A 3D analog of problem M for a third-order hyperbolic equation

Differential Equations, Vol. 40, No. 12, 2004, pp. 1781–1784. Translated from Differentsial’nye Uravneniya, Vol. 40, No. 12, 2004, pp. 1698–1700. c 2004 by Volkodavov, Rodionova, Bushkov. Original Russian Text Copyright

SHORT COMMUNICATIONS

A 3D Analog of Problem M for a Third-Order Hyperbolic Equation V. F. Volkodavov, I. N. Rodionova, and S. V. Bushkov Samara State University, Samara, Russia Samara State Space University, Samara, Russia Received April 15, 2002

We consider the equation uxyz +

2βy 2α(x − z) uxz − uyz = 0 2 2 (x − z) − y (x − z)2 − y 2

(1)

(0 < α, β < 1, α + β < 1) in the domain G bounded by the planes z = x − y, z = x + y, x = h, and z = 0. The plane y = 0 splits G into two subdomains, G1 (y > 0) and G2 (y < 0). ¯ belongs Problem MC. On the set G1 ∪ G2 , find a solution of Eq. (1) that is continuous in G, to the special classes Rh in G1 and G2 , and satisfies the conditions u(x, y, x − y) = τ1 (x, y), u(h, y, z) = φ1 (y, z), u(x, y, x + y) = τ2 (x, y), u(h, y, z) = φ2 (y, z), The matching conditions

¯ 1 = {(x, y) : (x, y) ∈ ∆ ¯ 2 = {(y, z) : (y, z) ∈ ∆ ¯ ∗1 = {(x, y) : (x, y) ∈ ∆ ¯ ∗2 = {(y, z) : (y, z) ∈ ∆

0 ≤ y ≤ x ≤ h}, 0 ≤ y ≤ h − z, 0 ≤ z ≤ h}, 0 ≤ −y ≤ x ≤ h}, 0 ≤ −y ≤ h − z, 0 ≤ z ≤ h}.

lim (uy − ux ) = lim (ux + uy )

y→0−0

y→0+0

(2) (3) (4) (5) (6)

must be satisfied on the plane y = 0. We assume that the following conditions hold:

∂ 2 τ1 ∂ 2 τ2 ¯1 , ¯∗ , ∈C ∆ ∈C ∆ 1 ∂x ∂y ∂x ∂y τ1 (h, y) = τ2 (h, y) = 0, τ1 (x, 0) = τ2 (x, 0), ∂τ1 (x, 0) ∂τ1 (x, 0) ∂τ2 (x, 0) ∂τ2 (x, 0) + = − , ∂x ∂y ∂y ∂x ∂ ∂ 2 φ1 (h − z, z) φ1 (h − z, z) = φ1 (h − z, z) = = 0, ∂z ∂z ∂y ∂ ∂ 2 φ2 (z − h, z) φ2 (z − h, z) = φ2 (z − h, z) = = 0. ∂z ∂z ∂y We set

and require that

  ∂ ∂ 2 φ1 (y, z) 1 = Φ1 (y, z), ∂y ∂z ∂y y

(7) (8) (9) (10) (11)

  ∂ ∂ 2 φ2 (−y, z) 1 = Φ2 (y, z) ∂y ∂z ∂y y


¯i , Φi (y, z) ∈ C ∆

i = 1, 2.

(12)

Just as in [1, pp. 19–25; 2, pp. 46–49], in the domains G1 and G2 , we obtain a solution of the modified Cauchy problem and introduce a special class Rh of solutions of Eq. (1). A solution of c 2004 MAIK “Nauka/Interperiodica” 0012-2661/04/4012-1781

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VOLKODAVOV et al.

this class in the domains G1 and G2 has the form x−y Z Zx β  α  u(x, y, z) = τ1 (x, y) − (x − s)2 − (ξ − s)2 dξ ds N1 (s, ξ) (ξ − s)2 − y 2 z

y+s

(13)

x−y Z Zh β  α  − (ξ − s)2 − (x − s)2 dξ, ds T1 (s, ξ) (ξ − s)2 − y 2 z

x

x+y Z Zx β  α  u(x, y, z) = τ2 (x, y) − (x − s)2 − (ξ − s)2 dξ ds N2 (s, ξ) (ξ − s)2 − y 2 z

s−y

(14)

x+y Z Zh β  α  − (ξ − s)2 − (x − s)2 dξ, ds T2 (s, ξ) (ξ − s)2 − y 2 z

x

respectively. The functions Ni and Ti are continuous in the respective open integration domains and absolutely integrable in the closed domains. The functions given by (13) and (14) satisfy conditions (2) and (4), respectively. From conditions (3) and (5), we find the functions N1 and N2 . By setting x = h in (13) and (14) and by taking account of condition (8), for N1 and N2 , we obtain the integral equations h−y Z Zh β  α  φ1 (y, z) = − (h − s)2 − (ξ − s)2 dξ, ds N1 (s, ξ) (ξ − s)2 − y 2 z

y+s

h+y Z

φ2 (y, z) = −

Zh

ds z

(15)

