DOI 10.1007/s10915-017-0580-y

**Analysis of the Finite Element Method for the**

**Laplace–Beltrami Equation on Surfaces with Regions of**

**High Curvature Using Graded Meshes**

**Johnny Guzman1** **· Alexandre Madureira2****,3****·**
**Marcus Sarkis4** **· Shawn Walker5**

Received: 11 May 2017 / Revised: 28 August 2017 / Accepted: 11 October 2017 © Springer Science+Business Media, LLC 2017

**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.

**Keywords Finite elements**· Laplace–Beltrami · Curvature

**1 Introduction**

Since the publication of the seminal paper [16], there has been a growing interest in the discretization of surface partial differential equations (PDEs) using finite element methods Johnny Guzman and Marcus Sarkis are supported by National Science Foundation (Grant Nos. DMS #1620100 and NSF-MPS 1522663).

### B

Johnny Guzman johnny_guzman@brown.edu Alexandre Madureira alm@lncc.br; alexandre.madureira@fgv.br Marcus Sarkis msarkis@wpi.edu Shawn Walker walker@math.lsu.edu1 _{Brown University, Providence, RI, USA}

2 _{Laboratório Nacional de Computação Científica, Petropolis, RJ, Brazil}
3 _{Fundacao Getulio Vargas, Rio de Janeiro, RJ, Brazil}

4 _{Mathematical Sciences Department, Worcester Polytechnic Institute, Worcester, MA, USA}
5 _{Department of Mathematics and Center for Computation and Technology (CCT), Louisiana State}

(FEMs). Such interest is motivated by important applications related to physical and biolog-ical phenomena, and also by the potential use of numerbiolog-ical methods to answer theoretbiolog-ical questions in geometry [16,17,33].

In this paper, we focus on linear finite element methods for the Poisson problem with the
Laplace–Beltrami operator on* ⊂ R*3*, a C*3two-dimensional compact orientable surface
without boundary. That is, we consider

*−u= f on .*

In order to motivate the results in our paper, we start by giving a short overview of
previ-ous results. A piecewise linear finite element method is proposed and analyzed in [16,17].
The basic idea is to consider a piecewise linear approximation of the surface, and pose a
finite element method over the discretized surface. Discretizing the surface, of course,
*cre-ates a geometric error, however, the advantage is that for a given discretization a surface*
parametrization is not necessary.

In [12] a generalization of the piecewise linear FEM is considered, based on higher order polynomials that approximate both the geometry and the PDE; the same paper proposes a variant of the method which employs parametric elements, and the method is posed on the surface, originating thus no geometric error. Discontinuous Galerkin schemes are considered in [1,11], and HDG and mixed versions are considered in [8]. Adaptive schemes are presented in [3,7,10,13,14]. An alternative approach, where a discretization of an outer domain induces the finite element spaces is proposed in [26,27]. See also [4,6,14,28,29]. In [2,7,9,30], the PDE itself is extended to a neighborhood of the surface before discretization.

Other problems and methods were considered as well, as a multiscale FEM for PDEs posed on rough surface [18], and stabilized methods [4,21,22,29]. In [23] the finite element exterior calculus framework was considered. Finally, transient and nonlinear PDEs were also subject of consideration, as reviewed in [17].

A common ground between all aforementioned papers is that the a-priori choices of
the surface discretization do not consider how to locally refine the mesh following some
optimality criterion. It is however reasonable to expect that some geometrical traits, as the
curvatures, have a local influence on the solution, and thus the mesh refinement could account
for that locally. Not surprisingly, numerical tests using adaptive schemes confirm that high
curvature regions require refined meshes [3,6,10,13]. This is no different from problems in
*nonconvex flat domains, where corner singularities arise, and meshes are used to tame the*
singularity at an optimal cost [31,32].

As far as we can tell, the development and analysis of a-priori strategies to deal with high
curvature regions have not been an object of investigation, so far. In this paper we consider
a simple setting: We suppose that the domain* = *1*∪ 2*, and assume that the maximum
of the principal curvatures in1 is much larger than those in2. We then seek a graded
mesh that gives us optimal error estimate. Of course, in the region1the triangles will be
much smaller than the mesh size in regions far from1. We consider the method originally
proposed by Dziuk [16].

To carry out the analysis, we first need to track the geometric constants carefully. This, as
far as we can tell, has not explicitly appeared in literature, although it is not a difficult task.
We do this by following [16,17] although in some cases we give different arguments while
trying to be as precise as possible. The estimate we obtain is found below in (41*). If u _{h}* is

*the finite element solution approximation to u then the result reads (see sections below for*precise notation):

**∇**
*u− u _{h}*

*L*2

_{()}≤ Ccp*(h+ h) f L*2

*+*

_{()}*f − fh*

*2*

_{L}

_{()}*+ C*⎛ ⎝

*T∈Th*

*h*2

*∇*

_{T}*2*

_{}*u*2

*2*

_{L}*⎞ ⎠ 1*

_{(T}_{)}*/2*

*.*

*Here f _{h}is an approximation to f , cp* is the Poincaré’s constant, and the numbers

*h*,

*h*

are geometric quantities. For instance,*h* = max*Tκ _{T}*2

*h*2

_{T}*where hT*is the diameter of the

*triangle T andκT* *measures the maximum principal curvatures on T* *(T* is the surface

*triangle corresponding to T ; see sections below). The important point here is thath+ h*

can be controlled locally. That is, if one wants to reach a certain tolerance, one needs to make

*hT* *small enough only depending on the geometry in T*. We point out that max*TκT* large is

*a necessary but not sufficient condition for cpto be large, this is related to a dumbbell shape*

of the surface. Surface with bumps for instance can allow large max*TκT* *and cp= O(1).*

On the other hand,∇* _{}*2

*u*

*2*

_{L}*does not depend only on the local geometry. In order*

_{(T}_{)}*to deal with this term, in the case of two sub-regions, we prove local H*2regularity results. Combining the local regularity and the a-priori error estimate (41) we are able to define a mesh grading strategy and prove Theorem16. The error estimate contained in Theorem16

is independent of the curvature in region1, and in some sense is the best error estimate one can hope for given the available information.

The paper is organized as follows. In Sect.2 we set the notation and derive several
fundamental estimates highlighting the influence of the curvatures. Section3 regards the
finite element and interpolation approximations. Finally, we present in Sect.4*a local H*2
estimate and a mesh grading scheme culminating in an error estimate that is independent of
*the bad curvature. The paper ends with numerical results in Sect.*5.

**2 Preliminaries**

As mentioned above, we assume that* is bounded, orientable, C*3surface without boundary.
Furthermore, we assume that there is a high curvature region1* , and define 2= \1*.
*For f* *∈ L*2_{() with}

*f d A= 0, let u ∈ ˚H*1*() be such that*

**∇***u***· ∇***v d A =*

*fv d A for all v ∈ ˚H*

1_{(),}_{(1)}

where ˚*H*1*() = {v ∈ H*1*():* * _{}v d A = 0}. We denote by ∇_{}*the tangential gradient [33],
and (1) is nothing but the weak formulation of the Poisson problem for the Laplace–Beltrami
operator. Existence and uniqueness of solution follows easily from the Poincaré’s
inequal-ity (Lemma8) and the Lax–Milgram theorem. Details on the definition of

**∇**

*v are given*

below. Consider now a triangulation *h* of the surface*. By that we mean that h* is a

*two-dimensional compact orientable polyhedral C*0 surface, and denoting by*Th* the set of

closed nonempty triangles such that∪*T∈ThT* *= h*, we assume that all vertices belong to
*, and that any two triangles have as intersection either the empty set, a vertex or an edge.*

*Let hT* *= diam(T ) and h = max{hT: T ∈* *Th}. For all T ∈* *Th* assume that there is a

three-dimensional neighborhood*NT* * of T where for every x*∈

*NT*there is a unique closest

* point a(x) ∈ (see Fig.*1) such that

### Γ

*N*

**ν**

**ν**

**x**

**a(x)**

*d > 0*

*d < 0*

**Fig. 1 Diagram of closest point map (**2). The exact surface is denoted by* which is contained in the “tubular”*
neighborhood*N . Since d is the signed distance function to , the zero level set of d coincides with *

*d(x) is the signed distance function from x to and define ν(x) = (∇d(x))t* _{and note}

that* ν(a(x)) = ν(x) and it is the unit normal to at a(x). Note that by using local tubular*
neighborhoods, we avoid unnecessary restrictions on the mesh size.

