Skip to content
• Research
• Open Access

# Some new nonlinear integral inequalities and their applications in the qualitative analysis of differential equations

Journal of Inequalities and Applications20112011:20

https://doi.org/10.1186/1029-242X-2011-20

• Received: 9 February 2011
• Accepted: 20 July 2011
• Published:

## Abstract

In this paper, some new nonlinear integral inequalities are established, which provide a handy tool for analyzing the global existence and boundedness of solutions of differential and integral equations. The established results generalize the main results in Sun (J. Math. Anal. Appl. 301, 265-275, 2005), Ferreira and Torres (Appl. Math. Lett. 22, 876-881, 2009), Xu and Sun (Appl. Math. Comput. 182, 1260-1266, 2006) and Li et al. (J. Math. Anal. Appl. 372, 339-349 2010).

MSC 2010: 26D15; 26D10

## Keywords

• integral inequality
• global existence
• integral equation
• differential equation
• bounded

## 1 Introduction

During the past decades, with the development of the theory of differential and integral equations, a lot of integral inequalities, for example , have been discovered, which play an important role in the research of boundedness, global existence, stability of solutions of differential and integral equations.

In , the following two theorems for retarded integral inequalities were established.

Theorem A: R+ = [0, ∞). Let u, f, g be nondecreasing continuous functions defined on R+ and let c be a nonnegative constant. Moreover, let ω C(R+, R+) be nondecreasing with ω(u) > 0 on (0, ∞) and α C1(R+, R+) be nondecreasing with α(t) ≤ t on R+. m, n are constants, and m > n > 0. If

where , r > 0,Ω - 1 is the inverse of Ω, Ω (∞) = ∞, and ξ R+ is chosen so that .

Recently, in , the author provided a more general result.

Theorem C: , R+ = (0, ∞). Let f(t, s) and be nondecreasing in t for every s fixed. Moreover, let be a strictly increasing function such that and suppose that is a nondecreasing function. Further, let be nondecreasing with {η, ω}(x) > 0 for x (0, ∞) and , with x0 defined as below. Finally, assume that is nondecreasing with α(t) ≤ t. If satisfies

where with xc(0) > x0 > 0 if and xc(0) > x0 ≥ 0 if

Here G -1 and ψ -1 are inverse functions of G and ψ, respectively.

In , Xu presented the following two theorems:

Theorem D: R+ = [0, ∞). Let u, f, g be real-valued nonnegative continuous functions defined for x ≥ 0, y ≥ 0 and let c be a nonnegative constant. Moreover, let ω C(R+, R+) be nondecreasing with ω(u) > 0 on (0, ∞) and α, β, C1(R+, R+) be nondecreasing with α(x) ≤ x, β(y) ≤ y on R+. m, n are constants, and m > n > 0. If

In this paper, motivated by the above work, we will prove more general theorems and establish some new integral inequalities. Also we will give some examples so as to illustrate the validity of the present integral inequalities.

## 2 Main results

In the rest of the paper we denote the set of real numbers as R, and R+ = [0, ∞) is a subset of R. Dom(f) and Im(f) denote the definition domain and the image of f, respectively.

Theorem 2.1: Assume that x, a C(R+, R+) and a(t) is nondecreasing. f i , g i , h i , t f i , t g i , t h i C(R+ × R+, R+), i = 1, 2. Let ω C(R+, R+) be nondecreasing with ω(u) > 0 on (0, ∞). p, q are constants, and p > q > 0. If α C1(R+, R+) is nondecreasing with α(t) ≤ t on R+, and
then there exists such that for where Proof: The proof for the existence of can be referred to Remark 1 in . We notice (3) obviously holds for t = 0. Now given an arbitrary number , for t (0, T], we have
Let the right-hand side of (7) be y(t). Then we have , , and

Considering is arbitrary, substituting T with t, and then the proof is complete.

Remark 1 : We note that the right-hand side of (2) is well defined since Ω (∞) = ∞.

Remark 2 : If we take p = 2, q = 1, ω(u) = u, h1(s, t) = h2(s, t) ≡ 0 or p = 2, q = 1, h1(s, t) = h2(s, t) ≡ 0, respectively, then our Theorem 2.1 reduces to [12, Theorems 2.1, 2.2].

Corollary 2.1: Assume that x, a, α, ω, Ω are defined as in Theorem 2.1. f i , g i , h i C(R+, R+), m i , n i , l i C1(R+, R+), i = 1, 2. If
then we can find some such that for Remark 3: If , m1(t) = n1(t) ≡ 1, l1(t) ≡ 0, m2(t) = n2(t) = l2(t) ≡ 0 for t R+, then Corollary 1 reduces to Theorem A [9, Theorem 2.1]. If , m1(t) ≡ 1, g1(t) ≡ 0, l1(t) ≡ 0, m2(t) ≡ 1, n2(t) = l2(t) ≡ 0 for t R + , then Corollary 2.1 reduces to Theorem B [9, Theorem 2.2].

