1. Introduction
In this paper, we consider the second-order difference equation
(1)
where
is a set of all integers,
α > 0 and
λ > 0 are constants,
Δxi =
xi+1 −
xi,
Δ2xi−1 =
Δ(
Δxi−1), and
will be specified later.
This equation is associated with the famous discrete logistic equation with diffusion:
(2)
for
and
, where the parameter
d > 0 corresponds to the rate at which the population diffuses, and the unknown function
x corresponds to the density of a population. The term −
x1+α in the equation corresponds to the fact that the population is self-limiting and the
p corresponds to the birth rate of population if the self-limitation is ignored. At points where
pi > 0 (<0) the population, ignoring self-limitation, has positive (negative) birth rate. Hence, we assume throughout this paper that
p takes on both positive and negative values on
; we further assume that the sequence
is bounded and there exists an integer
such that
.
If we write
λ = 1/
d, we see that the steady-state solutions of (
2) must satisfy (
1). Since the sequence
x represents a population density, it is nonnegative. Such solutions correspond to possible steady-state distributions of population. So our purpose in this paper is to establish the existence of positive solutions or the existence and uniqueness of positive solitons for (
1). A discrete soliton is a spatially localized solution of (
1) and decays to 0 at infinity; that is,
(3)
Recently, the existence of nontrivial discrete solitons for the general equation
(4)
has been extensively established by a number of authors [
1–
12]. Among the methods used are the principle of anticontinuity [
1–
3], variational methods [
4–
10], center manifold reduction [
11], Nehari manifold approach [
12], and so forth. On the other hand, a soliton of (
4) is also its homoclinic orbit or homoclinic solution. By using the variational methods, the existence of homoclinic orbits or homoclinic solutions has also been extensively discussed by a number of authors; see [
4–
10,
13–
16]. However, as far as we know, there are few papers concerned with the existence of positive solitons.
Equation (
2) is a discrete analogue of the well-known logistic equation of the form
(5)
Indeed, by means of standard finite difference methods, we set up a grid in the
x,
t plane with grid spacings
Δx and
Δt and then replace the second derivative
uxx with a central difference and
ut with a forward difference. By writing
xi =
iΔx,
tn =
nΔt,
pi =
p(
xi), and
, a finite difference scheme for (
5) is obtained:
(6)
or
(7)
which has the steady-state equation:
(8)
We note that the steady-state equation of (
5) is
(9)
or
(10)
where
μ = 1/
D. When
α = 1, (
10) is reduced to
(11)
In [
17], the authors considered the existence and uniqueness of positive solitons of (
11) by using sub- and upper-solution method. The present work is motivated by [
17].
To obtain a positive subsolution of (
1), we need a positive eigenvalue and its corresponding positive eigenfunction of the eigenvalue problem
(12)
Note that
p takes on both positive and negative values; thus, it is indefinite in a very strong sense. The corresponding problem for ordinary differential equation had been considered in [
18] by using variational method. However, such problem is new for the above discrete problem; see [
19–
21]. In Section
2, we will consider the above discrete eigenvalue problem by using the matrix and vector method.
As far as we know, for the discrete problem (1) there is no general sub- and upper-solution theorem on the set . Thus, a general sub- and upper-solution theorem will be firstly obtained in Section 3 and such theorem will be used for (1). On the other hand, our solutions are classical; however, [17] cannot insure this fact because some points of their subsolution are not derivative.
Our results will give some theory groundwork for the numerical calculation of (11). Note that λ = (Δx) 2/D ≠ μ when (Δx) 2 ≠ 1. This will lead to different results between (1) and (11).
2. Preliminaries
For any
with
a <
b, we consider the eigenvalue problem of the form
(13)
where [
a,
b] = {
a,
a + 1, …,
b},
is a real sequence, and there exists
i0 ∈ [
a,
b] such that
.
Denote
(14)
(15)
Then problem (
13) can be rewritten by matrix and vector in the form
(16)
Let H be a set of all real sequences with xa−1 = 0 = xb+1. For any u, v ∈ H, the inner product is defined as and the norm ‖·‖ is defined as .
