GEORGIAN MATHEMATICAL JOURNAL: Vol. 2, No. 3, 1995, 299-311

ON A SPATIAL PROBLEM OF DARBOUX TYPE FOR A SECOND-ORDER HYPERBOLIC EQUATION

S. KHARIBEGASHVILI

Abstract. The theorem of unique solvability of a spatial problem of Darboux type in Sobolev space is proved for a second-order hyperbolic equation.

In the space of variables x_{1}, x_{2}, t let us consider the second order hyperbolic equation

Lu ≡ □u + au_{x_{1}} + bu_{x_{2}} + cu_{t} + du = F,  (1)

where □ ≡ \frac{∂^{2}}{∂t^{2}} − \frac{∂^{2}}{∂x_{1}^{2}} − \frac{∂^{2}}{∂x_{2}^{2}} is a wave operator; the coefficients a, b, c, d and the right-hand side F of equation (1) are given real functions, and u is an unknown real function.

Denote by D : kt < x_{2} < t, 0 < t < t_{0}, −1 < k = const < 1, the domain lying in a half-space t > 0, which is bounded by a time-type plane surface S_{1} : kt − x_{2} = 0, 0 ≤ t ≤ t_{0}, a characteristic surface S_{2} : t − x_{2} = 0, 0 ≤ t ≤ t_{0} of equation (1), and a plane t = t_{0}.

Let us consider the Darboux type problem formulated as follows: find in the domain D the solution u(x_{1}, x_{2}, t) of equation (1) under the boundary conditions

u|_{S_{i}} = f_{i}, i = 1, 2,  (2)

where f_{i}, i = 1, 2, are given real functions on S_{i}; moreover (f_{1} −f_{2})|_{S_{1}∩S_{2}} = 0.

Note that in the class of analytic functions the problem (1),(2) is considered in [1]. In the case where S_{1} is a characteristic surface t + x_{2} = 0, 0 ≤ t ≤ t_{0}, the problem (1),(2) is studied in [1–3]. Some multidimensional analogues of the Darboux problems are treated in [4–6]. In the present paper the problem (1),(2) is investigated in the Sobolev space W_{2}^{1}(D).

299

1991 Mathematics Subject Classification. 35L20.

Key words and phrases. Characteristic, spatial problem of Darboux type, hyperbolic equation, a priori estimate.

1072-947X/95/0500-0299$07.50/0 \begin{matrix}c \\ ◯ \end{matrix} 1995 Plenum Publishing Corporation

Below we shall obtain first the solution of problem (1),(2) when equation (1) is a wave equation

□u ≡ \frac{∂^{2}u}{∂t^{2}} − \frac{∂^{2}u}{∂x_{1}^{2}} − \frac{∂^{2}u}{∂x_{2}^{2}} = F  (3)

and then using the estimates for that solution we shall prove the solvability of the problem (1),(2) in the Sobolev space W_{2}^{1}(D).

Using the method suggested in [7], we can get an integral representation of the regular solution of the problem (3),(2). Moreover, without loss of generality we can assume that for the domain D the value k = 0, i.e., D : 0 < x_{2} < t, 0 < t < t_{0}, since the case k ≠ 0 is reduced to the case k = 0 by a suitable Lorentz transform for which the wave operator □ is invariant. To this end we denote by D_{εδ} a part of the domain D : 0 < x_{2} < t, 0 < t < t_{0}, bounded by the surfaces S_{1} and S_{2}, the circular cone K_{ε} : r^{2} = (t − t^{0})(1 − ε) with vertex at the point (x^{0}, t^{0}) ∈ D, and the circular cylinder H_{δ} : r^{2} = δ^{2}, where r^{2} = (x_{1} − x_{1}^{0})^{2} + (x_{2} − x_{2}^{0})^{2} while ε and δ are sufficiently small positive numbers.

For any two twice continuously differentiable functions u and v we have an obvious identity

u□v − v□u = ∑_{i=1}^{2} \frac{∂}{∂x_{i}} (v \frac{∂u}{∂x_{i}} − u\frac{∂v}{∂x_{i}} ) − \frac{∂}{∂t}(v \frac{∂u}{∂t} − u\frac{∂v}{∂t} ).  (4)

Integrating equality (4) with respect to D_{εδ}, where u ∈ C^{1}(D̄) ∩ C^{2}(D) is a regular solution of the equation (3) and

v = E(r, t, t^{0}) = \frac{1}{2π} log \frac{t − t^{0} − \sqrt{(t − t^{0})^{2} − r^{2}}}{r} ,

we have

∫_{∂D_{εδ}} [E(r, t, t^{0})\frac{∂u}{∂N} − \frac{∂E(r, t, t^{0})}{∂N}u]ds + ∫_{D_{εδ}} F · E(r, t, t^{0})dxdt = 0, (5)

where N is the unit conormal vector at the point (x, t) = (x_{1}, x_{2}, t) ∈ ∂D_{εδ} with direction cosines cos N̂x_{1} = cos n̂x_{1}, cos N̂x_{2} = cos n̂x_{2}, cos N̂t = − cos n̂t and n is a unit vector of an outer normal to ∂D_{εδ}.

