Electronic Journal of Differential Equations, Vol. 2015 (2015), No. 09,... ISSN: 1072-6691. URL: or

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Electronic Journal of Differential Equations, Vol. 2015 (2015), No. 09, pp. 1–10.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
PARTICULAR SOLUTIONS OF GENERALIZED
EULER-POISSON-DARBOUX EQUATION
RAKHILA B. SEILKHANOVA, ANVAR H. HASANOV
Abstract. In this article we consider the generalized Euler-Poisson-Darboux
equation
2γ
2α
2β
utt +
ut = uxx + uyy +
ux +
uy , x > 0, y > 0, t > 0.
t
x
y
We construct particular solutions in an explicit form expressed by the Lauricella hypergeometric function of three variables. Properties of each constructed
solutions have been investigated in sections of surfaces of the characteristic
cone. Precisely, we prove that found solutions have singularity 1/r at r → 0,
where r2 = (x − x0 )2 + (y − y0 )2 − (t − t0 )2 .
1. Introduction
Many problems of modern mathematics and theoretical physics lead to the investigation of hypergeometric functions of many variables. In particular, problems
of superstring theory [8], analytical continuations of Mellin-Barnes integrals [14]
and algebraic geometry [13]. Systems of hypergeometric type differential equations
have numerous applications as nontrivial model examples in realization of algorithms for symbolic calculations, which are used in modern systems of computer
algebra [16]. Hypergeometric functions of many variables appear in quantum field
theory as solutions of Knizhnik-Zamolodchikov equation [22]. These equations can
be considered as generalized hypergeometric type equations and their solutions have
integral representations, which generalize classic Euler integrals for hypergeometric
functions of one variable. This approach allows us to link the special functions of
hypergeometric type and challenges the theory of representations of Lie algebras
and quantum groups [22].
Initially hypergeometric functions introduced by many authors with different
methods, which are not related with each other. Their occurrence is determined,
as a rule, by the need to solve problems, that led to a differential equation (or
system of equations), insoluble in the class of elementary functions. Thus arose
2000 Mathematics Subject Classification. 35Q05, 35L80, 35C65.
Key words and phrases. Generalized Euler-Poisson-Darboux equation; hyperbolic equation;
Lauricelli hypergeometric functions.
c
2015
Texas State University - San Marcos.
Submitted October 21, 2014. Published January 5, 2015.
1
2
R. B. SEILKHANOVA, A. H. HASANOV
EJDE-2015/09
Bessel functions, Hermite functions, Gauss hypergeometric function. Hypergeometric functions occupy an important place among the special functions of mathematical physics. Many problems of gas dynamics are reduced to boundary value
problems for degenerate equations of mixed type.
It is known that the degenerate equation of mixed type in the hyperbolic part
of the domain is reduced to the generalized equation of Euler-Poisson-Darboux
utt +
2α
2β
2γ
ut = uxx + uyy +
ux +
uy ,
t
x
y
x > 0, y > 0, t > 0,
(1.1)
where α > 0, β > 0 and γ > 0 are constants.
We note that Euler-Poisson-Darboux equation
uξη −
β
α
uξ +
uη = 0,
ξ−η
ξ−η
α > 0, β > 0, α + β < 1,
(1.2)
was considered in [21], where the Cauchy problem for (1.2) was solved. In [17, 18,
19], non-local boundary problems for (1.2) were investigated in characteristic triangles. In [11], two confluent hypergeometric functions of three variables were introduced. Further, for introduced hypergeometric functions authors prove formulas of
analytical continuation. Using the introduced confluent hypergeometric functions,
they constructed the Riemann function for the generalized Euler -Poisson-Darboux
equation
α
β
α
β
uξη + [
+
]uξ + [
−
]uη + γu = 0.
