# Some New Discrete Gronwall-Bellman Type Inequalities with Three Independent Variables and Applications

Authors: Weihua Liu; Haisong Huang
Journal: Journal of Function Spaces
 (2018)

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## Abstract

Some new discrete Gronwall-Bellman type inequalities with three independent variables which generalize some existing results and can be used as handy tools in the study of qualitative and quantitative properties of solutions of certain classes of difference equation are presented. As applications, some difference equations with the initial boundary conditions are also considered.


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## Body

## 1. Introduction

Finite difference inequalities which exhibit explicit bounds on unknown functions in general provide a very useful and important tool in the development of the theory of finite difference equations. Due to the fact that finite difference inequalities have been applied in various branches of finite difference equations, many such inequalities with in one or two variables have been established [1–4]. For example, Li [5, 6] investigated some new discrete inequalities in two independent variables which provide explicit bounds on unknown functions. Cheung [7–9] generalized some existing discrete Gronwall-Bellman-type inequalities to more general situations and applied them to study the boundedness, uniqueness, and continuous dependence of solutions of certain discrete boundary value problem for difference equations.Wang [10–13] established some nonlinear retarded difference inequalities and gave some applications of the obtained inequalities to the estimation of finite difference equations. Pachpatte [14–16] obtained certain inequalities arising in the theory of differential equations and provided new estimates for these types of inequalities. Ma et al. [17–19] obtained some power nonlinear Volterra-Fredholm type discrete inequalities. However, to the best of the authors' knowledge, few papers are only published on discrete inequalities with three independent variables. Hussain et al. [20] discussed some generalized Volterra-Fredholm type nonlinear discrete inequalities involving four iterated infinite sums and applied these results to new explicit bounds for solutions of certain difference equations.In 1979, Pachpatte and Singare [21] gave discrete Gronwall type inequalities in three independent variables as follows.Theorem 1.
Letu(x,y,z) and p(x,y,z) be real-valued nonnegative functions defined for (x,y,z)∈N03(N0≔{0,1,2,…}) for which the inequality (1)ux,y,z≤ax+by+cz+∑r=0x-1∑s=0y-1∑t=0z-1pr,s,tur,s,t,holds for (x,y,z)∈N03, where a(x),b(y),c(z)>0; Δa(x),Δb(y),Δc(z)>0(Δ is the operator defined by Δu(n)=u(n+1)-u(n)) are real-valued functions defined on N0. Then (2)ux,y,z≤a0+by+cz∏r=0x-11+Δaxar+b0+cz+∑s=0y-1∑t=0z-1pr,s,t,for (x,y,z)∈N03.In 1980, Singare and Pachpatte [22] established generalization of discrete Gronwall type inequalities in three independent variables in Theorem 1 in the following.Theorem 2.
Letu(x,y,z) and b(x,y,z) be real-valued nonnegative functions defined for (x,y,z)∈N03, and let a(x,y,z) be positive nondecreasing in the three variables, defined for (x,y,z)∈N03 for which the inequality (3)ux,y,z≤ax,y,z+∑r=0x-1∑s=0y-1∑t=0z-1br,s,tur,s,t,holds for (x,y,z)∈N03. Then (4)ux,y,z≤ax,y,z∏r=0x-11+∑s=0y-1∑t=0z-1pr,s,t,for (x,y,z)∈N03.Motivated by the results mentioned above, the aim of this paper is to establish some new and more general discrete Gronwall-Bellman type inequalities involving functions of three independent variables, which generalize and extend some existing results in [8, 17–19, 21, 22] and can be readily used as handy and powerful tools in the analysis of certain classes of partial finite difference and sum-difference equations. We also give some applications to convey the importance of these new inequalities to study various properties of the solutions of a class of initial value problems involving difference equations.


## 2. Main Results

In this section, we present some new and more general discrete Gronwall-Bellman type inequalities in three independent variables which might be useful tools in the analysis of certain classes of partial finite difference and sum-difference equations.In what follows,I≔[x0,X)∩Z, J≔[y0,Y)∩Z and K≔[z0,Z)∩Z are three fixed lattices of integral points in R, where x0,y0,z0∈Z,X,Y,Z∈Z∪{∞}. Let Ω≔I×J×K⊂Z3, R+≔(0,∞), R0≔[0,∞) and Ω(r,s,t) denote the sublattice [l0,r]×[m0,s]×[n0,t]∩Ω of Ω for any (r,s,t)∈Ω. We also use the following notions of the operators Δxu(x,y,z)≔u(x+1,y,z)-u(x,y,z), Δyu(x,y,z)≔u(x,y+1,z)-u(x,y,z), Δzu(x,y,z)≔u(x,y,z+1)-u(x,y,z), and Δxy2u(x,y,z)≔Δxu(x,y+1,z)-Δxu(x,y,z), and Δxyz3u(x,y,z)≔Δxy2u(x,y,z+1)-Δxy2u(x,y,z).Theorem 3.
