The theory of time scales had been begun in 1988 by Stefan Hilger [1], in order to develop a theory that can standardize a continuous and discrete analysis. Recently several authors in this field have investigated various forms of integral and integrodifferential equations under different hypotheses by using different ways, see [4] [5] [6] [7] [8]. In this article we consider the nonlinear integrodifferential equation of the following form

y Δ ( t ) = h ( t , y ( t ) , y Δ ( t ) , ∫ α t h 1 ( t , z , y ( z ) , y Δ ( z ) ) Δ z , ∫ α β h 2 ( t , z , y ( z ) , y Δ ( z ) ) Δ z ) ,                         t ∈ J T     with the initial condition   y ( α ) = y 0 , (1.1)

where y is unknown function and h : J T × ℝ n × ℝ n × ℝ n × ℝ n → ℝ n , h 1 , h 2 : J T 2 × ℝ n × ℝ n → ℝ n and h , h 1 , h 2 are given functions, assuming them to be rd-continuous functions, α < β , z ≤ t and J T = J ∩ T , J = [ α , ∞ ) . We denote a time scale by T which is nonempty closed subset of ℝ . ℝ n denotes Euclidean space with a suitable norm defined by |   .   | .

We can investigate the existence and uniqueness results for (1.1) by using the technique present in [6].

The operators σ ( t ) and ρ ( t ) denote the forward and backward operators respectively which are defined by σ ( t ) = inf { s ∈ T : s > t } ∈ T and ρ ( t ) = sup { s ∈ T : s < t } ∈ T , for all t ∈ T .

For t ∈ T , If t < sup T and σ ( t ) = t , then t is said to be right-dense; while If t > inf T and ρ ( t ) = t , then t is said to be left-dense. The graininess μ : T → [ 0 , ∞ ) is defined by μ ( t ) = σ ( t ) − t . The set T k is denoted by

T k = { T \ ( ρ ( sup T ) , sup T ]       if   sup T < ∞ T                                                               otherwise

Let z : T → ℝ , t ∈ T k , then z Δ ( t ) denotes the delta derivative of z at t which is exist with the property that given ε > 0 there is a neighbourhood U of t such that | z ( σ ( t ) , τ ) − z ( s , τ ) − z Δ ( t , τ ) ( σ ( t ) − s ) | ≤ ε | σ ( t ) − s | for all s ∈ U . Then g ( t ) = ∫ α t z ( t , τ ) Δ τ implies g Δ ( t ) = ∫ α t z Δ ( t , τ ) Δ τ + z ( σ ( t ) , t ) . If a function g : T → ℝ is continuous at any right-dense point t ∈ T and the left-hand limits exists (finite) at any left-dense point t ∈ T , then g is said to be rd-continuous. C r d denotes the class of all rd-continuous functions. We denote the class of all regressive functions by R which is defined by

R = { p ∈ C r d ( T , ℝ )     and     1 + p ( t ) μ ( t ) ≠ 0 , ∀ t ∈ T }

For p ∈ R , we define e p ( t , s ) = exp ( ∫ s t ξ μ ( τ ) ( p ( τ ) ) Δ τ ) for t , s ∈ T , with the cylinder transformation ξ h ( τ ) = { log ( 1 + h z ) h     if   h ≠ 0 z     if   h = 0 .

For more basic information about time scales calculus, see [1] [3].

We need the following result given in [2].

Lemma 2.1. suppose ν , b ∈ C r d and a ∈ R + . Then

ν Δ ( t ) ≤ a ( t ) ν ( t ) + b ( t ) , for all t ∈ T

Implies ν ( t ) ≤ ν ( α ) e a ( t , α ) + ∫ α t e a ( t , σ ( τ ) ) b ( τ ) Δ τ , for all t ∈ T .

In the following result we establish an integral inequality on time scales.

