NODDEA2012_VANKOVA

Some nonlinear difference inequalities
with “maxima” of Bihari type
L. V. Vankovaa
a Plovdiv University, Tzar Asen 24, 4000 Plovdiv, Bulgaria
ABSTRACT
In the paper several new types of nonlinear discrete inequalities are solved. The
main characteristic of the considered inequalities is the presence of the maximum of
the unknown function over a past time interval. These inequalities are discrete gener-
alizations of Bihari’s inequality. They are base of studying qualitative properties of the
solutions of nonlinear difference equations with “maxima”.
Keywords: Discrete Bihari type inequality, difference equations with maxima,
bounds
1. INTRODUCTION
In the recent years great attention has been paid to finite difference equations
and their applications in modeling of real world problems ([1], [2], [3], [4], [5]).
At the same time there are many real world processes in which the present state
depends significantly on its maximal value on a past time interval. Adequate
mathematical models of these processes are so called difference equations with
“maxima”. Meanwhile, this type of equations is not widely studied yet and there
are only some isolated results ([6], [7]).
The development of the theory of difference equations with “maxima” requires
solving of finite difference inequalities that involve the maximum value of the un-
known function. Note that many different types of inequalities in the continuous
case are solved in the papers [8], [9], [10], [11], [12], [13], [14], [15].
The main purpose of the paper is solving of a new type of nonlinear discrete
inequalities containing the maximum over a past time interval. Some of the solved
inequalities are applied to difference equations with “maxima” and bounds of their
solutions are obtained.
Further author information: (Send correspondence to L. V. Vankova.)
L. V. Vankova: E-mail: lilqna.v@gmail.com

2 L. V. Vankova
2. PRELIMINARY NOTES AND DEFINITIONS
Let R+ = [0, +∞), N be the set of all integers, N[α, β] = {α, α + 1, . . . β},
where α, β ∈ N : α < β and h ≥ 0 is a given fixed integer. Denote by ∆u(n) =
u(n + 1) − u(n). We will assume that m
s=n = 0 and m
s=n = 1 for m < n.
Definition 2.1. The function ω ∈ C(R+, R+) is said to be from the class Ω1 if
ω is a nondecreasing function and ω(u) > 0 for u > 0 and
∞
du
ω(u)
= ∞.
Definition 2.2. The function ω ∈ Ω1 is said to be from the class Ω2 if tω(u) ≤
ω(tu) for t ∈ [0, 1];
Definition 2.3. The function ω ∈ Ω2 is said to be from the class Ω3 if ω(u+v) ≤
ω(u) + ω(v).
Remark 2.4. For example, ω(u) =
√
u
1+
√
u
, ω(u) = u
1+u
∈ Ω3.
3. MAIN RESULTS
Theorem 3.1. Let the following conditions be fulfilled:
1. The functions fi, gj : N[0, T] → R+, i ∈ N[1, l ], j ∈ N[1, m], l, m < +∞.
2. The function ϕ : N[−h, 0] → R+ and the constants p ≥ 0, a > 0.
3. The functions ωi, ˜ωj ∈ Ω1, i ∈ N[1, l ], j ∈ N[1, m], l, m < +∞.
4. The function u : N[−h, T] → R+ satisfies the inequalities
u(n) ≤ a +
n−1
s=0
up
(s)
l
i=1
fi(s)ωi u(s) +
+
m
j=1
gj(s)˜ωj max
ξ∈N[s−h, s]
u(ξ) for n ∈ N[0, T],
(1)
u(n) ≤ ϕ(n) for n ∈ N[−h, 0]. (2)
Then for n ∈ N[0, β1] the inequality
u(n) ≤ Ψ−1
W−1
W Ψ(M) + A(n) (3)
holds, where Ψ−1
and W−1
are the inverse functions of
Ψ(r) =
r
r0
ds
sp
, 0 < r0 < M, r0 ≤ r, (4)

Some nonlinear difference inequalities with “maxima” ... 3
W(r) =
r
r1
ds
¯w Ψ−1(s)
, 0 < r1 < Ψ(M), r1 ≤ r, (5)
A(n) =
n−1
s=0
l
i=1
fi(s) +
m
j=1
gj(s) , (6)
¯w(t) = max max
1≤i≤l
ωi(t), max
1≤j≤m
˜ωj(t) for t ∈ R+, (7)
M = max a, max
n∈N[−h, 0]
ϕ(n) , (8)
β1 = sup n1 ∈ N[0, T] : W Ψ(M) + A(n) ∈ Dom W−1
,
W−1
W Ψ(M) + A(n) ∈ Dom Ψ−1
for all n ∈ N[0, n1] .
Proof. Define a nondecreasing function V : N[−h, T] → [M, +∞) by
V (n) =