Letu, a, c, and f be real-valued nonnegative functions defined for Ω and a and c be nondecreasing in each variables. Suppose that φ∈C(R0,R0) is a strictly increasing function with φ(0)=0 and ψ(u) is a nondecreasing continuous function with ψ(u)>0 for u∈R+. If(5)φux,y,z≤ax,y,z+cx,y,z∑r=x0x-1∑s=y0y-1∑t=z0z-1fr,s,tψur,s,t,for all (x,y,z)∈Ω, then(6)ux,y,z≤φ-1Φ-1Φax,y,z+cx,y,z∑r=x0x-1∑s=y0y-1∑t=z0z-1fr,s,t,for all (x,y,z)∈[x0,x1]×[y0,y1]×[z0,z1], where (7)Φr=∫r0rdsψφ-1s,r≥r0>0,φ-1, Φ-1 denote the inverse functions of φ, Φ and (x1,y1,z1)∈Ω is chosen so that (8)Φax,y,z+cx,y,z∑r=x0x-1∑s=y0y-1∑t=z0z-1fr,s,t∈domΦ-1.Proof.
It suffices to consider the casea(x,y,z)>0, since the case a(x,y,z)=0 can be arrived at by continuity argument. Let us first assume that a(x,y,z)>0. Fixing any numbers x¯, y¯, and z¯ with x0≤x≤x¯, y0≤y≤y¯ and z0≤z≤z¯, define a positive function ω(x,y,z) by(9)ωx,y,z≔ax¯,y¯,z¯+cx¯,y¯,z¯∑r=x0x-1∑s=y0y-1∑t=z0z-1fr,s,tψur,s,t.Then ω(x,y,z)>0, ω(x0,y,z)=ω(x,y0,z)=ω(x,y,z0)=a(x¯,y¯,z¯) and (5) can restated by(10)ux,y,z≤φ-1ωx,y,zon⁡Ωx¯,y¯,z¯.It is easy observe that ω(x,y,z) is a nondecreasing function in each variables defined for Ω(x¯,y¯,z¯) and(11)Δxωx,y,z=ωx+1,y,z-ωx,y,z=cx¯,y¯,z¯∑s=y0y-1∑t=z0z-1fx,s,tψux,s,t,Δxy2ωx,y,z=Δxωx,y+1,z-Δxωx,y,z=cx¯,y¯,z¯∑t=z0z-1fx,y,tψux,y,t.From (11) we have(12)Δxy2ωx,y,z+1-Δxy2ωx,y,z=cx¯,y¯,z¯fx,y,zψux,y,z≤cx¯,y¯,z¯fx,y,zψφ-1ωx,y,z.From (12) we observe that(13)Δxy2ωx,y,z+1-Δxy2ωx,y,zψφ-1ωx,y,z≤cx¯,y¯,z¯fx,y,z.From (13) we obtain(14)Δxy2ωx,y,z+1ψφ-1ωx,y,z+1-Δxy2ωx,y,zψφ-1ωx,y,z≤cx¯,y¯,z¯fx,y,z.Now keeping x, y fixed in (14), setting z=t, and substituting t=z0,z0+1,…,z-1 in (14) we obtain the estimate for Δxy2ω(x,y,z)/ψ[φ-1(ω(x,y,z))] such that(15)Δxy2ωx,y,zψφ-1ωx,y,z≤cx¯,y¯,z¯∑t=z0z-1fx,y,t.From (15) we observe that(16)Δxωx,y+1,zψφ-1ωx,y+1,z-Δxωx,y,zψφ-1ωx,y,z≤cx¯,y¯,z¯∑t=z0z-1fx,y,t.Keeping x, z fixed in (16), setting y=s, and substituting s=y0,y0+1,…,y-1 in (16) we obtain the estimate for Δxω(x,y,z)/ψ[φ-1(ω(x,y,z))] such that(17)Δxωx,y,zψφ-1ωx,y,z≤cx¯,y¯,z¯∑s=y0y-1∑t=z0z-1fx,s,t.From the definition of Φ and (17) we obtain(18)Φωx+1,y,z-Φωx,y,z=∫ωx,y,zωx+1,y,zdsψφ-1s≤Δxωx,y,zψφ-1ωx,y,z≤cx¯,y¯,z¯∑s=y0y-1∑t=z0z-1fx,s,t.Again keeping y, z fixed in (18), setting x=r, and substituting r=x0,x0+1,…,x-1 in (18), we obtain(19)Φωx,y,z≤Φax¯,y¯,z¯+cx¯,y¯,z¯∑r=x0x-1∑s=y0y-1∑t=z0z-1fx,s,t.Now using inequality (19) in (10), we get(20)ux,y,z≤φ-1Φ-1Φax¯,y¯,z¯+cx¯,y¯,z¯∑r=x0x-1∑s=y0y-1∑t=z0z-1fr,s,t.Taking x=x¯, y=x¯, and z=z¯ in inequality (20), since x¯, y¯, and z¯ are arbitrary, we get the required inequality.Corollary 4.
