On a Problem for an elliptic type equation of the second kind with a conormal and integral condition Текст научной статьи по специальности «Математика»

On a problem for an elliptic type equation of the second kind with a conormal and integral condition

B.I. Islomov1, A.A. Abdullaev2

1 National University of Uzbekistan named after Mirzo Ulugbek, Differential equations and Mathematical physics, 100174, Tashkent, Uzbekistan 2 Tashkent Institute of Irrigation and Agricultural Mechanization Engineers, Higher mathematics, Str. Kari Niyaziy, 39, 100000, Tashkent, Uzbekistan

islomovbozor@yandex.ru, akmal09.07.85@mail.ru

DOI 10.17586/2220-8054-2018-9-3-307-318

In this paper, we prove the uniqueness of a solution of the boundary value problem for an elliptic type equation of the second kind with the conormal and integral condition.

Keywords: elliptic type equation of the second kind, boundary value problems, boundary problems with Poincare condition, integral equation.

Received: 11 January 2018 Revised: 3 March 2018

1. Introduction

The Poincare problem with the conormal derivative are studied by Tricomi, Lavrentyev-Bitsadze and Geller-stedt [1-5].

Boundary value problems with the conormal derivative for the elliptic type equation with one and two lines of degeneration of the first kind is considered when on the lines of degeneration a function or its derivative is given by M. A. Usanatashvili [6], M. S. Salokhitdinov and B. Islomov [7,8], H. Islomov [9].

In this paper, we prove the unique solvability of a boundary value problem with the conormal and integral condition for the elliptic type equation of the second kind.

Stationary processes of a various physical nature (oscillations, heat conductivity, diffusion, electrostatics, etc.) are described by equations of the elliptic type [10]. In particular, in some nanophysical models such as hydrodynamics and gas dynamics elliptic equations are considered.

2. Boundary value problem with a conormal and integral condition

We consider the boundary value problem with a conormal and integral conditions for the following elliptic type equation of the second kind:

ymUxx + Uyy = 0, (1)

where -1 <m< 0 is a real number. Let D be a simply connected domain in the plane (x, y) bounded by a curve a at the first quadrant (x > 0, y > 0) with its end points A(0,0), B(1,0) and with the line segment AB of the real axis Ox.

Let us introduce the following notations:

__mm

J = {(x, y) : 0 < x < 1, y = 0}, dD = a U AB, 20 =--,

m + 2

note that as it is defined, we have:

-1 < 0 < 0. (2)

We denote by C (D) the space of continuous functions defined on a set D on the (x, y) plane or on the real line and Ck (D) denotes the space of k continuously differentiable functions on D.

In the domain D, we consider the following Problem Conormal (Problem CN) for the equation (1): Problem CN. Find a function u(x, y) with the following properties:

(1) u(x, y) G C (D) U C1 (D U a U J) and ux, uy can tend to infinity of the order less than -20 at points A (0,0) and B (1,0);

(2) u(x, y) G C2 (D) is a solution of the equation (1) in D;

(3) u(x, y) - satisfies the following boundary conditions: for all 0 < s < l

{¿(s)As[u] + p(s)u} = <(s) (3)

for all 0 < x < 1

n

ao(x)uy (x, 0) + a-(x)Daxju(x, 0) + an+i(x)u(x, 0) = b(x), (4)

j=i

where ¿(s), p(s), <(s), a-(x), and b(x) are given functions with the following conditions:

b(0) = 0, ao(1)=0, (5)

a0(x) =0, Vx € J, (6)

¿2(s) + p2(s) = 0, Vs € [0,l], (7)

n+1

^ a2 (x) =0, Vx € J, (8)

j=o

<S(s), p()), <(s) € C[0, l], __(9)

aj(x), b(x) € C (J) n C2 (J), (j = 0, n + 1), (10)

and

^ m dy du dx du

s ds dx ds dy'

dx dy

— = — cos(n, y), — = cos(n, x), where n is the external normal to the curve a, l is the length of

ds ds

the curve a, s is the length of an arc of the curve a, starting from the point B (1,0), and Daj [*] is the Riemann-Liouville integral operator of a fractional order a [11]:

x

D«*> = r(—a-y/tx—j. —1 <aj < o. o.)

o

Let a = max {|a-1}, and

1<j<n

a < —2^. (12)

We assume that the curve a satisfies the following conditions:

(1) functions x(s), y(s), which are the parametric representations of the curve a, have continuous derivatives x'(s), y'(s), and don't tend to zero at the same time, moreover, they have the second derivatives that satisfy Holder condition of an exponent k (0 < k < 1) in the interval 0 < s < l;

(2) near the endpoints of the curve a, they satisfy inequalities:

^ < const • ym+1(s), (13)

ds

and x(l) = y(0) = 0, x(0) = 1, y(l) = 0.