Passing in the equality (5) to the limit for ε → 0, δ → 0, we get

∫_{x_{2}^{0}}^{t^{0}} u(x_{1}^{0}, x_{2}^{0}, t)dt = ∫_{S_{1}^{∗}∪S_{2}^{∗}} [\frac{∂E(r, t, t^{0})}{∂N}u − E(r, t, t^{0})\frac{∂u}{∂N} ]ds −

− ∫_{D^{∗}} F · E(r, t, t^{0})dxdt,

where D^{∗} is a domain of D_{εδ} for ε = δ = 0, and S_{i}^{∗} = S_{i} ∩ ∂D^{∗}, i = 1, 2. Differentiation gives

u(x_{1}^{0}, x_{2}^{0}, t^{0}) = \frac{d}{dt^{0}} [ ∫_{S_{1}^{∗}∪S_{2}^{∗}} [\frac{∂E(r, t, t^{0})}{∂N}u − E(r, t, t^{0})\frac{∂u}{∂N} ]ds −

− ∫_{D^{∗}} F · E(r, t, t^{0})dxdt].  (6)

Remark. Since on the characteristic surface S_{2}^{∗} the direction of the conormal N coincides with that of a bicharacteristic lying on S_{2}^{∗}, we can, along with u|_{S_{2}^{∗}} = f_{2}, calculate also \frac{∂u}{∂N} over S_{2}^{∗}. At the same time, since the surface S_{1}^{∗} is a part of the plane x_{2} = 0, the direction of the conormal N coincides with that of an outer normal to ∂D^{∗}, i.e., \frac{∂}{∂N} = −\frac{∂}{∂x_{2}} . Therefore, to obtain an integral representation of the regular solution of the problem (3),(2), we should eliminate the value \frac{∂u}{∂N} |_{S_{1}^{∗}} on the right-hand side of the representation (6).

For this let us introduce a point P′(x_{1}^{0}, −x_{2}^{0}, t^{0}) symmetric to the point P (x_{1}^{0}, x_{2}^{0}, t^{0}) with respect to the plane x_{2} = 0. Denote by D_{ε} a part of the domain D bounded by the cone K_{ε}^{0} : (x_{1} −x_{1}^{0})^{2} +(x_{2} +x_{2}^{0})^{2} = (t−t^{0})^{2}(1−ε) with vertex at P′ and a boundary ∂D. Obviously, ∂D_{ε} ∩S_{1} ⊂ S_{1}^{∗} and ∂D_{0} ∩ S_{1} = S_{1}^{∗}. Put ∂D_{0} ∩ S_{2} = S̃_{2}, r̃ = \sqrt{(x_{1} − x_{1}^{0})^{2} + (x_{2} + x_{2}^{0})^{2}}. Integrating now the equality (4) with respect to D_{ε}, where u ∈ C^{1}(\overline{D}) ∩ C^{2}(D) is a regular solution of equation (3) and

v = E(r̃, t, t^{0}) = \frac{1}{2π} log \frac{t − t^{0} − \sqrt{(t − t^{0})^{2} − r^{2}}}{r̃} ,

and taking into account the fact that the function E(r̃, t, t^{0}) in D_{0} is nonsingular, after passing to the limit for ε → 0 we get the equality

\frac{d}{dt^{0}} [ ∫_{S_{1}^{∗}∪S_{2}^{∗}} [\frac{∂E(r̃, t, t^{0})}{∂N}u − E(r̃, t, t^{0})\frac{∂u}{∂N} ]ds −

− ∫_{D_{0}} F · E(r̃, t, t^{0})dxdt] = 0.  (7)

Since r = r̃ for x_{2} = 0, we have E(r̃, t, t^{0}) = E(r, t, t^{0}) on S_{1}^{∗}. Therefore, eliminating the value \frac{∂u}{∂N} |_{S_{1}^{∗}} from equalities (6) and (7), we finally get the integral representation of the regular solution of the problem (3),(2):

u(x_{1}^{0}, x_{2}^{0}, t^{0}) = \frac{d}{dt^{0}} [ ∫_{S_{1}^{∗}} [\frac{∂E(r, t, t^{0})}{∂N} − \frac{∂E(r̃, t, t^{0})}{∂N}]uds +

