Let \(D^{\prime}\) be one of the parallelograms covering the ‘nonsingular’ part of the polygon *D* defined in Section 2. The boundaries of the parallelogram \(D^{\prime } \) are denoted by \(\gamma_{j}^{\prime}\), enumerated counterclockwise starting from left, including the ends, \(\dot{\gamma}_{j}^{\prime }=\gamma_{j-1}^{\prime}\cap\gamma_{j}^{\prime}\), \(j=1,2,3,4\), denotes the vertices of \(D^{\prime}\), \(\gamma^{\prime}=\bigcup _{j=1}^{4}\gamma _{j}^{\prime}\), and \(\overline{D}^{\prime}=D^{\prime}\cup\gamma ^{\prime } \). Furthermore \(\gamma\cap\gamma^{\prime}\neq\emptyset\), but the vertices \(\dot{\gamma}_{m}^{\prime}\) with an interior angle of \(\alpha_{m}\pi\neq\pi/3\) are located either inside of *D*, or on the interior of a side \(\gamma_{m}\) of *D*, \(1\leq m\leq N\). We define the open parallelogram \(D^{\prime}\) as \(D^{\prime}= \{ ( x,y ) :0< y<\sqrt{3}a/2, d-y/\sqrt{3} <x<e-y/\sqrt{3} \}\). The boundary value problem (1)-(4) is considered on \(D^{\prime}\):

$$\begin{aligned}& \Delta v = 0\quad\text{on }D^{\prime}, \end{aligned}$$

(13)

$$\begin{aligned}& v = \psi_{j}\quad\text{on }\gamma_{j}^{\prime}, j=1,2,3,4, \end{aligned}$$

(14)

where \(\psi_{j}\) are the values of the solution of problem (1)-(4) on \(\gamma^{\prime}\).

Let \(h>0\), where \((e-d)/h\geq2\), \(a/\sqrt{3}h \geq2\) are integers. We assign to \(D^{\prime}\) a hexagonal grid of the form \(D_{h}^{\prime}= \{ (x,y)\in D^{\prime}:x=\frac{h}{2}(1-l)+kh, y=l\frac{\sqrt{3}h}{2}, k,l=0,\pm1,\pm2,\pm3,\ldots\} \). Let \(\gamma_{jh}^{\prime}\) be the set of nodes on the interior of \(\gamma _{j}^{\prime}\), and

$$\begin{aligned}& \dot{\gamma}_{jh}^{\prime} = \gamma_{j-1}^{\prime} \cap\gamma_{j}^{\prime},\qquad\gamma_{h}^{\prime}= \bigcup_{j=1}^{4}\gamma_{jh}^{\prime}, \quad j=1,2,3,4, \\& \overline{D}_{h}^{\prime} = D_{h}^{\prime}\cup \gamma_{h}^{\prime}. \end{aligned}$$

We consider the system of finite-difference equations:

$$\begin{aligned}& v_{h} = Sv_{h}\quad\text{on }D_{h}^{\prime}, \end{aligned}$$

(15)

$$\begin{aligned}& v_{h} = \psi_{j}\quad\text{on }\gamma _{jh}^{\prime}, j=1,2,3,4, \end{aligned}$$

(16)

where

$$\begin{aligned} Sv(x,y) =&\frac{1}{6}\biggl( v ( x+h,y ) +v \biggl( x+ \frac{h}{2},y+ \frac{\sqrt{3}}{2}h \biggr) +v \biggl( x- \frac{h}{2},y+\frac{\sqrt{3}}{2} h \biggr) \\ &{} +v ( x-h,y ) +v \biggl( x-\frac{h}{2},y-\frac{\sqrt{3}}{2} h \biggr) +v \biggl( x+\frac{h}{2},y-\frac{\sqrt{3}}{2}h \biggr) \biggr) . \end{aligned}$$

(17)

Since (17) has nonnegative coefficients and their sum is equal to 1, the solution of system (15), (16) exists and is unique (see [13]).

Everywhere below we will denote constants which are independent of *h* and of the cofactors on their right by \(c,c_{0},c_{1},\ldots\) , generally using the same notation for different constants for simplicity.