Letu, a, c, f, and φ be as defined in Theorem 3. If (21)φux,y,z≤ax,y,z+cx,y,z∑r=x0x-1∑s=y0y-1∑t=z0z-1fr,s,tφur,s,t,for all (x,y,z)∈Ω, then (22)ux,y,z≤φ-1ax,y,zexp⁡cx,y,z∑r=x0x-1∑s=y0y-1∑t=z0z-1fr,s,t,for all (x,y,z)∈Ω.Proof.
Suppose first thata(x,y,z)>0. Taking ψ=φ, we have (23)Φr=∫r0rdsφφ-1s=∫r0rdss=ln⁡rr0,r≥r0>0,and so Φ-1(r)=r0exp⁡r, in particular, it is defined everywhere on R. Hence, by Theorem 3, (24)ux,y,z≤φ-1r0exp⁡ln⁡ax,y,zr0+cx,y,z∑r=x0x-1∑s=y0y-1∑t=z0z-1fr,s,t=φ-1ax,y,zexp⁡cx,y,z∑r=x0x-1∑s=y0y-1∑t=z0z-1fr,s,tfor all (x,y,z)∈Ω. Finally, by continuity, this should also hold for the case c=0.Theorem 5.
Letu, a, c, and fi, gi(i=1,2,…,m) be real-valued nonnegative functions defined for Ω and a and c be nondecreasing in each variables. Suppose that φ,ϕ∈C(R0,R0) are strictly increasing functions with φ(0)=0 and ϕ(0)=0 and ψi(u) is a nondecreasing continuous function with ψi(u)>0(i=1,2,…,m) for u∈R+. If(25)φux,y,z≤ax,y,z+cx,y,z∑r=x0x-1∑s=y0y-1∑t=z0z-1ϕur,s,t∑i=1mfir,s,tψiur,s,t+gir,s,t,for all (x,y,z)∈Ω, let ψ=max1≤i≤m{ψi}, then(26)ux,y,z≤φ-1G-1Φ-1Φkx0,y,z+cx,y,z∑r=x0x-1∑s=y0y-1∑t=z0z-1∑i=1mfix,y,z,for all (x,y,z)∈[x0,x1]×[y0,y1]×[z0,z1], where (27)kx0,y,z=Gax,y,z+cx,y,z∑r=x0x-1∑s=y0y-1∑t=z0z-1∑i=1mgix,y,z,Gr=∫r0rdsϕφ-1s,Φr=∫r0rdsψφ-1G-1s,r≥r0>0,φ-1, G-1, Φ-1 denote the inverse functions of φ, G, Φ and (x1,y1,z1)∈Ω is chosen so that (28)Φax,y,z+cx,y,z∑r=x0x-1∑s=y0y-1∑t=z0z-1fr,s,t∈domΦ-1.Proof.
It suffices to consider the casea(x,y,z)>0, since the case a(x,y,z)=0 can be arrived at by continuity argument. Let us first assume that a(x,y,z)>0. Fixing any numbers x¯, y¯, and z¯ with x0≤x≤x¯,y0≤y≤y¯ and z0≤z≤z¯, define a positive function ω(x,y,z) by(29)ωx,y,z≔ax¯,y¯,z¯+cx¯,y¯,z¯∑r=x0x-1∑s=y0y-1∑t=z0z-1ϕur,s,t∑i=1mfir,s,tψiur,s,t+gir,s,t.Then ω(x,y,z)>0, ω(x0,y,z)=ω(x,y0,z)=ω(x,y,z0)=a(x¯,y¯,z¯) and (5) can restated by(30)ux,y,z≤φ-1ωx,y,zonΩx¯,y¯,z¯.It is easy observe that ω(x,y,z) is a nondecreasing function in each variable defined for Ω(x¯,y¯,z¯) and(31)Δxωx,y,z=ωx+1,y,z-ωx,y,z=cx¯,y¯,z¯∑s=y0y-1∑t=z0z-1ϕux,s,t∑i=1mfix,s,tψiux,s,t+gix,s,t,Δxy2ωx,y,z=Δxωx,y+1,z-Δxωx,y,z=cx¯,y¯,z¯∑t=z0z-1ϕux,y,t∑i=1mfix,y,tψiux,y,t+gix,y,t.From (31) we have(32)
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.From (32) we observe that(33)Δxy2ωx,y,z+1-Δxy2ωx,y,zϕφ-1ωx,y,z≤cx¯,y¯,z¯∑i=1mfix,y,zψiφ-1ωx,y,z+gix,y,z.From (33) we obtain (34)Δxy2ωx,y,z+1ϕφ-1ωx,y,z+1-Δxy2ωx,y,zϕφ-1ωx,y,z≤cx¯,y¯,z¯∑i=1mfix,y,zψiφ-1ωx,y,z+gix,y,z.
