Enforce the Dirichlet boundary condition by volume constraint in Point Integral method

arXiv:1506.02343v1 [math.NA] 8 Jun 2015

Enforce the Dirichlet boundary condition by volume constraint in Point Integral method

Zuoqiang Shi

June 9, 2015

Abstract

Recently, Shi and Sun proposed Point Integral method (PIM) to discretize Laplace-Beltrami operator on point cloud [16, 19]. In PIM, Neumann boundary is nature, but Dirichlet boundary needs some special treatment. In our previous work, we use Robin boundary to approximate Dirichlet boundary. In this paper, we introduce another approach to deal with the Dirichlet boundary condition in point integral method using the volume constraint proposed by Du et.al. [7].

1 Introduction

Partial differential equations on manifold appear in a wide range of applications such as material science [5, 9], fluid flow [12, 13], biology and biophysics [3, 10, 18, 2] and machine learning and data analysis [4, 6].

Due to the complicate geometrical structure of the manifold, it is very chanlleging to solve PDEs on manifold.

In recent years, it attracts more and more attentions to develop efficient numerical method to solve PDEs on manifold. In case of that the manifold is a 2D surface embedding in R³, many methods were proposed include level set methods [1, 21], surface finite elements [8], finite volume methods [15], diffuse interface methods [11] and local mesh methods [14].

In this paper, we focus on following Poisson equation with Dirichlet boundary condition −∆_Mu(x) = f(x), x∈ M

u(x) = 0, x∈∂M (1.1)

whereMis a smooth manifold isometrically embedded in R^d with the standard Euclidean metric and∂M is the boundary. ∆_M is the Laplace-Beltrami operator on manifold M. Let g be the Riemannian metric tensor ofM. Given a local coordinate system (x¹, x²,· · ·, x^k), the metric tensorg can be represented by a matrix [gij]k×k,

gij=< ∂

∂xⁱ, ∂

∂x^j >, i, j= 1,· · · , k.

Let [g^ij]k×k is the inverse matrix of [gij]k×k, then it is well known that the Laplace-Beltrami operator is

∆_M= 1

√detg

∂

∂xⁱ(g^ijp detg ∂

∂x^j).

In this paper, the metric tensor g is assumed to be inherited from the ambient space R^d, that is, M isometrically embedded in R^d with the standard Euclidean metric. If M is an open set inR^d, then ∆_M becomes standard Laplace operator, i.e., ∆_M=Pd

i=1

∂²

∂xⁱ².

∗Mathematical Sciences Center, Tsinghua University, Beijing, China, 100084. Email: zqshi@math.tsinghua.edu.cn.

In our previous papers, [16, 19], Point Integral method was developed to solve Poisson equation in point cloud. The main observation of the Point Integral method is that the solution of the Poisson equation can be approximated by an integral equation,

1 t

Z

M

Rt(x,y)(u(x)−u(y))dy−2 Z

∂M

R¯t(x,y)∂u

∂n(y)dµy = Z

M

R¯t(x,y)f(y)dy (1.2) wherenis the out normal ofMat∂M. The kernel functions

Rt(x,y) = 1 (4πt)^k/2R

kx−yk² 4t

, R¯t(x,y) = 1 (4πt)^k/2R¯

kx−yk² 4t

(1.3) and ¯R(r) =R+∞

r R(s)ds. t is a parameter, which is determined by the desensity of the point cloud in the real computations.

The kernel functionR(r) :R⁺→R⁺is assumed to beC²smooth and satisfies some mild conditions (see Section 1.1).

The integral approximation (1.2) is natural to solve the Poisson equation with Neumann boundary condition. To enforce the Dirichlet boundary condition, in our previous work [16, 19], we used Robin boundary condition to approximate the Dirichlet boundary condition. More specifically, we solve following problem instead of (1.1) with 0< β ≪1,

−∆_Mu(x) = f(x), x∈ M,

u(x) +β^∂u_∂n = 0, x∈∂M. (1.4)

Using (1.2), we have an integral equation to approximate the above Robin problem, 1

t Z

M

Rt(x,y)(u(x)−u(y))dy+ 2 β

Z

∂M

R¯t(x,y)u(y)dµy= Z

M

R¯t(x,y)f(y)dy. (1.5) We can prove that this approach converge to the original Dirichlet problem [19]. In the real computations, small β may give some trouble. The overcome this problem, we also introduced an itegrative method to enforce the Dirichlet boundary condition based on the Augmented Lagrangian Multiplier (ALM) method.

However, we can not prove the convergence of this iterative method, although it always converges in the numerical tests.

Recently, Du et.al. [7] proposed volume constraint to deal with the boundary condition in the nonlocal diffusion problem. They found that in the nonlocal diffusion problem, since the operator is nonlocal, only enforce the boundary condition on the boundary is not enough, we have to extend the boundary condition to a small region close to the boundary. Borrowing this idea, in nonlocal diffusion problem to handle the Dirichlet boundary. This idea gives us following integral equation with volume constraint:





 1 t Z

M

Rt(x,y)(u(x)−u(y))dy = Z

M

R¯t(x,y)f(y)dy, x∈ M^′t

u(x) = 0, x∈ V^t

(1.6)

Here,M^′t andV^tare subsets ofMwhich are defined as M^′t=n

x∈ M:B x,2√

t

∩∂M=∅o

, V^t=M\M^′t. (1.7)

The thickness ofV^tis 2√

twhich implies that|V^t|=O(√

t). The relation ofM,∂M,M^′tandV^tare sketched in Fig. 1.

