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INTEGRAL DESCRIPTION AND SOLUTION OF THE TILE TYPE PROBLEM FOR ONE OF THE FOURTH ORDER DIFFERENTIAL EQUATIONS OF MIXED GENERATION

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UDC 517.53.517.945
INTEGRAL DESCRIPTION AND SOLUTION OF THE TILE TYPE PROBLEM FOR ONE OF THE FOURTH ORDER DIFFERENTIAL EQUATIONS OF MIXED GENERATION
Sattorov A.S., Narzulloev Sh.Kh.
Tajik National University
The study of partial differential equations mainly developed at the end of the first half of the 20th century. in connection with solving physical problems. Many foreign and Soviet scientists are studying this class of equations [1-8].
A.V. Bitsadze [2] was the first to study the extremum principle for differential equations of mixed type when solving the Tricomi problem. M.V. Keldysh [3] proved the need to study the Tricomi problem for equations of mixed type depending on the accepted values of the coefficients of the lower order derivatives. E.I. Moiseev [4] solved the Tricomi problem for equations of mixed type in a special domain. MM. Smirnov [6] Tricomi problem for equations of mixed type in the researcher plane. In [7.8], boundary value problems for differential equations with singular coefficients of the second and higher orders are studied. In [9-14], an integral presented solution was found for one class of degenerate differential equations of the second and fourth orders. Some of these integral solution representations are used to solve Cauchy type problems in the characteristic domain.

In this article, for one of the degenerate differential equations of the fourth order of mixed kind in the plane, an integral representation of the solution is found using arbitrary functions. The found integral representations are used to solve a Cauchy type problem in the characteristic domain.
Let D be some region lying in the plane. The part of the region in which y>0 and y<0 is denoted by D^+ and D^- respectively. Equation (1) is of elliptic type in the region D^+ and of hyperbolic type in the region D^-.
In the region D^- consider the equation
L_(μ,ϑ) [N(x,y)L_(p,q) u]=0, (1)
Where
L_(p,q)≡xy ∂^2/(∂x^2 )+∂^2/(∂y^2 )+(1+2p)/2 y ∂/∂x+(6q-1)/2y ∂/∂y , (y<0)
N(x,y)=(x^((2+μ-p)/2) 〖(-y)〗^((4+3υ-3q)/2))/(9(q-υ)(q+υ-1)x+(p-μ)(p+μ-1) (-y)^3 ) ,
μ, ϑ, p and q are real numbers.
Let us study equation (1) in the region D^-.
Theorem 1. Let L_(μ,ϑ) u_(μ,ϑ)=0 and L_(p,q) u_(p,q)=0, then the regular solution of equation (1) in the domain D^- for μ < p , ϑ <q is described as follows
u(x,y)=u_(μ,ν) +x^((p-μ)/2) 〖(-y)〗^(3(q-ν)/2) u_(p,q). (2)
Proof. In equality (2) we apply the operator L_(μ,ϑ).
L_(μ,ϑ) u(x,y)=L_(μ,ϑ) u_(μ,ϑ)+L_(μ,ϑ) [x^((p-μ)/2) (-y)^(3(q-ν)/2) u_(p,q) ]=
=L_(μ,ϑ) [x^((p-μ)/2) (-y)^(3(q-ν)/2) u_(p,q) ]=xy (∂^2 [x^((p-μ)/2) (-y)^(3(q-ν)/2) u_(p,q) ])/(∂x^2 )+
+(∂^2 [x^((p-μ)/2) (-y)^(3(q-ν)/2) u_(p,q) ])/(∂y^2 )+(1+2μ)/2 y ∂[x^((p-μ)/2) (-y)^(3(q-ν)/2) u_(p,q) ]/∂x+
+(6ν-1)/2y ∂[x^((p-μ)/2) 〖(-y)〗^(3(q-ν)/2) u_(p,q) ]/∂y ,
After simplification, we calculate the corresponding derivatives with respect to x and y and find
L_(μ,ϑ) u(x,y)=1/4 x^((p-μ)/2-1) (-y)^(3(q-ν)/2-2)*
*[9(q-ν)(q+ν-1)x-(p-μ)(p+μ-1) (-y)^3 ] u_(p,q)
From the last equality we find
(x^((2+μ-p)/2) (-y)^((4+3ν-3q)/2))/(9(q-ν)(q+ν-1)x-(p-μ)(p+μ-1) (-y)^3 ) L_(μ,ϑ) u(x,y)=1/4 u_(p,q)
Let us apply the operator L_(p,q) to both sides of the last equality
L_(p,q) [(x^((2+μ-p)/2) (-y)^((4+3ν-3q)/2))/(9(q-ν)(q+ν-1)x-(p-μ)(p+μ-1) (-y)^3 ) L_(μ,ϑ) u(x,y)]=
=1/4 L_(p,q) u_(p,q)=0
Using Theorem 1, for equation (1), depending on the values of its coefficients, the integral representation of the solution is found using arbitrary functions.
Theorem 2. The regular solution of equation (1) in the domain D^- for 2μ>1,2ν>1,2p>1,2q>1 is described as follows
u(x,y)=A_μ A_ν T_(1-μ,1-ν) φ_1+A_p A_q x^(1/2 (p-μ) ) (-y)^(3/2 (q-ν)) T_(1-p,1-q) ψ_1, (4)
where φ_1,ψ_1 are arbitrary functions of one argument from the class C(D^-), and A_μ,A_ν,A_p,A_q are known constants,
T_(α.β) ψ_1=∫_0^1▒dσ ∫_0^1▒(ψ_1 [ω]dτ)/([σ(1-σ)]^α [τ(1-τ)]^β ),
ω=2√x (1-2σ)-(-y)^(3/2) (1-2τ).
Proof of Theorem 2. Here we show that the last equality (4) satisfies equation (1). In equality (4) we take the operator L_(μ,ν) we get
L_(μ,ν) u=L_(μ,ν) [T_(1-μ,1-ν) φ_1 ]+L_(μ,ν) [x^(1/2 (p-μ) ) (-y)^(3/2 (q-ν)) T_(1-p,1-q) ψ_1 ]=I_1+I_2,
Where

I_1=L_(μ,ν) [T_(1-μ,1-ν) φ_1 ]=0,
I_2=L_(μ,ν) [x^(1/2 (p-μ) ) (-y)^(3/2 (q-ν)) T_(1-p,1-q) ψ_1 ]=
=xy ∂^2/(∂x^2 ) [x^(1/2 (p-μ) ) (-y)^(3/2 (q-ν)) T_(1-p,1-q) ψ_1 ]+∂^2/(∂y^2 ) [x^(1/2 (p-μ) ) (-y)^(3/2 (q-ν)) T_(1-p,1-q) ψ_1 ]+
+(1+2μ)/2 y ∂/∂x [x^(1/2 (p-μ) ) (-y)^(3/2 (q-ν)) T_(1-p,1-q) ψ_1 ]+
+(6ν-1)/2y ∂/∂y [x^(1/2 (p-μ) ) (-y)^(3/2 (q-ν)) T_(1-p,1-q) ψ_1 ]. (5)
Let us calculate the partial derivatives of the first and second order from equality (5) and substitute them into (1):
u_x=1/2 (p-μ) x^(1/2 (p-μ)-1) (-y)^(3/2 (q-ν)) ∫_0^1▒dσ ∫_0^1▒(ψ_1 [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(1-q) )+
+x^(1/2 (p-μ)-1/2) (-y)^(3/2 (q-ν)) ∫_0^1▒dσ ∫_0^1▒(ψ_1^' [ω](1-2σ)dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(1-q) ),
I_x=∫_0^1▒dσ ∫_0^1▒(ψ_1^' [ω](1-2σ)dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(1-q) )=|■(u=ψ_1^' [ω]@du=-4√x ψ_1^'' [ω] )┤
├ ■(dϑ=((1-2σ))/[σ(1-σ)]^(1-p) @ϑ=1/p 1/[σ(1-σ)]^(-p) )┤|=4/p √x ∫_0^1▒dσ ∫_0^1▒(ψ_1^'' [ω]dτ)/([σ(1-σ)]^(-p) [τ(1-τ)]^(1-q) )
u_x=1/2 (p-μ) x^(1/2 (p-μ)-1) (-y)^(3/2 (q-ν)) ∫_0^1▒dσ ∫_0^1▒(ψ_1 [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(1-q) )+
+4/p x^(1/2 (p-μ) ) (-y)^(3/2 (q-ν)) ∫_0^1▒dσ ∫_0^1▒(ψ_1^'' [ω]dτ)/([σ(1-σ)]^(-p) [τ(1-τ)]^(1-q) ),
u_xx=1/2 (p-μ)[1/2 (p-μ)-1] x^(1/2 (p-μ)-2) (-y)^(3/2 (q-ν)) ∫_0^1▒dσ ∫_0^1▒(ψ_1 [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(1-q) )-
-2/p (2μ+1) x^(1/2 (p-μ)-1) (-y)^(3/2 (q-ν)) ∫_0^1▒dσ ∫_0^1▒(ψ_1^'' [ω]dτ)/([σ(1-σ)]^(-p) [τ(1-τ)]^(1-q) )+
+x^(1/2 (p-μ)-1) (-y)^(3/2 (q-ν)) ∫_0^1▒dσ ∫_0^1▒(ψ_1^'' [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(1-q) ),