For any
v ∈
H, we define
(17)
Consider the Rayleigh quotient
(18)
and let
(19)
For such defined λ1, we have the following important result.
Lemma 1. λ1 is a positive eigenvalue of the problem (13) or (16). Moreover λ1 is simple and the corresponding eigenfunction φ can be chosen such that φi > 0 for all i ∈ [a, b].
Proof. First, we show λ1 > 0. Clearly
(20)
for all
v ∈
H. Moreover, by the spectral theorem, (
Av,
v) ≥
γ1(
v,
v), where
(21)
is the first eigenvalue of the eigenvalue problem
(22)
Hence, if
v ∈
H and
, then
(23)
Hence
λ1 > 0.
Second, consider the linear eigenvalue problem
(24)
and define (
Su)
i = −
Δ2ui−1 −
λ1piui. It is easy to see that
λ1 is an eigenvalue for (
13) with corresponding eigenfunction
φ if and only if 0 is an eigenvalue of
S and so of (
24) with corresponding eigenfunction
φ. The minimal eigenvalue
α1 of
S is given by
(25)
Note that
for all
v ∈
H; thus, we have
α1 ≥ 0. Because of how we defined
λ1, there exists a sequence
v(n) ∈
H and
such that
(26)
which implies that
(27)
Thus, we have
α1 ≤ 0. In this case, we know that
α1 = 0 is the minimal eigenvalue of (
24). By Lemma
A.1 in the Appendix,
α1 is simple and the corresponding eigenfunction can be chosen to be positive on
H. Thus, the statement in this lemma is right.
Let
λ >
λ1; consider the eigenvalue problem of the form
(28)
We have the following result.
Lemma 2. If λ > λ1, then the minimal eigenvalue μ1 of (28) is negative and the corresponding eigenfunction φ(1) can be chosen so that for all i ∈ [a, b].
Proof. Let φ be the eigenvector obtained in Lemma 1 and ∥φ∥ = 1; then we have
(29)
Consequently,
(30)
In view of Lemma
A.1 in the Appendix, the eigenvector corresponding to
μ1 can be chosen to be positive. This completes the proof.
3. Main Results
First of all, we introduce the definitions of the subsolution and upper-solution and give a general sub- and upper-solution theorem.
For any
with
a <
b, consider the Dirichlet boundary value problem of the form
(31)
We will denote by (
31)
n the problem (
31) when
a = −
n and
b =
n.
Definition 3. A sequence is said to be an upper-solution of (31) if
(32)
A sequence
is said to be a subsolution of (
31) if
(33)
Theorem 4. Suppose that are functions with (where is defined by for ) and , are, respectively, a subsolution and an upper-solution of (31)n for all large n. Assume that f(i, z) is continuous with z ∈ [α, β] and there exists a constant k > 0 such that
(34)
for all
and
s2 >
s1 with
s1,
s2 ∈ [
α,
β] (where
and
). Then problem (
4) has a solution
x such that
.
Proof. Since and provide sub- and upper-solutions for (31)n , there exists a solution φ of (31)n such that for all |i | ≤ n.
In fact, for any sequence , clearly, the problem
(35)
has a unique solution
w. This defines a mapping
T :
u →
w. We claim that
T is increasing on
. In fact, for any sequences
,
with
, we have
(36)
where
Γ is an operator defined as (
Γu)
i = −
Δ2ui−1 +
kui for
i ∈ [−
n,
n]. Since
,
, by the strong monotonicity of
Γ, we obtain
; see [
21].
Let u(m) = Tu(m−1), ; v(m) = Tv(m−1), . In the following, we claim that
(37)
First of all, by using (
35) and the definition of subsolution, we see
(38)
and
,
. This implies that
u(0) −
u(1) ≤ 0 by the strong monotonicity of
Γ. A similar argument gives
v(1) ≤
v(0). The monotonicity of
Γ and
T gives the rest. So there exist
u and
v such that
(39)
By the definition of
u and
v, we see that
u and
v satisfy (
31)
n and
. Let
x(n) denote such solution corresponding to (
31)
n .
