Optimal control, flexural-torsional vibrations, necessary and sufficient condition, 49j20


EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 12, No. 1, 2019, 25-38
ISSN 1307-5543 – www.ejpam.com
Published by New York Business Global

On Determining Initial Conditions of Equations
Flexural-Torsinal Vibrations of a Bar

Aysel T. Ramazanova

University Duisburg-Essen, Germany

Abstract. The problem of finding the initial conditions in the boundary-value problem for the
system of flexural-torsional vibrations of a bar with additional conditions on the straight line is
reduced to an optimal control problem and studied by the methods of optimal control theory. The
gradient of the functional is calculated and using the gradient expression a necessary and sufficient
optimality condition are proved.

2010 Mathematics Subject Classifications: 49j20
Key Words and Phrases: Optimal control, flexural-torsional vibrations, necessary and sufficient
condition

1. Introduction

It is known that some problems of mathematical physics, mechanics, are described by
fourth order partial equations. A tuning fork, a bar vibrations equation, a rotary shaft,
oscillating motions equation and plate vibrations equation are among these equations give
some references. It is imperative optimal control problems in processes described by these
equations. The control connected with flexural-torsional vibrations of a bar has a great
signifficance in dynamics of aircraft constructions. Therefore, the study of bar vibrations
problems controls described by differential equations is necessary both from practical and
theoretical point of view.

2. Problem statement

We consider a boundary value problem for equations of flexural-torsional vibrations
of a bar, described by the system of two differential equations in the domain Q =
{0 < x < l, 0 < t < T} with boundary and initial conditions

∂2

∂x2

(
E (x) I (x)

∂2y

∂x2

)
+ ρ (x)A (x)

∂2y

∂t2
− ρ (x)A (x) e (x)

∂2θ

∂t2
= f1 (x, t) , (1)

DOI: https://doi.org/10.29020/nybg.ejpam.v12i1.3350

Email addresses: ramazanova-aysel@mail.ru

http://www.ejpam.com 25 c© 2019 EJPAM All rights reserved.



A.T. Ramazanova / Eur. J. Pure Appl. Math, 12 (1) (2019), 25-38 26

∂2

∂x2

(
E (x)Cw (x)

∂2θ

∂x2

)
−G (x)C (x)

∂2θ

∂x2
− ρ (x)A (x) e (x)

∂2y

∂t2
+

+ρ (x)
(
I (x) +A (x) e2 (x)

) ∂2θ
∂t2

= f2 (x, t) , (x, t) ∈ Q, (2)

yx=0 = y|x=l = 0,
∂y

∂x

∣∣∣∣
x=0

=
∂y

∂x

∣∣∣∣
x=l

= 0, 0≤t≤T, (3)

θ|x=0 = θ|x=l = 0,
∂θ

∂x

∣∣∣∣
x=0

=
∂θ

∂x

∣∣∣∣
x=l

= 0, 0≤t≤T, (4)

y|t=0 = 0,
∂y

∂t

∣∣∣∣
t=0

= v1 (x) , 0≤x≤l, (5)

θ|t=0 = 0,
∂θ

∂t

∣∣∣∣
t=0

= w1 (x) , 0≤x≤l, (6)

where l > 0, T > 0 are given numbers, y(x, t) is the lateral displacement of the bar, θ(x, t)
is the turning angle of the bar cross-section, E(x) is the Young modulus, I(x) is a polar
inertia moment of the cross section with respect to its gravity center, ρ(x) is a density
of the bar material, A(x) is the area cross section, e(x) is the distance from the gravity
center to the center of torsion, Cw(x) is the sectional moment of inertia of the cross-
section, G(x) a shear modulus, C(x) is geometrical rigidity of free torsion, E(x)Cw(x) is
the rigidity of flexural functions, G(x)C(x) is the rigidity of free torsion, the functions
(v1 (x) , w1 (x)) ∈ L2 (0, l)× L2 (0, l)- to be defined.

