Siberian Mathematical Journal, Vol. 48, No. 4, pp. 707–717, 2007 c 2007 Salakhitdinov M. S. and Urinov A. K. Original Russian Text Copyright
EIGENVALUE PROBLEMS FOR A MIXED-TYPE EQUATION WITH TWO SINGULAR COEFFICIENTS M. S. Salakhitdinov and A. K. Urinov
UDC 517.95
Abstract: We find the eigenvalues and eigenfunctions of two nonlocal problems for a mixed-type equation with the singular coefficients of lower-order terms. Keywords: mixed-type equation, singular coefficient, fundamental solution, Cauchy–Goursat problem, hypergeometric function, fractional operator
Consider the finite simply connected domain Ω bounded for y > 0 by the arc σ0 = {(x, y) : x2 + y 2 = 1, x > 0} and the straight line segment OA = {(x, y) : x = 0, 0 < y < 1}, and for y < 0, by the characteristics OC, x + y = 0, and BC, x − y = 1, of the equation uxx + sgn yuyy +
2α 2β ux + uy + λu = 0, x |y|
(1)
where λ is some numerical parameter and α, β ∈ R are such that 0 < 2α < 1, 0 < 2β < 1, α < β. Introduce the following notation: Ω1 = Ω ∩ (y < 0),
Ω2 = Ω ∩ (y > 0), OB = Ω ∩ (y = 0); x √ t ∂ 1, λ J0 [ λx(x − t)] dt; A0x [g(x)] ≡ g(x) − g(t) x ∂t
F0x
1 a, b g(x) ≡ c, xk Γ(c)
x
0
xk − tk ktk−1 dt, F a, b, c; xk
g(t)(x − t ) k
k c−1
0
where a, b, c ∈ R, k ∈ N; also, Γ(z) is the Euler gamma function, Jm (z) is the Bessel function of the first kind of order m, and F (a, b, c; z) is the Gauss hypergeometric function. √
1, λ and F0x are the operators introduced in [1] and [2] respectively. Note that A0x Finding the eigenvalues and eigenfunctions of boundary value problems for various equations of mixed type is the topic of many studies among which we should mention the articles by Moiseev, Ponomarev, Kal menov, and others; for instance, see [3–9]. In this article we find the eigenvalues and the corresponding eigenfunctions of two nonlocal problems for (1) in Ω. Similar problems for the Lavrent ev–Bitsadze equation and an equation of mixed type with one singular coefficient were previously studied in [10, 11]. Henceforth by a regular solution to (1) in Ω1 we understand some function u(x, y) with the properties
u(x, y) ∈ C(Ω1 ) ∩ C 2 (Ω1 ), u(x, 0) ∈ C[0, 1] ∩ C 2 (0, 1),
(−y)2β uy ∈ C(Ω1 ∪ OB), lim (−y)2β uy ∈ C 2 (0, 1)
y→0
that meets (1) in Ω1 ; here limy→0 (−y)2β uy could have singularity of order less than 1 − 2β as x → 0 or x → 1. Tashkent. Translated from Sibirski˘ı Matematicheski˘ı Zhurnal, Vol. 48, No. 4, pp. 882–893, July–August, 2007. Original article submitted June 8, 2006. c 2007 Springer Science+Business Media, Inc. 0037-4466/07/4804–0707
707
If u(x, y) is a regular solution to (1) in Ω1 then u(x, y) can be represented [12] as 1
y α τ (x − ty)(1 − t2 )β−1 1 − t Q1 (x, y, t) dt x
u(x, y) = γ1 −1 1−2β
1
−γ2 (−y)
−1
y α ν(x − ty)(1 − t2 )−β 1 − t Q2 (x, y, t) dt, x
(2)
√ √ where γ1 = Γ(β + 1/2)/[ πΓ(β)], γ2 = Γ( 12 − β)/[2 πΓ(1 − β)]; Q1 = Ξ2 (α, 1 − α, β; σ1 , σ2 ), Q2 = Ξ2 (α, 1 − α, 1 − β; σ1 , σ2 ); σ1 = y 2 (t2 − 1)/[4x(x − ty)], σ2 = −λy 2 (t2 − 1)/4; Ξ2 (a, b, c; x, y) is the Humbert hypergeometric function [13]. Henceforth by a regular solution to (1) in Ω2 we understand some function u(x, y) with the properties u(x, y) ∈ C(Ω2 ) ∩ C 1 (Ω2 ∪ σ0 ) ∩ C 2 (Ω2 ),
y 2β uy ∈ C(Ω2 ∪ OB)
that meets (1) in Ω2 , and by a regular solution to (1) in Ω we understand some function u(x, y) that is a regular solution to (1) in Ω1 and Ω2 and satisfies the conditions u(x, −0) = u(x, +0), 0 ≤ x ≤ 1;
lim (−y)2β uy = lim y 2β uy ,
y→−0
y→+0
0 < x < 1.
