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16Bulletin KRASEC. Phys. & Math. Sci, 2014, vol. , no. 2, pp. 3-12. ISSN 2313-0156
MSC 35M10
ABOUT A METHOD OF RESEARCH OF THE NON-LOCAL PROBLEM FOR THE LOADED MIXED TYPE EQUATION IN DOUBLE-CONNECTED
DOMAIN
O.Kh. Abdullayev
National University of Uzbekistan by Mirzo Ulugbeka, 100174, Uzbekistan,
Tashkent c., VUZ gorodok st.
E-mail: obidjon.mth@gmail.com
In the present paper an existence and uniqueness of solution of the non-local boundary value problem for the loaded elliptic-hyperbolic type equation on the third order in double-connected domain was investigated. The uniqueness of solution was proved by the extremum principle for the mixed type equations, and existence was proved by the method of integral equations.
Key words: loaded equation, elliptic-hyperbolic type, double-connected domain, an extremum principle, existence of solution, uniqueness of solution, method of integral equations
Introduction
We shall notice that with intensive research on problem of optimal control of the agroeconomical system, regulating the level of ground waters and soil moisture, it has become necessary to investigate a new class of equations called "LOADED EQUATIONS". Such equations were investigated in first in the works by N.N. Nazarov and N.Kochin, but they didn't use the term "LOADED EQUATIONS". For the first time, the most general definition of the LOADED EQUATIONS was given and various loaded equations were classified in detail by A.M. Nakhushev [1].
Let's notice that non-local problems for the loaded elliptic-hyperbolic type equations in double-connected domains have not been investigated. In the given paper, uniqueness of solution of the non-local boundary value problem for the loaded elliptic-hyperbolic type equation in double-connected domain was proved, and the method of the solvability of the investigated problem was presented.
Abdullayev Odidjon Khayrullaevich - Ph.D. (Phys. & Math.) Associate Professor Dept. of Differential equations of the National University of Uzbekistan named after Ulugbek, Uzbekistan. ©Abdullayev O.Kh., 2014.
The statement of the Problem
Let's consider the equation
Wxx + sgn (xy) + ^ (x, y) = 0 (1)
in double-connected domain of Q, bounded with to(two) lines:
a1 : x2 + y2 = 1 ; a2 : x2 + y2 = q2; atx > 0, y > 0;
a" : x2 + y2 = 1 ; : x2 + y2 = q2, atx < 0, y < 0; and characteristics:
AjA" : x - y = (-1)j+1; : x - y = (-1)j-1 ■ q, (j = 1,2)
at x■ y < 0, of the equations (1), where A1 (1;0), A2 (0;1), A1 (0;-1), A2(-1;0), B1 (q;0), B2 (0;q), B1 (0; -q), B (-q;0), 0 < q < 1, (j = 1,2) .
ft(x,y) = 1 - s^^(xy) ■ [A1 ■ ^(x,y) + ¿2 ■ ^2(x,y)], ¿1, A2 = const > 0; »1(x, y) = 1 + sgn2xy + x2) u(x, 0), »2(x, y) = 1 + sgn2xy + y2) u(0, y).
Let's enter designations:
E (q - (-1)j; q +(-1)j \ E ((-1)j-1 - q (-1)j - q
J V 2 ' 2 jV 2 ' 2
Cj ((-1)j-12; (-1)jD , Q = Qn (x > 0) n (y > 0),Q = Qn (x < 0) n (y < 0),
Ax = Qn (x + y > q) n (y < 0);A1 = Qn (x + y < -q) n (y < 0), A2 = nn (x + y > q) n (x < 0), A2 = Qn (x + y < -q) n (y > 0), Dl = Qn (-q < x + y < q) n (y < 0), D2 = Q n (-q < x + y < q) n (y > 0), D0 = Q0 U Ax U A2, D0 = Q0 U A1 U A2, Ij = {t: 0 < t < q}; I2+j = {t: 0 < (-1)j-1t < 1} , x at J = 1,
where t = , . „ y at j = 2.
Let's designate, through x + 1;, 02 ^y 2 1;y + ^ points of intersections
of characteristics of the equation (1) with leaving points (x, 0) e A1B1 and (0,y) e A2B2 with characteristics A1A1 and A2A2, accordingly.