Along the lines of proof of Theorem3, we obtain(35)Gωx,y,z≤Gax¯,y¯,z¯+cx¯,y¯,z¯∑r=x0x-1∑s=y0y-1∑t=z0z-1∑i=1mfix,y,zψiφ-1ωx,y,z+gix,y,z.Now, define function k(x,y,z) by (36)kx,y,z≔Gax¯,y¯,z¯+cx¯,y¯,z¯∑r=x0x¯-1∑s=y0y¯-1∑t=z0z¯-1∑i=1mgix,y,z+cx¯,y¯,z¯∑r=x0x-1∑s=y0y-1∑t=z0z-1∑i=1mfix,y,zψiφ-1ωx,y,z.Then k(x0,y,z)=G(a(x¯,y¯,z¯))+c(x¯,y¯,z¯)∑r=x0x¯-1‍∑s=y0y¯-1‍∑t=z0z¯-1‍∑i=1m‍gi(x,y,z) and (35) can be restated as(37)ωx,y,z≤G-1kx,y,zonΩx¯,y¯,z¯.It is easy observe that k(x,y,z) is a nondecreasing function in each variable defined for Ω(x¯,y¯,z¯) and(38)Δxkx,y,z=kx+1,y,z-kx,y,z=cx¯,y¯,z¯∑s=y0y-1∑t=z0z-1∑i=1mfix,s,tψiux,s,t,Δxy2kx,y,z=Δxkx,y+1,z-Δxkx,y,z=cx¯,y¯,z¯∑t=z0z-1∑i=1mfix,y,tψiux,y,t.From (38) we have(39)Δxy2kx,y,z+1-Δxy2kx,y,z=cx¯,y¯,z¯∑i=1mfix,y,zψiux,y,z≤cx¯,y¯,z¯ψφ-1G-1kx,y,z∑i=1mfix,y,z.From (39) we observe that(40)Δxy2kx,y,z+1-Δxy2kx,y,zψφ-1G-1kx,y,z≤cx¯,y¯,z¯∑i=1mfix,y,z.From (40) we obtain (41)Δxy2kx,y,z+1ψφ-1G-1kx,y,z+1-Δxy2kx,y,zψφ-1G-1kx,y,z≤cx¯,y¯,z¯∑i=1mfix,y,z.
Along the lines of proof of Theorem3, we obtain(42)Φkx,y,z≤Φkx0,y,z+cx¯,y¯,z¯∑r=x0x-1∑s=y0y-1∑t=z0z-1∑i=1mfix,y,z.Now using inequalities (37) and (42) in (30), we get(43)ux,y,z≤φ-1G-1Φ-1Φkx0,y,z+cx¯,y¯,z¯∑r=x0x-1∑s=y0y-1∑t=z0z-1∑i=1mfix,y,z.Taking x=x¯, y=x¯, and z=z¯ in inequality (43), since x¯, y¯, and z¯ are arbitrary, we get the required inequality.Corollary 6.
Letu, a, c, φ, ψ and fi, gi, ψi(i=1,2,…,m) be as defined in Theorem 5. If(44)φux,y,z≤ax,y,z+cx,y,z∑r=x0x-1∑s=y0y-1∑t=z0z-1φur,s,t∑i=1mfir,s,tψiur,s,t+gir,s,t,for all (x,y,z)∈Ω, let ψ=max1≤i≤m{ψi}, then (45)ux,y,z≤φ-1ax,y,zexpΦ1-1Φ1k1x0,y,z+cx,y,z∑r=x0x-1∑s=y0y-1∑t=z0z-1∑i=1mfix,y,z,for all (x,y,z)∈[x0,x1]×[y0,y1]×[z0,z1], where (46)k1x0,y,z=cx,y,z∑r=x0x-1∑s=y0y-1∑t=z0z-1∑i=1mgix,y,z,Φ1r=∫r0rdsψφ-1aes,r≥r0>0,where φ-1, Φ1-1 denote the inverse functions of φ, Φ1 and (x1,y1,z1)∈Ω is chosen so that (47)Φ1k1x0,y,z+cx,y,z∑r=x0x-1∑s=y0y-1∑t=z0z-1∑i=1mfix,y,z∈domΦ1-1.Proof.
