GEORGIAN MATHEMATICAL JOURNAL: Vol. 6, No. 6, 1999, 537-552

ON THE DIRICHLET PROBLEM IN A CHARACTERISTIC RECTANGLE FOR FOURTH ORDER LINEAR SINGULAR

HYPERBOLIC EQUATIONS

T. KIGURADZE

Abstract. In the rectangle D = (0, a) × (0, b) with the boundary Γ the Dirichlet problem

\frac{∂^{4}u}{∂x^{2}∂y^{2}} = p(x, y)u + q(x, y),

u(x, y) = 0 for (x, y) ∈ Γ

is considered, where p and q : D → ℝ are locally summable functions and may have nonintegrable singularities on Γ. The effective conditions guaranteeing the unique solvability of this problem and the stability of its solution with respect to small perturbations of the coefficients of the equation under consideration are established.

§ 1. Formulation of the Problem and Main Results

In the open rectangle D = (0, a) × (0, b) consider the linear hyperbolic equation

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

where p and q are real functions, Lebesgue summable on [δ, a − δ] × [δ, b − δ] for any small δ > 0. We do not exclude the case, where p and q are not summable on D and have singularities on the boundary of D. In this sense equation (1.1) is singular.

Let Γ be the boundary of D. In the present paper for equation (1.1) we study the homogeneous Dirichlet problem

u(x, y) = 0 for (x, y) ∈ Γ.  (1.2)

1991 Mathematics Subject Classification. 35L55. Key words and phrases. Fourth order linear singular hyperbolic equation, Dirichlet

problem, existence, uniqueness and stability of a solution.

537

1072-947X/99/1100-0537$16.00/0 \begin{matrix}c \\ ◯ \end{matrix} 1999 Plenum Publishing Corporation

In the regular case, i.e., when p and q are summable on D, problem (1.1), (1.2) was studied in [1].

Before formulating the main results, we introduce several notations. ℝ is the set of real numbers. \overline{D} = [0, a] × [0, b]. For any z ∈ ℝ set [z]_{+} = \frac{|z|+z}{2} . C(\overline{D}) is the space of continuous functions z : \overline{D} → ℝ. L_{loc}(D) is the space of functions z : D → ℝ which are Lebesgue summable

on [δ, a − δ] × [δ, b − δ] for any arbitrarily small δ > 0.

C̃_{loc}^{1,2}(D) is the space of functions z : D → ℝ, absolutely continuous on [δ, a − δ] × [δ, b − δ] for any arbitrarily small δ > 0 together with \frac{∂z}{∂x}, \frac{∂z}{∂y} and

\frac{∂^{2}z}{∂x∂y} and satisfying the condition ∫_{0}^{a} ∫_{0}^{b} [\frac{∂^{2}z(x,y)}{∂x∂y}]^{2} dx dy < +∞.

A function u ∈ C̃_{loc}^{1,2}(D) will be called a solution of equation (1.1) if it satisfies (1.1) almost everywhere in D.

A solution of problem (1.1), (1.2) will be sought in the class C̃_{loc}^{1,2}(D) ∩ C(\overline{D}).

Along with (1.1), we have to consider the equation

\frac{∂^{4}u}{∂x^{2}∂y^{2}} = \overline{p}(x, y)u + \overline{q}(x, y),  (1.3)

where \overline{p} and \overline{q} ∈ L_{loc}(D).

Definition 1.1. A solution of problem (1.1), (1.2) will be called stable with respect to small perturbations of the coefficients of equation (1.1) if there exist positive numbers δ and r such that for any \overline{p} and \overline{q} ∈ L_{loc}(D) satisfying the conditions

η_{1}(\overline{p} − p) =^{def} ∫_{0}^{a} ∫_{0}^{b} xy(a − x)(b − y)|\overline{p}(x, y) − p(x, y)| dx dy ≤ δ, (1.4)

η_{2}(\overline{q} − q) =^{def} ∫_{0}^{a} ∫_{0}^{b}[xy(a − x)(b − y)]^{\frac{1}{2}} |\overline{q}(x, y) − q(x, y)| dx dy < +∞, (1.5)

problem (1.3), (1.2) has a unique solution \overline{u} in C̃_{loc}^{1,2}(D) ∩ C(\overline{D}) and

[ ∫_{0}^{a} ∫_{0}^{b} (\frac{∂^{2}(\overline{u}(x, y) − u(x, y))}{∂x∂y})^{2} dx dy]^{\frac{1}{2}} < r[η_{1}(\overline{p} − p) + η_{2}(\overline{q} − q)]. (1.6)

∗For the definition of absolutely continuous functions in a rectangle see [2, §570] or

Remark 1.1. If inequality (1.6) holds, then, by Lemma 2.1 proved below, in the rectangle D the difference \overline{u} − u admits the estimate

|\overline{u}(x, y) − u(x, y)| ≤ 2r[xy(1 − \frac{x}{a})(1 − \frac{y}{b} )]^{\frac{1}{2}} [η_{1}(\overline{p} − p) + η_{2}(\overline{q} − q)].

Definition 1.2. We say that a function p ∈ L_{loc}(D) belongs to U(D) if there exists a number α ∈ [0, 1) such that for any function u ∈ C̃_{loc}^{1,2}(D) ∩ C(\overline{D}) satisfying the boundary condition (1.2), the estimate

∫_{0}^{a} ∫_{0}^{b}[p(x, y)]_{+}u^{2}(x, y) dx dy ≤ α ∫_{0}^{a} ∫_{0}^{b} (\frac{∂^{2}u(x, y)}{∂x∂y})^{2} dx dy (1.7)

is valid.

