GEORGIAN MATHEMATICAL JOURNAL: Vol. 4, No. 4, 1997, 341-354 ON THE SOLVABILITY OF THE MULTIDIMENSIONAL VERSION OF THE FIRST DARBOUX PROBLEM FOR A MODEL SECOND-ORDER DEGENERATING HYPERBOLIC EQUATION S. KHARIBEGASHVILI Abstract. A multidimensional version of the first Darboux problem is considered for a model second order degenerating hyperbolic equation. Using the technique of functional spaces with a negative norm, the correct formulation of this problem in the Sobolev weighted space is proved. In a space of variables x1 , x2 , t let us consider a second-order degenerating hyperbolic equation of the type Lu ≡ utt − |x2 |m ux1 x1 − ux2 x2 + a1 ux1 + a2 ux2 + a3 ut + a4 u = F, (1) where ai , i = 1, . . . , 4, F are given real functions and u is an unknown real function, m = const > 0. Denote by D : x2 < t < 1 − x2 , 0 < x2 < 12 , an unbounded domain lying in a space x2 > 0 and bounded by characteristic surfaces S1 : t − x2 = 0, 0 < x2 < 12 , S2 : t + x2 − 1 = 0, 0 < x2 < 12 , of equation (1) and by a plane surface S0 : x2 = 0, 0 < t < 1, of time type with an equation degenerating on it. The coefficients ai , i = 1, . . . , 4, of equation (1) in the domain D are assumed to be bounded functions of the class C 1 (D). For equation (1) let us consider a multidimensional version of the first Darboux problem formulated as follows: find in the domain D a solution u(x1 , x2 , t) of equation (1) satisfying the boundary condition (2) uS ∪S = 0. 0 1 1991 Mathematics Subject Classification. 35L80. Key words and phrases. Degenerating hyperbolic equation, multidimensional version of the first Darboux problem, Sobolev’s weighted space, functional space with negative norm. 341 c 1997 Plenum Publishing Corporation 1072-947X/97/0700-0341$12.50/0 342 S. KHARIBEGASHVILI The problem in the domain D for the equation L∗ v ≡ vtt − xm 2 vx1 x1 − vx2 x2 − (a1 v)x1 − (a2 v)x2 − (a3 v)t + a4 v = F (3) is posed analogously by the boundary condition v S0 ∪S2 = 0, (4) ∗ where L is a formally conjugate operator of L. Note that similar problems for m = 0, when equation (1) is not degenerating and contains, in its principal part, a wave operator, were studied in [1–6]. Other versions of multidimensional Darboux problems can be found in [7–9]. Denote by E, E ∗ the classes of functions from the Sobolev space W22 (D) satisfying respectively the boundary condition (2) and the boundary condition (4). Let W+ (W+∗ ) be the weighted Hilbert space obtained by closing the space E(E ∗ ) in the norm Z 2 2 2 2 kuk1,+ = (u2t + xm 2 ux1 + ux2 + u )dD. D W− (W−∗ ) the space with a negative norm constructed with reDenote by spect to L2 (D) and W+ (W+∗ ) [10]. Consider the condition −m M = sup x2 2 a1 (x1 , x2 , t) < ∞ (5) D imposed an the lowest