# Estimates of Certain Integral Inequalities on Time Scales

Authors: Deepak B. Pachpatte
Journal: Journal of Mathematics
 (2013)

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## Abstract

The main objective of this paper is to establish explicit bounds on certain integral inequalities on time scales, which can be used as tools in the study of certain classes of integral equations on time scales. Some applications of our results are also given.


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## Body

## 1. Introduction

Recently, many authors have studied various aspects of dynamic inequalities on time scales using various techniques [1–4]. In this paper, we obtain explicit bounds on certain integral inequalities on time scales. In the present paper we offer some fundamental dynamic integral inequalities on time scales which can be used as tools for handling the qualitative behavior of solutions of certain dynamic equations on time scales. Excellent information about introduction to time scales can be found in [5, 6]. In what follows, ℝ denotes the set of real numbers, ℤ the set of integers, and 𝕋 denotes arbitrary time scales. Now following [3, 4] we give some basic definitions about calculus on time scales in two variables.We say thatf:𝕋→ℝ is rd-continuous provided f is continuous at each right-dense point of 𝕋 and has a finite left sided limit at each left-dense point of 𝕋 and will be denoted by Crd. Let 𝕋1 and 𝕋2 be two time scales with at least two points and consider the time scales intervals 𝕋¯1=[x0,∞)∩𝕋1 and 𝕋¯2=[y0,∞)∩𝕋2 for x0∈𝕋1 and y0∈𝕋2 and Ω=𝕋1×𝕋2. Let σ1,  ρ1,  Δ1 and σ2,  ρ2,  Δ2 denote the forward jump operators, backward jump operators and the delta differentiation operator, respectively, on 𝕋1 and 𝕋2. Let a<b are points in 𝕋1,  c<d are points in 𝕋2, [a,b) is the half closed bounded interval in 𝕋1, and [c,d) is the half closed bounded interval in 𝕋2.We say that a real-valued functionf on 𝕋1×𝕋2 at (t1,t2)∈𝕋¯1×𝕋¯2 has a Δ1 partial derivative fΔ1(t1,t2) with respect to t1 if for each ϵ>0 there exists a neighborhood Ut1 of t1 such that
(1)|f(σ1(t1),t2)-f(s,t2)-fΔ1(t1,t2)(σ1(t1)-s)|≤ε|σ1(t1)-s|,
for all s∈Ut1. We say that f on 𝕋1×𝕋2 at (t1,t2)∈𝕋¯1×𝕋¯2 has a Δ2 partial derivative fΔ2(t1,t2) with respect to  t2 if for each η>0 there exists a neighborhood Ut2 of t2 such that
(2)|f(t1,σ2(t2))-f(t1,l)-fΔ2(t1,t2)(σ2(t2)-l)|≤η|σ2(t2)-l|,
for all  l∈Ut2. The function f is called rd-continuous in t2 if for every  α1∈𝕋1, the function f(α1,t2) is rd-continuous on 𝕋2. The function f is called rd-continuous in  t1 if for every α2∈𝕋2  the function f(t1,α2) is rd-continuous on 𝕋1.We require that the following lemma proved in [1, 2] holds.Lemma 1.
Letu,  f∈Crd(𝕋,ℝ),  k(t,s),  kΔ1(t,s)∈Crd(Ω,ℝ) and  c nonnegative constants. If
(3)u(t)≤c+∫t0tf(s)[u(s)+∫s0sk(s,τ)u(τ)Δτ]Δs,
for t∈𝕋, then
(4)u(t)≤c[1+∫t0tf(s)ef+A(s,s0)Δs],
for t∈𝕋, where
(5)A(t)=k(t,σ(t))+∫t0tkΔ1(t,η)Δη,
for t∈𝕋.Lemma 2 (see [1, Lemma]).
Letu,  a,  f∈Crd′(𝕋¯1×𝕋¯2,ℝ+) and a is nondecreasing in each of its variables. If
(6)u(x,y)≤a(x,y)+∫x0x∫y0yf(s,t)u(s,t)ΔtΔs,
for (x,y)∈𝕋¯1×𝕋¯2, then
(7)u(x,y)≤a(x,y)+e∫y0yf(x,t)Δt(x,x0),
for (x,y)∈𝕋¯1×𝕋¯2.


