Liu Advances in Difference Equations (2018) 2018:22 https://doi.org/10.1186/s13662-017-1450-5
RESEARCH
Open Access
Some new integral inequalities with mixed nonlinearities for discontinuous functions Haidong Liu* *
Correspondence:
[email protected] School of Mathematical Sciences, Qufu Normal University, Qufu, 273165, P.R. China
Abstract In this paper, we establish some new integral inequalities with mixed nonlinearities for discontinuous functions, which provide a handy tool in deriving the explicit bounds for the solutions of impulsive differential equations and differential-integral equations with impulsive conditions. Keywords: integral inequalities; discontinuous functions; mixed nonlinearities; impulsive differential equations
1 Introduction In recent years, the theory of impulsive differential systems has been attracting the attention of many mathematicians, and the interest in the subject is still growing. This is partly due to broad applications of it in many areas including threshold theory in biology, ecosystems management and orbital transfer of satellite, see [1]. One effective method for investigating the properties of solutions to impulsive differential systems is related to the integral inequalities for discontinuous functions (integro-sum inequalities). Up to now, a lot of integro-sum inequalities (for example, [2–18] and the references therein) have been discovered. For example, in 2003, Borysenko [3] considered the following integro-sum inequality:
t
x(t) ≤ a(t) +
q(τ )xm (τ ) dτ + t0
βi xm (ti – 0),
m > 0, m = 1.
t0
In 2009, Gallo and Piccirillo [8] further discussed the following nonlinear integro-sum inequality: x(t) ≤ c(t) + h(t)
t
f (s)w x b(s) ds + βi xm (ti – 0),
t0
m > 0.
t0
In 2012, Wang et al. [17] considered the nonlinear integro-sum inequality as follows: xm (t) ≤ c(t) + 2 +2
α(t)
m M1 f1 (t, s)u 2 (s) + N1 g1 (t, s)um (s) ds
α(t0 )
m M2 f2 (t, s)u 2 (s) + N2 g2 (t, s)um (s) ds + βi x(ti – 0),
t t0
m > 0.
t0
© The Author(s) 2018. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
Liu Advances in Difference Equations (2018) 2018:22
Page 2 of 16
Very recently, in 2016, Zheng et al. [18] considered the following nonlinear integro-sum inequality under the condition p > q > 0: p–q x (t) ≤ a0 (t) + p i=1 N
t
p
+
L
t
t0 s
bj (s)
t0
j=1
gi (s)xq φi (s) ds
cj (θ )xq wj (s) dθ ds + βi xm (ti – 0).
t0
t0
Motivated by [3, 8, 17, 18], in this paper, we investigate some new integro-sum inequality with mixed nonlinearities under the condition p > 0, q > 0 (p = q): x (t) ≤ a(t) +
t
p
f1 (s)x (s) ds + t0
t
s
g1 (τ )xp (τ ) dτ ds
f2 (s) t0
t0
s
g2 (τ )xq (τ ) dτ ds + c(t)
f3 (s)
+
t
q
t0
t0
βi xm (ti – 0)
t0
and the more general form
t
xp (t) ≤ a(t) +
f (s)xq (s) ds + t0
+
M
j=1
t
t
s
gj (τ )xp (τ ) dτ ds
bj (s) t0
t0
s
θk (τ )xq (τ ) dτ ds + d(t)
ck (s)
t0
k=1
L
t0
βi xm (ti – 0).
t0
We also discuss some nonlinear integro-sum inequality with positive and negative coefficients under the condition 0 < q < p < r: x (t) ≤ a(t) + b(t) p
+ c(t)
t
f (s)xp (s) + g(s)xq (s) – h(s)xr (s) ds
t0
βi xm (ti – 0),
t0
and the more general form under the condition 0 < qj < p < rj (j = 1, 2, . . . , L): x (t) ≤ a(t) +
t
p
p
f (s)x (s) ds + t0
+ c(t)
L j=1
t
t0
qj
gj (s)x (s) ds –
L j=1
t
hj (s)xrj (s) ds t0
βi xm (ti – 0).
t0
Based on these inequalities, we provide explicit bounds for unknown functions concerned and then apply the results to research the qualitative properties of solutions of certain impulsive differential equations.
2 Preliminaries Throughout the present paper, R denotes the set of real numbers; R+ = [0, +∞) is the subset of R; C(D, E) denotes the class of all continuous functions defined on the set D with range in the set E.
Liu Advances in Difference Equations (2018) 2018:22
Page 3 of 16
Lemma 2.1 ([19]) Assume that the following conditions for t ≥ t0 hold: (i) x0 is a nonnegative constant, (ii) t x(t) ≤ x0 + e(s)x(s) + l(s)xα (s) ds, t0
where x, e and l are nonnegative continuous functions and α = 1 is a positive constant. If 1 + (1 – α)x(α–1) 0
s l(s) exp (α – 1) e(τ ) dτ ds > 0
t
t0
t0
holds, then t e(s) ds x(t) ≤ x0 exp t0 1
1–α t s (α–1) × 1 + (1 – α)x0 l(s) exp (α – 1) e(τ ) dτ ds ,
t0
t ≥ t0 .
