Boundary control for pseudoparabolic equations in space Текст научной статьи по специальности «Математика»

Математические заметки СВФУ Июль—сентябрь, 2019. Том 26, № 3

UDC 517.9

BOUNDARY CONTROL FOR PSEUDOPARABOLIC EQUATIONS IN SPACE Z. K. Fayazova

Abstract. Let u(x, y, t) be a solution to the pseudoparabolic equation that satisfies the initial and boundary conditions. The value of the solution is given on the part of boundary of the considered region which contains the control parameter. It is required to choose the control parameter so that on a part of the regularity domain the solution takes the specified mean value. First, we consider an auxiliary boundary value problem for a pseudoparabolic equation. We prove the existence and uniqueness of the generalized solution from the corresponding class. The restriction for the admissible control is given in the integral form. By the separation variables method, the desired problem is reduced to the Volterra integral equation. The latter is solved by the Laplace transform method. The theorem on the existence of an admissible control is proved.

DOI: 10.25587/SVFU.2019.20.57.008 Keywords: pseudoparabolic equation, boundary control, admissible control, integral equation, Laplace transform

0. Introduction

We consider the boundary control problem with integral constraint associated with pseudo-parabolic equation in the domain Q = {(x, y) g R2 : |x| < n, |y| < n}. Let function u(t, x, y) be solution of the equation

ut = Aut + Au, (x,y,t) g qt = {0 <t<T, |x| <n, |y| < n}, (1)

and satisfies initial conditions

u|t=o = 0, -n < x, y < n (2)

and boundary conditions

u|x=n = 0, u|x=-n = M(t)^(y), 0 < t < T, |y| < n;

u|y=-n = 0, u|y=n = 0, 0 < t < T, |x| < n; (3)

0(-n) = 0(n) = 0,

where

dAu

Au = wxx + Aut =

One of the models theory of incompressible simple liquids with decaying memory can be described by equation (1) (see [1]). The questions of stability, uniqueness, and existence of solutions of some classical problems for equation which are under

© 2019 Z. K. Fayazova

consideration were studied in [2] (see also [3,4]). The point control problems for a parabolic and pseudo-parabolic equation were considered in [5]. The impulse control problems for some systems with distributed parameters were investigated in [6-8]. Controlling systems problem with distributed parameters, whose evolution is modeled by using partial differential equations, is set forth in the monographs [9,10].

In recent years, the interest in the study of distributed parameter systems has increased significantly, and therefore it should be noted the works V.A. Ilin and E.I. Moiseev, in which the questions of boundary control of various systems described by the wave equation are studied in details (see [11,12]). Works described in [1315] can be noted as works, whose results are related to problem of controlling the processes described by equations of a parabolic type and in particular, the heat exchange process.

Definition 1. Control /j(t) is called admissible if function ^(t) satisfies condition

Kt)| < 1.

For an arbitrary Banach space B and T > 0 by the symbol C([0,T] ^ B) we denote the Banach space of all continuous maps u : [0, T] ^ B with norm

iMi = max ||u(t)||.

0 <t<T

Further, by the symbol W2(D) we denote the subspace Sobolev space W2(D) formed by functions, which trace on dD is equal to zero. Note that due to the closure W2!(D) the sum of a series of functions from W2(D), converging in the metric W2(D) also belongs to W2(D) .

Definition 2. Under the solution of the problem (1)-(3) we understand a function u(x, y, t) presented in the form

n — x

u(x, y, t) = n(t)4>(y)—--v(x, y, t),

where function v(x,y,t) is generalized solution from C([0,T] ^ W2X (D)) the following problem:

n — x

Vt(x, y, t) - Av(x, y, t) = (p'imy) - (p'(t) + with uniform initial and boundary conditions:

v|t=o = v| on = 0.

Note that the class C([0,T] ^ W1(D)) is a subset of the class W21,0(DT), considered in the monograph [16] in order to determine a solution with homogeneous boundary conditions (for the corresponding uniqueness theorem, see Chap. III, Theorem 3.2, pp. 173-176).Thus, the generalized solution introduced above is a generalized solution in the sense of a monograph [16]. However, unlike a class solution W2'°(Dt), which is guaranteed to have a trace only for almost everyone t G [0,T], a solution from a class C([0, T] ^ W2(D)) continuously depends of t g [0, T] on the metric L2(D). Thus, the condition (3) is correctly defined.

Problem. For a given mean value of the solution 0(t), find the admissible control ^(t) that ensures this equality:

o o

u(x, y,t) dxdy = 0(t), t> 0. (4)

Theorem 1. Let the non-negative function ^(y) be twice continuously differen-tiable, has the third piecewise continuous derivative and decreases on the [—n, n].

