SOLVABILITY OF NONLOCAL INVERSE BOUNDARY-VALUE ...

Electronic Journal of Differential Equations, Vol. 2017 (2017), No. 125, pp. 1–14. ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu

SOLVABILITY OF NONLOCAL INVERSE BOUNDARY-VALUE PROBLEM FOR A SECOND-ORDER PARABOLIC EQUATION WITH INTEGRAL CONDITIONS ELVIN AZIZBAYOV, YASHAR MEHRALIYEV Communicated by Ludmila Pulkina

Abstract. This article studies a nonlocal inverse boundary-value problem for a second-order parabolic equation on the rectangular domain. First, we introduce a definition of a classical solution, and then the original problem is reduced to an equivalent problem. Existence and uniqueness of the solution of the equivalent problem is proved using a contraction mapping. Finally, using the equivalency, the existence and uniqueness of classical solution is obtained.

1. Introduction Practical requirements often lead to the problem of determining the coefficients or the right hand side of the differential equations for some known data about its solutions. Such problems are called inverse problems in mathematical physics. Inverse problems arise in various fields of human activity, such as seismology, mineral exploration, biology, medical visualization, computed tomography, Earth remote sensing, spectral analysis, nondestructive control, etc. Fundamentals of the theory and practice of research of inverse problems were established and developed in the pioneering works by Tikhonov [19], Lavrent’ev [11], [12], Ivanov [7], Romanov [18], Denisov [3]. Subsequently, the methods developed by them were applied to investigate a wide scale of inverse problems by their pupils and followers. Recently, there have been many studies of inverse problems for parabolic and other types of equations. A more detailed bibliography and a classification of problems are found in [1, 5, 6, 7, 8, 14, 15, 16]. Recently, problems with nonlocal conditions for partial differential equations have been of great interest. We note that most of the publications about problems with spatially nonlocal conditions and integral conditions for partial differential equations are found in[9, 4, 17]. In [10], a problem of time nonlocal integral conditions for hyperbolic conditions is investigated. In this article we study an inverse boundary-value problem for second-order parabolic equations with nonlocal conditions. A distinctive feature of this article 2010 Mathematics Subject Classification. 35R30, 35K10, 35A09, 35A02. Key words and phrases. Second order parabolic equation; finite domain; nonlocal integral condition; inverse value problem; classical solution. c

2017 Texas State University. Submitted February 17, 2017. Published May 8, 2017. 1

2

E. AZIZBAYOV, Y. MEHRALIYEV

EJDE-2017/125

is the consideration of a parabolic equation with both spatial and time non-local conditions. 2. Formulation of the problem Let T > 0 be a fixed number and DT := {(x, t) : 0 ≤ x ≤ 1, 0 ≤ t ≤ T }. We consider the equation c(t)ut (x, t) = uxx (x, t) + a(t)u(x, t) + b(t)g(x, t) + f (x, t)

(2.1)

in the rectangular domain DT . The inverse problem has nonlocal initial condition Z T p(t)u(x, t)dt = ϕ(x) (0 ≤ x ≤ 1), (2.2) u(x, 0) + δu(x, T ) + 0

periodic boundary condition (0 ≤ t ≤ T ),

u(0, t) = u(1, t) nonlocal integral condition Z

(2.3)

1

u(x, t)dx = 0

(0 ≤ t ≤ T ),

(2.4)

0

and the additional conditions u(xi , t) = hi (t)

(i = 1, 2; 0 ≤ t ≤ T ),

(2.5)

where δ ≥ 0, xi ∈ (0, 1) (i = 1, 2; x1 6= x2 ) are fixed numbers, 0 < c(t), g(x, t), f (x, t), 0 ≤ p(t), ϕ(x), hi (t) (i = 1, 2) are given functions, u(x, t), a(t), b(t) are the sought functions. Definition 2.1. The triplete {u(x, t), a(t), b(t)} is said to be a classical solution of problem (2.1)–(2.5), if the functions u(x, t), a(t) and b(t) satisfy the following conditions: (1) The function u(x, t) and its derivatives ut (x, t), ux (x, t), uxx (x, t) are continuous in the domain DT ; (2) the functions a(t) and b(t) are continuous on the interval [0, T ]; (3) equation (2.1) and conditions (2.2)–(2.5) are satisfied in the classical (usual) sense. Lemma 2.2. Suppose that δ ≥ 0, 0 < c(t) ∈ C[0, T ], a(t) ∈ C[0, T ] and 0 ≤ p(t) ∈ C[0, T ] hold. Then the problem c(t)y 0 (t) = a(t)y(t) (0 ≤ t ≤ T ), Z T y(0) + δy(T ) + p(t)y(t)dt = 0

(2.6) (2.7)

0

has a unique trivial solution. Proof. Obviously, the general solution of equation (2.6) has the form: a(τ ) dτ 0 c(τ )

Rt

y(t) = ce Using (2.7) we obtain RT c 1 + δe 0

a(t) dt c(t)

Z +

T

p(t)e 0

.

a(τ ) dτ 0 c(τ )

Rt

(2.8)

dt = 0.

