GEORGIAN MATHEMATICAL JOURNAL: Vol. 6, No. 5, 1999, 447-470

ON THE CORRECTNESS OF THE DIRICHLET PROBLEM IN A CHARACTERISTIC RECTANGLE FOR FOURTH

ORDER LINEAR HYPERBOLIC EQUATIONS

T. KIGURADZE

Abstract. It is proved that the Dirichlet problem is correct in the characteristic rectangle D_{ab} = [0, a] × [0, b] for the linear hyperbolic equation

\frac{∂^{4}u}{∂x^{2}∂y^{2}} = p_{0}(x, y)u + p_{1}(x, y) \frac{∂u}{∂x} + p_{2}(x, y) \frac{∂u}{∂y} +

+p_{3}(x, y) \frac{∂^{2}u}{∂x∂y} + q(x, y)

with the summable in D_{ab} coefficients p_{0}, p_{1}, p_{2}, p_{3} and q if and only if the corresponding homogeneous problem has only the trivial solution. The effective and optimal in some sense restrictions on p_{0}, p_{1}, p_{2} and p_{3} guaranteeing the correctness of the Dirichlet problem are established.

§ 1. Formulation of the Problem and Main Results

The Dirichlet problem for second order hyperbolic equations and some higher order linear hyperbolic equations with constant coefficients has long been attracting the attention of mathematicians. The problem has been the subject of numerous studies (see [1–20] and the references therein) but still remains investigated very little for a wide class of hyperbolic equations. This class includes the fourth order hyperbolic equation

\frac{∂^{4}u}{∂x^{2}∂y^{2}}=p_{0}(x, y)u+p_{1}(x, y)\frac{∂u}{∂x} +p_{2}(x, y)\frac{∂u}{∂y} +p_{3}(x, y)\frac{∂^{2}u}{∂x∂y} +q(x, y) (1.1)

for which the Dirichlet problem is considered here in the characteristic rectangle D_{ab} = [0, a] × [0, b].

447

1991 Mathematics Subject Classification. 35L55.

Key words and phrases. Fourth order linear hyperbolic equation, Dirichlet problem in the characteristic rectangle, correctness.

1072-947X/99/0900-0447$16.00/0 \begin{matrix}c \\ ◯ \end{matrix} 1999 Plenum Publishing Corporation

Throughout the paper it will be assumed that 0 < a < +∞, 0 < b < +∞, and p_{k} : D_{ab} → ℝ (k = 0, 1, 2, 3) and q : D_{ab} → ℝ are Lebesgue summable functions.

A function u : D_{ab} → ℝ will be called a solution of system (1.1) if it is absolutely continuous together with \frac{∂u}{∂x}, \frac{∂u}{∂y}, \frac{∂^{2}u}{∂x∂y} and satisfies (1.1) almost everywhere in D_{ab}.

Let Γ_{ab} be the boundary of the rectangle D_{ab}, and φ : Γ_{ab} → ℝ be a continuous function having the absolutely continuous partial derivatives \frac{∂φ(x,0)}{∂x}, \frac{∂φ(x,b)}{∂x} and \frac{∂φ(0,y)}{∂y}, \frac{∂φ(a,y)}{∂y} in [0, a] and [0, b], respectively.

The Dirichlet problem for equation (1.1) in D_{ab} consists in finding a solution u : D_{ab} → ℝ of (1.1) satisfying the boundary condition

u(x, y) = φ(x, y) for (x, y) ∈ Γ_{ab}.  (1.2)

Along with (1.1),(1.2), we shall consider the corresponding homogeneous problem

\frac{∂^{4}u}{∂x^{2}∂y^{2}} = p_{0}(x, y)u + p_{1}(x, y)\frac{∂u}{∂x} + p_{2}(x, y)\frac{∂u}{∂y} + p_{3}(x, y)\frac{∂^{2}u}{∂x∂y} , (1.1_{0})
u(x, y) = 0 for (x, y) ∈ Γ_{ab},  (1.2_{0})

and also the perturbed problem

\frac{∂^{4}u}{∂x^{2}∂y^{2}}=\overline{p}_{0}(x, y)u+\overline{p}_{1}(x, y)\frac{∂u}{∂x} +\overline{p}_{2}(x, y)\frac{∂u}{∂y} +\overline{p}_{3}(x, y)\frac{∂^{2}u}{∂x∂y} +\overline{q}(x, y), (1.3)
u(x, y) = \overline{φ}(x, y) for (x, y) ∈ Γ_{ab}.  (1.4)

Before formulating the main results, we introduce some notation and a definition.

ℝ is the set of real numbers.

L(D_{ab}) is the space of Lebesgue summable functions z : D_{ab} → ℝ.

C̃(D_{ab}) is the space of absolutely continuous functions z : D_{ab} → ℝ (see the definition in [21] or [22]).

C̃^{1}(D_{ab}) is the space of absolutely continuous functions z : D_{ab} → ℝ having the absolutely continuous partial derivatives \frac{∂z}{∂x}, \frac{∂z}{∂y} and \frac{∂^{2}z}{∂x∂y}.

C̃^{1}(Γ_{ab}) is the space of absolutely continuous functions ζ : Γ_{ab} → ℝ having the absolutely continuous partial derivatives \frac{∂ζ(x,0)}{∂x}, \frac{∂ζ(x,b)}{∂x} in [0, a] and the absolutely continuous partial derivatives \frac{∂ζ(0,y)}{∂y}, \frac{∂ζ(a,y)}{∂y} in [0, b].

Definition 1.1. Problem (1.1), (1.2) will be called correct if it has a unique solution u and for an arbitrary positive number ε and nonnegative function γ ∈ L(D_{ab}) there exists a positive number δ such that for arbitrary functions \overline{p}_{k} ∈ L(D_{ab}) (k = 0, 1, 2, 3), \overline{q} ∈ L(D_{ab}) and \overline{φ} ∈ C̃^{1}(Γ_{ab}) satisfying the conditions

|\overline{p}_{k}(x, y) − p_{k}(x, y)| ≤ γ(x, y) almost everywhere in D_{ab} (k = 0, 1, 2, 3),

| ∫_{0}^{x} ∫_{0}^{y}[\overline{p}_{k}(s, t) − p_{k}(s, t)] ds dt| ≤ δ for (x, y) ∈ D_{ab} (k = 0, 1, 2, 3),

| ∫_{0}^{x} ∫_{0}^{y}[\overline{q}(s, t) − q(s, t)] ds dt| ≤ δ for (x, y) ∈ D_{ab},

|\overline{φ}(x, y) − φ(x, y)| + |\frac{∂\overline{φ}(x, 0)}{∂x} − \frac{∂φ(x, 0)}{∂x}| + |\frac{∂\overline{φ}(x, b)}{∂x} − \frac{∂φ(x, b)}{∂x}| +

+|\frac{∂\overline{φ}(0, y)}{∂y} − \frac{∂φ(0, y)}{∂y}| + |\frac{∂\overline{φ}(a, y)}{∂y} − \frac{∂φ(a, y)}{∂y}| < δ for (x, y) ∈ Γ_{ab},

problem (1.3), (1.4) has a unique solution \overline{u} and

|\overline{u}(x, y) − u(x, y)| + |\frac{∂\overline{u}(x, y)}{∂x} − \frac{∂u(x, y)}{∂x}| + |\frac{∂\overline{u}(x, y)}{∂y} − \frac{∂u(x, y)}{∂y}| +

+|\frac{∂^{2}\overline{u}(x, y)}{∂x∂y} − \frac{∂^{2}u(x, y)}{∂x∂y}| < ε for (x, y) ∈ D_{ab}.