Here we would like to explain some notational conventions that we use. From now on,
the gradient of a scalar function will be a row vector. The normal vector* ν is a column vector*
(as well as

*which is defined below).*

**ν**h*We can now define, for every T* ∈*Th, the surface triangle T* **= {a(x): x ∈ T }. Then*** = ∪T∈ThT . For x* ∈

*NT*, define the self-adjoint extended Weingarten operator [15]

*H*3 → R3

**(x) : R***by H*2

**(x) = ∇***d(x) and let κi*

**(x), i = 1, 2, and 0 be the eigenvalues of***H*1

**(x). We note that κ***2*

**(a(x)) and κ**

**(a(x)) are the principal curvatures. We note that***κi (x) =*

*κi*

**(a(x))**1**+ d(x)κ**i**(a(x))**

(3) for curvature of parallel surfaces [20, Lemma 14.17] (also see [13]).

We assume the natural geometrical condition (see also [12] for a similar condition).
* |d(x)| max{|κ1(a(x))|, |κ*2

*0*

**(a(x))|} ≤ c***(4)*

**< 1 for all x ∈**NT,*where c*0is sufficiently small.

*At the estimates that follow in this paper we denote by C a generic constant that might*
*not assume the same value at all occurrences, but that does not depend on hT, u, f or on.*

*It might however depend for instance on the shape regularity of T* ∈*Th*.

*Given T* ∈*Th*, let

*κT* *= HL*∞*(T )*:= max

*i j* *Hi j**L*∞*(T ),* (5)

and* νh* ∈ R3

*be unit-normal vector to T such that*

**ν**h

**· ν > 0. We note that since H**is symmetric,*κT* *is also equivalent to the L*∞ *norm of the spectral radius of H (a(·)), or*

max*{|κ1 (a(·))|L*∞

*(T ), |κ*2

*∞*

**(a(·))|**L*(T )*}, since from (4) we have

1

1*+ c0|κi (a(x))| ≤ |κi(x)| ≤*
1

1*− c0|κi (a(x))|.*

**Assumption Throughout the paper we will assume**

*h*2* _{T}κ_{T}*2

*≤ c1< 1 for all T ∈Th,*(6)

It is easy to see that

*dL*∞*(T )≤ Ch*2*TκT,* (7)

**ν − ν**h*L*∞*(T )≤ ChTκT.* (8)

To show (7*), assume without loss of generality that T* ⊂ R2*× {0} and that ν3,ν*3*,h> 0. Let*

*Ihd be the Lagrange linear interpolant of d in T . Since d vanishes at the vertices, Ihd*≡ 0.

By [19] we have

*dL*∞*(T )+ hT∂d/∂xi**L*∞*(T )= d − Ihd**L*∞*(T )+ hT∂(d − Ihd)/∂xi**L*∞*(T )*

*≤ Ch*2

*T*∇2*d**L*∞*(T ),* (9)

*for i* *= 1, 2. Next, to prove (*8), start by noting that**ν**t_{h}*= (0, 0, 1), and then the estimate*
for the first two components*νi**L*∞*(T )= ∂d/∂xi**L*∞*(T )*of* ν follow from (*9) and (5). The

third component estimate follows from

*ν3*− 1*L*∞*(T )≤ (ν3− 1)(ν3+ 1)L*∞*(T )*=*ν*32− 1*L*∞*(T )≤ ν1*
2
*L*∞*(T )+ ν2*2*L*∞*(T )*
*≤ Ch*2
*TκT*2 *≤ ChTκT.*
Here we used (6).

From (7) and (5) we see that

*d HL*∞*(T )≤ Ch*2*TκT*2*.* (10)

*Therefore, making c1* sufficiently small in (6*) so that the eigenvalues of d (x)H(x) are*
smaller or equal to 1

*, then we will have*

**/2 for every x ∈**h*(I − d H)*−1_{}

*L*∞*(T )≤ C.* (11)

We define tangential projections onto* and h, respectively, as P* **= I − ν ⊗ ν and***Ph* **= I − ν**h**⊗ ν**h**, where q****⊗ r = qr**t**for two column vectors q and r. We recall that the**

tangential derivatives for a functions defined on a neighborhood of* (or h*) are given by

**∇*** v = (∇v)P, ∇hv = (∇v)Ph.* (12)

By using that* ν · ν = 1, we have*
0= 1

2* ∇(ν · ν) = (∇ν)ν = H(x)ν(x) for all x ∈ T,* (13)
Hence, we, of course, have

*P H* *= H = H P,* (14)

which we use repeatedly. Also, we can show that

*P(I − d H)*−1* ν = 0.* (15)

Indeed,* ν = (I −d H)ν by (*13

*) and so P(I −d H)*−1

*−1*

**ν = P(I −d H)**

**(I −d H)ν = Pν = 0.****2.1 Local Parametrization**

Let ˆ*T* *= {(θ1, θ*2*): 0 ≤ θ*1*, θ*2*≤ 1, 0 ≤ θ1+ θ2* *≤ 1} be the reference triangle. Fix T ∈Th*,

* let x0be one of the vertices, and let x1and x2*be vectors inR3

*representing two edges of T*

*(i.e. T*

*1*

**= {x0**+θ1**x***+θ2*2: 0 ≤ θ1

**x***, θ*2

*≤ 1, 0 ≤ θ1+θ2*

**≤ 1}). Let X : ˆT → T be given by*** X(θ*1

*, θ*2

*0*

**) = x***+ θ1*1

**x***+ θ2*2

**x**

**. We also define Y***: ˆT → T*1

**by Y**(θ*, θ*2

*1*

**) = a(X(θ***, θ*2

*)).*Since

*2*

**∇ X = [x1****, x**

**] we have (∇ X)**t**ν**h

**= 0. From the definition of a we have*** ∇a(x) = P(x) − d(x)H(x),*
and, hence

* ∇Y = (P − d H)∇ X.* (16)

*Therefore, using that P and H are symmetric and (*13) we have* (∇Y)tν = 0. Collecting the*
two results we have

**(∇ X)**t_{ν}

*h* * = 0, (∇Y)tν = 0.* (17)

Given a function*η ∈ L*1*(T) we define the pullback lift η _{}∈ L*1

*(T ) as*

*η (x) = η(a(x)),*

and for*η ∈ L*1*(T ) we define the push-forward lift η∈ L*1*(T) as*

*η (a(x)) = η(x),* (18)

and associate *ˆη : ˆT → R defined by*

*ˆη(θ1, θ*2* ) = η(X(θ*1

*, θ*2

*)) = η*1

**(Y(θ***, θ*2

*)).*Note that

*(η*1

_{})= η for η ∈ L*(T) and (η)*1

_{}= η for η ∈ L*(T ).*

Consider also the metric tensors

*G X(θ*1

*, θ*2

*) =*

*2*

**∇ X(θ1**, θ*)*

*t*

*2*

**∇ X(θ1**, θ*), G*1

**Y**(θ*, θ*2

*) =*

*2*

**∇Y(θ1**, θ*)*

*t*

*2*

**∇Y(θ1**, θ*).*From the definition of tangential derivative it is possible to show [33, Section 4.2.1] (see also (2.2) in [17]) that for a function