EJDE-2017/125

SOLVABILITY OF NONLOCAL INVERSE BVP

3

By δ ≥ 0, and p(t) ≥ 0, from the latter relation it is clear that c = 0. Substituting the value of c = 0 in (2.8), we obtain that problem (2.6), (2.7) has only the trivial solution. The proof is complete. Theorem 2.3. Assume the following conditions are satisfied: δ ≥ 0, 0 < c(t) ∈ R1 C[0, T ], 0 ≤ p(t) ∈ C[0, T ], f (x, t) ∈ C(DT ), ϕ(x) ∈ C[0, 1], 0 f (x, t)dx = 0 (0 ≤ t ≤ T ), g(x, t) ∈ C(DT ), Z 1 g(x, t)dx = 0 (0 ≤ t ≤ T ), 0

hi (t) ∈ C 1 [0, T ] (i = 1, 2), h(t) ≡ h1 (t)g(x2 , t) − h2 (t)g(x1 , t) 6= 0 (0 ≤ t ≤ T ), and the compatibility conditions Z 1 ϕ(x)dx = 0, (2.9) Z hi (0) + δhi (T ) +

0 T

p(t)hi (t)dt = ϕ(xi )

(i = 1, 2) .

(2.10)

0

Then the problem of finding a classical solution of (2.1)–(2.5) is equivalent to the problem of determining functions u(x, t) ∈ C 2,1 (DT ), a(t) ∈ C[0, T ], and b(t) ∈ C[0, T ], satisfying equation (2.1), conditions (2.2) and (2.3), and the conditions (0 ≤ t ≤ T ),

ux (0, t) = ux (1, t) c(t)h0i (t)

= uxx (xi , t) + a(t)hi (t) + b(t)g(xi , t) + f (xi , t)

(2.11) (2.12)

for i = 1, 2; 0 ≤ t ≤ T . Proof. Suppose that {u(x, t), a(t), b(t)} is a classical solution of (2.1)–(2.5). Integrating both sides of (2.1) with respect to x from 0 to 1 gives Z Z 1 d 1 c(t) u(x, t)dx = ux (1, t) − ux (0, t) + a(t) u(x, t)dx dt 0 0 (2.13) Z 1 Z 1 + b(t) g(x, t)dx + f (x, t)dx (0 ≤ t ≤ T ). 0

0

R1

Under the assumptions 0 f (x, t)dx = 0 and obtain (2.11). Setting x = xi in (2.1) we obtain

R1 0

g(x, t)dx = 0 (0 ≤ t ≤ T ), we

c(t)ut (xi , t) = u(xi , t) + a(t)u(xi , t) + b(t)u(xi , t) + f (xi , t)

(i = 1, 2; 0 ≤ t ≤ T ).

(2.14)

Further, assuming hi (t) ∈ C 1 [0, T ] (i = 1, 2) and differentiating (2.5), we have ut (xi , t) = hi (t)

(i = 1, 2).

(2.15)

From (2.14), by (2.5) and (2.15), we conclude that the relation (2.12) is fulfilled. Now, assume that {u(x, t), a(t), b(t)} is the solution of (2.1)–(2.3), (2.11), (2.12). Then from (2.13), taking into account (2.11), we find Z Z 1 d 1 u(x, t)dx = a(t) u(x, t)dx (0 ≤ t ≤ T ). (2.16) c(t) dt 0 0

4


By (2.2) and (2.9), it is easy to see that Z Z 1 Z 1 u(x, T )dx + u(x, 0)dx + δ 1

Z =

T

Z

Z

T

u(x, 0) + δu(x, T ) +

0

1

u(x, t)dxdt

p(t)

0

0

0

EJDE-2017/125

0

p(t)u(x, t)dt dx

(2.17)

0 1

Z

ϕ(x)dx = 0.

= 0

Since, by Lemma 2.2, problem (2.16), (2.17) has only a trivial solution, it follows that Z 1

(0 ≤ t ≤ T ),

u(x, t)dx = 0 0

i.e. the condition (2.4) holds. Moreover, from (2.12) and (2.14) we find d (u(xi , t) − hi (t)) = a(t)(u(xi , t) − hi (t)) (i = 1, 2; 0 ≤ t ≤ T ). dt Using (2.2) and the compatibility conditions (2.10) we have Z T u(xi , 0) − hi (0) + δ(u(xi , T ) − hi (0)) + p(t)(u(xi , t) − hi (t))dt c(t)

(2.18)

0 T

Z

Z p(t)u(xi , t)dt − hi (0) + δhi (0) +

= u(xi , 0) + δu(xi , T ) + 0

T

p(t)hi (t)dt

0 T

Z

= ϕ(xi ) − hi (0) + δhi (0) +

p(t)hi (t)dt = 0

(i = 1, 2).

