# Monotone-Iterative Method for the Initial Value Problem with Initial Time Difference for Differential Equations with “Maxima”

Authors: S. Hristova; A. Golev
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

The object of investigation of the paper is a special type of functional differential equations containing the maximum value of the unknown function over a past time interval. An improved algorithm of the monotone-iterative technique is suggested to nonlinear differential equations with “maxima.” The case when upper and lower solutions of the given problem are known at different initial time is studied. Additionally, all initial value problems for successive approximations have both initial time and initial functions different. It allows us to construct sequences of successive approximations as well as sequences of initial functions, which are convergent to the solution and to the initial function of the given initial value problem, respectively. The suggested algorithm is realized as a computer program, and it is applied to several examples, illustrating the advantages of the suggested scheme.


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## Body

## 1. Introduction

Equations with “maxima” find wide applications in the theory of automatic regulation. As a simple example of mathematical simulations by means of such equations, we shall consider the system of regulation of the voltage of a generator of constant current ([1]). The object of regulation is a generator of constant current with parallel stimulation, and the quantity regulated is the voltage on the clamps of the generator feeding an electric circuit with different loads. An equation with maxima is used if the regulator is constructed such that the maximal deviation of the quantity is regulated on the segment [t-r,t]. The equation describing the work of the regulator has the form(1.1)Tu′(t)+u(t)+qmax⁡s∈[t-r,t]⁡u(s)=f(t),
where T and q are constant characterizing the object, u(t) is the voltage regulated, and f(t) is the perturbing effect.Note that the above given model as well as all types of differential equations with “maxima” could be considered as delay functional differential equations. Differential equations with delay are studied by many authors (see, e.g., [2–14]). At the same time, the presence of maximum function in the equations causes impossibilities of direct application of almost all known results for delay functional differential equations (see the monograph [15]). Also, in the most cases, differential equations with “maxima,’’ including the some linear scalar equations, are not possible to be solved in an explicit form. That requires the application of approximate methods. In the recent years, several effective approximate methods, based on the upper and lower solutions of the given problem, was proved for various problems of differential equations in [16–21].In the current paper, an approximate method for solving the initial value problem for nonlinear differential equations with “maxima” is considered. This method is based on the method of lower and upper solutions. Meanwhile, in studying the initial value problems the authors usually keep the initial time unchanged. But, in the real repeated experiments it is difficult to keep this time fixed because of all kinds of disturbed factors. It requires the changing of initial time to be taken into consideration. Note that several qualitative investigations of the solutions of ordinary differential equations with initial time difference are studied in [22–30]. Also some approximate methods for various types of differential equations with initial time difference are proved in [23, 26, 31–33].In this paper, an improved algorithm of monotone-iterative techniques is suggested to nonlinear differential equations with “maxima.” The case when upper and lower solutions of the given problem are known at different initial time is studied. The behavior of the lower/upper solutions of the initial value problems with different initial times is studied. Also, some other improvements in the suggested algorithm are given. The main one is connected with the initial functions. In the known results for functional differential equations, the initial functions in the corresponding linear problems for successive approximations are the same (see, e.g., [20, 34–42]). In our case, it causes troubles with obtaining the solutions, which are successive approximations. To avoid these difficulties, we consider linear problems with different initial functions at any step. It allows us to chose appropriate initial data. Additionally, in connection with the presence of the maximum function in the equation and computer application of the suggested algorithm, an appropriate computer program is realized and it is applied to some examples illustrating the advantages and the practical application of the considered algorithm.