Note that for each fixed vector function (v1 (x) , w1 (x)) ∈ L2 (0, l) × L2 (0, l) problem
(1)-(6) has a unique generalized solution from the spaces W 2,1

2 (Q) [3,5,6].
To determine v (x) = (v1 (x) , w1 (x)), we give the additional conditions

y (x, T ; v) = ϕ1 (x) , 0 ≤ x ≤ l, (7)

θ (x, T ; v) = ϕ2 (x) , 0 ≤ x ≤ l, (8)

where ϕ1 (x) , ϕ2 (x) – are given functions.
We reduce this problem to the following optimal control problem: it is required to find

such a vector-function (v1 (x) , w1 (x)) ∈ L2 (0, l)×L2 (0, l), that minimizes the functional

J0 (v) =
1

2

∫ l

0

[
((y (x, T ; v)− ϕ1 (x))2 + (θ (x, T ; v)− ϕ2 (x))2

]
dx (9)

together with the solution of boundary value problem (1)-(6).
The function v (x) = (v1 (x) , w1 (x))- is called a control. We call problem (1)-(6),(9) a
reduced problem.The problem (1) - (6), (9) is regularized as follows. We introduce the
functional



A.T. Ramazanova / Eur. J. Pure Appl. Math, 12 (1) (2019), 25-38 27

Jα (v) = J0 (v) +
α

2

(
‖v1‖2L2(0,l)

+ ‖w1‖2L2(0,l)

)
, α = const > 0 (10)

Now for a class of admissible controls we take a convex, closed set Uad∈ L2 (0, l) ×
L2 (0, l) of vector-functions v (x) = (v1 (x) , w1 (x)).
Suppose that data of problem (1)-(6) satisfy the following conditions:

1) E (x) , I (x) , ρ (x) , A (x) , e (x) , Cω (x) , G (x) , C (x) , are mesaurable, bounded
and positive functions on the interval [0, l];

2) f1, f2 ∈ L2 (Q) , ϕ1, ϕ2 ∈ L2 (0, l)- are given functions.

3. Differentiability of functional (10)

We show that the functional (10) is differentiable in L2 (0, l)× L2 (0, l).
Introduce the following problem adjoint to the problem (1)-(6), (10):

∂2

∂x2

(
E (x) I (x)

∂2Ψ1

∂x2

)
+ ρ (x)A (x)

∂2Ψ1

∂t2
− ρ (x)A (x) e (x)

∂2Ψ2

∂t2
= 0,

(x, t) ∈ Q, (11)

∂2

∂x2

(
E (x)Cw (x)

∂2Ψ2

∂x2

)
−G (x)C (x)

∂2Ψ2

∂x2
− ρ (x)A (x) e (x)

∂2Ψ1

∂t2
+

+ρ (x)
(
I (x) +A (x) e2 (x)

) ∂2Ψ2

∂t2
= 0, (x, t) ∈ Q, (12)

Ψ1|x=0 = Ψ1|x=l = 0,
∂Ψ1

∂x

∣∣∣∣
x=0

=
∂Ψ1

∂x

∣∣∣∣
x=l

= 0, 0≤t≤T, (13)

Ψ2|x=0 = Ψ2|x=l = 0,
∂Ψ2

∂x

∣∣∣∣
x=0

=
∂Ψ2

∂x

∣∣∣∣
x=l

= 0, 0≤t≤T,

Ψ1|t=T = 0, Ψ2|t=T = 0, 0≤x≤l, (14)

∂Ψ1

∂t

∣∣∣∣
t=T

=

= −(I (x) +A (x) e2 (x))(ϕ1 (x)− y (x, T ; v) +A (x) e(x)(ϕ2 (x)− θ (x, T ; v))

ρ (x)A (x) I (x)
,

∂Ψ2

∂t

∣∣∣∣
t=T

= −(ϕ1 (x)− y (x, T ; v) e(x) + ϕ2 (x)− θ (x, T ; v))

ρ (x) I (x)
. (15)



A.T. Ramazanova / Eur. J. Pure Appl. Math, 12 (1) (2019), 25-38 28

We take the two admissible controls and assign them the increments
δv1 ∈ L2 (0, l) and δw1 ∈ L2 (0, l) in such a way that, (v1 (x) + δv1 (x) , w1 (x) + δw1 (x) ∈
Uad),

v (x) = (v1 (x) , w1 (x)) , v (x) + δv (x) =

= (v1 (x) + δv1 (x) , w1 (x) + δw1 (x)) ∈ L2 (0, l) .

Then the increment of the functional (10) is computed as

∆Jα (v) = Jα (v + δv)− Jα (v) =

=
1

2

l∫
0

[
(y (x, T ; v1 + δv1, w1 + δw1)− ϕ1 (x))2−

− (y (x, T ; v1, w1)− ϕ1 (x))2 + (θ (x, T ; v1 + δv1, w1 + δw1)− ϕ2 (x))2−

− (θ (x, T ; v1, w1)− ϕ2 (x))2
]
dx+

+
α

2

(
‖v1 + δv1‖2L2(0,l)

− ‖v1‖2L2(0,l)

)
+
α

2

(
‖w1 + δw1‖2L2(0,l)

)
, (16)

where
y (x, t; v (x) + δv (x)) = y (x, t; v) + δy(x, t)

θ (x, t; v (x) + δv (x)) = θ (x, t; v) + δθ(x, t).

