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

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 [2–18]. 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 ℝ₊=[0,∞), ℤ be the set of all integers, let ℎ≥0 be a given fixed integer and 𝑎,𝑏∈ℤ be such that a<𝑏. Denote by ℤ[𝑎,𝑏]={𝑧∈ℤ∶𝑎≤𝑧≤𝑏}.

Note that for any function 𝑄∶ℤ[𝑚,𝑛]→ℝ, 𝑚<𝑛, the equalities ∑^{𝑚𝑖=𝑛}𝑄(𝑖)=0 and ∏^{𝑚𝑖=𝑛}𝑄(𝑖)=1 hold.

Consider the following nonlinear difference equation with “maxima”: Δ𝑢(𝑘−1)=𝑓𝑘,𝑢(𝑘),max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝑢(𝑠)for𝑘∈ℤ[𝑎+1,𝑇](2.1) with an initial condition 𝑢(𝑘)=𝜑(𝑘)for𝑘∈ℤ[𝑎−ℎ+1,𝑎],(2.2) where the functions 𝑢∶ℤ[𝑎−ℎ+1,𝑇]→ℝ, Δ𝑢(𝑘−1)=𝑢(𝑘)−𝑢(𝑘−1), 𝑓∶ℤ[𝑎+1,𝑇]×ℝ²→ℝ, and 𝜑∶ℤ[𝑎−ℎ+1,𝑎]→ℝ, the points 𝑎,𝑇∈ℤ are such that 0≤𝑎<𝑇.

Definition 2.1. One will say that the function 𝛼∶ℤ[𝑎+1−ℎ,𝑇]→ℝ is a lower (upper) solution of the problem (2.1), (2.2), if Δ𝛼(𝑘−1)≤𝛼((𝑘≥))≤𝑓(≥𝑘),𝛼𝜑(𝑘(𝑘)),fmorax𝑘∈_{[𝑠∈ℤ𝑘−ℎ,k]}𝛼ℤ(𝑠[)𝑎+1fo−rℎ𝑘,∈𝑎]ℤ.[𝑎+1,𝑇],(2.3)

Let 𝛼,𝛽∶ℤ[𝑎+1−ℎ,𝑇]→ℝ be given functions such that 𝛼(𝑘)≤𝛽(𝑘) for 𝑘∈ℤ[𝑎+1−ℎ,𝑇]. Define the following sets: 𝑆(𝛼,𝛽)=Ω{(𝑢𝛼,∶𝛽ℤ)=[𝑎⎧⎪⎨⎪⎩+𝑘1,−𝑥₁ℎ,𝑇]→ℝ∶𝛼(𝑘)≤𝑢(𝑘)≤𝛽(𝑘),𝑘∈ℤ[𝑎+1−ℎ,𝑇]},,𝑥₂m∈aℤx[𝑎_{𝑠∈ℤ[𝑘−ℎ,𝑘]}+1,𝑇]×ℝ²∶𝛼(𝑘)≤𝑥₁≤𝛽(𝑘),𝛼(𝑠)≤𝑥₂≤max_{𝑠∈ℤ[𝑘−ℎ,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∶ℤ[𝑎+1,𝑇]→ℝ.(2)The functions 𝑀∶ℤ[𝑎+1,𝑇]→ℝ, 𝑁∶ℤ[𝑎+1,𝑇]→ℝ₊ are such that 𝑀(𝑘)+𝑁(𝑘)<1𝑓𝑜𝑟𝑘∈ℤ[𝑎+1,𝑇].(3.1)
Then the initial value problem for linear difference equation with “maxima” Δ𝑢(𝑘−1)=𝑄(𝑘)+𝑢(𝑀𝑘)(=𝑘)𝜑𝑢((𝑘𝑘))+𝑓𝑁𝑜𝑟(𝑘𝑘)∈ℤma[𝑎x+_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝑢1(−𝑠)ℎ,𝑎𝑓]𝑜𝑟𝑘∈ℤ[𝑎+1,𝑇],(3.2) has an unique solution on the interval ℤ[𝑎+1−ℎ,𝑇].

Proof. We will use the step method to solve the initial value problem (3.2). Assume for a fixed 𝑘∈ℤ[𝑎+1,𝑇] all values 𝑢(𝑗), 𝑗∈ℤ[a+1−ℎ,𝑘−1] are known. Then from (3.2), we obtain (1−𝑀(𝑘))𝑢(𝑘)=𝑢(𝑘−1)+𝑄(𝑘)+𝑁(𝑘)max_{𝑠∈ℤ[𝑘−ℎ,𝑘]}𝑢(𝑠).
Consider the following two possible cases.
Case  1. Let max_{𝑙∈ℤ[1,ℎ]}𝑢(𝑘−𝑙)≤(𝑄(𝑘)+𝑢(𝑘−1))/(1−𝑀(𝑘)−𝑁(𝑘)), or (𝑄(𝑘)+𝑢(𝑘−1))/(1−𝑀(𝑘)−𝑁(𝑘))≥𝑢(𝑘−𝑙) for 𝑙∈ℤ[1,ℎ].
Therefore, max_{𝑠∈ℤ[𝑘−ℎ,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(𝑘−1)1−𝑀(𝑘)−𝑁(𝑘).(3.3)
Case  2. Let (𝑄(𝑘)+𝑢(𝑘−1))/(1−𝑀(𝑘)−𝑁(𝑘))<max_{𝑙∈ℤ[1,ℎ]}𝑢(𝑘−𝑙)=𝑢(𝑘−𝑚) where 𝑚∈ℤ[1,ℎ].
If we assume 𝑢(𝑘)≥𝑢(𝑘−𝑚) then from (3.2), we obtain 𝑢(𝑘−𝑚)≤𝑢(𝑘)=(𝑄(𝑘)+𝑢(𝑘−1))/(1−𝑀(𝑘)−𝑁(𝑘))<𝑢(𝑘−𝑚). The obtained contradiction proves 𝑢(𝑘)<𝑢(𝑘−𝑚) and max_{𝑠∈ℤ[𝑘−ℎ,𝑘]}𝑢(𝑠)=𝑢(𝑘−𝑚).
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(𝑘−1)+𝑄(𝑘)+𝑁(𝑘)𝑢(𝑘−𝑚)1−𝑀(𝑘).(3.4)
Thus, we receive the value 𝑢(𝑘), 𝑘∈𝑍[𝑎+1,𝑇].