Theorem 1.1. Problem (1.1), (1.2) is correct if and only if the corresponding homogeneous problem (1.1_{0}), (1.2_{0}) has only the trivial solution.

Now consider the sequence of boundary value problems

\frac{∂^{4}u}{∂x^{2}∂y^{2}} = p_{0m}(x, y)u+p_{1m}(x, y)\frac{∂u}{∂x} + p_{2m}(x, y)\frac{∂u}{∂y} +
+p_{3m}(x, y)\frac{∂^{2}u}{∂x∂y} + q_{m}(x, y),  (1.5_{m})
u(x, y) = φ_{m}(x, y) for (x, y) ∈ Γ_{ab},  (1.6_{m})

under the following assumptions on the functions p_{km}, q_{m} and φ_{m}:

(i) p_{km} ∈ L(D_{ab}), q ∈ L(D_{ab}) (k = 0, 1, 2, 3; m = 1, 2, …) and the equalities

lim_{m→∞} ∫_{0}^{x} ∫_{0}^{y}[p_{km}(s, t) − p_{k}(s, t)] ds dt = 0 (k = 0, 1, 2, 3),

lim_{m→∞} ∫_{0}^{x} ∫_{0}^{y}[q_{m}(s, t) − q(s, t)] ds dt = 0

hold uniformly on D_{ab};

(ii) There exists a nonnegative function p ∈ L(D_{ab}) such that the inequalities

|p_{km}(x, y)| ≤ p(x, y) (k = 0, 1, 2, 3; m = 1, 2, …)

hold almost everywhere in D_{ab}; (iii) φ_{m} ∈ C̃^{1}(Γ_{ab}) (m = 1, 2, …) and the equalities

lim_{m→∞} \frac{∂^{k}φ_{m}(x, 0)}{∂x^{k}} = \frac{∂^{k}φ(x, 0)}{∂x^{k}}, lim_{m→∞} \frac{∂^{k}φ_{m}(x, b)}{∂x^{k}} = \frac{∂^{k}φ(x, 0)}{∂x^{k}} (k = 0, 1),

lim_{m→∞} \frac{∂^{k}φ_{m}(0, y)}{∂y^{k}} = \frac{∂^{k}φ(0, y)}{∂y^{k}}, lim_{m→∞} \frac{∂^{k}φ_{m}(a, y)}{∂y^{k}} = \frac{∂^{k}φ(a, y)}{∂y^{k}} (k = 0, 1)

hold uniformly on [0, a] and [0, b] respectively.

Theorem 1.1 can be reformulated in the manner as follows.

Theorem 1.1′. If the homogeneous problem (1.1_{0}), (1.2_{0}) has only the trivial solution, then problem (1.1), (1.2) has the unique solution u. Moreover, if the conditions (i)–(iii) hold, then beginning from some sufficiently large m_{0} problem (1.5_{m}), (1.6_{m}) has the unique solution u_{m} and

lim_{m→∞} \frac{∂^{j+k}u_{m}(x, y)}{∂x^{j}∂y^{k}} = \frac{∂^{j+k}u(x, y)}{∂x^{j}∂y^{k}} uniformly on D_{ab} (j, k = 0, 1).^{∗}

Remark 1.1. The conditions (i) and (ii) can be fulfilled even in the case where the differences p_{km}(x, y)−p_{k}(x, y) (k = 0, 1, 2, 3) and q_{m}(x, y)−q(x, y) have no limits as m → ∞. For example, the functions

p_{km}(x, y) = p_{k}(x, y) + sin mx sin my,
q_{m}(x, y) = q(x, y) + m sin mx sin my (k = 0, 1, 2, 3; m = 1, 2, …)

satisfy the conditions (i) and (iii).

Finally, we introduce the effective conditions for the correctness of problem (1.1),(1.2).

Theorem 1.2. Let the inequalities

|p_{k}(x, y) − p_{k0}(x, y)| ≤ l_{k} (k = 1, 2), |p_{3}(x, y)| ≤ l_{3},  (1.7)
p_{0}(x, y) − \frac{1}{2} \frac{∂p_{10}(x, y)}{∂x} − \frac{1}{2} \frac{∂p_{20}(x, y)}{∂y} ≤ l_{0}  (1.8)

hold almost everywhere in D_{ab}, where l_{k} (k = 0, 1, 2, 3) are constants and the functions p_{10} ∈ L(D_{ab}) and p_{20} ∈ L(D_{ab}) are absolutely continuous in the first and in the second argument, respectively, and have the partial derivatives \frac{∂p_{10}(x,y)}{∂x} and \frac{∂p_{20}(x,y)}{∂y} summable in D_{ab}. Moreover, let either l_{0} ≥0 and

∗Here and in what follows it is assumed that \frac{∂^{0}u(x,y)}{∂x^{0}} ≡ \frac{∂^{0}u(x,y)}{∂y^{0}} ≡ \frac{∂^{0}u(x,y)}{∂x^{0}∂y^{0}} ≡ u(x, y).

\frac{a^{2}b^{2}}{π^{4}} l_{0} + \frac{ab^{2}}{π^{3}} l_{1} + \frac{a^{2}b}{π^{3}}l_{2} + \frac{ab}{π^{2}} l_{3} < 1,  (1.9)

or l_{0} < 0 and

\frac{b}{π} l_{1} + \frac{a}{π} l_{2} + l_{3} < 2\sqrt{|l_{0}|}.  (1.10)

Then problem (1.1), (1.2) is correct.

Remark 1.2. The strict inequality (1.9) in Theorem 1.2 cannot be replaced by the nonstrict one

\frac{a^{2}b^{2}}{π^{4}} l_{0} + \frac{ab^{2}}{π^{3}} l_{1} + \frac{a^{2}b}{π^{3}}l_{2} + \frac{ab}{π^{2}} l_{3} ≤ 1.  (1.9′)

To make sure that is so, let us consider the equation

\frac{∂^{4}u}{∂x^{2}∂y^{2}} = \frac{π^{4}}{a^{2}b^{2}} u  (1.11)

with the boundary condition (1.2_{0}). For equation (1.11) conditions (1.7) and (1.8) hold, where p_{0}(x, y) ≡ l_{0} = \frac{π^{4}}{a^{2}b^{2}} , p_{k}(x, y) ≡ 0, l_{k} = 0 (k = 1, 2, 3) and p_{k0}(x, y) ≡ 0 (k = 1, 2), and instead of (1.9) condition (1.9′) holds. And yet problem (1.11), (1.2_{0}) is incorrect, since it has the infinite set of solutions of the form u(x, y) = c sin \frac{πx}{a} sin \frac{πy}{b}, where c is an arbitrary real constant.

§ 2. Auxiliary Statements

We introduce some additional notation to be used in this section.

If φ ∈ C̃^{1}(Γ_{ab}), then

η(φ)(x, y) = (1 − \frac{y}{b} )φ(x, 0) + \frac{y}{b} φ(x, b) + (1 − \frac{x}{a})φ(0, y) +

+\frac{x}{a}φ(a, y) − (1 − \frac{y}{b} )(1 − \frac{x}{a})φ(0, 0) − \frac{y}{b} (1 − \frac{x}{a})φ(0, b) −

−(1 − \frac{y}{b} )\frac{x}{a}φ(a, 0) − \frac{xy}{ab} φ(a, b) for (x, y) ∈ D_{ab}.  (2.1)

It is obvious that η(φ) ∈ C̃^{1}(D_{ab}) and

η(φ)(x, y) = φ(x, y) for (x, y) ∈ Γ_{ab}.  (2.2)

We denote by g_{1} and g_{2} the Green’s functions of the boundary value problems

\frac{d^{2}z}{dx^{2}} = 0, z(0) = z(a) = 0 and \frac{d^{2}z}{dy^{2}} = 0, z(0) = z(b) = 0.