3. Uniqueness of the solution of Problem CN

Assume we have:

— 1 < aj < 0, ¿(s)=0, Vs € [0,l], (14)

then the uniqueness of a solution of Problem CN can be proved by the method of integral energy. We have the following.

Theorem 1. If conditions (2), (5), (6), (7), (8) are satisfied and:

¿(s)p(s) > 0, 0 < s < l, (15)

an+i(x)

ao(x)

< 0, (16)

,ao(x)/x ' ao(1) then Problem CN in the domain D can't have more than one solution.

a, (x)\ a,- (1) ( -) , ,

-( M > 0, < 0, j = 1,^, (17)

Proof of Theorem 1. Let (x, y) be a point inside the domain D. Consider a domain Df g D, bounded by a curve af, parallel to a, and a segment of the straight line y = S (S > e > 0). We choose S, e small enough such that the point (x, y) belongs to the domain Df and u (x, y) G C2 (D|). We use the following identity:

d d

u[ymUxx + Uyy ] = — [ymu Ux] + dy [uuy ] - ymuX - uy. (18)

Integrating (18) on domain Df results in:

0 = J J u[ymuxx + uyy ]dxdy = JJ [ymuux ] + Jy [uuy ] j dxdy - J J [ymuX + u^ ]dxdy,

and applying Gauss-Ostrogradsky formula (the Green's theorem) (see [11]):

dP dQl f

—---— ^ dxdy = Qdx + Pdy

dx dy J J

we obtain:

[ymuxx + Uyy ]dxdy = -JJ [ymu2 + ]dxdy + J u [ymuxdy - uy dx] . From here on, by considering AB: y = 0 ^ dy = 0, and dy = cos(n, x)ds, dx = — cos(n, y)ds we have:

22

0= —ll [ymuX + uy]dxdy —j u(x,,)uy (x,,)dx + / uA,[u]d,, „„

Df xi

where x1, x2 - are the abscissas of the points of the intersection of the straight line y = S with the curve af. Taking into account the conditions (1) of Problem CN and y(s) = b(x) = 0 from (19) with (3) at S(s) = 0 and e ^ 0, S ^ 0 we get the following:

1

[ymu2 + uy] dxdy + / t(x)v(x)dx + / J(S))P(.S) u2ds = 0, (20)

J J ö2(s)

D 0

where:

u(x, 0) = t(x), (x, 0) G J, uy (x, 0) = v(x), (x, 0) G J. (21)

Due to the condition (15), the third integral of equality (20) implies that:

R = f ^TT1 u2ds > 0. (22)

J ö2(s)

a

Now, we show that the second term of the left-hand side of (20) is nonnegative. By (8) and (21), taking into account (4), we get:

(x) = —

Eaj(x) 7-1 o, / N . an+1(x) / \

jr D , t (x) +-— t (x)

ao(x) ao(x)

Using (12), we rewrite the second term of (20) in the form:

1 1 n x 1

R = f T(x)v(x)dx = - f Vp a(x)T(x) ^ dx f , T(t)d+ a- - f T2(x)dx = R21 + R22. (23)

J J r-fr(-aj) ao(x) J (x - i)1+a J ao(x)

0 0 J-1 0 0

Using the following formula (see [1]):

|x - t|-Y =-nY / z7-1 cos[z(x - t)]dz, 0 < Y < 1

I (y ) cos -2

20

and from the equality (23), we obtain:

i „

R

21

- E

Aj (x)t(x)dx

j=i

ao(x)r (—aj) r(1 + aj) cosn(1 + aj)/2

x CO

- J t (t)d^ z a j cos z(x — t)dz

oo

c 1

na, f „. , f Aj (x)t(x)