Lemma 3.2. Let the following conditions be fulfilled:(1) The functions 𝑀,𝑁∶ℤ[𝑎+1,𝑇]→ℝ satisfy the inequality 𝑀(𝑘)+𝑁(𝑘)<1𝑓𝑜𝑟𝑘∈ℤ[𝑎+1,𝑇].(3.5)(2) The function 𝑢∶ℤ[𝑎−ℎ+1,𝑇]→ℝ satisfies the inequalities Δ𝑢(𝑘−1)≤𝑀(𝑢𝑘)(𝑘𝑢)(𝑘≤)0+,𝑁𝑘(𝑘∈)ℤm[𝑎a−xℎ_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝑢+(𝑠1),,𝑎].𝑘∈ℤ[𝑎+1,𝑇],(3.6)
Then 𝑢(𝑘)≤0 for 𝑘∈ℤ[𝑎−ℎ+1,𝑇].

Proof. Assume the claim of Lemma 3.2 is not true. Then there exists 𝑗∈ℤ[𝑎+1,𝑇] such that 𝑢(𝑗)>0 and 𝑢(𝑘)≤0 for 𝑘∈ℤ[𝑎−ℎ+1,𝑗−1]. Therefore, max_{𝑠∈ℤ[𝑗−ℎ,𝑗]}𝑢(𝑠)=𝑢(𝑗) and according to inequality (3.6), we get 𝑢(𝑗)≤𝑢(𝑗)−𝑢(𝑗−1)≤(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∈ℤ[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∶ℤ(𝑎,∞)→ℝ₊, 𝑝,𝑢∶ℤ(𝑎,∞)→ℝ and 𝑢(𝑘)≤𝑝(𝑘)+𝑞(𝑘)^𝑘−1_𝑙=𝑎𝑓(𝑙)𝑢(𝑙)𝑓𝑜𝑟𝑘∈ℤ(𝑎,∞).(3.8)
Then for all 𝑘∈ℤ(𝑎,∞), the following inequality is valid: 𝑢(𝑘)≤𝑝(𝑘)+𝑞(𝑘)^𝑘−1_𝑙=𝑎𝑝(𝑙)𝑓(𝑙)^𝑘−1_𝜏_=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∶ℤ[𝑎+1,𝑇]→ℝ₊ and 𝑞(𝑘)+𝑄(𝑘)<1𝑓𝑜𝑟𝑘∈ℤ[𝑎+1,T].(3.10)(2) The function 𝑢∶ℤ[𝑎+1−ℎ,𝑇]→ℝ₊ and satisfies the inequalities 𝑢(𝑘)≤𝐶+^𝑘_𝑙=_𝑎+1𝑞(𝑙)𝑢(𝑙)+^𝑘_𝑙=_𝑎+1𝑢(𝑘)≤𝐶,𝑘𝑄∈(ℤ𝑙)[𝑎m+a1x−_{[]𝑠∈ℤ𝑙−ℎ,𝑙}𝑢ℎ(,𝑠𝑎)],,𝑘∈ℤ[𝑎+1,𝑇],(3.11) where 𝐶=const≥0.
Then, 𝑢(𝑘)≤𝐶∏^{𝑘𝜏=𝑎+1}(1−𝑞(𝜏)−𝑄(𝜏))𝑓𝑜𝑟𝑘∈ℤ[𝑎+1,𝑇].(3.12)

