NONLINEAR INTEGRAL INEQUALITIES INVOLVING ... - Ele-Math

M athematical 

Inequalities 

& Applications 

Volume 12, Number 4 (2012), 811–825 

NONLINEAR INTEGRAL INEQUALITIES INVOLVING 

MAXIMA OF THE UNKNOWN SCALAR FUNCTIONS 

M. BOHNER, S. HRISTOVA AND K. STEFANOVA 

(Communicated by J. Pečarić) 

Abstract. This paper deals with some nonlinear integral inequalities that involve the maximum 

of the unknown scalar function of one variable. The considered inequalities are generalizations 

of the classical integral inequality of Gronwall–Bellman. The importance of these integral inequalities 

is due to their wide applications in qualitative investigations of differential equations 

with “maxima”, and it is illustrated by some direct applications. 

1. Introduction 

In the past few years, a number of integral inequalities was established by many 

scholars, which are motivated by certain applications such as existence, uniqueness, 

continuous dependence, comparison, perturbation, boundedness and stability of solutions 

of differential and integral equations (see, for example, [4, 7, 12, 13, 15] and the 

references cited therein). Among these integral inequalities, we cite the famous Gronwall 

inequality and its various generalizations [1, 2, 5, 10, 9, 8]. 

In the last few decades, great attention has been paid to automatic control systems 

and their applications to computational mathematics and modeling. Many problems 

in control theory correspond to the maximal deviation of the regulated quantity (see 

[14]). Such kind of problems could be adequately modeled by differential equations 

that contain the maxima operator. Note that such equations involving “maxima” of the 

unknown function are called differential equations with “maxima”, see [3, 6]. In his 

survey [11], A. D. Mishkis also points out the necessity to study differential equations 

with “maxima”. 

The purpose of this paper is to establish some new nonlinear integral inequalities in 

the case when the “maxima” of the unknown scalar function is involved in the integral. 

Several cases depending on the type of nonlinearity are considered. These inequalities 

are mathematical tools in the theory of differential equations with “maxima”. Their 

importance is illustrated by some direct applications obtaining bounds of the solutions. 

Mathematics subject classification (2010): 26D10, 34D40. 

Keywords and phrases: Integral inequalities, maxima, scalar functions of one variable, differential 

equations with “maxima”. 

c○Ð, Zagreb 

811 

Paper MIA-15-69

812 M. BOHNER, S. HRISTOVA AND K. STEFANOVA 

∞. 

2. Main results 

Let h > 0 be a constant and suppose t0 and T are fixed points with 0 t0 < T 

DEFINITION 2.1. The function α ∈ C 1 ([t0,T),R+) is said to be from the class 

F if it is nondecreasing and satisfies α(t) t for t ∈ [t0,T). 

Let αi,βj ∈ F for i = 1,2,...,n and j = 1,2,...,m. Denote 

 

J = min min 

1in αi(t0), min 

1 jm β 

j(t0) . 

2.1. Constant additive term 

In this subsection, we discuss the case of a constant k in the inequality (2.1) below. 

For this purpose, we define the following functions. 

DEFINITION 2.2. The function ω ∈ C(R+,R+) is said to be from Ω1 if 

(i) ω(x) > 0 for x > 0 and ω is a nondecreasing function; 

(ii) ∞ dx 

ω(x) = ∞. 

In the case when the nonlinear functions under the integrals in the inequality (2.1) 

below are from the set Ω1 , we obtain the following result. 

THEOREM 2.3. Let the following conditions be fulfilled: 

(A 1 ) φ ∈ C([J − h,t0],R+). 

(A 2 ) k > 0. 

(A 3 ) αi,βj ∈ F for i = 1,2,...,n, j = 1,2,...,m. 

(A 4 ) fi,gj ∈ C([J,T),R+) for i = 1,2,...,n, j = 1,2,...,m. 

(A 5 ) ωi, ˜ω j ∈ Ω1 for i = 1,2,...,n, j = 1,2,...,m. 

(A 6 ) ψ ∈ C 1 (R+,R+) is increasing, ψ(0) = 0, and limt→∞ ψ(t) = ∞. 