In the domain of Q the following problem is investigated
Task. (Problem A.) To find function u(x,y) with following properties:
1) u(x,y) e C(Q);
2) u(x,y) is the regular solution of the equation (1) in the domain of Q\(xy = 0)\(x+y = ±q), besides, Uy e C(A1B1 UA^), Ux e C(A2B2UA\B\) at that ux(0,t), Uy(t, 0) can tend infinity of an order of less unit at t ^ ±q, and finite at t ^ ±1;
3) u(x,y) satisfies gluing conditions on lines of changing type:
uy(x, —0) = uy(x, +0), in regular intervals on(x,0) e A1B1 UA^B^,
ux(—0,y) = ux(+0,y), in regular intervals on (0,y) e A\B\ UA2B2, 4) u(x,y) satisfies to boundary conditions:
u(x,y) \a. = 0j(x,y); (x,y) e ô], (2)
u(x, y)
aJ = jx,y); (x,y) G aj\ (3)
d
—u(01 (x)) = a\(x)uy(x;0) + bi(x), q < x < 1, (4)
dx
d
—u (d2(y))= a2(y)ux(0;y) + b2(y), q < y < 1, (5) dy
u(x,y) BjBj = gj(t), t e Ij, (6) where 0j(x,y), 0j(x,y), gj(t), aj(t), bj(t), (j = 1,2) is given function, at that:
gj (q) = gj (!) , 02(q, 0)= g1 (q), 02(0, q)= g2(q), \ (7)
02(0, —q)= g1(—q), 0j(—q, 0) = g2(—q),f ( )
0j(x,y) = (xy)Y0j(x,y); 0j(x,y) e C (ôj), 2 < y< 3 (8)
0j(x,y) = (xy)Y0j(x,y); 0(x,y) e C (ôj) , 2 < Y< 3, (9)
gj (t ) e C (j n C2 (Ij ), (10)
aj (t), bj (t) e C [q, 1] n C2 (p, q) ; (11)
Theorem. If conditions (7) - 11) and
a1(x) > 2, a2(y) > 2 ; (12) are satisfied, then the solution of the Problem A exists and it is unique.
Proof. Note, that the solution the equation (1) in hyperbolic domains looks like:
y
u(x,y) = fi(x + y) + f2(y - x) + faj (y - t)z(0, t)dt, aty > 0, x < 0; (13)
q
-q
u(x,y)= fi (x + y) + f2(x - y) + xj (t - y)z(0, t)dt, atx > 0, y < 0; (14)
y
Owing to (13) we will receive, that the solution of Cauchy problem of the equation (1) satisfies conditions u(0,y) = t2(y) and ux(0,y) = v2(y) in the domain of A2, looks like:
y+x y+x
u(x,y)= T2(x+y) + T2(y - x) + 1 j V2(t)dt - X j (x+y - t)T2(t)dt-
2
y—X
y—x
X
-2 J (y-x-1)T2(t)dt + xj(y-1)T2(t)dt. (15)
q q
From here, by virtue, (5) we will receive a functional relation between T2(y) and v2(y) from the domain of A2 on the piece A2B2:
y+1
y ^r
11 i 2 2
2t2(y) + 2 V2(y) - ¿2J T2(t)dt + y J T2(t)dt = V2(y)a2(y) + &26O,
i.e.,
y+1 2
(2a2(y) - 1) V2(y) = t2 (y)+ X2 j T2(t)dt + T2(t)dt - ^(y). (16)
q q
Confirm, that a functional relation between T1(x) and v1(x)we will obtain by the same method, from the solution of Cauchy problem for the equation (1) satisfying to conditions u(x, 0) = T1(x); uy(x, 0) = v1(x) in the domain of A1, and with the account of conditions (4), which presented on the form:
x+1
2
(2a1 (x) — 1) v1 (x) = tJ (x) + h J t1 (t)dt + 2X1 J t1 (t)dt — b1 (x). (17)
□
The uniqueness of solution of the Problem A
It is known that, if the homogeneous problem has only trivial solution than a solution of the accordingly non-uniform problem is unique, therefore, we need to prove that the homogeneous problem has only trivial solution.
y
y
y
Let bi(y) = b2(x) = 0 then, from (16) and (17), accordingly, we will receive:
y+1
~T y
(2a2(y) - 1) V2(y) = t2(y) + h J T2(t)dt + 2^/ T2(t)dt, (18)
and
x+1
(2ai (x) - 1) Vi (x) = t1 (x) + Xi j Ti (t)dt + n (t)dt. (19)
q q
Lemma 1. // b1(y) = b2(x) = 0 and satisfied conditions (12) that solution u(x,y) of the equation (1) reaches the positive maximum, and the negative minimum in closed domain of D0 only on and o2.