(1.3)
η+ξ
η−ξ
η+ξ
η−ξ
Further, by the Riemann-function method the Cauchy problem for (1.3) was solved
in characteristic triangle. Solution is written in an explicit form. Note that Euler Poisson-Darboux equations (1.2) and (1.3) are written in characteristic coordinates.
In [20], the unique solvability of the Darboux problem with deviation for the
Euler -Poisson-Darboux equation was proved outside of characteristic cone. Other
type of the Euler -Poisson-Darboux equations were investigated in works [5, 7, 6,
23, 10, 15, 2, 3, 4].
2. Reduction of the Euler-Poisson-Darboux equation to a system of
Lauricella hypergeometric functions
Solution of the Euler-Poisson-Darboux equation (1.1) is searched in the form
u = P ω(ξ, η, ζ),
(2.1)
where ω(ξ, η, ζ) is unknown function and
−α−β−γ− 12
P = (r2 )
2
r12
2
r22
,
(2.2)
2
r32
r −
r −
r −
, η=
, ζ=
,
r2
r2
r2
2
2
2
2
2
2
r2 = (x − x0 ) + (y − y0 ) − (t − t0 ) , r12 = (x + x0 ) + (y − y0 ) − (t − t0 ) ,
ξ=
2
2
2
r22 = (x − x0 ) + (y + y0 ) − (t − t0 ) ,
2
2
2
r32 = (x − x0 ) + (y − y0 ) − (t + t0 ) .
Calculating necessary derivatives from (2.1) and substituting them into (1.1), we
obtain
A1 ωξξ + A2 ωηη + A3 ωζζ + B1 ωξη + B2 ωξζ + B3 ωηζ
(2.3)
+ C1 ωξ + C2 ωη + C3 ωζ + Dω = 0,
EJDE-2015/09
GENERALIZED EULER-POISSON-DARBOUX EQUATION
3
where
A1 = P ξx2 + P ξy2 − P ξt2 ,
A2 = P ηx2 + P ηy2 − P ηt2 ,
B1 = 2P ξx ηx + 2P ξy ηy − 2P ξt ηt ,
A3 = P ζx2 + P ζy2 − P ζt2 ,
B2 = 2P ξx ζx + 2P ξy ζy − 2P ξt ζt ,
B3 = 2P ηx ζx + 2P ηy ζy − 2P ηt ζt ,
2α
2β
2γ C1 = P (ξxx + ξyy − ξtt ) + 2(Px ξx + Py ξy − Pt ξt ) + P
ξx +
ξy −
ξt ,
x
y
t
2α
2β
2γ C2 = P (ηxx + ηyy − ηtt ) + 2(Px ηx + Py ηy − Pt ηt ) + P
ηx +
ηy −
ηt ,
x
y
t
2α
2β
2γ C3 = P (ζxx + ζyy − ζtt ) + 2(Px ζx + Py ζy − Pt ζt ) + P
ζx +
ζy −
ζt ,
x
y
t
2α
2β
2γ
D = Pxx + Pyy − Ptt + Px
+ Py
−
Pt .
x
y
t
Now we consider A1 . Since
2
2
ξx2 =
2
2
4(x + x0 ) (r2 ) − 8(x + x0 )(x − x0 )r2 r12 + 4(x − x0 ) (r12 )
4
(r2 )
2
2
2
2
4(y − y0 ) (r2 ) − 8(y − y0 ) r2 r12 + 4(y − y0 ) (r12 )
ξy2 =
2
,
4
(r2 )
2
ξt2 =
2
2
,
2
2
4(t − t0 ) (r2 ) − 8(t − t0 ) r2 r12 + 4(t − t0 ) (r12 )
4
(r2 )
,
we obtain
−3
A1 = 4P r12 (r2 )
−3
or A1 = 4P r12 (r2 )
2
2
[(r2 ) + (r12 ) − 2(x + x0 )(x − x0 ) − 2(y − y0 ) + 2(t − t0 ) ],
4x20 . By the equality 4x0 = −x−1 r2 ξ, we obtain
−1
A1 = −4P x−1 x0 (r2 )
ξ(1 − ξ).
