# Existence of Optimal Control for a Nonlinear-Viscous Fluid Model

Authors: Evgenii S. Baranovskii; Mikhail A. Artemov
Journal: International Journal of Differential Equations
 (2016)

---

## Abstract

We consider the optimal control problem for a mathematical model describing steady flows of a nonlinear-viscous incompressible fluid in a bounded three-dimensional (or a two-dimensional) domain with impermeable solid walls. The control parameter is the surface force at a given part of the flow domain boundary. For a given bounded set of admissible controls, we construct generalized (weak) solutions that minimize a given cost functional.


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## Body

## 1. Introduction

The control and optimization problems in hydrodynamics have been the focus of attention of the control theory specialists for a long time. Flow boundary control problems have attracted increasing interest in recent years (see, e.g., [1–7]). Such problems are of interest from a theoretical perspective and are beneficial to applications as boundary control is easy to implement in practice.In this paper, we study the optimal boundary control problem for a mathematical model describing steady flows of a nonlinear-viscous incompressible fluid in a bounded domain of spaceR
n, n
=
2,3, with impermeable solid walls. A distinguishing feature of the problem under consideration is that the surface force at the flow domain boundary is used as a control parameter instead of the nonhomogeneous Dirichlet boundary condition for the velocity field. Such an approach makes it possible to consider the case of flow control in a domain with impermeable solid walls without using external body forces as control parameters.It should be mentioned at this point that a lot of studies have been conducted towards mathematical models of nonlinear-viscous fluids (see monograph [8] and [9–12]). Nevertheless, there are very few results on the existence and properties of solutions of control problems for nonlinear-viscous fluid flows. To the best of our knowledge, some results have only been obtained for the two-dimensional case (see [13, 14]).Also, we would mention that there are many mathematical results concerning optimal control problems for the classical Navier-Stokes equations (see [15–17] and the references therein).The aim of this paper is to prove the solvability of the optimal control problem, which is discussed above. More precisely, for a given bounded set of admissible boundary controls, we will construct generalized (weak) solutions that minimize a given lower weakly semicontinuous cost functional.


