On the Positivity of Discrete Harmonic Functions and the Discrete - - PowerPoint PPT Presentation

On the Positivity of Discrete Harmonic Functions and the Discrete Harnack Inequality for Piecewise Linear Finite Elements

Presented by Andrew Miller

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Introduction

• Let Ω ⊂ RN, for N = 2, be a convex bounded domain with

sufficiently smooth boundary. We condsider the following Dirichlet problem for Laplace’s equation: −∆u = 0, in Ω u = b, on ∂Ω, where b ∈ C(∂Ω) and b ≥ 0.

• Goal: Approximate the solution using piecewise linear

conforming finite elements in order to investigate pointwise error estimates, the positivity of the discrete Green’s function, and the discrete Harnack inequality.

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Introduction

• Recall properties for the continuous Green’s function G(x, z)
• f the Laplace Equation. That is, the Green’s function with

singularity at z is the function G z(x) = G(x, z) given by −∆G z = δz, in Ω G z = 0, on ∂Ω, where δz is the Dirac delta distribution at z.

• We have the special property of being able to represent

solutions to our original problem using the Green’s function.

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Introduction Theorem If u ∈ C2(¯ Ω) solves the Laplace problem, then u(y) = −

• ∂Ω

b(x)∂G ∂ν (x, y) dS(x).

• From here we can see that since b(x) ≥ 0 and if G(x, y) is

positive for all x, y then ∂G

∂ν (x, y) ≤ 0 which gives that the

solution u(x) is positive.

• This is not quite the case when we look at the discrete case as

we shall soon see.

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Introduction

• We also have the continuous Harnack’s Inequality.

Theorem Assume u ≥ 0 is a C2 solution of −∆u = 0 in Ω, and suppose Ω0 ⋐ Ω is connected. Then there exists a constant C such that sup

Ω0

u ≤ C inf

Ω0 u.

The constant only depending on Ω0.

• This theorem essentially says that any two values of u on the

subdomain are comparable. We wish to do this in the discrete case.

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What is the Finite Element Method?

• Goal: Provide an introduction to the Finite Element Method

and give some preliminaries specific to our paper.

• The Finite Element Method is a numerical method for solving

partial differential equations in which we discretize Ω, interpolate functions on the discrete space using basis functions, and then numerically solve the PDE.

• How do we do this?

UConn (General Examination) FEM (Andrew Miller) 6 / 28

What is the Finite Element Method?

• Consider the variational formulation for our Laplace equation.

◮

• Ω

∇u · ∇v dx = 0 where v ∈ V = {v : v is continuous on Ω and all partial derivatives are piecewise continuous on Ω and v = 0 on the boundary}.

• We wish to construct a finite dimensional subspace Vh ⊂ V

and solve the PDE on Vh.

• Note: We need a polygonal computational domain Ωh ⊂ Ω

that approximates Ω with distx∈Ω(x, ∂Ωh) ≤ Ch2, h to be defined momentarily.

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What is the Finite Element Method?

• We first triangulate Ωh by subdividing Ωh into a set

Th = τ1, ..., τp of non-overlapping triangles, called elements, τi such that Ωh = ∪

τ∈Thτ = τ1 ∪ · · · ∪ τp,

where no vertex of one triangle lies on the edge of another triangle (this is what we consider conforming).

• The mesh parameter h represents the maximum length of all

edges in Th.

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What is the Finite Element Method?

• We now define Vh to be the set of all continuous functions on

Ωh that are linear (affine) when restricted to each triangle in Th and also define V 0

h (Ωh) = {v ∈ Vh : v|∂Ωh = 0}.

• We use parameters to descibe functions in Vh by choosing the

values v(xi) where xi, i = 1, ..., n is an interior node (resp. j = n + 1, ...n + m is a boundary node) of Th.

• In addition we have the standard nodal basis functions

{φk}n+m

k=1 for Vh(Ωh) defined by

φk(xl) = δlk ≡

• if l = k

1 if l = k, i, k = 1, ..., n + m.

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What is the Finite Element Method?

• An example of 1D basis functions on the interval [0, 1].
• An example of an arbitrary 2D basis function.

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What is the Finite Element Method?

• Therefore a function vh ∈ Vh has the following representation

vh(x) = n

i=1 αiφi(x) + n+m j=n+1 αjφk(x),

αi = v(xi) and αj = v(xj) for x ∈ Ωh.

• We then have the following discrete respresentation of the
• riginal problem:
• Define uh ∈ Vh(Ωh) to be the solution of the problem

(∇uh, ∇χ)Ωh = 0, ∀χ ∈ V 0

h (Ωh),

uh = Ihb, on ∂Ωh, where the interpolant Ihb is given by Ihb = n+m

j=n+1 b(xj)φj.

UConn (General Examination) FEM (Andrew Miller) 11 / 28

Preliminaries

• We can then represent uh(x) in matrix form as

U = −A−1HB.