u_y=-3/2 (q-ν) x^(1/2 (p-μ) ) (-y)^(3/2 (q-ν)-1) ∫_0^1▒dσ ∫_0^1▒(ψ_1 [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(1-q) )+
+x^(1/2 (p-μ) ) (-y)^(3/2 (q-ν)+1/2) ∫_0^1▒dσ ∫_0^1▒(ψ_1^' [ω](1-2τ)dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(1-q) ),
I_y=∫_0^1▒dσ ∫_0^1▒(ψ_1^' [ω](1-2τ)dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(1-q) )=|■(u=ψ_1^' [ω]@du=4/3 (-y)^(3/2) ψ_1^'' [ω] )┤
├ ■(dϑ=((1-2τ))/[τ(1-τ)]^(1-q) @ϑ=1/q 1/[τ(1-τ)]^(-q) )┤|=-4/3q (-y)^(3/2) ∫_0^1▒dσ ∫_0^1▒(ψ_1^'' [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(-q) ),
u_y=-3/2 (q-ν) x^(1/2 (p-μ) ) (-y)^(3/2 (q-ν)-1) ∫_0^1▒dσ ∫_0^1▒(ψ_1 [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(1-q) )-
-4/3q x^(1/2 (p-μ) ) (-y)^(3/2 (q-ν)+2) ∫_0^1▒dσ ∫_0^1▒(ψ_1^'' [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(-q) ),
u_yy=3/2 (q-ν)[3/2 (q-ν)-1] x^(1/2 (p-μ) ) (-y)^(3/2 (q-ν)-2)*
*∫_0^1▒dσ ∫_0^1▒(ψ_1 [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(1-q) )-2/3q (6ν-1) x^(1/2 (p-μ) ) (-y)^(3/2 (q-ν)+1)*
*∫_0^1▒dσ ∫_0^1▒(ψ_1^'' [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(-q) )+x^(1/2 (p-μ) ) (-y)^(3/2 (q-ν)+1)*
*∫_0^1▒dσ ∫_0^1▒(ψ_1^'' [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(1-q) ),
I_2=L_(μ,ν) [x^(1/2 (p-μ) ) (-y)^(3/2 (q-ν)) T_(1-p,1-q) ψ_1 ]=
=-1/2 (p-μ)[1/2 (p-μ)-1] x^(1/2 (p-μ)-1) (-y)^(3/2 (q-ν)+1) ∫_0^1▒dσ ∫_0^1▒(ψ_1 [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(1-q) )+
+2/p (2μ+1) x^(1/2 (p-μ) ) (-y)^(3/2 (q-ν)+1) ∫_0^1▒dσ ∫_0^1▒(ψ_1^'' [ω]dτ)/([σ(1-σ)]^(-p) [τ(1-τ)]^(1-q) )-
-x^(1/2 (p-μ) ) (-y)^(3/2 (q-ν)+1) ∫_0^1▒dσ ∫_0^1▒(ψ_1^'' [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(1-q) )+
+3/2 (q-ν)[3/2 (q-ν)-1] x^(1/2 (p-μ) ) (-y)^(3/2 (q-ν)-2) ∫_0^1▒dσ ∫_0^1▒(ψ_1 [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(1-q) )-
-2/3q (6ν-1) x^(1/2 (p-μ) ) (-y)^(3/2 (q-ν)+1) ∫_0^1▒dσ ∫_0^1▒(ψ_1^'' [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(-q) )+
+x^(1/2 (p-μ) ) (-y)^(3/2 (q-ν)+1) ∫_0^1▒dσ ∫_0^1▒(ψ_1^'' [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(1-q) )-
-(1+2μ)/2 1/2 (p-μ) x^(1/2 (p-μ)-1) (-y)^(3/2 (q-ν)+1) ∫_0^1▒dσ ∫_0^1▒(ψ_1 [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(1-q) )-
-2(1+2μ)/p x^(1/2 (p-μ) ) (-y)^(3/2 (q-ν)+1) ∫_0^1▒dσ ∫_0^1▒(ψ_1^'' [ω]dτ)/([σ(1-σ)]^(-p) [τ(1-τ)]^(1-q) )+
+(6ν-1)/2 3/2 (q-ν) x^(1/2 (p-μ) ) (-y)^(3/2 (q-ν)-2) ∫_0^1▒dσ ∫_0^1▒(ψ_1 [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(1-q) )+
+(6ν-1)/2 4/3q x^(1/2 (p-μ) ) (-y)^(3/2 (q-ν)+1) ∫_0^1▒dσ ∫_0^1▒〖(ψ_1^'' [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(-q) )=〗
=-1/4 (p-μ)(p+μ-1) x^(1/2 (p-μ)-1) (-y)^(3/2 (q-ν)+1) ∫_0^1▒dσ ∫_0^1▒(ψ_1 [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(1-q) )+
+9/4 (q-ν)(q+ν-1) x^(1/2 (p-μ) ) (-y)^(3/2 (q-ν)-2) ∫_0^1▒dσ ∫_0^1▒(ψ_1 [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(1-q) ),
We simplify the found partial derivatives in I_2 and get
I_2=L_(μ,ν) [x^(1/2 (p-μ) ) (-y)^(3/2 (q-ν)) T_(1-p,1-q) ψ_1 ]=
=-1/4 (p-μ)(p+μ-1) x^(1/2 (p-μ)-1) (-y)^(3/2 (q-ν)+1) ∫_0^1▒dσ ∫_0^1▒(ψ_1 [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(1-q) )+
+9/4 (q-ν)(q+ν-1) x^(1/2 (p-μ) ) (-y)^(3/2 (q-ν)-2) ∫_0^1▒dσ ∫_0^1▒(ψ_1 [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(1-q) )=
=1/4 [9(q-ν)(q+ν-1)x-(p-μ)(p+μ-1) (-y)^3 ]*
*x^(1/2 (p-μ)-1) (-y)^(3/2 (q-ν)-2) ∫_0^1▒dσ ∫_0^1▒(ψ_1 [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(1-q) ),
I_2=1/4 [9(q-ν)(q+ν-1)x-(p-μ)(p+μ-1) (-y)^3 ]*
*x^(1/2 (p-μ)-1) (-y)^(3/2 (q-ν)-2) ∫_0^1▒dσ ∫_0^1▒(ψ_1 [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(1-q) ).
After simplification we obtain the following equality
L_(μ,ϑ) u(x,y)=1/4 [9(q-ν)(q+ν-1)x-(1-p-μ)(p-μ) (-y)^3 ]*
*x^(1/2 (p-μ)-1) (-y)^(3/2 (q-ν)-2) ∫_0^1▒dσ ∫_0^1▒(ψ_1 [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(1-q) ).
(x^((2+μ-p)/2) (-y)^((4+3ν-3q)/2))/[9(q-ν)(q+ν-1)x-(1-p-μ)(p-μ) (-y)^3 ] L_(μ,ϑ) u(x,y)=
=1/4 ∫_0^1▒dσ ∫_0^1▒(ψ_1 [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(1-q) )
Let us apply the operator L_(p,q)- to both parts of the latter the equality
L_(p,q) [(x^((2+μ-p)/2) (-y)^((4+3q-3ν)/2))/[9(q-ν)(q+ν-1)x-(1-p-μ)(p-μ) (-y)^3 ] L_(μ,ϑ) u(x,y)]=
=L_(p,q) [1/4 ∫_0^1▒dσ ∫_0^1▒(ψ_1 [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(1-q) )]
It is easy to see that the right side of the last equality is equal to zero, i.e.
L_(p,q) [1/4 ∫_0^1▒dσ ∫_0^1▒(ψ_1 [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(1-q) )]=
=xy/4 ∂^2/(∂x^2 ) [T_(1-p,1-q) ψ_1 ]+1/4 ∂^2/(∂y^2 ) [T_(1-p,1-q) ψ_1 ]+
+(1+2p)/8 y ∂/∂x [T_(1-p,1-q) ψ_1 ]+(6q-1)/8y ∂/∂y [T_(1-p,1-q) ψ_1 ] (6)
Let us calculate the partial derivatives of the first and second order from equality (6):
u_x=x^(-1/2) ∫_0^1▒dσ ∫_0^1▒〖(ψ_1^' [ω](1-2σ)dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(1-q) ),〗
I_x=∫_0^1▒dσ ∫_0^1▒(ψ_1^' [ω](1-2σ)dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(1-q) )=|■(u=ψ_1^' [ω]@du=-4√x ψ_1^'' [ω] )┤
├ ■(dϑ=((1-2σ))/[σ(1-σ)]^(1-p) @ϑ=1/p 1/[σ(1-σ)]^(-p) )┤|=4/p √x ∫_0^1▒dσ ∫_0^1▒(ψ_1^'' [ω]dτ)/([σ(1-σ)]^(-p) [τ(1-τ)]^(1-q) )
u_x=4/p ∫_0^1▒dσ ∫_0^1▒(ψ_1^'' [ω]dτ)/([σ(1-σ)]^(-p) [τ(1-τ)]^(1-q) ),
u_xx=-2(1+2p)/p x^(-1) ∫_0^1▒dσ ∫_0^1▒(ψ_1^'' [ω]dτ)/([σ(1-σ)]^(-p) [τ(1-τ)]^(1-q) )+
+x^(-1) ∫_0^1▒dσ ∫_0^1▒(ψ_1^'' [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(1-q) )
u_y=(-y)^(1/2) ∫_0^1▒dσ ∫_0^1▒(ψ_1^' [ω](1-2τ)dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(1-q) ),
I_y=∫_0^1▒dσ ∫_0^1▒(ψ_1^' [ω](1-2τ)dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(1-q) )=|■(u=ψ_1^' [ω]@du=4/3 (-y)^(3/2) ψ_1^'' [ω] )┤
├ ■(dϑ=((1-2τ))/[τ(1-τ)]^(1-q) @ϑ=1/q 1/[τ(1-τ)]^(-q) )┤|=-4/3q (-y)^(3/2) ∫_0^1▒dσ ∫_0^1▒(ψ_1^'' [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(-q) ),
u_y=-(4y^2)/3q ∫_0^1▒dσ ∫_0^1▒(ψ_1^'' [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(-q) ),
u_yy=2y(6q-1)/3q ∫_0^1▒dσ ∫_0^1▒(ψ_1^'' [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(-q) )-y∫_0^1▒dσ ∫_0^1▒(ψ_1^'' [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(1-q) ).