*η : h*→ R,

**∇***h η(X) = ∇ ˆηG*
−1

**X**

**∇ X***t*

_{, ∇}*η*−1

**(Y) = ∇ ˆηG**

**Y**

**∇Y***t*

_{,}and multiplying by**∇ X and ∇Y we gather that**

* ∇ ˆη = ∇hη(X)∇ X, ∇ ˆη = ∇η(Y)∇Y.* (19)

Hence,

**∇***h η(X) = ∇η(Y)∇Y G*−1

**X***(20)*

**∇ X**t,**∇***η (Y) = ∇hη(X)∇ XG*−1

**Y***(21)*

**∇Y**t.Note that we can also write

*P = ∇YG*−1

_{Y}*−1*

**∇Y**t, Ph**= ∇ XG**

**X***(22)*

**∇ X**t.To see that this is the case, note first from (17) that* ∇Y G*−1

_{Y}

**∇Y**t**ν = 0. Next, consider for**** > 0 an arbitrary differentiable curve s: (−, ) → ˆT and α(t) = Y(s(t)). Then**

* ∇YG*−1

**Y****∇Y**t**α*** = ∇YG*−1

**Y**

**∇Y**t**∇Y s**

**= α**

**= Pα***,*

since* α*is tangent to

*. The same arguments hold for the identity regarding Ph*.

The following identities have appeared in the literature under different forms; see for example [12,13]. Again, we give a proof for completeness and to show the independence of

* Lemma 1 Letη : h→ R be differentiable, and define its forward lift ηas in (*18

*). It then*

*holds that*

**∇**

*hη =*

**∇**

*η*

**◦ a***Q onh,*(23)

*where Q= (I − d H)Ph, and*

**∇**

*η*(24)

**◦ a = (∇**hη)R on h,*where R*=

*I*−

**(ν**h**−ν)⊗(ν−ν**h)

**ν**h**·ν***(I − d H)*−1_{P. Moreover, there exists a constant C such that}*QL*∞*(T )+ RL*∞*(T )≤ C.* (25)
*Proof Using (*20) and (16) we get

**∇***h η(X) = ∇η(Y)[I − d(X)H(X)]∇ XG*−1

**X**

**∇ X**twhere we used that**∇*** _{}ην ⊗ ν = 0. Then (*23) follows from (22).

To prove (24), we use (16) and (14) to get**∇Y = P[I − d(X)H(X)]∇ X. Hence, we have**

**(∇Y)**t** _{= (∇ X)}**t

_{[I − d(X)H(X)]P.}_{(26)}Using (14) we get

*(I − d H)P(I − d H)*−1

_{I}_{−}

**ν**h**⊗ ν**

**ν**h**· ν***= P(I − d H)(I − d H)*−1

_{I}_{−}

**ν**h**⊗ ν**

**ν**h**· ν***= P*

*I*−

**ν**h**⊗ ν**

**ν**h**· ν***.*However,

*P*

*I*−

**ν**h**⊗ ν**

**ν**h**· ν**

**= I − ν ⊗ ν −****ν**h**⊗ ν***+*

**ν**h**· ν**

**(ν ⊗ ν)ν**h**⊗ ν**

**ν**h**· ν***= I −*

**ν**h**⊗ ν**

**ν**h**· ν***.*This gives

*(I − d H)P(I − d H)*−1

*I*−

**ν**h**⊗ ν**

**ν**h**· ν***= I −*

**ν**h**⊗ ν***(27) So, from (27), (26) and (17) we have*

**ν**h**· ν*** (∇Y)t_{(I − d H)}*−1

*I*−

**ν**h**⊗ ν**

**ν**h**· ν**

**= (∇ X)**t*I*−

**ν**h**⊗ ν**

**ν**h**· ν**

**= (∇ X)**t_{.}Thus, using (21) and the above identity we gather that
**∇***η (Y) = ∇hη(X)*

*I*−

**ν ⊗ ν**h

**ν**h**· ν***(I − d H)*−1

*−1*

_{∇YG}

**Y**

**∇Y***t*

**= ∇**

*h*

**η(X)***I*−

**ν ⊗ ν**h

**ν**h**· ν***(I − d H)*−1

_{P,}from (22). Clearly we have**(∇**_{}_{h}**η) ν**h**⊗ν**h* = 0 = (∇hη) νh⊗ν, and P(I −d H)*−1

**ν⊗ν = 0**follows from (15). So we get
**∇***η (Y) = ∇hη*

*I*−

**(ν**h**− ν) ⊗ (ν − ν**h)

**ν**h**· ν***(I − d H)*−1

*P.*

Here we used that**(ν**h**− ν) ⊗ (ν − ν**h**) = −ν**h**⊗ ν**h**− ν ⊗ ν + ν**h* ⊗ ν + ν ⊗ νh*. This

Next, we write an integration identity.

* Lemma 2 Letη ∈ L*1

*(T ). Then, if d A is the surface measure in Tand d Ah*

*is the surface*

*measure in T it follows that*

_{ }

*Tη*
_{d A}_{=}
*TηδT*
*d Ah,* (28)
*where*
*δT* =
det
*G YG*−1

**X***.*(29)

*Proof The result follows from the change of variables formulas [*15]
*T*
*η _{d A}*

_{=}

*ˆT*

*ˆη*

*det G*1

**Y**dθ*dθ*2

*,*

*ˆT*

*ˆη*

*det G*−1

_{X}*dθ*1

*dθ*2=

*T*

*η d Ah.*Combining (28), (23) and (24), we have that

*T*
**∇***h η · ∇hψ d Ah* =

*T*

**∇**

*ηQ*

**· ∇**

*1*

_{}ψQ*δ*

*T*

*d A,*(30)

*T*

**∇**

*η*

_{· ∇}*ψd A*=

*T*

**∇**

*h*

**ηR · ∇**hψ RδT*d Ah.*(31)

Next, we prove some bounds for*δT*.

* Lemma 3 Assuming that (*6

*) holds and definingδT*

*by (*29

*) we have that*

*δT*− 1*L*∞*(T )≤ Ch*2*TκT*2*,* (32)
* _{δ}*1

*− 1*

_{T}*L*∞

*(T )*

*≤ Ch*2

*TκT*2

*.*(33)

*Proof From (*16) and (17) we have

*G Y*

**= ∇ X**t**(I − d H − ν ⊗ ν)∇Y**

**= ∇ X**t_{(I − d H)∇Y}*2*

**= ∇ X**t_{(I − d H)}*Using (13) we get*

**t**_{∇ X − ∇ X}_{(I − d H)ν ⊗ ν∇ X.}*By (17),*

**(∇ X)**t**t**_{(I − d H)ν ⊗ ν∇ X = ∇ X}_{ν ⊗ ν∇ X.}

**(∇ X)**t_{ν}*h*Hence,

**⊗ ν**h**∇ X = (∇ X)**t**ν**h**⊗ ν∇ X = (∇ X)**t**ν ⊗ ν**h**∇ X = 0.**

**(∇ X)**t**t**_{ν ⊗ ν∇ X = (∇ X)}_{(ν − ν}*h*Therefore, we get

**) ⊗ (ν − ν**h**)∇ X.***G*

**Y***2*

**= (∇ X)**t[(I − d H)

**− (ν − ν**h**) ⊗ (ν − ν**h**)]∇ X,***or G*2

**Y****= (∇ X)**t**(I + B)∇ X where B = −2d H + d***H*2

**− (ν − ν**h**) ⊗ (ν − ν**h). Therefore,*G*−1

**Y**G

**X***−1*

**= I + M where M = ∇ X**tB**∇ XG**

**X***.*

It is clear that* ∇ XL*∞

*(T )*

*≤ ChT*. Also, not difficult to see that

*G*−1

**X***L*∞

*(T )*

*≤ Ch*−2

*T*.

