Electronic Journal of Differential Equations, Vol. 2009(2009), No. 141, pp. 1–10.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
EXISTENCE OF POSITIVE SOLUTIONS FOR BOUNDARY
VALUE PROBLEMS OF SECOND-ORDER NONLINEAR
DIFFERENTIAL EQUATIONS ON THE HALF LINE
XINGQIU ZHANG
Abstract. In this article, we study the existence of positive solutions for
Sturm-Liouville boundary-value problems of a second-order nonlinear differential equation on the half line. Our approach is based on the fixed point
theorem and the monotone iterative technique. Without assuming the existence of lower and upper solution, we obtain the existence of positive solutions,
and establish iterative schemes for approximating the solutions.
1. Introduction
In this article, we prove the existence positive solutions, and establish a an iterative scheme for their approximation, for the following Sturm-Liouville boundary
value problem of second-order differential equation on the half line
x00 (t) + q(t)f (t, x(t), x0 (t)) = 0,
0
αx(0) − βx (0) = 0,
t ∈ J+ ,
0
x (∞) = x∞ ≥ 0,
(1.1)
where J = [0, +∞), J+ = (0, +∞), α > 0, β ≥ 0, x0 (∞) = limt→+∞ x0 (t) and
f (t, u, v) : J × J × J → J is continuous. Throughout this paper, we assume the
following conditions.
(H1) f (t, u, v) ∈ C(J × J × J, J), f (t, 0, 0) 6≡ 0 on any subinterval of J and, when
u, v are bounded, f (t, (1 + t)u, v) is bounded on J;
(H2) q(t) is a nonnegative measurable function defined in J+ and q(t) does not
identically vanish on any subinterval of J+ and
Z +∞
Z +∞
0<
q(t)dt < +∞, 0 <
tq(t)dt < +∞.
0
0
Boundary value problems on half-line arise quite naturally in the study of radially
symmetric solutions of nonlinear elliptic equations and models of gas pressure in
a semi-infinite porous medium; see for example [1, 6, 7, 11, 15]. In the past few
years, the existence and multiplicity of bounded or unbounded positive solutions to
2000 Mathematics Subject Classification. 34B10, 34B15, 34B18, 34B40.
Key words and phrases. Successive iteration; singular differential equation; half line;
positive solution.
c
2009
Texas State University - San Marcos.
Submitted October 22, 2008. Published October 30, 2009.
Supported by grant 10671167 from the National Natural Science Foundation of China,
and 31805 from the the Natural Science Foundation of Liaocheng University.
1
2
X. ZHANG
EJDE-2009/141
nonlinear differential equations on the half line have been studied by different type
of techniques, we refer the reader to [1, 3, 5, 6, 7, 8, 9, 11, 14, 15] and references
therein. Most of papers only considered the existence of positive solutions of various
boundary value problems. Seeing such a fact, we cannot but ask “How can we find
the solutions when they are known to exist?” More recently, Ma, Du and Ge [10],
Sun and Ge [12, 13] proved the existence of positive solutions for some second-order
p-Laplacian boundary value problems which are defined on finite intervals by virtue
of the iterative technique.
To the best of our knowledge, up to now, it seems that no results in the literature
are available for the computation of positive solutions for boundary value problems
on the half line. Motivated by above papers, the purpose of this paper is to fill this
gap. As we know, it is very important to check the compactness of the corresponding
operator when we use the monotone iterative technique and Ascoli-Arzela theorem
plays a very important role. However, Ascoli-Arzela theorem is not suitable for
operators on the half line. So, it is needed to list some new conditions to meet the
requirement of compactness.
2. Preliminaries
First, we give some definitions.
Definition 2.1. Let E be a real Banach space. A nonempty closed set P ⊂ E is
said to be a cone provided that
(1) au + bv ∈ P for all u, v ∈ P and all a ≥ 0, b ≥ 0;
(2) u, −u ∈ P implies u = 0.
A map α : P → [0, +∞) is said to be concave on P , if
α(tu + (1 − t)v) ≥ tα(u) + (1 − t)α(v)
for all u, v ∈ P and t ∈ [0, 1].
