave functional
on K. Define the convex sets Pr and P (α, a, b) by
Pr := {x ∈ K : kxk < r},
P (α, a, b) := {x ∈ K : a ≤ α(x), kxk ≤ b}.
To prove our main result, we need the following Leggett-Williams fixed point theorem.
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THREE POSITIVE SOLUTIONS
3
Lemma 2.1 (see [3]). Let T : Pc → Pc be a completely continuous operator and let
α be a nonnegative continuous concave functional on K such that α(x) ≤ kxk for
all x ∈ Pc . Suppose there exist 0 < a < b < d ≤ c such that
(C1) {x ∈ P (α, b, d) : α(x) > b} =
6 ∅, and α(T x) > b for x ∈ P (α, b, d),
(C2) kT xk < a for kxk ≤ a, and
(C3) α(T x) > b for x ∈ P (α, b, c) with kT xk > d.
Then T has at least three fixed points x1 , x2 , and x3 such that kx1 k < a, b < α(x2 )
and kx3 k > a with α(x3 ) < b.
For fixed nonnegative constants a, b, c, d with ρ := ac + ad + bc > 0, we let
(
1 (b + as)(c + d − ct), 0 ≤ s ≤ t ≤ 1,
(2.1)
k(t, s) :=
ρ (b + at)(c + d − cs), 0 ≤ t ≤ s ≤ 1,
By definition, k ∈ C([0, 1] × [0, 1], R+ ) has the following properties:
(i) k(t, s) ≤ k(s, s) for all t, s ∈ [0, 1].
θc+d
(ii) For any θ ∈ (0, 1/2), there exists γ ∈ (0, min{ θa+b
a+b , c+d }] such that
k(t, s) ≥ γk(s, s), ∀t ∈ [θ, 1 − θ], s ∈ [0, 1].
R1
Lemma 2.2. Let f ∈ C[0, 1], h(t) ∈ C(0, 1) and 0 k(s, s)h(s)ds < +∞. The
boundary value problem
−u00 = h(t)f (t),
au(0) − bu0 (0) = 0,
cu(1) + du0 (1) = 0,
has a unique solution
Z
1
u(t) =
k(t, s)h(s)f (s)ds,
0
where k(t, s) is given by (2.1).
Let
(
(b1 + a1 s)(c1 + d1 − c1 t),
(b1 + a1 t)(c1 + d1 − c1 s),
(
1 (b2 + a2 s)(c2 + d2 − c2 t),
g1 (t, s) :=
ρ2 (b2 + a2 t)(c2 + d2 − c2 s),
1
k1 (t, s) :=
ρ1
For i, j = 2, . . . , define
Z 1
ki (t, s) :=
k1 (t, τ )ki−1 (τ, s)dτ,
Z
gj (t, s) :=
0
0 ≤ s ≤ t ≤ 1,
0 ≤ t ≤ s ≤ 1,
(2.2)
0 ≤ s ≤ t ≤ 1,
0 ≤ t ≤ s ≤ 1.
(2.3)
1
g1 (t, τ )gj−1 (τ, s)dτ
0
and the operators Ai : C[0, 1] → C[0, 1] and Bj : C[0, 1] → C[0, 1] by
Z 1
(Ai u)(t) :=
ki (t, s)u(s)ds, i = 1, 2, . . . , m,
0
Z
(Bj v)(t) :=
1
gj (t, s)v(s)ds,
0
j = 1, 2, . . . , n.
(2.4)
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J. XU, Z. YANG
EJDE-2009/163
For 1 ≤ i ≤ m − 1 and 1 ≤ j ≤ n − 1, let
Z 1
ξi := min
ki (t, s)ds,
θ≤t≤1−θ
ξ :=
ηi := max
1
0
1
Z
min
gj (t, s)ds,
θ≤t≤1−θ
ki (t, s)ds,
0≤t≤1
0
Z
µj :=
1
Z
νj := max
0≤t≤1
0
min
1≤i≤m−1,1≤j≤n−1
{ξi , µj },
η :=
gj (t, s)ds,
0
max
1≤i≤m−1,1≤j≤n−1
{ηi , νj }.
3. Main result
Let u(t) = (−1)m−1 x(2m−2) and v(t) = (−1)n−1 y (2n−2) . It is easy to see that
(1.1) is equivalent to the system of integro-ordinary differential equations
Z 1
Z 1