Theorem 1.1. Let

\begin{matrix}∫_{0}^{a} ∫_{0}^{b}[xy(a − x)(b − y)]^{\frac{3}{2}} |p(x, y)| dx dy < +∞, \\ ∫_{0}^{a} ∫_{0}^{b}[xy(a − x)(b − y)]^{\frac{1}{2}} |q(x, y)| dx dy < +∞, \end{matrix}  (1.8)

and p ∈ U(D). Then problem (1.1), (1.2) has a unique solution in C̃_{loc}^{1,2}(D)∩ C(\overline{D}), stable with respect to small perturbations of the coefficients of equation (1.1).

Theorem 1.2. Let conditions (1.8) be fulfilled and the inequality

p(x, y) ≤ \frac{λ_{0}(x, y) + λ_{1}}{xy(a − x)(b − y)} + \frac{λ_{2}}{x^{2}y^{2}(a − x)^{2}(b − y)^{2}}  (1.9)

hold almost everywhere in D, where λ_{0} is a nonnegative summable function, and λ_{1} and λ_{2} are nonnegative numbers such that

\frac{4}{ab} ∫_{0}^{a} ∫_{0}^{b} λ_{0}(x, y) dx dy + \frac{1}{4}λ_{1} + \frac{16}{a^{2}b^{2}} λ_{2} < 1.  (1.10)

Then the statement of Theorem 1.1 is valid.

As an example, consider the differential equation

\frac{∂^{4}u}{∂x^{2}∂y^{2}} = [\frac{l_{1}}{xy(a − x)(b − y)} + \frac{l_{2}}{x^{2}y^{2}(a − x)^{2}(b − y)^{2}} ]u +

+ l_{3}x^{μ_{1}}y^{μ_{2}}(a − x)^{ν_{1}}(b − y)^{ν_{2}},  (1.11)

where l_{i} (i = 1, 2, 3), μ_{j} and ν_{j} (j = 1, 2) are some real constants. By Theorem 1.2, if μ_{j} > −\frac{3}{2}, ν_{j} > −\frac{3}{2} (j = 1, 2) and

\frac{1}{4}[l_{1}]_{+} + \frac{16}{a^{2}b^{2}} [l_{2}]_{+} < 1,  (1.12)

then problem (1.11), (1.2) has a unique solution in C̃_{loc}^{1,2}(D) ∩ C(\overline{D}) and this solution is stable with respect to small perturbations of the coefficients of equation (1.11).

Note that condition (1.12) is sharp, since for l_{1} = 4, l_{2} = l_{3} = 0 problem (1.11), (1.2) has an infinite set of solutions. More precisely, for any c ∈ ℝ, the function u(x, y) = cxy(x−a)(y−b) is a solution of problem (1.11), (1.2).

§ 2. Auxiliary Statements

In this section, along with the notations introduced in §1, we shall make use of the following notations also.

A_{0}^{1,2} = \{u ∈ C̃_{loc}^{1,2}(D) ∩ C(\overline{D}) : u(x, y) = 0 for (x, y) ∈ Γ\}.

C̃^{1}(\overline{D}) is the space of functions z : \overline{D} → ℝ, absolutely continuous together with \frac{∂z}{∂x}, \frac{∂z}{∂y} and \frac{∂^{2}z}{∂x∂y}.

We introduce

Definition 2.1. Let α > 0. We say that a function p ∈ L_{loc}(D) belongs to U_{α}(D) if inequality (1.7) holds for any u ∈ A^{1,2}.

2.1. Properties of the functions from A_{0}^{1,2}.

Lemma 2.1. If u ∈ A^{1,2}, then

u^{2}(x, y) ≤ \frac{4}{ab}xy(a − x)(b − y)ρ^{2} for (x, y) ∈ D,  (2.1)

∫_{0}^{a} ∫_{0}^{b} \frac{u^{2}(x, y)}{xy(a − x)(b − y)} dx dy ≤ \frac{1}{4}ρ^{2},  (2.2)

∫_{0}^{a} ∫_{0}^{b} [\frac{u(x, y)}{xy(a − x)(b − y)}]^{2} dx dy ≤ \frac{16}{a^{2}b^{2}} ρ^{2},  (2.3)

where

ρ = [ ∫_{0}^{a} ∫_{0}^{b} (\frac{∂^{2}u(x, y)}{∂x∂y})^{2} dx dy]^{\frac{1}{2}} .  (2.4)

Proof. First, let us prove estimate (2.1). The condition u ∈ A^{1,2} yields the representations

u(x, y) = ∫_{ia}^{x} ∫_{jb}^{y} \frac{∂^{2}u(s, t)}{∂s∂t} ds dt (i, j = 0, 1) for (x, y) ∈ D.

Hence, by the Schwartz inequality and notation (2.4) it follows that

u^{2}(x, y) ≤ ρ^{2}|x − ia||y − jb| (i, j = 0, 1) for (x, y) ∈ D.

