INVERSE PROBLEM FOR FRACTIONAL ORDER PSEUDO-PARABOLIC EQUATION WITH INVOLUTION Текст научной статьи по специальности «Математика»

ISSN 2074-1871 Уфимский математический журнал. Том 12. № 4 (2020). С. 122-138.

INVERSE PROBLEM FOR FRACTIONAL ORDER PSEUDO-PARABOLIC EQUATION WITH INVOLUTION

D. SERIKBAEV

Abstract. In this paper, we consider an inverse problem on recovering the right-hand side of a fractional pseudo-parabolic equation with an involution operator. The major obstacle for considering the inverse problems is related with the well-posedness of the problem. Inverse problems are often ill-posed. For example, the inverse heat equation, deducing a previous distribution of temperature from final data, is not well-posed since the solution is highly sensitive to variations in the final data.

The advantage of this paper is two-fold. On the one hand, we investigate the solvability of the direct problem and prove the solvability to this problem. On the other hand, we study the inverse problem based on this direct problem and prove the solvability results in this problem, too.

First, we investigate the Cauchv problem for the time-fractional pseudo-parabolic equation with the involution operator, and secondly, we consider the inverse problem on recovering the right-hand side from an overdetermined final condition and prove that it is solvable.

To achieve our goals, we use methods corresponding to the different areas of mathematics such as the theory of partial differential equations, mathematical physics, and functional analysis. In particular, we use the ^-Fourier analysis method to establish the existence and uniqueness of solutions to this problem on the Sobolev space.

The classical and generalized solutions of the inverse problem are studied.

Keywords: fractional differential equation, inverse problem, involution, pseudo-parabolic equation.

Mathematics Subject Classification 2010: 35R30, 34K37

1. Introduction

In this paper we study the inverse problem for the nonlocal pseudo-parabolic equation with an involution of the space variable x. We investigate the equation

V*[u(t,x) - Uxx(t,x) + £Uxx(t,-K - x)] - Uxx(t,x) + £Uxx(t,-K - x) = f (x), (1.1)

for

(t,x) e tt = {0 <t<T< to, 0 <x<k], 0 <a ^ 1,

D. Serikbaev, Inverse problem for fractional order pseudo-parabolic equation with

involution.

© D. Serikbaev 2020.

The author was supported by the Science Committee of the Ministry of Education and Science of the Republic of Kazakhstan Grant AP08052425 and by the FWO Odysseus 1 grant G.0H94.18N: Analysis and Partial Differential Equations. Submitted October 16, 2020.

where is the Caputo derivative (see [14]) for 0 < a < 1 and := dt for a = 1, In L2(0,n) we consider a second order differential operator generated by the differential expression

C(u) = -u"(x) + eu"(n - x), 0 <x <n, (1.2)

subject to the boundary conditions

u(0) = 0, u(n) = 0, (1.3)

where |e| < 1, e G R. We can check easily that the introduced operator is self-adjoint (see [1,13,28]). For all |e| < 1, nonlocal problem (1.2), (1.3) has the following eigenvalues:

\2k = 4(1+ e)k2,k g N and X2k+i = (1 - e)(2k + 1)2, k g Z+,

and the corresponding system of eigenfunctions

2

u2k(x) = y — sin 2kx, k g N,

u2k+l(x) = J-sin(2k + 1)x, k G N u {0}.

(1.4)

As a motivation, we mention that pseudo-parabolic equations have numerous applications in sciences and engineering. For example, the energy functions of the isotropic materials can be expressed as solutions of pseudo-parabolic equations [5]. Some wave processes [3], filtration of the two-phase flow in porous media with the dynamic capillary pressure [4] are also modeled by pseudo-parabolic equations. The time-fractional pseudo-parabolic equation (1.1) occurs in studying flows of the Oldrovd-B fluid, one of the most important classes of dilute solutions of polymers [7,27].

The study of inverse problems for pseudo-parabolic equations was initiated in the 1980s. The first result obtained by Eundell [22] refers to the inverse identification problems for an unknown source function f in the following equation

d

— [u(t, x) + Lu(t, + Lu(t, x) = f, (1.5)

where L is an even order linear differential operator. Rundell proved global existence and uniqueness theorems for then cases when f depends either only on x or only on t.

The inverse problems on identifying the right hand sides of the pseudo-parabolic equations from a focal over-determination condition have important applications in various areas of applied sciences and engineering. Inverse source problems for the diffusion, sub-diffusion and for other types of equations are well studied. In this area some recent progresses have been done in the series of articles, see, for example, [1,2,9,10,12,13,15,21,23,24,26,28]. However, inverse problems for pseudo-parabolic equations and for their fractional analogues have been studied relatively less, see [8,11,17-19,22]. In our paper, we aim to fill this gap.