It suffices to consider the casea(x,y,z)>0, since the case a(x,y,z)=0 can be arrived at by continuity argument. Let us first assume that a(x,y,z)>0. Fixing any numbers x¯, y¯, and z¯ with x0≤x≤x¯, y0≤y≤y¯, and z0≤z≤z¯, define a positive function ω(x,y,z) by(48)ωx,y,z≔ax¯,y¯,z¯+cx¯,y¯,z¯∑r=x0x-1∑s=y0y-1∑t=z0z-1φur,s,t∑i=1mfir,s,tψiur,s,t+gir,s,t,Then ω(x,y,z)>0, ω(x0,y,z)=ω(x,y0,z)=ω(x,y,z0)=a(x¯,y¯,z¯) and (44) can restated by(49)ux,y,z≤φ-1ωx,y,zonΩx¯,y¯,z¯.Along the lines of proof of Theorem 5, we obtain (50)Δxy2ωx,y,z+1-Δxy2ωx,y,zωx,y,z≤cx¯,y¯,z¯∑i=1mfix,y,zψiφ-1ωx,y,z+gix,y,z.From the above inequality and Theorem 3, G(r)=∫r0r‍ds/φφ-1s=∫r0r‍ds/s=ln⁡r/r0, r≥r0>0, we observe that(51)ln⁡ωx,y,z≤ln⁡ax¯,y¯,z¯+cx¯,y¯,z¯∑r=x0x-1∑s=y0y-1∑t=z0z-1∑i=1mfix,y,zψiφ-1ωx,y,z+gix,y,z.Now, define a function k1(x,y,z) by (52)k1x,y,z≔cx¯,y¯,z¯∑r=x0x¯-1∑s=y0y¯-1∑t=z0z¯-1∑i=1mgix,y,z+cx¯,y¯,z¯∑r=x0x-1∑s=y0y-1∑t=z0z-1∑i=1mfix,y,zψiφ-1ωx,y,z.Then k1(x0,y,z)=c(x¯,y¯,z¯)∑r=x0x¯-1‍∑s=y0y¯-1‍∑t=z0z¯-1‍∑i=1m‍gi(x,y, z) and (51) can be restated as(53)ωx,y,z≤ax¯,y¯,z¯exp⁡k1x,y,zonΩx¯,y¯,z¯.Along the lines of proof of Theorem 5, we obtain (54)Δxy2k1x,y,z+1-Δxy2k1x,y,zψφ-1ax¯,y¯,z¯expk1x,y,z≤cx¯,y¯,z¯∑i=1mfix,y,z.From the above inequality and Theorem 3, we observe that(55)Φ1k1x,y,z≤Φ1k1x0,y,z+cx¯,y¯,z¯∑r=x0x-1∑s=y0y-1∑t=z0z-1∑i=1mfix,y,z.Now using inequalities (53) and (55) in (49), we get (56)ux,y,z≤φ-1ax¯,y¯,z¯exp⁡Φ1-1Φ1k1x0,y,z+cx¯,y¯,z¯∑r=x0x-1∑s=y0y-1∑t=z0z-1∑i=1mfix,y,z.Taking x=x¯, y=x¯, and z=z¯ in the above inequality, since x¯, y¯, and z¯ are arbitrary, we get the required inequality.For the special caseφ(u)=up (p>0 is constant), Corollary 6 gives the following retarded discrete inequality for nonlinear functions.Corollary 7.
Letu, a, c and fi, gi(i=1,2,…,m) be as defined in Theorem 5, and let ψ be as defined in Theorem 3. Suppose that p>q>0 are constants. If (57)upx,y,z≤ax,y,z+cx,y,z∑r=x0x-1∑s=y0y-1∑t=z0z-1upr,s,t∑i=1mfir,s,tψur,s,t+gir,s,t,for all (x,y,z)∈Ω, then (58)ux,y,z≤a1/px,y,zexp⁡1pΨ1-1Ψ1k1x0,y,z+cx,y,z∑r=x0x-1∑s=y0y-1∑t=z0z-1∑i=1mfix,y,z,for all (x,y,z)∈[x0,x1]×[y0,y1]×[z0,z1], where k1(x0,y,z) is defined in Corollary 6, (59)Ψ1r=∫r0rdsψa1/pexp⁡s/p,r≥r0>0,φ-1, Ψ1-1 denote the inverse functions of φ, Ψ1, and (x1,y1,z1)∈Ω is chosen so that (60)
Ψ
1
k
1
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0
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y
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+
c
x
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y
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∑
r
=
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1
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=
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0
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1
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i
x
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y
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∈
d
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m
Ψ
1
-
1
.Proof.
The proof is followed by an argument similar to that in the proof of Corollary6 with suitable modification. We omit the details here.For the special caseφ(u)=up and ϕ(u)=uq (p>q>0 are constants), Theorem 5 gives the following retarded discrete inequality for nonlinear functions.Corollary 8.
Letu, a, c and fi, gi(i=1,2,…,m) be as defined in Theorem 5, and let ψ be as defined in Theorem 3. Suppose that p>q>0 are constants. If (61)upx,y,z≤ax,y,z+cx,y,z∑r=x0x-1∑s=y0y-1∑t=z0z-1uqr,s,t∑i=1mfir,s,tψur,s,t+gir,s,t,for all (x,y,z)∈Ω, then (62)ux,y,z≤Φ2-1Φ2k2x0,y,z+cx,y,z∑r=x0x-1∑s=y0y-1∑t=z0z-1∑i=1mfix,y,z1/p-q,for all (x,y,z)∈[x0,x1]×[y0,y1]×[z0,z1], where (63)k2x0,y,z=ax,y,zp-q/p+p-qpcx,y,z∑r=x0x-1∑s=y0y-1∑t=z0z-1∑i=1mgix,y,z,Φ2r=∫r0rdsψs1/p-q,r≥r0>0,φ-1, Φ1-1 denote the inverse functions of φ, Φ1, and (x1,y1,z1)∈Ω is chosen so that (64)Φ2k2x0,y,z+cx,y,z∑r=x0x-1∑s=y0y-1∑t=z0z-1∑i=1mfix,y,z∈domΦ2-1.Proof.