L_(p,q) [1/4 ∫_0^1▒dσ ∫_0^1▒(ψ_1 [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(1-q) )]=
=-y(1+2p)/2p ∫_0^1▒dσ ∫_0^1▒(ψ_1^'' [ω]dτ)/([σ(1-σ)]^(-p) [τ(1-τ)]^(1-q) )+y/4 ∫_0^1▒dσ ∫_0^1▒(ψ_1^'' [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(1-q) )+
+y(6q-1)/6q ∫_0^1▒dσ ∫_0^1▒(ψ_1^'' [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(-q) )-y/4 ∫_0^1▒dσ ∫_0^1▒(ψ_1^'' [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(1-q) )+
+(1+2p)/2 y/p ∫_0^1▒dσ ∫_0^1▒(ψ_1^'' [ω]dτ)/([σ(1-σ)]^(-p) [τ(1-τ)]^(1-q) )-
(6q-1)/2 y/3q ∫_0^1▒dσ ∫_0^1▒(ψ_1^'' [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(-q) )=0,
L_(p,q) [1/4 ∫_0^1▒dσ ∫_0^1▒(ψ_1 [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(1-q) )]=0
The theorem has been proven.
Theorem 3. The regular solution of equation (1) in the domain D^- for 0<2μ<1,2ν>1,2p>1,2q>1 is assigned as follows
u(x,y)=A_μ A_ν T_(1-μ,1-ν) φ_1+A_(1-μ) A_ν x^(1/2 (1-2μ) ) T_(μ,1-ν) φ_2+
+A_p A_q x^(1/2 (p-μ) ) (-y)^(3/2 (q-ν)) T_(1-p,1-q) ψ_1, (6)
where φ_1,φ_2,ψ_1 are arbitrary functions of one argument from the class C(D^-), and A_μ,A_(1-μ),A_ν,A_p,A_q are known constants.
Theorem 4. The regular solution of equation (1) in the domain D^- for 2μ>1,0<2ν<1,2p>1,2q>1 is assigned as follows
u(x,y)=A_μ A_ν T_(1-μ,1-ν) φ_1+A_μ A_(1-ν) (-y)^(3/2 (1-2ν)) T_(1-μ,ν) (φ_2 ) ̃+
+A_p A_q x^(1/2 (p-μ) ) (-y)^(3/2 (q-ν)) T_(1-p,1-q) ψ_1, (7)
where φ_1,(φ_2 ) ̃,ψ_1 are arbitrary functions of one argument from the class C^2(D^-), and A_μ,〖A_ν,A〗_(1-ν),A_p,A_q are known constants.
Theorem 5. The regular solution of equation (1) in the domain D^- for 2μ>1,2ν>1,0<2p<1,2q>1 is described as follows
u(x,y)=A_μ A_ν T_(1-μ,1-ν) φ_1+A_p A_q x^(1/2 (p-μ) ) (-y)^(3/2 (q-ν)) T_(1-p,1-q) ψ_1+
+A_(1-p) A_q x^(1/2 (1-p-μ) ) (-y)^(3/2 (q-ν)) T_(p,1-q) ψ_2, (8)
where φ_1,ψ_1,ψ_2 are arbitrary functions of one argument from the class C^2 (D^-), and A_μ,A_ν,A_p,A_(1-p),A_q are known constants.
Proof of Theorem 5. Here we show that the last integral (8) satisfies equation (1).
Let us apply the operator L_(μ,ν) to both sides of equality (8).
L_(μ,ν) u=L_(μ,ν) [T_(1-μ,1-ν) φ_1 ]+L_(μ,ν) [x^(1/2 (p-μ) ) (-y)^(3/2 (q-ν)) T_(1-p,1-q) ψ_1 ]+
+L_(μ,ν) [x^(1/2 (1-p-μ) ) (-y)^(3/2 (q-ν)) T_(p,1-q) ψ_2 ]=I_1+I_2+I_3
Expressions I_1, I_2, I_3 are calculated separately:
I_1=L_(μ,ν) [T_(1-μ,1-ν) φ_1 ]=0
I_2=L_(μ,ν) [x^(1/2 (p-μ) ) (-y)^(3/2 (q-ν)) T_(1-p,1-q) ψ_1 ]=0
I_3=L_(μ,ν) [x^(1/2 (1-p-μ) ) (-y)^(3/2 (q-ν)) T_(p,1-q) ψ_2 ]=xy ∂^2/(∂x^2 ) [x^(1/2 (1-p-μ) ) (-y)^(3/2 (q-ν)) T_(p,1-q) ψ_2 ]+
+∂^2/(∂y^2 ) [x^(1/2 (1-p-μ) ) (-y)^(3/2 (q-ν)) T_(p,1-q) ψ_2 ]+(1+2μ)/2 y ∂/∂x [x^(1/2 (1-p-μ) ) (-y)^(3/2 (q-ν)) T_(p,1-q) ψ_2 ]+
+(6ν-1)/2y ∂/∂y [x^(1/2 (1-p-μ) ) (-y)^(3/2 (q-ν)) T_(p,1-q) ψ_2 ] (9)
Let us calculate the partial derivatives of the first and second order in equalities (9) and substitute them into (1) to obtain:
u_x=1/2 (1-p-μ) x^(1/2 (1-p-μ)-1) (-y)^(3/2 (q-ν)) ∫_0^1▒dσ ∫_0^1▒(ψ_2 [ω]dτ)/([σ(1-σ)]^p [τ(1-τ)]^(1-q) )+
+x^(1/2 (1-p-μ)-1/2) (-y)^(3/2 (q-ν)) ∫_0^1▒dσ ∫_0^1▒(ψ_2^' [ω](1-2σ)dτ)/([σ(1-σ)]^p [τ(1-τ)]^(1-q) ),
I_x=∫_0^1▒dσ ∫_0^1▒(ψ_2^' [ω](1-2σ)dτ)/([σ(1-σ)]^p [τ(1-τ)]^(1-q) )=|■(u=ψ_2^' [ω]@du=-4√x ψ_2^'' [ω] )┤
├ ■(dϑ=((1-2σ))/[σ(1-σ)]^p @ϑ=1/(1-p) 1/[σ(1-σ)]^(p-1) )┤|=4/(1-p) √x ∫_0^1▒dσ ∫_0^1▒(ψ_1^'' [ω]dτ)/([σ(1-σ)]^(p-1) [τ(1-τ)]^(1-q) )
u_x=1/2 (1-p-μ) x^(1/2 (1-p-μ)-1) (-y)^(3/2 (q-ν)) ∫_0^1▒dσ ∫_0^1▒(ψ_2 [ω]dτ)/([σ(1-σ)]^p [τ(1-τ)]^(1-q) )+
+4/(1-p) x^(1/2 (1-p-μ) ) (-y)^(3/2 (q-ν)) ∫_0^1▒dσ ∫_0^1▒(ψ_2^' [ω]dτ)/([σ(1-σ)]^(p-1) [τ(1-τ)]^(1-q) ),
u_xx=1/2 (1-p-μ)[1/2 (1-p-μ)-1] x^(1/2 (1-p-μ)-2) (-y)^(3/2 (q-ν))*
*∫_0^1▒dσ ∫_0^1▒(ψ_2 [ω]dτ)/([σ(1-σ)]^p [τ(1-τ)]^(1-q) )+
+2/(1-p) (1+2μ) x^(1/2 (1-p-μ)-1) (-y)^(3/2 (q-ν)) ∫_0^1▒dσ ∫_0^1▒(ψ_2^'' [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(1-q) )+
+x^(1/2 (1-p-μ)-1) (-y)^(3/2 (q-ν)) ∫_0^1▒dσ ∫_0^1▒(ψ_2^'' [ω]dτ)/([σ(1-σ)]^p [τ(1-τ)]^(1-q) ),
u_y=-3/2 (q-ν) x^(1/2 (1-p-μ) ) (-y)^(3/2 (q-ν)-1) ∫_0^1▒dσ ∫_0^1▒(ψ_2 [ω]dτ)/([σ(1-σ)]^p [τ(1-τ)]^(1-q) )+
+x^(1/2 (1-p-μ) ) (-y)^(3/2 (q-ν)-1/2) ∫_0^1▒dσ ∫_0^1▒(ψ_2^' [ω](1-2τ)dτ)/([σ(1-σ)]^p [τ(1-τ)]^(1-q) ),
I_y=∫_0^1▒dσ ∫_0^1▒(ψ_2^' [ω](1-2τ)dτ)/([σ(1-σ)]^p [τ(1-τ)]^(1-q) )=|■(u=ψ_2^' [ω]@du=4/3 (-y)^(3/2) ψ_2^'' [ω] )┤
├ ■(dϑ=((1-2τ))/[τ(1-τ)]^(1-q) @ϑ=1/q 1/[τ(1-τ)]^(-q) )┤|=-4/3q (-y)^(3/2) ∫_0^1▒dσ ∫_0^1▒(ψ_2^'' [ω]dτ)/([σ(1-σ)]^p [τ(1-τ)]^(-q) ),