## 2. Problem Formulation and Main Result

LetΩ be a bounded domain in R
n (n
=
2 or 3) with boundary Γ
∈
C
2. Consider the following optimal boundary control problem:(1)
v
·
∇
v
-
div
⁡
S
+
∇
p
=
f
in
Ω
,
(2)
∇
·
v
=
0
in
Ω
,
(3)
S
=
ψ
I
2
v
D
v
in
Ω
,
(4)
v
·
n
=
0
on
⁡
Γ
,
(5)
v
=
0
on
⁡
Γ
∖
Γ
c
,
(6)
S
n
τ
=
u
on
⁡
Γ
c
,
(7)
u
∈
U
,
(8)
J
v
,
S
,
u
⟶
i
n
f
,where v is the velocity field, p is the pressure function, S is the extra-stress tensor, f is the body force, the symbol ∇ denotes the gradient with respect to the spatial variables x
1
,
…
,
x
n, the divergence div
⁡
S is the vector with coordinates (9)
div
⁡
S
i
=
∑
j
=
1
n
∂
S
j
i
∂
x
j
,
D
(
v
) is the rate of deformation tensor, (10)
D
i
j
v
=
1
2
∂
v
i
∂
x
j
+
∂
v
j
∂
x
i
,
I
2
(
v
) is the second invariant of D
(
v
), (11)
I
2
v
=
∑
i
,
j
=
1
n
D
i
j
v
2
,
ψ is a given function, n is the unit vector of the outer normal to Γ, u is the control, v
·
n is the scalar product of the vectors v and n in space R
n, the symbol [
·
]
τ denotes the tangential component of a vector, that is, (12)
w
τ
=
w
-
w
·
n
n
,
Γ
c is a part of Γ from which the control is realized, U is the set of admissible controls, and J is a given cost functional.From here on, the following notations will be used.M
s
n
×
n denotes the space of symmetric n
×
n-matrices with the norm (13)
A
M
s
n
×
n
=
∑
i
,
j
=
1
n
A
i
j
2
1
/
2
.We use the standard notationsL
q
(
Ω
,
E
) and W
m
,
q
(
Ω
,
E
) for the Lebesgue and Sobolev spaces of vector functions defined on Ω with values in a finite-dimensional space E (for details, see [18]). The scalar product in the space L
2
(
Ω
,
E
) is denoted by (
·
,
·
).By definition, put(14)
L
τ
2
Γ
c
,
R
n
=
w
∈
L
2
Γ
c
,
R
n
:
w
·
n
=
0
.Moreover, we introduce the space(15)
X
Ω
,
R
n
=
v
∈
W
1,2
Ω
,
R
n
:
∇
·
v
=
0
,
v
Γ
·
n
=
0
,
v
Γ
∖
Γ
c
=
0with the following norm: (16)
v
X
Ω
,
R
n
=
D
v
L
2
Ω
,
M
s
n
×
n
.In the right-hand side of (15), the restriction of a vector function v
:
Ω
→
R
n to Γ is defined by the formula (17)
v
Γ
=
γ
0
v
,where γ
0
:
W
1,2
(
Ω
,
R
n
)
→
L
2
(
Γ
,
R
n
) is the trace operator.It follows from Korn’s inequality (see [8]) that the norm ·
X
(
Ω
,
R
n
) is equivalent to the norm induced from W
1,2
(
Ω
,
R
n
). Furthermore, we have the following estimates:(18)
v
L
2
Γ
,
R
n
≤
C
1
v
X
Ω
,
R
n
,
v
L
2
Ω
,
R
n
≤
C
2
v
X
Ω
,
R
n
,where C
1 and C
2 are positive constants.Suppose the following:(i)
the functionψ is measurable and there exist constants a
1 and a
2 such that (19)
0
<
a
1
≤
ψ
t
≤
a
2
,
t
∈
0
,
+
∞
,
(ii)
for anyA
,
B
∈
M
s
n
×
n, we have(20)
∑
i
,
j
=
1
n
ψ
A
M
s
n
×
n
2
A
i
j
-
ψ
B
M
s
n
×
n
2
B
i
j
A
i
j
-
B
i
j
≥
0
,
(iii)
the setU is bounded and sequentially weakly closed in L
τ
2
(
Γ
c
,
R
n
),
(iv)
the functionalJ
:
X
(
Ω
,
R
n
)
×
L
2
(
Ω
,
M
s
n
×
n
)
×
L
τ
2
(
Γ
c
,
R
n
)
→
R is lower weakly semicontinuous; that is, for any sequence {