## 2. Main Results

Our main results are given in the following theorems.Theorem 3.
Letu,  f,    a∈Crd(𝕋,R+),  k(t,s),   kΔ(t,s)∈Crd(Ω,R+). If
(8)u(t)≤a(t)+∫t0tf(s)[u(s)+∫s0sk(s,τ)u(τ)Δτ]Δs,
for t∈𝕋, then
(9)u(t)≤a(t)+h(t)[1+∫t0tf(s)ef+A(s,s0)Δs],
for t∈𝕋, where
(10)h(t)=∫t0tf(s)[a(s)+∫s0sk(s,τ)a(τ)Δτ]Δs,
for t∈𝕋 and A(t) defined by (5).Proof.
Define a functionz(t) by
(11)z(t)=∫t0tf(s)[u(s)+∫s0sk(s,τ)u(τ)Δτ]Δs,
then from (8), u(t)≤a(t)+z(t), and using this in (11) we get
(12)z(t)≤∫t0tf(s)[a(s)+z(s)+∫s0sk(s,τ)(a(τ)+z(τ))Δτ]Δs=h(t)+∫t0tf(s)[z(s)+∫s0sk(s,τ)z(τ)Δτ]Δs,
where h(t) is defined by (10). Clearly h(t) is nonnegative, rd-continuous and nondecreasing in t, t∈𝕋 for a(t) nondecreasing. First we assume that h(t)>0 for t∈𝕋. From (12), we have
(13)z(t)h(t)≤1+∫t0tf(s)[z(s)h(t)+∫s0sk(s,τ)z(τ)h(τ)Δτ]Δs.
Now as an application of Lemma 1 we have
(14)z(t)h(t)≤[1+∫t0tf(s)ef+A(s,s0)Δs].
The desired inequality (9) follows from (14) and the fact that u(t)≤a(t)+z(t). If h(t)≥0, we carry out the above procedure with h(t)+ϵ instead of h(t), where ϵ>0 is an arbitrary small constants, and then subsequently pass to the limit as ϵ→0 to obtain (9). The following theorem deals with two independent versions of inequalities established in Theorem 3.Theorem 4.
Letu,f∈Crd(Ω,R+), k(x,y,s,t), kΔ1(x,y,s,t),  kΔ2(x,y,s,t), kΔ2Δ1(x,y,s,t)∈  Crd(Ω×Ω,R+), and c be a nonnegative constants. If
(15)u(x,y)≤c+∫x0x∫y0yf(s,t)×[u(s,t)+∫s0s∫t0tk(s,t,τ,ξ)u(τ,ξ)ΔξΔτ]ΔtΔs,
for (x,y)∈Ω, then
(16)u(x,y)≤c[1+∫x0x∫y0yf(s,t)e∫t0t[f(τ,ξ)+A(τ,ξ)]ΔξΔτ(x,x0)]ΔtΔs,
where
(17)A(x,y)=k(x,y,σ(x),σ(y))+∫x0xkΔ1(x,y,τ,σ(y))Δτ+∫y0ykΔ1(x,y,σ(x),η)Δη+∫x0x∫y0ykΔ2Δ1(x,y,τ,η)ΔηΔτ,
for (x,y)∈Ω.Proof.
Letc>0 and define a function z(x,y) by the right-hand side of (15); then z(x0,y)=z(x,y0)=c,  u(x,y)≤z(x,y), and
(18)zΔ2Δ1(x,y)=f(x,y)[u(x,y)+∫x0x∫y0yk(x,y,τ,ξ)u(τ,ξ)ΔξΔτ]≤f(x,y)[z(x,y)+∫x0x∫y0yk(x,y,τ,ξ)z(τ,ξ)ΔξΔτ].
Define a function  v(x,y) by
(19)v(x,y)=z(x,y)+∫x0x∫y0yk(x,y,τ,ξ)z(τ,ξ)ΔξΔτ,
then v(x0,y)=z(x0,y)=c, v(x,y0)=z(x,y0)=c, z(x,y)≤v(x,y), zΔ1Δ2(x,y)≤f(x,y)v(x,y),   v(x,y) is nondecreasing for (x,y)∈Ω,  and