Proof. Define a function 𝑧∶ℤ[𝑎+1−ℎ,𝑇]→ℝ₊ by the equalities 𝑧(𝑘)=⎧⎪⎨⎪⎩𝐶+^𝑘_𝑙=𝑎+1𝑞(𝑙)𝑢(𝑙)+^𝑘_𝑙=_𝑎+1𝐶,𝑄(𝑙)max_{𝑠∈ℤ[𝑙−ℎ,𝑙]}𝑢(𝑠),𝑘𝑘∈∈ℤℤ[[𝑎𝑎++11,−𝑇]ℎ,,𝑎].(3.13)
It is easy to check that for any 𝑘∈ℤ[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∈ℤ[𝑎+1,𝑇], and max_{𝑠∈ℤ[𝑘−ℎ,𝑘]}𝑢(𝑠)≤max_{𝑠∈ℤ[𝑘−ℎ,𝑘]}𝑧(𝑠)=𝑧(𝑘) for 𝑘∈ℤ[𝑎+1,𝑇]. Therefore, for any 𝑘∈ℤ[𝑎+1,𝑇], we obtain 𝑧(𝑘)≤𝐶+^𝑘_𝑙=_𝑎+1(𝑞(𝑙)+𝑄(𝑙))𝑧(𝑙).(3.14)
From inequality (3.14), it follows 𝑧(𝑘)≤𝐶(1−𝑞(𝑘)−𝑄(𝑘))+(1−𝑞(𝑘1)−𝑄(𝑘))^𝑘−1_𝑙=𝑎+1(𝑞(𝑙)+𝑄(𝑙))𝑧(𝑙).(3.15)
According to Lemma 3.4 from inequality (3.15), we get for 𝑘∈ℤ[𝑎+1,𝑇]𝑧(𝑘)≤𝐶(×1−𝑞1(+𝑘)−𝑄_𝑘−1(𝑘))_𝑙=_𝑎+1𝑞(𝑙)+𝑄(𝑙)=𝐶1−𝑞(𝑙)−𝑄(𝑙)^𝑘−1_𝜏=𝑙+11+1−𝑞(𝑞𝜏()𝜏+)−𝑄𝑄(𝜏()𝜏)(×1−𝑞1(+𝑘)−𝑄_𝑘−1(𝑘))≤𝐶_𝑙=𝑎+11−𝑞(𝑞𝑙)(𝑙+)−𝑄𝑄(𝑙)(𝑙)∏^{𝑘−1𝜏=𝑙+1}(1−1𝑞(𝜏)−𝑄(𝜏))∏^{𝑘𝜏=𝑎+1}(1−𝑞(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∶ℤ[𝑎+1−ℎ,𝑇]→ℝ are a lower and an upper solutions of the problem (2.1), (2.2), respectively, and such that 𝛼₀(𝑘)≤𝛽₀(𝑘) for 𝑘∈ℤ[𝑎+1−ℎ,𝑇].(2) The function 𝑓∶ℤ[𝑎+1,𝑇]×ℝ²→ℝ 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,𝐺∶Ω(𝛼₀,𝛽₀)→ℝ are continuous and twice continuously differentiable with respect to their second and third arguments and the following inequalities are valid for 𝑘∈ℤ[𝑎+1,𝑇], (𝑘,𝑥,𝑦)∈Ω(𝛼₀,𝛽₀): 𝐺𝐹_𝑥𝑥(𝑘,𝑥,𝑦)≥0,𝐹_𝑥𝑦(𝑘,𝑥,𝑦)≥0,𝐹_𝑦𝑦(𝑘,𝑥,𝑦)≥0,_𝑥𝑥𝐺(𝑘,𝑥,𝑦)≥0,𝐺_𝑥𝑦(𝑘,𝑥,𝑦)≥0,𝐺_𝑦𝑦(𝑘,𝑥,𝑦)≥0,(4.2)_𝑥𝑘,𝛽⁰(𝑘),max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛽₀(𝑠)≤𝐹_𝑥𝑘,𝛼⁰𝐺(𝑘),max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛼₀(𝑠),(4.3)_𝑦𝑘,𝛽⁰(𝑘),max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛽₀(𝑠)≤𝐹_𝑦𝑘,𝛼⁰𝑀(𝑘)+𝑁(𝑘)<1,(𝑘),max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛼₀(𝑠),((44..45)) where 𝑀(𝑘)=𝐹_𝑥𝑘,𝛽⁰(𝑘),max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛽₀(𝑠)−𝐺_𝑥𝑘,𝛼⁰𝑁(𝑘)=𝐹_𝑦(𝑘),max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛼₀(𝑠),𝑘,𝛽⁰(𝑘),max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛽₀(𝑠)−𝐺_𝑦𝑘,𝛼⁰(𝑘),max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛼₀(𝑠).(4.6)(3) The function 𝜑∶ℤ[𝑎−ℎ+1,𝑎]→ℝ. Then there exist two sequences {𝛼_𝑛(𝑘)}^∞𝑛=0 and {𝛽_𝑛(𝑘)}^∞𝑛=0, 𝑘∈ℤ[𝑎+1−ℎ,𝑇], such that:(a) The functions 𝛼_𝑛∶ℤ[𝑎+1−ℎ,𝑇]→ℝ, (𝑛=1,2,…) are lower solutions of IVP (2.1), (2.2).(b) The functions 𝛽_𝑛∶ℤ[𝑎+1−ℎ,𝑇]→ℝ, (𝑛=1,2,…) are upper solutions of IVP (2.1), (2.2).(c) The following inequalities hold for 𝑘∈[𝑎+1−ℎ,𝑇]𝛼₀(𝑘)≤𝛼₁(𝑘)≤⋯≤𝛼_𝑛(𝑘)≤⋯≤𝛽_𝑛(𝑘)≤⋯≤𝛽₁(𝑘)≤𝛽₀(𝑘).(4.7)(d) Both sequences are convergent on ℤ[𝑎+1−ℎ,𝑇] and their limits 𝑉(𝑘)=lim_𝑛→∞𝛼_𝑛(𝑘) and 𝑊(𝑘)=lim_𝑛→∞𝛽_𝑛(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 𝜆_𝑖,𝜇_𝑖>0,𝑖=1,2 such that for the solution 𝑥(𝑘) of IVP (2.1), (2.2) in 𝑆(𝛼₀,𝛽₀), the inequalities ‖‖‖‖𝑥𝑥−−𝛼𝛽_𝑛+1𝑛+1‖‖≤𝜆₁‖‖𝑥−𝛼_𝑛‖‖²+𝜆₂‖‖𝑥−𝛽_𝑛‖‖²,‖‖≤𝜇₁‖‖𝑥−𝛼_𝑛‖‖²+𝜇₂‖‖𝑥−𝛽_𝑛‖‖²(4.8) hold, where ||𝑢||=max_{𝑠∈ℤ[𝑎+1−ℎ,𝑇]}|𝑢(𝑠)| for any function 𝑢∶ℤ[𝑎+1−ℎ,𝑇]→ℝ.