## 2. Preliminary Notes and Definitions

Consider the following initial value problem for the nonlinear differential equation with “maxima’’ (IVP):(2.1)x′=f(t,x(t),max⁡s∈[t-r,t]⁡x(s))fort∈[t0,∞),(2.2)x(t)=φ(t-t0)fort∈[t0-r,t0],
where x∈ℝ, t0≥0, f:ℝ+×ℝ×ℝ→ℝ, φ(t):[-r,0]→ℝ, and r>0 is a fixed constant.In the paper, we will study the differential equation with “maxima” (2.1) with an initial condition at two different initial points. For this purpose, we consider the differential equation (2.1) with the initial condition(2.3)x(t)=φ(t-τ0)fort∈[τ0-r,τ0],
where τ0≥0.Assume there exist a solutionx(t;t0,φ) of the IVP (2.1), (2.2) defined on [t0-r,t0+T], and there exists a solution y(t;τ0,φ) of the IVP (2.1), (2.3) defined on [τ0-r,τ0+T], where T>0 is a given constant. In the paper, we will compare the behavior of both solutions.Definition 2.1.
The functionα∈C([t0-r,t0+T),ℝ)∪C1([t0,t0+T),ℝ) is called a lower (upper) solution of the IVP (2.1), (2.2) for t≥t0-h if the following inequalities are satisfied:
(2.4)α′(t)≤(≥)f(t,α(t),max⁡s∈[t-r,t]⁡α(s))fort∈[t0,t0+T],α(t)≤(≥)φ(t-t0)fort∈[t0-r,t0].Note that the functionα(t) is a lower (upper) solution of the IVP (2.1), (2.3) for t∈[τ0-r,τ0+T] if the point t0 in the inequalities (2.4) is replaced by τ0.


## 3. Comparison Results

Often in the real world applications, the lower and upper solutions of one and the same differential equation are obtained at different initial time intervals.The following result is a comparison result for lower and upper solutions with initial conditions given on different initial time intervals.Theorem 3.1.
Let the following conditions be satisfied.(1)
Lett0,τ0≥0 be fixed such that η=τ0-t0>0.(2)
The functionα∈C([t0-r,t0+T],ℝ)∪C1([t0,t0+T],ℝ) is a lower solution of the IVP (2.1), (2.2).(3)
The functionβ∈C([τ0-r,τ0+T],ℝ)∪C1([τ0,τ0+T],ℝ) is an upper solution of the IVP (2.1), (2.3).(4)
The functionf(t,u,v):ℝ+×ℝ×ℝ→ℝ is nondecreasing in its both first and third argument t for any u∈ℝ, and there exist positive constants L1,L2 such that for t∈ℝ+ and u1≥u2 and v1≥v2, the inequality
(3.1)f(t,u1,v1)-f(t,u2,v2)≤L1(u1-u2)+L2(v1-v2)
holds.(5)
The functionφ(t)∈C([-r,0],ℝ).
Thenα(t)≤β(t+η) for t∈[t0,t0+T].Proof.
Choose a positive numberM such that M>L1+L2 and a positive number ϵ. Define a function w(t)=β(t+η)+ϵeMt for t∈[t0,t0+T] and w(t)=β(t+η)+ϵeMt0fort∈[t0-r,t0). Then for t∈[t0,t0+T], the following inequalities
(3.2)w(t)>β(t+η),max⁡s∈[t-r,t]⁡w(s)=max⁡s∈[t-r,t]⁡(β(s+η)+ϵeMs)≥max⁡s∈[t-r,t]⁡(β(s+η))=max⁡s∈[t+η-r,t+η]⁡β(s),max⁡s∈[t-r,t]⁡w(s)≤max⁡s∈[t-r,t]⁡(β(s+η))+ϵeMt=max⁡s∈[t+η-r,t+η]β(s)+ϵeMt
hold.