t0
Lemma 2.2 ([20]) Let x be a nonnegative function, 0 < q < p < r, c1 ≥ 0, k2 ≥ 0, c2 > 0 and k1 > 0. Then c1 xq – c2 xr ≤ (k1 – k2 )xp + θ1 (p, q, c1 , k1 ) + θ2 (p, r, c2 , k2 ), where p p –q p – q q p–q p–q p–q θ1 (p, q, c1 , k1 ) := c1 k1 , q p
p –p r r – p p r–p r–p r–p θ2 (p, r, c2 , k2 ) := c2 k2 . r r
3 Main results Theorem 3.1 Suppose that x is a nonnegative piecewise continuous function defined on [t0 , ∞) with discontinuities of the first kind in the points ti (i = 1, 2, . . .) and satisfies the integro-sum inequality xp (t) ≤ a(t) +
t
t0
t
t0
s
g1 (τ )xp (τ ) dτ ds
f2 (s) t0
t0
s
g2 (τ )xq (τ ) dτ ds + c(t)
f3 (s)
+
t
f1 (s)xq (s) ds +
t0
βi xm (ti – 0),
t ≥ t0 ,
(3.1)
t0
where 0 ≤ t0 < t1 < t2 < · · · , limi→∞ ti = ∞, functions a(t) ≥ 0 and c(t) ≥ 0 are defined on [t0 , ∞), f1 , f2 , f3 , g1 , g2 ∈ C(R+ , R+ ), βi ≥ 0 (i = 1, 2, . . .), p > 0, q > 0, p = q and m > 0 are constants. If
1+
p – q q–p p r (t) p i
q–p s l(s) exp e(τ ) dτ ds > 0, p ti–1 ti–1 t
i = 1, 2, . . . ,
Liu Advances in Difference Equations (2018) 2018:22
Page 4 of 16
then, for t ≥ t0 , the following estimates hold: x(t) ≤ v1 (t),
t ∈ [t0 , t1 ],
(3.2)
x(t) ≤ vi (t),
t ∈ (ti–1 , ti ], i = 2, 3, . . . ,
(3.3)
where t 1 e(s) ds vi (t) = ri (t) exp p ti–1 1
p–q t p – q q–p q–p s p × 1+ ri (t) l(s) exp e(τ ) dτ ds , p p ti–1 ti–1 t t e(t) = f2 (t) g1 (τ ) dτ , l(t) = f1 (t) + f3 (t) g2 (τ ) dτ , 1 p
t0
r1 (t) = max a(τ ) , t0 ≤τ ≤t
ti
ri+1 (t) = ri (t) + ti–1
f3 (s)
t
+
q f1 (s)vi (s) ds + ti
ti–1
ti–1
(3.5)
t
ti
f2 (s) ti–1
ti–1
q g2 (τ )vi (τ ) dτ
(3.4)
t0
h(t) = max c(τ ) , t0 ≤τ ≤t
i = 1, 2, . . . ,
p g1 (τ )vi (τ ) dτ
ds + h(t)βi vm i (ti – 0),
ds i = 1, 2, . . . .
Proof From (3.1) and (3.5), we have, for t ∈ I0 = [t0 , t1 ],
t
x (t) ≤ r1 (t) + p
f1 (s)x (s) ds + t0
s
g1 (τ )xp (τ ) dτ ds
f2 (s) t0
t
t0
s
g2 (τ )xq (τ ) dτ ds
f3 (s)
+
t
q
t0
(3.6)
t0
and r1 (t) is non-decreasing on [t0 , ∞). Take any fixed T ∈ [t0 , t1 ], and for arbitrary t ∈ [t0 , T], we have
t
xp (t) ≤ r1 (T) +
t0
s
g2 (τ )xq (τ ) dτ ds.
f3 (s)
+
g1 (τ )xp (τ ) dτ ds
t0
t
s
f2 (s)
t0
t
f1 (s)xq (s) ds +
t0
(3.7)
t0
Let u(t) = xp (t). Inequality (3.7) is equivalent to
t
u(t) ≤ r1 (T) +
q
t0
s
f3 (s)
+
s
f2 (s) t0
t
t
f1 (s)u p (s) ds +
g1 (τ )u(τ ) dτ ds t0
q
g2 (τ )u p (τ ) dτ ds.
t0
(3.8)
t0
Let
t
V (t) = r1 (T) +
t0 t
t0
t
s
g1 (τ )u(τ ) dτ ds t0
q
g2 (τ )u p (τ ) dτ ds. t0
s
f2 (s) t0
f3 (s)
+
q
f1 (s)u p (s) ds +
(3.9)
Liu Advances in Difference Equations (2018) 2018:22
Page 5 of 16
It follows from (3.8) and (3.9) that u(t) ≤ V (t),
V (t0 ) = r1 (T),
(3.10)
V (t) is non-decreasing and q
V (t) = f1 (t)u p (t) + f2 (t)
t
t
q
g1 (τ )u(τ ) dτ + f3 (t) t0
g2 (τ )u p (τ ) dτ .
(3.11)
t0
Since V (t) is non-decreasing, from (3.11) we have q
V (t) ≤ f1 (t)V p (t) + f2 (t)
t
t0
q p
≤ f1 (t)V (t) + f2 (t)
t
g1 (τ )V (τ ) dτ + f3 (t)
t
t0 t
g1 (τ ) dτ V (t) + f3 (t) t0
q
g2 (τ )V p (τ ) dτ q
g2 (τ ) dτ V p (t) t0
q p
≤ e(t)V (t) + l(t)V (t),
(3.12)
where e(t) and l(t) are defined as in (3.4). Integrating (3.12) from t0 to t yields V (t) ≤ r1 (T) +
t
q e(s)V (s) + l(s)V p (s) ds.