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: 𝑓𝑘,𝑥₂,𝑦₂−𝑓𝑘,𝑥₁,𝑦₁≤𝐹−_𝑥𝐺𝑘𝑘,,𝑥𝑥₂,₂𝑦₂𝑥₂−𝑥₁+𝐹_𝑦𝑘,𝑥₂,𝑦₂𝑦₂−𝑦₁−𝐺𝑘,𝑥₂,𝑦₂+𝐺𝑘,𝑥₁,𝑦₁,,𝑦₂+𝐺𝑘,𝑥₁,𝑦₁≤𝐺_𝑥𝑘,𝑥₁,𝑦₁𝑥₁−𝑥₂+𝐺_𝑦𝑘,𝑥₁,𝑦₁𝑦₁−𝑦₂.(4.9)
Consider the initial value problem for the linear difference equation with “maxima” Δ𝑥(𝑘−1)=𝑄₀(𝑘)𝑥𝑥((𝑘𝑘))=+𝜑𝑞₀(𝑘(𝑘)),𝑘ma∈xℤ_{[]𝑠∈ℤ𝑘−ℎ,𝑘}[𝑥𝑎(+𝑠)1+−𝑃ℎ₀(,𝑘𝑎)],,𝑘∈ℤ[𝑎+1,𝑇],((44..1101)) where the functions 𝑃₀,𝑄₀,𝑞₀∶ℤ[𝑎+1,𝑇]→ℝ are defined by the equalities 𝑃₀𝑄(𝑘)=𝑓𝑘,𝛼₀(𝑘),max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛼₀(𝑠)−𝑄₀(𝑘)𝛼₀(𝑘)−𝑞₀(𝑘)max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛼₀(𝑠),₀(𝑘)=𝐹_𝑥𝑘,𝛼⁰(𝑘),max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛼₀(𝑠)−𝐺_𝑥𝑘,𝛽⁰𝑞(𝑘),max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛽₀(𝑠)₀,(𝑘)=𝐹_𝑦𝑘,𝛼⁰(𝑘),max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛼₀(𝑠)−𝐺_𝑦𝑘,𝛽⁰(𝑘),max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛽₀(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∈ℤ[𝑎+1,𝑇]. According to Lemma 3.1 the IVP (4.10), (4.11) has an unique solution 𝛼₁(𝑘), defined on the interval ℤ[𝑎+1−ℎ,𝑇].
Define a function 𝑝₁∶ℤ[𝑎+1−ℎ,𝑇]→ℝ by the equality 𝑝₁(𝑘)=𝛼₀(𝑘)−𝛼₁(𝑘). Then we get 𝑝₁(𝑘)=0 for 𝑘∈ℤ[𝑎+1−ℎ,𝑎].
Let 𝑘∈ℤ[𝑎+1,𝑇]. From the choice of the function 𝛼₀(𝑘) and (4.10) for the function 𝛼₁(𝑘), we get Δ𝑝₁(𝑘−1)≤𝑄₀(𝑘)𝑝₁(𝑘)+𝑞₀(𝑘)max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝑝₁(𝑠).(4.13)
According to Lemma 3.2 for the function 𝑝₁(𝑘), it follows that 𝑝₁(𝑘)≤0 for 𝑘∈ℤ[𝑎+1−ℎ,𝑇]. Therefore, 𝛼₀(𝑘)≤𝛼₁(𝑘) for 𝑘∈ℤ[𝑎+1−ℎ,𝑇].
Consider the linear difference equation with “maxima” Δ𝑥(𝑘−1)=𝑄₀(𝑘)𝑥𝑥((𝑘𝑘))=+𝜑𝑞₀((𝑘𝑘)),m𝑘a∈xℤ_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝑥[𝑎(𝑠+)1+−𝑅ℎ₀,(𝑘𝑎]),,𝑘∈ℤ[𝑎+1,𝑇],((44..1145)) where the functions 𝑞₀(𝑘) and 𝑄₀(𝑘) are defined by equalities (4.12) and 𝑅₀(𝑘)=𝑓𝑘,𝛽₀(𝑘),max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛽₀(𝑠)−𝑄₀(𝑘)𝛽₀(𝑘)−𝑞₀(𝑘)max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛽₀(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 ℤ[𝑎+1−ℎ,𝑇].
Define a function 𝑝₂∶ℤ[𝑎+1−ℎ,𝑇]→ℝ by the equality 𝑝₂(𝑘)=𝛽₁(𝑘)−𝛽₀(𝑘). Then 𝑝₂(𝑘)=0 for 𝑘∈ℤ[𝑎+1−ℎ,𝑎].
Now, let 𝑘∈ℤ[𝑎+1,𝑇]. From the choice of the function 𝛽₀(𝑘) and (4.14) for the function 𝛽₁(𝑘), we get Δ𝑝₂(𝑘−1)≤𝑄+₀𝑞(₀𝑘)𝛽₁(𝑘)−𝛽₀(𝑘)≤𝑄₀(𝑘)max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛽₁(𝑠)−max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛽₀(𝑠)(𝑘)𝑝₂(𝑘)+𝑞₀(𝑘)max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝑝₂(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∈ℤ[𝑎+1−ℎ,𝑇]. Therefore, 𝛽₁(𝑘)≤𝛽₀(𝑘) for 𝑘∈ℤ[𝑎+1−ℎ,𝑇].
Define a function 𝑝₃∶ℤ[𝑎+1−ℎ,𝑇]→ℝ by the equality 𝑝₃(𝑘)=𝛼₁(𝑘)−𝛽₁(𝑘). Then 𝑝₃(𝑘)≤0 for 𝑘∈ℤ[𝑎+1−ℎ,𝑎].
Let 𝑘∈ℤ[𝑎+1,𝑇]. Then for the function 𝑝₃(𝑘), we get Δ𝑝₃(𝑘−1)=𝑓+𝑄𝑘₀,𝛼₀(𝑘),max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛼₀(𝑠)−𝑓𝑘,𝛽₀(𝑘),max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛽₀(𝑠)+𝑞₀(𝑘)𝛼₁(𝑘)−𝛼₀(𝑘)−𝑄₀(𝑘)𝛽₁(𝑘)−𝛽₀(𝑘)−𝑞₀(𝑘)max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛼₁(𝑠)−max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛼₀(𝑠)≤𝑄₀(𝑘)max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛽₁(𝑠)−max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛽₀(𝑠)(𝑘)𝑝₃(𝑘)+𝑞₀(𝑘)max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝑝₃(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∈ℤ[𝑎+1−ℎ,𝑇]. Therefore, 𝛼₁(𝑘)≤𝛽₁(𝑘) for 𝑘∈ℤ[𝑎+1−ℎ,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 ℤ[𝑎+1−ℎ,𝑇].