−u00 (t) = a(t)f1 t,
km−1 (t, s)u(s)ds, . . . ,
k1 (t, s)u(s)ds, u(t),
0
Z
0
1
Z
1
gn−1 (t, s)v(s)ds, . . . ,
0
0
Z
−v 00 (t) = b(t)f2 t,
1
Z
1
km−1 (t, s)u(s)ds, . . . ,
0
Z
g1 (t, s)v(s)ds, v(t) ,
k1 (t, s)u(s)ds, u(t),
0
1
Z
gn−1 (t, s)v(s)ds, . . . ,
0
1
g1 (t, s)v(s)ds, v(t) ,
0
subject to the boundary conditions
a1 u(0) − b1 u0 (0) = c1 u(1) + d1 u0 (1) = 0,
a2 v(0) − b2 v 0 (0) = c2 v(1) + d2 v 0 (1) = 0.
Furthermore, the above system is equivalent to the system
Z 1
Z 1
Z 1
u(t) =
k1 (t, s)a(s)f1 s,
km−1 (s, τ )u(τ )dτ, . . . ,
k1 (s, τ )u(τ )dτ, u(s),
Z
0
1
0
0
1
Z
gn−1 (s, τ )v(τ )dτ, . . . ,
0
Z
v(t) =
0
Z
0
1
Z
g1 (t, s)b(s)f2 s,
1
Z
0
1
1
km−1 (s, τ )u(τ )dτ, . . . ,
k1 (s, τ )u(τ )dτ, u(s),
0
Z
gn−1 (s, τ )v(τ )dτ, . . . ,
0
g1 (s, τ )v(τ )dτ, v(s) ds,
1
g1 (s, τ )v(τ )dτ, v(s) ds.
0
(3.1)
Let E := C([0, 1], R) × C([0, 1], R) endowed with the norm k(u, v)k := kuk + kvk,
where kuk := max0≤t≤1 |u(t)|, and define the cone K ⊂ E by
n
o
K := (u, v) ∈ E : u(t) ≥ 0, v(t) ≥ 0, t ∈ [0, 1], min (u(t) + v(t)) ≥ γk(u, v)k .
θ≤t≤1−θ
Clearly, (E, k · k) is a real Banach space and P is a cone on E. Define the operator
T : K → K by
T (u, v)(t) := (T1 (u, v)(t), T2 (u, v)(t)),
EJDE-2009/163
THREE POSITIVE SOLUTIONS
5
where
1
Z
k1 (t, s)a(s)f1 s, (Am−1 u)(s), . . . , (A1 u)(s),
0
u(s), (Bn−1 v)(s), . . . , (B1 v)(s), v(s) ds,
Z 1
T2 (u, v)(t) :=
g1 (t, s)b(s)f2 s, (Am−1 u)(s), . . . , (A1 u)(s),
0
u(s), (Bn−1 v)(s), . . . , (B1 v)(s), v(s) ds.
T1 (u, v)(t) :=
m+n
Now f ∈ C([0, 1] × R+
, R+ ) and g ∈ C([0, 1] × Rm+n
, R+ ) imply that T : K →
+
K is a completely continuous operator. In our setting, the existence of positive
solutions for (3.1) is equivalent to that of positive fixed points of T .
Lemma 3.1. The operator T maps K into K.
Proof. If (u, v) ∈ K, then
(T1 (u, v)(t))00
= −a(t)f1 t, (Am−1 u)(t), . . . , (A1 u)(t), u(t), (Bn−1 v)(t), . . . , (B1 v)(t), v(t)
≤ 0,
(T2 (u, v)(t))00
= −b(t)f2 t, (Am−1 u)(t), . . . , (A1 u)(t), u(t), (Bn−1 v)(t), . . . , (B1 v)(t), v(t)
≤ 0.
So T1 (u, v)(t) and T2 (u, v)(t) are concave on [0,1]. If (u, v) ∈ K, then from the
properties of k1 (t, s) and g1 (t, s), we have
Z
1
Z
1
k1 (s, s)a(s)f1 s, (Am−1 u)(s), . . . , (A1 u)(s),
0
u(s), (Bn−1 v)(s), . . . , (B1 v)(s), v(s) ds,
kT1 (u, v)(t)k ≤
and
g1 (s, s)b(s)f2 s, (Am−1 u)(s), . . . , (A1 u)(s),
0
u(s), (Bn−1 v)(s), . . . , (B1 v)(s), v(s) ds.
kT2 (u, v)(t)k ≤
On the other hand,
Z
min
θ≤t≤1−θ
1
k1 (s, s)a(s)f1 s, (Am−1 u)(s), . . . , (A1 u)(s), u(s),
0
(Bn−1 v)(s), . . . , (B1 v)(s), v(s) ds
T1 (u, v)(t) ≥ γ