The main advantage of the integral equation (1.6) is that there is not any differential operator in the integral equation. Then it is easy to discretized on point cloud. Assume we are given a set of sample points P ={p_i :p_i ∈ M, i= 1,· · ·, n}sampling the submanifoldM and one vectorV= (V1,· · ·, Vn)^t whereVi

2√ t

V^t M^′t M=M^′t∪ V^t

∂M

Figure 1: Computational domain for volume constraint

is the volume weight ofp_i inM. In addition, we assume that the point setP is a good sample of manifold Min the sense that the integral onMcan be well approximated by the summation overP, see Section 1.1.

Then, (1.6) can be easily discretized to get following linear system





 1 t

X

pj∈M

Rt(pi,p_j)(ui−uj)Vj = X

pj∈M

R¯t(pi,p_j)f(pj)Vj, p_i∈ M^′t,

ui = 0, p_i∈ V^t.

(1.8)

This is the discretization of the Poisson equation (1.1) given by Point Integral Method with volume constraint on point cloud.

Similarly, the eigenvalue problem

−∆_Mu(x) = λu, x∈ M

u(x) = 0, x∈∂M (1.9)

can be approximated by an integral eigenvalue problem





 1 t

Z

M

Rt(x,y)(u(x)−u(y))dy = λ Z

M

R¯t(x,y)u(y)dy, x∈ M^′t

u(x) = 0, x∈ V^t

(1.10)

And corresponding discretization is given as following





 1 t

X

pj∈M

Rt(pi,p_j)(ui−uj)Vj = λ X

pj∈M

R¯t(pi,p_j)ujVj, p_i∈ M^′t,

ui = 0, p_i∈ V^t.

(1.11)

1.1 Assumptions and main results

One of the main contribution of this paper is that, under some assumptions, we prove that the solution of the discrete system (1.8) converges to the solution of the Poisson equation (1.1) and the spectra of the eigen problem (1.11) converge to the spectra of the Laplace-Beltrami operator with Dirichlet boundary (1.9).

The assumptions we used are listed as following.

Assumption 1.1. • Assumptions on the manifold: M, ∂Mare both compact and C^∞ smooth.

• Assumptions on the sample points (P,V): (P,V)ish-integrable approximation ofM, i.e.

For any function f ∈C¹(M), there is a constant C independent ofhandf so that

Z

M

f(y)dy− X

pi∈M

f(pi)Vi

< Ch|supp(f)|kfkC¹(M).

• Assumptions on the kernel function R(r):

(a) R∈C²(R⁺);

(b) R(r)≥0 andR(r) = 0 for ∀r >1;

(c) ∃δ0>0 so thatR(r)≥δ0 for 0≤r≤¹2.

These assumptions are default in this paper and they are omitted in the statement of the theoretical results. And in the analysis, we always assume that t and h/√

t are small enough. Here, ”small enough”

means that they are less than a generic constant which only depends onM.

Under above assumptions, we have two theorems regarding the convergence of the Poisson equation and corresponding eigenvaule problem.

Theorem 1.1. Let u(x)be solution of (1.1)and u= [u1,· · ·, un]^tbe solution of (1.8) andf ∈C¹(M)in both problems. There exists C >0 only depends on Mand∂M, such that

ku−ut,hkH¹(M^′t)≤C

t^1/4+ h t^3/2

kfkC¹(M)

where

ut,h(x) =









 1 wt,h(x)



 X

pj∈M

Rt(x,pj)ujVj+t X

pj∈M

R¯t(x,pj)f(pj)Vj



, x∈ M^′t,

0, x∈ V^t.

(1.12)

andwt,h(x) =P

pj∈MRt(x,p_j)Vj.

Theorem 1.2. Letλi be theith largest eigenvalue of eigenvalue problem (1.9). And letλ^t,h_i be theith largest eigenvalue of discrete eigenvalue problem (1.11), then there exists a constantC such that

|λ^t,h_i −λi| ≤Cλ²_i

t^1/4+ h t^d/4+3

,

and there exist another constantC such that, for anyφ∈E(λi, T)X andX =H¹(M^′t), kφ−E(σ_i^t,h, Tt,h)φk^H1(M^′t)≤C

t^1/4+ h t^d/4+2

.

where σ_i^t,h ={λ^t,h_j ∈σ(Tt,h) : j ∈Ii} and Ii ={j ∈N: λj =λi},E(λ, T) is the Riesz spectral projection associated withλ.

2 Stability analysis

To prove the convergence, we need some stability results which are listed in this section. The first lemma is about the coercivity of the integral operator and the proof can be found in [19].

Lemma 2.1. For any functionu∈L²(M), there exists a constantC >0 only depends onM, such that Z

M

Z

M

Rt(x,y)(u(x)−u(y))²dxdy≥C Z

M|∇v|²dx,

where

v(x) = 1 wt(x)

Z

M

Rt(x,y)u(y)dy, andwt(x) =

Z

M

Rt(x,y) dy.

Next corollary directly follows from Lemma 2.1.