Moreover, using (8) and (10) we gather that*BL*∞*(T )* *≤ Ch*2*TκT*2. Hence,*ML*∞*(T )* ≤
*C h*2

*TκT*2*. Since M is symmetric, consider the spectral decomposition M= V V*−1, where
* = diag(λ*1*, λ*2*) and V is orthogonal. Denoting the ith column of V by vi*, we have that
*λi* *= v _{i}tMvi* and then

*λi*

*L*∞

*(T )≤ Ch*2

*TκT*2

*(for i= 1, 2). We also note that*

*G YG*−1

_{X}*= I + V V*−1

*= V (I + )V*−1

*.*Therefore, we obtain

*δ*2

*T*= det

*G*−1

**Y**G

_{X}*= det(V ) det(I + ) det(V*−1_{) = (1 + λ}

1*)(1 + λ*2*),*
which yields
*δ*2
*T* − 1*L*∞*(T )≤ Ch*
2
*TκT*2*.*

Using that*δT* *− 1 = (δ*2_{T}*− 1)/(δT* *+ 1), we obtain (*32). The inequality (33) follows from

the previous inequality and the fact*(δ _{T}*−1

*− 1) = δ*−1

_{T}*(1 − δT).*

*Remark 4 An alternative way to prove Lemma*3is to use [13, Proposition 2.1] given by

*δT (x) = ν · νh*

1* − d(x)κ1(x))(1 − d(x)κ*2

**(x))***,*

the identity 1* − ν · νh* = 12

*|2and (3), (4), (7) and (8).*

**|ν − ν**hWe can now state the following result which follows from Lemmas 1 and 3, and Eqs. (30), (31).

* Lemma 5 Assuming the hypotheses of Lemmas*1

*and*3

*, we have that*

**∇***η*

*L*2* _{(T}_{)}≤ C∇hηL*2

_{(T )}≤ C**∇**

*η*

*L*2_{(T}_{)}.

*In the following, we use the notation D _{i}u= (∇_{}u)i*, and write∇

*2*

_{}*u as the 3*× 3 matrix

*with entries D _{i}D_{j}u (also denoted by D_{i j}u).*

* Lemma 6 Assuming that (*6

*) holds, we have*∇2

*hη*

*L*2

*∇ 2*

_{(T )}≤ C*η*

*2*

_{L}

_{(T}_{)}+ C*hTκT*2

*+ h*2

*TκTγT*

**∇**

*η*

*L*2

_{(T}_{)},*whereγT*= max

*i j*

**∇H**i j*L*∞

*(T ).*

*Proof Let w = ∇_{}_{h}η. Using (*23

**) we see that for x**∈ T ,*wi (x) = (Ph)i k(I − d(x)H(x))k jDjη(a(x)),*

where we use Einstein summation convention.

*Using the product rule and the fact that Ph*is constant we have

**∇***hwi (x) = J*1

*2*

**(x) + J***3*

**(x) + J***where*

**(x),***1*

**J***= −(Ph)i kHk j*

*D*

_{j}η**◦ a****∇**

*hd,*

*2*

**J***= −(Ph)i kd*

*D*

_{j}η**◦ a****∇**

*hHk j,*

*3*

**J***= (Ph)i k(I − d H)k j*

**∇**

*h*

*D*

_{j}η**◦ a***.*

* We start with J*3. Using (23) we have

**∇**

*h*

*D*

_{j}η**◦ a**

**(x) = ∇**D_{j}η**(a(x))Q**Hence, using (25), (28), (32) and (6) we get
**∇***h*
*D _{j}η◦ a*

*L*2

_{(T )}≤ C**∇**2

*η*

*L*2

_{(T}_{)}.If we combine this inequality with (10) and (6), we have
* J*3

*L*2

_{(T ))}≤ C**∇**

*2*

_{}*η*

*L*2_{(T}_{)}.

Next, using (12) and the fact* νt_{h}Ph*= 0, we obtain

**∇***hd= (∇d)Ph* **= ν**

*t _{P}*

*h* **= (ν − ν**h)tPh.

Hence, using (8), (5), (28), (32) and (6) yields

* J*1

*L*2

*2*

_{(T )}≤ ChTκT**∇**

*η*

*L*2

_{(T}_{)}.Similarly, (7), (28), (32) and (6) yield

* J*2

*L*2

*2*

_{(T )}≤ Ch

_{T}κ_{T}γ_{T}**∇**

_{}η*L*2_{(T}_{)}.

Combining the above estimates gives the desired result.
We note that a similar result, with the*γT*dependence, can also be found in [5, Lemma 2.1]
*Remark 7 Note that the C*3 regularity imposed on * is needed in Lemma*6. That seems
unavoidable, at least under the choice of the local mapping (2) used in the analysis. The same
regularity is used in [13].

**3 Finite Element Spaces and Local Approximations**

We introduce the following finite dimensional approximation of (1). The finite element space
is given by
*Sh* =
*vh∈ C*0*(h):*
*h*

*vhd Ah* *= 0, vh*|*T* *is linear for all T* ∈*Th*

*,*
*S _{h}*=

*vh: vh*

*∈ Sh*

*.*

*For fh∈ L*2

*(h) with*

*h* *fhd Ah* *= 0, let uh∈ Sh*such that

*h*
**∇***huh***· ∇***hvhd Ah* =
*h*
*fhvhd Ah* for all*vh* *∈ Sh.* (34)

Existence and uniqueness of the finite-dimensional problem (34) follows from noting that if

*uhis a solution with fh= 0, then uh* *must be constant with zero average. Thus uh* = 0.

* Lemma 8 Assuming ⊂ R*3

*a C*3

*two-dimensional compact orientable surface without*

*boundary, there exists a constant cpsuch that*

*φL*2_{()}≤ cp**∇***φL*2_{()}*for allφ ∈ ˚H*1*().* (35)
Then we can state a simple energy estimate.

* Lemma 9 Let u solve (*1

*), then*

**∇***u** _{L}*2

_{()}*≤ cp f L*2

*(36)*

_{()}.*Before proving an a-priori estimate for u*that measures the inconsistency in going from

_{h}− u we will need to prove an important lemma*to h*. First, we need to develop notation to

use in the next proof. Since*δTd Ah* *= d A with respect to a given triangle T and its lifting*
*T*, let us define

*δh (x) = δT, if x ∈ T.*

**Lemma 10 Let**vhand zhbelong to Sh. Then the following holds

_{}*h*
**∇***hvh***· ∇***hzhd Ah*−
**∇***v*
*h***· ∇***zhd A*
* ≤ Ch***∇***vh** _{L}*2

_{()}**∇**

*z*

_{h}*L*2

*(37)*

_{()}

_{}*h*

*vhzhd Ah*−

*v*

*hzhd A*

*≤ Ch*

*vh*

*L*2

_{()}*zh*

*L*2

*(38)*

_{()},*whereh*= max

*T∈Thκ*2

*Th*2

*T.*

*Proof Using (*28) we get

_{}*h*
*vhzhd Ah*−
*v*
*hzhd A*
=_{}
* _{}vhzh*
1

*δ*

*h*− 1

*d A*Hence, (38) follows from (33).