• Here U represents the solution uh at the interior nodes with

U = (uh(x1), ..., uh(xn))T ∈ Rn.

• The matrix A ∈ Rn×n is the interior stiffness matrix, with

entries given by Aij = (∇φi, ∇φj)Ωh for i, j ∈ {1, ..., n}. The matrix H ∈ Rn×m is the boundary stiffness matrix, with entries given by Hjk = (∇φj, ∇φk)Ωh for j ∈ {1, ..., n} and k ∈ {n + 1, ..., n + m}. The vector B contains the boundary data with B = (b(xn+1), ..., b(xn+m))T ∈ Rm.

UConn (General Examination) FEM (Andrew Miller) 12 / 28

Preliminaries

• By reinterpreting matrix multiplication as a sum we have

uh(xi) = −

n

• j=1

n+m

• k=m+1

A−1

ij HjkBk.

• Now we use an interesting fact about the Discrete Green’s

Function with singularity at z which is the function G z

h (x) ∈ V 0 h (Ωh) satisfying

(∇G z

h , ∇χ)Ωh = (δz, χ)Ωh = χ(z)

∀χ ∈ V 0

h (Ωh).

• We substitute G xi

h (xj) = A−1 ij .

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Preliminaries

• Why is this true?
• First we know that the vector G (i) = (G xi

h (x1), ..., G xi h (xn))T

solves AG (i) = B(i), where B(i)

j

= (δxi, φj) = φj(xi) = δji.

• Therefore if we interpret the vectors as column matrices then

we can compactly write the equations as A[G (1), ..., G (n)] = [B(1), ..., B(n)] = In.

• Which shows that

G xi

h (xj) = A−1 ij .

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Preliminaries

• This gives

uh(xi) = −

n

• j=1

n+m

• k=n+1

Gh(xj, xi)HjkBk.

• Therefore positivity of the discrete Green’s function and

positivity of the boundary data is not sufficient to ensure positivity of the discrete solution.

• In fact, an explicit example was constructed showing the

discrete Green’s function is negative when the singularity and the node at which the negative value is obtained are both a distance O(h) from the boundary leading to nonpositivity of uh.

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Preliminaries Theorem (Dr˘ ag˘ anescu, Dupont, and Scott, 2004) There exists an ǫ0 > 0 such that for all h > 0 sufficiently small, G h,ǫ0

R

(Q) < 0. (That is, the Green’s function corresponding to a singularity at R is negative at Q).

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Preliminaries Theorem (Dr˘ ag˘ anescu, Dupont, and Scott, 2004) There exists an ǫ0 > 0 such that for all h > 0 sufficiently small, G h,ǫ0

R

(Q) < 0. (That is, the Green’s function corresponding to a singularity at R is negative at Q).

UConn (General Examination) FEM (Andrew Miller) 17 / 28

Pointwise Error Estimates for the Green’s Functions

• Goal: In this section we wish prove the following result;

Discrete Green’s Positivity Theorem (Leykekhman and Pruitt, 2016) Suppose D ⋐ Ω ⋐ R2 is smooth. Then there exists h0 > 0 such that for all 0 < h ≤ h0, we have G x0

h (x) > 0 for all x ∈ int Ωh and

x0 ∈ D.

• Loosely stated; this theorem says that the discrete Green’s

function on a smooth domain must eventually be positive as the mesh is refined (i.e. as the mesh parameter gets smaller we triangulate Ωh with more elements).

• This results turns out to be very difficult and technical to

prove requiring four theorems, three lemmas, and one proposition.

• Let’s try to get an idea of the build up we need.

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Pointwise Error Estimates for the Green’s Functions

• The following theorem provides an estimate between the

discrete Green’s function and the continuous Green’s function. Theorem (Leykekhman and Pruitt, 2016) Let x, y ∈ Ω with |x − y| ≥ d with Bd(x) ⋐ Ω. Then there exists a constant C independent of h, x, y, and d such that |G x

h (y) − G x(y)| ≤ Cℓhh2d−2,

where ℓh = | ln h|.

• To prove this we apply a best approximation theorem to get

|G x

h (y) − G x(y)| ≤ Cℓh

inf

χ∈V 0

h (Ωh)G x − χL∞(Bd/4(y)∩Ωh)

+ Cd−1G x − G x

h L2(Bd/4(y)∩Ωh).

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On the Positivity of the Discrete Green’s Function

• Using the pointwise error estimate we can piece together our

desired result. Discrete Green’s Positivity Theorem (Leykekhman and Pruitt, 2016) Suppose D ⋐ Ω ⋐ R2 is smooth. Then there exists h0 > 0 such that for all 0 < h ≤ h0, we have G x0

h (x) > 0 for all x ∈ int Ωh and

x0 ∈ D.

• The proof requires two additional lemmas, their results are:

Lemma 1: G x0

h (x0) ≥ C(| ln h| + 1).

