# Note on Some Nonlinear Integral Inequalities  and Applications to Differential Equations

Authors: Khaled Boukerrioua
Journal: International Journal of Differential Equations
 (2011)

---

## Abstract

Using ideas from Boukerrioua and Guezane-Lakoud (2008), some nonlinear integral inequalities are established.


---

## Body

## 1. Introduction

Integral inequalities provide a very useful and handy device for the study of qualitative as well as quantitative properties of solutions of differential equations. The Gronwall-Bellman type (see, e.g., [1–4]) is particularly useful in that they provide explicit bounds for the unknown functions. One of the most useful inequalities in the development of the theory of diferential equations is given in the following theorem.Theorem 1.1 (see [3]).
Ifu(t) and  f(t) are non-nonnegative continuous functions on [0,∞[ satisfying
(1.1)u2(t)≤c2+2∫0tf(s)u(s)ds,
for some constant c≥0, then
(1.2)u(t)≤c+∫0tf(s)ds,t∈[0,∞[.The importance of this inequality lies in its successful utilization of the situation for which the other available inequalities do not apply directly. It has been frequently used to obtain global existence, uniqueness, stability, boundedness, and other properties of the solution for wide classes of nonlinear differential equations. The aim of this paper is to give other results on nonlinear integral inequalities and their applications.