Therefore, we obtain(3.3)w′(t)≥f(t+η,β(t+η),max⁡s∈[t+η-r,t+η]⁡β(s))+ϵMeMt≥-L1(w(t)-β(t+η))-L2(max⁡s∈[t-r,t]⁡w(s)-max⁡s∈[t+η-r,t+η]⁡β(s))+f(t+η,w(t),max⁡s∈[t-r,t]⁡w(s))+ϵMeMt≥f(t+η,w(t),max⁡s∈[t-r,t]w(s))+ϵeMt(M-L1-L2)>f(t+η,w(t),max⁡s∈[t-r,t]⁡w(s))≥f(t,w(t),max⁡s∈[t-r,t]⁡w(s)).
Note that fort∈[t0-r,t0] we have α(t)≤φ(t-t0)=φ(t+η-τ0)≤β(t+η)<w(t). We will prove that
(3.4)α(t)≤w(t)fort∈[t0-r,t0+T].
Assume the contrary, that is, there exists a pointt1>t0 such that α(t)<w(t) for t∈[t0-r,t1), α(t1)=w(t1) and α(t)≥w(t) for t∈(t1,t2), where t2 is sufficiently close to t1. Therefore α′(t1)≥w′(t1). Then w′(t1)≤α′(t1)≤f(t1,α(t1),maxs∈[t1-r,t1]α(s))≤f(t1,w(t1),maxs∈[t1-r,t1]w(s))<w′(t1). The obtained contradiction proves the claim.
Therefore,α(t)≤w(t)<β(t+η) for t∈[t0-r,t0+T].Corollary 3.2.
Let the conditions of Theorem3.1 be fulfilled. Then α(t-η)≤β(t) for t∈[τ0-r,τ0+T].The comparison results are true if the inequalityτ0<t0 holds.Theorem 3.3 (comparison result).
Let the following conditions be satisfied.(1)
The conditions 2, 3, 5 of Theorem3.1 are satisfied.(2)
Lett0,τ0≥0 be fixed such that η=τ0-t0<0.(3)
The functionf(t,u,v):ℝ+×ℝ×ℝ→ℝ is nonincreasing in its first argument and is nondecreasing in its third argument for any u∈ℝ, and there exist positive constants L1,L2 such that for t∈ℝ+ and u1≥u2 and v1≥v2 the inequality(3.5)f(t,u1,v1)-f(t,u2,v2)≤L1(u1-u2)+L2(v1-v2)
holds.
Thenα(t)≤β(t+η) for t≥t0.The proof of Theorem3.3 is similar to the proof of Theorem 3.1 and we omit it.In what follows, we shall need that the functionsα,β∈C([t0-r,T+t0],ℝ) are such that α(t)≤β(t+η).Consider the sets:(3.6)S(α,β,t0,η)={u∈C([t0-r,T+t0],R):α(t)≤u(t)≤β(t+η)fort∈[t0-r,T+t0]},S̃(α,β,τ0,η)={u∈C([τ0-r,T+τ0],R):α(t-η)≤u(t)≤β(t)fort∈[τ0-r,T+τ0]},Ω(α,β,t0)={(t,x,y)∈[t0,T+t0]×R2:α(t)≤x≤β(t),max⁡s∈[t-r,t]⁡α(s)≤y≤max⁡s∈[t-r,t]⁡β(s)}.In our further investigations, we will need the following comparison result on differential inequalities with “maxima.’’Lemma 3.4 (see [43, Lemma  2.1]).
Let the functionm∈C([t0-r,T+t0],ℝ)∪C1([t0,T+t0],ℝ) satisfies the inequalities
(3.7)m′≤-L1m(t)-L2min⁡s∈[t-r,t]⁡m(s)fort∈[t0,T+t0],m(t)≤0fort∈[t0-r,t0],
where the positive constants L1,L2 are such that (L1+L2)T≤1.
Then the inequalitym(t)≤0 holds on [t0-r,T+t0].In our further investigations, we will use the following result, which is a partial case of Theorem3.1 [44].Lemma 3.5 (existence and uniqueness result).
Let the following conditions be fulfilled.(1)
The functionsQ∈C([t0,t0+T],ℝ),q∈C([t0,t0+T],ℝ).(2)
The functionsφ∈C([-r,0],ℝ).
Then the initial value problem for the linear scalar equation(3.8)u′=L1u(t)+L2max⁡s∈[t-r,t]⁡u(s)+Q(t)fort∈[t0,T+t0],u(t)=φ(t-t0)fort∈[t0-r,t0]
has a unique solution u(t-t0) on the interval [t0-r,T+t0].


