# Existence and Uniqueness of Solutions for a Discrete Fractional Mixed Type Sum-Difference Equation Boundary Value Problem

Authors: Weidong Lv
Journal: Discrete Dynamics in Nature and Society
 (2015)

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## Abstract

By means of Schauder’s fixed point theorem and contraction mapping principle, we establish the existence and uniqueness of solutions to a boundary value problem for a discrete fractional mixed type sum-difference equation with the nonlinear term dependent on a fractional difference of lower order. Moreover, a suitable choice of a Banach space allows the solutions to be unbounded and two representative examples are presented to illustrate the effectiveness of the main results.


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## Body

## 1. Introduction

Fora,b∈R, such that b-a is a nonnegative integer, we denote Na={a,a+1,a+2,…} and Nab={a,a+1,…,b} throughout this paper. It is also worth noting that, in what follows, for any function u defined on Na, we appeal to the convention ∑s=k1k2u(s)=0, when k1,k2∈Na with k1>k2.In this paper, we will consider the following discrete fractional mixed type sum-difference equation boundary value problem:(1)Δαut+ft+α-1,ut+α-1,Δα-1ut,Tut,Sut=0,t∈N0,uα-2=0,Δα-1u∞=u∞,where α∈(1,2], Δα and Δα-1 denote the discrete Riemann-Liouville fractional differences of order α and α-1, respectively, f:Nα-1×R×R×R×R→R, Δα-1u(∞)=limt→+∞Δα-1u(t)=u∞∈R, and (2)Tut=∑s=0tkt,sus+α-1,Sut=∑s=0∞ht,sus+α-1,where k:D→R, D={(t,s)∈N0×N0:s≤t}, and h:N0×N0→R.Continuous fractional calculus is a generalization of ordinary differentiation and integration on an arbitrary order that can be noninteger. This subject, as old as the problem of ordinary differential calculus, can go back to the times when Leibniz and Newton invented differential calculus. The theory of fractional differential equations has received a lot of attention and now constitutes a new important mathematical branch due to its extensive applications in various fields of science and engineer. For more details, see [1–13] and references therein.It is well known that discrete analogues of differential equations can be very useful in applications [14], in particular for using computer to simulate the behavior of solutions for certain dynamic equations. However, compared to the long and rich history of continuous fractional calculus, discrete fractional calculus attracted mathematicians and scientists into its fairly new research area in a short period of time. In this time period, the theory of discrete calculus has been developed in many directions parallel to the theory in continuous fractional calculus such as initial value problems and boundary value problems for fractional difference equations, discrete Mittag-Leffler functions, and inequalities with discrete fractional operators; see [15–39] and the references therein. At the same time, in [27], Atıcı and Şengül have shown the usefulness of discrete Gompertz fractional difference equation for tumor growth model, which implies that discrete fractional difference calculus will provide a new excellent tool to model real world phenomena in the future.Although, among all recently research topics, the branch of discrete finite fractional difference boundary value problems is currently undergoing active investigation [16, 31–38], significantly less is known about discrete infinite fractional difference boundary value problems with the nonlinear term dependent on a fractional difference operator. Here, we should point out that in [39], Lv and Feng, by simple analogy with the ordinary case, introduced some basic definitions of discrete fractional calculus for Banach-valued functions and initially studied a class of discrete infinite fractional mixed type sum-difference equation boundary value problems in abstract spaces by using contracting mapping principle. Furthermore, as far as we know, the theory of discrete fractional mixed type sum-difference equations boundary value problems is still a new research area. So in this paper, we continue to focus on this topic for real-valued functions and provide some sufficient conditions for the existence and uniqueness of solutions to problem (1). Particularly note that problem (1) is not like the problem in [39] and the biggest difference is the nonlinear term f in (1) explicitly dependent on the discrete fractional difference operator of lower order. Hence, these differences that cause the main difficulties that we have to deal with in this paper are those of constructing a special Banach space and establishing an appropriate compactness criterion in it.The outline for the remainder of this paper is as follows. In Section2, we recall some useful preliminaries for discrete fractional calculus and present the basic space and its compactness criterion for studying problem (1). In Section 3, by employing Schauder’s fixed point theorem and contraction mapping principle, we establish the existence and uniqueness results of problem (1). In Section 4, two concrete examples are provided to illustrate the possible applications of the obtained analytical results.