0

From this equality and (2.18), by Lemma 2.2, we conclude that conditions (2.5) are satisfied. The proof is complete. 3. Solvability of inverse boundary-value problem In [2], it is known that the system 1, cos λ1 x, sin λ1 x, . . . , cos λk x, sin λk x, . . . ,

(3.1)

where λk = 2kπ (k = 0, 1, . . .), is a basis for L2 (0, 1). Since the system (3.1) form a basis in L2 (0, 1), we shall seek the first component u(x, t) of classical solution {u(x, t), a(t), b(t)} of the problem (2.1)–(2.3), (2.11), (2.12) in the form u(x, t) =

∞ X

u1k (t) cos λk x +

k=0

∞ X

u2k (t) sin λk x

(λ = 2kπ),

k=1

where Z u10 (t) =

1

u(x, t)dx, 0

1

Z

u(x, t) cos λk x dx (k = 1, 2, . . .),

u1k (t) = 2 0

Z u2k (t) = 2

1

u(x, t) sin λk x dx (k = 1, 2, . . .). 0

(3.2)

EJDE-2017/125


5

Then applying the formal scheme of the Fourier method, for determining of unknown coefficients u1k (t) (k = 0, 1, . . . ) and u2k (t) (k = 1, 2, . . . ) of function u(x, t) from (2.1) and (2.2) we have

c(t)u010 (t) = F10 (t; u, a, b) c(t)u0ik (t)

(0 ≤ t ≤ T ),

λ2ik uik (t)

= Fik (t; u, a, b) (i = 1, 2; k = 1, 2 . . . ; 0 ≤ t ≤ T ), Z T uik (0) + δuik (T ) + p(t)uik (t)dt = ϕik (i = 1, 2; k = 0, 1, 2 . . .), +

(3.3) (3.4) (3.5)

0

where

Fik (t; u, a, b) = fik (t) + b(t)gik (t) + a(t)uik (t) (i = 1, 2; k = 0, 1, 2 . . .), Z 1 Z 1 f10 (t) = f (x, t)dx, g10 (t) = g(x, t)dx, 0

0

1

Z f1k (t) = 2

Z f (x, t) cos λk x dx,

f (x, t) sin λk x dx,

0

Z

1

f2k (t) = 2 0

1

g1k (t) = 2

Z g(x, t) cos λk x dx,

1

g2k (t) = 2

0

g(x, t) sin λk x dx, 0

Z ϕ10 =

1

Z ϕ(x)dx,

ϕ1k (t) = 2

0

1

ϕ(x) cos λk x dx, 0

Z

1

ϕ2k (t) = 2

ϕ(x) sin λk x dx 0

for k = 1, 2, . . . . Solving problem (3.3)–(3.5) we obtain

Z u10 (t) = (1 + δ)−1 ϕ10 − 0

Z

t

T

Z p(t)u10 (t)dt − δ 0

T

1 F10 (t; u, a, b)dt c(t)

1 + F10 (τ ; u, a, b)dτ , c(τ ) 0 R t λ2k Z T e− 0 c(s) ds uik (t) = ϕ − p(t)u (t)dt ik ik R T λ2k 0 1 + δe− 0 c(s) ds R T λ2k Z T R λ2k δe− 0 c(s) ds 1 − τt c(s) ds − F (τ ; u, a, b)e dτ ik R λ2 k ds − 0T c(s) 0 c(τ ) 1 + δe Z t R t λ2k 1 + Fik (τ ; u, a, b)e− τ c(s) ds dτ (i = 1, 2; k = 1, 2, . . .). 0 c(τ )

(3.6)

(3.7)

6


EJDE-2017/125

After substituting expressions u1k (t) (k = 0, 1, . . . ) and u2k (t) (k = 1, 2, . . . ) in (3.2), we obtain T

Z u(x, t) = (1 + δ)−1 ϕ10 −

Z 0

0

Z

t

+ 0

+

e−

λ2 k ds 0 c(s)

Rt

R λ2 k ds − 0T c(s)


Z

δe

T


e−

Rt 1 F1k (τ ; u, a, b)e− τ c(τ )

λ2 k ds 0 c(s)

RT


δe

−

RT

λ2 k ds c(s)

Rt

1 + δe−

k=1

λ2 k

0

λ2 k ds c(s)

Z

T

p(t)u1k (t)dt

0

0 1 + δe Z t Rt 1 + F1k (τ ; u, a, b)e− τ 0 c(τ ) ∞ n X

T

Z ϕ1k −

1 + δe

k=1

+

1 F10 (t; u, a, b)dt c(t)

1 F10 (τ ; u, a, b)dτ c(τ )

∞ n X

−

T

p(t)u10 (t)dt − δ

dτ (3.8)

o dτ cos λk x T

Z ϕ2k −

p(t)u2k (t)dt

0

Rt 1 F2k (τ ; u, a, b)e− τ c(τ )

1 + δe− 0 c(s) ds 0 Z t Rt 1 F2k (τ ; u, a, b)e− τ + 0 c(τ )

λ2 k ds c(s)

λ2 k ds c(s)

λ2 k ds c(s)

dτ

o dτ sin λk x.