Then there is a constant M > 0 such that for any function 0(t) g w21(—to, +œ), 0(t) = 0 with t < 0 that satisfies condition

l|0(t)llw2i(R+) < M an admissible control ^(t) that ensures condition (4) exists.

1. Main result

Lemma 1. Let the functions ^(t) and ^(y) satisfy the following conditions:

1) M g W21(R), m(0) = 0,

2) 0(y) g Wf[—n,n] and satisfies condition

n) =

Then solution of the problem (1)-(3) has the form

1 ^ . n(x + n) . m(y + n) u(x,y,t) = — > -sin-sin-li(t)

y ' 2tt ^ 1 + Xnm 2 2 ^ '

n,m=1

1 ^^ пфт ( q (t T ) / ч , + n) m(y + n)

+ ^ E e о sin о ' (5)

2п ^ (1+ A„m)2

where

n,m=1 Q

Anm , ч2 /m\2 i о

înm= 7—T-, Anm = (-) + 71, TO = 1,2,....

1 + Anm 2 2

vnm n

/ 1 [ ,, N • т(У + П)

Фт = - Ф{У) Sin

2

dy, m =1, 2,....

Proof. Solution of the problem one can write in the form

П _ x 2 f (t-T)

u{x,y,t)= ц{г)ф{у)—---> / -г

2п П n(1 + A„m)

n,m= 1 о

r i, s, , ,, , ^ , n(x + n) m(y + n)

x {м (т)(1 + (to/2)2) +M(r)(TO/2)2}iirsm- v ; —

2n 2n

We will prove that function y, t) represented by the indicated Fourier series,

21!

belongs to the class C([0, T] ^ W21(D)). It suffices to prove that the gradient of this

function taken with respect to (x, y) £ D, continuously depends from t £ [0, T] on the norm of the space L2(D). According to Parseval's equality, norm of this gradient is equal to

n,m=l

12

where

Notice, that

therefore

_ ^-nm ,

?nm 1 _i_ \ — '

1 + "nm

|pnm(t)| < Cim .

Consequently

4 ^ ^ i2

<n <m — 1 " '

<C!~ £ = X) \^a\2m2 = CW(y)\\l2[_vM.

"mi 2 ^ V"^ I j. |2 2 /illJ.'/ Ml2

n ^^ n2 3 ^^

n,m=l n,m=l

Note that the class C([0,T] ^ W1(D)) is a subset of the class W2'°(DT), considered in the monograph [16] in order to determine a solution with homogeneous boundary conditions (for the corresponding uniqueness theorem, see Chap. III, Theorem 3.2, pp. 173-176).Thus, the generalized solution introduced above is a generalized solution in the sense of a monograph [16]. However, unlike a class solution W2'°(Dt), which is guaranteed to have a trace only for almost everyone t £ [0,T], a solution from a class C([0,T] ^ W21(D)) continuously depends of t £ [0,T] on the metric L2(D). Thus, the condition (3) is correctly defined.

Lemma 1 proved.

Lemma 2. Let

n

1 f ^^(y + n) <t}™ = - <t>{y) sin---dy, m = 1, 2, 3, ....

—n

Then, under the conditions of Theorem 1 are valid inequalities

> 0, m = 1, 2, 3,.... (6)

Proof. We will introduce function

^(x) = ^(2x — n), 0 < x < n.

Then

n kn/m

2 i m 2 f

<f>m = — / ^¿(x) sin ma; da: = — / i/i(a;) sinriia; da;

n J '—' n

0 k=1 (k-1)n/m

kn

2

' 2 /*

V^- / ij){t/m) sini dt.

nm 7

k=1 (k-1)n

The terms in the last sum form an alternating sequence, decreasing in absolute value.Therefore, the sign of this sum coincides with the sign of the first term, i.e. condition (6) is satisfied. Lemma 1 proved. Using the integral constraint (4) we have

0 0

j j u(x,y,t) dxdy

Lu(t) f f V smn{x + lr) smm{y + 7r) dxdv

/ / n+*__'\Sln 9 Sm 9 Üxay

2n J J ^ (1 + A„m) 2

n,m=1

n,m=1 o

0 0

x / |sin^^sin^±^da;dy = 0(i) (7)

Qi+\ (4-\ 8 ^ ^m -2 -2

6(t =Mi- V n , x-rsin —sin —

n m(1 + Anm) 4 4

n,m=1

8 ^ ^m . 2 nn . 2 mn t _n (t )

+ - > —:—1-—r sill —sm - / e gnm[t t>u(t)(1t.

