A NONLOCAL PROBLEM FOR A THIRD ORDER PARABOLIC-HYPERBOLIC EQUATION WITH A SINGULAR COE CIENT Текст научной статьи по специальности «Математика»

DOI: 10.17516/1997-1397-2022-15-4-467-481 УДК 517.956

A Nonlocal Problem for a Third Order Parabolic-Hyperbolic Equation with a Singular Coefficient

Akhmadjon K. Urinov* Kobiljon S. Khalilov"

Fergana State University Fergana, Uzbekistan

Received 10.02.2022, received in revised form 03.04.2022, accepted 24.05.2022 Abstract. Non-classical problem with an integral condition for parabolic-hyperbolic equation of the third order is formulated and studied in this paper. The unique solvability of the problem was proved using the method integral equations. To do this the problem is equivalently reduced to a problem for a parabolic-hyperbolic equation of the second order with an unknown right-hand side. To study the obtained problem the formula of the Cauchy problem for hyperbolic equation with a singular coefficient and a spectral parameter was used. The solution of the first boundary value problem for the Fourier equation was also used.

Keywords: parabolic-hyperbolic equation, integral condition, uniqueness of the solution, existence of the solution, singular coefficient.

Citation: A.K. Urinov, K.S. Khalilov, A Nonlocal Problem for a Third Order Parabolic-Hyperbolic Equation with a Singular Coefficient, J. Sib. Fed. Univ. Math. Phys., 2022, 15(4), 467-481. DOI: 10.17516/1997-1397-2022-15-4-467-481.

Recently, problems with integral conditions for partial differential equations attract considerable interest. This is primarily due to the fact that problems with integral conditions have numerous applications in science and technology. Integral conditions appear when boundary conditions may not be available but the average value of the sought quantity is known. Conditions of this kind can appear in the mathematical modelling of phenomena in plasma physics, heat propagation, moisture transfer in capillary-porous media, demography processes and mathematical biology.

A problem with an integral condition was first considered by Cannon [1] and Kamynin [2] for the heat equation. Following these works, numerous problems with an integral condition for partial differential equations of the second order of parabolic, hyperbolic, elliptic types on the plane There are a number of works devoted to the study of problems with an integral condition for second order partial differential equations of mixed type. For example, problems with an integral condition for an elliptic-parabolic equation in a domain consisting of a rectangle and a semicircle were formulated and studied [26]. Problems for an elliptic-hyperbolic equation in a rectangular domain were considered [27].

Problems that are close to the subject of this work were considered in [3-14, 28-34]. Problems with an integral condition for a second order parabolic-hyperbolic equation with characteristic

* urinovak@mail.ru https://orcid.org/0000-0002-9586-1799 txalilov_q@mail.ru https://orcid.org/0000-0001-6934-6008 © Siberian Federal University. All rights reserved

line of type changing were considered in a rectangular domain [28]. Problems with an integral condition for a second order parabolic-hyperbolic equation with non-characteristic line of type changing were considered [29,30]. Parabolic-hyperbolic equations of the second order with characteristic line of type changing in the domain consisting of a rectangle and a characteristic triangle was considered and some problems with an integral condition in the domain of parabol-icity of the equation were studied [31,32]. Problems similar to equations of parabolic-hyperbolic type of the third order were studied [33,34] where a model equation [33] and an equation with a spectral parameter [34] were considered.

A non-local problem with an integral condition for an equation of mixed parabolic-hyperbolic type of the third order with a singular coefficient in the hyperbolic part is formulated and studied in this paper.

1. Formulation of the problem

Let D be a finite simply connected domain bounded for y > 0 by lines x = 0, x = 1, y =1 and for y < 0 it is bounded by straight lines x + y = 0, xy = 1, Di = D n {(x, y) : y > 0}, D2 = D n{(x,y) : y< 0}, Do = D n {(x,y) : y = 0}. Let us consider in domain D the following equation

where

L

(d/dx) Lu = 0, (1)

' Li = (d2/dx2 ) - (d/dy ) -A2, (x, y) G Di,

L = (d2/dx2 ) - (d2/dy 2) - (2ß/y) (d/dy) + A2, (x,y) G D2,

¡3, X1 and A2 are given real numbers such that 0 < 3 < (1/2).