Therefore u^{2}(x, y) ≤ ρ^{2} min{x, a − x} min{y, b − y} for (x, y) ∈ D. But min\{x, a − x\} ≤ \frac{2}{a}x(a − x) and min\{y, b − y\} ≤ \frac{2}{b}y(b − y) for (x, y) ∈ D. Consequently, estimate (2.1) is valid.

Now pass to proving estimate (2.2). By Hardy–Littlewood theorem (see [4], Theorem 262), we have

∫_{0}^{b} \frac{u^{2}(x, y)}{y(b − y)} dy ≤ \frac{1}{2} ∫_{0}^{b} (\frac{∂u(x, y)}{∂y})^{2} dy,

∫_{0}^{a} \frac{1}{x(a − x)}(\frac{∂u(x, y)}{∂y})^{2} dx ≤ \frac{1}{2} ∫_{0}^{a} (\frac{∂^{2}u(x, y)}{∂x∂y})^{2} dx

almost everywhere in (0, a) and (0, b), respectively. Therefore

∫_{0}^{a} [ ∫_{0}^{b} \frac{u^{2}(x, y)}{x(a − x)y(b − y)} dy] dx ≤ ∫_{0}^{b} [ ∫_{0}^{a} \frac{1}{x(a − x)}(\frac{∂u(x, y)}{∂y})^{2} dx] dy ≤

≤ \frac{1}{4} ∫_{0}^{a} ∫_{0}^{b} (\frac{∂^{2}u(x, y)}{∂x∂y})^{2} dx dy = \frac{1}{4}ρ^{2}.

Consequently, estimate (2.2) is valid.

As for estimate (2.3), it follows from V. I. Levin’s inequality (see [5] or [4, D.79]). Indeed,

∫_{0}^{a} [ ∫_{0}^{b} [\frac{u(x, y)}{x(a − x)y(b − y)}]^{2} dy] dx ≤

≤ \frac{4}{b^{2}} ∫_{0}^{b} [ ∫_{0}^{a} \frac{1}{x^{2}(a − x)^{2}} (\frac{∂u(x, y)}{∂y}) dy] dx ≤

≤ \frac{16}{a^{2}b^{2}} ∫_{0}^{a} ∫_{0}^{b} (\frac{∂^{2}u(x, y)}{∂x∂y})^{2} dx dy = \frac{16}{a^{2}b^{2}} ρ^{2}.

Lemma 2.2. Let u ∈ A^{1,2}. Then there exist sequences (x_{ik})_{k=1}^{+∞} and (y_{ik})_{k=1}^{+∞} such that

0 < x_{1k} < x_{2k} < a, 0 < y_{1k} < y_{2k} < b (k = 1, 2, …),  (2.5)
lim_{k→+∞} x_{1k} = 0, lim_{k→+∞} x_{2k} = a, lim_{k→+∞} y_{1k} = 0, lim_{k→+∞} y_{2k} = b (2.6)

and

lim_{k→+∞} ∫_{x_{1k}}^{x_{2k}}∫_{y_{1k}}^{y_{2k}} \frac{∂^{4}u(x, y)}{∂x^{2}∂y^{2}}u(x, y) dx dy = ρ^{2},  (2.7)

where ρ is the number given by (2.4). Proof. Let

w(x, y) = \frac{∂^{2}u(x, y)}{∂x∂y}.  (2.8)

For any natural k set

α_{0k} = \frac{a}{2k + 4}, α_{k} = \frac{a}{k + 2}, a_{0k} = \frac{(k + 1)a}{k + 2}, a_{k} = \frac{(2k + 3)a}{2k + 4},

β_{0k} = \frac{b}{2k + 4}, β_{k} = \frac{b}{k + 2}, b_{0k} = \frac{(k + 1)b}{k + 2}, b_{k} = \frac{(2k + 3)b}{2k + 4},

w_{1k}(x) = ∫_{β_{0k}}^{b_{k}} w^{2}(x, y) dy, w_{2k}(y) = ∫_{α_{0k}}^{a_{k}} w^{2}(x, y) dx,  (2.9)
∫_{α_{0k}}^{α_{k}} w_{1k}(x) dx + ∫_{a_{0k}}^{a_{k}} w_{1k}(x) dx + ∫_{β_{0k}}^{β_{k}} w_{2k}(y) dy + ∫_{b_{0k}}^{b_{k}} w_{2k}(y) dy = ε_{k}. (2.10)

Then lim_{k→+∞} ε_{k} = 0.

In view of the continuity of w, w_{1k} and w_{2k}, for any natural k there exist

x_{1k} ∈ [α_{0k}, α_{k}], x_{2k} ∈ [a_{0k}, a_{k}], y_{1k} ∈ [β_{0k}, β_{k}], y_{2k} ∈ [b_{0k}, b_{k}] such that

w_{1k}(x_{1k}) = min{w_{1k}(x) : α_{0k} ≤ x ≤ α_{k}},

w_{1k}(x_{2k}) = min{w_{1k}(x) : a_{0k} ≤ x ≤ a_{k}},

\frac{a}{2k + 4}[w^{2}(x_{1k}, y_{1k}) + w^{2}(x_{2k}, y_{1k})] + w_{2k}(y_{1k}) =

= min \{\frac{a}{2k + 4}[w^{2}(x_{1k}, y) + w^{2}(x_{2k}, y)] + w_{2k}(y) : β_{0k} ≤ y ≤ β_{k}\},

\frac{a}{2k + 4}[w^{2}(x_{1k}, y_{2k}) + w^{2}(x_{2k}, y_{2k})] + w_{2k}(y_{2k}) =

= min \{\frac{a}{2k + 4}[w^{2}(x_{1k}, y) + w^{2}(x_{2k}, y)] + w_{2k}(y) : b_{0k} ≤ y ≤ b_{k}\}.

