Below, everywhere we assume that index \(\alpha=1, 2\) and \(\beta= 3-\alpha\).
Let \(\bar{\Omega}=\{x=(x_{1},x_{2}): 0\leq x_{\alpha}\leq l_{\alpha}\}\) represent a rectangle with boundary Γ.
Consider the problem
$$\begin{aligned}& \begin{aligned} &(\lambda+2\mu)\frac{\partial^{2} u^{1}}{\partial x_{1}^{2}} +(\lambda+\mu) \frac{\partial^{2} u^{2}}{\partial x_{1}\,\partial x_{2}} + \mu \frac{\partial^{2} u^{1}}{\partial x_{2}^{2}}+f^{1}(x) =0,\\ &\mu\frac{\partial^{2} u^{2}}{\partial x_{1}^{2}}+ (\lambda+\mu) \frac{\partial^{2} u^{1}}{\partial x_{1}\,\partial x_{2}}+ (\lambda+2\mu) \frac{\partial^{2} u^{2}}{\partial x_{2}^{2}} +f^{2}(x) =0,\quad x\in\Omega, \end{aligned} \end{aligned}$$
(2)
$$\begin{aligned}& u^{\alpha}(x)=0, \qquad\frac{\partial u^{\beta}(x)}{\partial x_{\alpha}}=0,\quad x\in\Gamma, x_{\alpha}=0, l_{\alpha}. \end{aligned}$$
(3)
Here \(\lambda, \mu=\mbox{const}\), \(\mu>0\). \(\lambda\geq-\mu\) are Lamé coefficients, \(\mathbf {u}=(u^{1},u^{2})^{T}\) is an unknown displacement vector, and \(\mathbf {f}=(f^{1},f^{2})^{T}\) is the given vector.
As usual, by the symbol \(W_{2}^{s}(\Omega)\), \(s\geq0\), we denote the Sobolev space. For integer s the norm in \(W_{2}^{s}(\Omega)\) is given by the formula
$$\|u\|^{2}_{W_{2}^{s}(\Omega)}=\sum_{j=0}^{s}|u|^{2}_{W_{2}^{j}(\Omega)},\qquad |u|^{2}_{W_{2}^{j}(\Omega)}=\sum_{|\nu|=j} \bigl\| D^{\nu}u\bigr\| ^{2}_{L_{2}(\Omega)}, $$
where \(D^{\nu}=\partial^{|\nu|}/ (\partial x_{1}^{\nu_{1}}\partial x_{2}^{\nu _{2}} )\), and \(\nu=(\nu_{1},\nu_{2}) \) is a multi-index with non-negative integer components, \(|\nu|=\nu_{1}+\nu_{2}\).
If \(s=\bar{s}+\varepsilon\), where s̄ is an integer part of s and \(0<\varepsilon<1\), then
$$\|u\|^{2}_{W_{2}^{s}(\Omega)}=\|u\|^{2}_{W_{2}^{\bar{s}}(\Omega)}+|u|^{2}_{W_{2}^{ s}(\Omega)}, $$
where
$$|u|_{W_{2}^{s}(\Omega)}=\sum_{|\nu|=\bar{s}} \int_{\Omega}\int_{\Omega}\frac {|D^{\nu}u(x)-D^{\nu}u(y)|^{2}}{|x-y|^{2+2\varepsilon}}\,dx \,dy. $$
Particularly, for \(s=0\) we have \(W_{2}^{0}=L_{2}\).
We assume that solution of the problem (2), (3) belongs to \(W_{2}^{m}(\Omega)\), \(m\geq2\).
Let us introduce the mesh domains
$$\begin{aligned}& \bar{\omega}_{\alpha}=\{x_{\alpha}=i_{\alpha}h_{\alpha}: i_{\alpha}=0,1,\ldots ,n_{\alpha}, n_{\alpha}h_{\alpha}=l_{\alpha}, n_{\alpha}\geq4\},\\& \omega_{\alpha}= \bar{\omega}_{\alpha}\setminus\{0,l_{\alpha}\} ,\qquad \omega_{\alpha}^{+}= \omega_{\alpha}\cup\{l_{\alpha}\}, \\& \gamma_{-\alpha}= \bigl\{ x=(x_{1},x_{2}): x_{\alpha}=0, x_{\beta}\in\omega_{\beta}\bigr\} ,\qquad \gamma_{+\alpha}= \bigl\{ x=(x_{1},x_{2}): x_{\alpha}=l_{\alpha}, x_{\beta}\in \omega_{\beta}\bigr\} , \\& \gamma_{\beta}=\gamma\setminus(\gamma_{-\alpha}\cup \gamma_{+\alpha }),\qquad \omega=\omega_{1}\times\omega_{2},\qquad \bar{\omega}= \bar{\omega}_{1}\times\bar{\omega}_{2},\\& \omega^{+}= \omega_{1}^{+}\times\omega _{2}^{+},\qquad \gamma=\bar{\omega}\setminus \omega, \\& \bar{\gamma}_{\pm\alpha} = \bigl\{ x=(x_{1},x_{2})\in \gamma: x_{\alpha}=(l_{\alpha}\pm l_{\alpha})/2 \bigr\} ,\qquad \omega_{(1)}=\omega_{1} \times\bar{\omega}_{2},\qquad \omega_{(2)}=\bar{\omega}_{1} \times\omega_{2}, \end{aligned}$$
\(|h|^{2}=h_{1}^{2}+h_{2}^{2}\), \(\hbar_{\alpha}=h_{\alpha}\) when \(x_{\alpha}\in\omega_{\alpha}\), \(\hbar_{\alpha}=h_{\alpha}/2 \) when \(x_{\alpha}=0, l_{\alpha}\).