Now, using (3.2) and (2.12) we have n a(t) = [h(t)]−1 (c(t)h01 (t) − f (x1 , t))g(x2 , t) − (c(t)h02 (t) − f (x2 , t))g(x1 , t) + +

∞ X k=1 ∞ X

λ2k u1k (t)(g(x2 , t) cos λk x1 − g(x1 , t) cos λk x2 )

(3.9)

o λ2k u2k (t)(g(x2 , t) sin λk x1 − g(x1 , t) sin λk x2 ) ,

k=1

n b(t) = [h(t)]−1 h1 (t)(c(t)h02 (t) − f (x2 , t)) − h2 (t)(c(t)h01 (t) − f (x1 , t)) + +

∞ X k=1 ∞ X

λ2k u1k (t)(h1 (t) cos λk x2 − h2 (t) cos λk x2 )

(3.10)

o λ2k u2k (t)(h1 (t) sin λk x2 − h2 (t) sin λk x1 ) ,

k=1

where h(t) ≡ h1 (t)g(x2 , t) − h2 (t)g(x1 , t) 6= 0

(0 ≤ t ≤ T ).

(3.11)

EJDE-2017/125


7

Taking into account (3.7), from (3.9) and (3.10) we obtain a(t) = [h(t)]−1 {(c(t)h01 (t) − f (x1 , t))g(x2 , t) − (c(t)h02 (t) − f (x2 , t))g(x2 , t) +

∞ X

λ2k

e−

h

λ2 k ds 0 c(s)

Rt

1 + δe−

k=1

RT 0

R λ2 k ds − 0T c(s) RT

λ2 k ds c(s)

T

Z

δe

−

∞ X

λ2k

Rt 1 F1k (τ ; u, a, b)e− τ c(τ )

λ2 k

k=1

e−

h

λ2 k ds 0 c(s)

RT 0

R λ2 k ds − 0T c(s) RT

λ2 k ds c(s)

T

Z

δe

−

λ2 k ds c(s)

dτ

(3.12) T

Z ϕ2k −

p(t)u2k (t)dt

0

Rt 1 F2k (τ ; u, a, b)e− τ c(τ )

λ2 k

λ2 k ds c(s)

i dτ (g(x2 , t) cos λk x1

Rt

1 + δe−

p(t)u1k (t)dt −

0

1 + δe− 0 c(s) ds 0 Z t Rt 1 + F1k (τ ; u, a, b)e− τ 0 c(τ ) − g(x1 , t) cos λk x2 ) +

T

Z ϕ1k −

1 + δe− 0 c(s) ds 0 Z t Rt 1 + F2k (τ ; u, a, b)e− τ 0 c(τ ) o − g(x1 , t) sin λk x2 ) ,

λ2 k ds c(s)

λ2 k ds c(s)

dτ

i dτ (g(x2 , t) sin λk x1

n b(t) = [h(t)]−1 h1 (t)(c(t)h02 (t) − f (x2 , t)) − h2 (t)(c(t)h01 (t) − f (x1 , t)) +

∞ X

λ2k

δe−

λ2 k ds 0 c(s)

Rt

1 + δe−

k=1

−

e−

h

RT 0

−

RT

λ2 k ds c(s)

RT 0

λ2 k ds c(s)

T

Z

λ2 k ds c(s)

0

Z

+

∞ X

λ2k

k=1

−

δe−

e−

h

Rt 1 F1k (τ ; u, a, b)e− τ c(τ )

λ2 k ds 0 c(s)

1 + δe− RT 0

−

λ2 k ds c(s)

RT

λ2 k ds c(s)

λ2 k ds c(s)

0

Z 0

T

λ2 k ds c(s)

dτ

i dτ (h1 (t) cos λk x2 (3.13) Z

ϕ2k −

T

p(t)u2k (t)dt

0

Rt 1 F2k (τ ; u, a, b)e− τ c(τ )

1 + δe Z t Rt 1 + F2k (τ ; u, a, b)e− τ 0 c(τ ) o − h2 (t) sin λk x1 ) . 0

λ2 k ds c(s)

Rt

RT

p(t)u1k (t)dt

0

1 + δe Z t Rt 1 + F1k (τ ; u, a, b)e− τ 0 c(τ ) − h2 (t) cos λk x1 ) 0

T

ϕ1k −

λ2 k ds c(s)

λ2 k ds c(s)

dτ

i dτ (h1 (t) sin λk x2

Analogously, the following lemma was proved in [13].

8


EJDE-2017/125

Lemma 3.1. If {u(x, t), a(t), b(t)} is a solution of (2.1)–(2.3), (2.11), (2.12), then the functions Z 1 u(x, t)dx, u10 (t) = 0 1

Z u1k (t) = 2

u(x, t) cos λk x dx

(k = 1, 2, . . .),

u(x, t) sin λk x dx

(k = 1, 2, . . .)