u_y=-3/2 (q-ν) x^(1/2 (1-p-μ) ) (-y)^(3/2 (q-ν)-1) ∫_0^1▒dσ ∫_0^1▒(ψ_2 [ω]dτ)/([σ(1-σ)]^p [τ(1-τ)]^(1-q) )-
-4/3q x^(1/2 (1-p-μ) ) (-y)^(3/2 (q-ν)-1) ∫_0^1▒dσ ∫_0^1▒(ψ_2^'' [ω]dτ)/([σ(1-σ)]^p [τ(1-τ)]^(-q) ),
u_yy=3/2 (q-ν)[3/2 (q-ν)-1] x^(1/2 (1-p-μ) ) (-y)^(3/2 (q-ν)-2)*
*∫_0^1▒dσ ∫_0^1▒(ψ_2 [ω]dτ)/([σ(1-σ)]^p [τ(1-τ)]^(1-q) )-
-2(6ν+1)/3q x^(1/2 (1-p-μ) ) (-y)^(3/2 (q-ν)-1) ∫_0^1▒dσ ∫_0^1▒(ψ_2^'' [ω]dτ)/([σ(1-σ)]^p [τ(1-τ)]^(-q) )+
+x^(1/2 (1-p-μ) ) (-y)^(3/2 (q-ν)-1) ∫_0^1▒dσ ∫_0^1▒(ψ_2^'' [ω]dτ)/([σ(1-σ)]^p [τ(1-τ)]^(1-q) ),
We obtain partial derivatives in I_3
I_3=L_(μ,ν) [x^(1/2 (1-p-μ) ) T_(p,1-q) ψ_2 ]=
=-1/2 (1-p-μ)[1/2 (1-p-μ)-1] x^(1/2 (1-p-μ)-1) (-y)^(3/2 (q-ν)+1)*
*∫_0^1▒dσ ∫_0^1▒(ψ_2 [ω]dτ)/([σ(1-σ)]^p [τ(1-τ)]^(1-q) )-
-2/(1-p) (1+2μ) x^(1/2 (1-p-μ) ) (-y)^(3/2 (q-ν)+1) ∫_0^1▒dσ ∫_0^1▒(ψ_2^'' [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(1-q) )-
-x^(1/2 (1-p-μ) ) (-y)^(3/2 (q-ν)+1) ∫_0^1▒dσ ∫_0^1▒(ψ_2^'' [ω]dτ)/([σ(1-σ)]^p [τ(1-τ)]^(1-q) )+
+3/2 (q-ν)[3/2 (q-ν)-1] x^(1/2 (1-p-μ) ) (-y)^(3/2 (q-ν)-2) ∫_0^1▒dσ ∫_0^1▒(ψ_2 [ω]dτ)/([σ(1-σ)]^p [τ(1-τ)]^(1-q) )-
-2(6ν+1)/3q x^(1/2 (1-p-μ) ) (-y)^(3/2 (q-ν)-1) ∫_0^1▒dσ ∫_0^1▒(ψ_2^'' [ω]dτ)/([σ(1-σ)]^p [τ(1-τ)]^(-q) )+
+x^(1/2 (1-p-μ) ) (-y)^(3/2 (q-ν)-1) ∫_0^1▒dσ ∫_0^1▒(ψ_2^'' [ω]dτ)/([σ(1-σ)]^p [τ(1-τ)]^(1-q) )-
-(1+2μ)/2 1/2 (1-p-μ) x^(1/2 (1-p-μ)-1) (-y)^(3/2 (q-ν)+1) ∫_0^1▒dσ ∫_0^1▒(ψ_1 [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(1-q) )-
-(1+2μ)/2 4/(1-p) x^(1/2 (1-p-μ) ) (-y)^(3/2 (q-ν)+1) ∫_0^1▒dσ ∫_0^1▒(ψ_1^'' [ω]dτ)/([σ(1-σ)]^(-p) [τ(1-τ)]^(1-q) )+
+(6ν-1)/2 3/2 (q-ν) x^(1/2 (1-p-μ) ) (-y)^(3/2 (q-ν)-2) ∫_0^1▒dσ ∫_0^1▒(ψ_2 [ω]dτ)/([σ(1-σ)]^p [τ(1-τ)]^(1-q) )-
-(6ν-1)/2 4/3q x^(1/2 (1-p-μ) ) (-y)^(3/2 (q-ν)-2) ∫_0^1▒dσ ∫_0^1▒(ψ_2^'' [ω]dτ)/([σ(1-σ)]^p [τ(1-τ)]^(-q) )=
=-1/4 (1-p-μ)(p-μ) x^(1/2 (1-p-μ)-1) (-y)^(3/2 (q-ν)+1) ∫_0^1▒dσ ∫_0^1▒(ψ_2 [ω]dτ)/([σ(1-σ)]^p [τ(1-τ)]^(1-q) )+
+9/4 (q-ν)(q+ν-1) x^(1/2 (1-p-μ) ) (-y)^(3/2 (q-ν)-2) ∫_0^1▒dσ ∫_0^1▒(ψ_2 [ω]dτ)/([σ(1-σ)]^p [τ(1-τ)]^(1-q) ),
I_3=L_(μ,ν) [x^(1/2 (1-p-μ) ) T_(p,1-q) ψ_2 ]=
=-1/4 (1-p-μ)(p-μ) x^(1/2 (1-p-μ)-1) (-y)^(3/2 (q-ν)+1) ∫_0^1▒dσ ∫_0^1▒(ψ_2 [ω]dτ)/([σ(1-σ)]^p [τ(1-τ)]^(1-q) )+
+9/4 (q-ν)(q+ν-1) x^(1/2 (1-p-μ) ) (-y)^(3/2 (q-ν)-2) ∫_0^1▒dσ ∫_0^1▒(ψ_2 [ω]dτ)/([σ(1-σ)]^p [τ(1-τ)]^(1-q) )=
=1/4 [9(q-ν)(q+ν-1)x-(1-p-μ)(p-μ) (-y)^3 ]*
*x^(1/2 (p-μ-2) ) (-y)^(3/2 (q-ν)-2) x^(1/2 (1-2p) ) ∫_0^1▒dσ ∫_0^1▒(ψ_2 [ω]dτ)/([σ(1-σ)]^p [τ(1-τ)]^(1-q) ),
In (9) substituting I_1, I_2, I_3 and we get
L_(μ,ϑ) u(x,y)=1/4 [9(q-ν)(q+ν-1)x-(1-p-μ)(p-μ) (-y)^3 ]*
*x^(1/2 (1-p-μ) ) (-y)^(3/2 (q-ν)-2) x^(1/2 (1-2p) ) ∫_0^1▒dσ ∫_0^1▒(ψ_2 [ω]dτ)/([σ(1-σ)]^p [τ(1-τ)]^(1-q) ).
(x^((2+μ-p)/2) (-y)^((4-3q+3ν)/2))/[9(q-ν)(q+ν-1)x-(1-p-μ)(p-μ) (-y)^3 ] L_(μ,ϑ) u(x,y)=
=1/4 x^(1/2 (1-2p) ) ∫_0^1▒dσ ∫_0^1▒(ψ_2 [ω]dτ)/([σ(1-σ)]^p [τ(1-τ)]^(1-q) )
Apply the operator L_(p,q) to the last two equalities
L_(p,q) [(x^((2+μ-p)/2) (-y)^((4-3q+3ν)/2))/[9(q-ν)(q+ν-1)x-(1-p-μ)(p-μ) (-y)^3 ] L_(μ,ϑ) u(x,y)]=
=L_(p,q) [1/4 x^(1/2 (1-2p) ) ∫_0^1▒dσ ∫_0^1▒(ψ_2 [ω]dτ)/([σ(1-σ)]^p [τ(1-τ)]^(1-q) )]
We write the right side of the resulting equation separately
L_(p,q) [1/4 x^(1/2 (1-2p) ) ∫_0^1▒dσ ∫_0^1▒(ψ_2 [ω]dτ)/([σ(1-σ)]^p [τ(1-τ)]^(1-q) )]=
=xy/4 ∂^2/(∂x^2 ) [x^(1/2 (1-2p) ) T_(p,1-q) ψ_2 ]+1/4 ∂^2/(∂y^2 ) [x^(1/2 (1-2p) ) T_(p,1-q) ψ_2 ]+
+(1+2p)/8 y ∂/∂x [x^(1/2 (1-2p) ) T_(p,1-q) ψ_2 ]+(6q-1)/8y ∂/∂y [x^(1/2 (1-2p) ) T_(p,1-q) ψ_2 ] (10)
Let us calculate the partial derivatives of the first and second order from equality (10):
u_x=1/2 (1-2p) x^(1/2 (1-2p)-1) ∫_0^1▒dσ ∫_0^1▒(ψ_2 [ω]dτ)/([σ(1-σ)]^p [τ(1-τ)]^(1-q) )+
+x^(1/2 (1-2p)-1/2) ∫_0^1▒dσ ∫_0^1▒(ψ_2^' [ω](1-2σ)dτ)/([σ(1-σ)]^p [τ(1-τ)]^(1-q) ),
I_x=∫_0^1▒dσ ∫_0^1▒(ψ_2^' [ω](1-2σ)dτ)/([σ(1-σ)]^p [τ(1-τ)]^(1-q) )=|■(u=ψ_2^' [ω]@du=-4√x ψ_2^'' [ω] )┤
├ ■(dϑ=((1-2σ))/[σ(1-σ)]^p @ϑ=1/(1-p) 1/[σ(1-σ)]^(p-1) )┤|=4/(1-p) √x ∫_0^1▒dσ ∫_0^1▒(ψ_1^'' [ω]dτ)/([σ(1-σ)]^(p-1) [τ(1-τ)]^(1-q) )

u_x=1/2 (1-2p) x^(1/2 (1-2p)-1) ∫_0^1▒dσ ∫_0^1▒(ψ_2 [ω]dτ)/([σ(1-σ)]^p [τ(1-τ)]^(1-q) )+
+4/(1-p) x^(1/2 (1-2p) ) ∫_0^1▒dσ ∫_0^1▒(ψ_1^'' [ω]dτ)/([σ(1-σ)]^(p-1) [τ(1-τ)]^(1-q) ),
u_xx=1/2 (1-2p)[1/2 (1-2p)-1] x^(1/2 (1-2p)-2) ∫_0^1▒dσ ∫_0^1▒(ψ_2 [ω]dτ)/([σ(1-σ)]^p [τ(1-τ)]^(1-q) )+
-2(1-2p)/(1-p) 1/2 x^(1/2 (1-2p)-1) ∫_0^1▒dσ ∫_0^1▒(ψ_1^'' [ω](1-2σ)dτ)/([σ(1-σ)]^(p-1) [τ(1-τ)]^(1-q) )+
+x^(1/2 (1-2p)-1) ∫_0^1▒dσ ∫_0^1▒(ψ_1^'' [ω]dτ)/([σ(1-σ)]^p [τ(1-τ)]^(1-q) ).