## 2. Preliminaries

In this section, we firstly present here some necessary definitions and basic results about discrete fractional calculus.Definition 1 (see [30]).
For anyt and ν, the falling factorial function is defined as (3)tν_=Γt+1Γt+1-ν,provided that the right-hand side is well defined. We appeal to the convention that if t+1-ν is a pole of the Gamma function and t+1 is not a pole, then tν_=0.Definition 2 (see [40]).
Theνth discrete fractional sum of a function f:Na→R, for ν>0, is defined by(4)Δa-νft=1Γν∑s=at-νt-s-1ν-1_fs,t∈Na+ν. Also, we define the trivial sum Δa-0f(t)=f(t),t∈Na.Definition 3 (see [30]).
Theνth discrete Riemann-Liouville fractional difference of a function f:Na→R, for ν>0, is defined by (5)Δaνft=ΔnΔa-n-νft,t∈Na+n-ν,where n is the smallest integer greater than or equal to ν and Δn is the nth order forward difference operator. If ν=n∈N1, then Δanf(t)=Δnf(t).Remark 4.
From Definitions2 and 3, it is easy to see that Δa-ν maps functions defined on Na to functions defined on Na+ν and Δaν maps functions defined on Na to functions defined on Na+n-ν, where n is the smallest integer greater than or equal to ν. For ease of notation, we throughout this paper omit the subscript a in Δaνf(t) and Δa-νf(t) when it is not to lead to domains confusion and general ambiguity.Lemma 5 (see [30]).
Letf:Na→R and ν,μ>0. Then (6)Δa+μ-νΔa-μft=Δa-ν-μft=Δa+ν-μΔa-νft,t∈Na+μ+ν.Lemma 6 (see [31]).
Letn∈N1 and f:Nν-n→R with ν∈(n-1,n]. Then (7)Δ0-νΔν-nνft=ft+c1tν-1_+c2tν-2_+⋯+cntν-n_,fort∈Nν-n, ci∈R, i∈N1n.Lemma 7 (see [15]).
Leta∈R and μ>0 be given. Then, for ν∈(n-1,n], n∈N1, (8)Δa+μνt-aμ_=μν_t-aμ-ν_,t∈Na+μ+n-ν.Lemma 8 (see [28]).
Letf:Na→R, p∈N1 and ν>p. Then (9)ΔpΔa-νft=Δa-ν-pft.Next, we define the space,(10)X=u:Nα-2⟶R∣supt∈Nα-2⁡ut1+tα-1_<∞,supt∈N0⁡Δα-1ut<∞equiped with the norm (11)uX=max⁡supt∈Nα-2⁡ut1+tα-1_,supt∈N0⁡Δα-1ut.Furthermore, using the linear functional analysis theory, we can easily verify that(X,·X) is a Banach space, and then we present the following compactness criterion in it.Lemma 9.