0

Z u2k (t) = 2

1

0

satisfy system (3.6), (3.7) on the interval [0, T ]. From Lemma 3.1 it follows that to prove the uniqueness of the solution of problem (2.1)–(2.3), (2.11), (2.12), it suffices to prove the uniqueness of the solution of (3.8), (3.12), (3.13). 3 Now, consider the space B2,T [13] consisting of functions of the form u(x, t) =

∞ X

u1k (t) cos λk x +

k=0

∞ X

u2k (t) sin λk x (λ = 2kπ)

k=1

in domain DT , where the functions u1k (t) (k = 0, 1, . . . ), u2k (t) (k = 1, 2, . . . ), are continuous on [0, T ] and satisfy the condition ∞ ∞ 2 1/2 X 2 1/2 X + λ3k ku2k (t)kC[0,T ] < +∞. ku10 (t)kC[0,T ] + λ3k ku1k (t)kC[0,T ] k=1

k=1 3 B2,T

The norm in the space

is

3 ku(x, t)kB2,T = ku10 (t)kC[0,T ] +

∞ X

λ3k ku1k (t)kC[0,T ]

2 1/2

k=1

+

∞ X

λ3k ku2k (t)kC[0,T ]

2 1/2

.

k=1 3 We denote by ET3 , the Banach space B2,T × C[0, T ] × C[0, T ] of vector functions z(x, t) = {u(x, t), a(t), b(t)} with norm 3 3 kz(x, t)kB2,T = ku(x, t)kB2,T + ka(t)kC[0,T ] + kb(t)kC[0,T ] .

3 It is known that B2,T and ET3 are Banach spaces Now consider the operator

Φ(u, a, b) = {Φ1 (u, a, b), Φ2 (u, a, b), Φ3 (u, a, b)} in the space ET3 , where Φ1 (u, a, b) = u ˜(x, t) ≡

∞ X

u ˜1k (t) cos λk x +

k=0

Φ2 (u, a, b) = a ˜(t),

∞ X

u ˜2k (t) sin λk x,

k=1

Φ3 (u, a, b) = ˜b(t)

and the functions u ˜10 (t), u ˜ik (t) (i = 1, 2; k = 0, 1, 2, . . . ), a ˜(t) and ˜b(t) are equal to the right-hand sides of (3.6), (3.7), (3.12), and (3.13) respectively.

EJDE-2017/125


9

Using simple transformations from (3.6), (3.7), (3.12), and (3.13) we obtain k˜ u10 (t)kC[0,T ] h ≤ (1 + δ)−1 |ϕ10 | + T kp(t)kC[0,T ] ku10 (t)kC[0,T ] 1/2 √ Z T 1 + δk |f10 (τ )|2 dτ + T ka(t)kC[0,T ] ku10 (t)kC[0,T ] kC[0,T ] T c(t) 0 Z T 1/2 i √ (3.14) + T kb(t)kC[0,T ] |g10 (τ )|2 dτ 0

h√ 1 +k kC[0,T ] T c(t)

Z

T

|f10 (τ )|2 dτ

1/2

0

+ T ka(t)kC[0,T ] ku10 (t)kC[0,T ] +

√

T kb(t)kC[0,T ]

Z

T

|g10 (τ )|2 dτ

1/2 i

,

0 ∞ X

Big(λ3k k˜ uik (t)kC[0,T ]

2 1/2

k=1 ∞ ∞ 2 1/2 X 2 1/2 √ X √ 3 ≤2 2 + 2 2T kp(t)kC[0,T ] λk |ϕik | λ3k kuik kC[0,T ] k=1

k=1

√ 1 + 2 2(1 + δ)k kC[0,T ] T c(t) √

Z 0

∞ T X

2 1/2 λ3k |fik (τ )| dτ

(3.15)

k=1

∞ X 2 1/2 √ 1 kC[0,T ] T ka(t)kC[0,T ] λ3k kuik (t)kC[0,T ] + 2 2(1 + δ)k c(t) k=1 Z TX ∞ 2 1/2 √ 1 + 2 2T (1 + δ)k λ3k |gik (τ )| dτ , kC[0,T ] kb(t)kC[0,T ] c(t) 0 k=1

k˜ a(t)kC[0,T ] n ≤ k[h(t)]−1 kC[0,T ] (c(t)h01 (t) − f (x1 , t))g(x2 , t) − (c(t)h02 (t)

− f (x2 , t))g(x1 , t) C[0,T ] +

∞ X

λ−2 k

1/2

k|g(x2 , t)| + |g(x1 , t)|kC[0,T ]

i=1

k=1

+ T kp(t)kC[0,T ]

2 X ∞ X i=1

+ (1 + δ)k

2 X ∞ hX

λ3k kuik (t)kC[0,T ]

λ3k |ϕik |

2 1/2

k=1

2 1/2

k=1

2 Z T X ∞ 2 1/2 √ X 1 kC[0,T ] T λ3k |fik (τ )| dτ c(t) 0 i=1

(3.16)

k=1

∞ 2 X 2 1/2 X 1 kC[0,T ] T ka(t)kC[0,T ] λ3k kuik (t)kC[0,T ] c(t) i=1 k=1 ∞ 2 Z T X X √ 2 1/2 io 1 + (1 + δ)k kC[0,T ] T kb(t)kC[0,T ] λ3k |gik (τ )dτ | , c(t) 0 i=1

+ (1 + δ)k

k=1

10


EJDE-2017/125

k˜b(t)kC[0,T ] n ≤ k[h(t)]−1 kC[0,T ] k(h1 (t)(c(t)h02 (t) − f (x2 , t)) − h2 (t)(c(t)h01 (t) − f (x1 , t))kC[0,T ] +

