Introduction to Finite Elements and the Discrete Green's ... · 4/10/2019 · engineering, and...

Introduction to Finite Elements and theDiscrete Green’s Function

Andrew Miller

April 10, 2019

UConn (Sigma Seminar) (Andrew Miller) 1 / 29

Outline

1 Introduce Elliptic PDEs.

2 Introduce FEM for solving Elliptic PDEs.

3 Discuss Advantages to FEM.

4 Introduce and discuss Green’s Functions.

5 A discrete Harnack Inequality and 2D versus 3D.

6 Future Work.


Introduction

• Let Ω ⊂ RN be a convex bounded domain with sufficientlysmooth boundary. We condsider the following Dirichletproblem for Laplace’s equation:

−∆u = f , in Ω

u = b, on ∂Ω,

where f ∈ L2(Ω), b ∈ C(∂Ω), and b ≥ 0.

• The standard procedure is to multiply through by a testfunction v ∈ H1

0 (Ω) and integrate by parts to arrive at thefollowing form,

a(u, v) = (∇u,∇v)Ω =

∫Ω∇u · ∇v dx =

∫Ωfv dx = (f , v)Ω,

where ( , )Ω denotes the standard L2-inner product on Ω.


Introduction

• For

a(u, v) = (∇u,∇v)Ω =

∫Ω∇u · ∇v dx =

∫Ωfv dx = (f , v)Ω;

a(u, v) is refered to as a bilinear form on H1 × H1 → R and(f , v) is a linear functional on H1. The well knownLax-Milgram Theorem guarantees existence of a solution toour problem.


Introduction

Theorem (Lax and Milgram, 1954)

Assume that a(u, v) : H × H → R is a bilinear mapping, for whichthere exists constants α, β > 0 such that,

|a(u, v)| ≤ α‖u‖‖v‖ bounded,

andβ‖u‖2 ≤ a(u, u) coercive.

Also, let (f , v) be a continuous linear functional on H. Then thereexists a unique element u ∈ H such that,

a(u, v) = (f , v),

for all v ∈ H.


Introduction

• Existence of solutions is awesome!

• Actual solutions are hard to find...

• In many applications, like fluid mechanics, aeronauticalengineering, and mathematical physics, the domain may bedifficult to work with and the actual solutions orapproximations to actual solutions may be desirable.

• Instead of working with the continuous problem, we wish todiscrete version of the same problem.


What is the Finite Element Method?

• Goal: Provide an introduction to the Finite Element Methodand give some preliminaries specific to my research.

• The Finite Element Method is a numerical method for solvingpartial differential equations in which we discretize Ω,interpolate functions on the discrete space using basisfunctions, and then numerically solve the PDE.

• How do we do this?



• Consider the variational formulation for our Laplace equation.

I∫Ω

∇u · ∇v dx = (f , v)Ω where v ∈ H10 (Ω).

• We wish to construct a finite dimensional subspace Vh ⊂ H10

and solve the PDE on Vh.



• We first triangulate Ω by subdividing Ω into a setTh = τ1, ..., τp of non-overlapping triangles, called elements, τisuch that,

Ω = ∪τ∈Th

τ = τ1 ∪ · · · ∪ τp,

where no vertex, called nodes, of one triangle lies on the edgeof another triangle (this is what we consider conforming).

• The mesh parameter h represents the maximum length of alledges in Th and we refer to this ”new” domain as Ωh.



• We now define Vh to be the set of all continuous functions onΩh that are linear (affine) when restricted to each triangle inTh and also define V 0

h (Ωh) = v ∈ Vh : v |∂Ωh= 0.

• We use parameters to descibe functions in Vh by choosing thevalues 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 functionsφkn+m

k=1 for Vh(Ωh) defined by

φk(xl) = δlk ≡

0 if l 6= k

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



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

• An example of an arbitrary 2D basis function.



• Therefore a function vh ∈ Vh has the following representation

vh(x) =∑n

i=1 viφi (x) +∑n+m

j=n+1 vjφk(x),

vi = v(xi ) and vj = v(xj) for x ∈ Ωh.