The proof is followed by an argument similar to that in the proof of Theorem5 with suitable modification. We omit the details here.Theorems3 and 5 can easily be applied to generate other useful discrete inequalities in more effective tools of the study of certain partial finite difference and sum-difference equations. For example, we have the following results.Theorem 9.
Letu, a, c, φ, ψ, and f be as defined in Theorem 3, and let L:Ω×R0→R0 be real-valued function which satisfies the following condition:(65)0≤Lx,y,z,v-Lx,y,z,w≤Mx,y,z,wv-w,for x,y,z∈R0 and v≥w≥0, where M(x,y,z,w):Ω×R0→R0 be real-valued function. If(66)φux,y,z≤ax,y,z+cx,y,z∑r=x0x-1∑s=y0y-1∑t=z0z-1fr,s,tLr,s,t,ψur,s,t,for all (x,y,z)∈Ω, then(67)ux,y,z≤φ-1Φ-1Φωx0,y,z+cx,y,z∑r=x0x-1∑s=y0y-1∑t=z0z-1fr,s,tMx,y,z,for all (x,y,z)∈[x0,x1]×[y0,y1]×[z0,z1], where Φ(r) is defined in Theorem 3, (68)ωx0,y,z=ax,y,z+cx,y,z∑r=x0x-1∑s=y0y-1∑t=z0z-1fr,s,tLr,s,t,φ-1, Φ-1 denote the inverse functions of φ, Φ, and (x1,y1,z1)∈Ω is chosen so that (69)Φωx0,y,z+cx,y,z∑r=x0x-1∑s=y0y-1∑t=z0z-1fr,s,tMx,y,z∈domΦ-1.Proof.
It suffices to consider the casea(x,y,z)>0, since the case a(x,y,z)=0 can be arrived at by continuity argument. Let us first assume that a(x,y,z)>0. Due to the property of L, inequality (66) becomes(70)φux,y,z≤ax,y,z+cx,y,z∑r=x0x-1∑s=y0y-1∑t=z0z-1fr,s,tLr,s,t+Mr,s,tψur,s,t.Fixing any numbers x¯, y¯, and z¯ with x0≤x≤x¯, y0≤y≤y¯, and z0≤z≤z¯, define a positive function ω(x,y,z) by(71)ωx,y,z≔ax¯,y¯,z¯+cx¯,y¯,z¯∑r=x0x¯-1∑s=y0y¯-1∑t=z0z¯-1fr,s,tLr,s,t+cx¯,y¯,z¯∑r=x0x-1∑s=y0y-1∑t=z0z-1fr,s,tMr,s,tψur,s,tThen ω(x,y,z)>0, ω(x0,y,z)=a(x¯,y¯,z¯)+c(x¯,y¯,z¯)∑r=x0x¯-1‍∑s=y0y¯-1‍∑t=z0z¯-1‍f(r,s,t)L(r,s,t) and (70) can restated by(72)ux,y,z≤φ-1ωx,y,zonΩx¯,y¯,z¯.Along the lines of proof of Theorem 3, we obtain (73)Δxy2ωx,y,z+1-Δxy2ωx,y,zψφ-1ωx,y,z≤cx¯,y¯,z¯fx,y,zMr,s,t.From the above inequality and Theorem 3, we observe that(74)Φωx,y,z≤Φωx0,y,z+cx¯,y¯,z¯∑r=x0x-1∑s=y0y-1∑t=z0z-1fx,y,zMr,s,t.Now using inequalities (74) in (72), we get (75)ux,y,z≤φ-1Φ-1Φωx0,y,z+cx¯,y¯,z¯∑r=x0x-1∑s=y0y-1∑t=z0z-1fr,s,tMx,y,zTaking x=x¯, y=x¯, and z=z¯ in the above inequality, since x¯, y¯, and z¯ are arbitrary, we get the required inequality.Corollary 10.
Letu, a, c, φ, ψ, and f be as defined in Theorem 3, and let L, M be as defined in Theorem 9. If (76)φux,y,z≤ax,y,z+cx,y,z∑r=x0x-1∑s=y0y-1∑t=z0z-1fr,s,tLr,s,t,φur,s,t,for all (x,y,z)∈Ω, then (77)ux,y,z≤φ-1ωx0,y,zexp⁡cx,y,z∑r=x0x-1∑s=y0y-1∑t=z0z-1fr,s,tMx,y,z,for all (x,y,z)∈Ω, where ω(x0,y,z) is defined in Theorem 9.Proof.
The proof is followed by an argument similar to that in the proof of Corollary4 and Theorem 9 with suitable modification. We omit the details here.Theorem 11.