Problem A0λ . Find the values of λ and the corresponding nontrivial regular solutions to (1) in Ω satisfying the conditions ∂u(x, y) (3) = 0, (x, y) ∈ σ0 , α0 u(x, y) + β0 ∂n u(0, y) = 0, 0 ≤ y ≤ 1, (4)
√ 1+α−β β−α d , α−β 1, λ 2 2 x 2 (x2 ) 2 F0x (x2 )α+β−1 u(θ0 ) + γ0 u(x, 0) = 0, 0 ≤ x ≤ 1, (5) x1−α−β A0x 1 − β, x2 dx where α0 , β0 , γ0 ∈ R with α02 + β02 = 0; n is the exterior normal to σ0 ; and θ0 is the point of
intersection of the characteristic OC with the characteristic issuing from (x, 0) ∈ OB; i.e., θ0 = x2 , − x2 . Before turning to Problem A0λ , we consider the Cauchy–Goursat problem for (1) in Ω1 and some properties that follow from representation of its solution. The Cauchy–Goursat Problem. Find a regular solution to (1) in Ω1 satisfying the conditions lim (−y)2β uy = ν(x), 0 < x < 1;
y→0
u(x, −x) = ϕ(2x), 0 ≤ x ≤ 1/2,
where ν(x) and ϕ(x) are given functions such that ν(x) belongs to C 2 (0, 1) and can have singularity of order less than 1 − 2β as x → 0 or x → 1, and ϕ(x) ∈ C 1 [0, 1] ∩ C (2,δ) (0, 1) for some δ > 0. Using the results of [12, 13], it is not difficult to verify that a solution to this problem exists, is unique, and can be represented as
u(x, y) = χ ¯2
2β−1
x+y
−β ν(t) r02 (t/x)α Ξ2 (α, 1 − α, 1 − β; r1 , r2 ) dt
0 x+y
Φ0 (t)H(x, y; t/2, −t/2) dt +
+ 0
708
x−y
x+y
Φ0 (t)R(x, y; t/2, −t/2) dt,
(6)
where Φ0 (t) = ϕ (t) + (α + β)t−1 ϕ(t), χ ¯ = Γ(β)/[Γ(1 − β)Γ(2β)]; r02 = (x − t)2 − y 2 , r1 = −r02 /(4xt), 2 r2 = −λr0 /4; H(x, y; x0 , y0 ) and R(x, y; x0 , y0 ) are respectively the Green–Hadamard and Riemann functions [12] of (1);
β H(x, y; x0 , y0 ) = χ ¯ (x0 /x)α (−y0 )2β 4/R02 ∞ λR02 n 1 − × H2 (β − n, β, α, 1 − α, 2β; R1 , R2 ), (1 − β)n n! 4 n=0
R(x, y; x0 , y0 ) = (x0 /x)α (y0 /y)β n ∞ λR02 1 − × F3 (β, α, 1 − β, 1 − α, 1 + n; 1/R1 , −R2 ); (n!)2 4 n=0
R02
= (x − x0 )2 − (y − y0 )2 ,
R1 = 4yy0 /R02 , ∞
Ξ2 (a, b, c; x, y) =
m,k=0
H2 (a, b, c, d, e; x, y) =
R2 = R02 /(4xx0 );
(a)m (b)m m k x y , (c)m+k m!k!