### Lemma 3.1

*Let*

$$ \psi_{j}(s)\in C^{4,\lambda}\bigl(\gamma_{j}^{\prime} \bigr),\quad0< \lambda<1 $$

(18)

*and*

$$ \psi_{j-1}^{ ( 3p ) }(s_{j})=\psi_{j}^{ ( 3p ) }(s_{j}), \quad p=0,1, $$

(19)

*be satisfied on the vertices whose interior angles are*
\(\alpha_{j}\pi =\pi/3\), *where*
\(j=1,2,3,4\). *Then the solution of problem* (13), (14)

$$ v\in C^{4,\lambda}\bigl(\overline{D}^{\prime}\bigr). $$

(20)

### Proof

The closed parallelogram \(\overline{D}^{\prime}\) lies inside the polygon *D* defined in Section 2, and the vertices \(\dot{\gamma} _{m}^{\prime}\) with an interior angle of \(\alpha_{m}\pi\neq\pi/3\) are located either inside of *D* or on the interior of a side \(\gamma_{m}\) of *D*, \(1\leq m\leq N\). Since the boundary functions (14), by the definition of the boundary functions \(\varphi_{j}\) in problem (1), (2) satisfy conditions (3), (4), from the results in [14], (20) follows. □

Let \(D_{h,k}^{\prime}\) be the set of nodes whose distance from the point \(P\in D_{h}^{\prime}\) to \(\gamma_{h}^{\prime}\) is \(\frac{\sqrt{3}}{2} kh\), \(1\leq k\leq a^{\ast}\), where \(a^{\ast}= [ \frac{2d_{t}}{\sqrt{3} h} ] \), \([ c ] \) denotes the integer part of *c*, and \(d_{t}\) is the minimum of the half-lengths of the sides of the parallelogram.

### Lemma 3.2

*Let*
\(w_{h}^{k}\neq\mathrm{const.}\)
*be the solution of the system of equations*

$$\begin{aligned}& w_{h}^{k} = Sw_{h}^{k}+f_{h}^{k} \quad\textit{on }D_{h,k}^{\prime}, \\& w_{h}^{k} = Sw_{h}^{k}\quad \textit{on }D_{h}^{\prime}\backslash D_{h,k}^{\prime }, \\& w_{h}^{k} = 0\quad\textit{on }\gamma_{h}^{\prime}, \end{aligned}$$

*and*
\(z_{h}^{k}\neq\mathrm{const.}\)
*be the solution of the system of equations*

$$\begin{aligned}& z_{h}^{k} = Sz_{h}^{k}+g_{h}^{k} \quad\textit{on }D_{h,k}^{\prime}, \\& z_{h}^{k} = Sz_{h}^{k}\quad \textit{on }D_{h}^{\prime}\backslash D_{h,k}^{\prime }, \\& z_{h}^{k} = 0\quad\textit{on }\gamma_{h}^{\prime}, \end{aligned}$$

*where*
\(1\leq k\leq a^{\ast}\). *If*
\(\vert f_{h}^{k}\vert \leq g_{h}^{k}\), *then*

$$ \bigl\vert w_{h}^{k}\bigr\vert \leq z_{h}^{k}, \quad1\leq k\leq a^{\ast}. $$

(21)

### Proof

The proof follows analogously to the proof of the comparison theorem given in [13]. □

### Lemma 3.3

*Let*
*v*
*be the trace of the solution of problem* (13), (14) *on*
\(\overline{D}_{h}^{\prime}\), *and*
\(v_{h}\)
*be the solution of system* (15), (16). *If*

$$ \psi_{j}(s)\in C^{4,\lambda}\bigl(\gamma_{j}^{\prime} \bigr),\quad0< \lambda<1, j=1,2,3,4 $$

*and*

$$ \psi_{j-1}^{ ( 3p ) }(s_{j})=\psi_{j}^{ ( 3p ) }(s_{j}), \quad p=0,1, $$

*on the vertices with an interior angle of*
\(\alpha_{j}\pi=\pi/3\), \(j=1,2,3,4\), *then*

$$ \max_{\overline{D}_{h}^{\prime}} \vert v-v_{h}\vert \leq ch^{4}. $$

(22)

