1. Introduction and Problem Formulation
In this work, we consider the following optimal control problem for the model describing steady non-isothermal creeping flows of an incompressible fluid through a bounded domain
with the locally Lipschitz boundary
:
where
is the velocity of the fluid at a point
,
is the deviation from the average temperature value,
denotes the external forces,
is the deformation rate tensor,
, the function
represents the pressure field,
is the viscosity,
is the thermal conductivity,
is a coefficient characterizing the heat transfer on solid walls of the flow domain,
stands for the heat source intensity,
is the unit outward normal to the surface
,
S is a flat (straight for
) portion of
or the union of several such portions. Functions
and
play the role of controls,
is the set of admissible controls, while
is a given cost functional. By the symbol
we denote the tangential component of a vector, i.e.,
.
The main scope of the present paper is to prove the solvability of problem (
1) in the weak formulation. The proof is based on the Galerkin procedure, methods of the topological degree theory, compactness arguments, and the well-known theorem of Krasnoselskii on the continuity of a superposition operator acting in Lebesgue spaces. In addition, for control system (
1), we introduce the concept of the marginal function, which shows how the optimal value of the cost functional changes with a change in the set of admissible controls. In this paper, it is proved that the marginal function of (
1) is lower semi-continuous.
It should be mentioned that many mathematical works are devoted to optimization and control problems for non-isothermal flows of a viscous fluid. The results from the existing literature mainly deal with the case where the flow occurs inside a bounded domain and external body forces and/or heat sources are used as control parameters (see, for instance, [
1,
2,
3,
4,
5,
6,
7,
8]). Moreover, the Dirichlet-type boundary control by the velocity field is quite well studied (for details, see References [
9,
10,
11,
12,
13]). However, for the analysis of real applications, it is very important to consider models of pressure driven flows, which typically occur in control problems for the heat and mass transfer in pipeline networks. For this purpose, following the approach from [
14], we employ mixed boundary conditions, including the inhomogeneous Dirichlet condition for the pressure on a portion of the boundary of the flow domain; but at the same time, we formulate the optimal control problem without using the
curl operator and boundary conditions associated with this operator. Another distinguishing feature of the present paper is that it takes into account the dependence of both the viscosity and the thermal conductivity coefficient on the temperature. This expands the range of possible applications of our results.
2. Preliminaries
In this section, we present some notations and function spaces utilized in the paper.
Let
,
be Banach spaces. By
we denote the space of all bounded linear mappings from
to
. The space
is equipped with the norm
The strong (weak) convergence in a Banach space is denoted by → (⇀).
Let
and
be subsets of a Banach space
E. By definition, put
This quantity is termed as the directed Hausdorff distance (or one-sided Hausdorff distance) from the set to the set .
Throughout this paper, boldface symbols denote vector- and matrix-valued quantities. For vectors
and matrices
, by
and
, we denote the scalar products, respectively:
As usual,
denotes the Euclidean norm:
The symbol
∇ stands for the gradient with respect to the variables
. The divergence operator is defined as follows:
for
and
.
We employ the standard notation for Lebesgue spaces, such as , , where , and the Sobolev space . When it comes to classes of -valued functions, we use bold face letters, for instance, , , etc.
Recall that the restriction of a function
to the surface
is defined by the rule
, where
is the trace operator (see, e.g., [
15] Section 2.4.2),
if
, and
q is an arbitrary number from
if
.
Let us introduce two sets:
and two basic spaces for handling problem (
1):
In our study, it is convenient to use the following scalar product and norm in
:
From ([
16] Chap. I, Theorems 2.2 and 2.3) it follows that the scalar product
is well defined and the norm
is equivalent to the standard
-norm if the set
has the positive (
)-dimensional measure.
In the space
, we introduce the scalar product and the norm as follows:
By the Friedrichs inequality (see, e.g., [
15] Section 1.1.8, Theorem 1.9), it is easy to show that the scalar product
is well defined and the norm
is equivalent to the standard
-norm.
7. Proof of Theorem 1
Let us prove the existence result (a). Our first step is to show that .