Hence it follows that

∆Jα (v) =

∫ l

0
[(y (x, T ; v)− ϕ1 (x)) δy (x, T ) + (θ (x, T ; v)− ϕ2 (x)) δθ (x, T )] dx+

+α

∫ l

0
(v1δv1+w1δw1)dx+R, (17)

where

R =
1

2

∫ l

0

[
(δy (x, T ))2 + (δθ (x, T ))2

]
dx+

α

2

(∫ l

0
((δv1)

2+(δw1)
2)dx

)
and

(δy (x, t) , δθ (x, t)) ∈W 2,1
2 (Q)×W 2,1

2 (Q) is the generalized solution of the

∂2

∂x2

(
E (x) I (x)

∂2δy

∂x2

)
+ ρ (x)A (x)

∂2δy

∂t2
−

−ρ (x)A (x) e (x)
∂2δθ

∂t2
= 0, (x, t) ∈ Q, (18)



A.T. Ramazanova / Eur. J. Pure Appl. Math, 12 (1) (2019), 25-38 29

∂2

∂x2

(
E (x)Cw (x)

∂2δθ

∂x2

)
−G (x)C (x)

∂2δθ

∂x2
−

−ρ (x)A (x) e (x)
∂2δy

∂t2
+ ρ (x)

(
I (x) +A (x) e2 (x)

) ∂2δθ
∂t2

= 0, (x, t) ∈ Q , (19)

δy|x=0 = δy|x=l = 0,
∂δy

∂x

∣∣∣∣
x=0

=
∂δy

∂x

∣∣∣∣
x=l

= 0, 0 ≤ t ≤ T, (20)

δθ|x=0 = δθ|x=l = 0,
∂δθ

∂x

∣∣∣∣
x=0

=
∂δθ

∂x

∣∣∣∣
x=l

= 0, 0 ≤ t ≤ T, (21)

δy|t=0 = 0,
∂δy

∂t

∣∣∣∣
t=0

= δv1 (x) ,

δθ|t=0 = 0,
∂δθ

∂t
|t=0 = δw1 (x) , 0 ≤ x ≤ l (22)

i.e. for any function

∀ η1 = η1 (x, t) , η2 = η2 (x, t) ∈W 2,1
2 (Q) ,

η1|x=0 = η1|x=l = 0,
∂η1
∂x

∣∣∣∣
x=0

=
∂η1
∂x

∣∣∣∣
x=l

= 0, 0≤t≤T,

η2|x=0 = η2|x=l = 0,
∂η2
∂x

∣∣∣∣
x=0

=
∂η2
∂x

∣∣∣∣
x=l

= 0, 0≤t≤T,

the following integral identities are fulfilled∫∫
Q

(
E (x) I (x)

∂2δy

∂x2
∂2η1
∂x2

− ρ (x)A (x)
∂δy

∂t

∂η1
∂t

+

ρ (x)A (x) e (x)
∂δθ

∂t

∂η1
∂t

)
dxdt+

+

∫ l

0
ρ (x)A (x)

∂δy

∂t
η1|T0 dx−

∫ l

0
ρ (x)A (x) e (x)

∂δθ

∂t
η1|T0 dx= 0, (23)

∫∫
Q

(
E (x)Cw (x)

∂2δθ

∂x2
∂2η2
∂x2

− G (x)C (x)
∂2η2
∂x2

δθ+

+ ρ (x)A (x) e (x)
∂δy

∂t

∂η2
∂t
− ρ (x)

(
I (x) +A (x) e2 (x)

) ∂δθ
∂t

∂η2
∂t

)
dxdt−

∫ l

0
ρ (x)A (x) e (x)

∂δy

∂t
η2|T0 dx+



A.T. Ramazanova / Eur. J. Pure Appl. Math, 12 (1) (2019), 25-38 30

+

∫ l

0
ρ (x)

(
I (x) +A (x) e2 (x)

) ∂δθ
∂t

η2|T0 dx= 0. (24)

As the functions (Ψ1 (x, t) ,Ψ2 (x, t))are the generalized solutions of problem (11)- (15),
for any functions g1, g2 ∈W