(
v
k
,
S
k
,
u
k
)
}
k
=
1
∞ such that v
k
→
v weakly in X
(
Ω
,
R
n
), S
k
→
S weakly in L
2
(
Ω
,
M
s
n
×
n
), and u
k
→
u weakly in L
τ
2
(
Γ
c
,
R
n
), we have (21)
J
v
,
S
,
u
≤
lim
inf
k
→
∞
⁡
J
v
k
,
S
k
,
u
k
.Example 1.
Let us consider the following cost functionals:(22)
J
1
v
,
S
,
u
=
λ
1
v
-
v
~
L
2
Ω
,
R
n
2
+
λ
2
S
-
S
~
L
2
Ω
,
M
s
n
×
n
2
+
λ
3
u
-
u
~
L
τ
2
Γ
c
,
R
n
2
,
J
2
v
,
S
,
u
=
-
λ
1
v
-
w
~
L
2
Ω
,
R
n
2
+
λ
2
S
-
S
~
L
2
Ω
,
M
s
n
×
n
2
+
λ
3
u
-
u
~
L
τ
2
Γ
c
,
R
n
2
,where v
~ is a favorable velocity field; w
~ is an unfavorable velocity field, that is, a velocity field whose appearance is undesirable; S
~ is a favorable extra-stress tensor; u
~ is a favorable surface force at Γ
c; and λ
1, λ
2, and λ
3 are positive cost parameters. It is obvious that condition (iv) holds for the functionals J
=
J
i, i
=
1,2.Remark 2.
We do not assume that the set of admissible controls is convex. As is known, the convexity condition is widely used in studying of optimal control problems (see, e.g., [17]). However, this condition does not always hold in applications. Obviously, condition (iii) is weaker than the convexity condition. For example, (iii) is satisfied if the set U can be represented as the union of finite number of convex closed sets in the space L
τ
2
(
Γ
c
,
R
n
).Now we introduce the concept of admissible triplets of (1)–(8) by analogy with the definition of generalized (weak) solutions to hydrodynamic models with slip boundary conditions (see, e.g., [8, 19, 20]).Letf
∈
L
2
(
Ω
,
R
n
).Definition 3.
One says that a triplet(
v
,
S
,
u
)
∈
X
(
Ω
,
R
n
)
×
L
2
(
Ω
,
M
s
n
×
n
)
×
L
τ
2
(
Γ
c
,
R
n
) is anadmissible triplet of control system (1)–(8) if the equality(23)
-
∑
i
=
1
n
v
i
v
,
∂
φ
∂
x
i
+
ψ
I
2
v
D
v
,
D
φ
=
f
,
φ
+
∫
Γ
c
u
·
φ
d
Γ
cholds for any φ
∈
X
(
Ω
,
R
n
) and if conditions (3) and (7) hold.Remark 4.
Equation (23) appears for the following reasons. Let us assume that (
v
,
S
,
p
,
u
) is a classical solution of (1)–(7). We take the L
2-scalar product of (1) with φ
∈
X
(
Ω
,
R
n
). By integrating by parts, we obtain (24)
-
∑
i
=
1
n
v
i
v
,
∂
φ
∂
x
i
+
S
,
D
φ
-
∫
Γ
c
S
n
·
φ
d
Γ
c
=
f
,
φ
.Combining this with (3) and (6), we get (23).
On the other hand, it is not difficult to prove that if an admissible triplet(
v
,
S
,
u
) is sufficiently smooth, then there exists a function p such that (
v
,
S
,
p
,
u
) is a classical solution to (1)–(7).LetM be the set of admissible triplets to problem (1)–(8).Definition 5.
A triplet(
v
∗
,
S
∗
,
u
∗
)
∈
M is called asolution of optimization problem (1)–(8) if the equality (25)
J
v
∗
,
S
∗
,
u
∗
=
inf
v
,
S
,
u
∈
M
⁡
J
v
,
S
,
uholds.Our main result provides existence of solutions to (1)–(8).Theorem 6.
If conditions (i), (ii), (iii), and (iv) hold, then optimization problem (1)–(8) has at least one solution.


