In the rest of this paper, for the sake of convenience, we denote T0 = [t0, ∞) ∩T, and always assume T0 ⊂ T κ.
Lemma 2.1 [15]: Assume that a ≥ 0, p ≥ q ≥ 0, and p ≠ 0, then for any K > 0
Lemma 2.2: Suppose u, a ∈ C
rd
,
, m ≥ 0, and a is nondecreasing. Then,
implies
where e
m
(t, t0) is the unique solution of the following equation
Proof: From [[16], Theorem 5.6], we have
, t ∈ T0. Since a(t) is nondecreasing on T0, then
. On the other hand, from [[14], Theorem 2.39 and 2.36 (i)], we have
. Combining the above information, we can obtain the desired inequality.
Theorem 2.1: Suppose u, a, b, f ∈ C
rd
(T0, R+), and a, b are nondecreasing. ω ∈ C(R+, R+) is nondecreasing. τ ∈ (T0, T), τ (t) ≤ t, -∞ < α = inf{τ(t), t ∈ T0} ≤ t0, ϕ ∈ C
rd
([α, t0] ∩T, R+). p > 0 is a constant. If u(t) satisfies, the following integral inequality
with the initial condition
then
where G is an increasing bijective function, and
Proof: Let T ∈ T0 be fixed, and
Then considering a, b are nondecreasing, we have
Furthermore, for t ∈ [t0, T ] ∩T, if τ(t) ≥ t0, considering τ (t) ≤ t, then τ
i
(t) ∈ [t0, T ] ∩T, and from (6) we obtain
If τ(t) ≤ t0, from (2) we obtain
So from (7) and (8), we always have
Moreover,
that is,
On the other hand, for t ∈ [t0, T ] ∩T, if σ(t) > t, then
If σ(t) = t, then
where ξ lies between v(s) and v(t). So we always have
.
Using the statements above, we deduce that
Replacing t with s in the inequality above, and an integration with respect to s from t0 to t yields
where G is defined in (4).
Considering G is increasing, and v(t0) = a(T ), it follows that
Combining (6) and (12), we get
Taking t = T in (12), yields
Since T ∈ T0 is selected arbitrarily, then substituting T with t in (13) yields the desired inequality (3).
Remark 2.1: Since T is an arbitrary time scale, then if we take T for some peculiar cases in Theorem 2.1, then we can obtain some corollaries immediately. Especially, if T = R, t0 = 0, then Theorem 2.1 reduces to [[17], Theorem 2.2], which is the continuous result. However, if we take T = Z, we obtain the discrete result, which is given in the following corollary.
Corollary 2.1: Suppose T = Z, n0 ∈ Z, and Z0 = [n0, ∞) ∩ Z. u, a, b, f ∈ (Z0, R+), and a, b are decreasing on Z0. τ ∈ (Z0, Z), τ (n) ≤ n, -∞ < α = inf{τ(n), n ∈ Z0} ≤ n0, ϕ ∈ C
rd
([α, n0] ∩ Z, R+). ω is defined the same as in Theorem 2.1. If for n ∈ Z0, u(n) satisfies
with the initial condition
then
In Theorem 2.1, if we change the conditions for a, b, ω p; then, we can obtain another bound for the function u(t).
Theorem 2.2: Suppose u, a, b, f ∈ C
rd
(T0, R+), ω ∈ C(R+, R+) is nondecreasing, subadditive, and submultiplicative, that is, for ∀α ≥ 0, β ≥ 0 we always have ω(α + β) ≤ ω(α) + ω (β) and ω(αβ) ≤ ω(α)ω(β). τ, α, ϕ are the same as in Theorem 2.1. If u(t) satisfies the inequality (1) with the initial condition (2), then for ∀K > 0, we have
where
is an increasing bijective function, and
Proof: Let
Then,
Similar to the process of (7)-(9), we have
Considering ω is nondecreasing, subadditive, and submultiplicative, Combining (16), (18), and Lemma 2.1, we obtain
where A(t) is defined in (15).
Let T be fixed in T0, and t ∈ [t0, T] ⋂ T. Denote
Considering A(t) is nondecreasing, then we have
Furthermore,
Similar to Theorem 2.1, we have
Substituting t with s in (22), and an integration with respect to s from t0 to t yields
which is followed by
Combining (17), (21), and (23), we obtain
Taking t = T in (24), yields
Since T is selected from T0 arbitrarily, then substituting T with t in (25), we can obtain the desired inequality (14).
Remark 2.2: Theorem 2.2 unifies some known results in the literature. If we take T = R, t0 = 0, τ(t) = t, K = 1, then Theorem 2.2 reduces to [[18], Theorem 2(b3)], which is one case of continuous inequality. If we take T = Z, t0 = 0, τ(t) = t, K = 1, then Theorem 2.2 reduces to [[18], Theorem 4(d3)], which is the discrete analysis of [[18], Theorem 2(b3)].
