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 UConn (General Examination) FEM (Andrew Miller) 1 / 28

Introduction • Let Ω ⊂ R N , 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. UConn (General Examination) FEM (Andrew Miller) 2 / 28

Introduction • Recall properties for the continuous Green’s function G ( x , z ) of 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 = δ z , in Ω G z = 0 , on ∂ Ω , where δ z 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. UConn (General Examination) FEM (Andrew Miller) 3 / 28

Introduction Theorem If u ∈ C 2 (¯ Ω) solves the Laplace problem, then � b ( x ) ∂ G u ( y ) = − ∂ν ( x , y ) dS ( x ) . ∂ Ω • 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. UConn (General Examination) FEM (Andrew Miller) 4 / 28

Introduction • We also have the continuous Harnack’s Inequality. Theorem Assume u ≥ 0 is a C 2 solution of − ∆ u = 0 in Ω , and suppose Ω 0 ⋐ Ω is connected. Then there exists a constant C such that sup u ≤ C inf Ω 0 u . Ω 0 The constant only depending on Ω 0 . • This theorem essentially says that any two values of u on the subdomain are comparable. We wish to do this in the discrete case. UConn (General Examination) FEM (Andrew Miller) 5 / 28

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. � ∇ u · ∇ v dx = 0 where v ∈ V = { v : v is continuous on Ω ◮ Ω and all partial derivatives are piecewise continuous on Ω and v = 0 on the boundary } . • We wish to construct a finite dimensional subspace V h ⊂ V and solve the PDE on V h . • Note: We need a polygonal computational domain Ω h ⊂ Ω that approximates Ω with dist x ∈ Ω ( x , ∂ Ω h ) ≤ Ch 2 , h to be defined momentarily. UConn (General Examination) FEM (Andrew Miller) 7 / 28

What is the Finite Element Method? • We first triangulate Ω h by subdividing Ω h into a set T h = τ 1 , ..., τ p of non-overlapping triangles, called elements, τ i such that Ω h = τ ∈T h τ = τ 1 ∪ · · · ∪ τ p , ∪ 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 T h . UConn (General Examination) FEM (Andrew Miller) 8 / 28

What is the Finite Element Method? • We now define V h to be the set of all continuous functions on Ω h that are linear (affine) when restricted to each triangle in T h and also define V 0 h (Ω h ) = { v ∈ V h : v | ∂ Ω h = 0 } . • We use parameters to descibe functions in V h by choosing the values v ( x i ) where x i , i = 1 , ..., n is an interior node (resp. j = n + 1 , ... n + m is a boundary node) of T h . • In addition we have the standard nodal basis functions { φ k } n + m k =1 for V h (Ω h ) defined by � 0 if l � = k φ k ( x l ) = δ lk ≡ 1 if l = k , i , k = 1 , ..., n + m . UConn (General Examination) FEM (Andrew Miller) 9 / 28

What is the Finite Element Method? • An example of 1 D basis functions on the interval [0 , 1]. • An example of an arbitrary 2 D basis function. UConn (General Examination) FEM (Andrew Miller) 10 / 28

What is the Finite Element Method? • Therefore a function v h ∈ V h has the following representation v h ( x ) = � n i =1 α i φ i ( x ) + � n + m j = n +1 α j φ k ( x ), α i = v ( x i ) and α j = v ( x j ) for x ∈ Ω h . • We then have the following discrete respresentation of the original problem: • Define u h ∈ V h (Ω h ) to be the solution of the problem ( ∇ u h , ∇ χ ) Ω h = 0 , ∀ χ ∈ V 0 h (Ω h ) , u h = I h b , on ∂ Ω h , where the interpolant I h b is given by I h b = � n + m j = n +1 b ( x j ) φ j . UConn (General Examination) FEM (Andrew Miller) 11 / 28

Preliminaries • We can then represent u h ( x ) in matrix form as U = − A − 1 HB . • Here U represents the solution u h at the interior nodes with U = ( u h ( x 1 ) , ..., u h ( x n )) T ∈ R n . • The matrix A ∈ R n × n is the interior stiffness matrix, with entries given by A ij = ( ∇ φ i , ∇ φ j ) Ω h for i , j ∈ { 1 , ..., n } . The matrix H ∈ R n × m is the boundary stiffness matrix, with entries given by H jk = ( ∇ φ j , ∇ φ k ) Ω h for j ∈ { 1 , ..., n } and k ∈ { n + 1 , ..., n + m } . The vector B contains the boundary data with B = ( b ( x n +1 ) , ..., b ( x n + m )) T ∈ R m . UConn (General Examination) FEM (Andrew Miller) 12 / 28

Preliminaries • By reinterpreting matrix multiplication as a sum we have n n + m � � A − 1 u h ( x i ) = − ij H jk B k . j =1 k = m +1 • 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 (Ω h ) satisfying ∀ χ ∈ V 0 ( ∇ G z h , ∇ χ ) Ω h = ( δ z , χ ) Ω h = χ ( z ) h (Ω h ) . • We substitute G x i h ( x j ) = A − 1 ij . UConn (General Examination) FEM (Andrew Miller) 13 / 28

Preliminaries • Why is this true? • First we know that the vector G ( i ) = ( G x i h ( x 1 ) , ..., G x i h ( x n )) T solves AG ( i ) = B ( i ) , where B ( i ) = ( δ x i , φ j ) = φ j ( x i ) = δ ji . j • 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 ) ] = I n . • Which shows that G x i h ( x j ) = A − 1 ij . UConn (General Examination) FEM (Andrew Miller) 14 / 28

Preliminaries • This gives n + m n � � u h ( x i ) = − G h ( x j , x i ) H jk B k . j =1 k = n +1 • 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 u h . UConn (General Examination) FEM (Andrew Miller) 15 / 28

Preliminaries Theorem (Dr˘ ag˘ anescu, Dupont, and Scott, 2004) There exists an ǫ 0 > 0 such that for all h > 0 sufficiently small, G h ,ǫ 0 ( Q ) < 0 . R (That is, the Green’s function corresponding to a singularity at R is negative at Q). UConn (General Examination) FEM (Andrew Miller) 16 / 28

Preliminaries Theorem (Dr˘ ag˘ anescu, Dupont, and Scott, 2004) There exists an ǫ 0 > 0 such that for all h > 0 sufficiently small, G h ,ǫ 0 ( Q ) < 0 . R (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 ⋐ R 2 is smooth. Then there exists h 0 > 0 such that for all 0 < h ≤ h 0 , we have G x 0 h ( x ) > 0 for all x ∈ int Ω h and x 0 ∈ 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 Ω h 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. UConn (General Examination) FEM (Andrew Miller) 18 / 28