## 3. Proof of Theorem6

The proof of Theorem6 is based on the Galerkin method and monotonicity methods [21], as well as the following lemma.Lemma 7.
LetB
R
=
{
x
∈
R
m
:
x
R
m
≤
R
} be a closed ball. Suppose the continuous mapping F
:
B
R
×
[
0,1
]
→
R
m satisfies the following conditions: (a)
F
(
x
,
λ
)
≠
0 for any (
x
,
λ
)
∈
∂
B
R
×
[
0,1
],
(b)
F
(
x
,
0
)
=
A
x for any x
∈
B
R,
whereA
:
R
m
→
R
m is an isomorphism;then for any λ
∈
[
0,1
] the equation F
(
x
,
λ
)
=
0 has at least one solution x
λ
∈
B
R.Lemma7 can be proved by methods of topological degree theory (see, e.g., [22]).Proof of Theorem6.
First we show that the set of admissible triplets is nonempty. Let us fix an elementu
0
=
(
u
1
0
,
…
,
u
n
0
)
∈
U. Suppose {
φ
j
}
j
=
1
∞ is an orthonormal basis of the space X
(
Ω
,
R
n
).
For an arbitrary fixed numberm
∈
N, we consider the following auxiliary problem.
Find a vector(
α
m
1
,
…
,
α
m
m
)
∈
R
m such that(26)
-
λ
∑
i
=
1
n
v
i
m
v
m
,
∂
φ
j
∂
x
i
+
ψ
λ
I
2
v
m
D
v
m
,
D
φ
j
=
λ
f
,
φ
j
+
λ
∫
Γ
c
u
0
·
φ
j
d
Γ
c
,
j
=
1
,
…
m
,
(27)
v
m
=
∑
j
=
1
m
α
m
j
φ
j
,where λ is a parameter, λ
∈
[
0,1
].
First we prove some a priori estimates of solutions to problem (26) and (27). Let (
α
m
1
,
…
,
α
m
m
) be a solution of system (26) and (27) with a fixed parameter λ
∈
[
0,1
]. We multiply (26) by α
m
j and add the corresponding equalities for j
=
1
,
…
,
m. Taking into account the equality (28)
∑
i
=
1
n
v
i
m
v
m
,
∂
v
m
∂
x
i
=
0
,we obtain(29)
ψ
λ
I
2
v
m
D
v
m
,
D
v
m
=
λ
f
,
v
m
+
λ
∫
Γ
c
u
0
·
v
m
d
Γ
c
.Using (18) and (19), from (29) we obtain the estimate (30)
a
1
v
m
X
Ω
,
R
n
2
≤
f
L
2
Ω
,
R
n
v
m
L
2
Ω
,
R
n
+
u
0
L
τ
2
Γ
c
,
R
n
v
m
L
τ
2
Γ
c
,
R
n
≤
C
2
f
L
2
Ω
,
R
n
+
C
1
u
0
L
τ
2
Γ
c
,
R
n
v
m
X
Ω
,
R
n
.This yields that(31)
v
m
X
Ω
,
R
n
≤
a
1
-
1
C
2
f
L
2
Ω
,
R
n
+
C
1
u
0
L
τ
2
Γ
c
,
R
n
.
Applying Lemma7 to system (26) and (27), we see that problem (26) and (27) is solvable for any λ
∈
[
0,1
] and m
∈
N.
Let{
v
m
}
m
=
1
∞ be a sequence of vector functions that satisfy (26) and (27) with λ
=
1. It is clear that(32)
-
∑
i
=
1
n
v
i
m
v
m
,
∂
φ
j
∂
x
i
+
ψ
I
2
v
m
D
v
m
,
D
φ
j
=
f
,
φ
j
+
∫
Γ
c
u
0
·
φ
j
d
Γ
c
,
j
=
1
,
…
,
m
.
Note that estimate (31) is independent of m. This shows the existence of a vector function v
0
∈
X
(
Ω
,
R
n
) and a subsequence m
′
→
∞ such that v
m
′
→
v
0 weakly in X
(
Ω
,
R
n
). For the sake of simplicity, we assume that(33)
v
m
⟶
v
0
weakly  in
X
Ω
,
R
n
as
m
⟶
∞
.Moreover, by the Sobolev embedding theorems, we have(34)
v
m