≥ γkT1 (u, v)(t)k,
6
J. XU, Z. YANG
EJDE-2009/163
and
Z
min
θ≤t≤1−θ
1
g1 (s, s)b(s)f2 s, (Am−1 u)(s), . . . , (A1 u)(s), u(s),
0
(Bn−1 v)(s), . . . , (B1 v)(s), v(s) ds
T2 (u, v)(t) ≥ γ
≥ γkT2 (u, v)(t)k.
Combining the preceding inequalities, we arrive at
min (T1 (u, v)(t) + T2 (u, v)(t)) ≥
θ≤t≤1−θ
min (T1 (u, v)(t)) +
θ≤t≤1−θ
min (T2 (u, v)(t))
θ≤t≤1−θ
≥ γ(kT1 (u, v)(t)k + kT2 (u, v)(t)k).
This completes the proof.
In this paper, we use the following assumptions:
((H1) a(t) and b(t) do not vanish identically on any subinterval of (0, 1), and there
R1
exists t0 ∈ (0, 1) such that a(t0 ) > 0, b(t0 ) > 0 and 0 < 0 k1 (s, s)a(s)ds <
R1
+∞, 0 < 0 g1 (s, s)b(s)ds < +∞.
Finally, we define the nonnegative continuous concave functional
α(u, v) :=
min (u(t) + v(t)).
θ≤t≤1−θ
We observe here that, for each (u, v) ∈ K, α(u, v) ≤ k(u, v)k. Let
Z 1
Z 1
ξe1 := min
k1 (t, s)a(s)ds, ηe1 := max
k1 (t, s)a(s)ds,
θ≤t≤1−θ
µ
e1 :=
0≤t≤1
0
Z
min
1
g1 (t, s)b(s)ds,
θ≤t≤1−θ
0
0
Z
νe1 := max
0≤t≤1
1
g1 (t, s)b(s)ds .
0
Theorem 3.2. Let 1 ≤ i ≤ m − 1, and 1 ≤ j ≤ n − 1.Assume there exist nonnegative numbers a, b, c such that 0 < a < b ≤ min{γ, ξe1 /e
η1 , ξi /ηi , µ
e1 /e
ν1 , µj /νj }c,
and f (t, x1 , . . . , xm−1 , xm , y1 , . . . , yn−1 , yn ), g(t, x1 , . . . , xm−1 , xm , y1 , . . . , yn−1 , yn )
satisfy the following growth conditions:
(H2) f1 (t, x1 , . . . , xm−1 , xm , y1 , . . . , yn−1 , yn ) ≤ c/2e
η1 ,
f2 (t, x1 , . . . , xm−1 , xm , y1 , . . . , yn−1 , yn ) ≤ c/2e
ν1 , for all t ∈ [0, 1], xi + yj ∈
[0, ηc], xm + yn ∈ [0, c].
(H3) f1 (t, x1 , . . . , xm−1 , xm , y1 , . . . , yn−1 , yn ) < a/2e
η1 ,
f2 (t, x1 , . . . , xm−1 , xm , y1 , . . . , yn−1 , yn ) < a/2e
ν1 , for all t ∈ [0, 1], xi + yj ∈
[0, ηa], xm + yn ∈ [0, a].
(H4) f1 (t, x1 , . . . , xm−1 , xm , y1 , . . . , yn−1 , yn ) ≥ b/2ξe1 ,
f2 (t, x1 , . . . , xm−1 , xm , y1 , . . . , yn−1 , yn ) ≥ b/2e
µ1 , for all t ∈ [θ, 1 − θ], xi +
yj ∈ [ξb, ηb/γ], xm + yn ∈ [b, b/γ].
Then (1.1) has at least three positive fixed points (u1 , v1 ), (u2 , v2 ), (u3 , v3 ) such
that k(u1 , v1 )k < a, b < minθ≤t≤1−θ (u2 (t) + v2 (t)), and k(u3 , v3 )k > a, with
minθ≤t≤1−θ (u3 (t) + v3 (t)) < b.
Proof. We note first that T : Pc → Pc is a completely continuous operator. If
(u, v) ∈ Pc , then k(u, v)k ≤ c. For 1 ≤ i ≤ m − 1 and 1 ≤ j ≤ n − 1,
Z 1
Z 1
(Ai u)(t) + (Bj v)(t) =
ki (t, s)u(s)ds +
gj (t, s)v(s)ds ≤ ηi kuk + νj kvk ≤ ηc.
0
0
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THREE POSITIVE SOLUTIONS
7
From (H2), we have
kT (u, v)(t)k = max |T1 (u, v)(t)| + max |T2 (u, v)(t)|