t0
From the above and Lemma 2.1, we get p t
p–q t p – q q–p q–p s p V (t) ≤ r1 (T) exp r (T) e(s) ds 1+ l(s) exp e(τ ) dτ ds , p 1 p t0 t0 t0
and then from (3.10) and the assumption u(t) = xp (t), we have t 1 1 p e(s) ds x(t) ≤ r1 (T) exp p t0 1 p–q
t p – q q–p q–p s p r (T) l(s) exp e(τ ) dτ ds . × 1+ p 1 p t0 t0 Since the above inequality is true for any t ∈ [t0 , T], we obtain T 1 e(s) ds x(T) ≤ r1 (T) exp p t0 1 p–q
T q–p s p – q q–p r1 p (T) l(s) exp e(τ ) dτ ds . × 1+ p p t0 t0 1 p
Replacing T by t yields t 1 x(t) ≤ r1 (t) exp e(s) ds p t0 1 p–q
t p – q q–p q–p s p r (t) l(s) exp e(τ ) dτ ds × 1+ p 1 p t0 t0 1 p
= v1 (t),
t ∈ I0 = [t0 , t1 ].
This means that (3.1) is true.
(3.13)
Liu Advances in Difference Equations (2018) 2018:22
Page 6 of 16
For t ∈ I1 = (t1 , t2 ], from (3.1), (3.2), (3.5) and (3.13), we get
t
xp (t) ≤ r1 (t) +
t0 t
t0
g2 (τ )xq (τ ) dτ ds + h(t)β1 xm (t1 – 0)
= r1 (t) +
t1
t1 t
t
t1
t0 t
t1
t1
t0 t
t1 t1
t1
t0 s
t1 t1
= r1 (t) +
t0 t
t1 t
t0
s
g1 (τ )xp (τ ) dτ ds
f2 (s) t1
t1
t
t1
f3 (s) t1
q
g2 (τ )v1 (τ ) dτ ds
t0
t
t1
t
t1
f2 (s) t0 t1
f3 (s) t0
t
f1 (s)xq (s) ds +
g1 (τ )xp (τ ) dτ ds +
t0
t0
p
g1 (τ )v1 (τ ) dτ ds
q
g2 (τ )v1 (τ ) dτ ds
g2 (τ )xq (τ ) dτ ds + h(t)β1 vm 1 (t1 – 0)
t
f1 (s)x (s) ds +
t0
p
g1 (τ )v1 (τ ) dτ ds
g2 (τ )xq (τ ) dτ ds + h(t)β1 vm 1 (t1 – 0)
q
t
t1
f2 (s)
t
s
t1
t1
s
g1 (τ )xp (τ ) dτ ds
f2 (s) t1
t1
s
g2 (τ )xq (τ ) dτ ds.
f3 (s) t1
t1
t1
f1 (s)x (s) ds +
q
t
= r2 (t) + +
t1 s
t1
g2 (τ )xq (τ ) dτ ds
t0
t
g2 (τ )v1 (τ ) dτ ds +
q
f3 (s)
+
t1
q
f1 (s)v1 (s) ds +
f2 (s)
+
p
t1
f3 (s)
+
t
f3 (s)
g1 (τ )v1 (τ ) dτ ds +
t0
t0 t
g1 (τ )xp (τ ) dτ ds
t1
t1
f3 (s)
+
q f1 (s)v1 (s) ds +
f2 (s)
+
s
g2 (τ )xq (τ ) dτ ds + h(t)β1 xm (t1 – 0)
t1
≤ r1 (t) +
g1 (τ )xp (τ ) dτ ds
t0
t1
g2 (τ )xq (τ ) dτ ds +
t1
t0
f2 (s)
t0 s
t1
f2 (s)
t
t1
f3 (s)
+
t0
f3 (s)
+
f1 (s)x (s) ds +
g1 (τ )xp (τ ) dτ ds +
t1
q
t1
f2 (s) t1 t1
t
q
f1 (s)x (s) ds + t0
g2 (τ )xq (τ ) dτ ds
g2 (τ )xq (τ ) dτ ds + h(t)β1 xm (t1 – 0)
t1
≤ r1 (t) +
t0
t0
t0
t0
g1 (τ )xp (τ ) dτ ds
t0 s
t1
s
f2 (s)
f3 (s)
t0 s
t1
+
g1 (τ )xp (τ ) dτ ds +
f3 (s)
+
f1 (s)x (s) ds +
s
f2 (s)
t1
q
t1
t
+
t
q
f1 (s)x (s) ds + t0
g1 (τ )xp (τ ) dτ ds t0
s
t0
s
f2 (s) t0
f3 (s)
+
t
f1 (s)xq (s) ds +
(3.14)
t1
Inequality (3.14) is the same as (3.6) if we replace r1 (t) and t0 with r2 (t) and t1 in (3.6), respectively. Thus, by (3.14), we have, for t ∈ I1 = (t1 , t2 ], t 1 x(t) ≤ r2 (t) exp e(s) ds p t1 1
p–q t p – q q–p q–p s p × 1+ r2 (t) l(s) exp e(τ ) dτ ds = v2 (t). p p t1 t1 1 p
Liu Advances in Difference Equations (2018) 2018:22
Page 7 of 16
Suppose that t 1 1 p e(s) ds x(t) ≤ ri (t) exp p ti–1 1
p–q t p – q q–p q–p s p × 1+ r (t) l(s) exp e(τ ) dτ ds = vi (t) p i p ti–1 ti–1
(3.15)
holds for t ∈ Ii–1 = (ti–1 , ti ], i = 2, 3, . . . . Then, for t ∈ Ii = (ti , ti+1 ], from (3.1), (3.2), (3.5) and (3.15) we obtain
t
xp (t) ≤ r1 (t) +
t0 t
t0
= r1 (t) +
s
g2 (τ )xq (τ ) dτ ds + h(t)
i–1
tk+1
s
t0
t
q
f3 (s)
g1 (τ )xp (τ ) dτ ds
ti
s
s
f2 (s)
t0
t
g1 (τ )xp (τ ) dτ ds +
f2 (s)
tk+1
+ h(t)
f1 (s)xq (s) ds ti