—2^cos nay/ z a dWAj

j= 2 J J ao(x)

o

CO

cos nai/ za j dz/AM • «

2 7 7 ao(x) ox j = 1 0 0

—E

dxj [cos zx cos zt + sin zx sin zt]T(t)dt

0

x \ 2 / x \ 2'

/ t(t) cos ztdt I + I / t(t) sin ztdt

0 / vo

Integrating the last integrals by parts on x, we have:

aj(x)

R21 = — E cos j z a dzl j( ) ^^ 21 I a0 (x)

j=1 1

j(x)

j t(t) cos ztdt I + jy t(t) sin ztdt

x=0

— E

j=1

a«(x)

naW a. aj (1) cos —z a ' j

dx

+ y^ cos —^ I z a j dz j=1 0

Hence, by (17) we have

00 2 / x \ 2'

t(t) cos ztdt I + ^y t(t) sin ztdt 00

x \ 2 / x \ 2'

/ t(t) cos ztdt I + I / t(t) sin ztdt 00

x \ 2 / x \ 2'

/ t(t) cos ztdt I + I / t(t) sin ztdt

00

a0 (1)

dz

1

0

aj(x) a0(x)

dx.

Furthermore, by (16) and (23), we have: Finally, using (24) and (25), by (23), we obtain:

R21 > 0. R22 > 0.

(24)

(25)

(26)

R2 > 0.

Using the relations (22) and (26), by (20) it follows that ux = uy =0 in D, that is, u = const for all (x, y) € D. The fact that each term in (20) tends to zero concludes u = 0 on a. Thus, u = 0 in D for £(s) = 0.

Remark. The uniqueness of a solution of Problem CN for p(s) =0, Vs € [0, l] is proved using the maximum principle [7].

Theorem 1 is proved.

4. Existence of a solution of Problem CN for S(s) = 0

We consider the following auxiliary problem. Problem DK. Find a solution u(x, y) € C(D) n C 1(D U a U J) n C2(D) of the equation (1) in the domain D that satisfies conditions (3) and the following:

u| o = t(x), 0 < x < 1,

(27)

where t(x) is a continuous function that satisfies Holder condition with the exponent y0 > 1 — 2,0 in the interval (0,1) and it have the following representation:

1

(28)

t (x) = J (t — x)-2^ T (t)dt,

where a function T(t) is continuous in (0,1) and it is integrated in [0,1]. The uniqueness of a solution of Problem DK follows from identity (20).

x

x

2

2

x

x

The solution of Problem DK that satisfies conditions (3) and (27) for the equation (1) in the domain D exists and unique, moreover it has the following representation (see [11, eq. (10.78)]:

u(x,y) = J t(00; x,y)d£ +y ||SyGi&n; x,y)ds, (29)

0 0

where G2(£, n; x, y) is the Green's function of Problem DK for equation (1), and it has the following form (see, [11]):

g2 (e n; x y) = g02 n; x y) + H2 (e n;^ y), (30)

where G02(£, n; x, y) is the Green's function of Problem DK for equation (1) on the normal domain D0 bounded

with the segment AB and the normal curve a0 : ( x - - ) + -—4—-r ym+2 = 1

\ 2/ (m + 2)2 4

H2 (e n;x, y) = g2 (e n;x, y) - G02(e n;x, y) i

= j A2(s; e,n){ As[Gq2 (e (s) (s); x,y)] + Gq2 (C (s) ,n (s) ; x,y)j ds

where A2 (s; e, n) is a solution of the integral equation:

i

A2 (s; e, n) + 2 J A2 (t; e, n) { A [«2 (C (t), n (t) ; x(s), y(s))] + P|s)«2 (C (t), n (t) ; x(s), y(s)) jdt (32)

= —2q2 (e(s),n(s); e,n),

where q2 (x, y, xo, yo) is the fundamental solution of the equation (1) and it has the following form:

/ 4 x 4ß-2

«2 (x, y,xo,yo) = k2 (r2)-ß (1 - a)1-2ß F (1 - ß, 1 - ß, 2 - 2ß; 1 - a), (33)

\m + 2/

,m + 2

where:

r2 1 , ,9 4 / -±2 ^i2 \2

r2 \ = (x — xo) + "-—2 ( y 2 T yo 2

r1 (m + 2)

m / 4 \2-2^ r2 (1 — ff)

a = r2 , ^ = 2(m + 2) < 0, k2 = 4nvmT2; r (2 — 2£), where F (a, b, c; z) is the Gauss hypergeometric function [7].

