Curved geometry, mesh generation and conservative transfer — Examples

The derivative matrix is $DF=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, whose determinant is $-1$.

Since $-1$ is not bounded below by any positive $j_0$, this map is rejected by the admissibility certificate.

$g^{\mathrm{\prime }}(t)=a(\pi \mathrm{/}4){\sec }^2(\pi t\mathrm{/}4)>0$, so $H$ maps $[-1,1{]}^2$ bijectively onto $[-a,a{]}^2$, since $g(-1)=-a$ and $g(1)=a$. Its Gram determinant gives the same volume factor by [F2]. On the top sector, ${\theta }^{\mathrm{\prime }}=-\pi \mathrm{/}4$ and $r_s=1-a\mathrm{/}\sin \theta \ge 1-a\sqrt{2}>0$ because $\sin \theta \ge \sqrt{2}\mathrm{/}2$. Polar-coordinate differentiation gives $\det DF=r{r}_s(-{\theta }^{\mathrm{\prime }})>0$; the Gram determinant satisfies $\sqrt{\det (D{F}^{T}DF)}=\det DF$. Rotations preserve these determinants.

At $s=0$, $F(t,0)=(a\cot \theta (t),a)=(g(t),a)$, which is exactly the trace of $H$ on its top edge; at $s=1$, $r=1$, so the outer trace is the unit-circle arc from angle $3\pi \mathrm{/}4$ to $\pi \mathrm{/}4$. The two radial side traces match those of adjacent rotated sectors. Hence all ten pieces meet only along identical mapped edges.

For each angle $\theta \in [\pi \mathrm{/}4,3\pi \mathrm{/}4]$, the square boundary on that ray has radius $a\mathrm{/}\sin \theta$ and the unit-circle boundary has radius $1$. The sector map covers exactly the radii between them. Four rotated angular ranges cover the annulus outside the central square, with shared corner rays only; $H$ covers the central square. Therefore the ten mapped triangles cover the unit disk without interior overlap.

The central area is $4a^2=1$. By [F3], the sector area is the polar integral $\frac{1}{2}{\int }_{\pi \mathrm{/}4}^{3\pi \mathrm{/}4}(1-a^2{\csc }^2\theta )d\theta$. Since $[-\cot \theta {]}_{\pi \mathrm{/}4}^{3\pi \mathrm{/}4}=2$, this equals $\pi \mathrm{/}4-a^2=(\pi -1)\mathrm{/}4$. Four rotated sectors have total area $\pi -1$, so the union area is $\pi$, consistent with the unit disk.

The shear is a global $C^1$ diffeomorphism with inverse $(X,Y)\mapsto (X,Y-X^2\mathrm{/}4)$ and derivative $\left(\begin{array}{cc}1& 0\\ x\mathrm{/}2& 1\end{array}\right)$ of determinant $1$. Apply [F4] to each overlap to see that its physical area equals the reference overlap area. In the reference square, the two diagonals meet at the center; the two diagonals divide the square into four corner triangles, each with base $2$ along a side and altitude $1$ to the center, so each nonempty source-target overlap has area $1$.

Each target triangle has area $2$ and receives source mass $2(1)+0(1)=2$, so its target average is $1$. The source mass is $2\cdot 2+0\cdot 2=4$, and target mass is $2(1)+2(1)=4$.

The exact remap therefore has source values $(2,0)$, four nonempty overlaps of area $1$, target cell areas $2$, and target averages $(1,1)$. This explicitly verifies the remap formula.

Differentiate $\mathrm{\Phi }$ to obtain $D\mathrm{\Phi }=\left(\begin{array}{cc}1& 0\\ x\mathrm{/}2& 1\end{array}\right)$ and $\det D\mathrm{\Phi }=1$. The quadratic term makes the image of a nonvertical straight edge a parabolic arc, while vertical edges remain vertical.

Apply [F2] on the shape-regular mesh to get the stated value error, derivative error, and positive-Jacobian certificate.

On each shared reference edge, the two affine traces of $I_h(x^2\mathrm{/}4)$ agree at its endpoints and therefore agree everywhere on that edge. Thus ${\mathrm{\Phi }}_h$ has matching traces; the exact map is globally smooth and has matching traces as well. On each boundary edge, ${\mathrm{\Phi }}_h$ is the straight segment joining the exact images of its vertices. The vertical boundary traces agree exactly with $\mathrm{\Phi }$; on each horizontal boundary segment, ${\mathrm{\Phi }}_h(x,\pm 1)=(x,\pm 1+I_h(x^2\mathrm{/}4))$, so the boundary discrepancy is bounded by $C{h}^2$ from step 1.2. Compatibility here does not assert equality of the exact parabolic arcs and their straight approximations. By [F2], ${\mathrm{\Phi }}_h$ is continuous and injective on the compact square, hence by [F4] is a homeomorphism onto its image, as is $\mathrm{\Phi }$. Its cell restrictions are invertible affine maps extending to neighborhood diffeomorphisms; the exact restrictions extend by the global shear. These facts verify matching and boundary-trace compatibility and common image topology in [F1].

Step 1.1 bounds $D\mathrm{\Phi }$ uniformly, and step 1.2 gives $\mathrm{\parallel }D{\mathrm{\Phi }}_h{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }D\mathrm{\Phi }{\mathrm{\parallel }}_{\mathrm{\infty }}+Ch$ and determinant lower bounds $J=1$, $J_h\ge 1\mathrm{/}2$. Apply [F3] cellwise, with the compatible maps from step 2.1, and substitute $e_1\le Ch$ from step 1.2. This yields $\mathrm{\parallel }B-B_h{\mathrm{\parallel }}_{\mathrm{\infty }}\le Ch$ with constants uniform over the mesh family.

For every $n\ge 1$, ${\epsilon }_n\in \mathbb{D}$ and the four sites therefore satisfy the dyadic-coordinate input condition of [F1]. The first three points are noncollinear in $z=0$ and $p_4$ has positive height, so the four points are affinely independent and form a nondegenerate tetrahedron. Its volume is one third of base area $1\mathrm{/}2$ times height ${\epsilon }_n$, namely $V_n={\epsilon }_n\mathrm{/}6$. With only these four sites, its circumsphere contains no other site in its interior, so [F1] makes it an empty-sphere Delaunay tetrahedron.

The total face area $S_n$ is at least the base area $1\mathrm{/}2$ and is bounded above uniformly in $n$, since every edge length is at most $2$ and each of the four face areas is at most half the product of two such edge lengths; in particular $S_n\le 8$. The inradius is $r_n=3V_n\mathrm{/}{S}_n={\epsilon }_n\mathrm{/}(2S_n)$, so ${\epsilon }_n\mathrm{/}16\le r_n\le {\epsilon }_n$ and $r_n\to 0$. The diameter is at least $\mathrm{\mid }{p}_1-p_2\mathrm{\mid }=1$. Thus empty-sphere Delaunay membership imposes no positive uniform inradius-to-diameter bound in three dimensions.