Then it is obvious that the sequences (x_{ik})_{k=1}^{+∞} and (y_{ik})_{k=1}^{+∞} (i = 1, 2) satisfy conditions (2.5) and (2.6). On the other hand, from (2.9) and (2.10) we have

ε_{k} ≥ \frac{a}{2k + 4}[w_{1k}(x_{1k}) + w_{1k}(x_{2k})] + ∫_{β_{0k}}^{β_{k}} w_{2k}(y) dy + ∫_{b_{0k}}^{b_{k}} w_{2k}(y) dy ≥

≥ ∫_{β_{0k}}^{β_{k}} [\frac{a}{2k + 4}w^{2}(x_{1k}, y) + \frac{a}{2k + 4}w^{2}(x_{2k}, y) + w_{2k}(y)] dy +

+ ∫_{b_{0k}}^{b_{k}} [\frac{a}{2k + 4}w^{2}(x_{1k}, y) + \frac{a}{2k + 4}w^{2}(x_{2k}, y) + w_{2k}(y)] dy ≥

≥ \frac{ab}{(2k + 4)^{2}} [w^{2}(x_{1k}, y_{1k}) + w^{2}(x_{2k}, y_{1k}) + w^{2}(x_{1k}, y_{2k}) + w^{2}(x_{2k}, y_{2k})] +

+\frac{b}{2k + 4}[w_{2k}(y_{1k}) + w_{2k}(y_{2k})] (k = 1, 2, …).

Therefore

w_{1k}(x_{ik})≤(k + 2)ε_{0k}, w_{2k}(y_{ik})≤(k + 2)ε_{0k} (i = 1, 2; k = 1, 2, …), (2.11)
|w(x_{ik}, y_{jk})| ≤ (k + 2)ε_{0k} (i, j = 1, 2; k = 1, 2, …),  (2.12)

where ε_{0k} = max \{\frac{2ε_{k}}{a} , \frac{2ε_{k}}{b} , 2(\frac{ε_{k}}{ab} )^{\frac{1}{2}} \} and

lim_{k→+∞} ε_{0k} = 0.  (2.13)

Moreover, inequality (2.1) implies that

|u(x_{ik}, y_{jk})| ≤ \frac{γ}{k + 2} (i, j = 1, 2; k = 1, 2, …),  (2.14)

where γ = 2ρ(ab)^{\frac{1}{2}} .

544 T. KIGURADZE

For every natural k consider the integral

I_{k} = ∫_{x_{1k}}^{x_{2k}}∫_{y_{1k}}^{y_{2k}} \frac{∂^{4}u(x, y)}{∂x^{2}∂y^{2}}u(x, y) dx dy.  (2.15)

By equality (2.8) and the formula of integration by parts, we have

I_{k} = ∫_{x_{1k}}^{x_{2k}}( ∫_{y_{1k}}^{y_{2k}} \frac{∂^{2}w(x, y)}{∂x∂y}u(x, y) dy) dx = ∫_{x_{1k}}^{x_{2k}}[\frac{∂w(x, y_{2k})}{∂x}u(x, y_{2k}) −

−\frac{∂w(x, y_{1k})}{∂x}u(x, y_{1k})] dx − ∫_{y_{1k}}^{y_{2k}}( ∫_{x_{1k}}^{x_{2k}} \frac{∂w(x, y)}{∂x} \frac{∂u(x, y)}{∂y} dx) dy =

= w(x_{2k}, y_{2k})u(x_{2k}, y_{2k}) − w(x_{1k}, y_{2k})u(x_{1k}, y_{2k}) − −w(x_{2k}, y_{1k})u(x_{2k}, y_{1k}) + w(x_{1k}, y_{1k})u(x_{1k}, y_{1k}) −

− ∫_{x_{1k}}^{x_{2k}}[w(x, y_{2k})\frac{∂u(x, y_{2k})}{∂x} − w(x, y_{1k})\frac{∂u(x, y_{1k})}{∂x}] dx −

− ∫_{y_{1k}}^{y_{2k}}[w(x_{2k}, y)\frac{∂u(x_{2k}, y)}{∂y} − w(x_{1k}, y)\frac{∂u(x_{1k}, y)}{∂y}] dy + I_{0k}, (2.16)

where

I_{0k} = ∫_{x_{1k}}^{x_{2k}}∫_{y_{1k}}^{y_{2k}}(\frac{∂^{2}u(x, y)}{∂x∂y})^{2} dx dy.

Moreover, as it follows from (2.4) and (2.6),

lim_{k→+∞} I_{0k} = ρ^{2}.  (2.17)

By virtue of the condition u ∈ A^{1,2} and equality (2.8) we have

\frac{∂u(x, y_{ik})}{∂x} = ∫_{(i−1)b}^{y_{ik}} w(x, y) dy, \frac{∂u(x_{ik}, y)}{∂y} = ∫_{(i−1)a}^{x_{ik}} w(x, y) dx (i = 1, 2).