Proof. According to the extremum principle for the hyperbolic [2] and elliptic equations [3], the solution u(x,y) of the equation (1) can reach the positive maximum and the negative minimum in closed domain of D0 only on A2B2 UA1B1 and ÔT U We need to prove, that the solution u(x,y) of the equation (1) can't reach the positive maximum and the negative minimum in A2B2 UA1B1.
Let the function u(x,y) (u(0,y) = T2(y)) reach the positive maximum (the negative minimum) on some point ofy0 g A2B2, then from (18) and based on (12) we will receive
v2(yo) =
2a2(yo) - 1
yo+1
yo
T2(t )dt + 2 T2(t )dt
> 0 (v2(yo) < 0).
Therefore, in the view of to the gluing condition, we have v+(y0) > 0 (v+(y0) < 0), and it contradicts the known Zareba-Zero principle [3], according to which in a point of a positive maximum (a negative minimum) should be v2(y0) < 0 (v2(y0) > 0). Thus, u(x,y) does not reach the positive maximum (the negative minimum) on the point of y0 e A2B2.
Owing to (19) and (12), it is similarly proved that the function u(x,y) does not reach the positive maximum and the negative minimum in the interval of A1B1.
The Lemma 1 is proved. □
As T2(y) does not reach the positive maximum and the negative minimum in the interval of A2B2 then, T2(y) = const in the interval of A2B2. Further by virtue u(x,y) e C(G), we will receive T2(l) = 01(0,1) = 0, T2(q) = 02(0,q) = 0 at the ^(x,y) = 0. From here, as T2(y) e C (A2£2) we will conclude that, T2(y) = 0 in A2B2. Therefore, if b2(y) = 0 and a2(y) > 1, then from (18) we will get v2(y) = 0. Also, by the same method, we can get T1(x) = 0 in A1B1 and v1(x) = 0.
Hence, owing to uniqueness of solution of the Cauchy problem for the equation (1), we will have u(x,y) = 0 in domains A1 and A2.
Consequently, in the view of the Lemma 1, at 01(x,y) = 02(x,y) = 0 we will deduce, that u(x,y) = 0 in closed domain of Q0. Thus, u(x,y) = 0 in D0.
Further, owing to condition 7 and considering the uniqueness of solution of the Gaursat problem atg(.t) = 0, t e I. we will receive u(x,y) = 0 in D. (j = 1,2).
Takes place:
Lemma 2. The solution u(x,y) of the equation (1) reach the positive maximum and the negative minimum in closed domain D*0 only on A*2B*2 UA'lBp o* and o*. ( Lemma 2 will be proved similarly as Lemma 1)
On the basis of Lemma 2, on account of 3 at 0*(x,y) = 0, we will receive u(x,y) = 0 in the domain of D0. Thus, the solution of homogeneous Problem A is identically equal to zero in the domain of Q. (The uniqueness of solution of the Problem A is proved.)
The existence of solution of the problem I
It's known, that the solution u(x,y) of the Nyman problem (Problem N) [2] satisfying conditions:
1) u(x,y) e C(D0) HC2(D0\xy = 0) nC1(D0) is the solution of the equation (1);
2) uy e C(A1B1), ux e C(A2B2), at that ux(0,t), uy(t,0) can tend to infinity of an order of less unit at t ^ q, and finite at t ^ 1;
3) satisfies to boundary conditions (2) (j=1,2) and ux(0,y) = v2(y), uy(x,0) = v1(x) is exists and is unique and represented in the form [2]
u(x, y) =J 01(1, n) dn G(%, n; x, y)dS -J , n) dn G(%, n; x, y)dS+
O 02
1 1
+ J v+(t)G(t, 0; x, y)dt + f v+(t)G(0, t; x, y)dt, (20)
q q
where G(%, n;x,y) is Green's function of the problem N for the Laplase equation in the domain of which looks like:
0(£,, n; x, y) = In
a (\nv+lna Anv+lna An(-v)+lnn /ln(-v)+lnn 61 l 2nir ) 2Kir ) 6 I -2Wr-) 6 I-2nTr-
M ^ M1^ M M1
(21)
where v = % + in, v = % - in, M = x + iy, M = x - iy, r = 1 ln q, i2 = -1, 01 (% ) = 01 (% |-^ is theta function.