We will use the following space to study (1.1)
|x(t)|
< ∞,
E = x ∈ C 1 [0, +∞) : sup
t∈J t + 1
lim x0 (t) exists .
t→+∞
Then E is a Banach space equipped with the norm kxk = max{kxk1 , kx0 k∞ }, where
0
0
kxk1 = supt∈J |x(t)|
t+1 , kx k∞ = supt∈J |x (t)|. Let E+ = {x ∈ E : x(t) ≥ 0}. Define
the cone P ⊂ E by
P = x ∈ E : x(t) ≥ 0, t ∈ [0, +∞), x is concave on [0, +∞),
αx(0) − βx0 (0) = 0, and lim x0 (t) exists .
t→+∞
00
Remark 2.2. If x satisfies (1.1), then x (t) = −q(t)f (t, x(t), x0 (t)) ≤ 0 on [0, +∞),
which implies that x is concave on [0, +∞). Moreover if x0 (∞) = x∞ ≥ 0, then
x0 (t) ≥ 0, t ∈ [0, +∞) and so x is monotone increasing on [0, +∞).
Lemma 2.3. Assume that (H1)-(H2) hold. Then x ∈ E+ ∩ C 2 [J+ , J] is a solution
of (1.1) if and only if x ∈ C[J, E] is a solution of the integral equation
Z
β +∞
x(t) =
q(s)f (s, x(s), x0 (s))ds + x∞
α 0
(2.1)
Z t Z +∞
0
+
q(τ )f (τ, x(τ ), x (τ ))dτ ds + tx∞ .
0
s
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EXISTENCE OF POSITIVE SOLUTIONS
3
Proof. Suppose that x ∈ E+ ∩C 2 [J+ , J] is a solution of (1.1). For t ∈ J, integrating
(1.1) from t to +∞, we have
Z +∞
x0 (t) = x∞ +
f (s, x(s), x0 (s))ds.
(2.2)
t
Integrating from 0 to t,
Z tZ
+∞
q(τ )f (τ, x(τ ), x0 (τ ))dτ ds.
x(t) = x(0) + tx∞ +
(2.3)
s
0
Thus, by (2.2) we obtain
Z
0
+∞
f (s, x(s), x0 (s))ds,
x (0) = x∞ +
0
which together with the boundary value condition implies
Z +∞
β
x(0) =
x∞ +
f (s, x(s), x0 (s))ds .
α
0
(2.4)
Substituting the above expression into (2.3), we have
Z
β +∞
x(t) =
q(s)f (s, x(s), x0 (s))ds + x∞
α 0
Z t Z +∞
+
q(τ )f (τ, x(τ ), x0 (τ ))dτ ds + tx∞ .
0
s
R t R +∞
Next, we show that the integral 0 s q(τ )f (τ, x(τ ), x0 (τ ))dτ ds are convergent.
Since x ∈ E+ , then there exists r0 such that kxk < r0 . Set Br0 = sup{f (t, (1 +
t)u, v)|(t, u, v) ∈ J × [0, r0 ] × [0, r0 ]}, and we have by exchanging the integral order
Z t Z +∞
Z +∞ Z +∞
q(τ )f (τ, x(τ ), x0 (τ ))dτ ds ≤
q(τ )f (τ, x(τ ), x0 (τ ))dτ ds
0
s
0
s
(2.5)
Z +∞
≤
sq(s)ds · Br0 .
0
R t R +∞
By (H2), we know that 0 s q(τ )f (τ, x(τ ), x0 (τ ))dτ ds is convergent. Thus, we
have proved that the right term in (2.1) is well defined. Conversely, if x is a solution
of integral equation, then direct differentiation gives the proof.
Now, we define an operator A : P → C 1 [0, +∞) by
Z
β +∞
q(s)f (s, x(s), x0 (s))ds + x∞
(Ax)(t) =
α 0
Z t Z +∞
+
q(τ )f (τ, x(τ ), x0 (τ ))dτ ds + tx∞ .
0
(2.6)
s
To obtain the complete continuity of A, the following lemma is needed.
Lemma 2.4 ([2, 8]). Let W be a bounded subset of P . Then W is relatively compact
in E if {W (t)/(1+t)} and {W 0 (t)} are both equicontinuous on any finite subinterval
of [0, +∞) and for any ε > 0, there exists N > 0 such that
x(t )
x(t2 ) 1
−
< ε, |x0 (t1 ) − x0 (t2 )| < ε, ∀ t1 , t2 ≥ N,
1 + t1
1 + t2
4
X. ZHANG
EJDE-2009/141
uniformly with respect to x ∈ W as t1 , t2 ≥ N , where W (t) = {x(t)|x ∈ W },
W 0 (t) = {x0 (t)|x ∈ W }, t ∈ [0, +∞).
Lemma 2.5. Assume that (H1)-(H2) hold. Then A : P → P is completely continuous.