Corollary 2.1. For any function u∈L2(M^′t), there exists a constant C >0 only depneds onM, such that 1

t Z

M^′t

Z

M^′t

Rt(x,y)(u(x)−u(y))²dxdy+1 t

Z

M^′t

u²(x) Z

Vt

Rt(x,y)dy

dx≥C Z

M^′t

|∇v|²dx, where

v(x) = 1 wt(x)

Z

M^′t

Rt(x,y)u(y)dy,

andwt(x) = Z

M

Rt(x,y) dy.

Proof. Let

˜ u(x) =

u(x), x∈ M^′t, 0, x∈ V^t. Using Lemma 2.1,

Z

M^′t

|∇v|²dx≤ Z

M|∇v|²dx

≤ C t

Z

M

Z

M

Rt(x,y)(˜u(x)−u(y))˜ ²dxdy

= C t

Z

M^′t

Z

M^′t

Rt(x,y)(u(x)−u(y))²dxdy+C t

Z

M^′t

u²(x) Z

Vt

Rt(x,y)dy

dx.

Using Lemma 2.1, we can also get following lemma regarding the stability inL²(M).

Lemma 2.2. For any function u∈L2(M)with u(x) = 0in V^t, there exists a constantC >0 independent ont

1 t

Z

M

Z

M

Rt(x,y)(u(x)−u(y))²dxdy≥Ckuk²L2(M), as long ast small enough.

Proof. Let

v(x) = 1 wt(x)

Z

M

Rt(x,y)u(y)dy.

Sinceu(x) = 0, x∈ V^t, we have

v(x) = 0, x∈∂M.

By Lemma 2.1 and the Poincare inequality, there exists a constantC >0, such that Z

M|v(x)|²dx≤ Z

M|∇v(x)|²dx ≤ C t

Z

M

Z

M

Rt(x,y) (u(x)−u(y))²dµxdµy

Letδ= _2wmax^wmin_+wmin. Ifuis smooth and close to its smoothed versionv, in particular, Z

M

v²(x)dµx≥δ² Z

M

u²(x)dµx, (2.1)

then the proof is completed.

Now consider the case where (2.1) does not hold. Note that we now have ku−vkL²(M) ≥ kukL2(M)− kvkL2(M)>(1−δ)kukL2(M)

> 1−δ

δ kvk^L2(M)= 2wmax

wmin kvk^L2(M). Then we have

Ct

t Z

M

Z

M

R

|x−y|² 4t

(u(x)−u(y))²dµxdµy

= 2Ct

t Z

M

u(x) Z

M

R

|x−y|² 4t

(u(x)−u(y))dµydµx

= 2 t

Z

M

u²(x)wt(x)dµx− Z

M

u(x)v(x)wt(x)dµx

= 2 t

Z

M

(u(x)−v(x))²wt(x)dµx+ Z

M

(u(x)−v(x))v(x)wt(x)dµx

≥ 2 t Z

M

(u(x)−v(x))²wt(x)dµx−2 t

Z

M

v²(x)wt(x)dµx

1/2Z

M

(u(x)−v(x))²wt(x)dµx

1/2

≥ 2wmin

t Z

M

(u(x)−v(x))²dµx−2wmax

t Z

M

v²(x)dµx

1/2Z

M

(u(x)−v(x))²dµx

1/2

≥ wmin

t Z

M

(u(x)−v(x))²dµx≥wmin

t (1−δ)² Z

M

u²(x)dµx. This completes the proof for the theorem.

Corollary 2.2. For any functionu∈L2(M^′t), there exists a constantC >0 independent ont, such that 1

t Z

M^′t

Z

M^′t

Rt(x,y)(u(x)−u(y))²dxdy+ Z

M^′t

u²(x) Z

V^t

Rt(x,y)dy

dx≥Ckuk²L2(M^′t), as long ast small enough.

Proof. Consider

˜ u(x) =

u(x), x∈ M^′t, 0, x∈ V^t. and apply Lemma 2.2.

Now, we can prove one important theorem.

Theorem 2.1. Let u(x)∈L²(M)be solution of following integral equation





 1

t Z

M

Rt(x,y)(u(x)−u(y))dy = r(x), x∈ M^′t

u(x) = 0, x∈ V^t

(2.2) There existsC >0only depends on Mand∂M, such that

kuk^H1(M^′t)≤Ckrk^L2(M^′t)+Ctk∇rk^L2(M^′t)

Proof. First of all, we have 1

t Z

M^′t

u(x) Z

M

Rt(x,y)(u(x)−u(y))dydx

= 1 t

Z

M^′t

u(x) Z

M^′t

Rt(x,y)(u(x)−u(y))dydx+1 t

Z

M^′t

u(x) Z

Vt

Rt(x,y)(u(x)−u(y))dydx

= 1 2t

Z

M^′t

Z

M^′t

Rt(x,y)(u(x)−u(y))²dxdy+1 t

Z

M^′t

u²(x) Z

Vt

Rt(x,y)dydx.

Now we can get L² estimate ofu. Using Corollary 2.2, we have kuk²2,M^′t ≤ C

t Z

M^′t

Z

M^′t

Rt(x,y)(u(x)−u(y))²dxdy+C t

Z

M^′t

|u(x)|² Z

V^t

Rt(x,y)dy

dx

≤

C t

Z

M^′t

u(x) Z

M

Rt(x,y)(u(x)−u(y))dydx

≤ Ckuk^2,M^′tkrk^2,M^′t

This gives that

kuk^L2(M^′t)≤Ckrk^L2(M^′t). (2.3) Next, we turn to estimate theL² norm of∇etin M^′t. Using the integral equation (2.2),uhas following expression

ut(x) = 1 wt(x)

Z

M^′t

Rt(x,y)ut(y)dy+ t

wt(x)r(x), x∈ V^t. (2.4) Thenk∇utk²2,M^′t can be bounded as following

k∇utk²2,M^′t ≤C

∇ 1

wt(x) Z

M^′t

Rt(x,y)ut(y)dy

!