Therefore

g_{1}(x, s) = \begin{cases}(x − a)\frac{s}{a} for s ≤ x, \\ (s − a)\frac{x}{a} for s > x, \end{cases}  (2.3)

g_{2}(y, t) = \begin{cases}(y − b)\frac{t}{b} for t ≤ y, \\ (t − b)\frac{y}{b} for t > y. \end{cases}  (2.4)

Under C^{1,1}(D_{ab}) we understand the Banach space of continuous functions z : D_{ab} → ℝ having the continuous partial derivatives \frac{∂z}{∂x}, \frac{∂z}{∂y}, \frac{∂^{2}z}{∂x∂y}, with the norm

∥z∥_{C^{1,1}} = max \{ ∑_{j,k=0}^{1} |\frac{∂^{j+k}z(x, y)}{∂x^{j}∂y^{k}} | : (x, y) ∈ D_{ab}\}.

It is obvious that C̃^{1}(D_{ab}) ⊂ C^{1,1}(D_{ab}). On the other hand, Arcela– Ascoli lemma gives rise to

Lemma 2.1. Let Z ⊂ C̃^{1}(D_{ab}) and there exist summable functions h_{1} : [0, a] → [0, +∞) and h_{2} : [0, b] → [0, +∞) such that every function z ∈ Z satisfies the inequality

|\frac{∂^{3}z(x, y)}{∂x^{2}∂y}| ≤ h_{1}(x)

for every y ∈ [0, b] and almost every x ∈ [0, a] and the inequality

|\frac{∂^{3}z(x, y)}{∂x∂y^{2}} | ≤ h_{2}(y)

for every x ∈ [0, a] and almost every y ∈ [0, b]. Moreover, if Z is bounded in the topology of the space C^{1,1}(D_{ab}), then it is the relative compact of this space.

In C^{1,1}(D_{ab}) consider the operator equation

u(x, y) = η(φ)(x, y) + ∫_{0}^{a} ∫_{0}^{b} g_{1}(x, s)g_{2}(y, t)[Q(u)(s, t) + q(s, t)]dsdt, (2.5)

where

Q(u)(x, y) = p_{0}(x, y)u(x, y) + p_{1}(x, y)\frac{∂u(x, y)}{∂x} +
+ p_{2}(x, y)\frac{∂u(x, y)}{∂y} + p_{3}(x, y)\frac{∂^{2}u(x, y)}{∂x∂y}.  (2.6)

Under a solution of equation (2.5) we understand a function u ∈ C^{1,1}(D_{ab}) satisfying (2.5) at every point of D_{ab}.

2.1. Lemma on Equivalence of Problem (1.1),(1.2) and Equation (2.5).

Lemma 2.2. Problem (1.1), (1.2) is equivalent to equation (2.5). More precisely, problem (1.1), (1.2) is solvable if and only if equation (2.5) is solvable and the sets of their solutions coincide.

Proof. First we show that if problem (1.1), (1.2) is solvable, then its arbitrary solution u is a solution of equation (2.5). Put

v(x, y) = u(x, y) − η(φ)(x, y).  (2.7)

Then in view of (2.1) and (2.6)

\frac{∂^{2}}{∂x^{2}} (\frac{∂^{2}v(x, y)}{∂y^{2}}) = Q(u)(x, y) + q(x, y) almost for every (x, y) ∈ D_{ab}, (2.8)

since

\frac{∂^{4}η(φ)(x, y)}{∂x^{2}∂y^{2}} = 0 for (x, y) ∈ D_{ab}.

On the other hand, by virtue of (1.2) and (2.2) we have

\frac{∂^{2}v(0, y)}{∂y^{2}} = \frac{∂^{2}v(a, y)}{∂y^{2}} = 0 for 0 ≤ y ≤ b,  (2.9)
v(x, 0) = v(x, b) = 0 for 0 ≤ x ≤ a.  (2.10)

By the condition Q(u) ∈ L(D_{ab}) it follows from (2.8) and (2.9) that the function v satisfies the equality

\frac{∂^{2}v(x, y)}{∂y^{2}} = ∫_{0}^{a} g_{1}(x, s)[Q(u)(s, y) + q(s, y)] ds

for every x ∈ [0, a] and almost for every y ∈ [0, b]. Hence, taking into the account (2.9) and (2.7), we get

u(x, y) − η(φ)(x, y) = ∫_{0}^{a} ∫_{0}^{b} g_{1}(x, s)g_{2}(y, t)[Q(u)(s, t) + q(s, t)] ds dt

for (x, y) ∈ D_{ab}. Thus u is a solution of equation (2.5).

To complete the proof, we have to show that if equation (2.5) is solvable, then its arbitrary solution u is a solution of problem (1.1), (1.2). Indeed, since η(φ) ∈ C̃^{1}(D_{ab}) and u ∈ C^{1.1}(D_{ab}), it follows from (2.3)–(2.6) that u ∈ C̃^{1}(D_{ab}) and

\frac{∂^{2}u(x, y)}{∂y^{2}} = \frac{∂^{2}η(φ)(x, y)}{∂y^{2}} + ∫_{0}^{a} g_{1}(x, s)[Q(u)(s, y) + q(s, y)] ds,

\frac{∂^{4}u(x, y)}{∂x^{2}∂y^{2}} = Q(u)(x, y) + q(x, y) almost for every (x, y) ∈ D_{ab},

i.e., u is the solution of equation (1.1). On the other hand, by (2.2)–(2.4) it follows from (2.5) that u satisfies the boundary condition (1.2).

2.2. Lemma on the Fredholm Property of Problem (1.1),(1.2).

Lemma 2.3. Problem (1.1), (1.2) is uniquely solvable if and only if the corresponding homogeneous problem (1.1_{0}), (1.2_{0}) has only the trivial solution.

Proof. For an arbitrary u ∈ C^{1,1}(D_{ab}) put

A(u)(x, y) = ∫_{0}^{a} ∫_{0}^{b} g_{1}(x, s)g_{2}(y, t)Q(u)(s, t) ds dt,  (2.11)

where Q : C^{1,1}(D_{ab}) → L(D_{ab}) is the operator given by (2.6). In view of (2.3), (2.4) and (2.6) it is clear that A : C^{1,1}(D_{ab}) → C^{1,1}(D_{ab}) is a linear bounded operator.

By Lemma 2.1 problem (1.1), (1.2) is equivalent to the equation

u(x, y) = u_{0}(x, y) + A(u)(x, y),  (2.12)

where

u_{0}(x, y) = η(φ)(x, y) + ∫_{0}^{a} ∫_{0}^{b} g_{1}(x, s)g_{2}(y, t)q(s, t) ds dt

and u_{0} ∈ C̃^{1}(D_{ab}). As for problem (1.1_{0}), (1.2_{0}), it is equivalent to the equation

u(x, y) = A(u)(x, y).  (2.12_{0})

Now, we have to show that equation (2.12) is uniquely solvable if and only if equation (2.12_{0}) has only the trivial solution.

By Fredholm’s alternative for operator equations ([2.3], Theorem XII. 2.3) we need only to show that A is a completely continuous operator, which is equivalent to the fact that A(B_{ρ}) = {v = A(u) : u ∈ B_{ρ}} is a relative compact in C^{1,1}(D_{ab}), where ρ is an arbitrary positive number and B_{ρ} = {u ∈ C^{1,1}(D_{ab}) : ∥u∥_{C^{1,1}} ≤ ρ}.