For more information on pseudo-parabolic equations, we refer to a book by Demidenko and Uspenskii [6] and the references therein.

2. Direct problem

Before formulating a problem, we first introduce fractional differentiation operators.

Definition 2.1. The Riemann-Liouville fractional integral Ia of order 0 < a < 1 for an integrable function f is defined by the formula

1 r*

V{a) Jc

where Г is the Euler gamma function.

The Caputo fractional derivative of order 0 < a < 1 of a differentiable function f is defined by the formula

V?if](t) — 11-a[f(t)] — Г ds, t e [c,d].

Г(1 - a) Jc [t - s)a

Further information on fractional derivatives can be found in [14].

In what follows, we widely use the properties of the Mittag-Leffler type function (see [16]), which is introduced as

Ea,f (z) —

—/ r(am + d)

m=0 y

In [25], the following estimate for the Mittag-Leffler function was proved:

-, , w!-r- ^ Ea,i(-z) ^ \———, z > °

1 + 1(1 — a)z 1 + 1(1 + a) 1z

as 0 < a < 1 and this estimate fails for a ^ 1. Thus, it follows that

0 <Ea1(—z) < 1, z> 0. (2.1)

Definition 2.2. The space W2'2[0, -n] is a Hilbert one consisting of all elements of L2[0,n] having generalized derivatives up to order 2 in L2, i.e.

W2>2[0,n] = [f e L2[0,n]\ f',f e L2[0,n]}

and the norm is defined by

W 2,2 [0,7] — ^

г=0

d%f\ 2

dxl

L2 [0,7]

, *k] is the subspace W2'2[0, -n] defined as the closure with respect to || ■ \\w2,2of all twice continuously differentiable in [0, -n] functions vanishing at the points 0 and n. The L2-scalar product of two functions f,g : [0, -n] ^ R is defined by

(2.2)

(f,9)L2 = f(x)g(x)dx. 0

In this section, we study the Cauchv problem for the pseudo-parabolic equation V"[u(t, x) — uxx(t, x) + euxx(t, -k — x)]

— Uxx(t,X) + £Uxx(t,K — X) = f (t,x), (t,x) e Q, with the initial data

u(0,x) = <p(x), x e [0,n], (2.3)

and the homogeneous Dirichlet boundary conditions

u(t, 0) = u(t,n) = 0, (2.4)

where Q := [0 <t <T < <x, 0 < x < n}, f (t,x) and p(x) are given functions. The following statement holds true.

Theorem 2.3. Let |e| < 1, f e C([0,T]; C2[0,n]) and f (t, 0) = f (t,n) = 0, <p e C4[0,n] and p(l\0) = (k) = 0 i = 0,2. Then there exists a unique regular solution u e C 1([0,T]; C2\0,^}) of problem (2.2)-(2.4) and this solution can be written in the form

<x <x

u(t, x) = ^^ wk(t) sin(2fc + 1)x + ^^ vk(t) sin 2kx, k=0 k=1

zm

7

where

- w =- E-1 (- ш+Ь- ¿11 i0' I (-ife')) ^ -

щ(t) (-r+L-e) - £/| (-r+L-s*)) m -

for t E [0,T] and for all к E N, where

Plk = (f, U2k+1 )l2 , <^2k = (f, U2k)l2 ,

fl(t) = (f(t, •),U2k+l)L2 , f2(t) = (f(t, •),U2k )L2 .

Доказательство. We begin with proving the existence. Since the system of eigenfunctions (1,4) is an orthonormal basis in L2(0, к), we seek the function u(t,x) in the form

те те

u(t, x) = ^^ wk(t) sin(2k + l)x + ^^ vk(t) sin 2kx, (2,5)

k=0 k=1

where wk(t) and vk(t) are unknown functions. Substituting equation (2,5) into equations (2,2), (2,3), we obtain the following equations corresponding to the functions wk(t) and vk(t):

VatWk (t) ,, Wk (t) = k.w , k E No, (2.6)

1 + -2k+1 1 + -2k+1

A2k+1 w (f) _ fk(t)

+A-Wk (t) _ 1 + A

+ A2k+1 1 + A2k

Wk(0) _ Pik, k E No, (2.7)

and

V?vk(t) + j+^Vk, k E N, (2.8) 1 + A2k 1 + A2k

Vk(0) _ V2k, kE N. (2.9)

According to [16], the solutions of equations (2.6) and (2.8) satisfying initial conditions (2.7) and (2.9) can be represented in the form

^- IT- f* fci {-TT,

1 + A2k+1 ) Mk+1 Jo ds V V 1 + A2

and

*(i) _(-r^--hHi (-глЬ;s*)) M-s)ds- tm

for t E [0,T] and for all k E N. Substituting (2.10) and (2.11) into (2.5), we obtain a solution of problem (2.2)-(2.4).