Let 𝑘∈ℤ[𝑎+1,𝑇]. From the inequalities 𝛼₀(𝑘)≤𝛼₁(𝑘)≤𝛽₀(𝑘) for 𝑘∈ℤ[𝑎+1−ℎ,𝑇], max_{𝑠∈ℤ[𝑘−ℎ,𝑘]}𝛼₀(𝑠)≤max_{𝑠∈ℤ[𝑘−ℎ,𝑘]}𝛼₁(𝑠)≤max_{𝑠∈ℤ[𝑘−ℎ,𝑘]}𝛽₀(𝑠) for 𝑘∈ℤ[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(𝑘−1)=𝑓+𝑞₀𝑘,𝛼₀(𝑘),max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛼₀(𝑠)+𝑄₀(𝑘)𝛼₁(𝑘)−𝛼₀(𝑘)=𝑓𝑘,(𝑘𝛼)₁max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛼₁(𝑠)−max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛼₀(𝑠)+𝑓𝑘(,𝑘𝛼)₀,max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛼₁(𝑠)+𝑄₀(𝑘)𝛼₁(𝑘)−𝛼₀(𝑘)+𝑞₀(𝑘),max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛼₀(𝑠)−𝑓𝑘,𝛼₁(𝑘),max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛼₁(𝑠)≤𝑓𝑘,(𝑘𝛼)₁max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛼₁(𝑠)−max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛼₀(𝑠)(𝑘),max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛼₁(𝑠).(4.19)
Thus, the function 𝛼₁(𝑘) is a lower solution of (2.1), (2.2) on ℤ[𝑎+1−ℎ,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 ℤ[𝑎+1−ℎ,𝑇].
Analogously, we can construct two sequences of functions {𝛼_𝑛(𝑘)}^∞𝑛=1 and {𝛽_𝑛(𝑘)}^∞𝑛=1. If the functions 𝛼_𝑗(𝑘) and 𝛽_𝑗(𝑘), 𝑗=1,2,…,𝑛−1,𝑛, are obtained such that 𝛼_𝑗,𝛽_𝑗∈𝑆(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(𝑘−1)=𝑄_𝑛(𝑘)𝑥𝑥((𝑘𝑘))=+𝜑𝑞_𝑛(𝑘(𝑘)),𝑘ma∈xℤ_{[]𝑠∈ℤ𝑘−ℎ,𝑘}[𝑥𝑎(+𝑠)1+−𝑃ℎ_𝑛(,𝑘𝑎)],,𝑘∈ℤ[𝑎+1,𝑇],((44..2201)) and the initial value problem for the linear difference equation with “maxima’’ Δ𝑥(𝑘−1)=𝑄_𝑛(𝑘)𝑥𝑥((𝑘𝑘))=+𝜑𝑞_𝑛((𝑘𝑘)),m𝑘a∈xℤ_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝑥[𝑎(𝑠+)1+−𝑅ℎ_𝑛,(𝑘𝑎]),,𝑘∈ℤ[𝑎+1,𝑇],(4.22) where 𝑅𝑃_𝑛(𝑘)=𝑓𝑘,𝛼_𝑛(𝑘),max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛼_𝑛(𝑠)−𝑄_𝑛(𝑘)𝛼_𝑛(𝑘)−𝑞_𝑛(𝑘)max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛼_𝑛(𝑠),_𝑛𝑄(𝑘)=𝑓𝑘,𝛽_𝑛(𝑘),max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛽_𝑛(𝑠)−𝑄_𝑛(𝑘)𝛽_𝑛(𝑘)−𝑞_𝑛(𝑘)max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛽_𝑛(𝑠),_𝑛(𝑘)=𝐹_𝑥𝑘,𝛼^𝑛(𝑘),max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛼_𝑛(𝑠)−𝐺_𝑥𝑘,𝛽^𝑛𝑞(𝑘),max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛽_𝑛(𝑠)≥0,_𝑛(𝑘)=𝐹_𝑦𝑘,𝛼^𝑛(𝑘),max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛼_𝑛(𝑠)−𝐺_𝑦𝑘,𝛽^𝑛(𝑘),max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛽_𝑛(𝑠)≥0.(4.23)
Since 𝛼_𝑗,𝛽_𝑗∈𝑆(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 𝑄_𝑛(𝑘)≤𝑀(𝑘), 𝑞_𝑛(𝑘)≥𝑞₀(𝑘)≥0 and 𝑞_𝑛(𝑘)≤𝑁(𝑘), that is, 𝑄_𝑛(𝑘)+𝑞_𝑛(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 𝛼_𝑛+1(𝑘) and 𝛽_𝑛+1(𝑘), 𝑘∈ℤ[𝑎+1−ℎ,𝑇], correspondingly.
The proof that the functions 𝛼_𝑛+1,𝛽_𝑛+1∈𝑆(𝛼_𝑛,𝛽_𝑛), 𝛼_𝑛+1(𝑘)≤𝛽_𝑛+1(𝑘), and they are lower/upper solutions of (2.1), (2.2) on the interval ℤ[𝑎+1−ℎ,𝑇] is the same as in the case of 𝑛=1 and we omit it.
For any fixed 𝑘∈ℤ[𝑎+1−ℎ,𝑇], the sequences {𝛼_𝑛(𝑘)}^∞𝑛=0 and {𝛽_𝑛(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 ℤ[𝑎+1−ℎ,𝑇], that is, there exist functions 𝑉,𝑊∶ℤ[𝑎+1−ℎ,𝑇]→ℝ such that lim_𝑛→∞𝛼_𝑛(𝑘)=𝑉(𝑘),lim_𝑛→∞𝛽_𝑛(𝑘)=𝑊(𝑘).(4.24)
From inequalities (4.7), it follows that 𝑉,𝑊∈𝑆(𝛼₀,𝛽₀).
Now, we will prove that for any 𝑘∈ℤ[𝑎+1,𝑇] the following equality holds: lim_𝑛→∞max_{[]𝜉∈ℤ𝑘−ℎ,𝑘}𝛼_𝑛(𝜉)=max_{𝜉∈ℤ[𝑘−ℎ,𝑘]}lim_𝑛→∞𝛼_𝑛(𝜉).(4.25)
Let 𝑘∈ℤ[𝑎+1,𝑇] be fixed. We denote max_{𝜉∈ℤ[𝑘−ℎ,𝑘]}𝛼_𝑛(𝜉)=𝐴_𝑛. From inequalities (4.7) for every 𝜉∈ℤ[𝑘−ℎ,𝑘], the inequalities 𝛼_𝑛−1(𝜉)≤𝛼_𝑛(𝜉)≤𝐴_𝑛 hold and thus, 𝐴_𝑛−1≤𝐴_𝑛, that is, the sequence {𝐴_𝑛}^∞𝑛=0 is monotone nondecreasing and bounded from above by 𝛽₀(𝑘). Therefore, there exists the limit 𝐴=lim_𝑛→∞𝐴_𝑛.
From the monotonicity of the sequence of the lower solutions 𝛼_𝑛(𝑘), we get that for 𝜉∈ℤ[𝑘−ℎ,𝑘] it is fulfilled 𝛼_𝑛(𝜉)≤𝑉(𝜉). Let 𝜂∈ℤ[𝑘−ℎ,𝑘] be such that max_{𝜉∈ℤ[𝑘−ℎ,𝑘]}𝑉(𝜉)=𝑉(𝜂). From the inequalities 𝛼_𝑛(𝜂)≤𝐴_𝑛≤𝐴 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)<𝐴_𝑁≤𝐴 hold. Therefore, there exists 𝜉∈ℤ[𝑘−ℎ,𝑘] such that 𝛼_𝑁(𝜉)=max_{𝜉∈ℤ[𝑘−ℎ,𝑘]}𝛼_𝑁(𝜉)=𝐴_𝑁 or 𝑉(𝜂)<𝛼_𝑁(𝜉)≤𝑉(𝜉). The obtained contradiction proves the validity of the required inequality (4.25).