⟶
v
0
strongly  in
L
4
Ω
,
R
n
as
m
⟶
∞
.
Using (34), we get (35)
∑
i
=
1
n
v
i
m
v
m
,
∂
φ
j
∂
x
i
⟶
∑
i
=
1
n
v
i
0
v
0
,
∂
φ
j
∂
x
i
as
m
⟶
∞
.Therefore we can pass to the limit m
→
∞ in equality (32) and obtain(36)
lim
m
→
∞
⁡
ψ
I
2
v
m
D
v
m
,
D
φ
j
=
∑
i
=
1
n
v
i
0
v
0
,
∂
φ
j
∂
x
i
+
f
,
φ
j
+
∫
Γ
c
u
0
·
φ
j
d
Γ
cfor any j
∈
N. Since {
φ
j
}
j
=
1
∞ is a basis of the space X
(
Ω
,
R
n
), it follows that equality (36) remains valid if we replace φ
j by an arbitrary vector function φ
∈
X
(
Ω
,
R
n
):(37)
lim
m
→
∞
⁡
ψ
I
2
v
m
D
v
m
,
D
φ
=
∑
i
=
1
n
v
i
0
v
0
,
∂
φ
∂
x
i
+
f
,
φ
+
∫
Γ
c
u
0
·
φ
d
Γ
c
.
Now we multiply (32) by α
m
j and add the corresponding equalities for j
=
1
,
…
,
m. The result is (38)
ψ
I
2
v
m
D
v
m
,
D
v
m
=
f
,
v
m
+
∫
Γ
c
u
0
·
v
m
d
Γ
c
.Hence we find in the limit(39)
lim
m
→
∞
⁡
ψ
I
2
v
m
D
v
m
,
D
v
m
=
f
,
v
0
+
∫
Γ
c
u
0
·
v
0
d
Γ
c
.Taking into account (20), (33), (37), and (39), we obtain the estimate (40)
-
μ
∑
i
=
1
n
v
i
0
v
0
,
∂
φ
j
∂
x
i
+
μ
ψ
I
2
v
0
-
μ
φ
j
D
v
0
-
μ
φ
j
,
D
φ
j
-
μ
f
,
φ
j
-
μ
∫
Γ
c
u
0
·
φ
j
d
Γ
c
=
-
f
,
v
0
-
∫
Γ
c
u
0
·
v
0
d
Γ
c
+
∑
i
=
1
n
v
i
0
v
0
,
∂
v
0
-
μ
φ
j
∂
x
i
+
f
,
v
0
-
μ
φ
j
+
∫
Γ
c
u
0
·
v
0
-
μ
φ
j
d
Γ
c
+
μ
ψ
I
2
v
0
-
μ
φ
j
D
v
0
-
μ
φ
j
,
D
φ
j
=
-
lim
m
→
∞
⁡
ψ
I
2
v
m
D
v
m
,
D
v
m
+
lim
m
→
∞
⁡
ψ
I
2
v
m
D
v
m
,
D
v
0
-
μ
φ
j
+
lim
m
→
∞
⁡
ψ
I
2
v
0
-
μ
φ
j
D
v
0
-
μ
φ
j
,
D
v
m
-
D
v
0
-
μ
φ
j
=
-
lim
m
→
∞
⁡
ψ
I
2
v
m
D
v
m
-
ψ
I
2
v
0
-
μ
φ
j
D
v
0
-
μ
φ
j
,
D
v
m
-
D
v
0
-
μ
φ
j
≤
0for any number μ
>
0. Multiplying the obtained inequality by μ
-
1, we get(41)
-
∑
i
=
1
n
v
i
0
v
0
,
∂
φ
j
∂
x
i
+
ψ
I
2
v
0
-
μ
φ
j
D
v
0
-
μ
φ
j
,
D
φ
j
-
f
,
φ
j
-
∫
Γ
c
u
0
·
φ
j
d
Γ
c
≤
0for any j
∈
N and μ
>
0.
Using Krasnoselskii’s theorem [22] on continuity of Nemytskii operators, we can pass to the limit μ
→
0 in (41):(42)
-
∑
i
=
1
n
v
i
0
v
0
,
∂
φ
j
∂
x
i
+
ψ
I
2
v
0
D
v
0
,
D
φ
j
-
f
,
φ
j
-
∫
Γ
c
u
0
·
φ
j
d
Γ
c
≤
0
.Since {
φ
j
}
j
=
1
∞ is a basis of the space X
(
Ω
,
R
n
), it follows that inequality (42) remains valid if we replace φ
j by an arbitrary vector function φ
∈
X
(
Ω
,
R
n
). Furthermore, since φ is an arbitrary vector function from the space X
(
Ω
,
R
n
), we have (43)
-
∑
i
=
1
n
v
i
0
v
0
,
∂
φ
∂
x
i
+
ψ
I
2
v
0
D
v
0
,
D
φ
-
f
,
φ
-
∫
Γ
c
u
0
·
φ
d
Γ
c
=
0
.This implies that the triplet (
v
0
,
ψ
(
I
2
(
v
0
)
)
D
(
v
0
)
,
u
0
) is an admissible triplet of problem (1)–(7) and thus M
≠
∅.
We will show thatM is bounded in the space X
(
Ω
,
R
n
)
×
L
2
(
Ω
,
M
s
n
×
n
)
×
L
τ
2
(
Γ