+ ∫_{S_{2}^{∗}} [\frac{∂E(r, t, t^{0})}{∂N}u − E(r, t, t^{0})\frac{∂u}{∂N} ]ds − ∫ ; _{S_{2}} [\frac{∂E(r̃, t, t^{0})}{∂N}u −

−E(r̃, t, t^{0})\frac{∂u}{∂N} ]ds + ∫_{D_{0}} F · E(r̃, t, t^{0})dxdt − ∫_{D^{∗}} F · E(r, t, t^{0})dxdt]. (8)

Denote by C_{∗}^{∞}(\overline{D}) the space of functions of the class C^{∞}(\overline{D}) having bounded supports, i.e.,

C_{∗}^{∞}(\overline{D}) = \{u ∈ C^{∞}(\overline{D}) : diam supp u < ∞\}.

The spaces C_{∗}^{∞}(S_{i}), i = 1, 2, are defined analogously.

According to the remark above and using the formula (8), the solution u(x_{1}, x_{2}, t) of the problem (3),(2) will be defined uniquely; moreover, as is easily seen, for any F ∈ C_{∗}^{∞}(\overline{D}), f_{i} ∈ C_{∗}^{∞}(S_{i}), i = 1, 2, this solution belongs to the class C_{∗}^{∞}(\overline{D}).

Denote by W_{2}^{1}(D), W_{2}^{2}(D) and W_{2}^{1}(S_{i}), i = 1, 2, the well-known Sobolev spaces.

Definition. Let f_{i} ∈ W_{2}^{1}(S_{i}), i = 1, 2, F ∈ L_{2}(D). The function u ∈ W_{2}^{1}(D) is said to be a strong solution of the problem (3),(2) of the class W_{2}^{1} if there is a sequence u_{n} ∈ C_{∗}^{∞}(\overline{D}) such that u_{n} → u, □u_{n} → F and u_{n}|_{S_{i}} → f_{i} in the spaces W_{2}^{1}(D), L_{2}(D) and W_{2}^{1}(S_{i}), i = 1, 2, respectively, i.e., for n → ∞

∥u_{n} − u∥_{W_{2}^{1}(D)} → 0, ∥□u_{n} − F ∥_{L_{2}(D)} → 0,
∥u_{n}|_{S_{i}} − f_{i}∥_{W_{2}^{1}(S_{i})} → 0, i = 1, 2.

Lemma 1. For −1 < k < 0 the a priori estimate

∥u∥_{W_{2}^{1}(D)} ≤ C( ∑_{i=1}^{2} ∥f_{i}∥_{W_{2}^{1}(S_{i})} + ∥F ∥_{L_{2}(D)})  (9)

is valid for any u ∈ C_{∗}^{∞}(\overline{D}), where f_{i} = u|_{S_{i}}, i = 1, 2, F = □u, and the positive constant C does not depend on u.

Proof. Introduce the notations:

D_{τ} = {(x, t) ∈ D : t < τ}, D_{0τ} = ∂D_{τ} ∩ {t = τ}, 0 < τ ≤ t_{0},

S_{iτ} = ∂D_{τ} ∩ S_{i}, i = 1, 2, S_{τ} = S_{1τ} ∪ S_{2τ} , α_{1} = cos { \atop (n, x_{1})},

α_{2} = cos { \atop (n, x_{2})}, α_{3} = cos { \atop (n, t)}.

Here n = (α_{1}, α_{2}, α_{3}) is the unit vector of an outer normal to ∂D_{τ}; moreover, as is easily seen,

n|_{S_{1τ}} =(0, \frac{−1}{\sqrt{1 + k^{2}}} , \frac{k}{\sqrt{1 + k^{2}}} ), n|_{S_{2τ}} =(0, \frac{\sqrt{2}}{2}, \frac{−\sqrt{2}}{2}), n|_{D_{0τ}} =(0, 0, 1).