Proof. It is clear that (Ax)(t) ≥ 0 for any x ∈ P , t ∈ J. By (2.6), we have
Z +∞
0
(Ax) (t) =
q(s)f (s, x(s), x0 (s))ds + x∞ ≥ 0,
(2.7)
t
(Ax)00 (t) = −q(t)f (t, x(t)) ≤ 0.
(2.8)
These two inequalities imply that (T P ) ⊂ P . Now, we prove that A is continuous
and compact respectively. Let xn → x as n → ∞ in P , then there exists r0 such that
supn∈N \{0} kxn k < r0 . Let Br0 = sup{f (t, (1 + t)u, v)|(t, u, v) ∈ J × [0, r0 ] × [0, r0 ]}.
By (H2), we have
Z +∞
Z +∞
0
0
q(τ )|f (τ, xn (τ ), xn (τ )) − f (τ, x(τ ), x (τ ))|dτ ≤ 2Br0 ·
q(s)ds < +∞,
0
0
(2.9)
and
Z tZ
+∞
q(τ )|f (τ, xn (τ ), x0n (τ )) − f (τ, x(τ ), x0 (τ ))|dτ ds
0
s
Z +∞ Z +∞
≤
q(τ )|f (τ, xn (τ )) − f (τ, x(τ ), x0 (τ ))|dτ ds
0
s
Z +∞
≤ 2Br0
sq(s)ds < +∞.
(2.10)
0
It follows from (2.6), (2.9), (2.10), and the Lebesgue dominated convergence theorem that
n 1 β Z +∞
q(s)(f (s, xn (s), x0n (s)) − f (s, x(s), x0 (s)))ds
kAxn − Axk1 = sup
1+t α 0
t∈J
Z t Z +∞
o
+
q(τ )(f (τ, xn (τ ), x0n (τ )) − f (τ, x(τ ), x0 (τ )))dτ ds
0
s
o
n 1 β Z +∞
q(s)|f (s, xn (s), x0n (s)) − f (s, x(s), x0 (s))|ds
≤ sup
1+tα 0
t∈J
n Z t Z +∞
o
+ sup
q(τ )|f (τ, xn (τ )) − f (τ, x(τ ), x0 (τ ))|dτ ds → 0,
t∈J
0
s
as n → ∞. Also
0
0
k(Axn ) − (Ax) k∞ = sup
nZ
t∈J
Z
+∞
o
q(s)|f (s, xn (s), x0n (s)) − f (s, x(s), x0 (s))|ds
t
+∞
≤ 2Br0
q(s)ds < +∞.
0
Therefore, A is continuous.
Let Ω be any bounded subset of P . Then, there exists r > 0 such that kxk ≤ r
for any x ∈ Ω. Therefore, we have
kAxk1
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EXISTENCE OF POSITIVE SOLUTIONS
5
n 1 β Z +∞
q(s)f (s, x(s), x0 (s))ds + x∞
1+t α 0
t∈J
Z t Z +∞
o
q(τ )f (τ, x(τ ), x0 (τ ))dτ ds + tx∞ +
s
0
Z
n t Z +∞
o
β +∞
≤
q(s)ds · Br + x∞ + sup
q(s)f (s, x(s), x0 (s))ds + x∞
α 0
1+t 0
t∈J
β
Z +∞
≤
+1
q(s)ds · Br + x∞ ,
α
0
= sup
and
0
k(Ax) k∞
n Z
= sup t∈J
t
+∞
o Z
q(s)f (s, x(s), x (s))ds + x∞ ≤
0
+∞
q(s)ds · Br + x∞ .
0
R +∞ R +∞
So, T Ω is bounded. Remembering that the integral 0
q(τ )dτ ds is convers
gent, so for any T ∈ J+ and t1 , t2 ∈ [0, T ], by the absolute continuity of the integral,
we have
(Ax)(t ) (Ax)(t ) 1
2 −
1 + t1
1 + t2
Z
1
β +∞
1 ≤
q(s)f (s, x(s), x0 (s))ds + x∞ · −
α 0
1 + t1
1 + t2
1 Z t1 Z +∞
+
q(τ )f (τ, x(τ ), x0 (τ ))dτ ds
1 + t1 0
s
Z t2 Z +∞
1
−
q(τ )f (τ, x(τ ), x0 (τ ))dτ ds
1 + t2 0
s
t
t1 1
+
−
x∞
1 + t1
1 + t1
Z
Z Z
1
1 1 t2 +∞
β +∞
−
+
≤
q(s)ds · Br + x∞ q(τ )dτ ds · Br α 0
1 + t1
1 + t2
1 + t1 t1 s
Z t2 Z +∞
1
1 t1
t1 −
−
+
q(τ )dτ ds · Br +
x∞
1 + t1
1 + t2
1 + t1
1 + t1
0
s
which approaches zero, uniformly as t1 → t2 . Also
Z t2
Z
0
0
0
|(Ax) (t1 ) − (Ax) (t2 )| = q(s)f (s, x(s), x (s))ds ≤ Br · t1
t2
t1
q(s)ds
which approaches zero, uniformly as t1 → t2 . Thus, we have proved that T Ω is
equicontinuous on any finite subinterval of [0, +∞).