β  α  (h − s)2 − (ξ − s)2 dξ. N2 (s, ξ) (ξ − s)2 − y 2

(16)

s−y

Under the assumption that Eq. (15) has a solution, we differentiate both sides with respect to z and then with respect to y. Next, we replace y by t and apply the operator h−z Z
−β 2 . . . t2 − y 2 dt y

to both sides of the resulting identity. On the right-hand side, we change the order of integration, and on the left-hand side, we perform integration by parts taking ∂ 2 φ1 (t, z) 1 ∂z ∂t 2t

u=

with regard to conditions (10). After this, we differentiate both sides with respect to y. Then we obtain the function N1 . The function N2 can be found in a similar way. With regard to the notation introduced above, we rewrite these functions in the form −α

y [(h − z)2 − y 2 ] Ni (z, y + z) = B(1 + β, 1 − β)

h−z Z
−β Φi (t, z) t2 − y 2 dt,

i = 1, 2.

(17)

y

The existence of solutions of Eqs. (15) and (16) can be proved by a straightforward computation with regard to conditions (10) and (11). DIFFERENTIAL EQUATIONS

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A 3D ANALOG OF PROBLEM M FOR A THIRD-ORDER HYPERBOLIC EQUATION

1783

The continuity of the solution on the plane y = 0, the matching condition (6), and condition (9) imply the integral equations Zh

Zh (x − s)ds

z

α+i−2  %i (s, ξ) (ξ − s)2 − (x − s)2 dξ

x

Zx

Zx (x − s)ds

= z

(18i )

α+i−2 µi (s, ξ) (x − s)2 − (ξ − s)2 dξ,

i = 1, 2,

µi = (ξ − s)2β [N1 (s, ξ) ± N2 (s, ξ)] ,

i = 1, 2.



s

where %i = (ξ − s)2β [T1 (s, ξ) ± T2 (s, ξ)] ,

(19)

(The plus sign corresponds to i = 1, and the minus sign corresponds to i = 2.) To ensure the solvability of Eq. (182 ) for %2 , we subject its right-hand side to a condition that results in an additional condition on the functions φi : h−s Z [Φ1 (t, s) − Φ2 (t, s)] t2 dt = 0.

(20)

0

The unique solvability of Eqs. (18i ) for %i can be proved in the same way as for Eq. (15). Then we obtain the solutions %i (z, x + z) = cos(πα)µi (z, x + z)  α Zh sin(πα) (h − z)2 − (ξ − z)2 2x − µi (z, ξ) dξ, π (ξ − z)2 − x2 (h − z)2 − x2

i = 1, 2.

z

From (19), (21), and (17), we find T1 and T2 : Ti (s, ξ) =

 −α sin(πα) (ξ − s)−1 (h − s)2 − (ξ − s)2 π ξ−s  Z   2 2 2 −β × Φk (y, s)y (ξ − s) − y F 1, 1 − β, 2; 0

y2 (ξ − s)2

 dy

h−s −α Z (ξ − s) sin(πα) [(h − s)2 − (ξ − s)2 ] − Φk (y, s) πB(1 + β, 1 − β) ξ−s (     (ξ − s)2 1 (ξ − s)2 × D β − 1, β, + y 2−2β F β, β, β + 1; y2 1+β y2

y 2 − (ξ − s)2 β (ξ − s)2 [y 2 − (ξ − s)2 ] −β )  2  2 2 − (ξ − s) y − (ξ − s)2 y dy − (ξ − s)−2β D −β, 1 − β; y2 y2 +

1

ln

−α

(ξ − s) cos(πα) [(h − s)2 − (ξ − s)2 ] + B(1 + β, 1 − β) where i = 1, 2 and k = 2, 1. DIFFERENTIAL EQUATIONS

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h−s Z
−β Φk (y, s) y 2 − (ξ − s)2 dy, ξ−s

(21)

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The function D(a, b, x) =

∞ X

((a)n /n!) Ψ(b + n)xn

n=0

was introduced and analyzed in [3, pp. 28–29]. The fact that Ti and Ni belong to the abovementioned class can be proved by a straightforward computation. By substituting the Ti and Ni thus found into (13) and (14), we obtain a solution of Problem MC. The uniqueness of the solution of Problem MC follows from the uniqueness of the solution of the Cauchy problem, which was used for deriving the estimates (13) and (14), and from the uniqueness of the solution of all integral equations obtained in the course of the solution of the problem. Theorem. If conditions (7)–(12) and (20) are satisfied, then Problem MC for Eq. (1) has a unique solution. REFERENCES 1. Volkodavov, V.F. and Rodionova, I.N., Integral’nye uravneniya i ikh prilozheniya (Integral Equations and Their Applications), Kiev, 1995, issue 10. 2. Volkodavov, V.F., Rodionova, I.N., and Saltuganov, N.M., Integral’nye uravneniya Vol’terra pervogo roda s parametrom i ikh prilozheniya (Volterra Integral Equations of the First Kind with Parameter and Their Applications), Ioshkar-Ola, 1997. 3. Volkodavov, V.F., Lerner, Sh.E., Nikolaev, N.Ya., and Nosov, V.A., Tablitsy nekotorykh funktsii Rimana, integralov i ryadov (Tables of Some Riemann Functions, Integrals, and Series), Kuibyshev, 1982.

DIFFERENTIAL EQUATIONS

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2004