Now we are going to prove the convergence of the obtained infinite series corresponding to the functions u(t,x), V"u(t,x), uxx(t,x) and Vj*uxx(t,x). By the assumptions of the theorem we have

<^w(0) = 0, ^l)(n) = 0, x = 0, 2, f(t, 0) = f(t,n) = 0.

This implies that

2 n

<£1k =\ ~ V(x) sin(2k + 1)xdx ui 0

1 F2 1

= (2kT1T4V 21 ^(x)sm(2k +1)xdx = WTW^,

[2 r

<£2k =\~ ^(x)sin2kxdx ui 0

1 F2 1

= iW 21 ^(4)(x)sm2kxdx =

f!(t) = f(t,x)sin(2k + 1)xdx

= — w+w vf [ f"(t-x) sm(2k +1)xdx = — wrr?fr(t)-

fk(t) = \ — I f(t ,x)sin2kxdx V k j 0

= — W2\fl if" (* ,x)sin2kxdx = — fk(2\t).

Here ^ ^ f!(2)(t)^d fk(2)(t) are

(4)

<P1k

(4) ^2k

14? = y 2 J V(4)(x) sin(2k + 1)xdx,

2k) = \ 2 I p(4\x) sin2kxdx, V k j 0

k

fl(2)(t) = \J2 J ,x) sin(2k + 1)xdx, fk(2)(t) = JQ f"(t,x)sin2kxdx,

where

f(t ,x) = 0 (t,x). Our next step is to calculate u(t,x), uxx(tV™ux:x:(t,x). We have:

<x <x

By (2.6) and (2.8) we find Vfwk(i) and V*vk(t):

t i^kW ^t

fk(t) A2 k+1

t W 1 + A2 k+1 1 + k+1 and

(2.12)

(2.13)

V*u(t, x) = Y^ V*Wk(t) sin(2k + 1)x + Y v?Vk(t) sin 2kx. (2.14)

k=0 k=1

V?Wk(t) = n k" — Y+^^k(t), (2.15)

Mk+1 1 +

^ (^ = ¡+L — T+rk^ (2'16)

Substituting (2,15), (2,16) into (2,14) and taking into consideration formulae (2,10), (2,11), (2.12), (2.13), we get:

V-u(t,x) = -

ft (2)(t)

k=0 ro

E

k=0 oo

(1 + X2k+l)(2k + 1)2 1 -£

(1 + \2k+l)(2k+1)2

sin

ro f2(2) (+) (2k +1)x - E (i+x^lw

sin 2 k x

k=1

<p<£EaJ- A sin(2k + 1)x

V 1 + X2k+l J

E n^„2 [' £ (E.(-^s) ) fl'»(t - syu (,17)

k=0 ro

£

(1 + \2k+i)(2k + 1)2 Jo ds

(1 + e) sin 2kx

a,l

(__h^fa\

V 1 + X2k ) X2k

^ sin 2 k x r±(F (

(1T\2k)4k2J0 dS \ a,lv 1 + x^2k

S a))f2r(t - s )d s.

Applying the operator -X to (2.5) and taking into consideration formulae (2.10), (2.11), (2.12), (2.13), we have

uxx(ix) ^ ^

k=0 oo

EaJta) sin(2k + 1)x V 1 + X2k+l J

(2 k + 1)2~a'l\ 1 + \2k+i

t , S / x

k=0

ro (4)

2 k+l

ro (4) /

^ V2k E k=l v ^ sin 2 kx f' d ( ^ f

+ I slM

ta\ sin 2 kx

1 + X2k

~T ( Ea,l (—A^s-)) fk(2\t - s)d s.

