A Symmetric Model of Viscous Relaxing Fluid. An Evolution Problem
An evolution problem on small motions of the viscous rotating relaxing fluid in a bounded domain is studied. The problem is reduced to the Cauchy problem for the first-order integro-differential equation in a Hilbert space. Using this equation, we prove a strong unique solvability theorem for the co...
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Zakora, D. 2016-10-03T14:57:29Z 2016-10-03T14:57:29Z 2012 A Symmetric Model of Viscous Relaxing Fluid. An Evolution Problem / D. Zakora // Журнал математической физики, анализа, геометрии. — 2012. — Т. 8, № 2. — С. 190-206. — Бібліогр.: 13 назв. — англ. 1812-9471 https://nasplib.isofts.kiev.ua/handle/123456789/106718 An evolution problem on small motions of the viscous rotating relaxing fluid in a bounded domain is studied. The problem is reduced to the Cauchy problem for the first-order integro-differential equation in a Hilbert space. Using this equation, we prove a strong unique solvability theorem for the corresponding initial-boundary value problem. Исследована эволюционная задача о малых движениях вязкой вращающейся релаксирующей жидкости в ограниченной области. Задача приведена к задаче Коши для интегро-дифференциального уравнения первого порядка в гильбертовом пространстве. С использованием этой задачи Коши доказана теорема об однозначной сильной разрешимости соответствующей начально-краевой задачи. The author expresses his gratitude to Prof. N.D. Kopachevsky for fruitful discussions of the paper. en Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України Журнал математической физики, анализа, геометрии A Symmetric Model of Viscous Relaxing Fluid. An Evolution Problem Article published earlier |
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A Symmetric Model of Viscous Relaxing Fluid. An Evolution Problem |
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A Symmetric Model of Viscous Relaxing Fluid. An Evolution Problem |
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A Symmetric Model of Viscous Relaxing Fluid. An Evolution Problem |
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symmetric model of viscous relaxing fluid. an evolution problem |
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Журнал математической физики, анализа, геометрии |
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Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
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An evolution problem on small motions of the viscous rotating relaxing fluid in a bounded domain is studied. The problem is reduced to the Cauchy problem for the first-order integro-differential equation in a Hilbert space. Using this equation, we prove a strong unique solvability theorem for the corresponding initial-boundary value problem.
Исследована эволюционная задача о малых движениях вязкой вращающейся релаксирующей жидкости в ограниченной области. Задача приведена к задаче Коши для интегро-дифференциального уравнения первого порядка в гильбертовом пространстве. С использованием этой задачи Коши доказана теорема об однозначной сильной разрешимости соответствующей начально-краевой задачи.
|
| issn |
1812-9471 |
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https://nasplib.isofts.kiev.ua/handle/123456789/106718 |
| citation_txt |
A Symmetric Model of Viscous Relaxing Fluid. An Evolution Problem / D. Zakora // Журнал математической физики, анализа, геометрии. — 2012. — Т. 8, № 2. — С. 190-206. — Бібліогр.: 13 назв. — англ. |
| work_keys_str_mv |
AT zakorad asymmetricmodelofviscousrelaxingfluidanevolutionproblem AT zakorad symmetricmodelofviscousrelaxingfluidanevolutionproblem |
| first_indexed |
2025-11-26T01:42:40Z |
| last_indexed |
2025-11-26T01:42:40Z |
| _version_ |
1850604689448501248 |
| fulltext |
Journal of Mathematical Physics, Analysis, Geometry
2012, vol. 8, No. 2, pp. 190�206
A Symmetric Model of Viscous Relaxing Fluid.
An Evolution Problem
D. Zakora
Taurida National V.I. Vernadsky University
4 Vernadsky Ave., Simferopol 95007, Crimea, Ukraine
E-mail: dmitry−@crimea.edu
Received April 12, 2011
An evolution problem on small motions of the viscous rotating relaxing
�uid in a bounded domain is studied. The problem is reduced to the Cauchy
problem for the �rst-order integro-di�erential equation in a Hilbert space.
Using this equation, we prove a strong unique solvability theorem for the
corresponding initial-boundary value problem.
Key words: viscous �uid, compressible �uid, existence, uniqueness,
integro-di�erential equation.
Mathematics Subject Classi�cation 2000: 45K05, 58C40, 76R99.
Introduction
As it is known, a barotropic �uid is modelled by an equation of the state relat-
ing �uid pressure and density. In particular, it is supposed that this relationship
is linear. The problem of motions of an ideal rotating �uid with compressibility
being taken into account was �rst studied in V.N. Maslennikova's works. In the
paper [1] the spectral problem on normal oscillations of the �uid (viscous or ideal)
�lling a rotating bounded domain was studied.
In [2, p. 390�410] (see also [3]) the problem on small motions of an ideal
relaxing �uid in a bounded region, excluding rotation, gravity force, and at some
model restrictions on the boundary conditions for dynamical density, was studied.
The model of the relaxing �uid is a generalization of the barotropic �uid in the
sense that the �uid pressure and the �uid density are coupled through the integral
Volterra operator. In the monograph [2] the theorem on the strong solvability
of the appropriate initial-boundary value problem was proved, and a spectral
problem on normal oscillations was studied.
In [4, 5], the problem on small motions of an ideal relaxing �uid �lling a
bounded region and being in�uenced by a gravitational �eld was studied. In this
c© D. Zakora, 2012
A Symmetric Model of Viscous Relaxing Fluid. An Evolution Problem
case the �uid density in the state of relative equilibrium was assumed to be con-
stant. It turns out that the neglect of a change of the stationary density causes
both the breaking of a symmetry of the problem and the noncompact perturba-
tion of the operator pencil corresponding to the spectral problem (even in the case
of a barotropic model). In [6] there was proposed a modi�ed rheological relation
leading to the symmetric model of the ideal relaxing �uid with consideration for
the exact steady state of a �uid. For this model, the evolution and the spectral
problems were studied.