7T m(l + Xnm) 4 A J ^ '

*rt •Wl--v '

n,m=1 0

We introduce notations

8 ^ ,2 nn .2 mn

= — > —n-;-r sm —sin ——,

n m(1 + Anm) 4 4

n,m=1

8 -A ^me-qnmt . 2 nn . 2 mn

K2 = ~ V -,, , .-rr Sill —Sin ——.

n ,m=i m(1 + Anm)2 4 4

n,m=1

Then (7) we can write in the following form

t

K^(i) + J K(t - r)M(r) dr = 0(i). (8)

By definition of the Laplace transform we have

?(p) = J e—pt M(t) dt. 0

Applying the Laplace transform to the second kind Volterra integral equation (8) and taking into account properties of the transform convolution we get

CO t CO

0(p) = J e—pt dt J K(t — s)^(s) ds + Ki J e—PV(t) dt = Ki/Z(p) + K(p)/Z(p). 0 0 0 Consequently,

№) = -%-• 0) Ki + K (p)

From here we get with some a > 0

a+ic ^ + o ^

M(i) = JL f dp = _L f „ + e(^)t ^ (10)

K(p) + Ki 2n J K(a + + Ki

a—io — o

Now we estimate the series for Ki and K(a + ¿£) by the conditions 0m > 0, m = 1,2,...,

8 ^ 0m . 2 mn . 2 nn ill = — > —----- sm -sin

n ^—^ 1 A

n ^ m(1 + A„m) 4 4 '

n,m=i

We transform this series

oo ,

E^m . 2 mn . 2 nn

im(l + (f)2 + (|)2)Sm TSm T

n,m=i 2 2

=,Tl2 ^T_ \ "" ym_0111 ~4

1 \ - fflm sm —__\ - 0m_Sin_

2 ^ TO 1 + (21^1)2 + (m)2 + Z^ TO 1 + (2n- 1)2 + n,m=i v 2 7 v 2 7 n,m=i 2

_ V^ _02m-1_

~ 4 (2m - 1)(1 + (2n - 1)V4 + (2to - 1)2/4)

1 'C 04m —2

+ 7 E

2 n-m=i 2(2m — 1)(1 + (2n — 1)2/4 + (2m — 1)2)

02m-1

2 n^i (2m — 1)(1 + (2n — 1)2 + (2m — 1)2/4)

o /

04m —2

2(2m — 1)(1 + (2n — 1)2 + (2m — 1)2)'

n,m=i

Let us prove the convergence of these series. Consider the series

04m-2

E

(2m — 1)(1 + (2n — 1)2 + (2m — 1)2)'

n,m=i

Using obvious inequality of

1 1

<

1 + (2n - 1)2 + (2m - 1)2 " 1 + (2n - 1)2

we get that

o , o 1 oo ,

04m—2 ^ V^ 04m—2

(2m - 1)(1 + (2n - l)2 + (2m - l)2) " ¿^ (1 + (2n - l)2) ¿^ (2m - 1) '

° é

The last series (2m-i) converges, since 0m is the Fourier coefficients of the twice

m=l

continuously differentiable function 0(y). Similarly, we can prove the convergence of the series for K(p).

Evaluate the expression for the Kl from below

8 0m . 2 nn . 2 mi

K i = — > —--- sin — sin -

n ¿-= m(1 + Anm) 4 4

n,m=l

8 0rn . 2 nn . 2 mn — _ \ _sin _sin _

7rn^im(l + (m/2)2 + (n/2)2) 4 4

(because 1 + (m/2)2 + (n/2)2 < (1 + (m/2)2)(1 + (n/2)2)))

64 ^^ 1 ^^ 0m 2 mn 2 m > — / 1-7 / —Ta-sm —— sin —.

n ^ 4 + n2 ^^ m(4 + m2) 4 4

n,m=l n=l

Estimating each of these series separately, we have

o n o n oo n

E1 • 2 «I 1 „ „ V"^ 1

--— sm — = > ---—0, 5 + > ■

(4 + n2) 4 4 + (2n _ 1)2 '

(4 + n2) 4 ^4 + (2n - 1)2 ' ^4(1 + (2n - 1)2)

n=l n=l ' n=l v v ' '

= — (thTT + th(Tr/2). (a) 16

o

E0m . 2

—--sin

m(4 + n2) 4 (without loss of generality, we assume that ^ « -^j-)

oo

E02m-1 „ _ \ - 04m —2

m=l (2m - 1)(4 + (2m - 1)2) ' ^ (4m - 2)(1 + (2m - 1)2)

