Hindawi Publishing Corporation
Journal of Applied Mathematics
Volume 2012, Article ID 528719, 13 pages
doi:10.1155/2012/528719
Research Article
Positive Solutions for Nonlinear Differential
Equations with Periodic Boundary Condition
Shengjun Li,1 Li Liang,1 and Zonghu Xiu2
1
2
College of Information Sciences and Technology, Hainan University, Haikou 570228, China
College of Science and Information Science, Qingdao Agricultural University, Qingdao 266109, China
Correspondence should be addressed to Shengjun Li, shjli626@126.com
Received 22 February 2012; Accepted 23 March 2012
Academic Editor: Yeong-Cheng Liou
Copyright q 2012 Shengjun Li et al. This is an open access article distributed under the Creative
Commons Attribution License, which permits unrestricted use, distribution, and reproduction in
any medium, provided the original work is properly cited.
We study the existence of positive solutions for second-order nonlinear differential equations with
nonseparated boundary conditions. Our nonlinearity may be singular in its dependent variable.
The proof of the main result relies on a nonlinear alternative principle of Leray-Schauder. Recent
results in the literature are generalized and significantly improved.
1. Introduction
In this paper, we establish the positive periodic solutions for the following singular
differential equation:
− pxy qxy f x, y, y ,
0 ≤ x ≤ T,
1.1
and boundary conditions
y0 yT ,
y1 0 y1 T ,
1.2
where p, q ∈ CR/T Z, the nonlinearity f ∈ CR/T Z × 0, ∞ × R, R, and y1 x pxy x denotes the quasi-derivative of yx. We call boundary conditions 1.2 the periodic
boundary conditions which are important representatives of nonseparated boundary
2
Journal of Applied Mathematics
conditions. In particular, the nonlinearity may have a repulsive singularity at y 0, which
means that
lim f x, y, z ∞,
y→0
uniformly in x, z ∈ R2 .
1.3
Also f may take on negative values. Electrostatic or gravitational forces are the most
important examples of singular interactions.
During the last few decades, the study of the existence of periodic solutions for
singular differential equations have deserved the attention of many researchers 1–8. Some
classical tools have been used to study singular differential equations in the literature,
including the degree theory 7–9, the method of upper and lower solutions 5, 10,
Schauder’s fixed point theorem 2, 11, 12, some fixed point theorems in cones for completely
continuous operators 13–15, and a nonlinear Leray-Schauder alternative principle 16–18.
However, the singular differential equation 1.1, in which the nonlinearity is
dependent on the derivative and does not require f to be nonnegative, has not attracted
much attention in the literature. There are not so many existence results for 1.1 even when
the nonlinearity is independent of the derivative. In this paper, we try to fill this gap and
establish the existence of positive T -periodic solutions of 1.1; proof of the existence of
positive solutions is based on an application of a nonlinear alternative of Leray-Schauder,
which has been used by many authors 16–18.
The rest of this paper is organized as follows. In Section 2, some preliminary results
will be given, including a famous nonlinear alternative of Leray-Schauder type. In Section 3,
we will state and prove the main results.
2. Preliminaries
Let us denote ut and vt by the solutions of the following homogeneous equations:
− pxy qxy 0,
0 ≤ x ≤ T,
2.1
satisfying the initial conditions
u0 1,
u1 0 0,
v0 0,
v1 0 1,
2.2
and set
D uT v1 T − 2.
2.3
Throughout this paper, we assume that 1.1 satisfies the following condition 2.4:
T
px > 0,
qx > 0,
0
1
dx < ∞,
px
T
0
qxdx < ∞.
2.4
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3
Lemma 2.1 see 19. For the solution yx of the boundary value problem
− pxy qxy hx,
y0 yT ,
0 ≤ x ≤ T,
2.5
1
y 0 y1 T ,
the formula
yx T
Gx, shsds,
2.6
x ∈ 0, T ,
0
holds, where
Gx, s u1 T vT uxus −
vxvs
D
D
⎧ 1
uT − 1
v T − 1
⎪
⎪
uxvs −
usvx, 0 ≤ s ≤ x ≤ T,
⎪
⎪
⎨
D
D
⎪
⎪
1
⎪
⎪
⎩ v T − 1 usvx − uT − 1 uxvs, 0 ≤ x ≤ s ≤ T,
D
D
2.7
is the Green’s function, and the number D is defined by 2.3.