(A 7 ) u ∈ C([J − h,T),R+) satisfies for some p 0 the inequalities 

 

ψ u(t) 

k+ 

n 

∑ 

αi(t) 

i=1 αi(t0) 

m β j(t) 

+ 

∑ 

j=1 

β j(t0) 

fi(s)u p (s)ωi 

g j(s)u p (s) ˜ω j 

 

u(s) ds (2.1) 

 

max 

ξ ∈[s−h,s] u(ξ) 

 

ds for t ∈ [t0,T), 

u(t) φ(t) for t ∈ [J − h,t0]. (2.2)

Then for t0 t t1 , the inequality 

holds, where 

t1 = sup 

 

NONLINEAR INTEGRAL INEQUALITIES 813 

u(t) ψ −1 

 

Ψ −1 

 

W −1 

W Ψ(M) 

+ A(t) 

 

r 

Ψ(r) = 

r0 

r 

W(r) = 

r1 

(2.3) 

ds 

ψ −1 (s) p, 0 < r0 < k, (2.4) 

ds 

q ψ−1 (Ψ−1 (s)) , 0 < r1 

 

< Ψ(M), (2.5) 

q(t) = max max 

1in ωi(t), max 

1 jm ˜ω 

j(t) , (2.6) 

A(t) = 

n 

∑ 

i=1 

αi(t) 

αi(t0) 

M = max 

fi(s)ds+ 

 

k, ψ 

 

m β j(t) 

∑ 

j=1 β j(t0) 

max 

s∈[J−h,t0] φ(s) 

τ ∈ [t0,T) : W Ψ(M) + A(t) ∈ Dom W −1 , 

W −1 

W Ψ(M) 

+ A(t) ∈ Dom Ψ −1 and 

Ψ −1 

 

W −1 

W Ψ(M) 

+ A(t) 

 

∈ Dom ψ −1 

Proof. Define a function z : [J − h,T) → R+ by 

g j(s)ds, (2.7) 

, (2.8) 

for t ∈ [t0,τ] 

⎧ 

n αi(t) 

M + ∑ fi(s)u 

⎪⎨ i=1 αi(t0) 

z(t) = 

⎪⎩ 

p 

(s)ωi u(s) ds 

m β j(t) 

+ ∑ g j(s)u 

j=1 β j(t0) 

p t ∈ [t0,T), 

(s) ˜ω j max u(ξ) ds, 

ξ ∈[s−h,s] 

M, t ∈ [J − h,t0]. 

The function z is nondecreasing. Since ψ(u(t)) ψ(max s∈[J−h,t0] φ(s)) M = z(t) 

for t ∈ [J −h,t0] by (2.2) and (2.8) and ψ(u(t)) z(t) for t ∈ [t0,T) by (2.1) and (2.8), 

the inequality 

u(t) ψ −1 z(t) holds for t ∈ [J − h,T). (2.9) 

 

.


Note that maxξ ∈[s−h,s] ψ−1z(ξ) = ψ−1z(s) for s ∈ [β j(t0),βj(T)), j = 1,2,...,m. 

Then from inequality (2.1) and the definition of the function q, we get for t ∈ [t0,T) 

n αi(t) 

z(t) M + fi(s) 

ψ −1 z(s) 

ds 

+ 

M + 

+ 

∑ 

i=1 αi(t0) 

m β j(t) 

∑ 

j=1 β j(t0) 

n αi(t) 

∑ 

i=1 αi(t0) 

m β j(t) 

∑ 

j=1 β j(t0) 

ψ −1 z(s) p 

ωi 

 

g j(s) ψ −1 z(s) p 

˜ω j ψ −1 z(s) 

ds 

 

fi(s) ψ −1 z(s) p 

q ψ −1 z(s) 

ds 

 

g j(s) ψ −1 z(s) p 

q ψ −1 z(s) 

ds =: K(t), 

(2.10) 

where the function K : [t0,T) → [M,∞) is nondecreasing and satisfies K(t0) = M . Differentiate 

the function K and use its monotonicity (observe Definition 2.1) and (2.10) 

to obtain 

K ′ (t) = 

n 

∑ 

i=1 

m 

+ ∑ 

j=1 

n 

∑ fi 

i=1 

m 

+ ∑ 

j=1 

 

 

+ 

 

fi αi(t) 

ψ −1 z(αi(t)) p 

q ψ −1 z(αi(t)) 

α ′ i(t) 

 

g j β j(t) 

ψ −1 z(β j(t)) p 

q ψ −1 z(β j(t)) 

β ′ j (t) 

αi(t) 

ψ −1 K(αi(t)) p 

q 

 

ψ −1 z(αi(t)) 

α ′ i (t) 

 

g j β j(t) 

ψ −1 K(β j(t)) p 

q ψ −1 z(β j(t)) 

β ′ j (t) 

ψ −1 K(t) p n 

∑ 

m 

∑ 

j=1 

i=1 

From (2.4) and (2.11), we have that 

(Ψ ◦ K) ′ (t) = 

 

fi(αi(t))q ψ −1 z(αi(t)) 

α ′ i (t) 

 

g j(β j(t))q ψ −1 z(β j(t)) 

β ′ 

j(t) . 