1 + X

2 k

(2.18)

Finally, applying the operator -Xz to (2.17), we have

Vtauxx(t ,x) =

ro fl(2)

k=0

fl(2'(t) 1 + X

sin

2 k+l

(2 k + 1)x + Y,

ro .2(2)

k=l

fr(t)

1 + X

sin 2 x

2 k

ro

+ ZT-^viiEa.l

k=0 oo

1 + X2 k+1 1

+ ^ 1 + X2k+llo

k=0 oo

f X2k+l ta\

V 1 + X2k+l J

(Ea,l (

a

sin(2 k + 1)x

— ( E- l

X2 k+1 a 1(2)

I V^ 1 + g J4) E + 1 , X V2k Ea,l

k=1 oo

(__h^fa\

V 1 + X2k )

2 k+1

a sin 2 x

sa fk(2)(t - s) ds

sin(2 k + 1)x (2.19)

1 + X2 k

" -i-(EaJ -

1 + X2k Jo ds V 'V 1 + X2k

+ ^ 11 + X2k Jo

1 + X2k

d ( „ ( X2k - W ¿2(2)

sa fk(2)(t - s)ds

sin 2 x.

Let us to show the convergence of the integral term:

[ ds{Ea,1{ 1 + X^

? Sa ) ) fk(t - s)ds

«

is {Ea'1{- ÎT\,Sa

« I fk (t - s )|

« max I fk (t)I

0<t<TlJkK n

T ( E»,i

I fk (t - s )ids 1 + X?"

is (- IT\,sa

ds.

(2.20)

Hereinafter we suppose that i = 1, 2 and if i = 1, then Ç = 2k + 1, k E N0, while if i = 2, then £ = 2k, k E N. In the above estimates, the integrand is the derivative of the Mittag-Leffler function, which is a non-positive function since

T I Ea,i

X

i + \

? s»

X?

1 + X?

lEa,a{ i + Xtsa)

« 0.

(2.21)

It is known that

Ea,a(-z) > 0, z > 0,

see, for instance, [20]. Hence,

(Ea,i( -

Is 1 ^

X

1 + X.

? sa

ds

fo ds ' E»1

X

1 + X,

? sa

ds

(2.22)

Substituting the latter relations into (2.20) and employing (2.1), we obtain:

( Ea,i (

fo ds ' Ea

X?

1 + X?

sa) fk(t - s )ds

SSTIfk (t)I{1 -E*+{ - ïV))

«C max I fk (i)I.

0«t«T k

Using the estimates (2.1), (2.23) and taking into consideration the formulae (2.12), we arrive at the following inequality:

(2.23) (2.13),

n1

(4) |

ct,x(n)

k=0

ro | (4) | (2 k + 1)4 + C 16k4

ro |

«c e ^+CE-

J2k

k=i

+ C max y A iti™(t)I N2 +C max V j^2^) 0««T k=0 X2k+i(2k + 1)2 0«t«Tt=i X2k4k2

(2.24)

By formula (2.17), we have

t a\\ct,x(n)

« c E

i№iïi

k=0

(1 + X2M)(2k + 1)2

+ c E

k=i

+ C max

0<t.<T —'

I fl(2)(t) I

0«MTÏ=0 (1 + X2k+i)(2k + 1)2 0«t«T^ (1 + X2k)4k2'

(1 + X2k )4k2

ro

+ C max

0<t.<T —'

I f2k(2\t) I

(2.25)

0

0

0

a —

0

It follows from formula (2,18) that

+ C ± » + c max ± J»

(2.26)

and by the formula (2.19), we have

k=o 1 + W 1 + ^k

+ C mx ± JÄ + c max ± Ä

(2.27)

Hence, series (2.24), (2.25), (2.26) and (2.27) converge absolutely and uniformly in the domain

H. " "

Now we are in a position to show the uniqueness of the solutions. Let w(t ,x) and v(x, t) be solutions of problem (2.2)-(2.4), that is,

V%[w(t, x) - wxx(t, x) + ewxx(t ,k - x)] - wxx(t, x) + ewxx(t ,k - x) = f(t, x), w(0,x) = <(x), V"[v(t ,x) - Vxx{t ,x) + £Vxx{t ,K - x)] - Vxx(t, x) + £Vxx{t ,K - x) = f (t, x) , v(0,x) = <(x).

By subtracting these equations one from the other and denoting u(t,x) = w(t,x) - v(t,x), we obtain that

V"[u(t, x) - Uxx(t, x) + £Uxx(t ,K - x)] - Uxx(t, x) + £Uxx(t - x) = 0, (2.28)

u(0, x) = 0. (2.29)

We also have

wk(t) = \ — u(t,x)sin(2k + 1)xdx, ke N0,

V W o

vk(t) = \ — f u(t,x)sin—kxdx, k e N.