Hence, for −1 < k < 0

\begin{matrix}α_{3}|_{S_{iτ}} < 0 i = 1, 2, α_{3}^{−1}(α_{3}^{2} − α_{1}^{2} − α_{2}^{2})|_{S_{1}} > 0, \\ (α_{3}^{2} − α_{1}^{2} − α_{2}^{2})|_{S_{2}} = 0. \end{matrix}  (10)

Multiplying both parts of equation (3) by 2u_{t}, where u ∈ C_{∗}^{∞}(\overline{D}), F = □u, integrating the obtained expression over the region to D_{τ}, and taking into account (10), we get

2 ∫_{D_{τ}} F u_{τ} dxdt = ∫_{D_{τ}} (\frac{∂u_{t}^{2}}{∂t} + 2u_{x_{1}}u_{tx_{1}} + 2u_{x_{2}}u_{tx_{2}})dxdt −

−2 ∫_{S_{τ}} (u_{x_{1}}u_{t}α_{1} + u_{x_{2}}u_{t}α_{2})ds = ∫_{D_{0τ}} (u_{t}^{2} + u_{x_{1}}^{2} + u_{x_{2}}^{2})dx +

+ ∫_{S_{τ}} [(u_{t}^{2} + u_{x_{1}}^{2} + u_{x_{2}}^{2})α_{3} − 2(u_{x_{1}}u_{t}α_{1} + u_{x_{2}}u_{t}α_{2})]ds =

= ∫_{D_{0τ}} (u_{t}^{2} + u_{x_{1}}^{2} + u_{x_{2}}^{2})dx + ∫_{S_{τ}} α_{3}^{−1}[(α_{3}u_{x_{1}} − α_{1}u_{t})^{2} + (α_{3}u_{x_{2}} − α_{2}u_{t})^{2} +

+(α_{3}^{2} − α_{1}^{2} − α_{2}^{2})u_{t}^{2}]ds ≥ ∫_{D_{0τ}} (u_{t}^{2} + u_{x_{1}}^{2} + u_{x_{2}}^{2})dx +

+ ∫_{S_{τ}} α_{3}^{−1}[(α_{3}u_{x_{1}} − α_{1}u_{t})^{2} + (α_{3}u_{x_{2}} − α_{2}u_{t})^{2}]ds.  (11)

Putting

W (τ) = ∫_{D_{0τ}} (u_{t}^{2} + u_{x_{1}}^{2} + u_{x_{2}}^{2})dx, ũ_{i} = α_{3}u_{x_{i}} − α_{i}u_{t}, i = 1, 2,

304 S. KHARIBEGASHVILI

from (11) we have

W (τ) ≤ \frac{\sqrt{1 + k^{2}}}{|k|} ∫_{S_{1τ}} (u_{1}^{2} + u_{2}^{2})ds + \sqrt{2} ∫_{S_{2τ}} (u_{1}^{2} + u_{2}^{2})ds +

+ ∫_{D_{τ}} (F^{2} + u_{t}^{2})dxdt ≤ \frac{\sqrt{1 + k^{2}}}{|k|} ∫_{S_{1τ}} (u_{1}^{2} + u_{2}^{2})ds + \sqrt{2} ∫_{S_{2τ}} (u_{1}^{2} + u_{2}^{2})ds +

+ ∫_{0}^{τ} dξ ∫_{D_{0ξ}} u_{t}^{2}dx + ∫_{D_{τ}} F^{2}dxdt ≤ \frac{\sqrt{1 + k^{2}}}{|k|} ∫_{S_{τ}} (u_{1}^{2} + u_{2}^{2})ds +

+ ∫_{0}^{τ} W (ξ)dξ + ∫_{D_{τ}} F^{2}dxdt.  (12)

Let (x, τ_{x}) be a point of intersection of the surface S_{1} ∪S_{2} with a straight line parallel to the axis t and passing through the point (x, 0). We have

u(x, τ) = u(x, τ_{x}) + ∫_{τ_{x}}^{τ} u_{t}(x, t)dt,

whence it follows that

∫_{D_{0τ}} u^{2}(x, τ)dx ≤ 2 ∫_{D_{0τ}} u^{2}(x, τ_{x})dx +

+2|τ − τ_{x}| · ∫_{D_{0τ}} dx ∫_{τ_{x}}^{τ} u_{t}^{2}(x, t)dt = 2 ∫_{S_{τ}} α_{3}^{−1}u^{2}ds +

+2|τ − τ_{x}| ∫_{D_{τ}} u_{t}^{2}dxdt ≤ C_{k}( ∫_{S_{τ}} u^{2}ds + ∫_{D_{τ}} u_{t}^{2}dxdt),  (13)

where C_{k} = 2 max (\frac{\sqrt{1+k^{2}}}{|k|} , t_{0}).