LetV⊆X be a bounded set. If for any given ϵ>0, there exists a positive integer T=T(ϵ) such that (12)ut21+t2α-1_-ut11+t1α-1_<ϵ,(13)Δα-1us2-Δα-1us1<ϵwhenever t1,t2∈Nα+T, s1,s2∈NT+1, and u∈V; then V is relatively compact in X.Proof.
Evidently, it is sufficient to prove thatV is totally bounded. In what follows we divide this proof into two steps.
Step  1. Let us consider the case t∈Nα-2α+T.
Denote byVNα-2α+T the restriction of V on Nα-2α+T. Then VNα-2α+T, equipped with the norm u∞=supt∈Nα-2α+T(ut/(1+tα-1_)), is a finite dimension Banach space. So we know that VNα-2α+T is relatively compact from the boundness of V; hence VNα-2α+T is totally bounded; namely, for any ϵ>0, there exist finitely many ball Bϵ(ui), ui∈VNα-2α+T, i∈N1n, such that (14)VNα-2α+T⊂⋃i=1nBϵui,where Bϵ(ui)={u∈VNα-2α+T:u-ui∞=supt∈Nα-2α+Tu(t)/(1+tα-1_)-ui(t)/(1+tα-1_)<ϵ}.
Similarly, denoteVN0T+1α-1={Δα-1u:N0T+1→R∣u∈VNα-2α+T}. Then VN0T+1α-1 is also a Banach space with the norm Δα-1u∞=supt∈N0T+1|Δα-1u(t)| and it can be covered by finitely many balls Bϵ(Δα-1vj); that is, (15)VN0T+1α-1⊂⋃j=1mBϵΔα-1vj,vj∈VNα-2α+T,where Bϵ(Δα-1vj)={Δα-1u∈VN0T+1α-1:Δα-1u-Δα-1vj∞=supt∈N0T+1Δα-1u(t)-Δα-1vj(t)<ϵ}.
Step  2. Define Vij={u∈V:uNα-2α+T∈Bϵ(ui),Δα-1uN0T+1∈Bϵ(Δα-1vj)}.
Let us consider the caset∈Nα-2α+T. It is obvious that VNα-2α+T⊂⋃1≤i≤n,1≤j≤m‍VijNα-2α+T. Now, let us take uij∈Vij; then V can be covered by the balls B4ϵ(uij),i∈N1n,j∈N1m, where (16)B4ϵuij=u∈V:u-uijX<4ϵ.
In fact, for anyu∈V, the argument in Step 1 implies that there exist i and j such that uNα-2α+T∈Bϵ(ui), Δα-1uN0T+1∈Bϵ(Δα-1vj). Hence, for t∈Nα-2α+T and s∈N0T+1, we have (17)ut1+tα-1_-uijt1+tα-1_≤ut1+tα-1_-uit1+tα-1_+uit1+tα-1_-uijt1+tα-1_<2ϵ,(18)Δα-1us-Δα-1uijs≤Δα-1us-Δα-1vjs+Δα-1vjs-Δα-1uijs<2ϵ.For arbitrary t∈Nα+T, (12) and (17) yield that (19)ut1+tα-1_-uijt1+tα-1_≤ut1+tα-1_-uα+T1+α+Tα-1_+uα+T1+α+Tα-1_-uijα+T1+α+Tα-1_+uijα+T1+α+Tα-1_-uijt1+tα-1_<ϵ+2ϵ+ϵ=4ϵ, and for any s∈NT+1, (13) and (18) ensure that (20)Δα-1us-Δα-1uijs≤Δα-1us-Δα-1uT+1+Δα-1uT+1-Δα-1uijT+1+Δα-1uijT+1-Δα-1uijs<ϵ+2ϵ+ϵ=4ϵ.Relations (17)–(20) show that u-uijX<4ϵ. Therefore, V is totally bounded and this lemma is proved.