(2.4)
η(1 − η),
(2.5)
ζ(1 − ζ).
(2.6)
Similarly we have
−1
A2 = −4P y −1 y0 (r2 )
2 −1
A3 = −4P t−1 t0 (r )
Further, we calculate representation of B1 . Finding necessary derivatives from the
arguments and substituting them we obtain
n 2(x + x )r2 − 2(x − x )r2 2(x − x )r2 − 2(x − x )r2
0
0 1
0
0 2
B1 = 2P
2
2
2
2
(r )
(r )
2(y − y0 )r2 − 2(y − y0 )r12 2(y + y0 )r2 − 2(y − y0 )r22
+
2
2
(r2 )
(r2 )
2(t − t0 )r2 − 2(t − t0 )r12 2(t − t0 )r2 − 2(t − t0 )r22 o
−
,
2
2
(r2 )
(r2 )
or
n
−4
B1 = 8P (r2 )
[(x + x0 )r2 − (x − x0 )r12 ][(x − x0 )r2 − (x − x0 )r22 ]
+ [(y − y0 )r2 − (y − y0 )r12 ][(y + y0 )r2 − (y − y0 )r22 ]
o
− [(t − t0 )r2 − (t − t0 )r12 ][(t − t0 )r2 − (t − t0 )r22 ] .
4
R. B. SEILKHANOVA, A. H. HASANOV
EJDE-2015/09
After some evaluations, we deduce
B1 = 4P y −1 y0 (r2 )
−1
ξη + 4P x−1 x0 (r2 )
−1
ξη.
(2.7)
ξζ,
(2.8)
ηζ.
(2.9)
Similarly one obtains
B2 = 4P t−1 t0 (r2 )
−1
−1
ξζ + 4P x−1 x0 (r2 )
2 −1
B3 = 4P t−1 t0 (r )
2 −1
ηζ + 4P y −1 y0 (r )
Further, considering the following expressions
−1
(2x−1 x0 ξ − ξ),
1
α + β + γ + (x−1 x0 ξ + ξ),
2
ξxx + ξyy − ξtt = 2(r2 )
−1
Px ξx + Py ξy − Pt ξt = 2P (r2 )
2β
2γ
2α
−1 ξx +
ξy −
ξt = −4(r2 ) αx−1 x0 + αξ + αx−1 x0 (1 − ξ) − βy −1 y0
x
y
t
+ βξ + βy −1 y0 (1 − ξ) − γt−1 t0 + γξ + γt−1 t0 (1 − ξ) ,
we define
1
+ 1)ξ]
2
−1
−1
+ 4P (r2 ) y −1 y0 βξ + 4P (r2 ) t−1 t0 γξ.
−1 −1
C1 = −4P (r2 )
x
x0 [2α − (2α + β + γ +
(2.10)
Similarly, we define
1
+ 1)η]
2
−1
−1
+ 4P (r2 ) x−1 x0 αη + 4P (r2 ) t−1 t0 γη,
1
−1
C3 = −4P (r2 ) t−1 t0 [2γ − (α + β + 2γ + + 1)ζ]
2
2 −1 −1
2 −1 −1
+ 4P (r ) x x0 γζ + 4P (r ) y y0 βζ.
−1 −1
C2 = −4P (r2 )
y
y0 [2β − (α + 2β + γ +
(2.11)
(2.12)
After simple calculations we obtain
1
1
−1
−1
D = 4(α + β + γ + )P (r2 ) αx−1 x0 + 4(α + β + γ + )P (r2 ) βy −1 y0
2
2
(2.13)
1
2 −1 −1
+ 4(α + β + γ + )P (r ) γt t0 .