2,1
2 (Q) ,

g1|x=0 = g1|x=l = 0,
∂g1
∂x

∣∣∣∣
x=0

=
∂g1
∂x

∣∣∣∣
x=l

= 0, 0≤t≤T,

g2|x=0 = g2|x=l = 0,
∂g2
∂x

∣∣∣∣
x=0

=
∂g2
∂x

∣∣∣∣
x=l

= 0, 0≤t≤T,

the following integral identities are fulfilled∫∫
Q

(
E (x) l (x)

∂2Ψ1

∂x2
∂2g1
∂x2

− ρ (x)A (x)
∂Ψ1

∂t

∂g1
∂t

+

+ ρ (x)A (x) e (x)
∂Ψ2

∂t

∂g1
∂t

)
dxdt+

∫ l

0
ρ (x)A (x)

∂Ψ1

∂t
g1|T0 dx−

−
∫ l

0
ρ (x)A (x) e (x)

∂Ψ2

∂t
g1|T0 dx= 0, (x, t) ∈ Q, (25)

∫∫
Q

(
E (x)Cw (x)

∂2Ψ2

∂x2
∂2g2
∂x2

−G (x)C (x)
∂2Ψ2

∂x2
g2+

+ρ (x)A (x) e (x)
∂Ψ1

∂t

∂g2
∂t
− ρ (x)

(
I (x) +A (x) e2 (x)

) ∂Ψ2

∂t

∂g2
∂t

)
dxdt−

−
∫ l

0
ρ (x)A (x) e (x)

∂Ψ1

∂t
g2|T0 dx+

+

∫ l

0
ρ (x)

(
I (x) +A (x) e2 (x)

) ∂Ψ2

∂t
g2|T0 dx= 0, (x, t) ∈ Q. (26)

In equalities (23) and (24) instead of η1 (x, t) and η2 (x, t) we take Ψ1 (x, t) and Ψ2 (x, t),
in the identities (25) and (26) instead of g1 (x, t) and g2 (x, t) we take δy (x, t) and δθ (x, t)
respectively, subtract the obtained relations and sum them. Then we have,∫ l

0
[(y (x, T ; v)− ϕ1 (x)) δy (x, T ) dx+ (θ (x, T ; v)− ϕ2 (x)) δθ (x, T ) dx =

=

∫ l

0
[ρ (x)A (x) Ψ1 (x, 0)−ρ (x)A (x) e (x) Ψ2 (x, 0)] δv1dx+ (27)

+

∫ l

0
[−ρ (x)A (x) e (x)Ψ1 (x, 0) +ρ (x) (I (x) +A(x) e2 (x)

)
Ψ2 (x, 0)

]
δw1dx



A.T. Ramazanova / Eur. J. Pure Appl. Math, 12 (1) (2019), 25-38 31

Therefore from formulas (17) and (27) it follows that

∆Jα (v) =

∫ l

0
[ρ (x)A (x) Ψ1 (x, 0)− ρ (x)A (x) e (x) Ψ2 (x, 0) + αv1] δv1dx+

+

∫ l

0

[
−ρ (x)A (x) e (x) Ψ1 (x, 0) + ρ (x)

(
I (x) +A (x) e2 (x)

)
·

· Ψ2 (x, 0) + αw1] δw1dx+R (28)

Next we show that
‖δy (x, T )‖2L2(0,l)

≤ c‖δv1‖2L2(0,l)
, (29)

‖δθ (x, T )‖2L2(0,l)
≤ c‖δw1‖2L2(0,l)

. (30)

For this purpose first we show that

‖δy (x, t)‖2
W 2,1

2 (Q)
≤ c‖δv1‖2L2(0,l)

, (31)

‖δθ (x, t)‖2
W 2,1

2 (Q)
≤ c‖δw1‖2L2(0,l)

(32)

For proving estimations (31) and (32) we apply the Faedo-Galerkin method. Let {ωi (x)}∞i=1

be a fundamental system in
0

W 2
2 (0, l) and∫ l

0
ωi (x)ωk (x) dx =

{
1, i = k,
0, i 6= k.