Next, we prove that for any ε > 0, there exits sufficiently large N > 0 such that
(Ax)(t ) (Ax)(t ) 2 1
−
(2.11)
< ε, |(Ax)0 (t1 ) − (Ax)0 (t2 )| < ε,
1 + t1
1 + t2
for all t1 , t2 ≥ N and all x ∈ Ω. For any x ∈ Ω, we have
Z t Z +∞
(Ax)(t) 1
q(τ )f (τ, x(τ ), x0 (τ ))dτ ds + x∞ .
lim = lim
t→∞
t→∞ 1 + t 0
1+t
s
(2.12)
6
X. ZHANG
EJDE-2009/141
Similar to (2.5), we get
Z t Z +∞
Z
q(τ )f (τ, x(τ ), x0 (τ ))dτ ds ≤ Br
0
s
+∞
τ q(τ )dτ < +∞,
0
which shows that
1
t→∞ 1 + t
+∞
Z tZ
q(τ )f (τ, x(τ ), x0 (τ ))dτ ds = 0.
lim
0
(2.13)
s
On the other hand, we arrive at
Z
0
+∞
q(s)f (s, x(s), x0 (s))ds + x∞
lim |(Ax) (t)| = lim
t→∞
t→∞
t
Z
≤ Br · lim
t→∞
(2.14)
+∞
q(s)ds + x∞ = x∞ .
t
and |(Ax)0 (t)| tend to x∞
It follows from (2.12), (2.13), and (2.14) that (Ax)(t)
1+t
uniformly as t → ∞. So, for any ε > 0, there exists N1 > 0 such that
(Ax)(t)
ε
− x∞ < , ∀ t ≥ N1 .
1+t
2
Consequently, for any t1 , t2 ≥ N1 , we
(Ax)(t )
1
− x∞ <
1 + t1
have
ε
,
2
(Ax)(t )
ε
2
− x∞ < .
1 + t2
2
Similarly, we can prove that there exists N2 > 0 such that
ε
ε
(Ax)0 (t1 ) − x∞ < , (Ax)0 (t2 ) − x∞ < , ∀ t1 , t2 ≥ N2 .
2
2
(2.15)
(2.16)
Choose N = max{N1 , N2 }, then (2.11) can be easily seen by (2.15) and (2.16). By
Lemma 2.4, we know that A : P → P is completely continuous.
3. Main results
For notational convenience, we denote
Z
Z +∞
β
n Z +∞
o
β +∞
m=
+ 1 x∞ , n =
q(τ )dτ + max
q(τ )dτ,
τ q(τ )dτ .
α
α 0
0
0
We will prove the following existence results.
Theorem 3.1. Assume that (H1)-(H2) hold, and there exists a > 2m such that
(S1) f (t, x1 , y1 ) ≤ f (t, x2 , y2 ) for any 0 ≤ t < +∞, 0 ≤ x1 ≤ x2 , 0 ≤ y1 ≤ y2 ;
a
, (t, u, v) ∈ [0, +∞) × [0, a] × [0, a].
(S2) f (t, (1 + t)u, v) ≤ 2n
Then the boundary value problem (1.1) has two positive nondecreasing on [0, +∞)
and concave solutions w∗ and v ∗ , such that 0 < kw∗ k ≤ a, and limn→∞ wn =
limn→∞ An w0 = w∗ , where
w0 (t) =
a
β
(t + 1) + x∞ + tx∞ , t ∈ J,
2
α
and 0 < kv ∗ k ≤ a, limn→∞ vn = limn→∞ An v0 = v ∗ , where v0 (t) = 0, t ∈ J.
