GEORGIAN MATHEMATICAL JOURNAL: Vol. 5, No. 1, 1998, 71-90

NUMERICAL SOLUTIONS TO THE DARBOUX PROBLEM WITH THE FUNCTIONAL DEPENDENCE

Z. KAMONT AND H. LESZCZYŃSKI

Abstract. The paper deals with the Darboux problem for the equation D_{xy}z(x, y) = f(x, y, z_{(x,y)}) where z_{(x,y)} is a function defined by z_{(x,y)}(t, s) = z(x + t, y + s), (t, s) ∈ [−a_{0}, 0] × [−b_{0}, 0]. We construct a general class of difference methods for this problem. We prove the existence and uniqueness of solutions to implicit functional difference equations by means of a comparison method; moreover we give an error estimate. The convergence of explicit difference schemes is proved under a general assumption that given functions satisfy nonlinear estimates of the Perron type. Our results are illustrated by a numerical example.

§ 1. Introduction

Given any two metric spaces X and Y, we denote by C(X, Y) the class of all continuous functions from X into Y. Take a, b > 0 and a_{0}, b_{0} ∈ R_{+}, R_{+} = [0, +∞). Define

E = (0, a] × (0, b], E^{0} = ([−a_{0}, a] × [−b_{0}, b])\E

and B = [−a_{0}, 0] × [−b_{0}, 0]. Given a function z : E^{0} ∪ E → R and a point (x, y) ∈ E, we define the function z_{(x,y)} : B → R by

z_{(x,y)}(t, s) = z(x + t, y + s), (t, s) ∈ B.  (1)

Suppose that f : \overline{E} × C(B, R) → R and ϕ : E^{0} → R are given functions. (Here \overline{E} is the closure of E.) Consider the Darboux problem

D_{xy}z(x, y) = f(x, y, z_{(x,y)}), (x, y) ∈ E,  (2)
z(x, y) = ϕ(x, y) for (x, y) ∈ E^{0},  (3)

1991 Mathematics Subject Classification. 65N06, 35R10. Key words and phrases. Volterra condition, differential-integral equation, implicit

functional-difference equation, comparison method, nonlinear estimate.

71

1072-947X/98/0100-0071$12.50/0 \begin{matrix}c \\ ◯ \end{matrix} 1998 Plenum Publishing Corporation

where D_{xy}z = \frac{∂^{2}z}{∂x∂y}. We consider classical solution to problem (2), (3). A function v ∈ C(E^{0} ∪E, R) is regarded as a solution of (2), (3) if D_{xy}v exists on E, D_{xy}v ∈ C(E, R), and v satisfies (2), (3). Sufficient conditions for the existence and uniqueness of a solution to (2), (3) are given in [1], see also [2].

For a few recent years, a certain number of papers concerning numerical methods for functional partial differential equations have been published.

Difference methods for nonlinear parabolic functional differential problems were considered in [3]–[5]. The main problem in these investigations is to find a difference approximation which is stable and satisfies consistency conditions with respect to the original problem. A method of difference inequalities or simple theorems on linear recurrent inequalities are used in the investigation of stability.

The semidiscretizations in space variables of linear parabolic Volterra integral-differential equations (the method of lines, Galerkin or collocation techiques) lead to large systems of stiff ordinary integral-functional equations. The analysis of spatial and temporal disretizations of linear integralfunctional parabolic problems has received considerable attention during the last years [6]–[11]. Most of these contributions seem to focus on the convergence theory. There are very few numerical studies.

Difference methods for first order functional differential equations with initial or initial-boundary conditions were studied in [12], [13]. The proofs of the convergence were based either on functional difference inequalities or on a general theorem on error estimates for approximate solutions to functional difference equations of the Volterra type with initial or initialboundary conditions and with unknown function of several variables.

The convergence of difference methods for functional hyperbolic systems in the Schauder canonic form was studied in [14].

For further biblioghaphical information concerning numerical methods for functional partial differential equations we suggest to see the survey papers [15] and [13].

The paper is organized as follows. In Section 2 we construct a general class of difference schemes for (2), (3). This leads to implicit functional difference problems. The existence and uniqueness of solutions to such problems are considered in Section 3. The comparison method of investigation of functional difference equations is used. The next section deals with a theorem on the convergence of explicit difference schemes with nonlinear estimates for given functions. We assume that increment functions satisfy nonlinear estimates of the Perron type with respect to the functional variable. In Section 5, we establish an error estimate for implicit difference methods. We give a numerical example.

Differential equations with a deviated argument and integral-differential problems can be obtained from (2), (3) by a specification of given operators.

§ 2. Discretization

Given any two sets X and Y, we denote by F [X, Y] the class of all functions defined on X and taking values in Y. We will denote by N and Z the sets of natural numbers and integers, respectively.

We construct a mesh in E^{0} ∪ E in the following way. Suppose that (h, k) ∈ (0, a] × (0, b] stand for steps of the mesh. Write

x_{i} = ih, y_{j} = jk, x_{i+ \frac{1}{2}} = ih + \frac{h}{2}, y_{j+ \frac{1}{2}} = jk + \frac{k}{2} , i, j ∈ Z.

Denote by I_{0} the set of all (h, k) ∈ (0, a] × (0, b] such that there exist M_{0}, N_{0} ∈ N such that M_{0}h = a_{0}, N_{0}k = b_{0}. We assume that I_{0} ≠ ∅ and there is a sequence {(h_{n}, k_{n})}, (h_{n}, k_{n}) ∈ I_{0}, such that lim_{n→∞}(h_{n}, k_{n}) = (0, 0). For (h, k) ∈ I_{0} we put Z_{hk} = {(x_{i}, y_{j}) : i, j ∈ Z}, and E_{hk}^{0} = Z_{hk} ∩ E^{0}, E_{hk} = Z_{hk} ∩ E. There are M, N ∈ N such that Mh ≤ a < (M + 1)h, Nk ≤ b < (N + 1)k. Let

A_{hk} = {(x_{i}, y_{j}) : 0 ≤ i ≤ M − 1, 0 ≤ j ≤ N − 1}.

Using the above definitions of M and N, we have

E_{hk} = {(x_{i}, y_{j}) : 1 ≤ i ≤ M, 1 ≤ j ≤ N}.

Let K, L ∈ Z be fixed and assume that −M_{0} ≤ K ≤ 1, −N_{0} ≤ L ≤ 1. Write

D_{hk} = {(x_{i}, y_{j}) : −M_{0} ≤ i ≤ K, −N_{0} ≤ j ≤ L}.

For z ∈ F [E_{hk}^{0} ∪ E_{hk}, R] we write z^{(i,j)} = z(x_{i}, y_{j}), (x_{i}, y_{j}) ∈ E_{hk}^{0} ∪ E_{hk}. In the same way we define w^{(i,j)} for w ∈ F [D_{hk}, R].

