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Kyungpook Mathematical Journal 2020; 60(1): 211-222

Published online March 31, 2020

Copyright © Kyungpook Mathematical Journal.

On Some Fractional Quadratic Integral Inequalities

Ahmed M. A. El-Sayed, Hind H. G. Hashem∗

Department of Mathematics and Computer Science, Faculty of Science, Alexandria University, Alexandria, Egypt
e-mail : amasayed@alexu.edu.eg
Department of Mathematics, Faculty of Science, Qassim University, Buraidah, Saudi Arabia
e-mail : hendhghashem@yahoo.com

Received: September 5, 2018; Revised: March 17, 2019; Accepted: March 18, 2019

Integral inequalities provide a very useful and handy tool for the study of qualitative as well as quantitative properties of solutions of differential and integral equations. The main object of this work is to generalize some integral inequalities of quadratic type not only for integer order but also for arbitrary (fractional) order. We also study some inequalities of Pachpatte type.

Keywords: fractional-order integral inequality, fractional quadratic integral equations, Bellman-Gronwall’s inequality, inequality of Pachpatte type

In the theory of differential and integral equations, Gronwall’s lemma has been widely used in various applications since its first appearance in the article by Bellman in 1943. In this article the author gave a fundamental lemma, known as Gronwall-Bellman Lemma, which is used to study the stability and asymptotic behavior of solutions of differential equations. Gronwall’s lemma has seen several generalizations to various forms [3, 11, 28].

The literature on these inequalities and their applications is vast; see [1, 2, 4] and the references given therein [19, 20, 23]. In addition, as the theory of calculus on time scales has developed over the last few years, some Gronwall-type integral inequalities on time scales have been established by many authors [17, 21, 24].

In this work, we shall study some fractional integral inequalities which can be used to prove the uniqueness of solutions for differential and integral equations of fractional-order. These inequalities are similar to Bellman-Gronwall type inequalities which play a fundamental role in the qualitative as well as quantitative study of differential equations.

Let L1 = L1[a, b] be the class of Lebesgue integrable functions on [a, b] with the standard norm.

Now, we shall introduce the definitions of the fractional-order integral operators (see [18, 25, 26, 27]). Let β be a positive real number.

Definition 1.1

The left-sided fractional integral of order β of the function f is defined on [a, b] by

Ia+βf(t)=at(t-s)β-1Γ(β)f(s)ds,t>a

and when a = 0, we have I0+βf(t), t>0.

Definition 1.2

The right-sided fractional integral of order β of the function f is defined on [a, b] by

Ib-βf(t)=tb(s-t)β-1Γ(β)f(s)ds,         t<b

and when b = 0, we have I0-βf(t), t<0.

For further properties of fractional calculus see [18, 25, 26, 27].

The existence of solutions of nonlinear quadratic integral equations and some properties of their solutions have been well studied recently (see [5, 6, 7, 8, 9, 10] and [12, 13, 14, 15, 16]).

Let α, β > 0. The existence of solutions of the quadratic integral equation of arbitrary (fractional) orders α and β

x(t)=h(t)+Ia+αf(t,x(t))Ia+βg(t,x(t)),         t>a

have been studied in [16] and [14] under the following assumptions.

  • The functions f, g : [a, b] × R+R+ satisfy the Carathèodory condition (i.e. are measurable in t for all xR+ and continuous in x for all t ∈ [a, b]). Moreover, there exist two functions m1,m2L1, and a positive constant k such that

    f(t,x)m1(t)+kx(t),         g(t,x)m2(t),(t,x)[a,b]×R+.

  • There exists a positive constant M2 such that Ia+γm2(t)M2, γ <β.

  • There exists a positive constant M1 such that Ia+σm1(t)M1, σ<α.

  • hL1.

