Analysis of the Finite Element Method for the Laplace–Beltrami Equation on Surfaces with Regions of High Curvature Using Graded Meshes

Abstract

We derive error estimates for the piecewise linear finite element approximation of the Laplace–Beltrami operator on a bounded, orientable, $C^3$, surface without boundary on general shape regular meshes. As an application, we consider a problem where the domain is split into two regions: one which has relatively high curvature and one that has low curvature. Using a graded mesh we prove error estimates that do not depend on the curvature on the high curvature region. Numerical experiments are provided.

Appendix A: One Technical Result

Lemma 19

Assume that $\Gamma$ is a $C^3$ two-dimensional compact orientable surface without boundary, and that $u\in H^2(\Gamma )$. Then

$$\begin{aligned}&\sum _{i,j=1}^3\int _{\Gamma _2}\underline{D}_{ij}u\underline{D}_i\left( \rho ^2\underline{D}_ju\right) \,dA \\&\quad =\sum _{j=1}^3\int _{\Gamma _2}\Delta _\Gamma u\underline{D}_j\left( \rho ^2\underline{D}_ju\right) \,dA -\int _{\Gamma _2}\rho ^2\bigl ({\text {tr}}(H)H-2H^2\bigr ){\varvec{\nabla }_\Gamma }u\cdot {\varvec{\nabla }_\Gamma }u\,dA. \end{aligned}$$

Proof

In what follows, we use the two identities [17]

$$\begin{aligned} \int _\Gamma \underline{D}_iuv\,dA=-\int _\Gamma u\underline{D}_iv\,dA+\int _\Gamma uv{\text {tr}}(H)\nu _i\,dA, \end{aligned}$$

(52)

and

$$\begin{aligned} \underline{D}_{ij}u=\underline{D}_{ji}u+(H{\varvec{\nabla }_\Gamma }u)_j\nu _i-(H{\varvec{\nabla }_\Gamma }u)_i\nu _j, \end{aligned}$$

(53)

for all $C^3(\Gamma )$ functions u, v. Also we will use that of course

$$\begin{aligned} {\varvec{\nabla }_\Gamma }u\cdot {\varvec{\nu }}=0. \end{aligned}$$

(54)

We assume for the proof that $u\in C^3(\Gamma )$, and the general result follows from density arguments. Following [17, Lemma 3.2], and using the Einstein summation convention,

$$\begin{aligned}&\int _\Gamma \underline{D}_{ij}u\underline{D}_i\left( \rho ^2\underline{D}_ju\right) \,dA \\&\quad =-\int _\Gamma \underline{D}_{iij}u\left( \rho ^2\underline{D}_ju\right) \,dA +\int _\Gamma \underline{D}_{ij}u\rho ^2\underline{D}_ju{\text {tr}}(H)\nu _i\,dA&\text { by } (52) \\&\quad =-\int _\Gamma \underline{D}_{iij}u \left( \rho ^2\underline{D}_ju\right) \,dA\quad&\text { by } (55) \\&\quad =-\int _\Gamma \underline{D}_i\bigl [\underline{D}_{ji}u+(H{\varvec{\nabla }_\Gamma }u)_j\nu _i-(H{\varvec{\nabla }_\Gamma }u)_i\nu _j\bigr ]\rho ^2\underline{D}_ju\,dA \quad&\text { by } (53) \\&\quad =-\int _\Gamma \underline{D}_i\bigl [\underline{D}_{ji}u+H_{jk}\underline{D}_ku\nu _i-H_{ik}\underline{D}_ku\nu _j\bigr ]\rho ^2\underline{D}_ju\,dA \\&\quad =-\int _\Gamma \bigl (\underline{D}_{iji}u+H_{jk}H_{ii}\underline{D}_ku-H_{ik}H_{ij}\underline{D}_ku\bigr )\rho ^2\underline{D}_ju\,dA&\text { by } (54) \\&\quad =-\int _\Gamma \left( \underline{D}_{iji}u \rho ^2\underline{D}_ju + \rho ^2({\text {tr}}(H) H- H^2) {\varvec{\nabla }_\Gamma }u \cdot {\varvec{\nabla }_\Gamma }u \right) \,dA. \end{aligned}$$

To handle the first term on the right hand side, we use (53) and the fact that $\underline{D}_{ii} u= \Delta _{\Gamma } u$ to write:

$$\begin{aligned} -\int _\Gamma \underline{D}_{iji}u\rho ^2\underline{D}_ju\,dA&=-\int _\Gamma \bigl [\underline{D}_j\Delta _\Gamma u+(H{\varvec{\nabla }_\Gamma }\underline{D}_iu)_j\nu _i-(H{\varvec{\nabla }_\Gamma }\underline{D}_iu)_i\nu _j\bigr ]\rho ^2\underline{D}_ju\,dA \\&=-\int _\Gamma \underline{D}_j\Delta _\Gamma u\rho ^2\underline{D}_ju+H_{jk}\underline{D}_{ki}u\nu _i\rho ^2\underline{D}_ju\,dA, \end{aligned}$$

where we used (54) in the last equation. But $\underline{D}_{ki}u\nu _i=\underline{D}_k(\underline{D}_iu\nu _i)-\underline{D}_iuH_{ki}=-\underline{D}_iuH_{ki}$, and then

$$\begin{aligned} -\int _\Gamma \underline{D}_{iji} u\rho ^2 \underline{D}_ju\,dA&=\int _\Gamma -\underline{D}_j\Delta _\Gamma u\rho ^2\underline{D}_ju+H_{jk}\underline{D}_iuH_{ki}\rho ^2\underline{D}_ju\,dA \\&=\int _\Gamma \Delta _\Gamma u\underline{D}_j\left( \rho ^2\underline{D}_ju\right) +\rho ^2H^2{\varvec{\nabla }_\Gamma }u\cdot {\varvec{\nabla }_\Gamma }u\,dA. \end{aligned}$$

In the last equation we used (52) and (54). This completes the proof. $\square$