u_y=x^(1/2 (1-2p) ) (-y)^(1/2) ∫_0^1▒dσ ∫_0^1▒(ψ_1^' [ω](1-2τ)dτ)/([σ(1-σ)]^p [τ(1-τ)]^(1-q) ),
I_y=∫_0^1▒dσ ∫_0^1▒(ψ_2^' [ω](1-2τ)dτ)/([σ(1-σ)]^p [τ(1-τ)]^(1-q) )=|■(u=ψ_2^' [ω]@du=4/3 (-y)^(3/2) ψ_2^'' [ω] )┤
├ ■(dϑ=((1-2τ))/[τ(1-τ)]^(1-q) @ϑ=1/q 1/[τ(1-τ)]^(-q) )┤|=-4/3q (-y)^(3/2) ∫_0^1▒dσ ∫_0^1▒(ψ_2^'' [ω]dτ)/([σ(1-σ)]^p [τ(1-τ)]^(-q) ).
u_y=-(4y^2)/3q x^(1/2 (1-2p) ) ∫_0^1▒dσ ∫_0^1▒(ψ_1^'' [ω]dτ)/([σ(1-σ)]^p [τ(1-τ)]^(-q) ),
u_yy=2y(6q-1)/3q x^(1/2 (1-2p) ) ∫_0^1▒dσ ∫_0^1▒(ψ_1^'' [ω]dτ)/([σ(1-σ)]^p [τ(1-τ)]^(-q) )-
-yx^(1/2 (1-2p) ) ∫_0^1▒dσ ∫_0^1▒(ψ_1^'' [ω]dτ)/([σ(1-σ)]^p [τ(1-τ)]^(1-q) ).
We put the results obtained on the right side of (10).
y/8 (1-2p)[1/2 (1-2p)-1] x^(1/2 (1-2p)-1) ∫_0^1▒dσ ∫_0^1▒(ψ_2 [ω]dτ)/([σ(1-σ)]^p [τ(1-τ)]^(1-q) )+
-y(1-2p)/4(1-p) x^(1/2 (1-2p) ) ∫_0^1▒dσ ∫_0^1▒(ψ_1^'' [ω](1-2σ)dτ)/([σ(1-σ)]^(p-1) [τ(1-τ)]^(1-q) )+
+y/4 x^(1/2 (1-2p) ) ∫_0^1▒dσ ∫_0^1▒(ψ_1^'' [ω]dτ)/([σ(1-σ)]^p [τ(1-τ)]^(1-q) )+
+y(6q-1)/6q x^(1/2 (1-2p) ) ∫_0^1▒dσ ∫_0^1▒(ψ_1^'' [ω]dτ)/([σ(1-σ)]^p [τ(1-τ)]^(-q) )-
-y/4 x^(1/2 (1-2p) ) ∫_0^1▒dσ ∫_0^1▒(ψ_1^'' [ω]dτ)/([σ(1-σ)]^p [τ(1-τ)]^(1-q) )+
+(1+2p)/8 y 1/2 (1-2p) x^(1/2 (1-2p)-1) ∫_0^1▒dσ ∫_0^1▒(ψ_2 [ω]dτ)/([σ(1-σ)]^p [τ(1-τ)]^(1-q) )+
+(1+2p)/8 y 4/(1-p) x^(1/2 (1-2p) ) ∫_0^1▒dσ ∫_0^1▒(ψ_1^'' [ω]dτ)/([σ(1-σ)]^(p-1) [τ(1-τ)]^(1-q) )-
-(6q-1)/8y (4y^2)/3q x^(1/2 (1-2p) ) ∫_0^1▒dσ ∫_0^1▒(ψ_1^'' [ω]dτ)/([σ(1-σ)]^p [τ(1-τ)]^(-q) )=0
L_(p,q) [(x^((2+μ-p)/2) (-y)^((4+3q-3ν)/2))/[9(q-ν)(q+ν-1)x-(1-p-μ)(p-μ) (-y)^3 ] L_(μ,ϑ) u(x,y)]=0
The theorem has been proven.
Theorem 6. The regular solution of equation (1) in the domain D^- for 2μ>1,2ν>1,2p>1,0<2q<1 is described as follows
u(x,y)=A_μ A_ν T_(1-μ,1-ν) φ_1+A_p A_q x^(1/2 (p-μ) ) (-y)^(3/2 (q-ν)) T_(1-p,1-q) ψ_1+
+A_p A_(1-q) x^(1/2 (p-μ) ) (-y)^(3/2 (1-q-ν)) T_(1-p,q) (ψ_2 ) ̃, (11)
where φ_1,ψ_1,(ψ_2 ) ̃ are arbitrary functions of one argument from the class C^2(D^-), and A_μ,A_ν,A_p,〖A_q,A〗_(1-q) are known constants.
Proof of Theorem 6. Here we show that the last integral (11) satisfies equation (1). In equality (11) applying the operator L_(μ,ν) we obtain
L_(μ,ν) u=L_(μ,ν) [T_(1-μ,1-ν) φ_1 ]+L_(μ,ν) [x^(1/2 (p-μ) ) (-y)^(3/2 (q-ν)) T_(1-p,1-q) ψ_1 ]+
+L_(μ,ν) [x^(1/2 (p-μ) ) (-y)^(3/2 (1-q-ν)) T_(1-p,q) ψ_2 ]=I_1+I_2+I_3
I_1=L_(μ,ν) [T_(1-μ,1-ν) φ_1 ]=0
I_2=L_(μ,ν) [x^(1/2 (p-μ) ) (-y)^(3/2 (q-ν)) T_(1-p,1-q) ψ_1 ]=0
I_3=L_(μ,ν) [x^(1/2 (p-μ) ) (-y)^(3/2 (1-q-ν)) T_(1-p,q) ψ_2 ]=
=xy ∂^2/(∂x^2 ) [x^(1/2 (p-μ) ) (-y)^(3/2 (1-q-ν)) T_(1-p,q) ψ_2 ]+
+∂^2/(∂y^2 ) [x^(1/2 (p-μ) ) (-y)^(3/2 (1-q-ν)) T_(1-p,q) ψ_2 ]+
+(1+2μ)/2 y ∂/∂x [x^(1/2 (p-μ) ) (-y)^(3/2 (1-q-ν)) T_(1-p,q) ψ_2 ]+
+(6ν-1)/2y ∂/∂y [x^(1/2 (p-μ) ) (-y)^(3/2 (1-q-ν)) T_(p,1-q) ψ_2 ] (12)
Let us calculate the singular derivatives of the first and second order from equation (12) and substitute them into (1) to obtain:
u_x=1/2 (p-μ) x^(1/2 (p-μ)-1) (-y)^(3/2 (1-q-ν)) ∫_0^1▒dσ ∫_0^1▒(ψ_1 [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^q )+
+x^(1/2 (p-μ)-1/2) (-y)^(3/2 (1-q-ν)) ∫_0^1▒dσ ∫_0^1▒(ψ_1^' [ω](1-2σ)dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^q ),
I_x=∫_0^1▒dσ ∫_0^1▒(ψ_1^' [ω](1-2σ)dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^q )=|■(u=ψ_1^' [ω]@du=-4√x ψ_1^'' [ω] )┤
├ ■(dϑ=((1-2σ))/[σ(1-σ)]^(1-p) @ϑ=1/p 1/[σ(1-σ)]^(-p) )┤|=4/p √x ∫_0^1▒dσ ∫_0^1▒(ψ_1^'' [ω]dτ)/([σ(1-σ)]^p [τ(1-τ)]^(1-q) )
u_x=1/2 (p-μ) x^(1/2 (p-μ)-1) (-y)^(3/2 (1-q-ν)) ∫_0^1▒dσ ∫_0^1▒(ψ_1 [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^q )+
+4/p x^(1/2 (p-μ) ) (-y)^(3/2 (1-q-ν)) ∫_0^1▒dσ ∫_0^1▒(ψ_1^'' [ω]dτ)/([σ(1-σ)]^(-p) [τ(1-τ)]^q ),
u_xx=1/2 (p-μ)[1/2 (p-μ)-1] x^(1/2 (p-μ)-2) (-y)^(3/2 (1-q-ν)) ∫_0^1▒dσ ∫_0^1▒(ψ_1 [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^q )-
-2/p (2μ+1) x^(1/2 (p-μ)-1) (-y)^(3/2 (1-q-ν)) ∫_0^1▒dσ ∫_0^1▒(ψ_1^'' [ω]dτ)/([σ(1-σ)]^(-p) [τ(1-τ)]^q )+
+x^(1/2 (p-μ)-1) (-y)^(3/2 (1-q-ν)) ∫_0^1▒dσ ∫_0^1▒(ψ_1^'' [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^q ),
u_y=-3/2 (1-q-ν) x^(1/2 (p-μ) ) (-y)^(3/2 (1-q-ν)-1) ∫_0^1▒dσ ∫_0^1▒(ψ_2 [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^q )+
+x^(1/2 (p-μ) ) (-y)^(3/2 (1-q-ν)+1/2) ∫_0^1▒dσ ∫_0^1▒(ψ_1^' [ω](1-2τ)dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^q )
I_y=∫_0^1▒dσ ∫_0^1▒(ψ_2^' [ω](1-2τ)dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^q )=|■(u=ψ_2^' [ω]@du=4/3 (-y)^(3/2) ψ_2^'' [ω] )┤
├ ■(dϑ=((1-2τ))/[τ(1-τ)]^q @ϑ=1/(1-q) 1/[τ(1-τ)]^(q-1) )┤|=-4/3(1-q) (-y)^(3/2) ∫_0^1▒dσ ∫_0^1▒(ψ_2^'' [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(q-1) ),
u_y=-3/2 (1-q-ν) x^(1/2 (p-μ) ) (-y)^(3/2 (1-q-ν)-1) ∫_0^1▒dσ ∫_0^1▒(ψ_2 [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^q )-
-4/3(1-q) x^(1/2 (p-μ) ) (-y)^(3/2 (1-q-ν)+2) ∫_0^1▒dσ ∫_0^1▒(ψ_2^'' [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(q-1) ).