If along with this we take into account equalities (2.4) and (2.9) and inequality (2.11), then we get

| ∫_{x_{1k}}^{x_{2k}}w(x, y_{ik})\frac{∂u(x, y_{ik})}{∂x} dx| = | ∫_{x_{1k}}^{x_{2k}}w(x, y_{ik})( ∫_{(i−1)b}^{y_{ik}} w(x, y) dy) dx| ≤

≤[ ∫_{x_{1k}}^{x_{2k}}w^{2}(x, y_{ik}) dx]^{\frac{1}{2}} [ ∫_{x_{1k}}^{x_{2k}}( ∫_{(i−1)b}^{y_{ik}} w(x, y) dy)^{2} dx]^{\frac{1}{2}} ≤[w_{2k}(y_{ik})]^{\frac{1}{2}} ×

×[\frac{b}{k + 2} ∫_{0}^{a} ∫_{0}^{b} w^{2}(x, y) dx dy]^{\frac{1}{2}} ≤(bε_{0k})^{\frac{1}{2}} ρ (i = 1, 2; k = 1, 2, …), (2.18)

| ∫_{y_{1k}}^{y_{2k}}w(x_{ik}, y)\frac{∂u(x_{ik}, y)}{∂y} dy| ≤ (aε_{0k})^{\frac{1}{2}} ρ (i = 1, 2; k = 1, 2, …). (2.19)

Using conditions (2.12)–(2.14), (2.18), and (2.19), from (2.16) we find

|I_{k} − I_{0k}| ≤ 4γε_{0k} + 2(bε_{0k})^{\frac{1}{2}} ρ + 2(aε_{0k})^{\frac{1}{2}} ρ → 0 for k → +∞.

Hence, according to (2.15) and (2.17), there follows equality (2.7).

2.2. On one property of the set U_{α}(D).

Lemma 2.3. Let α > 0, δ > 0, p ∈ U_{α}(D) and the function \overline{p} ∈ L_{loc}(D) satisfy the inequality

∫_{0}^{a} ∫_{0}^{b} xy(a − x)(b − y)[\overline{p}(x, y) − p(x, y)]_{+} dx dy ≤ δ.  (2.20)

Then

\overline{p} ∈ U_{β}(D),  (2.21)

where β = α + \frac{4}{ab}δ.

Proof. Let u be an arbitrary function from A^{1,2}, and ρ be the number given by (2.4). Then by Definition 2.1 and Lemma 2.1 inequalities (1.7) and (2.1) are valid. Moreover, if we take into account inequalities (2.20) and

[\overline{p}(x, y)]_{+} ≤ [p(x, y)]_{+} + [\overline{p}(x, y) − p(x, y)]_{+},

then we get

∫_{0}^{a} ∫_{0}^{b}[\overline{p}(x, y)]_{+}u^{2}(x, y) dx dy ≤ ∫_{0}^{a} ∫_{0}^{b}[p(x, y)]_{+}u^{2}(x, y) dx dy +

+ ∫_{0}^{a} ∫_{0}^{b}[\overline{p}(x, y) − p(x, y)]_{+}u^{2}(x, y) dx dy ≤ αρ^{2} +

+\frac{4ρ^{2}}{ab} ∫_{0}^{a} ∫_{0}^{b}xy(a − x)(b − y)[\overline{p}(x, y) − p(x, y)]_{+}dx dy ≤ (α+ \frac{4}{ab}δ)ρ^{2} =βρ^{2}.

Consequently, inclusion (2.21) is true.

2.3. Lemmas on a priori estimates.

Lemma 2.4. Let p ∈ U_{α}(D), where 0 < α < 1, and the function q ∈ L_{loc}(D) satisfy the condition

η_{2}(q) =^{def} ∫_{0}^{a} ∫_{0}^{b}[xy(a − x)(b − y)]^{\frac{1}{2}} |q(x, y)| dx dy < +∞. (2.22)

Moreover, if problem (1.1), (1.2) has a solution u ∈ C̃_{loc}^{1,2}(D) ∩ C(\overline{D}), then

ρ ≤ \frac{2}{(1 − α)\sqrt{ab}}η_{2}(q),  (2.23)

where ρ is a number given by (2.4).

Proof. By virtue of Definition 2.1 and Lemmas 2.1 and 2.2 the function u satisfies conditions (1.7) and (2.1), and there exist sequences (x_{ik})_{k=1}^{+∞} and (y_{ik})_{k=1}^{+∞} (i = 1, 2) satisfying conditions (2.5) and (2.6) such that equality (2.7) is true.

Multiply both sides of (1.1) by u(x, y) and integrate them over [x_{1k}, x_{2k}]× [y_{1k}, y_{2k}] for any natural k. Then with regard to (1.7),(2.1) and (2.22) we find

∫_{x_{1k}}^{x_{2k}}∫_{y_{1k}}^{y_{2k}}u(x, y)\frac{∂^{4}u(x, y)}{∂x^{2}∂y^{2}} dx dy = ∫_{x_{1k}}^{x_{2k}}∫_{y_{1k}}^{y_{2k}}p(x, y)u^{2}(x, y) dx dy +

+ ∫_{x_{1k}}^{x_{2k}}∫_{y_{1k}}^{y_{2k}}q(x, y)u(x, y) dx dy ≤ ∫_{0}^{a} ∫_{0}^{b}[p(x, y)]_{+}u^{2}(x, y) dx dy +

+ ∫_{0}^{a} ∫_{0}^{b} |q(x, y)||u(x, y)| dx dy ≤ αρ^{2} + \frac{2}{\sqrt{ab}}η_{2}(q)ρ (k = 1, 2, …).