We will need a discrete version of the restriction operator given by (1). If z : E_{hk}^{0} ∪ E_{hk} → R and 0 ≤ i ≤ M − 1, 0 ≤ j ≤ N − 1 then the function z_{[i,j]} : D_{hk} → R is defined as follows:

z_{[i,j]}(t, s) = z(x_{i} + t, y_{j} + s), (t, s) ∈ D_{hk}.

We consider the difference operator δ given by

δz^{(i,j)} = \frac{1}{hk} [z^{(i+1,j+1)} − z^{(i+1,j)} − z^{(i,j+1)} + z^{(i,j)}].

Suppose that

F_{hk} : A_{hk} × F [D_{hk}, R] → R, ϕ_{hk} : E_{hk}^{0} → R

are given functions. Consider the problem

δz^{(i,j)} = F_{hk}(x_{i}, y_{j}, z_{[i,j]}), (x_{i}, y_{j}) ∈ A_{hk},  (4)
z^{(i,j)} = ϕ_{hk}^{(i,j)} for (x_{i}, y_{j}) ∈ E_{hk}^{0}.  (5)

Remark 2.1. If K = 1 and L = 1 then problem (4), (5) turns out to be an implicit difference method. If K ≤ 0 or L ≤ 0, then problem (4), (5) represents a simple functional difference equation of the Volterra type. It is obvious that in this case there exists exactly one solution u_{hk} : E_{hk}^{0} ∪E_{hk} → R of (4), (5).

Example 1. Suppose that F : \overline{E} × R → R and φ : [0, a] → R, ψ : [0, b] → R are given functions. We assume that φ(0) = ψ(0). Put a_{0} = 0, b_{0} = 0 and f(x, y, w) = F (x, y, w(0, 0)), (x, y, w) ∈ \overline{E} × C(B, R). Then (2),

(3) reduces to the classical Darboux problem

D_{xy}z(x, y) = F (x, y, z(x, y)), (x, y) ∈ E,  (6)
z(x, 0) = φ(x) for x ∈ [0, a], z(0, y) = ψ(y) for y ∈ [0, b].  (7)

One of the implicit difference schemes for (6), (7) takes the form

δz^{(i,j)} = F (x_{i+ \frac{1}{2}} , y_{j+ \frac{1}{2}} , \frac{1}{4}(z^{(i,j)} + z^{(i+1,j)} + z^{(i,j+1)} + z^{(i+1,j+1)})),

(x_{i}, y_{j}) ∈ A_{hk},

z^{(i,0)} = φ(x_{i}) for 0 ≤ i ≤ M, z^{(0,j)} = ψ(y_{j}) for 0 ≤ j ≤ N.

The most natural explicit difference method for (6), (7) takes the form

δz^{(i,j)} = F (x_{i}, y_{j}, z^{(i,j)}), (x_{i}, y_{j}) ∈ A_{hk},

with the above boundary condition.

Example 2. Suppose that a_{0} < 0, b_{0} < 0. For the same F we put

f(x, y, w) = F (x, y, ∫_{B} w(t, s)dt ds), (x, y, w) ∈ \overline{E} × C(B, R).

Then problem (2), (3) is equivalent to the integral-differential equation

D_{xy}z(x, y) = F (x, y, ∫_{B} z(x + t, y + s)dt ds), (x, y) ∈ E,  (8)

with boundary condition (3). Now, we construct an explicit difference method for (8), (3). Let

B_{hk} = {(x_{i}, y_{j}) : −M_{0} ≤ i ≤ 0, −N_{0} ≤ j ≤ 0, }.

We define the operator T_{hk} : F [B_{hk}, R] → F [B, R] in the following way. Suppose that w ∈ F [B_{hk}, R] and (t, s) ∈ B. Then there is (x_{i}, y_{j}) ∈ B_{hk} such that i < 0, j < 0 and x_{i} ≤ t ≤ x_{i+1}, y_{j} ≤ s ≤ y_{j+1}. We put

T_{hk}w(t, s) = w^{(i,j)}[1 − \frac{t − x_{i}}{h} ] [1 − \frac{s − y_{j}}{k} ] +

+ w^{(i+1,j+1)} \frac{t − x_{i}}{h} \frac{s − y_{j}}{k} +

+ w^{(i+1,j)} \frac{t − x_{i}}{h} [1 − \frac{s − y_{j}}{k} ] +

+ w^{(i,j+1)} [1 − \frac{t − x_{i}}{h} ] \frac{s − y_{j}}{k} .  (9)

It is easy to see that T_{hk} : F [B_{hk}, R] → C(B, R). We assume that D_{hk} = B_{hk}, i.e., K = 0, L = 0. We will approximate solutions to problem (8), (3) by means of solutions to the equation

δz^{(i,j)} = F (x_{i}, y_{j}, ∫_{B} T_{hk}z_{[i,j]}(t, s)dt ds), (x_{i}, y_{j}) ∈ A_{hk},  (10)

with boundary condition (5).

Example 3. Suppose that

F : \overline{E} × R → R, φ : \overline{E} → R, ψ : \overline{E} → R, ϕ : E^{0} → R

are given functions, and

−a_{0} ≤ φ(x, y) − x ≤ 0, −b_{0} ≤ ψ(x, y) − y ≤ 0 for (x, y) ∈ \overline{E}.

Let

f(x, y, w) = F (x, y, w(φ(x, y) − x, ψ(x, y) − y)), (x, y, w) ∈ \overline{E} × C(B, R).

Then problem (2), (3) reduces to the differential equation with a deviated argument

D_{xy}z(x, y) = F (x, y, z(φ(x, y), ψ(x, y))big), (x, y) ∈ E,

with boundary condition (3). It is easy to construct difference methods for the above equation using the ideas from Example 2. Further examples of the operator F_{hk} are given in Sections 4 and 5.

§ 3. Existence and Uniqueness of Solutions of Functional Difference Problems

The existence and uniqueness of a solution of problem (4), (5) is investigated by the comparison method. This method is based on the association of the operator F_{hk} with another operator σ_{hk}, which is followed by thorough analysis of a comparison equation. If the latter equation possesses adequate properties, then problem (4), (5) has exactly one solution which is the limit of a sequence of successive approximations. We obtain the simplest case of the operator σ_{hk} corresponding to equation (4) if the function F_{hk} satisfies the Lipschitz condition. The comparison problem is linear in this case.

The following property of the operator δ is important:

Lemma 3.1. Problem (4), (5) is equivalent to

z^{(i,j)} = hk ∑_{μ=0}^{i−1} ∑_{ν=0}^{j−1} F_{hk}(x_{μ}, y_{ν}, z_{[μ,ν]}) + ϕ_{hk}^{(i,0)} + ϕ_{hk}^{(0,j)} − ϕ_{hk}^{(0,0)}, (11)
1 ≤ i ≤ M, 1 ≤ j ≤ N,
z^{(i,j)} = ϕ_{hk}^{(i,j)} for (x_{i}, y_{j}) ∈ E_{hk}^{0}.  (12)

We omit the simple proof of the lemma.

