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Hindawi Publishing Corporation
Advances in Difference Equations
Volume 2010, Article ID 101959, 19 pages
doi:10.1155/2010/101959
Research Article
A Mixed Problem for Quasilinear Impulsive
Hyperbolic Equations with Non Stationary
Boundary and Transmission Conditions
Akbar B. Aliev
1
and Ulviya M . Mamedova
2
1
Azerbaijan Technical University, AZ 1073, Baki, Azerbaijan
2
Institute of Mathematics and Mechanics of NAS of Azerbaijan, AZ 1141, Baku, Azerbaijan
Correspondence should be addressed to Akbar B. Aliev,
Received 10 March 2010; Revised 13 June 2010; Accepted 26 October 2010
Academic Editor: Toka Diagana
Copyright q 2010 A. B. Aliev and U. M. Mamedova. This is an open access article distributed
under the Creative Commons Attribution License, which permits unrestricted use, distribution,
and reproduction in any medium, provided the original work is properly cited.
The initial-boundary value problem for a class of linear and nonlinear equations in Hilbert space
is considers. We prove the existence and uniqueness of solution of this problem. The results of
this investigation are applied to solvability of initial-boundary value problems for quasilinear
impulsive hyperbolic equations with non-stationary transmission and boundary conditions.
1. Abstract Model Initial Boundary Value Problem with
Non Stationary Boundary and Transmission Conditions for
the Impulsive Linear Hyperbolic Equations
In paper 1 there is given an abstract scheme of investigation of mixed problems for
hyperbolic equations with non stationary boundary conditions. In this direction, some results
were obtained in 2.
In this paper, we offer the analogues abstract model of investigation of mixed
problem with non stationary boundary and transmission conditions for impulsive linear and
semilinear hyperbolic equations.
1.1. Statement of the Problem and Main Theorem
Let H
i
,H
i
0
,X
i
ν
,Y
j
μ
ν 1, 2, ,s
i
; i 1, 2, ,m; μ 1, 2, ,r
j
; j 1, 2, ,m be Hilbert
Spaces. Consider the following abstract initial-boundary value problem:
2AdvancesinDifference Equations
¨u
i
t
A
i
t
u
i
t
f
i
t
,
hyperbolic equations
, 1.1
B
iν
¨u
i
t
m
k1
C
i
kν
t
u
k
t
g
iν
t
,
non stationary boundary and transmission conditions
,
m
k1
D
i
kμ
u
k
t
0,
stationary boundary and transmission conditions
,
u
i
0
u
0
i
, ˙u
i
0
u
1
i
,
initial conditions
,
1.2
where t ∈ 0,T,¨u
i
d
2
u
i
/dt
2
,˙u
i
du
i
/dt, A
i
t are the linear closed operators in H
i
; B
iν
are
the linear operators from H
i
to X
i
ν
; C
i
kν
t are the linear operators from H
k
to X
i
ν
; D
j
kμ
are the
linear operators from H
k
to Y
j
μ
; ν 1, ,s
i
, i 1, ,m, μ 1, ,r
j
, j 1, ,m,k 1, ,m.
We will investigate this problem under the following conditions.
i Let H
i
0
⊂ H
i
,andletH
i
0
be densely in H
i
and continuously imbedded into it, i
1, 2, ,m.
In the Hilbert space H
i
, it was defined the system of the inner products ·, ·
H
i
t
,which
generate uniform equivalent norms, that is,
c
−1
1
u
2
H
i
≤
u
2
H
i
t
≤ c
1
u
2
H
i
,c
1
> 0,
u
2
H
i
t
u, u
H
i
t
,t∈
0,T
,i 1, 2, ,m.
1.3
For each u ∈ H
i
, the function t →u
2
H
i
t
: 0,T → R
is continuously differentiable,
i 1, 2, ,m.
In the Hilbert space X
i
ν
, it w as defined the system of the inner products ·, ·
X
ν
i
,which
generate uniform equivalent norms, that is,
c
−1
2
v
2
X
i
ν
≤
v
2
X
i
ν
t
≤ c
2
v
2
X
i
ν
,c
2
> 0,
v
2
X
i
ν
t
v, v
X
i
ν
t
,t∈
0,T
,ν 1, 2, ,s
i
,i 1, 2, ,m.
1.4
For each v ∈ X
i
ν
, the function t →v
2
X
i
ν
t
: 0,T → R
is continuously differentiable.
ii For each t ∈ 0,T and i 1, 2, ,m, A
i
t is a linear closed operator in H
i
whose
domain is H
i
0
; A
i
t acts boundedly from H
i
0
to H
i
; A
i
t is strongly continuously
differentiable.
iii The linear operators B
iν
,thatactfromH
i
1/2
to X
i
ν
, bounded, where H
i
1/2
H
i
0
,H
i
1/2
is interpolation space between H
i
0
and H
i
of order 1/2 ν 1, ,s
i
,i
1, ,msee 3.
iv For each t ∈ 0,T, the linear operators C
i
kν
t,thatactfromH
k
to X
i
ν
,are
bounded; C
i
kν
t is strongly continuously differentiable ν 1, ,s
i
, i 1, ,m;
k 1, ,m.
Advances in Difference Equations 3
v The linear operators D
j
kμ
,fromH
k
1/2
into Y
j
μ
, act boundedly μ 1, ,r
j
, j
1, ,m; k 1, ,m.