In the present paper the rheological relation from [6] is applied to construct a
model of the viscous relaxing �uid. We reduce the initial-boundary value problem
describing this model to the Cauchy problem for an integro-di�erential equation
in some Hilbert space. It should be noted that there are some versions of the
initial-boundary value problem in an abstract form. Here the way leading to the
equation well-adapted for further studying of the spectral problem is chosen.
1. Small Motions of a Rotating Viscous Relaxing Fluid
1.1. Statement of the problem. Consider a container that uniformly
rotates around the axis parallel to the gravity force and is completely �lled by
a viscous inhomogeneous �uid. The �uid is said to occupy a bounded region
Ω ⊂ R3. Let ~n be a unit vector perpendicular to the boundary S := ∂Ω and
directed out of the region Ω. We introduce a system of the coordinates Ox1x2x3
which is toughly connected with the container so that the Ox3 axis coincides
with the rotating axis and is directed opposite the gravity force, and the origin
of coordinates is in the region Ω. In this case, the uniform velocity of rotation
of the container takes the form ~ω0 := ω0~e3, where ~e3 is the unit vector along the
rotating axis Ox3, and ω0 > 0 for de�niteness. The external stationary �eld of
forces ~F0 is considered to be a gravitational �eld acting along the rotating axis,
i.e., ~F0 = −g~e3, g > 0.
Let us consider the relative equilibrium state of the �uid. From the Navier�
Stokes equation of motion of a viscous �uid, written in the moving system of
coordinates, we obtain the formula for the stationary pressure gradient
∇P0 = ρ0(−~ω0 × (~ω0 × ~r)− g~e3) = ρ0∇(2−1|~ω0 × ~r|2 − gx3), (1.1)
where ~r is the radius-vector of a moving point of the region Ω, and ρ0 is the
stationary density of the �uid.
In the state of relative equilibrium, the dynamic components of pressure and
density responsible for relaxing e�ects in the �uid are leaking. Therefore, we will
consider that the �uid is barotropic in the relative equilibrium state and satis�es
the equation of the state ∇P0 = a2∞∇ρ0, where a∞ is the sound velocity in the
�uid. This equation and relation (1.1) allow us to conclude that ρ0 and a2∞ can be
Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 2 191
D. Zakora
functions of the parameter z := 2−1ω2
0(x
2
1 + x2
2)− gx3. Further we will consider
that the sound velocity a∞ is de�ned for the �uid and it is constant. Then the
stationary density can be determined as a function of the parameter z. In this
case, the stationary density ρ0 is constant only when there is neither rotation
nor gravitational �eld in the system. The function ρ0(z) satis�es the conditions
0 < α1 ≤ ρ0(z) ≤ α2.
We now represent the total pressure and the density of the �uid in the form
P̂ (t, x) = P0(z)+p(t, x), ρ̂(t, x) = ρ0(z) + ρ̃(t, x), where p(t, x) and ρ̃(t, x) are the
dynamical pressure and the density, respectively, arising at small motions of the
�uid relative to its steady state. Assume that the dynamic components satisfy
the following rheological relation:
Pm
( ∂
∂t
)
∇p(t, x) = a2
∞
(
Pm
( ∂
∂t
)
+ ρ0(z)Qm−1
( ∂
∂t
))
∇ρ̃(t, x), (1.2)
where Pm(λ) and Qm−1(λ) are polynomials with the degrees m and m−1, respec-
tively. In this case we can obviously consider that the coe�cient of the highest
degree in the polynomial Pm(λ) is equal to 1. Following the reasoning and ideas
from [7], we will assume that all roots of the polynomial Pm(λ) are real, di�erent
from one another, and negative (we denote them by −bl (l = 1,m)) while the
roots of the polynomial Qm−1(λ) are real, negative and alternate with the roots
of Pm(λ). Thus from (1.2) we obtain the equation of the state
∇p(t, x) = a2
∞
(
∇ρ̃(t, x)− ρ0(z)
t∫
0
∇K̃(t− s)ρ̃(s, x) ds
)
, (1.3)
where K̃(t) :=
∑m
l=1 kl exp(−blt). The numbers b−1
l are used as the times of re-
laxation in the system, and kl > 0 (l = 1,m) are some structural constants. As
a mathematical generalization of the presented constructions, we assume that
K̃ = K̃(t, x) is a su�ciently smooth positive kernel in the evolution problem.
Let us linearize the Navier�Stokes equation written in the moving system of
coordinates with respect to the relative equilibrium state. Using the equation of
the state (1.3), we obtain the problem of small motions of a viscous relaxing �uid
�lling a uniformly rotating solid body
∂~u(t, x)
∂t
− 2ω0
(
~u(t, x)× ~e3
)
= ρ−1
0 (z)
(
µ∆~u(t, x) + (η + 3−1µ)∇div~u(t, x)
)
−∇(
a2
∞ρ−1
0 (z)ρ̃(t, x)
)
+
t∫
0
∇(
a2
∞K̃(t− s, x)ρ̃(s, x)
)
ds + ~f(t, x) (in Ω),
∂ρ̃(t, x)
∂t
+ div
(
ρ0(z)~u(t, x)
)
= 0 (in Ω), ~u(t, x) = ~0 (on S),
192 Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 2
A Symmetric Model of Viscous Relaxing Fluid. An Evolution Problem
where ~u(t, x) is the �eld of velocities in the �uid, ρ̃(t, x) is the dynamic density
of the �uid, µ and η are the dynamic and the second viscosity of the �uid, and
~f(t, x) is a weak �eld of external forces imposed on the gravitational �eld.
