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

Outline 2

Outline  When is a differential equation (D.E.) singularly perturbed? 2

Outline  When is a differential equation (D.E.) singularly perturbed?  Examples 2

Outline  When is a differential equation (D.E.) singularly perturbed?  Examples  Properties of the solution to such problems 2

Outline  When is a differential equation (D.E.) singularly perturbed?  Examples  Properties of the solution to such problems  Numerical methods for the approximation of these solutions 2

Classical Example:         2 ( ) ( ) 1, ( 1,1) u x u x x I      ( 1) (1) 0 u u Here 0 < ε ≤ 1, multiplies the highest derivative in the D.E. and is a given parameter that can approach 0. 3

Classical Example:         2 ( ) ( ) 1, ( 1,1) u x u x x I      ( 1) (1) 0 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? 3

Solution for various values of  1  =1/4 0.9 0.8 0.7  =1/2 0.6 u(x) 0.5  =3/4 0.4  =1 0.3 0.2 0.1 0 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 x 4

In the general one-dimensional case          2 ( ) ( ) ( ) ( ), ( , ) u x b x u x f x x I       ( ) ( ) 0 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 layers 1 . 5

In the general one-dimensional case          2 ( ) ( ) ( ) ( ), ( , ) u x b x u x f x x I       ( ) ( ) 0 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 layers 1 . 1 Ludwig Prandtl, On the motion of a fluid with very small viscosity , Third World Congress of Mathematicians, August 1903. 5

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Note: The following D.E. is not singularly pertubed             2  1 ( ) ( ) 1, ( 1,1) u x u x x I       ( 1) (1) 0 u u 8

Two-dimensional example      2 2 u u        2 2    ( , ) ( , ) ( , ), (0,1) b x y u x y f x y x   2 2    x y   0 u   Source: Niall Madden (NUI Galway) 9

One more two-dimensional example             2 D a S  t Source: Niall Madden (NUI Galway) 10

Source: Niall Madden (NUI Galway) 11

Example with boundary and interface layers               2 ( ) ( ) ( ) ( ) in ( 1,0), ( ) ( ) ( ) 0 in (0,1), u x u x f x u x u x                     2 ( 1) (1) 0, (0) (0) 0, ( ) (0) ( ) (0) 0 u u u u u u       Exact solution for f(x) = 1 and  = 0.1 Exact solution for f(x) = 1 and  = 0.05 1 1 0.9 0.9 0.8 0.8 0.7 0.7 0.6 0.6 u exact u exact 0.5 0.5 0.4 0.4 0.3 0.3 0.2 0.2 0.1 0.1 0 0 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 x x 12

Two-dimensional example with boundary and interface layers         2 in , u u f            in , u u f          0 on \ , u          0 on \ u          0 on , u u        u u        2 on , h       13

What can mathematics tell us? 14

What can mathematics tell us? The solution to such problems may be decomposed as   u u u S BL boundary layer smooth 14

What can mathematics tell us? The solution to such problems may be decomposed as   u u u S BL boundary layer smooth 14

What can mathematics tell us? The solution to such problems may be decomposed as   u u u S BL boundary layer smooth = 14

What can mathematics tell us? The solution to such problems may be decomposed as   u u u S BL boundary layer smooth = 14

What can mathematics tell us? The solution to such problems may be decomposed as   u u u S BL boundary layer smooth – = Source: Niall Madden (NUI Galway) 14

Bounds of the form              ( ) ( ) / (1 )/ m q m m m x x , ( ) u C u x C e e S BL may be proven. 15

Bounds of the form              ( ) ( ) / (1 )/ m q m m m x x , ( ) u C u x C e e S BL may be proven. How? 15

Bounds of the form              ( ) ( ) / (1 )/ m q m m m x x , ( ) u C u x C e e S BL may be proven. How? We express u as     j u u j  0 j with u j unknown (for the moment) functions and we substitute in the D.E. 15

For example, for the boundary value problem            2 : ( ) ( ) ( ) ( ), ( 1,1) L u u x b x u x f x x I       ( 1) ( ) 1 0 u u we have 16

For example, for the boundary value problem            2 : ( ) ( ) ( ) ( ), ( 1,1) L u u x b x u x f x x I       ( 1) ( ) 1 0 u u we have               2 j j ( ) ( ) ( ) ( ) u x b x u x f x x I j j   0 0 j j 16

For example, for the boundary value problem            2 : ( ) ( ) ( ) ( ), ( 1,1) L u u x b x u x f x x I       ( 1) ( ) 1 0 u u we have               2 j j ( ) ( ) ( ) ( ) u x b x u x f x x I j j   0 0 j j We equate like powers of ε on both sides of the above equation. 16

                          2 0 3 4 2 0 ... u bu u bu u bu f 0 0 1 1 2 2 17

                          2 0 3 4 2 0 ... u bu u bu u bu f 0 0 1 1 2 2  f u       k , , 0,2,4..., 0, 1,3,5,... u u k u k  0 2 k k b b 17

                          2 0 3 4 2 0 ... u bu u bu u bu f 0 0 1 1 2 2  f u       k , , 0,2,4..., 0, 1,3,5,... u u k u k  0 2 k k b b Thus we define the smooth part of the solution as M    2 M j ( ) u x u 2 S j  0 j 17

                          2 0 3 4 2 0 ... u bu u bu u bu f 0 0 1 1 2 2  f u       k , , 0,2,4..., 0, 1,3,5,... u u k u k  0 2 k k b b Thus we define the smooth part of the solution as M    2 M j ( ) u x u 2 S j  0 j and (after some calculations) we find        2 2 M M ( ) ( ) ( ) L u x u x u x  2 S M 17

This shows that    M as 0 u u S but     M 1 0. u S 18

This shows that    M as 0 u u S but     M 1 0. u S To correct this we define the boundary layers u  through the equations BL         0, L u x I 0, L u x I   BL  BL           ( 1) 0  M u ( 1) ( 1) u u BL BL S        M (1) (1) u u (1) 0  u  BL S BL 18

and we observe that          2 2 M M L u u u u  2 S BL M as well as   ( 1)       M 0 u u u S BL BL 19

and we observe that          2 2 M M L u u u u  2 S BL M as well as   ( 1)       M 0 u u u S BL BL Finally we define the remainder r M by         2 2 M , ( 1,1) L r u x I   M    ( 1) 0 r M and we have the decomposition 19