K ′ (t) 

ψ−1K(t) n 

p ∑ 

i=1 

+ 

 

fi αi(t) 

q ψ −1 z(αi(t)) 

α ′ i(t) 

m 

∑ 

j=1 

 

g j β j(t) 

q ψ −1 z(β j(t)) 

β ′ j (t). 

Integrate (2.12) from t0 to t ∈ [t0,t1] and change the variables to get 

 

n αi(t) 

Ψ K(t) Ψ(M)+ fi(s)q ψ −1 z(s) 

ds 

+ 

∑ 

i=1 αi(t0) 

m β j(t) 

∑ 

j=1 β j(t0) 

 

g j(s)q ψ −1 z(s) 

ds =: K1(t), 

(2.11) 

(2.12) 

(2.13)


where the function K1 is nondecreasing and satisfies K1(t0) = Ψ(M) and, due to (2.10) 

and (2.13), 

z(t) K(t) Ψ −1 K1(t) holds for t ∈ [t0,t1). (2.14) 

Differentiate the function K1 and use its monotonicity (observe Definition 2.1) and 

(2.14) to obtain 

K ′ 

1(t) = fi(αi(t))q ψ −1 z(αi(t)) 

α ′ i(t) 

n 

∑ 

i=1 

+ 

 

q 

m 

∑ 

j=1 

 

g j(β j(t))q ψ −1 z(β j(t)) 

β ′ j (t) 

ψ −1 

Ψ −1 K1(t) n 

∑ fi(αi(t))α 

i=1 

′ i(t)+ 

From (2.15) and (2.5), we get 

(W ◦ K1) ′ (t) = 

 

K ′ 1 (t) 

 

q ψ−1 

Ψ−1K1(t) 

n 

∑ 

i=1 

fi(αi(t))α ′ i(t)+ 

m 

∑ 

j=1 

m 

∑ 

j=1 

g j(β j(t))β ′ 

j(t) . 

g j(β j(t))β ′ j(t). 

Integrate (2.16) from t0 to t ∈ [t0,t1] and change the variables to get 

(2.15) 

(2.16) 

W K1(t) = W Ψ(M) + A(t), (2.17) 

where the function A is defined by (2.7). Since W −1 is increasing and since, due to 

(2.9) and (2.14), u(t) ψ−1 (z(t)) ψ−1 

Ψ−1K1(t) 

, (2.17) implies the required 

inequality (2.3). 

COROLLARY 2.4. Let k = 0 and φ(t) ≡ 0. Suppose (A 3 )–(A 7 ) hold. Then for 

t0 t t2 , the inequality 

u(t) ψ −1 

 

Ψ −1 

 

W −1 

 

A(t) 

 

holds, where 

t2 = sup 

 

τ ∈ [t0,T) : A(t) ∈ Dom W −1 , W −1 A(t) ∈ Dom Ψ −1 

and Ψ −1 

W −1 A(t) 

∈ Dom ψ −1 

for t ∈ [t0,τ] . 

Proof. This claim follows from Theorem 2.3 by choosing an arbitrary ε > 0, letting 

k = φ(t) = ε , and taking the limit as ε → 0. 

Note that the inequalities (2.1), (2.2) could have another type of solution as follows, 

which is simpler than (2.3) but the used integral function is more complicated.


THEOREM 2.5. Suppose (A 1 )–(A 7 ) hold. Then for t0 t t3 , the inequality 

u(t) ψ −1 

Ψ −1 

 

Ψ1(M)+A(t) 

(2.18) 

holds, where Ψ −1 

1 is the inverse function of 

Ψ1(r) = 

r 

r3 

1 

ds 

ψ −1 (s) pq ψ −1 (s) , 0 < r3 < k, (2.19) 

the functions q and A and the constant M are defined by (2.6), (2.7), and (2.8), respectively, 

and 

 

t3 = sup τ ∈ [t0,T) : Ψ1(M)+A(t) ∈ Dom Ψ −1 

1 and 

Ψ −1 

 

1 Ψ1(M)+A(t) ∈ Dom ψ −1 

for t ∈ [t0,τ] . 