By condition (2.6) every function u ∈ B_{ρ} satisfies the inequality

|Q(u)(x, y)| ≤ ρp(x, y)

almost everywhere in D_{ab}, where

p(x, y) = ∑_{k=0}^{3} |p_{k}(x, y)|.

With regard to this fact and equalities (2.3), (2.4) and (2.6), we conclude from (2.11) that A(B_{ρ}) ⊂ C̃^{1}(D_{ab}). Moreover, for any v ∈ A(B_{ρ}) the inequality

|\frac{∂^{3}v(x, y)}{∂x^{2}∂y}| ≤ h_{1}(x)

holds for every y ∈ [0, b] and almost for every x ∈ [0, a] and the inequality

|\frac{∂^{3}v(x, y)}{∂x∂y^{2}} | ≤ h_{2}(y)

holds for every x ∈ [0, a] and almost for every y ∈ [0, b], where

h_{1}(x) = ρ ∫_{0}^{b} p(x, t) dt, h_{2}(y) = ρ ∫_{0}^{a} p(s, y) ds.

Moreover, h_{1} : [0, a] → [0, +∞) and h_{2} : [0, b] → [0, +∞) are summable functions. Hence Lemma 2.1 implies that A(B_{ρ}) is a relative compact of the space C^{1,1}(D_{ab}).

2.3. Krasnosel’skii–Krein Type Lemmas.

Lemma 2.4. Let f ∈ L(D_{ab}), z ∈ C(D_{ab}),

f_{m} ∈ L(D_{ab}), z_{m} ∈ C̃(D_{ab}) (m = 1, 2, …),  (2.13)
lim_{m→∞} ∫_{0}^{x} ∫_{0}^{y}[f_{m}(s, t) − f(s, t)] ds dt = 0 uniformly on D_{ab}, (2.14)

and

lim_{m→∞} z_{m}(x, y) = z(x, y) uniformly on D_{ab}.  (2.15)

456 T. KIGURADZE

Moreover, if

lim sup_{m→∞} ( ∫_{0}^{a} |\frac{∂z_{m}(s, 0)}{∂s}| ds + ∫_{0}^{b} |\frac{∂z_{m}(0, t)}{∂t}| dt +

+ ∫_{0}^{a} ∫_{0}^{b} |\frac{∂^{2}z_{m}(s, t)}{∂s∂t}| ds dt) < +∞,  (2.16)

then

lim_{m→∞} ∫_{0}^{x}∫_{0}^{y}[f_{m}(s, t)z_{m}(s, t)−f(s, t)z(s, t)]dsdt=0 uniformly on D_{ab}^{∗}. (2.17)

Proof. By virtue of condition (2.16) and the boundedness of the sequence (z_{m})_{m=1}^{∞} there exists a positive constant ρ such that

|z_{m}(x, y)| + ∫_{0}^{a} ∫_{0}^{b} |f(s, t)| ds dt + ∫_{0}^{a} |\frac{∂z_{m}(s, y)}{∂s}| ds + ∫_{0}^{b} |\frac{∂z_{m}(x, t)}{∂t}| dt +

+ ∫_{0}^{a} ∫_{0}^{b} |\frac{∂^{2}z_{m}(s, t)}{∂s∂t}| ds dt < ρ for (x, y) ∈ D_{ab}.  (2.18)

For an arbitrary natural m set

γ_{m}(x, y) = ∫_{0}^{x} ∫_{0}^{y}[f_{m}(s, t) − f(s, t)] ds dt,  (2.19)
δ_{m}(x, y) = ∫_{0}^{x} ∫_{0}^{y}[f_{m}(s, t)z_{m}(s, t) − f(s, t)z(s, t)] ds dt.  (2.20)

Then

δ_{m}(x, y)=∫_{0}^{x} ∫_{0}^{y} \frac{∂^{2}γ_{m}(s, t)}{∂s∂t}z_{m}(s, t)dsdt+∫_{0}^{x} ∫_{0}^{y} f(s, t)[z_{m}(s, t)−z(s, t)]dsdt=

=γ_{m}(x, y)z_{m}(x, y)+∫_{0}^{x} ∫_{0}^{y} \frac{∂^{2}z_{m}(s, t)}{∂s∂t}γ_{m}(s, t)dsdt−∫_{0}^{x} \frac{∂z_{m}(s, y)}{∂s}γ_{m}(s, y)ds−

∗An analogous statement for functions of one variable was proved for the first time in [24].

−∫_{0}^{y} \frac{∂z_{m}(x, t)}{∂t}γ_{m}(x, t) dt+∫_{0}^{x}∫_{0}^{y} f(s, t)[z_{m}(s, t)−z(s, t)] ds dt. (2.21)

On the other hand, by (2.14) and (2.15) there exists a sequence of positive numbers (ε_{m})_{m=1}^{∞} such that

γ_{m}(x, y)| < ε_{m}, |z_{m}(x, y) − z(x, y)| < ε_{m} for (x, y) ∈ D_{ab}, (2.22)

and

lim_{m→∞} ε_{m} = 0.  (2.23)

Taking into account (2.18) and (2.22), from (2.21) we find

|δ_{m}(x, y)| < ρε_{m} for (x, y) ∈ D_{ab} (m = 1, 2, …).

Hence, by virtue of equalities (2.20) and (2.23), we obtain condition (2.17).

Lemma 2.5. Let f ∈ L(D_{ab}), z ∈ C(D_{ab}) and along with (2.13)–(2.15) the conditions

z_{m}(x, y) = z_{0m}(x, y) + \overline{z}_{m}(x, y), \overline{z}_{m} ∈ C̃(D_{ab}) (m = 1, 2, …), (2.24)

lim sup_{m→∞} ( ∫_{0}^{a} |\frac{∂\overline{z}_{m}(s, 0)}{∂s}| ds + ∫_{0}^{b} |\frac{∂\overline{z}_{m}(0, t)}{∂t}| dt +

+ ∫_{0}^{a} ∫_{0}^{b} |\frac{∂^{2}\overline{z}_{m}(s, t)}{∂s∂t}| ds dt) < +∞, (2.25)
lim_{m→∞} z_{0m}(x, y) = z_{0}(x, y) uniformly on D_{ab}  (2.26)

hold, where

z_{0} ∈ C̃(D_{ab}).  (2.27)

Moreover, let there exist a nonnegative function f^{∗} ∈ L(D_{ab}) such that the inequality

|f_{m}(x, y)| ≤ f^{∗}(x, y) (m = 1, 2, …)  (2.28)

holds almost everywhere in D_{ab}. Then condition (2.17) is fulfilled.

Proof. By conditions (2.13), (2.15), (2.24) and (2.26)

lim_{m→∞} \overline{z}_{m}(x, y) = \overline{z}(x, y) uniformly on D_{ab},

458 T. KIGURADZE

where \overline{z} ∈ C(D_{ab}). But by Lemma 2.4 the latter condition, along with (2.14) and (2.25), yields

lim_{m→∞} ∫_{0}^{x} ∫_{0}^{y}[f_{m}(s, t)\overline{z}_{m}(s, t) − f(s, t)\overline{z}(s, t)] ds dt = 0 uniformly on D_{ab}.

Thus to complete the proof we have only to show that

lim_{m→∞} δ_{0m}(x, y) = 0 uniformly on D_{ab},  (2.29)

where

δ_{0m}(x, y) = ∫_{0}^{x} ∫_{0}^{y}[f_{m}(s, t)z_{0m}(s, t) − f(s, t)z_{0}(s, t)] ds dt.