2
Substituting (2.4)-(2.13) in (2.3), we obtain
4P x0 n
−
ξ(1 − ξ)ωξξ − ξηωξη − ξζωξζ − αηωη − γζωζ
xr2
o
1
1
+ [2α − (α + β + γ + + α + 1)ξ]ωξ − (α + β + γ + )αω
2
2
4P y0 n
−
η(1 − η)ωηη − ξηωξη − ηζωηζ − βξωξ − βζωζ
yr2
(2.14)
o
1
1
+ [2β − (α + β + γ + + β + 1)η]ωη − (α + β + γ + )βω
2
2
4P t0 n
−
ζ(1 − ζ)ωζζ − ξζωξζ − ηζωηζ − γξωξ − γηωη
tr2
o
1
1
+ [2γ − (α + β + γ + + γ + 1)ζ]ωζ − (α + β + γ + )γω = 0.
2
2
EJDE-2015/09
GENERALIZED EULER-POISSON-DARBOUX EQUATION
5
Hence, the following equality is valid
ξ(1 − ξ)ωξξ − ξηωξη − ξζωξζ + [2α − (α + β + γ +
1
+ α + 1)ξ]ωξ
2
1
− αηωη − γζωζ − (α + β + γ + )αω = 0,
2
η(1 − η)ωηη − ξηωξη − ηζωηζ + [2β − (α + β + γ +
1
+ β + 1)η]ωη
2
(2.15)
1
− βξωξ − βζωζ − (α + β + γ + )βω = 0,
2
ζ(1 − ζ)ωζζ − ξζωξζ − ηζωηζ + [2γ − (α + β + γ +
1
+ γ + 1)ζ]ωζ
2
1
− γξωξ − γηωη − (α + β + γ + )γω = 0.
2
Thus, the Euler-Poisson-Darboux equation (1.1) equivalently reduced to the system
(2.15).
3. Particular solutions of the Euler-Poisson-Darboux equation
In [1], system (2.15) was considered for the n-dimensional case
(n)
xj (1 − xj )
− bj
∂ 2 FA
∂x2j
n
X
k=1, k6=j
xk
− xj
n
X
(n)
xk
k=1, k6=j
(n)
∂FA
∂xk
(n)
(n)
∂F
∂ 2 FA
+ [cj − (a + bj + 1)xj ] A
∂xk ∂xj
∂xj
− abj FA = 0,
j = 1, 2, . . . , n.
There were found 2n particular solutions of this system. All of them are expressed
(n)
by Lauricella hypergeometric functions FA . In particular case, system of hypergeometric functions (2.15) has the following solutions [1]
1
(3)
ω1 = FA α + β + γ + ; α, β, γ; 2α, 2β, 2γ; ξ, η, ζ ,
(3.1)
2
3
(3)
ω2 = ξ 1−2α FA − α + β + γ + ; 1 − α, β, γ; 2 − 2α, 2β, 2γ; ξ, η, ζ ,
(3.2)
2
3
(3)
(3.3)
ω3 = η 1−2β FA α − β + γ + ; α, 1 − β, γ; 2α, 2 − 2β, 2γ; ξ, η, ζ ,
2
3
(3)
(3.4)
ω4 = ζ 1−2γ FA α + β − γ + ; α, β, 1 − γ; 2α, 2β, 2 − 2γ; ξ, η, ζ ,
2
5
(3)
ω5 = ξ 1−2α η 1−2β FA − α − β + γ + ; 1 − α, 1 − β, γ;
2
(3.5)
2 − 2α, 2 − 2β, 2γ; ξ, η, ζ ,
5
(3)
ω6 = ξ 1−2α ζ 1−2γ FA − α + β − γ + ; 1 − α, β, 1 − γ; 2 − 2α, 2β,
2
(3.6)
2 − 2γ; ξ, η, ζ ,
5
(3)
ω7 = η 1−2β ζ 1−2γ FA α − β − γ + ; α, 1 − β, 1 − γ; 2α, 2 − 2β,
2
(3.7)
2 − 2γ; ξ, η, ζ ,
6
R. B. SEILKHANOVA, A. H. HASANOV
EJDE-2015/09
7
− α − β − γ + ; 1 − α, 1 − β, 1 − γ;
2
2 − 2α, 2 − 2β, 2 − 2γ; ξ, η, ζ ,
(3)
ω8 = ξ 1−2α η 1−2β ζ 1−2γ FA
(3.8)
where the Lauricella hypergeometric function is defined as
∞
X
(a)m+n+p (b1 )m (b2 )n (b3 )p
m,n,p=0
(c1 )m (c2 )n (c3 )p m!n!p!