Proof. Following the proof of Theorem 2.3, we obtain the inequalities (2.10) and 

(2.11). From (2.11) and Definition 2.1, we conclude 

K ′ 

(t) ψ −1 K(t) p 

q 

+ 

m 

∑ 

j=1 

From (2.19) and (2.20), we have that 

g j(β j(t))β ′ j (t) 

 

. 

∑ 

ψ −1 K(t) n 

i=1 

(Ψ1 ◦ K) ′ K 

(t) = 

′ (t) 

 

ψ−1K(t) p 

q ψ−1K(t) 

 

n 

∑ 

i=1 

 

fi αi(t) α ′ i(t)+ 

m 

∑ 

j=1 

fi(αi(t))α ′ i(t) 

 

g j β j(t) β ′ j(t). 

Integrate inequality (2.21) from t0 to t ∈ [t0,t3] and change the variables to get 

Ψ1 

 

K(t) 

Ψ1(M)+ 

n 

∑ 

i=1 

αi(t) 

αi(t0) 

fi(s)ds+ 

m β j(t) 

∑ 

j=1 β j(t0) 

(2.20) 

(2.21) 

g j(s)ds = Ψ1(M)+A(t). (2.22) 

By (2.9), (2.10), and (2.22), we obtain the required inequality (2.18). 

In the case when p = 0, both solutions of inequalities (2.1), (2.2) given in Theorem 

2.3 and Theorem 2.5 coincide. 

COROLLARY 2.6. Suppose (A 1 )–(A 6 ) hold and assume


(A ′ 7 ) u ∈ C([J − h,T),R+) satisfies the inequalities 

 

ψ u(t) 

k+ 

n 

∑ 

αi(t) 

i=1 αi(t0) 

m β j(t) 

+ 

∑ 

j=1 

β j(t0) 

fi(s)ωi 

g j(s) ˜ω j 

 

u(s) ds 

u(t) φ(t) for t ∈ [J − h,t0]. 


u(t) ψ −1 

holds, where W −1 is the inverse function of 

r 

W(r) = 

r4 

 

max u(ξ) 

ξ ∈[s−h,s] 


 

W −1 

 

W(M)+A(t) 

 

ds 

q ψ −1 (s) , 0 < r4 < k, 


and 

 

t4 = sup 

τ ∈ [t0,T) : W(M)+A(t) ∈ Dom W −1 and 

W −1 

 

W(M)+A(t) ∈ Dom ψ −1 

for t ∈ [t0,τ] . 

In the case when the left part of the considered inequality is linear, i.e., ψ(x) = x, 

we obtain the following particular case of Theorem 2.3. 

COROLLARY 2.7. Suppose (A 1 )–(A 5 ) hold and assume 

(A ′′ 

7 ) u ∈ C([J − h,T),R+) satisfies the inequalities 

u(t) k+ 

n 

∑ 

αi(t) 

i=1 αi(t0) 

m β j(t) 

+ 

∑ 

j=1 

β j(t0) 

fi(s)ωi 

g j(s) ˜ω j 

 

u(s) ds 

u(t) φ(t) for t ∈ [J − h,t0]. 


u(t) W −1 

 

 

max u(ξ) 

ξ ∈[s−h,s] 


W(M)+A(t)


holds, where W −1 is the inverse function of 

W(r) = 

r 

r 5 

ds 

q(s) , r5 > 0, (2.23) 


and 

 

 

t5 = sup 

2.2. Monotone additive term 

τ ∈ [t0,T) : W(M)+A(t) ∈ Dom W −1 for t ∈ [t0,τ] 

In this subsection, we solve an inequality in which the constant k from Subsection 

2.1 is replaced by a monotonic function. For this purpose, we introduce the following 

set of functions. 

DEFINITION 2.8. The function ψ ∈ C 1 (R+,R+) is said to be from Λ if 

(i) ψ is an increasing function; 

(ii) tψ(x) ψ tx for 0 t 1. 