Analogously, we can show that the functions 𝛽_𝑛(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 ℤ[𝑎+1−ℎ,𝑇].
Let 𝑘∈ℤ[𝑎+1−ℎ,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∈ℤ[𝑎+1−ℎ,𝑎].
Let 𝑘∈ℤ[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∈ℤ[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∈ℤ[𝑎−ℎ+1,𝑇]. In the case of nonuniqueness, let 𝑢∈𝑆(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 {𝛼_𝑛(𝑘)}^∞𝑛=0, {𝛽_𝑛(𝑘)}^∞𝑛=0 is quadratic. Let 𝑥(𝑘) be a solution of (2.1), (2.2) in 𝑆(𝛼₀,𝛽₀).
Define the functions 𝐴_𝑛+1,𝐵_𝑛+1∶ℤ[𝑎+1−ℎ,𝑇]→ℝ₊, 𝑛=0,1,… by the equalities 𝐴_𝑛+1(𝑘)=𝑥(𝑘)−𝛼_𝑛+1(𝑘),𝐵_𝑛+1(𝑘)=𝛽_𝑛+1(𝑘)−𝑥(𝑘).(4.26)
It is obvious that 𝐴_𝑛+1(𝑘)=0 for 𝑘∈ℤ[𝑎+1−ℎ,𝑎].
Let 𝑘∈[𝑎+1,𝑇]. According to the definitions of the functions 𝐴_𝑛+1(𝑘), 𝛼_𝑛+1(𝑘) and the condition (2) of Theorem 4.1, we get Δ𝐴_𝑛+1(𝑘−1)=𝑓+𝑄𝑘_𝑛,𝑥(𝑘),max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝑥(𝑠)−𝑓𝑘,𝛼_𝑛(𝑘),max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛼_𝑛(𝑠)−𝑄_𝑛(𝑘)𝐴_𝑛+1(𝑘)+𝑞_𝑛(𝑘)max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝑥(𝑠)−max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛼_𝑛+1(𝑠)≤𝑄_𝑛(𝑘)𝐴_𝑛(𝑘)−𝑞_𝑛(𝑘)max_{[𝑠∈ℤk]−ℎ,𝑘}𝑥(𝑠)−max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛼_𝑛(𝑠)+(𝐹𝑘)𝐴_𝑛+1(𝑘)+𝑞_𝑛(𝑘)max_{𝑠∈ℤ[𝑘−ℎ,𝑘]}𝐴_𝑛+1(𝑠)_𝑥𝑘,𝑥(𝑘),max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝑥(𝑠)−𝐺_𝑥𝑘,𝛼^𝑛+𝐹(𝑘),max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛼_𝑛(𝑠)−𝑄_𝑛(𝑘)𝐴_𝑛(𝑘)_𝑦𝑘,x(𝑘),max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝑥(𝑠)−𝐺_𝑦𝑘,𝛼^𝑛×max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}(𝑘),max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛼_𝑛(𝑠)−𝑞_𝑛(𝑘)𝑥(𝑠)−max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛼_𝑛(𝑠).(4.27)
According to the mean value theorem, there exist points 𝜁_𝑖 and 𝜉_𝑗,𝑖=1,3,𝑗=2,4 such that 𝛼𝛼_𝑛(𝑘)≤𝜁_𝑖≤𝑥(𝑘),max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛼_𝑛(𝑠)≤𝜁_𝑗≤max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝑥(𝑠),_𝑛𝐹(𝑘)≤𝜉_𝑖≤𝛽_𝑛(𝑘),max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛼_𝑛(𝑠)≤𝜉_𝑗≤max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛽_𝑛(𝑠),(4.28)_𝑥𝑘,𝑥(𝑘),max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝑥(𝑠)−𝐺_𝑥𝑘,𝛼^𝑛=𝐹_𝑥(𝑘),max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛼_𝑛(𝑠)−𝑄_𝑛(𝑘)𝑘,𝑥(𝑘),max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝑥(𝑠)−𝐹_𝑥𝑘,𝛼^𝑛+𝐹_𝑥(𝑘),max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝑥(𝑠)𝑘,𝛼^𝑛(𝑘),max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝑥(𝑠)−𝐹_𝑥𝑘,𝛼^𝑛+𝐺_𝑥(𝑘),max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛼_𝑛(𝑠)𝑘,𝛽^𝑛(𝑘),max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛽_𝑛(𝑠)−𝐺_𝑥𝑘,𝛼^𝑛+𝐺_𝑥(𝑘),max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛽_𝑛(𝑠)𝑘,𝛼^𝑛(𝑘),max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛽_𝑛(𝑠)−𝐺_𝑥𝑘,𝛼^𝑛=𝐹_𝑥𝑥(𝑘),max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛼_𝑛(𝑠)𝑘,𝜁¹+𝐹_𝑥𝑦,max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝑥(𝑠)𝐴_𝑛(𝑘)+𝐺_𝑥𝑥𝑘,𝛼_𝑛(𝑘),𝜁₂max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝑥(𝑠)−max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛼_𝑛(𝑠)𝑘,𝜁¹+𝐺_𝑥𝑦,max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛽_𝑛(𝑠)𝛽_𝑛(𝑘)−𝛼_𝑛(𝑘)𝐹𝑘,𝛼_𝑛(𝑘),𝜁₂max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛽_𝑛(𝑠)−max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛼_𝑛(𝑠),(4.29)_𝑦𝑘,𝑥(𝑘),max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝑥(𝑠)−𝐺_𝑦𝑘,𝛼^𝑛=𝐹_𝑦𝑥(𝑘),max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛼_𝑛(𝑠)−𝑞_𝑛(𝑘)𝑘,𝜉³+𝐹_𝑦𝑦,max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝑥(𝑠)𝐴_𝑛(𝑘)+𝐺_𝑦𝑥𝑘,𝛼_𝑛(𝑘),𝜉₄max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝑥(𝑠)−max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛼_𝑛(𝑠)𝑘,𝜉³+𝐺_𝑦𝑦,max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛽_𝑛(𝑠)𝛽_𝑛(𝑘)−𝛼_𝑛(𝑘)𝑘,𝛼_𝑛(𝑘),𝜉₄max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛽_𝑛(𝑠)−max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛼_𝑛(𝑠).(4.30)