u_yy=3/2 (1-q-ν)[3/2 (1-q-ν)-1] x^(1/2 (p-μ) ) (-y)^(3/2 (1-q-ν)-2)*
*∫_0^1▒dσ ∫_0^1▒(ψ_2 [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^q )-
-2(6ν-1)/3(1-q) x^(1/2 (p-μ) ) (-y)^(3/2 (1-q-ν)+1) ∫_0^1▒dσ ∫_0^1▒(ψ_2^'' [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(q-1) )+
+x^(1/2 (p-μ) ) (-y)^(3/2 (1-q-ν)+1) ∫_0^1▒dσ ∫_0^1▒(ψ_2^'' [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^q ).
I_3=L_(μ,ν) [x^(1/2 (p-μ) ) (-y)^(3/2 (1-q-ν)) T_(1-p,q) ψ_2 ]=
=-1/2 (p-μ)[1/2 (p-μ)-1] x^(1/2 (p-μ)-1) (-y)^(3/2 (1-q-ν)+1)*
*∫_0^1▒dσ ∫_0^1▒(ψ_1 [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^q )+
+2/p (2μ+1) x^(1/2 (p-μ) ) (-y)^(3/2 (1-q-ν)+1) ∫_0^1▒dσ ∫_0^1▒(ψ_1^'' [ω]dτ)/([σ(1-σ)]^(-p) [τ(1-τ)]^q )-
-x^(1/2 (p-μ) ) (-y)^(3/2 (1-q-ν)+1) ∫_0^1▒dσ ∫_0^1▒(ψ_1^'' [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^q )+
+3/2 (1-q-ν)[3/2 (1-q-ν)-1] x^(1/2 (p-μ) ) (-y)^(3/2 (1-q-ν)-2)*
*∫_0^1▒dσ ∫_0^1▒(ψ_2 [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^q )-
-2(6ν-1)/3(1-q) x^(1/2 (p-μ) ) (-y)^(3/2 (1-q-ν)+1) ∫_0^1▒dσ ∫_0^1▒(ψ_2^'' [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(q-1) )+
+x^(1/2 (p-μ) ) (-y)^(3/2 (1-q-ν)) ∫_0^1▒dσ ∫_0^1▒(ψ_2^'' [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^q )-
-(1+2μ)/2 1/2 (p-μ) x^(1/2 (p-μ)-1) (-y)^(3/2 (1-q-ν)+1)*
*∫_0^1▒dσ ∫_0^1▒(ψ_1 [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^q )-
-(1+2μ)/2 4/p x^(1/2 (p-μ) ) (-y)^(3/2 (1-q-ν)+1) ∫_0^1▒dσ ∫_0^1▒(ψ_1^'' [ω]dτ)/([σ(1-σ)]^(-p) [τ(1-τ)]^q )+
+(6ν-1)/2 3/2 (1-q-ν) x^(1/2 (p-μ) ) (-y)^(3/2 (1-q-ν)) ∫_0^1▒dσ ∫_0^1▒(ψ_2 [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^q )+
+(6ν-1)/2 4/3(1-q) x^(1/2 (p-μ) ) (-y)^(3/2 (1-q-ν)-2) ∫_0^1▒dσ ∫_0^1▒(ψ_2^'' [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(q-1) )=
I_3=-1/2 (p-μ)[1/2 (p-μ)-1] x^(1/2 (p-μ)-1) (-y)^(3/2 (1-q-ν)+1)*
*∫_0^1▒dσ ∫_0^1▒(ψ_2 [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^q )+
+3/2 (1-q-ν)[3/2 (1-q-ν)-1] x^(1/2 (p-μ) ) (-y)^(3/2 (1-q-ν)-2)*
*∫_0^1▒dσ ∫_0^1▒(ψ_2 [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^q )-
-(1+2μ)/2 1/2 (p-μ) x^(1/2 (p-μ)-1) (-y)^(3/2 (1-q-ν)+1) ∫_0^1▒dσ ∫_0^1▒(ψ_2 [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^q )+
+(6ν-1)/2 3/2 (1-q-ν) x^(1/2 (p-μ) ) (-y)^(3/2 (1-q-ν)-2) ∫_0^1▒dσ ∫_0^1▒(ψ_2 [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^q )=
=-1/2 (p-μ) (p+1)/2 x^(1/2 (p-μ)-1) (-y)^(3/2 (1-q-ν)+1) ∫_0^1▒dσ ∫_0^1▒(ψ_2 [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^q )+
+9/4 (1-q-ν)(ν-q) x^(1/2 (p-μ) ) (-y)^(3/2 (1-q-ν)-2) ∫_0^1▒dσ ∫_0^1▒(ψ_2 [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^q )=
=1/4 [9(q-ν)(q+ν-1)x-(1-p-μ)(p-μ) (-y)^3 ]*
*x^(1/2 (p-μ) ) (-y)^(3/2 (1-q-ν)-2) ∫_0^1▒dσ ∫_0^1▒(ψ_2 [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^q ),
In (12) substituting I_1, I_2, I_3 and we get
L_(μ,ϑ) u(x,y)=1/4 [9(q-ν)(q+ν-1)x-(1-p-μ)(p-μ) (-y)^3 ]*
*x^(1/2 (p-μ) ) (-y)^(3/2 (1-q-ν)-2) ∫_0^1▒dσ ∫_0^1▒(ψ_2 [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^q ).
(x^((2+μ-p)/2) (-y)^((4+3ν-3q)/2))/[9(q-ν)(q+ν-1)x-(1-p-μ)(p-μ) (-y)^3 ] L_(μ,ϑ) u(x,y)=
=1/4 (-y)^(3/2 (1-2q)) ∫_0^1▒dσ ∫_0^1▒(ψ_2 [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^q )
Apply the operator L_(p,q) to the last two equalities
L_(p,q) [(x^((2+μ-p)/2) (-y)^((4+3ν-3q)/2))/[9(q-ν)(q+ν-1)x-(1-p-μ)(p-μ) (-y)^3 ] L_(μ,ϑ) u(x,y)]=
=L_(p,q) [1/4 (-y)^(3/2 (1-2q)) ∫_0^1▒dσ ∫_0^1▒(ψ_2 [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^q )]
We write the right side of the resulting equation separately
L_(p,q) [1/4 (-y)^(3/2 (1-2q)) ∫_0^1▒dσ ∫_0^1▒(ψ_2 [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^q )]=
=xy/4 ∂^2/(∂x^2 ) [(-y)^(3/2 (1-2q)) T_(1-p,q) ψ_2 ]+1/4 ∂^2/(∂y^2 ) [(-y)^(3/2 (1-2q)) T_(1-p,q) ψ_2 ]+
+(1+2p)/8 y ∂/∂x [(-y)^(3/2 (1-2q)) T_(1-p,q) ψ_2 ]+(6q-1)/8y ∂/∂y [(-y)^(3/2 (1-2q)) T_(1-p,q) ψ_2 ] (13),
Let us calculate the partial derivatives of the first and second order of equality (13):
u_x=x^(-1/2) 1/4 (-y)^(3/2 (1-2q)) ∫_0^1▒dσ ∫_0^1▒(ψ_2^' [ω](1-2σ)dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^q ),
I_x=∫_0^1▒dσ ∫_0^1▒(ψ_2^' [ω](1-2σ)dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^q )=|■(u=ψ_2^' [ω]@du=-4√x ψ_1^'' [ω] )┤
├ ■(dϑ=((1-2σ))/[σ(1-σ)]^(1-p) @ϑ=1/p 1/[σ(1-σ)]^(-p) )┤|=4/p √x ∫_0^1▒dσ ∫_0^1▒(ψ_2^'' [ω]dτ)/([σ(1-σ)]^(-p) [τ(1-τ)]^q )
u_x=1/p (-y)^(3/2 (1-2q)) ∫_0^1▒dσ ∫_0^1▒(ψ_2^'' [ω]dτ)/([σ(1-σ)]^(-p) [τ(1-τ)]^q ).
u_xx=-(1+2p)/2p x^(-1) (-y)^(3/2 (1-2q)) ∫_0^1▒dσ ∫_0^1▒(ψ_2^'' [ω]dτ)/([σ(1-σ)]^(-p) [τ(1-τ)]^q )
+x^(-1) 1/4 (-y)^(3/2 (1-2q)) ∫_0^1▒dσ ∫_0^1▒(ψ_2^'' [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^q ).