The equation Lu = 0 belongs to parabolic type in domain D1 and it belongs to hyperbolic type in domain D2 and segment D0 is the line of type changing of the equation. The following problem with an integral condition for equation (1) is studied in domain D.

Problem 1. Find a function u (x,y) with the following properties: 1) u (x,y) G C (D), ux, uy G C (D U D3); 2) u (x, y) satisfies equation (1) in D1UD2; 3) u (x, y) satisfies the following conditions

u (0, y) = yi (y), u (1,y) = y (y), 0 < y < 1; (2)

/ u (x,y) dx = (y), 0 < y < 1; (3)

■Jo

= ^2 (x), 0 < x < (1/2); (4)

u(x,y)\D3 = (x), dn

D3

lim uy (x, y) = lim (— y)2ß uy (x,y), 0 <x< 1, (5)

y^+0 y^-0

where D3 = {(x, y) : y = —x, 0 ^ x ^ 1/2 }, n is the inner normal to D3, and <fj (y), (x), ^2 (x) are given functions such that ^ (y) G C1 [0,1], j = 1,3; ^i(x) G C1 [0,1/2] n C2 (0,1/2), ^2 (x) G C [0,1/2] n C1 (0,1/2), ^(0) = ^ (0) and ^1' (x), ^2 (x) G L1 [0,1/2].

2. Pleminaries

To study the considered problem the following operators are used [35], [36]:

1

Dlx [f (x)] =

= Jr(-7)

fx

/ (x - t)-Y-1f (t) dt, y< 0,

Jo

lDY-lf (x) ^

y e (0,1),

A 0X 2 [g (x)] = g (x) -I g (t)( H g-J

t\m d

\MW x (x - t)

B,

m,\2

x

fx ,x\ l-m d r /_

[g (x)] = g (x) + yo g (t)[t) oxJ0 [\A2^t (t - x)

dt, m = 0,1,

dt, m = 0, 1,

where Jv(x) is the Bessel function of the first kind [37], r(z) is the gamma function [38]. These operators have the following properties

Lemma 1 ([35]). For all f (x) G C (0,1) fl L [0,1] the following equality is valid:

DY0xD- f (x) = f (x), Y > 0. (6)

Lemma 2 ([35]). If 0 < 2f3 < 1 and x-? f (x) G C (0,1) f L [0,1] then the equality

Dßx x2ß-1Dß-1x-ß f (x) = xß-1Dlß-1f (x)

(7)

is valid.

Lemma 3 ([36,39]). For all functions g(x) G C(0,1) f L[0,1] the following equalities are hold:

Am/2{BmX'X2 [g (x)]}= g (x), BmXM {Am/2 [g (x)]}= g (x), m = 0,1.

(8)

Lemma 4 ([39,40]). If v (x) e C(0a) (0,1), a> ß > 0, [x (1 - x)] 2ß v (x) e L [0,1] then the following equality

( \ xß—i rx

A^lxß-1Dlß-1xßBlxX2 [v (x) x-^ = r(1 - 2ß) yo v (t) (x - t)-2ß J-ß [\A2 \ (x - t)] dt (9)

is valid.

In addition to the above lemmas the following statements are also used.