We look for approximate solutions
(
δyN (x, t) , δθN (x, t)

)
of problem (18), (19) in the form

δyN (x, t) =
∑N

i=1 c
N
1i (t)ωi (x) and δθN (x, t) =

∑N
i=1 c

N
2i (t)ωi (x) from the following rela-

tions ∫ l

0
(E (x) I (x)

∂2δyN

∂x2
d2ωp(x)

∂x2
+ ρ (x)A (x)

∂2δyN

∂t2
ωp(x)−

− ρ (x)A (x) e (x)
∂2δθN

∂t2
ωp(x)

)
dx = 0, p = 1, N, (33)

∫ l

0
(E (x)Cw (x)

∂2δθN

∂x2
d2ωp(x)

∂x2
− G (x)C (x) δθN

d2ωp(x)

∂x2
−

−ρ (x)A (x) e (x)
∂2δyN

∂t2
ωp(x)+

ρ (x)
(
I (x) +A (x) e2 (x)

) ∂2δθN
∂t2

ωp(x)

)
dx = 0 p = 1, N, (34)



A.T. Ramazanova / Eur. J. Pure Appl. Math, 12 (1) (2019), 25-38 32

cN1i
∣∣
t=0

= 0,
dcN1i (t)

dt

∣∣∣∣
t=0

= (v1, ωi) , (35)

cN2i
∣∣
t=0

= 0,
dcN2i (t)

dt

∣∣∣∣
t=0

= (w1, ωi) , i = 1, N. (36)

Equalities (33) and (34) are the system of linear ordinary differential equations of
second order with the unknowns cN1i (t) and cN2i (t), i = 1, N solved with respect to d2cN1i

/
dt2

and d2cN2i
/
dt2. Under the conditions on the problem data, this system is uniquely solvable

under initial conditions (35) and (36), moreover d2cN1i
/
dt2, d2cN2i

/
dt2 ∈ L2 (0, T ), i = 1, N.

Multiplying each of the equalities (33) and (34) by its own dcN1p
/
dt, dcN2p

/
dt and sum-

ming over p from 1 to N we come to the equalities∫ l

0
(E (x) I (x)

∂2δyN

∂x2
∂3δyN

∂x2∂t
+ ρ (x)A (x)

∂2δyN

∂t2
∂δyN

∂t
−

−ρ (x)A (x) e (x)
∂2δθN

∂t2
∂δyN

∂t

)
dx = 0, (37)

∫ l

0
(E (x)Cw (x)

∂2δθN

∂x2
∂3δθN

∂x2∂t
− G (x)C (x) δθN

∂3δθN

∂x2∂t
−

ρ (x)A (x) e (x)
∂2δyN

∂t2
∂δθN

∂t
+

+ ρ (x)
(
I (x) +A (x) e2 (x)

) ∂2δθN
∂t2

∂δθN

∂t

)
dx = 0 (38)

Suppose that G (x)C (x) are independent of x.
Then it follows that

1

2

d

dt

∫ l

0

[
(E (x) I (x)

(
∂2δyN

∂x2

)2

+ ρ (x)A (x)

(
∂δyN

∂t

)2

+ E (x)Cw (x)

(
∂2δθN

∂x2

)2

+

+GC

(
∂δθN

∂x

)2

+ ρ (x)
(
I (x) +A (x) e2 (x)

)(∂δθN
∂t

)2

−

− 2ρ (x)A (x) e (x)

(
∂δyN

∂t

∂δθN

∂t

)]
dx = 0. (39)

We integrate the last equality with respect to t from 0 to t :

∫ l

0

[
(E (x) I (x)

(
∂2δyN

∂x2

)2

+ ρ (x)A (x)

(
∂δyN

∂t

)2

+ E (x)Cw (x)

(
∂2δθN

∂x2

)2

+



A.T. Ramazanova / Eur. J. Pure Appl. Math, 12 (1) (2019), 25-38 33

+GC

(
∂δθN

∂x

)2

+ ρ (x)
(
I (x) +A (x) e2 (x)

)(∂δθN
∂t

)2

−

−2ρ (x)A (x) e (x)

(
∂δyN

∂t

∂δθN

∂t

)]
dx =

=

∫ l

0

[
ρ (x)A (x)

(
∂δyN (x, 0)

∂t

)2

+ ρ (x)
(
I (x) +A (x) e2 (x)

)(∂δθN (x, 0)

∂t

)2

−

− 2ρ (x)A (x) e (x)

(
∂δyN (x, 0)

∂t

∂δθN (x, 0)

∂t

)]
dx. (40)

In equality (40) we make some transformations:

∫ l

0

[
(E (x) I (x)

(
∂2δyN

∂x2

)2

+ ρ (x)A (x)

(
∂δyN

∂t

)2

+ E (x)Cw (x)

(
∂2δθN

∂x2

)2

+

+GC

(
∂δθN

∂x

)2

+ ρ (x)
(
I (x) +A (x) e2 (x)