Now we present a more general result than Theorem 2.1. We study the following delay integral inequality on time scales.
where u, a, b, f, g, h ∈ Crd(T0, R+), ω ∈ C(R+, R+), and a, b, ω are nondecreasing, η ∈ C(R+, R+) is increasing, τ
i
∈ (T0, T) with τ
i
(t) ≤ t, i = 1, 2, and -∞ < α = inf{min{τ
i
(t), i = 1, 2}, t ∈ T0} ≤ t0.
Theorem 2.3: Define a bijective function
such that
, with
. If
is increasing, and for t ∈ T0, u(t) satisfies the inequality (26) with the initial condition
where ϕ ∈ C
rd
([α, t0] ⋂ T, R+), then
Proof: Let the right side of (26) be v(t), then
For t ∈ T0, if τ
i
(t) ≥ t0, considering τ
i
(t) ≤ t, then τ
i
(t) ∈ T0, and from (29), we have
If τ
i
(t) ≤ t0, from (27), we obtain
So from (30) and (31), we always have
Furthermore, considering η is increasing, we get that
Fix a T ∈ T0, and let t ∈ [t0, T] ⋂ T. Define
Since a, b are nondecreasing on T0, it follows that
On the other hand,
Similar to Theorem 2.1, we have
Replacing t with s, and an integration for (36) with respect to s from t0 to t yields
Since c(t0) = a(T), and G is increasing, it follows that
Combining (29), (35), (38), we have
Taking t = T in (39), yields
Since T ∈ T0 is selected arbitrarily, then substituting T with t in (40) yields the desired inequality (28).
Remark 2.3: If we take η(u) = up, g(t) ≡ 0, then Theorem 2.3 reduces to Theorem 2.1.
Next, we consider the delay integral inequality of the following form.
where u, f, g, h, a, τ
i
, i = 1, 2 are the same as in Theorem 2.3, m ∈ C(R+, R+), p > 0 is a constant, ω ∈ C(R+, R+) is nondecreasing, and ω is submultitative, that is, ω(αβ) ≤ ω(α)ω(β) holds for ∀α ≥ 0, β ≥ 0.
Theorem 2.4: Suppose G ∈ (R+, R) is an increasing bijective function defined as in Theorem 2.1. If u(t) satisfies, the inequality (41) with the initial condition
then
Proof: Let the right side of (41) be v(t). Then,
and similar to the process of (30)-(32) we have
Furthermore,
A suitable application of Lemma 2.2 to (46) yields
Fix a T ∈ T0, and let t ∈ [t0, T] ⋂ T. Define
Then,
and
Similar to Theorem 2.1, we have
An integration for (50) from t0 to t yields
Considering G is increasing and
, it follows
Combining (44), (49), and (51), we have
Taking t = T in (52), yields
Since T ∈ T0 is selected arbitrarily, after substituting T with t in (53), we obtain the desired inequality (43).
Remark 2.4: If we take ω(u) = u, τ1(t) = t, h(t) ≡ 0, then Theorem 2.4 reduces to [[11], Theorem 3]. If we take m(t) = f(t) = h(t) ≡ 0, then Theorem 2.4 reduces to Theorem 2.1 with slight difference.
Finally, we consider the following integral inequality on time scales.
where u, f, g, ω, τ1, τ2 are the same as in Theorem 2.3, p, q, C are constants, and p > q > 0, C > 0.
Theorem 2.5: If u(t) satisfies (54) with the initial condition (42), then
where
, H are two increasing bijective functions, and
Proof: Let the right side of (54) be v(t). Then,
and similar to the process of (30)-(32) we have
Furthermore,
Similar to Theorem 2.1, we have
An integration for (59) from t0 to t yields
Considering
is increasing, and v(t0) = C, then (60) implies
Given a fixed number T in T0, and t ∈ [t0, T]. Let
Then,
and furthermore,
that is,
Integrating (64) from t0 to t yields
Since H is increasing, and
, then (65) implies
Combining (57), (63), and (66), we obtain
Taking t = T in (67), and since T is an arbitrary number in T0, then the desired inequality can be obtained after substituting T with t.
Remark 2.5: If we take T = R, τ1(t) = τ2(t), then we can obtain a new bound of for the unknown continuous function u(t), which is different from the result using the method in [[19], Theorem 2.1].
Remark 2.6: If we take T = R in Theorem 2.3-2.4, or take T = Z in Theorem 2.3-2.5, then immediately we obtain a number of corollaries on continuous and discrete analysis, which are omitted here.