## 3. Main Result

In this section, we will establish the existence and uniqueness of solutions for problem (1) by using Schauder’s fixed point theorem and contraction mapping principle. For the sake of convenience and to abbreviate our presentation, for any function u∈X, we denote (21)gut=ft+α-1,ut+α-1,Δα-1ut,Tut,Sut,t∈N0in the sequel discussion and list here the following conditions: (C1)
k∗=supt∈N0⁡∑s=0tkt,s<∞ and h∗=supt∈N0⁡∑s=0∞(|h(t,s)|[1+(s+α-1)α-1_]/(1+t+α-1)α-1_) < ∞.(C2)
There exist functionspi:Nα-1→[0,∞), i∈N15, with(22)p∗=∑t=α-1∞1+tα-1_p1t+p3tk∗+p4th∗+p2t<Γαandp5∗=∑t=α-1∞p5(t)<∞ such that (23)ft,u,v,w,ϖ≤p1tu+p2tv+p3tw+p4tϖ+p5t,for(t,u,v,w,ϖ)∈Nα-1×R×R×R×R.(C2′)
f∗=∑t=α-1∞‍|f(t,0,0,0,0)|<∞, and there exist nonnegative numbers ai,i∈N14, and a function q:Nα-1→[0,+∞) with q∗=∑t=α-1∞q(t)[(1+tα-1_)(a1+a3k∗+a4h∗)+a2]<Γ(α) such that (24)ft,u,v,w,ϖ-ft,u¯,v¯,w¯,ϖ¯≤qta1u-u¯+a2v-v¯+a3w-w¯+a4ϖ-ϖ¯fort∈Nα-1,u,v,w,ϖ,u¯,v¯,w¯,ϖ¯∈R.Lemma 10.
If(C1) and (C2) hold, then, for any u∈X, (25)∑t=0∞gut≤p∗uX+p5∗.Proof.
For anyu∈X, t∈N0, using (C1), (C2), and the monotonicity of ια-1_, ι∈Nα-1 produces(26)gut=ft+α-1,ut+α-1,Δα-1ut,Tut,Sut≤p1t+α-1ut+α-1+p2t+α-1Δα-1ut+p3t+α-1Tut+p4t+α-1Sut+p5t+α-1≤1+t+α-1α-1_p1t+α-1+p3t+α-1∑s=0tkt,s1+s+α-1α-1_1+t+α-1α-1_+p4t+α-1∑s=0∞ht,s1+s+α-1α-1_1+t+α-1α-1_supt∈N0⁡ut+α-11+t+α-1α-1_+p2t+α-1Δα-1ut+p5t+α-1≤1+t+α-1α-1_p1t+α-1+p3t+α-1k∗+p4t+α-1h∗+p2t+α-1uX+p5t+α-1.Summating both sides of (26), we can get (25). The proof is completed.Lemma 11.
If(C1) and (C2) hold, then the unique solution of problem (1) is (27)ut=∑s=0∞Gt,sgus+u∞Γαtα-1_,t∈Nα-2,where (28)Gt,s=1Γαtα-1_-t-s-1α-1_,s∈N0t-α,tα-1_,s∈Nt-α+1.Proof.
Ifu:Nα-2→R satisfies the equation of problem (1), then Lemma 6 implies that (29)ut=-1Γα∑s=0t-αt-s-1α-1_gus+c1tα-1_+c2tα-2_,for some ci∈R,i∈N12, t∈Nα-2. By u(α-2)=0, we get c2=0.
Therefore,(30)ut=-1Γα∑s=0t-αt-s-1α-1_gus+c1tα-1_,t∈Nα-2.By virtue of Lemmas 5, 7, and 8, we have (31)Δα-1ut=-∑s=0t-1gus+c1Γα,t∈N0.Using the condition Δα-1u(∞)=u∞ in (31), we obtain (32)c1=1Γα∑s=0∞gus+u∞. Now, substitution of c1 into (30) gives (33)ut=-1Γα∑s=0t-αt-s-1α-1_gus+1Γα∑s=0∞tα-1_gus+u∞Γαtα-1_=∑s=0∞Gt,sgus+u∞Γαtα-1_,t∈Nα-2,where G(t,s) is defined by (28). The proof is completed.Remark 12.