To prove (37) we use (30) to get
*h*
**∇***hvh***· ∇***hzhd Ah* =
**∇***vhQ*
·**∇***z _{h}Q*

_{1}

*δh*

*d A.*

Using that**∇**_{}**(·)P = ∇**_{}(·) we have*h*
**∇***hvh***· ∇***hzhd Ah* =
**∇***v*
*hM***· ∇***zhd A,*
where
*M*= 1
*δ*
*h*
*(P Q)(P Q)t _{.}*

On the other hand, again using that**∇**_{}**(·)P = ∇**_{}(·) we get**∇***v*
*h***· ∇***zhd A*=
**∇***v*
*hP***· ∇***zhd A.*
Hence,
_{}*h*
**∇***hvh***· ∇***hzhd Ah*−
**∇***v*
*h***· ∇***zhd A*
=
_{}**∇***vh (M − P) · ∇zhd A*

*.*(39)

We now proceed to bound*(M − P). We first use (*14), to get on*h*
*P Q ◦ a = P Q = P(I − d H)Ph= (I − d H)P Ph.*

Hence, we get

*M ◦ a =* 1

*δh(I − d H)S(I − d H),*

*where S= P PhP, and then the triangle inequality yields*

*M − PL*∞*(T)*=*δ**h*−1*(I − d H)S(I − d H) − P** _{L}*∞

_{(T )}≤*δ*−1*h* *(I − d H)S(I − d H) − S** _{L}*∞

*∞*

_{(T )}+ S − PL*(T ).*Now using (10), (6) and (33

*) and the fact that S is bounded we obtain*

*δ*−1*h* *(I − d H)S(I − d H) − S** _{L}*∞

*2*

_{(T )}≤ Ch*TκT*2*.*

*Since P νh⊗ ν P = Pν ⊗ νhP= Pν ⊗ ν P = 0 we have that on h*,

*S*

**= P(I − ν**h**⊗ ν**h**)P = P(I − (ν**h**− ν) ⊗ (ν**h**− ν))P.***Finally using that P*2_{= P we have}

*S − P = −P(νh− ν) ⊗ (νh− ν)P.*

Therefore, it follows from (8) that*S − PL*∞*(T )≤ Ch*2*TκT*2. Using the previous inequalities

we obtain

*M − PL*∞*(T)≤ Ch*2*Tκ*2*T,*

and hence

*M − PL*∞*()≤ Ch,*

and thus (37) follows from this inequality and (39).
* Theorem 11 Let u*∈ ˚

*H*1

*() be the solution of (*1

*) and let uh∈ Shthat solves (*34

*). Assume*

*that f*

_{h}satisfies f_{h}*2*

_{L}*2*

_{()}≤ C f_{L}

_{()}. Then there exists a constant C such that
**∇**
*u− u _{h}*

*L*2

_{()}≤ C min_{φ}*h∈Sh*

**∇**

*u− φ*

_{h}*L*2

_{()}*+ Ccp*

*h f L*2

*+*

_{()}*f − f*

_{h}*L*2

_{()}*,*

*whereh*= max

*T∈ThκT*2

*h*2

*Tis as in Lemma*10

*.*

*Proof For an arbitraryφh* *∈ Sh*, set*ξh* *= uh− φh*and*ξh= uh− φh*. By Lemma5we have

**∇***ξh*

*Then, we write for an arbitrary constant c*
**∇***h(uh− φh)*
2
*L*2_{(}*h)*=
*h*
**∇***huh***· ∇***hξhd Ah*−
*h*
**∇***hφh***· ∇***hξhd Ah*
=
*h*
*fhξhd Ah*−
*h*
**∇***hφh***· ∇***hξhd Ah* by (34)
=
*h*
*fhξhd Ah*−
*f* *· ξ*
*hd A*
+
**∇***u***· ∇***ξ*
*hd A*−
*h*
**∇***hφh***· ∇***hξhd Ah,* by (1)
*= J1+ J2+ J3+ J4,*

where after using that_{}*f d A*= 0 =_{}

*h* *fhd Ah*,
*J*1=
*h*
*fh(ξh− c) d Ah*−
*f*
*h*
*ξh− c*
*d A, J*2=
*f _{h}− f*

*ξh− c*

*d A,*

*J*3=

**∇**

*u− φ*

_{h}**· ∇**

*ξhd A, J*4=

**∇**

*φ*

*h*

**· ∇**

*ξhd A*−

*h*

**∇**

*hφh*

**· ∇**

*hξhd Ah.*

By applying (38) and the Poincaré’s inequality (35) we get

*J*1*≤ C cph** fh*
*L*2_{()}

**∇***ξh*

*L*2_{()},

*where we chose c*= * _{||}*1

_{}ξ_{h}d A. Using the Cauchy–Schwarz inequality and the Poincaré’sinequality (35) we have
*J*2*≤ cp* * fh− f*
*L*2_{()}**∇***ξh*
*L*2* _{()}.*
Using the Cauchy–Schwarz inequality we get

*J*3≤**∇***(u − φh)** _{L}*2

_{()}**∇**

*ξh*

*2*

_{L}*Using (37) gives*

_{()}.*J*4

*≤ C h*

**∇**

*φh*

*L*2

_{()}**∇**

*ξh*

*L*2

*Hence, combining the above results we get*

_{()}
**∇***ξh*
*L*2_{()}≤ Ccp*h f L*2* _{()}*+

*f*

_{h}− f*L*2

_{()}*+ C*

**∇**

*(u − φh)*

*L*2

_{()}+ h**∇**

*φh*

*L*2

_{()}*,*

where we used that* f _{h}*

*2*

_{L}

_{()}*≤ f*2

_{L}*. If we use the triangle inequality and (36) we*

_{()}obtain
**∇***φh** _{L}*2

*≤*

_{()}**∇**

*u− φ*

_{h}*L*2

*2*

_{()}+ cp f L*().*

The result now follows after taking the minimum over*φh∈ Sh*and use the fact that*h* ≤ 1

which follows from (6).

*Let Ih,*be the standard Lagrange interpolant in*honto Shand define Ihη= (Ih,η)*∈
*S _{h}*. We then have the following estimate.

* Lemma 12 Let u be the solution of (*1

*). Then,*

**∇**

*(u − Ihu)L*2

*2*

_{()}≤ Ccp(h+ h) f L*+ ⎛ ⎝*

_{()}*T∈Th*

*h*2

*∇*

_{T}*2*

_{}*u*2

*2*

_{L}*⎞ ⎠ 1*

_{(T}_{)}*/2*

*,*

*whereh*= max

*T∈Thh*3

*TγTκT* *andγT* *was defined in Lemma*6*.*

*Proof Recall that u _{}is the pullback lift of u. Using Lemma*5, and approximation properties
of the Lagrange interpolant, we have

**∇***(u − Ihu)L*2* _{(T}_{)}≤ C∇h(u− Ih,u)L*2

*∇*

_{(T )}≤ ChT*2*

_{}

_{h}u*2*

_{L}*We get from Lemma6that*

_{(T )}.**∇***(u − Ihu)L*2_{(T}_{)}≤ C*hT*∇* _{}*2

*u*

*2*

_{L}*+*

_{(T}_{)}*h*2

*2*

_{T}κ_{T}*+ h*3

_{T}κTγT**∇**

*u*

*2*

_{L}

_{(T}_{)}*. (40)*

*The result follows easily by summing over T and applying (*36).
We can combine Theorem11and Lemma12to get the following theorem.