Differentiating the equation (29) by y, then by tending y to zero and by (30) and (33), we obtain a functional relation between t(x) and v(x), transferred from the domain D to J:

111 1

r \ 1 fu 12^-2 , / T(t)d^ f ,,^d2H2(t,0;x,0),,, / ,sdq2(t,n;x,

v(x) = k2J |t — x| t(t)dt — k2J + ^ 2tx)2-2, + J t(t)-dndy-dt+y x(s)-dy-ds,

o o ( ) o o

(34)

where x(s) is a solution of the integral equation:

x(s) + 2 J x(t){ As[q2 (e(t),n(t); x(s),y(s)) ] + q2 (£(t),n(t); x(s),y(s))j dt = ^. (35)

o

Lemma 1. Let a function t (x) belongs to the class C (1'7l)(0,1), y1 > —2,0, then the following identities hold on (0,1):

x x

/ (x — t)2^-2T(t)dt = 2mbry dx2^/ (x — t)2' t(t)dt, (36)

oo 1 1

/ (t — x)2^-2T(t)dt =2^—1)&J (t — x)2^ t(t)dt, (37)

xx

where 0 < —2^ < 1.

Proof. We rewrite identities (36) and (37) as follows:

and

T1(x) = limT1f(x) = limi i (x - t)2? t(t)dt

f^0 f^0 | dx2 J

0

T2(x) = limT2f(x) = limi i (t - x)2? t(t)dt /. f^0 f^0 | dx2 J I

It follows that:

T1f(x) = dx

e2?t(x - e) + 2^ y (x - t)2?-1 t(t)dt 0

x — f

= e2?T'(x - e) + 20e2?-V(x - e) + 20 (20 - 1) J (x - t)2?-2 t(t)dt,

2f dx

-e2?t(x + e) - 20 j (t - x)2?-V(t)dt x+f

x1

-e2?T'(x + e)+20e2?-V(x + e) + 20 (20 - 1) J (t - x)2?-2 t(t)dt.

(38)

(39)

The conditions of the Lemma 1 and relations (38) and (39) at e ^ 0 imply the identities (36) and (37). This completes the proof.

Lemma 2. Let t (x ) G C(1'Yl) (0,1), Y1 > -20 and it is representable in the form of (28), then the following identities hold on (0,1):

xx

J (x - t)2?-2 t(t)dt =

d2

20 (20 - 1) dx2

xx

J (x - t)2? T(t)dt

= r (1 + 20) r (1 - 20) D1-2,D2,- 1T(x) = 20 (20 - 1) D0x Dx1 T (x)

n cot n20 1 - 20

1 1-2, 1 f it\ T(t)

t <x) - rWG)

dt, tx

(40)

J (t - x)2?-2 t(t)dt =

d2

20 (20 - 1) dx2

x1

J (t - x)2^ T(t)dt

= r(1 + 20)r(1 - 20)D1-2,DX?-1T(x) 20 (20 - 1) x1 x1 ( )

- T (x),

(20 - 1) sin 20n

x1 x1 x1 J (t + x - 2tx)2?-2 t(t)dt = J (t + x - 2tx)2?-2 dtj (z - t)-2? T(z)dz

1

t \ 1-2? T(t)dt

1 - 20 x x + t - 2xt

(41)

where (0 < -20 < 1).

xf

xf

1

x+f

1

1

1

1

Proof. Using definitions of the integro-differential operator with a fractional order (see [11, §4, (4.1), (4.6)]):

1

J (x - t)-CT-1 f (t)dt, x G (p, q), a < 0,

Df (x) =

r(—a),

p

f (x), a = 0,

dn

d^ D-f (x)] , n - 1 < a < n, n G N,

and representation (28) and the identity r(z)r(1 - z) = n/sin nz from the right-hand side of (40) implies that:

xx xx

A1 = /(x -1)2?-2 T (t)dt = 20(20-1) (x -1)2? T (t)dt 00 xx x1

1 d2 2, -2, (43)

20 (20 - 1) dx2

0 t

J (x - t)2^ dtj (z - t)-2^ T(z)dz

= r (1 + 20) r (1 - 20) (1+2,)D2,- 1T (x) = n D1-2, D2,-1

= 20 (20 - 1) dx2D0x Dx1 T (x) = (20 - 1)sin20nD°x Dx1

Using the formula (see [11, p. 24, lemma 4.5]):

b

a T / n , » sin na /" / t — a \ a $(t) ,

DaxD-ba$(x) = cos na $(x) +- - —^-dt, 0 < a < 1

n J Vx — <2 y t — x

and the equality (43) we have:

Ai = nCOt20nT(x) f (tY^dt.