(3)
FA (a; b1 , b2 , b3 ; c1 , c2 , c3 ; x, y, z) =
xm y n z p .
Further, substituting (3.1) - (3.8) in (2.1), we obtain
q1 (x, y, t; x0 , y0 , t0 )
−α−β−γ− 21
= k1 (r2 )
1
FA α + β + γ + ; α, β, γ; 2α, 2β, 2γ; ξ, η, ζ ,
2
(3.9)
q2 (x, y, t; x0 , y0 , t0 )
α−β−γ− 32
= k2 (r2 )
3
− α + β + γ + ; 1 − α, β, γ;
2
(3.10)
3
α − β + γ + ; α, 1 − β, γ; 2α,
2
(3.11)
3
α + β − γ + ; α, β, 1 − γ; 2α, 2β,
2
(3.12)
5
− α − β + γ + ; 1 − α,
2
(3.13)
5
− α + β − γ + ; 1 − α, β,
2
(3.14)
5
α − β − γ + ; α, 1 − β,
2
(3.15)
(3)
1−2α
FA
1−2β
FA
(xx0 )
2 − 2α, 2β, 2γ; ξ, η, ζ ,
q3 (x, y, t; x0 , y0 , t0 )
−α+β−γ− 32
= k3 (r2 )
(yy0 )
2 − 2β, 2γ; ξ, η, ζ ,
(3)
q4 (x, y, t; x0 , y0 , t0 )
−α−β+γ− 23
= k4 (r2 )
(tt0 )
1−2γ
(3)
FA
2 − 2γ; ξ, η, ζ ,
q5 (x, y, t; x0 , y0 , t0 )
α+β−γ− 52
= k5 (r2 )
1−2α
(xx0 )
1−2β
(yy0 )
(3)
FA
1 − β, γ; 2 − 2α, 2 − 2β, 2γ; ξ, η, ζ ,
q6 (x, y, t; x0 , y0 , t0 )
α−β+γ− 52
= k6 (r2 )
(xx0 )
1−2α
1−2γ
(tt0 )
1 − γ; 2 − 2α, 2β, 2 − 2γ; ξ, η, ζ ,
(3)
FA
q7 (x, y, t; x0 , y0 , t0 )
= k7 (r2 )
−α+β+γ− 52
(yy0 )
1−2β
1−2γ
(tt0 )
1 − γ; 2α, 2 − 2β, 2 − 2γ; ξ, η, ζ ,
(3)
FA
q8 (x, y, t; x0 , y0 , t0 )
7
(3)
FA − α − β − γ + ;
2
1 − α, 1 − β, 1 − γ; 2 − 2α, 2 − 2β, 2 − 2γ; ξ, η, ζ
1−2α
= k8 (xx0 )
1−2β
(yy0 )
(tt0 )
1−2γ
α+β+γ− 27
(r2 )
(3.16)
EJDE-2015/09
GENERALIZED EULER-POISSON-DARBOUX EQUATION
7
where ki , i = 1, 8 are constants.
4. Some properties of particular solutions
Theorem 4.1. If α, β, γ > 0, then particular solutions (3.9)–(3.16) tends to infinity
of the order 1/r at r → 0.