To symmetrize the system, let us replace a∞ρ
−1/2
0 (z)ρ̃(t, x) = ρ(t, x). As a
result, we arrive at the basic problem
∂~u(t, x)
∂t
− 2ω0
(
~u(t, x)× ~e3
)
= ρ−1
0 (z)
(
µ∆~u(t, x) + (η + 3−1µ)∇div~u(t, x)
)
−∇(
a∞ρ
−1/2
0 (z)ρ(t, x)
)
+
t∫
0
∇(
K(t− s, x)ρ(s, x)
)
ds + ~f(t, x) (in Ω), (1.4)
∂ρ(t, x)
∂t
+ a∞ρ
−1/2
0 (z)div
(
ρ0(z)~u(t, x)
)
= 0 (in Ω), ~u(t, x) = ~0 (on S), (1.5)
where K(t, x) := a∞ρ
1/2
0 (z)K̃(t, x).
For the completeness of the statement of the problem, we set the initial con-
ditions
~u(0, x) = ~u0(x), ρ(0, x) = ρ0(x). (1.6)
1.2. Auxiliary operators and their properties. Let us introduce a vector
Hilbert space ~L2(Ω, ρ0) with the scalar product and the norm
(~u,~v)~L2(Ω,ρ0) :=
∫
Ω
ρ0(z)~u(x) · ~v(x) dΩ, ‖~u‖2
~L2(Ω,ρ0)
=
∫
Ω
ρ0(z)|~u(x)|2 dΩ.
We introduce a scalar Hilbert space L2(Ω) of functions square summable in
the region Ω, and also its subspace L2,ρ0(Ω) := {f ∈ L2(Ω) | (f, ρ
1/2
0 )L2(Ω) = 0}.
We de�ne an orthogonal projector
Πf := f − (f, ρ
1/2
0 )L2(Ω)‖ρ1/2
0 ‖−2
L2(Ω)ρ
1/2
0 (z).
Obviously, the formula ΠL2(Ω) = L2,ρ0(Ω) is valid.
To pass to the operator formulation of problem (1.4)�(1.6), we introduce a
number of operators and study their properties.
We introduce an operator S~u(t, x) := i
(
~u(t, x)× ~e3
)
, D(S) = ~L2(Ω, ρ0). The
following lemma, whose proof is similar to that of an analogous lemma on prop-
erties of the Coriolis operator from [8], is valid.
Lemma 1.1. The operator S is self-adjoint and bounded in ~L2(Ω, ρ0): S = S∗,
S ∈ L(~L2(Ω, ρ0)); moreover, ‖S‖ = 1.
Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 2 193
D. Zakora
In what follows, we will consider that the function K̃(t, x) is continuously dif-
ferentiable with respect to spatial variables and twice continuously di�erentiable
with respect to time, and the boundary S of the region Ω belongs to the class C2.
Lemma 1.2. We introduce the space HA := {~u ∈ ~W 1
2 (Ω)
∣∣∣ ~u = ~0 (on S)} with
the following scalar product and norm:
(~u,~v)A :=
∫
Ω
E(~u,~v) dΩ, ‖~u‖2
A :=
∫
Ω
E(~u, ~u) dΩ,
E(~u,~v) :=
(
η − 2
3
µ
)
div~udiv~v +
1
2
µ
3∑
j,k=1
(∂uj
∂xk
+
∂uk
∂xj
)( ∂vj
∂xk
+
∂vk
∂xj
)
.
The space HA is a Hilbert space; it is compactly embedded in the space ~L2(Ω, ρ0):
HA ⊂→⊂→ ~L2(Ω, ρ0). A generating operator A of a Hilbert pair (HA; ~L2(Ω, ρ0)),
which is self-adjoint and positive de�nite in ~L2(Ω, ρ0), possesses a discrete spec-
trum and is de�ned on D(A) = ~W 2
2 (Ω) ∩HA. For every �eld ~w ∈ ~L2(Ω, ρ0), the
generalized solution of the problem
−ρ−1
0 (z)
(
µ∆~u + (η + 3−1µ)∇div~u
)
= ~w (in Ω), ~u = ~0 (on S),
given by the formula ~u = A−1 ~w, exists and it is unique.
P r o o f. We now show that HA is a Hilbert space. Let us introduce a new
equivalent norm in the space ~W 1
2 (Ω) by the formula
‖~u‖2
~W 1
2,S(Ω)
:=
∫
Ω
3∑
k=1
|∇uk|2 dΩ +
∫
S
~u dS
2
.
Let ~W 1
2,S(Ω) denote the space ~W 1
2 (Ω) with a new norm.
For any �eld ~u ∈ ~W 1
2,S(Ω) with the condition ~u = ~0 on the boundary S, the
following Korn inequality is valid (see [9, p. 23, Theorem 2.7]):
∫
Ω
E(~u, ~u) dΩ ≥ c1
∫
Ω
3∑
k=1
|∇~uk|2 dΩ = c1‖~u‖2
~W 1
2,S(Ω)
,
where c1 is positive constant depending only on the domain Ω.
By using the Korn inequality, for any �eld ~u ∈ HA, we can get the following
inequalities:
c1‖~u‖2
~W 1
2,S(Ω)
≤ ‖~u‖2
A ≤ max{3η, 2µ}‖~u‖2
~W 1
2,S(Ω)
. (1.7)
194 Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 2
A Symmetric Model of Viscous Relaxing Fluid. An Evolution Problem
The inequalities imply that HA is a Hilbert space.
The Hilbert space HA is dense in ~L2(Ω, ρ0). Taking into account that the
inequality ‖~u‖~L2(Ω,ρ0) ≤ c2‖~u‖ ~W 1
2,S(Ω) is valid for any �eld ~u ∈ ~W 1
2,S(Ω), from the
left-hand inequality in (1.7) we obtain that HA and ~L2(Ω, ρ0) form a Hilbert pair
(HA; ~L2(Ω, ρ0)).