g2 (τ )x (τ ) dτ ds +
tk
k=0
βi xm (ti – 0)
t
f1 (s)xq (s) ds +
tk
i–1
t0
tk
i–1 tk+1 k=0
+
g1 (τ )xp (τ ) dτ ds t0
t0
k=0
+
s
f2 (s) t0
f3 (s)
+
t
f1 (s)xq (s) ds +
s
g2 (τ )xq (τ ) dτ ds
f3 (s)
t0
ti
t0
βi xm (ti – 0)
t0
≤ r1 (t) +
i–1 k=0
+
i–1 tk+1
t f2 (s)
+
f2 (s)
tk+1
i–1
+ h(t)
i–1
t
p
k
tj+1
q
g2 (τ )x (τ ) dτ ds
tj
tj+1
q
g2 (τ )x (τ ) dτ ds +
t
ti
tj
k=0
+
i–1 k=0
tk+1
tk
tk+1
tk
q
f1 (s)vk+1 (s) ds +
f2 (s)
k j=0
tj
tj+1
t
f1 (s)xq (s) ds ti
p g1 (τ )vj+1 (τ )
dτ ds
s
g2 (τ )xq (τ ) dτ ds
f3 (s)
βi xm (ti – 0)
i–1
g1 (τ )xp (τ ) dτ ds ti
t0
≤ r1 (t) +
s
f2 (s) ti
j=0
j=0
g1 (τ )x (τ ) dτ ds
tj
tj
ti
p
g1 (τ )x (τ ) dτ ds +
f3 (s)
f3 (s)
+
f1 (s)xq (s) ds
tj+1
tj+1
tk
t
k j=0
j=0
i–1
t
ti
tk
ti
k=0
f1 (s)xq (s) ds +
tk
k=0
+
tk+1
ti
Liu Advances in Difference Equations (2018) 2018:22
t +
i–1
f2 (s) ti
+
Page 8 of 16
tj+1 tj
j=0
i–1 tk+1
k
f3 (s)
tk
k=0
i–1
f3 (s) ti
+ h(t)
tj
j=0
tj+1
tj+1
tj
j=0
t +
g1 (τ )xp (τ ) dτ ds
ti
ti
q g2 (τ )vj+1 (τ )
dτ ds
t
ti
t
f1 (s)xq (s) ds +
t
s
f2 (s)
t
ds +
s
g2 (τ )xq (τ ) dτ ds
f3 (s) ti
ti
βi v m i (ti – 0)
= ri+1 (t) + +
t
ds +
q g2 (τ )vj+1 (τ ) dτ
t0
p g1 (τ )vj+1 (τ ) dτ
s
g1 (τ )xp (τ ) dτ ds
f2 (s) ti
ti
s
g2 (τ )xq (τ ) dτ ds.
f3 (s) ti
(3.16)
ti
Inequality (3.16) is the same as (3.6) if we replace r1 (t) and t0 with ri+1 (t) and ti in (3.6), respectively. Thus, by (3.16), we have, for t ∈ Ii = (ti , ti+1 ], t 1 1 p x(t) ≤ ri+1 (t) exp e(s) ds p ti 1 p–q
t q–p s p – q q–p p r (t) l(s) exp e(τ ) dτ ds . × 1+ p i+1 p ti ti By induction, we know that (3.3) holds for t ∈ (ti , ti+1 ], for any nonnegative integer i. This completes the proof of Theorem 3.1. Theorem 3.2 Suppose that x is a nonnegative piecewise continuous function defined on [t0 , ∞) with discontinuities of the first kind in the points ti (i = 1, 2, . . .) and satisfies the integro-sum inequality
t
xp (t) ≤ a(t) +
f (s)xq (s) ds + t0
+
M k=1
L j=1
t
s
gj (τ )xp (τ ) dτ ds
bj (s)
t0
t0
s
θk (τ )xq (τ ) dτ ds + d(t)
ck (s)
t0
t
t0
βi xm (ti – 0),
t ≥ t0 ,
(3.17)
t0
where 0 ≤ t0 < t1 < t2 < · · · , limi→∞ ti = ∞, a(t) ≥ 0 is defined on [t0 , ∞), f ∈ C(R+ , R+ ), bj , gj ∈ C(R+ , R+ ) (j = 1, 2, . . . , L), cj , θj ∈ C(R+ , R+ ) (j = 1, 2, . . . , M), βi ≥ 0 (i = 1, 2, . . .), p > 0, q > 0, p = q, and m > 0 are constants. If 1+
p – q q–p p r (t) p i
q–p s l(s) exp e(τ ) dτ ds > 0, p ti–1 ti–1 t
i = 1, 2, . . . ,
then, for t ≥ t0 , the following estimates hold: x(t) ≤ v1 (t),
t ∈ [t0 , t1 ],
(3.18)
x(t) ≤ vi (t),
t ∈ (ti–1 , ti ], i = 2, 3, . . . ,
(3.19)
Liu Advances in Difference Equations (2018) 2018:22
Page 9 of 16
where t 1 e(s) ds vi (t) = ri (t) exp p ti–1 1 p–q
t q–p s p – q q–p p ri (t) l(s) exp e(τ ) dτ ds , × 1+ p p ti–1 ti–1 1 p
i = 1, 2, . . . , e(t) =
L
t
bj (t)
gj (τ ) dτ ,
l(t) = f (t) +
M
t0
j=1
t0 ≤τ ≤t
ti
q f (s)vi (s) ds +
ri+1 (t) = ri (t) + ti–1 M k=1
θk (τ ) dτ , t0
h(t) = max d(τ ) ,
t0 ≤τ ≤t
+
L j=1
ck (s)
t
t
ck (t)
k=1
r1 (t) = max a(τ ) ,
ti–1
ti
ti–1
t
ti
bj (s) ti–1
ti–1
p gj (τ )vi (τ ) dτ
ds
q θk (τ )vi (τ ) dτ ds + h(t)βi vm i (ti – 0),
i = 1, 2, . . . .