∞ X

λ−2 k

1/2

k|h1 (t)| + |h2 (t)|kC[0,T ]

2 X ∞ hX i=1

k=1

+ T kp(t)kC[0,T ]

2 X ∞ X i=1

+ (1 + δ)k

λ3k kuik (t)kC[0,T ]

2 1/2 λ3k |ϕik |

k=1

2 1/2

k=1

2 Z T X ∞ 2 1/2 √ X 1 kC[0,T ] T λ3k |fik (τ )| dτ c(t) 0 i=1

(3.17)

k=1

2 X ∞ 2 1/2 X 1 kC[0,T ] T ka(t)kC[0,T ] λ3k kuik (t)kC[0,T ] c(t) i=1 k=1 √ 1 + (1 + δ)k kC[0,T ] T kb(t)kC[0,T ] c(t) 2 Z T X ∞ 2 1/2 io X × λ3k |gik (τ )dτ | .

+ (1 + δ)k

i=1

0

k=1

Assume that the data for problem (2.1)–(2.3), (2.11) and (2.12) satisfy the following conditions (A1) ϕ(x) ∈ C 2 [0, 1], ϕ000 (x) ∈ L2 (0, 1), ϕ(0) = ϕ(1), ϕ0 (0) = ϕ0 (1), ϕ00 (0) = ϕ00 (1); (A2) f (x, t), fx (x, t), fxx (x, t) ∈ C 2 [0, 1], fxxx (x, t) ∈ L2 (DT ), f (0, t) = f (1, t), fx (0, t) = fx (1, t), fxx (0, t) = fxx (1, t) = 0 (0 ≤ t ≤ T ); (A3) g(x, t), gx (x, t), gxx (x, t) ∈ C 2 [0, 1], gxxx (x, t) ∈ L2 (DT ), g(0, t) = g(1, t), gx (0, t) = fx (1, t), gxx (0, t) = gxx (1, t) = 0 (0 ≤ t ≤ T ); (A4) δ ≥ 0, 0 ≤ p(t) ∈ C[0, T ], hi (t) ∈ C 1 [0, T ] (i = 1, 2), h(t) = h1 (t)g(x2 , t) − h2 (t)g(x1 , t) 6= 0 (0 ≤ t ≤ T ). Then from (3.14)–(3.17) we find that 3 3 ≤ A1 (T ) + B1 (T )ka(t)kC[0,T ] ku(x, t)kB2,T k˜ u(x, t)kB2,T 3 + C1 (T )ku(x, t)kB2,T + D1 (T )kb(t)kC[0,T ] , 3 k˜ a(t)kC[0,T ] ≤ A2 (T ) + B2 (T )ka(t)kC[0,T ] ku(x, t)kB2,T 3 + C2 (T )ku(x, t)kB2,T + D2 (T )kb(t)kC[0,T ] , 3 k˜b(t)kC[0,T ] ≤ A3 (T ) + B3 (T )ka(t)kC[0,T ] ku(x, t)kB2,T 3 + C3 (T )ku(x, t)kB2,T + D3 (T )kb(t)kC[0,T ] ,

(3.18)

(3.19)

(3.20)

EJDE-2017/125


11

where

1 A1 (T ) = (1 + δ)−1 2kϕ(x)kL2 (0,1) + 2δk kC[0,T ] kf (x, t)kL2 (DT ) c(t) √ 1 kC[0,T ] kf (x, t)kL2 (DT ) + 4 2kϕ000 (x)kL2 (0,1) +k c(t) √ 1 kC[0,T ] kfxxx (x, t)kL2 (DT ) , + 4 2T k c(t) 1 B1 (T ) = δ(1 + δ)−1 + 1 T k kC[0,T ] , c(t) √ C1 (T ) = (1 + δ)−1 + 2 2 T kp(t)kC[0,T ] , √ √ 1 Tk D1 (T ) = δ(1 + δ)−1 + 1 + 2 2 kC[0,T ] kgxxx (x, t)kL2 (DT ) , c(t) n A2 (T ) = k[h(t)]−1 kC[0,T ] (c(t)h01 (t) − f (x1 , t))g(x2 , t) − (c(t)h02 (t) ∞ 1/2 X

λ−2 k|g(x1 , t)| + |g(x2 , t)|kC[0,T ] − f (x2 , t))g(x1 , t) C[0,T ] + k k=1

h io √ 1 × 2kϕ000 (x)kL2 (0,1) + 2(1 + δ)k kC[0,T ] T kfxxx (x, t)kL2 (DT ) , c(t) ∞ X 1/2 (λ−2 B2 (T ) = k[h(t)]−1 kC[0,T ] k ) k=1

× k|g(x1 , t)| + |g(x2 , t)|kC[0,T ] (1 + δ)T k −1

C2 (T ) = k[h(t)]

kC[0,T ]

∞ X

1 kC[0,T ] , c(t)