In view of (2.19) and (2.27)

δ_{0m}(x, y) = ∫_{0}^{x} ∫_{0}^{y} \frac{∂^{2}γ_{m}(s, t)}{∂s∂t}z_{0}(s, t) ds dt +

+ ∫_{0}^{x} ∫_{0}^{y} f_{m}(s, t)[z_{0m}(s, t) − z_{0}(s, t)] ds dt = γ_{m}(x, y)z_{0}(x, y) +

+ ∫_{0}^{x} ∫_{0}^{y} \frac{∂^{2}z_{0}(s, t)}{∂s∂t}γ_{m}(s, t) ds dt − ∫_{0}^{x} \frac{∂z_{0}(s, y)}{∂s}γ_{m}(s, y) ds −

− ∫_{0}^{y} \frac{∂z_{0}(x, t)}{∂t}γ_{m}(x, t) dt + ∫_{0}^{x} ∫_{0}^{y} f_{m}(s, t)[z_{0m}(s, t) − z_{0}(s, t)] ds dt,

which with regard to (2.28) gives

|δ_{0m}(x, y)| ≤ ρε_{m} for (x, y) ∈ D_{ab} (m = 1, 2, …),  (2.30)

where

ρ = max \{|z_{0}(x, y)| + ∫_{0}^{a} |\frac{∂z_{0}(s, y)}{∂s}| ds + ∫_{0}^{b} |\frac{∂z_{0}(x, t)}{∂t}| dt : (x, y) ∈ D_{ab}\} +

+ ∫_{0}^{a} ∫_{0}^{b} |\frac{∂^{2}z_{0}(s, t)}{∂s∂t}| ds dt + ∫_{0}^{a} ∫_{0}^{b} f^{∗}(s, t) ds dt,

ε_{m} = max{|γ_{m}(x, y)| + |z_{0m}(x, y) − z_{0}(x, y)| : (x, y) ∈ D_{ab}}.

But by (2.14) and (2.26) equality (2.23) holds. (2.23) and (2.30) yield condition (2.29).

Lemma 2.6. Let

v_{m}(x, y) = ∫_{0}^{a} ∫_{0}^{b} g_{1}(x, s)g_{2}(y, t)f_{m}(s, t)z_{m}(s, t) ds dt,  (2.31)

v(x, y) = ∫_{0}^{a} ∫_{0}^{b} g_{1}(x, s)g_{2}(y, t)f(s, t)z(s, t) ds dt,  (2.32)

where g_{1} and g_{2} are the functions given by (2.3) and (2.4), and f_{m}, z_{m} (m = 1, 2, …), f and z are the functions satisfying either the conditions of Lemma 2.4, or the conditions of Lemma 2.5. Then

lim_{m→∞} ∥v_{m} − v∥_{C^{1,1}} = 0.  (2.33)

Proof. By (2.31) and (2.32)

v_{m} ∈ C̃^{1}(D_{ab}) (m = 1, 2, …), v ∈ C̃^{1}(D_{ab})

and the equalities

\frac{∂^{j+k}v_{m}(x, y)}{∂x^{j}∂y^{k}} =

= ∫_{0}^{a} ∫_{0}^{b} g_{1j}(x, s)g_{2k}(y, t)f_{m}(s, t)z_{m}(s, t) ds dt (j, k = 0, 1), (2.34)

\frac{∂^{j+k}v(x, y)}{∂x^{j}∂y^{k}} = ∫_{0}^{a} ∫_{0}^{b} g_{1j}(x, s)g_{2k}(y, t)f(s, t)z(s, t) ds dt (j, k = 0, 1), (2.35)

hold, where g_{10}(x, s) = g_{1}(x, s), g_{11}(x, s) = \frac{∂g_{1}(x,s)}{∂x}, g_{20}(y, t) = g_{2}(y, t) and g_{21}(y, t) = \frac{∂g_{1}(y,t)}{∂y}.

Let σ(x, y) be an arbitrary polynomial of two variables. If the functions f_{m}, z_{m} (m = 1, 2, …), f and z satisfy the conditions of Lemma 2.4 (Lemma 2.5), then the functions f_{m}, σz_{m} (m = 1, 2, …), f and σz satisfy the conditions of this lemma too. Therefore for arbitrary fixed a_{0} ∈ [0, a] and b_{0} ∈ [0, b] we have

lim_{m→∞} ∫_{a_{0}}^{x} ∫_{b_{0}}^{y} σ(s, t)[f_{m}(s, t)z_{m}(s, t) − f(s, t)z(s, t)] ds dt = 0

uniformly on D_{ab}. Hence by (2.34) and (2.35) we conclude that

lim_{m→+∞} \frac{∂^{j+k}v_{m}(x, y)}{∂x^{j}∂y^{k}} = lim_{m→+∞} \frac{∂^{j+k}v(x, y)}{∂x^{j}∂y^{k}} (j, k = 0, 1) uniformly on D_{ab}.

Therefore condition (2.33) holds.

When z_{m}(x, y) ≡ z(x, y) ≡ 1 (m = 1, 2, …) Lemma 2.6 takes the form of

Lemma 2.7. Let

v_{m}(x, y) = ∫_{0}^{a} ∫_{0}^{b} g_{1}(x, s)g_{2}(y, t)f_{m}(s, t) ds dt (m = 1, 2, …),

v(x, y) = ∫_{0}^{a} ∫_{0}^{b} g_{1}(x, s)g_{2}(y, t)f(s, t) ds dt,

where f_{m} ∈ L(D_{ab}) (m = 1, 2, …) and f ∈ L(D_{ab}) are the functions satisfying condition (2.14). Then condition (2.33) holds.

2.4. Lemma on Sequences of Solutions of Problems (1.5_{m}),(1.6_{m}) (m=1,2, …).

Lemma 2.8. Let the functions p_{km}, q_{m} and φ_{m} (k = 0, 1, 2, 3; m = 1, 2, …) satisfy the conditions (i)–(iii) and let, for every natural m, problem (1.5_{m}), (1.6_{m}) have a solution u_{m}. Moreover, let

ρ = sup{∥u_{m}∥_{C^{1,1}} : m = 1, 2, … } < +∞.  (2.36)

Then there exist a subsequence (u_{m_{n}})_{n=1}^{∞} of the sequence (u_{m})_{m=1}^{∞} and a solution u of problem (1.1), (1.2) such that

lim_{n→∞} ∥u_{m_{n}} − u∥_{C^{1,1}} = 0.  (2.37)

Proof. By Lemma 2.2 and the conditions (ii) and (2.36) for any natural m the function u_{m} admits the representation

u_{m}(x, y) = v_{m}(x, y) + \overline{v}_{m}(x, y),  (2.38)

where

v_{m}(x, y) = η(φ_{m})(x, y) + ∫_{0}^{a} ∫_{0}^{b} g_{1}(x, s)g_{2}(y, t)q_{m}(s, t) ds dt, (2.39)

\overline{v}_{m}(x, y) = ∫_{0}^{a} ∫_{0}^{b} g_{1}(x, s)g_{2}(y, t)Q_{m}(s, t) ds dt,  (2.40)

ON THE CORRECTNESS OF THE DIRICHLET PROBLEM 461

Q_{m}(x, y) = p_{0m}(x, y)u_{m}(x, y) + p_{1m}(x, y)\frac{∂u_{m}(x, y)}{∂x} +

+p_{2m}(x, y)\frac{∂u_{m}(x, y)}{∂y} + p_{3m}(x, y)\frac{∂^{2}u_{m}(x, y)}{∂x∂y}  (2.41)

and

|Q_{m}(x, y)| ≤ ρp(x, y) almost everywhere in D_{ab}.  (2.42)

According to the conditions (i), (iii) and Lemma 2.7

lim_{m→∞} ∥v_{m} − v∥_{C^{1,1}} = 0,  (2.43)

where

v(x, y) = η(φ)(x, y) + ∫_{0}^{a} ∫_{0}^{b} g_{1}(x, s)g_{2}(y, t)q(s, t) ds dt  (2.44)

and

v ∈ C̃^{1}(D_{ab}).  (2.45)