To �nd the generating operator A of the Hilbert pair (HA; ~L2(Ω, ρ0)), we use
the identity (see [8, p. 33])
(A~u,~v)~L2(Ω,ρ0) = (~u,~v)A, ~u ∈ D(A), ~v ∈ HA = D(A1/2). (1.8)
The identity of Betty is valid for any ~u ∈ ~W 2
2 (Ω), ~v ∈ ~W 1
2,S(Ω):
∫
Ω
L~u · ~v dΩ =
∫
Ω
E(~u,~v) dΩ−
∫
S
σ(~u)~n · ~v dS,
L~u := −(
µ∆~u + (η + 3−1µ)∇div~u
)
, σ(~u) := {σj,k(~u)}3
j,k=1,
σj,k(~u) :=
(
η − 2
3
µ
)
δjkdiv~u + µ
(∂uj
∂xk
+
∂uk
∂xj
)
, j, k = 1, 2, 3.
By using the identities above, for the twice di�erentiable �eld ~u we can trans-
form identity (1.8) as follows:
(A~u,~v)~L2(Ω,ρ0) =
∫
Ω
E(~u,~v) dΩ =
∫
Ω
L~u · ~v dΩ +
∫
S
σ(~u)~n · ~v dS
=
∫
Ω
L~u · ~v dΩ = (−ρ−1
0 (z)
(
µ∆~u + (η + 3−1µ)∇div~u
)
, ~v)~L2(Ω,ρ0).
This implies that any twice di�erentiable solution ~u of the equation A~u = ~w is
a solution of the problem −ρ−1
0 (z)
(
µ∆~u + (η + 3−1µ)∇div~u
)
= ~w (in Ω), ~u = ~0
(on S). This problem has a unique generalized solution ~u = A−1 ~w for any �eld
~w ∈ ~L2(Ω, ρ0). It should be noticed that the �eld ~u is called a generalized solution
of the described problem if the identity
∫
Ω
E(~u,~v) dΩ =
∫
Ω
ρ0 ~w · ~v dΩ
[
= (~w,~v)~L2(Ω,ρ0)
]
∀~v ∈ HA
holds.
It follows from the smoothness of the boundary S that D(A) = ~W 2
2 (Ω) ∩HA.
It follows from the left-hand inequality in (1.7) and the compactness of the
embedding of the space ~W 1
2,S(Ω) in ~L2(Ω, ρ0) that HA is compactly embedded
in the space ~L2(Ω, ρ0): HA ⊂→⊂→ ~L2(Ω, ρ0). This implies the compactness of the
operator A−1; hence the spectrum of the operator A is discrete.
Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 2 195
D. Zakora
De�ne an operator B~u(t, x) := a∞ρ
−1/2
0 (z)div(ρ0(z)~u(t, x)), D(B) := HA.
Lemma 1.3. The adjoint operator B∗ρ = −∇(a∞ρ
−1/2
0 ρ), D(B∗) = W 1
2,ρ0
(Ω),
W 1
2,ρ0
(Ω) := W 1
2 (Ω) ∩ L2,ρ0(Ω). The following inequality takes place:
∃ c > 0 : ‖B~u‖W 1
2,ρ0
(Ω) ≤ c‖A~u‖~L2(Ω,ρ0) ∀~u ∈ D(A).
P r o o f. Let ~u ∈ D(B) := D(A1/2) = HA. Then we have
(B~u, ρ)L2,ρ0(Ω) =
∫
Ω
a∞ρ
−1/2
0 div(ρ0~u)ρ dΩ = −
∫
Ω
ρ0~u · ∇
(
a∞ρ
−1/2
0 ρ
)
dΩ
+
∫
S
a∞ρ
1/2
0 ρ~u·~n dS = −
∫
Ω
ρ0~u·∇
(
a∞ρ
−1/2
0 ρ
)
dΩ = (~u,−∇(
a∞ρ
−1/2
0 ρ
)
)~L2(Ω,ρ0).
The above implies the formula for the operator B∗.
Consider the problem
L~u := −µ∆~u− (η + 3−1µ)∇div~u = ~f (in Ω), BL~u := ~u = ~g (on S). (1.9)
It can be shown in the usual way that the matrix di�erential expression L is
nonsingular and properly elliptic, and the boundary condition BL covers it (see
[10]). It follows from the theorem of normal solvability (see [10, p. 241]) that
there exist constants c3 > 0, c4 > 0 (not depending on the �eld ~u) such that the
following inequalities are valid:
c3‖~u‖2
~W 2
2 (Ω)
≤ ‖L~u‖2
~L2(Ω)
≤ c4‖~u‖2
~W 2
2 (Ω)
∀~u ∈ ~W 2
2 (Ω, BL), (1.10)
~W 2
2 (Ω, BL) := {~u ∈ ~W 2
2 (Ω)
∣∣BL~u = ~u = ~0 (on S)} = D(A),
‖~u‖2
~W 2
2 (Ω)
:=
3∑
k=1
[
‖uk‖2
L2(Ω) +
∑
|α|=2
‖Dαuk‖2
L2(Ω)
]
.
Further, from the Erling�Nirenberg inequality (see [11, p. 33]) it follows that
there exists a constant c5 > 0 (not depending on the �eld ~u) such that the equality
∥∥∂uk
∂xj
∥∥2
L2(Ω)
≤ c5‖~u‖2
~W 2
2 (Ω)
∀~u ∈ ~W 2
2 (Ω), k, j = 1, 2, 3 (1.11)
is valid.
Let now ~u ∈ D(A) = ~W 2
2 (Ω, BL). By using inequalities (1.10), (1.11), we get
‖B~u‖2
W 1
2,ρ0
(Ω) = a2
∞
∫
Ω
(|∇ρ
−1/2
0 div(ρ0~u)|2 + |ρ−1/2
0 div(ρ0~u)|2) dΩ ≤ c6‖~u‖2
~W 2
2 (Ω)
≤ c6c
−1
3 ‖L~u‖2
~L2(Ω)
≤ c6c
−1
3 max
x∈Ω
ρ0
∫
Ω
ρ0|ρ−1
0 L~u|2 dΩ = c‖A~u‖2
~L2(Ω,ρ0)
,
196 Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 2
A Symmetric Model of Viscous Relaxing Fluid. An Evolution Problem
where c = c(c3, c5, ρ0, a∞, Ω) > 0 is some absolute constant.