From the expression ofG(t,s), we can easily find that G(t,s)≥0 and G(t,s)/(1+tα-1_)<1/Γ(α) for (t,s)∈Nα-2×N0.For anyu∈X, define an operator F by (34)Fut=∑s=0∞Gt,sgus+u∞Γαtα-1_,t∈Nα-2 and due to Lemma 10 and Remark 12, we have (35)Fut1+tα-1_≤∑s=0∞Gt,s1+tα-1_gus+u∞tα-1_Γα1+tα-1_≤1Γαp∗uX+p5∗+u∞,t∈Nα-2.On the other hand, by virtue of Lemmas 5, 7, 8, and 10, we get (36)Δα-1Fut=∑s=t∞gus+u∞,(37)Δα-1Fut≤p∗uX+p5∗+u∞which hold for t∈N0. So (35) and (37) imply that F:X→X is well defined and bounded. Furthermore, from Lemma 11, we can transform problem (1) into an operator equation u=Fu and it is clear to see that u is a solution of problem (1) which is equivalent to a fixed point of F.Remark 13.
Settingu¯=v¯=w¯=ϖ¯=0 in (C2′), we have (38)ft,u,v,w,ϖ≤qta1u+a2v+a3w+a4ϖ+ft,0,0,0,0for (t,u,v,w,ϖ)∈Nα-1×R×R×R×R, which implies that condition (C2′) is stronger than (C2). So under assumptions (C1) and (C2′), the operator F:X→X defined by (34) is also well defined.Now, we are in the position to give the main results of this work.Theorem 14.
Assume thatf:Nα-1×R×R×R×R→R is continuous, and suppose that conditions (C1) and (C2) hold. Then problem (1) has at least one solution u∈X.Proof.
In what follows, we divide this proof into three steps.
Step  1. Choose (39)R≥u∞+p5∗Γα-p∗and let (40)U=u∈X:uX≤R.Then, for any u∈U, by (35), (37), and the fact Γ(α)∈(0,1], we can verify that FuX≤R, which implies F:U→U.
Step  2. Let V be s subset of U. We employ Lemma 9 to verify that FV is relatively compact.
In view of Lemma10 and the boundness of V, there exists M>0 such that (41)∑t=0∞gut≤Mforanyu∈V.By (34) and (36), we have (42)Fut21+t2α-1_-Fut11+t1α-1_≤1Γαu∞+∑τ=0∞guτt2α-1_1+t2α-1_-t1α-1_1+t1α-1_+1Γα∑τ=0t2-αt2-τ-1α-1_1+t2α-1_guτ-∑τ=0t1-αt1-τ-1α-1_1+t1α-1_guτ,t1,t2∈Nα-2,Δα-1Fus2-Δα-1Fus1≤∑ϱ=s1s2-1guϱ,fors1,s2∈N0withs1<s2.Observing (42), together with limt→+∞(tα-1_/(1+tα-1_))=1 and the conditions of Lemma 9, we only need to show that, for any ϵ>0, there exists sufficiently large positive integer T such that, for any t1,t2∈Nα+T, (43)∑τ=0t2-αt2-τ-1α-1_1+t2α-1_guτ-∑τ=0t1-αt1-τ-1α-1_1+t1α-1_guτ<ϵ,and for any s1,s2∈NT+1 with s2>s1, (44)∑ϱ=s1s2-1guϱ≤ϵ.