Let us introduce the following designations:
H H
1
⊕···⊕H
m
,
H
0
u : u
u
1
, ,u
m
,u
i
∈ H
i
0
,i 1, ,m;
m
k1
D
j
kμ
u
k
0,μ 1, ,r
j
,j 1, ,m
,
H
1/2
u : u
u
1
, ,u
m
,u
i
∈ H
i
1/2
,i 1, ,m;
m
k1
D
j
kμ
u
k
0,μ 1, ,r
j
,j 1, ,m
,
H
1
w : w
w
1
, ,w
m
,w
i
u
i
,B
i1
u
i
, ,B
is
i
u
i
,i 1, ,m,
where
u
1
, ,u
m
∈
H
0
,
H
i
H
i
⊕ X
i
1
⊕···⊕X
i
s
i
, H
m
i1
H
i
, H
1/2
H
1
, H
1/2
.
1.5
From condition v, it follows that the space
H
1/2
with the norm
u
H
1/2
m
i1
u
i
H
i
1/2
1.6
is a subspace of
H
1/2
u : u
u
1
, ,u
m
,u
i
∈ H
i
1/2
,i 1, ,m
H
1
1/2
×···×H
m
1/2
. 1.7
vi Let the linear manifold
H
0
be dense in
H
1/2
, and let linear manifold H
1
be dense in
H.
viiGreen’s Identity. For arbitrary
u, v ∈
H
0
and t ∈ 0,T, the following identity is
valid:
m
i1
⎡
⎣
A
i
t
u
i
,v
i
H
i
t
s
i
ν1
m
k1
C
i
kν
tu
k
,B
iν
v
i
X
i
ν
t
⎤
⎦
m
i1
⎡
⎣
u
i
,A
i
t
v
i
H
i
t
s
i
ν1
B
iν
u
i
,
m
k1
C
i
kν
tv
k
X
i
ν
t
⎤
⎦
.
1.8
4AdvancesinDifference Equations
viii For all
u u
1
, ,u
m
∈
H
0
, the following inequality is fulfilled:
c
1
m
i1
u
i
2
H
i
s
i
ν1
B
iν
u
i
2
X
i
ν
≤
m
i1
⎡
⎣
A
i
t
u
i
,u
i
H
i
t
s
i
ν1
m
k1
C
i
kν
tu
k
,B
iν
u
i
X
i
ν
t
⎤
⎦
≤ c
2
m
i1
u
i
2
H
i
1/2
,
1.9
where c
1
∈ R, c
2
> 0.
ix For each t ∈ 0,T, an operator pencil
L
t
λ
:
u
u
1
, ,u
m
−→ L
t
λ
u
L
t
10
λ
u, L
t
11
λ
u, ,L
t
1s
1
λ
u, ,L
t
m0
λ
u, L
t
m1
λ
u, ,L
t
ms
m
λ
u
,
1.10
which acts boundedly from
H
0
to H, has a regular point λ λ
0
∈ R,where
L
t
i0
λ
u λu
i
A
i
t
u
i
,i 1, 2, ,m,
L
t
iν
λ
u λB
iν
u
i
m
k1
C
i
kν
t
u
k
,ν 1, 2, ,s
i
,i 1, 2, ,m.
1.11
x u
0
i
∈ H
i
0
,u
1
i
∈ H
i
1/2
,
m
k1
D
j
kμ
u
0
k
0,
m
k1
D
j
kμ
u
1
k
0
i 1, 2, ,m, μ 1, 2, ,r
j
,j 1, 2, ,m
. 1.12
xi f
i
· ∈ W
1
p
0,T; H
i
,p≥ 1,i 1, ,m,
g
iν
·
∈ W
1
p
0,T; X
i
ν
,p≥ 1,ν 1, ,s
i
,i 1, ,m. 1.13
Definition 1.1. The function t → u
1
t, ,u
m
t is called a solution of problem 1.1-1.2 if
the function t →
utu
1
t, ,u
m
t from 0,T to
H
0
is continuous, and the function
t −→
u
1
t
,B
11
u
1
t
, ,B
1s
1
u
1
t
, ,u
m
t
,B
m1
u
m
t
, ,B
ms
m
u
m
t
1.14
from 0,T to H is twice continuously differentiable and 1.1-1.2 are satisfied.
Theorem 1.2. Let conditions (i)–(xi) are satisfied, then the problem 1.1-1.2 has a unique solution.
Advances in Difference Equations 5
Proof. We define the op erator At in the Hilbert space H in the following way:
D
A
t
H
1
,
A
t
w
A
1
t
u
1
,
m
k1
C
1
k1
t
u
k
, ,
m
k1
C
1
ks
1
t
u
k
, ,A
t
m
u
m
,
m
k1
C
m
k1
t
u
k
, ,
m
k1
C
m
ks
m
t
u
k
,t∈
0,T
,w∈H
1
.
1.15
Then the problem 1.1-1.2 is represented as the Cauchy problem
¨w A
t
w Φ
t
,
w
0
w
0
, ˙w
0
w
1
,
1.16
where wtu
1
t,B
11
u
1
t, ,B
1s
1
u
1
t, ,u
m
t,B
m1
u
m
t, ,B
ms
m
u
m
t,
Φ
t
f
1
t
,g
11
t
, ,g
1s
1
t
, ,f
m
t
,g
m1
t
, ,g
ms
m
t
,
w
0
u
0
1
,B
11
u
0
1
, ,B
1s
1
u
0
1
, ,u
0
m
,B
m1
u
0
m
, ,B
ms
m
u
0
m
,
w
1
u
1
1
,B
11
u
1
1
, ,B
1s
1
u
1
1
, ,u
1
m
,B
m1
u
1
m
, ,B
ms
m
u
1
m
.
1.17
It is obvious that if u
1
t, ,u
m
t is the solution of problem 1.1-1.2,thenwt is
the solution of the problem 1.16. On the contrary, if
w
t
∈ C
2
0,T
; H
∩ C
1
0,T
;
H
1
, H
1/2
∩ C
0,T
; H
1
1.18
is the solution of problem 1.16,thenwtu
1
t,B
11
u
1
t, ,B
1s
1
u
1
t, ,u
m
t,
B
m1
u
m
t, ,B
ms
m
u
m
t and u
1
t, ,u
m
t is the solution of problem 1.1-1.2.
