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Journal of Applied Mathematics
Volume 2012 (2012), Article ID 159031, 17 pages
http://dx.doi.org/10.1155/2012/159031
Research Article

Quasilinearization of the Initial Value Problem for Difference Equations with “Maxima”

Faculty of Mathematics and Informatics, Plovdiv University, Tzar Asen 24, 4000 Plovdiv, Bulgaria

Received 26 May 2012; Accepted 6 July 2012

Academic Editor: Mehmet Sezer

Copyright © 2012 S. Hristova 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.

Abstract

The object of investigation of the paper is a special type of difference equations containing the maximum value of the unknown function over a past time interval. These equations are adequate models of real processes which present state depends significantly on their maximal value over a past time interval. An algorithm based on the quasilinearization method is suggested to solve approximately the initial value problem for the given difference equation. Every successive approximation of the unknown solution is the unique solution of an appropriately constructed initial value problem for a linear difference equation with “maxima,” and a formula for its explicit form is given. Also, each approximation is a lower/upper solution of the given mixed problem. It is proved the quadratic convergence of the successive approximations. The suggested algorithm is realized as a computer program, and it is applied to an example, illustrating the advantages of the suggested scheme.

1. Introduction

In the last few decades, great attention has been paid to automatic control systems and their applications to computational mathematics and modeling. Many problems in the control theory correspond to the maximal deviation of the regulated quantity. In the case when the dynamic of these problems is modeled discretely, the corresponding equations are called difference equations with “maxima”. The presence of the maximum function in the equation requires not only more complicated calculations but also a development of new methods for qualitative investigations of the behavior of their solutions(see, e.g., the monograph [1]). The character of the maximum function leads to variety of the types of difference equations. Some special types of difference equations are studied in [218]. At the same time, when the unknown function at any point is presented in both sides of the equation nonlinearly as well as it is involved in the maximum function, the given equation is not possible to be solved in an explicit form. It requires development of some approximate methods for their solving.

In the present paper, a nonlinear difference equation of delayed type is considered. The equation contains the maximum of the unknown function over a discrete past time interval. The main goal of the paper is suggesting an algorithm for an approximate solving of an initial value problem for the given difference equation.

2. Preliminary Notes

Let +=[0,), be the set of all integers, let 0 be a given fixed integer and 𝑎,𝑏 be such that a<𝑏. Denote by [𝑎,𝑏]={𝑧𝑎𝑧𝑏}.

Note that for any function 𝑄[𝑚,𝑛], 𝑚<𝑛, the equalities 𝑚𝑖=𝑛𝑄(𝑖)=0 and 𝑚𝑖=𝑛𝑄(𝑖)=1 hold.

Consider the following nonlinear difference equation with “maxima”: Δ𝑢(𝑘1)=𝑓𝑘,𝑢(𝑘),max[]𝑠𝑘,𝑘[]𝑢(𝑠)for𝑘𝑎+1,𝑇(2.1) with an initial condition [],𝑢(𝑘)=𝜑(𝑘)for𝑘𝑎+1,𝑎(2.2) where the functions 𝑢[𝑎+1,𝑇], Δ𝑢(𝑘1)=𝑢(𝑘)𝑢(𝑘1), 𝑓[𝑎+1,𝑇]×2, and 𝜑[𝑎+1,𝑎], the points 𝑎,𝑇 are such that 0𝑎<𝑇.

Definition 2.1. One will say that the function 𝛼[𝑎+1,𝑇] is a lower (upper) solution of the problem (2.1), (2.2), if Δ𝛼(𝑘1)()𝑓𝑘,𝛼(𝑘),max[𝑠𝑘,k][],𝛼[].𝛼(𝑠)for𝑘𝑎+1,𝑇(𝑘)()𝜑(𝑘)for𝑘𝑎+1,𝑎(2.3)

Let 𝛼,𝛽[𝑎+1,𝑇] be given functions such that 𝛼(𝑘)𝛽(𝑘) for 𝑘[𝑎+1,𝑇]. Define the following sets: [][]𝑆(𝛼,𝛽)={𝑢𝑎+1,𝑇𝛼(𝑘)𝑢(𝑘)𝛽(𝑘),𝑘𝑎+1,𝑇},Ω(𝛼,𝛽)=𝑘,𝑥1,𝑥2[]𝑎+1,𝑇×2𝛼(𝑘)𝑥1𝛽(𝑘),max𝑠[𝑘,𝑘]𝛼(𝑠)𝑥2max𝑠[𝑘,𝑘].𝛽(𝑠)(2.4)

3. Comparison Results

In our further investigations, we will use the following results for difference inequalities with “maxima”.

Lemma 3.1 (existence and uniqueness). Let the following conditions be fulfilled:(1)The function 𝑄[𝑎+1,𝑇].(2)The functions 𝑀[𝑎+1,𝑇], 𝑁[𝑎+1,𝑇]+ are such that [].𝑀(𝑘)+𝑁(𝑘)<1𝑓𝑜𝑟𝑘𝑎+1,𝑇(3.1)
Then the initial value problem for linear difference equation with “maxima” Δ𝑢(𝑘1)=𝑄(𝑘)+𝑀(𝑘)𝑢(𝑘)+𝑁(𝑘)max[]𝑠𝑘,𝑘[],𝑢[]𝑢(𝑠)𝑓𝑜𝑟𝑘𝑎+1,𝑇(𝑘)=𝜑(𝑘)𝑓𝑜𝑟𝑘𝑎+1,𝑎(3.2) has an unique solution on the interval [𝑎+1,𝑇].

Proof. We will use the step method to solve the initial value problem (3.2). Assume for a fixed 𝑘[𝑎+1,𝑇] all values 𝑢(𝑗), 𝑗[a+1,𝑘1] are known. Then from (3.2), we obtain (1𝑀(𝑘))𝑢(𝑘)=𝑢(𝑘1)+𝑄(𝑘)+𝑁(𝑘)max𝑠[𝑘,𝑘]𝑢(𝑠).
Consider the following two possible cases.
Case  1. Let max𝑙[1,]𝑢(𝑘𝑙)(𝑄(𝑘)+𝑢(𝑘1))/(1𝑀(𝑘)𝑁(𝑘)), or (𝑄(𝑘)+𝑢(𝑘1))/(1𝑀(𝑘)𝑁(𝑘))𝑢(𝑘𝑙) for 𝑙[1,].
Therefore, max𝑠[𝑘,𝑘]𝑢(𝑠)=𝑢(𝑘). Then applying the inequality (3.1), we obtain the unique solution 𝑢(𝑘) of the problem (3.2) given by the equality 𝑢(𝑘)=𝑄(𝑘)+𝑢(𝑘1).1𝑀(𝑘)𝑁(𝑘)(3.3)
Case  2. Let (𝑄(𝑘)+𝑢(𝑘1))/(1𝑀(𝑘)𝑁(𝑘))<max𝑙[1,]𝑢(𝑘𝑙)=𝑢(𝑘𝑚) where 𝑚[1,].
If we assume 𝑢(𝑘)𝑢(𝑘𝑚) then from (3.2), we obtain 𝑢(𝑘𝑚)𝑢(𝑘)=(𝑄(𝑘)+𝑢(𝑘1))/(1𝑀(𝑘)𝑁(𝑘))<𝑢(𝑘𝑚). The obtained contradiction proves 𝑢(𝑘)<𝑢(𝑘𝑚) and max𝑠[𝑘,𝑘]𝑢(𝑠)=𝑢(𝑘𝑚).
From the inequalities (3.1) and 𝑁(𝑘)0, it follows that 𝑀(𝑘)<1 and therefore, the unique solution of problem (3.2) is given by the equality 𝑢(𝑘)=𝑢(𝑘1)+𝑄(𝑘)+𝑁(𝑘)𝑢(𝑘𝑚).1𝑀(𝑘)(3.4)
Thus, we receive the value 𝑢(𝑘), 𝑘𝑍[𝑎+1,𝑇].