Theorem 3.1. Let ν , r , b 1 , b 2 , p , q , g , d ∈ C r d ( J T , ℝ   + ) and assume that

ν ( t ) ≤ r ( t ) + b 1 ( t ) ∫ α t { [ b 2 ( t ) p ( τ ) + 1 ] ν ( τ ) + b 2 ( τ ) ∫ α τ p ( z ) ν ( z ) Δ z                   + q ( τ ) ∫ α β g ( z ) ν ( z ) Δ z } Δ τ + d ( t ) ∫ α β g ( z ) ν ( z ) Δ z ,   t ∈ J T , (3.1)

If

N = ∫ α β g ( γ ) N 2 ( γ ) Δ γ < 1 , (3.2)

Implies

ν ( t ) ≤ N 1 ( t ) + A N 2 ( t ) , t ∈ J T , (3.3)

where

N 1 ( t ) = r ( t ) + b 1 ( t ) ∫ α t [ b 2 σ ( τ ) p ( τ ) + 1 ] [ r ( τ ) + b ( τ ) ∫ α τ e ( b 2 σ p + p + 1 ) b ( τ , σ ( z ) ) × ( b 2 σ ( z ) p ( z ) + p ( z ) + 1 ) r ( z ) Δ z ] Δ τ , (3.4)

N 2 ( t ) = d ( t ) + b 1 ( t ) ∫ α t { [ b 2 σ ( τ ) p ( τ ) + 1 ] [ d ( τ ) + b ( τ ) ∫ α τ e ( b 2 σ p + p + 1 ) b ( τ , σ ( z ) ) × ( [ b 2 σ ( z ) p ( z ) + p ( z ) + 1 ] d ( z ) + q ( z ) ) Δ z ] + q ( τ ) } Δ τ , (3.5)

for t ∈ J T ,

b 2 σ ( t ) = b 2 ( σ ( t ) ) = b 2 ∘ σ , b ( t ) = max t ∈ J T { b 1 ( t ) , b 2 ( t ) + b 2 Δ ( t ) } , (3.6)

A = 1 1 − N ∫ α β g ( γ ) N 1 ( γ ) Δ γ , (3.7)

Proof. Let

λ = ∫ α β g ( z ) ν ( z ) Δ z , (3.8)

we shall define functions B 1 ( t ) and B 2 ( t ) by

B 1 ( t ) = ∫ α t { [ b 2 ( t ) p ( τ ) + 1 ] ν ( τ ) + b 2 ( τ ) ∫ α τ p ( z ) ν ( z ) Δ z     + q ( τ ) ∫ α β g ( z ) ν ( z ) Δ z } Δ τ , (3.9)

B 2 ( t ) = B 1 ( t ) + ∫ α t p ( z ) [ r ( z ) + b 1 ( z ) B 1 ( z ) + d ( z ) λ ] Δ z , (3.10)

then B 1 ( α ) = 0 , B 2 ( α ) = 0 , B 1 ( t ) ≤ B 2 ( t ) and we have

ν ( t ) ≤ r ( t ) + b 1 ( t ) B 1 ( t ) + d ( t ) λ , (3.11)

from (3.9), we get

B 1 Δ ( t ) = ∫ α t b 2 Δ ( t ) p ( τ ) ν ( τ ) Δ τ + [ b 2 σ ( t ) p ( t ) + 1 ] ν ( t )                         + b 2 ( t ) ∫ α t p ( z ) ν ( z ) Δ z + q ( t ) ∫ α β g ( z ) ν ( z ) Δ z ≤ [ b 2 σ ( t ) p ( t ) + 1 ] r ( t ) + [ b 2 σ ( t ) p ( t ) + 1 ] b 1 ( t ) B 1 ( t )     + [ b 2 σ ( t ) p ( t ) + 1 ] d ( t ) λ + [ b 2 ( t ) + b 2 Δ ( t ) ] ∫ α t p ( z ) ν ( z ) Δ z + q ( t ) λ

≤ [ b 2 σ ( t ) p ( t ) + 1 ] r ( t ) + b 2 σ ( t ) p ( t ) b ( t ) B 1 ( t )     + b ( t ) [ B 1 ( t ) + ∫ α t p ( z ) ν ( z ) Δ z ] + [ b 2 σ ( t ) p ( t ) + 1 ] d ( t ) λ + q ( t ) λ ≤ [ b 2 σ ( t ) p ( t ) + 1 ] [ r ( t ) + b ( t ) B 2 ( t ) + d ( t ) λ ] + q ( t ) λ , (3.12)