Lemma 2.2 see 19. Under condition 2.4, the Green’s function Gx, s of the boundary value
problem 2.5 is positive, that is, Gx, s > 0, for x, s ∈ 0, T .
One denotes
B max Gx, s,
A min Gx, s,
0≤s,x≤T
σ
0≤s,x≤T
A
.
B
2.8
Thus B > A > 0 and 0 < σ < 1.
Remark 2.3. If px 1, qx m2 > 0, then the Green’s function Gx, s of the boundary
value problem 2.5 has the form
Gx, s ⎧ mx−s
emT −xs
e
⎪
⎪
,
⎪
⎪
mT
⎪
⎨ 2m e − 1
⎪
⎪
⎪
ems−x emT x−s
⎪
⎪
⎩
,
2m emT − 1
0 ≤ s ≤ x ≤ T,
2.9
0 ≤ x ≤ s ≤ T.
It is obvious that Gx, s > 0 for 0 ≤ s, x ≤ T , and a direct calculation shows that
A
emT/2
mT
,
m e −1
B
1 emT
,
2m emT − 1
σ
2emT/2
< 1.
1 emT
2.10
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Journal of Applied Mathematics
Let X C0, T , and we suppose that F : 0, T × R × R → 0, ∞ is a continuous
function. Define an operator:
T y x T
Gx, sF s, ys, zs ds
2.11
0
for y ∈ X and x ∈ 0, T . It is easy to prove that T is continuous and completely continuous.
3. Main Results
In this section, we state and prove the new existence results for 1.1. In order to prove our
main results, the following nonlinear alternative of Leray-Schauder is needed, which can be
T
found in 20. Let us define the function ωx 0 Gx, sds. The usual L1 -norm over 0, T is denoted by · 1 , and the supremum norm of C0, T is denoted by · .
Lemma 3.1. Assume Ω is a relatively compact subset of a convex set E in a normed space X. Let
T : Ω → E be a compact map with 0 ∈ Ω. Then one of the following two conclusions holds:
i T has at least one fixed point in Ω;
ii there exist u ∈ ∂Ω and 0 < λ < 1 such that u λT u.
Now we present our main existence result of positive solution to problem 1.1.
Theorem 3.2. Suppose that 1.1 satisfies 2.4. Furthermore, assume that there exists a constant
r > 0 such that
H1 there exists a constant M > 0 such that Fx, y, z fx, y, z M ≥ 0 for all x, y, z ∈
0, T × 0, r × R;
H2 there exist continuous, nonnegative functions gy, hy, and y such that
F x, y, z ≤ g y h y |z|,
∀ x, y, z ∈ 0, T × 0, r × R,
3.1
where gy > 0 is nonincreasing, hy/gy is nondecreasing in 0, r and y is
nondecreasing in 0, ∞;
H3 there exist a nonincreasing positive continuous function g0 y on 0, ∞, and a constant
R0 > 0 such that Fx, y, z ≥ g0 y for x, y, z ∈ 0, T × 0, R0 × ∞, where g0 y
R
satisfies limy → 0 g0 y ∞ and limy → 0 y 0 g0 udu ∞;
H4 the following inequalities hold:
σr > Mω,
r
> ω,
gσr − Mω 1 hr/gr L1 r L2 T where L1 2q1 /min0≤x≤T px, L2 2M/min0≤x≤T px.
Then 1.1 has at least one positive T -periodic solution y with 0 < y Mω ≤ r.
3.2
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5
Proof. Since H4 holds, let N0 {n0 , n0 1, . . .}, and we can choose n0 ∈ {1, 2, . . .} such that
1/n0 < σr − Mω and
1
hr
L1 r L2 T < r.
ωgσr − Mω 1 gr
n0
3.3
To show 1.1 has a positive solution, we should only show that
− pxy qxy F x, yx − Mωx, y x − Mω x
3.4
has a positive solution y satisfying 1.2. If it is right, then kx yx − Mωx is a solution
of 1.1 since
qx yx − Mωx
− pxk qxk − px yx − Mωx
F x, yx − Mωx, y x − Mω x − M
f x, yx − Mωx, y x − Mω x
f x, kx, k x ,
3.5
where −pxω x qxωx 1 is used.
