HEJ, HU ISSN 1418-7108 Manuscript no.: ANM-980205-A

A Finite Element Representation

Let us assume that the intervals $[0,a]$ and $[0,b]$ are endowed with equidistant meshes corresponding to the grid sizes $h_a$ and $h_b$, respectively and let $N_a= {a}/{h_a}$ and $N_b={b}/{h_b}$. We discretize $Q = [0,a] \times [0,b]$ as a direct product of the discretizations defined on the intervals $[0,a]$ and $[0,b]$. The elements of the piecewise bilinear finite element space $V_h(Q)$ corresponding to this discretization can be expressed as

$$g(x_1,x_2) = \sum_{i=0}^{N_a} \sum_{j=0}^{N_b} g_{ij} \varphi_i(x_1) \psi_j(x_2), \quad (x_1,x_2) \in Q$$ (14)

where $g_{ij}$ denote the function values at the grid points $(i h_a,j h_b)$ and $\varphi_i$ and $\psi_j$ are the one-dimensional piecewise linear shape funcions corresponding to the discretization of the intervals $[0,a]$, $[0,b]$, respectively.

In this space the 'partial' seminorms $\mid . \mid^2_{H_a^{1/2}(Q)}$ and $\mid . \mid^2_{H_b^{1/2}(Q)}$ can be simplified to a sum of one-dimensinal seminorms:

Theorem 3.1   Let $g \in V_h(Q)$. Then

$$2 \mid g \mid^2_{H_a^{1/2}(Q)} \le h_a \sum_{i=0}^{N_a} \... ...2 -y_2 \mid^2} dy_2 dx_2 \le 12 \mid g \mid^2_{H_a^{1/2}(Q)},$$ (15)

and

$$2 \mid g \mid^2_{H_b^{1/2}(Q)} \le h_b \sum_{j=0}^{N_b} \... ... -y_1 \mid^2} dy_1 dx_1 \le 12 \mid g \mid^2_{H_b^{1/2}(Q)}.$$ (16)

The proof of the theorem is based on the following lemma.

Lemma 3.2  

$$2 \int_0^h({c\frac{t}{h}+d({1-\frac{t}{h}})})^2dt \le h(c^2+d^2) \le 6 \int_0^h({c\frac{t}{h}+d({1-\frac{t}{h}})})^2dt,$$ (17)

$$\mbox{for all } c,d \in (-\infty,infty) \mbox{ and } h > 0$$

Proof 3.3   Since

$$6 \int_0^h({c\frac{t}{h}+d({1-\frac{t}{h}})})^2dt = h(c^2+d^2+cd)$$

the inequality

$$\frac{1}{2}(c^2+d^2) \le c^2+d^2+cd \le \frac{3}{2}(c^2+d^2)$$

proves our lemma.

Proof 3.4 (Proof of Theorem 3.1.)   In the case of $g \in V_h(Q)$ for the seminorm $\mid . \mid_{H_a^{1/2}(Q)}$ the following indentities hold:

$$\mid g \mid^2_{H_a^{1/2}(Q)} =$$

$$= \int_0^a \int_0^b \int_0^b \frac{\mid \sum_{i=0}^{N_a}\... ...-\psi_j(y_2)) \mid^2} {\mid x_2 - y_2 \mid^2} dy_2 dx_2 dx_1$$ (18)

$$= \sum_{i=0}^{N_a-1} \int_0^b \int_0^b \frac{1}{\mid x_2 - ... ...hi_i(x_1) \sum_{j=0}^{N_a} g_{ij}(\psi_j(x_2)-\psi_j(y_2)) +$$

$$+ \varphi_{i+1}(x_1) \sum_{j=0}^{N_a} g_{i+1,j}(\psi_j(x_2)-\psi_j(y_2)) \mid^2 dx_1 dy_2 dx_2$$ (19)

$$= \sum_{i=0}^{N_a-1} \int_0^b \int_0^b \frac{1}{\mid x_2 - ... ..._1}{h_a} \sum_{j=0}^{N_a} g_{ij}(\psi_j(x_2)-\psi_j(y_2)) +$$

$$+ ({1-\frac{x_1}{h_a}}) \sum_{j=0}^{N_a} g_{i+1,j}(\psi_j(x_2)-\psi_j(y_2)) \mid^2 dx_1 dy_2 dx_2$$ (20)

Using Lemma 3.2 with $t=x_1$, $h=h_a$, $c=\sum_{j=0}^{N_a}g_{ij}(\psi_j(x_2)-\psi_j(y_2))$ and $d=\sum_{j=0}^{N_a}g_{i+1,j}(\psi_j(x_2)-\psi_j(y_2))$ one can get

$$2\int_{0}^{h_a} \mid \frac{x_1}{h_a} \sum_{j=0}^{N_a} g_{... ...0}^{N_a} g_{i+1,j}(\psi_j(x_2)-\psi_j(y_2)) \mid^2 dx_1 \le$$

$$\le h_a({ \mid \sum_{j=0}^{N_a}g_{ij}(\psi_j(x_2)-\psi_j(... ..._{j=0}^{N_a}g_{i+1,j}(\psi_j(x_2)-\psi_j(y_2)) \mid^2 }) \le$$ (21)

$$\le 6\int_{0}^{h_a} \mid \frac{x_1}{h_a} \sum_{j=0}^{N_a} ... ..._{j=0}^{N_a} g_{i+1,j}(\psi_j(x_2)-\psi_j(y_2)) \mid^2 dx_1.$$

Consider

$$\mid \sum_{j=0}^{N_a}g_{ij}(\psi_j(x_2)-\psi_j(y_2)) \mid^2 = \mid g(i h_a,x_2) - g(i h_a,y_2)\mid^2$$

and summing up (21) for $i$, (20) implies the inequality (15). The inequality (16) can be proved analogously.

HEJ, HU ISSN 1418-7108 Manuscript no.: ANM-980205-A