Then standard a priori estimates and a diagonalization argument show that there exists a subsequence of {x(n)} which converges to solution x of (4) on every bounded subset of . Moreover, since for all n, it follows that on . The proof is complete.
In the following, we use the sub- and upper-solution theorem to establish the existence of positive solutions or the existence and uniqueness of positive solitons for (1).
For problem (
1), we have assumed that there exists
such that
. For any
n>|
i0|, consider the eigenvalue problem (
13) when
a = −
n and
b =
n and denote the corresponding positive eigenvalue
λ1 defined in (
19) by
. Then
(40)
Theorem 5. For any , (1) has a positive solution, where is defined in (40).
Proof. For any , clearly, there exists a positive integer n1≥|i0| such that i0 ∈ [−n1, n1] and . In view of Lemma 2, the eigenvalue problem
(41)
has a negative eigenvalue
and its corresponding eigenfunction
with
can be chosen so that
for all
i ∈ [−
n1,
n1]. For any
ɛ > 0, we define
(42)
At this time, we have
(43)
for
i ∈ [−
n1,
n1] and sufficiently small
ϵ. When
i =
n1 + 1,
(44)
and
for
i >
n1 + 1. Similarly, we can also prove that
for
i ≤ −
n1 − 1. Thus, the sequence
defined in (
42) is a subsolution of (
1).
Note that α > 0 and the sequence {pi} is bounded. Thus, the sufficiently large positive constant is a supersolution of (1).
For , we have
(45)
where
ξ ∈ (
s1,
s2). Note that the sequences {
pi},
, and
M are bounded; thus, (
1) satisfies condition (
34). By Theorem
4, problem (
1) has a solution
u such that
. We claim that
u > 0. Suppose there exists
such that
; then we obtain
since
. Thus we obtain
. So
u ≡ 0, which contradicts
u≢0. The proof is complete.
Theorem 6. Assume that and there exists a constant C > 6/λ and n0 > |i0| such that
(46)
for |
i | >
n0. Then (
1) has a positive soliton and there exists a constant
M > 0 and
n2 ≥
n0 such that
(47)
Proof. Let be the subsolution of (1) obtained in Theorem 5 and
(48)
where
M > 0 is a constant determined later. By simple calculation, we have
(49)
By the assumption on
p, there exist
n2 ≥
n0 and
M1 > 0 such that
(50)
for
M ≥
M1 and |
i | ≥
n2.
For any M ≥ M1, define
(51)
We will show that
ψ is upper-solution of (
1) by appropriate choice of
M. In fact,
- (a)
for |i | > n2, we have by the choice of M;
- (b)
for |i | < n2 − 1, we have
(52)
-
by choosing M > M1 large enough since α > 0 and p is bounded;
- (c)
similarly, for i = n2, n2 − 1, −n2, −n2 + 1, one can show that for M > M1 large enough.
In view of (a)–(c), we construct upper-solution ψ of (1) with . Using the sub- and upper-solution Theorem 4, we complete the proof.
Theorem 7. Assume that there exists a positive integer m > |i0| such that pi ≤ 0 for |i| ≥ m. Let x be a bounded positive solution of (1); then
(53)
Proof. At this time, we have
(54)
Hence
Δxi−1 is an increasing sequence for |
i| ≥
m and so there exists a positive integer
n1 such that
Δxi is one sign for |
i| >
n1. Thus
x is eventually a monotone sequence for |
i | >
n1. Note that
x is bounded. Thus, lim
n→+∞xn and lim
n→−∞xn exist.
If limn→+∞xn ≠ 0, we have
(55)
which implies that
(56)
This is impossible since
x is bounded and so we must have lim
n→+∞xn = 0. Similarly, we also have lim
n→−∞xn = 0. The proof is complete.
Theorem 8. For any λ ≠ 0, there exists at most one positive solution of (1) such that lim|i|→∞xi = 0.