Letψ be as defined in Theorem 3, let u, a, c, φ, ϕ and fi, gi(i=1,2,…,m) be as defined in Theorem 5, and let L, M be as defined in Theorem 9. If (78)φux,y,z≤ax,y,z+cx,y,z∑r=x0x-1∑s=y0y-1∑t=z0z-1ϕur,s,t∑i=1mfir,s,tLr,s,t,ψur,s,t+gir,s,t,for all (x,y,z)∈Ω, then (79)
u
x
,
y
,
z
≤
φ
-
1
G
-
1
Φ
-
1
Φ
k
3
x
0
,
y
,
z
+
c
x
,
y
,
z
∑
r
=
x
0
x
-
1
∑
s
=
y
0
y
-
1
∑
t
=
z
0
z
-
1
∑
i
=
1
m
f
i
x
,
y
,
z
M
r
,
s
,
t
,for all (x,y,z)∈[x0,x1]×[y0,y1]×[z0,z1], where G, Φ(r) are defined in Theorem 5, (80)k3x0,y,z=Gax,y,z+cx,y,z∑r=x0x-1∑s=y0y-1∑t=z0z-1∑i=1mfix,y,zLr,s,t+gix,y,z,φ-1, G-1, Φ-1 denote the inverse functions of φ, G, Φ, and (x1,y1,z1)∈Ω is chosen so that (81)Φk3x0,y,z+cx,y,z∑r=x0x-1∑s=y0y-1∑t=z0z-1∑i=1mfix,y,zMr,s,t∈domΦ-1.Proof.
The proof is followed by an argument similar to that in the proof of Theorems5 and 9 with suitable modification. We omit the details here.


## 3. Applications

In this section, we will show that our results are useful in establishing the boundedness and uniqueness of solutions to certain partial finite difference and sum-difference equations. Consider the partial finite difference equations with the initial boundary conditions:(82)Δxyz3φux,y,z=Fx,y,z,ux,y,zwith(83)ux0,y,z=e1y,z,ux,y0,z=e2x,z,ux,y,z0=e3x,y,e2x0,z=e3x0,y=0,e1y0,z=e3x,y0=0,e1y,z0=e2x,z0=0.The following two results deal with the boundedness on the solutions of problem (82).Proposition 12.
LetF be real-valued function such that(84)Fx,y,z,u≤fx,y,zψu,φe1y,z+φe2x,z+φe3x,y≤ax,y,z,where a and f are real-valued nonnegative functions defined for Ω and a is nondecreasing in each variables. Suppose that φ∈C(R0,R0) is a strictly increasing function with φ(0)=0 and ψ(u) is a nondecreasing continuous function with ψ(u)>0 for u∈R+. If u(x,y,z) is any solution of problem (82) with the condition (83), then(85)ux,y,z≤φ-1Φ-1Φax,y,z+∑r=x0x-1∑s=y0y-1∑t=z0z-1fr,s,t,for all (x,y,z)∈[x0,x1]×[y0,y1]×[z0,z1], where Φ(r) is defined in Theorem 3, φ-1, Φ-1 denote the inverse functions of φ, Φ, and (x1,y1,z1)∈Ω is chosen so that (86)Φax,y,z+∑r=x0x-1∑s=y0y-1∑t=z0z-1fr,s,t∈domΦ-1.Proof.
It is easy to see that the solutionu(x,y,z) of problem (82) satisfies equivalent sum-difference equation:(87)φux,y,z=φe1y,z+φe2x,z+φe3x,y+∑r=x0x-1∑s=y0y-1∑t=z0z-1Fr,s,t,ur,s,t.From (84) and (87), we have (88)φux,y,z≤φe1y,z+φe2x,z+φe3x,y+∑r=x0x-1∑s=y0y-1∑t=z0z-1Fr,s,t,ur,s,t≤ax,y,z+∑r=x0x-1∑s=y0y-1∑t=z0z-1fx,y,zψur,s,t.Now, a suitable application of inequality given in Theorem 3 to the above yields the desired result (85).Proposition 13.
LetF be real-valued function such that (89)Fx,y,z,u≤fx,y,zLx,y,z,ψu,φe1y,z+φe2x,z+φe3x,y≤ax,y,z,where a and f are real-valued nonnegative functions defined for Ω, a is nondecreasing in each variables, and L:Ω×R0→R0 is real-valued function which satisfies the condition (90)0≤Lx,y,z,v-Lx,y,z,w≤Mx,y,z,wv-w,for x,y,z∈R0 and v≥w≥0, M(x,y,z,w):Ω×R0→R0 be real-valued function. Suppose that φ∈C(R0,R0) is a strictly increasing function with φ(0)=0 and ψ(u) is a nondecreasing continuous function with ψ(u)>0 for u∈R+. If u(x,y,z) is any solution of problem (82) with condition (83), then (91)ux,y,z≤φ-1Φ-1Φωx0,y,z+∑r=x0x-1∑s=y0y-1∑t=z0z-1fr,s,tMx,y,z,for all (x,y,z)∈[x0,x1]×[y0,y1]×[z0,z1], where Φ(r) is defined in Theorem 3, (92)ωx0,y,z=ax,y,z+∑r=x0x-1∑s=y0y-1∑t=z0z-1fr,s,tLr,s,t,φ-1, Φ-1 denote the inverse functions of φ, Φ, and (x1,y1,z1)∈Ω is chosen so that (93)Φωx0,y,z+∑r=x0x-1∑s=y0y-1∑t=z0z-1fr,s,tMx,y,z∈domΦ-1.Proof.