Let us define the system of inner product in Hilbert space H in the following way:
w
1
,w
2
Ht
m
i1
w
1
i
,w
2
i
H
i
t
m
i1
s
i
ν1
B
iν
u
1
i
,B
iν
u
2
i
X
i
ν
t
,t∈
0,T
,
1.19
where w
l
w
l
1
, ,w
l
m
,w
l
i
u
l
i
,B
i1
u
l
i
, ,B
is
i
u
l
i
,i 1, 2, ,m,u
l
1
, ,u
l
m
∈
H
0
,l 1, 2.
We denote space H with inner product 1.19 by Ht.
We will prove later the following auxiliary results.
Statement 1.3. There exists such c
3
> 0, that
c
−1
3
w
2
H
≤
w
2
Ht
≤ c
3
w
2
H
,t∈
0,T
, 1.20
6AdvancesinDifference Equations
and the function t →w
2
Ht
: 0,T → R
is continuously differentiable, where w
2
Ht
w, w
Ht
.
Statement 1.4. At is a symmetric operator in Ht for each t ∈ 0,T.
Statement 1.5. At has a regular point for each t ∈ 0,T in R.
At is symmetric and RAtλIHt,forsomeλ ∈ R; therefore, for each t ∈
0,T, At is a selfadjoint operator in Htsee 4,chapterx.
Taking into account viii and Statement 1.3,weget
Atw, w
Ht
m
i1
⎡
⎣
A
i
t
u
i
,u
i
H
i
t
s
i
ν1
m
k1
C
i
kν
tu
k
,B
iν
u
i
X
i
ν
t
⎤
⎦
≥ c
1
w
2
Ht
,
1.21
that is, At is a lower semibounded selfadjoint op erator in Ht.
Thus, the operator A
0
tAtλ
0
I is selfadjoint and positive definite, where λ
0
>c
1
.
Problem 1.16 canberewrittenas
¨w
t
A
0
t
w
t
− λ
0
w
t
F
t
,
w
0
w
0
, ˙w
0
w
1
.
1.22
It is known that if w
0
∈H
1
and w
1
∈H
1/2
, then the problem 1.22 has a unique
solution w ∈ C
2
0,T; H ∩ C
1
0,T; H
1/2
∩C0,T; H
1
see 5, 6.
To complete the proof of the theorem, we need to show that w
0
∈H
1
and w
1
∈H
1/2
.
By conditions of the theorem u
0
i
∈ H
i
0
,
m
k1
D
j
kμ
u
0
k
0i 1, 2, ,m; μ 1, 2, ,r
j
,
j 1, 2 , ,m and B
iν
are bounded operators from H
i
1/2
to X
i
ν
,ν 1, 2, ,s
i
,i 1, 2, ,m.
Therefore,
w
0
u
0
1
,B
11
u
0
1
, ,B
1s
1
u
0
1
, ,u
0
m
,B
m1
u
0
m
, ,B
ms
m
u
0
m
∈H
1
. 1.23
On the other hand, u
1
i
∈ H
i
1/2
and
m
k1
D
j
kμ
u
1
k
0 i 1, 2, ,m,μ 1, 2, ,r
j
,j
1, 2, ,m, therefore, B
iν
u
1
i
∈ X
i
ν
ν 1, 2, ,s
i
,i 1, 2, ,m.Consequently,
w
1
u
1
1
,B
11
u
1
1
, ,B
1s
1
u
1
1
, ,u
1
m
,B
m1
u
1
m
, ,B
ms
m
u
1
m
∈J,
J
w : w
w
1
, ,w
m
,w
i
u
i
,B
i1
u
i
, ,B
is
i
u
i
,u
i
∈ H
i
1/2
,
m
k1
D
j
kμ
u
k
0,i 1, ,m,μ 1, ,r
j
,j 1, ,m
.
1.24
Advances in Difference Equations 7
From the definition of interpolation spaces see 3,chapter1, 7,chapter1,weget
the following inclusion:
H
1
⊂H
1/2
⊂
H
1/2
m
i1
H
i
1/2
⊕ X
i
1
⊕···⊕X
i
s
i
.
1.25
By virtue of definition, the powers of positive selfadjoint operator see 8,chapter2,
7,chapter1,wehavethatDA
1/2
0
t H
1/2
and
c
−1
w
H
1/2
≤
A
1/2
0
tw
Ht
≤ c
w
H
1/2
,c>0.
1.26
Assume that w ∈ DA
0
H
1
,then
A
1/2
0
tw
2
Ht
A
0
tw, w
Ht
m
i1
⎡
⎣
A
i
t
u
i
,u
i
H
i
t
s
i
ν1
m
k1
C
i
kν
t
u
k
,B
iν
u
i
X
i
ν
t
⎤
⎦
λ
0
m
i1
u
i
,u
i
H
i
t
s
i
ν1
B
iν
u
i
,B
iν
u
i
X
i
ν
t
.
1.27
By virtue of conditions ii, viii, 1.26,and1.27,weget
w
2
H
1/2
≤ c
m
i1
u
i
2
H
i
1/2
.
1.28
Let w
1
∈J. By virtue of condition vi,
H
0
is dense in
H
1/2
; therefore, there exists a
sequence
u
p
u
p
1
, ,u
p
m
,suchthatu
p
∈
H
0
and
u
p
− u
1
H
1
1/2
⊕···⊕H
m
1/2
−→ 0, at p −→ ∞.