We de�ne the operators Q := BA−1/2, Q+ := A−1/2B∗.
Lemma 1.4. The operator Q is bounded Q ∈ L(~L2(Ω, ρ0), L2,ρ0(Ω)).
The operator Q∗ admits an extension in a continuity to the bounded operator Q+,
Q+ = Q∗, Q+ = Q∗|D(B∗).
P r o o f. Let ~u ∈ D(B) := D(A1/2) = HA. We have
‖B~u‖2
L2,ρ0(Ω) =
∫
Ω
a2
∞ρ−1
0 |div(ρ0~u)|2 dΩ ≤ 2a2
∞
∫
Ω
(
ρ−1
0 |∇ρ0|2|~u|2
+ ρ0|div~u|2
)
dΩ ≤ c7
∫
Ω
E(~u, ~u) dΩ = c7‖A1/2~u‖2
~L2(Ω,ρ0)
,
where c7 = c7(a∞, ρ0, Ω) > 0 is some absolute constant. After replacement
A1/2~u = ~v, ~v ∈ ~L2(Ω, ρ0), it follows from the last inequality that the operator
Q is bounded. This implies the boundedness of the operator Q∗ and the simply
checked relation Q+ = Q∗|D(B∗).
We de�ne an operator function M(t)ρ(t, x) := Πρ0(z)K̃(t, x)Πρ(t, x). Obvi-
ously, the operator function M(t) is bounded self-adjoint and positive de�nite in
L2,ρ0(Ω).
1.3. Reduction to the �rst-order integro-di�erential equation.
Solvability of the initial-boundary value problem. Using the operators
introduced above, we can write problem (1.4)�(1.6) as a system of two equations
with initial-value conditions in the Hilbert space H = ~L2(Ω, ρ0)⊕ L2,ρ0(Ω),
d~u
dt
+ (2ω0iS + A)~u−B∗ρ +
t∫
0
B∗M(t− s)ρ(s) ds = ~f(t),
dρ
dt
+ B~u = 0, (~u; ρ)τ (0) := (~u0; ρ0)τ .
(1.12)
Notice that the kernel K̃(t, x) is continuously di�erentiable with respect to
spatial variables, and twice continuously di�erentiable with respect to time. This
implies that the operator function M(t) is twice continuously di�erentiable with
values in the space L(W 1
2,ρ0
(Ω)).
Let us change the variables in system (1.12) according to the formula
ρ̂(t) :=
t∫
0
M(t− s)ρ(s) ds, ρ̂(0) = 0.
Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 2 197
D. Zakora
Considering this di�erentiated relation as a di�erential equation associated with
system (1.12), we arrive at a Cauchy problem for a �rst-order integro-di�erential
equation in the Hilbert space H := ~L2(Ω, ρ0)⊕ L2,ρ0(Ω)⊕ L2,ρ0(Ω),
dζ
dt
+ Bζ =
t∫
0
M(t− s)ζ(s) ds + F(t), ζ(0) = ζ0, (1.13)
where ζ(t) := (~u(t); ρ(t); ρ̂(t))τ , ζ0 := (~u0; ρ0; 0)τ , F(t) := (~f(t); 0; 0)τ . The nonzero
components of the operator blocks B and M(t) have the following representa-
tion: M3,2(t) := M ′(t), B1,1 := 2ω0iS + A, B1,2 := −B∗, B1,3 := B∗, B2,1 := B,
B3,2 := −M(0).
Thus, if ~u and ρ are a solution of problem (1.4)�(1.6) (the problem on small
motions of the viscose rotating relaxing �uid in a bounded domain) such that
all the reasonings above are applicable, then the function ζ is a solution of the
Cauchy problem for the �rst-order integro-di�erential equation (1.13).
De�nition 1.1. If the function ζ is a strong solution of the Cauchy prob-
lem (1.13), then the corresponding functions ~u, ρ are called a strong solution of the
initial-boundary value problem (1.4)�(1.6). A function ζ(t) is called a strong so-
lution of the Cauchy problem (1.13) (see [12, p. 38]) if ζ(t) ∈ D(B) for any t from
R+ := [0, +∞), Bζ(t) ∈ C(R+;H), ζ(t) ∈ C1(R+;H), ζ(0) = ζ0, and the equation
from (1.13) is satis�ed for any t ∈ R+.
Let us change the sough function in problem (1.13), ζ(t) = eatξ(t) :=
eat(~v(t); q(t); q̂(t))τ ,
dξ
dt
+ (A+ S)ξ =
t∫
0
e−a(t−s)M(t− s)ξ(s) ds + e−atF(t), ξ(0) = ζ0, (1.14)
where nonzero components of the operator blocksA and S have the formA1,1 := A,
A1,2 := −B∗, A1,3 := B∗, A2,1 := B, A2,2 = A3,3 := aI, S3,2 := −M(0), S1,1 :=
2ω0iS + aI (here I is a unit operator in the corresponding space). The operator
S is bounded in H. The domain of de�nition of the operator A has the form
D(A) := D(A) ⊕ D(B∗) ⊕ D(B∗). The operator A appears to be non closed.
This fact complicates the using of theorems of solvability of abstract di�erential
equations. In this connection we will prove the following lemma.
Lemma 1.5. Let a > 4−1‖Q‖2. Then the operator A admits a closure to the
maximal uniformly accretive operator A which can be presented in the symmetric
form
A = diag(A1/2, I, I)
I −Q∗ Q∗
Q aI 0
0 0 aI
diag(A1/2, I, I)
198 Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 2
A Symmetric Model of Viscous Relaxing Fluid. An Evolution Problem
and in the Schur�Frobenius form A = (I + D1)diag(A, aI + QQ∗, aI)(I + D2).