coefficient a1 in equation (1). The uniqueness theorem required for solving problem (1), (2) of the class W22 (D) follows from Lemma 1. Let condition (5) be fulfilled. Then for every u ∈ W22 (D) satisfying the homogeneous boundary condition uS0 = 0 (6) the a priori estimate kuk1,+ ≤ c kf k1,∗ + kF kL2 (D) (7) is valid, where the positive constant c does not depend on u; f = u|S1 , F = Lu, Z h ∂f 2 i 2 xm ds, kf k21,∗ = 2 fx1 + ∂N S1 1 ∂ ∂ ∂ + =− is the derivative with respect to the conormal and ∂N 2 ∂x2 ∂t which is the inner differential operator on the characteristic surface S1 . MULTIDIMENSIONAL VERSION OF THE FIRST DARBOUX PROBLEM 343 Proof. Let n = (ν1 , ν2 , ν0 ) be the outer unit vector to ∂D, i.e., ν1 = ct). For the function u ∈ W22 (D) cos([ n, x1 ), ν2 = cos([ n, x2 ), ν0 = cos(n, satisfying the boundary condition (6) and λ = const > 0 a simple integration by parts gives 2 Z −λt e utt ut dD = D −2 Z D = −2 + Z e−λt u2t ν0 De −λt λe−λt u2t dD, (8) D ∂D e−λt (xm 2 ux1 x1 ut + ux2 x2 ut )dD = Z e−λt (xm 2 ux1 ut ν1 + ux2 ut ν2 )ds + ∂D Z ds + Z 2 (xm 2 u x1 + u2x2 )ν0 ds + Z 2 2 e−λt (λxm 2 ux1 + λux2 )dD. (9) D It can be easily seen that 1 1 uS = ut S0 = ν0 S0 = 0, nS1 = 0, √ , − √ , 0 2 2 2 2 ν0 S > 0, (ν02 − xm − ν ) ν = 0. 2 S ∪S 2 1 1 2 (10) 2 Taking into account (8)–(10), multiplying both parts of equation (1) by 2e−λt ut , where F = Lu, and integrating the obtained expression with respect to D, we obtain 2(Lu, e −λt ut )L2 (D) = Z D 2 2 e−λt λ(ut2 + xm 2 ux1 + ux2 ) + +2(a1 ux1 + a2 ux2 + a3 ut + a4 u)ut dD + Z S1 ∪S2 e−λt ν0−1 xm 2 (ν0 ux1 − 2 2 2 − xm −ν1 ut ) + (ν0 ux2 − ν2 ut ) + 2 ν1 − ν2 )ut ds ≥ Z 2 2 ≥ e−λt λ(u2t + xm 2 ux1 + ux2 ) + 2(a1 ux1 + a2 ux2 + a3 ut + 2 2 D √ +a4 u)ut dD − 2 Z S1 (ν02 h ∂u 2 i 2 ds. e−λt xm 2 ux1 + ∂N (11) 344 S. KHARIBEGASHVILI Owing to condition (6) and the structure of D, one can easily verify that the inequality Z Z 2 u dD ≤ c0 ux2 2 dD (12) D D is valid for some c0 = const > 0 not depending on u ∈ W22 (D). Using inequality (5), we can show that m |2a1 ux1 ut | ≤ 2M (x22 ux1 )ut ≤ M (x2m u2x1 + u2t ). (13) By virtue of (12) and (13), from (11) for sufficiently large λ we get 2(Lu, e−λt ut )L2 (D) ≥ c1 Z 2 2 2 (u2t + xm 2 ux1 + ux2 + u )dD − D − c2 Z h 2 xm 2 ux1 + S1 ∂u 2 i ds, ∂N (14) where the positive constants c1 and c2 do not depend on u; note that depending on λ, the constant c1 can be chosen arbitrarily large. Therefore estimate (7) follows obviously from (14). Remark 1. Since for the principal part of the operator L the derivative ∂ ∂ ∂ ∂ = ν0 − xm with respect to the conormal − ν2 is the inner 2 ν1 ∂N ∂t ∂x1 ∂x2 differential operator on the characteristic surfaces of equation (1), by (2) and (4) we have ∂u = 0, ∂N S1 for the functions