(20)vΔ1Δ2(x,y)=zΔ1Δ2(x,y)+k(x,y,σ(x),σ(y))z(x,y)+∫x0xkΔ1(x,y,τ,σ(y))z(τ,y)Δτ+∫y0ykΔ2(x,y,σ(x),ξ)z(x,ξ)Δξ+∫x0x∫y0ykΔ2Δ1(x,y,τ,ξ)z(τ,ξ)ΔξΔτ≤f(x,y)v(x,y)+k(x,y,σ(x),σ(y))v(x,y)+∫x0xkΔ1(x,y,τ,σ(y))v(τ,y)Δτ+∫y0ykΔ2(x,y,σ(x),ξ)v(x,ξ)Δξ+∫x0x∫y0ykΔ2Δ1(x,y,τ,ξ)v(τ,ξ)ΔξΔτ≤|f(x,y)+A(x,y)|v(x,y),
where A(x,y) is defined by (17).
Using the fact thatz(x,y)>0,   zΔ1(x,y)≥0,   zΔ2(x,y)≥0, and we have
(21)vΔ2Δ1(x,y)v(x,y)≤[f(x,y)+A(x,y)]+vΔ1(x,y)vΔ2(x,y)v(x,y)v(x,σ(y)).
Now keepingx fixed and delta integrating with respect to t from y0 to y, we obtain estimates
(22)vΔ1(x,y)v(x,y)≤∫y0y[f(x,ξ)+A(x,y)]Δ2ξ,
keeping y fixed in (22) and delta integrating s from x0 to x, we get
(23)v(x,y)≤c+∫x0x∫y0y[f(τ,ξ)+A(τ,ξ)]v(x,y)Δ2ξΔ1τ.
Then Lemma 2 implies
(24)v(x,y)≤c+e∫y0y[f(τ,ξ)+A(τ,ξ)]Δξ(x,x0).
Using (24) in (18) and integrating the resulting inequality from 0 to y and then from 0 to x for  (x,y)∈Ω, we get
(25)z(x,y)≤c[1+∫x0x∫y0yf(s,τ)e∫t0t[f(τ,ξ)+A(τ,ξ)]ΔξΔτ(s,s0)ΔtΔs].
Using (25) in  u(x,y)≤z(x,y), we get the required inequality in (16).Theorem 5.
Letu,   f,   a∈Crd(Ω×Ω,R+) and k(x,y,s,t),   kΔ1(x,y,s,t),   kΔ2(x,y,s,t),  kΔ2Δ1(x,y,s,t)∈Crd(Ω,R+). If
(26)u(x,y)≤a(x,y)+∫x0x∫y0yf(s,t)≤a(x,y)+×[u(s,t)+∫s0s∫t0tk(s,t,τ,ξ)u(τ,ξ)ΔξΔτ]ΔtΔs,
For  (x,y)∈Ω, then
(27)u(x,y)≤a(x,y)+h¯(x,y)×[1+∫x0x∫y0yf(s,t)e∫t0t[f(τ,ξ)+A(τ,ξ)]ΔξΔτ(x,x0)]ΔtΔs,
where
(28)h¯(x,y)=∫x0x∫y0yf(s,t)×[a(s,t)+∫s0s∫t0tk(s,t,σ(x),ξ)a(τ,ξ)ΔξΔτ]ΔtΔs,
for (x,y)∈Ω  and A(x,y)  as defined in (17).Proof.
Define a functionz(x,y)  by the right-hand side of (26)
(29)z(x,y)≤∫x0x∫y0yf(s,t)×[u(s,t)+∫s0s∫t0tk(s,t,τ,ξ)u(τ,ξ)ΔξΔτ]ΔtΔs,
then from (26) we have u(x,y)≤a(x,y)+z(x,y), and using this in (28) we get
(30)z(x,y)≤∫x0x∫y0yf(s,t)[∫s0s∫t0tk(s,t,τ,ξ)[a(τ,ξ)+z(τ,ξ)]ΔξΔτa(s,t)+z(s,t)+∫s0s∫t0tk(s,t,τ,ξ)×[a(τ,ξ)+z(τ,ξ)]ΔξΔτ∫s0s∫t0tk(s,t,τ,ξ)]ΔtΔs≤h¯(x,y)+∫x0x∫y0yf(s,t)[∫s0s∫t0tk(s,t,τ,ξ)z(τ,ξ)ΔξΔτz(s,t)+∫s0s∫t0tk(s,t,τ,ξ)×z(τ,ξ)ΔξΔτ∫s0s∫t0tk(s,t,τ,ξ)]ΔtΔs,
where h¯(x,y) is defined by (28). Here h¯(x,y) nondecreasing for (x,y)∈Ω, then from (30) we have
(31)z(x,y)h¯(x,y)≤1+∫x0x∫y0yf(s,t)×[z(s,t)h¯(s,t)+∫s0s∫t0tk(s,t,τ,ξ)z(τ,ξ)h¯(τ,ξ)ΔξΔτ]ΔtΔs.
Now as an application of inequality in Theorem 3 to (31) we get
(32)z(x,y)h¯(x,y)≤[1+∫x0x∫y0yf(s,t)e[f(τ,ξ)+A(τ,ξ)]ΔξΔτ(x,x0)]ΔtΔs.
The required inequality in (27) follows from (32) and using the fact that u(x,y)≤a(x,y)+z(x,y).