* Theorem 13 Assume that the hypothesis of Theorem*11

*holds. Then,*

**∇**
*u− u _{h}*

*L*2

_{()}≤ Ccp*(h+ h) f L*2

*+*

_{()}*f − f*

_{h}*L*2

_{()}*+C*⎛ ⎝

*T∈Th*

*h*2

*∇*

_{T}*2*

_{}*u*2

*2*

_{L}*⎞ ⎠ 1*

_{(T}_{)}*/2*

*.*(41)

**4 Graded Meshes for Two Subdomains**

In this section we consider the case where there is a high curvature region1 * , and*

2 *= \1*. We let*κ(i)* *= HL*∞*(i)* and assume that*κ(1)* * κ(2)*. We also let*κ =*

*HL*∞*()= κ(1)*. We use our above results from the previous sections and local regularity

results in order to grade a mesh so that the error is balanced.

We start by just stating the global regularity result which is found in [17]. We note that
this result does not fit our graded meshing strategy; instead, we establish Lemma15.
* Lemma 14 Assume that is a C*3

*orientable compact surface without boundary, and that*

*u*∈ ˚

*H*1

*() solves (*1

*). Then u∈ H*2

*(), and there exists a constant C that is independent of*

*the curvatures of such that*

∇2

*u**L*2* _{()}≤ (1 + Ccpκ) f L*2

*Applying Theorem13and Lemma14it easily follows that*

_{()}.**∇**

*u− u*

_{h}*L*2

_{()}≤ Ccp*f − fh*

*L*2

_{()}+ C*h+ cp(h+ h+ κh)*

*f L*2

*Then, requiring*

_{()}. (42)*h+ h*

*≤ h(1 + κ) we get*

**∇**
*u− u _{h}*

*L*2

_{()}≤ Ccp f − fh*L*2

*()+ Ch*1

*+ cp(1 + κ)*

*f L*2

*(43)*

_{()}.Of course, this is not a very good estimate in the case*κ = κ(1)* * κ(2)*since we would
have the mesh size far away from1still depending on*κ(1)*. Instead, we would like to have
the mesh size to only depend on*κ(2)*a distance order one away from1. In order to do this
we will need a local regularity result, as follows.

* Lemma 15 (Weighted-H*2

*regularity) Let u∈ H*2

*() be the solution of (*1

*), and consider*

*the subset*2 *⊆ . Let ρ ∈ W*1*,∞() be such that supp ρ ⊆ *2*. Then there is an universal*

*constant C such that*

*ρ∇*2
*u**L*2_{(}_{2}* _{)}≤ CρW*1

*,∞*1

_{()}*+ cp(1 + κ(2))*

*f L*2

*(44)*

_{()}.*Proof Note that D _{i}(ρ*2

*D*2

_{j}u) = 2ρ D_{i}ρ D_{j}u+ ρ*D*

_{i j}u, and
2
*ρ*2_{D}*i ju Di ju d A*=
2
*D _{i j}u D_{i}*

*ρ*2

_{D}*ju*

*d A*− 2 2

*ρ Di ju Diρ Dju d A*≤ 2

*D*

_{i j}u D_{i}*ρ*2

_{D}*ju*

*d A*

**+ 2∇**

*∞*

_{}ρL*()ρ Di ju*

*L*2

*2*

_{(}*)Dju*

*L*2

*(*2

*)*≤ 2

*D*

_{i j}u D_{i}*ρ*2

_{D}*ju*

*d A*

**+ 2∇**

*2*

_{}ρ*∞*

_{L}*2*

_{()}Dju*L*2

*2*

_{(}*)*+ 1 2

*ρ Di ju*2

*L*2

*2*

_{(}*).*Then 2

*ρ*2

_{D}*i ju Di ju d A*≤ 2 2

*D*

_{i j}u D_{i}*ρ*2

_{D}*ju*

*d A*

**+ 4|∇**

*2*

_{}ρ|*∞*

_{L}*2*

_{()}Dju*L*2

*2*

_{(}*).*From Lemma19, 3

*i, j=1*2

*D*

_{i j}u D_{i}*ρ*2

_{D}*ju*

*d A*= 3

*j*=1 2

*u D*

_{j}*ρ*2

_{D}*ju*

*d A*− 2

*ρ*2

_{tr}

*2*

_{(H)H − 2H}

_{∇}*u*

**· ∇**

*≤ 2*

_{}u d A*u*2

*2*

**ρ∇**_{}**ρ · ∇**_{}u+ ρ

_{}u*d A+ C(κ(2))*2 2

*ρ*2

_{|∇}*u*|2

*d A*

*≤ C*

**∇**

*ρL*∞

*()ρu*

*L*2

*2*

_{(}*)*

**∇**

*u*

*L*2

*(*2

*)+ ρu*2

*L*2

*2*

_{(}*)*

*+ (κ(2)*2

_{)}*2*

_{ρ}*L*∞

*()*

**∇**

*u*2

*L*2

*2*

_{(}*)*

*.*Thus, 3

*i, j=1*2

*(ρ Di ju)*2

*d A*

*≤ C*

*ρu*2

*2*

_{L}*2*

_{(}*)*+

**∇**

*ρL*∞

*()+ κ(2)ρL*∞

*()*2

**∇**

*u*2

*2*

_{L}*2*

_{(}*)*

*.*

The result now follows after using the energy estimate (36).
Lemma15holds with a generic function* ρ, but we will apply the result with ρ(x) =*
dist

*1*

**(x,***) which is a 1-Lipschitz function.*

**4.1 The Graded Mesh**
*We start by defining d*1 =

1*+ cp(1 + κ(2))*

*/(1 + cpκ(1)). We then define the region*

*D*1 * = {x ∈ : dist(1, x) ≤ d*1}. We also let ˜

*T*1

_{h}*= {T ∈*

*Th: T*∩

*D*1 = ∅} and define ˜

*T*2

*h* *= {T ∈Th: T /∈ ˜Th*1*}. We set h1* *= hd1*. Finally, we let* ρ(x) = dist(x, *1

*) and also set*

*ρT* *= dist(T, *1*) for all T ∈Th*, where*ρ(·) is defined using the geodesic distance [*25] and

*∇ρL*∞*(*2*)*≤ 1.

Our graded mesh will then satisfy:
(M1) *h+ h≤ h(1 + κ(2))*

*(M2) hT* *≤ h1for every T* ∈ ˜*T _{h}*1

*(M3) hT* *≤ min{ρT, 1}h for every T ∈ ˜T _{h}*2

We recall that*h*and*h* were respectively defined in Theorem11and Lemma12.

A few comments are in order. First, note that condition (M1) is completely local, and in
*the case O(κ(1)) = O(κ(2)), condition (M1) would be necessary to get an estimate of the*
form (43), as the argument above (43) shows.

*Note that if O(κ(1)) = O(κ(2)) then O(h*1*) = O(h). The mesh size for triangles that*
are unit distance from1 (i.e.*ρT* *= O(1)) can be chosen so that O(hT) = O(h). In the*

intermediate region a grading giving by (M3) needs to be satisfied. Finally, note that there
is a smooth transition for triangles in the border of*D*1. Indeed, if*ρT* *= d1*then by (M3),
*hT* *≤ d1h= h1*

We now state and prove our main result.