## 4. Main Results

DenoteJ1=[t0-r,T+t0] and J2=[τ0-r,T+τ0], where t0,τ0≥0 are fixed numbers.Case 1.
Lett0≤τ0.
Theorem 4.1.
Let the following conditions be fulfilled.(1)
The pointst0,τ0≥0 are such that η=τ0-t0≥0.(2)
The functionα0(t)∈C([t0-r,T+t0],ℝ)∪C1([t0,T+t0],ℝ) is a lower solution of the IVP (2.1), (2.2) in J1.(3)
The functionβ0(t)∈C([τ0-r,T+τ0],ℝ)∪C1([τ0,T+τ0],ℝ) is an upper solution of the IVP (2.1), (2.3) in J2.(4)
The functionφ∈C([-r,0],ℝ).(5)
The functionf∈C(Ω(α0,β0,t0),ℝ) is nondecreasing in its first argument and satisfies the one side Lipschitz condition
(4.1)f(t,u1,v1)-f(t,u2,v2)≥L1(u2-u1)+L2(v2-v1),
for u1≤u2 and v1≤v2, where L1,L2>0 are such that(4.2)(L1+L2)T≤1.
Then there exist two sequences of functions{αn(t)}0∞ and {βn(t)}0∞ such that(a)
the functionsαn(t)=1,2,…  are lower solutions of the IVP (2.1), (2.2) on [t0-r,T+t0];(b)
the functionsβn(t)=1,2,…  are upper solutions of the IVP (2.1), (2.3) on [τ0-r,T+τ0];(c)
the sequence{αn(t)}0∞ is increasing;(d)
the sequence{βn(t)}0∞ is decreasing;(e)
the inequalities(4.3)α0(t)≤⋯≤αn(t)≤βn(t+η)≤⋯≤β0(t+η),t∈[t0-r,T+t0],α0(t-η)≤⋯≤αn(t-η)≤βn(t)≤⋯≤β0(t),t∈[τ0-r,T+τ0]
hold;(f)
both sequences uniformly converge andx(t)=limn→∞αn(t) is a solution of the IVP (2.1), (2.2) in S(α0,β0,t0,η), and y(t)=limn→∞βn(t) is a solution of the IVP (2.1), (2.3) in S̃(α0,β0,τ0,η).Proof.
According to Theorem3.1, the inequality α0(t)≤β0(t+η) holds on [t0-r,T+t0].
LetL0=mins∈[t0-r,t0](φ(s-t0)-α0(s))≥0. Choose a number k0∈[0,1) such that
(4.4)k0≤L0.
Therefore,φ(t-t0)-α0(t)≥L0>k0L0 or α0(t)≤φ(t-t0)-k0L0 on [t0-r,t0].
We consider the linear differential equation with “maxima’’(4.5)x′(t)=f(t,α0(t),max⁡s∈[t-r,t]⁡α0(s))-L1(x(t)-α0(t))-L2(max⁡s∈[t-r,t]⁡x(s)-max⁡s∈[t-r,t]⁡α0(s))fort∈[t0,T+t0],
with the initial condition
(4.6)x(t)=φ(t-t0)-k0L0fort∈[t0-r,t0].
According to Lemma 3.5, the linear initial value problem (4.5), (4.6) has a unique solution α1(t) in J1.
We will prove thatα0(t)≤α1(t) on J1.
From the choice ofL0,k0 and equality (4.6), it follows that α0(t)≤φ(t-t0)-k0L0=α1(t) for t∈[t0-r,t0].
Now, lett∈[t0,T+t0]. Consider the function u(t)=α0(t)-α1(t) defined on J1. It is clear that u(t)≤0 on [t0-r,t0]. From the definition of the function α0(t) and (4.5), we have
(4.7)u′(t)≤-L1u(t)-L2(max⁡s∈[t-r,t]⁡α0(s)-max⁡s∈[t-r,t]⁡α1(s))fort∈[t0,T+t0].
Now from inequality (4.7) and
(4.8)max⁡s∈[t-h,t]⁡α0(s)-max⁡s∈[t-r,t]⁡α1(s)=max⁡s∈[t-r,t]⁡α0(s)-α1(ξ)≥α0(ξ)-α1(ξ)≥min⁡s∈[t-r,t]⁡(α0(s)-α1(s))=min⁡s∈[t-r,t]⁡u(s),
we obtain
(4.9)u′(t)≤-L1u(t)-L2min⁡s∈[t-r,t]⁡u(s),t∈[t0,T+t0].
According to Lemma 3.4, the inequality u(t)≤0 holds on J1, that is, α0(t)≤α1(t).
We will prove that the functionα1(t) is a lower solution of the IVP (2.1), (2.2) on J1.
From equality (4.6), it follows the validity of the inequality α1(t)≤φ(t-t0) on [t0-r,t0]. Now, let t∈[t0,T+t0]. Then, since α0(t)≤α1(t), according to the one side Lipschitz condition 2 of Theorem 4.1, we have
(4.10)α1′(t)=Q(t)-L1(α1(t)-α0(t))-L2(max⁡s∈[t-r,t]⁡α1(s)-max⁡s∈[t-r,t]⁡α0(s))=f(t,α1(t),max⁡s∈[t-r,t]⁡α1(s))+f(t,α0(t),max⁡s∈[t-r,t]⁡α0(s))-f(t,α1(t),max⁡s∈[t-r,t]⁡α1(s))-L1(α1(t)-α0(t))-L2(max⁡s∈[t-r,t]⁡α1(s)-max⁡s∈[t-r,t]⁡α0(s))≤f(t,α1(t),max⁡s∈[t-r,t]⁡α1(s))fort∈[t0,T+t0],
where Q(t)=f(t,α0(t),maxs∈[t-r,t]α0(s)).
Next, according to Lemma3.4 for the function u(t)=α1(t)-β0(t-η), the inequality α1(t)≤β0(t+η) holds on J1, that is, the inclusion α1∈S(α0,β0,t0,η) is valid.
LetC0=mins∈[τ0-r,τ0](β0(s)-φ(s-τ0))≥0. Choose a number p0∈[0,1) such that
(4.11)p0≤C0.