Now, consider the nonlinear quadratic integral inequality of fractional orders

x(t)h(t)+Ia+αf(t,x(t))Ia+βg(t,x(t)),t>a,   α,   β>0

Theorem 2.1

Let assumptions (i)–(iv) be satisfied. Let x(t) ∈ L1and satisfies inequality(2.2)for almost all t ∈ [a, b]. IfkM2bα+β-γΓ(β-γ+1)Γ(α+1)<1, then

x(t)n=0(kM2bβ-γΓ(β-γ+1)Ia+α)n[h(t)+M1M2bβ-γ+α-σΓ(β-γ+1)Γ(α-σ+1)].
Proof

Using assumptions (i)–(iv) and inequality (2.2), we have

x(t)h(t)+Ia+α(m1(t)+kx(t)).Ia+βm2(t),t>a,α,β>0x(t)h(t)+Ia+α-σIa+σm1(t).Ia+β-γIa+γm2(t)+kIa+αx(t).Ia+β-γIa+γm2(t),x(t)h(t)+M1M2bβ-γ+α-σΓ(β-γ+1)Γ(α-σ+1)+kM2bβ-γΓ(β-γ+1)Ia+αx(t),t>a,α>0(I-kM2bβ-γΓ(β-γ+1)Ia+α)x(t)h(t)+M1M2bβ-γ+α-σΓ(β-γ+1)Γ(α-σ+1)

Since

kM2bβ-γΓ(β-γ+1)Ia+αx(t)=abkM2bβ-γΓ(β-γ+1)Ia+αx(t)dtabkM2bβ-γΓ(β-γ+1)at(t-s)α-1Γ(α)x(s)dsdtkM2bβ-γΓ(β-γ+1)absb(t-s)α-1Γ(α)dtx(s)dskM2bα+β-γΓ(β-γ+1)Γ(α+1)abx(s)ds

and kM2bα+β-γΓ(β-γ+1)Γ(α+1)<1, then

kM2bβ-γΓ(β-γ+1)Ia+αx(t)<x(t)         and         (I-kM2bβ-γΓ(β-γ+1)Ia+α)-1exists.

Then

x(t)n=0(kM2bβ-γΓ(β-γ+1)Ia+α)n[h(t)+M1M2bβ-γ+α-σΓ(β-γ+1)Γ(α-σ+1)].

Remark 2.2

If we replace assumptions (ii) and (iii) by

  • (ii*) There exists a positive constant M2 such that Ia+βm2(t)M2;

  • (iii*) There exists a positive constant M1 such that Ia+αm1(t)M1,

then we can obtain the following result.

Theorem 2.3

Let assumptions (i), (ii*), (iii*) and(iv) be satisfied. Let x(t) ∈ L1satisfies the inequality(2.2)for almost all t ∈ [a, b]. IfkM2bα+βΓ(β+1)Γ(α+1)<1, then

x(t)n=0(kM2bβΓ(β+1)Ia+α)n[h(t)+M1M2bβ+αΓ(β+1)Γ(α+1)].

Inequality (2.2) involves many integral inequalities of fractional order. So, some particular cases can be obtained as follows.

Corollary 2.4

Let f, g : [a, b] × R+R+are continuous functions, and hL1. If x(t) ∈ L1and satisfies the inequality (2.2) for almost all t ∈ [a, b]. Then

x(t)h(t)+M1M2bβ+αΓ(β+1)Γ(α+1),

whereM1=supt[a,b]g(t,x),M2=supt[a,b]f(t,x).

Corollary 2.5

Let assumptions (i)–(iv) (with α = β, f = g) be satisfied. Let x(t) ∈ L1and satisfies the inequality

x(t)h(t)+(Ia+αf(t,x(t)))2,   t>a,α>0

for almost all t ∈ [a, b]. IfkM2b2α-γΓ(α-γ+1)Γ(α+1)<1, then

x(t)n=0(kM2bα-γΓ(α-γ+1)Ia+α)n[h(t)+M22b2α-γΓ(α-γ+1)Γ(α+1)].

Corollary 2.6

Let assumptions (i)–(iv) (with α = β, f = g and h = 0) be satisfied. Let x(t) ∈ L1and satisfies the inequality

x(t)Ia+αf(t,x(t)),t>a,         α>0

for almost all t ∈ [a, b]. IfkM2b2α-γΓ(α-γ+1)Γ(α+1)<1, then

x(t)n=0(kM2bα-γΓ(α-γ+1)Ia+α)n[M22b2α-γΓ(α-γ+1)Γ(α+1)].

Letting α, β → 1, then we have

Corollary 2.7

Let h(t), f(t, x) and g(t, x) satisfy the assumptions of Theorem 2.3. Let x(t) ∈ L1and satisfies the quadratic integral inequality

x(t)h(t)+atf(t,x(t))dt.atg(t,x(t))dt,t>a

for almost all t ∈ [a, b]. If kM2b2 < 1, then

x(t)n=0(kM2bIa+α)n[h(t)+M1M2b2].