2

2,M^′t

+Ct² ∇

r(x) wt(x)

2 2,M^′t

(2.5) Corollary 2.1 gives a bound the first term of (2.5).

∇ 1

wt(x) Z

M^′t

Rt(x,y)ut(y)dy

!

2

2,M^′t

(2.6)

≤C t

Z

M^′t

Z

M^′t

Rt(x,y)(ut(x)−ut(y))²dydx+C t

Z

M^′t

|ut(x)|² Z

Vt

Rt(x,y)dy

dx.

The second terms of (2.5) can be bounded by direct calculation.

∇

r(x) wt(x)

2

2,M^′t

≤ C

∇r(x) wt(x)

2

2,M^′t

+C

r(x)∇wt(x) (wt(x))²

2

2,M^′t

(2.7)

≤ Ck∇r(x)k²2,M^′t+C

t kr(x)k²2,M^′t. Now we have the bound ofk∇utk^2,M^′t by combining (2.5), (2.6), and (2.7)

k∇utk²2,M^′t ≤ C t

Z

M^′t

Z

M^′t

Rt(x,y)(ut(x)−ut(y))²dxdy (2.8) +C

t Z

M^′t

|ut(x)|² Z

Vt

Rt(x,y)dy

dx+Ct²k∇r(x)k²2,M^′t+Ctkr(x)k²2,M^′t.

Then the bound ofk∇utk^2,M^′t can be obtained also from (2.8) k∇utk²2,M^′t

≤ C t

Z

M^′t

Z

M^′t

Rt(x,y)(ut(x)−ut(y))²dxdy+Ctkr(x)k²2,M^′t

+C t

Z

M^′t

|ut(x)|² Z

Vt

Rt(x,y)dy

dx+Ct²k∇r(x)k²2,M^′t

≤ C

t Z

M^′t

ut(x) Z

M

Rt(x,y)(ut(x)−ut(y))dydx +Ct²k∇r(x)k²2,M^′t+Ctkr(x)k²2,M^′t

≤ kutk2,M^′tkrk2,M^′t+Ct²k∇r(x)k²2,M^′t+Ctkr(x)k²2,M^′t

≤ Ckrk²L²(M^′t)+Ct²k∇r(x)k²2,M^′t. Then we have

k∇utk^2,M^′t ≤Ckrk^L2(M^′t)+Ctk∇r(x)k2,M^′t. (2.9) The proof is completed by putting (2.3) and (2.9) together.

3 Convergence analysis

The main purpose of this section is to prove that the solution of (1.8) converges to the solution of the original Poisson equation (1.1), i.e. Theorem 1.1 in Section 1.1. To prove this theorem, we split it to two parts. First, we prove that the solution of the integral equation (1.6) converges to the solution of the Poisson equation (1.1), which is given in Theorem 3.2. Then we prove Theorem 3.3 to show that the solution of (1.8) converges to the solution of (1.6).

3.1 Integral approximation of Poisson equation

To prove the convergence of the integral equation (1.6), we need following theorem about the consistency which is proved in [20].

Theorem 3.1. Let u(x) be the solution of the problem (1.1). Let u∈H³(M)and r(x) = 1

t Z

M

Rt(x,y)(u(x)−u(y))dy− Z

M

R¯t(x,y)f(y)dy.

There exists constants C, T0 depending only onM and∂M, so that for anyt≤T0,

kr(x)kL²(M^′t) ≤ Ct^1/2kuk^H3(M), (3.1) k∇r(x)kL²(M^′t) ≤ CkukH³(M). (3.2)

Using the consistency result, Theorem 3.1 and the stability results presented in Section 2, we can get following theorem which shows the convergence of the integral equation (1.6).

Theorem 3.2. Letu(x)be solution of (1.1)andut(x)be solution of (1.6). There existsC >0only depends onM and∂M, such that

ku−utk^H1(M^′t)≤Ct^1/4kfk^H1(M)

Proof. Letet(x) =u(x)−ut(x), first of all, we have 1

t Z

M^′t

et(x) Z

M

Rt(x,y)(et(x)−et(y))dydx (3.3)

=1 t

Z

M^′t

et(x) Z

M^′t

Rt(x,y)(et(x)−et(y))dydx+1 t

Z

M^′t

et(x) Z

V^t

Rt(x,y)(et(x)−et(y))dydx

=1 2t

Z

M^′t

Z

M^′t

Rt(x,y)(et(x)−et(y))²dxdy+1 t Z

M^′t

et(x) Z

Vt

Rt(x,y)(et(x)−et(y))dydx. The second term can be calculated as

1 t

Z

M^′t

et(x) Z

V^t

Rt(x,y)(et(x)−et(y))dydx (3.4)

= 1 t

Z

M^′t

|et(x)|² Z

Vt

Rt(x,y)dy

dx−1 t

Z

M^′t

et(x) Z

Vt

Rt(x,y)u(y)dy

dx.