Relation (41) yields that there exists a positive number L∈N0 such that (45)∑t=L+1∞gut≤ϵ3uniformlywithrespecttou∈V.On the other hand, from the monotonicity of ια-1_ and limt→+∞((t-L-1)α-1_/(1+tα-1_))=1, there exist T∈NL+1 such that, for any t1,t2∈Nα+T and τ∈N0L, (46)t2-τ-1α-1_1+t2α-1_-t1-τ-1α-1_1+t1α-1_≤1-t2-τ-1α-1_1+t2α-1_+1-t1-τ-1α-1_1+t1α-1_≤1-t2-L-1α-1_1+t2α-1_+1-t1-L-1α-1_1+t1α-1_<ϵ3M.Now taking t1,t2∈Nα+T, by virtue of (41), (45), and (46), we get(47)∑τ=0t2-αt2-τ-1α-1_1+t2α-1_guτ-∑τ=0t1-αt1-τ-1α-1_1+t1α-1_guτ≤∑τ=0Lt2-τ-1α-1_1+t2α-1_-t1-τ-1α-1_1+t1α-1_guτ+∑τ=L+1t2-αt2-τ-1α-1_1+t2α-1_guτ+∑τ=L+1t1-αt1-τ-1α-1_1+t1α-1_guτ<ϵ3M∑τ=0∞guτ+2∑τ=L+1∞guτ<ϵ.Moreover, from (45), we have (48)∑ϱ=s1s2-1guϱ≤∑ϱ=L+1∞guϱ<ϵwhich holds for any s1,s2∈NT+1 with s2>s1 and arbitrary u∈V. Moreover, it follows from (47) and (48) that (43) and (44) hold. Consequently, by Lemma 9, FV is relatively compact.
Step  3. F:U→U is a continuous operator.
Letun,u∈U,n∈N1 such that un-uX→0 as n→∞. Then by (C2), for any ϵ>0 there exists a positive integer L such that (49)∑t=α+L∞1+tα-1_p1t+p3tk∗+p4th∗+p2t<Γα6Rϵ,∑t=α+L∞p2t<Γα6ϵ.On the other hand, from the continuity of f, we know that there exists N∈N1 such that, for any n>N and t∈N0L, (50)gunt-gut≤Γα3L+1ϵ. Therefore, for t∈Nα-2 and n>N, by (49)-(50) and Remark 12, we can obtain that (51)Funt-Fut1+tα-1_≤∑s=0∞Gt,s1+tα-1_guns-gus<1Γα∑s=0Lguns-gus+∑s=L+1∞guns-gus≤1Γα∑s=0Lguns-gus+2R∑s=α+L∞1+sα-1_p1s+p3sk∗+p4sh∗+p2s+2∑s=α+L∞p5s<ϵ3+ϵ3+ϵ3=ϵ.Meanwhile, for t∈N0 and n>N, applying (49)-(50) again, we can easily verify that (52)Δα-1Funt-Δα-1Fut≤∑s=t∞gunt-gut≤∑s=0∞gunt-gut<Γαϵ<ϵ.Then, by virtue of (51) and (52), we conclude that Fun-FuX≤ϵ as n>N, which asserts the continuity of F.
Therefore, by Schauder’s fixed point theorem, we obtain that problem (1) has at least one solution in U and the proof is finished.Theorem 15.
Suppose that conditions(C1) and (C2′) hold. Then problem (1) has a unique solution u∈X.Proof.
For anyu,v∈X, in view of (C2′) and Remark 12, we have (53)Fut-Fvt1+tα-1_≤∑s=0∞Gt,s1+tα-1_gus-gvs≤1Γα∑s=0∞qs+α-1a1us+α-1-vs+α-1+a2Δα-1us-Δα-1vs+a3Tus-Tvs+a4Sus-Svs≤1Γαq∗u-vX,t∈Nα-2.On the other hand, by (36) and using (C2′) again, we have (54)Δα-1Fut-Δα-1Fvt≤∑s=t∞gus-gvs≤q∗u-vX,t∈N0.So, from (53), (54) and the facts that q∗<Γ(α) and Γ(α)∈(0,1] when α∈(1,2], we know that F is a contraction mapping. By means of Banach contraction mapping principle, we get that F has a unique fixed point in X; that is, problem (1) has a unique solution. This completes the proof.


