An hp-FEM for a singularly perturbed problem of reaction ...xenophon/slides/BAIL_2018.pdf · An...

An hp-FEM for a singularly perturbed

reaction-convection-diffusion problem

with two small parameters

Christos Xenophontos

Department of Mathematics and Statistics

University of Cyprus

joint work with

Irene Sykopetritou (Ph.D. student)

1

► The model problem

► Analytic regularity of the solution

► Variational formulation

► Discretization by an hp FEM – error estimates

► Numerical example

► Closing remarks

Outline

2

The Model Problem

1 2( ) ( ) ( ) ( ) ( ) ( ) in (0,1)

(0) (1) 0

u x b x u x c x u x f x I

u u

Find u(x) such that

2

The Model Problem

1 2( ) ( ) ( ) ( ) ( ) ( ) in (0,1)

(0) (1) 0


u u

where 0 < ε1, ε2 << 1 are given constants and b(x), c(x),

f(x) are given (analytic) functions, which satisfy

Find u(x) such that

0 ,n

2

The Model Problem

1 2( ) ( ) ( ) ( ) ( ) ( ) in (0,1)

(0) (1) 0


u u

Find u(x) such that

( ) ( ) ( )

, , ,!, ! , !n n n n n n

f f b b c cI I If C n b C n c C n

0 ,n

for some constants , , .b c fC C

where 0 < ε1, ε2 << 1 are given constants and b(x), c(x),

f(x) are given (analytic) functions, which satisfy

3

In addition, we assume there exist constants β, γ, ρ such

that

2( ) 0 , ( ) 0 , ( ) ( ) 02

b x c x c x b x

3

In addition, we assume there exist constants β, γ, ρ such

that

2( ) 0 , ( ) 0 , ( ) ( ) 02

b x c x c x b x

Example: u(x) for

ε1 = 0.0005

ε2 = 0.05

b = c = f = 1

4

There exist constants K, C > 0, independent of ε and n such

that the solution u(x) satisfies

Proposition:

( ) 1 1

1 2( )max , ,

nn n

L Iu CK n

4

There exist constants K, C > 0, independent of ε and n such

that the solution u(x) satisfies

( ) 1 1

1 2( )max , ,

nn n

L Iu CK n

Remark:

This classical differentiability result is sufficient in the so-

called asymptotic range of the (approximating polynomial)

degree p, but not for practical values of p.

Proposition:

5

Let λ0(x), λ1(x) be the solutions of the characteristic equation

and set0 0 1 1

[0,1][0,1]max ( ) , min ( )

xxx x

or equivalently2 2

2 2 1

0,1[0,1]

1

( ) ( ) 4 ( )min

2x

b x b x c x

5

Let λ0(x), λ1(x) be the solutions of the characteristic equation

and set0 0 1 1

[0,1][0,1]max ( ) , min ( )

xxx x

or equivalently2 2

2 2 1

0,1[0,1]

1

( ) ( ) 4 ( )min

2x

b x b x c x

The values of μ0 , μ1 determine the strength of the

boundary layer and since |λ0(x)| < |λ1(x)|, the boundary layer

near x = 1 is stronger.

6

Essentially, there are three regimes [Linß, 2010]:

μ0 μ1

Convection-Diffusion ε1 << ε2 = 1 1 1/ε1

Reaction-Convection-Diffusion ε1 << (ε2)2 << 1 1 /ε2 ε2 /ε1

Reaction-Diffusion 1 >> ε1 >> (ε2)2 1/ (ε1)

1/2 1/ (ε1)1/2

6


We will focus on the case

μ0 μ1




1/2 1/ (ε1)1/2

2

1 2 1

and we anticipate layers of width O(1/ε2) at the left endpoint

and width O(ε2 /ε1) at the right endpoint.

6


μ0 μ1




1/2 1/ (ε1)1/2

2

1 2 1

and we anticipate layers of width O(1/ε2) at the left endpoint

and width O(ε2 /ε1) at the right endpoint.

For simplicity we will assume that ( ) , ( ) .b x b c x c

We will focus on the case

7

Define the stretched variables

2 2 1/ , (1 ) /x x x x

and make the formal ansatz

7

Define the stretched variables

and make the formal ansatz

2, ,2 1 2 ,

0 0

~ / ( ) ( ) ( )BLj BLi

i j i ji j

i j

u u x u x u x

with the functions to be determined., ,, , ,BLBL

i j i ji ju u u

2 2 1/ , (1 ) /x x x x

8

Substituting in the DE and equating like powers of ε1, ε2 we

obtain the following:

8

0,0

,0 1,0

0, 1,

, 2, 1 1,

/

, 1

0, 1

1, 2, 1

i i

j j

i j i j i j

u f c

bu u i

c

u u j

u u bu i jc

Smooth part:

Substituting in the DE and equating like powers of ε1, ε2 we

obtain the following:

,0 ,0

, , , 1

0, 0

, 0, 1

BL BL

i i

BL BL BL

i j i j i j

b u cu i

b u cu u i j

Left boundary layer:

9

,0 ,0

, 1 , 1 ,

0, 0

, , 0

BL BL

i i

BL BL BL

i j i j i j

u b u i

u b u cu i j

Left boundary layer:

Right boundary layer:

9

,0 ,0

, , , 1

0, 0

, 0, 1

BL BL

i i

BL BL BL

i j i j i j

b u cu i

b u cu u i j

10

There exist constants

independent of ε and n such that

Proposition:

( )

, 1 2( )! ! 0,1,2,...n n n

i j L Iu Cn K i K n

2

( )/

, 2

( )

( )n n i jBL xi j n

i j

L I

u C K i j e

2 1

( )

(1 ) /, 2 1

( )

( ) /

n n i jBLn xi j

i j

L I

u C K i j e

, 0, 0,1,2,...i j n

1 2, , , , , , , , 0C K K C K C K

11

Next, we define for some M

2

2 1 2 ,

0 0

( ) / ( )M M

ji

M i j

i j

u x u x

(Smooth Part)

2,2 1 2

0 0

( ) / ( )M M

jBL BLi

M i j

i j

u x u x

(Left BL)

2, ,2 1 2

0 0

( ) / ( )M MBL BLji

i j i j

i j

u x u x

(Right BL)

11


2

2 1 2 ,

0 0

( ) / ( )M M

ji

M i j

i j

u x u x

(Smooth Part)

2,2 1 2

0 0

( ) / ( )M M

jBL BLi

M i j

i j

u x u x

(Left BL)

2, ,2 1 2

0 0

( ) / ( )M MBL BLji

i j i j

i j

u x u x

(Right BL)

(Remainder)BLBL

M MM Mr u u u u

11


2

2 1 2 ,

0 0

( ) / ( )M M

ji

M i j

i j

u x u x

(Smooth Part)

2,2 1 2

0 0

( ) / ( )M M

jBL BLi

M i j

i j

u x u x

(Left BL)

2, ,2 1 2

0 0

( ) / ( )M MBL BLji

i j i j

i j

u x u x

(Right BL)

BLBL

M MM Mr u u u u

(Remainder)

and we have the decomposition

BLBL

M MM Mu u u u r

12

By construction, u satisfies the BVP.

12

By construction, u satisfies the BVP.

Moreover, we assume that

(otherwise the previous definitions do not make sense)

12 2

2

1, 1M M

(§)

13

Assume (§) holds. Then there exist positive constants K1, K2,

K3, γ, δ, independent of ε and n such that for any n = 0, 1, 2, …

Proposition:

( )

1( )!n n

M L Iu Cn K

2

( )( , )/

2 2( )n

BL dist x In nMu x CK e

2 1

( )

( , ) /23

1

( )

nnBL

dist x InMu x CK e

2 1 22 2

/ /1/2

1( ) ( ) ( )max ,M M ML I L I L I

r r r e e

14

Variational Formulation

1

0,B u v F v v H I

1

0Find such thatu H I

1 2, , , ,B u v u v bu v cu v

where, with the usual L2(Ι) inner product,

,F v f v

,

It follows that the bilinear form is coercive, i.e.

2 1

0, ,E

B u u C u u H I C

15

denotes the energy norm.

2 2 2

1 1, 0,E I Iu u u

where

16

Discretization

As usual, we seek 1

0 s. t.N Nu V H I

,N NB u v F v v V

16

Discretization

As usual, we seek 1

0 s. t.N Nu V H I

,N NB u v F v v V

where the space VN is defined as follows: Let IST = [–1, 1] be

the reference element and Πp(IST) the space of polynomials on

IST , of degree p. With an arbitrary partition of I,

we define

0

N

i ix

16

Discretization

As usual, we seek 1

0 s. t.N Nu V H I

,N NB u v F v v V

where the space VN is defined as follows: Let IST = [–1, 1] be

the reference element and Πp(IST) the space of polynomials on

IST , of degree p. With an arbitrary partition of I,

we define

0

N

i ix

1

0( ) ( ) : ( ( )) ( ), 1,...,p

N j p STV S u H I u Q I j N

1 1( ) (2 )( )j j j j jQ x x x x where

17

Definition: Spectral Boundary Layer Mesh

1 20, and 0 1p

2

1

0, : withpS p S H

define

1

1

1 1 1

0 1 0

0,1 if 1/ 2

0, ,1 ,1 if 1/ 2

p

p p p

For

The polynomial degree is taken to be uniformly p over all

elements.

0 κp/μ0 1– κp/μ1 1

18

The following result from [Schwab, 1998] is the main tool in the

analysis.

2 2

2

2

222 ( 1)

2( ) ( )

22

2 ( 1)

( )( )

( ) ( ) , ( ) ( )

1 ( )!( )

( )!

( )!( )

( )!

p p

s s

p L I L I

s s

p L IL I

u a I a u b I b

p su I u b a u

p p s

p su I u b a u

p s

Theorem: Let I = [a, b]. Then for any u C(I) there exists

Ipu Πp(I) such that, for any 0 s p,

19

where u is the exact solution and Ipu is the interpolant on the

Spectral Boundary Layer Mesh.

p

p Eu I u Ce

Proposition: There exist constants C, > 0, depending only

on the input data, such that

Using the previous theorem, we are able to prove the following

19

where u is the exact solution and Ipu is the interpolant on the

Spectral Boundary Layer Mesh.

p

p Eu I u Ce



Using the previous theorem, we are able to prove the following

Sketch of Proof: Separated into two cases.