The proof is similar to that of Theorem 3.1, and we omit these details. Theorem 3.3 Suppose that x is a nonnegative piecewise continuous function defined on [t0 , ∞) with discontinuities of the first kind in the points ti (i = 1, 2, . . .) and satisfies the integro-sum inequality: xp (t) ≤ a(t) + b(t) + c(t)
f (s)xp (s) + g(s)xq (s) – h(s)xr (s) ds
t
t0
βi xm (ti – 0),
t ≥ t0 ,
(3.20)
t0
where 0 ≤ t0 < t1 < t2 < · · · , limi→∞ ti = ∞, a(t) is defined on [t0 , ∞) and a(t0 ) = 0, b(t) ≥ 0 and c(t) ≥ 0 are defined on [t0 , ∞), f , g ∈ C(R+ , R+ ), h ∈ C(R+ , (0, +∞)), 0 < q < p < r, βi ≥ 0 (i = 1, 2, . . .) and m > 0 are constants. Then, for any continuous functions k1 (t) > 0 and k2 (t) ≥ 0 on [t0 , ∞) satisfying k(t) = k1 (t) – k2 (t) ≥ 0, the following estimates hold: x(t) ≤ v1 (t),
t ∈ [t0 , t1 ],
(3.21)
x(t) ≤ vi (t),
t ∈ (ti–1 , ti ], i = 2, 3, . . . ,
(3.22)
where
t 1 f (s) + k(s) ds , vi (t) = ri (t) exp e(t) p ti–1 d(t) = max a(τ ) , e(t) = max b(τ ) , 1 p
t0 ≤τ ≤t
t0 ≤τ ≤t
i = 1, 2, . . . , l(t) = max c(τ ) , t0 ≤τ ≤t
r1 (t) = d(t) + e(t)w(t), t θ1 p, q, g(s), k1 (s) + θ2 p, r, h(s), k2 (s) ds, w(t) = t0
(3.23) (3.24)
(3.25)
Liu Advances in Difference Equations (2018) 2018:22
Page 10 of 16
p –q p p – q q p–q p–q θ1 p, q, g(s), k1 (s) = g (s)k1p–q (s), q p p r r – p p r–p –p θ2 p, r, h(s), k2 (s) = h r–p (s)k2r–p (s), r r ti p f (s) + k(s) vi (s) ds + l(t)βi vm ri+1 (t) = ri (t) + e(t) i (ti – 0),
(3.26)
(3.27) i = 1, 2, . . . .
(3.28)
ti–1
Proof From (3.20) and (3.24), we obtain, for t ∈ I0 = [t0 , t1 ],
t
x (t) ≤ d(t) + e(t) p
f (s)xp (s) + g(s)xq (s) – h(s)xr (s) ds.
(3.29)
t0
From Lemma 2.1, (3.24)-(3.27) and (3.29), we have
t
f (s) + k(s) xp (s) + θ1 p, q, g(s), k1 (s) + θ2 p, r, h(s), k2 (s) ds
xp (t) ≤ d(t) + e(t)
t0 t
= d(t) + e(t) + e(t)
θ1 p, q, g(s), k1 (s) + θ2 p, r, h(s), k2 (s) ds
t0 t
f (s) + k(s) xp (s) ds
t0
= d(t) + e(t)w(t) + e(t)
t
f (s) + k(s) xp (s) ds
t0
t
= r1 (t) + e(t)
f (s) + k(s) xp (s) ds,
(3.30)
t0
r1 (t) and e(t) are non-decreasing on [t0 , ∞). Take any fixed T ∈ [t0 , t1 ], and for arbitrary t ∈ [t0 , T], we have
t
f (s) + k(s) xp (s) ds.
x (t) ≤ r1 (T) + e(T) p
(3.31)
t0
Let u(t) = xp (t). Inequality (3.31) is equivalent to u(t) ≤ r1 (T) + e(T)
t
f (s) + k(s) u(s) ds.
(3.32)
t0
Define a function V (t) by the right-hand side of (3.32). Then V (t) is positive and V (t0 ) = r1 (T),
u(t) ≤ V (t),
V (t) = e(T) f (t) + k(t) u(t) ≤ e(T) f (t) + k(t) V (t),
(3.33) t ∈ [t0 , T].