## 3. Applications

In this section, we give application of inequality in Theorem4 to study certain properties of solution of nonlinear partial integrodifferential equation on time scales,
(33)uΔ2Δ1(x,y)=F(x,y,u(x,y),∫x0x∫y0yh(x,y,τ,ξ,u(τ,ξ))ΔξΔτ),
with the initial boundary condition
(34)u(x,y0)=α1(x),u(x0,y)=α2(y),α1(0)=α2(0)=0,
where u∈Crd(Ω,R+),  h∈Crd(Ω×R,R+), and  F∈Crd(Ω×Ω,R+).The following theorem deals with the estimate on the solution of (33)-(34).Theorem 6.
Assume that(35)|h(x,y,s,t,u(s,t))|≤k(x,y,s,t)|u(s,t)|,|F(x,y,u,v)|≤f(x,y)[|u|+|v|],|α1(x)+α2(y)|≤c,
where k, f, and c are as defined in (9). If u(x,y),  (x,y)∈Ω is any solution of (33)-(34), then
(36)|u(x,y)|≤c[1+∫x0x∫y0yf(s,t)e∫t0t[f(τ,ξ)+A(τ,ξ)]Δt(s,s0)ΔtΔs]
for (x,y)∈Ω,  where A(x,y) is defined by (17).Proof.
The solutionu(x,y)  of (33)-(34) can be written as
(37)u(x,y)=α1(x)+α2(y)+∫x0x∫y0yF(∫s0s∫t0th(s,t,τ,ξ,u(τ,ξ))s,t,u(s,t),∫s0s∫t0th(s,t,τ,ξ,u(τ,ξ))ΔξΔτ)ΔtΔs.
Using (35) in (37), we have
(38)|u(x,y)|≤c+∫x0x∫y0yf(s,t)×[∫s0s∫t0tk(s,t,τ,ξ,|u(τ,ξ)|)ΔξΔτ|u(s,t)|+(∫s0s∫t0tk(s,t,τ,ξ,|u(τ,ξ)|)×ΔξΔτ∫s0s∫t0tk(s,t,τ,ξ,|u(τ,ξ)|))∫s0s∫t0tk(s,t,τ,ξ,|u(τ,ξ)|)]ΔtΔs.
Now an application of inequality in Theorem 4 yields the estimate in (36).In the next result, we give uniqueness of solutions of (33)-(34).Theorem 7.
Suppose that the functionsh, F in (33) satisfy the conditions
(39)|h(x,y,s,t,u1)-h(x,y,s,t,u2)|≤k(x,y,s,t)|u1-u2|,|F(x,y,u1,u2)-F(x,y,v1,v2)|≤f(x,y)[|u1-v1|+|u2-v2|],
where k and f are as in Theorem 4. Then the problem (33)-(34) has at most one solution on Ω.Proof.
Letu1(x,y) and u2(x,y) be two solutions of (33)-(34) on  Ω, then we have (40)u1(x,y)-u2(x,y)=∫x0x∫y0y[F(s,t,u1(s,t),∫s0s∫t0th(s,t,τ,ξ,u1(τ,ξ))ΔξΔτ)-F(s,t,u2(s,t),∫s0s∫t0th(s,t,τ,ξ,u2(τ,ξ))×ΔξΔτ∫s0s∫t0th(s,t,τ,ξ,u2(τ,ξ)))∫s0s∫t0th(s,t,τ,ξ,u2(τ,ξ))]ΔtΔs.
From (39) and (40), we have
(41)|u1(x,y)-u2(x,y)|≤∫x0x∫y0y[∫s0s∫t0tk(s,t,τ,ξ)|u1(τ,ξ)-u2(τ,ξ)|ΔξΔτf(s,t)|u1(s,t)-u2(s,t)|+∫s0s∫t0tk(s,t,τ,ξ)|u1(τ,ξ)-u2(τ,ξ)|×ΔξΔτ∫s0s∫t0tk(s,t,τ,ξ)|u1(τ,ξ)-u2(τ,ξ)|]ΔtΔs.
As an application of inequality in Theorem 4 with  c=0 yields |u1(x,y)-u2(x,y)|≤0, therefore u1(x,y)=u2(x,y); that is, there is at most one solution of (33)-(34) on Ω.



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