integrating the inequality (3.12) and using B 1 ( α ) = 0 , we have

B 1 ( t ) ≤ ∫ α t { [ b 2 σ ( τ ) p ( τ ) + 1 ] [ r ( τ ) + b ( τ ) B 2 ( τ ) + d ( τ ) λ ] + q ( τ ) λ } Δ τ , (3.13)

therefore

B 2 Δ ( t ) = B 1 Δ ( t ) + p ( t ) [ r ( t ) + b 1 ( t ) B 1 ( t ) + d ( t ) λ ] ≤ [ b 2 σ ( t ) p ( t ) + 1 ] [ r ( t ) + b ( t ) B 2 ( t ) + d ( t ) λ ] + q ( t ) λ     + p ( t ) [ r ( t ) + b ( t ) B 2 ( t ) + d ( t ) λ ] = [ b 2 σ ( t ) p ( t ) + p ( t ) + 1 ] b ( t ) B 2 ( t )     + [ b 2 σ ( t ) p ( t ) + p ( t ) + 1 ] [ r ( t ) + d ( t ) λ ] + q ( t ) λ , (3.14)

now applying lemma 2.1, we get

B 2 ( t ) ≤ ∫ α t e ( b 2 σ p + p + 1 ) b ( t , σ ( z ) )                       × ( [ b 2 σ ( z ) p ( z ) + p ( z ) + 1 ] [ a ( z ) + d ( z ) λ ] + q ( z ) λ ) Δ z , (3.15)

from (3.11), (3.13) and (3.15), we obtain that

ν ( t ) ≤ N 1 ( t ) + λ N 2 ( t ) , (3.16)

and from (3.8) and (3.16) we observe that

λ ≤ A , (3.17)

using (3.17) in (3.16) we obtain (3.3). □

We provide the result that includes the estimate on the solutions of (1.1) as follows.

Theorem 3.2. Assume that the following conditions satisfied

| h ( t , ν 1 , ν 2 , ν 3 , ν 4 ) | ≤ L [ | ν 1 | + | ν 2 | + | ν 3 | + | ν 4 | ] , (3.18)

| h 1 ( t , z , u , ν ) | ≤ c 1 ( t ) s 1 ( z ) [ | u | + | ν | ] , (3.19)

| h 2 ( t , z , u , ν ) | ≤ c 2 ( t ) s 2 ( z ) [ | u | + | ν | ] , (3.20)

for the functions h , h 1 , h 2 in (1.1), where 0 ≤ L < 1 is a constant and

c 1 , s 1 , c 2 , s 2 ∈ C r d ( J T , ℝ + )

If y ( t ) is a solution of (1.1) on J T , then

| y ( t ) | + | y Δ ( t ) | ≤ M 1 ( t ) + D 1 M 2 ( t ) , t ∈ J T , (3.21)

where

M 1 ( t ) = | y 0 | 1 − L + L 1 − L ∫ α t [ c 1 σ ( τ ) s 1 ( τ ) + 1 ] [ | y 0 | 1 − L + b ( τ ) ∫ α τ e ( c 1 σ s 1 + s 1 + 1 ) b ( τ , σ ( z ) )                           × ( c 1 σ ( z ) s 1 ( z ) + s 1 ( z ) + 1 ) | y 0 | 1 − L Δ z ] Δ τ , t ∈ J T (3.22)

M 2 ( t ) = L 1 − L c 2 ( t )                         + L 1 − L ∫ α t { [ c 1 σ ( τ ) s 1 ( τ ) + 1 ] [ L 1 − L c 2 ( τ ) + b ( τ ) ∫ α τ e ( c 1 σ s 1 + s 1 + 1 ) b ( τ , σ ( z ) )                         × ( [ c 1 σ ( z ) s 1 ( z ) + s 1 ( z ) + 1 ] L 1 − L c 2 ( z ) + c 2 ( z ) ) Δ z ] + c 2 ( τ ) } Δ τ , t ∈ J T

(3.23)

Assume that

b ( t ) = max t ∈ J T { L 1 − L , c 1 ( t ) + c 1 Δ ( t ) } , (3.24)

λ = ∫ α β s 2 ( γ ) M 2 ( γ ) Δ γ < 1 , (3.25)

D 1 = 1 1 − λ ∫ α β s 2 ( γ ) M 1 ( γ ) Δ γ , (3.26)