Let η_{hk} : E_{hk}^{0} ∪ E_{hk} → R be a function given by

η_{hk}(x, y) = ϕ_{hk}(x, y) for (x, y) ∈ E_{hk}^{0} and η_{hk}(x, y) = 0 for (x, y) ∈ E_{hk}.

We define a sequence {z_{n}}, z_{n} : E_{hk}^{0} ∪ E_{hk} → R, in the following way:

z_{0}^{(i,j)} = η_{hk}^{(i,j)} for (x_{i}, y_{j}) ∈ E_{hk}^{0} ∪ E_{hk},  (13)

and

z_{n+1}^{(i,j)} = hk ∑_{μ=0}^{i−1} ∑_{ν=0}^{j−1} F_{hk}(x_{μ}, y_{ν}, (z_{n})_{[μ,ν]}) + ϕ_{hk}^{(i,0)} + ϕ_{hk}^{(0,j)} − ϕ_{hk}^{(0,0)}, (14)
1 ≤ i ≤ M, 1 ≤ j ≤ N,
z_{n+1}^{(i,j)} = ϕ_{hk}^{(i,j)} for (x_{i}, y_{j}) ∈ E_{hk}^{0},  (15)

where n = 0, 1, 2, . . ..

We prove that, under suitable assumptions on the function F_{hk}, the sequence {z_{n}} converges to the unique solution of problem (4), (5).

For w ∈ F (D_{hk}, R) we put

∥w∥_{hk} = max {|w^{(i,j)}| : (x_{i}, y_{j}) ∈ D_{hk}}.

For the above w we define the function |w|_{hk} : D_{hk} → R_{+} by

|w|_{hk}(x, y) = |w(x, y)|, (x, y) ∈ D_{hk}.

We will consider a comparison function σ_{hk} : A_{hk} × F [D_{hk}, R_{+}] → R_{+} coresponding to the function F_{hk}.

Assumption H_{1}. Suppose that

1^{0} for each (x, y) ∈ A_{hk} the function σ_{hk}(x, y, ·) : F [D_{hk}, R_{+}] → R_{+} is nondecreasing, and σ_{hk}(x, y, Θ_{hk}) = 0 for (x, y) ∈ A_{hk}, where Θ_{hk}(x, y) = 0 for (x, y) ∈ D_{hk}, 2^{0} for (x, y, w) ∈ A_{hk} × F [D_{hk}, R], \overline{w} ∈ F [D_{hk}, R] we have

|F_{hk}(x, y, w) − F_{hk}(x, y, \overline{w})| ≤ σ_{hk}(x, y, |w − \overline{w}|_{hk}),  (16)

3^{0} there exists a function \overline{g}_{hk} : E_{hk}^{0} ∪ E_{hk} → R_{+} which is a solution to the problem

ω^{(i,j)} ≥ hk ∑_{μ=0}^{i−1} ∑_{ν=0}^{j−1} σ_{hk}(x_{μ}, y_{ν}, ω_{[μ,ν]}) + \overline{η}_{hk}^{(i,j)},  (17)
1 ≤ i ≤ M, 1 ≤ j ≤ N,
ω^{(i,j)} = |ϕ_{hk}^{(i,j)}| for (x_{i}, y_{j}) ∈ E_{hk}^{0},  (18)

with the function \overline{η}_{hk} satisfying the condition

\overline{η}_{hk}^{(i,j)} ≥ hk ∑_{μ=0}^{i−1} ∑_{ν=0}^{j−1} |F_{hk}(x_{μ}, y_{ν}, (\overline{η}_{hk})_{[μ,ν]})| +
+ |ϕ_{hk}^{(i,0)}| + |ϕ_{hk}^{(0,j)}| + |ϕ_{hk}^{(0,0)}|,  (19)

where 1 ≤ i ≤ M, 1 ≤ j ≤ N,

4^{0} the function ω(x, y) = 0, (x, y) ∈ E_{hk}^{0} ∪ E_{hk}, is the unique solution to the problem

ω^{(i,j)} = hk ∑_{μ=0}^{i−1} ∑_{ν=0}^{j−1} σ_{hk}(x_{μ}, y_{ν}, ω_{[μ,ν]}), 1 ≤ i ≤ M, 1 ≤ j ≤ N, (20)
ω^{(i,j)} = 0 for (x_{i}, y_{j}) ∈ E_{hk}^{0}  (21)

in the class of functions satisfying the condition 0 ≤ ω(x, y) ≤ \overline{g}_{hk}(x, y), (x, y) ∈ E_{hk}.

Theorem 3.2. If Assumption H_{1} is satisfied then there exists a solution \overline{z} : E_{hk}^{0} ∪ E_{hk} → R of problem (4), (5). The solution is unique in the class of functions z : E_{hk}^{0} ∪ E_{hk} → R satisfying the condition

|z(x, y)| ≤ \overline{g}_{hk}(x, y) for (x, y) ∈ E_{hk}.

Proof. Consider the sequence {g_{n}}, g_{n} : E_{hk}^{0} ∪ E_{hk} → R_{+} given by

g_{0} = \overline{g}_{hk},

and

g_{n+1}^{(i,j)} = hk ∑_{μ=0}^{i−1} ∑_{ν=0}^{j−1} σ_{hk}(x_{μ}, y_{ν}, (g_{n})_{[μ,ν]}), 1 ≤ i ≤ M, 1 ≤ j ≤ n, (22)
g_{n+1}^{(i,j)} = 0 for (x_{i}, y_{j}) ∈ E_{hk}^{0},  (23)

where n = 0, 1, 2, . . .. We prove that

g_{n+1}(x, y) ≤ g_{n}(x, y) for (x, y) ∈ E_{hk}, n = 0, 1, 2, …,  (24)
lim_{n→∞} g_{n}(x, y) = 0, (x, y) ∈ E_{hk},  (25)
|z_{n}(x, y)| ≤ \overline{g}_{hk}(x, y), (x, y) ∈ E_{hk}, n = 0, 1, 2, …,  (26)
|z_{n+r}(x, y) − z_{n}(x, y)| ≤ g_{n}(x, y), (x, y) ∈ E_{hk}, n, r = 0, 1, 2, …. (27)

It follows from condition 3^{0} of Assumption H_{1} that g_{1}(x, y) ≤ g_{0}(x, y) for (x, y) ∈ E_{hk}. Assume that, for fixed n ∈ N, we have g_{n}(x, y) ≤ g_{n−1}(x, y), (x, y) ∈ E_{hk}. It follows from the monotonicity of σ_{hk} with respect to the functional variable and from (22) that g_{n+1}(x, y) ≤ g_{n}(x, y) for (x, y) ∈ E_{hk}. Then we have (24) by induction on n ∈ N.