Letting β → 0, then we have

Corollary 2.8

Let h(t), f(t, x) and g(t, x) satisfy the assumptions of Theorem 2.3. Let x(t) ∈ L1and satisfies the quadratic integral inequality

x(t)h(t)+g(t,x(t))Ia+αf(t,x(t)),t>a,         α>0

for almost all t ∈ [a, b]. Ifk.M2bαΓ(α+1)<1t[a,b], then

x(t)n=0(km2(t)Ia+α)n[h(t)+M1m2(t)].
Proof

Using assumptions of Theorem 2.3, then inequality (2.3) becomes

x(t)h(t)+m2(t).Ia+α(m1(t)+kx(t)),t>a,α,β>0x(t)h(t)+m2(t)Ia+αm1(t)+km2(t)Ia+αx(t),x(t)h(t)+M1m2(t)+km2(t).Ia+αx(t),   t>a,α>0         (I-km2(t)Ia+α)x(t)h(t)+M1.m2(t).

Similarly

km2(t)Ia+αx(t)k.M2bαΓ(α+1)x(t)

Then

x(t)n=0(km2(t)Ia+α)n[h(t)+M1m2(t)].

In the same fashion, we can prove the following corollary.

Corollary 2.9

Let h(t), f(t, x) satisfy assumptions of Corollary 2.8 and g(t, x) = 1. Let x(t) ∈ L1and satisfies the inequality

x(t)h(t)+Ia+αf(t,x(t)),t>a,α>0

for almost all t ∈ [a, b]. IfkbαΓ(α+1)<1, then

x(t)n=0(kIa+α)n[h(t)+M1].

Letting f(t, x(t)) = m(t)x(t), β → 0 and g(t, x(t)) = 1, then we have the following result.

Corollary 2.10

Let h(t),m(t) ∈ L1and m(t) > 0. Let x(t) ∈ L1and satisfies the inequality

x(t)h(t)+Ia+αm(t)x(t),t>a,α>0

for almost all t ∈ [a, b]. If Mbα < Γ(α + 1), where M is a positive constant such thatsupt[a,b]m(t)=M, then

x(t)1m(t)j=0(m(t)Ia+α)jm(t)h(t),t>a.

When m(t) is continuous, then we get the following corollary.

Corollary 2.11

Let h(t) ∈ L1, m(t) > 0 and m(t) is continuous on [a, b]. Let x(t) ∈ L1and satisfies(2.5)for almost all t ∈ [a, b]. If bαM <Γ(α + 1), then

x(t)i=0MiIa+αih(t),   t>a

wheresupt[a,b]m(t)=M.

Some special cases will be considered, when m(t) = K,K ≠ 0.

Corollary 2.12

Let h(t) ∈ L1and x(t) ∈ L1satisfying the inequality

x(t)h(t)+KIa+αx(t),   t>a,         α>0

for almost all t ∈ [a, b]. If bαK <Γ(α + 1), then

x(t)i=0KiIa+αih(t),   t>a.

Corollary 2.13

Let 0 ≤ βα,A ≥ 0 and let x(t) ∈ L1satisfying the inequality

x(t)At-β+KI0+αx(t),   t>0,         α>0

for almost all t ∈ [0, b]. If bαK <Γ(α + 1), then

x(t)CAt-β,   t>0

where C depends only on K, α and b.

For h(t) = 0, we have the following corollary.

Corollary 2.14

Let x(t) ∈ L1and x(t) > 0 satisfying

x(t)KIa+αx(t),t>a,         α>0

for almost all t ∈ [a, b]. If bαK <Γ(α + 1), then

x(t)=0,   t>a.

Now, if we reverse the inequality (2.5) we shall obtain the next lemma.