Proof. Let a ( t ) = | y ( t ) | + | y Δ ( t ) | , t ∈ J T , since y ( t ) is a solution of (1.1), then by using this and the hypotheses, we get

a ( t ) = | y 0 + ∫ α t h ( τ , y ( τ ) , y Δ ( τ ) , ∫ α τ h 1 ( τ , z , y ( z ) , y Δ ( z ) ) Δ z ,     ∫ α β h 2 ( τ , z , y ( z ) , y Δ ( z ) ) Δ z ) Δ τ |     + | h ( t , y ( t ) , y Δ ( t ) , ∫ α t h 1 ( t , z , y ( z ) , y Δ ( z ) ) Δ z , ∫ α β h 2 ( t , z , y ( z ) , y Δ ( z ) ) Δ z ) | ≤ | y 0 | + ∫ α t L [ a ( τ ) + ∫ α τ c 1 ( τ ) s 1 ( z ) a ( z ) Δ z + ∫ α β c 2 ( τ ) s 2 ( z ) a ( z ) Δ z ] Δ τ     + L [ a ( t ) + ∫ α t c 1 ( t ) s 1 ( z ) a ( z ) Δ z + ∫ α β c 2 ( t ) s 2 ( z ) a ( z ) Δ z ]

from the above inequality, we have

a ( t ) ≤ | y 0 | 1 − L + L 1 − L ∫ α t { [ c 1 ( t ) s 1 ( τ ) + 1 ] a ( τ ) + c 1 ( τ ) ∫ α τ s 1 ( z ) a ( z ) Δ z     + c 2 ( τ ) ∫ α β s 2 ( z ) a ( z ) Δ z } Δ τ + L 1 − L c 2 ( t ) ∫ α β s 2 ( z ) a ( z ) Δ z , (3.27)

Now applying theorem 3.1 in (3.27) we obtain (3.21). □

Remark 3.3. Since y ( t ) is a solution of (1.1). Then (3.21) yields the bounds on y ( t ) and y Δ ( t ) . If the estimate in (3.21) is bounded, implies the solution y ( t ) and y Δ ( t ) are also bounded on J T .

Consider (1.1) with the following corresponding equation

Y Δ ( t ) = H ( t , Y ( t ) , Y Δ ( t ) , ∫ α t h 1 ( t , z , Y ( z ) , Y Δ ( z ) ) Δ z ,                                 ∫ α β h 2 ( t , z , Y ( z ) , Y Δ ( z ) ) Δ z ) , t ∈ J T ,

with the initial condition

Y ( α ) = Y 0 (3.28)

where H ∈ C r d ( J T × ℝ n × ℝ n × ℝ n × ℝ n , ℝ n ) , h 1 , h 2 as in (1.1).

The next result concerning the closeness of solution of (1.1) and (3.28).

Theorem 3.4. Suppose that the following conditions satisfied

| h ( t , ν 1 , ν 2 , ν 3 , ν 4 ) − h ( t , ν ¯ 1 , ν ¯ 2 , ν ¯ 3 , ν ¯ 4 ) | ≤ L [ | ν 1 − ν ¯ 1 | + | ν 2 − ν ¯ 2 | + | ν 3 − ν ¯ 3 | + | ν 4 − ν ¯ 4 | ] , (3.29)

| h 1 ( t , z , u , ν ) − h 1 ( t , z , u ¯ , ν ¯ ) | ≤ c 1 ( t ) s 1 ( z ) [ | u − u ¯ | + | ν − ν ¯ | ] , (3.30)

| h 2 ( t , z , u , ν ) − h 2 ( t , z , u ¯ , ν ¯ ) | ≤ c 2 ( t ) s 2 ( z ) [ | u − u ¯ | + | ν − ν ¯ | ] , (3.31)

where the functions h , h 1 , h 2 in (1.1), and 0 ≤ L < 1 is a constant.

Also c 1 , s 1 , c 2 , s 2 ∈ C r d ( J T , ℝ   + ) , and

| h ( t , ν 1 , ν 2 , ν 3 , ν 4 ) − H ( t , ν 1 , ν 2 , ν 3 , ν 4 ) | ≤ ε , (3.32)

| y 0 − Y 0 | ≤ δ , (3.33)

where y 0 and H , Y 0 as in (1.1) and (3.28) respectively.