Corollary 2:2: Assume that x, a, α, ω, Ω are defined as in Theorem 2.1. f, g, h, t f, t g, t h C(R+ × R+, R+). If
then for Corollary 2:3: Assume that x, a, α, ω, Ω are defined as in Theorem 2.1. f, g, h C(R+, R+), m, n, l C1(R+, R+). If
then for Motivated by Corollary 2.2 and Theorem C , we will give the following more general theorem:

Theorem 2:2: Assume that f(s, t), g(s, t), h(s, t) C(R+ × R+, R+) are nondecreasing in t for each s fixed, and ϕ C(R+, R+) is a strictly increasing function with . ψ, ω C(R+, R+) are nondecreasing with ψ (x) > 0, ω(x) > 0 for x (0, ∞) and , a(t), α(t) are defined as in Theorem 2.1, and a(0) > t0 > 0. If x C(R+, R+) satisfies the following integral inequality containing multiple integrals
then we can find some such that for Proof: The proof for the existence of can be referred to Remark 1 in . We notice (22) obviously holds for t = 0. Now given an arbitrary number T > 0, . Define
Integrating (27) from 0 to t, considering J is increasing, we can obtain
Define , then
Integrating (31) from 0 to t, considering and Y is increasing, it follows

Considering is arbitrary, substituting T with t we have completed the proof.

Remark 4: If h(s, t) ≡ 0, α1(t) = α2(t) = α(t), then Theorem 2.2 becomes Theorem C [10, Theorem 1].

Now we will apply the concept of establishing Theorem 2.2 to the situation with two independent variables.

Theorem 2:3: Assume that f i (x, y), g i (x, y), h i (x, y) C(R+ × R+, R+), i = 1, 2, and ϕ C(R+, R+) is a strictly increasing function with . a(x, y) C(R+ × R+, R+) is nondecreasing in x for every fixed y and nondecreasing in y for every fixed x. α(x), β(y) C1(R+, R+) are nondecreasing with α(x) ≤ x, β(y) ≤ y. ψ, ω C(R+, R+) are nondecreasing with ψ(x) > 0, ω(x) > 0 for x (0, ∞) and , where 0 < t0 < a(0, 0).

If u C(R+ × R+, R+) satisfies the following integral inequality containing multiple integrals
then we can find some , so that for all , Proof: The process for seeking for , can also be referred to Remark 1 in .

If we take x = 0 or y = 0, then (35) holds trivially. Now fix , , and x (0, x0], y (0, y0]. Let
Considering a(x, y) is nondecreasing, we have u(x, y) ≤ ϕ -1(z(x, y)) ≤ ϕ -1(z(x0, y)). Moreover,
Integrating (42) from 0 to y, considering we have

Since , are arbitrary, substitute x0, y0 with x, y and the proof is complete.

Corollary 2.4: Assume that f(x, y), g(x, y), h(x, y) C(R+ × R+, R+), and a, ϕ, ψ, ω, α, β, J, Y are defined as in Theorem 2.3. If u C(R+ × R+, R+) satisfies the following integral inequality containing multiple integrals
then we can find some , such that for all , Remark 5: If we take h(x, y) ≡ 0, ψ (u(x, y)) = u n (x, y), ϕ(x, y) = u m (x, y), m > n > 0, then Corollary 2.4 reduces to Theorem D [11, Theorem 2.1].

Corollary 2.5: Assume that f i , g i (x, y) C(R+ × R+, R+), i = 1, 2, and a, ϕ, ψ, ω, J, Y are defined as in Theorem 2.3. If u C(R+ × R+, R+) satisfies the following integral inequality containing multiple integrals
then we can find some , such that for all , Remark 6: If we take f1(x, y) = f2(x, y) ≡ 0, ψ(u(x, y)) = u n (x, y), ϕ(x, y) = u m (x, y), m > n, then Corollary 8 reduces to Theorem E [11, Theorem 2.2].

## 3 Applications

In this section, we will present two examples in order to illustrate the validity of the above results. In the first example, we will try to prove the global existence of the solutions of a delay differential equation, while in the second example, we will obtain the bound of the solutions of an integral equation.

For the sake of proving the global existence of solutions of differential equations, we first recall some basic facts. Consider the following equation

with X0 R n , H C(R+ × R2n, R n ), α C1(R+, R+) satisfying α(t) ≤ t for t ≥ 0. A result in  guarantees that for every X0 R n , Equation 45 has a solution, but the uniqueness of solutions cannot be guaranteed. Furthermore, every solution of (45) has a maximal time of existence T > 0, and if T < ∞, then .

where p is an even number. α(t) is a nondecreasing function, α(t) C1(R+, R+), α(t) ≤ t, t ≥ 0. p > q > 0. Assume , , , where , , and v is nondecreasing.

If (x(t), y(t)) is a solution of (46) defined on the maximal existence interval [0, T), integrating (47) from 0 to t, we have
where . From Theorem 2.1 we have

Obviously we have {|x(t)|, |y(t)|} ≤ |u(t)|. So x(t), y(t) do not blow up in finite time. Then T = ∞, and the solutions of (46) are global.

where u C(R+ × R+, R+), |F (x, y, u(x, y))| ≤ f(x, y)u(x, y),

|G(x, y, u(x, y))| ≤ g(x, y)u2(x, y), f, g C(R+ × R+, R+), a(x, y), α(x), β(y) are defined as in Theorem 2.3.

Let ϕ(u) = uln(u + 1), ω (u) = u, η(u) = u. Then one can easily see the conditions of Theorem 2.3 are satisfied. So we can obtain the bound of u(x, y) as
where , are determined similar to the process in Theorem 2.3, and

Remark 7: we note that the methods in  are not available for the estimate of bound for the solution of Equation 48.

## 5 Authors'contributions

BZ carried out the main part of this article. Both of the two authors read and approved the final manuscript.

## Declarations

### Acknowledgements

The authors thank the referees very much for their valuable suggestions on this paper.

## Authors’ Affiliations

(1)
School of Science, Shandong University of Technology, Zibo, Shandong, 255049, China
(2)
School of Mathematical Sciences, Qufu Normal University, Qufu, Shandong, 273165, China

## References

Advertisement 