Further, we define difference quotients in the \(x_{\alpha}\) direction as follows:
$$V_{x_{\alpha}}= \bigl(V^{(+1_{\alpha})}-V \bigr)/h_{\alpha},\qquad V_{\bar{x}_{\alpha}}= \bigl(V-V^{(-1_{\alpha})} \bigr)/h_{\alpha},\qquad V_{\stackrel {\circ }{x}_{\alpha}}= ( V_{x_{\alpha}} + V_{\bar{x}_{\alpha}} )/2, $$
where
$$V^{(\pm1_{1})}(x)=V(x_{1}\pm h_{1},x_{2}),\qquad V^{(\pm1_{2})}(x)=V(x_{1}, x_{2}\pm h_{2}). $$
The notations \(V^{(\pm0.5_{\alpha})}\) will have a similar meaning. Let
$$I_{\alpha}V(x):=\frac{V(x)+V^{(-1_{\alpha})}(x)}{2}. $$
Denote
$$\Lambda_{\alpha\alpha}V:= \left \{ \textstyle\begin{array}{@{}l@{\quad}l} \frac{2}{h_{\alpha}}V_{x_{\alpha}}, & x \in\gamma_{-\alpha}, \\ V_{\bar{x}_{\alpha}x_{\alpha}}, & x\in\bar{\omega}\setminus\gamma _{\alpha}, \\ -\frac{2}{h_{\alpha}}V_{\bar{x}_{\alpha}}, & x\in\gamma_{+\alpha}, \end{array}\displaystyle \right .\qquad \Lambda_{12}V:= \left \{ \textstyle\begin{array}{@{}l@{\quad}l} V_{x_{\alpha} \stackrel {\circ }{x}_{\beta}}, & x\in\gamma_{-\alpha}, \\ V_{\stackrel {\circ }{x}_{1} \stackrel {\circ }{x}_{2}}, & x\in\omega, \\ V_{\bar{x}_{\alpha} \stackrel {\circ }{x}_{\beta}}, & x\in\gamma _{+\alpha}. \end{array}\displaystyle \right . $$
On the set of mesh functions we define scalar products and norms:
$$\begin{aligned}& (Y,V)_{\tilde{\omega}}=\sum_{x\in\tilde{\omega}}h_{1}h_{2}Y(x)V(x),\qquad \| Y\|_{\tilde{\omega}}=(Y,Y)^{1/2}_{\tilde{\omega}},\quad \tilde{\omega}\subseteq \bar{\omega}, \\& (Y,V)_{(\alpha)}=\sum_{x\in\omega_{(\alpha)}} h_{\alpha}\hbar_{\beta} Y(x)V(x), \qquad\|Y\|_{(\alpha)}=(Y,Y)^{1/2}_{(\alpha)}. \end{aligned}$$
By \(\stackrel {\circ }{H}_{h} (\omega)\) we denote a set of two-dimensional vector-functions \(\mathbf{ V}=(V^{1}, V^{2})^{T}\), whose components are defined on ω̄ and equal to zero on \(\gamma_{\alpha}\), respectively. Let \(H_{h}\) be the set of two-dimensional vector-functions, whose components are defined on the meshes \(\omega_{(\alpha)}\), respectively.
Define in \(\stackrel {\circ }{H}_{h} (\omega)\) the inner product and the norms:
$$\begin{aligned}& (\mathbf {U}, \mathbf {V})= \bigl(U^{1},V^{1} \bigr)_{(1)}+ \bigl(U^{2},V^{2} \bigr)_{(2)},\qquad \| \mathbf {U}\|=(\mathbf {U}, \mathbf {U})^{1/2}, \\& \|\mathbf {U}\|^{2}_{W_{2}^{2}(\omega)}:=\sum_{\alpha=1}^{2} \bigl( \bigl\| \Lambda_{11}U^{\alpha}\bigr\| ^{2}_{(\alpha)}+ \bigl\| \Lambda_{22}U^{\alpha}\bigr\| ^{2}_{(\alpha)}+ 2 \bigl\| U^{\alpha}_{\bar{x}_{1} \bar{x}_{2}}\bigr\| ^{2}_{\omega^{+}} \bigr). \end{aligned}$$
For functions, defined on Ω, we need the following averaging operators:
$$\begin{aligned}& S_{1}^{-} u(x):= \frac{1}{h_{1}} \int_{x_{1}-h_{1}}^{x_{1}}u(t_{1},x_{2}) \,dt_{1},\qquad T_{1}u(x):=\frac{1}{h_{1}^{2}} \int _{x_{1}-h_{1}}^{x_{1}+h_{1}}\bigl(h_{1}-|x_{1}-t_{1}|\bigr)u(t_{1},x_{2}) \,dt_{1}, \\& T_{1}u(0,x_{2}):=\frac{2}{h_{1}^{2}} \int_{0}^{h_{1}}(h_{1}-t_{1})u(t_{1},x_{2}) \,dt_{1}, \\& T_{1}u(l_{1},x_{2}):=\frac{2}{h_{1}^{2}} \int _{l_{1}-h_{1}}^{l_{1}}(h_{1}-l_{1}+t_{1})u(t_{1},x_{2}) \,dt_{1}. \end{aligned}$$
Analogously are defined operators \(T_{2}\), \(S_{2}^{-}\). Note that these operators are commutative and
$$T_{\alpha}\frac{\partial^{2} u}{\partial x_{\alpha}^{2}}=u_{\bar{x}_{\alpha}x_{\alpha}},\qquad T_{\alpha}\frac{\partial u}{\partial x_{\alpha}}= \bigl( S_{\alpha}^{-} u \bigr)_{ x_{\alpha}}. $$