Since 0 ≤ g_{n}(x, y) for (x, y) ∈ E_{hk}, n ∈ N, there exists

\overline{g}(x, y) = lim_{n→∞} g_{n}(x, y), (x, y) ∈ E_{hk}.

It follows from (22), (23) that \overline{g} is a solution to problem (20), (21). Condition 4^{0} of Assumption H_{1} implies \overline{g}(x, y) = 0 for (x, y) ∈ E_{hk}.

Now we prove (26). This relation is obvious for n = 0. If we assume that |z_{n}(x, y)| ≤ \overline{g}_{hk}(x, y), (x, y) ∈ E_{hk}, then we deduce from (16), (19) and the monotonicity of σ_{hk} that

|z_{n+1}^{(i,j)}| ≤ hk ∑_{μ=0}^{i−1} ∑_{ν=0}^{j−1} |F_{hk}(x_{μ}, y_{ν}, (z_{n})_{[μ,ν]}) − F_{hk}(x_{μ}, y_{ν}, (η_{hk})_{[μ,ν]})| +

+ hk ∑_{μ=0}^{i−1} ∑_{ν=0}^{j−1} |F_{hk}(x_{μ}, y_{ν}, (η_{hk})_{[μ,ν]})| + |ϕ_{hk}^{(i,0)}| + |ϕ_{hk}^{(0,j)}| + |ϕ_{hk}^{(0,0)}| ≤

≤ hk ∑_{μ=0}^{i−1} ∑_{ν=0}^{j−1} σ_{hk}(x_{μ}, y_{ν}, (\overline{g}_{hk})_{[μ,ν]}) +

+ |ϕ_{hk}^{(i,0)}| + |ϕ_{hk}^{(0,j)}| + |ϕ_{hk}^{(0,0)}| + \overline{η}_{hk}^{i,j)} ≤ \overline{g}_{hk}^{(i,j)},

where 1 ≤ i ≤ M, 1 ≤ j ≤ N; and it is seen that inequality (26) is obtained by induction on n ∈ N.

We prove (27) by induction on n ∈ N. Estimate (27) for n = 0 follows from (26). If we assume that for fixed n ∈ N we have

|z_{n+r}^{(i,j)} − z_{n}^{(i,j)}| ≤ g_{n}^{(i,j)}, (x_{i}, y_{j}) ∈ E_{hk}, r = 0, 1, 2, …,

then, applying (14), (16), (22) and the monotonicity of σ_{h}, we get

|z_{n+r+1}^{(i,j)} − z_{n+1}^{(i,j)}| ≤ hk ∑_{μ=0}^{i−1} ∑_{ν=0}^{j−1} |F_{hk}(x_{μ}, y_{ν}, (z_{n+r})_{[μ,ν]}) −

− F_{hk}(x_{μ}, y_{ν}, (z_{n})_{[μ,ν]})| ≤

≤ hk ∑_{μ=0}^{i−1} ∑_{ν=0}^{j−1} σ_{hk}(x_{μ}, y_{ν}, |(z_{n+r})_{[μ,ν]} − (z_{n})_{[μ,ν]}|_{hk}) ≤

≤ hk ∑_{μ=0}^{i−1} ∑_{ν=0}^{j−1} σ_{hk}(x_{μ}, y_{ν}, (g_{n})_{[μ,ν]}) = g_{n+1}^{(i,j)},

which completes the proof of assertion (27).

It follows from (25), (27) that there is \overline{z} : E_{hk}^{0} ∪ E_{hk} → R such that

\overline{z}(x, y) = lim_{n→∞} z_{n}(x, y), (x, y) ∈ E_{hk}^{0} ∪ E_{hk}.

Relations (14), (15) imply the function \overline{z} is a solution to problem (11), (12).

Suppose that \overline{u} : E_{hk}^{0} ∪ E_{hk} → R is another solution to problem (11), (12), and that |\overline{u}(x, y)| ≤ \overline{g}_{hk}(x, y) for (x, y) ∈ E_{hk}. Then we obtain by induction on n ∈ N the relation

|\overline{u}(x, y) − z_{n}(x, y)| ≤ g_{n}(x, y) for (x, y) ∈ E_{hk}, n = 0, 1, 2, ….

It follows from (25) that \overline{u} = \overline{z}, which completes the proof of Theorem 1.2.

Now, we prove a result on the global uniqueness of solution to (11), (12).

Lemma 3.3. Suppose that Assumption H_{1} is satisfied and the function \overline{ω}(x, y) = 0 for (x, y) ∈ E_{hk}^{0} ∪ E_{hk} is the only solution of the problem

ω^{(i,j)} ≤ hk ∑_{μ=0}^{i−1} ∑_{ν=0}^{j−1} σ_{hk}(x_{μ}, y_{ν}, ω_{[μ,ν]}), 1 ≤ i ≤ M, 1 ≤ j ≤ N, (28)
ω^{(i,j)} = 0 for (x_{i}, y_{j}) ∈ E_{hk}^{0}.  (29)

Then the solution \overline{z} : E_{hk}^{0} ∪ E_{hk} → R to problem (11), (12) is unique.

Proof. If \overline{z}, \overline{u} : E_{hk}^{0} ∪ E_{hk} → R are solutions to (11), (12), then ω̃ = \overline{z} − \overline{u} satisfies (28), (29), and the assertion follows.

Now, we give sufficient conditions for the uniqueness of the solution \overline{ω} = 0 to problem (28), (29).

Lemma 3.4. Suppose that the function σ_{hk} satisfies the conditions:

1^{0} for each function λ_{hk} : E_{hk} → R_{+} there exists a solution to the problem

ω^{(i,j)} ≥ hk ∑_{μ=0}^{i−1} ∑_{ν=0}^{j−1} σ_{hk}(x_{μ}, y_{ν}, ω_{[μ,ν]}) + λ_{hk}^{(i,j)},  (30)
1 ≤ i ≤ M, 1 ≤ j ≤ N,
ω^{(i,j)} = 0 for (x_{i}, y_{j}) ∈ E_{hk}^{0},  (31)

2^{0} the function \overline{ω} = 0 is a unique solution to problem (20), (21). Under these assumptions, the function \overline{ω}(x, y) = 0, (x, y) ∈ E_{hk}^{0} ∪ E_{hk},

is the only solution to problem (28), (29).

Proof. Suppose that ω̃ : E_{hk}^{0} ∪ E_{hk} → R_{+} is a solution to problem (28), (29). Consider the sequence {ω_{n}}, ω_{n} : E_{hk}^{0} ∪ E_{hk} → R_{+} given by

(i) ω_{0} is a solution of (30), (31) for λ_{hk} = ω̃, (ii) if ω_{n} is a given function then

ω_{n+1}^{(i,j)} = hk ∑_{μ=0}^{i−1} ∑_{ν=0}^{j−1} σ_{hk}(x_{μ}, y_{ν}, (ω_{n})_{[μ,ν]}), 1 ≤ i ≤ M, 1 ≤ j ≤ N, (32)
ω_{n+1}^{(i,j)} = 0 for (x_{i}, y_{j}) ∈ E_{hk}^{0}.  (33)

We obtain

ω̃(x, y) ≤ ω_{n}(x, y) for (x, y) ∈ E_{hk}, n = 0, 1, 2, …,
0 ≤ ω_{n+1}(x, y) ≤ ω_{n}(x, y) for (x, y) ∈ E_{hk}, n = 0, 1, 2, ….