EJDE-2009/141
EXISTENCE OF POSITIVE SOLUTIONS
7
Proof. By Lemma 2.5, we know that A : P → P is completely continuous. For any
x1 , x2 ∈ P with x1 ≤ x2 , x01 ≤ x02 , from the definition of A and (S1), we can easily
get that Ax1 ≤ Ax2 . We denote
P a = {x ∈ P : kxk ≤ a}.
First, we prove that A : P a → P a . If x ∈ P a , then kxk ≤ a. By (2.1), (S1) and
(S2 ), we get
n 1 β Z +∞
q(s)f (s, x(s), x0 (s))ds + x∞
kAxk1 = sup
1+t α 0
t∈J
Z t Z +∞
o
q(τ )f (τ, x(τ ), x0 (τ ))dτ ds + tx∞ +
s
0
β
Z +∞
β
a
≤
+ 1 x∞ +
+1
q(τ )dτ ·
α
α
2n
0
a
≤ a,
≤m+n·
2n
and
o
n Z +∞
q(s)f (s, x(s), x0 (s))ds + x∞ k(Ax)0 k∞ = sup t∈J
+∞
t
Z
≤
q(s)ds ·
0
a
+ x∞ ≤ a.
2n
β
Hence, we have proved that A : P a → P a . Let w0 (t) = a2 (t + 1) + α
x∞ + tx∞ , 0 ≤
2
t < +∞, then w0 (t) ∈ P a . Let w1 = Aw0 , w2 = A w0 , then by Lemma 2.5, we
have that w1 ∈ P a and w2 ∈ P a . We denote wn+1 = Awn = An w0 , n = 0, 1, 2, . . . .
Since A : P a → P a , we have wn ∈ A(P a ) ⊂ P a , n = 1, 2, 3, . . . . It follows from the
complete continuity of A that {wn }∞
n=1 is a sequentially compact set.
By (2.1) and (S2), we get
Z
β +∞
w1 (t) =
q(s)f (s, ω0 (s), ω00 (s))ds + x∞
α 0
Z t Z +∞
+
q(τ )f (τ, ω0 (τ ), ω00 (τ ))dτ ds + tx∞
0
s
(3.1)
Z
Z +∞
β
β +∞
a
a
+ x∞ +
+ tx∞
q(s)ds ·
τ q(τ )dτ ·
≤
α 0
2n α
2n
0
a β
≤ + x∞ + tx∞ ≤ w0 (t), 0 ≤ t < +∞,
2 α
and
Z
+∞
ω10 (t) = (Aω0 )0 (t) =
q(s)f (s, ω0 (s), ω00 (s))ds + x∞
t
Z +∞
a
≤
q(s)ds ·
+ x∞
2n
0
a
≤ + x∞ = ω00 (t), 0 ≤ t < +∞.
2
So, by (3.1), (3.2) and (S1) we have
w2 (t) = (Aw1 )(t) ≤ (Aw0 )(t) = w1 (t),
w20 (t)
0
0
0
0 ≤ t < +∞,
= (Aw1 ) (t) ≤ (Aw0 ) (t) = (w1 ) (t), 0 ≤ t < +∞.
(3.2)
8
X. ZHANG
EJDE-2009/141
By induction, we get
wn+1 (t) ≤ wn (t),
0
wn+1
(t) ≤ wn0 (t),
0 ≤ t < +∞, n = 0, 1, 2, . . . .
Thus, there exists w∗ ∈ P a such that wn → w∗ as n → ∞. Applying the continuity
of A and wn+1 = Awn , we get that Aw∗ = w∗ .
Let v0 (t) = 0, 0 ≤ t < +∞, then v0 (t) ∈ P a . Let v1 = Av0 , v2 = A2 v0 , then by
Lemma 2.5, we have that v1 ∈ P a and v2 ∈ P a . We denote vn+1 = Avn = An v0 , n =
0, 1, 2, . . . . Since A : P a → P a , we have vn ∈ A(P a ) ⊂ P a , n = 1, 2, 3, . . . . It follows
from the complete continuity of A that {vn }∞
n=1 is a sequentially compact set.
Since v1 = Av0 ∈ P a , we have
v1 (t) = (Av0 )(t) = (A0)(t) ≥ 0,
v10 (t)
0
0
= (Av0 ) (t) = (A0) (t) =
v00 (t)
0 ≤ t < +∞,
≥ 0,
0 ≤ t < +∞.
So, that we have
v2 (t) = (Av1 )(t) ≥ (A0)(t) = v1 (t),
0 ≤ t < +∞,
v20 (t)
0 ≤ t < +∞.