By (2.40) and (2.42) for any natural m we have

\overline{v}_{m} ∈ C̃^{1}(D_{ab}), ∥\overline{v}_{m}∥_{C^{1,1}} ≤ (ab + a + b + 1)ρ ∫_{0}^{a} ∫_{0}^{b} p(s, t)ds dt, (2.46)
|\frac{∂^{3}\overline{v}_{m}(x, y)}{∂x^{2}∂y}| ≤ h_{1}(x) for x ∈ [0, a]\I_{1y}, y ∈ [0, b],  (2.47)
|\frac{∂^{3}\overline{v}_{m}(x, y)}{∂x∂y^{2}} | ≤ h_{2}(y) for x ∈ [0, a], y ∈ [0, b]\I_{2x},  (2.48)

and

|\frac{∂^{4}\overline{v}_{m}(x, y)}{∂x^{2}∂y^{2}} | ≤ ρp(x, y) almost everywhere in D_{ab},  (2.49)

where

h_{1}(x) = ρ ∫_{0}^{b} p(x, t) dt, h_{2}(y) = ρ ∫_{0}^{a} p(s, y) ds,

and I_{1y} ⊂ [0, a] and I_{2x} ⊂ [0, b] are some sets of zero measure.

By virtue of Lemma 2.1 conditions (2.46)–(2.49) guarantee that the sequence (\overline{v}_{m})_{m=1}^{∞} is a relative compact in the topology of the space C^{1,1}(D_{ab}).

Therefore there exist a function \overline{v}_{0} ∈ C^{1,1}(D_{ab}) and a subsequence (\overline{v}_{m_{n}})_{n=1}^{∞} of this sequence such that

lim_{m→∞} ∥\overline{v}_{m_{n}} − \overline{v}∥_{C^{1,1}} = 0.

This and (2.43) imply condition (2.37), where

u(x, y) = v(x, y) + \overline{v}(x, y)

and u ∈ C^{1,1}(D_{ab}). By Lemma 2.2, to complete the proof, it remains to show that u is a solution of equation (2.5).

Put

z_{0n}(x, y) = u_{m_{n}}(x, y), z_{1n}(x, y) = \frac{∂z_{0n}(x, y)}{∂x}, z_{2n}(x, y) = \frac{∂z_{0n}(x, y)}{∂y},

z_{3n}(x, y) = \frac{∂^{2}z_{0n}(x, y)}{∂x∂y};

w_{0n}(x, y) = v_{m_{n}}(x, y), w_{1n}(x, y) = \frac{∂w_{0n}(x, y)}{∂x}, w_{2n}(x, y) = \frac{∂w_{0n}(x, y)}{∂y},

w_{3n}(x, y) = \frac{∂^{2}w_{0n}(x, y)}{∂x∂y};

\overline{w}_{0n}(x, y) = \overline{v}_{m_{n}}(x, y), \overline{w}_{1n}(x, y) = \frac{∂\overline{w}_{0n}(x, y)}{∂x}, \overline{w}_{2n}(x, y) = \frac{∂\overline{w}_{0n}(x, y)}{∂y},

\overline{w}_{3n}(x, y) = \frac{∂^{2}\overline{w}_{0n}(x, y)}{∂x∂y};

w_{0}(x, y) = v(x, y), w_{1}(x, y) = \frac{∂v(x, y)}{∂x}, w_{2}(x, y) = \frac{∂v(x, y)}{∂y},

w_{3}(x, y) = \frac{∂^{2}v(x, y)}{∂x∂y}.

Then by (2.37), (2.43) and (2.45) we have

lim_{n→∞} z_{0n}(x, y) = u(x, y), lim_{n→∞} z_{1n}(x, y) = \frac{∂u(x, y)}{∂x},

lim_{n→∞} z_{2n}(x, y) = \frac{∂u(x, y)}{∂y}, lim_{n→∞} z_{3n}(x, y) = \frac{∂^{2}u(x, y)}{∂x∂y},

(2.50)

lim_{n→∞} w_{kn}(x, y) = w_{k}(x, y)

uniformly on D_{ab} and

w_{k} ∈ C̃(D_{ab}) (k = 0, 1, 2, 3).  (2.51)

On the other hand, it follows from (2.46)–(2.49) that

lim sup_{n→∞} ( ∫_{0}^{a} |\frac{∂\overline{w}_{kn}(s, 0)}{∂s}| ds + ∫_{0}^{b} |\frac{∂\overline{w}_{kn}(0, t)}{∂t}| dt +

+ ∫_{0}^{a} ∫_{0}^{b} |\frac{∂^{2}\overline{w}_{kn}(s, t)}{∂s∂t}| ds dt) < +∞ (k = 0, 1, 2, 3).  (2.52)

According to (2.38), (2.40) and (2.41) we have

z_{kn}(x, y) = w_{kn}(x, y) + \overline{w}_{kn}(x, y) (k = 0, 1, 2, 3),  (2.53)
z_{0n}(x, y) = w_{0n}(x, y) + ∑_{k=0}^{3} ∫_{0}^{a} ∫_{0}^{b} g_{1}(x, s)g_{2}(y, t)p_{kn}(s, t)z_{kn}(s, t)dsdt.(2.54)

But by Lemma 2.6 and the conditions (i), (ii), and (2.50)–(2.53) from (2.54) we obtain

u(x, y) = v(x, y) + ∫_{0}^{a} ∫_{0}^{b} g_{1}(x, s)g_{2}(y, t)Q(u)(s, t) ds dt,

where Q is the operator given by equality (2.6). Taking this and equality (2.44) into account, it becomes clear that u is the solution of equation (2.5).

2.5. Lemma on the Stability of the Unique Solvability Property of Problem (1.1_{0}),(1.2_{0}).

Lemma 2.9. Let problem (1.1_{0}), (1.2_{0}) have only the trivial solution. Then for any nonnegative function γ ∈ L(D_{ab}) there exists a positive number δ_{0} such that for arbitrary functions \overline{p}_{k} ∈ L(D_{ab}) (k = 0, 1, 2, 3) satisfying the inequalities

|\overline{p}_{k}(x, y) − p_{k}(x, y)| ≤ γ(x, y) almost for every (x, y) ∈ D_{ab} (k = 0, 1, 2, 3),

| ∫_{0}^{x} ∫_{0}^{y}[\overline{p}_{k}(s, t) − p_{k}(s, t)] ds dt| ≤ δ for (x, y) ∈ D_{ab} (k = 0, 1, 2, 3),

the differential equation

\frac{∂^{4}u}{∂x^{2}∂y^{2}} = \overline{p}_{0}(x, y)u + \overline{p}_{1}(x, y)\frac{∂u}{∂x} + \overline{p}_{2}(x, y)\frac{∂u}{∂y} + \overline{p}_{3}(x, y)\frac{∂^{2}u}{∂x∂y}

has only the trivial solution satisfying the boundary conditions (1.2_{0}).