Then the following inequalities are valid: 𝐴_𝑛(𝑘)𝛽_𝑛(𝑘)−𝛼_𝑛(𝑘)≤𝐴_𝑛(𝑘)𝐴_𝑛(𝑘)+𝐵_𝑛(𝑘)≤32𝐴^2𝑛(𝑘)+12𝐵^2𝑛𝐴(𝑘),_𝑛(𝑘)max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛽_𝑛(𝑠)−max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛼_𝑛(𝑠)≤32‖‖𝐴^𝑛‖‖²+12‖‖𝐵^𝑛𝐴_𝑛‖‖²,(𝑘)max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛽_𝑛(𝑠)−max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛼_𝑛(𝑠)≤32‖‖𝐴^𝑛‖‖²+12‖‖𝐵^𝑛max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}‖‖²,𝐴_𝑛(𝑠)𝛽_𝑛(𝑘)−𝛼_𝑛(𝑘)≤32‖‖𝐴^𝑛‖‖²+12‖‖𝐵^𝑛max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}‖‖²,𝐴_𝑛(𝑠)max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛽_𝑛(𝑠)−max_{s[]∈ℤ𝑘−ℎ,𝑘}𝛼_𝑛(𝑠)≤32‖‖𝐴^𝑛‖‖²+12‖‖𝐵^𝑛‖‖².(4.31)
From inequalities (4.29) and (4.31), we obtain 𝐹^𝑥𝑘,𝑥(𝑘),max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝑥(𝑠)−𝐺_𝑥𝑘,𝛼^𝑛≤𝐹_𝑥𝑥(𝑘),max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛼_𝑛(𝑠)−𝑄_𝑛(𝑘)𝐴_𝑛(𝑘)𝑘,𝜁¹+𝐹_𝑥𝑦,max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝑥(𝑠)𝐴^2𝑛(𝑘)+𝐺_𝑥𝑥𝑘,𝛼_𝑛(𝑘),𝜁₂32|||||𝐴^𝑛|||²+12|||||𝐵^𝑛|||²𝑘,𝜁¹,max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛽_𝑛(𝑠)32‖‖𝐴^𝑛‖‖²+12‖‖𝐵^𝑛‖‖²+𝐺_𝑥𝑦𝑘,𝛼_𝑛(𝑘),𝜁₂32‖‖𝐴^𝑛‖‖²+12‖‖𝐵^𝑛‖‖².(4.32)
From inequalities (4.30), (4.31), we get 𝐹^𝑦𝑘,𝑥(𝑘),max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝑥(𝑠)−𝐺_𝑦𝑘,𝛼^𝑛×max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}(𝑘),max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛼_𝑛(𝑠)−𝑞_𝑛(𝑘)≤𝐹_𝑦𝑥𝑥(𝑠)−max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛼_𝑛(𝑠)𝑘,𝜉³+𝐹_𝑦𝑦,max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝑥(𝑠)𝐴^2𝑛(𝑘)+𝐺_𝑦𝑥𝑘,𝛼_𝑛(𝑘),𝜉₄32‖‖𝐴^𝑛‖‖²+12‖‖𝐵^𝑛‖‖²𝑘,𝜉³,max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝛽_𝑛(𝑠)32‖‖𝐴^𝑛‖‖²+12‖‖𝐵^𝑛‖‖²+𝐺_𝑦𝑦𝑘,𝛼_𝑛(𝑘),𝜉₄32‖‖𝐴^𝑛‖‖²+12‖‖𝐵^𝑛‖‖².(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 𝐿_𝑘,𝑁_𝑘 such that Δ𝐴_𝑛+1(𝑘−1)≤𝑄+_𝑛𝐿(𝑘_𝑘)𝐴_𝑛+1(𝑘)+𝑞_𝑛(𝑘)max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝐴_𝑛+1(𝑠)‖‖𝐴_𝑛‖‖²+𝑁_𝑘‖‖𝐵_𝑛‖‖².(4.34)
Therefore, 𝐴_𝑛+1(𝑘)≤𝐴+𝐿_𝑛+1𝑘(𝑘−1)+𝑄_𝑛(𝑘)𝐴_𝑛+1(𝑘)+𝑞_𝑛(𝑘)max_{[]𝑠∈ℤ𝑘−ℎ,𝑘}𝐴_𝑛+1(𝑠)𝐴…‖‖…𝐴_𝑛‖‖…²_𝑛+1+𝑁_𝑘‖‖𝐵_𝑛‖‖²,(𝑘−1)≤𝐴+𝑞_𝑛+1𝑛(𝑘−2)+𝑄_𝑛(𝑘−1)𝐴_𝑛+1(𝑘−1)𝐴_𝑛+1(𝑘−1)max_{[]𝑠∈ℤ𝑘−ℎ−1,𝑘−1}𝐴_𝑛+1(𝑠)+𝐿_𝑘−1‖‖𝐴_𝑛‖‖²+𝑁_𝑘−1‖‖𝐵_𝑛‖‖²,(𝑎+1)≤𝐴+𝑞_𝑛+1𝑛(𝑎)+𝑄_𝑛(𝑎+1)𝐴_𝑛+1(𝑎+1)(𝑎+1)max_{[]𝑠∈ℤ𝑎+1−ℎ,𝑎+1}𝐴_𝑛+1(𝑠)+𝐿_𝑎+1‖‖𝐴_𝑛‖‖²+𝑁_𝑎+1‖‖𝐵_𝑛‖‖².(4.35)
From inequalities in (4.35), we obtain 𝐴_𝑛+1(𝑘)≤𝐿_𝑛‖‖𝐴^𝑛‖‖²+‖‖𝐵_𝑛‖‖²+∑^𝑘_𝑙=𝑎+1𝑄^𝑛(𝑙)𝐴_𝑛+1(𝑙)+𝑞_𝑛(𝑙)max_{[]𝜂∈𝑙−ℎ,𝑙}𝐴𝑘∈ℤ[𝑎+1,𝑇_𝑛+1],𝐴_𝑛+1(𝜂),(𝑘)=0≤𝐿_𝑛‖‖𝐴^𝑛‖‖²+‖‖𝐵_𝑛‖‖²,𝑘∈ℤ[𝑎−ℎ+1,𝑎],(4.36) where 𝐿_𝑛=∑^{𝑇𝑗=𝑎+1}(𝐿_𝑗+𝑁_𝑗).
According to Lemma 3.5 from inequalities in (4.36), it follows 𝐴_𝑛+1(𝑘)≤𝐿^𝑛‖‖𝐴^𝑛‖‖²+‖‖𝐵_𝑛‖‖²∏^{𝑘𝑙=𝑎+1}1−𝑞_𝑛(𝑙)−𝑄_𝑛(𝑙).(4.37)
From (4.37) and the condition (2) of Theorem 4.1, it follows that there exist positive constants 𝜆_𝑖, where 𝑖=1,2, such that 𝐴_𝑛+1(𝑘)≤𝜆₁‖‖𝐴_𝑛‖‖²+𝜆₂‖‖𝐵_𝑛‖‖²,𝑘∈ℤ[𝑎+1,𝑇].(4.38)
In a similar way, we can prove that there exist positive constants 𝜇_𝑗, where 𝑗=1,2, such that 𝐵_𝑛+1(𝑘)≤𝜇₁‖‖𝐵_𝑛‖‖²+𝜇₂‖‖𝐴_𝑛‖‖²,𝑘∈ℤ[𝑎+1,𝑇].(4.39)
Inequalities (4.38), (4.39) and the definitions of the functions 𝐴_𝑛+1(𝑘), 𝐵_𝑛+1(𝑘) imply the validity of (4.8), that is, the convergence of the monotone sequences {𝛼_𝑛(𝑘)}^∞𝑛=0 and {𝛽_𝑛(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”: Δ𝑢(𝑘−1)=12−0.5𝑢(𝑘)−2+0.5max1_{𝑠∈ℤ[𝑘−2,𝑘]}𝑢(𝑠)−𝑢(𝑘−1),𝑘∈ℤ[1,3],(5.1) with an initial condition 𝑢(𝑘)=0,𝑘∈ℤ[−1,0].(5.2)