∞ (a)m−k (b)m (c)k (d)k m k x y , (e)m m!k!
m,k=0
∞ (a)m (b)k (c)m (d)k m k F3 (a, b, c, d, e; x, y) = x y , (e)m+k m!k! m,k=0
(a)m = a(a + 1)(a + 2) . . . (a + m − 1) = Γ(a + m)/Γ(a). Lemma 1. If u(x, y) is a regular solution to (1) in Ω1 with the properties u(x, −x) ∈ C 1 [0, 1] ∩ C (2,δ) (0, 1) for some δ > 0, u(x, 0) = τ (x), and limy→0 (−y)2β uy = ν(x), then 1+α−β α−β d , , 2 β−α 2 α+β−1 2 2 (x ) Γ(1 − β)x (x ) 2 F0x . u(θ0 ) τ (x) = χ ¯2 1 − β, x2 dx2 α x t (x − t)2 λ 2β−1 −2β 2 ν(t)(x − t) , Ξ2 α, 1 − α, 1 − β; − , − (x − t) dt. +¯ χ2 x 4xt 4 1−α−β
β−α
√
1, λ x A0x
0
Proof. Take some regular solution u(x, y) to (1) in Ω1 such that limy→0 (−y)2β uy = ν(x) and u(x, −x) = ϕ(2x) ∈ C 1 [0, 1] ∩ C (2,δ) (0, 1) for some δ > 0. Then (6) holds. Hence, using the expansions [12, 14] of H2 and Ξ2 , we obtain Ξ2 (a, b, c; x, y) =
∞ k=0
H2 (a, b, c, d, e; x, y) =
yk F (a, b, c + k; x), (c)k k!
∞ (a)m (b)m m x F (c, d, 1 − a − m; −y), (e)m m!
m=0
and writing Φ1 (t) =
d α+β ϕ(t)] dt [t
as y → 0 we have x τ (x) = χ ¯2
−2β
ν(t)(x − t)
β−1 0
α t x 709
(x − t)2 λ 2 ¯ 2−α x−α−β , − (x − t) dt + χ ×Ξ2 α, 1 − α, 1 − β; − 4xt 4 x t−x λ −β β , (t − x) dt. × Φ1 (t)(x − t) t Ξ2 α, 1 − α, 1 − β; 2t 4
(8)
0
Denoting by Φ2 (x) the second integral on the right-hand side of (8) and taking into account the following formula [14]: ∞ (a)m (b)m m √ x J c+m−1 (2i y), Ξ2 (a, b, c; x, y) = (c)m m! m=0
where J m (z) = Γ(m + 1)(z/2)−m Jm (z) is the Bessel–Clifford function, we have x Φ2 (x) = 0
m ∞ 1 (α)m (1 − α)m − Φ1 (t) tβ−m (x − t)m−β J m−β [ λx(x − t)] dt. (1 − β)m m! 2 m=0
Hence, using the easy identity ∂ m−β (x − t) J m−β [ λx(x − t)] = (s − t)m−β J0 [ λx(x − s)] ds, ∂t t
x
after some manipulations we obtain √
1, λ −1 [x Φ3 (x)], Φ2 (x) = xA0x
where
x
−β
Φ1 (t)t (x − t) β
Φ3 (x) = 0
(9)
t−x F α, 1 − α, 1 − β; dt. 2t
Using the well-known properties of the hypergeometric function [13], c−a c+a−1 c−1 , , c; 4z(1 − z) , F (a, 1 − a, c; z) = (1 − z) F 2 2 F (a, b, c; z) = (1 − z)−b F (c − a, b, c; z/(z − 1)), d [xα+β ϕ(zx)], we have and the equality Φ1 (xz) = z α+β−1 dx β−α d (x2 ) 2 F0x Φ3 (x) = 2 Γ(1 − β)x 2 dx
β
2
1+α−β
, 1 − β, 2
α−β 2 x2
(x2 )α+β−1 ϕ(x).