## 2. Main Results

In this section, we begin by giving some material necessary for our study. We denote byℝ, the set of real numbers and ℝ+ the nonnegative real numbers.Lemma 2.1.
Forx∈ℝ+,y∈ℝ+,1/p+1/q=1, one has
(2.1)x1/py1/q≤xp+yq.Lemma 2.2 (see [1]).
Letb(t) and f(t) be continuous functions for t≥α, let v(t) be a differentiable function for t≥α and suppose
(2.2)v′(t)≤b(t)v(t)+f(t),t≥α,v(α)≤v0.
Then for t≥α,
(2.3)v(t)≤v0exp⁡∫αtb(s)ds+∫αtf(s)(exp⁡∫stb(σ)dσ)ds.Now we state the main results of this workTheorem 2.3.
Letu,a,b,hi(i=1,…,n) be real-valued nonnegative continuous functions and there exists a series of positive real numbers p1,p2,…,pn and u(t) satisfy the following integral inequality,
(2.4)up(t)≤a(t)+b(t)∫0t∑i=1i=nhi(s)upi(s)ds,
for t∈ℝ+ then
(2.5)u(t)≤{a(t)+b(t)∫0t∑i=1i=nhi(s)(pipa(s)+p-pip)×exp⁡(∫stb(σ)∑i=1i=npiphi(σ)dσ)ds}1/P,
for p≥p⋆=max{pi,i=1,…,n}.Proof.
Define a functionv(t) by
(2.6)v(t)=∫0t∑i=1i=nhi(s)upi(s)ds,
then v(0)=0 and (2.4) can be written as
(2.7)up(t)≤a(t)+b(t)v(t).
By (2.7) and Lemma 2.1, we get
(2.8)upi(t)=(up(t))pi/p≤(a(t)+b(t)v(t))pi/p≤pip(a(t)+b(t)v(t))+p-pip.
Differentiating (2.6), we get
(2.9)v′(t)=∑i=1i=nhi(t)upi(t).
Using (2.8) and (2.9), it yields
(2.10)v′(t)≤∑i=1i=nhi(t)(pip(a(t)+b(t)v(t))+p-pip)=A(t)+B(t)v(t),
where
(2.11)B(t)=b(t)∑i=1i=npiphi(t),A(t)=∑i=1i=nhi(t)(pipa(t)+p-pip).
By Lemma 2.2, we have
(2.12)v(t)≤∫0tA(s)exp⁡[∫stB(σ)dσ]ds.
Using (2.7) and (2.12), we get
(2.13)u(t)≤{a(t)+b(t)∫0t∑i=1i=nhi(s)(pipa(s)+p-pip)×exp⁡(∫stb(σ)∑i=1i=npiphi(σ)dσ)ds}1/P.
This achieves the proof of the theorem.Remark 2.4.
if we taken=2,p≥1,p1=p,andp2=1, then the inequality  established in Theorem 2.3 become the inequality given in [4,Theorem 1(a1)].Theorem 2.5.
Suppose that the hypothesis of Theorem2.3 holds. Assume that the function (a(t)+p⋆/p)/b(t) is nondecreasing and
(2.14)up(t)≤a(t)+b(t)∫0t∑i=1i=nhi(s)upi(s)ds,
for t∈ℝ+ then
(2.15)u(t)≤(a(t)+p⋆p)1/p×(1-(p⋆p-1)∫0tb(s)∑i=1i=nhi(t)(a(t)+p⋆p)(p⋆/p)-1ds)1/p-p⋆,
for p⋆=min{pi,i=1,…,n}≤p<p⋆ and t<βp,p⋆,where
(2.16)βp,p⋆=sup⁡{t∈R+:((p⋆/p)-1)∫0tb(s)∑i=1i=nhi(t)(a(t)+p⋆p)(p⋆/p)-1ds≤1}.Proof.
Forp⋆≤p<p⋆
By (2.7) and the fact that (p⋆/p)>1, one gets:
(2.17)upi(t)=(up(t))pi/p≤(a(t)+b(t)v(t))(pi/p)≤(a(t)+b(t)v(t)+p⋆/p)p⋆/p.
Differentiating (2.6) and using (2.17), we obtain
(2.18)v′(t)≤∑i=1i=nhi(t)(a(t)+b(t)v(t)+p⋆p)p⋆/p,
then
(2.19)v′(t)≤(b(t)∑i=1i=nhi(t))(a(t)+(p⋆/p)b(t)+v(t))(a(t)+p⋆p+b(t)v(t))(p⋆/p)-1.
Since the function (a(t)+p⋆/p)/b(t) is nondecreasing, for 0≤t≤τ then,
(2.20)v′(t)≤M(t)(a(τ)+(p⋆/p)b(τ)+v(t)),
where
(2.21)M(t)=b(t)∑i=1i=nhi(t)(a(t)+p⋆p+b(t)v(t))(p⋆/p)-1.
Consequently
(2.22)v(t)+a(τ)+(p⋆/p)b(τ)≤a(τ)+(p⋆/p)b(τ)exp⁡∫0tM(s)ds.
For τ=t, we can see that
(2.23)a(t)+p⋆p+b(t)v(t)≤(a(t)+p⋆p)exp⁡∫0tM(s)ds,
then the function M(t) can be estimated as
(2.24)M(t)≤b(t)∑i=1i=nhi(t)(a(t)+p⋆p)(p⋆/p)-1⋅exp⁡∫0t(p⋆p-1)M(s)ds.
Let
(2.25)L(t)=(p⋆p-1)M(t).
Now we estimate the expression L(t)exp(-∫0tL(s)ds) by using (2.24) to get
(2.26)L(t)exp⁡(∫0t-L(s)ds)≤(p⋆p-1)b(t)∑i=1i=nhi(t)(a(t)+p⋆p)(p⋆/p)-1.
Remarking that
(2.27)L(t)exp⁡(∫0t-L(s)ds)=ddt(-exp⁡(∫0t-L(s)ds))≤(p⋆p-1)b(t)∑i=1i=nhi(t)(a(t)+p⋆p)(p⋆/p)-1,
we integrate (2.27) from 0 to t to get
(2.28)(1-exp⁡∫0t-L(s)ds)≤∫0t(p⋆p-1)b(t)∑i=1i=nhi(t)(a(t)+p⋆p)(p⋆/p)-1ds,
replacing L(t) by its value in (2.28), we obtain
(2.29)(1-exp⁡∫0t(1-p⋆p)M(s)ds)≤∫0t(p⋆p-1)b(t)∑i=1i=nhi(t)(a(t)+p⋆p)(p⋆/p)-1ds,
then
(2.30)exp⁡∫0tM(s)ds≤{1-(∫0t(p⋆p-1)b(t)∑i=1i=nhi(t)(a(t)+p⋆p)(p⋆/p)-1)ds}1/(1-(p⋆/p)).
Using (2.7), (2.23), and (2.30) we have,
(2.31)u(t)≤(a(t)+p⋆p)1/p⋅(1-(p⋆p-1)∫0tb(s)∑i=1i=nhi(t)(a(t)+p⋆p)(p⋆/p)-1ds)1/(p-p⋆).
This achieves the proof of the theorem.Theorem 2.6.
Suppose that the hypothesis of Theorem2.3 holds. Assume that the function (a(t)+p⋆/p)/b(t), is nondecreasing and
(2.32)up(t)≤a(t)+b(t)∫0t∑i=1i=nhi(s)upi(s)ds,
for t∈ℝ+ then
(2.33)u(t)≤(a(t)+p⋆p)1/p×(1-(p⋆p-1)∫0tb(s)∑i=1i=nhi(t)(a(t)+p⋆p)(p⋆/p)-1ds)1/(p-p⋆),
for p<p⋆ and t<βp,p⋆,
(2.34)βp,p⋆=sup⁡{t∈R+:((p⋆/p)-1)∫0tb(s)∑i=1i=nhi(t)(a(t)+p⋆p)(p⋆/p)-1ds≤1}.Proof.
forp<p⋆=min{pi,i=1,…,n}.
Using (2.7), the fact that the function (a(t)+p⋆/p)/b(t) is nondecreasing, and p⋆/p≻1, we have
(2.35)upi(t)=(up(t))pi/p≤(a(t)+b(t)v(t))pi/p≤(a(t)+b(t)v(t)+p⋆p)p⋆/p.
Differentiating (2.6) and using (2.35), we obtain
(2.36)v′(t)≤∑i=1i=nhi(t)(a(t)+b(t)v(t)+p⋆p)p⋆/p,
then