1.29
Hence it follows, that
u
p
−
u
q
H
1
1/2
⊕···⊕H
m
1/2
−→ 0atp, q −→ ∞.
1.30
Then from 1.28 and 1.30 it follows that {w
p
} is fundamental in H
1/2
,thatis,
w
p
− w
q
H
1/2
−→ 0, at p, q −→ ∞,
1.31
where w
p
u
p
1
,B
11
u
p
1
, ,B
1s
1
u
p
1
, ,u
p
m
,B
m1
u
p
m
, ,B
ms
m
u
p
m
,p 1, 2,
8AdvancesinDifference Equations
Thus, there exists w ∈H
1/2
such that
w
p
− w
H
1/2
−→ 0, at p −→ ∞.
1.32
On the other hand, H
1/2
⊂
H
1/2
, therefore,
w
p
− w
H
1/2
−→ 0, at p −→ ∞.
1.33
Hence,
u
p
− u
H
1
1/2
⊕···⊕H
m
1/2
−→ 0, at p −→ ∞,
1.34
where
u u
1
, ,u
m
. From this, by virtue of 1.29, u u
1
,thatis,
w
u
1
1
,B
11
u
1
1
, ,B
1s
1
u
1
1
, ,u
1
m
,B
m1
u
1
m
, ,B
ms
m
u
1
m
w
1
. 1.35
Thus, w
1
∈H
1/2
. The theorem is proved.
1.2. Proof of Auxiliary Results
Validity of Statement 1.3 follows from condition i, the Statement 1.4 from condition vii.
Proof of Statement 3. Consider in Hilbert space H the equation
λw A
t
w F,t∈
0,T
, 1.36
where F f
1
,f
11
, ,f
1s
1
, ,f
m
,f
m1
, ,f
ms
m
∈H,λ∈ R.
Equation 1.36 is equivalent to the following system of differential-operator
equations:
L
t
i0
λ
u λu
i
A
i
t
u
i
f
i
,t∈
0,T
,i 1, 2, ,m,
L
t
iν
λ
u λB
iν
u
i
m
k1
C
i
kν
t
u
k
g
iν
,t∈
0,T
,ν 1, 2, ,s
i
,i 1, 2, ,m,
m
k1
D
j
kμ
u
k
0,μ 1, 2 , ,r
j
,j 1, 2, ,m.
1.37
By virtue of ix,problem1.37 has a solution
u u
1
, ,u
m
∈
H
0
for some λ ∈ R.Thus,
for each t ∈ 0,T,
R
λI A
t
H
t
, 1.38
where I is an identity operator in Ht,thatis,A has a regular point.
Advances in Difference Equations 9
2. Abstract Model of Initial Boundary Value Problem with
Non Stationary Boundary and Transmission Conditions for
the Impulsive Semilinear Hyperbolic Equations
Consider the following initial boundary value problem:
¨u
i
t
A
i
t
u
i
t
f
i
t,
u
t
,
˙
u
t
,
B
iν
¨u
i
t
m
k1
C
i
kν
t
u
k
t
g
iν
t,
u
t
,
¨
u
t
,
m
k1
D
i
kμ
u
k
t
0,
u
i
0
u
0
i
, ˙u
i
0
u
1
i
,
2.1
where t ∈ 0,T, ν 1, ,s
i
, μ 1, ,r
i
, i 1, ,m,
˙
u u
1
, ,u
m
,
¨
u ˙u
1
, , ˙u
m
, A
i
t,
B
iν
, C
i
kν
t and D
i
kμ
satisfy all conditions of Theorem 1.2.
Assume, that the nonlinear operators f
i
and g
iν
satisfy the following conditions.
xi
Suppose that the nonlinear operators
t,
u,
˙
u
−→ f
i
t,
u,
˙
u
:
0,T
×
m
i1
H
i
1/2
×
m
i1
H
i
−→ H
i
,
t,
u,
˙
u
−→ g
iν
t,
u,
˙
u
:
0,T
×
m
i1
H
i
1/2
×
m
i1
H
i
−→ X
i
ν
2.2
satisfy the local Lipschitz conditions in the following sense: for arbitrary t
1
,t
2
∈
0,T,
u
1
, v
1
, u
2
, v
2
∈
H
1/2
×
H,
f
i
t
1
, u
1
, v
1
− f
i
t
2
, u
2
, v
2
H
i
≤ c
i
r
|
t
1
− t
2
|
m
i1
u
1
i
− u
2
i
H
i
1/2
v
1
i
− v
2
i
H
i
,
g
iν
t
1
, u
1
, v
1
− g
iν
t
2
, u
2
, v
2
X
i
ν
≤ c
iν
r
|
t
1
− t
2
|
m
i1
u
1
i
− u
2
i
H
i
1/2
v
1
i
− v
2
i
H
i
,
2.3
where c
i
·,c
iν
∈ CR
,R
,ν 1, ,s
i
,i 1, ,m,
r
m
i1
2
l1
u
l
i
H
i
1/2
v
l
i
H
i
. 2.4
10 Advances in Difference Equations
Theorem 2.1. Let conditions (i)–(x) and (xi
) be satisfied, then there exists T
∈ 0,T, such that the
problem 2.1 has a unique solution
u
u
1
, ,u
m
∈ C
0,T
,
H
0
∩ C
1
0,T
,
H
1/2
∩ C
2
0,T
,
H
. 2.5
Additionally, if
E
t
m
i1
u
i
t
H
i
1/2
˙u
i
t
H
i
≤ ϕ
m
i1
u
0
i
H
i
1/2
u
1
i
H
i
,t∈
0,T
, 2.6
where ϕ· ∈ CR
,R
,thenT
T. Otherwise, there exists T
0
∈ 0,T,suchthat
lim
t →T
0
−0
E
t
∞.