u ∈ E and v ∈ E ∗ . ∂v =0 ∂N S2 Lemma 2. Let condition (5) be fulfilled. u ∈ E, v ∈ E ∗ the inequalities kLukW−∗ ≤ c1 kukW+ , ∗ kL vkW− ≤ c2 kvk ∗ W+ (15) Then for all functions (16) (17) are fulfilled, where the positive constants c1 and c2 do not depend respectively on u and v, k · kW+ = k · kW+∗ = k · k1,+ . MULTIDIMENSIONAL VERSION OF THE FIRST DARBOUX PROBLEM 345 Proof. According to the definition of a negative norm and because of (2), (4), and (15), we have −1 −1 kLukW−∗ = sup kvkW ∗ (Lu, v)L2 (D) = sup kvkW ∗ (Lu, v)L2 (D) = = sup kvk−1 W∗ + v∈E ∗ Z ∗ v∈W+ + + v∈E ∗ −1 [ut vν0 − xm 2 ux1 vν1 − ux2 vν2 ]ds + sup kvkW ∗ v∈E ∗ ∂D + Z [−ut vt + D +x2m ux1 vx1 + ux2 vx2 + a1 ux1 v + a2 ux2 v + a3 ut v + a4 uv]dD = Z Z ∂ −1 −1 = sup kvkW ∗ v ds + sup kvkW ∗ [−ut vt + xm 2 ux1 vx1 + + + ∂N v∈E ∗ v∈E ∗ S1 ∪S2 D Z −1 [−ut vt + +ux2 vx2 + a1 ux1 v + a2 ux2 v + a3 ut v + a4 uv]dD = sup kvkW ∗ v∈E ∗ + D +xm 2 ux1 vx1 + ux2 vx2 + a1 ux1 v + a2 ux2 v + a3 ut v + a4 uv]dD. (18) Due to condition (5) and the Schwartz inequality we obtain Z hZ i 21 2 2 × (u2t + xm )dD u + u [−ut vt + x2m ux1 vx1 + ux2 vx2 ]dD ≤ 3 2 x1 x2 D × hZ 2 2 (vt2 + xm 2 vx1 + vx2 )dD D i 12 D ≤ 3kukW+ kvkW+∗ , (19) Z Z 1 x2m u2x1 dD) 2 kvkL2 (D) + (a1 ux1 v + a2 ux2 v + a3 ut v + a4 uv]dD ≤ M D D + sup |a2 | kux2 kL2 (D) kvkL2 (D) + sup |a3 | kut kL2 (D) kvkL2 (D) + D D + sup |a4 | kukL2 (D) kvkL2 (D) ≤ M + D 4 X sup |ai | kukW+ kvkW+∗ . (20) i=1 D From (18)–(20) it follows that 4 X ∗ kLukW−∗ ≤ 3 + M + sup |ai | sup kvk−1 W ∗ kukW+ kvkW+ = i=2 D v∈E ∗ + 4 X = 3+M + sup |ai | kukW+ , i=2 D which proves inequality (16). Thus Lemma 2 is completely proved, since the proof of inequality (17) repeats that of the inequality (16). 346 S. KHARIBEGASHVILI Remark 2. By inequalities (16) and (17), the operator L : W+ → W−∗ (L : W+∗ → W− ) with a dense domain of definition E(E ∗ ) admits a closure which is a continuous operator from W+ (W+∗ ) to W−∗ (W− ). Denoting this closure as previously by L(L∗ ), we note that it is defined on the whole Hilbert space W+ (W+∗ ). Lemma 3. Problems (1), (2) and (3), (4) are self-conjugate, i.e., the equality (Lu, v) = (u, L∗ v) (21) holds for every u ∈ W+ and v ∈ W+∗ . Proof. By Remark 2 it suffices to prove equality (21) when u ∈ E and v ∈ E ∗ . We have Z (Lu, v) = (Lu, v)L2 (D) = [ut vν0 − xm 2 ux1 vν1 − ux2 vν2 ]ds + + Z ∂D [a1 ν1 + a2 ν2 + a3 ν0 ]uv ds + Z D ∂D − ut vt + xm 2 ux1 vx1 + ux2 vx2 − −u(a1 v)x1 − u(a2 v)x2 − u(a3 v)t + a4 uv dD = −ux2 vν2 ]ds + Z [a1 ν1 + a2 ν2 + a3 ν0 ]uv ds − ∂D −uvx2 ν2 ]ds + Z D Z [ut vν0 − xm 2 ux1 vν1 − ∂D Z [uvt ν0 − xm 2 uvx1 ν1 − ∂D uvtt − xm 2 uvx1 x1 − uvx2 x2 − u(a1 v)x1 − −u(a2 v)x2 − u(a3 v)t + a4 uv dD = Z h ∂v ∂u −u + v ∂N ∂N ∂D i +(a1 ν1 + a2 ν2 + a3 ν3 ]uv ds + (u, L∗ v)L2 (D) . (22) By (2), (4), and (15), equality (21) follows directly from (22), which proves Lemma 3. Lemma 4. Let condition (5) be fulfilled. Then for every u ∈ W+ we have the inequality ckukL2 (D) ≤ kLukW−∗ with the positive constant c not depending on u. (23) MULTIDIMENSIONAL VERSION OF THE FIRST DARBOUX PROBLEM 347 Proof. By Remark 2 it suffices to prove the inequality (23) when u ∈ E. If u ∈ E, then it can be easily verified that the function v(x1 , x2 , t) = ϕ2 (x Z 1 ,x2 ) e−λτ u(x1 , x2 , τ )dτ, λ = const > 0, t where t = ϕ2 (x1 , x2 ) is the equation of the characteristic surface S2 , belongs to the space E ∗ , and the equalities vt (x1 , x2 , t) = −e−λt u(x1 , x2 , t), u(x1 , x2 , t) = −eλt vt (x1 , x2 , t) (24) are valid. By (10), (15) and (24) we have Z h i ∂u + (a1 ν1 + a2 ν2 + a3 ν0 )uv ds + v (Lu, v)L2 (D) = ∂N D Z + [−ut vt + xm 2 ux1 vx1 + ux2 vx2 − ua1x1 v − ua1 vx1 − ua2x2 v − D −ua2 vx2 − ua3t v − ua3 vt + a4 uv]dD = + Z Z e−λt ut u dD + D eλt [−xm 2 vx1 t vx1 − vx2 t vx2 + a1x1 vt v + a1 vt vx1 + a2x2 vt v + D +a2 vt vx2 + a3t vt v + a3 vt2 − a4 vt v]dD. (25) Analogously to (8) and (9), because of (2) we have Z Z Z 1 1 e−λt u2 ν0 ds + e−λt λu2 dD = e−λt ut u dD = 2 2 D D ∂D Z Z 1 1 = e−λt u2 ν0 ds + eλt λvt2 dD = 2 2 S2 D Z Z 1 1 eλt λvt2 dD, = eλt vt2 ν0 ds + (26) 2 2 S2 D Z Z 1 v v eλt [−xm eλt [x2m vx21 + vx22 ]ν0 ds + − v v ]dD = − x t x x t x 2 1 1 2 2 2 D ∂D Z 1 m 2 λt (27) + e λ[x2 vx1 + vx22 ]dD. 2 D 348 S. KHARIBEGASHVILI Since v S = 0, for some α on S2 we have 2 vx1 = αν1 , vt = αν0 , vx2 = αν2 . Therefore, since the surface S2 is characteristic, we have (vt2 − x2m vx21 − vx22 )S = α2 (ν02 − xm ν12 − ν22 )S = 0. 2 2 2 (28) Due to the fact that ν0 S0 = 0, ν0 S1 < 0 and owing to equalities (4) and (28), we find that Z Z 1 1 2 2 eλt [xm eλt vt2 ν0 ds − 2 vx1 + vx2 ]ν0 ds = 2 2 S2 ∂D Z Z 1 1 2 2 eλt vt2 ν0 ds − eλt [xm = 2 vx1 + vx2 ]ν0 ds − 2 2 S2 S1 Z Z Z 1 1 1 λt m 2 λt 2 2 2 − e [x2 vx1 + vx2 ]ν0 ds ≥ e vt ν0 ds − eλt [xm 2 vx1 + 2 2 2 S2 S2 S2 Z 1 2 2 (29) +vx22 ]ν0 ds = eλt [vt2 − xm 2 vx1 − vx2 ]ν0 ds = 0. 2 S2 Taking into consideration (26), (27), and (29), from (25) we get Z Z 1 1 λt 2 (Lu, v)L2 (D) = eλt λvt2 dD − e vt ν0 ds + 2 2 S2 D Z Z 1 1 2 2 eλt [xm eλt λ[x2m vx21 + vx22 ]dD + − 2 vx1 + vx2 ]ν0 ds + 2 2 D ∂D Z + eλt a1 vt vx1 + a2 vt vx2 + a3 vt2 + (a1x1 + a2x2 + a3t − a4 )vt v dD ≥ D ≥ λ 2 Z D Z 2 2 v eλt [vt2 + xm + v − ]dD eλt [a1 vt vx1 + x2 2 x1 D +a2 vt vx2 + Putting a3 vt2 + (a1x1 + a2x2 + a3t − a4 )vt v]dD. µ = max sup |a2 |, sup |a3 |, sup |a1x1 + a2x2 + a3t − a4 | D D D (30) MULTIDIMENSIONAL VERSION OF THE FIRST DARBOUX PROBLEM 349 and taking into account (5), we find that Z eλt [a1 vt vx1 + a2 vt vx2 + a3 vt2 + (a1x1 + a2x2 + a3t − a4 )vt v]dD ≤ D ≤ Z D h 1 i µ 2 µ 2 2 eλt M (xm (vx2 + vt2 ) + µvt2 + (v 2 + vt2 ) dD = 2 vx 1 + v t ) + 2 2 2 Z h 1 1 µ 2 µ i 2 M + 2µ vt2 + M xm vx2 + v 2 dD ≤ = eλt 2 vx1 + 2 2 2 2 D Z 1 ≤ (31) M + 2µ eλt [vt2 + x2m vx21 + vx22 + v 2 ]dD. 2 D λ Since the function e 2 t v S = 0, by virtue of inequality (12) we have 0 Z Z Z λt 2 λt 2 2 2 e v dD ≤ c0 e vx2 dD ≤ c0 eλt [vt2 + xm 2 vx1 + vx2 ]dD D D D and, consequently, Z 2 2 eλt[vt2 + xm 2 vx1 + vx2]dD ≥ D 1 1 + c0 Z 2 2 2 eλt[vt2 + xm 2 vx1 + vx2 + v ]dD. (32) D By (31), (32), and (24) from (30) we obtain Z λ 2 2 2 (Lu, v)L2 (D) ≥ eλt [vt2 + xm 2 vx1 + vx2 + v ]dD − 2(1 + c0 ) D Z 1 2 2 2 λt 2 e [vt + xm − M + 2µ 2 vx1 + vx2 + v ]dD = 2 D Z λ 1 2 2 2 = − M − 2µ eλt [vt2 + xm 2 vx1 + vx2 + v ]dD ≥ 2(1 + c0 ) 2 D i 12 i 12 h Z hZ eλt vt2 dD ≥σ [vt2 + x2m vx21 + vx22 + v 2 ]dD = D =σ hZ D D kvkW+∗ ≥ σ · inf e−λt kukL2 (D) kvkW+∗ , D (33) 1 λ − M − 2µ > 0 for sufficiently large λ, while 2(1 + c0 ) 2 = const > 0 owing to the structure of the domain D. where σ = inf D e−λt e−λt u2 dD i 21 350 S. KHARIBEGASHVILI If now we apply the generalized Schwartz inequality (Lu, v)L2 (D) ≤ kLukW−∗ kvkW−∗ to the left-hand side of (33), then after reduction by kvkW+∗ we obtain inequality (23), where c = σ inf D e−λt = const > 0. Lemma 5. Let condition (5) be fulfilled. Then for every v ∈ W+∗ the inequality ckvkL2 (D) ≤ kL∗ vkW− (34) is valid for some c = const > 0 not depending on v ∈ W+∗ . Proof. As in Lemma 4, by Remark 2 it suffices to prove the validity of inequality (34) for v ∈ E ∗ . Let v ∈ E ∗ and introduce into consideration the function u(x1 , x2 , t) = Zt eλτ v(x1 , x2 , τ )dτ, λ = const > 0, ϕ1 (x1 ,x2 ) where t = ϕ1 (x1 , x2 ) is the equation of the characteristic surface S1 . It is easily seen that the function u(x1 , x2 , t) belongs to the class E, and we have the equalities ut (x1 , x2 , t) = eλt v(x1 , x2 , t), v(x1 , x2 , t) = e−λt ut (x1 , x2 , t). (35) Because of (10), (15), and (35) we have Z h i ∂v ∗ (L v, u)L2 (D) = u − (a1 ν1 + a2 ν2 + a3 ν0 )uv ds + ∂N ∂D Z + [−vt ut + xm 2 vx1 ux1 + vx2 ux2 + a1 vux1 + a2 vux2 + a3 vut + a4 uv]dD = D =− Z D + eλt vt v dD + Z Z e−λt [xm 2 ux1 t ux1 + ux2 t ux2 ]dD + D e−λt [a1 ux1 + a2 ux2 + a3 ut + a4 u]ut dD. D Similarly to (26)–(29) we can prove the equalities Z Z Z 1 1 λt 2 λt e v ν0 ds + eλt λv 2 dD = − e vt v dD = − 2 2 D ∂D D (36) MULTIDIMENSIONAL VERSION OF THE FIRST DARBOUX PROBLEM 351 Z 1 e−λt λu2t dD, 2 S1 D Z Z 1 2 2 e−λt [xm e−λt [xm 2 ux1 t ux1 + ux2 t ux2 ]dD = 2 ux1 + ux2 ]ν0 ds + 2 D ∂D Z 1 −λt m 2 e λ[x2 ux1 + u2x2 ]dD, + 2 D 2 2 2 (ut − xm 2 ux1 − ux2 ) S1 = 0, Z Z 1 1 − e−λt ut2 ν0 ds + e−λt [x2m u2x1 + u2x2 ]ν0 ds = 2 2 S1 ∂D Z Z 1 1 2 2 −λt 2 =− e ut ν0 ds + e−λt [xm 2 ux1 + ux2 ]ν0 ds + 2 2 S1 S1 Z 1 −λt m 2 + e [x2 ux1 +u2x2 ]ν0 ds ≥ 2 S Z 2 1 2 2 ≥− e−λt [u2t − xm 2 ux1 − ux2 ]ν0 ds = 0. 