On the finite element approximation of 4 th order singularly - PowerPoint PPT Presentation

On the finite element approximation of 4 th order singularly perturbed eigenvalue problems Christos Xenophontos Department of Mathematics and Statistics University of Cyprus joint work with D. Savvidou (UCY) and H.-G. Roos (TU-Dresden)

The Model Problem Singularly perturbed 4 th order eigenvalue problem :     4 Find such that 0 u x ( ) C ( ) , I  ( )   −  +  =  = 2 (4) ( ) u x ( ) ( ) x u x ( ) ( ) x u x u ( ) x in I (0, 1)   = = = = u (0) u (1) u (0) u (1) 0 2

The Model Problem Singularly perturbed 4 th order eigenvalue problem :     4 Find such that 0 u x ( ) C ( ) , I  ( )   −  +  =  = 2 (4) ( ) u x ( ) ( ) x u x ( ) ( ) x u x u ( ) x in I (0, 1)   = = = = u (0) u (1) u (0) u (1) 0 where ε  (0, 1] is a given small parameter and α ( x ), β ( x ) ≥ 0 , are given sufficiently smooth functions. 2

Remarks: [ MOSER , 1955] 3

Remarks: [ MOSER , 1955] • The problem is self-adjoint. 3

Remarks: [ MOSER , 1955] • The problem is self-adjoint. • For all positive eigenvalues λ k ( ε ), we have   =  lim ( ) (0) k k  → 0 3

Remarks: [ MOSER , 1955] • The problem is self-adjoint. • For all positive eigenvalues λ k ( ε ), we have   =  lim ( ) (0) k k  → 0 where λ k (0) are the eigenvalues of the reduced/limiting problem. If λ k (0) is real, then so is λ k ( ε ) 3

Remarks: [ MOSER , 1955] • The problem is self-adjoint. • For all positive eigenvalues λ k ( ε ), we have   =  lim ( ) (0) k k  → 0 where λ k (0) are the eigenvalues of the reduced/limiting problem. If λ k (0) is real, then so is λ k ( ε ) • λ k ( ε ) can be expanded as a power series in ε . 3

Theorem: [ MOSER , 1955] Each eigenfunction u k can be decomposed as + − = + + S BL , BL , u u u u k k k k + S BL , where u denotes the smooth part, denotes the left u k k − BL , boundary layer and denotes the right boundary layer. u k Moreover, for n = 0, 1, 2, … 4

Theorem: [ MOSER , 1955] Each eigenfunction u k can be decomposed as + − = + + S BL , BL , u u u u k k k k + S BL , where u denotes the smooth part, denotes the left u k k − BL , boundary layer and denotes the right boundary layer. u k Moreover, for n = 0, 1, 2, … ( ) ( ) ( ) n  S u x C , k k ( ) ( ) ( ) n ( ) n + − e  −  − − e  − −      BL , 1 n x / BL , 1 n (1 x )/ u ( ) x C , u ( ) x C k k k k 4

Variational Formulation Find   ( ) ( )    =  = = = = 2 2 u H I u H I : (0) u u (0) u (1) u (1) 0 k 0   and such that k ( ) ( ) =    2 B u v , u v , v H I k k k 0 where is the usual L 2 ( Ι ) inner product and   , ( )     =  +   2 B u v , u v , u v , + u v , 5

Discretization ( )     h 2 h We seek u V H I , s. t. k h 0 k ( ) =    h h B u , v u v , v V k k h 6

Discretization ( )     h 2 h We seek u V H I , s. t. k h 0 k ( ) =    h h B u , v u v , v V k k h We define the energy norm as   2 =  2 + 2 + 2   2 2 u u u u u H ( ) I 2 2 2 0 E L ( ) I L ( ) I L ( ) I 6

Discretization ( )     h 2 h We seek u V H I , s. t. k h 0 k ( ) =    h h B u , v u v , v V k k h We define the energy norm as   2 =  2 + 2 + 2   2 2 u u u u u H ( ) I 2 2 2 0 E L ( ) I L ( ) I L ( ) I and we have ( )   2   2 B u u , u u H ( ) I 0 E 6

In order to define the finite element space V h let    = =    = 0 x x ... x 1 0 1 N be an arbitrary mesh on I = (0, 1) and set ( ) = = − = I x , x , h x x , j 1,..., N − − j j 1 j j j j 1 7

In order to define the finite element space V h let    = =    = 0 x x ... x 1 0 1 N be an arbitrary mesh on I = (0, 1) and set ( ) = = − = I x , x , h x x , j 1,..., N − − j j 1 j j j j 1 P   With the space of polynomials on ( α , β ) of degree ( , ) p less than or equal to p  2 N + 1, we define 7