1/2 k|g(x1 , t)| + |g(x2 , t)|kC[0,T ] kp(t)kC[0,T ] T, (λ−2 k )

k=1

D2 (T ) = k[h(t)]−1 kC[0,T ]

∞ X

1/2 (λ−2 ) k|g(x1 , t)| + |g(x2 , t)|kC[0,T ] k

k=1

√ 1 × (1 + δ)k kC[0,T ] T kgxxx (x, t)kL2 (DT ) , c(t) n A3 (T ) = k[h(t)]−1 kC[0,T ] h1 (t)(c(t)h02 (t) − f (x2 , t)) − h2 (t)(c(t)h01 (t) − f (x1 , t))kC[0,T ] +

∞ X

λ−2 k

1/2

k|h1 (t)| + |h2 (t)|kC[0,T ]

k=1

io h √ 1 × 2kϕ000 (x)kL2 (0,1) + 2(1 + δ)k kC[0,T ] T kfxxx (x, t)kL2 (DT ) , c(t) ∞ X 1/2 1 B3 (T ) = k[h(t)]−1 kC[0,T ] (λ−2 k|h1 (t)| + |h2 (t)|kC[0,T ] (1 + δ)T k kC[0,T ] , k ) c(t) k=1

C3 (T ) = k[h(t)]−1 kC[0,T ]

∞ X k=1

1/2 (λ−2 k|h1 (t)| + |h2 (t)|kC[0,T ] kp(t)kC[0,T ] T, k )

12


D3 (T ) = k[h(t)]−1 kC[0,T ]

∞ X

EJDE-2017/125

1/2 (λ−2 k|h1 (t)| + |h2 (t)|kC[0,T ] k )

k=1

√ 1 kC[0,T ] T kgxxx (x, t)kL2 (DT ) . (1 + δ)k c(t) From (3.18)–(3.20) we conclude that 3 k˜ u(x, t)kB2,T + k˜ a(t)kC[0,T ] + k˜b(t)kC[0,T ] 3 ≤ A(T ) + B(T )ka(t)kC[0,T ] ku(x, t)kB2,T

(3.21)

3 + C(T )ku(x, t)kB2,T + D(T )kb(t)kC[0,T ] ,

where A(T ) = A1 (T ) + A2 (T ) + A3 (T ), B(T ) = B1 (T ) + B2 (T ) + B3 (T ), C(T ) = C1 (T ) + C2 (T ) + C3 (T ), D(T ) = D1 (T ) + D2 (T ) + D3 (T ). Theorem 3.2. If conditions (A1)–(A4) and the condition (B(T )(A(T ) + 2) + C(T ) + D(T ))(A(T ) + 2) < 1

(3.22)

hold, then problem (2.1)–(2.3), (2.11), (2.12) has a unique solution in the ball K = KR (kzkET3 ≤ R ≤ A(T ) + 2) of the space ET3 . Proof. In the space ET3 , we consider the equation z = Φz,

(3.23)

where z = {u, a, b}, the components Φi (u, a, b) (i = 1, 2, 3), of operator Φ(u, a, b), defined by the right side of equations (3.8), (3.12) and (3.13). Consider the operator Φ(u, a, b), in the ball K = KR of the space ET3 . Similarly, with the aid of (3.21) we obtain that for any z1 , z2 , z3 ∈ KR the following inequalities hold kΦzkET3 3 3 ≤ A(T ) + B(T )ka(t)kC[0,T ] ku(x, t)kB2,T + C(T )ku(x, t)kB2,T + D(T )kb(t)kC[0,T ]

≤ A(T ) + B(T )(A(T ) + 2)2 + C(T )(A(T ) + 2) + D(T )(A(T ) + 2) < A(T ) + 2, (3.24) kΦz1 − Φz2 kET3 3 ≤ B(T )R ka1 (t) − a2 (t)kC[0,T ] + ku1 (x, t) − u2 (x, t)kB2,T

(3.25)

3 + C(T )ku1 (x, t) − u2 (x, t)kB2,T + D(T )kb1 (t) − b2 (t)kC[0,T ] .

Then by (3.22), from (3.24) and (3.25) it is clear that the operator Φ on the set K = KR satisfy the conditions of the contraction mapping principle. Therefore the operator Φ has a unique fixed point {z} = {u, a, b}, in the ball K = KR , which is a solution of equation (3.23); i.e. in the sphere K = KR is the unique solution of the systems (3.8), (3.12), (3.13). Then the function u(x, t), as an element of space 3 B2,T , is continuous and has continuous derivatives ux (x, t) and uxx (x, t) in DT .

EJDE-2017/125


13

Next, from (3.4) it follows that u0ik (t) (i = 1, 2; k = 1, 2 . . .) is continuous on [0, T ] and consequently we have ∞ X 1/2 (λk ku0ik (t)kC[0,T ] )2 k=1 ∞ √ hX 2 1/2 1 kC[0,T ] 2 λ3k kuik (t)kC[0,T ] c(t) k=1 i

+ kfx (x, t) + a(t)ux (x, t) + b(t)gx (x, t)kC[0,T ] L2 (0,1) < +∞ (i = 1, 2).