u_y=-3/2 (1-2q) 1/4 (-y)^(3/2 (1-2q)-1) ∫_0^1▒dσ ∫_0^1▒(ψ_2 [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^q )+
+1/4 (-y)^(3/2 (1-2q)+1/2) ∫_0^1▒dσ ∫_0^1▒(ψ_2^' [ω](1-2τ)dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^q ),
I_y=∫_0^1▒dσ ∫_0^1▒(ψ_2^' [ω](1-2τ)dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^q )=|■(u=ψ_2^' [ω]@du=4/3 (-y)^(3/2) ψ_2^'' [ω] )┤
├ ■(dϑ=((1-2τ))/[τ(1-τ)]^q @ϑ=1/(1-q) 1/[τ(1-τ)]^(q-1) )┤|=-4/3(1-q) (-y)^(3/2) ∫_0^1▒dσ ∫_0^1▒(ψ_2^'' [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(q-1) ),
u_y=-3/2 (1-2q) 1/4 (-y)^(3/2 (1-2q)-1) ∫_0^1▒dσ ∫_0^1▒(ψ_2 [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^q )-
-1/3(1-q) (-y)^(3/2 (1-2q)+2) ∫_0^1▒dσ ∫_0^1▒(ψ_2^'' [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(q-1) ).
u_yy=3/2 (1-2q)[3/2 (1-2q)-1] 1/4 (-y)^(3/2 (1-2q)-2)*
*∫_0^1▒dσ ∫_0^1▒(ψ_2 [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^q )-
-1/6(1-q) (6q-1) (-y)^(3/2 (1-2q)+1) ∫_0^1▒dσ ∫_0^1▒(ψ_2^'' [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(q-1) )+
+1/4 (-y)^(3/2 (1-2q)+1) ∫_0^1▒dσ ∫_0^1▒(ψ_2^'' [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^q ),
We substitute the obtained results into the right side of (13)
(1+2p)/2p (-y)^(3/2 (1-2q)+1) ∫_0^1▒dσ ∫_0^1▒(ψ_2^'' [ω]dτ)/([σ(1-σ)]^(-p) [τ(1-τ)]^q )-
-1/4 (-y)^(3/2 (1-2q)+1) ∫_0^1▒dσ ∫_0^1▒(ψ_2^'' [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^q )+
+3/2 (1-2q)[3/2 (1-2q)-1] 1/4 (-y)^(3/2 (1-2q)-2)*
*∫_0^1▒dσ ∫_0^1▒(ψ_2 [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^q )-
-1/6(1-q) (6q-1) (-y)^(3/2 (1-2q)+1) ∫_0^1▒dσ ∫_0^1▒(ψ_2^'' [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(q-1) )+
+1/4 (-y)^(3/2 (1-2q)+1) ∫_0^1▒dσ ∫_0^1▒(ψ_2^'' [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^q )-
-(1+2p)/2 1/p (-y)^(3/2 (1-2q)+1) ∫_0^1▒dσ ∫_0^1▒(ψ_2^'' [ω]dτ)/([σ(1-σ)]^(-p) [τ(1-τ)]^q )+
+(6q-1)/2 3/2 (1-2q) 1/4 (-y)^(3/2 (1-2q)-2) ∫_0^1▒dσ ∫_0^1▒(ψ_2 [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^q )+
+(6q-1)/2 1/3(1-q) (-y)^(3/2 (1-2q)+1) ∫_0^1▒dσ ∫_0^1▒(ψ_2^'' [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^(q-1) )=0
L_(p,q) [1/4 (-y)^(3/2 (1-2q)) ∫_0^1▒dσ ∫_0^1▒(ψ_2 [ω]dτ)/([σ(1-σ)]^(1-p) [τ(1-τ)]^q )]=0.
The theorem has been proven.
Theorem 7. The regular solution of equation (1) in the domain D^- for 0<2μ<1,0<2ν<1,2p>1,2q>1 and q>ν is described as follows
u(x,y)=A_μ A_ν T_(1-μ,1-ν) φ_1+A_(1-μ) A_ν x^(1/2 (1-2μ) ) T_(μ,1-ν) φ_2+
+A_μ A_(1-ν) (-y)^(3/2 (1-2ν)) T_(1-μ,ν) (φ_2 ) ̃+A_(1-μ) A_(1-ν) x^(1/2 (1-2μ) ) (-y)^(3/2 (1-2ν)) T_(μ,ν) φ_3+
+A_p A_q x^(1/2 (p-μ) ) (-y)^(3/2 (q-ν)) T_(1-p,1-q) ψ_1, (14)
where φ_1,φ_2,(φ_2 ) ̃,ψ_1 are arbitrary functions of one argument from the class С^2(D^-) and A_μ,A_(1-μ),A_ν,A_(1-ν),A_p,A_q are given constants.
Theorem 8. The regular solution of equation (1) in the domain D^- for 0<2μ<1,2ν>1,0<2p<1,2q>1 and p>μ,q>ν is described as follows
u(x,y)=A_μ A_ν T_(1-μ,1-ν) φ_1+A_(1-μ) A_ν x^(1/2 (1-2μ) ) T_(μ,1-ν) φ_2+
+A_p A_q x^(1/2 (p-μ) ) (-y)^(3/2 (q-ν)) T_(1-p,1-q) ψ_1+
+A_(1-p) A_q x^(1/2 (1-p-μ) ) (-y)^(3/2 (q-ν)) T_(p,1-q) ψ_2,(15)
where φ_1,φ_2,ψ_1,ψ_2 are arbitrary functions of one argument from the class С^2(D^-) and A_μ,A_(1-μ),A_ν,A_p,A_(1-p),A_q are given constants.
Problem K. It is required to find a regular solution to equation (1) for 2μ>1,
0<2ν<1,2p>1,2q>1 in the region D^- with initial conditions
lim┬(y→-0)⁡U(x,y)=f_1 (x), 〖 lim┬(y→-0) (-y)^(3ν-1/2)〗⁡〖∂U/∂y〗=f_2 (x),
lim┬(y→-0)⁡[(-y)^((5-3q-3ν)/2) ∂/∂y [(-y)^(3ν-1/2) ∂U/∂y]]=g(x) (K)
where f_1 (x), f_2 (x), g(x) —are given continuous functions in the interval 0<x<1.
Solution of problem K. To solve this problem we apply Theorem 4. From equality (7) using the initial condition (K) we obtain.
A_μ ∫_0^1▒(φ_1 [2√x (1-2σ)]dτ)/[σ(1-σ)]^(1-μ) =f_1 (x),
∫_0^1▒(φ_1 [2√x (1-2σ)]dτ)/[σ(1-σ)]^(1-μ) =1/B(ν;ν) f_1 (x),
φ_1 (x)=4^(2μ-k-1)/(B(1-α;α) A_μ [(k+α-1)(k+α-2)…α] )*
*d/dx ∫_0^y▒(d/((s/2)^(3/2) d(s/2)^2 ))^k ((s/2)^(7/2-μ) f_1 (s/2)^2 ) sds/(x^2-S^2 )^a . (16)
lim┬(y→-0)⁡〖(-y)^(3ν-1/2) 〗 ∂U/∂y=f_2 (x),
-3/2 (1-2ν) A_μ ∫_0^1▒(φ_2 [2√x (1-2σ)]dτ)/[σ(1-σ)]^(1-μ) =f_2 (x),
Performing the substitution 2√x (1-2σ)=η we get,
2√x (1-2σ)=η; (1-σ)=1-(1/2-η/(4√x))=
1-2σ=η/(2√x); =1/2+η/(4√x)
2σ=1-η/(2√x); σ=0; η=2√x
σ=1/2-η/(4√x); σ=1; η=-2√x
dσ=-dη/(4√x)
∫_(2√x)^(-2√x)▒(φ(η) dη/(4√x))/[(1/2-η/(4√x))(1/2+η/(4√x))]^(1-μ) =-2/(3BA_μ (1-2ν) ) f_2 (x);
-∫_(2√x)^(-2√x)▒(φ(η) dη/(4√x))/[(1/4-η^2/16x)]^(1-μ) =-2/(3A_μ (1-2ν) ) f_2 (x);
-1/(4√x) ∫_(2√x)^(-2√x)▒φ(η)dη/[(1/4-η^2/16x)]^(1-μ) =-2/(3A_μ (1-2ν) ) f_2 (x);
-1/(4√x) ∫_(2√x)^(-2√x)▒φ(η)dη/〖(1/4)^(1-μ) [(1-η^2/4x)]〗^(1-μ) =-2/(3A_μ (1-2ν) ) f_2 (x);
-1/(4√x) (1/4)^(μ-1) ∫_(2√x)^(-2√x)▒φ(η)dη/[(1-η^2/4x)]^(1-μ) =-2/(3A_μ (1-2ν) ) f_2 (x);
-1/(4√x) (1/4)^(μ-1) ∫_(2√x)^(-2√x)▒φ(η)dη/[((4x-η^2)/4x)]^(1-μ) =-2/(3A_μ (1-2ν) ) f_2 (x);
-1/(4√x) (1/4)^(μ-1) ∫_(2√x)^(-2√x)▒φ(η)dη/〖(1/4x)^(1-μ) [4x-η^2 ]〗^(1-μ) =-2/(3A_μ (1-2ν) ) f_2 (x);
-1/(4√x) (1/16)^(μ-1) (1/x)^(μ-1) ∫_(2√x)^(-2√x)▒φ(η)dη/[4x-η^2 ]^(1-μ) =-2/(3A_μ (1-2ν) ) f_2 (x);
-(x^(1-μ) 4^(1-2μ))/√x ∫_(2√x)^(-2√x)▒φ(η)dη/[4x-η^2 ]^(1-μ) =-2/(3A_μ (1-2ν) ) f_2 (x);
-∫_(2√x)^(-2√x)▒φ(η)dη/[(2√x)^2-η^2 ]^(1-μ) =-(2x^(μ-3/2) 4^(2μ-1))/(3A_μ (1-2ν) ) f_2 (x);
Changing the limits of the integral
∫_(-2√x)^(2√x)▒φ(η)dη/[(2√x)^2-η^2 ]^(1-μ) =-(2x^(μ-3/2) 4^(2μ-1))/(3A_μ (1-2ν) ) f_2 (x);
Let us divide the integral into two parts
∫_(-2√x)^0▒φ(η)dη/[(2√x)^2-η^2 ]^(1-μ) +∫_0^(2√x)▒φ(η)dη/[(2√x)^2-η^2 ]^(1-μ) =
=-(2x^(μ-3/2) 4^(2μ-1))/(3A_μ (1-2ν) ) f_2 (x);
In the first integral, we change η to -η and get. η=-η
-∫_(2√x)^0▒φ(η)dη/[(2√x)^2-η^2 ]^(1-μ) +∫_0^(2√x)▒φ(η)dη/[(2√x)^2-η^2 ]^(1-μ) =
