Boundary (and other) layers and how to approximate them Christos - - PowerPoint PPT Presentation

Boundary (and other) layers and how to approximate them

Christos Xenophontos Department of Mathematics & Statistics University of Cyprus

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Outline

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Outline

 When is a differential equation (D.E.) singularly perturbed?

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Outline

 Examples  When is a differential equation (D.E.) singularly perturbed?

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Outline

 Properties of the solution to such problems  When is a differential equation (D.E.) singularly perturbed?  Examples

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Outline

 Numerical methods for the approximation of these solutions  When is a differential equation (D.E.) singularly perturbed?  Examples  Properties of the solution to such problems

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Classical Example:

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( ) ( ) 1, ( 1,1) ( 1) (1) u x u x x I u u              Here 0 < ε ≤ 1, multiplies the highest derivative in the D.E. and is a given parameter that can approach 0.

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Classical Example:

2

( ) ( ) 1, ( 1,1) ( 1) (1) u x u x x I u u              Here 0 < ε ≤ 1, multiplies the highest derivative in the D.E. and is a given parameter that can approach 0. Note that as ε  0, the solution u  1, but what happens with the boundary conditions?

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• 1
• 0.8
• 0.6
• 0.4
• 0.2

0.2 0.4 0.6 0.8 1 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 x u(x) Solution for various values of 

=1 =1/4 =1/2 =3/4

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In the general one-dimensional case

2

( ) ( ) ( ) ( ), ( , ) ( ) ( ) u x b x u x f x x I u u                with b > 0, f given functions, the solution u  f / b when ε  0, and if (f / b)|I ≠ 0, then u will contain boundary layers1.

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In the general one-dimensional case

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( ) ( ) ( ) ( ), ( , ) ( ) ( ) u x b x u x f x x I u u                with b > 0, f given functions, the solution u  f / b when ε  0, and if (f / b)|I ≠ 0, then u will contain boundary layers1.

1Ludwig Prandtl, On the motion of a fluid with very small viscosity, Third World

Congress of Mathematicians, August 1903.

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7

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Note: The following D.E. is not singularly pertubed

 

2

1 ( ) ( ) 1, ( 1,1) ( 1) (1) u x u x x I u u                 

9

Two-dimensional example

2 2 2 2 2 2

( , ) ( , ) ( , ), (0,1) u u b x y u x y f x y x x y u 



                     

Source: Niall Madden (NUI Galway)

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One more two-dimensional example

 

2D

a S t           

Source: Niall Madden (NUI Galway)

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Source: Niall Madden (NUI Galway)

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Example with boundary and interface layers

2 2

( ) ( ) ( ) ( ) in ( 1,0), ( ) ( ) ( ) 0 in (0,1), ( 1) (1) 0, (0) (0) 0, ( ) (0) ( ) (0) u x u x f x u x u x u u u u u u

         

 

         

                 

• 1
• 0.8
• 0.6
• 0.4
• 0.2

0.2 0.4 0.6 0.8 1 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 x uexact Exact solution for f(x) = 1 and  = 0.1

• 1
• 0.8
• 0.6
• 0.4
• 0.2

0.2 0.4 0.6 0.8 1 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 x uexact Exact solution for f(x) = 1 and  = 0.05

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Two-dimensional example with boundary and interface layers

2 2

in , in , 0 on \ , 0 on \ 0 on ,

• n ,

u u f u u f u u u u u u h

         

   

               

                                   

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What can mathematics tell us?

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What can mathematics tell us?

S BL boundary layer smooth

u u u   The solution to such problems may be decomposed as

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What can mathematics tell us?

S BL boundary layer smooth

u u u   The solution to such problems may be decomposed as

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What can mathematics tell us?

S BL boundary layer smooth

u u u   The solution to such problems may be decomposed as =

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What can mathematics tell us?

S BL boundary layer smooth

u u u   The solution to such problems may be decomposed as =

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What can mathematics tell us?

S BL boundary layer smooth

u u u   The solution to such problems may be decomposed as = –

Source: Niall Madden (NUI Galway)

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Bounds of the form

 

( ) ( ) / (1 )/

, ( )

m q m m m x x S BL

u C u x C e e

 

 

   

   may be proven.