Lemma 3.2. Let the following conditions be fulfilled:(1) The functions 𝑀,𝑁[𝑎+1,𝑇] satisfy the inequality [].𝑀(𝑘)+𝑁(𝑘)<1𝑓𝑜𝑟𝑘𝑎+1,𝑇(3.5)(2) The function 𝑢[𝑎+1,𝑇] satisfies the inequalities Δ𝑢(𝑘1)𝑀(𝑘)𝑢(𝑘)+𝑁(𝑘)max[]𝑠𝑘,𝑘[],𝑢[].𝑢(𝑠),𝑘𝑎+1,𝑇(𝑘)0,𝑘𝑎+1,𝑎(3.6)
Then 𝑢(𝑘)0 for 𝑘[𝑎+1,𝑇].

Proof. Assume the claim of Lemma 3.2 is not true. Then there exists 𝑗[𝑎+1,𝑇] such that 𝑢(𝑗)>0 and 𝑢(𝑘)0 for 𝑘[𝑎+1,𝑗1]. Therefore, max𝑠[𝑗,𝑗]𝑢(𝑠)=𝑢(𝑗) and according to inequality (3.6), we get 𝑢(𝑗)𝑢(𝑗)𝑢(𝑗1)(𝑀(𝑗)+𝑁(𝑗))𝑢(𝑗).(3.7)
The above inequality contradicts (3.5).

Remark 3.3. Note if both 𝑀(𝑘),𝑁(𝑘)0 for 𝑘[𝑎+1,𝑇], then inequality (3.5) is satisfied.

Now, we will prove a linear difference inequality in which at any point both the unknown function and its maximum over a past time interval are involved also in the right side of the inequality.

In the proof of our preliminary results, we will need the following lemma.

Lemma 3.4 (see [2], Theorem  4.1.1). Let 𝑓,𝑞(𝑎,)+, 𝑝,𝑢(𝑎,) and 𝑢(𝑘)𝑝(𝑘)+𝑞(𝑘)𝑘1𝑙=𝑎𝑓(𝑙)𝑢(𝑙)𝑓𝑜𝑟𝑘(𝑎,).(3.8)
Then for all 𝑘(𝑎,), the following inequality is valid: 𝑢(𝑘)𝑝(𝑘)+𝑞(𝑘)𝑘1𝑙=𝑎𝑝(𝑙)𝑓(𝑙)𝑘1𝜏=𝑙+1(1+𝑞(𝜏)𝑓(𝜏)).(3.9)

Now, we will solve a generalized linear difference inequality with “maxima.”

Lemma 3.5. Let the following conditions be fulfilled:(1) The functions 𝑞,𝑄[𝑎+1,𝑇]+ and [].𝑞(𝑘)+𝑄(𝑘)<1𝑓𝑜𝑟𝑘𝑎+1,T(3.10)(2) The function 𝑢[𝑎+1,𝑇]+ and satisfies the inequalities 𝑢(𝑘)𝐶+𝑘𝑙=𝑎+1𝑞(𝑙)𝑢(𝑙)+𝑘𝑙=𝑎+1𝑄(𝑙)max[]𝑠𝑙,𝑙[],[],𝑢(𝑠),𝑘𝑎+1,𝑇𝑢(𝑘)𝐶,𝑘𝑎+1,𝑎(3.11) where 𝐶=const0.
Then, 𝐶𝑢(𝑘)𝑘𝜏=𝑎+1[].(1𝑞(𝜏)𝑄(𝜏))𝑓𝑜𝑟𝑘𝑎+1,𝑇(3.12)

Proof. Define a function 𝑧[𝑎+1,𝑇]+ by the equalities 𝑧(𝑘)=𝐶+𝑘𝑙=𝑎+1𝑞(𝑙)𝑢(𝑙)+𝑘𝑙=𝑎+1𝑄(𝑙)max𝑠[𝑙,𝑙][],[].𝑢(𝑠),𝑘𝑎+1,𝑇𝐶,𝑘𝑎+1,𝑎(3.13)
It is easy to check that for any 𝑘[𝑎+1,𝑇], the inequality 𝑧(𝑘+1)𝑧(𝑘) holds. Also, from the definition of 𝑧(𝑘), it follows that 𝑢(𝑘)𝑧(𝑘),𝑘[𝑎+1,𝑇], and max𝑠[𝑘,𝑘]𝑢(𝑠)max𝑠[𝑘,𝑘]𝑧(𝑠)=𝑧(𝑘) for 𝑘[𝑎+1,𝑇]. Therefore, for any 𝑘[𝑎+1,𝑇], we obtain 𝑧(𝑘)𝐶+𝑘𝑙=𝑎+1(𝑞(𝑙)+𝑄(𝑙))𝑧(𝑙).(3.14)
From inequality (3.14), it follows 𝐶𝑧(𝑘)+1(1𝑞(𝑘)𝑄(𝑘))(1𝑞(𝑘)𝑄(𝑘))𝑘1𝑙=𝑎+1(𝑞(𝑙)+𝑄(𝑙))𝑧(𝑙).(3.15)
According to Lemma 3.4 from inequality (3.15), we get for 𝑘[𝑎+1,𝑇]𝐶𝑧(𝑘)×(1𝑞(𝑘)𝑄(𝑘))1+𝑘1𝑙=𝑎+1𝑞(𝑙)+𝑄(𝑙)1𝑞(𝑙)𝑄(𝑙)𝑘1𝜏=𝑙+1𝑞1+(𝜏)+𝑄(𝜏)=𝐶1𝑞(𝜏)𝑄(𝜏)×(1𝑞(𝑘)𝑄(𝑘))1+𝑘1𝑙=𝑎+1𝑞(𝑙)+𝑄(𝑙)11𝑞(𝑙)𝑄(𝑙)𝑘1𝜏=𝑙+1𝐶(1𝑞(𝜏)𝑄(𝜏))𝑘𝜏=𝑎+1.(1𝑞(𝜏)𝑄(𝜏))(3.16) Inequality (3.16) implies the validity of the required inequality (3.12).