If we pass in this inequality to the limit as k → +∞, then by (2.7) we get

ρ^{2} ≤ αρ^{2} + \frac{2}{\sqrt{ab}}η_{2}(q)ρ.

Consequently, estimate (2.23) is true.

Lemma 2.5. Let 0 < α < 1, δ > 0,

β = α + \frac{4}{ab}δ < 1,  (2.24)

and the functions p ∈ U_{α}(D), \overline{p}, q and \overline{q} ∈ L_{loc}(D) satisfy conditions (1.4) and (1.5). Moreover, let problem (1.1), (1.2) have a solution u ∈ C̃_{loc}^{1,2}(D) ∩ C(\overline{D}), and problem (1.3), (1.2) have a solution \overline{u} ∈ C̃_{loc}^{1,2}(D) ∩ C(\overline{D}). Then inequality (1.6) is valid, where

r = max \{\frac{4ρ}{(1 − β)ab}, \frac{2}{(1 − β)\sqrt{ab}}\}  (2.25)

and ρ is the number given by (2.4).

Proof. Note that by Lemma 2.3 inclusion (2.21) is true. Set

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

Then from (1.1)–(1.3) we have

\frac{∂^{4}v(x, y)}{∂x^{2}∂y^{2}} = \overline{p}(x, y)v(x, y) + [\overline{p}(x, y) − p(x, y)]u(x, y) + \overline{q}(x, y) − q(x, y),

v(x, y) = 0 for (x, y) ∈ Γ.

Hence, by Lemma 2.4 and conditions (2.21) and (2.24), there follows the estimate

ρ^{∗} =^{def} [ ∫_{0}^{a} ∫_{0}^{b} (\frac{∂^{2}v(x, y)}{∂x∂y})^{2} dx dy]^{\frac{1}{2}} ≤ \frac{2}{(1 − β)\sqrt{ab}}η_{2}(\overline{q} − q) +

+\frac{2}{(1 − β)\sqrt{ab}} ∫_{0}^{a} ∫_{0}^{b}[xy(a − x)(b − y)]^{\frac{1}{2}} |\overline{p}(x, y) − p(x, y)||u(x, y)| dx dy.

Now if we apply conditions (1.5) and (2.1), then it becomes clear that

ρ^{∗} ≤ \frac{4ρ}{(1 − β)ab}η_{1}(\overline{p} − p) + \frac{2}{(1 − β)\sqrt{ab}}η_{2}(\overline{q} − q).

Consequently, estimate (1.6) is true, where the constant r is given by (2.25).

2.4. Lemmas on the existence and uniqueness of solutions of problem (1.1),(1.2).

Lemma 2.6. If p ∈ U_{α}(D), where 0 < α < 1, then problem (1.1), (1.2) has at most one solution in C̃_{loc}^{1,2}(D) ∩ C(\overline{D}).

Proof. Let u_{i} ∈ C̃_{loc}^{1,2}(D) ∩ C(\overline{D}) (i = 1, 2) be arbitrary solutions of problem (1.1), (1.2). Set u(x, y) = u_{2}(x, y)−u_{1}(x, y). It is obvious that u is a solution of the homogeneous problem

\frac{∂^{4}u}{∂x^{2}∂y^{2}} = p(x, y)u  (2.26)

and u ∈ A^{1,2}. Hence by Lemma 2.4 it follows that \frac{∂^{2}u(x,y)}{∂x∂y} ≡ 0 and

u(x, y) ≡ ∫_{0}^{x} ∫_{0}^{y} \frac{∂^{2}u(s, t)}{∂s∂t} ≡ 0.

Consequently, u_{1}(x, y) ≡ u_{2}(x, y).

Lemma 2.7. If the functions p and q are summable on D and p ∈ U_{α}(D), where 0 < α < 1, then problem (1.1), (1.2) has one and only one solution in C̃_{loc}^{1,2}(D) ∩ C(\overline{D}).

Proof. By Lemma 2.6, problem (2.26), (1.2) has only the trivial solution in the space C̃_{loc}^{1,2}(D) ∩ C(\overline{D}). Consequently, this problem has only the trivial solution in C̃^{1}(\overline{D}), since

C̃^{1}(\overline{D}) ⊂ C̃_{loc}^{1,2}(D) ∩ C(\overline{D}).  (2.27)

But by Theorem 1.1 from [1] the summability of p and q on D and the unique solvability of problem (2.26),(1.2) in the space C̃^{1}(\overline{D}) guarantee the existence and uniqueness of a solution u ∈ C̃^{1}(\overline{D}) of problem (1.1),(1.2). Hence Lemma 2.6 and condition (2.27) imply that u is the unique solution of problem (1.1),(1.2) in C̃_{loc}^{1,2}(D) ∩ C(\overline{D}).