Proof. By the expansion for the Lauricella hypergeometric function [12]
(3)
FA (a; b1 , b2 , b3 ; c1 , c2 , c3 ; x, y, z)
∞
X
(a)i+j+k (b1 )i+j (b2 )i+k (b3 )j+k
=
i,j,k=0
(c1 )i+j (c2 )i+k (c3 )j+k i!j!k!
xi+j y i+k z j+k
× F (a + i + j, b1 + i + j; c1 + i + j; x)F (a + i + j + k, b2 + i + k; c2 + i + k; y)
× F (a + i + j + k, b3 + j + k; c3 + j + k; z),
(4.1)
the particular solution (3.9) is rewritten as follows
−α−β−γ− 12
q1 (x, y, t; x0 , y0 , t0 ) = k1 (r2 )
×
∞
X
(α + β + γ + 21 )i+j+k (α)i+j (β)i+k (γ)j+k
(2α)i+j (2β)i+k (2γ)j+k i!j!k!
i,j,k=0
i+j
i+k
j+k
r2 − r12 r2 − r22 r2 − r32 r2
r2
r2
1
r2 − r12 × F α + β + γ + + i + j, α + i + j; 2α + i + j;
2
r2
1
r2 − r22 × F α + β + γ + + i + j + k, β + i + k; 2β + i + k;
2
r2
2
r − r32 1
.
× F α + β + γ + + i + j + k, γ + j + k; 2γ + j + k;
2
r2
×
−b
Using the formula F (a, b; c; x) = (1 − x)
obtain
q1 (x, y, t; x0 , y0 , t0 ) = k1 (r2 )
− 21
−α
(r12 )
(4.2)
F (c − a, b; c; x/(x − 1)) [9], from (4.2) we
(r22 )
−β
−γ ∗
q1 (x, y, t; x0 , y0 , t0 ),
(r32 )
(4.3)
where
q1∗ (x, y, t; x0 , y0 , t0 ) =
∞
X
(α + β + γ + 21 )i+j+k (α)i+j (β)i+k (γ)j+k
i,j,k=0
(2α)i+j (2β)i+k (2γ)j+k i!j!k!
i+j
i+k
j+k
r12 − r2 r22 − r2 r32 − r2 r12
r22
r32
1
r2 − r2 (4.4)
× F α − β − γ − , α + i + j; 2α + i + j; 1 2
2
r1
1
r2 − r2 × F β − α − γ − − j, β + i + k; 2β + i + k; 2 2
2
r2
2
1
r − r2 × F γ − α − β − − i, γ + j + k; 2γ + j + k; 3 2
.
2
r3
×
8
R. B. SEILKHANOVA, A. H. HASANOV
EJDE-2015/09
Now we show that q1∗ (x, y, t; x0 , y0 , t0 ) is bounded aa r → 0. Since
r12 − r2
r22 − r2
r32 − r2
=
lim
=
lim
= 1,
r→0
r→0
r→0
r12
r22
r32
lim
hence, one can find
1
F α − β − γ − , α + i + j; 2α + i + j; 1
2
Γ(2α)Γ(β + γ + 21 )(2α)i+j
=
,
Γ(α)Γ(α + β + γ + 12 )(α + β + γ + 12 )i+j
1
F β − α − γ − − j, β + i + k; 2β + i + k; 1
2
Γ(2β)Γ(α + γ + 21 )(2β)i+k (α + γ + 12 )2j
,
=
Γ(β)Γ(α + β + γ + 21 )(α + β + γ + 21 )i+j+k
1
F γ − α − β − − i, γ + j + k; 2γ + j + k; 1
2
Γ(2γ)(2γ)j+k Γ(α + β + 21 )(α + β + 21 )i
.