The above relations can be proved by induction on n ∈ N. Let \overline{ω} : E_{hk}^{0} ∪ E_{hk} → R_{+} be defined by

\overline{ω}(x, y) = lim_{n→∞} ω_{n}(x, y).

It follows from (32), (33) that \overline{ω} = 0. Since ω̃ ≤ \overline{ω}, the assertion follows.

§ 4. Convergence of Difference Methods with Nonlinear Estimates for Increment Functions

In this section we consider the particular case of the set D_{hk}. We assume that K = 0, L = 0. We will use the following comparison lemma:

Lemma 4.1. Suppose that K = 0 and L = 0 in the definition of D_{hk} and

1^{0} the function G_{hk} : A_{hk} ×F [B_{hk}, R] → R is non-decreasing with respect to the functional variable, 2^{0} the functions u, v : E_{hk}^{0} ∪ E_{hk} → R satisfy the relations

u^{(i,j)} − hk ∑_{μ=0}^{i−1} ∑_{ν=0}^{j−1} G_{hk}(x_{μ}, y_{ν}, u_{[μ,ν]}) ≤

≤ v^{(i,j)} − hk ∑_{μ=0}^{i−1} ∑_{ν=0}^{j−1} G_{hk}(x_{μ}, y_{ν}, v_{[μ,ν]}),

whenever 1 ≤ i ≤ M, 1 ≤ j ≤ N; and u^{(i,j)} ≤ v^{(i,j)} on E_{hk}^{0}.

Then u^{(i,j)} ≤ v^{(i,j)} for (x_{i}y_{j}) ∈ E_{hk}.

We omit the simple proof of the lemma. Denote by Ξ the class of all functions α : I_{0} → R_{+} such that

lim_{(h,k)→(0,0)} α(h, k) = 0.

Assumption H_{2}. Suppose that K = 0, L = 0 and 1^{0} conditions 1^{0}, 2^{0}, 4^{0} of Assumption H_{1} are satisfied, 2^{0} the solution \overline{ω}(x_{i}, y_{j}) = 0, (x_{i}, y_{j}) ∈ E_{hk}^{0} ∪ E_{hk}, of the problem

\begin{matrix}δω^{(i,j)} = σ_{hk}(x_{i}, y_{j}, ω_{[i,j]}), (x_{i}, y_{j}) ∈ A_{hk}, \\ ω^{(i,j)} = 0 for (x_{i}, y_{j}) ∈ E_{hk}^{0}, \end{matrix}  (34)

is stable in the following sense: if ω_{hk} : E_{hk}^{0} ∪ E_{hk} → R_{+} is the solution of the problem

δω^{(i,j)} = σ_{hk}(x_{i}, y_{j}, ω_{[i,j]}) + α(h, k), (x_{i}, y_{j}) ∈ A_{hk},  (35)
ω^{(i,j)} = α_{0}(h, k) for (x_{i}, y_{j}) ∈ E_{hk}^{0},  (36)

where α, α_{0} ∈ Ξ then there is β ∈ Ξ such that ω_{hk}^{(i,j)} ≤ β(h, k) for (x_{i}, y_{j}) ∈ E_{hk}.

Theorem 4.2. Suppose that Assumption H_{2} is satisfied, and 1^{0} u_{h} : E_{hk}^{0} ∪ E_{hk} → R is a solution to problem (4), (5) and there is

α_{0} ∈ Ξ such that

|ϕ^{(i,j)} − ϕ_{hk}^{(i,j)}| ≤ α_{0}(h, k) for (x_{i}, y_{j}) ∈ E_{hk}^{0};  (37)

2^{0} v : E^{0} ∪ E → R is a solution to problem (2), (3) and v is of class C^{3}, on \overline{E};

3^{0} the following compatibility condition is satisfied: there is α̃ ∈ Ξ such that

|F_{hk}(x_{i}, y_{j}, (v_{hk})_{[i,j]}) − f(x_{i}, y_{j}, v_{(x_{i},y_{j})})| ≤ α̃(h, k),  (38)
(x_{i}, y_{j}) ∈ A_{hk},

where the function v_{hk} is the restriction of the function v to the set E_{hk}^{0} ∪ E_{hk}.

Under these assumptions there exists β ∈ Ξ such that

|u_{hk}^{(i,j)} − v_{hk}^{(i,j)}| ≤ β(h, k), (x_{i}, y_{j}) ∈ E_{hk}.  (39)

Proof. Let Γ_{hk} : A_{hk} → R be defined by

δv_{hk}^{(i,j)} = F_{hk}(x_{i}, y_{j}, (v_{hk})_{[i,j]}) + Γ_{hk}^{(i,j)}, (x_{i}, y_{j}) ∈ A_{hk}.  (40)

It follows from assumption 2^{0} that there is α_{1} ∈ Ξ such that

|δv_{hk}^{(i,j)} − D_{xy}v^{(i,j)}| ≤ α_{1}(h, k), (x_{i}, y_{j}) ∈ A_{hk}.  (41)

From the above inequality and from the compatibility condition (39) we deduce that there is α ∈ Ξ such that |Γ_{hk}^{(i,j)}| ≤ α(h, k) for (x_{i}, y_{j}) ∈ A_{hk}. Let ω_{hk} = |u_{hk} − v_{hk}|. Then the function ω_{hk} satisfies the relations

ω_{hk}^{(i,j)} ≤ hk ∑_{μ=0}^{i−1} ∑_{ν=0}^{j−1} σ_{hk}(x_{μ}, y_{ν}, (ω_{hk})_{[μ,ν]}) + ih jk α(h, k),  (42)
1 ≤ i ≤ M, 1 ≤ j ≤ N,
ω_{hk}^{(i,j)} ≤ α_{0}(h, k) for (x_{i}, y_{j}) ∈ E_{hk}^{0}.  (43)

Let ω̃ : E_{hk}^{0} ∪ E_{hk} → R_{+} be a solution of the problem

ω^{(i,j)} = hk ∑_{μ=0}^{i−1} ∑_{ν=0}^{j−1} σ_{hk}(x_{μ}, y_{ν}, (ω)_{[μ,ν]}) + abα(h, k),  (44)
1 ≤ i ≤ M, 1 ≤ j ≤ N,
ω^{(i,j)} = α_{0}(h, k) for (x_{i}, y_{j}) ∈ E_{hk}^{0}.  (45)

Relations (42), (43) and Lemma 4.1 imply ω_{hk}^{(i,j)} ≤ ω_{hk}^{(i,j)} for (x_{i}, y_{j}) ∈ E_{hk}. Now we obtain the assertion of our theorem from the stability of the functional difference problem (34).

