Existence and Uniqueness of Positive Solitons for a Second-Order Difference Equation

A discrete logistic steady-state equation with both positive and negative birth rate of population will be considered. By using suband upper-solution method, the existence of bounded positive solutions and the existence and uniqueness of positive solitons will be established. To this end, the Dirichlet eigenvalue problem with positive and negative coefficients is considered, and a general suband upper-solution theorem is also obtained.


Introduction
In this paper, we consider the second-order difference equation where Z is a set of all integers,  > 0 and  > 0 are constants, Δ  =  +1 −   , Δ 2  −1 = Δ(Δ −1 ), and {  } ∈Z will be specified later.This equation is associated with the famous discrete logistic equation with diffusion: for  ∈ Z and  ∈ Z + = {0, 1, 2, . ..},where the parameter  > 0 corresponds to the rate at which the population diffuses, and the unknown function  corresponds to the density of a population.The term − 1+ in the equation corresponds to the fact that the population is self-limiting and the  corresponds to the birth rate of population if the self-limitation is ignored.At points where   > 0 (<0) the population, ignoring self-limitation, has positive (negative) birth rate.Hence, we assume throughout this paper that  takes on both positive and negative values on Z; we further assume that the sequence {  } ∈Z is bounded and there exists an integer  0 ∈ Z such that   0 > 0.
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][5][6][7][8][9][10][13][14][15][16].However, as far as we know, there are few papers concerned with the existence of positive solitons.

2
Discrete Dynamics in Nature and Society Equation ( 2) is a discrete analogue of the well-known logistic equation of the form Indeed, by means of standard finite difference methods, we set up a grid in the ,  plane with grid spacings Δ and Δ and then replace the second derivative   with a central difference and   with a forward difference.By writing   = Δ,   = Δ,   = (  ), and    ≈ (  ,   ), a finite difference scheme for ( 5) is obtained: or which has the steady-state equation: We note that the steady-state equation of ( 5) is or where  = 1/.When  = 1, (10) is reduced to 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 Note that  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][20][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 Z. 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.

Preliminaries
For any ,  ∈ Z with  < , we consider the eigenvalue problem of the form where [, ] = {,  + 1, . . ., }, {  }  = is a real sequence, and there exists Then problem ( 13) can be rewritten by matrix and vector in the form Let  be a set of all real sequences {  } +1 =−1 with  −1 = 0 =  +1 .For any , V ∈ , the inner product is defined as Consider the Rayleigh quotient and let For such defined  1 , we have the following important result.
Lemma 1.  1 is a positive eigenvalue of the problem (13) or (16).Moreover  1 is simple and the corresponding eigenfunction  can be chosen such that   > 0 for all  ∈ [, ].
Second, consider the linear eigenvalue problem and define ( It is easy to see that  1 is an eigenvalue for (13) with corresponding eigenfunction  if and only if 0 is an eigenvalue of  and so of (24) with corresponding eigenfunction .The minimal eigenvalue  1 of  is given by Note that   1 (V) ≥ 0 for all V ∈ ; thus, we have  1 ≥ 0.
Because of how we defined  1 , there exists a sequence V () ∈  and Thus, we have  1 ≤ 0. In this case, we know that  1 = 0 is the minimal eigenvalue of (24).By Lemma A.1 in the Appendix,  1 is simple and the corresponding eigenfunction can be chosen to be positive on .Thus, the statement in this lemma is right.
Let  >  1 ; consider the eigenvalue problem of the form We have the following result.
Proof.Let  be the eigenvector obtained in Lemma 1 and ‖‖ = 1; then we have Consequently, In view of Lemma A.1 in the Appendix, the eigenvector corresponding to  1 can be chosen to be positive.This completes the proof.

Main Results
First of all, we introduce the definitions of the subsolution and upper-solution and give a general sub-and upper-solution theorem.
Then standard a priori estimates and a diagonalization argument show that there exists a subsequence of { () } which converges to solution  of (4) on every bounded subset of Z.Moreover, since  ≤  () ≤  for all , it follows that  ≤  ≤  on Z.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).
Theorem 6. Assume that  >  (∞) and there exists a constant  > 6/ and  0 > | 0 | such that for || >  0 .Then (1) has a positive soliton and there exists a constant  > 0 and  2 ≥  0 such that Proof.Let  be the subsolution of (1) obtained in Theorem 5 and where  > 0 is a constant determined later.By simple calculation, we have By the assumption on , there exist  2 ≥  0 and  1 > 0 such that for  ≥  1 and || ≥  2 .
For any  ≥  1 , define We will show that  is upper-solution of (1) by appropriate choice of .In fact, This is impossible since  is bounded and so we must have lim  → +∞   = 0. Similarly, we also have lim  → −∞   = 0.The proof is complete.