=
Γ(γ)Γ(α + β + γ + 21 )(α + β + γ + 21 )i+j+k
(4.5)
(4.6)
(4.7)
Due to (4.5)–(4.7) at r → 0 from (4.4) we obtain
lim q1∗ (x, y, t; x0 , y0 , t0 )
r→0
=
Γ(2α)Γ(2β)Γ(2γ)Γ(α + β + 12 )Γ(α + γ + 21 )Γ(β + γ + 21 )
Γ(α)Γ(β)Γ(γ)Γ3 (α + β + γ + 21 )
×
(4.8)
(α + β + 21 )i (α + γ + 21 )j (α)i+j (β)i+k (γ)j+k
.
(α + β + γ + 21 )i+j (α + β + γ + 12 )i+j+k i!j!k!
i,j,k=0
∞
X
It is easy to show that
∞
X
(α + β + 12 )i (α + γ + 12 )j (α)i+j (β)i+k (γ)j+k
i,j,k=0
(α + β + γ + 12 )i+j (α + β + γ + 12 )i+j+k i!j!k!
Γ( 12 )Γ2 (α + β + γ + 21 )
=
.
Γ(α + β + 12 )Γ(α + γ + 12 )Γ(β + γ + 21 )
Thus, from (4.4) we deduce
lim
r→0
q1∗ (x, y, t; x0 , y0 , t0 )
(4.9)
√
=
πΓ(2α)Γ(2β)Γ(2γ)
.
Γ(α)Γ(β)Γ(γ)Γ(α + β + γ + 21 )
(4.10)
From here considering (4.10), from (4.3) at r → 0 we have the estimate
c0
|q1 (x, y, t; x0 , y0 , t0 )| ≤ ,
(4.11)
r
where
k1 22α+2β+2γ−1 Γ(α + β + γ)Γ(2α)Γ(2β)Γ(2γ)
c0 =
.
α
β
γ
Γ(α)Γ(β)Γ(γ)Γ(2α + 2β + 2γ)(r12 ) (r22 ) (r32 )
Estimate (4.11) states that function q1 (x, y, t; x0 , y0 , t0 ) at r → 0 tends to infinity
of the order 1/r. Similarly one can prove that every function qi (x, y, t; x0 , y0 , t0 ),
i = 2, 3, . . . , 8 has singularity 1/r as r → 0.
EJDE-2015/09
GENERALIZED EULER-POISSON-DARBOUX EQUATION
9
Theorem 4.2. If α, β, γ > 0, then the found particular solutions have the following
properties
∂ ∂ ∂ q1
= 0, y 2β q1 = 0,
= 0, t2γ q1 ∂x x=0
∂y y=0
∂t t=0
∂ ∂ q2 x=0 = 0, y 2β q2 = 0, t2γ q2 = 0,
∂y y=0
∂t t=0
∂ ∂ x2α q3 = 0, q3 y=0 = 0, t2γ q3 = 0,
∂x x=0
∂t t=0
∂ ∂ = 0, y 2β q4 = 0, q4 t=0 = 0,
x2α q4 ∂x x=0
∂y y=0
∂ q5 x=0 = 0, q5 y=0 = 0, t2γ q5 = 0,
∂t t=0
∂ = 0, q6 t=0 = 0,
q6 x=0 = 0, y 2β q6 ∂y y=0
∂ x2α q7 = 0, q7 y=0 = 0, q7 t=0 = 0,
∂x x=0
q8 x=0 = 0, q8 y=0 = 0, q8 t=0 = 0.
x2α
Proofs of the above equalities are based on elementary calculations. These properties could be used in studying various boundary problems for the equation (1.1).
Acknowledgements. We are grateful to the Professor H. M. Srivastava for his
suggestion to consider this problem.
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Rakhila B. Seilkhanova
S. Baishev University Aktobe, Department for Information Systems and Applied Mathematics, 030000 Aktobe, Zhubanov str. 302, Kazakhstan
E-mail address: srahila@inbox.ru
Anvar H. Hasanov
S. Baishev University Aktobe, Department for Information Systems and Applied Mathematics, 030000 Aktobe, Zhubanov str. 302, Kazakhstan
E-mail address: anvarhasanov@yahoo.com
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