0≤t≤1
0≤t≤1
Z 1
= max
k1 (t, s)a(s)f1 s, (Am−1 u)(s), . . . , (A1 u)(s),
0≤t≤1 0
u(s), (Bn−1 v)(s), . . . , (B1 v)(s), v(s) ds
Z 1
+ max
g1 (t, s)b(s)f2 s, (Am−1 u)(s), . . . , (A1 u)(s),
0≤t≤1 0
u(s), (Bn−1 v)(s), . . . , (B1 v)(s), v(s) ds
Z 1
Z 1
g1 (t, s)b(s)ds ≤ c.
k1 (t, s)a(s)ds + c/2e
ν1 max
≤ c/2e
η1 max
0≤t≤1
0≤t≤1
0
0
Therefore, T : Pc → Pc . In an analogous argument, one can verify that (H3) implies
condition (C2) of Lemma 2.1. Clearly, {(u, v) ∈ P (α, b, b/γ) : α(u, v) > b} =
6 ∅. If
(u, v) ∈ P (α, b, b/γ), then b ≤ u(t) + v(t) ≤ b/γ, t ∈ [θ, 1 − θ].
For 1 ≤ i ≤ m − 1 and 1 ≤ j ≤ n − 1, we have
Z 1
Z 1
(Ai u)(t) + (Bj v)(t) =
ki (t, s)u(s)ds +
gj (t, s)v(s)ds ≥ ξb,
0
1
Z
(Ai u)(t) + (Bj v)(t) =
Z
gj (t, s)v(s)ds ≤ ηb/γ.
ki (t, s)u(s)ds +
0
0
1
0
By (H4),
α(T (u, v)(t)) =
≥
min (T1 (u, v)(t) + T2 (u, v)(t))
θ≤t≤1−θ
min
T1 (u, v)(t) + min T2 (u, v)(t)
θ≤t≤1−θ
Z 1
≥ min
k1 (t, s)a(s)f1 s, (Am−1 u)(s), . . . , (A1 u)(s),
θ≤t≤1−θ 0
u(s), (Bn−1 v)(s), . . . , (A1 v)(s), v(s) ds
Z 1
+ min
g1 (t, s)b(s)f2 s, (Am−1 u)(s), . . . , (A1 u)(s),
θ≤t≤1−θ 0
u(s), (Bn−1 v)(s), . . . , (B1 v)(s), v(s) ds
Z 1
Z 1
e
≥ b/2ξ1 min
k1 (t, s)a(s)ds + b/2e
µ1 min
g1 (t, s)b(s)ds
θ≤t≤1−θ
θ≤t≤1−θ
0
θ≤t≤1−θ
0
= b.
Therefore, condition (C1) of Lemma 2.1 is satisfied.
Finally, we show that (C3) is also satisfied. If (u, v) ∈ P (α, b, c) and kT (u, v)k >
b/γ, then
α(T (u, v)(t)) =
min (T1 (u, v)(t) + T2 (u, v)(t)) ≥ γkT (u, v)k > b.
θ≤t≤1−θ
Therefore, (C3) of Lemma 2.1 is also satisfied. By Lemma 2.1, the system (1.1) has
at least three positive fixed points (u1 , v1 ), (u2 , v2 ), (u3 , v3 ) such that k(u1 , v1 )k < a,
8
J. XU, Z. YANG
EJDE-2009/163
b < minθ≤t≤1−θ (u2 (t)+v2 (t)), and k(u3 , v3 )k > a, with minθ≤t≤1−θ (u3 (t)+v3 (t)) <
b.
An example. Consider a system of nonlinear second-order and fourth-order ordinary differential equations (with m = 1, n = 2):
−x00 = a(t)f1 (t, x, y, −y 00 ) = 0,
y (4) = b(t)f2 (t, x, y, −y 00 ) = 0,
x(0) − x0 (0) = x(1) + x0 (1) = 0,
0
(3.2)
0
y(0) − y (0) = y(1) + y (1) = 0,
00
y (0) − y 000 (0) = y 00 (1) + y 000 (1) = 0.
Let u := x and v := −y 00 , then the problem (3.2) is equivalent to the following
system of nonlinear integral equations
Z 1
Z 1
u(t) =
k1 (t, s)a(s)f1 (s, u(s),
g1 (s, τ )v(τ )dτ, v(s)),
0
0
(3.3)
Z 1
Z 1
v(t) =
g1 (t, s)b(s)f2 (s, u(s),
0
g1 (s, τ )v(τ )dτ, v(s)),
0
where
1
k1 (t, s) = g1 (t, s) :=
3
(
(1 + s)(2 − t), 0 ≤ s ≤ t ≤ 1,
(1 + t)(2 − s), 0 ≤ t ≤ s ≤ 1.
We choose a(t) := (1 − t)−1/2 , b(t) := t−1/2 ,
 t
1
+ 10
(x1 + x3 )2 ,