If y ( t ) and Y ( t ) be solutions of (1.1) and (3.28) on J T , then

| y ( t ) − Y ( t ) | + | y Δ ( t ) − Y Δ ( t ) | ≤ M 3 ( t ) + D 2 M 2 ( t ) , t ∈ J T , (3.34)

where M 3 ( t ) is described by the right side of (3.22) by substituting m ( t ) = δ + ε ( 1 + t − α ) instead of | y 0 | , M 2 ( t ) , b ( t ) and λ be as in (3.23), (3.24) and (3.25) respectively and

D 2 = 1 1 − λ ∫ α β r 2 ( γ ) M 3 ( γ ) Δ γ , (3.35)

Proof. Let w ( t ) = | y ( t ) − Y ( t ) | + | y Δ ( t ) − Y Δ ( t ) | , t ∈ J T , we have

w ( t ) ≤ | y 0 − Y 0 |     + ∫ α t | h ( τ , y ( τ ) , y Δ ( τ ) , ∫ α τ h 1 ( τ , z , y ( z ) , y Δ ( z ) ) Δ z , ∫ α β h 2 ( τ , z , y ( z ) , y Δ ( z ) ) Δ z )     − h ( τ , Y ( τ ) , Y Δ ( τ ) , ∫ α τ h 1 ( τ , z , Y ( z ) , Y Δ ( z ) ) Δ z , ∫ α β h 2 ( τ , z , Y ( z ) , Y Δ ( z ) ) Δ z ) | Δ τ     + ∫ α t | h ( τ , Y ( τ ) , Y Δ ( τ ) , ∫ α τ h 1 ( τ , z , Y ( z ) , Y Δ ( z ) ) Δ z , ∫ α β h 2 ( τ , z , Y ( z ) , Y Δ ( z ) ) Δ z )     − H ( τ , Y ( τ ) , Y Δ ( τ ) , ∫ α r h 1 ( τ , z , Y ( z ) , Y Δ ( z ) ) Δ z , ∫ α β h 2 ( τ , z , Y ( z ) , Y Δ ( z ) ) Δ z ) | Δ τ

+ | h ( t , y ( t ) , y Δ ( t ) , ∫ α t h 1 ( t , z , y ( z ) , y Δ ( z ) ) Δ z , ∫ α β h 2 ( t , z , y ( z ) , y Δ ( z ) ) Δ z ) − h ( t , Y ( t ) , Y Δ ( t ) , ∫ α t h 1 ( t , z , Y ( z ) , Y Δ ( z ) ) Δ z , ∫ α β h 2 ( t , z , Y ( z ) , Y Δ ( z ) ) Δ z ) | + | h ( t , Y ( t ) , Y Δ ( t ) , ∫ α t h 1 ( t , z , Y ( z ) , Y Δ ( z ) ) Δ z , ∫ α β h 2 ( t , z , Y ( z ) , Y Δ ( z ) ) Δ z ) − H ( t , Y ( t ) , Y Δ ( t ) , ∫ α t h 1 ( t , z , Y ( z ) , Y Δ ( z ) ) Δ z , ∫ α β h 2 ( t , z , Y ( z ) , Y Δ ( z ) ) Δ z ) |

≤ δ + ∫ α t L [ w ( τ ) + ∫ α τ c 1 ( τ ) s 1 ( z ) w ( z ) Δ z + ∫ α β c 2 ( τ ) s 2 ( z ) w ( z ) Δ z ] Δ τ     + ∫ α t ε Δ τ + L [ w ( t ) + ∫ α t c 1 ( t ) s 1 ( z ) w ( z ) Δ z + ∫ α β c 2 ( t ) s 2 ( z ) w ( z ) Δ z ] + ε = m ( t ) + L ∫ α t [ w ( τ ) + c 1 ( τ ) ∫ α τ s 1 ( z ) w ( z ) Δ z + c 2 ( τ ) ∫ α β s 2 ( z ) w ( z ) Δ z ] Δ τ     + L [ w ( t ) + ∫ α t c 1 ( t ) s 1 ( z ) w ( z ) Δ z + c 2 ( t ) ∫ α β s 2 ( z ) w ( z ) Δ z ]

then we get

w ( t ) ≤ m ( t ) 1 − L + L 1 − L ∫ α t { [ c 1 ( t ) s 1 ( τ ) + 1 ] w ( τ ) + c 1 ( τ ) ∫ α τ s 1 ( z ) w ( z ) Δ z     + c 2 ( τ ) ∫ α β s 2 ( z ) w ( z ) Δ z } Δ τ + L 1 − L c 2 ( t ) ∫ α β s 2 ( z ) w ( z ) Δ z , (3.36)

