A New Generalized Gronwall-Bellman Type Inequality

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2012 International Conference on Image, Vision and Computing (ICIVC 2012)
IPCSIT vol. 50 (2012) © (2012) IACSIT Press, Singapore
DOI: 10.7763/IPCSIT.2012.V50.46
A New Generalized Gronwall-Bellman Type Inequality
Qinghua Feng+
School of Science, Shandong University of Technology, Zhangzhou Road 12, Zibo, Shandong, China, 255049
Abstract. In this paper, a new nonlinear integral inequality is established, which provide a handy tool for
analyzing the global existence of solutions of differential and integral equations.
Keywords: Integral inequality; Global existence; Integral equation; Differential equation
1. Introduction
During the past decades, with the development of the theory of differential and integral equations, a lot of
integral inequalities, for example [1-13], have been discovered, which play an important role in the research of
boundedness, global existence, stability of solutions of differential and integral equations.
In [12], Jiang proved the following theorem:
Theorem A: R+ = [0, ∞) . Suppose that x (t ), f (t ), h(t ) ∈ C ( R+ , R+ ) . Then the following form of delay integral
inequality:
t
x p (t ) ≤ C + ∫ [ f ( s ) x p ( s ) + h ( s ) x p (σ ( s ))]ds, t ∈ R+
0
with the initial condition
1
x (t ) = φ (t ), t ∈ [α , 0], φ (σ (t )) ≤ ( ρ (t )) p , t ∈ R+ , σ (t ) ≤ 0
where p, q are constants, p > 0, q > 0, p ≠ q . σ (t) ∈C(R+ , R) , σ (t ) ≤ t, −∞ < α = inf{σ (t ), t ∈ R+ } ≤ 0,φ ∈C([α ,0], R+ ) ,
implies that
t
x (t ) ≤ exp( ∫
0
t
s
p −q
1
f (s)
p−q
p−q
h ( s ) exp( ∫
f (τ ) dτ ) ds ] p−q
ds ) [C p + ∫
p
p
p
0
0
for t ∈ [0, t0 ] , where t0 is a positive number satisfying
inf {C
t∈[0, t0 ]
p−q
p
t
+∫
0
s
p−q
p−q
h ( s ) exp( ∫
f (τ ) dτ ) ds} > 0 .
p
p
0
In this paper, motivated by the above work, we will prove more general theorem and establish a new integral
inequality. Also we will give one example so as to illustrate the validity of the present integral inequality.
+
Corresponding author.
E-mail address: fqhua@sina.com
2. Main Results
Theorem 2.1: Assume that x, a ∈ C ( R+ , R+ ) , a (t ) is non-decreasing. f , g , ∂ t f , ∂ t g ∈ C ( R+ × R+ , R+ ) .
ω ∈ C ( R+ , R+ ) be nondecreasing with ω (u ) > 0 on (0, ∞) . If x (t ) satisfies the following delay integral
inequality:
t
x p (t ) ≤ a(t ) + ∫ [ f ( s, t ) x p (σ 1 ( s)) + g ( s, t ) x q (σ 2 ( s))ω ( x(σ 3 ( s)))]ds, t ∈ R+
(1)
0
with the initial condition
1
x(t ) = φ (t ), t ∈ [α , 0], φ (σ i (t )) ≤ (a (t )) p , t ∈ R+ , σ i (t ) ≤ 0, i = 1, 2,3
(2)
where p, q are constants, p > q > 0 , σi ∈C(R+ , R) , σi (t) ≤ t, −∞ < α = inf{min{σi (t), i = 1,2,3}, t ∈ R+} ≤ 0,φ ∈C([α,0], R+ ) ,
then for t ∈ R+ ,
x (t ) ≤ {Ω −1[Ω(exp(
t
p−q
1
p−q
p−q
p−q
p−q
F1 (t ))a p (t )) + exp(
F1 (t )) ∫
F2 '( s ) exp( −
F1 ( s )) ds ]} p−q
p
p
p
p
0
(3)
where
t
t
0
r
0
F1 (t ) = ∫ f ( s , t ) ds, F2 (t ) = ∫ g ( s, t ) ds ,
Ω(r ) = ∫
1
1
1
p −q
ω (s )
(4)
ds, Ω −1 is the inverse of Ω .