§ 3. Proofs of the Main Results

Proof of Theorem 1.1. Note that by Definitions 1.2 and 2.1 there exists α ∈ (0, 1) such that

p ∈ U_{α}(D).  (3.1)

For any natural m set

D_{m} = (\frac{a}{4m}, \frac{(4m − 1)a}{4m}) × (\frac{b}{4m}, \frac{(4m − 1)}{4m}),

p_{m}(x, y) = \begin{cases}p(x, y) & for (x, y) ∈ D_{m}, \\ 0 & for (x, y) ∈ D\D_{m}, \end{cases}

q_{m}(x, y) = \begin{cases}q(x, y) & for (x, y) ∈ D_{m}, \\ 0 & for (x, y) ∈ D\D_{m} \end{cases}

and consider the differential equation

\frac{∂^{4}u}{∂x^{2}∂y^{2}} = p_{m}(x, y)u + q(x, y).  (3.2_{m})

It is clear that p_{m} and q_{m} (m = 1, 2, …) are summable on D and the conditions

|p_{m}(x, y)| ≤ |p(x, y)|, |q_{m}(x, y)| ≤ |q(x, y)| (m = 1, 2, …), (3.3)
lim_{m→∞} p_{m}(x, y) = p(x, y), lim_{m→∞} q_{m}(x, y) = q(x, y)  (3.4)

hold almost everywhere in D. Moreover, by (3.1),

p_{m} ∈ U_{α}(D) (m = 1, 2, …).  (3.5)

By Lemmas 2.4 and 2.7, for any natural m problem (3.2_{m}), (1.2) has the unique solution u_{m} in C̃_{loc}^{1,2}(D) ∩ C(\overline{D}) and

ρ_{m} ≤ \frac{2}{(1 − α)\sqrt{ab}} ∫_{0}^{a} ∫_{0}^{b}[xy(a − x)(b − y)]^{\frac{1}{2}} |q_{m}(x, y)| dx dy,

where ρ_{m} = [ ∫_{0}^{a} ∫_{0}^{b} (\frac{∂^{2}u_{m}(x,y)}{∂x∂y})^{2} dx dy]^{\frac{1}{2}} . Hence, taking into account (1.8) and (3.3), we find

ρ_{m} ≤ γ (m = 1, 2, …),  (3.6)

where γ = \frac{2}{(1−α)\sqrt{ab}} ∫_{0}^{a} ∫_{0}^{b}[xy(a − x)(b − y)]^{\frac{1}{2}} |q(x, y)| dx dy.

By Lemma 2.1 and condition (3.6), we have

|u_{m}(x, y)| ≤ γ_{0}[xy(a − x)(b − y)]^{\frac{1}{2}} for (x, y) ∈ \overline{D} (m = 1, 2, …), (3.7)

\begin{matrix}|u_{m}(\overline{x}, \overline{y}) − u_{m}(x, y)| = | ∫_{x}^{\overline{x}} ∫_{0}^{\overline{y}} \frac{∂^{2}u_{m}(s, t)}{∂s∂t} ds dt + ∫_{0}^{x} ∫_{y}^{\overline{y}} \frac{∂^{2}u_{m}(s, t)}{∂s∂t} ds dt| ≤ \\ ≤ (\overline{y}|\overline{x} − x|)^{\frac{1}{2}} ρ_{m} + (x|\overline{y} − y|)^{\frac{1}{2}} ρ_{m} ≤ γ_{0}(|\overline{x} − x|)^{\frac{1}{2}} + |\overline{y} − y|)^{\frac{1}{2}} ) \\ for 0 ≤ x ≤ \overline{x} ≤ a, 0 ≤ y ≤ \overline{y} ≤ b (m = 1, 2, …), \end{matrix} (3.8)

where γ_{0} = max\{2(ab)^{− \frac{1}{2}} , a^{\frac{1}{2}} , b^{\frac{1}{2}} \}γ.

By the Arzela–Ascoli lemma, conditions (3.7) and (3.8) guarantee the existence of a subsequence (u_{m_{k}})_{k=1}^{+∞} of the sequence (u_{m})_{m=1}^{+∞}, uniformly convergent on \overline{D}. Set

lim_{k→+∞} u_{m_{k}}(x, y) = u(x, y).  (3.9)

Then from (3.7) we get

|u(x, y)| ≤ γ_{0}[xy(a − x)(b − y)]^{\frac{1}{2}} for (x, y) ∈ \overline{D}.  (3.10)

For any natural k the function u_{m_{k}} admits the representation

u_{m_{k}}(x, y) = ∫_{0}^{a} ∫_{0}^{b} g_{1}(x, s)g_{2}(y, t)[p_{m_{k}}(s, t)u_{m_{k}}(s, t) + q_{m_{k}}(s, t)] ds dt, (3.11)

where

g_{1}(x, s) = \begin{cases}s(\frac{x}{a} − 1) for s ≤ x, \\ x(\frac{s}{a} − 1) for s > x, \end{cases} g_{2}(y, t) = \begin{cases}t(\frac{y}{b} − 1) for t ≤ y, \\ y(\frac{t}{b} − 1) for t > y. \end{cases}

Moreover, it is obvious that the functions g_{1} and g_{2} admit the estimates

|g_{1}(x, s)| ≤ (1 − \frac{s}{a})s, |g_{2}(y, t)| ≤ (1 − \frac{t}{b})t,  (3.12)

\begin{matrix}|\frac{∂g_{1}(x, s)}{∂x}| ≤ [x(1 − \frac{x}{a})]^{−1}s(1 − \frac{s}{a}), \\ |\frac{∂g_{2}(y, t)}{∂y}| ≤ [y(1 − \frac{y}{b} )]^{−1}t(1 − \frac{t}{b}). \end{matrix}  (3.13)