Lemma 2.15

Let h(t), k(t) ∈ L1and k(t) > 0. Let x(t) ∈ L1and satisfies the inequality

x(t)h(t)-Ia+αk(t)x(t),t>a,         α>0

for almost all t ∈ [a, b]. If there exist a function mL1and a positive constant M such thatIb-βm(t)M, β < α. Moreover, Mbαβ < Γ(αβ + 1). Then

x(t)1k(t)i=0(-k(t)Ia+α)ik(t)h(t),t>a.
Proof

Multiply both sides of (2.6) by k(t), then

k(t)x(t)k(t)h(t)-k(t)Ia+αk(t)x(t)

and setting y(t) = k(t) x(t), we get

y(t)k(t)h(t)-k(t)Ia+αy(t)         (I-(-k(t)Ia+α))y(t)k(t)h(t)

Since

(-k(t)Ia+αy(t))=ab-k(t)Ia+αy(t)dtab-k(t)at(t-s)α-1Γ(α)y(s)dsdtabsb(t-s)α-1Γ(α)k(t)dty(s)dsabIb-α-βIb-βm(t)dty(s)dsMabsb(t-s)α-β-1Γ(α-β)dty(s)dsMbα-βΓ(α-β+1)aby(s)ds.

Then

(-k(t))Ia+αy(t)<y(t)         (-k(t))Ia+α<1      and         (I-(-k(t))Ia+α)-1         exists.

Then

y(t)i=0(-k(t))Ia+α)ik(t)h(t),

and therefore x(t) is estimated by

x(t)1k(t)i=0(-k(t))Ia+α)ik(t)h(t).

The next corollaries give particular cases for inequality (2.6).

Corollary 2.16

Let h(t) ∈ L1, k(t) > 0 and k(t) is continuous on [a, b]. Let x(t) ∈ L1and satisfies(2.6)for almost all t ∈ [a, b]. If bαM <Γ(α + 1), then

x(t)i=0MiIa+αih(t),   t>a,

wheresupt[a,b]k(t)=M.

Corollary 2.17

Let h(t) ∈ L1and x(t) ∈ L1satisfying the inequality

x(t)h(t)-kIa+αx(t),   t>a,         α>0

for almost all t ∈ [a, b]. If bαk < Γ(α + 1), then

x(t)i=0(-k)iIa+αih(t),   t>a,.

Corollary 2.18

Let 0 ≤ βα,A ≥ 0 and let x(t) ∈ L1satisfying the inequality

x(t)At-β-kI0+αx(t),   t>a,         α>0

for almost all t ∈ [0, b]. If bαk < Γ(α + 1), then

x(t)CAt-β,   t>0.

where C depends only on k,α and b.

Corollary 2.19

Let x(t) ∈ L1and satisfies the inequality

x(t)kIa+αx(t),         α>0

for almost all t ∈ [a, b]. If bαk < Γ(α + 1), then

x(t)0,t>0.

Remark 2.20

Clearly, we can obtain similar results if we replace the left-sided fractional-order integralIa+α by the right-sided fractional-order integralIb-α in inequality (2.2).

Over the years integral inequalities have become an important tool in the analysis of various differential and integral equations. These inequalities are useful in investigating the asymptotic behavior and the stability on the solutions of integral equations. Pachpatte [22] gave a new integral inequality and studied the boundedness, asymptotic behavior and growth of the solutions of an integral equation using the inequality. We introduce this inequality as follows.

Theorem 3.1.([22])

Let u, f, g be real-valued nonnegative continuous functions defined on R+, and c1, c2be nonnegative constants. If

u(t)(c1+0tf(s)u(s)ds)(c2+0tg(s)u(s)ds)

andc1c20tR(s)Q(s)ds<1for all tR+, then

u(t)c1c2Q(t)1-c1c20tR(s)Q(s)ds,tR+,

where

R(t)=0t[f(t)g(s)+f(s)g(t)]ds,         Q(t)=exp(0t[c1g(s)+c2f(s)]ds.)

Now consider the quadratic inequality of fractional order

x(t)(c1+Ia+αf(t,x(t)))(c2+Ia+βg(t,x(t))),         t>a,   α,   β>0

Theorem 3.2

Let assumptions (i), (ii*), (iii*) and (iv) be satisfied. Let x(t) ∈ L1satisfies the inequality(3.1)for almost all t ∈ [a, b]. Ifk(c2+M1)bαΓ(α+1)<1, then

x(t)n=0(k(c2+M1)Ia+α)n(c1c2+c1M1+M2cs2+M1M2).x(t)n=0(k(c2+M1))ntnα(c1c2+c1M1+M2cs2+M1M2)Γ(αn+1).
Proof

The proof can straight forward as in Theorem 2.1.

When f(t, x) = m(t)x(t) and g(t, x) = k(t)x(t), then we obtain Pachpatte inequality which is studied in [22].

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