)(∂δθN
∂t

)2

−

− ρ (x)A (x) e (x)

((
∂δyN

∂t

)2

+

(
∂δθN

∂t

)2
)]

dx ≤

≤
∫ l

0

[
ρ (x)A (x)

(
∂δyN (x, 0)

∂t

)2

+ ρ (x)
(
I (x) +A (x) e2 (x)

)(∂δθN (x, 0)

∂t

)2

+

+ ρ (x)A (x) e (x)

((
∂δyN (x, 0)

∂t

)2

+

(
∂δθN (x, 0)

∂t

)2
)]

dx. (41)

It’s clear that,

∫ l

0

∣∣∣∣∂δyN (x, 0)

∂t

∣∣∣∣2 dx ≤ ∫ l

0

∣∣∣∣∣
N∑
i=1

(δv1, ωi)ωi (x)

∣∣∣∣∣
2

dx ≤

≤ c
N∑
i=1

|δv1i|2 ≤ c
∞∑
i=1

|δv1i|2 ≤ c ‖δv1‖2L2(0,l)
, (42)

∫ l

0

∣∣∣∣∂δθN (x, 0)

∂t

∣∣∣∣2 dx ≤ ∫ l

0

∣∣∣∣∣
N∑
i=1

(δw1, ωi)ωi (x)

∣∣∣∣∣
2

dx ≤



A.T. Ramazanova / Eur. J. Pure Appl. Math, 12 (1) (2019), 25-38 34

≤ c
N∑
i=1

|δw1i|2 ≤ c
∞∑
i=1

|δw1i|2 ≤ c ‖δw1‖2L2(0,l)
, (43)

where δv1i =
∫ l
0 δv1 (x)ωi (x) dx, δw1i =

∫ l
0 δw1 (x)ωi (x) dx are Fourier coefficients of the

functions δv1 (x) , δw1 (x). From (41), (53), (43) we get:

∫ l

0

[
(E (x) I (x)

(
∂2δyN

∂x2

)2

+ ρ (x)A (x)

(
∂δyN

∂t

)2

+ E (x)Cw (x)

(
∂2δθN

∂x2

)2

+

+GC

(
∂δθN

∂x

)2

+ ρ (x)
(
I (x) +A (x) e2 (x)

)(∂δθN
∂t

)2

−

− ρ (x)A (x) e (x)

((
∂δyN

∂t

)2

+

(
∂δθN

∂t

)2
)]

dx ≤

≤
∫ l

0

[
ρ (x)A (x)

(
∂δyN (x, 0)

∂t

)2

+ ρ (x) (I (x) +

+A (x) e2 (x)

(
∂δθN (x, 0)

∂t

)2

+

+ ρ (x)A (x) e (x)

((
∂δyN (x, 0)

∂t

)2

+

(
∂δθN (x, 0)

∂t

)2
)]

dx ≤

≤ c
(
‖δv1‖2L2(0,l)

+ ‖δw1‖2L2(0,l)

)
(44)

Assume that 1 − e (x) ≥ α0 > 0, I (x) + A (x) e (x) (e (x)− 1) ≥ α1 > 0,
∀x ∈ [0, l], where α0, α1 > 0- are the given numbers.
Since E (x) , I (x) , A (x) , Cω (x) , ρ (x) are positive functions on the segment [0, l], by

equivalence of the norms in the space
0

W 2
2 (0, l), from the last inequality

by means of elementary transformations we get:

∫ l

0

(δyN (x, t)
)2

+

(
∂δyN (x, t)

∂t

)2

+

(
∂δyN (x, t)

∂x

)2

+

(
∂2δyN (x, t)

∂x2

)2

+

+
(
δθN (x, t)

)2
+

(
∂δθN (x, t)

∂t

)2

+

(
∂δθN (x, t)

∂x

)2

+

(
∂2δθN (x, t)

∂x2

)2]
dx ≤



A.T. Ramazanova / Eur. J. Pure Appl. Math, 12 (1) (2019), 25-38 35

≤ c
∫ t

0

∫ l

0

(δyN (x, s)
)2

+

(
∂δyN (x, s)

∂t

)2

+

(
∂δyN (x, s)

∂x

)2

+

(
∂2δyN (x, s)

∂x2

)2

+

+
(
δθN (x, s)

)2
+

(
∂δθN (x, s)

∂t

)2

+

(
∂δθN (x, s)

∂x

)2

+

(
∂2δθN (x, s)

∂x2

)2]
dxds+

+c
(
‖δv1‖2L2(0,l)