4. Quasilinearization

We will apply the method of quasilinearization to obtain approximate solution of the IVP for the nonlinear difference equation with “maxima’’ (2.1), (2.2). We will prove the convergence of the sequence of successive approximations is quadratic.

Theorem 4.1. Let the following conditions be fulfilled:(1) The functions 𝛼0,𝛽0[𝑎+1,𝑇] are a lower and an upper solutions of the problem (2.1), (2.2), respectively, and such that 𝛼0(𝑘)𝛽0(𝑘) for 𝑘[𝑎+1,𝑇].(2) The function 𝑓[𝑎+1,𝑇]×2 satisfies for (𝑘,𝑥,𝑦)Ω(𝛼0,𝛽0) the equality 𝑓(𝑘,𝑥,𝑦)=𝐹(𝑘,𝑥,𝑦)𝐺(𝑘,𝑥,𝑦),(4.1)where the functions 𝐹,𝐺Ω(𝛼0,𝛽0) are continuous and twice continuously differentiable with respect to their second and third arguments and the following inequalities are valid for 𝑘[𝑎+1,𝑇], (𝑘,𝑥,𝑦)Ω(𝛼0,𝛽0): 𝐹𝑥𝑥(𝑘,𝑥,𝑦)0,𝐹𝑥𝑦(𝑘,𝑥,𝑦)0,𝐹𝑦𝑦𝐺(𝑘,𝑥,𝑦)0,𝑥𝑥(𝑘,𝑥,𝑦)0,𝐺𝑥𝑦(𝑘,𝑥,𝑦)0,𝐺𝑦𝑦𝐺(𝑘,𝑥,𝑦)0,(4.2)𝑥𝑘,𝛽0(𝑘),max[]𝑠𝑘,𝑘𝛽0(𝑠)𝐹𝑥𝑘,𝛼0(𝑘),max[]𝑠𝑘,𝑘𝛼0𝐺(𝑠),(4.3)𝑦𝑘,𝛽0(𝑘),max[]𝑠𝑘,𝑘𝛽0(𝑠)𝐹𝑦𝑘,𝛼0(𝑘),max[]𝑠𝑘,𝑘𝛼0(𝑠),(4.4)𝑀(𝑘)+𝑁(𝑘)<1,(4.5)where 𝑀(𝑘)=𝐹𝑥𝑘,𝛽0(𝑘),max[]𝑠𝑘,𝑘𝛽0(𝑠)𝐺𝑥𝑘,𝛼0(𝑘),max[]𝑠𝑘,𝑘𝛼0,(𝑠)𝑁(𝑘)=𝐹𝑦𝑘,𝛽0(𝑘),max[]𝑠𝑘,𝑘𝛽0(𝑠)𝐺𝑦𝑘,𝛼0(𝑘),max[]𝑠𝑘,𝑘𝛼0.(𝑠)(4.6)(3) The function 𝜑[𝑎+1,𝑎].Then there exist two sequences {𝛼𝑛(𝑘)}𝑛=0 and {𝛽𝑛(𝑘)}𝑛=0, 𝑘[𝑎+1,𝑇], such that:(a) The functions 𝛼𝑛[𝑎+1,𝑇], (𝑛=1,2,) are lower solutions of IVP (2.1), (2.2).(b) The functions 𝛽𝑛[𝑎+1,𝑇], (𝑛=1,2,) are upper solutions of IVP (2.1), (2.2).(c) The following inequalities hold for 𝑘[𝑎+1,𝑇]𝛼0(𝑘)𝛼1(𝑘)𝛼𝑛(𝑘)𝛽𝑛(𝑘)𝛽1(𝑘)𝛽0(𝑘).(4.7)(d) Both sequences are convergent on [𝑎+1,𝑇] and their limits 𝑉(𝑘)=lim𝑛𝛼𝑛(𝑘) and 𝑊(𝑘)=lim𝑛𝛽𝑛(𝑘) are the minimal and the maximal solutions of IVP (2.1), (2.2) in 𝑆(𝛼0,𝛽0). In the case, IVP (2.1), (2.2) has an unique solution in 𝑆(𝛼0,𝛽0) both limits coincide, that is, 𝑉(𝑘)=𝑊(𝑘).(e) The convergence is quadratic, that is, there exist constants 𝜆𝑖,𝜇𝑖>0,𝑖=1,2 such that for the solution 𝑥(𝑘) of IVP (2.1), (2.2) in 𝑆(𝛼0,𝛽0), the inequalities 𝑥𝛼𝑛+1𝜆1𝑥𝛼𝑛2+𝜆2𝑥𝛽𝑛2,𝑥𝛽𝑛+1𝜇1𝑥𝛼𝑛2+𝜇2𝑥𝛽𝑛2(4.8)hold, where ||𝑢||=max𝑠[𝑎+1,𝑇]|𝑢(𝑠)| for any function 𝑢[𝑎+1,𝑇].