Proof: We notice (3) holds for t = 0 obviously. Let the right side of (1) be ϕ p (t ) . Then
x (t ) ≤ ϕ (t )
(5)
When σ i (t ) ≥ 0 , we have
x(σ i (t )) ≤ ϕ (σ i (t )) ≤ ϕ (t )
(6)
When σ i (t ) ≤ 0 , we have
1
x (σ i (t )) = φ (σ i (t )) ≤ a p (t ) ≤ ϕ (t )
(7)
So from (6), (7) we always have x (σ i (t )) ≤ ϕ (t ), i = 1, 2, 3 . Fix T > 0 . Then for t ∈ (0, T ] , we have
t
ϕ p (t ) ≤ a(T ) + ∫ [ f ( s, t )ϕ p (s) + g (s, t )ϕ q (s)ω (ϕ (s))]ds
(8)
0
Let the right side of (8) be u p (t ) . Then
ϕ (t ) ≤ u (t ), t ∈ (0, T ]
(9)
and
u
p −1
t
∂g ( s, t ) q
1 ∂f ( s, t ) p
ϕ ( s)ω (ϕ (s))ds + g (t , t )ϕ q (t )ω (ϕ (t ))]
(t )u '(t ) = [ ∫
ϕ ( s )ds + f (t , t )ϕ p (t ) + ∫
t
∂
p 0 ∂t
0
t
t
≤
d ∫ f ( s, t )ds
0
dt
t
1 p
ϕ (t ) +
p
d ∫ g ( s, t )ω (u ( s))ds
t
≤
0
dt
1 q
ϕ (t )
p
t
d ∫ f ( s, t )ds
1 p
u (t ) +
p
0
dt
d ∫ g ( s, t )ω (u ( s ))ds
0
dt
1 q
u (t )
p
(10)
Then
t
u '(t ) ≤
d ∫ f (s, t )ds
0
dt
t
1
u(t ) +
p
d ∫ g(s, t)ω(u(s))ds
0
dt
1 1+q− p
u
(t )
p
(11)
Let v(t ) = u p − q (t ) . Then
t
v '(t ) ≤
d ∫ f (s, t )ds
0
dt
t
p−q
v(t ) +
p
d ∫ g(s, t )ω(u(s))ds
0
dt
t
p−q
≤
p
d ∫ f (s, t )ds
0
dt
t
p−q
v(t ) +
p
d ∫ g(s, t )ds
0
dt
1
p−q
ω(v p−q (t))
p
that is,
v '(t ) −
1
p−q
p−q
F1 '(t )v(t ) ≤
F2 '(t )ω(v p−q (t ))
p
p
(1
2)
Multiplying exp(− p − q F1 (t )) on both sides of (12), we have
p
d[exp(−
p−q
F1(t))v(t)]
1
p−q
p−q
p
≤
F2 '(t)ω(v p−q (t))exp(−
F1(t))
dt
p
p
(13)
Integrating (13) from 0 to t , it follows
exp(−
t
p −q
p−q
p−q
F1(t))v(t) − v(0) ≤ ∫
F2 '(s)ω(v (s))exp(−
F1 (s))ds
p
p
p
0
1
p−q
(14)
1
p−q
p−q
p −q
F2 '(s)ω(v p−q (s))exp(−
F1(s))ds}exp(
F1(t))
p
p
p
(15)
p −q
p
Since v(0) = u p − q (0) = a (T ) , we have
p−q
t
v(t) ≤ {a p (T ) + ∫
0
and
t
p−q
1
1
p−q
p−q
u(t) ≤ exp( F1(t)) {a p (T ) + ∫
F2 '(s)ω(u(s))exp(−
F1(s))ds}p−q
p
p
p
0
(16)
t
p−q
Let k(t) = a p (T ) + ∫
0
1
p−q
p−q
F2 '(s)ω(v p−q (s))exp(−
F1(s))ds .