(2.37)v′(t)≤(b(t)∑i=1i=nhi(t))(a(t)+(p⋆/p)b(t)+v(t))(a(t)+p⋆p+b(t)v(t))(p⋆/p)-1,
from the proof of Theorem 2.5, we get the required inequality in (2.33).Remark 2.7.
if we taken=2, then the inequalities established in Theorems 2.5 and 2.6 become the inequalities given in [5, Theorem 1.2].Theorem 2.8.
Suppose that the hypothesis of Theorem2.3 holds and moreover the function  b(t) is decreasing. Let c be a real valued nonnegative continuous and nondecreasing function for t∈ℝ+. If
(2.38)up(t)≤cp(t)+b(t)∫0t∑i=1i=nhi(s)upi(s)ds,
then(1)(2.39)u(t)≤c(t){1+b(t)∫0t∑i=1i=nHi(s)exp⁡(∫stb(σ)∑i=1i=npipHi(σ)dσ)ds}1/p,
for p≥p⋆, where
(2.40)Hi(s)=hi(s)c(s)pi-p.(2)(2.41)u(t)≤c(t)(1+p⋆p)1/p×(1-(p⋆p-1)∫0tb(s)∑i=1i=nHi(s)(1+p⋆p)(p⋆/p)-1ds)1/(p-p⋆),
for p⋆≤p<p⋆ andt<βp,p⋆where
(2.42)βp,p⋆=sup⁡{t∈R+:((p⋆/p)-1)∫0tb(s)∑i=1i=nHi(s)(1+p⋆p)(p⋆/p)-1ds≤1}.(3)(2.43)u(t)≤c(t)(1+p⋆p)1/p×(1-(p⋆p-1)∫0tb(s)∑i=1i=nHi(s)(1+p⋆p)(p⋆/p)-1ds)1/(p-p⋆),
for  and t<βp,p⋆where
(2.44)βp,p⋆=sup⁡{t∈R+:((p⋆/p)-1)∫0tb(s)∑i=1i=nHi(s)(1+p⋆p)(p⋆/p)-1ds≤1}.Proof.
Sincec(t) is a nonnegative, continuous, and nondecreasing function, for t∈ℝ+, from (2.38) we observe that
(2.45)(u(t)c(t))p≤1+b(t)∫0t∑i=1i=nhi(s)c(s)pi-p(u(s)c(s))pids,
we put
(2.46)w(t)=u(t)c(t),Hi(t)=hi(t)c(t)pi-p,
then, we have
(2.47)wp(t)≤1+b(t)∫0t∑i=1i=nHi(s)wpi(s)ds.
Then a direct application of the inequalities established in Theorems2.3, 2.5, and 2.6 gives the required results.Theorem 2.9.
Suppose that the hypothesis of Theorem2.3 holds. Assume that the functions (a(t)+p⋆/p)/b(t),(a(t)+p⋆/p)/b(t) are nondecreasing and let k(t,s) and its derivative partial ∂/∂sk(t,s) be real-valued nonnegative continuous functions, for 0≤s≤t≤∞. If
(2.48)up(t)≤a(t)+b(t)∫0tK(t,s)∑i=1i=nhi(s)upi(s)ds,
then(1)(2.49)u(t)≤{a(t)+b(t)∫0tA(σ)(exp⁡∫σtB(τ)dτ)dσ)}1/p,
where
(2.50)A(t)=K(t,t){∑i=1i=nhi(t)(pipa(t)+p-pip)}+∫0t∂∂sk(t,s){∑i=1i=nhi(s)(pipa(s)+p-pip)}ds,(2.51)B(t)=K(t,t)b(t)∑i=1i=npiphi(t)+∫0t∂∂sk(t,s)b(s)∑i=1i=npiphi(s)ds,
for p≥p⋆.
(2)(2.52)u(t)≤(a(t)+p⋆p)1/p{1-(p⋆p-1)∫0tl(s)(a(s)+p⋆p)(p⋆/p)-1ds}1/(p-p⋆).
For p⋆≤p≤p⋆andt<βp,p⋆,where(2.53)βp,p⋆=sup⁡{t∈R+:((p⋆/p)-1)∫0tl(s)(a(s)+p⋆p)(p⋆/p)-1ds≤1}.(3)(2.54)u(t)≤(a(t)+p⋆p)1/p{1-(p⋆p-1)∫0tl(s)(a(s)+p⋆p)(p⋆/p)-1ds}1/(p-p⋆).