Here we use the definition ofetand the volume constraint condition ut(x) = 0, x∈ V^t to get thatet(x) = u(x), x∈ V^t.

The first term is positive which is good for us. We only need to bound the second term of (3.4) to show that it can be controlled by the first term. First, the second term can be bounded as following

1 t Z

M^′t

et(x) Z

Vt

Rt(x,y)u(y)dy

dx

(3.5)

≤ 1 t

Z

M^′t

|et(x)| Z

Vt

Rt(x,y)dy

1/2Z

Vt

Rt(x,y)|u(y)|²dy 1/2

dx

≤ 1 t

Z

M^′t

1 2|et(x)|²

Z

Vt

Rt(x,y)dy

dx+ 2 Z

M^′t

Z

Vt

Rt(x,y)|u(y)|²dy

dx

!

≤ 1 2t

Z

M^′t

|et(x)|² Z

Vt

Rt(x,y)dy

dx+2 t

Z

Vt

|u(y)|² Z

M^′t

Rt(x,y)dx

! dy

≤ 1 2t

Z

M^′t

|et(x)|² Z

Vt

Rt(x,y)dy

dx+C t

Z

Vt

|u(y)|²dy

≤ 1 2t

Z

M^′t

|et(x)|² Z

V^t

Rt(x,y)dy

dx+C√

tkfk²H¹(M). Here we use Lemma A.1 in Appendix A to get the last inequality.

By substituting (3.5), (3.4) in (3.3), we get

1 t Z

M^′t

et(x) Z

M

Rt(x,y)(et(x)−et(y))dydx

(3.6)

≥ 1 2t

Z

M^′t

Z

M^′t

Rt(x,y)(et(x)−et(y))²dxdy +1

2t Z

M^′t

|et(x)|² Z

Vt

Rt(x,y)dy

dx−Ckfk²H¹(M)

√t.

This is the key estimate we used to get convergence.

Notice thatet(x) satisfying an integral equation, 1

t Z

M

Rt(x,y)(et(x)−et(y))dy=r(x), ∀x∈ M^′t, (3.7)

wherer(x) = ¹_tR

MRt(x,y)(u(x)−u(y))dy−R

MR¯t(x,y)f(y)dy.

From Theorem 3.1, we know that

kr(x)kL²(M^′t)≤Ct^1/2kuk^H3(M)≤C√

tkfk^H1(M), (3.8) k∇r(x)kL²(M^′t)≤Ckuk^H3(M)≤Ckfk^H1(M). (3.9) Now we can getL²estimate ofet. Using Corollary 2.2, we have

ketk²2,M^′t ≤C t

Z

M^′t

Z

M^′t

Rt(x,y)(et(x)−et(y))²dxdy (3.10) +C

t Z

M^′t

|et(x)|² Z

Vt

Rt(x,y)dy

dx (from (3.6)) ≤

C t

Z

M^′t

et(x) Z

M

Rt(x,y)(et(x)−et(y))dydx

+Ckfk²H¹(M)

√t

(from (3.7)) ≤Cketk^2,M^′tkrk^2,M^′t+Ckfk²H¹(M)

√t (from (3.8)) ≤Ckfk^H1(M)ketk^2,M^′t

√t+Ckfk²H¹(M)

√t.

This gives that

ketk2,M^′t ≤Ct^1/4kfkH¹(M). (3.11) Next, we turn to estimate the L²norm of∇etin M^′t. Using the integral equation (3.7),ethas following expression

et(x) = 1 wt(x)

Z

M

Rt(x,y)et(y)dy+ t

wt(x)r(x) (3.12)

= 1

wt(x) Z

M^′t

Rt(x,y)et(y)dy+ 1 wt(x)

Z

Vt

Rt(x,y)u(y)dy+ t wt(x)r(x).

Thenk∇etk²2,M^′t can be bounded as following

k∇etk²2,M^′t ≤C

∇ 1

wt(x) Z

M^′t

Rt(x,y)et(y)dy

!

2

2,M^′t

(3.13)

+C ∇

1 wt(x)

Z

Vt

Rt(x,y)u(y)dy

2

2,M^′t

+Ct² ∇

r(x) wt(x)

2 2,M^′t

.

Corollary 2.1 gives a bound the first term of (3.13).

∇ 1

wt(x) Z

M^′t

Rt(x,y)et(y)dy

!

2

2,M^′t

(3.14)

≤ C t

Z

M^′t

Z

M^′t

Rt(x,y)(et(x)−et(y))²dydx+C t

Z

M^′t

|et(x)|² Z

Vt

Rt(x,y)dy

dx.

The second and third terms of (3.13) can be bounded by direct calculation.

∇

r(x) wt(x)

2

2,M^′t

≤ C

∇r(x) wt(x)

2

2,M^′t

+C

r(x)∇wt(x) (wt(x))²

2

2,M^′t

(3.15)

≤ Ck∇r(x)k²2,M^′t+C

t kr(x)k²2,M^′t

≤ Ckfk²H¹(M),

and

∇

1 wt(x)

Z

Vt

Rt(x,y)u(y)dy

(3.16)

≤

∇wt(x) (wt(x))²

Z

Vt

Rt(x,y)u(y)dy

+

1 wt(x)

Z

Vt

∇Rt(x,y)u(y)dy

≤ Ckfk^H1(M)

Z

Vt

Rt(x,y)dy+Ckfk^H1(M)

Z

Vt

|R^′_t(x,y)|dy.