We have
t V (t) ≤ V (t0 ) exp e(T) f (s) + k(s) ds t0
t f (s) + k(s) ds , = r1 (T) exp e(T) t0
(3.34)
Liu Advances in Difference Equations (2018) 2018:22
Page 11 of 16
and then, from (3.33), (3.34) and the assumption u(t) = xp (t), we get
t 1 f (s) + k(s) ds . x(t) ≤ r1 (T) exp e(T) p t0 1 p
Since the above inequality is true for any t ∈ [t0 , T], we obtain
T 1 1 x(T) ≤ r1p (T) exp e(T) f (s) + k(s) ds . p t0 Replacing T by t yields
t 1 x(t) ≤ r1 (t) exp e(t) f (s) + k(s) ds = v1 (t), p t0 1 p
t ∈ I0 = [t0 , t1 ].
(3.35)
This means that (3.21) is true for t ∈ [t0 , t1 ]. For t ∈ I1 = (t1 , t2 ], from Lemma 2.1 and (3.20), (2.24)-(2.27) and (3.35), we obtain
t
xp (t) ≤ d(t) + e(t)
f (s)xp (s) + g(s)xq (s) – h(s)xr (s) ds + l(t)β1 xm (t1 – 0)
t0 t
f (s) + k(s) xp (s) + θ1 p, q, g(s), k1 (s) + θ2 p, r, h(s), k2 (s) ds
≤ d(t) + e(t)
t0
+ l(t)β1 vm 1 (t1
= d(t) + e(t) + e(t)
t0 t
– 0)
t
θ1 p, q, g(s), k1 (s) + θ2 p, r, h(s), k2 (s) ds + l(t)β1 vm 1 (t1 – 0)
f (s) + k(s) xp (s) ds
t0
≤ d(t) + e(t)w(t) + l(t)β1 vm 1 (t1 – 0) + e(t) + e(t)
t
t1 t0
p f (s) + k(s) v1 (s) ds
f (s) + k(s) xp (s) ds
t1
= r1 (t) + l(t)β1 vm 1 (t1 – 0) t t1 p f (s) + k(s) v1 (s) ds + e(t) f (s) + k(s) xp (s) ds + e(t) t0
= r2 (t) + e(t)
t1
t
f (s) + k(s) xp (s) ds.
(3.36)
t1
Inequality (3.36) is the same as (3.30) if we replace r1 (t) and t0 with r2 (t) and t1 in (3.36), respectively. Thus, by (3.35) and (3.36), we get, for t ∈ I1 = (t1 , t2 ],
t 1 1 p f (s) + k(s) ds = v2 (t). x(t) ≤ r2 (t) exp e(t) p t1 Suppose that
t 1 1 p f (s) + k(s) ds x(t) ≤ ri (t) exp e(t) p ti–1 = vi (t) holds for t ∈ Ii–1 = (ti–1 , ti ], i = 2, 3, . . . .
(3.37)
Liu Advances in Difference Equations (2018) 2018:22
Page 12 of 16
Then, for t ∈ Ii = (ti , ti+1 ], from Lemma 2.1 and (3.20), (3.24)-(3.27) and (3.37), we have
t
x (t) ≤ d(t) + e(t) p
f (s)xp (s) + g(s)xq (s) – h(s)xr (s) ds + l(t) βi xm (ti – 0)
t0
f (s) + k(s) xp (s) + θ1 p, q, g(s), k1 (s) + θ2 p, r, h(s), k2 (s) ds
≤ d(t) + e(t) + l(t)
t0
t t0
βi v m i (ti – 0)
t0
= d(t) + e(t)w(t) + e(t) + l(t)
i–1
f (s) + k(s) xp (s) ds + e(t)
tk
k=0
tk+1
t
f (s) + k(s) xp (s) ds
ti
βi v m i (ti – 0)
t0
≤ r1 (t) + e(t) + l(t)
i–1
k=0
tk+1
tk
p f (s) + k(s) vk+1 (s) ds + e(t)
t
f (s) + k(s) xp (s) ds
ti
βi v m i (ti – 0)
t0
t
≤ ri+1 (t) + e(t)
f (s) + k(s) xp (s) ds.
(3.38)
ti
Inequality (3.38) is the same as (3.30) if we replace r1 (t) and t0 with ri+1 (t) and ti in (3.38), respectively. Thus, by (3.35) and (3.38), we have, for t ∈ Ii = (ti , ti+1 ],
t 1 1 p f (s) + k(s) ds . x(t) ≤ ri+1 (t) exp e(t) p ti By induction, we know that (3.30) holds for t ∈ (ti , ti+1 ], for any nonnegative integer i. This completes the proof of Theorem 3.3. Theorem 3.4 Suppose that x is a nonnegative piecewise continuous function defined on [t0 , ∞) with discontinuities of the first kind in the points ti (i = 1, 2, . . .) and satisfies the integro-sum inequality xp (t) ≤ a(t) +
t
f (s)xp (s) ds + t0
+ c(t)
L j=1
βi xm (ti – 0),
t
gj (s)xqj (s) ds –
t0
L j=1
t
hj (s)xrj (s) ds t0
t ≥ t0 ,
t0
where 0 ≤ t0 < t1 < t2 < · · · , limi→∞ ti = ∞, a(t) is defined on [t0 , ∞) and a(t0 ) = 0, b(t) ≥ 0 and c(t) ≥ 0 are defined on [t0 , ∞), f , g ∈ C(R+ , R+ ), h ∈ C(R+ , (0, +∞)), 0 < qj < p < rj (j = 1, 2, . . . , L), βi ≥ 0, i = 1, 2, . . . , and m > 0 are constants. Then, for any continuous functions k1 (t) > 0 and k2 (t) ≥ 0 on [t0 , ∞) satisfying k(t) = k1 (t) – k2 (t) ≥ 0, the following estimates hold: x(t) ≤ v1 (t),
t ∈ [t0 , t1 ],
x(t) ≤ vi (t),
t ∈ (ti–1 , ti ], i = 2, 3, . . . ,
Liu Advances in Difference Equations (2018) 2018:22
Page 13 of 16
where
t 1 1 p vi (t) = ri (t) exp f (s) + Lk(s) ds , p ti–1 l(t) = max c(τ ) , d(t) = max a(τ ) , t0 ≤τ ≤t
w(t) =
i = 1, 2, . . . ,
t0 ≤τ ≤t
r1 (t) = d(t) + w(t),
L t θj p, qj , gj (s), k1 (s) + θj p, rj , hj (s), k2 (s) ds, j=1
t0
p p –q p – qj qj p–qj p–qj p–qj θj p, qj , gj (s), k1 (s) = gj (s)k1 (s), j = 1, 2, . . . , L, qj p rj p rj – p p rj –p r–p rj –p j –p hj (s)k2 (s), j = 1, 2, . . . , L, θj p, rj , hj (s), k2 (s) = rj rj ti p ri+1 (t) = ri (t) + f (s) + Lk(s) vi (s) ds + l(t)βi vm i = 1, 2, . . . . i (ti – 0),
ti–1
The proof is similar to that of Theorem 3.3, and we omit these details.