Here I is a unit operator in H, and the operators I +D1, I +D2 are lower and
upper triangular blocks, respectively. The distinct from zero components of the op-
erator blocks D1 and D2 have the form (D1)2,1 := QA−1/2, (D2)1,2 := −A−1/2Q∗,
(D2)1,3 := A−1/2Q∗, (D2)2,3 := −(aI + QQ∗)−1QQ∗. The domain of the operator
A has the form D(A) = {(~v; q; q̂)τ ∈ H| q, q̂ ∈ L2,ρ0(Ω), ~v − A−1/2Q∗(q − q̂) ∈
D(A)}.
P r o o f. From the formulas for the operator A it is seen that it is presented
as a product of three closed operators. It follows from the Schur�Frobenius fac-
torization that there exists the operator (A)−1, and the domain of the operator
(A)−1 is the whole space H. Thus, A is the closed operator.
Let us check that the operatorA is accretive onD(A). Let ξ = (~v; q; q̂)τ ∈ D(A).
We �x ‖Q‖(2a)−1 < ε < 2‖Q‖−1. Using the Cauchy inequality, we get
Re(Aξ, ξ) = ‖A1/2~v‖2 + Re(Q∗q̂, A1/2~v) + a‖q‖2 + a‖q̂‖2
≥ ‖A1/2~v‖2 − ‖Q∗‖‖q̂‖‖A1/2~v‖+ a‖q‖2 + a‖q̂‖2
≥ ‖A1/2~v‖2 − ‖Q∗‖(ε2−1‖A1/2~v‖2 + (2ε)−1‖q̂‖2
)
+ a‖q‖2 + a‖q̂‖2
≥ min{(1− ‖Q∗‖ε2−1)γ(A), a− ‖Q∗‖(2ε)−1}‖ξ‖2 =: γ(A)‖ξ‖2,
where γ(A) > 0 is an exact lower boundary of the operator A. It follows from
the estimations above that γ(A) > 0 if only the number a satis�es the lemma
conditions. Hence the operator A is uniformly accretive. The range of the values
of the operator A coincides with the whole space H. Hence it is a maximal
uniformly accretive operator.
It can be shown that the factorizations from this lemma are valid for the
operator A when Q∗ is replaced by Q+. By direct computations it is also checked
that the operator A is uniformly accretive on the domain D(A). In this case the
operator A admits a closure which coincides with A.
Notice that in the lemma the number a > 0 can be chosen to be so large that
A+ S would be a maximal uniformly accretive operator. Everywhere below we
will assume that it is.
Let us consider along with (1.14) a Cauchy problem with the closed operator
dξ
dt
+ (A+ S)ξ =
t∫
0
e−a(t−s)M(t− s)ξ(s) ds + e−atF(t), ξ(0) = ζ0. (1.15)
The following solvability theorem is valid for problem (1.15).
Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 2 199
D. Zakora
Theorem 1.1. Let ζ0 ∈ D(A) and function F(t) satis�es the Gelder condition,
∀ τ ∈ R+ ∃ K = K(τ) > 0, k(τ) ∈ (0, 1],
‖F(t)−F(s)‖H ≤ K|t− s|k at 0 ≤ s, t ≤ τ.
Then the strong solution of the Cauchy problem (1.15) exists and it is unique.
P r o o f. Using the Schur�Frobenius factorization for the operator A, we
change the sought function in problem (1.15), z(t) = (I +D2)ξ(t). After a series
of simple transformations we get the following Cauchy problem:
dz
dt
+ B̂z =
t∫
0
M̂(t− s)z(s) ds + F̂(t), z(0) = (I +D2)ζ0 =: z0, (1.16)
where B̂ := (I + T )A0 +D, M̂(t) := e−at(I +D2)M(t)(I +D2)−1, F̂(t) :=
e−at(I +D2)F(t). The operator A0 := diag(A, aI + QQ∗, aI) is the central block
in the Schur�Frobenius factorization of the operator A. The nonzero compo-
nents of the operator T have the form T1,2 := −A−1/2Q∗, T1,3 := A−1/2Q∗,
T2,1 := (I −A−1/2Q∗)QA−1/2. The nonzero component of the operator D3 has
the form (D3)2,3 := −a(aI + QQ∗)−1QQ∗, and D := (I +D2)S(I +D2)−1 +D3.
It follows from the lemmas about auxiliary operators and the introduced no-
tations that the operator D is bounded, and T is a compact operator in H.
The domain of the operator A0 is D(A0) = D(A)⊕ L2,ρ0(Ω)⊕ L2,ρ0(Ω). The
operator A0 is self-adjoint and positive de�nite, and the operator −A0 is a gen-
erator of a strongly continuous analytical semigroup of operators. This implies
(see [12]) that the operator −B̂ is a generator of a strongly continuous semigroup
U(t) := exp(−tB̂) in H analytical in some sector containing a positive semiaxis.
It follows from M(t) ∈ C2(R+;L(W 1
2,ρ0
(Ω))) and formulas for the operator
function M̂(t) that M̂(t) ∈ C1(R+;L(H)).
The further proof follows the ideas from monograph [2].
By the conditions of the theorem, the function F(t) satis�es the Gelder condi-
tion, ζ0 ∈ D(A), and hence z0 := (I +D2)ζ0 ∈ D(A0). We will assume that the
Cauchy problem (1.16) has a strong solution z(t). Taking into account Theorem
1.4 (see [13, p. 130]), we obtain
z(t) = U(t)z0 +
t∫
0
U(t− s)F̂(s) ds +
t∫
0
U(t− s)
s∫
0
M̂(s− τ)z(τ) dτ
ds
= U(t)z0 +
t∫
0
U(t− s)F̂(s) ds +
t∫
0
dτ
∫ t
τ
U(t− s)M̂(s− τ)z(τ) ds. (1.17)
200 Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 2
A Symmetric Model of Viscous Relaxing Fluid. An Evolution Problem
Consider the internal integral in (1.17). It follows from M̂(t) ∈ C1(R+;L(H))
that there exists the partial derivative
∂
∂s
(
U(t−s)B̂−1M̂(s−τ)z(τ)
)
= U(t−s)M̂(s−τ)z(τ)+U(t−s)B̂−1 ∂
∂s
M̂(s−τ)z(τ).