Therefore,β0(t)-φ(t-τ0)≥C0>p0C0 or β0(t)≥φ(t-τ0)+p0C0 on [τ0-r,τ0].
We consider the linear differential equation with “maxima’’(4.12)x′(t)=f(t,β0(t),max⁡s∈[t-r,t]⁡β0(s))-L1(x(t)-β0(t))-L2(max⁡s∈[t-r,t]⁡x(s)-max⁡s∈[t-r,t]⁡β0(s))fort∈[τ0,T+τ0]
with the initial condition
(4.13)x(t)=φ(t+τ0)+p0C0fort∈[τ0-r,τ0].
There exists a unique solutionβ1(t) of the IVP (4.12), (4.13), which is defined on J2.
The functionβ1(t) is an upper solution of the IVP (2.1), (2.3) on J2, and the inclusion β1∈S̃(α0,β0,τ0,η) is valid. The proofs are similar to the ones about the function α1. We omit the proofs.
Functionsα1(t),β1(t) are a lower and an upper solution of the IVP (2.1), (2.2), and (2.1), (2.3) correspondingly. According to Lemma 3.4 the inequality α1(t)≤β1(t+η) on J1 holds.
Similarly, recursively, we can construct two sequences of functions{αn(t)}0∞ and {βn(t)}0∞. In fact, if the functions αn(t) and βn(t) are known, and Ln=mins∈[-r,0](φ(s)-αn(s)),Cn=mins∈[-r,0](βn(s)-φ(s)), and the numbers kn,pn∈[0,1) are such that
(4.14)kn≤Ln,pn≤Cn,
then the function αn+1(t) is the unique solution of the initial value problem for the linear differential equation with “maxima’’
(4.15)x′=f(t,αn(t),max⁡s∈[t-r,t]⁡αn(s))-L1(x-αn(t))-L2(max⁡s∈[t-r,t]⁡x(s)-max⁡s∈[t-r,t]⁡αn(s))fort∈[t0,T+t0],(4.16)x(t)=φ(t)-knLnfort∈[t0-r,t0],
and the function βn+1(t) is the unique solution of the initial value problem
(4.17)x′(t)=f(t,βn(t),max⁡s∈[t-r,t]⁡βn(s))-L1(x-βn(t))-L2(max⁡s∈[t-h,t]⁡x(s)-max⁡s∈[t-r,t]βn(s))fort∈[τ0,T+τ0],(4.18)x(t)=φ(t)+pnCnfort∈[τ0-r,τ0].
Now following exactly as for the casen=0, it can be proved that the function αn+1(t) is a lower solution of the IVP (2.1), (2.2) on J1, the function βn+1(t) is an upper solution of the IVP (2.1), (2.3) on J2, the inclusions αn+1∈S(αn,βn,t0,η), βn+1∈S̃(αn,βn,τ0,η) are valid, and the inequalities (4.3) hold.
Therefore, the sequence{αn(t)}0∞ is uniformly convergent on J1 and {βn(t)}0∞ is uniformly convergent on J2.
Denote(4.19)lim⁡n→∞⁡αn(t)=u(t),t∈[t0-r,T+t0],lim⁡n→∞⁡βn(t)=v(t),t∈[τ0-r,T+τ0].
From the uniform convergence and the definition of the functions αn(t) and βn(t), it follows that the following inequalities hold
(4.20)α0(t)≤u(t)≤v(t+η)≤β0(t+η),t∈[t0-r,T+t0].Case 2.
Lett0>τ0.
In this case, we could approximate again the solution of the given initial value problem, starting from lower and upper solutions given at two different initial points. Since the proofs are similar, we will set up only the results.
Theorem 4.2.
Let the following conditions be fulfilled.(1)
The pointst0,τ0≥0 are such that η=t0-τ0≥0.(2)
The conditions 2, 3, 4 of Theorem4.1 are satisfied.(3)
The functionf∈C(Ω(α0,β0,τ0),ℝ) is nonincreasing in its first argument and satisfies the one side Lipschitz condition
(4.21)f(t,u1,v1)-f(t,u2,v2)≥L1(u2-u1)+L2(v2-v1)
for u1≤u2 and v1≤v2, where L1,L2>0 are such that
(4.22)(L1+L2)T≤1.
Then there exist two sequences of functions {αn(t)}0∞ and {βn(t)}0∞ such that(a)
the functionsαn(t)=1,2,…  are lower solutions of the IVP (2.1), (2.2) on [t0-r,T+t0];(b)
the functionsβn(t)=1,2,…  are upper solutions of the IVP (2.1), (2.3) on [τ0-r,T+τ0];(c)
the sequence{αn(t)}0∞ is increasing;(d)
the sequence{βn(t)}0∞ is decreasing;(e)
the inequalities(4.23)α0(t)≤⋯≤αn(t)≤βn(t+η)≤⋯≤β0(t+η),t∈[t0-r,T+t0],α0(t-η)≤⋯≤αn(t-η)≤βn(t)≤⋯≤β0(t),t∈[τ0-r,T+τ0]
hold;(f)
both sequences uniformly converge andx(t)=limn→∞αn(t) is a solution of the IVP (2.1), (2.2) in S(α0,β0,t0,η) and y(t)=limn→∞βn(t) is a solution of the IVP (2.1), (2.3) in S̃(α0,β0,τ0,η).