REMARK 2.9. Note that the functions ψ(x) = x and ψ(x) = x n , where n > 1, are 

from Λ. 

DEFINITION 2.10. The function ω ∈ Ω1 is said to be from Ω2 if 

ω(tx) tω(x) for 0 t 1. 


(B1 ) φ ∈ C([J − h,t0],[0, ˜k]), where ˜k = k(t0). 

(B2 ) k ∈ C([t0,T),[1,∞)) is nondecreasing. 

(B3 ) αi,βj ∈ F for i = 1,2,...,n, j = 1,2,...,m. 

(B4 ) fi,gj ∈ C([J,T),R+) for i = 1,2,...,n, j = 1,2,...,m. 

(B5 ) ωi, ˜ω j ∈ Ω2 for i = 1,2,...,n, j = 1,2,...,m. 

(B6 ) ψ ∈ Λ. 

(B7 ) u ∈ C([J − h,T),R+) satisfies for some p 0 the inequalities 

 

ψ u(t) 

k(t)+ 

n 

∑ 

αi(t) 

i=1 αi(t0) 

m β j(t) 

+ 

∑ 

j=1 

β j(t0) 

fi(s)u p (s)ωi 

g j(s)u p (s) ˜ω j 

 

u(s) ds (2.24) 

 

max 

ξ ∈[s−h,s] u(ξ) 

 

ds, t ∈ [t0,T), 

u(t) φ(t), t ∈ [J − h,t0]. (2.25) 

.



u(t) k(t)ψ −1 

 

Ψ −1 

 

W −1 

 

W Ψ(1) + A1(t) 

(2.26) 

holds, where the functions Ψ and W are defined by (2.4) and (2.5), respectively, and 

A1(t) = 

n αi(t) p m 

fi(s) k(s) ds+ 

β j(t) p g j(s) k(s) ds, (2.27) 

∑ 

i=1 

αi(t0) 

∑ 

j=1 

β j(t0) 

 

t6 = sup τ ∈ [t0,T) : W Ψ(1) + A1(t) ∈ Dom W −1 , 

W −1 

 

W Ψ(1) + A1(t) ∈ Dom Ψ −1 and 

Ψ −1 

W −1 

W Ψ(1) 

+ A1(t) ∈ Dom ψ −1 

for t ∈ [t0,τ] . 

Proof. From (2.24), (2.25), (B2 ), (B6 ), and 0 1 

k(t) 1, we obtain 

n 

u(t) 

αi(t) 

ψ 1+ 

k(t) ∑ fi(s)u 

i=1 αi(t0) 

p 

u(s) 

(s)ωi ds (2.28) 

k(s) 

m β j(t) 

+ ∑ g j(s)u p 

maxξ∈[s−h,s] u(ξ) 

(s) ˜ω j 

ds, t ∈ [t0,T), 

k(s) 

j=1 

β j(t0) 

u(t) φ(t) 

 

k(t0) k(t0) 1, t ∈ [J − h,t0]. (2.29) 

Let s ∈ [β j(t0),βj(T)), where 1 j m is arbitrary. From the monotonicity of the 

function k in [t0,T), we obtain the inequality 

max ξ ∈[s−h,s] u(ξ) 

k(s) 

= u(ξ1) 

k(s) 

u(ξ1) 

max 

k(ξ1) 

ξ ∈[s−h,s] 

where ξ1 ∈ [s − h,s]. Define a function v ∈ C([J − h,T),R+) by 

⎧ 

⎪⎨ 

u(t) 

for t ∈ [t0,T) 

k(t) 

v(t) = 

⎪⎩ 

u(t) 

for t ∈ [J − h,t0]. 

k(t0) 

u(ξ) 

k(ξ) , 

Then inequalities (2.28) and (2.29) can be rewritten as 

n 

ψ v(t) 1+ 

αi(t) 

fi(s)k p (s)v p 

(s)ωi v(s) ds (2.30) 

∑ 

i=1 αi(t0) 

m β j(t) 

+ 

∑ 

j=1 

β j(t0) 

g j(s)k p (s)v p (s) ˜ω j 

 

max v(ξ) 

ξ ∈[s−h,s] 

ds, t ∈ [t0,T), 

v(t) 1, t ∈ [J − h,t0]. (2.31)


Theorem 2.3, applied to the inequalities (2.30) and (2.31), implies the validity of inequality 

(2.26). 

