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). □