2 =− 1 2 Z e−λt u2t ν0 ds + (37) (38) (39) (40) S1 To obtain inequality (40) we have used the fact that ν0 |S2 > 0. Owing to (37)–(40), from (36) we get Z 1 2 2 (L∗ v, u)L2 (D) ≥ e−λt λ[ut2 + xm 2 ux1 + ux2 ]dD + 2 D Z Z λ −λt + e [a1 ux1 + a2 ux2 + a3 ut + a4 u]ut dD ≥ e−λt [ut2 + x2m u2x1 + 2 D D Z +u2x2 ]dD − e−λt [a1 ux1 + a2 ux2 + a3 ut + a4 u]ut dD. D whence, as in obtaining inequality (33) from (30), we have h 1 i λ (L∗ v, u)L2 (D) ≥ − M + max sup |ai | × i=2,3,4 D 2(1 + c0 ) 2 × inf e−λt kvkL2 (D) kukW+ . D Inequality (34) follows directly from the above inequality for sufficiently large λ. Definition 1. If F ∈ L2 (D), then the function u will be called a strong generalized solution of problem (1), (2) of the class W+ if u ∈ W+ , and 352 S. KHARIBEGASHVILI there exists a sequence of functions un ∈ E such that un → u and Lun → F respectively in the spaces W+ and W−∗ as n → ∞, i.e., lim kun − ukW+ = 0. n→∞ lim kLun − F kW−∗ = 0. n→∞ Definition 2. If F ∈ W−∗ , then the function u will be called a strong generalized solution of problem (1), (2) of the class L2 if u ∈ L2 (D), and there exists a sequence of functions un ∈ E such that un → u and Lun → F respectively in the spaces L2 (D) and W−∗ as n → ∞, i.e., lim kun − ukL2 (D) = 0. n→∞ lim kLun − F kW−∗ = 0. n→∞ According to the results of [11], the theorems below are consequences of Lemmas 2–5. Theorem 1. Let condition (5) be fulfilled. Then for every F ∈ W−∗ there exists a unique strong generalized solution u of problem (1), (2) of the class L2 for which the estimate kukL2 (D) ≤ ckF kW−∗ , (41) with a positive constant c not depending on F, is valid. Theorem 2. Let condition (5) be fulfilled. Then for every F ∈ L2 (D) there exists a unique strong generalized solution u of problem (1), (2) of the class W+ for which estimate (41) is valid. Proof. The existence of a solution of problem (1), (2) in Theorem 2 follows, for example, from the arguments as follows. By virtue of inequality (34), the functional (F, v)L2 (D) can be regarded as a linear continuous functional of L∗ v, where v ∈ E ∗ , F ∈ L2 (D). Indeed, using this inequality, we have |(F, v)L2 (D) | ≤ kF kL2 (D) kvkL2 (D) ≤ c∗ kL∗ vkW− , c∗ = const > 0. By the Khan–Banach theorem, this functional can be linearly and continuously extended into the whole space W− . Following the theorem on a general type of a linear continuous functional over W− , there exists a function u ∈ W+ such that (u, L∗ )L2 (D) = (F, v)L2 (D) , v ∈ E ∗ . (42) Equality (42) means that u is a weak generalized solution of the problem (1), (2). Let us now show that this solution is also a strong