Proof. Suppose that u and v are such two distinct solutions. As before we can construct an arbitrarily small subsolution and so there must exist a solution w of (1) such that w ≤ u and w ≤ v. Multiplying the u-equation by w and the w-equation by u, then we have
(57)
or
(58)
Let
n → +
∞; we have
(59)
which implies that
u =
w. Similarly, we can also prove that
v =
w. The proof is complete.
Corollary 9. Assume that all conditions of Theorem 8 hold; then (1) has a unique positive soliton.
4. Conclusion
This paper studied a discrete logistic steady-state equation with both positive and negative birth rate of population. By using sub- and upper-solution method (Theorem 4), the existence of positive solution and positive soliton is obtained (Theorems 5 and 6). Uniqueness of homoclinic type solution is also obtained (Theorem 8).
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
Acknowledgment
The authors would like to thank the anonymous referees for their comments and suggestions on the paper.
Appendix
To obtain a positive eigenfunction of (13) or a positive eigenvector of (16), we need the following lemma. We give the proof by the method of [22].
Lemma A.1. For any (b − a + 1) × (b − a + 1) diagonal matrix
(A.1)
let
μ1 be the minimal eigenvalue of
A +
Q, where
A is defined in (
14). Then
μ1 is simple and the corresponding eigenvector
φ(1) can be chosen to be positive.
Proof. Without loss of generality, we assume qi > 0 for i ∈ [a, b]. If this is not the case, choose a constant C > 0 large enough such that qi + C > 0. Consider the matrix A + Q + CE instead, where E is a (b − a + 1)×(b − a + 1) identity matrix. Denote ; then for any i ∈ [a, b].
Since A + Q is a real (b − a + 1)×(b − a + 1) symmetric positive definite matrix, A + Q has b − a + 1 real eigenvalues. We repeat each eigenvalue according to its multiplicity as follows:
(A.2)
and we choose eigenvectors
φ(1),
φ(2), …,
φ(b−a+1) for
A +
Q such that
(A.3)
is an orthonormal basis for
. In the following we give the proof of Lemma
A.1 in four steps.
Step 1. For any , define B[u, v] = vT(A + Q)u. It is obvious that
(A.4)
Step 2. We claim that if and ∥u∥ = 1, then u is a solution of
(A.5)
if and only if
(A.6)
Obviously (
A.5) implies (
A.6). On the other hand, suppose (
A.6) is valid. Then, writing
dk = (
u,
φ(k)), we have
(A.7)
Hence
(A.8)
Consequently
dk = (
u,
φ(k)) = 0, if
μk >
μ1. It follows that
(A.9)
for some
m, where (
A +
Q)
φ(k) =
μ1φ(k) for
k = 1,2, …,
m. Therefore
(A.10)
This proves (
A.5).
Step 3. We will show that if satisfies (A + Q)u = μ1u, u ≠ 0, then either u > 0 or u < 0.
To see this, let us assume without loss of generality that ∥u∥ = 1 and note α + β = 1 for α = ∥u+∥2, β = ∥u−∥2, where and . In this case, we have
(A.11)
Thus
(A.12)
By Step 2,
u+ and
u− satisfy (
A +
Q)
u+ =
μ1u+, (
A +
Q)
u− =
μ1u−. Thus (
A +
Q)
u+ ≥ 0, (
A +
Q)
u− ≥ 0. That is,
(A.13)
If there exists an
i0 ∈ [
a,
b] such that
, then
u+ = 0. Similar arguments apply to
u−, and so either
u > 0 or
u < 0. Thus, we can choose a positive eigenvector
φ(1) corresponding to the minimal eigenvalue
μ1.
Step 4. Finally assume that u and are two eigenvectors corresponding to μ1; in view of Step 3,
(A.14)
and so there exists a real constant
k such that
(A.15)
or
(A.16)
But since
also satisfies
, by Step 3,
or
or
. In view of (
A.16), we have
. Hence the eigenvalue
μ1 is simple. This completes the proof of Lemma
A.1.