Now applying theorem 3.1, yields (3.34). □

The following theorem provide the continuous depends of solutions of (1.1) on given initial values.

Theorem 3.5. Assume that the conditions (3.29), (3.30) and (3.31) are satisfied for the functions h , h 1 , h 2 in (1.1). Let y 1 ( t ) and y 2 ( t ) be the solutions of equation

y Δ ( t ) = h ( t , y ( t ) , y Δ ( t ) , ∫ α t h 1 ( t , z , y ( z ) , y Δ ( z ) ) Δ z ,                               ∫ α β h 2 ( t , z , y ( z ) , y Δ ( z ) ) Δ z ) , t ∈ J T ,

with the given initial values

y 1 ( α ) = c 1 and y 2 ( α ) = c 2 , (3.37)

where h , h 1 , h 2 as in (1.1), c 1 and c 2 are constants. Then

| y 1 ( t ) − y 2 ( t ) | + | y 1 Δ ( t ) − y 2 Δ ( t ) | ≤ M 4 ( t ) + D 3 M 2 ( t ) , t ∈ J T , (3.38)

where M 4 ( t ) is described by the right side of (3.22) by substituting | c 1 − c 2 | instead of | y 0 | , M 2 ( t ) , b ( t ) and λ be as in (3.23), (3.24) and (3.25) respectively and

D 3 = 1 1 − λ ∫ α β r 2 ( γ ) M 4 ( γ ) Δ γ , (3.39)

Proof. Let n ( t ) = | y 1 ( t ) − y 2 ( t ) | + | y 1 Δ ( t ) − y 2 Δ ( t ) | , t ∈ J T , we get

n ( t ) ≤ | c 1 − c 2 |   + ∫ α t | h ( τ , y 1 ( τ ) , y 1 Δ ( τ ) , ∫ α τ h 1 ( τ , z , y 1 ( z ) , y 1 Δ ( z ) ) Δ z , ∫ α β h 2 ( τ , z , y 1 ( z ) , y 1 Δ ( z ) ) Δ z )   − h ( τ , y 2 ( τ ) , y 2 Δ ( τ ) , ∫ α τ h 1 ( τ , z , y 2 ( z ) , y 2 Δ ( z ) ) Δ z , ∫ α β h 2 ( τ , z , y 2 ( z ) , y 2 Δ ( z ) ) Δ z ) | Δ τ   + | h ( t , y 1 ( t ) , y 1 Δ ( t ) , ∫ α t h 1 ( t , z , y 1 ( z ) , y 1 Δ ( z ) ) Δ z , ∫ α β h 2 ( t , z , y 1 ( z ) , y 1 Δ ( z ) ) Δ z )   − h ( t , y 2 ( t ) , y 2 Δ ( t ) , ∫ α t h 1 ( t , z , y 2 ( z ) , y 2 Δ ( z ) ) Δ z , ∫ α β h 2 ( t , z , y 2 ( z ) , y 2 Δ ( z ) ) Δ z ) |

≤ | c 1 − c 2 | + L ∫ α t [ n ( τ ) + c 1 ( τ ) ∫ α τ s 1 ( z ) n ( z ) Δ z + c 2 ( τ ) ∫ α β s 2 ( z ) n ( z ) Δ z ] Δ τ   + L [ n ( t ) + ∫ α t c 1 ( t ) s 1 ( z ) n ( z ) Δ z + c 2 ( t ) ∫ α β s 2 ( z ) n ( z ) Δ z ]

then

n ( t ) ≤ | c 1 − c 2 | 1 − L + L 1 − L ∫ α t { [ c 1 ( t ) s 1 ( τ ) + 1 ] n ( τ ) + c 1 ( τ ) ∫ α τ s 1 ( z ) n ( z ) Δ z                   + c 2 ( τ ) ∫ α β s 2 ( z ) n ( z ) Δ z } Δ τ + L 1 − L c 2 ( t ) ∫ α β s 2 ( z ) n ( z ) Δ z , (3.40)