* Theorem 16 Suppose that u*∈ ˚

*H*1

*() solves (*1

*) and uh*

*∈ Sh*

*solves (*34

*). Assume that the*

*mesh satisfies (M1–M3). Then we have*

**∇**
*u− u _{h}*

*L*2

*1*

_{()}≤ C*+ cp(1 + κ(2))*

*h f*2

_{L}

_{()}+ Ccp*f − fh*

*L*2

*Before proving the result let us state a few comments. Note that the right hand side of (45) looks like the right hand side of (43) with*

_{()}. (45)*κ(2)*instead of

*κ. Therefore, with the available*information, (45) is essentially the best result we can hope for. So, we found a graded mesh where one has a fine mesh in the region where the curvature is high to get the best error estimate.

*Proof (of Theorem*16) By (41) and our assumption (M1) we have
**∇**
*u− u _{h}*

*L*2

_{()}≤ Ccp*(1 + κ(2)*

_{)h f }*L*2

*+*

_{()}*f − f*

_{h}*L*2

_{()}*+C*⎛ ⎝

*T∈Th*

*h*2

*∇*

_{T}*2*

_{}*u*2

*2*

_{L}*⎞ ⎠ 1*

_{(T}_{)}*/2*

*.*Next, we estimate

*T∈Th*

*h*2

*∇*

_{T}*2*

_{}*u*2

*2*

_{L}*=*

_{(T}_{)}*T∈ ˜T*1

*h*

*h*2

*∇*

_{T}*2*

_{}*u*2

*2*

_{L}*+*

_{(T}_{)}*T∈ ˜T*2

*h*

*h*2

*∇*

_{T}*2*

_{}*u*2

*2*

_{L}*By our (M2) we get*

_{(T}_{)}.*T∈ ˜T*1

*h*

*h*2

*∇*

_{T}*2*

_{}*u*2

*2*

_{L}*21∇2*

_{(T}_{)}≤ Ch*u*2

*L*2

_{()},and we gather from Lemma14that
*T∈ ˜T _{h}*1

*h*2

*∇*

_{T}*2*

_{}*u*2

*2*

_{L}*21 1*

_{(T}_{)}≤ Ch*+ cpκ(1)*2

*f*2

*L*2

_{()}*= Ch*2 1+c

*p(1+κ(2))*2

*f*2

*L*2

_{()}.The other term we bound in the following way by using (M3):
*T∈ ˜T _{h}*2

*h*2

*∇*

_{T}*2*

_{}*u*2

*2*

_{L}*≤*

_{(T}_{)}*T∈ ˜T*2

_{h}*h*2

*T*

*ρ*2

*T*

*ρ∇*2

*u*2

*L*2

*2*

_{(T}_{)}≤ h*ρ∇*2

*u*2

*L*2

*2*

_{(}*).*

Now using (44), we have
*T∈ ˜T _{h}*2

*h*2

*∇*

_{T}*2*

_{}*u*2

*2*

_{L}*2 1*

_{(T}_{)}≤ Ch*+ cp(1 + κ(2))*2

*2*

_{ f }*L*2

_{()}.The result now follows.

**4.2 An L****2****() Estimate**

*We now derive an error estimate in the L*2*() norm, based on the usual duality argument.*
*We note that the conditions (M1–M3) are no longer enough to guarantee a h*2convergence
that is independent of*κ(1)*. Actually, (M1) is reinforced by imposing that

(M4) *h≤ h*2*(1 + κ(2))*

* Theorem 17 Suppose that u*∈ ˚

*H*1

*() solves (*1

*) and uh*

*∈ Shsolves (*34

*). Assume that f*

_{h}*satisfies f*

_{h}*2*

_{L}*2*

_{()}≤ C f_{L}

_{()}, and that the mesh satisfies (M1–M4). Then we have
*u − uh*
*L*2* _{()}≤ Ch*
2

_{ϒ}

_{ϒ + c}*p+ h(1 + κ(2))(cp+ hϒ)*

*f L*2

_{()}*+ Ccp(cp+ hϒ)*

*f − fh*

*L*2

_{()}.*Hereϒ = 1 + cp(1 + κ(2)).*

*Proof First, letv ∈ ˚H*1_{() be the weak solution of}*−v = u − ˜uh* on
where
*˜uh= uh*−
1
*||*
*u*
*hd A.*
Then
*u − ˜uh*
2
*L*2* _{()}* =

**∇**

**v · ∇***u− ˜u*

_{h}*d A*=

**∇**

*(v − Ih*

**v) · ∇***u− ˜u*

_{h}*d A*−

**∇**

*Ih*

**v · ∇**˜u*hd A*+

*f Ihv d A,*

where we used (1). Then, using (34), the fact that derivatives of constants are zero, and that

*f d A*= 0 =

*h* *fhd Ah*, we can show that

*u − ˜uh*
2
*L*2* _{()}= J1+ J2+ J3+ J4,*
where

*J*1=

*f*

*h(Ihv − c)d A −*

*h*

*fh(Ihv − c)d Ah*

*J*2=

*f*

*− f*

_{h}*(Ihv − c) d A*

*J*3=

**∇**

*(v − Ih*

**v) · ∇***u− u*

_{h}*d A,*

*J*4=

*h*

**∇**

*h((Ihv)*−

**) · ∇**huhd Ah**∇**

*Ih*

**v · ∇**u*hd A.*

*Here we choose c* = * _{||}*1

*38), the Poincaré’s inequality (35), and that*

_{}Ihv d A. Using (* f*

*h**L*2* _{()}≤ C f _{L}*2

*,*

_{()}*J*1 *≤ Ccph f L*2_{()}**∇*** _{}(Ihv)L*2

*Using the Poincaré’s inequality (35) we get*

_{()}.*J*2*≤ Ccp** f − fh*

*L*2_{()}**∇***(Ihv)L*2*().*
Using (37) we get

*J*4*≤ Ch***∇***uh** _{L}*2

_{()}**∇**

*(Ihv)L*2

*().*Using the triangle inequality and the energy estimate (36) we have

*J*4*≤ Ch*
**∇**
*u _{h}− u*

*L*2

*2*

_{()}+ cp f L*()*

**∇**

*(Ihv)L*2

_{()}.*To bound J*3we use the Cauchy–Schwarz inequality

*J*3≤**∇**

*u _{h}− u*

_{}

*L*2_{()}**∇***(Ihv − v)L*2*().*
Using Lemma12we get the estimate

**∇***(v − Ihv)*2* _{L}*2

*2*

_{()}≤C c*p(h+ h)*2

*u − ˜uh*2

*L*2

*+*

_{()}*T∈Th*

*h*2

*∇*

_{T}*2*

_{}*v*2

*2*

_{L}

_{(T}_{)}.As we did in the proof of Theorem16[using (M1–M3)] we can show that

*T∈Th*

*h*2* _{T}*∇

*2*

_{}*v*2

*2*

_{L}*2*

_{(T}_{)}≤ C h*ϒ*2

*u − ˜uh*

2

*L*2* _{()}.*
Hence, using (M1) we have

**∇***(v − Ihv)L*2_{()}≤ Ch ϒ*u − ˜u _{h}*

Therefore, we get the bound
*J*3*≤ Chϒ***∇**
*u _{h}− u*

*L*2

_{()}*u − ˜uh*

*L*2

*Using (46), the triangle inequality and a energy estimate we have*

_{()}.**∇***(Ihv)L*2_{()}≤ C(cp+ hϒ)*u − ˜uh*
*L*2* _{()}.*
It follows then that

*J*1*+ J2+ J4≤ C(cp+ hϒ)*
*cph f L*2_{()}+ cp* f − fh** _{L}*2

_{()}+ h**∇**

*u*

_{h}− u*L*2

*×*

_{()}*u − ˜uh*

*L*2

*Therefore,*

_{()}.*u − ˜uh*

*L*2

_{()}*≤ C(cp+ hϒ)*

*cph f L*2

_{()}+ cp*f − fh*

*L*2

_{()}*+ C*

*(cp+ hϒ)h+ hϒ*

**∇**

*u*

_{h}− u*L*2

*We gather from (M4) and (45) that*

_{()}.
*u − ˜uh*
*L*2_{()}*≤ C(cp+ hϒ)h*
2_{ϒ f }*L*2_{()}+ Ccp(cp+ hϒ)* f − fh*
*L*2_{()}*+ C**(cp+ hϒ)(1 + κ(2))h*2*+ hϒ*
*hϒ f L*2* _{()}.* (47)
The triangle inequality yields