In the case when Theorem 2.5 is applied instead of Theorem 2.3 in the last part of 

the proof of Theorem 2.11, we obtain the following result. 

THEOREM 2.12. Suppose (B1 )–(B7 ) hold. Then for t0 t t7 , the inequality 

u(t) k(t)ψ −1 

Ψ −1 

 

Ψ1(1)+A1(t) 

holds, where Ψ1 and A1 are defined by (2.19) and (2.27), respectively, and 

t7 = sup 

 

1 

τ ∈ [t0,T) : Ψ1(1)+A1(t) ∈ Dom Ψ −1 

1 and 

Ψ −1 

 

1 Ψ1(1)+A(t) ∈ Dom ψ −1 

for t ∈ [t0,τ] . 

In the case when p = 0 and the left part of the considered inequality is linear, i.e., 

ψ(x) = x, the results of Theorem 2.11 and Theorem 2.12 coincide, and we obtain the 

following result. 

COROLLARY 2.13. Suppose (B1 )–(B5 ) hold and assume 

(B ′ 7 ) u ∈ C([J − h,T),R+) satisfies the inequalities 

u(t) k(t)+ 

n 

∑ 

αi(t) 

i=1 αi(t0) 

m β j(t) 

+ 

∑ 

j=1 

β j(t0) 

fi(s)ωi 

g j(s) ˜ω j 

u(t) φ(t), t ∈ [J − h,t0]. 


u(t) k(t)W −1 

 

 

u(s) ds 

 

max 

ξ ∈[s−h,s] u(ξ) 

 

ds, t ∈ [t0,T), 

W(1)+A(t) 

holds, where the functions W and A are defined by (2.23) and (2.7), respectively, and 

t8 = sup 

 

τ ∈ [t0,T) : W(1)+A(t) ∈ Dom W −1 for t ∈ [t0,τ] 

 

 

.

2.3. Arbitrary additive term 


In this subsection, we solve a nonlinear inequality in which the constant k of 

Subsection 2.1 is replaced by an arbitrary function. In this case, we define the following 

set of functions. 

DEFINITION 2.14. The function ω ∈ Ω2 is said to be from Ω3 if 

ω(x)+ω(y) ω(x+y). 

REMARK 2.15. Note that the functions ω(x) = √ x and ω(x) = x are from Ω3 . 


(C2 ) k ∈ C([J − h,T),R+). 

(C3 ) αi,βj ∈ F for i = 1,2,...,n, j = 1,2,...,m. 

(C4 ) fi,gj ∈ C([J,T),R+) for i = 1,2,...,n, j = 1,2,...,m. 

(C5 ) ωi, ˜ω j ∈ Ω3 for i = 1,2,...,n, j = 1,2,...,m. 

(C6 ) µ ∈ C([t0,T),[1,∞)) is nondecreasing. 

(C7 ) u ∈ C([J − h,T),R+) satisfies the inequalities 

u(t) k(t)+ µ(t) 

+ 

 

n αi(t) 

∑ 

i=1 αi(t0) 

m β j(t) 

∑ 

j=1 

β j(t0) 

fi(s)ωi 

 

u(s) ds (2.32) 

 

g j(s) ˜ω j max u(ξ) 

ξ ∈[s−h,s] 

ds , t ∈ [t0,T), 

u(t) k(t), t ∈ [J − h,t0]. (2.33) 


u(t) k(t)+M(t)e(t)W −1 

 

W(1)+A2(t) 

holds, where the function W is defined by (2.23), 

e(t) = 1+ 

n 

∑ 

αi(t) 

i=1 αi(t0) 

m β j(t) 

+ 

∑ 

j=1 

β j(t0) 

fi(s)ωi 

g j(s) ˜ω j 

(2.34) 

 

k(s) ds (2.35) 

 

max k(ξ) 

ξ ∈[s−h,s] 

ds, t ∈ [t0,T),


A2(t) = 

n 

∑ 

i=1 

αi(t) 

αi(t0) 

fi(s)M(s)ds+ 

m 

∑ 

j=1 

β j(t) 

β j(t0) 

g j(s)M(s)ds, 

 

µ(t) for t ∈ [t0,T), 

M(t) = 

µ(t0) for t ∈ [J − h,t0], 

 

t9 = sup τ ∈ [t0,T) : W(1)+A2(t) ∈ Dom W −1 

for t ∈ [t0,τ] . 