Proof. Assume the contrary, i.e., that the lemma is not true. Then there exist a nonnegative function γ ∈ L(D_{ab}) and a sequence of summable in D_{ab} functions (p_{km})_{m=1}^{∞} (k = 0, 1, 2, 3) such that

|p_{km}(x, y) − p_{k}(x, y)| ≤ γ(x, y) almost for every (x, y) ∈ D_{ab}
(k = 0, 1, 2, 3; m = 1, 2, …),

| ∫_{0}^{x} ∫_{0}^{y}[p_{k}(s, t) − p_{k}(s, t)] ds dt| ≤ \frac{1}{m} for (x, y) ∈ D_{ab}

(k = 0, 1, 2, 3; m = 1, 2, …)

and for an arbitrary natural m the differential equation

\frac{∂^{4}u}{∂x^{2}∂y^{2}} = p_{0m}(x, y)u + p_{1m}(x, y)\frac{∂u}{∂x} + p_{2m}(x, y)\frac{∂u}{∂y} + p_{3m}(x, y)\frac{∂^{2}u}{∂x∂y}

has a solution u_{m} such that u_{m}(x, y) = 0 for (x, y) ∈ Γ_{ab} and

∥u_{m}∥_{C^{1,1}} = 1.  (2.55)

On the other hand, by Lemma 2.8 the sequence (u_{m})_{m=1}^{∞} contains a subsequence (u_{m_{n}})_{n=1}^{∞} converging to the solution of problem (1.1_{0}), (1.2_{0}) in the norm of the space C^{1,1}. But problem (1.1_{0}), (1.2_{0}) has only the trivial solution. Therefore

lim_{n→∞} ∥u_{m_{n}}∥_{C^{1,1}} = 0,

which contradicts condition (2.55). The obtained contradiction proves the lemma.

2.6. On an Integral Identity for Functions from the Space C̃^{1}(D_{ab}) Satisfying the Boundary Condition (1.2_{0}).

Lemma 2.10. If the function u ∈ C̃^{1}(D_{ab}) satisfies the boundary condition (1.2_{0}), then

∫_{0}^{a} ∫_{0}^{b} u(s, t)\frac{∂^{4}u(s, t)}{∂s^{2}∂t^{2}} ds dt = ∫_{0}^{a} ∫_{0}^{b} [\frac{∂^{2}u(s, t)}{∂s∂t}]^{2} ds dt.  (2.56)

Proof. By the formula of integration by parts and condition (1.2_{0}) we have

∫_{0}^{b} u(s, t)\frac{∂^{4}u(s, t)}{∂s^{2}∂t^{2}} dt = − ∫_{0}^{b} \frac{∂u(s, t)}{∂t} \frac{∂^{3}u(s, t)}{∂s^{2}∂t} dt  (2.57)

and

∫_{0}^{a} \frac{∂u(s, t)}{∂t} \frac{∂^{3}u(s, t)}{∂s^{2}∂t} ds = − ∫_{0}^{a} [\frac{∂^{2}u(s, t)}{∂s∂t}]^{2} ds.  (2.58)

almost everywhere in [0, a] and [0, b].

Now, by integrating (2.57) from 0 to a and taking (2.58) into account we obtain equality (2.56).

2.7. On an Analogue of Wirtinger’s Lemma for Functions of Two Variables.

Lemma 2.11. If a function u ∈ C^{1,1}(D_{ab}) satisfies the boundary condition (1.2_{0}), then

∫_{0}^{a} ∫_{0}^{b} [\frac{∂u(s, t)}{∂s}]^{2} ds dt ≤ \frac{b^{2}}{π^{2}} ∫_{0}^{a} ∫_{0}^{b} [\frac{∂^{2}u(s, t)}{∂s∂t}]^{2} ds dt,  (2.59)

∫_{0}^{a} ∫_{0}^{b} [\frac{∂u(s, t)}{∂t}]^{2} ds dt ≤ \frac{a^{2}}{π^{2}} ∫_{0}^{a} ∫_{0}^{b} [\frac{∂^{2}u(s, t)}{∂s∂t}]^{2} ds dt,  (2.60)

∫_{0}^{a} ∫_{0}^{b} u^{2}(s, t) ds dt ≤ \frac{a^{2}b^{2}}{π^{4}} ∫_{0}^{a} ∫_{0}^{b} [\frac{∂^{2}u(s, t)}{∂s∂t}]^{2} ds dt.  (2.61)

Proof. By Wirtinger’s lemma (see [25], Lemma 257) and condition (1.2_{0}) we have the inequalities

∫_{0}^{b} [\frac{∂u(s, t)}{∂s}]^{2} dt ≤ \frac{b^{2}}{π^{2}} ∫_{0}^{b} [\frac{∂^{2}u(s, t)}{∂s∂t}]^{2} dt,  (2.62)

∫_{0}^{a} [\frac{∂u(s, t)}{∂t}]^{2} ds ≤ \frac{a^{2}}{π^{2}} ∫_{0}^{a} [\frac{∂^{2}u(s, t)}{∂s∂t}]^{2} ds,  (2.63)

∫_{0}^{b} u^{2}(s, t) dt ≤ \frac{b^{2}}{π^{2}} ∫_{0}^{b} [\frac{∂u(s, t)}{∂t}]^{2} dt.  (2.64)

If we integrate (2.62) and (2.63) from 0 to a and from 0 to b, respectively, then we get inequalities (2.59) and (2.60). If we integrate (2.64) from 0 to a, then with regard to (2.60) we obtain inequality (2.61).

§ 3. Proofs of the Main Results

Proof of Theorem 1.1′. By virtue of Lemmas 2.3 and 2.9 problem (1.1), (1.2) is uniquely solvable, and beginning from some sufficiently large m_{0} problem (1.1_{m}), (1.2_{m}) is also uniquely solvable. Denote the solutions of these problems by u and u_{m}, respectively. Our goal is to prove that

lim_{m→∞} ∥u_{m} − u∥_{C^{1,1}} = 0.  (3.1)

Assume the opposite, i.e., that condition (3.1) does not hold. Then without loss of generality it can be assumed that

∥u_{m} − u∥_{C^{1,1}} ≥ ε (m = m_{0}, m_{0} + 1, …),  (3.2)

where ε is a positive constant independent of m. There exist two possibilities: either (u_{m})_{m=1}^{∞} satisfies condition (2.36), or

lim sup_{m→∞} ∥u_{m}∥_{C^{1,1}} = +∞.  (3.3)

Assume first that (2.36) holds. Then by Lemma 2.8 there exists a subsequence (u_{m_{n}})_{n=1}^{∞} of the sequence (u_{m})_{m=1}^{∞} satisfying condition (2.37). But this is impossible on account of condition (3.2).

To complete the proof, we have to show that the assumption that condition (3.3) is fulfilled leads us to a contradiction.

Choose a subsequence (u_{m_{n}})_{n=1}^{∞} from (u_{m})_{m=1}^{∞} such that

ρ_{n} = ∥u_{m_{n}}∥_{C^{1,1}} > 0 (n = 1, 2, …), lim_{n→∞} ρ_{n} = +∞.

Put v_{n}(x, y) = \frac{1}{ρ_{n}} u_{m_{n}}(x, y). Then

∥v_{n}∥_{C^{1,1}} = 1 (n = 1, 2, …)  (3.4)

and for every natural n the function v_{n} is a solution of the problem

\frac{∂^{4}v}{∂x^{2}∂y^{2}} = p_{0m_{n}}(x, y)v + p_{1m_{n}}(x, y)\frac{∂v}{∂x} + p_{2m_{n}}(x, y)\frac{∂v}{∂y} +
+p_{3m_{n}}(x, y)\frac{∂^{2}v}{∂x∂y} + q̃_{n}(x, y),
v(x, y) = \overline{φ}_{n}(x, y) for (x, y) ∈ Γ_{ab},

where q̃_{n}(x, y) = \frac{1}{ρ_{n}} q_{m_{n}}(x, y), φ̃_{n}(x, y) = \frac{1}{ρ_{n}} φ_{m_{n}}(x, y). Moreover,

lim_{n→∞} ∫_{0}^{x} ∫_{0}^{y} q̃_{n}(s, t) ds dt = 0 uniformly on D_{ab}  (3.5)

and the equalities

lim_{n→∞} \frac{∂^{k}φ̃_{n}(x, 0)}{∂x^{k}} = 0, lim_{n→∞} \frac{∂^{k}φ̃_{n}(x, b)}{∂x^{k}} = 0 (k = 0, 1),  (3.6)
lim_{n→∞} \frac{∂^{k}φ̃_{n}(0, y)}{∂y^{k}} = 0, lim_{n→∞} \frac{∂^{k}φ̃_{n}(a, y)}{∂y^{k}} = 0 (k = 0, 1)  (3.7)

hold uniformly on [0, a] and [0, b], respectively.