1 1 - 20 v 7 1 - 207 Vx/ t - x

The last equality implies the identity (40).

Now we prove the identity (41). Due to the representation (28) we have:

x1 x1 x1 1 d2 x1

A2 = / (t - x)2-2 T(t)dt = 20T20-T) d-/(t - x)2" T(t)dt

x

1

d

20 (20 - 1) dx2

I (t - x)2^ t(t)d^y (z - t)-2^ T(z)dz.

Using definitions of the integro-differential operator with a fractional order (see, [11, §4, see (4.13), (4.14)]):

b

1, , ,-„-1

1(-a) y (t - x)-ff-1f (t)dt, x G (p, q), a < 0,

Dqf(x) = \ r, s x (44)

r

f(x), a = 0

-( b

dn

n — 1 <a<n, n G N,

(-1)7din Dl-V(x)

and the formula DabD-baf (x) = f (x), we have:

A = n n1-2^ n-(1-2«T(x) = ^ T(x)

A2 =(20 - 1) sin20n 1 Dx1 T (x) = (20 - 1)sin20nT (x).

The last equality implies the identity (41). We consider the expression:

x1

A3 = y (t + x - 2tx)2^ 2 t(t)dt. (45)

0

Substituting (28) into (42), and changing the order of the integrations, we have:

1

1

A3 = J (t + x — 2tx)2^-2 dtj (z — t)-2^ T(z)dz = J T(z)dzy (z — t)-2^ (t + x — 2tx)2^-2 dt. o t o o

Putting t = z(1 — s) in internal integral and using the formula (see [11, p.8, (2.10)])

1

f-,-a-1/1 AC-«-^ r (a) r (c — a) ta 1 (1 — t) (1 — zt) dt =— t^/n- F (a, b, c; z),

J r(c)

0 < Re a < Re c, |arg(1 — z)| < n. (46)

We obtain:

1 1

A3 = J T (z) z-2'+1 (z + x — 2x z)2'-2 dz| s-2' 00

' _ (1 — 2x) zs l2'-2 (z + x — 2x z)

ds

r(1 — 20)r(1^ z V"' T(z) -F (1 — 20, 2 — 20, 2 — 20 ; (1 — 2x)z ) dz.

r (2 — 20) J \x + z — 2xz/ x + z — 2x z y ' ' 'x + z — 2xz

o

Using the formulas F (a, b, b; z) = (1 — z)-a and r(1 + a) = ar(a), from the last expression we have:

1 1 (t V-2^ T (t)dt

A3 —

1 — 20J \x ) x +1 — 2xt

o

The last equality implies the identity (42). Lemma 2 is proved.

Thus, putting (40), (41), (42) and (28) into (34), we get a functional relation between T(x) and v(x), transferred from the domainD to J:

k2ntan 0n v (x) =--1 — 20 T (x)

1 1-2^ 1 t

+ 1—*/;> t«

11

d'lTm i x,0) <' — z)-2'dz (47)

x t x I t 2xt

0 "0 0

i

+ / ^ x, 0) x(s)ds, (x, 0) G J.

0

By the conditions (21) and the relations (8), (11), (28), from (4) on the interval J, we get a functional relation between t(x) and v(x):

n x 1

V(x) = — V • , 1 , f (x — t)-aj-1 dt / (z — t)-2' T(z)dz

A0 (x) r(—aj )J V y J y ' W

j=1 1 0 4 (48)

an+1 (x) ^ \-2' \ j, , b(x)

, , /(t —x)-2' T(t)dt + , (x, 0) G J. a0 (x) J a0 (x)

1

1

1

z

Theorem 2. f the conditions (2), (5)-(10) and (12)-(14) are satisfied, then there exists the solution of Problem CN in the domain D.