Lemma 5. Any solution of the equation (d/dx) L2u = 0 in domain D2 can be represented in the form

u (x,y) = v (x,y) + u (y), (10)

where v (x, y) is the general solution of the equation

2ß

Vxx - Vyy - yVy + A2V = 0, (x,y) e D2, and u (y) is an arbitrary function from the class C [-1/2 , 0] n C2 (-1/2 , 0).

x

Proof. Let u(x,y) be a solution of the equation (d/dx) L2u = 0. Integrating this equation in domain D2 with respect to x, we obtain

23 2

uxx - uyy - yuy + A\u = uo (y), (x,y) G D2, (12)

where u0 (y) is an arbitrary function from the class C (-1/2,0) n L [-1/2,0]. It is easy to verify that any function of the following form

, N iy f1 (y) f2 (n) - fi (n) f2 (y) , w + ^ r 1 /O m

u (y) = J -An)-uo (n) dn, a = const G [-1/2, 0] ,

satisfies equation (12) and f (y) = (-y)1/2-lJ1/2-i(-|A2|y), f2(y) = (-y)1/2-l3J/3-1/2(-\A2\y) are linearly independent solutions of the homogeneous equation

u" (y) + (23/y) u' (y) - A2u (y) = 0, y G (-1/2, 0), (13)

A (y) = f1 (y) f '2 (y) - f2 (y) f ' 1 (y) =0 is the Wronskian of functions f1(y) and f2(y). Therefore, equality (10) is true.

Now, let the function u(x,y) be representable in form (10). Then, substituting (10) into (d/dx) L2u and taking into account that v(x,y) is a solution of equation (11), we immediately obtain the equality (d/dx) L2u = 0. Lemma 5 is proved. □

The following lemma can be proved in a similar way.

Lemma 6. Any solution of the equation (d/dx) L1u = 0 in D1 can be represented as

u (x,y)= w (x,y)+ 5 (y), (14)

where w (x, y) is the general solution of the equation

wxx - Wy - Xfw = 0, (15)

and 5 (y) is an arbitrary function from the class C [0,1] n C1 (0,1).

3. Study of the problem

Let us prove the unique solvability of the problem 1. To do this representation (10) of the solution of the equation (d/dx) L2u = 0 is used. Obviously, the function

"l(y) = (-y)1/2-^ [AJ-1/2-?(-№) + BJ^-!/2(-\\2\y)]

is a solution of equations (11) and (13) where A and B are arbitrary constants. Taking this into account when considering Problem 1, one can assume without loss of generality that arbitrary function w(y) in representation (10) satisfies the following conditions

" (0) = 0, lim(-y)2^"' (y) = 0. (16)

Otherwise, rewriting function (10) in the form u(x,y) = [v(x,y) + "i(y)j + ["(y) — "i(y)j, one can distribute A and B so that the new function " (y) = " (y) — "1 (y) satisfies conditions (16). Let u (x,y) be a solution of Problem 1. Taking into account the conditions of the problem, the following notation and assumptions are we introduced

lim u (x, y) = lim u (x, y) = t (x), 0 ^ x ^ 1; (17)

y^+0 y^-0

lim uy (x,y)= lim (-y)2ß uy (x,y) = v (x), 0 <x< 1; (18)

y^+0 "" y^-0

t (x) G C1 [0,1] f C3 (0,1), v (x) G C [0,1] f C2 (0,1), v' (x) G L [0,1]. (19)

If equalities (10), (16)-(19) are taken into account then function v (x,y) in domain D2 can be treated as a solution of the modified Cauchy problem for equation (11) [39,41]:

v (x, y) = Yi t t [x + y (1 - 2t)] T23-213-1 [-2\\2 \yT] dt-Jo

- Y2(-y)1-2? i v [x + y (1 - 2t)] T-2?~I-? [-2\\2\yT] dt, (20) Jo

where T ), yi = r(20)/r2(0), Y2 =r(1-20)/r2(1 -0), I3 (x) = r(1+ 0) (x/2)-3I3 (x),

and I3 (x) is the modified Bessel function [37].