2.7
In the Hilbert space H,theproblem2.1 is represented as the Cauchy problem
¨w A
0
t
w F
t, w, ˙w
,
w
0
w
0
, ˙w
0
w
1
,
2.8
where w u
1
,B
11
u
1
, ,B
1s
1
u
1
, ,u
m
,B
m1
u
m
, ,B
ms
m
u
m
,
w
0
u
0
1
,B
11
u
0
1
, ,B
1s
1
u
0
1
, ,u
0
m
,B
m1
u
0
m
, ,B
ms
m
u
0
m
,
w
1
u
1
1
,B
11
u
1
1
, ,B
1s
1
u
1
1
, ,u
1
m
,B
m1
u
1
m
, ,B
ms
m
u
1
m
,
F
t, w, ˙w
λ
0
w F
1
t, w, ˙w
,
F
1
t, w, ˙w
f
1
t,
u,
˙
u
,g
11
t,
u,
˙
u
, ,g
1s
1
t,
u,
˙
u
, ,
f
m
t,
u,
˙
u
,g
m1
t,
u,
˙
u
, ,g
ms
m
t,
u,
˙
u
.
2.9
From xi’, it follows that, for arbitrary t
1
,t
2
∈ 0 ,T,w
1
,w
2
∈H
1/2
,z
1
,z
2
∈H,
F
t
1
,w
1
,z
1
−F
t
2
,w
2
,z
2
H
≤ c
r
|
t
1
− t
2
|
w
1
− w
2
H
1/2
z
1
− z
2
H
, 2.10
where c· ∈ CR
,R
,r
2
l1
w
l
H
1/2
z
l
H
.
Advances in Difference Equations 11
Thus, the nonlinear operator F satisfies the condition of local solvability of the Cauchy
problem for the quasilinear hyperbolic equations in Hilbert space see 6, 9. Taking this into
account, the problem 2.8 has a unique solution
w ∈ C
2
0,T
; H
∩ C
1
0,T
; H
1/2
∩ C
0,T
; H
1
.
2.11
3. Initial Boundary Value Problem with
Non Stationary Boundary and Transmission Condition for
the Impulsive Semilinear Hyperbolic Equations
Let a
1
<a
2
< ···<a
m1
. We consider in the domain 0,T ×
m
i1
a
i
,a
i1
the following mixed
problem
¨u
i
t, x
− p
i
t
u
i
t, x
f
i
t, x, u
i
t, x
,u
i
t, x
, ˙u
i
t, x
,ϕ
i
u,
˙
u
,
t, x
∈
0,T
×
a
i
,a
i1
,i 1, 2, ,m,
u
i
t, a
i1
u
i1
t, a
i1
,i 1, 2, ,m− 1,t>0,
¨u
1
t, a
1
− q
0
t
u
1
t, a
1
g
0
t, ψ
0
u,
˙
u
,t>0,
¨u
i
t, a
i1
q
i
t
u
i
t, a
i1
− u
i1
t, a
i1
g
i
t, ψ
i
u,
˙
u
,
i 1, 2, ,m− 1,t>0,
¨u
m
t, a
m1
q
m
t
u
m
t, a
m1
g
m
t, ψ
m
u,
˙
u
,t>0,
u
i
0,x
u
0
i
x
, ˙u
i
0,x
u
1
i
x
,x∈
a
i
,b
i
,i 1, 2, ,m,
3.1
where ˙u
i
∂u
i
/∂t, u
i
∂u
i
/∂x, ¨u
i
∂
2
u
i
/∂t
2
,u
i
∂
2
u
i
/∂x
2
, u u
1
, ,u
m
,
˙
u
˙u
1
, , ˙u
m
, p
i
, q
j
,f
i
,g
j
,u
0
i
,u
1
i
are some functions, ϕ
i
and ψ
j
are some functionals, which will
be specified below, i 1, ,m,j 0, 1, ,m.
Recently, differential equations with impulses are great interest because of the
needs of modern technology, where impulsive automatic control systems and impulsive
computing systems are very important and intensively develop broadening the scope of their
applications in technical problems, heterogeneous by their physical nature and functional
purpose see 10,chapter1.
Assume that the following conditions are held:
1
0
p
i
t ∈ C
1
0,T,q
j
t ∈ C
1
0,T; p
i
t > 0,q
j
t > 0,t∈ 0,T, i 1, ,m,j
0, 1, ,m,
2
0
f
i
· ∈ C
1
0,T × a
i
,a
i1
×R
4
,i 1, 2, ,m,
3
0
g
j
· ∈ C
1
0,T,R,j 0, 1, ,m,
12 Advances in Difference Equations
4
0
ϕ
i
· are nonlinear functionals acting from
m
k1
W
1
2
a
k
,a
k1
× L
2
a
k
,a
k1
3.2
to R and for arbitrary
u
1
, v
1
, u
2
, v
2
∈
m
k1
W
1
2
a
k
,a
k1
× L
2
a
k
,a
k1
the following
inequality holds
ϕ
i
u
1
, v
1
− ϕ
i
u
2
, v
2
≤ c
i
r
m
k1
u
1
k
− u
2
k
W
1
2
a
k
,a
k1
v
1
k
− v
2
k
L
2
a
k
,a
k1
,
i 1, 2, ,m,
3.3
where r
m
k1
u
1
k
W
1
2
a
k
,a
k1
u
2
k
W
1
2
a
k
,a
k1
v
1
k
L
2
a
k
,a
k1
v
1
k
L
2
a
k
,a
k1
,
c
i
·
∈ C
R
,R
,R
0, ∞
,i 1, 2, ,m, 3.4
5
0
ψ
j
· are nonlinear functionals acting from
m
k1
W
1
2
a
k
,a
k1
× L
2
a
k
,a
k1
3.5
to R and for arbitrary
u
1
, v
1
, u
2
, v
2
∈
m
k1
W
1
2
a
k
,a
k1
× L
2
a
k
,a
k1
the following
inequality holds
ψ
j
u
1
, v
1
− ψ
j
u
2
, v
2
≤ c
j
r
m
k1
u
1
k
− u
2
k
W
1
2
a
k
,a
k1
v
1
k
− v
2
k
L
2
a
k
,a
k1
,
3.6
where c
j
· ∈ CR
,R
,j 0, 1, ,m,andr—is defined as in 3.3,
6
0
u
0
i
∈ W
2
2
a
i
,a
i1
,u
1
i
∈ W
1
2
a
i
,a
i1
, i 1, 2, ,m,where
u
0
j
a
j1
u
0
j1
a
j1
,
u
1
j
a
j1
u
1
j1
a
j1
,j 1, 2, ,m− 1.