The proof is followed by an argument similar to that in the proof of Theorem3, using Theorem 9 with suitable modification. We omit the details here.The following theorem deal with the uniqueness on the solutions of problem (82).Proposition 14.
LetF be real-valued function such that(94)Fx,y,z,v-Fx,y,z,w≤fx,y,zφv-φw,where f is real-valued nonnegative function on Ω. Then problem (82) with condition (83) has at most one positive solution on Ω.Proof.
Letu(x,y,z) and u¯(x,y,z) be two solutions of problem (82) on Ω. By (94), we obtain (95)φux,y,z-φu¯x,y,z≤∑r=x0x-1∑s=y0y-1∑t=z0z-1Fr,s,t,ur,s,t-∑r=x0x-1∑s=y0y-1∑t=z0z-1Fr,s,t,u¯r,s,t≤∑r=x0x-1∑s=y0y-1∑t=z0z-1fr,s,tφur,s,t-φu¯r,s,t.An application of Corollary 4 with suitable modification to function φ-1(|φ(u(x,y,z))-φ(u¯(x,y,z))|) in the last inequality yields (96)φux,y,z-φu¯x,y,z≤0,for (x,y,z)∈Ω. Hence u=u¯ on Ω.Finally, we investigate the continuous dependence of the solutions of problem (82) on the function F and the boundary data e1, e2, and e3. For this we consider the following variation of problem (82).(97)Δxyz3φux,y,z=F¯x,y,z,ux,y,zwith(98)ux0,y,z=e¯1y,z,ux,y0,z=e¯2x,z,ux,y,z0=e¯3x,y,e¯2x0,z=e¯3x0,y=0,e¯1y,z=e¯3x,y=0,e¯1y,z0=e¯2x,z0=0.Proposition 15.
Consider problem (82) and problem (97). Assume that(99)Fx,y,z,v-Fx,y,z,w≤fx,y,zφv-φw,(100)φe1y,z-φe¯1y,z+φe2x,z-φe¯2x,z+φe3x,y-φe¯3x,y≤ε2and(101)∑r=x0x-1∑s=y0y-1∑t=z0z-1Fr,s,t,ξ-F¯r,s,t,ξ≤ε2,where (x,y,z)∈Ω and v,w,ξ∈R. Then (102)φux,y,z-φu¯x,y,z≤εexp⁡∑r=x0x-1∑s=y0y-1∑t=z0z-1fr,s,t,for all (x,y,z)∈Ω. Hence, φ(u) depends continuously on F, e1, e2, and e3.Proof.
Letu(x,t,z) and u¯(x,t,z) be solutions of problem (82) and problem (97), respectively. Then u satisfies (87) and u¯ satisfies the corresponding equation: (103)φu¯x,y,z=φe¯1y,z+φe¯2x,z+φe¯3x,y+∑r=x0x-1∑s=y0y-1∑t=z0z-1F¯r,s,t,u¯r,s,t.
Hence, we obtain(104)φux,y,z-φu¯x,y,z≤φe1y,z-φe¯1y,z+φe2x,z-φe¯2x,z+φe3x,y-φe¯3x,y+∑r=x0x-1∑s=y0y-1∑t=z0z-1Fr,s,t,ur,s,t-∑r=x0x-1∑s=y0y-1∑t=z0z-1F¯r,s,t,u¯r,s,t≤ε2+∑r=x0x-1∑s=y0y-1∑t=z0z-1Fr,s,t,ur,s,t-∑r=x0x-1∑s=y0y-1∑t=z0z-1Fr,s,t,u¯r,s,t+∑r=x0x-1∑s=y0y-1∑t=z0z-1Fr,s,t,u¯r,s,t-∑r=x0x-1∑s=y0y-1∑t=z0z-1F¯r,s,t,u¯r,s,t≤ε+∑r=x0x-1∑s=y0y-1∑t=z0z-1fr,s,tφur,s,t-φu¯r,s,t.As an application of Corollary 4 with suitable modification, function φ-1(|φ(u(x,y,z))-φ(u¯(x,y,z))|) in the last inequality yields by assumptions (99)-(101) (105)φux,y,z-φu¯x,y,z≤εexp⁡∑r=x0x-1∑s=y0y-1∑t=z0z-1fr,s,t,for all (x,y,z)∈Ω. When restricted to any compact sublattice ∑r=x0x-1‍∑s=y0y-1‍∑t=z0z-1‍f(r,s,t) that is bounded, (106)φux,y,z-φu¯x,y,z≤ε·K,for some K>0 for all (x,y,z) in this compact sublattice. Hence, φ(u) depends continuously on F, e1, e2, and e3.


## 4. Conclusions

By using the inequality analyze technique, some new discrete Gronwall-Bellman type inequalities with three independent variables have been obtained. Our main theorems generalize and extend some existing results in the literature. Furthermore, the boundedness and uniqueness of solutions to certain partial finite difference and sum-difference equations have been established by applying our obtained results.



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