Remark 4.3. If the assumptions of Theorem 4.2 are satisfied then we have the following estimate for the error of approximate solutions to problem (2), (3):

|u_{hk}^{(i,j)} − v_{hk}^{(i,j)}| ≤ ω_{hk}^{(i,j)}, (x_{i}, y_{j}) ∈ E_{hk},

where the function ω̃_{h} is the only solution to problem (44), (45) with α_{0} given by (37), α = α̃ + α_{1}, and α̃, α_{1} are defined by (39), (41).

Now, we consider problem (2), (3) and the difference method

\begin{matrix}δz^{(i,j)} = f(x_{i}, y_{j}, T_{hk}z_{[i,j]}), (x_{i}, y_{j}) ∈ A_{hk}, \\ z^{(i,j)} = ϕ_{hk}^{(i,j)} for (x_{i}, y_{j}) ∈ E_{hk}^{0}, \end{matrix}  (46)

where the operator T_{hk} is defined in Example 2. It is obvious that there exists exactly one solution to problem (46).

Assumption H_{3}. Suppose that the function f : \overline{E} × C(B, R) → R is continuous, and there is a function σ : \overline{E} × R_{+} → R_{+} such that

1^{0} σ is continuous, and σ(x, y, 0) = 0 for (x, y) ∈ \overline{E};

2^{0} σ is nondecreasing with respect to all variables and the function \overline{ω}(x, y) = 0, (x, y) ∈ \overline{E} is the unique solution to the problem

D_{xy}z(x, y) = σ(x, y, z(x, y)), (x, y) ∈ \overline{E},
z(x, 0) = 0 for x ∈ [0, a], z(0, y) = 0 for y ∈ [0, b];

3^{0} the estimate

|f(x, y, w) − f(x, y, \overline{w})| ≤ σ(x, y, ∥w − \overline{w}∥_{B})

is satisfied on \overline{E} × C(B, R).

Theorem 4.4. Suppose that Assumption H_{3} is satisfied, and 1^{0} u_{hk} : E_{hk}^{0} ∪E_{hk} → R is a solution to problem (46), and there is α_{0} ∈ Ξ

such that estimate (37) holds;

2^{0} v : E^{0} ∪ E → R is a solution to problem (2), (3), and v is of class C^{3} on \overline{E}.

Then there is β ∈ Ξ such that

|u_{hk}^{(i,j)} − v_{hk}^{(i,j)}| ≤ β(h, k), (x_{i}, y_{j}) ∈ E_{hk},  (47)

where v_{hk} is the restriction of the function v to the set E_{hk}^{0} ∪ E_{hk}.

Proof. We apply Theorem 4.2 in the proof of assertion (47). Put D_{hk} = B_{hk} and

F_{hk}(x, y, w) = f(x, y, T_{hk}w), (x, y, w) ∈ A_{hk} × F [D_{hk}, R]. (48)

Then we have

|F_{hk}(x, y, w) − F_{hk}(x, y, \overline{w})| ≤ σ(x, y, ∥T_{hk}(w − \overline{w})∥_{B}) =
= σ(x, y, ∥w − \overline{w}∥_{hk}) on A_{hk} × F [D_{hk}, R].

Consider the problem

δω^{(i,j)} = σ(x_{i}, y_{j}, ω^{(i,j)}), 0 ≤ i ≤ M − 1, 0 ≤ j ≤ N − 1, (49)
ω^{(i,j)} = 0 for (x_{i}, y_{j}) ∈ E_{hk}^{0}.  (50)

We prove that the solution \overline{ω}(x, y) = 0, (x, y) ∈ E_{hk}^{0} ∪ E_{hk}, to the problem

δω^{(i,j)} = σ(x_{i}, y_{j}, ω^{(i,j)}), (x_{i}, y_{j}) ∈ A_{hk}, ω^{(i,j)} = 0 for (x_{i}, y_{j}) ∈ E_{hk}^{0},

is stable in the sense of condition 2^{0} of Assumption H_{2}.

Let \overline{ω}_{hk} : E_{hk}^{0} ∪ E_{hk} → R_{+} be a solution to the problem

δω^{(i,j)} = σ(x_{i}, y_{j}, ω^{(i,j)}) + α(h, k), (x_{i}, y_{j}) ∈ A_{hk},  (51)
ω^{(i,j)} = α_{0}(h, k) for (x_{i}, y_{j}) ∈ E_{hk}^{0},  (52)

where α α_{0} ∈ Ξ. Consider the Darboux problem

D_{xy}z(x, y) = σ(x, y, z(x, y)) + α(h, k), (x, y) ∈ \overline{E},
z(x, 0) = α_{0}(h, k) for x ∈ [0, a], z(0, y) = α_{0}(h, k) for y ∈ [0, b].

Since there is ε_{0} > 0 such that if h + k ≤ ε_{0}, there is also a solution z_{hk} : \overline{E} → R to the above problem, and

lim_{(h,k)→(0,0)} z_{hk}(x, y) = 0 uniformly on \overline{E}.  (53)

It follows from the monotonicity of σ that for h + k ≤ ε_{0}, (x_{i}, y_{j}) ∈ A_{hk}, we have the relations

z_{hk}^{(i,j)} = ∫_{0}^{x_{i}} ∫_{0}^{y_{j}}[σ(t, s, z_{hk}(t, s)) + α(h, k)]dt ds + α_{0}(h, k) ≥

≥ hk ∑_{μ=0}^{i−1} ∑_{ν=0}^{j−1}[σ(x_{μ}, y_{ν}, z_{hk}^{(μ,ν)}) + α(h, k)] + α_{0}(h, k).

Then the function z_{hk} satisfies the difference inequality

z_{hk}^{(i,j)} ≥ hk ∑_{μ=0}^{i−1} ∑_{ν=0}^{j−1}[σ(x_{μ}, y_{ν}, z_{hk}^{(μ,ν)}) + α(h, k)] + α_{0}(h, k),

1 ≤ i ≤ M, 1 ≤ j ≤ N.

The function \overline{ω}_{hk} satisfies the equation

\overline{ω}_{hk}^{(i,j)} = hk ∑_{μ=0}^{i−1} ∑_{ν=0}^{j−1}[σ(x_{μ}, y_{ν}, \overline{ω}_{hk}^{(μ,ν)}) + α(h, k)] + α_{0}(h, k),

1 ≤ i ≤ M, 1 ≤ j ≤ N.

It follows from Lemma 4.1 that \overline{ω}_{hk}^{(i,j)} ≤ z_{hk}^{(i,j)} for (x_{i}, y_{j}) ∈ E_{hk}. Thus the stability of problem (49), (50) follows from condition (53).