By Lemma 2.8 the conditions (i), (ii), (3.5)–(3.7) and the unique solvability of the homogeneous problem (1.1_{0}), (1.2_{0}) guarantee that there exists a subsequence (v_{n_{j}})_{j=1}^{∞} of the sequence (v_{n})_{n=1}^{∞} such that

lim_{j→∞} ∥v_{n_{j}}∥_{C^{1,1}} = 0.

But this contradicts condition (3.4). The obtained contradiction proves the theorem. □

Proof of Theorem 1.1. Assume that the theorem is not true. Then by virtue of Lemmas 2.3 and 2.9 there exist a positive number ε, a nonnegative function γ ∈ L(D_{ab}) and sequences of functions (p_{km})_{m=1}^{∞} (k = 0, 1, 2, 3), (q_{m})_{m=1}^{∞} and (φ_{m})_{m=1}^{∞} such that the conditions (i)–(iii) hold, where

p(x, y) = ∑_{k=0}^{3} |p_{k}(x, y)| + γ(x, y),

and for any natural m problem (1.1_{m}), (1.2_{m}) has a unique solution u_{m} satisfying the inequality

∥u_{m} − u∥_{C^{1,1}} > ε.

But this is impossible, since equality (3.1) is valid by Theorem 1.1′. □

Proof of Theorem 1.2. By Theorem 1.1 we have only to show that in the conditions of Theorem 1.2 the homogeneous problem (1.1_{0}), (1.2_{0}) has only the trivial solution. Assume the opposite, i.e., that this problem has a nontrivial solution u. Then according to Lemma 2.10 we have

∫_{0}^{a} ∫_{0}^{b} [\frac{∂^{2}u(s, t)}{∂s∂t}]^{2} ds dt = ∫_{0}^{a} ∫_{0}^{b} p_{0}(s, t)u^{2}(s, t) ds dt +

+ ∫_{0}^{a} ∫_{0}^{b} p_{1}(s, t)u(s, t)\frac{∂u(s, t)}{∂s} ds dt + ∫_{0}^{a} ∫_{0}^{b} p_{2}(s, t)u(s, t)\frac{∂u(s, t)}{∂t} ds dt +

+ ∫_{0}^{a} ∫_{0}^{b} p_{3}(s, t)u(s, t)\frac{∂^{2}u(s, t)}{∂s∂t} ds dt.  (3.8)

468 T. KIGURADZE

Applying the Schwartz inequality and condition (1.7), we find

∫_{0}^{a} ∫_{0}^{b} p_{1}(s, t)u(s, t)\frac{∂u(s, t)}{∂s}dsdt ≤ ∫_{0}^{a} ∫_{0}^{b} p_{10}(s, t)u(s, t)\frac{∂u(s, t)}{∂s}dsdt + l_{1}ρ_{0}ρ_{1},

∫_{0}^{a} ∫_{0}^{b} p_{2}(s, t)u(s, t)\frac{∂u(s, t)}{∂t}dsdt ≤ ∫_{0}^{a} ∫_{0}^{b} p_{20}(s, t)u(s, t)\frac{∂u(s, t)}{∂t}dsdt + l_{2}ρ_{0}ρ_{2},

∫_{0}^{a} ∫_{0}^{b} p_{3}(s, t)u(s, t)\frac{∂^{2}u(s, t)}{∂s∂t} ds dt ≤ l_{3}ρ_{0}ρ_{3},

where ρ_{0}, ρ_{1}, ρ_{2} and ρ_{3} are the positive constants given by the equalities

ρ_{0}^{2} = ∫_{0}^{a} ∫_{0}^{b} u^{2}(s, t) ds dt, ρ_{1}^{2} = ∫_{0}^{a} ∫_{0}^{b} [\frac{∂u(s, t)}{∂s}]^{2} ds dt,

ρ_{2}^{2} = ∫_{0}^{a} ∫_{0}^{b} [\frac{∂u(s, t)}{∂t}]^{2} ds dt, ρ_{3}^{2} = ∫_{0}^{a} ∫_{0}^{b} [\frac{∂^{2}u(s, t)}{∂s∂t}]^{2} ds dt.

On the other hand, according to the formula of integration by parts and condition (1.2_{0}) we have

∫_{0}^{a} ∫_{0}^{b} p_{10}(s, t)u(s, t)\frac{∂u(s, t)}{∂s} ds dt = −\frac{1}{2} ∫_{0}^{a} ∫_{0}^{b} \frac{∂p_{10}(s, t)}{∂s}u^{2}(s, t) ds dt,

∫_{0}^{a} ∫_{0}^{b} p_{20}(s, t)u(s, t)\frac{∂u(s, t)}{∂t} ds dt = −\frac{1}{2} ∫_{0}^{a} ∫_{0}^{b} \frac{∂p_{20}(s, t)}{∂t}u^{2}(s, t) ds dt.

Taking this and inequality (1.8) into account, from (3.8) we get

ρ_{3}^{2} ≤ ∫_{0}^{a} ∫_{0}^{b} [p_{0}(s, t) − \frac{1}{2} \frac{∂p_{10}(s, t)}{∂s} − \frac{1}{2} \frac{∂p_{20}(s, t)}{∂t}]u^{2}(s, t) ds dt +

+l_{1}ρ_{0}ρ_{1} + l_{2}ρ_{0}ρ_{2} + l_{3}ρ_{0}ρ_{3} ≤ l_{0}ρ_{0}^{2} + l_{1}ρ_{0}ρ_{1} + l_{2}ρ_{0}ρ_{2} + l_{3}ρ_{0}ρ_{3}. (3.9)

By Lemma 2.11 equalities (2.59)–(2.61) are valid, i.e.,

ρ_{1} ≤ \frac{b}{π} ρ_{3}, ρ_{2} ≤ \frac{a}{π} ρ_{3}, ρ_{0} ≤ \frac{ab}{π^{2}} ρ_{3}.  (3.10)

Hence for l_{0} ≥ 0 (3.9) implies

0 < ρ_{3}^{2} ≤ (\frac{a^{2}b^{2}}{π^{4}} l_{0} + \frac{ab^{2}}{π^{3}} l_{1} + \frac{a^{2}b}{π^{3}}l_{2} + \frac{ab}{π^{2}} l_{3})ρ_{3}^{2},

which contradicts inequality (1.9).

When l_{0} < 0, by (1.10) and (3.10), inequality (3.9) again gives the contradiction

ρ_{3}^{2} + |l_{0}|ρ_{0}^{2} ≤ (\frac{b}{π} l_{1} + \frac{a}{π} l_{2} + l_{3})ρ_{0}ρ_{3} < 2\sqrt{|l_{0}|}ρ_{0}ρ_{3} ≤ ρ_{3}^{2} + |l_{0}|ρ_{0}^{2}.

Acknowledgement

This work was supported by INTAS Grant 96–1060.

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(Received 12.06.1998)

Author’s address: I. Javakhishvili Tbilisi State University Faculty of Physics, Chair of Higher Mathematics 3, Chavchavadze Ave., Tbilisi 380028 Goergia