Proof. Define a function z : [J − h,T) → R+ by 

⎧ n αi(t) 

∑ fi(s)ωi u(s) ds 

⎪⎨ i=1 αi(t0) 

m β 

z(t) = 

j(t) 

+ ∑ g j(s) ˜ω j max 

⎪⎩ 

j=1 β j(t0) 

ξ ∈[s−h,s] u(ξ) 

 

ds, t ∈ [t0,T), 

0, t ∈ [J − h,t0]. 

From (2.32) and the definition of the function z, we obtain for s ∈ [β j(t0),βj(T)) and 

1 j m 

u(t) k(t)+M(t)z(t) for t ∈ [J − h,T]. (2.36) 

Since the function M is nondecreasing on [J − h,T), we obtain 

max u(ξ) max k(ξ)+M(s) max z(ξ). (2.37) 

ξ ∈[s−h,s] ξ ∈[s−h,s] 

ξ ∈[s−h,s] 

From (2.36), (2.37), (C3 ), and (C5 ), we get for t ∈ [t0,T) 

αi(t) 

αi(t0) 

fi(s)ωi 

αi(t) 

u(s) ds 

 

αi(t0) 

αi(t) 

αi(t0) 

fi(s)ωi 

fi(s)ωi 

 

k(s)+M(s)z(s) ds (2.38) 

αi(t) 

k(s) ds+ fi(s)M(s)ωi z(s) ds 

αi(t0) 

and 

 

β j(t) 

g j(s) ˜ω j max 

β j(t0) 

ξ ∈[s−h,s] u(ξ) 

 

ds 

 

β j(t) 

g j(s) ˜ω j max 

β j(t0) 

ξ ∈[s−h,s] k(ξ) 

 

β j(t) 

ds+ g j(s)M(s) ˜ω j max 

β j(t0) 

ξ ∈[s−h,s] z(ξ) 

 

ds. 

(2.39) 

From the definition of the function z and (2.36), (2.38), and (2.39), it follows that 

n αi(t) 

z(t) e(t)+ ∑ fi(s)M(s)ωi z(s) ds (2.40) 

i=1 αi(t0) 

m 

β j(t) 

+ ∑ g j(s)M(s) ˜ω j max z(ξ) ds, t ∈ [t0,T), 

j=1 β j(t0) 

ξ ∈[s−h,s] 

z(t) 0, t ∈ [J − h,t0], (2.41)


where the function e is defined by (2.35). Note that e : [t0,T) → [1,∞) is nondecreasing 

and e(t0) = 1. Corollary 2.13 applied to inequalities (2.40) and (2.41) implies the 

required inequality (2.34). 

3. Applications 

In this section, we consider the differential equation with “maxima” 

with the initial condition 

px p−1 x ′ 

= F t,x(t), max 

s∈[σ(t),τ(t)] x(s) 

 

for t ∈ [t0,T), (3.1) 

x(t) = ϕ(t) for t ∈ [τ(t0) − h,t0], (3.2) 

where ϕ : [τ(t0) − h,t0] → R, F : [t0,T)×R×R → R, and p is a natural number. 

THEOREM 3.1. (Upper bound) Let the following conditions be fulfilled: 

(H 1 ) ϕ ∈ C([τ(t0) − h,t0],R). 

(H 2 ) τ,σ ∈ F and there exists a constant h with 0 < τ(t) − σ(t) h for t t0 . 

(H 3 ) F ∈ C([t0,T)×R×R,R) satisfies 

 

F(t,u,v) Q(t) u q + R(t) v q 

where Q,R ∈ C([t0,T),R+) and q ∈ (0, p). 

for u,v ∈ R and t t0, 

(H 4 ) x : [τ(t0) − h,T) → R is a solution of the initial value problem (3.1), (3.2). 

Then x satisfies the inequality 

 

 

x(t) p−q M p−q 

p + 

where M = sup s∈[τ(t0)−h,t0] 

p − q 

p 

p ϕ(s) . 

t 

Proof. The function x satisfies the integral problem 

t0 

 

Q(s)+R(s) ds for t ∈ [t0,T), (3.3) 

x(t) p = ϕ(t0) p + 

t 

t0 

F 

 

s,x(s), max 

ξ ∈[σ(s),τ(s)] x(ξ) 

 


x(t) = ϕ(t) for t ∈ [τ(t0) − h,t0].