V W o

a

(2.30)

Applying the operator V™ to (2.30), we obtain

V'wk(t) = \/— [ V'u(t, x) sin(2k + 1)xdx, ke N0,

k .1 o

2 ^ I 'n

(2.31)

V'vk (t) = \ - u(t ,x)sin2kxdx, ke N.

0

We multiply both sides of equations (2.28), (2.29) by the functions sin (2k + l)x, sin 2 kx and integrate in the variable x torn 0 to k. Taking into consideration the self-adjointness of the nonlocal operator c, and by (2.30)-(2.31), we have

Vwk(t) + (t) = 0, k E No, (2.32)

1 + *2k+1

wk(0) = 0, k E N0, (2.33)

and

V?vk(t) + -^~vk(t) = 0, k E N, (2.34)

1 + M k

vk(0) = 0, k E N. (2.35)

By formulae (2.10), (2.11) with p1k = 0, fl(t) = 0, p2k = 0, fk(t) = 0, the solutions of problem (2.32), (2.33) and (2.34), (2.35) read as wk(t) = vk(t) = 0. Hence, by the completeness of the system u2k(x), u2k+1(x) in L2, we obtain u(x, t) = 0. □

2.1. Generalized solution. In this section we provide a solution to the direct problem with data from Sobolev spaces.

Theorem 2.4. Let |e| < 1, f E C([0,T];L2[0,n]) and p E W02'2[0,'k]. Then there exists a unique generalized solution u E C :([0,T]; Wq'2[0,k}) of problem (2.2)-(2.4) and it can be written as

<x <x

u(t, x) = ^^ wk(t) sin(2k + 1)x + ^^ vk(t) sin 2kx, k=0 k=1

where

iTx^* ) + Mk+Jo MM-TTA^S

and

Wk(t) = <PikEa,i (-^A + f ± (Еа,1 (-^sA) fl(t - s)ds,

\ 1 + \2k+i J Mk+i Jо ds \ \ 1 + \2k+i J J

- W = ^- I+к, + Ъ Li (- r+k ^ I fi - ^

for all t E [0,T] and for each, к E N, where

<Pik = ('P,U2k+i )l2 , <^2k = (<f,U2k )l2 ,

fl(t) = (f(t, •),U2k+i)L2 , fk(t) = (f(t, •),U2k) L2 . Доказательство. Using property (2.1), we get the following estimates:

те Ifi(t)12 те \f2(t)12

llUllCФМО^ J СMl Ьм + С ^£ lJkrJL + С max £ Щг11-

0JtJ1 k=0 A2k+i 0JJT k=i A2k

тпгсфП,2Мjc imI12M +c .J- g j^b +С ssst £ (Щу

12 ^ r<\\. „ 112 С

xx nC([0,T],L2 [0,-]) J СH<p xxl|L2[0,-] + (1 _ jg-j )2 11 11 C([0,T],L2 [0,-]

11 ^t Uxxl|C([0,T],L2[0,-]) xx 11 jj2 [0,-]

+ С11Ш

C([0,T ],Ь2[0,ж]).

I | uxx 11 (

\ uxx 1 |C([0,T],L2[0,^])

^CIIp xx 11 j2 [0,^]

+ CIIfII Cl

Arguing as in the proof of Theorem 2.3, we complete the proof. □

3. Inverse problem In this Section, we study an inverse problem for the pseudo-parabolic equation (1.1).

( u( , x) , ( x))

homogeneous Dirichlet boundary conditions

u(t, 0) = u(t,n) = 0, (3.1)

and an initial condition

u(0,x) = p(x), x E [0,n], (3.2)

with an additional condition

u(T,x) = ^>(x), x E [0,k], (3,3)

where <(x) and ip(x) are sufficiently smooth, given functions.

Using the Fourier method, one can cheek the unique solvability of this problem, A regular solution of problem 3,1 is a pair of functions (u(t,x), f (x)), where u E C 1([0,T],C2([0,k])) and f E C2([0, k]), A generalized solution of problem 3,1 is a pair of functions (u(t,x), f(x)), where u E C 1([0,T],W2'2([0,k])) and f E L2([0,k]).

3.1. Main results. For equation (1,1) with conditions (3.1)-(3.3) the following theorem holds true.

Theorem 3.2. Let T > 0 N < 1, E Ca[0,k] and

<l)(0) = <i)(K) = ^ (0) = ^(K) = 0, i = 0, 2.