=-(2x^(μ-3/2) 4^(2μ-1))/(3A_μ (1-2ν) ) f_2 (x);
∫_0^(2√x)▒φ(η)dη/[(2√x)^2-η^2 ]^(1-μ) +∫_0^(2√x)▒φ(η)dη/[(2√x)^2-η^2 ]^(1-μ) =
=-(2x^(μ-3/2) 4^(2μ-1))/(3A_μ (1-2ν) ) f_2 (x);
2∫_0^(2√x)▒φ(η)dη/[(2√x)^2-η^2 ]^(1-μ) =-(2x^(μ-3/2) 4^(2μ-1))/(3A_μ (1-2ν) ) f_2 (x);
We include the following symbols: k=[μ] is the integer part, λ=(μ) is the fractional part, then it is written like this
∫_0^(2√x)▒〖φ(η) [(2√x)^2-η^2 ]^(k+λ-1) dη〗=-(2x^(μ-3/2) 4^(2μ-1))/(3A_μ (1-2ν) ) f_2 (x); (17)
From equation (17) we take the derivative k times
2x^(-1/2) (k+λ-1) ∫_0^(2√x)▒〖φ(η) [(2√x)^2-η^2 ]^(k+λ-2) dη〗=
=-d/dx ((2x^(μ-3/2) 4^(2μ-1))/(3A_μ (1-2ν) ) f_2 (x));
∫_0^(2√x)▒〖φ(η) [(2√x)^2-η^2 ]^(k+λ-2) dη〗=
=-(2*4^(2μ-1))/(6x^(-1/2) (k+λ-1) A_μ (1-2ν) ) d/dx (x^(μ-3/2) f_2 (x));
∫_0^(2√x)▒〖φ(η) [(2√x)^2-η^2 ]^(k+λ-2) dη〗=
=-4^(2μ-1)/(3(k+λ-1) A_μ (1-2ν) ) d/(x^(-1/2) dx) (x^(μ-3/2) f_2 (x));
____ ____ ____ ____ ____ ____ ____ ____ ____ ____
∫_0^(2√x)▒〖φ(η) [(2√x)^2-η^2 ]^(k+λ-1) dη〗=
=-4^(2μ-k-1)/(3A_μ (1-2ν)[(k+λ-1)(k+λ-2)…λ] ) (d/(x^(-1/2) dx))^k (x^(μ-3/2) f_2 (x));
(d/(x^(-1/2) dx))^k=d/(x^(-1/2) dx),…,d/(x^(-1/2) dx) (k-temperature)
∫_0^(2√x)▒〖φ(η) [(2√x)^2-η^2 ]^(λ-1) dη〗=
=-4^(2μ-k-1)/(3A_μ (1-2ν)[(k+λ-1)(k+λ-2)…λ] ) (d/(x^(-1/2) dx))^k (x^(3/2-μ) f_2 (x));
In the last equality, we change 2√x to S, multiply both sides of the equation by SdS/(x^2-S^2 )^a and integrate over S in the limit from 0 to y.
∫_0^y▒sds/(x^2-S^2 )^a ∫_0^y▒〖φ(η)/[s^2-η^2 ]^(1-a) dη〗=
=-4^(2μ-k-1)/(3A_μ (1-2ν)[(k+λ-1)(k+λ-2)…λ] )*
*∫_0^y▒(d/((s/2)^(3/2) d(s/2)^2 ))^k ((s/2)^(7/2-μ) f(s/2)^2 ) sds/(x^2-S^2 )^a ;
Using the Dirichlet formula, we change the order of the integral.
∫_0^y▒φ(η) dη∫_0^y▒sds/〖(x^2-S^2 )^a [s^2-η^2 ]〗^(1-a) =
=-4^(2μ-k-1)/(3A_μ (1-2ν)[(k+λ-1)(k+λ-2)…α] )*
*∫_0^y▒(d/((s/2)^(3/2) d(s/2)^2 ))^k ((s/2)^(7/2-μ) f(s/2)^2 ) sds/(x^2-S^2 )^a ;
I=∫_η^y▒sds/〖(x^2-S^2 )^a [s^2-η^2 ]〗^(1-a) =|■(S^2=η^2+t(x^2-S^2 )@1) S=η; t=0 2) S=x; t=1@2Sds=(x^2-η^2 )dt)|=
=1/2 ∫_0^1▒(x^2-η^2 )dt/(〖〖(1-t)^a (x^2-η^2 )〗^a [x^2-η^2 ]〗^(1-a) t^(1-a) )=1/2 ∫_0^1▒(x^2-η^2 )dt/((1-t)^a (x^2-η^2 ) t^(1-a) )=
=1/2 ∫_0^1▒dt/((1-t)^a t^(1-a) )=1/2 B(1-α;α),
1/2 B(1-α;α) ∫_0^y▒φ(η) dη=
=-4^(2μ-k-1)/(3A_μ (1-2ν)[(k+λ-1)(k+λ-2)…λ] )*
*∫_0^y▒(d/((s/2)^(3/2) d(s/2)^2 ))^k ((s/2)^(7/2-μ) f(s/2)^2 ) sds/(x^2-S^2 )^a ;
∫_0^y▒φ(η) dη=-4^(2μ-k-1)/(3B(1-λ;λ) A_μ (1-2ν)[(k+λ-1)(k+λ-2)…λ] )*
*∫_0^y▒(d/((s/2)^(3/2) d(s/2)^2 ))^k ((s/2)^(7/2-μ) f(s/2)^2 ) sds/(x^2-S^2 )^a ;
Differentiating the last equality with respect to x we get,
φ_2 (x)=-4^(2μ-k-1)/(3B(1-λ;λ) A_μ (1-2ν)[(k+λ-1)(k+λ-2)…λ] )*
*d/dx ∫_0^y▒(d/((s/2)^(3/2) d(s/2)^2 ))^k ((s/2)^(7/2-μ) f(s/2)^2 ) sds/(x^2-S^2 )^a , (18)
Taking into account the second condition of problem K, from equality (7) we find, i.e.
lim┬(y→-0)⁡[x^(1/2 (p-μ) ) (-y)^((5+3ν-3q)/2) ∂/∂y [(-y)^(3ν-1/2) ∂U/∂y]]=3/2 (1-2ν) A_μ*
*[3/2 (q+ν-1)] x^(1/2 (p-μ) ) ∫_0^1▒ψ[2√x (1-2σ)]dτ/[σ(1-σ)]^(1-p) =g(x)
∫_0^1▒ψ[2√x (1-2σ)]dτ/[σ(1-σ)]^(1-p) =(4x^(1/2 (μ-p) ))/(9(q+ν-1)(1-2ν) A_μ ) g(x)
g(x)=(x^(1/2 (μ-p) ) 4^(2μ-k))/(27B(1-λ;λ) A_μ (q+ν-1)(1-2ν)[(k+λ-1)(k+λ-2)…λ] )*
*d/dx ∫_0^y▒(d/((s/2)^(3/2) d(s/2)^2 ))^k ((s/2)^(7/2-μ) f(s/2)^2 ) sds/(x^2-S^2 )^a (19)
Theorem 9. If 2μ>1,0<2ν<1,2p>1,2q>1 and q> ϑ q+ ϑ>1. Then the solution to problem K in the domain (D^- ) ̃ is given by formula (7), where arbitrary functions φ_1, φ_2 and φ_3 are determined from equations (16), (18) and (19).
REVIEWER: Odinaev R.N.,
Doctor of Physical and Mathematical Sciences,
Professor

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INTEGRAL REPRESENTATION AND SOLUTION OF A CAUCHY TYPE PROBLEM FOR ONE OF THE FOURTH ORDER MIXED RAD DIFFERENTIAL EQUATIONS
Partial differential equations were developed mainly at the end and beginning of the twentieth century in connection with the solution of physical problems. Many domestic and foreign scientists have been studying this scientific direction. Second-order partial differential equations have been widely studied, but higher-order partial differential equations have been studied relatively little. Higher order differential equations of mixed type are comparatively less studied. From this point of view, the study and investigation of differential equations of mixed type of higher order is relevant. In this article, for the first time, the solution to a degenerate fourth-order mixed type differential equation is expressed using solutions to two second-order differential equations. After this, for degenerate differential equations of mixed type (1) of the fourth order, the integral representation of the solution is determined explicitly using arbitrary functions. Some found integral representations of the solution are used to solve Cauchy type problems in the characteristic domain. The solution to the Cauchy type problem is found in explicit form.
Key words: degeneration, integral image, mixed genus, Cauchy type problem, regular solution, optional functions.
Information about the authors: Sattorov Abdumanon Sattorovich – Tajik National University, Professor of the Department of Higher Mathematics. Address: 734025, Dushanbe, Tajikistan, Rudaki Avenue, 17. E-mail: This email address is being protected from spambots. You need JavaScript enabled to view it..
Narzulloev Shukhrat Kholmakhmadovich-Tajik National University, doctoral student of the Department of of Higher Mathematics. Address: 734025, Dushanbe, Tajikistan, Rudaki Avenue, 17. Phone: (+992) 988-08-84-78. E-mail: shuxratnarzulloev1997@ gmail.com.
Article received 08.05.2024
Approved after review 03.09.2024
Accepted for publication 21.10.2024

   
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