## 5. Computer Realization

Since the set of differential equations, which could be solved in an explicit form, is very narrow, we will realize the above suggested algorithm numerically. At present, we cannot solve numerically differential equations with “maxima” by the existing ready-made systems, such asMathematica, Mathlab, and so forth, because of the maximum of the unknown function over a past time interval. It requires proposed and computer realized new algorithms for solving such kind of equations. In [43], an algorithm for solving a class of differential equations with “maxima,” based on the trapezoid’s method, is presented. The main problem in the suggested algorithm is obtaining the maximum of the unknown function over a past time interval. It is based on the idea that the local maximum depends on values that are initially inserted and later they are removed within the range. A special structure is applied to help quicker obtaining the local maximum ([43]). The suggested scheme could be based on any numerical method for solving differential equations. In this paper, we will use Euler method in the application of the algorithm ([43]) for solving differential equations with “maxima.” We combine the algorithm of Euler method and obtaining of maximum of the function, and we get the following modification:
(5.1)tk+1=tk+h,yn+1(tk+1)=yn+1(tk)+hf(tk,yn+1(tk),yn(tk),max⁡⁡yn,k,max⁡⁡yn+1,k),
where the values maxyn,k=maxs∈[tk-r,tk]yn(s) and maxyn+1,k=maxs∈[tk-r,tk]yn+1(s) are obtained by the algorithm suggested in [43], and the initial values of yn+1(t0) are found by initial conditions (4.16) and (4.18).