Substituting the function v(x,y) from (20) into (10), we find the function u(x,y) as

'(x,y) = Yi f t [x + y (1 - 2t)] T23-21?-i [-2\\2\yT] dt-

o

- Y2(-y)1-23 i v [x + y (1 - 2t)] T-23~I-3 [-2\\2\yT] dt + w (y). (21)

o

After satisfying condition u (x, y)\D3 = u (x, -x) = ^i(x), x G [0,1/2], we obtain

1 i t (2xt) T23-2I3-i [2\A2\xT] dt-

o

- Y2xi-23 i v (2xt) T-23I-3 [2\A2\xT] dt + w (-x) = ^(x), x G [0,1/2] . (22)

o

Differentiating equality (22) with respect to x and using the equality (d/dx) JY (x) = - (x/2(y + 1)) JY+i (x), we obtain

Y1 t t' (2xt) T2ß-2Iß-1 (2\A2\xT) 2tdt-Y1 t t (2xt) T2ß-2 -ß (2\A2\xT) 2\A2\Tdt-

0 0 2ß

-Y2 (1 - 2ß) x-2ß i v (2xt) T-2ßI-ß (2\A2\xT) dt-l2x1-2ß i v' (2xt) 2tT-2ßI-ß (2\A2\xT) dt+

00

+ Y2x1-2ß 11 v (2xt) T-2ß 1-ß (2\A2\xT) dt - u' (-x) = ^(x), x e [0,1/2]. (23)

Jo 2 - 2ß

o

Let us calculate now (d/dn ) u\D3. First, we find ux and uy:

r-i

ux = Y1 f t' [x + y (1 - 2t)] T2ß-2Iß-1 -2\A2 \yT] -

0

- Y2(-y)1-2ß i v' [x + y (1 - 2t)] T-2ßI-ß (-2\A2\yT) dt,

0

uy = Y1 i t' [x + y (1 - 2t)] (1 - 2t) T2ß-2Iß-1 (-2\A2\yT) dt-1

- yJ t [x + y (1 - 2t)] T2ß-2 (-2]A2lyT)~Iß (-2\A2\yT)(-2\A2\T) dt+

0 2ß

+ 72 (1 — 2£) (—y)-2p f v [x + y (1 — 2t)] T-2pI-p (—2\\2\yT) dt—

0

— 72(—y)1-2P [ v' [x + y (1 — 2t)](1 — 2t) T-2pI-p (—2\\2\T) dt+

0

+72(—y)1-2P 0 v [x + y (1 — 2t)] T-2p (—Ii-p (—2\\2\yT) (—2\\2\T) dt + " ' (y). Then, according to the formula

(d/dn ) u\D3 = (ux cos (n, x) + uy cos (n, y)) \D3 = (V2/2j (ux + uy) D3 and the second of boundary conditions (4), we obtain

Yi

i t' (2xt) (2 - 2t) T2ß-2Iß-1 (2\A2\xT) dt+ Jo

+yi f t (2xt) t2ß-2 (2AlxT)I-ß (2\A2\xT)(2\A2\T) dt+ jo 2ß

+72 (1 — 23) x-2p i v (2xt) T-2pI-p (2\X2\xT) dt—

—72x1-2p i v' (2xt) (2 — 2t) T-2pI-p (2\A2\xT) dt—

—72x1-2p0 v (2xt) T-2p|(A—T))1— (2\A2\xT) (2\A2 \T) dt + " ' (—x) =

(x), x G [0,1/2] . Combining this relation term by term with relation (23), we obtain

271 i t' (2xt) T2p-2Ip-1 (2\A2\xT) dt — 2l2x1-2p i v' (2xt) T-2pI-p (2\A2\xT) dt =

00

= V1(x) + V2Mx), x G [0, 1/2] . Using the change of variable 2x = z G [0,1] in the last relation, we obtain

(24)

Yi

/i i 2 ß Çi

t' (zt) T2ß-2Iß-i (\A2\zT) dt - Y2(iy J v' (zt) T-2ßI-ß (\A2\zT)

dt =

= V1(z/2) + V2^2(z/2), z G [0,1] . If we replace zt by £ then £ G [0, z], t = £/z, 1 — t = (z — £)/z, dt = d£/z. Then, taking into account T =y/t (1 — t) = (1/z)V £ (z — £), I7 (\ A2\zT ) = A W £ (z — £)

we have

Yi z

i-2ß t ' (£)[£ (z - £)]ß-i~I ß-i

\A2\V£ (z - d d£-

-Y222ß-i v' (£)[£ (z - £)]-ß I-ß \\A2W £ (z - £)] d£ = $ (z), z G [0,1],

(25)

where $ (z) = (z/2) + V2^2 (z/2).