Proof. From Taylor formula and condition (2) of Theorem 4.1 for (𝑘,𝑥1,𝑦1), (𝑘,𝑥2,𝑦2)Ω(𝛼0,𝛽0), the following inequalities are valid: 𝑓𝑘,𝑥2,𝑦2𝑓𝑘,𝑥1,𝑦1𝐹𝑥𝑘,𝑥2,𝑦2𝑥2𝑥1+𝐹𝑦𝑘,𝑥2,𝑦2𝑦2𝑦1𝐺𝑘,𝑥2,𝑦2+𝐺𝑘,𝑥1,𝑦1,𝐺𝑘,𝑥2,𝑦2+𝐺𝑘,𝑥1,𝑦1𝐺𝑥𝑘,𝑥1,𝑦1𝑥1𝑥2+𝐺𝑦𝑘,𝑥1,𝑦1𝑦1𝑦2.(4.9)
Consider the initial value problem for the linear difference equation with “maxima” Δ𝑥(𝑘1)=𝑄0(𝑘)𝑥(𝑘)+𝑞0(𝑘)max[]𝑠𝑘,𝑘𝑥(𝑠)+𝑃0[],𝑥[],(𝑘),𝑘𝑎+1,𝑇(4.10)(𝑘)=𝜑(𝑘),𝑘𝑎+1,𝑎(4.11) where the functions 𝑃0,𝑄0,𝑞0[𝑎+1,𝑇] are defined by the equalities 𝑃0(𝑘)=𝑓𝑘,𝛼0(𝑘),max[]𝑠𝑘,𝑘𝛼0(𝑠)𝑄0(𝑘)𝛼0(𝑘)𝑞0(𝑘)max[]𝑠𝑘,𝑘𝛼0𝑄(𝑠),0(𝑘)=𝐹𝑥𝑘,𝛼0(𝑘),max[]𝑠𝑘,𝑘𝛼0(𝑠)𝐺𝑥𝑘,𝛽0(𝑘),max[]𝑠𝑘,𝑘𝛽0,𝑞(𝑠)0(𝑘)=𝐹𝑦𝑘,𝛼0(𝑘),max[]𝑠𝑘,𝑘𝛼0(𝑠)𝐺𝑦𝑘,𝛽0(𝑘),max[]𝑠𝑘,𝑘𝛽0.(𝑠)(4.12)
From inequality (4.4), it follows that 𝑞0(𝑘)0 and from inequalities (4.2), (4.5), we get 𝑄0(𝑘)+𝑞0(𝑘)<1 for 𝑘[𝑎+1,𝑇]. According to Lemma 3.1 the IVP (4.10), (4.11) has an unique solution 𝛼1(𝑘), defined on the interval [𝑎+1,𝑇].
Define a function 𝑝1[𝑎+1,𝑇] by the equality 𝑝1(𝑘)=𝛼0(𝑘)𝛼1(𝑘). Then we get 𝑝1(𝑘)=0 for 𝑘[𝑎+1,𝑎].
Let 𝑘[𝑎+1,𝑇]. From the choice of the function 𝛼0(𝑘) and (4.10) for the function 𝛼1(𝑘), we get Δ𝑝1(𝑘1)𝑄0(𝑘)𝑝1(𝑘)+𝑞0(𝑘)max[]𝑠𝑘,𝑘𝑝1(𝑠).(4.13)
According to Lemma 3.2 for the function 𝑝1(𝑘), it follows that 𝑝1(𝑘)0 for 𝑘[𝑎+1,𝑇]. Therefore, 𝛼0(𝑘)𝛼1(𝑘) for 𝑘[𝑎+1,𝑇].
Consider the linear difference equation with “maxima” Δ𝑥(𝑘1)=𝑄0(𝑘)𝑥(𝑘)+𝑞0(𝑘)max[]𝑠𝑘,𝑘𝑥(𝑠)+𝑅0[],𝑥[],(𝑘),𝑘𝑎+1,𝑇(4.14)(𝑘)=𝜑(𝑘),𝑘𝑎+1,𝑎(4.15) where the functions 𝑞0(𝑘) and 𝑄0(𝑘) are defined by equalities (4.12) and 𝑅0(𝑘)=𝑓𝑘,𝛽0(𝑘),max[]𝑠𝑘,𝑘𝛽0(𝑠)𝑄0(𝑘)𝛽0(𝑘)𝑞0(𝑘)max[]𝑠𝑘,𝑘𝛽0(𝑠).(4.16)
According to Lemma 3.1, the linear initial value problem (4.14), (4.15) has a unique solution 𝛽1(𝑘), defined on the interval [𝑎+1,𝑇].
Define a function 𝑝2[𝑎+1,𝑇] by the equality 𝑝2(𝑘)=𝛽1(𝑘)𝛽0(𝑘). Then 𝑝2(𝑘)=0 for 𝑘[𝑎+1,𝑎].
Now, let 𝑘[𝑎+1,𝑇]. From the choice of the function 𝛽0(𝑘) and (4.14) for the function 𝛽1(𝑘), we get Δ𝑝2(𝑘1)𝑄0𝛽(𝑘)1(𝑘)𝛽0(𝑘)+𝑞0(𝑘)max[]𝑠𝑘,𝑘𝛽1(𝑠)max[]𝑠𝑘,𝑘𝛽0(𝑠)𝑄0(𝑘)𝑝2(𝑘)+𝑞0(𝑘)max[]𝑠𝑘,𝑘𝑝2(𝑠).(4.17)
Inequality (4.17) proves the function 𝑝2(𝑘) satisfies inequality (3.6). According to Lemma 3.2, it follows that 𝑝2(𝑘)0 for 𝑘[𝑎+1,𝑇]. Therefore, 𝛽1(𝑘)𝛽0(𝑘) for 𝑘[𝑎+1,𝑇].
Define a function 𝑝3[𝑎+1,𝑇] by the equality 𝑝3(𝑘)=𝛼1(𝑘)𝛽1(𝑘). Then 𝑝3(𝑘)0 for 𝑘[𝑎+1,𝑎].
Let 𝑘[𝑎+1,𝑇]. Then for the function 𝑝3(𝑘), we get Δ𝑝3(𝑘1)=𝑓𝑘,𝛼0(𝑘),max[]𝑠𝑘,𝑘𝛼0(𝑠)𝑓𝑘,𝛽0(𝑘),max[]𝑠𝑘,𝑘𝛽0(𝑠)+𝑄0𝛼(𝑘)1(𝑘)𝛼0(𝑘)𝑄0𝛽(𝑘)1(𝑘)𝛽0(𝑘)+𝑞0(𝑘)max[]𝑠𝑘,𝑘𝛼1(𝑠)max[]𝑠𝑘,𝑘𝛼0(𝑠)𝑞0(𝑘)max[]𝑠𝑘,𝑘𝛽1(𝑠)max[]𝑠𝑘,𝑘𝛽0(𝑠)𝑄0(𝑘)𝑝3(𝑘)+𝑞0(𝑘)max[]𝑠𝑘,𝑘𝑝3(𝑠).(4.18)
Inequality (4.18) proves the function 𝑝3(𝑘) satisfies inequality (3.6). According to Lemma 3.2, it follows that 𝑝3(𝑘)0 for 𝑘[𝑎+1,𝑇]. Therefore, 𝛼1(𝑘)𝛽1(𝑘) for 𝑘[𝑎+1,𝑇].
Furthermore, the functions 𝛼1(𝑘) and 𝛽1(𝑘)𝑆(𝛼0,𝛽0).