M +
n−1
s=0
up
(s)
l
i=1
fi(s)ωi u(s)
+
m
j=1
gj(s)˜ωj max
ξ∈N[s−h,s]
u(ξ) for n ∈ N[0, T],
M for n ∈ N[−h, 0],
where M is defined by (8). From the definition of V (n) it follows that
u(n) ≤ V (n) for n ∈ N[−h, T], (9)
max
ξ∈N[n−h, n]
u(ξ) ≤ max
ξ∈N[n−h, n]
V (ξ) = V (n) for n ∈ N[0, T]. (10)
From (7), (9) and (10) we get for n ∈ N[0, T]
V (n) ≤ M +
n−1
s=0
V (s)
p
¯w V (s)
l
i=1
fi(s) +
m
j=1
gj(s) := K(n), (11)
where K : N[0, T] → [M, +∞) is a nondecreasing function and K(0) = M and
V (n) ≤ K(n) holds for n ∈ N[0, T]. Therefore, from (11) we get for n ∈ N[0, T]
∆K(n)
K(n)
p ≤ ¯w K(n)
l
i=1
fi(n) +
m
j=1
gj(n) . (12)

4 L. V. Vankova
According to the mean value theorem we get for some ξ ∈ K(n), K(n + 1)
∆Ψ K(n) = Ψ K(n + 1) − Ψ K(n)
= Ψ (ξ)∆K(n) =
∆K(n)
ξp
≤
∆K(n)
K(n)
p ,
(13)
where Ψ is defined by (4). Then from (12) and (13) it follows for n ∈ N[0, T]
∆Ψ K(n) ≤ ¯w K(n)
l
i=1
fi(n) +
m
j=1
gj(n) . (14)
Summing up inequality (14) from 0 to n − 1 for n ∈ N[0, T], we get
Ψ K(n) ≤ Ψ(M) +
n−1
s=0
¯w K(s)
l
i=1
fi(s) +
m
j=1
gj(s) := K1(n), (15)
where the function K1 : N[0, T] → [Ψ(M), +∞) is nondecreasing and K1(0) =
Ψ(M) and V (n) ≤ K(n) ≤ Ψ−1
K1(n) holds for n ∈ N[0, β1]. Therefore,
∆K1(n)
¯w Ψ−1 K1(n)
≤
l
i=1
fi(n) +
m
j=1
gj(n) for n ∈ N[0, β1]. (16)
According to the mean value theorem it follows for ξ ∈ K1(n), K1(n + 1)
∆W K1(n) = W K1(n + 1) − W K1(n)
= W ξ ∆K1(n) =
∆K1(n)
¯w Ψ−1 ξ
≤
∆K1(n)
¯w Ψ−1 K1(n)
, (17)
where W is defined by (5). Therefore, for n ∈ N[0, β1] we get
∆W K1(n) ≤
l
i=1
fi(n) +
m
j=1
gj(n). (18)
Summing up inequality (18) from 0 to n − 1 for n ∈ N[0, β1], we obtain
W K1(n) ≤ W Ψ(M) + A(n), (19)
where A(n) is defined by (6). Since W−1
is an increasing function from inequalities
u(n) ≤ V (n) ≤ K(n) ≤ Ψ−1
K1(n) and (19) we obtain (3).

Note that the inequalities (1), (2) could have another type of solution as fol-
lows, which is simpler than (3) but the used integral function is more complicated.
Theorem 3.2. Let the conditions of Theorem 3.1 be satisfied.
u(n) ≤ Ψ−1
1 Ψ1(M) + A(n) (20)
holds, where A(n), ¯w(s) and M are defined by (6), (7) and (8), respectively, and
Ψ−1
1 is the inverse function of
Ψ1(r) =
r
r2
ds
sp ¯w(s)
, 0 < r2 < M, r2 ≤ r, (21)
β2 = sup n2 ∈ N[0, T] : Ψ1(M) + A(n) ∈ Dom Ψ−1
1 for all n ∈ N[0, n2] .
Proof. Following the proof of Theorem 3.1, we obtain inequality (12). Ac-
cording to the mean value theorem it follows for some ξ ∈ K(n), K(n + 1)
∆Ψ1 K(n) = Ψ1 K(n + 1) − Ψ1 K(n)
= Ψ1(ξ)∆K(n) =
∆K(n)
ξp ¯w ξ
≤
∆K(n)
K(n)
p
¯w K(n)
,
(22)
where Ψ1 is defined by (21). Therefore, for n ∈ N[0, T] we get
∆Ψ1 K(n) ≤
l
i=1
fi(n) +
m
j=1
gj(n). (23)
Summing up inequality (23) from 0 to n − 1 for n ∈ N[0, T] and obtain
Ψ1 K(n) ≤ Ψ1 M + A(n), (24)
where A(n) is defined by (6). Since Ψ−1
1 is an increasing function from inequalities
u(n) ≤ V (n) ≤ K(n) and (24) we obtain the required inequality (20).
In the case when the functions ωi, ˜ωj ∈ Ω2 the following results are valid.
3. The function a : N[0, T] → [1, +∞) is nondecreasing, p = constant ≥ 0.