If along with this we take into account conditions (1.8), (3.3), and (3.7), then we obtain the inequalities

|g_{1}(x, s)g_{2}(y, t)[p_{m_{k}}(s, t)u_{m_{k}}(s, t)+q_{m_{k}}(s, t)]|≤q^{∗}(s, t) (k =1, 2, …), (3.14)

|\frac{∂g_{1}(x, s)}{∂x} \frac{∂g_{2}(y, t)}{∂y}[p_{m_{k}}(s, t)u_{m_{k}}(s, t) + q_{m_{k}}(s, t)]| ≤

≤ [xy(1 − \frac{x}{a})(1 − \frac{y}{b} )]^{−1}q^{∗}(s, t) (k = 1, 2, …),  (3.15)

where q^{∗}(s, t) = γ_{0}[st(1 − \frac{s}{a})(1 − \frac{t}{b})]^{\frac{3}{2}} |p(s, t)| + st(1 − \frac{s}{a})(1 − \frac{t}{b})|q(s, t)| and q^{∗} is summable on D.

Now if we apply the Lebesgue’s theorem on the passage to the limit under the integral, then, with regard to (3.4), (3.9), and (3.14), from (3.11) we get

u(x, y) = ∫_{0}^{a} ∫_{0}^{b} g_{1}(x, s)g_{2}(y, t)[p(s, t)u(s, t) + q(s, t)] ds dt. (3.16)

By virtue of (3.15), equalities (3.11) and (3.16) yield

lim_{k→+∞} \frac{∂^{2}u_{m_{k}}(x, y)}{∂x∂y} = \frac{∂^{2}u(x, y)}{∂x∂y}

uniformly on every closed subset of D. Taking into account this fact, from (3.6) we get

∫_{0}^{a} ∫_{0}^{b} (\frac{∂^{2}u(x, y)}{∂x∂y})^{2} dx dy ≤ γ.  (3.17)

By virtue of (1.8), (3.12), (3.13) it follows from (3.16) and (3.17) that u ∈ C̃_{loc}^{1,2}(D) ∩ C(\overline{D}) and u is a solution of equation (1.1). On the other hand, it is clear from (3.10) that u satisfies the boundary condition (1.2).

By Lemma 2.6, problem (1.1),(1.2) has no solution different from u in C̃_{loc}^{1,2}(D) ∩ C(\overline{D}).

To complete the proof, we have to show the stability of the solution u with respect to small perturbation of the coefficients of equation (1.1).

Let δ be an arbitrary positive number satisfying inequality (2.24), and ρ and r be numbers given by equalities (2.4) and (2.25). Consider arbitrary functions \overline{p} and \overline{q} ∈ L_{loc}(D) satisfying conditions (1.4) and (1.5). Then by conditions (1.8),(3.1) and Lemma 2.3,

∫_{0}^{a} ∫_{0}^{b}[xy(a − x)(b − y)]^{\frac{3}{2}} |\overline{p}(x, y)| dx dy < +∞,

∫_{0}^{a} ∫_{0}^{b}[xy(a − x)(b − y)]^{\frac{1}{2}} |\overline{q}(x, y)| dx dy < +∞

and \overline{p} ∈ U_{β}(D). But according to the above-said, these conditions guarantee the existence and uniqueness of a solution \overline{u} ∈ C̃_{loc}^{1,2}(D) ∩ C(\overline{D}) of problem (1.1),(1.3). On the other hand, by Lemma 2.5, the solutions u and \overline{u} satisfy conditions (1.6).

Proof of Theorem 1.2. Let

α = \frac{4}{ab} ∫_{0}^{a} ∫_{0}^{b} λ_{0}(x, y) dx dy + \frac{1}{4}λ_{1} + \frac{16}{a^{2}b^{2}} λ_{2}.

By Theorem 1.1 and inequality (1.10), to prove Theorem 1.2 it is sufficient to establish that the function p satisfies condition (3.1).

Let u be an arbitrary funciton from A^{1,2}. Then by Lemma 2.1, inequalities (2.1)–(2.3) are valid, where ρ is the number given by (2.4). If along with this we take into account inequality (1.9), then we get

∫_{0}^{a} ∫_{0}^{b}[p(x, y)]_{+}u^{2}(x, y) dx dy ≤ ∫_{0}^{a} ∫_{0}^{b} λ_{0}(x, y)\frac{u^{2}(x, y)}{xy(a − x)(b − y)} dx dy +

+λ_{1} ∫_{0}^{a} ∫_{0}^{b} \frac{u^{2}(x, y)}{xy(a − x)(b − y)} dx dy + λ_{2} ∫_{0}^{a} ∫_{0}^{b} [\frac{u(x, y)}{xy(a − x)(b − y)}]^{2} dx dy ≤

≤ [\frac{4}{ab} ∫_{0}^{a} ∫_{0}^{b} λ_{0}(x, y) dx dy + \frac{λ_{1}}{4} + \frac{16}{a^{2}b^{2}} λ_{2}]ρ^{2} = αρ^{2}.

Hence, in view of the arbitrariness of u there follows inclusion (3.1).

Acknowledgement

This work was supported by INTAS Grant 96–1060.

References

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(Received 12.06.1998)

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