Introducing the notation

W_{0}(τ) = ∫_{D_{0τ}} (u^{2} + u_{t}^{2} + u_{x_{1}}^{2} + u_{x_{2}}^{2})dx

and adding the inequalities (12) and (13) we obtain

W_{0}(τ) ≤ C_{k}[ ∫_{S_{τ}} (u^{2} + u_{1}^{2} + u_{2}^{2})ds + ∫_{0}^{τ} W_{0}(ξ)dξ + ∫_{D_{τ}} F^{2}dxdt]

from which by Gronwall’s lemma we find that

W_{0}(τ) ≤ C_{1k}[ ∫_{S_{τ}} (u^{2} + u_{1}^{2} + u_{2}^{2})ds + ∫_{D_{τ}} F^{2}dxdt].  (14)

We can easily see that α_{3} \frac{∂}{∂x_{i}} − α_{i} \frac{∂}{∂t} is the interior differential operator on the surface S_{τ}. Therefore, by virtue of (2), the inequality

∫_{S_{τ}} (u^{2} + u_{1}^{2} + u_{2}^{2})ds ≤ C̃_{3} ∑_{i=1}^{2} ∥f∥_{W_{2}^{1}(S_{iτ})}^{2}  (15)

is valid.

It follows from (14) and (15) that

W_{0}(τ) ≤ C_{2k}( ∑_{i=1}^{2} ∥f_{i}∥_{W_{2}^{1}(S_{iτ})}^{2} + ∥F ∥_{L_{2}(D_{τ})}^{2}).  (16)

Integrating both parts of the inequality (16) with respect to τ, we obtain the estimate (9).

Remark. It is easy see that the a priori estimate (9) is also valid for a function u of the class W_{2}^{2}(D), since the space C_{∗}^{∞}(\overline{D}) is everywhere a dense subset of the space W_{2}^{2}(D). It should be noted that the constant C in (9) tends to infinity for k → 0 and it becomes, generally speaking, invalid in the limit for k = 0, i.e. for S_{1} : x_{2} = 0, 0 ≤ t ≤ t_{0}. At the same time, following the proof of Lemma 1, we can see that the estimate (9) is also valid for k = 0 if f_{1} = u|_{S_{1}} = 0.

The following theorem holds.

Theorem 1. Let −1 < k < 0. Then for every f_{i} ∈ W_{2}^{1}(S_{i}), i = 1, 2, F ∈ L_{2}(D) there exists a unique strong solution of the problem (3), (2) of the class W_{2}^{1} for which the estimate (9) is valid.

Proof. It is known that the spaces C_{∗}^{∞}(\overline{D}) and C_{∗}^{∞}(S_{i}), i = 1, 2, are dense everywhere in the spaces L_{2}(D) and W_{2}^{1}(S_{i}), i = 1, 2, respectively. Therefore there exist sequences F_{n} ∈ C_{∗}^{∞}(D) and f_{in} ∈ C_{∗}^{∞}(S_{i}), i = 1, 2, such that

lim_{n→∞} ∥F − F_{n}∥_{L_{2}(D)} = lim_{n→∞} ∥f_{i} − f_{in}∥_{W_{2}^{1}(S_{i})} = 0, i = 1, 2. (17)

Moreover, because of the condition (f_{1} − f_{2})|_{S_{1}∩S_{2}} = 0, the sequences f_{1n} and f_{2n} can be chosen so that

(f_{1n} − f_{2n})|_{S_{1}∩S_{2}} = 0, n = 1, 2, ….

According to the integral representation (8) of the regular solutions of the problem (3),(2), there exists a sequence u_{n} ∈ C_{∗}^{∞}(\overline{D}) of solutions of that problem for F = F_{n}, f_{i} = f_{in}, i = 1, 2.

By virtue of the inequality (9) we have

∥u_{n} − u_{m}∥_{W_{2}^{1}(D)} ≤
≤ C( ∑_{i=1}^{2} ∥f_{in} − f_{im}∥_{W_{2}^{1}(S_{i})} + ∥F_{n} − F_{m}∥_{L_{2}(D)}).  (18)

It follows from (17) and (18) that the sequence u_{n} of the functions is fundamental in the space W_{2}^{1}(D). Therefore, since the space W_{2}^{1}(D) is complete, there exists a function u ∈ W_{2}^{1}(D) such that u_{n} → u, □u_{n} → F, and u_{n}|_{S_{i}} → f_{i} in W_{2}^{1}(D), L_{2}(D), and W_{2}^{1}(S_{i}), i = 1, 2, respectively, for n → ∞. Hence the function u is the strong solution of the problem (3),(2) of the class W_{2}^{1}. The uniqueness of the strong solution of the problem (3),(2) of the class W_{2}^{1} follows from the inequality (9).

Remark. Theorem 1 remains also valid for k = 0, i.e., for S_{1} : x_{2} = 0, 0 ≤ t ≤ t_{0} if f_{1} = u|_{S_{1}} = 0.