(10)
Taking into account the equality ϕ(x) = u(θ0 ) and inserting (10) into (9) and (9) into (8), we obtain (7). Lemma 1 is proved. Lemma 2. We have the equality x
−2β β−1
(x − t) 0
710
t
√ √ (x − t)2 λ 2 , − (x − t) dt = l0 x−β Jp ( λx), Jp ( λt)Ξ2 α, 1 − α, 1 − β; − 4xt 4
(11)
where l0 = 22β−1 Γ(l1 )Γ(l3 − l2 )Γ(1 − 2β)/[Γ(1 − l2 )Γ(1 + l1 − l3 )], 2l1 = α + β + p, 2l2 = α + β − p, 2l3 = 1 + 2β, p = (α + β)2 + q 2 , Re q 2 > 0, and we pick the branch of the root that corresponds to the arithmetic root. Proof. Take some regular solution u(x, y) to (1) in Ω1 satisfying the conditions u(x, −x) = 0, τ (x) = u(x, 0), and ν(x) = limy→0 (−y)2β uy . Then Lemma 1 implies the equality x (x − t)2 λ 2β−1 −2β α 2 , − (x − t) dt. ν(t)(x − t) (t/x) Ξ2 α, 1 − α, 1 − β; − (12) τ (x) = χ ¯2 4xt 4 0
On the other hand, seeking a regularsolution to (1) in Ω1 satisfying the condition u(x, −x) = 0 in the form u(x, y) = P (ρ)Q(θ), where ρ = x2 − y 2 and θ = −y 2 /ρ2 , we find √ 1 , u(x, y) = c0 ρ−α−β (−θ)−l1 Jp ( λρ)F l1 , 1 + l1 − l3 , 1 + l1 − l2 , θ which implies that √ τ (x) = lim u(x, 0) = l4 x−α−β Jp ( λx), y→0 (13) √ ν(x) = lim (−y)2β uy = l4 (1 − 2β)k0 xβ−α−1 Jp ( λx). y→0
Here c0 = 0 is an arbitrary number, l4 = c0 Γ(1 + l1 − l2 )Γ(1 − l3 )/[Γ(1 − l2 )Γ(1 + l1 − l2 )], and k0 = Γ(l3 )Γ(1 − l2 )Γ(1 + l1 − l3 )/[Γ(l1 )Γ(l3 − l2 )Γ(2 − l3 )]. Inserting (13) into (12) and taking into account the uniqueness of a solution to the Cauchy–Goursat problem, we see that (11) holds. Lemma 2 is proved. Remark. The equality (11) with α = 0 implies a well-known formula 6.581 (5) on p. 712 of [15]. We now address Problem A0λ . Take some solution u(x, y) to A0λ such that τ (x) = u(x, 0) and ν(x) = limy→0 (−y)2β uy . Then Lemma 1 implies that (7) holds. We then deduce from (5) a functional relation between τ (x) and ν(x) on OB which is inherited from Ω1 : 2β−1
x ν(t)(x − t)
¯2 γ3 τ (x) = χ
−2β
0
α t (x − t)2 λ 2 , − (x − t) dt, Ξ2 α, 1 − α, 1 − β; − x 4xt 4
(14)
where γ3 = 1 + γ0 2β−α Γ(β)/Γ(2β). Therefore, Problem A0λ is reduced to the following equivalent elliptic Problem Cλ0 : Find the values of λ and the corresponding nontrivial regular solutions to (1) in Ω2 satisfying (3), (4), and(14). 2 2 We seek y for a solution to this problem in the form u(x, y) =0 P (r)Q(ϕ), where r = x + y and −1 ϕ = tan x with 0 ≤ r ≤ 1 and 0 ≤ ϕ ≤ π/2. Then Problem Cλ splits into two:
r2 P (r) + (1 + 2α + 2β)rP (r) + (λr2 − μ2 )P (r) = 0, |P (0) < +∞|,
0 < r < 1,
α0 P (r) + β0 P (r) = 0 for r = 1;
(15) (16)
and
Q (ϕ) + (2β cot ϕ − 2α tan ϕ)Q (ϕ) + μ2 Q(ϕ) = 0,
0 < ϕ < π/2,
Q(π/2) = 0, ¯2 γ3 P (x)Q(0) = χ
2β−1
(18) x
2β
[Q (ϕ)(sin ϕ) ]|ϕ→0
×Ξ2 α, 1 − α, 1 − β; −
(17)
(x − t)−2β (t/x)α P (t)t2β−1
0
t)2
(x − 4xt
λ , − (x − t)2 dt, 4
(19)
where μ is the separation constant. 711
Making the substitution ψ = sin2 ϕ, we rewrite (17) as 1 ψ(1 − ψ)Qψψ + [1/2 + β − (1 + a + β)ψ]Qψ + μ2 Q = 0. 4 Using solutions to the Gauss hypergeometric equation [13], we find the general solution to (17): 0
Q(ϕ) = c1 F (a0 , b0 , c0 ; sin2 ϕ) + c2 (sin2 ϕ)1−c F (a0 − c0 + 1, b0 − c0 + 1, 2 − c0 ; sin2 ϕ), where c1 , c2 are arbitrary numbers, 2a0 = α+β+w, 2b0 = α+β−w, 2c0 = 1+2β, and w = This implies that Q(0) = c1 , lim Q (sin ϕ)2β = (1 − 2β)c2 .