Consider the family of equations:
qx
,
− pxy qxy λFn x, yx − Mωx, y x − Mω x n
3.6
where λ ∈ 0, 1, n ∈ N0 , and
Fn x, y, z ⎧
⎪
⎪
F x, y, z
⎪
⎪
⎨
if y ≥
⎪
⎪
1
⎪
⎪
⎩F x, , z
n
1
if y ≤ .
n
1
,
n
3.7
Problem 3.6−1.2 is equivalent to the following fixed point of the operator equation:
y Tn y,
3.8
where Tn is a continuous and completely continuous operator defined by
Tn yx λ
T
0
1
Gx, sFn s, ys − Mωs, y s − Mω s ds ,
n
3.9
and we used the fact
T
0
Gx, sqsds ≡ 1
see Lemma 2.1 with h q .
3.10
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Journal of Applied Mathematics
Now we show y /
r for any fixed point y of 3.8. If not, assume that y is a fixed
point of 3.8 such that y r. Note that
1
yx − λ
n
T
Gx, sFn s, ys − Mωs, y s − Mω s ds
0
≥ λA
T
Fn s, ys − Mωs, y s − Mω s ds
0
T
Fn s, ys − Mωs, y s − Mω s ds
0
T
Gx, sFn s, ys − Mωs, y s − Mω s ds
≥ σ max λ
σBλ
x∈0,T 1
.
σ y − n
3.11
0
So we have
1
1 ≥ σ y − 1 1 ≥ σr,
y
−
yx ≥ σ n n
n
n
for 0 ≤ x ≤ T.
3.12
In order to pass the solutions of the truncation equation 3.6 with λ 1 to that of the
original equation 3.4, we need the fact that y is bounded. Now we show that
y ≤ L1 r
3.13
for a solution yx of 3.6.
Integrating 3.6 from 0 to T with λ 1, we obtain
T
0
T
qx
dx.
Fn x, yx − Mωx, y x − Mω x qxyxdx n
0
3.14
Since y0 yT , there exists x0 ∈ 0, T such that y x0 0; therefore,
x
pxy x psy s ds
x0
x
qs
qsys − Fn s, ys − Mωs, y s − Mω s −
ds
x0
n
T
qs
≤
ds
qsys Fn s, ys − Mωs, y s − Mω s n
0
T
2
qsysds
0
≤ 2r
T
qsds.
0
3.15
Journal of Applied Mathematics
7
So,
2r
y x ≤
T
0
qsds
px
2r q1
≤
: L1 r.
min0≤x≤T px
3.16
Similarly we have
ω x ≤
2T
min0≤x≤T px
.
3.17
By 3.12, we obtain yx − Mωx ≥ σr − Mωx ≥ σr − Mω > 1/n0 ≥ 1/n. Thus
from condition H2 T
1
yx λ
Gx, sFn s, ys − Mωs, y s − Mω s ds n
0
T
1
Gx, sF s, ys − Mωs, y s − Mω s ds λ
n
0
T
1
Gx, sF s, ys − Mωs, y s − Mω s ds ≤
n
0
T
h ys − Mωs
1
≤
Gx, sg ys − Mωs 1 y s Mω s ds n
g ys − Mωs
0
T
hr
1
≤ gσr − Mω 1 L1 r L2 T Gt, s ds gr
n
0
hr
1
L1 r L2 T ω .
≤ gσr − Mω 1 gr
n0
3.18
Therefore,
hr
1
r y ≤ gσr − Mω 1 L1 r L2 T ω .
gr
n0
3.19
This is a contradiction, so y /
r.
Using Lemma 3.1, we know that
y Tn y
3.20
has a fixed point, denoted by yn , that is, equation
qx
− pxy qxy Fn x, yx − Mωx, y x − Mω x n
3.21
has a periodic solution yn with yn < r. Using similar procedure to that of the proof of 3.13,
we can prove that
yn ≤ L1 r.
3.22
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Journal of Applied Mathematics
In the next lemma, we will show that yn x − Mωx have a uniform positive lower bound,
that is, there exists a constant δ > 0, independent of n ∈ N0 , such that
yn x − Mωx ≥ δ
3.23
for all n ∈ N0 .