For p<p⋆ and t<βp,p⋆, where
(2.55)βp,p⋆=sup⁡{t∈R+:((p⋆/p)-1)∫0tl(s)(a(s)+p⋆p)(p⋆/p)-1ds≤1},(2.56)l(t)=b(t)K(t,t)∑i=1i=nhi(t)+b(t)1-(p⋆/p)∫0t∂∂sk(t,s)b(s)p⋆/p∑i=1i=nhi(s)ds.Proof.
Let(2.57)v(t)=∫0tK(t,s)∑i=1i=nhi(s)upi(s)ds,
(1) for p≥p⋆.
Differentiating(2.57) we get
(2.58)v′(t)=K(t,t)∑i=1i=nhi(t)upi(t)+∫0t∂∂sk(t,s)∑i=1i=nhi(s)upi(s)ds.
Using (2.8) and (2.58) and the fact that v(t) is nondecreasing, we obtain, for 0≤s≤t,
(2.59)v′(t)≤(K(t,t)b(t)∑i=1i=npiphi(t)+∫0t∂∂sk(t,s)b(s)∑i=1i=npiphi(s)ds)v(t)+K(t,t){∑i=1i=nhi(t)(pipa(t)+p-pip)}+∫0t∂∂sk(t,s){∑i=1i=nhi(s)(pipa(s)+p-pip)}ds,
then
(2.60)v′(t)≤B(t)v(t)+A(t).
By Lemma2.2, we have
(2.61)v(t)≤∫0tA(s)(exp⁡∫stB(σ)dσ)ds,
where
(2.62)A(t)=K(t,t){∑i=1i=nhi(t)(pipa(t)+p-pip)}+∫0t∂∂sk(t,s){∑i=1i=nhi(s)(pipa(s)+p-pip)}ds,B(t)=K(t,t)b(t)∑i=1i=n(pip)hi(t)+∫0t∂∂sk(t,s)b(s)∑i=1i=n(pip)hi(s)ds.
Finally using (2.61) in (2.7), we get the required inequality.(2)
Forp⋆≤p<p⋆. Using (2.17) and (2.58), we get
(2.63)v′(t)≤K(t,t)∑i=1i=nhi(t)(a(t)+p⋆p+b(t)v(t))p⋆/p+∫0t∂∂sk(t,s)(a(s)+p⋆p+b(s)v(s))p⋆/pds,(2.64)v′(t)≤b(t)K(t,t)∑i=1i=nhi(t)(a(t)+(p⋆/p)b(t)+v(t))(a(t)+p⋆p+b(t)v(t))(p⋆/p)-1+∫0t∂∂sk(t,s)×{b(s)p⋆/p∑i=1i=nhi(s)(a(s)+(p⋆/p)b(s)+v(s))(a(s)+(p⋆/p)b(s)+v(s))(p⋆/p)-1}ds.
For 0≤s≤t≤τ, and the fact that (a(t)+p⋆/p)/b(t)is nondecreasing, we have
(2.65)(a(s)+(p⋆/p)b(s)+v(s))≤(a(t)+(p⋆/p)b(t)+v(t))≤(a(τ)+(p⋆/p)b(τ)+v(t)),(a(s)+(p⋆/p)b(s)+v(s))(p⋆/p)-1≤(a(t)+(p⋆/p)b(t)+v(t))(p⋆/p)-1=b(t)1-(p⋆/p)(a(t)+p⋆p+b(t)v(t))(p⋆/p)-1,
then,
(2.66)v′(t)≤M(t)[a(τ)+(p⋆/p)b(τ)+v(t)],
where
(2.67)M(t)=(b(t)K(t,t)∑i=1i=nhi(t)+b(t)1-(p⋆/p)∫0t∂∂sk(t,s)b(s)p⋆/p∑i=1i=nhi(s)ds)×(a(t)+p+b(t)v(t))(p⋆/p)-1.
From the proof of Theorem 2.5, we get the required inequality.(3)
Using (2.35) and (2.58), we get(2.68)v′(t)≤K(t,t)∑i=1i=nhi(t)(a(t)+p⋆p+b(t)v(t))p⋆/p+∫0t∂∂sk(t,s)(a(s)+p⋆p+b(s)v(s))p⋆/pds.
taking account the fact that (a(t)+(p⋆/p))/b(t) is nondecreasing and from the proof of Theorem 2.6, we get the required inequality.Remark 2.10.
if we taken=2,p≥1,p1=p,p2=1, then the inequality  established in Theorem 2.9 (part 1) becomes the inequality given in [4, Theorem 1(a3)].