+ ‖δw1‖2L2(0,l)

)
. (45)

Applying the Gronuoll lemma, we have:

∫ l

0

(δyN (x, t)
)2

+

(
∂δyN (x, t)

∂t

)2

+

(
∂δyN (x, t)

∂x

)2

+

(
∂2δyN (x, t)

∂x2

)2

+

+
(
δθN (x, t)

)2
+

(
∂δθN (x, t)

∂t

)2

+

(
∂δθN (x, t)

∂x

)2

+

(
∂2δθN (x, t)

∂x2

)2]
dx ≤

≤ c
(
‖δv1‖2L2(0,l)

+ ‖δw1‖2L2(0,l)

)
∀t ∈ [0, T ] . (46)

From the last inequality it follows that∫ T

0

∫ l

0

[(
δyN (x, t)

)2
+

(
∂δyN (x, t)

∂t

)2

+

(
∂δyN (x, t)

∂x

)2

+

(
∂2δyN (x, t)

∂x2

)2

+

+
(
δθN (x, t)

)2
+

(
∂δθN (x, t)

∂t

)2

+

(
∂δθN (x, t)

∂x

)2

+

(
∂2δθ (x, t)

∂x2

)2
]
dx ≤

≤ c
(
‖δv1‖2L2(0,l)

+ ‖δw1‖2L2(0,l)

)
. (47)

From the sequence
(
δyN , δθN

)
we can choose a subsequence weakly convergent inW 2,1

2 (Q)×
W 2,1

2 (Q) to some element (δy, δθ) ∈W 2,1
2 (Q)×W 2,1

2 (Q). By virtue of weak lower semicon-
tinuity of the norm in the Hilbert space, we get from (47) that for δy (x, t) and δθ (x, t) the
following estimation is valid

‖δy‖2
W 2,1

2 (Q)
+ ‖δθ‖2

W 2,1
2 (Q)

≤ c
(
‖δv1‖2L2(0,l)

+ ‖δw1‖2L2(0,l)

)
.

Hence the estimate (31) and (32).
Since W 2,1

2 (Q) is boundedly imbedded in L2 (0, T ) [6, pp. 73-74], hence it follows that



A.T. Ramazanova / Eur. J. Pure Appl. Math, 12 (1) (2019), 25-38 36

‖δy (x, T )‖2L2(0,l)
≤ c1 ‖δy‖2W 2,1

2 (Q)
≤ c

(
‖δv1‖2L2(0,l)

+ ‖δw1‖2L2(0,l)

)
, (48)

‖δθ (x, T )‖2L2(0,l)
≤ c2 ‖δθ‖2W 2,1

2 (Q)
≤ c

(
‖δv1‖2L2(0,l)

+ ‖δw1‖2L2(0,l)

)
.

It is easy to show that (δy, δθ) is the generalized solution of problem
(18)-(19) [6, pp. 210-215].

Thus, from (48) we find that

R =
1

2

∫ l

0

[
(δy (x, T ))2 + (δθ (x, T ))2

]
dx+

α

2

(∫ l

0
((δv1)

2+(δw1)
2)dx

)
≤

≤ c
(
‖δv1‖2L2(0,l)

+ ‖δw1‖2L2(0,l)

)
. (49)

Thus, from (28) and (49) it follows that the differential of the functional J (v) is equal
to 〈

J
′
(v) , δv

〉
=

∫ l

0
[ρ (x)A (x) Ψ1 (x, 0)−ρ (x)A (x) e (x) Ψ2 (x, 0) +αv1] δv1dx+

+
[
−ρ (x)A (x) e (x) Ψ1 (x, 0) +ρ (x) (I (x) +A(x) e2 (x)

)
Ψ2 (x, 0) +αw1

]
δw1dx.

3. Necessary and sufficient condition of optimality

Theorem 1. For the control v (x) = (v01 (x), w0
1 (x)) to be an optimal control in

problem (1)-(6), (10) it is necessary and sufficient that

∫ l

0
[ρ (x)A (x) Ψ1 (x, 0)−ρ (x)A (x) e(x)Ψ2 (x, 0) +αv1]

(
v1 (x)− v01(x)

)
dx+

+

[∫ l

0
−ρ (x)A (x) e (x)Ψ1 (x, 0) +ρ (x) (I (x) +A(x) e2 (x)

)
Ψ2 (x, 0) +αw1

]
×

×
(
w1 (x)− w0

1(x)
)

dx≥0 ,∀v = (v1, w1) ∈ Uad. (50)

Proof. Let v0 (x) =
(
v01 (x) , w0

1 (x)
)
- to be an optimal control in problem (1)-(6),

(10). As Uad- is a convex set in L2 (0, l)× L2 (0, l), by virtue of the known theorem from
[7, pp. 28], 〈

J
′
(v) , v − v0

〉
≥ 0,∀v ∈ Uad.