Integrating this from τ to t with respect to s, we can see that
t∫
τ
U(t− s)M̂(s− τ)z(τ) ds = B̂−1
(
M̂(t− τ)z(τ)− U(t− τ)M̂(0)z(τ)
−
t∫
τ
U(t− s)
∂
∂s
M̂(s− τ)z(τ) ds
)
=: B̂−1M̂1(t, τ)z(τ). (1.18)
From (1.17) and (1.18) we get that any strong solution z(t) of the Cauchy
problem (1.16) satis�es the following Volterra integral equation:
z(t) = ẑ(t)+
t∫
0
B̂−1M̂1(t, s)z(s) ds, ẑ(t) := U(t)z0+
t∫
0
U(t−s)F̂(s) ds. (1.19)
Here ẑ(t) is the solution of the Cauchy problem (1.16), which does not contain
an integral term, and thus ẑ(t) ∈ C(R+;D(B̂)) ∩ C1(R+;H).
Let us show that equation (1.19) has a unique solution and this solution is the
strong solution of the Cauchy problem (1.16). Introduce the spaceH(B̂) := (D(B̂),
‖ · ‖H(B̂)
), where ‖z‖H(B̂)
:= ‖B̂z‖ for any z ∈ D(B̂) = D(A0). It is known that
H(B̂) is a Banach space.
It follows from (1.18) that B̂−1M̂1(t, s) ∈ C, 0 ≤ s ≤ t < +∞;L(H(B̂)).
Therefore, equation (1.19), considered in H(B̂), is a Volterra integral equation
of the second kind with continuous kernel. Therefore, taking into account the
inclusion ẑ(t) ∈ C(R+;H(B̂)), we may conclude that equation (1.19) has the
unique solution z(t) ∈ C(R+;H(B̂)).
From the inclusion ẑ(t) ∈ C1(R+;H), we obtain that z(t) is a continuously
di�erentiable function with values in the Hilbert space H. Direct computations
show that z(t) satis�es de�nition 1.1; thus, z(t) is the unique strong solution
of problem (1.16). Then ξ(t) = (I +D2)−1z(t) is the unique strong solution of
problem (1.15).
Using Theorem 1.1, we study the strong solutions of problem (1.4)�(1.6) (the
problem on small motions of the viscous rotating relaxing �uids in a bounded
domain).
Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 2 201
D. Zakora
Theorem 1.2. Let the �eld ~f(t, x) satisfy the Gelder condition
∀ τ ∈ R+ ∃ K = K(τ) > 0, k(τ) ∈ (0, 1]: ‖~f(t)− ~f(s)‖~L2(Ω,ρ0) ≤ K|t− s|k, 0 ≤
s, t ≤ τ . Then for any ~u0 ∈ D(A) and ρ0 ∈ D(B∗) there exists a unique strong
solution of the initial-boundary value problem (1.4)�(1.6).
P r o o f. By the de�nition 1.1, we are to prove that problem (1.13) has a
unique strong solution. By the theorem conditions, ζ0 := (~u0; ρ0; 0)τ ∈ D(A)
⊂ D(A), and the function F(t) := (~f(t); 0; 0)τ satis�es the Gelder condition.
From Theorem 1.1 we get that problem (1.15) has the unique strong solution
ξ(t) = (~v(t); q(t); q̂(t))τ . We represent the equation from (1.15) as the system
d~v
dt
+ (aI + 2ω0iS)~v + A(~v −A−1/2Q∗q + A−1/2Q∗q̂) = e−at ~f(t),
dq
dt
+ aq + QA1/2~v = 0,
dq̂
dt
+ aq̂ −M(0)q =
t∫
0
e−a(t−s)M ′(t− s)q(s) ds.
(1.20)
The function ξ(t) will be the unique strong solution of the Cauchy prob-
lem (1.14) if in the �rst equation of this system it is possible to remove the
brackets before the operator A.
Using the formula ξ(0) = ζ0 = (~u0; ρ0; 0)τ and Q = BA−1/2, from the second
equation of system (1.20) we obtain
q(t) = e−atρ0 −
t∫
0
e−a(t−s)QA1/2~v(s) ds = e−atρ0 −
t∫
0
e−a(t−s)B~v(s) ds.
From the above, by using the third equation from (1.20), we get
q̂(t) =
t∫
0
M(t− s)
e−asρ0 −
s∫
0
e−a(s−τ)B~v(τ) dτ
ds.
Lemma 1.4 implies that A−1/2Q∗|D(B∗) = A−1/2Q+ = A−1B∗. Using the for-
mulas for q(t) and q̂(t), we get that the inclusion ~v(t) − A−1/2Q∗(q(t) − q̂(t)) ∈
202 Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 2
A Symmetric Model of Viscous Relaxing Fluid. An Evolution Problem
D(A) is valid if and only if
~v(t) + A−1/2Q∗
t∫
0
e−a(t−s)B~v(s)−M(t− s)
s∫
0
e−a(s−τ)B~v(τ) dτ
ds
= ~v(t) +
t∫
0
A−1/2Q∗
e−a(t−s)I −
t∫
s
M(t− τ)e−a(τ−s) dτ
B~v(s) ds
= ~v(t) +
t∫
0
A−1/2Q∗R(t, s)B~v(s) ds =: ~w(t) ∈ D(A), (1.21)
R(t, s) := e−a(t−s)I −
t∫
s
e−a(τ−s)M(t− τ) dτ.
Analogously as in Theorem 1.1, introduce the space H(A) := (D(A), ‖·‖H(A)),
where ‖~v‖H(A) := ‖A~v‖~L2(Ω,ρ0). Equation (1.21), considered in H(A), is the
Volterra integral equation of the second kind with the right side of the equation
~w(t) which is continuous in H(A). We show that A−1/2Q∗R(t, s)B ∈ C, 0 ≤ s ≤
t < +∞;L(H(A)).
