In the rest of this paper, for the sake of convenience, we denote **T**_{0} = [*t*_{0}, ∞) ∩**T**, and always assume **T**_{0} ⊂ **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*, *t*_{0}) is the unique solution of the following equation

**Proof**: From [[16], Theorem 5.6], we have , *t* ∈ **T**_{0}. Since *a*(*t*) is nondecreasing on **T**_{0}, 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
}(**T**_{0}, **R**_{+}), and *a*, *b* are nondecreasing. *ω* ∈ *C*(**R**_{+}, **R**_{+}) is nondecreasing. *τ* ∈ (**T**_{0}, **T**), *τ* (*t*) ≤ *t*, -∞ < *α* = inf{*τ*(*t*), *t* ∈ **T**_{0}} ≤ *t*_{0}, *ϕ* ∈ *C*_{
rd
}([*α*, *t*_{0}] ∩**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* ∈ **T**_{0} be fixed, and

Then considering *a*, *b* are nondecreasing, we have

Furthermore, for *t* ∈ [*t*_{0}, *T* ] ∩**T**, if *τ*(*t*) ≥ *t*_{0}, considering *τ* (*t*) ≤ *t*, then *τ*_{
i
}(*t*) ∈ [*t*_{0}, *T* ] ∩**T**, and from (6) we obtain

If *τ*(*t*) ≤ *t*_{0}, from (2) we obtain

So from (7) and (8), we always have

Moreover,

that is,

On the other hand, for *t* ∈ [*t*_{0}, *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 *t*_{0} to *t* yields

where *G* is defined in (4).

Considering *G* is increasing, and *v*(*t*_{0}) = *a*(*T* ), it follows that

Combining (6) and (12), we get

Taking *t* = *T* in (12), yields

Since *T* ∈ **T**_{0} 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**, *t*_{0} = 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**, *n*_{0} ∈ **Z**, and **Z**_{0} = [*n*_{0}, ∞) ∩ **Z**. *u*, *a*, *b*, *f* ∈ (**Z**_{0}, **R**_{+}), and *a*, *b* are decreasing on **Z**_{0}. *τ* ∈ (**Z**_{0}, **Z**), *τ* (*n*) ≤ *n*, -∞ *< α* = inf{*τ*(*n*), *n* ∈ **Z**_{0}} ≤ *n*_{0}, *ϕ* ∈ *C*_{
rd
}([*α*, *n*_{0}] ∩ **Z**, **R**_{+}). *ω* is defined the same as in Theorem 2.1. If for *n* ∈ **Z**_{0}, *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
}(**T**_{0}, **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 **T**_{0}, and *t* ∈ [*t*_{0}, *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 *t*_{0} 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 **T**_{0} 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**, *t*_{0} = 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**, *t*_{0} = 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*(**T**_{0}, **R**_{+}), *ω* ∈ *C*(**R**_{+}, **R**_{+}), and *a*, *b*, *ω* are nondecreasing, *η* ∈ *C*(**R**_{+}, **R**_{+}) is increasing, *τ*_{
i
} ∈ (**T**_{0}, **T**) with *τ*_{
i
}(*t*) ≤ *t*, *i* = 1, 2, and -∞ < *α* = inf{min{*τ*_{
i
}(*t*), *i* = 1, 2}, *t* ∈ **T**_{0}*}* ≤ *t*_{0}.

**Theorem 2.3**: Define a bijective function such that , with . If is increasing, and for *t* ∈ **T**_{0}, *u*(*t*) satisfies the inequality (26) with the initial condition

where *ϕ* ∈ *C*_{
rd
}([*α*, *t*_{0}] ⋂ **T**, **R**_{+}), then

**Proof**: Let the right side of (26) be *v*(*t*), then

For *t* ∈ **T**_{0}, if *τ*_{
i
}(*t*) ≥ *t*_{0}, considering *τ*_{
i
}(*t*) ≤ *t*, then *τ*_{
i
}(*t*) ∈ **T**_{0}, and from (29), we have

If *τ*_{
i
}(*t*) ≤ *t*_{0}, from (27), we obtain

So from (30) and (31), we always have

Furthermore, considering *η* is increasing, we get that

Fix a *T* ∈ **T**_{0}, and let *t* ∈ [*t*_{0}, *T*] ⋂ **T**. Define

Since *a*, *b* are nondecreasing on **T**_{0}, 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 *t*_{0} to *t* yields

Since *c*(*t*_{0}) = *a*(*T*), and *G* is increasing, it follows that

Combining (29), (35), (38), we have

Taking *t* = *T* in (39), yields

Since *T* ∈ **T**_{0} is selected arbitrarily, then substituting *T* with *t* in (40) yields the desired inequality (28).

**Remark 2.3**: If we take *η*(*u*) = *u*^{p}, *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* ∈ **T**_{0}, and let *t* ∈ [*t*_{0}, *T*] ⋂ **T**. Define

Then,

and

Similar to Theorem 2.1, we have

An integration for (50) from *t*_{0} 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* ∈ **T**_{0} 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*, *τ*_{1}(*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*, *τ*_{1}, *τ*_{2} 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 *t*_{0} to *t* yields

Considering is increasing, and *v*(*t*_{0}) = *C*, then (60) implies

Given a fixed number *T* in **T**_{0}, and *t* ∈ [*t*_{0}, *T*]. Let

Then,

and furthermore,

that is,

Integrating (64) from *t*_{0} 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 **T**_{0}, then the desired inequality can be obtained after substituting *T* with *t*.

**Remark 2.5**: If we take **T** = **R**, *τ*_{1}(*t*) = *τ*_{2}(*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.