From (20) y = 0 andx = 0, we will get functional relations between T+(y), v+(y) and T+(x), v+(x), respectively, on the pieces A2B2 and A1B1 getting from the domain :
T+(x) = £ (-l)k-1 J0kß,n)YnG(k,n;x,0)dS + Jv+(t)G(t,0;x,0)dt+
k=l Ok q
1
+ J v+(t)G(0, t; x, 0)dt. (22)
2
T+(y) = t (-1 )k-1 f 0k£, n)d^Gtt, n;0,y)dS +f v+(t)G(t, 0;0,y)dt+
k=1 ok dn q
1
1
i
+ y v2+(t)G(o, t ;o, y)dt, (23)
After differentiating equalities (22) by x and (23) by y, we obtain
1 1
<+(*) = / v+(t) + /V2+(t) ^^^dt + (x), (24)
where F (x) = £ (-1)k-1 / , n) , n;x, o)dS, and
k=1 ok
T2+(y) = / v,+(t ) ^^^IMd, + j V2+(t ) ^^liMd, + F2(y), (25)
where F2(y)= j" (-1)k-1 f ,n)¿0«, n;0,y)dS.
k=1 ok
Further, the equations (16) and (17) we will rewrite in the form:
where
t)(t) + XjJ K(t,z)t¡(z)dz = Fj(t), (26)
q
Fj (t) = (2aj (t) - 1) Vj (t) + bj (t) + AjTj (q) (^ - 3q) , (27)
K(t, z) = ( 5++1 - 3z, q - z —t+1 t =( xatj = 1, (28)
V 7 I t++1 - z, t - z - ^, \yatj = 2. v 7
Note that we will search the function v.(t) from the class of C2(q, 1), which can tend to infinity of an order of less unit at t ^ q, and finite at t ^ 1. From here and owing to account (11), we will decide Fj(t) e C2(q, 1) and that Fj(t) can tend to infinity of an order of less unit at t ^ q, and finite at t ^ 1. Hence, by virtue
|Kj (t, z)| — const,
we will conclude that the equations (26) are the Volterra integral equations of the second kind which unequivocally solved by the method of consecutive approach [2] and its solution is represented in the form:
t+1
2
Tj(t) = Aj / Rj(t,z,Aj)Fj(z)dz + Fj(t), (29)
where Rj (t, z, Aj) are resolves of the kernels Kj (t, z), and that
Ir. (t, z, a. )| — const, (j = 1,2). (30)
1
Further, having excluded Tj(t) from the relations (24) and (29), we will receive system of the integral equations
x+1 2
(2ai(x) - 1 ) Vi (x) ± A f Ri (x, z, X1)v1(z)dz - f V±(t) dG(t^dt = ^(x)
y±1 2
(2a2(y) - 1)V2(y)±hf R2(y,z,A2)V2(z)dz-JV±(t)ï^^dt = ®2(y)
(31)
where
, x±1
1 2
®i(x) = jV±(t)dG(Q¿,xxQ)dt - A1 j R1(x,z, A1)b1(z)dz-b1(x)±F((x),
(32)
h(x) = b1(x)±
AMq, Q) 2
(5x - 6q ± 1) ;
1 ^ &2(y) = J V±(t) ^^^yQ^dt - hf R2(y, z, h)b2(z)dz - b2(y)± F±(y),
(33)
~b2(y)= b2 (y)± (5y - 6q ± 1).
G(t, Q; x, Q) = G(Q, t ;Q, x) = K1(x, t),
K1(x, t ) = 1 n
ln t ln x ln q
± k;(X, t )± K;(X, -t )
G(t, Q;Q, x) = G(Q, t ; x, Q) = K{2x, t ) = k[ x,-it ),
(34)
(35)
k;(X, t )= in
1 - tx
tx
± I (in
n=1
1 - q2ntx
t - q2nx
± in
1 - q-2ntx
t - q-2nx
(36)
Further, after some simplifications, from (31)-(36) we can get the system of singularity integral equation concerning v1(x) and v2(y) which represented on(in) the form [2]
( 1
V(x) = f V1(t )A1(x, t )dt ± c^1(x)
q
( f1
v2x) = f V2(t )A2(x, t )dt ± <I)2 (x)
(37)
where j (x) =
_ Jjx)
2a j (x)-1 '
Aj (x, t ) =
A Rj(x,z,Aj) K1(x,t) qsr x±1.