Lemma 2: ∇G x0

h L∞(Ωh) ≤ Ch−1.

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On the Positivity of the Discrete Green’s Function

• Idea of proof:
• It is sufficient to consider the case when D = Bd(x0) with

d ≥ ch and dist(∂Bd, ∂Ω) ≥ d0 for some arbitrary d0.

• Case 1: |x − x0| ≤ Kh| ln h|1/2.

◮ Apply lemmas (1) and (2).

• Case 2: Kh| ln h|1/2 ≤ |x − x0| and dist(x, ∂Ω) ≥ d0.

◮ We must apply a theorem from Schatz and Walhbin, 1977, that gives the following estimate, |G x0(x) − G x0

h (x)| ≤ CK h2 |x−x0|2 ln

• |x−x0|

h

• .
• Case 3: dist(x, ∂Ω) ≤ d0.

◮ We apply the earlier theorem giving an estimate between G x0

h

and G x0 and note that if xj is an interior node then |xj − x0| = O(1).

• Finally we interpolate between the nodes to complete the

proof.

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Discrete Harnack Inequality

• Goal: We wish to prove a Harnack type inequality for

piecewise linear finite elements in 2D.

• Recall that we can represent a solution uh to our discrete

variational problem in the following way, uh(xi) = − n

j=1

n+m

k=n+1 Gh(xj, xi)HjkBk.

• Now we use the support of the nodal basis functions to

simplify this sum.

• Let

N(xk) denote the set of all neighboring nodes to xk. Then we can rearrange and rewrite to get uh(xi) = −

n+m

• k=n+1
• xj∈

N(xk)

Gh(xj, xi) (∇φj, ∇φk)Ωh

• Hjk

b(xk)

Bk

.

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Discrete Harnack Inequality

• Now we need the following assumption of the boundary

stiffness matrix H: Assumption For every triangulation of Ωh, the associated boundary stiffness matrix H must satisfy H ≤ 0, i.e (∇φj, ∇φk)Ωh ≤ 0 for all j ∈ {1, ..., n} and k ∈ {n + 1, ..., n + m}.

• In 2D this is equivalent to the following edge condition: For

every edge in the triangulation with one node on the boundary and one node in the interior we must have the sum of the angles opposite the edge be no more than 180o.

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Discrete Harnack Inequality

• An immediate consequence of our theorem on the error

between G z

h and G z is the following Lemma;

Lemma (Leykekhman and Pruitt, 2016) Suppose Ω0 ⋐ Ω1 ⋐ Ω. There there exists h0 > 0 and a constant C∗ such that for all 0 < h ≤ h0, if x ∈ Ω0 and z ∈ Ω\Ω1, the estimate |G(x, z) − Gh(x, z)| ≤ C∗h2| ln h|, holds.

• This then gives the following Harnack type inequalty for the

discrete Green’s function.

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Discrete Harnack Inequality Lemma (Leykekhman and Pruitt, 2016) Suppose Ω0 ⋐ Ω. Let 0 < c1 < c2 be positive constants. Then there exits h0 > 0 and a constant C > 0 independent of h such that, for all 0 < h ≤ h0 and for all x, y ∈ Ω0 and for all z ∈ Ω with c1h ≤ dist(z, ∂Ω) ≤ c2h, we have G z

h (x) ≥ CG z h (y).

• Here the idea is to apply the previous lemma and the

continuous Harnack inequality on G(·, z) in order to compare both Gh(x, z) with G(x, z) and Gh(y, z) with G(y, z).

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Discrete Harnack Inequality

• Now we are ready to prove the discrete Harnack inequality for

all discrete harmonic functions. Theorem (Leykekhman and Pruitt, 2016) Suppose Ω0 ⋐ Ω. Then there exists h0 > 0 and a constant C > 0 such that for all 0 < h < h0, and for all discrete harmonic functions uh satisfying uh(x) ≥ 0 for x ∈ ∂Ω and for all nodes x∗, y∗ ∈ Ω0, we have uh(x∗) ≥ Cuh(y∗).

• The proof involves applying the Harnack’s inequality for the

discrete Green’s function as well as the symmetry of the discrete Green’s function.

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Discrete Harnack Inequality

• Finally as a corollary to the theorem, by interpolating at nodal

points, we obtain a Harnack inequality for piecewise linear finite elements valid for all points in a subdomain. Theorem (Leykekhman and Pruitt, 2016) Suppose Ω0 ⋐ Ω1 ⋐ Ω. Then there exists h0 > 0 and a constant C > 0, depending on Ω0, Ω1, such that for all 0 < h ≤ h0 and for all discrete harmonic functions uh satisfying uh(x) ≥ 0 for x ∈ ∂Ω, and for all x, y ∈ Ω0, we have uh(x) ≥ Cuh(y).

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Thank you!

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