Then the second term of (3.14) has following bound

∇

1 wt(x)

Z

Vt

Rt(x,y)u(y)dy

2

2,M^′t

(3.17)

≤ Ckfk²H¹(M)

Z

M^′t

Z

V^t

Rt(x,y)dy 2

dx+Ckfk²H¹(M)

Z

M^′t

Z

V^t|R^′_t(x,y)|dy 2

dx

≤ Ckfk²H¹(M)

Z

M^′t

Z

Vt

Rt(x,y)dydx+Ckfk²H¹(M)

Z

M^′t

Z

Vt

|R^′_t(x,y)|dydx

≤ Ckfk²H¹(M)|V^t| ≤Ckfk²H¹(M)

√t.

Now we have the bound ofk∇etk^2,M^′t by combining (3.13), (3.15), (3.14) and (3.17) k∇etk²2,M^′t ≤ C

t Z

M^′t

Z

M^′t

Rt(x,y)(et(x)−et(y))²dxdy (3.18) +C

t Z

M^′t

|et(x)|² Z

Vt

Rt(x,y)dy

dx+Ckfk²H¹(M)

√t.

Then the bound ofk∇etk^2,M^′t can be obtained also from (3.18) k∇etk²2,M^′t ≤C

t Z

M^′t

Z

M^′t

Rt(x,y)(et(x)−et(y))²dxdy (3.19) +C

t Z

M^′t

|et(x)|² Z

V^t

Rt(x,y)dy

dx+Ckfk²H¹(M)

√t

(from (3.6)) ≤

C t

Z

M^′t

et(x) Z

M

Rt(x,y)(et(x)−et(y))dydx

+Ckfk²H¹(M)

√t

(from (3.7)) ≤ketk^2,M^′tkrk^2,M^′t+Ckfk²H¹(M)

√t (from (3.8)) ≤CkfkH¹(M)ketk2,M^′t

√t+Ckfk²H¹(M)

√t (from (3.11)) ≤Ct^3/4kfk^H1(M)+Ckfk²H¹(M)

√t.

Then we have

k∇etk^2,M^′t ≤Ct^1/4kfk^H1(M). (3.20) The proof is completed by putting (3.11) and (3.20) together.

3.2 Discretization of the integral equation

Supposeu= [u1,· · ·, un]^tis the discrete solution which means that it solves (1.8). First, we interpolate the discrete solution from the point cloudP ={p₁,· · ·,p_n}to the whole manifoldM. Fortunately, the discrete

equation (1.8) gives a natural interpolation.

ut,h(x) =









 1 wt,h(x)



 X

pj∈M

Rt(x,pj)ujVj+t X

pj∈M

R¯t(x,pj)f(pj)Vj



, x∈ M^′t,

0, x∈ V^t.

(3.21)

Then, we have following theorem regarding the convergence fromut,hto ut.

Theorem 3.3. Letut(x)be the solution of the problem (1.6)andube the solution of the problem (1.8). If f ∈C¹(M) in both problems, then there exists constantsC >0 depending only onMand ∂M so that

kut,h−utk^H1(M^′t) ≤ Ch

t^3/2kfk^C1(M), as long ast and √^h

t are both small enough.

To prove this theorem, we need the stability result, Theorem 2.1, and the consistency result which is given in Theorem 3.4.

To simplify the notations, we introduce some operators here.

Ltu(x) = 1 t

Z

M^′t

Rt(x,y)(u(x)−u(y))dy+u(x) t

Z

Vt

Rt(x,y)dy, (3.22) L^′_tu(x) = 1

t Z

M^′t

Rt(x,y)(u(x)−u(y))dy, (3.23) (Lt,hu)(x) = 1

t X

pj∈M^′t

Rt(x,p_j)(u(x)−u(pj))Vj+u(x) t

X

pj∈Vt

Rt(pi,p_j)Vj, (3.24) (L^′_t,hu)(x) = 1

t X

pj∈M^′t

Rt(x,p_j)(u(x)−u(pj))Vj. (3.25)

It is easy to check thatut,hsatisfies following equation ifu= [u1,· · ·, un]^tsolves (1.8), Lt,hut,h= X

pj∈M

R¯t(x,pj)f(pj)Vj (3.26)

Now, we can state the consistency result as following.

Theorem 3.4. Letut(x)be the solution of the problem (1.6)andube the solution of the problem (1.8). If f ∈C¹(M) , in both problems, then there exists constantsC >0 depending only onMand∂Mso that

kLt(ut,h−ut)k^L2(M^′t) ≤ Ch

t^3/2kfk^C1(M), (3.27) k∇Lt(ut,h−ut)k^L2(M^′t) ≤ Ch

t² kfk^C1(M). (3.28)

as long ast and √^h

t are small enough.

4 Convergence of the eigenvalue problem

In this section, we investigate the convergence of the eigenvalue problem (1.11) to the eigenvalue prob- lem (1.9). First, we introduce some operators.

Denote the operatorT :L²(M)→H²(M) to be the solution operator of the following problem ∆_Mu(x) =f(x), x∈ M,

u(x) = 0, x∈∂M.

wherenis the out normal vector ofM.