Proof. Excluding v(x) from relation (47) and (48) we have:

T(x)-

cot 0n f f t

1-2,

n J \ x ,

0

(1 - 20 ) cot 0n

1

1

x t x | t 2xt

t

T (t)dt

T (t)dJ (t - z)-2, d2g2(z'0;x,0) dz J dndy

(1 - 20) cot 0n j (x) 1

T(t)dt / (x - z)-aj-1 (t - z)-2p dz

k2n

j=1

ao (x) r(-aj),

-2,

(1 - 20) cot 0n j aj (x) 1

T(t)dt / (x - z)-aj-1 (t - z)-2p dz

k2n

j=1

ao (x) r(-aj),

-2,

(1 - 20)cot0n an+1(x)

k2n

a0 (x)

1

J (t - x)-^ T(t)dt

(20 - 1) cot 0n b(x) (1 - 20) cot 0W , ,dq2(^(s),n(s); x, 0) , +--;- X(s)---ds,

k2n

a0 (x)

k2n

dy

or

T(x) - Y3

1

1

x t x | t 2xt

T(t)dt -J K (x,t)T(t)dt = F (x), 0

where, Y3 = - cot 0n, T(x) = x1-^T(x), n

K1 (x,t), 0 < t < 1,

K (x,t)={ K2 (x, t), 0 < t < x,

K3 (x,t) + K4 (x, t), x < t < 1,

(49)

(50)

K ( t) (1 - 20) ■ Y3 (x)1-20 i( )-2, d2H2(z, 0; x, 0) d

K1 (x,t) =-k2-IV J(t - z) ■-dndy-dz,

0

(51)

K2 (x,t) = (1 - 20) • 73 (x) 1-2' j f (x - z)-aj-1(t - z)-2,dz

k2

a0 (x) r(-«j)J j = 1 0

K3 (x,t) =

(1 - 20) ■ 73 (x) 1-2, aj (x) 1

k2

E

j=1

ao (x) r (-aj)

(x - z)-aj-1(t - z)-^dz

K4 (x, t) = (1 - 20) Y3 (x) 1-2' On^ (t - x)-2,. 4V ' ; k2 \t) a0(x) v 7 •

F (x)

(20 - 1) Yзx1-2, k2

b(x) a0(x)

i

-J X(s)

dq2 (g(s),n(s); x, 0) dy

ds

We investigate the kernel and the right-hand side of the singular integral equation (49). Lemma 3. Let 0 <x< 1, 0 <z< 1, then the following inequality holds:

d2H2(z, 0; x, 0)

< C1(x + z -

where C1 is a constant depending only on the domain D.

The proof of Lemma 3 is similar to that of Lemma 18.1(see [11, p. 133-136]).

(52)

(53)

(54)

(55)

t

x

1

x

1

1

x

t

Using (56) by (51) we have

|K1(x,t)| < C1

(1 — 20)73 (xV-2' k2

t

J (t — z)-2'(x + z — 2xz)2'-1dz

Changing variables z = t(1 — a) and using formulas (46) from (57) we have:

|K1(x,t)| < C1

(1 — 20)73 k2

x I t — 2xt

1-2'

2'

1

(1 — 2x)t

-c

x I t — 2xt

2'-1

dc.

(57)

(58)

Using (46) from (57), we get:

|K1(x,t)| < C1

(1 — 20) 73

1-2'

k2 V x + t — 2xt Since c — a — b = 2 — 20 — 2 + 40 = 20 < 0, using a formula [7]:

F 1 — 20, 1 — 20, 2 — 20;

t (1 — 2x) x +1 — 2xt

F (a, b, c, z) = (1 — z)c a b F (c — a, c — b, c; z), |arg(1 — z)| <n

and an estimate

const

F (a, b, c, z) < { const(1 — z)c-a-b const[1 + l (1 — z)]

at c — a — b > 0, 0 < z < 1, at c — a — b < 0, 0 < z < 1 , at c a b = 0, 0 < z < 1

(59)

(60)

(61)

from (59) we have the following: |K1(x,t)| < C1C2

(1 — 20) Y3

1-2'

2'

<

C3x

k2 \ x +1 — 2xt y y x +1 — 2xt y x + t — 2xt We consider the kernel (52). We make a replacement z = in (52) and taking into account (46) we get:

(62)