6 L. V. Vankova
4. The function ϕ : N[−h, 0] → R+ with maxs∈N[−h, 0]
ϕ(s) ≤ a(0).
u(n) ≤ a(n) +
n−1
s=0
up
(s)
l
i=1
fi(s)ωi u(s) +
+
m
j=1
gj(s)˜ωj max
ξ∈N[s−h, s]
u(ξ) for n ∈ N[0, T],
(25)
u(n) ≤ ϕ(n) for n ∈ N[−h, 0]. (26)
u(n) ≤ a(n)Ψ−1
W−1
W Ψ(1) + A1(n) (27)
holds, where Ψ and W are deﬁned by (4) and (5), respectively, and
A1(n) =
n−1
s=0
ap
(s)
l
i=1
fi(s) +
m
j=1
gj(s) , (28)
β3 = sup n3 ∈ N[0, T] : W Ψ(1) + A1(n) ∈ Dom W−1
,
W−1
W Ψ(1) + A1(n) ∈ Dom Ψ−1
for all n ∈ N[0, n3] .
Proof. From inequalities (25), (26) and 0 < 1
a(n)
≤ 1 it follows
u(n)
a(n)
≤ 1 +
n−1
s=0
up
(s)
l
i=1
fi(s)ωi
u(s)
a(s)
+
m
j=1
gj(s)˜ωj
maxξ∈N[s−h, s]
u(ξ)
a(s)
for n ∈ N[0, T],
(29)
u(n) ≤
ϕ(n)
a(0)
≤ 1 for n ∈ N[−h, 0]. (30)
From the monotonicity of the function a(n) it follows
maxξ∈N[n−h, n]
u(ξ)
a(n)
=
u(η)
a(n)
≤
u(η)
a(η)
≤ max
ξ∈N[n−h, n]
u(ξ)
a(ξ)
for n ∈ N[0, T], (31)

where η ∈ N[n−h, n]. Define a nondecreasing function v : N[−h, T] → R+ by
v(n) =



u(n)
a(n)
for n ∈ N[0, T],
u(n)
a(0)
for n ∈ N[−h, 0].
Use inequalities (29), (30), (31) and the definition of v(n) and get
v(n) ≤ 1+
n−1
s=0
ap
(s)vp
(s)
l
i=1
fi(s)ωi v(s) +
+
m
j=1
gj(s)˜ωj max
ξ∈N[s−h, s]
v(ξ) for n ∈ N[0, T],
(32)
v(n) ≤ 1 for n ∈ N[−h, 0]. (33)
According to Theorem 3.1 for a = M = 1 we obtain (27).
Theorem 3.4. Let the conditions of Theorem 3.3 be satisfied.
u(n) ≤ a(n)Ψ−1
1 Ψ1(1) + A1(n) (34)
holds, where Ψ1(r) and A1(n) are defined by (21) and (28), respectively, and
β4 = sup n4 ∈ N[0, T] : Ψ1(1) + A1(n) ∈ Dom Ψ−1
1 for all n ∈ N[0, n4] .
Proof. Following the proof of Theorem 3.3, we obtain inequalities (32) and
(33). According to Theorem 3.2 for a = M = 1 we obtain (34).
In the case when the functions ωi, ˜ωj ∈ Ω3 the following result is valid.
2. The function a : N[−h, T] → R+.
u(n) ≤ a(n) +
n−1
s=0
l
i=1
fi(s)ωi u(s) +
+
m
j=1
gj(s)˜ωj max
ξ∈N[s−h, s]
u(ξ) for n ∈ N[0, T],
(35)