Denote Tt:L²(M)→L²(M) to be the solution operator of the following problem







−1 t

Z

M

Rt(x,y)(u(x)−u(y))dy = Z

M

R¯t(x,y)f(y)dy, x∈ M^′t

u(x) = 0, x∈ V^t

The last solution operator isTt,h:C(M)→C(M) which is defined as follows.

Tt,h(f)(x) =









 1 wt,h(x)



 X

pj∈M

Rt(x,pj)ujVj−t X

pj∈M

R¯t(x,pj)f(pj)Vj



, x∈ M^′t,

0, x∈ V^t,

wherewt,h(x) =P

pj∈MRt(x,p_j)Vj andu= (u1,· · · , un)^twithuj= 0, p_j∈ V^tsolves the following linear system

−1 t

X

pj∈M

Rt(p_i,p_j)(ui−uj)Vj= X

pj∈M

R¯t(p_i,p_j)f(p_j)Vj.

We know thatT, TtandTt,hhave following properties.

Proposition 4.1. For any t >0,h >0,

1. T, Tt are compact operators on H¹(M) into H¹(M); Tt, Tt,h are compact operators on C¹(M) into C¹(M).

2. All eigenvalues of T, Tt, Tt,h are real numbers. All generalized eigenvectors of T, Tt, Tt,h are eigenvec- tors.

Proof. The proof of (1) is straightforward. First, it is well known thatT is compact operator. Tt,his actually finite dimensional operator, so it is also compact. To show the compactness of Tt, we need the following formula,

Ttu= 1 wt(x)

Z

M

Rt(x,y)Ttu(y)dy+ t wt(x)

Z

M

R¯t(x,y)u(y)dy, ∀u∈H¹(M).

Using the assumption that R ∈ C², direct calculation would gives that that Ttu∈C². This would imply the compactness ofTtboth inH¹andC¹.

For the operator T, the conclusion (2) is well known. The proof ofTtandTt,h are very similar, so here we only present the proof forTt.

Letλbe an eigenvalue ofTtanduis corresponding eigenfunction, then LtTtu=λLtu

which implies that

λ= R

M

R

MR¯t(x,y)u^∗(x)u(y)dxdy R

Mu^∗(x)(Ltu)(x)dx whereu^∗ is the complex conjugate ofu.

Using the symmetry ofLtand ¯R(x,y), it is easy to show thatλ∈R.

Let u be a generalized eigenfunction of Tt with multiplicity m > 1 associate with eigenvalue λ. Let v= (Tt−λ)^m−1u,w= (Tt−λ)^m−2u, thenvis an eigenfunction ofTtand

Ttv=λv, (Tt−λ)w=v

andv(x) = 0, w(x) = 0, x∈ V^t.

By applyingLton both sides of above two equations, we have λLtv = Lt(Ttv) =

Z

M

R¯t(x,y)v(y)dy= Z

M^′t

R¯t(x,y)v(y)dy, x∈ M^′t

Ltv = Lt(Ttw)−λLtw= Z

M^′t

R¯t(x,y)w(y)dy−λLtw, x∈ M^′t

Using above two equations and the fact thatLt is symmetric, we get 0 =

*

w, λLtv− Z

M^′t

R¯t(x,y)v(y)dy +

M^′t

=

*

λLtw− Z

M^′t

R¯t(x,y)w(y)dy, v +

M^′t

= hLtv, vi_M^′_t≥Ckvk²2

which implies that (Tt−λ)^m−1u = v = 0. This proves that u is a generalized eigenfunction of Tt with multiplicitym−1. Repeating above argument, we can show thatuis actually an eigenfunction ofTt.

Theorem 3.2 actually gives thatTt converges toT inH¹norm.

Theorem 4.1. Under the assumptions in Section 1.1, for t small enough, there exists a constant C > 0 such that

kT−Ttk^H1 ≤Ct^1/4.

Using the arguments in [20] and Theorem 3.3, we can get that Tt,hconverges toTtin C¹ norm.

Theorem 4.2. Under the assumptions in Section 1.1, for t, hsmall enough, there exists a constantC >0 such that

kTt,h−Ttk^C1 ≤ Ch t^d/4+2.

And we also have the bound ofTtandTt,h following the arguments in [20].

Theorem 4.3. Under the assumptions in Section 1.1, for t, h small enough, there exists a constant C independent ont andh, such that

kTtk^H1 ≤C, kTt,hk∞≤Ct^−d/4, kTt,hk^C1≤Ct^−(d+2)/4.

. Before state the main theorem of the spectral convergence, we need to introduce some notations. Let X be a complex Banach space andL:X →X be a compact linear operator. ρ(L) is the resolvent set ofL which is given byz∈Csuch thatz−Lis bijective. The spectrum ofLisσ(L) =C\ρ(L). If λis a nonzero eigenvalue ofL, the ascent multiplicityαofλ−Lis the smallest integer such that ker(λ−L)^α= ker(λ−L)^α+1. Given a closed smooth curve Γ⊂ρ(L) which encloses the eigenvalueλand no other elements of σ(L), the Riesz spectral projection associated withλis defined by

E(λ, L) = 1 2πi

Z

Γ

(z−L)⁻¹dz, wherei=√

−1 is the unit imaginary

Now we are ready to state the main theorem about the convergence of the eigenvalue problem. And its proof can be given from Theorems 4.1, 4.2 and 4.3 following same arguments as those in [20].