generalized solution of problem (1), (2) of the class W+ . Since the space E is dense in W+ , there exists a sequence un ∈ E of functions such that lim kun − ukW+ = 0. n→∞ (43) MULTIDIMENSIONAL VERSION OF THE FIRST DARBOUX PROBLEM 353 Using equalities (21) and (42), we have (un − u, L∗ v)L2 (D) = (Lun − F, v)L2 (D) . Now, according to the generalized Schwartz inequality, (un − u, L∗ v)L (D) ≤ kun − ukW+ kL∗ vkW− . 2 (44) (45) It follows from (43)–(45) that in the space W−∗ the sequence Lun of functions converges weakly to the function F . But since this sequence, because of (16) and (43), converges in the norm of the space W−∗ , we obtain lim kLun − F kW−∗ = 0. n→∞ Consequently, the function u is a strong generalized solution of problem (1), (2) of the class W+ . This fact can be proved in a different way. Indeed, using equalities (21) and (42) and inequality (17), we have kLun − F kW−∗ = sup kvk−1 W ∗ (Lun − F, v)L2 (D) = ∗ v∈W+ + −1 = sup kvkW ∗ (Lun , v)L2 (D) − (F, v)L2 (D) = + v∈E ∗ −1 = sup kvkW ∗ (un , L∗ v)L2 (D) − (u, L∗ v)L2 (D) = v∈E ∗ + ∗ = sup kvk−1 W ∗ (un − u, L v)L2 (D) ≤ v∈E ∗ ≤ sup v∈E ∗ + −1 kvkW ∗ kun + − ukW+ kL∗ vkW− ≤ ∗ ≤ sup kvk−1 W ∗ kun − ukW+ c2 kvkW+ = c2 kun − ukW+ , v∈E ∗ + whence limn→∞ kLun − F kW−∗ = 0. The uniqueness of a strong generalized solution of problem (1), (2) of the class W+ in Theorem 2 as well as estimate (41) follow from inequality (23). As for Theorem 1, it can be proved as follows. Since the space L2 (D) is dense in the space W−∗ , for every element F ∈ W−∗ there exists a sequence Fn ∈ L2 (D) of functions such that lim kFn − F kW−∗ = 0. According to n→∞ Theorem 2, for every function Fn ∈ L2 (D) there exists a unique strong generalized solution un of problem (1), (2) of the class W+ . Furthermore, using inequality (23) and passing to the limit, we obtain the existence and the uniqueness of a strong generalized solution of problem (1), (2) of the class L2 as well as estimate (41). 354 S. KHARIBEGASHVILI References 1. J. Hadamard, Lectures on Cauchy’s problem in partial differential equations. Yale University Press, New Haven, 1923. 2. J. Tolen, Probléme de Cauchy sur la deux hypersurfaces caracteristiques sécantes. C. R. Acad. Sci. Paris, Sér. 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(Russian) Izv. Akad. Nauk Kazakh. SSR. Ser. Fiz.–Mat., (1982), No. 3, 18–25. 10. Yu. M. Berezansky, Decomposition in eigenfunctions of self-conjugate operators. (Russian) Naukova Dumka, Kiev, 1965. 11. I. I. Ljashko, V. P. Didenko, and O. E. Tsitritskii, Filtration of noise. (Russian) Naukova Dumka, Kiev, 1979. (Received 23.11.1994) Author’s address: Department of Theoretical Mechanics (4) Georgian Technical University 77, M. Kostava Str., Tbilisi 380075 Georgia