Then there exists a unique regular solution u E C *([0, T], C2[0, k]), f E C2\0,k] of problem 3.1 and it can be written as

- I1 -E«A -Т+Й+ТH hin(2k + l)x u(t,x) =j(x) + £ -\-LLr-(J£ -

(1 -(2k + 1)2 - (l -EaA - Г)) sin2kx

+liT—у у—-^

k=T I1 -e«t{-T+tTa)) 4k2

f(x) = - Jxx(x) + Sjxx(n -x) + Y —- ( к Hк ) ч sin(2k + 1)x

n 1 ТП / ^2k + -\ глп i

k=o i-E A- X2k+1 Ta\ 1 l+\2k+1T J

+ £ (l+e)(f аш2кх

k=1 1 ~E«i{ - "

for (t,x) E П, where

Ры = (<Pxx,u2k+l)b2 , P^k = (Pxx,U2k )l2 , i>fk = (^xx,U2k+l)b2 , фЦ =(Фхх,Щк) L2 .

Доказательство. We begin with proving the existence. Since the eigenfunctions (1,4) form an orthonormal basis in L2[0, к], we seek functions u(t,x^d f(x) in the form

те те

u(t, x) = ^^ wk(t) sin(2k + l)x + "Yj yk(t) sin 2kx, (3,4)

k=0 k=1

and

те те

f(x) = Y ft sin(2k + l)x + Y fl Sin 2kx, (3.5)

k=0 k=1 where wk (t), vk (t), /¿and fl are unknown. Substituting equations (3.4) and (3.5) into equation (1.1), we obtain the following equations for the functions wk(t), vk(t) and the constants fl, fl:

T+f^^ (t) = УУ, v-Vk (t) + -^Vk (t) = -Ä-.

1 + *2k+T 1 + *2k+T 1 + *2k 1 + Л2к

Solving these equations, we obtain

wk (t) = ^ + CikEaJ - A

X

2 k+l

(__X2k+l

V 1 + X2k+l J

k2

Vk (t) = + C2kEa,l[ - , , X2 k 1 + X2 k

V 1 + X2k )

where the constants fl, fj2, C1k and C2k are unknown. To find these constants, we use conditions

k k

Pik = (P,u2k+l)j2 , P2k = (P,u2k )j2 , ^1k = 0>,u2k+l)j2 , ^2k = (tp,u2k) L2 .

C1 k

Wk (0) Wk (T)

X

l

+ Clk = Plk,

Then we have

2 k+l fk X2k+l

Cl k =

+ ClkEa,l\ —

-

X

2 k+l

1 + X2 k+l Plk - llk

T

llk.

1- E A - X2k+1 T- |

1 Ealy l+A2fc+i J

Va'M l+A2fc+1

k1

C2 k

fk = PlkX2k+l — ClkX2k+l.

k2

Vk (0) = + C2k = P2k, x2k

Vk (T ) = # + C2kEaJ - X2k

X

2 k

P2k — C2k + C2kEa,l[ ---k-T"

( X2k t-\

V 1 + X2k J

( X2k t-\ =

V 1 + X2k J

l2k,

Then we obtain

C2 k =

1 + X2k

P2k - l2k

l2k.

1 — E A__X2k Ta ]

1 Ea'l\ l+\2kT J

k2

k2 = P2 k X2 k - C2 k X

2 k.

Substituting wk (t), vk (t), /¿and /| into equations (3,4) and (3,5), we find

u(t, x) =p(x) +

Clk ( Ea,l ( k=0 ^ ^

EaA ta) - 1) sin(2k + 1)x

1 + X2 k+l

+ Y,C2k(Ea>l(^-Y+X^kta) - 0 sin2kx.

By the assumption of the theorem we have

p(0) = p(n) = ^(0) = ^(n) = 0.

This implies

Cl k =

Plk - l

l k

P(2) ilt2) plk - Wlk

1 - E-l( - (1 - E-,l( - 1+2+T^ ) (2k + 1)2

In the same way we obtain:

P(2) l (2) C =__P2k - r2k_

2k (1 -Ea'l( - 4k 2

Then we get

ro (1 -Ea,l( - l+t+^A) sin(2k+1)x u(t,x) = P(x) + £ --LL--(plk - 4<2)

k=0 (1 -Ea,l[ - (2k+1)2

1 - Eali -ta) I sin2kx

+ £ V - l+A2fe JI (p22k -i(2).