(20)
(a + β)2 + μ2 .
ϕ→0
(21)
The function √ P (r) = r−α−β Jw ( λr), where w = (α + β)2 + μ2 , Re μ2 > 0, λ = 0, is a solution to (15) bounded for r = 0. Inserting it into (19) and using (11) and (21), we find c2 = c1 γ3 Γ(1 − b0 )Γ(1 + a0 − c0 )Γ(c0 )/[Γ(a0 )Γ(c0 − b0 )Γ(2 − c0 )].
(22)
Requiring that (20) satisfies (18) and using (22), we find that nontrivial solutions to (17)–(19) exist in Ω2 only for μn = wn2 − (α + β)2 , where wn is the solution to the equation π π cos (α − β + w) + γ3 sin (α + β + w) = 0 2 2 that satisfies the condition w > α + β; i.e.,
for α + β + −α − β − π2 cotan−1 ζ + 2n wn = 2 −1 −α − β − π cotan ζ + 2(n + 1) for α + β +
1 π 1 π
cotan−1 ζ < 1, cotan−1 ζ ≥ 1,
(23)
where ζ = (γ3 + sin βπ)/ cos βπ, n = 1, 2, . . . . Consequently, by (20), (22), and (23), the eigenfunctions of (17)–(19) are the functions Qn (ϕ) = cn {F (an , bn , c0 ; sin2 ϕ) +γ3 kn (sin ϕ)1−2β F (1 + an − c0 , 1 + bn − c0 , 2 − c0 ; sin2 ϕ)},
n = 1, 2, . . . ,
where cn = 0 are arbitrary numbers, 2an = α + β + wn , 2bn = α + β − wn , and kn = Γ(1 − bn )Γ(1 + an − c0 )Γ(c0 )/[Γ(an )Γ(c0 − bn )Γ(2 − c0 )]. Inserting w = wn into the solution to (15) and fulfilling the second condition in (16), we have √ √ √ [α0 − (α + β)β0 ]Jwn ( λ) + β0 λ · Jw n ( λ) = 0, n = 1, 2, . . . . (24) √ If β0 = 0 then (24) take the form Jwn ( λ) = 0. Since wn > 0, these equations have only real roots [16]. If, however, β0 = 0 and the condition (α0 /β0 ) − α − β + w1 ≥ 0 is fulfilled then for n = 2, 3, . . . the conditions (α0 /β0 ) − α − β + wn > 0 are fulfilled. Then by the general theory of Bessel functions [16] all roots of (24) are real. Consequently, if β0 = 0 or β0 = 0 and (α0 /β0 ) − α − β + w1 ≥ 0 then each equation in (24) has (w ) countably many real roots. Denoting by θm k the mth root of (24) for n = k, we obtain the eigenvalues (wn ) 2 , where n, m = 1, 2, . . . , of Problem A0λ (Cλ0 ). λn,m = θm 712
The eigenfunctions corresponding to the eigenvalues found in Ω2 are given by the formula un,m (x, y) = cn,m r−α−β Jwn ( λn,m r){F (an , bn , c0 ; sin2 ϕ) +γ3 kn (sin ϕ)1−2β F (1 + an − c0 , 1 + bn − c0 , 2 − c0 ; sin2 ϕ)}, where cn,m = 0 are arbitrary numbers, n, m = 1, 2, . . . . The eigenfunctions of Problem A0λ in the domain Ω1 are defined as solutions to the equations uxx − uyy +