By Lemma 1.3, BA−1 ∈ L(~L2(Ω, ρ0),W 1
2,ρ0
(Ω)). It follows from Lemma 1.4
that A1/2Q∗|W 1
2,ρ0
(Ω)=D(B∗) = B∗ ∈ L(W 1
2,ρ0
(Ω), ~L2(Ω, ρ0)). Finally, it follows from
the properties of the operator function M(t) that the kernel R(t, s) is a continuous
operator function with values in L(W 1
2,ρ0
(Ω)). Now estimate the norm
‖A−1/2Q∗R(t, s)B~v‖H(A) = ‖A1/2Q∗R(t, s)B~v‖ = ‖B∗R(t, s)(BA−1)A~v‖
≤ ‖B∗‖L(W 1
2,ρ0
,~L2)‖R(t, s)‖L(W 1
2,ρ0
)‖BA−1‖L(~L2,W 1
2,ρ0
)‖~v‖H(A) =: const‖~v‖H(A).
This implies that Eq. (1.21) is the Volterra integral equation of the second kind
with continuous kernel. It follows from inclusion ~w(t) ∈ C(R+; H(A)) that equa-
tion (1.21) has the unique solution ~v(t) ∈ C(R+; H(A)). Thus, the solution com-
ponent ~v(t) ∈ D(A) and therefore, in system (1.20), the brackets before the ope-
rator A can be removed. As a result, ξ(t) is the solution of equation (1.14) with
non closed operator. Using the inverse replacement in (1.14), we obtain that
ζ(t) = eatξ(t) is the unique strong solution of the Cauchy problem (1.13).
1.4. Reduction to the �rst-order di�erential equation in the
case of the kernel of a special type. Let us consider the case when
K̃(t) =
∑m
l=1 kl(x) exp(−blt), and kl(x) > 0 (l = 1,m) are some structural func-
tions that are assumed to be continuously di�erentiable in the domain Ω. In this
Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 2 203
D. Zakora
case the operator function M(t) from (1.12) becomes M(t) =
∑m
l=1 exp(−blt)Ml,
where Mlρ(t, x) := Πρ0(z)kl(x)Πρ(t, x). Obviously, the operators Ml (l = 1,m)
are bounded self-adjoint and positive de�nite in L2,ρ0(Ω).
In system (1.12) make the changes
ρl(t) :=
t∫
0
exp(−bl(t− s))Mlρ(s) ds, ρl(0) = 0 (l = 1,m),
and represent the obtained system of di�erential equations with initial conditions
as a Cauchy problem for a �rst-order di�erential equation in the Hilbert space
H := ~L2(Ω, ρ0)⊕ L2,ρ0(Ω)⊕ Ĥ, where Ĥ := ⊕m
l=1L2,ρ0(Ω),
d
dt
~u
ρ
ρ̂
+
2ω0iS + A −B∗ B̂∗
B 0 0
0 −M̂ Îb
~u
ρ
ρ̂
=
~f
0
0
,
~u(0)
ρ(0)
ρ̂(0)
=
~u0
ρ0
0
. (1.22)
The following notations are used here: ρ̂ := (ρ1; . . . ; ρm)τ , B̂∗ := (B∗, . . . , B∗),
M̂ := (M1, . . . , Mm)τ , Îb := diag(b1I, . . . , bmI).
We change the sough function in problem (1.22): (~u(t); ρ(t); ρ̂(t))τ = eat(~v(t);
q(t); q̂(t))τ . Then this problem is transformed into the Cauchy problem
dξ
dt
+ (A+ S)ξ = F(t), ξ(0) = ξ0, (1.23)
where ξ(t) := (~v(t); q(t); q̂(t))τ , ξ0 := (~u0; ρ0; 0)τ , F(t) := (e−at ~f(t); 0; 0)τ ,
A :=
A −B∗ B̂∗
B aI 0
0 0 aÎ
, S :=
2ω0iS + aI 0 0
0 0 0
0 −M̂ Îb
.
For the operator A the lemma below is valid.
Lemma 1.6. Let a > 4−1‖Q‖2. Then the operator A admits a closure to a
maximal uniformly accretive operator A and can be presented in the symmetric
form
A = diag(A1/2, I, Î)
I −Q∗ Q̂∗
Q aI 0
0 0 aÎ
diag(A1/2, I, Î), Q̂∗ := (Q∗, . . . , Q∗)
and in the Schur�Frobenius form A = (I + D1)diag(A, aI + QQ∗, aÎ)(I + D2),
where I is a unit operator in H, and the operators I +D1 and I +D2 are
204 Journal of Mathematical Physics, Analysis, Geometry, 2012, vol. 8, No. 2
A Symmetric Model of Viscous Relaxing Fluid. An Evolution Problem
lower and upper triangular blocks, respectively. The distinct from zero com-
ponents of the operator blocks D1 and D2 have the forms (D1)2,1 := QA−1/2,
(D2)1,2 := −A−1/2Q∗, (D2)1,3 := A−1/2Q̂∗, (D2)2,3 := −(aI + QQ∗)−1QQ̂∗. The
domain of the operator A has the form D(A) = {q ∈ L2,ρ0(Ω), q̂ ∈ Ĥ : ~v −
A−1/2Q∗q + A−1/2Q̂∗q̂ ∈ D(A)}.
The proof of Lemma 1.6 is similar to that of Lemma 1.5.
Basing on Eq. (1.23), as in the previous point, it is possible to prove Theorem
1.2 of strong solvability of initial-boundary value problem (1.4)�(1.6).
Let us consider the homogeneous equation (1.23) with the closed operator A.
We search for its solution in the following form: ξ(t) = exp(−(λ + a)t)ξ. As a
result, we get the spectral problem (A + S − aI)ξ = λξ which we will associate
with the problem on normal oscillations of the viscous rotating relaxing �uid. The
last one will be studied in the next paper.
The author expresses his gratitude to Prof. N.D. Kopachevsky for fruitful
discussions of the paper.
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