Now for the problem (3),(2) let us introduce the notion of a weak solution of the class W_{2}^{1}. Put S_{3} = ∂D ∩ {t = t_{0}}, V = {v ∈ W_{2}^{1}(D) : v|_{S_{1}∪S_{3}} = 0}.

Definition. Let f_{i} ∈ W_{2}^{1}(S_{i}), i = 1, 2, F ∈ L_{2}(D). The function u ∈ W_{2}^{1}(D) is said to be a weak solution of the problem (3),(2) of the class W_{2}^{1} if it satisfies both the boundary conditions (2) and the identity

∫_{D} (u_{t}v_{t} − u_{x_{1}}v_{x_{1}} − u_{x_{2}}v_{x_{2}})dxdt + ∫_{S_{2}} \frac{∂f_{2}}{∂N} vds + ∫_{D} F vdxdt = 0 (19)

for any v ∈ V, where \frac{∂}{∂N} is a derivative with respect to a conormal to S_{2}.

Obviously, every strong solution of the problem (3),(2) of the class W_{2}^{1} is a weak solution of the same class.

Lemma 2. For k = 0, i.e., for S_{1} : x_{2} = 0, 0 ≤ t ≤ t_{0} the problem (3),

(2) cannot have more than one weak solution of the class W_{2}^{1}.

Proof. Let the function u ∈ W_{2}^{1}(D) satisfy the identity (19) for u|_{S_{i}} = f_{i} = 0, i = 1, 2, F = 0. In this identity we take as v the function

v(x_{1}, x_{2}, t) = \begin{cases}0 for t ≥ τ, \\ ∫_{τ}^{t} u(x_{1}, x_{2}, σ)dσ for |x_{2}| ≤ t ≤ τ, \end{cases}  (20)

where 0 < τ ≤ t_{0}.

Obviously, v ∈ V and

\begin{matrix}v_{t} = u, v_{x_{i}} = ∫_{τ}^{t} u_{x_{i}}(x_{1}, x_{2}, σ)dσ, i = 1, 2, \\ v_{tx_{i}} = u_{x_{i}}, v_{tt} = u_{t}. \end{matrix}  (21)

By virtue of (20) and (21), the identity (19) for f_{2} = 0, F = 0 takes the form

∫_{D_{τ}} (v_{tt}v_{t} − v_{tx_{1}}v_{x_{1}} − v_{tx_{2}}v_{x_{2}})dxdt = 0

or

∫_{D_{τ}} \frac{∂}{∂t}(v_{t}^{2} − v_{x_{1}}^{2} − v_{x_{2}}^{2})dxdt = 0,  (22)

where D_{τ} = D ∩ {t < τ}.

Using the Gauss–Ostrogradsky formula on the left-hand side of (22), we obtain

∫_{∂D_{τ}} (v_{t}^{2} − v_{x_{1}}^{2} − v_{x_{2}}^{2}) cos n̂tds = 0.  (23)

Since ∂D_{τ} = S_{1τ} ∪ S_{2τ} ∪ S_{3τ}, where S_{iτ} = ∂D_{τ} ∩ S_{i}, i = 1, 2, S_{3τ} = ∂D_{τ} ∩ {t = τ} and

cos n̂t|_{S_{1τ}} = 0, cos n̂t|_{S_{2τ}} = −\frac{1}{\sqrt{2}}, cos n̂t|_{S_{3τ}} = 1,

u|_{S_{iτ}} = f_{i} = 0, i = 1, 2, v_{x_{i}}|_{S_{3τ}} = 0, i = 1, 2, v_{t} = u,

it follows from (23) that

∫_{S_{3τ}} u^{2}dx_{1}dx_{2} + \frac{1}{\sqrt{2}} ∫_{S_{2τ}} (v_{x_{1}}^{2} + v_{x_{2}}^{2})ds = 0.

Hence, u|_{S_{3τ}} = 0 for any τ from the interval (0, t_{0}]. Therefore, u ≡ 0 in the domain D.

Due to the fact that the strong solution of the problem (3),(2) of the class W_{2}^{1} is at the same time a weak solution of the class W_{2}^{1}, from Lemma 2 and the remark following after Theorem 1 we have

Theorem 2. Let k = 0, i.e., S_{1} : x_{2} = 0, 0 ≤ t ≤ t_{0} and u|_{S_{1}} = f_{1} = 0. Then for any f_{2} ∈ W_{2}^{1}(S_{2}) and F_{2} ∈ L_{2}(D) there exists a unique weak solution u of the problem (3), (2) of the class W_{2}^{1} for which the estimate (9) is valid.