2α 2β ux − uy + λn,m u = 0, x y
n, m = 1, 2, . . . ,
satisfying the initial conditions
u(x, 0) = τn,m (x) = cn,m x−α−β Jwn ( λn,m x),
lim (−y)2β uy (x, y) = νn,m (x) = cn,m kn (1 − 2β)xβ−α−1 Jwn ( λn,m x);
y→0
they are given by (2). This completes our study of Problem A0λ . Problem A1λ . Find the values of λ and the corresponding nontrivial regular solutions to (1) in Ω satisfying (3), (4) and such that α+β √ α+β−1 1−α−β 2α−1 d 1, λ 2 , 2 (x2 ) 2 u(θ0 ) x 2 (x2 ) 2 F0x xβ−α A0x 2 β, x dx +γ0 lim (−y)2β uy (x, y) = 0, y→0
0 < x < 1.
(25)
Before addressing Problem A1λ , consider the Darboux problem for (1) in Ω1 and some corollaries of the representation of its solution. The Darboux Problem. Find a regular solution to (1) in Ω1 satisfying the conditions u(x, 0) = τ (x), 0 ≤ x ≤ 1,
u(x, −x) = ϕ(2x), 0 ≤ x ≤ 1/2,
where τ (x) and ϕ(x) are given functions such that τ (0) = ϕ(0), τ (x) ∈ C[0, 1] ∩ C 2 (0, 1), and ϕ(x) ∈ C 1 [0, 1] ∩ C (2,δ) (0, 1) for some δ > 0. Kapilevich constructed [12] the Riemann–Hadamard function for this problem. Using it, by analogy with [12, 14, 17] it is easy to verify that a solution to the Darboux problem exists, is unique, and can be represented as 1−2β
x+y
β−1 τ (t) r02 (t/x)α Ξ2 (α, 1 − α, β; r1 , r2 ) dt
u(x, y) = χ1 (−y)
0 x+y
x−y
Φ0 (t)H(x, y; t/2, −t/2)dt +
+ 0
Φ0 (t)R(x, y; t/2, −t/2) dt,
(26)
x+y
where χ1 = (1 − 2β)χ21−2β , χ = Γ(1 − β)/[Γ(β)Γ(2 − 2β)], H(x, y; x0 , y0 ) = χ(x0 /x)α R11−2β (2/R0 )2β ∞ λR2 n 1 − 0 H2 (1 − β − n, 1 − β, α, 1 − α, 2 − 2β; R1, R2 ), × (β)n n! 4 n=0
r0 , r1 , r2 , R0 , R1 , R2 , Φ0, (t), R(x, y; x0 , y0 ) are the same functions as in the Cauchy–Goursat problem. 713
Lemma 3. If u(x, y) is a regular solution to (1) in Ω1 with the properties u(x, −x) ∈ C 1 [0, 1] ∩ for some δ > 0, u(x, 0) = τ (x), and limy→0 (−y)2β uy = ν(x), then
C (2,δ) (0, 1)
−α
ν(x) = χ1 x
d dx
x 0
τ (t)tα dt + (2β − 1) (x − t)1−2β
x 0
(x − t)2 λ , − (x − t)2 × Ξ2 α, 1 − α, β; − 4xt 4 α+β √ d 1−α−β β−α 1, λ 2 1−α−β 2 , 2 −Γ(β)2 χ1 x A0x x 2 (x ) F0x β, dx
τ (t)tα (x − t)2β−2
− 1 dt α+β−1 2 x2
(x2 )
2α−1 2
u(θ0 ).