A.T. Ramazanova / Eur. J. Pure Appl. Math, 12 (1) (2019), 25-38 37

From the last inequality we get necessity. As problem (1)-(6), (10) is a linear-quadratic,
the obtained condition is a sufficient condition as well for the optimality of the control
v0 (x).

Conclusion: In this paper, the inverse problem of determining the right-hand sides
of the flexural-torsional vibrations of a rod is considered. This problem is reduced to the
problem of optimal control. The gradient of the functional is calculated and, using the
gradient expression, a necessary and sufficient optimality condition is proved.
Example 1. We consider a boundary value problem for equations of flexural-torsional
vibrations of a bar, described by the system of two differential equations in the domain
Q = {0 < x < 1, 0 < t < 1}

∂4y

∂x4
+ 4

∂2y

∂t2
− 2

∂2θ

∂t2
= f1 (x, t) , (51)

∂4θ

∂x4
− ∂2θ

∂x2
− 2

∂2y

∂t2
+ 3

∂2θ

∂t2
= f2 (x, t) , (x, t)∈Q (52)

yx=0 = y|x=1 = 0,
∂y

∂x

∣∣∣∣
x=0

=
∂y

∂x

∣∣∣∣
x=1

= 0, 0≤t≤T, (53)

θ|x=0 = θ|x=1 = 0,
∂θ

∂x

∣∣∣∣
x=0

=
∂θ

∂x

∣∣∣∣
x=1

= 0, 0≤t≤T, (54)

y|t=0 = 0,
∂y

∂t

∣∣∣∣
t=0

= v1 (x) , (55)

θ|t=0 = 0,
∂θ

∂t

∣∣∣∣
t=0

= v2 (x) , (56)

f1 (x, t) = 24t, f2 (x, t) = 44t+ 24tx− 24tx2.

In the special case, the coefficients of equations (51)-(52) were taken in the from:E =
1
2 , I = 2, ρ = 1, e = 1

2 , A = 4, Cw = 2, G = 1, C = 1.
In order to determine v (x) = (v1 (x) , v2 (x)), we give the additional conditions:

y

(
x,

1

2
, v

)
=
x2(1− x)2

2

θ

(
x,

1

2
, v

)
= x2(1− x)2.

In this special case the functional (9) has the form

J0 (v) =
1

2

∫ 1

0

(y(x, 1

2
, v

)
− x2(1− x)2

2

)2

+

(
θ

(
x,

1

2
, v

)
− x2(1− x)2

)2
]
dx.



REFERENCES 38

It is easy to verify that

y (x, t) = tx2(1− x)2, θ (x, t) = 2tx2(1− x)2

and
inf J (v)

v∈L2(0,1)×L2(0,1)

= min J (v)
v∈L2(0,1)×L2(0,1)

= 0,

and the minimum of the functionalJ (v) is attained for v = v0 (x) =
(
v01 (x) , v02 (x)

)
=
(
x2(1− x)2, 2x2(1− x)2

)
.

In this case necessary and sufficient condition (28) is fulfilled by itself, when α = 0.

References

[1] J.L. Arman, Application of theory of optimal control of distributed parameters systems
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[2] A.G. Butkovsky, A.I. Egorov, K.A. Lurie, Optimal Control of Distributed Sys-
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[3] A.Z. Ishmukhametov, Stability and approximation of optimal control of distributed
parameters systems. Computing center of RAS, 2001.

[4] V. Komkov, The Optimal control theory of a transverse vibrations of a beam. SIAMJ.
Control 6 (1968), no. 3, 401-421.

[5] F.E. Lomovtsev, N.A. Yurchuk, Cauchy problem for second order hyperbolic
differential-operator equations, Diff. uravn. 12 (1976), no.12, 2242-2250.

[6] O.A. Ladyzhenskaya, Boundary value problems of mathematical physics.
Moscow:Nauka, 1973.

[7] F.P. Vasil’ev, Methods for solving extrum problems. Moscow: Nauka 42, (2016), No
2, 174-187

[8] A.T.Ramazanova, G.F.Kuliev, On Finding the Right-Hand Sides of Equations of
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