20

Case 1: κp(μ1)–1 ≥ ½, asymptotic case, Δ = {0, 1}

20


2 2

22( 1)

2( ) ( )

11 1 1

1 22

11

2

1 ( )!

( )!

1 ( )! max 1, ,

( )!

1 ( )! 1 , (0, )

( )!

s

p L I L I

ss

ss

p su I u u

p p s

p sCK s

p p s

p sC K s s p

p p s

20


2 2

22( 1)

2( ) ( )

11 1 1

1 22

11

2

1 ( )!

( )!

1 ( )! max 1, ,

( )!

1 ( )! 1 , (0, )

( )!

s

p L I L I

ss

ss

p su I u u

p p s

p sCK s

p p s

p sC K s s p

p p s

Choosing s = λp, λ(0, 1) at our disposal, we get

2

2 11

2( )

1 ( )!1

( )!

pp

p L I

p pu I u C K p

p p p

21

2

2

( )

p

p L Iu I u Ce

Stirling’s formula allows us to obtain

21

2

2

( )

p

p L Iu I u Ce

Stirling’s formula allows us to obtain

Similarly

2

2

( )

p

p

L I

u I u Cpe

and this completes the proof in the asymptotic case.

22

Case 2: κp(μ0)–1 < 1/2, pre-asymptotic case

22


The mesh consists of 3 elements and we use the

decomposition

BLBL

M MM Mu u u u r

22



decomposition

BLBL

M MM Mu u u u r

Each term is approximated separately; the remainder is not

approximated since it is exponentially small.

22



decomposition

BLBL

M MM Mu u u u r

Each term is approximated separately; the remainder is not

approximated since it is exponentially small.

The approximation of the smooth part is the same as in

Case 1.

23

For left boundary layer, separate approximations are

constructed on I1 = [0, κp(μ0)–1] and I \ I1.

23



Here we use the linear

interpolant of BL

Mu

23




interpolant of BL

Mu

Here we construct the interpolant using the previous

theorem.

23




interpolant of BL

Mu

Here we construct the interpolant using the previous

theorem.

Similarly for the boundary layer on the right.

24

where uFEM is the finite element solution and Ipu is the

interpolant, both on the Spectral Boundary Layer Mesh.

p

FEM p Eu I u Ce



24





p

FEM p Eu I u Ce

Sketch of Proof: By coercivity and Galerkin orthogonality,

there holds, with ξ := Ipu – uFEM ,

24





2

,( , ) ( , )pE I

C B B u I u

p

FEM p Eu I u Ce

Sketch of Proof: By coercivity and Galerkin orthogonality,

there holds, with ξ := Ipu – uFEM ,

25

2

,

1 2

( , ) ( , )

, , ,

pE I

p p pI

I I

C B B u I u

u I u b u I u c u I u

25

2

,

1 2

( , ) ( , )

, , ,

pE I

p p pI

I I

C B B u I u


Cauchy-Schwarz

2

2, , ,,, max 1,p pE I I E IE I

I

C b u I u c u I u

25

2

,

1 2

( , ) ( , )

, , ,

pE I

p p pI

I I

C B B u I u


Cauchy-Schwarz

2

2, , ,,, max 1,p pE I I E IE I

I

C b u I u c u I u

For this term, we integrate by parts to get

26

2 2, ,p pI

I

b u I u b u I u

26

2 2, ,p pI

I

b u I u b u I u

Then, we consider each interval in the mesh separately and

we make use of the fact that

1

0 1/ 2p

26

2 2, ,p pI

I

b u I u b u I u

Then, we consider each interval in the mesh separately and

we make use of the fact that

1

0 1/ 2p

Ultimately, we are able to establish

2 2

,

, ,

,

p pI

I

p

E I

b u I u b u I u

Ce

27

p

FEM Eu u Ce



Proof:

p

FEM FEM p pE E Eu u u I u I u u Ce

28

Numerical Results

We consider the problem with b = c = f = 1 and we

are computing

,

,

100EXACT FEM E I

EXACT E I

u uError

u

33

Closing Remarks

• The hp FEM on the Spectral Boundary Layer Μesh

yields robust exponential convergence for all values of

0 ε1 ε2 1.

33

Closing Remarks

• The hp FEM on the Spectral Boundary Layer Μesh

yields robust exponential convergence for all values of

0 ε1 ε2 1.

• We are in the process of studying a two-dimensional

analog:

1 2( , ) ( ) ( ) ( , ) ( , ) in

0 on

uu x y b x c x u x y f x y

x

u

An hp-FEM for a singularly perturbed problem of reaction ...xenophon/slides/BAIL_2018.pdf · An...

Documents

Transcript of An hp-FEM for a singularly perturbed problem of reaction ...xenophon/slides/BAIL_2018.pdf · An...