0
0
= (Av1 ) (t) ≥ (A0) (t) =
v10 (t),
By induction, we get
vn+1 (t) ≥ vn (t),
0
vn+1
(t) ≥ vn0 (t),
0 ≤ t < +∞, n = 0, 1, 2, . . . .
Thus, there exists v ∗ ∈ P a such that vn → v ∗ as n → ∞. Applying the continuity
of A and vn+1 = Avn , we get that Av ∗ = v ∗ .
If f (t, 0, 0) 6≡ 0, 0 ≤ t < ∞, then the zero function is not the solution of (1.1).
Thus, v ∗ is a positive solution of (1.1). It is well known that each fixed point of
A in P is a solution of (1.1). Hence, we assert that w∗ and v ∗ are two positive,
nondecreasing on [0, +∞) and concave solutions of (1.1).
β
Remark 3.2. The iterative schemes in Theorem 3.1 are w0 (t) = a2 (t + 1) + α
x∞ +
n
n
tx∞ , wn+1 = Awn = A w0 , n = 0, 1, 2, . . . and v0 (t) = 0, vn+1 = Avn = A v0 , n =
0, 1, 2, . . . . They start off with a known simple linear function and the zero function
respectively. It is convenient in application. We can easily get that w∗ and v ∗ are
the maximal and minimal solutions of the boundary value problem (1.1). Of course
w∗ and v ∗ may coincide and then the boundary value problem (1.1) has only one
solution in P .
The following theorem can be obtained directly from Theorem 3.1.
Theorem 3.3. Assume that (H1)-(H2) hold and there exists 2m < a1 < a2 < · · · <
an such that
(S1) f (t, x1 , y1 ) ≤ f (t, x2 , y2 ) for any 0 ≤ t < +∞, 0 ≤ x1 ≤ x2 , 0 ≤ y1 ≤ y2 ;
ak
(S2) f (t, (1 + t)u, v) ≤ 2n
, (t, u, v) ∈ [0, +∞) × [0, ak ] × [0, ak ], k = 1, 2, . . . , n.
Then the boundary value problem (1.1) has 2n positive nondecreasing on [0, +∞)
and concave solutions wk∗ and vk∗ , such that 0 < kwk∗ k ≤ ak , and limn→∞ wkn =
limn→∞ An wk0 = wk∗ , where
ak
β
(t + 1) + x∞ + tx∞ , t ∈ J,
2
α
and 0 < kvk∗ k ≤ ak , limn→∞ vkn = limn→∞ An v0 = vk∗ , where vk0 (t) = 0, t ∈ J.
wk0 (t) =
EJDE-2009/141
EXISTENCE OF POSITIVE SOLUTIONS
9
4. Example
Consider the boundary-value problem
1
x00 (t) + √
f (t, x(t), x0 (t)) = 0, t ∈ J+ ,
t(1 + t)2
2x(0) − 3x0 (0) = 0, x0 (∞) = 0,
(4.1)
where

4

10−2 | cos(2t + 1)| + 10−2 u
+
1+t
4
f (t, u, v) =

3
10−2 | cos(2t + 1)| + 10−2 1+t
+
1
10
1
10
v
400
,
u ≤ 3,
v
400
,
u ≥ 3.
1
Set q(t) = √t(1+t)
2 . It is clear that (H1) and (S2) hold. Let α = 2, β = 3, x∞ = 0.
By direct computation, we can obtain that
Z +∞
Z 1
Z +∞
Z +∞
1
1
8
1
√
√ dt +
√
dt <
dt = ,
q(t)dt =
(4.2)
2
2
3
t(1 + t)
t
t·t
0
0
1
0
√
Z +∞
Z +∞
Z 1√
Z +∞
8
t
t
√
dt = .
(4.3)
tq(t)dt =
dt
<
tdt
+
2
2
t
3
t(1 + t)
0
0
0
1
By (4.2) and (4.3), we have m = 0, n < 20
3 . Choose a = 300 and check check (S2).
Since nonlinear term f satisfies
1
81
3
179
300
300
f (t, (1 + t)u, v) ≤ 2 +
+
=
<
,
<
10
100 40
200
2n
2 · 20
3
for t ∈ [0, +∞), u, v ∈ [0, 300]; then all the conditions in Theorem 3.1 are satisfied.
Therefore, the conclusion of Theorem 3.1 holds.
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Xingqiu Zhang
School of Mathematics, Liaocheng University, Liaocheng, Shandong, China
E-mail address: zhxq197508@163.com