Now, we will prove that the function 𝛼1(𝑘) is a lower solution of (2.1), (2.2) on the interval [𝑎+1,𝑇].
Let 𝑘[𝑎+1,𝑇]. From the inequalities 𝛼0(𝑘)𝛼1(𝑘)𝛽0(𝑘) for 𝑘[𝑎+1,𝑇], max𝑠[𝑘,𝑘]𝛼0(𝑠)max𝑠[𝑘,𝑘]𝛼1(𝑠)max𝑠[𝑘,𝑘]𝛽0(𝑠) for 𝑘[𝑎+1,𝑇], inequalities (4.9), definitions (4.12), and inequalities (4.2) which prove the monotonic property of the first derivatives of the functions 𝐹(𝑘,𝑥,𝑦) and 𝐺(𝑘,𝑥,𝑦) we get Δ𝛼1(𝑘1)=𝑓𝑘,𝛼0(𝑘),max[]𝑠𝑘,𝑘𝛼0(𝑠)+𝑄0𝛼(𝑘)1(𝑘)𝛼0(𝑘)+𝑞0(𝑘)max[]𝑠𝑘,𝑘𝛼1(𝑠)max[]𝑠𝑘,𝑘𝛼0(𝑠)=𝑓𝑘,𝛼1(𝑘),max[]𝑠𝑘,𝑘𝛼1(𝑠)+𝑄0𝛼(𝑘)1(𝑘)𝛼0+𝑓(𝑘)𝑘,𝛼0(𝑘),max[]𝑠𝑘,𝑘𝛼0(𝑠)𝑓𝑘,𝛼1(𝑘),max[]𝑠𝑘,𝑘𝛼1(𝑠)+𝑞0(𝑘)max[]𝑠𝑘,𝑘𝛼1(𝑠)max[]𝑠𝑘,𝑘𝛼0(𝑠)𝑓𝑘,𝛼1(𝑘),max[]𝑠𝑘,𝑘𝛼1.(𝑠)(4.19)
Thus, the function 𝛼1(𝑘) is a lower solution of (2.1), (2.2) on [𝑎+1,𝑇].
In a similar way, we can prove that the function 𝛽1(𝑘) is an upper solution of (2.1), (2.2) on the interval [𝑎+1,𝑇].
Analogously, we can construct two sequences of functions {𝛼𝑛(𝑘)}𝑛=1 and {𝛽𝑛(𝑘)}𝑛=1. If the functions 𝛼𝑗(𝑘) and 𝛽𝑗(𝑘), 𝑗=1,2,,𝑛1,𝑛, are obtained such that 𝛼𝑗,𝛽𝑗𝑆(𝛼0,𝛽0) and the claims (a), (b), (c) of Theorem 4.1 are satisfied, then we consider the initial value problem for the linear difference equation with “maxima” Δ𝑥(𝑘1)=𝑄𝑛(𝑘)𝑥(𝑘)+𝑞𝑛(𝑘)max[]𝑠𝑘,𝑘𝑥(𝑠)+𝑃𝑛[],𝑥[],(𝑘),𝑘𝑎+1,𝑇(4.20)(𝑘)=𝜑(𝑘),𝑘𝑎+1,𝑎(4.21) and the initial value problem for the linear difference equation with “maxima’’ Δ𝑥(𝑘1)=𝑄𝑛(𝑘)𝑥(𝑘)+𝑞𝑛(𝑘)max[]𝑠𝑘,𝑘𝑥(𝑠)+𝑅𝑛[],𝑥[],(𝑘),𝑘𝑎+1,𝑇(𝑘)=𝜑(𝑘),𝑘𝑎+1,𝑎(4.22) where 𝑃𝑛(𝑘)=𝑓𝑘,𝛼𝑛(𝑘),max[]𝑠𝑘,𝑘𝛼𝑛(𝑠)𝑄𝑛(𝑘)𝛼𝑛(𝑘)𝑞𝑛(𝑘)max[]𝑠𝑘,𝑘𝛼𝑛𝑅(𝑠),𝑛(𝑘)=𝑓𝑘,𝛽𝑛(𝑘),max[]𝑠𝑘,𝑘𝛽𝑛(𝑠)𝑄𝑛(𝑘)𝛽𝑛(𝑘)𝑞𝑛(𝑘)max[]𝑠𝑘,𝑘𝛽𝑛𝑄(𝑠),𝑛(𝑘)=𝐹𝑥𝑘,𝛼𝑛(𝑘),max[]𝑠𝑘,𝑘𝛼𝑛(𝑠)𝐺𝑥𝑘,𝛽𝑛(𝑘),max[]𝑠𝑘,𝑘𝛽𝑛𝑞(𝑠)0,𝑛(𝑘)=𝐹𝑦𝑘,𝛼𝑛(𝑘),max[]𝑠𝑘,𝑘𝛼𝑛(𝑠)𝐺𝑦𝑘,𝛽𝑛(𝑘),max[]𝑠𝑘,𝑘𝛽𝑛(𝑠)0.(4.23)
Since 𝛼𝑗,𝛽𝑗𝑆(𝛼0,𝛽0), the first derivatives of the function 𝐹(𝑘,𝑥,𝑦) and 𝐺(𝑘,𝑥,𝑦) are nondecreasing in Ω(𝛼0,𝛽0), and inequalities (4.7) hold, we obtain 𝑄𝑛(𝑘)𝑀(𝑘), 𝑞𝑛(𝑘)𝑞0(𝑘)0 and 𝑞𝑛(𝑘)𝑁(𝑘), that is, 𝑄𝑛(𝑘)+𝑞𝑛(𝑘)𝑀(𝑘)+𝑁(𝑘)<1. Therefore, according to Lemma 3.1, the initial value problems (4.20), (4.21), and (4.22) have unique solutions 𝛼𝑛+1(𝑘) and 𝛽𝑛+1(𝑘), 𝑘[𝑎+1,𝑇], correspondingly.
The proof that the functions 𝛼𝑛+1,𝛽𝑛+1𝑆(𝛼𝑛,𝛽𝑛), 𝛼𝑛+1(𝑘)𝛽𝑛+1(𝑘), and they are lower/upper solutions of (2.1), (2.2) on the interval [𝑎+1,𝑇] is the same as in the case of 𝑛=1 and we omit it.
For any fixed 𝑘[𝑎+1,𝑇], the sequences {𝛼𝑛(𝑘)}𝑛=0 and {𝛽𝑛(𝑘)}𝑛=0 are monotone nondecreasing and monotone nonincreasing, respectively, and they are bounded by 𝛼0(𝑘) and 𝛽0(𝑘). Therefore, they are convergent on [𝑎+1,𝑇], that is, there exist functions 𝑉,𝑊[𝑎+1,𝑇] such that lim𝑛𝛼𝑛(𝑘)=𝑉(𝑘),lim𝑛𝛽𝑛(𝑘)=𝑊(𝑘).(4.24)
From inequalities (4.7), it follows that 𝑉,𝑊𝑆(𝛼0,𝛽0).