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Bounds of the form

 

( ) ( ) / (1 )/

, ( )

m q m m m x x S BL

u C u x C e e

 

 

   

   may be proven. How?

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Bounds of the form

 

( ) ( ) / (1 )/

, ( )

m q m m m x x S BL

u C u x C e e

 

 

   

   may be proven. We express u as

j j j

u u 

 

 with uj unknown (for the moment) functions and we substitute in the D.E. How?

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2

: ( ) ( ), ( 1,1) ( ( ( ) 1) ( ) ) 1 L u b x f x x u x u x I u u



                For example, for the boundary value problem we have

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2

: ( ) ( ), ( 1,1) ( ( ( ) 1) ( ) ) 1 L u b x f x x u x u x I u u



                For example, for the boundary value problem we have

2

( ) ) ( ( ) ( )

j j j j j j

b x f x x u x x I u   

   

      

 

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2

: ( ) ( ), ( 1,1) ( ( ( ) 1) ( ) ) 1 L u b x f x x u x u x I u u



                For example, for the boundary value problem we have

2

( ) ) ( ( ) ( )

j j j j j j

b x f x x u x x I u   

   

      

 

We equate like powers of ε on both sides of the above equation.

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     

3 4 1 1 2 2 2 2

... u bu u bu u bu f                    

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     

3 4 1 1 2 2 2 2

... u bu u bu u bu f                    

2

, , 0,2,4..., 0, 1,3,5,...

k k k

u f u u k u k b b



      

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Thus we define the smooth part of the solution as

     

3 4 1 1 2 2 2 2

... u bu u bu u bu f                    

2

, , 0,2,4..., 0, 1,3,5,...

k k k

u f u u k u k b b



      

2 2

( )

M M j S j j

u x u 





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Thus we define the smooth part of the solution as

     

3 4 1 1 2 2 2 2

... u bu u bu u bu f                    

2

, , 0,2,4..., 0, 1,3,5,...

k k k

u f u u k u k b b



      

2 2

( )

M M j S j j

u x u 



 and (after some calculations) we find

 

2 2 2

( ) ( ) ( )

M M S M

L u x u x u x







  

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This shows that

 

1 0.

M S

u  

but as

M S

u u   

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This shows that

 

1 0.

M S

u  

To correct this we define the boundary layers through the equations 0, ( 1) ( 1) (1)

BL M BL S BL

L u x I u u u

   

            but

as

M S

u u   

0, ( 1) (1) (1)

BL BL M BL S

L u x I u u u

   

          

BL

u

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and we observe that

 

2 2 2 M M S BL M

L u u u u





 

   

 ( 1)

M S BL BL

u u u

 

    as well as

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and we observe that

 

2 2 2 M M S BL M

L u u u u





 

   

 ( 1)

M S BL BL

u u u

 

    as well as Finally we define the remainder rM by

2 2

, ( 1,1) ( 1)

M M M

L r u x I r







         

and we have the decomposition

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left BL remain M S BL BL rig s h m t d B

• oth

er L M

u u u u r

 

   

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left BL remain M S BL BL rig s h m t d B

• oth

er L M

u u u u r

 

    By construction and ( 1) L u f u



  

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left BL remain M S BL BL rig s h m t d B

• oth

er L M

u u u u r

 

    By construction and ( 1) L u f u



   The next step is to obtain estimates for each term in the decomposition of u in order to decide how to best approximate it.

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Proposition: There exist constants C, Κ > 0 independent of ε, u and n, such that

 

2

( ) 1 ( )

max , 0,1,2,...

n n n L I

u CK n n     

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Proposition: There exist constants C, Κ > 0 independent of ε, u and n, such that

 

2

( ) 1 ( )

max , 0,1,2,...

n n n L I

u CK n n      Proposition: There exist constants C, K1, K2 > 0 independent of ε and n, such that if 0 < 2ΜεΚ1 < 1, then

 

( ) 2 ( )

! 0,1,2,...

n M n S L I

u CK n n



  

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Proposition: There exist constants C, Κ > 0 independent of ε such that for each x[–1, 1] and n = 0, 1, 2, …

       

( ) (1 )/ 1 ( ) (1 )/ 1

( ) max , ( ) max ,

n n x n BL n n x n BL

u x Ce K n u x Ce K n

   

 

       

 

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Proposition: There exist constants C, Κ > 0 independent of ε such that for each x[–1, 1] and n = 0, 1, 2, …

       

( ) (1 )/ 1 ( ) (1 )/ 1

( ) max , ( ) max ,

n n x n BL n n x n BL

u x Ce K n u x Ce K n

   

 

       

  Proposition: There exist constants C, K > 0 independent of ε, such that

   

2

( ) 2 2 ( )

2 , 0,1,2

n M n M L I

r C M K n  



 

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What does all this mean?