(27)
Proof. Take some solution u(x, y) to (1) in Ω1 such that u(x, 0) = τ (t) and u(x, −x) = ϕ(2x) ∈ ∩ C (2,δ) (0, 1) for some δ > 0. Then (26) holds. Consider the function
C 1 [0, 1]
1−2β
x+y−ε
τ (t)(r02 )β−1 (t/x)α dt
uε (x, y) = χ1 (−y)
0
+χ1 (−y)1−2β
x+y−ε
τ (t)(r02 )β−1 (t/x)α [Ξ2 (α, 1 − α, β; r1 , r2 ) − 1] dt
0 x+y−ε
x−y
Φ0 (t)H(x, y; −t/2, −t/2) dt +
+ 0
Φ0 (t)R(x, y; t/2, −t/2) dt,
x+y+ε
where ε > 0 is some sufficiently small number. It is obvious that limε→0 uε (x, y) = u(x, y). By direct calculation we deduce from (28) that ∂uε = χ1 (−y)1−2β τ (x + y − ε)[ε(ε − 2y)]β−1 [(x + y − ε)/x]α ∂y x+y−ε β−2 −2β +χ1 (−y) τ (t) r02 (t/x)α (2β − 1)r02 + 2(β − 1)y 2 dt 0 1−2β
+χ1 (−y) τ (x + y − ε)[ε(ε − 2y)]β−1 [(x + y − ε)/x]α
ε(2y − ε) λ , ε(2y − ε) − 1 × Ξ2 α, 1 − α, β; 4x(x + y − ε) 4 x+y−ε β−1 −2β τ (t) r02 (t/x)α [Ξ2 (α, 1 − α, β; r1 , r2 ) − 1] dt −χ1 (1 − 2β)(−y) 0 2−2β
x+y−ε
β−2 τ (t) r02 (t/x)α
+2χ1 (−y)
0
2 ∂ × (β − 1)[Ξ2 (α, 1 − α, β; r1 ; r2 ) − 1] + r0 2 Ξ2 (α, 1 − α, β; r1 ; r2 ) dt ∂r0 x−y x−y ,− −Φ0 (x − y)R x, y; 2 2 714
x+y−ε x+y−ε ,− +[Φ0 (x + y − ε) − Φ0 (x + y + ε)]H x, y; 2 2 x+y+ε x+y+ε x+y−ε x+y−ε +Φ0 (x + y + ε) H x, y; ,− − R x, y; ,− 2 2 2 2 x+y−ε x−y ∂ ∂ Φ0 (t) H(x, y; t/2, −t/2)dt + Φ0 (t) R(x, y; t/2, −t/2) dt. + ∂y ∂y 0
x+y+ε
Multiplying both sides of the last equality by (−y)2β and passing to the limit as ε → 0 and y → 0, we have −α
ν(x) = χ1 x x +(2β − 1)
τ (t)tα (x − t)2β−2
0 −α β−α−1
−χ1 2
d dx
x
Φ0 (t)t
0
τ (t)tα dt (x − t)1−2β
(x − t)2 λ Ξ2 α, 1 − α, β − − (x − t)2 − 1 dt 4xt 4
x α+1
x
(x − t)
0
β−1
t−x λ , x(t − x) dt. Ξ2 α, 1 − α, β; 2t 4
(29)
Applying the above method to Φ2 (x), we can prove that the last integral in (29) is equal to α+β √ α+β−1 2α−1 d , 1, λ 1−β 2 1−α−β 2 2 (x2 ) 2 u(θ0 ). Γ(β)xA0x x 2 (x ) 2 F0x 2 2 β, x dx This implies the claim of Lemma 3. Lemma 4. If τ (x) ∈ C[0, 1] ∩ C 2 [0, 1] and τ (0) = 0 then the integral equation x
−2β
ν(t)(x − t) 0
(x − t)2 λ 2 , − (x − t) dt (t/x) Ξ2 α, 1 − α, 1 − β; − 4xt 4 α
= [21−2β Γ(2β)Γ(1 − β)/Γ(β)]τ (x)
(30)
has a unique solution ν(x) ∈ C 2 (0, 1), given by the formula ν(x) = (1 − 2β)χ21−2β x−α x +(2β − 1) 0
τ (t)tα (x − t)2β−2
d dx
x 0
τ (t)tα dt (x − t)1−2β
(x − t)2 λ 2 Ξ2 α, 1 − α, β; − , − (x − t) − 1 dt , 4xt 4
(31)
and, conversely, if ν(x) ∈ C 2 (0, 1) (and ν(x) can have singularity of order less than 1−2β as x → 0, x → 1) then the integrodifferential equation (31) has a unique solution τ (x) ∈ C[0, 1] ∩ C 2 [0, 1] satisfying the condition τ (0) = 0 and given by (30). The claims of this lemma follow from Lemmas 1 and 3 for u(x, −x) = 0 by the uniqueness of solutions to the Cauchy–Goursat and Darboux problems. 715