Now, we will prove that for any 𝑘[𝑎+1,𝑇] the following equality holds: lim𝑛max[]𝜉𝑘,𝑘𝛼𝑛(𝜉)=max𝜉[𝑘,𝑘]lim𝑛𝛼𝑛(𝜉).(4.25)
Let 𝑘[𝑎+1,𝑇] be fixed. We denote max𝜉[𝑘,𝑘]𝛼𝑛(𝜉)=𝐴𝑛. From inequalities (4.7) for every 𝜉[𝑘,𝑘], the inequalities 𝛼𝑛1(𝜉)𝛼𝑛(𝜉)𝐴𝑛 hold and thus, 𝐴𝑛1𝐴𝑛, that is, the sequence {𝐴𝑛}𝑛=0 is monotone nondecreasing and bounded from above by 𝛽0(𝑘). Therefore, there exists the limit 𝐴=lim𝑛𝐴𝑛.
From the monotonicity of the sequence of the lower solutions 𝛼𝑛(𝑘), we get that for 𝜉[𝑘,𝑘] it is fulfilled 𝛼𝑛(𝜉)𝑉(𝜉). Let 𝜂[𝑘,𝑘] be such that max𝜉[𝑘,𝑘]𝑉(𝜉)=𝑉(𝜂). From the inequalities 𝛼𝑛(𝜂)𝐴𝑛𝐴 for every 𝑛=0,1,2, it follows 𝑉(𝜂)𝐴. Assume that 𝑉(𝜂)<𝐴. Then there exists a natural number 𝑁 such that the inequalities 𝑉(𝜂)<𝐴𝑁𝐴 hold. Therefore, there exists 𝜉[𝑘,𝑘] such that 𝛼𝑁(𝜉)=max𝜉[𝑘,𝑘]𝛼𝑁(𝜉)=𝐴𝑁 or 𝑉(𝜂)<𝛼𝑁(𝜉)𝑉(𝜉). The obtained contradiction proves the validity of the required inequality (4.25).
Analogously, we can show that the functions 𝛽𝑛(𝑘) also satisfy (4.25).
Now, we will prove that the function 𝑉(𝑘) is a solution of the IVP (2.1), (2.2) on [𝑎+1,𝑇].
Let 𝑘[𝑎+1,𝑎]. Take a limit as 𝑛 in (4.21) and get 𝑉(𝑘)=𝜑(𝑘).
Therefore, the function 𝑉(𝑘) satisfies equality (2.2) for 𝑘[𝑎+1,𝑎].
Let 𝑘[𝑎+1,𝑇]. Taking a limit in (4.20) as 𝑛 and applying (4.25), we obtain the function 𝑉(𝑘) satisfies equality (2.1) for 𝑘[𝑎+1,𝑇].
In a similar way, we can prove that 𝑊(𝑘) is a solution of the IVP (2.1), (2.2).
Therefore, we obtain two solutions of (2.1), (2.2) in 𝑆(𝛼0,𝛽0).
In the case of uniqueness of the solution of (2.1), (2.2) in 𝑆(𝛼0,𝛽0), we have 𝑉(𝑘)=𝑊(𝑘) for 𝑘[𝑎+1,𝑇]. In the case of nonuniqueness, let 𝑢𝑆(𝛼0,𝛽0) be another solution of (2.1), (2.2). Then it is easy to prove that 𝑉(𝑘)𝑢(𝑘)𝑊(𝑘), that is, 𝑉 is the minimal solution and 𝑊 is the maximal solution of (2.1), (2.2) in 𝑆(𝛼0,𝛽0).
We will prove that the convergence of the sequences {𝛼𝑛(𝑘)}𝑛=0, {𝛽𝑛(𝑘)}𝑛=0 is quadratic. Let 𝑥(𝑘) be a solution of (2.1), (2.2) in 𝑆(𝛼0,𝛽0).
Define the functions 𝐴𝑛+1,𝐵𝑛+1[𝑎+1,𝑇]+, 𝑛=0,1, by the equalities 𝐴𝑛+1(𝑘)=𝑥(𝑘)𝛼𝑛+1(𝐵𝑘),𝑛+1(𝑘)=𝛽𝑛+1(𝑘)𝑥(𝑘).(4.26)
It is obvious that 𝐴𝑛+1(𝑘)=0 for 𝑘[𝑎+1,𝑎].
Let 𝑘[𝑎+1,𝑇]. According to the definitions of the functions 𝐴𝑛+1(𝑘), 𝛼𝑛+1(𝑘) and the condition (2) of Theorem 4.1, we get Δ𝐴𝑛+1(𝑘1)=𝑓𝑘,𝑥(𝑘),max[]𝑠𝑘,𝑘𝑥(𝑠)𝑓𝑘,𝛼𝑛(𝑘),max[]𝑠𝑘,𝑘𝛼𝑛(𝑠)+𝑄𝑛𝐴(𝑘)𝑛+1(𝑘)+𝑞𝑛(𝑘)max[]𝑠𝑘,𝑘𝑥(𝑠)max[]𝑠𝑘,𝑘𝛼𝑛+1(𝑠)𝑄𝑛𝐴(𝑘)𝑛(𝑘)𝑞𝑛(𝑘)max[𝑠k],𝑘𝑥(𝑠)max[]𝑠𝑘,𝑘𝛼𝑛(𝑠)𝑄𝑛𝐴(𝑘)𝑛+1(𝑘)+𝑞𝑛(𝑘)max𝑠[𝑘,𝑘]𝐴𝑛+1+𝐹(𝑠)𝑥𝑘,𝑥(𝑘),max[]𝑠𝑘,𝑘𝑥(𝑠)𝐺𝑥𝑘,𝛼𝑛(𝑘),max[]𝑠𝑘,𝑘𝛼𝑛(𝑠)𝑄𝑛𝐴(𝑘)𝑛+𝐹(𝑘)𝑦𝑘,x(𝑘),max[]𝑠𝑘,𝑘𝑥(𝑠)𝐺𝑦𝑘,𝛼𝑛(𝑘),max[]𝑠𝑘,𝑘𝛼𝑛(𝑠)𝑞𝑛×(𝑘)max[]𝑠𝑘,𝑘𝑥(𝑠)max[]𝑠𝑘,𝑘𝛼𝑛(.𝑠)(4.27)