To prove the solvability of the problem (1),(2) we shall use the solvability of the problem (3),(2) and the fact that in the specifically chosen equivalent norms of the spaces L_{2}(D), W_{2}^{1}(D), W_{2}^{1}(S_{i}), i = 1, 2, the lowest terms in equation (1) give arbitrarily small perturbations.

Introduce in the space W_{2}^{1}(D) an equivalent norm depending on the parameter γ,

∥u∥_{D,1,γ}^{2} = ∫_{D} e^{−γt}(u^{2} + u_{t}^{2} + u_{x_{1}}^{2} + u_{x_{2}}^{2})dxdt, γ > 0.

In the same manner we introduce the norms ∥F ∥_{D,0,γ}, ∥f_{i}∥_{S_{i},1,γ} in the spaces L_{2}(D), W_{2}^{1}(S_{i}), i = 1, 2.

Making use of the inequality (16), we obtain the a priori estimate for u ∈ C_{∗}^{∞}(D) with respect to the norms ∥ · ∥_{D,1,γ}, ∥ · ∥_{S_{i,1,γ}}, i = 1, 2. Multiplying both parts of the inequality (16) by e^{−γt} and integrating the obtained inequality with respect to τ from 0 to t_{0} we get

∥u∥_{D,1,γ}^{2} = ∫_{0}^{t_{0}} e^{−γτ} W_{0}(τ)dτ ≤ C_{2k}( ∑_{i=1}^{2} ∫_{0}^{t_{0}} e^{−γt}∥f_{i}∥_{W_{2}^{1}(S_{iτ})}^{2}dτ +

+ ∫_{0}^{t_{0}} e^{−γτ} ∥F ∥_{L_{2}(D_{τ})}^{2}dτ).  (24)

We have

∫_{0}^{t_{0}} e^{−γt}∥F ∥_{L_{2}(D_{τ})}^{2}dτ = ∫_{0}^{t_{0}} e^{−γt}[ ∫_{0}^{τ} ( ∫_{D_{0σ}} F^{2}dx)dσ]dτ =

= ∫_{0}^{t_{0}} [ ∫_{D_{0σ}} F^{2}dx ∫_{σ}^{t_{0}} e^{−γτ} dτ]dσ = \frac{1}{γ} ∫_{0}^{t_{0}}(e^{−γσ} − e^{−γt_{0}})[ ∫_{D_{0σ}} F^{2}dx]dσ ≤

≤ \frac{1}{γ} ∫_{0}^{t_{0}} e^{−γσ}[ ∫_{D_{0σ}} F^{2}dx]dσ = \frac{1}{γ} ∥F ∥_{D,0,γ}^{2},  (25)

where D_{0τ} = ∂D_{τ} ∩ {t = τ}, 0 < τ ≤ t_{0}.

Analogously we obtain

∫_{0}^{t_{0}} e^{−γτ} ∥f_{i}∥_{W_{2}^{1}(S_{iτ})}^{2}dτ ≤ \frac{C_{3}}{γ} ∥f_{i}∥_{S_{i},1,γ}^{2}, i = 1, 2,  (26)

where C_{3} is a positive constant independent of f_{i} and the parameter γ.

From the inequalities (24)–(26) we have the a priori estimate for u ∈ C_{∗}^{∞}(\overline{D})

∥u∥_{D,1,γ} ≤ \frac{C_{4}}{\sqrt{γ}} ( ∑_{i=1}^{2} ∥f_{i}∥_{S_{i},1,γ} + ∥F ∥_{D,0,γ})  (27)

for −1 < k < 0, where C_{4} = const > 0 does not depend on u and the parameter γ.

Below, the coefficients a, b, c, and d in equation (1) are assumed to be bounded measurable functions in the domain D.

Consider the space

V = L_{2}(D) × W_{2}^{1}(S_{1}) × W_{2}^{1}(S_{2}).

To the problem (1),(2) there corresponds an unbounded operator

T : W_{2}^{1}(D) → V

with the domain of definition Ω_{T} = C_{∗}^{∞}(D) ⊂ W_{2}^{1}(D), acting by the formula

T u = (Lu, u|_{S_{1}}, u|_{S_{2}}), u ∈ Ω_{T} .