c
,
R
n
). Suppose (
v
,
S
,
u
) is an arbitrary triplet from M and φ
=
v. It follows from (23) that (44)
ψ
I
2
v
D
v
,
D
v
=
f
,
v
+
∫
Γ
c
u
·
v
d
Γ
c
.This yields that(45)
v
X
Ω
,
R
n
≤
a
1
-
1
C
2
f
L
2
Ω
,
R
n
+
C
1
sup
w
∈
U
⁡
w
L
τ
2
Γ
c
,
R
n
.Moreover, taking into account (19), we obtain(46)
S
L
2
Ω
,
M
s
n
×
n
=
ψ
I
2
v
D
v
L
2
Ω
,
M
s
n
×
n
≤
a
2
D
v
L
2
Ω
,
M
s
n
×
n
=
a
2
v
X
Ω
,
R
n
≤
a
2
a
1
-
1
C
2
f
L
2
Ω
,
R
n
+
C
1
s
u
p
w
∈
U
w
L
τ
2
Γ
c
,
R
n
.Recall that the set U is bounded in L
τ
2
(
Γ
c
,
R
n
). Therefore from estimates (45) and (46) it follows that the set M is bounded in the space X
(
Ω
,
R
n
)
×
L
2
(
Ω
,
M
s
n
×
n
)
×
L
τ
2
(
Γ
c
,
R
n
).
Now we will show that the setM is sequentially weakly closed. Take a sequence {
(
v
k
,
S
k
,
u
k
)
}
k
=
1
∞
⊂
M such that v
k
→
v
^ weakly in X
(
Ω
,
R
n
), S
k
→
S
^ weakly in L
2
(
Ω
,
M
s
n
×
n
), and u
k
→
u
^ weakly in L
τ
2
(
Γ
c
,
R
n
) as k
→
∞. Let us check that (
v
^
,
S
^
,
u
^
)
∈
M.
By definition, we have(47)
-
∑
i
=
1
n
v
i
k
v
k
,
∂
φ
∂
x
i
+
ψ
I
2
v
k
D
v
k
,
D
φ
=
f
,
φ
+
∫
Γ
c
u
k
·
φ
d
Γ
cfor any φ
∈
X
(
Ω
,
R
n
). Arguing as above, we conclude that (48)
-
∑
i
=
1
n
v
^
i
v
^
,
∂
φ
∂
x
i
+
ψ
I
2
v
^
D
v
^
,
D
φ
=
f
,
φ
+
∫
Γ
c
u
^
·
φ
d
Γ
c
.From condition (iii), we get u
^
∈
U. Thus, it remains to show that(49)
S
^
=
ψ
I
2
v
^
D
v
^
.
Sincev
k
→
v
^ weakly in X
(
Ω
,
R
n
), we see that(50)
D
v
k
⟶
D
v
^
weakly  in
L
2
Ω
,
R
s
n
×
n
as
k
⟶
∞
.Note also that(51)
ψ
I
2
v
k
D
v
k
,
Φ
=
S
k
,
Φ
⟶
S
^
,
Φ
as
k
⟶
∞
,for any Φ
∈
L
2
(
Ω
,
R
s
n
×
n
).
Using the equalityS
k
=
ψ
(
I
2
(
v
k
)
)
D
(
v
k
), we rewrite (47) as follows: (52)
-
∑
i
=
1
n
v
i
k
v
k
,
∂
φ
∂
x
i
+
S
k
,
D
φ
=
f
,
φ
+
∫
Γ
c
u
k
·
φ
d
Γ
c
.Passing to the limit k
→
∞ in this equality, we obtain(53)
-
∑
i
=
1
n
v
^
i
v
^
,
∂
φ
∂
x
i
+
S
^
,
D
φ
=
f
,
φ
+
∫
Γ
c
u
^
·
φ
d
Γ
c
.Substituting v
^ for φ in (53), we get(54)
S
^
,
D
v
^
=
f
,
v
^
+
∫
Γ
c
u
^
·
v
^
d
Γ
c
.Further, substituting v
k for φ in (47), we find (55)
lim
k
→
∞
⁡
ψ
I
2
v
k
D
v
k
,
D
v
k
=
f
,
v
^
+
∫
Γ
c
u
^
·
v
^
d
Γ
c
.Combining this with (54), we obtain(56)
lim
k
→
∞
⁡
ψ
I
2
v
k
D
v
k
,
D
v
k
=
S
^
,
D
v
^
.By [21, Chapter III, Lemma 1.3] and (50), (51), and (56), we get (49).
Applying the generalized Weierstrass theorem (see [23]), we conclude that there exists an element (
v
∗
,
S
∗
,
u
∗
)
∈
M such that (57)
J
v
∗
,
S
∗
,
u
∗
=
inf
v
,
S
,
u
∈
M
⁡
J
v
,
S
,
u
.This proves Theorem 6.



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*Source: 9428128-2016-06-27.xml*