Now, we prove the consistency condition for equation (48). We will use the following property of the operator T_{hk} ([12]): if w ∈ F [B, R], w is of class C^{3} and w_{hk} is the restriction of w to the set B_{hk}, then there is C > 0 such that

∥T_{hk}w_{hk} − w∥_{B} ≤ C(h^{2} + k^{2}).

It follows from the above property of T_{hk} and from assumption 2^{0} that the operator F_{hk} given by (48) satisfies condition (39) with α̃ ∈ Ξ.

§ 5. Convergence of Implicit Difference Methods

In this section, we consider a general class of difference problems consistent with (2), (3) which satisfy Assumption H_{1} and are convergent. We formulate a functional difference equation.

Let K = 1, L = 1 in the definition of D_{hk}. We define the operator

T̃_{hk} : F [D_{hk}, R] → F [ [−a_{0}, h] × [−b_{0}, k], R]

in the following way. Let w ∈ F [D_{hk}, R] and (t, s) ∈ [−a_{0}, h]×[−b_{0}, k]. Then there is (x_{i}, y_{j}) such that −M_{0} ≤ i < 1, −N_{0} ≤ j < 1 and x_{i} ≤ t ≤ x_{i+1}, y_{j} ≤ j ≤ y_{j+1}. We define (T̃_{hk}w)(t, s) as the right-hand side of formula (9). Denote by S_{hk} : F [D_{hk}, R] → F [B, R] the operator given by

(S_{hk}w)(t, s) = (T̃_{hk}w)(t + \frac{h}{2}, s + \frac{k}{2} ), (t, s) ∈ B.

The function S_{hk}w is the restriction of the function T̃_{hk}w to the set [−a_{0} + \frac{h}{2}, \frac{h}{2}] × [−b_{0} + \frac{k}{2} , \frac{k}{2} ] which is shifted to the set B.

Consider problem (2), (3) and the difference equation

δz^{(i,j)} = f(x_{i+ \frac{1}{2}} , y_{j+ \frac{1}{2}} , S_{hk}z_{[i,j]}), 0 ≤ i ≤ M − 1, 0 ≤ j ≤ N − 1, (54)

with boundary condition (5).

Assumption H_{4}. Suppose that the function f : \overline{E} × C(B, R) → R is continuous and there is L̃ ∈ R_{+} such that

|f(x, y, w) − f(x, y, \overline{w})| ≤ L̃∥w − \overline{w}∥_{D} on \overline{E} × C(B, R).

Theorem 5.1. Suppose that Assumption H_{4} is satisfied and 1^{0} v : E^{0} ∪ E → R is a solution to problem (2), (3), and v is of class C^{4}

on \overline{E}, 2^{0} abL̃ < 1, and there is α_{0} ∈ Ξ such that inequality (37) holds true. Then there exists exactly one solution u_{hk} : E_{hk}^{0} ∪ E_{hk} → R to problem

(54), (5), and there is C ∈ R_{+} such that the following error estimate holds:

|u_{hk}^{(i,j)} − v^{(i,j)}| ≤ \frac{Cx_{i}y_{j}(h^{2} + k^{2}) + 2α_{0}(h, k)}{1 − x_{i}y_{j}L̃}, (x_{i}, y_{j}) ∈ E_{hk}. (55)

Proof. We put

F_{hk}(x_{i}, y_{j}, w) = f(x_{i+ \frac{1}{2}} , y_{j+ \frac{1}{2}} , S_{hk}w), (x_{i}, y_{j}, w) ∈ A_{hk} × F [D_{hk}, R].

Then F_{hk} satisfies the Lipschitz condition with respect to the functional variable with the constant L̃. Put

σ(x, y, w) = L̃∥w∥_{hk}, (x, y, w) ∈ A_{hk} × F [D_{hk}, R_{+}].  (56)

Then equation (20) is equivalent to

ω^{(i,j)} = hk ∑_{μ=1}^{i} ∑_{ν=1}^{j} L̃∥ω_{[μ−1,ν−1]}∥_{hk}, 1 ≤ i ≤ M, 1 ≤ j ≤ N.

The above equation with boundary condition (21) is equivalent to the problem

\begin{matrix}ω^{(i,j)} = hk ∑_{μ=1}^{i} ∑_{ν=1}^{j} L̃ω^{(μ,ν)}, 1 ≤ i ≤ M, 1 ≤ j ≤ N, \\ ω^{(i,j)} = 0 for (x_{i}, y_{j}) ∈ E_{hk}^{0}. \end{matrix}  (57)

It follows from assumption 2^{0} that problem (57) satisfies conditions 3^{0}, 4^{0} of Assumption H_{1} and that the unique solution to problem (28) (29) with σ given by (56) is ω(x, y) = 0. Then there exists exactly one solution u_{hk} : E_{hk}^{0} ∪ E_{hk} → R to problem (54), (5).

Let Γ_{hk} : A_{hk} → R be defined by

δv_{hk}^{(i,j)} = f(x_{i+ \frac{1}{2}} , y_{j+ \frac{1}{2}} , S_{hk}(v_{hk})_{[i,j]}) + Γ_{hk}^{(i,j)}, (x_{i}, y_{j}) ∈ A_{hk}.

There is C̃ ∈ R_{+} such that we have

|δv_{hk}^{(i,j)} − D_{xy}v(x_{i+ \frac{1}{2}} , y_{j+ \frac{1}{2}} )| ≤ C̃(h^{2} + k^{2}), (x_{i}, y_{j}) ∈ A_{hk},

and

|T̃_{hk}(v_{hk})_{[i,j]}(t, s) − v_{(x_{i+1},y_{j+1})}(t, s)| ≤ C̃(h^{2} + k^{2}),

where (t, s) ∈ [−a_{0}, h] × [−b_{0}, k], 0 ≤ i ≤ M − 1, 0 ≤ j ≤ N − 1. Then there is C ∈ R_{+} such that |Γ_{hk}^{(i,j)}| ≤ C(h^{2} + k^{2}) for (x_{i}, y_{j}) ∈ A_{hk}. Let ω_{hk} = u_{hk} − v_{hk}. Obviously, the function ω_{hk} satisfies the inequalities

|ω_{hk}^{(i,j)}| ≤ hk ∑_{μ=0}^{i−1} ∑_{ν=0}^{j−1} [L̃(∥ω_{hk})_{[μ,ν]}∥_{hk} + C(h^{2} + k^{2})] + 2α_{0}(h, k),
1 ≤ i ≤ M, 1 ≤ j ≤ N,
ω_{hk}^{i,j)} ≤ α_{0}(h, k) for (x_{i}, y_{j}) ∈ E_{hk}^{0}.