8 L. V. Vankova
u(n) ≤ a(n) for n ∈ N[−h, 0]. (36)
u(n) ≤ a(n) + B(n)Ψ−1
1 Ψ1(1) + A(n) (37)
holds, where A(n) and Ψ1(n) are defined by (6) and (21), respectively, and
B(n) = 1 +
n−1
s=0
l
i=1
fi(s)ωi a(s) +
m
j=1
gj(s)˜ωj max
ξ∈N[s−h, s]
a(ξ) , (38)
β5 = sup n5 ∈ N[0, T] : Ψ1(1) + A(n) ∈ Dom Ψ−1
1 for all n ∈ N[0, n5] .
Proof. Define a nondecreasing function z : N[−h, T] → R+ by the equations
z(n) =



n−1
s=0
l
i=1
fi(s)ωi u(s) +
m
j=1
gj(s)˜ωj max
ξ∈N[s−h, s]
u(ξ) for n ∈ N[0, T],
0 for n ∈ N[−h, 0].
From (35), (36) and the definition of the function z(n) it follows
u(n) ≤ a(n) + z(n) for n ∈ N[−h, T], (39)
max
ξ∈N[n−h, n]
u(ξ) ≤ max
ξ∈N[n−h, n]
a(ξ) + max
ξ∈N[n−h, n]
z(ξ) for n ∈ N[0, T]. (40)
From (39), (40) and the definition of z(n) we get for n ∈ N[0,T]
z(n) ≤ B(n) +
n−1
s=0
l
i=1
fi(s)ωi z(s) +
m
j=1
gj(s)˜ωj max
ξ∈N[s−h, s]
z(ξ) , (41)
where the nondecreasing function B(n) is defined by (38) and B(0) = 1.
According to Theorem 3.4 for p = 0 and a = M = 1 from (41) and the
definition of the function z(n) for n ∈ N[−h, 0] we get for n ∈ N[0, β5]
z(n) ≤ B(n)Ψ−1
1 Ψ1(1) + A(n) , (42)
where A(n) and Ψ1 are defined by (6) and (21), respectively. From (39) and (42)
it follows the validity of the required inequality (37).

4. APPLICATIONS
Example: Consider the following difference equation with “maxima”
∆v(n) = F n, v(n), max
s∈N[n−h, n]
v(s) for n ∈ N[0, T], (43)
with the initial condition
v(n) = ϕ(n) for n ∈ N[−h, 0], (44)
where v : N[−h, T] → R, ϕ : N[−h, 0] → R, F : N[0, T] × R2
→ R.
Theorem 4.1. (Upper bound) Let the following conditions be fulfilled:
1. The functions F : N[0, T] × R2
→ R, R, Q : N[0, T] → R+ satisfies
F n, x, y ≤ x
q
R(n) x
t
+ Q(n) y
t
for n ∈ N[0, T], x, y ∈ R,
where the constants q, t are such that 0 ≤ q < 1, 0 ≤ t < 1 and 1 − q − t > 0.
2. The function ϕ : N[−h, 0] → R and M = maxs∈N[−h, 0]
|ϕ(s)|.
3. The IVP (43), (44) has at least one solution, defined for n ∈ N[−h, T].
Then for n ∈ N[0, T] the solution of the IVP (43), (44) satisfies the inequality
|v(n)| ≤ 1−q−t
M1−q−t + (1 − q − t)
n−1
s=0
R(s) + Q(s) . (45)
Proof. For the solution of the IVP (43), (44) it follows for n ∈ N[0, T]
|v(n)| ≤ M +
n−1
s=0
|v(s)|q
R(s)|v(s)|t
+ Q(s) max
ξ∈N[s−h,s]
v(ξ)
t
, (46)
|v(n)| ≤ M for n ∈ N[−h, 0]. (47)
Set u(n) = |v(n)| for n ∈ N[−h, T]. According to Theorem 3.2 from (46),
(47) for a = M, l ≡ m ≡ 1, p = q, f(n) ≡ R(n), g(n) ≡ Q(n), n ∈ N[0, T],
ω(V ) ≡ ˜ω(V ) = V t
, Ψ1(r) =
r
0
ds
sq+t = r1−q−t
1−q−t
, Ψ1(M) = M1−q−t
1−q−t
, Ψ−1
1 (r) =
r(1 − q − t)
1
1−q−t
, Dom(Ψ−1
1 ) = R+ we obtain for n ∈ N[0, T] the required
inequality (45).

10 L. V. Vankova
Acknowledgments. Research was partially supported by Plovdiv Univer-
sity, Fund “Scientific Research” MU11FMI005/29.05.2011.
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