*u − uh*
*L*2* _{()}*≤

*u − ˜uh*

*L*2

*+ 1*

_{()}*||*1

*/2*

_{}u_{h}d A*.*(48) We now use that

_{}*huhd Ah* = 0 and (38) to get
1
*||*1*/2*
* _{}u_{h}d A* = 1

*||*1

*/2*

*−*

_{}u_{h}d A*h*

*uhd Ah*

*≤ Ch*

*uh*

*L*2

*Using the triangle inequality, (35) and (36) we gather that*

_{()}.1
*||*1*/2*
_{}u_{h}d A* ≤ C*
*h**uh− u** _{L}*2

*2*

_{()}+ hc*p f L*2

_{()}*.*

Finally, from (48), (6), and (M4) we have
*u − uh*
*L*2_{()}≤ C*u − ˜uh*
*L*2* _{()}+ Ccph*
2

_{ϒ f }*L*2

_{()}.The result follows from this inequality and (47).

*Remark 18 Although we only proved results for domains without boundaries, we anticipate*

that our analysis will carry over to surfaces with boundary. In fact, in the next section we will
provide numerical experiments for a surface* with a boundary.*

**Fig. 2 Visualization of the surface**

**5 Numerical Experiments**

We consider a simple example of a surface with a high curvature “ridge,” and show our adapted mesh as well as properties of the solution.

**5.1 Surface Parameterization**
Let* be the surface parameterized by*

**X**(x, y) =*x, y, 1 −*
*x*2_{+ 5 · 10}−2* _{y}*2

*−5*

_{+ 2.5 · 10}*, for (x, y) ∈ U,*(49)

*where U*

*= {(x, y) ∈ R*2

*: x*2

*+ y*2

*< 1} (see Fig.*2). The curvature regions1,2are defined by

*U*1=

*(x, y) ∈ R*2

_{:}

*x*0

*.05*2 +

*y*0

*.5*2 ≤ 1

*,*1

*2*

**= X(U1**),*= \ 1.*(50) This leads to the following maximum curvatures on1,2:

*κ(1)*= max
1
*κ = 199.970, κ(2)*= max
2
*κ = 8.701,* *κ(1)*
*κ(2)* *= 22.984.*
**5.2 “Exact” Solution**

We use zero boundary conditions on*∂ and choose the right-hand-side f : → R to be*

*f(x, y, z) =*
50*.0 exp*
1
*(x−0.2)*2* _{+y}*2

_{−0.2}*, if (x − 0.2)*2

*2*

_{+ y}*0*

_{< 0.2,}*,*else

*.*(51)

**Fig. 3 Plot of the “exact” solution u on**

*Note that f is similar to a “bump” function [*24*] and is C*∞*() and has compact support on*

*.*

*In lieu of an exact analytic solution, we compute a reference “exact” solution (denoted u)*
on a mesh consisting of 3,679,489 vertices and 7,356,928 triangles obtained from refining
an initial coarse mesh. The number of free degrees-of-freedom of the reference solution is
3,677,441 (after eliminating boundary degrees-of-freedom). See Fig.3*for a plot of u.*

We also plot an approximation of|∇* _{}*2

*u*| to illustrate how the hessian is influenced by the

high curvature of the domain, which is concentrated at the high curvature ridge of the surface (see Fig.4).

**5.3 Adapted Mesh and Solution**

Our adapted mesh is generated by first starting with a coarse mesh that satisfies (10), (6), (11). We then iteratively check the criteria in (M1), (M2), (M3) in Sect.4.1. At each iteration, if any triangle does not satisfy the criteria, then it is marked for refinement. We then refine all marked triangles using standard longest-edge bisection. Figure5shows a plot of our final adapted mesh.

Figure6shows the “pointwise” error**|∇*** _{}(u − uh)|, where uh* is the numerical solution

on the adapted mesh. Note that the graded mesh, essentially, eliminates the error in the high
*curvature region. However, the grading strategy does not specifically account for f , so the*
*error is larger where f is large.*

**Fig. 4 Plot of the hessian of u (viewed from the top). Note that it peaks in the high curvature region**

**Fig. 6 Plot of|∇**_{}(u − uh)| (viewed from the top). The adapted mesh essentially eliminates the error in the

high curvature region

**Appendix A: One Technical Result**

**Lemma 19 Assume that*** is a C*3 *two-dimensional compact orientable surface without*
*boundary, and that u∈ H*2_{(). Then}

3
*i, j=1*
2
*D _{i j}u D_{i}*

*ρ*2

_{D}*ju*

*d A*= 3

*j*=1 2

*u D*

_{j}*ρ*2

_{D}*ju*

*d A*− 2

*ρ*2

_{tr}

*2*

_{(H)H − 2H}

_{∇}*u*

**· ∇**

_{}u d A.*Proof In what follows, we use the two identities [*17]

*Diuv d A = −*
*u Div d A +*
*uv tr(H)νid A,* (52)
and
*D _{i j}u= D_{j i}u+ (H∇_{}u)jνi− (H∇u)iνj,* (53)

*for all C*3*() functions u, v. Also we will use that of course*

**∇***u · ν = 0.* (54)

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

*Di ju Di*
*ρ*2_{D}*ju*
*d A*
= −
*Dii ju*
*ρ*2_{D}*ju*
*d A*+
*Di juρ*
2_{D}*ju tr(H)νid A* by (52)
= −
*Dii ju*
*ρ*2_{D}*ju*
*d A* by (54)
= −
*Di*
*D _{j i}u+ (H∇u)jνi− (H∇u)iνj*

*ρ*2

_{D}*ju d A*by (53) = −

*Di*

*D*

_{j i}u+ Hj kDkuνi− Hi kDkuνj*ρ*2

_{D}*ju d A*= −

*D*

_{i j i}u+ Hj kHiiDku− Hi kHi jDku*ρ*2

_{D}*ju d A*by (54) = −

*D*2

_{i j i}uρ*D*2

_{j}u+ ρ*(tr(H)H − H*2

**)∇**_{}u**· ∇**

_{}u*d A.*

To handle the first term on the right hand side, we use (53*) and the fact that D _{ii}u= _{}u*

to write:
−
*Di j iuρ*
2_{D}*ju d A*= −
*D _{j}_{}u+ (H∇_{}D_{i}u)jνi− (H∇Diu)iνj*

*ρ*2

_{D}*ju d A*= −

*Djuρ*2

_{D}*ju+ Hj kDkiuνiρ*2

*Dju d A,*

where we used (54*) in the last equation. But D _{ki}uνi= Dk(Diuνi) − Diu Hki*

*= −Diu Hki*,

and then
−
*Di j iuρ*
2_{D}*ju d A*=
*−Djuρ*
2_{D}*ju+ Hj kDiu Hkiρ*2*Dju d A*
=
*u Dj*
*ρ*2_{D}*ju*
*+ ρ*2* _{H}*2

_{∇}*u*

**· ∇**

_{}u d A.In the last equation we used (52) and (54). This completes the proof.

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