• We then have the following discrete respresentation of theoriginal problem:

• Find uh ∈ Vh(Ωh) to be the solution of the problem

(∇uh,∇v)Ωh= (f , v)Ωh

, ∀v ∈ V 0h (Ωh),

uh = Ihb, on ∂Ωh,

where the interpolant Ihb is given by

Ihb =∑n+m

j=n+1 b(xj)φj .



• Now, for simplicity, let’s assume u = 0 on ∂Ωh. Then we wishto find the coefficients αi for uh(x) =

∑ni=1 αiφi (x) in,

(f , v)Ωh=

∫Ωh

∇uh · ∇v dx , ∀v ∈ V 0h (Ωh).

• Since each v ∈ V 0h is represented by the basis functions it

suffices to check for each j = 1, ..., n,

(f , φj)Ωh=

∫Ωh

∇uh · ∇φj dx .



• That is, for each j = 1, ..., n,

(f , φj)Ωh=

∫Ωh

∇uh · ∇φj dx

=

∫Ωh

∇

(n∑

i=1

αiφi

)· ∇φj dx

=n∑

i=1

αi

∫Ωh

∇φi · ∇φj dx .



• Set fj = (f , φj)Ωhand Aji =

∫Ωh∇φi · ∇φj dx and we arrive at

for each j = 1, ..., n,fj = Ajiαi ,

which is a n × n linear system for the unknowns αi .

• Therefore in matrix form,

A~α = ~f ,

where A is called the stiffness matrix and ~f is called the loadvector.

• Thus we need only to solve for ~α in order to find our solutionuh, the so called Finite Element Method (FEM).

• This we can do numerically! Programs like Matlab andFreeFEM.


Advantages to FEM

• In the Dirichlet case (zero boundary condition) we have thatA is a symmetric positive definite matrix, immediatelyimplying that A−1 exists and thus a solution exists.

• Recall the support of the basis functions, φi , is small. Thisimplies that (∇φi ,∇φj)Ωh

= 0 when the support of φi and φjdoes not share any elements.

I This means that most of the entries of A are zero, making A asparse matrix.

I This makes for a large reduction in memory needed to run aFEM solver.


Advantages to FEM

• We have the following a priori estimate on the FEM solution:

‖u − uh‖L2(Ω) ≤ Ch2‖D2u‖L2(Ω).

I In general limh→0‖u − uh‖ = 0 and abstractly lim

h→0V 0h → H1

0 .

• Galerkin Orthogonality; we have for both u and uh,∫Ω∇u·∇v dx =

∫Ωfv dx and

∫Ω∇uh·∇v dx =

∫Ωfv dx , ∀v ∈ V 0

h ,

since V 0h ⊂ H1

0 .Therefore we can subtract to get,∫Ω∇(u − uh) · ∇v dx = 0.


Green’s Functions

• Recall the continuous Green’s function G (x , z) of the LaplaceEquation. That is, the Green’s function with singularity at z isthe 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 .

• The Discrete Green’s Function with singularity at z which isthe function G z

h (x) ∈ V 0h (Ωh) satisfying

(∇G zh ,∇v)Ωh

= (δz , v)Ωh= v(z) ∀v ∈ V 0

h (Ωh).

• I.e. G zh (x) is the FEM solution to the continuous Green’s

function problem.


Green’s Functions

• In addition we can also regularize the continuous Green’sfunction and the Dirac distribution!

• Define

δz = 1σ(B(0,1))εN

1Bε ,

where Bε = B(z , ε). We will assume that the parameter ε issmall enough such that B(z , ε) ⊂ τ0, where τ0 is the elementsuch that z ∈ τ0.

• Then we define G z(x) as the solution to

−∆G z = δz , in Ω

G z = 0, on ∂Ω.

• Quick fun fact!


Green’s Functions

Theorem (Bertoluzza Et Al., 2017)

Consider the continuous Green’s function and regularized Green’sfunction Laplace problem;

−∆G z = δz , in Ω and −∆G z = δz , in Ω

G z = 0, on ∂Ω and G z = 0, on ∂Ω.

The piecewise linear finite element solution to both problems areequal.

• Essentially, G zh (x) is the finite element approximation to both

problems! Another huge advantage in the analysis.


Green’s Functions

• Let’s recall a special way to represent solutions to Laplace’sproblem using the continuous Green’s function.

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) ispositive for all x , y then ∂G

∂ν (x , y) ≤ 0 which gives that thesolution u(x) is positive.

