Finite-element meshes and interpolation — Examples

The first matrix is $A=\left(\begin{array}{cc}2& 0\\ 0& 3\end{array}\right)$, with determinant $6>0$. Applying it to the three vertices gives $(0,0),(2,0),(0,3)$. This is a right triangle with legs $2$ and $3$, so its area is $\frac{1}{2}\cdot 2\cdot 3=3$.

The second matrix is $B=\left(\begin{array}{cc}2& 1\\ 0& 3\end{array}\right)$, with determinant $6>0$. Applying it to the square's vertices gives $(0,0),(2,0),(1,3),(3,3)$, whose opposite edges are parallel because they are translates of the two column vectors $(2,0)$ and $(1,3)$. Hence the image is a parallelogram with base length $2$ and altitude $3$, so its area is $2\cdot 3=6$. It is an affine image of a square, not a simplicial cell.

Each cell in the stated triangular-grid subdivision has side $a=1\mathrm{/}n$. Its altitude is $\sqrt{3}a\mathrm{/}2$, so its area is $\sqrt{3}{a}^2\mathrm{/}4$ and its incenter lies one third of the altitude from each side; hence $\rho =\sqrt{3}a\mathrm{/}6$. Its diameter is $h=a$, so $h\mathrm{/}\rho =2\sqrt{3}$, independently of $n$. By [F1] this family is shape regular.

For $0<ϵ\le 1$, the sliver triangle's incenter joined to its three vertices partitions it into three triangles with common altitude ${\rho }_{ϵ}$, so its area is ${\rho }_{ϵ}$ times half its perimeter. Its side lengths are $1,ϵ,\sqrt{1+{ϵ}^2}$, its longest side is $h_{ϵ}=\sqrt{1+{ϵ}^2}$ and its area is $ϵ\mathrm{/}2$. Its perimeter is $1+ϵ+\sqrt{1+{ϵ}^2}$, so ${\rho }_{ϵ}=\frac{ϵ}{1+ϵ+\sqrt{1+{ϵ}^2}}\mathrm{.}$ The numerator of $h_{ϵ}\mathrm{/}{\rho }_{ϵ}$ tends to $1$, while its denominator is of order $ϵ$; hence the ratio tends to infinity. The second triangle in each strip is congruent, so every cell has the same ratio. Thus no uniform shape-regularity constant exists for this fixed-square mesh family.

Choose $R>0$ with $\overline{B(c,R)}\subset {\widehat{S}}_2$, and choose $\chi \in C_c^{\mathrm{\infty }}(B(c,R))$ equal to one on $B(c,R\mathrm{/}2)$. For $0<ϵ<R\mathrm{/}4$, put $L_{ϵ}=\log (R\mathrm{/}ϵ)$ and $u_{ϵ}(x)=\chi (x)\frac{\log \text{\hspace{0.17em}⁣}\left(R\mathrm{/}\sqrt{\mathrm{\mid }x-c{\mathrm{\mid }}^2+{ϵ}^2}\right)}{L_{ϵ}}\mathrm{.}$ This is smooth and compactly supported in the simplex, and $u_{ϵ}(c)=1$.

In polar coordinates about $c$, $r\ge 0$ and the stated range $0<ϵ<R\mathrm{/}4$ gives $r^2+{ϵ}^2\ge {ϵ}^2>0$ and the radial derivative has no zero denominator. The undivided logarithm has uniformly bounded squared $L^2$ integral on $B(c,R)$: its absolute value is bounded by $\log (R\mathrm{/}r)+C$ for $0<r<R$ and small $ϵ$, and ${\int }_0^{R}r(\log (R\mathrm{/}r)+C{)}^{2}dr<\mathrm{\infty }$. Its radial derivative has magnitude $r\mathrm{/}(r^2+{ϵ}^2)$, so ${\int }_0^R\frac{r^3}{(r^2+{ϵ}^2{)}^2}dr\le C+\log (R\mathrm{/}ϵ)=C+L_{ϵ},$ by splitting at $r=ϵ$. After division by $L_{ϵ}$, these estimates show the $L^2$ norm and the gradient contribution from $\chi \mathrm{\nabla }\log (R\mathrm{/}\sqrt{r^2+{ϵ}^2})$ tend to zero. The term $\mathrm{\nabla }\chi$ is supported where $R\mathrm{/}2\le r<R$, on which the logarithm is uniformly bounded; its divided $L^2$ norm also tends to zero. Therefore $\mathrm{\parallel }{u}_{ϵ}{\mathrm{\parallel }}_{W^{1,2}({\widehat{S}}_2)}\to 0$.

If point evaluation at $c$ were bounded, some finite $C$ would satisfy $\mathrm{\mid }u(c)\mathrm{\mid }\le C\mathrm{\parallel }u{\mathrm{\parallel }}_{W^{1,2}}$ for every smooth $u$. By [F2] this is the boundedness condition for the candidate nodal functional. Applying it to $u_{ϵ}$ gives $1\le C\mathrm{\parallel }{u}_{ϵ}{\mathrm{\parallel }}_{W^{1,2}}\to 0$, a contradiction. Hence evaluation is unbounded at the critical exponent.

Apply [F1] to the first triangle and the ordered triangle $[v_0,v_2,v_3]$ to obtain $\mathrm{\partial }[v_0,v_1,v_2]=[v_1,v_2]-[v_0,v_2]+[v_0,v_1],$ $\mathrm{\partial }[v_0,v_2,v_3]=[v_2,v_3]-[v_0,v_3]+[v_0,v_2]\mathrm{.}$ The edge $[v_0,v_2]$ appears with coefficient $-1$ in the first boundary and $+1$ in the second.

Adding the two formulas cancels the shared diagonal and leaves $[v_0,v_1]+[v_1,v_2]+[v_2,v_3]-[v_0,v_3]$, which is the oriented outer boundary since $-[v_0,v_3]=[v_3,v_0]$. This explicit calculation agrees with the general cancellation rule [F2].