Theorem 4.4. Under the assumptions in Section 1.1, let λi be the ith smallest eigenvalue of T counting multiplicity, andλ^t,h_i be theith smallest eigenvalue ofTt,h counting multiplicity, then there exists a constant C such that

|λ^t,h_i −λi| ≤C

t^1/4+ h t^d/4+3

,

and there exist another constantC such that, for anyφ∈E(λi, T)X andX =H¹(M), kφ−E(σ_i^t,h, Tt,h)φkH¹(M)≤C

t^1/4+ h t^d/4+2

. whereσ^t,h_i ={λ^t,h_j ∈σ(Tt,h) :j∈Ii} andIi={j∈N:λj=λi}.

The convergence result, Thorem 1.2, follows easily from the above theorem and Proposition 4.1.

5 Numerical results

In this section, we present several numerical results to show the convergence of the Point Integral method with volume constraint, PIM VC for short, from point clouds.

The numerical experiments were carried out in unit disk. We discretize unit disk with 684, 2610, 10191 and 40269 points respectively and check the convergence of the point integral method with volume constraint.In the experiments, the volume weight vector V is estimated using the method proposed in [17]. First, we locally approximate the tangent space at each point and then project the nearby points onto the tangent space over which a Delaunay triangulation is computed in the tangent space. The volume weight is estimated as the volume of the Voronoi cell of that point.

Table 1 gives thel²error of different methods with 684, 2610, 10191 and 40269 points. The exact solution is cos 2πp

x²+y². PIM Robin is the Point Integral method and using Robin boundary to approximate the Dirichlet boundary condition, i.e. solving the integral equation (1.5) and hereβis chosen to be 10⁻⁴. PIM VC is the Point Integral method and using volume constraint to enforce the Dirichlet boundary condition. These two methods both converge. The rates of convergence are very close and the error of PIM VC is a little larger than the error of PIM Robin.

|P| 684 2610 10191 40269

PIM Robin 0.1500 0.0428 0.0140 0.0052 PIM VC 0.3046 0.0747 0.0201 0.0067

Table 1: l²error with different number of points. FEM: Finite Element method; PIM Robin: Point Integral method with Robin boundary; PIM VC: Point Integral method with volume constraint. The exact solution is cos 2πp

x²+y².

Fig. 2 shows the result of the eigenvalues of Laplace-Beltrami operator with Dirichlet boundary in unit disk. Clearly, the eigenvalues also converge and the larger eigenvalues have larger errors which verify the theoretical result, Theorem 1.2.

Above numerical results in unit disk are just toy examples to demonstrate the convergence of the Point Integral method with volume constraint. However, our method applies in any point clouds which sample smooth manifolds. Fig. 3 shows the first two eigenfunctions on two complicated surfaces (left hand and head of Max Plank).

6 Conclusion

In this paper, we use the volume constraint [7] in the Point Integral method to handle the Dirichlet boundary condition. And the convergence is proved both for Poisson equation and eigen problem of Laplace-Beltrami

Figure 2: Eigenvalue given by Point Integral method with volume constraint in unit disk.

operator from point cloud. Our study shows that Point Integral method together with the volume constraint gives an efficient numerical approach to solve the Poisson equation with Dirichlet boundary on point cloud.

In this paper, we focus on the Poisson equation. For other PDEs, we can also use the idea of volume constraint to enforce the Dirichlet boundary condition. The progress will be reported in our subsequent papers.

Acknowledgments. This research was partially supported by NSFC Grant (11201257 to Z.S., 11371220 to Z.S. and J.S. and 11271011 to J.S.), and National Basic Research Program of China (973 Program 2012CB825500 to J.S.).

A One basic estimates

Lemma A.1. Let u(x)be the solution of (1.1)andf ∈H¹(M), then there is a generic constantC >0and T0>0 only depend on Mand∂M, for anyt < T0,

Z

Vt

|u(y)|²dy≤Ct^3/2kfk²H¹(M).

Proof. Both Mand ∂M are compact andC^∞ smooth. Consequently, it is well known that both M and

∂Mhave positive reaches, which means that there existsT0>0 only depends onMand∂M, ift < T0, V^t can be parametrized as (z(y), τ)∈∂M×[0,1], wherey=z(y)+τ(z^′(y)−z(y)) and

det

dy d(z(y),τ)

≤C√

t and C >0 is a constant only depends on M and ∂M. Herez^′(y) is the intersection point between∂M^′ and the line determined byz(y) andy. The parametrization is illustrated in Fig.4.

Figure 3: The first (upper row) and second (lower row) eigenfunctions with Dirichlet boundary.

y z^′(y)

z(y)

∂M

V_t M^′_t

Γs,τ

Figure 4: Parametrization ofV^t First, we have

Z

V^t|u(y)|²dy= Z

V^t|u(y)−u(z(y))|²dy

= Z

Vt

Z 1 0

d

dsu(y+s(z(y)−y))ds

2

dy

= Z

V^t

Z 1 0

(z(y)−y)· ∇u(y+s(z(y)−y))ds

2

dy

≤ Ct Z

Vt

Z 1

0 |∇u(y+s(z(y)−y))|²dsdy

≤ Ct sup

0≤s≤1

Z

V^t|∇u(y+s(z(y)−y))|²dy. Here, we use the fact thatkz(y)−yk²≤2√

tto get the second last inequality.