### Proof

Let \(\epsilon_{h}=v_{h}-v\) on \(\overline{D}_{h}^{\prime}\). Clearly

$$\begin{aligned}& \epsilon_{h} = S\epsilon_{h}+(Sv-v)\quad\text{on }D_{h}^{\prime}, \end{aligned}$$

(23)

$$\begin{aligned}& \epsilon_{h} = 0\quad\text{on }\gamma_{h}^{\prime}. \end{aligned}$$

(24)

Let \(D_{1h}^{\prime}\) contain the set of nodes whose distance from the boundary \(\gamma^{\prime}\) is \(\frac{\sqrt{3}h}{2}\), and hence for \((x,y)\in D_{1h}^{\prime}\), \((x+sH,y+sK)\in\overline{D}^{\prime}\) for \(0\leq s\leq1\), \(H=\pm\frac{h}{2}, \pm h\), \(K=0, \pm\frac{\sqrt{3}h}{2} \), \(H^{2}+K^{2}>0\), and \(D_{2h}^{\prime}=D_{h}^{\prime}\backslash D_{1h}^{\prime}\).

Moreover, let

$$ \epsilon_{h}=\epsilon_{h}^{1}+\epsilon _{h}^{2}. $$

(25)

We rewrite problem (23), (24) as

$$\begin{aligned}& \epsilon_{h}^{1} = S\epsilon_{h}^{1}+(Sv-v) \quad\text{on }D_{1h}^{\prime}, \\& \epsilon_{h}^{1} = S\epsilon_{h}^{1} \quad\text{on }D_{2h}^{\prime}, \\& \epsilon_{h}^{1} = 0\quad\text{on }\gamma _{h}^{\prime} \end{aligned}$$

(26)

and

$$\begin{aligned}& \epsilon_{h}^{2} = S\epsilon_{h}^{2} \quad\text{on }D_{1h}^{\prime}, \\& \epsilon_{h}^{2} = S\epsilon_{h}^{2}+(Sv-v) \quad\text{on }D_{2h}^{\prime}, \\& \epsilon_{h}^{2} = 0\quad\text{on }\gamma _{h}^{\prime}. \end{aligned}$$

(27)

In order to obtain an estimation for \(Sv-v\) on \(D_{1h}^{\prime}\), we use Taylor’s formula. On the basis of Lemma 3.1, we have

$$ \vert Sv-v\vert \leq c_{3}h^{4}\quad\text{on }D_{1h}^{\prime}. $$

(28)

Since at least two values of \(\epsilon_{h}^{1}\) in \(S\epsilon _{h}^{1}\) are lying on the boundary \(\gamma_{h}^{\prime}\), on which \(\epsilon _{h}^{1}=0\), from (26), (28), and the maximum principle (see [13]), we obtain

$$ \max_{\overline{D}_{h}^{\prime}}\bigl\vert \epsilon_{h}^{1} \bigr\vert \leq\frac{2}{3}\max_{\overline{D}_{h}^{\prime}}\bigl\vert \epsilon_{h}^{1}\bigr\vert +c_{3}h^{4}. $$

Hence

$$ \max_{\overline{D}_{h}^{\prime}}\bigl\vert \epsilon_{h}^{1} \bigr\vert \leq c_{4}h^{4}, $$

(29)

where \(c_{4}=3c_{3}\).