3.7
By applying Theorem 2.1, we obtain the fo llowing result.
Advances in Difference Equations 13
Theorem 3.1. Let conditions (1
0
)–(6
0
) be held, then there exists a T
∈ 0,T, such that the problem
3.1 has a unique solution
u u
1
, ,u
m
,where
u
i
∈ C
2
0,T
; L
2
a
i
,a
i1
∩ C
1
0,T
; W
1
2
a
i
,a
i1
∩ C
0,T
; W
2
2
a
i
,a
i1
,
u
i
t, a
i
,u
i
t, a
i1
∈ C
2
0,T
,R
,i 1, 2, ,m.
3.8
Proof. Let us denote H
i
L
2
a
i
,a
i1
, H
i
0
W
2
2
a
i
,a
i1
, X
i
ν
,Y
j
μ
, ν 1, 2, ,s
i
, i
1, 2, ,m, μ 1, 2, ,r
j
, j 1, 2, ,m,wheres
i
2,r
j
1.
In space H
i
and X
i
ν
are defined the following inner p roducts:
u, v
H
i
t
p
−1
i
t
a
i1
a
i
uvdx,
h
1
,h
2
X
1
1
t
q
−1
0
t
h
1
h
2
,
h
1
,h
2
X
i
2
t
q
−1
i
t
h
1
h
2
,
h
1
,h
2
∈ ,i 1, 2, ,m.
3.9
From differentiability of the functions p
i
t, i 1, 2, ,m,andq
j
t, j 0, 1, ,m it
follows that the condition i is sa tisfied.
Let us define the following operators:
A
i
tu
i
−p
i
tu
i
,u
i
∈ DA
i
t W
2
2
a
i
,a
i1
,
B
11
u
1
u
1
a
1
,B
j1
0,B
2i
u
i
u
i
a
i1
,i 1, 2, ,m,j 2, ,m,
C
1
11
tu
1
−q
0
tu
1
a
1
,C
m
m1
tu
m
q
m
tu
m
a
m1
,
C
i
k1
t0, for all other i, k,
C
i
i2
tu
i
q
i
tu
i
a
i1
,i 1, 2, ,m,
C
j
j2
tu
j1
−q
j
tu
j1
a
j1
,j 1, 2, ,m− 1,
C
i
k2
t0, for all other i, k,
D
i
i1
u
i
−u
i
a
i1
,D
i
i1,1
u
i1
u
i1
a
i1
,i 1, 2, ,m− 1,
D
i
k1
0,k
/
i, k
/
i 1.
We also define the nonlinear operators as follows:
F
i
t, u, vf
i
t, x, u
i
x,u
i
x,v
i
x,ϕ
i
u, v,i 1, 2, ,m,
G
11
t, u, vg
0
t, ψ
0
u, v,
G
i2
t, u, vg
i
t, ψ
i
u, v,i 1, 2, ,m,
G
i1
t, u, v0,i 2, 3, ,m.
14 Advances in Difference Equations
It is easy to verify that linear operators A
i
t,B
iν
,C
k
iν
t, and D
k
iμ
and the nonlinear operators
F
i
,G
i1
, and G
i2
, i 1, ,m satisfy the conditions of Theorem 2.1,andtheproblem3.1 is
represented as an abstract initial boundary-value problem in the following way:
¨u
i
t
A
i
t
u
i
t
F
i
t,
u,
˙
u
,
B
11
¨u
1
t
C
1
11
t
u
1
t
G
0
t,
u,
˙
u
,
B
i2
¨u
i
t
C
i
i2
t
u
i
t
C
i
i2,2
t
u
i
t
G
i
t,
u,
˙
u
,
B
m2
¨u
m
t
C
m
m2
t
u
m
t
G
m
t,
u,
˙
u
,
D
i
i1
u
i
D
i
i1
u
i1
0,i 1, 2, ,m− 1.
3.10
We will show that conditions of Theorem 2.1 are satisfied. Conditions i–v follow
immediately from definitions of spaces H
i
,X
i
ν
, and Y
j
μ
and operators A
i
t,B
iν
,C
i
kν
t,
and D
j
kμ
, and traces theorems see 3,chapter2,wherek 1, 2, ,m; ν 1, 2, ,s
i
;
i 1, 2, ,m; μ 1, 2, ,r
j
; j 1 , 2, ,m.
The linear manifolds
H
0
and H
1
are defined in the following way:
H
0
u, u
u
1
, ,u
m
,u
i
∈ W
2
2
a
i
,a
i1
,i 1, ,m,
u
j
a
j1
u
j1
a
j1
,j 1, ,m− 1
,
H
1
w, w
w
1
, ,w
m
,w
1
u
1
,u
1
a
2
,u
1
a
1
,
w
i
u
i
,u
i
a
i1
,i 2, ,m,
u ∈
H
0
.