Fix a pair . Let be an orthonormal basis of the space , an orthonormal basis of the space , and m an arbitrary fixed integer.
Consider the following one-parameter problem in space :
Find a vector such thatwhere and are defined as follows:and λ is a parameter, . We shall establish
a priori estimates of solutions to problem (
11)–(
13). Suppose a vector
satisfies (
11)–(
13). Let us multiply (
11) by
and add the obtained equalities for
. This yields
Then, using conditions
(C2) and
(C5), the Cauchy–Bunyakovsky–Schwarz inequality, and
, we derive
Note that
where
is the trace operator, the symbol
stands for the identity mapping. Combining (
14) with (
15)–(
17), we get
It immediately follows that
Now multiply (
12) by
and add the results for
; this gives
Using the integration by parts formula, it is easy to show that the term
vanishes. Indeed,
Therefore, equality (
19) can be rewritten as follows:
from which, by using conditions
(C2) and
(C5), the Cauchy–Bunyakovsky–Schwarz inequality, and
, we derive
whence
With this majoration, we deduce from (
18) that
Furthermore, since
is an orthonormal basis of the space
and
is an orthonormal basis of the space
, from (
13) it follows that
Taking into account (
20) and (
21), we obtain
Note that these estimates do not depend on
as well as
. Therefore, we can apply Proposition 1 to justify the solvability of problem (
11)–(
13). Here, of course, we used Proposition 2 and conditions
(C2)–
(C5) to establish the continuity of the corresponding mappings.
Let
be a solution to (
11)–(
13) with
. Setting
we have
Consider the sequence
. It is clear that estimate (
21) remains valid if we replace
with
. Consequently, the sequence
is bounded in the space
. Likewise, taking into account estimate (
20), we see that the sequence
is bounded in the space
. Therefore, there exists a pair
such that
converges to
weakly in
and
converges to
weakly in
, for some subsequence
as
. Without loss of generality, it can be assumed that
Since the trace operator
is compact (see [
15] Section 2.6.2, Theorem 6.2), we have
Moreover, from (24) and (25) and the compactness theorem for the identity mapping
(see [
15] Section 2.6.1, Theorem 6.1), it follows that
Using (24)–(28) and Proposition 2, we can pass to the limit
in (22) and (23); this gives
for each
. Because
is a basis of
and
is a basis of
, equalities (29) and (30) remain valid if we replace
and
with arbitrary vector function
and function
, respectively:
Thus, we established that is an admissible quadruplet to problem (1) and hence .
Now, consider a sequence
such that
Owing to coercivity condition
(C9), we deduce from (31) that the set
is bounded in the space
. Hence, there exists a subsequence
such that
for some quadruplet
.
The inclusion
implies that
from which, by (32)–(35), one can derive the following equalities:
Comparing this convergence with (35), we conclude that . Moreover, since and the set is closed in the space , we see that the function belongs to . Next, taking into account condition (C6), the inclusion , and convergence (34), we get . Thus, .
In addition, by condition
(C8) and (31)–(35), we can obtain
This means that the quadruplet belongs to the set , and hence . Therefore, we have shown that optimal control problem (1) is solvable in the sense of Definition 2.
We now turn to proving assertion (b). Assume the converse. Then, there exists a subsequence
such that
Consider a sequence
such that
Moreover, by Definition 3, we have
It follows from (36) and (39) that there exists a number
such that
for each
. Hence,
In view of condition
(C9), the set
is bounded in
. Therefore, without loss of generality, it can be assumed that
for some quadruplet
.
Besides, since both the identity mapping
and the trace operator
are compact, the following convergences hold:
Comparing (43) with (46), one can infer that
and
In view of (10), there exist sequences
and
such that
From (42), (47), and (48), it follows that
and, by conditions
(C6) and
(C7), we get
.
Next, taking into account the convergence results (40)–(42), (44), and (45), we pass to the limit
in equalities (37) and (38) and obtain
Therefore, we have established that the quadruplet
belongs to the set
. This is a key point of the proof of assertion (b). Indeed, using condition
(C8) and (39)–(43), we can deduce the following relations:
thus contradicting inequality (36). This contradiction concludes the proof.