≤k

Hence we conclude that the function ut (x, t) is continuous in the domain DT . Further, it is possible to verify that equation (2.1) and conditions (2.2), (2.3), (2.11), (2.12) are satisfied in the usual sense. Consequently, {u(x, t), a(t), b(t)} is a solution of (2.1)–(2.3), (2.11), (2.12), and by Lemma 3.1 it is unique in the ball K = KR . The proof is complete. From Theorem 3.2 and Theorem 2.3, it follows directly the following assertion. Theorem 3.3. Suppose that all assumptions of Theorem 3.2, and the compatibility conditions (2.9), (2.10) hold. If Z 1 Z 1 f (x, t)dx = 0, g(x, t)dx = 0 (0 ≤ t ≤ T ) 0

0

then problem (2.1)–(2.5) has a unique classical solution in the ball K = KR . Acknowledgements. The authors would like to express their deep gratitude to the editorial team and the anonymous referees for the careful reading of the manuscript as well as their valuable comments and suggestions which helped to improve the present paper. References [1] Yu. Ya. Belov, A. Sh. Lyubanova, S. V. Polyntseva, R. V.Sorokin, I. V. Frolenkov; Inverse Problems of Mathematical Physics (in Russian), Siberian Federal University, Krasnoyarsk, 2008. [2] B. M. Budak, A. A. Samarskii, A. N. Tikhonov; Collection of Problems on Mathematical Physics (in Russian), Nauka, Moscow, 1979. [3] A .M. Denisov; Introduction to the Theory of Inverse Problems (in Russian), Moscow University Press, Moscow, 1994. [4] D. G. Gordeziani, G. A. Avalishvili; On the constructing of solutions of the nonlocal initial boundary value problems for one-dimensional medium oscillation equations, Matematiheskoe Modelirovanie 12(2000), No. 1, 94-103 (in Russian). [5] M. S. Hussein, D. Lesnic, M. I. Ismailov; An inverse problem of finding the time-dependent diffusion coefficient from an integral condition, Mathematical Methods in the Applied Sciences 39 (2) (2016), 963-980. [6] M. I. Ivanchov; Inverse problem for equations of parabolic type, Mathematical Studies, vol. 10, Lviv: VNTL Publishers, Monograph Series, 2003. [7] V. K. Ivanov, V. V. Vasin, V. P. Tanana; Theory of Linear Ill-Posed Problems and Its Applications (in Russian), Nauka, Moscow, 1978. [8] A. I. Kozhanov; On the solvability of the inverse problem of determining the thermal conductivity coefficient, Siberian Mathematical Journal 40(5) (2005), 841-856. [9] A.I. Kozhanov and L.S. Pulkina; On the solvability of boundary value problems with a nonlocal boundary condition of integral form for multidimensional hyperbolic equations, Differential Equations 42(2006), No. 9, 1166-1179.

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[10] S. V. Kirichenko; On a boundary value problem for mixed type equation with nonlocal initial conditions in the rectangle, Vestn. Samar. Gos. Tekhn. Univ. Ser. Fiz. -Mat. Nauki 32 (2013), No. 3, 185-189. [11] M. M. Lavrent’ev, V. G. Romanov, S. T. Shishatskii; TIll-Posed Problems of Mathematical Physics and Analysis (in Russian), Moscow, 1980. [12] M. M. Lavrent’ev, V. G. Vasil’ev, V. G. Romanov; Multidimensional Inverse Problems for Differential Equations (in Russian), Novosibirsk, 1969. [13] Y. T. Mehraliyev; On solvability of an inverse boundary value problem for a second order elliptic equation (in Russian), Bulletin of Tver State University, Series: Applied mathematics (2011), No. 23, 25-38. [14] Shin-ichi Nakamura; An an inverse problem for quasi-linear parabolic equations, Tokyo Journal of Mathematics 21 (1998), No. 2, 463-469. [15] A. I. Prilepko, D. S. Tkachenko; Properties of solutions of a parabolic equation and the uniqueness of the solution of the inverse source problem with integral overdetermination, Computational Mathematics and Mathematical Physics 43 (2003), No. 4, 537-546. [16] L. S. Pulkina; Solution to nonlocal problems of pseudohyperbolic equations, Electronic journal of differential equations 2014 (2014), No. 116, 1-9. [17] L. S. Pulkina; Boundary-value problems for a hyperbolic equation with nonlocal conditions of the I and II kind, Russian Mathematics 56(2012), No 4, 62-69. [18] V. G. Romanov; Inverse Problems of Mathematical Physics (in Russian), Moscow, 1984. [19] A. N.T ikhonov; On stability of inverse problems (in Russian), Doklady Akademii Nauk SSSR 39 (1943), No. 5, 195-198. Elvin Azizbayov Department of Computational Mathematics, Baku State University, Z. Khalilov Str. 23, Baku, AZ1148, Azerbaijan E-mail address: [email protected] Yashar Mehraliyev Department of Differential and Integral equations, Baku State University, Z. Khalilov Str. 23, Baku, AZ1148, Azerbaijan E-mail address: yashar [email protected]