Next, we consider the estimation of \(\epsilon_{h}^{2}\). Let \(D_{2h,k}^{\prime}\) be the set of nodes whose distance from the point \(P\in D_{2h}^{\prime}\) to \(\gamma_{h}^{\prime}\) is \(\frac{\sqrt{3}}{2}kh\), \(2\leq k\leq a^{\ast}\), where \(a^{\ast}= [ \frac{2d_{t}}{ \sqrt{3}h} ] \), \([ c ] \) denotes the integer part of *c*, and \(d_{t}\) is the minimum of the half-lengths of the sides of the parallelogram. Furthermore, \(D_{2h,1}^{\prime}\equiv D_{1h}^{\prime}\) and \(D_{2h,0}^{\prime}\equiv\gamma_{h}^{\prime}\). Since the vertices with \(\alpha_{j}=\frac{1}{3}\) of the parallelogram \(D^{\prime}\) are never used as a node of the hexagonal grid for the estimation of \(\vert Sv-v\vert \) on \(D_{2h,k}^{\prime}\), \(2\leq k\leq a^{\ast}\), we use the inequality

$$ \max_{p+q=6}\biggl\vert \frac{\partial^{6}v(x,y)}{\partial x^{p}\, \partial y^{q}}\biggr\vert \leq c_{0}\rho^{\lambda-2}\quad\text{on }\overline{D}^{\prime } \backslash\gamma_{m}^{\prime}, $$

for the sixth order derivatives, where *ρ* is the distance from \((x,y)\in D^{\prime}\) to \(\gamma_{m}^{\prime}\). Hence, we obtain

$$ \vert Sv-v\vert \leq c_{5}h^{6}/(kh)^{2-\lambda} \quad\text{on } D_{2h,k}^{\prime}, 2\leq k\leq a^{\ast}. $$

(30)

Consider a majorant function of the form

$$ Y_{k}=\left \{ \begin{array}{l@{\quad}l} 3m&\text{if }P\in D_{2h,m}^{\prime}, 0\leq m\leq k, \\ 3k&\text{if }P\in D_{2h,m}^{\prime}, m>k. \end{array} \right . $$

(31)

Hence \(Y_{k}\) is a solution of the finite-difference problem

$$\begin{aligned}& Y_{k} = SY_{k}+\mu_{k}\quad\text{on }D_{2h,k}^{\prime}, \\& Y_{k} = SY_{k}\quad\text{on }D_{h}^{\prime} \backslash D_{2h,k}^{\prime}, \\& Y_{k} = 0\quad\text{on }\gamma_{h}^{\prime}, \end{aligned}$$

(32)

where \(1\leq\mu_{k}\leq3\), \(1\leq k\leq a^{\ast}\).

We represent the solution of system (27) as the sum of the solution of the following subsystems:

$$\begin{aligned}& \epsilon_{h,k}^{2} = S\epsilon_{h,k}^{2}+ \mu_{k}^{\prime}\quad\text{on } D_{2h,k}^{\prime}, \\& \epsilon_{h,k}^{2} = S\epsilon_{h,k}^{2} \quad\text{on }D_{h}^{\prime }\backslash D_{2h,k}^{\prime}, \\& \epsilon_{h,k}^{2} = 0\quad\text{on }\gamma _{h}^{\prime}, \end{aligned}$$

(33)

where \(1\leq k\leq a^{\ast}\), \(\mu_{k}^{\prime}=0\) when \(k=1\) and \(\vert \mu_{k}^{\prime} \vert \leq c_{6}\frac{h^{4+\lambda}}{k^{2-\lambda}}\) when \(k=2,3,\ldots,a^{\ast}\).

By (32), (33), and Lemma 3.2, it follows that

$$ \bigl\vert \epsilon_{h,k}^{2}\bigr\vert \leq c_{6}\frac{h^{4+\lambda}}{k^{2-\lambda}}Y_{k}. $$

(34)

Hence, by taking (33) and (34) into consideration, we have

$$\begin{aligned} \max_{D_{h}^{\prime}}\bigl\vert \epsilon_{h}^{2} \bigr\vert \leq&\sum_{k=1}^{a^{\ast}}\epsilon _{h,k}^{2}\leq\sum_{k=1}^{a^{\ast}}c_{6} \frac{h^{4+\lambda}}{k^{2-\lambda}}Y_{k} \\ \leq&3c_{6}h^{4+\lambda}\sum_{k=1}^{a^{\ast}} \frac{1}{k^{1-\lambda}} \leq c_{7}h^{4}. \end{aligned}$$

(35)

On the basis of (25), (29), and (35), we have estimation (22). □