 100

2
t
2
100 + 6[(x1 + x3 ) − 2(x1 + x3 )] + 5 ,
t

+ 20 log2 (x1 + x3 ) + 2(x1 + x3 ) + 25 ,


 100
t
1
542
100 + 4 (x1 + x3 ) + 5 ,
θ := 1/4, f1 (t, x1 , x2 , x3 ) equals
t ∈ [0, 1],
t ∈ [0, 1],
t ∈ [0, 1],
t ∈ [0, 1],
0 ≤ x1 + x3 ≤ 2, x2 ≥ 0,
2 < x1 + x3 < 4, x2 ≥ 0,
4 ≤ x1 + x3 ≤ 16, x2 ≥ 0,
x1 + x3 > 16, x2 ≥ 0.
and f2 (t, x1 , x2 , x3 ) equals
 t
1
+ 10
(x1 + x3 )2 ,
t ∈ [0, 1],


 100

t
13
2
2
t ∈ [0, 1],
100 + 8 [(x1 + x3 ) − 2(x1 + x3 )] + 5 ,
t
5
2

+
log
(x
+
x
)
+
2(x
+
x
)
+
,
t ∈ [0, 1],
3
1
3

2 1
2
5

 100
t
1
192
t ∈ [0, 1],
100 + 4 (x1 + x3 ) + 5 ,
0 ≤ x1 + x3 ≤ 2, x2 ≥ 0,
2 < x1 + x3 < 4, x2 ≥ 0,
4 ≤ x1 + x3 ≤ 16, x2 ≥ 0,
x1 + x3 > 16, x2 ≥ 0.
Then by direct calculation, we obtain
ξ ≈ 0.4120,
η = 0.5,
ξe1 ≈ 0.0417,
µ
e1 ≈ 0.9273,
ηe1 ≈ 0.8889,
νe1 ≈ 1.1111.
It is easy to check that (H1) holds. Choose γ = 41 , a = 1, b = 4, c = 800. Also, it is
easy to verify that f1 and f2 satisfy conditions (H2)-(H4). So the system (3.2) has
at least three positive solutions (u1 , v1 ), (u2 , v2 ), (u3 , v3 ) such that k(u1 , v1 )k < 1,
4 < min1/4≤t≤3/4 (u2 (t) + v2 (t)), and k(u3 , v3 )k > 1, with min1/4≤t≤3/4 (u3 (t) +
v3 (t)) < 4.
EJDE-2009/163
THREE POSITIVE SOLUTIONS
9
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Jiafa Xu
Department of Mathematics, Qingdao Technological University, No 11, Fushun Road,
Qingdao, China
E-mail address: xujiafa292@sina.com
Zhilin Yang
Department of Mathematics, Qingdao Technological University, No 11, Fushun Road,
Qingdao, China
E-mail address: zhilinyang@sina.com