Let us denote the first and second integrals in the right-hand side of equality (25) by l1 and l2 and transform them. By virtue of the equality

d

(z - £)ß-i I ß-i \A2\V£(z-£) = —I (z - t)ß-i Jo \A2\V£(£-t)

dt,

o

o

z

£

which can be easily proved using the expansion of functions Ip-i(x) and J0(x) in power series, we rewrite l\ as

h = £tdzf (z - t)p-i Jo [^i vûï-t] dt} ¿t.

Integrating by parts the integral over t and performing the external operation (d/dz), we have

"(z - t)3-i J

Hence, changing the order of integration in the integral and changing the specification of variables, we find

h = £ T'(o^-1 {(z - o13-1 + f (z - t)ß-1 dj [1A21 vW-ö] ^}

ii = J* (z - e)3-^ t'(oz3-1 + 0 t '(t)t3-i d-J [m v'ft-o] dt^dt

By virtue of notation Dqx and B^^2 we obtain from the last relation

li =r(3)D0-/B^2 [t'(z)z3-1 ]. (26)

Similarly, we find

I2 = r(1 - 0)D3-1 B0'ZX2 [v'(z)z-3]. (27)

Due to (26) and (27), relation (25) can be rewritten as

Yir (0) x1-23D-3B0xX2 [t' (x) x3-1] - Y2223-1r (1 - 0) D^B^2 [v' (x) x-3] = $ (x). (28)

From here, applying the operator Al'x2 D3xx23-1 and taking into account (6)-(9), we have

t' (z) = Y3 i v' (t) (z - t)-23 J-3 [\A2\ (z - t)] dt + F (z), (29)

■Jo

where Y3 = 223-1r (0) /r (1 - 0) r (20), F(x) = r (0) x1-3A1^2D3x [x23-1$ (x)] /r (20). Integrating (29) with respect to z from 0 to x, we obtain

t(x) = t(0) + Y3 i v' (t) M (x - t) dt + F1 (x), (30) Jo

x

where F1 (x) = f F (z) dz,

0

M x -1) = f' (z - t)-23>J-f [M (z - t)] dz = V r(1 - A'-')' ( A2 y " - t)1+'k°2'.

v ' Jt ' 3U 2K n f=0 k!r (1 + k - 0) V 2) 1 + 2k - 20

Applying the integration by parts from relation (30), we find

{x —

t (x) = (0) - Y3v (0) M (x) + F1 (x) + Y3 v (t) (x - t)-23 J-3 [A \ (x - t)] dt. (31)

Jo

Multiplying both sides of (23) by x23 and then setting the limit as x ^ 0, we obtain lim x23^1(x) = -y2(1 - 20)v(0) / T-23dt = -y2(1 - 20)v(0) / [t (1 - t)]-3dt = -v (0).

x^0 Jo Jo

Taking the last relation into account, we obtain from (31) that

f x

T (x)= Y3 v (t)(x - t)-2 J_p [|A2| (x - t)] dt + F2(x),

o

(32)

where F2(x) = ^1 (0) + y3M (x) lim x21 ^1 (x/2) + F1 (x).

Introducing the notation F3(x) = y-1 [t(x) - F2(x)], we obtain from (32) an integral equation with respect to v(x):

f x

/ v(t)(x — t)-2ß J_p [|A2| (x — t)] dt = F3(x).