K2 (x,t) =

(1 — 20) • y3 (x \ 1-2' ^ a, (x) x-a,-1t1-2'

k2

j=1

a0 (x) r(—ay )

(1 — s)

-2'

1-^

-a,- -1

ds

(1 — 20) • Y3 r(1 — 20) (x A1-2' E jx) x-a,-1t1-2' F 1 + 2 — 20; t_

k2 r(2 — 20) V t

Y3x^ ay (x) x-aj-2'

j=1

a0 (x) r(—ay )

(63)

t

()r( rF 1, 1 + a-, 2 — 20; - . k2 j=1 ao (x) i (—a-) V xy

Considering (6), (10), (12), (61) and taking into account c — a — b = 2 — 20 — 1 — 1 — a- = —20 — a- > —20 — a > 0, 0 < t < x < 1 from (63) we have an estimation:

|K (x,t)| < fY,

Y3^ C4C6 x-a,-2' ^ Y3C4C6

<

2'

k^^f C5 r(—ay ) k2C5 j=1

E

j=1

1

xa, r (—ay )

<C7x-

a-2',

or

|K (x,t)| < C7 x-(a+2« < Cg. (64)

Similarly, we estimate K3 (x, t). Due to (6), (10), (12), (61) and taking into account c—a—b = 1—a- — 1—20 = —20 — a-, 0 < x < t < 1 from (53) we have

Y3 (xx1-2'^ C4C9 x-a,t-2' ^ Y3C4C9

K(x_)|< Y3 Î E

<

2'

E

1

or

t) f-f C5 r(—ay ) " k2C5 t

IK (x,t)| < C11. Due to (6), (10) with x < t < 1, from (54) it follows that:

xaj r (—ay )

< C10 x-(a+2') < C11,

(65)

IK,(x.«)|< » (xf" < C13<' — x)-2'

or

IK4 (x,t)| < C13(t — x)-2' < C14.

(66)

1

x

7

x

x

x

1

t

x

x

x

Now we estimate the right-hand side of the equality (49). Differentiating (33) with respect to y and at y = 0, we obtain:

dq2 (e, n; t, 0)

dy

k2n

(e - t)2 +

4

(m + 2)

rn

m+2

,-1

Substituting (67) to (55) we have:

F(x)

(20 - 1) Yзx1-2, k2

b(x) a0(x)

- k2

n(s)x(s)

(e(s) -1)2 + ^ nm+2(s)

1-,

ds

(67)

(68)

It is clear that the function F (x) has derivatives of any order in the interval (0,1). Let us study the behavior of the function F (x) and its derivative at x ^ 0 and x ^ 1. In this regard, we consider the following expression:

F1(x) = k2 J

n(s)x(s)

(e(s) -1)2 + ^ nm+2(s)

1-,

ds.

By (9) for the sufficiently small x > 0, we have

i

|F1 (x)| < k^ |x(s)

i-E

(e - x)2 + <mw nm+2

1-,

ds + O(1)

(69)

<C15

i-E

_ x)2 + nm+2

(e - x)2 +

(m+2)

— rr' 2 '/

1-,

ds + O(1).

Hence, using (13) for sufficiently small e > 0, we get

|F1 (x)| <C16

i-E

™2 I 4 ,r)m+2

x + (m+2)2 n

1+,ds + O(1) <C17

dn

[x2 + n2]2 +,

Ï+, + O(1).

(70)

Substituting = w in (70) and by relations (46), (61) and formulas [11, (2.17), (2.14), (2.22), p. 10-13], we have:

|F1 (x)| <

S2

x2,+1 S

r(1, 5)r(0)

<

r(0, 5 + 0) r(1, 5)r(0)

S2i +

S

„2,+ 1

r(1, 5)r(-0)

x

.2,

r(0, 5 + 0)

+

S r(1,5)r(-0)

r(0, 5)r(1 - 0)

r(0, 5)r(1 - 0) (x2 + S2)^ <C18 x-^,

(x2 + S2 x2,+1F 0,^, 1+ 0

1

x2 + S2

or

|F1 (x)| ^^x-^. If 1 - x it is sufficiently small, then as before, we get:

|F1 (x)| = C19 (1 - x)-^ .