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

The function 𝛽₀(𝑘)=1, 𝑘∈ℤ[−1,3], is an upper solution of (5.1), (5.2) because the inequality 0>1/1.5−1/2.5−1=−11/15 holds.

The conditions of Theorem 4.1 are satisfied since 𝐹_𝑥𝑥(𝑘,𝑥,𝑦)=0.5/(2−0.5𝑥)³>0 and 𝐺_𝑦𝑦(𝑘,𝑥,𝑦)=0.5/(2+0.5𝑦)³ for −1≤𝑥,𝑦≤1. 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 𝛼_𝑛(𝑘) is a solution of (4.20), (4.21) which is reduced to the following initial value problem: Δ𝛼_𝑛(𝑘−1)=𝑄+_𝑛−1(𝑘)1𝛼_𝑛(𝑘)+𝑞_𝑛−1(𝑘)max_{𝑠∈ℤ[𝑘−2,𝑘]}𝛼_𝑛(𝑠)2−0.5𝛼_𝑛−1(𝑘)−𝛼^𝑛(𝑘−1)−1−𝑄_𝑛−12+0.5max_{𝑠∈ℤ[𝑘−2,𝑘]}𝛼_𝑛−1(𝑠)(𝑘𝛼)𝛼_𝑛−1(𝑘)−𝑞_𝑛−1(𝑘)max_{𝑠∈ℤ[𝑘−2,𝑘]}𝛼_𝑛−1(𝑠),𝑘∈ℤ[1,3_𝑛],(𝑘)=0,𝑘∈ℤ[−1,0],(5.3) and the successive approximation 𝛽_𝑛(𝑘) is a solution of (4.22) which is reduced to the following initial value problem: Δ𝛽_𝑛(𝑘−1)=𝑄+_𝑛−1(𝑘)1𝛽_𝑛(𝑘)+𝑞_𝑛−1(𝑘)max_{𝑠∈ℤ[𝑘−2,𝑘]}𝛽_𝑛(𝑠)2−0.5𝛽_𝑛−1(𝑘)−𝛽^𝑛(𝑘−1)−1−𝑄_𝑛−12+0.5max_{𝑠∈ℤ[𝑘−2,𝑘]}𝛽_𝑛−1(𝑠)(𝑘)𝛽𝛽_𝑛−1(𝑘)−𝑞_𝑛−1(𝑘)max_{𝑠∈ℤ[𝑘−2,𝑘]}𝛽_𝑛−1(𝑠),𝑘∈ℤ[1,3_𝑛],(𝑘)=0,𝑘∈ℤ[−1,0],(5.4) where 𝑄_𝑛−1(𝑘)=0.52−0.5𝛼_𝑛−1(𝑘)²,𝑞_𝑛−1(𝑘)=0.52+0.5max_{𝑠∈ℤ[𝑘−2,𝑘]}𝛽_𝑛−1(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.

Table 1 

Values of successive approximations 𝛼_𝑛(𝑘) and 𝛽_𝑛(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.