The fact yn < r and yn ≤ L1 r shows that {yn }n∈N0 is a bounded and equicontinuous family on 0, T . Thus the Arzela–Ascoli Theorem guarantees that {yn }n∈N0 has
a subsequence, {yni }i∈N converging uniformly on 0, T to a function y ∈ X. F is uniformly
continuous since yn satisfies δ Mωx ≤ yn x ≤ r for all x ∈ 0, T . Moreover, yni satisfies
the integral equation
yni x T
0
1
Gx, sFn s, yni s − Mωs, yn i s − Mω s ds .
ni
3.24
Letting i → ∞, we arrive at
yx T
Gx, sFn s, ys − Mωs, y s − Mω s ds.
3.25
0
Therefore, y is a positive periodic solution of 1.1 and satisfies 0 < y Mω ≤ r.
Lemma 3.3. There exists a constant δ > 0 such that any solution yn of 3.6 (with λ 1) satisfies
3.23 for all n large enough.
Proof. By condition H3 , there exist R1 ∈ 0, R0 and a continuous function g0 such that
F x, y, y − qxy ≥ g0 y ≥ max M, r q
3.26
for all x, y ∈ 0, T × 0, R1 , where g0 satisfies condition also like in H3 .
Choose n1 ∈ N0 such that 1/n1 ≤ R1 , and let N1 {n1 , n1 1, . . .}. For n ∈ N1 , let
αn min yn x − Mωx ,
0≤x≤T
βn max yn x − Mωx .
0≤x≤T
3.27
We first show that βn > R1 for all n ∈ N1 . If not, assume that βn ≤ R1 for some n ∈ N1 .
If 1/n ≤ yn x − Mωx ≤ R1 , we obtain from 3.26
Fn x, yn x − Mωx, yn x − Mω x F x, yn x − Mωx, yn x − Mω x
≥ qx yn x − Mωx g0 yn x − Mωx
≥ g0 yn x − Mωx > r q,
3.28
and, if yn x − Mωx ≤ 1/n, we obtain
Fn x, yn x −
Mωx, yn x
qx
1
1
1 g0
≥ g0
> r q. 3.29
F x, , yn x ≥
n
n
n
n
Journal of Applied Mathematics
9
So we have
Fn x, yn x − Mωx, yn x − Mω x > r q,
for βn ≤ R1 .
3.30
Integrating 3.6 with λ 1 from 0 to T , we deduce that
T
qx
dx
− px yn x qxyn x − Fn x, yn x − Mωx, yn x − Mω x −
n
0
T
T
1 T
qx yn x dx −
qxdx −
Fn x, yn x − Mωx, yn x − Mω x dx
n 0
0
0
T
<
qxyn xdx − r qT ≤ 0.
0
0
3.31
This is a contradiction. Thus βn > R1 , and we have
yn − Mω > R1 ,
∀n ∈ N1 .
3.32
To prove 3.23, we first show
yn x ≥ Mωx 1
,
n
0 ≤ x ≤ T for n ∈ N1 .
3.33
Let N1 P ∪ Q; here αn ≥ R1 if n ∈ P , and αn < R1 if n ∈ Q. If n ∈ P , it is easy to verify
3.33 is satisfied. We now show 3.33 holds if n ∈ Q. If not, suppose there exists n ∈ Q with
1
αn min yn x − Mωx yn cn − Mωcn <
0≤x≤T
n
3.34
for some cn ∈ 0, T . As αn yn cn − Mωcn < R1 , by βn > R1 , there exists cn ∈ 0, T without loss of generality, we assume an < cn such that yn an Mωan R1 and yn x ≤
Mωx R1 for an ≤ x ≤ cn .
From 3.26, we easily show that
Fn x, yx − Mωx, y x − Mω x > qx yn x − Mωx M
for x ∈ an , cn .