According to the mean value theorem, there exist points 𝜁𝑖 and 𝜉𝑗,𝑖=1,3,𝑗=2,4 such that 𝛼𝑛(𝑘)𝜁𝑖𝑥(𝑘),max[]𝑠𝑘,𝑘𝛼𝑛(𝑠)𝜁𝑗max[]𝑠𝑘,𝑘𝛼𝑥(𝑠),𝑛(𝑘)𝜉𝑖𝛽𝑛(𝑘),max[]𝑠𝑘,𝑘𝛼𝑛(𝑠)𝜉𝑗max[]𝑠𝑘,𝑘𝛽𝑛𝐹(𝑠),(4.28)𝑥𝑘,𝑥(𝑘),max[]𝑠𝑘,𝑘𝑥(𝑠)𝐺𝑥𝑘,𝛼𝑛(𝑘),max[]𝑠𝑘,𝑘𝛼𝑛(𝑠)𝑄𝑛(𝑘)=𝐹𝑥𝑘,𝑥(𝑘),max[]𝑠𝑘,𝑘𝑥(𝑠)𝐹𝑥𝑘,𝛼𝑛(𝑘),max[]𝑠𝑘,𝑘𝑥(𝑠)+𝐹𝑥𝑘,𝛼𝑛(𝑘),max[]𝑠𝑘,𝑘𝑥(𝑠)𝐹𝑥𝑘,𝛼𝑛(𝑘),max[]𝑠𝑘,𝑘𝛼𝑛(𝑠)+𝐺𝑥𝑘,𝛽𝑛(𝑘),max[]𝑠𝑘,𝑘𝛽𝑛(𝑠)𝐺𝑥𝑘,𝛼𝑛(𝑘),max[]𝑠𝑘,𝑘𝛽𝑛(𝑠)+𝐺𝑥𝑘,𝛼𝑛(𝑘),max[]𝑠𝑘,𝑘𝛽𝑛(𝑠)𝐺𝑥𝑘,𝛼𝑛(𝑘),max[]𝑠𝑘,𝑘𝛼𝑛(𝑠)=𝐹𝑥𝑥𝑘,𝜁1,max[]𝑠𝑘,𝑘𝐴𝑥(𝑠)𝑛(𝑘)+𝐹𝑥𝑦𝑘,𝛼𝑛(𝑘),𝜁2max[]𝑠𝑘,𝑘𝑥(𝑠)max[]𝑠𝑘,𝑘𝛼𝑛(𝑠)+𝐺𝑥𝑥𝑘,𝜁1,max[]𝑠𝑘,𝑘𝛽𝑛(𝛽𝑠)𝑛(𝑘)𝛼𝑛(𝑘)+𝐺𝑥𝑦𝑘,𝛼𝑛(𝑘),𝜁2max[]𝑠𝑘,𝑘𝛽𝑛(𝑠)max[]𝑠𝑘,𝑘𝛼𝑛,𝐹(𝑠)(4.29)𝑦𝑘,𝑥(𝑘),max[]𝑠𝑘,𝑘𝑥(𝑠)𝐺𝑦𝑘,𝛼𝑛(𝑘),max[]𝑠𝑘,𝑘𝛼𝑛(𝑠)𝑞𝑛(𝑘)=𝐹𝑦𝑥𝑘,𝜉3,max[]𝑠𝑘,𝑘𝐴𝑥(𝑠)𝑛(𝑘)+𝐹𝑦𝑦𝑘,𝛼𝑛(𝑘),𝜉4max[]𝑠𝑘,𝑘𝑥(𝑠)max[]𝑠𝑘,𝑘𝛼𝑛(𝑠)+𝐺𝑦𝑥𝑘,𝜉3,max[]𝑠𝑘,𝑘𝛽𝑛(𝛽𝑠)𝑛(𝑘)𝛼𝑛(𝑘)+𝐺𝑦𝑦𝑘,𝛼𝑛(𝑘),𝜉4max[]𝑠𝑘,𝑘𝛽𝑛(𝑠)max[]𝑠𝑘,𝑘𝛼𝑛.(𝑠)(4.30)
Then the following inequalities are valid: 𝐴𝑛𝛽(𝑘)𝑛(𝑘)𝛼𝑛𝐴(𝑘)𝑛𝐴(𝑘)𝑛𝐵(𝑘)+𝑛3(𝑘)2𝐴2𝑛1(𝑘)+2𝐵2𝑛𝐴(𝑘),𝑛(𝑘)max[]𝑠𝑘,𝑘𝛽𝑛(𝑠)max[]𝑠𝑘,𝑘𝛼𝑛3(𝑠)2𝐴𝑛2+12𝐵𝑛2,𝐴𝑛(𝑘)max[]𝑠𝑘,𝑘𝛽𝑛(𝑠)max[]𝑠𝑘,𝑘𝛼𝑛(3𝑠)2𝐴𝑛2+12𝐵𝑛2,max[]𝑠𝑘,𝑘𝐴𝑛𝛽(𝑠)𝑛(𝑘)𝛼𝑛3(𝑘)2𝐴𝑛2+12𝐵𝑛2,max[]𝑠𝑘,𝑘𝐴𝑛(𝑠)max[]𝑠𝑘,𝑘𝛽𝑛(𝑠)maxs[]𝑘,𝑘𝛼𝑛(3𝑠)2𝐴𝑛2+12𝐵𝑛2.(4.31)
From inequalities (4.29) and (4.31), we obtain 𝐹𝑥𝑘,𝑥(𝑘),max[]𝑠𝑘,𝑘𝑥(𝑠)𝐺𝑥𝑘,𝛼𝑛(𝑘),max[]𝑠𝑘,𝑘𝛼𝑛(𝑠)𝑄𝑛𝐴(𝑘)𝑛(𝑘)𝐹𝑥𝑥𝑘,𝜁1,max[]𝑠𝑘,𝑘𝐴𝑥(𝑠)2𝑛(𝑘)+𝐹𝑥𝑦𝑘,𝛼𝑛(𝑘),𝜁232|||||𝐴𝑛|||2+12|||||𝐵𝑛|||2+𝐺𝑥𝑥𝑘,𝜁1,max[]𝑠𝑘,𝑘𝛽𝑛3(𝑠)2𝐴𝑛2+12𝐵𝑛2+𝐺𝑥𝑦𝑘,𝛼𝑛(𝑘),𝜁232𝐴𝑛2+12𝐵𝑛2.(4.32)
From inequalities (4.30), (4.31), we get 𝐹𝑦𝑘,𝑥(𝑘),max[]𝑠𝑘,𝑘𝑥(𝑠)𝐺𝑦𝑘,𝛼𝑛(𝑘),max[]𝑠𝑘,𝑘𝛼𝑛(𝑠)𝑞𝑛×(𝑘)max[]𝑠𝑘,𝑘𝑥(𝑠)max[]𝑠𝑘,𝑘𝛼𝑛(𝑠)𝐹𝑦𝑥𝑘,𝜉3,max[]𝑠𝑘,𝑘𝐴𝑥(𝑠)2𝑛(𝑘)+𝐹𝑦𝑦𝑘,𝛼𝑛(𝑘),𝜉432𝐴𝑛2+12𝐵𝑛2+𝐺𝑦𝑥𝑘,𝜉3,max[]𝑠𝑘,𝑘𝛽𝑛3(𝑠)2𝐴𝑛2+12𝐵𝑛2+𝐺𝑦𝑦𝑘,𝛼𝑛(𝑘),𝜉432𝐴𝑛2+12𝐵𝑛2.(4.33)
Since the second derivatives of the functions 𝐹,𝐺 are continuous and bounded in Ω(𝛼0,𝛽0), it follows from inequalities in (4.27), (4.32), and (4.33), there exist positive constants 𝐿𝑘,𝑁𝑘 such that Δ𝐴𝑛+1(𝑘1)𝑄𝑛(𝐴𝑘)𝑛+1(𝑘)+𝑞𝑛(𝑘)max[]𝑠𝑘,𝑘𝐴𝑛+1(𝑠)+𝐿𝑘𝐴𝑛2+𝑁𝑘𝐵𝑛2.(4.34)
Therefore, 𝐴𝑛+1(𝐴𝑘)𝑛+1(𝑘1)+𝑄𝑛(𝐴𝑘)𝑛+1(𝑘)+𝑞𝑛(𝑘)max[]𝑠𝑘,𝑘𝐴𝑛+1(𝑠)+𝐿𝑘𝐴𝑛2+𝑁𝑘𝐵𝑛2,𝐴𝑛+1𝐴(𝑘1)𝑛+1(𝑘2)+𝑄𝑛𝐴(𝑘1)𝑛+1(𝑘1)+𝑞𝑛(𝑘1)max[]𝑠𝑘1,𝑘1𝐴𝑛+1(𝑠)+𝐿𝑘1𝐴𝑛2+𝑁𝑘1𝐵𝑛2,𝐴𝑛+1𝐴(𝑎+1)𝑛+1(𝑎)+𝑄𝑛𝐴(𝑎+1)𝑛+1(𝑎+1)+𝑞𝑛(𝑎+1)max[]𝑠𝑎+1,𝑎+1𝐴𝑛+1(𝑠)+𝐿𝑎+1𝐴𝑛2+𝑁𝑎+1𝐵𝑛2.(4.35)
From inequalities in (4.35), we obtain 𝐴𝑛+1𝐿(𝑘)𝑛𝐴𝑛2+𝐵𝑛2+𝑘𝑙=𝑎+1𝑄𝑛𝐴(𝑙)𝑛+1(𝑙)+𝑞𝑛(𝑙)max[]𝜂𝑙,𝑙𝐴𝑛+1,[],𝐴(𝜂)𝑘𝑎+1,𝑇𝑛+1𝐿(𝑘)=0𝑛𝐴𝑛2+𝐵𝑛2[],,𝑘𝑎+1,𝑎(4.36) where 𝐿𝑛=𝑇𝑗=𝑎+1(𝐿𝑗+𝑁𝑗).
According to Lemma 3.5 from inequalities in (4.36), it follows 𝐴𝑛+1𝐿(𝑘)𝑛𝐴𝑛2+𝐵𝑛2𝑘𝑙=𝑎+11𝑞𝑛(𝑙)𝑄𝑛.(𝑙)(4.37)
From (4.37) and the condition (2) of Theorem 4.1, it follows that there exist positive constants 𝜆𝑖, where 𝑖=1,2, such that 𝐴𝑛+1(𝑘)𝜆1𝐴𝑛2+𝜆2𝐵𝑛2[].,𝑘𝑎+1,𝑇(4.38)
In a similar way, we can prove that there exist positive constants 𝜇𝑗, where 𝑗=1,2, such that 𝐵𝑛+1(𝑘)𝜇1𝐵𝑛2+𝜇2𝐴𝑛2[].,𝑘𝑎+1,𝑇(4.39)
Inequalities (4.38), (4.39) and the definitions of the functions 𝐴𝑛+1(𝑘), 𝐵𝑛+1(𝑘) imply the validity of (4.8), that is, the convergence of the monotone sequences {𝛼𝑛(𝑘)}𝑛=0 and {𝛽𝑛(𝑘)}𝑛=0 is quadratic.