0

Solving this integral equation [39], we obtain the relation between unknown functions t (x) and v (x) which is brought to D0 from domain D2

v (x) = Y4C^2 [t (x) — F2 (x)], 0 < x < 1, where Y4 = Y3-1 r-1(1 — 2ß) = 21-2ßr(1 — ß)r(2ß)/r(ß)r(1 — 2ß),

C1^2 [q (x)] =

r(2ß)

<Af

dx 0

1 d rx Jß [|A2| (x — t)]

0 ( x — t)

1-2ß

q (t) dt +

A2

4(ß + ß2).

Jß+1 [|A2|(x — t)]

(x — t)-2ß

(33)

q ( t) dt .

Performing the same transformations that we do to obtain (28) from (24), we have from (22)

that

Y1r(3) x1-2pD-xB^2 [t (x) xl-1} - Y2221-1 r(1 - 3) Dl-B^2 [v (x) x-1 ] =

= ^1 (x/2) - u (-x/2), 0 < x < 1. (34)

Further, taking into account (8) and x1-21D-1 xl-1D2x,-1xl3g(x) = Dq-1 g(x) (which can be verified by using the operator Dlxx213-1) and introducing the notation x1 g(x) = f (x) (in this case it takes form (7)), we have

x1-21 D-xB^2 {A^2x/-1D20l-1xlB^ [v (x) x-]} =

= x1 -2ßD-ßxß-1 Dlß-1 xßBß2 [v (x) x-ß] = Dß-1 Bß2 [v (x) x-ß] .

(35)

Then, substituting the function t(x) from (32) into (34) and taking into account (9), (35) and (26), we find the unknown function u (x) in the form

x — 2

M 2) — Y1x1-2ß J F2 (£) [£ (x — 0]ß-1Iß-1

IA2W e (x — o'

d£, 0 < x < 1.

Setting the limit at y ^ +0 in equation (1) and in boundary conditions (2), (3) and taking into account notations (17), (18), we obtain the second relation between unknown functions t (x) and v (x), which is brought to D0 from domain D1, and conditions for the function t (x):

t'' (x) — A1t (x) — v (x) = k, 0 < x < 1,

t (0) = V1 (0), T (1) = ^2 (0), f T (x) dx = (0),

0

(36)

0

x

where k is an unknown number.

Substituting the expression for v (x) from (33) into (36), we obtain integro-differential equation for the unknown function t (x):

t '' (x) - A\t (x) - YiC1^2 [t (x)] = k - YiC1^2 [F2 (x)], 0 <x< 1. (38)

Therefore, the unknown function t (x) is a solution of problem {(38), (37)}. From this problem we find the function t (x). First, we prove uniqueness of the solution of problem {(38), (37)}. Let us consider the homogeneous problem

t'' (x) - X\t (x) - lACl£2 [t (x)] = k, (39)

t (0)=0,t (1) = 0, i t (x) dx = 0. (40)

Jo

Multiplying (39) by the function t (x) and integrating the obtained relation over segment [0,1], we obtain

/ t (x) t'' (x) dx - A2 t2 (x) dx - 74 / t (x) COx^2 [t (x)] dx = k t (x) dx.

Jo Jo Jo Jo

Hence, integrating the first integral by parts and then taking into account (40) and t' (x) e C [0,1], we have

[ [t' (x)fdx + A\ i t2 (x) dx + y4 [ t (x) C^2 [t (x)] dx = 0. (41)

Jo Jo Jo

Here, the notation

r- 1 (1 - 20)Clo'X2 [t (x)]= M (x). (42)

is introduced. Hence, taking into account the conditions t (0) =0 and t' (x) e C [0,1], we find the function t (x) as follows [39]:

i x

t (x)= (x - t)-213 J-j3 [|A2| (x - t)] n (t) dt. (43)

o

Substituting (42) and (43) into (41), we obtain

»1 x

r{[r' (x)]2 + Mr2 (x)\dx + »(z)dziXJ-ß [IX2j (x " ¡^ (t) dt =0. (44)

Jo k J J0 J0 (x — t) H

It was proved that the last integral in (44) is non-negative [39]. Then, this relation implies that t' (x) = 0, i.e., t (x) = const, x e (0,1). Taking into account that t (x) e C [0,1] and t (0) = t (1) = 0, we have t (x) = 0, x e [0,1]. Therefore, the homogeneous problem {(39),(40)} has only a trivial solution. It follows from this that if there exists solution of problem {(37),(38)} then it is unique.