We can easily prove that the operator T admits a closure \overline{T}. In fact, let u_{n} ∈ Ω_{T}, u_{n} → 0 in W_{2}^{1}(D) and T u_{n} → (F, f_{1}, f_{2}) in the space V. First we shall show that F = 0. For φ ∈ C_{0}^{∞}(D) we have

(Lu_{n}, φ) = (u_{n}, □φ) + (Ku, φ),  (28)

where Ku = au_{x_{1}} +bu_{x_{2}} +cu_{t} +du. Since u_{n} → 0 in W_{2}^{1}(D), it follows from (28) that (Lu_{n}, φ) → 0. On the other hand, by the definition of a strong solution, we have the convergence Lu_{n} → F in L_{2}(D). Therefore (f, φ) = 0 for any φ ∈ C_{0}^{∞}(D), and hence, F = 0. That f_{1} = f_{2} = 0 follows from the fact that u_{n} → 0 in W_{2}^{1}(D) and the contraction operator u → (u|_{S_{1}}, u|_{S_{2}}) acts boundedly from W_{2}^{1}(D) to L_{2}(S_{1}) × L_{2}(S_{2}).

To the problem (3),(2) there corresponds an unbounded operator T_{0} : W_{2}^{1}(D) → V obtained from the operator T for a = b = c = d = 0. As was shown above, the operator T_{0} also admits a closure \overline{T}_{0}. Obviously, the operator K_{0} : W_{2}^{1}(D) → V acting by the formula K_{0}u = (Ku, 0, 0) is bounded and

T = T_{0} + K_{0}.  (29)

Note that the domains of definition Ω\overline{T} and Ω_{\overline{T}_{0}} of the closed operators \overline{T} and \overline{T}_{0} coincide by virtue of (29) and the fact that the operator K_{0} is bounded.

We can easily see that the existence and uniqueness of the strong solution of the problem (1),(2) of the class W_{2}^{1} as well as the estimate (9) for this solution follow from the existence of the bounded right operator \overline{T}^{−1} inverse to \overline{T} and defined in a whole space V.

The fact that the operator \overline{T}_{0} has a bounded right inverse operator \overline{T} _{0}^{−1} : V → W_{2}^{1}(D) for −1 < k < 0 follows from Theorem 1 and the estimate (9) which, as we have shown above, can be written in equivalent norms in the form of (27). It is easy to see that the operator

K_{0}\overline{T} _{0}^{−1} : V → V

is bounded and by virtue of (27) its norm admits the following estimate

∥K_{0}\overline{T} _{0}^{−1}∥ ≤ \frac{C_{4}C_{5}}{\sqrt{γ}} ,  (30)

where C_{5} is a positive constant depending only on the coefficients a, b, c, and d of equation (1).

Taking into account (30), we note that the operator (I+K_{0}\overline{T} _{0}^{−1}) : V → V has a bounded inverse operator (I +K_{0}\overline{T} _{0}^{−1})^{−1} for sufficiently large γ, where I is the unit operator. Now it remains only to note that the operator

\overline{T} _{0}^{−1}(I + K_{0}\overline{T} _{0}^{−1})^{−1}

is a bounded operator right inverse to \overline{T} and defined in a whole space V. Thus the following theorem is proved.

Theorem 3. Let −1 < k < 0. Then for any f_{i} ∈ W_{2}^{1}(S_{i}), i = 1, 2, F ∈ L_{2}(D) there exists a unique strong solution u of the problem (1), (2) of the class W_{2}^{1} for which the estimate (9) is valid.

References

1. J. Hadamard, Lectures on Cauchy’s problem in partial differential equations. Yale University Press, New Haven, 1923.

2. J. Tolen, Probleme de Cauchy sur les deux hipersurfaces caracteristiques secantes. C.R.Acad.Sci. Paris Ser. A-B 291(1980), No. 1, 49-52.

3. S. Kharibegashvili, On a characteristic problem for the wave equation. Proc. I. Vekua Inst. Appl. Math. Tbilisi St. Univ. 47(1992), 76–82.

4. A. V. Bitsadze, On mixed type equations on three-dimensional domains. (Russian) Dokl. Akad. Nauk SSSR 143(1962), No. 5, 1017-1019.

5. A. M. Nakhushev, A multidimensional analogue of the Darboux problem for hyperbolic equations. (Russian) Dokl. Akad. Nauk SSSR 194(1970), No. 1, 31-34.

6. T. Sh. Kalmenov, On multidimensional regular bounded value problems for the wave equation. (Russian) Izv. Akad. Nauk Kazakh. SSR, Ser. Fiz.-Mat. (1982), No. 3, 18-25.

7. A. V. Bitsadze, Some classes of partial differential equations. (Russian) Nauka, Moscow, 1981.

(Received 25.10.1993)

Author’s address: I. Vekua Institute of Applied Mathematics Tbilisi State University 2, University St., Tbilisi 380043 Republic of Georgia