5. Application

Now, we will give an example to illustrate the suggested above scheme for approximate obtaining of a solution.

Consider the following nonlinear difference equation with “maxima”: 1Δ𝑢(𝑘1)=120.5𝑢(𝑘)2+0.5max𝑠[𝑘2,𝑘]𝑢[],(𝑠)𝑢(𝑘1),𝑘1,3(5.1) with an initial condition [].𝑢(𝑘)=0,𝑘1,0(5.2)

The function 𝛼0(𝑘)=1, 𝑘[1,3], is a lower solution of (5.1), (5.2) because the inequality 0<1/(20.5(1))1/(2+0.5(1))+1=4/15+1=11/15 holds.

The function 𝛽0(𝑘)=1, 𝑘[1,3], is an upper solution of (5.1), (5.2) because the inequality 0>1/1.51/2.51=11/15 holds.

The conditions of Theorem 4.1 are satisfied since 𝐹𝑥𝑥(𝑘,𝑥,𝑦)=0.5/(20.5𝑥)3>0 and 𝐺𝑦𝑦(𝑘,𝑥,𝑦)=0.5/(2+0.5𝑦)3 for 1𝑥,𝑦1. Also, the inequality (4.5) holds, because in this case 𝑀(𝑘)=0.5/(1.5)2=2/9, 𝑁(𝑘)=(0.5/(1.5)2)=2/9 and 𝑀(𝑘)+𝑁(𝑘)<1.

According to Theorem 4.1, the initial value problem (5.1), (5.2) has a solution which is between 𝛼0(𝑘)=1 and 𝛽0(𝑘)=1. It is obviously the problem (5.1), (5.2) has a zero solution. This solution also could be obtained by constructing two sequences of successive approximations.

The successive approximation 𝛼𝑛(𝑘) is a solution of (4.20), (4.21) which is reduced to the following initial value problem: Δ𝛼𝑛(𝑘1)=𝑄𝑛1(𝑘)𝛼𝑛(𝑘)+𝑞𝑛1(𝑘)max𝑠[𝑘2,𝑘]𝛼𝑛+1(𝑠)20.5𝛼𝑛1(𝑘)𝛼𝑛1(𝑘1)2+0.5max𝑠[𝑘2,𝑘]𝛼𝑛1(𝑠)𝑄𝑛1(𝑘)𝛼𝑛1(𝑘)𝑞𝑛1(𝑘)max𝑠[𝑘2,𝑘]𝛼𝑛1[],𝛼(𝑠),𝑘1,3𝑛[],(𝑘)=0,𝑘1,0(5.3) and the successive approximation 𝛽𝑛(𝑘) is a solution of (4.22) which is reduced to the following initial value problem: Δ𝛽𝑛(𝑘1)=𝑄𝑛1(𝑘)𝛽𝑛(𝑘)+𝑞𝑛1(𝑘)max𝑠[𝑘2,𝑘]𝛽𝑛+1(𝑠)20.5𝛽𝑛1(𝑘)𝛽𝑛1(𝑘1)2+0.5max𝑠[𝑘2,𝑘]𝛽𝑛1(𝑠)𝑄𝑛1(𝑘)𝛽𝑛1(𝑘)𝑞𝑛1(𝑘)max𝑠[𝑘2,𝑘]𝛽𝑛1[],𝛽(𝑠),𝑘1,3𝑛[],(𝑘)=0,𝑘1,0(5.4) where 𝑄𝑛1(𝑘)=0.520.5𝛼𝑛1(𝑘)2,𝑞𝑛1(𝑘)=0.52+0.5max𝑠[𝑘2,𝑘]𝛽𝑛1(𝑠)2.(5.5) Initial value problems (5.3) and (5.4) are solved by a computer program, using the algorithm given in the proof of Lemma 3.1 and the results are written in Table 1.

tab1
Table 1: Values of successive approximations 𝛼𝑛(𝑘) and 𝛽𝑛(𝑘), 𝑛=1,2,3,4,5.

Table 1 demonstrates both sequences monotonically approach the exact zero solution. This illustrates the application of the proved above procedure for approximately obtaining of the solution.

Acknowledgments

The research in this paper was partially supported by Fund Scientific Research MU11FMI005/29.05.2011, Plovdiv University, Bulgaria and BG051PO001/3.3-05-001 Science and Business, financed by the Operative Program “Development of Human Resources”, European Social Fund.

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