Now, we prove the existence of the solution of this problem. We rewrite (38) as t'' (x) = p (x), where

p (x) = k + A2t (x) + Y4CoXX2 [t (x) - F2 (x)] . (45)

The solution of this equation that satisfies the first two conditions of (37) is defined as follows [44]

t (x) = pi (0) (1 - x) + ^2 (0) x + i p (t) G (x, t) dt, (46)

o

where G (x,t) = x (t — 1) for x ^ t, G (x, t) = t (x — 1) for x ^ t.

Substituting (45) into (46) and then integrating the resulting relation over x by [0,1] and

1 1

taking into account the last of conditions (37) and f f G (x,t) dxdt = — (1/12), we find the

0 0

unknown number k

k = —12^3(0) + 6^i(0) + (0) + 12 J J G(x,t){X2t(t)+ jC?2 [t(t) — F2(t)]} dtdx. Substituting k into (45) and (46), we obtain after some transformations that

t (x) = 0 Q (x, t^ XIt (t) + lAChX2 [t (t)]} dt + pi (x), (47)

where Q (x, t) = G (x, t) + 3xt (x — 1) (t — 1),

pi(x) = ^i(0) (1 — 4x + 3x2) — v>2(0)x(2 — 3x) + 6^(0)x(1 — x) — f Q(x,t)Cif2 [F2(t)] dt.

Jo

Taking into account the form of the operator C^2 and the equality

J Q (x,t)dt Jq t (t)(t - z)2ß-\Jß [|A2| (t - z)] dzdt =

i' 1 d i't

= - J0 —Q (x>t) dt t (z)(t - z)2ß-1 Jß p2| (t - z)] dz =

0 JO

r 1 r 1 d

- t (z) dz (t - z)2ß-1 Jß [|A21 (t - z)] -Q (x,t) dt,

we obtain an integral equation for the unknown function t (x):

t (x) - Qi (x,z) t (z) dz = pi (x), x e (0,1), (48)

Jo

where

/1 f)

(t - z)2ß-1 Jß [| A21 (t - z)] -Q (x, t) dt+

iaa2

i Q (x,t)(t - z)2ß Jß [|A2| (t - z)] dt.

J z

2(1+ /3)r(1 + 20)

It is easy to verify that Qi (x,z) e C (0 < x,z < 1) n C2 (0 < x,z < 1, x = z) and pi (x) e C [0,1] n C2 (0,1). Therefore, (48) is the Fredholm integral equation of the second kind [45]. It is equivalent to problem {(37), (38)}. The homogeneous integral equation corresponding to equation (48) is equivalent to homogeneous problem {(39), (40)}. Since, the last problem has only a trivial solution the homogeneous integral equation corresponding to (48) has also only a trivial solution. Then, according to alternative of Fredholm [45], the solution of non-homogeneous integral equation (48) exists and it is unique.

Once the function t (x) is found from (48), the function v (x) can be found from (33). Substituting t (x), v (x) and w (x) into (21), we find a solution of Problem 1 in domain D2.

Now, we turn to the study of Problem 1 in domain Di. Here, we have the problem 1': find the function u (x,y) that satisfies equation (1) in domain Di and conditions (2), (3), u (x, 0) = t (x), 0 ^ x ^ 1, where t(x) is the function defined in (48).