~Cl ^ (2m-1)3(4 +(2m-l)2) > W

Consider the Laplace transform of the expression K(t):

oo

K(p) = f e-ptK(t) dt

8 ^^ 6m 1 2 nn 2 mn — > —--—-sm — sin -

n ^ m(1 + Anm)2 P + qnm 4 4

n,m=1 v

8 ^ ^m a + Qnm . 2 nn . 2 mn

x sm —— sm

n ^ m(1 + Anm)2 (a + Qnm)2 + C2 4 4

n,m=1 v

8 ^^ 1 . 2nn . 2 mn — it— > —--— --—--sin —sm -

n,m=1 v

= Re K (p) — ¿ImK (p). Consider the absolute value of the imaginary part if (a + ¿C)

IT ^ -Ml 8|C| ^ ^m 1 .2 nn -2 mn

n,m=1 v

because > 0, m =1, 2,.... It is easy to see the truth of the following inequalities for any n, m, C, a > 0

(a + qnm )2 + C2 < [1 + (a + qnm )2 ](1 + C2). Therefore, it is not difficult by using the following

11

>

(a + qnm )2 + C2 " [1 + (a + qnm )2](1+ C2) to prove the inequality

| Im K(a + ¿C)|

> 1^1 r V 1__t™ U7r m7r_L

-1 Z^ (1 i (H\2\2 n , fm)2)2m3 A A in -1-1)2

1+t2 nj?=1(l + (f)2)2(l + (f)2)2™3 4 4 (a + l)2 + l

;1 + C2'

Real part of expression K1 + K(a + ¿C) | ReK1 + Re K(a + ¿C)|

00

. „ , „ ^m a + qnm . 2 nn . 2 mn

> K1 + C > —~ ---777--r^-rr sm — sm -

1 + A nm 2 a + gnm 2+£2 4 4

n,m=1 v / w / /

^ „ 1 ^ <t>m a + 0, 5__1_ . 2 ror . 2 TO7T

" 1+ l+^nZliTO3(1 + (f)2)(1 + (f)2)((a + 1)2 + 1)Sm 4 Sm 4

>^1+C2a; 1

'1 + C2

where

oo

x - <j)m a + 0, 5 1 . 2 n7T . 2 TO7r

2a = Jhl (1 + (1 + (^)2)((a +1)2 +1} sm Tsm ~

Summarizing the obtained inequalities for Ki +K(a+i£), we obtain the estimate

\Ki +K(a + i£,)\ >Ci + ■

C2

>Ci.

(1+^)1/2

Let the image of a given function 6(t) satisfies the following condition

co

f |6(^)| dC<

(11)

Then, by passing to the limit at a ^ 0 from (10), we can obtain the equality

oo

= ¿ I

Kl + K (iC)

e*4 d£.

Next we need the following lemma.

Lemma 3. Let 6 £ Wi( —to, +to) , 6(t) = 0 with t < 0. Then for the image of the 6(t) function the inequality is true

/ |6(i^)l d^<

Proof. We use integration by parts in the integral representing the image of the given function 6(t)

-(a+i«)t

= / e^^^di = --0(t)

J -a - iC

+

1

t=0

a + iC

e-(a+ii)t0'(t) dt. (12)

Then using the obtained inequality and multiplying by the corresponding coefficient we get

(a + i£)0(a + iC)^ e-(a+liV(t) dt.

Hence, when a —>■ 0 we have

Similarly,

Consequently,

o

iC^(iC) = J e-iît0'(t) dt. 0

o

(9(iC) = J e-iit0(t) dt.

(13)

(14)

(15)

l^(iC)|2 (1+ C2) dC < Co||0(t)||2

W1 (R+ ).

t=oo

Using these inequalities, we can prove the following inequalities:

mM= I J 1 +

-oo —oo

1/2

<{ I mW*} < [ (jr^jä^

1/2 f oo 2 >> 1/2

+ { I em)\2d^ \ [ (1^552^

< / (1+ai«i2d^ < C0y0(i)yw2i(R+)-

Lemma 3 is proved.

According to the above,

l/n J,~2TT J \K1+K{ii)\ * " 2ttCi J 1 ^

Therefore

1 „ ,, „, . \ 11 2 . C0

As M we took

H«I < ^coiiofoii;,< = i-

C0

Acknowledgments. The author is grateful to Academician Sh. A. Alimov for the attention shown to this work.

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Submitted August 23, 2019 Revised August 28, 2019 Accepted September 3, 2019

Zarina Fayazova

Department of Higher Mathematics, Tashkent State Technical University 2A Almazarskaya Street, Tashkent 100095, Uzbekistan z.fayazova@yahoo.com