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
(
;
;
;
) be Hilbert Spaces. Consider the following abstract initial-boundary value problem:
where
,
,
,
are the linear closed operators in
;
are the linear operators from
to
;
are the linear operators from
to
;
are the linear operators from
to
;
,
,
,
.
We will investigate this problem under the following conditions.
-
(i)
Let
, and let
be densely in
and continuously imbedded into it,
.
In the Hilbert space
, it was defined the system of the inner products
, which generate uniform equivalent norms, that is,
For each
, the function
is continuously differentiable,
.
In the Hilbert space
, it was defined the system of the inner products
, which generate uniform equivalent norms, that is,
For each
, the function
is continuously differentiable.
-
(ii)
For each
and 
is a linear closed operator in
whose domain is
;
acts boundedly from
to
;
is strongly continuously differentiable.
-
(iii)
The linear operators
, that act from
to
, bounded, where
is interpolation space between
and
of order
(see [3]).
-
(iv)
For each
, the linear operators
, that act from
to
, are bounded;
is strongly continuously differentiable
,
;
.
-
(v)
The linear operators
, from
into
, act boundedly
,
;
.
Let us introduce the following designations:
From condition (v), it follows that the space
with the norm
is a subspace of
-
(vi)
Let the linear manifold
be dense in
, and let linear manifold
be dense in
.
-
(vii)
(Green's Identity). For arbitrary
and
, the following identity is valid:
-
(viii)
For all
, the following inequality is fulfilled:
where
.
-
(ix)
For each
, an operator pencil
which acts boundedly from
to
, has a regular point
, where
-
(x)
-
(xi)
,
Definition 1.1.
The function
is called a solution of problem (1.1)-(1.2) if the function
from
to
is continuous, and the function
from
to
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.
Proof.
We define the operator
in the Hilbert space
in the following way:
Then the problem (1.1)-(1.2) is represented as the Cauchy problem
where
,
It is obvious that if
is the solution of problem (1.1)-(1.2), then
is the solution of the problem (1.16). On the contrary, if
is the solution of problem (1.16), then
,
and
is the solution of problem (1.1)-(1.2).
Let us define the system of inner product in Hilbert space
in the following way:
where
,
.
We denote space
with inner product (1.19) by
.
We will prove later the following auxiliary results.
Statement 1.3.
There exists such
, that
and the function
is continuously differentiable, where
=
.
Statement 1.4.
is a symmetric operator in
for each
.
Statement 1.5.
has a regular point for each
in
.
is symmetric and
, for some
; therefore, for each
is a selfadjoint operator in
(see [4, chapter x]).
Taking into account (viii) and Statement 1.3, we get
that is,
is a lower semibounded selfadjoint operator in
.
Thus, the operator
is selfadjoint and positive definite, where
.
Problem (1.16) can be rewritten as
It is known that if
and
, then the problem (1.22) has a unique solution 



(see [5, 6]).
To complete the proof of the theorem, we need to show that
and
.
By conditions of the theorem 
;
,
and
are bounded operators from
to
. Therefore,
On the other hand,
and 
, therefore, 
. Consequently,
From the definition of interpolation spaces (see [3, chapter 1], [7, chapter 1]), we get the following inclusion:
By virtue of definition, the powers of positive selfadjoint operator (see [8, chapter 2], [7, chapter 1]), we have that
and
Assume that
, then
By virtue of conditions (ii), (viii), (1.26), and (1.27), we get
Let
. By virtue of condition (vi),
is dense in
; therefore, there exists a sequence
, such that
and
Hence it follows, that
Then from (1.28) and (1.30) it follows that
is fundamental in
, that is,
where
,….
Thus, there exists
such that
On the other hand,
, therefore,
Hence,
where
. From this, by virtue of (1.29),
, that is,
Thus,
. 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.
Consider in Hilbert space
the equation
where
.
Equation (1.36) is equivalent to the following system of differential-operator equations:
By virtue of (ix), problem (1.37) has a solution
for some
. Thus, for each
,
where
is an identity operator in
, that is,
has a regular point.