## 4. Examples

In this section, we will illustrate the possible applications of the above established analytical results with the following two concrete examples.Example 1.
Consider the discrete fractional difference boundary value problem:(55)Δ3/2ut+3-t+11+t+1/21/2_2sin⁡ut+1/2+4-t+1/21+t+1/21/2_×1+ut+12+1+t+121/2_Δ1/2ut+∑s=0t1t+s+22_us+12+∑s=0∞cos⁡t2ss+22_1+s+1/21/2_us+121/4=0,t∈N0,u-12=0,Δ1/2u∞=u∞.
Conclusion. Problem (55) has at least one solution u:N-1/2→R.Proof.
Corresponding to problem (1), we have α=3/2, (56)kt,s=1t+s+22_,ht,s=cos⁡t2ss+22_1+s+1/21/2_,ft,u,v,w,ϖ=3-t-1/21+t1/2_2sin⁡u+4-t1+t1/2_1+u+1+t1/2_v+w+ϖ1/4,(t,u,v,w,ϖ)∈N1/2×R×R×R×R.
From the expression off, it is easy to see that f is continuous. Furthermore, we can verify that (57)k∗=supt∈N0⁡∑s=0t1t+s+22_=supt∈N0⁡12t+1=12<∞,h∗=supt∈N0⁡11+t+1/21/2_∑s=0∞cos⁡t2s1+s+1/21/2_1+s+1/21/2_s+22_≤supt∈N0⁡11+t+1/21/2_∑s=0∞1s+22_≤11+Γ3/2<34<∞.So condition (C1) is satisfied.
On the other hand, by using a simple inequality(58)1+zγ≤1+γz,forz∈0,+∞,γ∈0,1, we have (59)ft,u,v,w,ϖ≤3-t+1/21+t1/2_2sin⁡u+4-t1+t1/2_1+u+1+t1/2_v+w+ϖ1/4≤4-t+11+t1/2_u+1+t1/2_v+w+ϖ+3-t+1/21+t1/2_2+4-t1+t1/2_,and therefore (60)ft,u,v,w,ϖ≤p1tu+p2tv+p3tw+p4tϖ+p5t,where (61)p1t=p3t=p4t=4-t+11+t1/2_,p2t=4-t+1,p5t=3-t+1/21+t1/2_2+4-t1+t1/2_.By directly calculation, we have (62)p∗<1324<Γ32,p5∗<76.Thus, condition (C2) holds. So, by Theorem 14, our conclusion follows.Example 2.
Consider the following problem:(63)Δ4/3ut+2-t+181+t+1/31/3_cos⁡ut+1/3+sin⁡Δ1/3ut+3-t+1e21+t+1/31/3_∑s=0t1t+s+22_us+13+e-t+1e32+cost+1/3+t+1/31/3_∑s=0∞sin⁡t+ess+22_1+s+1/31/3_us+13=0,t∈N0,u-23=0,Δ1/3u∞=u∞.
Conclusion. Problem (63) has a unique solution u:N-2/3→R.Proof.
It is easy to see that problem (63) is the form of problem (1), where α=4/3, (64)kt,s=1t+s+22_,ht,s=sin⁡t+ess+22_1+s+1/31/3_,ft,u,v,w,ϖ=2-t+2/381+t1/3_cos⁡u+sin⁡v+3-t+2/3e21+t1/3_w+e-t+2/3e32+cos⁡t+t1/3_ϖ,for (t,u,v,w,ϖ)∈N1/3×R×R×R×R.
Choosinga1=a2=1/8, a3=1/e2, a4=1/e3, and q(t)=2-(t+2/3)/(1+t1/3_),t∈N1/3, then we can verify that f∗<1/4, k∗=1/2, h∗<0.5283, q∗<0.3454<0.8930≈Γ(4/3), and (65)ft,u,v,w,ϖ-ft,u¯,v¯,w¯,ϖ¯≤qta1u-u¯+a2v-v¯+a3w-w¯+a4ϖ-ϖ¯holds for any t∈N1/3, u,v,w,ϖ,u¯,v¯,w¯,ϖ¯∈R.
Clearly, all conditions of Theorem15 are fulfilled. Therefore, we can conclude that problem (63) has a unique solution.



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*Source: 376261-2015-10-11.xml*