3.35
10
Journal of Applied Mathematics
Using 3.6 with λ 1 for yn x, we have, for x ∈ an , cn ,
−px yn x − Mω x − px yn x M px ω x −qxyn x Fn x, yn x − Mωx, yn x − Mω x
qx
− M 1 − qxωx
n
> −qxyn x qx yn x − Mωx M
qx
− M 1 − qxωx
n
qx
≥ 0.
n
3.36
As yn cn − Mω cn 0, px > 0, so yn x − Mω x < 0 for all x ∈ an , cn , and
the function νn : yn − Mω is strictly decreasing on an , cn . We use ηn to denote the inverse
function of yn restricted to an , cn . Thus there exists bn ∈ an , cn such that yn bn −Mωbn 1/n and
yn x − Mωx ≤
1
n
for cn ≥ x ≥ bn ,
1
≤ yn x − Mωx ≤ R1
n
for bn ≥ x ≥ an .
3.37
By using the method of substitution, we obtain
R1
F ηn ν, ν, ν dν 1/n
an
b
ann
b
F x, yn x − Mωx, yn x − Mω x yn x − Mω x dx
Fn x, yn x − Mωx, yn x − Mω x yn x − Mω x dx
ann qx yn x − Mω x dx
− px yn x qxyn x −
n
b
ann − px yn x
yn x − Mω x dx
bn
an qx yn x − Mω x dx.
qxyn x −
n
bn
3.38
By the facts yn < r, yn and ω are bounded, one can easily obtain that the second term
is bounded. The first term is
2
2
pbn yn bn − pan yn an Mpan yn an ω an − Mpbn yn bn ω bn an
an
pxyn xyn xdx − M
pxyn xω xdx,
bn
bn
3.39
Journal of Applied Mathematics
11
which is also bounded. As a consequence, there exists L > 0 such that
R1
F ηn y , y, y dy ≤ L.
3.40
1/n
On the other hand, by H3 , we can choose n2 ∈ N1 large enough such that
R1
F ηn y , y, y dy ≥
R1
1/n
g0 y dy > L
3.41
1/n
for all n ∈ N2 {n2 , n2 1, . . .}. This is a contradiction. So 3.33 holds.
Finally, we will show that 3.23 is right in n ∈ Q. Noticing estimate 3.33 and
employing the method of substitution, we obtain
R1
F ηn y , y, y dy αn
an
c
ann
c
F x, yn x − Mωx, yn x − Mω x yn x − Mω x dx
Fn x, yn x − Mωx, yn x − Mω x yn x − Mω x dx
ann cn
qx yn x − Mω x dx.
− pxyn x qxyn x −
n
3.42
Obviously, the right-hand side of the above equality is bounded. On the other hand, by H3 ,
R1
F ηn y , y, y dy ≥
αn
R1
g0 y dy −→ ∞
3.43
αn
if αn → 0 . Thus we know that αn ≥ δ for some constant δ > 0; the proof is completed.
Corollary 3.4. Let the nonlinearity in 1.1 be
f x, y, z 1 |z|γ y−α μyβ ex ,
0 ≤ x ≤ T,
3.44
where α > 0, β, γ ≥ 0, ex ∈ C0, T , and μ > 0 is a positive parameter,
i if β γ < 1, then 1.1 has at least one positive periodic solution for each μ > 0;
ii if β γ ≥ 1, then 1.1 has at least one positive periodic solution for each 0 < μ < μ∗ , where
μ∗ is some positive constant.
Proof. We will apply Theorem 3.2. Take
M e0 max |ex|,
0≤x≤T
g y y−α ,
h y μyβ 2e0 ,
z 1 |z|γ .
3.45
12
Journal of Applied Mathematics
Then conditions H1 –H3 are satisfied and the existence condition H4 becomes
rσr − Mωα − ω 1 L1 r L2 T γ − 2e0 ωr α 1 L1 r L2 T γ
μ<
r αβ ω 1 L1 r L2 T γ
3.46
for some r > 0. So 1.1 has at least one positive periodic solution for
rσr − Mωα − ω 1 L1 r L2 T γ − 2e0 ωr α 1 L1 r L2 T γ
0 < μ < μ : sup
.
r αβ ω 1 L1 r L2 T γ
r>0
3.47
∗
Note that μ∗ ∞ if β γ < 1 and μ∗ < ∞ if β γ ≥ 1. We have i and ii.
Acknowledgments
This work is supported by the National Natural Science Foundation of China Grant no.
11161017 and Hainan Natural Science Foundation Grant no. 111002.
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