3.11
We also define the spaces
H
1/2
u, u
u
1
, ,u
m
,u
i
∈ W
1
2
a
i
,a
i1
,i 1, ,m
,
H
1/2
u, u
u
1
, ,u
m
,u
i
∈ W
1
2
a
i
,a
i1
,i 1, ,m,
u
j
a
j1
u
j1
a
j1
,j 1, ,m− 1
.
3.12
Statement 3.2. H
1
is dense in
H
L
2
a
1
,b
1
⊕
⊕
⊕
m
i2
L
2
a
i
,b
i
⊕
.
3.13
Advances in Difference Equations 15
Proof. Assume that u
1
,α
1
,α
0
,u
2
,α
2
, ,u
m
,α
m
∈H. Consider the following functions:
u
0
i
x
a
i1
− x
a
i1
− a
i
α
i−1
x − a
i
a
i1
− a
i
α
i
,x∈
a
i
,a
i1
,i 1, ,m.
3.14
From definitions of u
0
i
x, i 1, ,m,wecanseethat
u
0
i
a
i1
u
0
i1
a
i1
α
i
,i 1, 2, ,m− 1.
3.15
Let
u u
1
, ,u
m
∈
H. Consider the function
z
z
1
, ,z
m
u
1
− u
0
1
, ,u
m
− u
0
m
. 3.16
It is obvious that z ∈
m
i1
L
2
a
i
,a
i1
. On the other hand,
m
i1
Da
i
,a
i1
,a
i1
m
i1
L
2
a
i
,a
i1
,whereDa
i
,a
i1
i 1, ,m is a space of infinitely differentiable finite
functions. Therefore, for an arbitrary ε>0, there exist the functions h
i
∈Da
i
,a
i1
,
i 1, ,m,suchthat
m
i1
z
i
− h
i
<ε.
3.17
By denoting
h
i
u
0
i
h
i
from 3.17,weget
m
i1
u
i
−
h
i
L
2
a
i
,a
i1
<ε,
3.18
where
h
i
∈ C
∞
a
i
,a
i1
,
h
i
a
i
α
i−1
,i 1, ,m.
Thus,
u
1
,u
1
a
2
,u
1
a
1
,u
2
,u
2
a
2
, ,u
m
,u
m
a
m1
−
h
1
,α
1
,α
0
,h
2
,α
1
, ,h
m
,α
m
H
<ε.
3.19
The following statement is proved in the same way.
Statement 3.3.
H
0
is dense
H
1/2
.
Now, we prove that the condition vi holds.
16 Advances in Difference Equations
Let
u u
1
, ,u
m
, v v
1
, ,v
m
∈
H
0
,then
m
i1
⎡
⎣
A
i
tu
i
,v
i
H
i
t
s
i
ν1
m
k1
C
i
kν
tu
k
,B
iν
v
i
X
i
ν
t
⎤
⎦
m
i1
−
a
i1
a
i
u
i
v
i
dx
−u
1
a
1
,v
1
a
1
m
i1
u
i
a
i1
− u
i1
a
i1
,v
i
a
i1
u
m
a
m1
,v
m
a
m
m
i1
u
i
a
i
,v
i
a
i
−
u
i
a
i1
,v
i
a
i1
−
m
i1
a
i1
a
i
u
i
v
i
dx −
u
1
a
1
,v
1
a
1
m−1
i1
u
i
a
i1
v
i
a
i1
−
m−1
i1
u
i1
a
i1
v
i
a
i1
u
m
a
m1
v
m
a
m1
m
i1
u
i
a
i
v
i
a
i
− u
i
a
i1
v
i
a
i1
− u
1
a
1
v
1
a
1
m−1
i1
u
i
a
i
v
i
a
i
−
m
i2
u
i
a
i
v
i
a
i
u
m
a
m1
v
m
a
m1
m
i1
a
i1
a
i
u
i
v
i
dx
m
i1
a
i1
a
i
u
i
v
i
dx.
3.20
Similary, we obtain the following identity:
m
i1
⎡
⎣
u
i
,A
i
t
v
i
H
i
t
s
i
ν1
B
iν
ν
i
,
m
k1
C
i
kν
t
u
k
X
i
ν
t
⎤
⎦
m
i1
a
i1
a
i
u
i
v
i
dx. 3.21
Thus, by virtue of 3.20-3.21, the condition vi holds.
From 3.20 or 3.21, putting v
i
u
i
, we also obtain the identity
m
i1
a
i1
a
i
u
2
i
dx
m
i1
⎡
⎣
A
i
tu
i
,u
i
H
i
t
s
i
ν1
m
k1
C
i
kν
tu
k
,B
iν
u
i
X
i
ν
t
⎤
⎦
, 3.22
that is, condition viii is satisfied, c
1
c
2
1.
Advances in Difference Equations 17
Now, we verify fulfillment of condition ix. To that end, we consider the mixed
problem
λu
i
− p
i
t
u
i
h
i
x
,i 1, 2, ,m, 3.23
λu
1
a
1
− q
0
t
u
1
a
1
h
10
,
λu
i
a
i1
q
i
t
u
i
a
i1
− u
i1
a
i1
h
i0
,i 1, 2, ,m− 1,
λu
m
a
m1
q
m
t
u
m
a
m1
h
m0
,
3.24
where h
i
∈ L
2
a
i
,a
i1
,i 1, ,m; h
j0
∈ R, j 0, 1, ,m, λ ∈ R.
Let h
i
x be the extend of function h
i
x to R. We consider the system of the differential
equations
λu
i
− p
i
t
u
i
xx
h
i
x
,i 1, 2, ,m.