In order to define the finite element space V h let    = =    = 0 x x ... x 1 0 1 N be an arbitrary mesh on I = (0, 1) and set ( ) = = − = I x , x , h x x , j 1,..., N − − j j 1 j j j j 1 P   With the space of polynomials on ( α , β ) of degree ( , ) p less than or equal to p  2 N + 1, we define   ( ) =   = 2 V u H I : u P I ( ), j 1, , N h 0 p j I j 7

Definition:   N Let be an arbitrary partition of the interval ( a , b ) x = i i 0 and suppose that for a sufficiently smooth function f ( x ), x  ( a , b ),   =  =  the values are given. Then, there f x ( ) y , f ( ) x y i i i i  I f P 1 ( , ) a b exists a unique polynomial , called the Hermite + 2 N interpolant of f , given by N  +  = I f ( ) x ( y H , ( ) x y H ( )) x i 0 i i 1, i = i 0 where, with L i ( x ) the Lagrange polynomial of degree N associated with node x i ,   dL = − − = − 2 2 i ( ) 1 2( ) ( ) ( ) , ( ) ( ) ( ) H x x x x L x H x x x L x   0, i i i i 1, i i i   dx 8

Theorem:   Let u  C 2 n +2 ([ a , b ]) and let be a mesh on [ a , b ],  = N x = i i 0 with maximum mesh size h and with N a multiple of n . If u I is the piecewise Hermite interpolant of u , having degree at most 2 n +1 on each subinterval [ x i -1 , x i ], i = 1,…, N then 9

Theorem:   Let u  C 2 n +2 ([ a , b ]) and let be a mesh on [ a , b ],  = N x = i i 0 with maximum mesh size h and with N a multiple of n . If u I is the piecewise Hermite interpolant of u , having degree at most 2 n +1 on each subinterval [ x i -1 , x i ], i = 1,…, N then ( ) ( ) k + − + −  = + I 2 n 2 k (2 n 2) u u Ch u , k 0,1,...,2 n 1   L ( ) I L ( ) I 9

Theorem:   Let u  C 2 n +2 ([ a , b ]) and let be a mesh on [ a , b ],  = N x = i i 0 with maximum mesh size h and with N a multiple of n . If u I is the piecewise Hermite interpolant of u , having degree at most 2 n +1 on each subinterval [ x i -1 , x i ], i = 1,…, N then ( ) ( ) k + − + −  = + I 2 n 2 k (2 n 2) u u Ch u , k 0,1,...,2 n 1   L ( ) I L ( ) I In our setting ( ) ( ) k + − + −  = I p 1 k ( p 1) u u Ch u , k 0,1,..., p   L ( ) I L ( ) I 9

Definition: Exponentially graded mesh With N > 4 a multiple of 4 we define  = − −  − ( ) t ln(1 4 C t ) , t [0,1/ 4 1/ N ]  p ,    = − −   C 1 exp  +  p ,   ( p 1) 10

Definition: Exponentially graded mesh With N > 4 a multiple of 4 we define  = − −  − ( ) t ln(1 4 C t ) , t [0,1/ 4 1/ N ]  p ,    = − −   C 1 exp  +  p ,   ( p 1)   +  = −  ( p 1) ( / j N ) , j 0,1,..., N / 4 1    −   x x = + − + = −  3 N /4 N /4 1   x x ( j N / 4 1) , j N / 4...,3 N / 4 − + j N /4 1   2 N / 2    −   N j − +  = +    1 ( 1) , 3 / 4 1,..., p j N N     N 10

Definition: Exponentially graded mesh With N > 4 a multiple of 4 we define  = − −  − ( ) t ln(1 4 C t ) , t [0,1/ 4 1/ N ]  p ,    = − −   C 1 exp  +  p ,   ( p 1)   +  = −  ( p 1) ( / j N ) , j 0,1,..., N / 4 1    −   x x = + − + = −  3 N /4 N /4 1   x x ( j N / 4 1) , j N / 4...,3 N / 4 − + j N /4 1   2 N / 2    −   N j − +  = +    1 ( 1) , 3 / 4 1,..., p j N N     N 10

Lemma: [X., CMAM 2017] I u Let u BL denote either boundary layer and let be its BL Hermite interpolant based on the exponential mesh. Then ( ) ( ) k − − + − −   = I 1 k ( p 1 k ) u u C N , k 0,1,..., p BL BL  L ( ) I and  − − + −  I 1/2 p 1 u u C N BL BL H 2 ( ) I 11