4 Application In this section, we will apply the results which we have established above to the estimates of solutions of certain impulsive differential equations. Example 4.1 Consider the following impulsive differential equation: ⎧ p t dx (t) ⎪ ⎪ dt = F(t, x(t), t0 G(s, t, x(s)) ds), ⎨ ⎪ ⎪ ⎩
t = ti ,
x|t=ti = d(t)βi xm (ti – 0),
(4.1)
x(t0 ) = x0 ,
where p > 0, m > 0 are constants, the functions d(t) ≥ 0, t ∈ [t0 , ∞), F ∈ C(R × R × R, R+ ) and G ∈ C(R × R × R, R+ ) satisfy the following conditions: F(t, u, v) ≤ f (t)|u|q + |v|,
(4.2)
L M G(s, t, w) ≤ bj (t)gj (s)|w|p + ck (t)θk (s)|w|q , j=1
(4.3)
k=1
where q > 0 (q = p) is a constant, and f (t), gj (t), bj (t) (j = 1, 2, . . . , L), cj (t), θj (t) (j = 1, 2, . . . , M) are defined as in Theorem 3.2. If p – q q–p r p (t) 1+ p i
q–p s l(s) exp e(τ ) dτ ds > 0, p ti–1 ti–1 t
i = 1, 2, . . . ,
then for t ≥ t0 , every solution x(t) of Eq. (4.1) satisfies the following estimates: x(t) ≤ v1 (t), t ∈ [t0 , t1 ], x(t) ≤ vi (t), t ∈ (ti–1 , ti ], i = 2, 3, . . . , where l(t), e(t), ri (t) and vi (t) (i = 1, 2, . . .) are defined as in Theorem 3.2.
(4.4) (4.5)
Liu Advances in Difference Equations (2018) 2018:22
Page 14 of 16
Proof The solution x(t) of Eq. (4.1) satisfies the following equivalent equation:
p
xp (t) = x0 +
F τ , x(τ ),
t
t0
τ
G s, τ , x(s) ds dτ + d(t) βi xm (ti – 0).
t0
t0
From conditions (4.2) and (4.3), it is easy to have t F τ , x(τ ),
x(t) p ≤ |x0 |p +
t0
t0
+ c(t)
τ
G s, τ , x(s) ds dτ
m βi x(ti – 0)
t0
t
≤ |x0 |p +
L q f (τ ) x(τ ) dτ +
t0
+
M
j=1
t t0
k=1
τ
ck (τ )
t
τ
bj (τ ) t0
p gj (s) x(s) ds dτ
t0
q m θk (s) x(s) ds dτ + d(t) βi x(ti – 0) ,
t0
t ≥ t0 .
t0
By using Theorem 3.2, we easily obtain estimates (4.4) and (4.5) of solutions of Eq. (4.1). Example 4.2 Consider the following impulsive differential equation: ⎧ 1 dx(t) 2 ⎪ ⎪ dt = f (t)x(t) + g(t)x 3 (t) – h(t)x (t), ⎨ ⎪ ⎪ ⎩
t = ti ,
x|t=ti = a(t)βi x3 (ti – 0),
(4.6)
x(t0 ) = x0 ,
where 0 ≤ t0 < t1 < t2 < · · · , limi→∞ ti = ∞, f , g ∈ C(R+ , R+ ), h ∈ C(R+ , (0, +∞)), a(t) ≥ 0 is defined on [t0 , ∞) and βi ≥ 0 (i = 1, 2, . . .) are constants. Then, for any continuous functions k1 (t) > 0 and k2 (t) ≥ 0 on [t0 , ∞) satisfying k(t) = k1 (t) – k2 (t) ≥ 0, the following estimates hold: x(t) ≤ v1 (t), t ∈ [t0 , t1 ], x(t) ≤ vi (t), t ∈ (ti–1 , ti ], i = 2, 3, . . . ,
(4.7) (4.8)
where
vi (t) = ri (t) exp
t
f (s) + k(s) ds ,
ti–1
r1 (t) = |x0 | + w(t),
t0 ti
ri+1 (t) = ri (t) + ti–1
and
t w(t) =
(4.9)
1 2 3 –1 √ g 2 (s)k1 2 (s) + h–1 (s)k22 (s) ds, 4 3 3
f (s) + k(s) vi (s) ds + l(t)βi v3i (ti – 0),
l(t) = max a(t) . t0 ≤τ ≤t
i = 1, 2, . . . ,
i = 1, 2, . . . ,
(4.10) (4.11) (4.12)
Liu Advances in Difference Equations (2018) 2018:22
Page 15 of 16
Proof The solution x(t) of Eq. (4.6) satisfies the following equivalent equation: x(t) = x0 +
t
1 f (s)x(s) + g(s)x 3 (s) – h(s)x2 (s) ds + a(t) βi x3 (ti – 0).