• This is not quite the case when we look at the discrete case aswe shall soon see.

• Why do we care?


Green’s Functions

• We also have the continuous Harnack’s Inequality.

Theorem

Assume u ≥ 0 is a C2 solution of

−∆u = 0 in Ω,

and suppose Ω0 b Ω is connected. Then there exists a constant Csuch that

supΩ0

u ≤ C infΩ0

u.

The constant only depending on Ω0.

• This theorem essentially says that any two values of u on thesubdomain are comparable. We wish to do this in the discretecase.


Discrete Harnack Inequality

• First consider the following Laplace problem:

−∆u = 0, in Ω

u = b, on ∂Ω,

where b ∈ C(∂Ω) and b ≥ 0.

• This problem has the following FEM variational formulation:

(∇uh,∇v)Ωh= 0, ∀v ∈ V 0

h (Ωh),

uh = Ihb, on ∂Ωh,

where the interpolant Ihb is given by

Ihb =∑n+m

j=n+1 b(xj)φj .



• We can then represent uh(x) in matrix form as~U = −A−1H ~B.

• 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. Thematrix H ∈ Rn×m is the boundary stiffness matrix. The vector~B contains the boundary data withB = (b(xn+1), ..., b(xn+m))T ∈ Rm.

• By reinterpreting matrix multiplication as a sum we have

uh(xi ) = −n∑

j=1

n+m∑k=m+1

A−1ij HjkBk .

• Now to prove a discrete Harnack Inequality we wish to showuh is positive.



• To do so we substitute G xih (xj) = A−1

ij . Why is this true?

• First we know that the vector ~G (i) = (G xih (x1), ...,G xi

h (xn))T

solves

A ~G (i) = ~B(i),

where B(i)j = (δxi , φj) = φj(xi ) = δji .

• Therefore if we interpret the vectors as column matrices thenwe can compactly write the equations as

A[ ~G (1), ..., ~G (n)] = [~B(1), ..., ~B(n)] = In.• Which shows that

G xih (xj) = A−1

ij .



• This gives

uh(xi ) = −∑n

j=1

∑n+mk=n+1 Gh(xj , xi )HjkBk .

• Therefore positivity of the discrete Green’s function andpositivity of the boundary data is not sufficient to ensurepositivity of the discrete solution.

• Now we need the following assumption of the boundarystiffness matrix H:

Assumption

For every triangulation of Ωh, the associated boundary stiffnessmatrix H must satisfy H ≤ 0, i.e (∇φj ,∇φk)Ωh

≤ 0 for allj ∈ 1, ..., n and k ∈ n + 1, ..., n + m.

• In 2D this is equivalent to the following edge condition: Forevery edge in the triangulation with one node on the boundaryand one node in the interior we must have the sum of theangles opposite the edge be no more than 180o .


Positivity of G zh (x)

• Under the edge condition, the discrete Harnack Inequality is astraight forward result. The challenge lies in showingG zh (x) ≥ 0.

Discrete Green’s Positivity Theorem (Leykekhman andPruitt, 2016)

Suppose D b Ω b R2 is smooth. Then there exists h0 > 0 suchthat for all 0 < h ≤ h0, we have G z

h (x) > 0 for all x ∈ int Ωh andz ∈ D.

• In 2D many of the results rely on the fact that G (x , z) scaleslike ln |x − z |.


Challenges to Positivity of G zh (x) in 3D

• This is not the case in 3D which is where my work lies, hereG (x , z) scales like 1

|x−z| .

• We believe that we need stronger conditions on the mesh,beyond the boundary stiffness matrix condition. Why?I A Delaunay triangulation is one of the most commonly used

mesh generating schemes.I In 2D Delaunay will bound the minimum angle in each

triangle. Good!!I In 3D Delaunay will not bound the minimum dihedral angle.

Bad!!

• This can lead to very thin tetrahedra which can then lead to abad structure of the basis functions and will cause negativityof the discrete Green’s function.


Future Work

• What’s next?

• Once we finish the positivity in 3D, can we loosen meshconditions in both 2D and 3D?

• Can we extend the Harnack Inequality to non-homogeneousPDE and Parabolic PDEs?

• Investigate the possibility of proving positivity and Harnackfor 2D surfaces.


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