## 3. Further Results

In this section, we investigate some Gronwall-type inequalities.Theorem 3.1.
Assume thatu(t) and  f(t) are non-nonnegative continuous functions on [0,∞[ and c≥0 is a constant. If k(t,s) is defined as in Theorem 2.9, then
(3.1)up(t)≤c+∫t0tf(τ)[uq(τ)+∫t0τk(τ,s)ur(s)ds]dτ,
implies
(3.2)u(t)≤{c+∫t0tf(τ)[(qpc+p-qp)exp⁡∫t0tA*(s)ds+∫t0tB*(s)(exp⁡∫stA*(σ)dσ)ds]dτ}1/p
where
(3.3)A*(t)={qpf(t)+rq(K(t,s)+∫t0t∂∂sK(t,s)ds)},B*(t)=p-rp{K(t,s)+∫t0t∂∂sK(t,s)ds},
where p≠0,  0≤q≤p and 0≤r≤p.Proof.
Define a functionz(t) by the right side of (3.1) then
(3.4)z′(t)=f(t)[uq(t)+∫t0tK(t,s)ur(s)ds]≤f(t)[qpZ(t)+p-qp+∫t0tk(t,s)(rpZ(t)+p-rp)ds].
Define a function v(t) by
(3.5)v(t)=qpZ(t)+p-qp+∫t0tK(t,s)(rpZ(s)+p-rp)ds.
Then
(3.6)v(t0)=qpc+p-qp,qpZ(t)≤v(t),z′(t)≤f(t)v(t),
and v(t) is nondecreasing for t∈ℝ+.
Then,(3.7)v′(t)=qpz′(t)+K(t,s)(rpZ(t)+p-rp)+∫t0t∂∂sk(t,s)(rpZ(s)+p-rp)ds,
its follow from (3.4), (3.6), and (3.7) that
(3.8)v′(t)≤[{qpf(t)+rq(K(t,s)+∫t0t∂∂sK(t,s)ds)}v(t)+p-rp{K(t,s)+∫t0t∂∂sK(t,s)ds}],
then (3.8) can be written as
(3.9)v′(t)≤A*(t)v(t)+B*(t),
where
(3.10)A*(t)={qpf(t)+rq(K(t,s)+∫t0t∂∂sK(t,s)ds)},B*(t)=p-rp{K(t,s)+∫0t∂∂sK(t,s)ds},
by Lemma 2.2  we obtain
(3.11)v(t)≤(qpc+p-qp)(exp⁡∫t0tA*(s)ds)+∫t0tB*(s)(exp⁡∫stA*(σ)dσ)ds,
from (3.6) and (3.11), it follows that
(3.12)z′(t)≤f(t)[(qpc+p-qp)exp⁡∫t0tA*(s)ds+∫t0tB*(s)(exp⁡∫stA*(σ)dσ)ds],
integrating (3.12), we obtain
(3.13)z(t)≤c+∫t0tf(τ)[(qpc+p-qp)exp⁡∫t0tA*(s)ds+∫t0tB*(s)(exp⁡∫stA*(σ)dσ)ds]dτ,
but
(3.14)u(t)≤(z(t))1/p,
then the result required is found.


## 4. Application

In this section we present some applications of Theorems2.3, 2.5, 2.6 and 3.1 to investigate certain properties of solutions of differential equation.Example 4.1.
We consider a nonlinear differential equation(4.1)(up)′(t)=H(t,u(t)),u(0)=u0.
Assume that p,pi(i=1,…,n)≥0, are fixed real numbers,u0 is a real constant, and H:ℝ+×ℝ+→ℝ+,
(4.2)|H(t,u(t))|≤l(t)+∑i=1i=nhi(t)upi(t),hi,l:ℝ+→ℝ+  are continuous functions.Integrating (4.1), from 0 to t and using (4.2) we obtain (4.3)|u(t)|p≤a(t)+∫0t∑i=1i=nhi(s)|upi(s)|(s)ds,
where a(t)=|u0|p+∫0tl(s)ds. By applying Theorems 2.3, 2.5, and 2.6, we estimate the solution u(t) of the equation, that gives us a bound of the solution.Example 4.2.
Consider the following initial value problem:(4.4)(up(t))′=f(t)[uq(t)+∫t0tK(t,s)ur(s)ds],u(t0)=c,
where f(t) and k(t,s) are as defined in Theorem 3.1, and p≠0,0≤q≤p and 0≤r≤p,andc is a constant.Theorem 4.3.
Assumeu(t) is a solution of (4.4), then
(4.5)|u(t)|≤{|c|p+∫t0tf(τ)[(qpc+p-qp)exp⁡∫t0tA*(s)ds+∫t0tB*(s)(exp⁡∫stA*(σ)dσ)ds]dτ}1/p
where A*(t) and B*(t) are defined in (3.10).Proof.
The solutionu(t) of (4.4) satisfies the following equivalent equation:
(4.6)up(t)=cp+∫t0tf(τ)[uq(τ)+∫t0tk(τ,s)ur(s)ds]dτ.
It follows from (4.6) that
(4.7)|u(t)|p≤|c|p+∫t0tf(τ)[|u(τ)|q+∫t0tk(τ,s)|u(τ)|rds]dτ.
Using Theorem3.1, we obtain (4.5).



---

*Source: 456216-2011-11-16.xml*