We now address Problem A1λ . Take some solution u(x, y) to Problem A1λ such that u(x, 0) = τ (x) and limy→0 (−y)2β uy = ν(x). Then, using (27) and (25), we find a functional relation between τ (x) and ν(x) on OB which is inherited from Ω1 : γ4 ν(x) = (1 − 2β)χ21−2β x−α x +(2β − 1)
τ (t)tα (x − t)2β−2
d dx
x 0
τ (t)tα dt (x − t)1−2β
(x − t)2 λ 2 , − (x − t) − 1 dt , Ξ2 α, 1 − α, β; − 4xt 4
0
where γ4 = 1 − γ0 22−a−3β Γ(1 − β)/Γ(2 − 2β). Hence, by Lemma 4 we have 2β−1
x
−2β
ν(t)(x − t)
¯2 τ (x) = γ4 χ
0
(x − t)2 λ 2 , − (x − t) dt. (t/x) Ξ2 α, 1 − α, 1 − β; − 4xt 4 α
(32)
Therefore, Problem A1λ is reduced to the following equivalent elliptic Problem Cλ1 : Find the values of λ and the corresponding nontrivial regular solutions to (1) in Ω2 satisfying (3), (4), and (32). This problem is settled by analogy with Problem Cλ0 . Arguing as in the case of Problem Cλ0 , we can ¯ n,m (n, m = 1, 2, . . . ) of Problem A1 (C 1 ) are determined as the roots of the verify that the eigenvalues λ λ λ equations √ √ √ [α0 − (α + β)β0 ]Jwn ( λ) + β0 λ · Jw n ( λ) = 0, n = 1, 2, . . . , where
wn =
− π2 tan−1 ζ¯ + β − α + 2(n − 1), − 2 tan−1 ζ¯ + β − α + 2n, π
if tan−1 ζ¯ < −απ, if tan−1 ζ¯ ≥ −απ,
ζ¯ = (γ4 + sin βπ)/ cos βπ, and the corresponding eigenfunctions in Ω2 are given by the formulas u ¯ n,m (x, y) = ¯cn,m r
−α−β
¯ n,m r){γ4 F (¯ Jwn ( λ an , ¯bn , c0 ; sin2 ϕ)
+k¯n (sin ϕ)1−2β F (1 + a ¯n − c0 , 1 + ¯bn − c0 , 2 − c0 ; sin2 ϕ)}, where ¯cn,m = 0 are arbitrary numbers. 2¯ an = α + β + wn ,
2¯bn = α + β − wn ,
2c0 = 1 + 2β,
¯n − c0 )Γ(c0 )/[Γ(¯ an )Γ(c0 − ¯bn )Γ(2 − c0 )], k¯n = Γ(1 − ¯bn )Γ(1 + a
n, m = 1, 2, . . . .
The eigenfunctions of Problem A1λ in Ω1 are determined in terms of the functions u ¯ n,m as in Problem A0λ . √
1, λ and F0x in Problems A0λ and A1λ are invertible, for β0 = γ0 = 0 Note that because the operators A0x we have the homogeneous Tricomi problem for (1) in Ω. In this case wn and wn simplify to wn = wn = 2n − α − 1/2 for n = 1, 2, . . . . Note in closing that the unique solvability of two nonlocal problems of type A0λ and A1λ for (1) in the domain of hyperbolicity is studied in a modified form in [18].
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[email protected];
[email protected]
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