## 6. Applications

Now we will illustrate the employment of the suggested above scheme to a particular nonlinear scalar differential equations with “maxima.’’Initially we will consider the case of one and the same initial points of both initial value problems. By this way, we will emphasize our considerations to the advantages of the involved in the initial conditions constants.Example 6.1.
Consider the following scalar nonlinear differential equation with “maxima’’:(6.1)x′=11-x(t)-2max⁡s∈[t-0.1,t]⁡x(s)-1,fort∈[0,0.35],
with the initial condition
(6.2)x(t)=0,t∈[-0.1,0],
where x∈ℝ. It is easy to check that the initial value problem (6.1), (6.2) has a zero solution.
In this casef(t,x,y)≡1/(1-x)-2y-1, t0=τ0=0, T=0.35, r=0.1, J1=J2=[-0.1,0.35]. Choose α0(t)=-1/4 and β0(t)=1/4. Then α0(t) is a lower solution and β0(t) is an upper solution of the IVP (6.1), (6.2). In this case, Ω(α0,β0,t0)={(t,u,v)∈[0,0.35]×ℝ2:-1/4≤u,v≤1/4}. Let (t,u1,v1),(t,u2,v2)∈Ω(α0,β0,t0) and u1≤u2 and v1≤v2. Then
(6.3)f(t,u1,v1)-f(t,u2,v2)=11-u1-11-u2+2(v2-v1)=1(1-u1)(1-u2)(u2-u1)+2(v2-v1)≥1625(u2-u1)+2(v2-v1).
That is, L1=0.64,L2=2, and the inequality (L1+L2)T<1 holds.
The successive approximationsαn+1(t) and βn+1(t) will be the unique solutions of the linear problems
(6.4)x′=11-αn(t)+0.64αn(t)-1-0.64x-2max⁡s∈[t-0.1,t]⁡x(s)fort∈[0,0.35],x(t)=-knLnfort∈[-0.1,0],x′(t)=11-βn(t)-1+0.64βn(t)-0.64x-2max⁡s∈[t-0.1,t]⁡x(s)fort∈[0,0.35],x(t)=pnCnfort∈[-0.1,0],
where Ln=mins∈[-0.1,0](-αn(s))≥0,Cn=mins∈[-0.1,0](βn(s))≥0, and the numbers kn,pn∈[0,1) are chosen such that kn≤Ln, pn≤Cn, n=1,2,3,….
Now we will construct an increasing sequence of lower solutions and a decreasing sequence of upper solutions, which will be convergent to the zero solution.
The first lower approximationα1(t) is a solution of the IVP
(6.5)x′=-925-0.64x-2max⁡s∈[t-0.1,t]⁡x(s)fort∈[0,0.35],x(t)=-k04,t∈[-0.1,0].
Choose k0=6/11. Then the IVP (6.5) has an exact solution α1(t)=-3/22.
The second lower approximationα2(t) is a solution of the IVP
(6.6)x′=-57275-0.64x-2max⁡s∈[t-0.1,t]⁡x(s)fort∈[0,0.35],x(t)=-k1322,t∈[-0.1,0].
Choose k1=19/33. Then the IVP (6.6) has an exact solution α2(t)=-57/725. It is clear that α0(t)≤α1(t)≤α2(t)<0=x(t).
The first upper approximationβ1(t) is a solution of the IVP
(6.7)x′(t)=3775-0.64x-2max⁡s∈[t-0.1,t]⁡x(s)fort∈[0,0.35],x(t)=p04,t∈[-0.1,0].
Choose p0=74/99. Then the IVP (6.7) has an exact solution β1(t)=37/198≈0.1869.
The second upper approximationβ2(t) is a solution of the IVP
(6.8)x′=-114481398475-0.64x-2max⁡s∈[t-0.1,t]⁡x(s)fort∈[0,0.35],x(t)=p137198,t∈[-0.1,0].
Choose p1=114481/196581. Then the IVP (6.8) has an exact solution β2(t)=114481/1051974≈0.108825.
It is obviouse thatβ0(t)>β1(t)>β2(t)>0=x(t)>α2(t)>α1(t)>α0(t) on the interval [-0.1,0.35].
Example6.1 illustrates that the presence of the constants pn and kn allow us easily to construct each successive approximation in an explicit form.Now we will illustrate the suggested method in the case when lower and upper solutions are defined for different initial points and on different intervals.Example 6.2.
Consider the following IVP for the scalar nonlinear differential equation with “maxima’’:(6.9)x′=t1+x(t)-0.5max⁡s∈[t-0.1,t]⁡x(s)-2fort∈[1.5,1.75],x(t)=0,t∈[1.4,1.5],(6.10)x′=t1+x(t)-0.5max⁡s∈[t-0.1,t]⁡x(s)-2fort∈[2.35,2.6],x(t)=0,t∈[2.25,2.35],
where x∈ℝ.
In this case,f(t,x,y)≡t/(1+x)-0.5y-2, t0=1.5, τ0=2.35, η=τ0-t0=0.85, T=0.25, r=0.1, J1=[1.4,1.75], J2=[2.25,2.6].