3.25
Hence, we have
λ
u
i
− k
2
p
i
t
u
i
xx
h
i
x
,i 1, 2, ,m,
3.26
where g Fg is a Fourier transformation of the function gx.From3.26,weobtain
u
h
i
/λ k
2
p
i
t, then functions u
i
F
−1
u
i
F
−1
h
i
/λ k
2
p
i
t satisfy 3.25,and
their constrictions on a
i
,a
i1
satisfy the 3.23. It is clear that u
i
∈ W
2
2
a
i
,a
i1
.Considering
linearity of the problem 3.23, 3.24, the solution can be represented in the form
u
i
v
i
u
i
, 3.27
where v
i
u
i
− u
i
is a solution of the following problem:
λv
i
x
− p
i
t
v
i
x
0, 3.28
λv
1
a
1
− q
0
t
v
1
a
1
h
10
,
λv
i
a
i1
q
i
t
v
i
a
i1
− v
i1
a
i1
h
i0
,i 1, 2, ,m− 1,
λv
m
a
m1
q
m
t
v
m
a
m1
h
m0
,
3.29
where
h
10
h
10
− λu
1
a
1
q
0
tu
1
a
1
,
h
i0
h
i0
− λu
i
a
i1
− q
i
t
u
i
a
i1
− u
i1
a
i1
,i 1, 2, ,m− 1,
h
m0
0
h
m
0
− λu
m
a
m1
q
m
t
u
m
a
m1
.
3.30
18 Advances in Difference Equations
A general solution of a system 3.28 is found in the following form:
v
i
x
c
i1
e
−x−a
i
√
λ/p
i
t
c
i2
e
−b
i
−x
√
λ/p
i
t
,i 1, 2, ,m.
3.31
Then, for determination of c
i1
,c
i2
,i 1, 2, ,m,from3.29,wegetthefollowing
system of the algebraic equations:
λ
c
11
c
12
e
−a
2
−a
1
√
λ/p
i
t
− q
0
t
λ
p
i
t
c
11
− c
12
e
−a
2
−a
1
√
λ/p
i
t
h
0
,
λ
c
i1
e
−a
i1
−a
i
√
λ/p
i
t
− c
i2
q
i
t
λ
p
i
t
c
i1
e
−a
i1
−a
i
√
λ/p
i
t
c
i2
−
λ
p
i1
t
c
i1,1
c
i1,2
e
−a
i2
−a
i1
√
λ/p
i1
t
h
i0
,i 1, 2, ,m− 1,
c
i1
e
−a
i1
−a
i
√
λ/p
i
t
− c
i2
−
c
i1,1
− c
i1,2
e
−a
i2
−a
i1
√
λ/p
i1
t
0,i 1, ,m− 1,
λ
c
m1
e
−a
m1
−a
m
√
λ/p
m
t
c
m2
−c
m1
e
−a
m1
−a
m
√
λ/p
m
t
c
m2
h
m
0
.
3.32
Let Rλ be a matrix of coefficients of system 3.32.From3.32, it is clear that Rλ
R
0
λR
1
λ,wheredetR
0
λ → ∞ and det R
1
λ → 0asλ → ∞.Thus,forsufficiently
large λ, Rλ is invertible and det Rλ → ∞. Therefore, the system 3.32 has a unique
solution.
Thus, for sufficiently positive large λ,theproblem3.23-3.24 has a unique solution
u u
1
, ,u
m
∈ H
0
.
Thus, the condition ix is satisfied. The fulfillment of other conditions follows from
1
0
–6
0
.
Now, let us consider a class of nonlinear equations, for which the large solvability
theorem takes place.
Let
f
i
t, x, u
i
,u
i
, ˙u
i
,ϕ
u,
˙
u
−
|
u
i
|
ρ
i
u
i
f
1i
t, x, u
i
,u
i
, ˙u
i
, ϕ
i
u,
˙
u
,
g
0
t, ψ
0
u,
˙
u
−
|
u
1
a
1
|
τ
0
u
1
a
1
g
01
t, ψ
0
u,
˙
u
,
g
i
t, ψ
i
u,
˙
u
−
|
u
i
a
i1
|
τ
i
u
i
a
i1
g
i1
t, ψ
i
u,
˙
u
,i 1, 2, ,m,
3.33
where ρ
i
≥ 0, τ
j
≥ 0, i 1, 2, ,m; j 0, 1, ,mand
7
0
f
1i
,g
1j,
ϕ
i
and ψ
j
, i 1, 2, ,m,j 1, 2, ,msatisfy the conditions 2
0
−−5
0
.
8
0
|f
i
t, x, u
i
,v
i
,ξ
i
,η|≤c1 |u
i
|
ρ
i
2/2
|v
i
| |ξ
i
| |η|,
Advances in Difference Equations 19
9
0
|g
0i
t, η|≤c1 |η|,
10
0
|ϕ
i
u, v|≤c1
n
i1
|u
i
|
ρ
i
2/2
|v
i
|
2
|u
i
y
i
|
τ
i
2/2
,
where y
i
a
i1
, i 0, 1, ,m, ρ maxmin
i1,2, ,m
ρ
i
, 2.
Theorem 3.4. Let conditions 7
0
–10
0
be held and initial data satisfy the condition 6
0
, then the
problem 3.1 has a unique solution
u u
1
, ,u
m
,where
u
i
∈ C
2
0,T
; L
2
a
i
,a
i1
∩ C
1
0,T
; W
1
2
a
i
,a
i1
∩ C
0,T
; W
2
2
a
i
,a
i1
,
u
i
t, a
i
,u
i
t, a
i1
∈ C
2
0,T
; R
,i 1, 2, ,m.
3.34
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