Aj 2aj(x)-1 2aj(x)-1, q - z - 2 ;
KK1 (x,z) x±1 <r z ^ 1
2aj(x)-1 , 2 - z - 1,
(38)
2 ln t
K (x, t) = ± K(x, t) ± K(x,-t), x in q
1
K (x, t ) = -n
E. n2nf —2nf f,—2n
q q i q i + q
l t _ q2nx 1 q2ntx 1 q-2ntx +
t - x 1 - tx n=1 V t - q2nx 1 - q2ntx 1 - q-2ntx t - q2nx Further, we will investigate the function $1(x):
. (40)
x+1 2
$1(x) = y v+(t)K1(x, it)dt - A1 y R1(x, z, A1)b1(z)dz - Mx)+
+/ *«, n ) dx ( ^^^^^ ) d n-/*«, n ) dx ( ^ ) d «+
-/ ««, n )dx (^) dn +./ ««, n) dx (^) d« =
o2 o2
= /*(, )KlM )dt - 2/*(«, n ) « dx ( ) d « -
>,(«, n ) d « + 2j m, n )
yi-«2 dx q(
« d f dG(«, n;x, oy, d« +
x/q2-«2 «M d«
. q ( )
+/««, n ) dx ( ^^ ) d «+ß (x)=
n
o
= a1 (x) + a2 (x) + a3 (x) + a4 (x) + a5 (x) + ß (x)
x+1 2
where ^(x) = -A1 / R1(x,z,A1)&1(z)dz-¿1 (x). q
Owing to account (9), (11), (21), (30), (39) and (40), we will get (see [2]): (x)| — const, |a1(x)| — const, |a2(x)| — const, |a3(x)| — const,
|a4(x)| — (x-q)Y-3const, |a5(x)| — (x-q)Y-3const, (2 < y< 3). (41)
Hence, by virtue (12) and (41), we will conclude that <I>1 (x) e C2(q, 1), and <i)1(x) can tend to infinity an order of less one at x ^ q, and at x ^ 1 it is limited.
It is similarly proved, that <i>2(y) e C2(q, 1), and $2(y) can tend to infinity an order of less one at y ^ q, and at y ^ 1 it is limited.
Thus, the system of integral equation (37) is reduced to the Fredholm integral equations of the second kind, by the known method Karleman-Vekua [4], just as in works [2], [3].
Note, that unique solvability of the Fredholm integral equations of the second kind follows from the uniqueness of solution of the Problem A and from the theory integral equations.
1
t
1
1
Solving the system of integral equations (37), we will found Vi(x) and v2(y) [2], further owing to account (24) and T1(q) = fa(q,0), T2(q) = 02(O,q) from (29) we will find Tj(t) (j=1,2).
Hence, after having found t1 (x) and v1(x) , t2(y) and v2(y), the solution of the Problem A can be restored in the domain of as the solution of the Problem N (20), and in domains Aj (j=1,2) as the solution of the Cauchy problem. The solution of the Problem A in domains of Dj(j= 1,2) can be restored as a solution of the Gaursat problem with conditions (6) and u(x,y) \BjEj = hj(t), where hj(t) (j=1,2) are traces of solution of the Cauchy problems in domains Aj (j=1,2), on the line x + y = q, and reciprocally in domains A*t (j=1,2) as the solution of the Cauchy-Gaursat problem with conditions
uy(x,Q) = Vj(x); ux(Q,y) = vj(y), -1 < x,y < -q and u(x,y)
jEj
= hj (t), where hj (t)
(j=1,2) are traces of solution of the Gaursat problems in domains Dj (j=1,2). And finally the solution of the Problem A can be restored in the domain of as the solution of the problem N, similarly as (20). The Theorem is proved.
References
1. Nakhushev A.M. Nagruzhennye uravneniya i ih prilozheniya [The loaded equations and their applications]. Moscow, Nauka Publ., 2012. 232 p.
2. Abdullaev O.Kh. Kraevaya zadacha dlya nagruzhennogo uravneniya 'elliptiko-giperbolicheskogo tipa v dvusvyaznoj oblasti [Boundary value problem for a loaded equation elliptic-hyperbolic type in double connected domain]. Vestnik KRAUNC. Fiziko-matematicheskie nauki - Bulletin of the Kamchatka Regional Association «Education-Scientific Center». Physical & Mathematical Sciences, 2014, vol. 8, no. 2, pp. 33-48.
3. Bitsadze A.V. Differential equations of Mixed type. New York, Pergamon Press, 1964. 160 p.
4. Iuskheleshvili N.I. Singulyarnye integral'nye uravneniya [Singularity integral equations]. Moscow, Nauka Publ., 1968. 512 p.
Original article submitted: 10.12.2014