Now applying theorem 3.1 in (3.40) we obtain (3.38). □

Remark 3.6. The inequality (3.38) gives the uniqueness of solutions of (3.37). If we have c 1 = c 2 = 0 , then we get M 5 ( t ) = 0 and D 3 = 0 , implies the right hand side of (3.37) is equal to zero.

Now consider the initial value problems

Y Δ ( t ) = h ( t , Y ( t ) , Y Δ ( t ) , ∫ α t h 1 ( t , z , Y ( z ) , Y Δ ( z ) ) Δ z ,                               ∫ α β h 2 ( t , z , Y ( z ) , Y Δ ( z ) ) Δ z , μ ) , t ∈ J T , Y ( α ) = Y 0 , (3.41)

Y Δ ( t ) = h ( t , Y ( t ) , Y Δ ( t ) , ∫ α t h 1 ( t , z , Y ( z ) , Y Δ ( z ) ) Δ z ,                               ∫ α β h 2 ( t , z , Y ( z ) , Y Δ ( z ) ) Δ z , μ 0 ) , t ∈ J T , Y ( α ) = Y 0 , (3.42)

where h ∈ C r d ( J T × ℝ n × ℝ n × ℝ n × ℝ n , ℝ n ) and μ , μ 0 are parameters.

The dependency of solutions of (3.41) and (3.42) on parameters follows in the next theorem.

Theorem 3.7. Suppose that the conditions (3.30) and (3.31) are satisfied and

| h ( t , ν 1 , ν 2 , ν 3 , ν 4 , μ ) − h ( t , ν ¯ 1 , ν ¯ 2 , ν ¯ 3 , ν ¯ 4 , μ ) | ≤ L [ | ν 1 − ν ¯ 1 | + | ν 2 − ν ¯ 2 | + | ν 3 − ν ¯ 3 | + | ν 4 − ν ¯ 4 | ] , (3.43)

| h ( t , ν 1 , ν 2 , ν 3 , ν 4 , μ ) − h ( t , ν 1 , ν 2 , ν 3 , ν 4 , μ 0 ) | ≤ k ( t ) | μ − μ 0 | , (3.44)

where 0 ≤ L < 1 is a constant and k ∈ C r d ( J T , ℝ   + ) . Let Y 1 ( t ) and Y 2 ( t ) be respectively, the solutions of (3.41) and (3.42) on J T , then

| Y 1 ( t ) − Y 2 ( t ) | + | Y 1 Δ ( t ) − Y 2 Δ ( t ) | ≤ M 5 ( t ) + D 4 M 2 ( t ) , t ∈ J T , (3.45)

where M 5 ( t ) is described by the right side of (3.22) by substituting | μ − μ 0 | k ¯ ( t ) instead of | y 0 | , M 2 ( t ) , b ( t ) and λ be as in (3.23), (3.24) and (3.25) respectively.

Let

k ¯ ( t ) = k ( t ) + ∫ α t k ( r ) Δ τ , (3.46)

D 4 = 1 1 − λ ∫ α β r 2 ( γ ) M 5 ( γ ) Δ γ , (3.47)

Proof. Let P ( t ) = | Y 1 ( t ) − Y 2 ( t ) | + | Y 1 Δ ( t ) − Y 2 Δ ( t ) | , t ∈ J T , we have