Then for the norm of the solution x, we obtain 

 

x(t) p 

ϕ(t0) p t 

 

+ F s,x(s), max 

ϕ(t0) p + 

t0 

t 

t0 

t 

ξ ∈[σ(s),τ(s)] 

 

ds 

x(ξ) 

 

Q(s) 

x(s) q 

+ R(s) max 

ξ ∈[σ(s),τ(s)] x(ξ) 

 

 

q 

ds 

ϕ(t0) p + Q(s) 

t0 

 

x(s) q ds 

 

t 

 

+ R(s) max x(ξ) 

t0 ξ ∈[σ(s),τ(s)] 

q ds for t ∈ [t0,T), (3.4) 

 

x(t) = ϕ(t) for t ∈ [τ(t0) − h,t0]. (3.5) 

Change the variable s = τ −1 (η) in the second integral of (3.4), use the inequality 

max ξ ∈[σ(s),τ(s)] |x(ξ)| max ξ ∈[τ(s)−h,τ(s)] |x(ξ)| for s ∈ [t0,T) that follows from (H 2 ), 

and obtain 

 

x(t) p 

ϕ(t0) p t 

+ Q(s) 

t0 

 

x(s) q ds 

τ(t) 

+ R(τ 

τ(t0) 

−1 (η))(τ −1 ) ′ 

(η) 

Note that the conditions of Corollary 2.6 are satisfied for 

max 

ξ ∈[η−h,η] 

 

x(ξ) 

u(t) = |x(t)|, n = 1, α1(t) = t, m = 1, β1 = τ, 

k = |ϕ(t0)| p , f1 = Q, g1 = (R ◦ τ −1 )(τ −1 ) ′ on [τ(t0),T), 

ψ(x) = x p , ψ −1 (x) = p√ x, Dom(ψ −1 ) = R+, 

ω1(x) = ˜ω1(x) = x q r q 

− 

, W(r) = s p ds = 

0 

p p−q 

r p , 

p − q 

W −1 (r) = p−q 

p − q 

p r 

p , Dom(W −1 ) = R+. 

According to Corollary 2.6, from (3.6) and (3.5), we obtain (3.3). 

q 

dη. 

(3.6) 

COROLLARY 3.2. Let ϕ(t) ≡ 0. Suppose (H 2 )–(H 4 ) hold. Then the solution x 

of the initial value problem (3.1), (3.2) satisfies the inequality 

 

 

x(t) p−q p − q t 

Q(s)+R(s) ds for t ∈ [t0,T). 

p 

t0 

REMARK 3.3. Note that in the case of p = q = 1 in (H 3 ), we use the functions 

ψ(x) = ψ−1 (x) = x and W(r) = lnr, W −1 (r) = er , apply Corollary 2.6, and obtain 

|x(t)| e lnM+ 

t 

t 

t0 Q(s)+R(s) ds t0 Q(s)+R(s) ds 

= Me 

for t ∈ [t0,T), 

which gives us the known result about uniqueness of the solution of a first-order differential 

equation with Lipschitz-continuous right-hand side.


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scales, JIPAM. J. Inequal. Pure Appl. Math. 6, 1 (2005), Article 6, 23 pp. (electronic). 

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[12] BABURAO G. PACHPATTE, Inequalities for differential and integral equations, volume 197 of Mathematics 

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“Nauka”, Moscow, 1978. 

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Rosenblatt and Lawrence Shampine. Ergebnisse der Mathematik und ihrer Grenzgebiete, Band 55. 

Springer-Verlag, New York, 1970. 

(Received February 22, 2011) M. Bohner 

Department of Mathematics and Statistics 

Missouri University of Science and Technology 

Rolla, MO 65409-0020 

USA 

e-mail: bohner@mst.edu 

Mathematical Inequalities & Applications 

www.ele-math.com 

mia@ele-math.com 

S. Hristova 

Department of Applied Mathematics and Modeling 

Plovdiv University, Plovdiv 4000 

Bulgaria 

e-mail: snehri@uni-plovdiv.bg 

K. Stefanova 

Department of Applied Mathematics and Modeling 

Plovdiv University, Plovdiv 4000 

Bulgaria

NONLINEAR INTEGRAL INEQUALITIES INVOLVING ... - Ele-Math

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