t0
t0
From the assumptions of f , g and h, it follows x(t) ≤ |x0 | +
1 2 f (s) x(s) + g(s) x(s) 3 – h(s) x(s) ds
t t0
+ a(t)
3 βi x(ti – 0) ,
t ≥ t0 .
t0
By using Theorem 3.3, we easily obtain estimates (4.7) and (4.8) of solutions of Eq. (4.6). Acknowledgements The author thanks the reviewers for their helpful and valuable suggestions and comments on this paper. This research was supported by the National Natural Science Foundation of China (No. 11671227) A Project of Shandong Province Higher Educational Science and Technology Program (China) (No. J14LI09). Competing interests The author declares that he has no competing interests. Authors’ contributions All authors read and approved the final manuscript.
Publisher’s Note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. Received: 20 July 2017 Accepted: 13 December 2017 References 1. Bainov, DD, Simeonov, PS: Impulsive Differential Equations: Periodic Solution and Applications. Longman, Harlow (1993) 2. Mitropolskiy, YuA, Iovane, G, Borysenko, SD: About a generalization of Bellman-Bihari type inequalities for discontinuous functions and their applications. Nonlinear Anal. 66, 2140-2165 (2007) 3. Borysenko, SD: About one integral inequality for piece-wise continuous functions. In: Proc. X Int. Kravchuk Conf., Kyiv, p. 323 (2004) 4. Iovane, G: Some new integral inequalities of Bellman-Bihari type with delay for discontinuous functions. Nonlinear Anal. 66, 498-508 (2007) 5. Gallo, A, Piccirillo, AM: About new analogies of Gronwall-Bellman-Bihari type inequalities for discontinuous functions and estimated solutions for impulsive differential systems. Nonlinear Anal. 67, 1550-1559 (2007) 6. Borysenko, SD, Ciarletta, M, Iovane, G: Integro-sum inequalities and motion stability of systems with impulse perturbations. Nonlinear Anal. 62, 417-428 (2005) 7. Borysenko, S, Iovane, G: About some new integral inequalities of Wendroff type for discontinuous functions. Nonlinear Anal. 66, 2190-2203 (2007) 8. Gallo, A, Piccirillo, AM: On some generalizations Bellman-Bihari result for integro-functional inequalities for discontinuous functions and their applications. Bound. Value Probl. 2009, Article ID 808124 (2009) 9. Iovane, G: On Gronwall-Bellman-Bihari type integral inequalities in several variables with retardation for discontinuous functions. Math. Inequal. Appl. 11, 599-606 (2008) 10. Gallo, A, Piccirillo, AM: About some new generalizations of Bellman-Bihari results for integro-functional inequalities with discontinuous functions and applications. Nonlinear Anal. 71, 2276-2287 (2009) 11. Deng, SF, Prather, C: Generalization of an impulsive nonlinear singular Gronwall-Bihari inequality with delay. J. Inequal. Pure Appl. Math. 9, Article ID 34 (2008) 12. Wang, WS, Zhou, XL: A generalized Gronwall-Bellman-Ou-Iang type inequality for discontinuous functions and applications to BVP. Appl. Math. Comput. 216, 3335-3342 (2010) 13. Shao, J, Meng, FW: Nonlinear impulsive differential and integral inequalities with integral jump conditions. Adv. Differ. Equ. 2016, 112 (2016) 14. Zheng, B: Some generalized Gronwall-Bellman type nonlinear delay integral inequalities for discontinuous functions. J. Inequal. Appl. 2013, 297 (2013) 15. Mi, YZ, Deng, SF, Li, XP: Nonlinear integral inequalities with delay for discontinuous functions and their applications. J. Inequal. Appl. 2013, 430 (2013) 16. Mi, YZ: Generalized integral inequalities for discontinuous functions with two independent variables and their applications. J. Inequal. Appl. 2014, 524 (2014)
Liu Advances in Difference Equations (2018) 2018:22
Page 16 of 16
17. Wang, WS, Li, ZZ, Tang, AM: Nonlinear retarded integral inequalities for discontinuous functions and their applications. In: Computer, Informatics, Cybernetics and Applications, vol. 107, pp. 149-159 (2012) 18. Zheng, ZW, Gao, X, Shao, J: Some new generalized retarded inequalities for discontinuous functions and their applications. J. Inequal. Appl. 2016, 7 (2016) 19. Li, YS: The bound, stability and error estimates of the solution of nonlinear different equations. Acta Math. Sin. 12(1), 28-36 (1962) (in Chinese) 20. Tian, YZ, Cai, YL, Li, LZ, Li, TX: Some dynamic integral inequalities with mixed nonlinearities on time scales. J. Inequal. Appl. 2015, 12 (2015)