(3.6)

k=l ^ -Ea,l(- l+tTa)) 4k2

As above, we also find that

f(x) = - Pxx(x) + SPxx(7T -x) + Y, (1 ^ x sin(2k + 1)x

^2k+l rpa

k=0 1-E 1I - X2k+1 T— I

1 Ea'ly l+\2k+l J

+ ± (1+£)fk sn2kx.

k=l 1 -Eal[ - J+^Ta)

(3.7)

The assumption T > 0 and the estimate (2.1) guarantee that the denominators in formulae (3.6) and (3.7) do not vanish. Moreover, there exists a constant C > 0 such that

1 - Ea,l(- - Y+^T^ >C> 0, (3.8)

for all £ E Z+

Let us calculate Vfu(t,x), uxx(t,x) and V"uxx(t,x). According to [16], we have

Vta (Ea,l(-Xta)) = -XEa,l(-Xta). (3.9)

Applying the operator V? to (3,6) and taking into consideration (3,9), we have

^ (1 - е)ЕаЛ - j+^r) sin(2k + 1)x V?u(t, x) = Y T-f-fr-- №)

k=° (1 -ш++;та)) a + W

те (1 + е)ЕаЛ - 1+--tA sin2kx

+ £ t-A—У-( - 4Ï),

k= l1 -Е«д(- 1+tTa)) (1 + ^2k)

те Î1 -E*,l( - A) Sln(2k+1)X

Uxx(t,x) =tpxx (x) -7-\-(Lpfk - Ф\к)

k=0 1- E Л__X2k+1 Ta 1

1 1+\2k+1T )

те [1 -ЕаЛ - T+t-tA) sin2kx

y—v ; ;л -№),

^2 k Ta I ^ 1+A2 kT j

(3.10)

(3.11)

k=1 1 — Е 1__Л2к Ta

1 Е»,1\ 1 + ^2k

and we also have

те \2к+1ЕаЛ(-I+f2+i7tA sin(2k + 1)x V?Uxx(t,x) = - Y T-f-(t-(^ - №)

k=° l1 -Е*,1{ - Ш++1Та)) (1 + ^2k+1)

те \2кЕа,А - ï+ff-ta Sin 2kx

£ T-> +2k <N-№ - №).

k=1 1-Е„1 -

(3.12)

(1 -Е*>{ - Ш-Та)) (1 + ^2k )

In view of estimate (2.1), in order to ensure the convergence of series (3.6), (3.10), (3.11), (3.12) and (3.7), we have the following estimate

I, II ^ || M + c ^ ^fk1 + \Ф\Yk 1 + С ^ W^k 1 + ^fh (О 1 o\

Wu\\ct,x(Щ < \M\c([0,7]) + c^ + + С^-—-. (3.13)

k=0 ( ) k=l

Bv the assumptions of the theorem we have

<^г)(0) = 0, р(г)(п) = 0, г = 0, 2, Ф(г)(0) = 0, ф(г)(п) = 0, г = 0, 2.

This implies that the coefficients of the function ф(х) satisfy identities (2.12). Using this fact for f(x), we arrive at the estimate

II f\\ II + С ^ ^ + ^Ы1 + С V4 ^k 1 + 1ф2к)1 /0 1Л

k=0 ( + ) k=l

We also have:

№u\\Cm^ ^

ro + № +CPii + iiffi

+ 1 + X2k

k=0 oo

1+ X

2 k+l

||uxx II

xx\\Ct,x(n)

<

C

k=0

P4I + uffi (2 k + 1)2

ro (4)

k=1 oo

+ C

+ ii2fi

4k 2

k=l

(4) ro (4)

I,™ M _<C ^lk1 + 1141 +C ^ ^k1 + 141

Wt uxx Hct^fi) ^ C (2k + 1)2 + C 4k2 .

k=0

k=l

4k2

This completes the proof of the existence of the solution.

We proceed to proving the uniqueness. We suppose that {u1(t, x), /^x)} and {u2(t, x), f2(x)} are solutions of Problem 3,1, Then u(t,x) = u1(t,x) -u2(t,x) and f(x) = f^x) — f2(x) satisfy the following problem:

V" [u(t ,x) — uxx(t ,x) + £uxx(t ,n — x)] — uxx(t, x) + £uxx(t ,n — x) = f (x) , (3.15)

u(0, x) = 0, u( T, x) = 0.

We also have

wk (t) = \ — I u(t ,x)sin(2k + 1)xdx, ui 0

k E N0,

and

k( )

kl

k2

- I u(t,x)sin2kxdx, K . ¡0

kEN,

f(x) sin(2k + 1) xdx, k E N0,

f(x)sin2kxdx, kE N0.