The function

ω_{hk}^{(i,j)} = \frac{Cx_{i}y_{j}(h^{2} + k^{2}) + 2α_{0}(h, k)}{1 − x_{i}y_{j}L̃} , (x_{i}, y_{j}) ∈ E_{hk},
ω_{hk}^{(i,j)} = 2α_{0}(h, k) for (x_{i}, y_{j}) ∈ E_{hk}^{0},

satisfies the inequalities

ω_{hk}^{(i,j)} ≥ hk ∑_{μ=0}^{i−1} ∑_{ν=0}^{j−1} [L̃∥(ω)_{[μ,ν]}∥_{hk} + C(h^{2} + k^{2})] + 2α_{0}(h, k),
1 ≤ i ≤ M, 1 ≤ j ≤ N,
ω^{(i,j)} ≥ α_{0}(h, k) for (x_{i}, y_{j}) ∈ E_{hk}^{0}.

Consequently, we obtain assertion (55) from Lemma 4.1.

Numerical example. Define E = (0, 1]×(0, 1], E^{0} = ([−\frac{1}{2}, 1]×[\frac{1}{2}, 1])\E and B = [−\frac{1}{2}, 0] × [−\frac{1}{2}, 0]. Consider the Darboux problem

D_{xy}z(x, y) = 2(x + y)(z(x − 0.25, y − 0.25) − z(x, y)) −
− (x + y) ∫_{− \frac{1}{2}}^{0} ∫_{− \frac{1}{2}}^{0} z(x + t, y + s)dt dx + f(x, y),  (58)

(x, y) ∈ \overline{E},
z(x, y) = sin(1 + x + y) for (x, y) ∈ E^{0},  (59)

where

f(x, y) = (x + y − 1) sin(1 + x + y) − (x + y) sin(x + y).

Let M_{0}, N_{0}, M, N be natural numbers which satisfy

M_{0}h = 0.5, N_{0}k = 0.5, M = 2M_{0}, N = 2N_{0}.

Assume that M_{0} and N_{0} are even numbers. Consider the difference equation corresponding to equation (58)

δz^{(i,j)} = 2(x_{i+ \frac{1}{2}} + y_{j+ \frac{1}{2}} )(S_{hk}z_{(i,j)}(−0.25, −0.25) − S_{hk}z_{(i,j)}(0, 0)) −

− (x_{i+ \frac{1}{2}} + y_{j+ \frac{1}{2}} ) ∫_{− \frac{1}{2}}^{0} ∫_{− \frac{1}{2}}^{0} T_{hk}z_{(i,j)}(t, s)dt ds + f^{(i,j)},

0 ≤ i ≤ M − 1, 0 ≤ j ≤ N − 1.

Let m_{0} = \frac{1}{2}M_{0} and n_{0} = \frac{1}{2}N_{0}. Then we have

S_{hk}z_{(i,j)}(−0.25, −0.25) = I^{(−)}[i, j], S_{hk}z_{(i,j)}(0, 0) = I^{(0)}[i, j],

88 Z. KAMONT AND H. LESZCZYŃSKI

where

I^{(−)}[i, j] = \frac{1}{4}(z^{(i−m_{0},j−n_{0})} + z^{(i−m_{0}+1,j−n_{0})} +
+ z^{(i−m_{0},j−n_{0}+1)} + z^{(i−m_{0}+1,j−n_{0}+1)}),
I^{(0)}[i, j] = \frac{1}{4}(z^{(i,j)} + z^{(i+1,j)} + z^{(i,j+1)} + z^{(i+1,j+1)}).

Let w ∈ F (B_{hk}, R). Then

∫_{x_{i}}^{x_{i+1}} ∫_{y_{j}}^{y_{j+1}}T_{hk}w(t, s)dt ds = \frac{hk}{4} (w^{(i,j)} + w^{(i+1,j)} + w^{(i,j+1)} + w^{(i+1,j+1)})

and consequently

∫_{− \frac{1}{2}}^{0} ∫_{− \frac{1}{2}}^{0} T_{hk}z_{(i,j)}(t, s)dt ds = I[i, j],

where

I[i, j] = \frac{hk}{4} (z^{(i−M_{0},j−N_{0})} + z^{(i,j−N_{0})} + z^{(i−M_{0},j)} + z^{(i,j)}) +

+ \frac{hk}{2} ∑_{i′=1}^{M_{0}−1} (z^{(i−M_{0}+i′,j−N_{0})} + z^{(i−M_{0}+i′,j)}) +

+ \frac{hk}{2} ∑_{j′=1}^{N_{0}−1} (z^{(i−M_{0},j−N_{0}+j′)} + z^{(i,j−N_{0}+j′)}) +

+ hk ∑_{i′=1}^{M_{0}−1} ∑_{j′=1}^{N_{0}−1} z^{(i−M_{0}+i′,j−N_{0}+j′)}.

We approximate the solution v : E^{0} ∪ E → R of problem (58), (59) by means of solutions of the implicit difference equation

z^{(i+1,j+1)} − z^{(i+1,j)} − z^{(i,j+1)} + z^{(i,j)} =

= −hk(x_{i+ \frac{1}{2}} + y_{j+ \frac{1}{2}} )I[i, j] +
+ 2hk(x_{i+ \frac{1}{2}} + y_{j+ \frac{1}{2}} )(I^{(−)}[i, j] − I^{(0)}[i, j]) + hkf^{(i,j)},  (60)
0 ≤ i ≤ M − 1, 0 ≤ j ≤ N − 1,

with the boundary condition

z^{(i,j)} = sin(1 + x_{i} + y_{j}), (x_{i}, y_{j}) ∈ E_{hk}^{0}.  (61)

The function v(x, y) = sin(1 + x + y), (x, y) ∈ E^{0} ∪ E, is the solution to problem (58), (59). Let u_{hk} : E_{hk}^{0} ∪ E_{hk} → R be a solution to problem (60), (61), and ε_{hk} = v_{hk} − u_{hk}, where u_{hk} is the restriction of the function v to the set E_{hk}^{0} ∪ E_{hk}. Some values of ε_{hk}^{(i,j)} are listed in the table for h = k = 10^{−3}.

TABLE OF ERRORS

x = 0.80 x = 0.85 x = 0.90 x = 0.95  x = 1
y = 0.80 1.325 10^{−4} 1.597 10^{−4} 1.898 10^{−4} 2.229 10^{−4} 2.593 10^{−4}
y = 0.85 1.587 10^{−4} 1.905 10^{−4} 2.244 10^{−4} 2.618 10^{−4} 3.028 10^{−4}
y = 0.90 1.898 10^{−4} 1.244 10^{−4} 2.627 10^{−4} 3.048 10^{−4} 3.510 10^{−4}
y = 0.95 2.229 10^{−4} 2.618 10^{−4} 3.048 10^{−4} 3.521 10^{−4} 4.041 10^{−4}
y = 1 2.593 10^{−4} 3.028 10^{−4} 3.510 10^{−4} 3.641 10^{−4} 4.625 10^{−4}

The computation was performed by the computer IBM AT.

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(Received 18.12.1995)

Authors’ address: Institute of Mathematics University of Gdańsk Wit Stwosz Str. 57 80-952 Gdańsk Poland