We will prove the existence and uniqueness of the solution of problem 1'. Let u (x,y) be a solution of problem 1'. To study this problem we use representation (14) of the solution of the equation (d/dx ) Liu = 0. In this case, without loss of generality, one can assume that ¿(0) = 0. If we temporarily assume that S (y) is a known function then problem 1', due to (14) and S (0) = 0, is equivalent to the problem of finding a solution of equation (15) in domain D\ that satisfies the conditions

w (0, y) = (y) - S (y), w (1,y) = ?2 (y) - S (y), 0 < y < 1;

w (x, 0) = t (x), 0 ^ x ^ 1,

/ w (x,y) dx = (y) - S (y), 0 < y < 1. Jo

(49)

(50)

(51)

Then function w (x,y) is a solution of the first boundary value problem for equation (15) in domain D\ with the boundary conditions (49) and (50), and it can be represented as [42]

t fy

•(x,y) = T (£) e-x2y G (x,y; £, 0) d£ + fa (n) - S (n)] e-xl(y-n)Gç (x,y;0,n) dV-

oo

iy 2 - / [<P2 (n) - S (n)] e-Xl(y-n)Gç (x, y; 1, n) dn

o

(52)

where G (x, y; £, n) is the Green's function of the first boundary value problem [43] for the equation

wxx - wy = 0:

G (x,y; £,n) =

1

2Vn (y - n) n=

J2 iexp

(x - £ + 2n)

exp

(x + £ + 2n) 4 (y - n)

». (53)

4 (y - n)

Substituting w (x,y) from (52) into condition (51), we obtain after some transformations that

where

f1 fy 2

s (y) -/ / S (n) e-Xl(y-n) [Gç (x,y;0,n) - Gç (x,y;1,n)] dndx = g(y), 0 < y < 1,

oo

g (y) = V3 (y) - 0 { 0 t (£) e-X2yG (x, y; £, 0) d£+

/y 2 ty 2 1

(n) e-Xi(y-n)Gç (x, y; 0, n) dn - J (n) e-Xi(y-n)Gç (x, y; 1, n) dn> dx.

(54)

Using (53), it is easy to verify that

/ Gç (x,y;0,n) dx = - Gç (x,y;1,n) dx = K (y,n),

oo

(55)

where

K (y, n) =

+

Vn (y - n) Vn (y - n)

exp

n=1

r 2 n2 ' (2n - 1)2

y - n_ - exp 4 (y - n)

After changing the order of integration over variables x and n, and then taking into account (55), we obtain from (54) the Volterra integral equation of the second kind with respect to S (y):

î y 2

S(y) - e-Xl(y-n)Ki (y,n) S (n) dn = gi(y),

o

1

2

where K1 (y, n) = 2K (y, n),

f y f 1 i'1

91(y) = (y) - [Mn)+ Mn)] e-x2(y-n)K(y,n)dn - / T^e-^ G(x,y; 0)dÇd Jo Jo Jo

oo

Obviously, the kernel Ki(y,n) has a weak singularity. Using the properties of functions t(x), yi(y), y2(y) and (y), it is easy to show that gi(y) e C [0,1] n Cl(0,1). Therefore, equation (56) has a unique solution in this class [45]. Solving it, we find function S(y). Thus, function w(x,y) is defined by (52) in domain Di. Then solution of Problem 1 (Problem 1') in domain Di is determined by expression (14). The study of Problem 1 is completed.

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Нелокальная задача для одного параболо-гиперболического уравнения третьего порядка с сингулярным коэффициентом

Ахмаджон К. Уринов Кобилжон С. Халилов

Ферганский государственный университет Фергана, Узбекистан

Аннотация. В настоящей работе сформулирована и исследована неклассическая задача с интегральным условием для параболо-гиперболического уравнения третьего порядка. Методом интегральных уравнений доказана однозначная разрешимость поставленной задачи. При этом поставленная задача эквивалентно сведена к задаче для параболо-гиперболического уравнения второго порядка с неизвестной правой частью. При исследовании последней задачи использованы формулы решения задачи Коши для гиперболического уравнения, имеющего сингулярный коэффициент и спектральный параметр, а также решения первой краевой задачи для параболического уравнения Фурье.

Ключевые слова: параболо-гиперболическое уравнение, интегральное условие, единственность решения, существование решения, сингулярный коэффициент.