 The smooth part of the solution is as smooth as the

data allows (e.g. analytic data gives analytic smooth part).

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What does all this mean?

 The smooth part of the solution is as smooth as the

data allows (e.g. analytic data gives analytic smooth part).

 Boundary layers behave like the one-dimensional

function exp(–x/ε) and “live” only in a region near the boundary of the domain.

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What does all this mean?

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So, we need to know how to approximate the one- dimensional function exp( – x/ε).

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So, we need to know how to approximate the one- dimensional function exp( – x/ε). In two-dimensions the boundary layer part of the solution behaves like the (one-dimensional) function exp( – ρ/ε), where ρ denotes the distance normal to the boundary of our domain.

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χ is a smooth cut-off function and S(θ) is a smooth function satisfying

/

( , ) ( ) ( )

BL

u S e       





( ) S C   

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Approximation

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It is obvious that the boundary layers cannot be

• ignored. In particular, the approximation must

account for their presence, otherwise the (overall) result will not be accurate.

Approximation

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For example, consider

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( ) ( ) 1, ( 1,1) ( 1) (1) u x u x x I u u              with exact solution cosh( / ) ( ) 1 cosh(1/ )

EXACT

x u x    

Suppose we use the Finite Element Method with piecewise polynomials of degree 1 on a uniform mesh.

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• 1
• 0.8
• 0.6
• 0.4
• 0.2

0.2 0.4 0.6 0.8 1 0.5 1 1.5 x y

 = 10-3, uniform mesh, p = 1

uFEM uEXACT

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Choice of the mesh

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Choice of the mesh

“Overkill”

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• Shishkin mesh

τ 1– τ 1

1 min ,( 1) ln 4 p N          

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• Shishkin mesh

τ 1– τ 1

uniform mesh

1 min ,( 1) ln 4 p N          

2N N N

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• Shishkin mesh

τ 1– τ 1

uniform mesh

1 min ,( 1) ln 4 p N          

Here p is the polynomial degree of the approximation. Increasing Ν (or equivalently, reducing the mesh size), we improve our approximation.

2N N N

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This mesh may be used in conjunction with Finite Differences, Finite Elements, etc.

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This mesh may be used in conjunction with Finite Differences, Finite Elements, etc. It may be shown that

 

ln ,

p p N L

u u CN N N





    (i.e. almost optimal algebraic convergence of order p.)

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• 1
• 0.8
• 0.6
• 0.4
• 0.2

0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1 x y

 = 10-3, Shishkin mesh, p = 1

uFEM uEXACT

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• Spectral Boundary Layer Mesh

pε 1– pε 1

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• Spectral Boundary Layer Mesh

pε 1– pε 1

Here p is the degree of the approximating polynomials which is increased in order to improve accuracy. The mesh remains constant with just 3 subintervals/elements.

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It may be shown that

, 0,

p N L

u u Ce p









    

(i.e. exponential convergence)

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• 1
• 0.8
• 0.6
• 0.4
• 0.2

0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1 x y

 = 10-3, Spectral Boundary Layer Mesh, p = 8

uFEM uEXACT

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From theory to practice …

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From theory to practice …

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Epilogue

Singularly perturbed problems arise in many different applications and research in this area is quite active. Currently, we are investigating 4th order problems:

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Epilogue

Singularly perturbed problems arise in many different applications and research in this area is quite active. Currently, we are investigating 4th order problems:  

2 (4)( )

( ) ( ) ( ) ( ) ( ) in (0,1) (0) (0) (1) (1) u x x u x x u x f x I u u u u                    

1-D:

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d2u/dx2

Example: f(x) = α(x) = β(x) = 1, ε = 0.01

u(x) du/dx

41

 

2 2

in 0 on u a u bu f u u n                   2-D:

with Ω a bounded, smooth domain and a, b and f given smooth (e.g. analytic) functions.