Applying the operator V" to (3,18), we get V?wk (t) =

— I V— u(t,x) sin(2k + 1)xdx

k J 0

2

V—vk(t) = \/~ V— u(t, x) sin 2kxdx,

, k E N0, k N.

(3.18)

(3.19)

(3.20)

We multiply both sides of equations (3.15)-(3.17) by the functions sin (2 k + 1)x, sin 2 kx and integrate in the variable x torn 0 to n. Taking into consideration the self-adjointness of the nonlocal operator c, and by (3.18)—(3.20), we have

V—Wk (t) + A2k+^wk (t) = fk

1 + X2 k+1

Wk (0) = 0, Wk (T) = 0,

1 + X2 k+1

k E N0, k E N0,

k E N0,

(3.21)

(3.22)

(3.23)

n

2

k J 0

n

0

and

V?vk (t) + T^vk (t)

fk

1 + X2k 1 + X2k Vk(0) = 0, k e N,

Vk(T) = 0, k e N.

ke N,

(3.24)

(3.25)

(3.26)

Arguing as in the proof of the existence result, we confirm easily that the solutions of equations (3.21)-(3.23) and (3.24)-(3.26) are wk(t) = vk(t) = ft = ft = 0. Then

0 0

— I u(t,x) sin(2k + 1)xdx, k e No,

* ■ Io

— I u(t,x)sin2kxdx, ke N,

* . to

and

0 0

— I f(x) sin(2k + 1)xdx, k e N0,

* Jo

— I f(x)sin2kxdx, ke N0.

* .to

Therefore, owing to the completeness of the system u2k(x), u2k+1(x)rn L2, we obtain u(x, t) = 0

and f(x) = 0. This prove the uniqueness of the solution of problem 3.1. The proof is complete.

□

3.2. Generalized solutions. Once we deal with a generalized solution instead of the regular one of problem 3.1, Theorem 3.2 can be modified as follows.

Theorem 3.3. Let T > 0 N < 1 and p,^ e WQ2^,*}. Then there exists a unique generalized solution u e C1([0,T},Wq'2\0,*}), f e L2[0,n] of problem 3.1 and it can be written as

- (l -EaJ - l+t+^A) sin(2k+1)x u(t,x) =p(x) + £ --¿Z^-(^ -

k=0 ^ -Ea,^- 1+2+T«)) (2k+1)2 - (l -EaJ-tA) sin2kx

+ £ P-— Pk - №),

k=l ^ - Ea,iy - J^Ta)) 4k 2

- (1- r)( P 2)-P,(2)) f(x) = - Pxx (x) + £pXx(* -x) + Y, —-p x sin(2k + 1)x

k=0 1 -Eal[ —

/ ^2fe+i Ta \

I l+A2fc+i T J

+ —--7—- k \ sm 2 kx,

k=l 1 Ea,l( -

for (t,x) E П, where

V1k = (<Pxx,U2k+l)l2 , Ры = (<Pxx,U2k)l2 ,

Фгк = (Фxx,U2k+l)L2 , = (Фхх,и2к)l2 .

Доказательство. Reproducing the proof of Theorem 3,2, we can prove the unique solvability in the considered case, too. We just need to clarify the functional classes. By using Plancherel identity and orthogonality of u2k, u2k+i, we get:

~ 2) 12 + u(2)\2 ™

Hull W2IM) i СMhM + c E ilk +Vf +c E

_ i4?I2 + I4?I2

(2k +!)' (2k)4 ■

WfWb [0,7т] ^ C\\VxA2L2[0,IГ] +C\\ФХА2Ь2[0,ТГ].

Moreover, we find

\Wu\\2Cаоздо,^ ^ l\\k + +2Ы)21 (1 + \2k)2

\v(2\2 + \^\2 + r ^ \v<ik\2 +

W^xxWC([0,T],l2[0,tT]) ^ &W^xxWL2^] + &||rlpxx || ¿2[Q,n] ,

xx \ \ L2 [0,^]

+ CU xx \ \ L2 [0,^]'

These estimates allow us to reproduce the arguing from the proof of Theorem 3,2, The proof

is complete, □

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Daurenbek Serikbaev,

A! Iambi Kazakh National University,

Al-Farabi av, 71,

050040, Almatv, Kazakhstan

Department of Mathematics: Analysis, Logic and Discrete Mathematics, Ghent University, Krijgslaan 281, B-9000, Gent, Belgium

Institute of Mathematics and Mathematical Modeling, Pushkin str,, 125, Almatv, Kazakhstan

E-mail: daurenbek. serikbaevSugent. be