By the same arguments, we obtain:

|F1 (x)| < C20x-2,-\ |F1 (x)| = C21 (1 - x)-2^1 . From the relations (6), (10), (71), (72), (73) from (68), we conclude that:

F (x) G C (J) n C 1(J).

The function F'(x) tends to infinity of the order less 20 + 1 at x ^ 1, and when x ^ 0 it is bounded. Introducing new variables:

z =

t2

1 - 2t + 2t2

1 - 2x + 2x2

(71)

(72)

(73)

(74)

(75)

n

n

à

n

2

2

x

x

2

x

z=

in the equation (49), we have:

i i

"(z) + Y3| ^ - j K(z, C)"(ZR = F(z), (76)

0 0

where:

"(z) = (1 - 2x + 2x2)T(x), F(z) = (1 - 2x + 2x2)F(x),

1 - 2t + 2t2 (1 - 2x + 2x2)(1 - 2t + 2t2)

K (Z,C) = 2t(1 - t)(1 - 2x + 2x2)K (x,t) + 73 (1 - t)(t + x - 2xt) ,

x = V* t = VC x yz+V1-T VC + V1-C'

Since 1 + y3 =0, the equation (49) is the normal type. Its index is equal to zero in a class of h2 functions "(z) g H(0,1), bounded at the ends of the segment J (see [12]).

We apply the method of regularization of Carleman-Vekua [12] to the equation (76). This method is developed by S. G. Mikhlin [13] and M. M. Smirnov [11, p. 258]. This results in the Fredholm's integral equation of the second kind, solvability of which follows from the uniqueness of a solution of Problem CN. By the notations T(x) = x1-2^T(x), "(z) = (1 - 2x + 2x2)T(x), we obtain a function T(x) that is continuous on (0,1) and is integrable in [0,1].

Substituting the solution T(x) of Fredholm's equation of the second kind in (28), we get t(x). Furthermore, knowing the function t(x), the solution of Problem CN for equation (1) in the domain D is defined as a solution of Problem DK for equation (1) with conditions (3) and (27).

Thus, the existence of a solution of Problem CN for £(s) = 0 is proved. Theorem 2 is proved.

References

[1] Vostrova L.K. Mixed boundary value problem for the general Lavrentyev-Bitsadze equation. Scientific notes Kuibyshev State pedagogical college, 1959, 29, P. 45-66.

[2] Melentyev B.V. On a boundary problem for the mixed type equations. Soviet Math. Doklady, 1964, 154 (6), P. 1262-1265.

[3] Salakhitdinov M.S., Kadyrov Z. A problem with a normal derivative for an equation of mixed type with nonsmooth lines of degeneracy. Differ. Uravn., 1986, 22 (1), P. 103-114

[4] Salakhitdinov M.S., Mingziyaev B. A boundary value problem with the conormal derivative for the mixed type equation with two lines of degeneration. In Differential equations and their applications, Tashkent, 1979, P. 3-14 (in Russian).

[5] Islomov B.I. Gellerstedt's boundary value problem for the mixed type equation of first kind. In Boundary value problems for the nonclassical equations of mathematical physics, Tashkent, 1986, P. 121-134.

[6] Usanatashvili M.A. A problem with an inclined fractional derivative for an equation of mixed type. Soobsheniye AN GSSR, 1978, 90 (34), P. 10-16.

[7] Salokhitdinov M.S., Islomov B.I. The equations of the mixed type with two lines of degeneration. Tashkent, "MUMTOZ SO'Z", 2009, 264 p.

[8] Salokhitdinov M.S., Islomov B.I. Nonlocal boundary value problem of a conormal derivative for the mixed type equation with two internal lines and various orders of degeneration. News of higher education institutions. Mathematics, 2011, 1, P. 49-58.

[9] Islomov H. A problem with a conormal derivative for the elliptic type equation with one line of degeneration. Uzbek mathematical journal, 2012, 1, P. 47-60.

[10] Tikhonov A.N., Samarskii A.A. Equations of Mathematical Physics. Nauka, Moscow, 1977, 742 p.

[11] Smirnov M.M. The mixed type equations. Moscow, 1985, 304 p.

[12] Muskhelishvili N.I. Singular integral equations. Nauka, 1968, 512 p.

[13] Mikhlin S.G. On the integral equations of F. Tricomi. Soviet Math. Doklady, 1948, 59 (6), P. 1053-1056.