p
p
Then
1
1
u (t ) ≤ exp( F1 (t ))k p−q (t )
p
(17)
and
k '(t) =
1
p−q
p−q
p−q
1
p−q
F2 '(t)ω(u(t))exp(−
F1(t)) ≤
F2 '(t)ω(exp( F1(T ))k p−q (t))exp(−
F1(t))
p
p
p
p
p
Then
k '(t)
p−q
p−q
≤
F2 '(t)exp(−
F1(t))
1
1
p
p
p−q
ω(exp( F1(T ))k (t))
p
(18)
Integrating (18) from 0 to t , it follows
Ω[exp(
t
p−q
p−q
p−q
p−q
p −q
p −q
F1 (T ))k (t )] −Ω[exp(
F1 (T ))a p (T )] ≤ exp(
F1(T ))∫
F2 '(s)exp(−
F1(s))ds
p
p
p
p
p
0
(19)
t
p−q
q− p
p−q
p−q
p−q
p−q
F1(T ))Ω−1[Ω(exp(
F1(T ))a p (T )) + exp(
F1(T ))∫
F2 '(s)exp(−
F1(s))ds]
p
p
p
p
p
0
(20)
Then
k(t) ≤ exp(
From (17) and (20), it follows
u(t) ≤ {Ω−1[Ω(exp(
t
p−q
1
p −q
p−q
p−q
p −q
F1(T ))a p (T )) + exp(
F1(T ))∫
F2 '(s)exp(−
F1(s))ds]}p−q , t ∈(0,T ]
p
p
p
p
0
(21)
Combining (5), (9), (21) we have
x(t) ≤ {Ω−1[Ω(exp(
t
p−q
1
p −q
p−q
p−q
p −q
F1(T ))a p (T )) + exp(
F1(T ))∫
F2 '(s)exp(−
F1(s))ds]}p−q , t ∈(0,T ]
p
p
p
p
0
(22)
Setting t = T and considering T ∈ R+ is arbitrary, we have completed the proof.
Corallary 2.2: Assume that x, a, f , g ∈ C(R+ , R+ ), m ∈ C1 (R+ , R+ ) . If x, a, p, q,σi (t ),α,φ, ω are the same as in
Theorem 2.1, and x(t) satisfies the following delay integral inequality:
t
x p (t ) ≤ a(t ) + m(t )∫ [ f (s) x p (σ1 (s)) + g (s) xq (σ 2 (s))ω( x(σ 3 (s)))]ds, t ∈ R+
(23)
0
with the initial condition (2), then for t ∈ R+ ,
t
p−q
p−q
p−q
p−q
x(t ) ≤ {Ω −1[Ω(exp(
F1 (t ))a (t )) + exp(
F1 (t )) ∫
F2 '( s ) exp( −
F1 ( s )) ds ]}
p
p
p
p
0
p −q
p
1
p −q
(24)
where
t
t
0
0
(25)
F1 (t ) = m (t ) ∫ f ( s ) ds , F2 (t ) = m (t ) ∫ g ( s ) ds
3.
Application
Example: We consider the following delay differential equation
( x p (t )) ' = F (t , x (σ 1 (t )), x(σ 2 (t )), x(σ 3 (t )))
(26)
with the initial condition
1
x(t ) = φ (t ), t ∈ [α , 0],| φ (σ i (t )) |≤ A p , t ∈ R+ , σ i (t ) ≤ 0, i = 1, 2,3
where
A>0
is
a
constants
and
A = x p (0) , F ∈ C( R+ × R3 , R) ,. σi (t) ∈C(R+ , R) , σi (t) ≤ t, −∞< α
= inf{min{σi (t), i =1,2,3}, t ∈R+} ≤ 0, φ ∈C([α,0], R+ ) , Assume | F(t, x, y, z) |≤ f (t) | x | p +g(t) | x |q v(| z |)
∞
where f , g , v ∈ C ( R+ , R+ ) , v is nondecreasing, and v ( s ) ds = ∞ . p > q > 0
∫
1
Integrating (26) from 0 to t , it follows
t
x p (t ) − x p (0) = ∫ F ( s , x (σ 1 ( s )), x (σ 2 ( s )), x (σ 3 ( s ))) ds
o
So
t
| x p (t ) | − | x p (0) |≤| x p (t ) − x p (0) | ≤ | F ( s , x (σ 1 ( s )), x (σ 2 ( s )), x (σ 3 ( s ))) | ds
∫
o
t
≤ ∫ [ f ( s ) | x (σ 1 ( s )) | p + g ( s ) | x (σ 2 ( s )) |q v (| x (σ 3 ( s )) |)]ds
o
Taking ω (u ) = v (u ) , from Theorem 2.1, we can reach the estimate
| x(t ) |≤ {Ω −1[Ω(exp(
t
p−q
1
p−q
p−q
p−q
p−q
F1 (t )) | A | p ) + exp(
F1 (t )) ∫
F2 '( s ) exp( −
F1 ( s )) ds ]} p−q
p
p
p
p
0
where
t
t
0
0
F1 (t ) = ∫ f ( s ) ds, F2 (t ) = ∫ g ( s ) ds ,
which shows x(t ) does not blow up in finite time. So the solution of (26) is global.
4. Conclusions
In this paper, we establish a new integral inequality, which provides a handy tool in the qualitative analysis
of solutions of integral equations and differential equations. Our result generalizes the result in [12].
5. References
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