P ( t ) ≤ ∫ α t | h ( τ , Y 1 ( τ ) , Y 1 Δ ( τ ) , ∫ α τ h 1 ( τ , z , Y 1 ( z ) , Y 1 Δ ( z ) ) Δ z , ∫ α β h 2 ( τ , z , Y 1 ( z ) , Y 1 Δ ( z ) ) Δ z , μ )   − h ( τ , Y 2 ( τ ) , Y 2 Δ ( τ ) , ∫ α τ h 1 ( τ , z , Y 2 ( z ) , Y 2 Δ ( z ) ) Δ z , ∫ α β h 2 ( τ , z , Y 2 ( z ) , Y 2 Δ ( z ) ) Δ z , μ ) | Δ τ   + ∫ α t | h ( τ , Y 2 ( τ ) , Y 2 Δ ( τ ) , ∫ α τ h 1 ( τ , z , Y 2 ( z ) , Y 2 Δ ( z ) ) Δ z , ∫ α β h 2 ( τ , z , Y 2 ( z ) , Y 2 Δ ( z ) ) Δ z , μ )   − h ( τ , Y 2 ( τ ) , Y 2 Δ ( τ ) , ∫ α τ h 1 ( τ , z , Y 2 ( z ) , Y 2 Δ ( z ) ) Δ z , ∫ α β h 2 ( τ , z , Y 2 ( z ) , Y 2 Δ ( z ) ) Δ z , μ 0 ) | Δ τ

  + | h ( t , Y 1 ( t ) , Y 1 Δ ( t ) , ∫ α t h 1 ( t , z , Y 1 ( z ) , Y 1 Δ ( z ) ) Δ z , ∫ α β h 2 ( t , z , Y 1 ( z ) , Y 1 Δ ( z ) ) Δ z , μ )   − h ( t , Y 2 ( t ) , Y 2 Δ ( t ) , ∫ α t h 1 ( t , z , Y 2 ( z ) , Y 2 Δ ( z ) ) Δ z , ∫ α β h 2 ( t , z , Y 2 ( z ) , Y 2 Δ ( z ) ) Δ z , μ ) |   + | h ( t , Y 2 ( t ) , Y 2 Δ ( t ) , ∫ α t h 1 ( t , z , Y 2 ( z ) , Y 2 Δ ( z ) ) Δ z , ∫ α β h 2 ( t , z , Y 2 ( z ) , Y 2 Δ ( z ) ) Δ z , μ )   − h ( t , Y 2 ( t ) , Y 2 Δ ( t ) , ∫ α t h 1 ( t , z , Y 2 ( z ) , Y 2 Δ ( z ) ) Δ z , ∫ α β h 2 ( t , z , Y 2 ( z ) , Y 2 Δ ( z ) ) Δ z , μ 0 ) |

≤ ∫ α t L [ P ( τ ) + ∫ α τ c 1 ( τ ) s 1 ( z ) P ( z ) Δ z + ∫ α β c 2 ( τ ) s 2 ( z ) P ( z ) Δ z ] Δ τ   + ∫ α t k ( τ ) | μ − μ 0 | Δ τ   + L [ P ( t ) + ∫ α t c 1 ( t ) s 1 ( z ) P ( z ) Δ z + ∫ α β c 2 ( t ) s 2 ( z ) P ( z ) Δ z ] + k ( t ) | μ − μ 0 | = | μ − μ 0 | k ¯ ( t ) + L ∫ α t [ P ( τ ) + c 1 ( τ ) ∫ α τ s 1 ( z ) P ( z ) Δ z + c 2 ( τ ) ∫ α β s 2 ( z ) P ( z ) Δ z ] Δ τ   + L [ P ( t ) + ∫ α t c 1 ( t ) s 1 ( z ) P ( z ) Δ z + ∫ α β c 2 ( t ) s 2 ( z ) P ( z ) Δ z ]

then we have

P ( t ) ≤ | μ − μ 0 | k ¯ ( t ) 1 − L     + L 1 − L ∫ α t { [ c 1 ( t ) s 1 ( τ ) + 1 ] P ( τ ) + c 1 ( τ ) ∫ α τ s 1 ( z ) P ( z ) Δ z     + c 2 ( τ ) ∫ α β s 2 ( z ) P ( z ) Δ z } Δ τ + L 1 − L c 2 ( t ) ∫ α β s 2 ( z ) P ( z ) Δ z , (3.48)

Now applying theorem 3.1, we have (3.45). □

The authors declare no conflicts of interest regarding the publication of this paper.

Noori, M.I. and Mahmood, A.H. (2020) On a Nonlinear Volterra-Fredholm Integrodifferential Equa- tion on Time Scales. Open Access Library Journal, 7: e6103. https://doi.org/10.4236/oalib.1106103