Chooseα0(t)=-1/4. Then for t∈[1.5,1.75] the inequality
(6.11)f(t,α0(t),max⁡s∈[t-0.1,t]⁡α0(s))=t1-1/4+18-2=4t3-158≥1.543-158=0.125>0
holds. Therefore, α0(t) is a lower solution of the IVP (6.9) in [1.4,1.75].
Letβ0(t)=1/4. Then for t∈[2.35,2.6] the inequality
(6.12)f(t,β0(t),max⁡s∈[t-0.1,t]⁡β0(s))=t1+1/4-18-2=4t5-178≤2.645-178=-0.045<0
holds. Therefore, β0(t) is an upper solution of the IVP (6.10) on [2.25,2.6].
In this case,τ0>t0 and Ω(α0,β0,τ0)={(t,u,v)∈[1.5,1.75]×ℝ2:-1/4≤u,v≤1/4}.
Lett∈[1.5,1.76], -1/4≤u1≤u2≤1/4 and -1/4≤v1≤v2≤1/4. Then from the inequality t/((1+u1)(1+u2))≤(16/9)t≤3.1111<3.2, we obtain
(6.13)f(t,u1,v1)-f(t,u2,v2)=t1+u1-t1+u2+0.5(v2-v1)=t(1+u1)(1+u2)(u2-u1)+0.5(v2-v1)≤3.2(u2-u1)+(v2-v1).
That is, L1=3.2,L2=0.5, and the inequality (L1+L2)T<1 holds.
According to Theorem4.1, there exist two convergent monotone sequences of functions {αn(t)}0∞ and {βn(t)}0∞ such that (4.3) holds and their limits are solutions of the IVP (6.9) and (6.10).
The successive approximationsαn+1(t) will be the unique solutions of the linear problems
(6.14)x′=t1+αn(t)-0.5αn(t)-2-3.2(x-αn(t))-0.5(max⁡s∈[t-0.1,t]⁡x(s)-max⁡s∈[t-0.1,t]⁡αn(s))=t1+αn(t)+2.7αn(t)+0.5max⁡s∈[t-0.1,t]⁡αn(s)-2-3.2x-0.5max⁡s∈[t-0.1,t]⁡x(s)fort∈[1.5,1.75],x(t)=-knLnfort∈[1.4,1.5],
and the successive approximations βn+1(t) will be the unique solutions of the linear problems
(6.15)x′(t)=t1+βn(t)+2.7βn(t)+0.5max⁡s∈[t-0.1,t]⁡βn(s)-2-3.2x-0.5max⁡s∈[t-0.1,t]⁡x(s)fort∈[2.35,2.6],x(t)=pnCnfort∈[2.25,2.35],
where α0(t)=-1/4, β0(t)=1/4, Ln=Cn=0.25, kn=0.25n, pn=0.25n, and n=1,2,3,….
Note the above IVPs are linear, but we are not able to obtain their solutions in explicit form because of the presence of maximum function.We will use the computer realization of the considered method, explained in Section4, to solve the initial value problems (6.14) and (6.15) for n=1,2,3,4,5. In this case at any step, the function yn(tk) in (5.1) is replaced by the functions αn(tk) or βn(tk), respectively, the function yn+1(tk) is the next approximation αn+1(tk) or βn+1(tk), respectively, and the function f is the right part of (6.14) or (6.15), respectively, and h=0.00001. Some of the numerical values of the successive approximations {αn(t)}0∞ and {βn(t)}0∞, which are lower/upper solutions of the given problem, are shown in the Table 1. Also, the obtained successive approximations are graped on the Figure 1. Both, Table 1 and Figure 1, illustrate us the monotonicity of the sequences {αn(t)}0∞ and {βn(t)}0∞ and the validity of the inequalities(6.16)α0(t)≤α1(t)≤⋯≤αn(t)≤βn(t+.85)≤⋯≤β0(t+.85),t∈[1.5,1.75].Table 1
Values of the successive lower/upper approximationsαk(t) and βk(t+.85).
t1.501.551.601.651.701.75β1(t+.85)0.25000000.23962850.23263880.22866080.22731770.2281468β2(t+.85)0.06250000.07954870.09548780.11388750.13071750.1464314β3(t+.85)0.01562500.03226340.04906840.06710410.08509140.1029515β4(t+.85)0.00390630.02111060.03835520.05588820.07342480.0909491β5(t+.85)0.00097660.01839820.03582320.05329960.07075230.0881732α5(t)−0.0009766−0.0236676−0.0416446−0.0552493−0.0648509−0.0708277α4(t)−0.0039063−0.0269991−0.0452915−0.0591247−0.0688717−0.0749110α3(t)−0.0156250−0.0399773−0.0590208−0.0731539−0.0828101−0.0884511α2(t)−0.0625000−0.0871172−0.1048556−0.1164338−0.1225992−0.1243204α1(t)−0.2500000−0.2427252−0.2336358−0.2230384−0.2111877−0.1982954Figure 1
(graphs ofαk(t) and βk(t+.85), k=1,2,3,4).



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