Energy Estimates and Weak Boundary Procedures for LAM Marco - PowerPoint PPT Presentation

Energy Estimates and Weak Boundary Procedures for LAM Marco Kupiainen 1 , 2 1 Department of Computational Mathematics University of Link¨ oping 2 SMHI, Rossby Centre marco.kupiainen@smhi.se ASM 2013 Reykjavik, Iceland

Outline Introduction/Motivation for this work Motivating examples Model problem Why do we impose Boundary Conditions? Short Description of Boundary methods Energy estimate for Davies-relaxation Spectral Radius of operators Computational results Conclusion Conclusion II

Introduction/Motivation for this work ◮ Boundary conditions are set using ad hoc intuition (Davies relaxation) ◮ Different effects in e.g. Arctic and Europe ◮ ”The apparent success of spectral nudging for one-way nesting is at least partly an artifact of very bad procedures for windowing and blending” 1 1 John P. Boyd ”Limited-Area Fourier Spectral Models and Data Analysis Schemes: Windows, Fourier Extension, Davies Relaxation and All That”, Mon. Weat. Rev. Vol. 133 2005

Motivating example ◮ Europe has ”standing waves” ◮ Similar problems over Arctic (mitigated with Spectral Nudging) 20 18 16 14 12 10 8 6 4 2 /home/sm_marku/test/Cordex/fc198909010000qq ◮ Analyze boundary procedures!

Model problem Simplification for purpose of illustration, without loosing generality! u t + � 3 i =1 ∇ � ◮ � F i ( � u ) = 0 N-S/Euler/primitive equations

Model problem Simplification for purpose of illustration, without loosing generality! u t + � 3 i =1 ∇ � ◮ � F i ( � u ) = 0 N-S/Euler/primitive equations ◮ ⇓ 1D system

Model problem Simplification for purpose of illustration, without loosing generality! u t + � 3 i =1 ∇ � ◮ � F i ( � u ) = 0 N-S/Euler/primitive equations ◮ ⇓ 1D system u t + � ◮ � f ( � u ) x = 0

Model problem Simplification for purpose of illustration, without loosing generality! u t + � 3 i =1 ∇ � ◮ � F i ( � u ) = 0 N-S/Euler/primitive equations ◮ ⇓ 1D system u t + � ◮ � f ( � u ) x = 0 ◮ ⇓ Semi-linear system

Model problem Simplification for purpose of illustration, without loosing generality! u t + � 3 i =1 ∇ � ◮ � F i ( � u ) = 0 N-S/Euler/primitive equations ◮ ⇓ 1D system u t + � ◮ � f ( � u ) x = 0 ◮ ⇓ Semi-linear system ◮ � u t + A ( � u ) � u x = 0

Model problem Simplification for purpose of illustration, without loosing generality! u t + � 3 i =1 ∇ � ◮ � F i ( � u ) = 0 N-S/Euler/primitive equations ◮ ⇓ 1D system u t + � ◮ � f ( � u ) x = 0 ◮ ⇓ Semi-linear system ◮ � u t + A ( � u ) � u x = 0 ◮ ⇓ Linear system

Model problem Simplification for purpose of illustration, without loosing generality! u t + � 3 i =1 ∇ � ◮ � F i ( � u ) = 0 N-S/Euler/primitive equations ◮ ⇓ 1D system u t + � ◮ � f ( � u ) x = 0 ◮ ⇓ Semi-linear system ◮ � u t + A ( � u ) � u x = 0 ◮ ⇓ Linear system � � ◮ � 0 1 u t + A � u x = 0 , e.g. A = 1 0

Model problem Simplification for purpose of illustration, without loosing generality! u t + � 3 i =1 ∇ � ◮ � F i ( � u ) = 0 N-S/Euler/primitive equations ◮ ⇓ 1D system u t + � ◮ � f ( � u ) x = 0 ◮ ⇓ Semi-linear system ◮ � u t + A ( � u ) � u x = 0 ◮ ⇓ Linear system � � ◮ � 0 1 u t + A � u x = 0 , e.g. A = 1 0 A = T Λ T − 1 ⇒ � u t + T Λ T − 1 � ◮ ⇓ u x = 0 Scalar PDE

Model problem Simplification for purpose of illustration, without loosing generality! u t + � 3 i =1 ∇ � ◮ � F i ( � u ) = 0 N-S/Euler/primitive equations ◮ ⇓ 1D system u t + � ◮ � f ( � u ) x = 0 ◮ ⇓ Semi-linear system ◮ � u t + A ( � u ) � u x = 0 ◮ ⇓ Linear system � � ◮ � 0 1 u t + A � u x = 0 , e.g. A = 1 0 A = T Λ T − 1 ⇒ � u t + T Λ T − 1 � ◮ ⇓ u x = 0 Scalar PDE ◮ u t + u x = 0

Why do we impose Boundary Conditions? (Energy point of view) u t + u x = 0 , x ∈ [0 , 1]

Why do we impose Boundary Conditions? (Energy point of view) u t + u x = 0 , x ∈ [0 , 1] ◮ Multiply with u and integrate!

Why do we impose Boundary Conditions? (Energy point of view) u t + u x = 0 , x ∈ [0 , 1] ◮ Multiply with u and integrate! � 1 � 1 ◮ 2 0 uu t dx + 2 0 uu x = 0

Why do we impose Boundary Conditions? (Energy point of view) u t + u x = 0 , x ∈ [0 , 1] ◮ Multiply with u and integrate! � 1 � 1 ◮ 2 0 uu t dx + 2 0 uu x = 0 ◮ Use integration by parts!

Why do we impose Boundary Conditions? (Energy point of view) u t + u x = 0 , x ∈ [0 , 1] ◮ Multiply with u and integrate! � 1 � 1 ◮ 2 0 uu t dx + 2 0 uu x = 0 ◮ Use integration by parts! � 1 dt � u � 2 + [ u 2 ] 1 � u � 2 = d 0 u 2 dx 0 = 0 , ◮

Why do we impose Boundary Conditions? (Energy point of view) u t + u x = 0 , x ∈ [0 , 1] ◮ Multiply with u and integrate! � 1 � 1 ◮ 2 0 uu t dx + 2 0 uu x = 0 ◮ Use integration by parts! � 1 dt � u � 2 + [ u 2 ] 1 � u � 2 = d 0 u 2 dx 0 = 0 , ◮ dt � u � 2 = ( u (0 , t )) 2 − ( u (1 , t )) 2 d ◮ � �� � � �� � Faster than exponential growth Decay

Why do we impose Boundary Conditions? (Energy point of view) u t + u x = 0 , x ∈ [0 , 1] ◮ Multiply with u and integrate! � 1 � 1 ◮ 2 0 uu t dx + 2 0 uu x = 0 ◮ Use integration by parts! � 1 dt � u � 2 + [ u 2 ] 1 � u � 2 = d 0 u 2 dx 0 = 0 , ◮ dt � u � 2 = ( u (0 , t )) 2 − ( u (1 , t )) 2 d ◮ � �� � � �� � Faster than exponential growth Decay ◮ We must set conditions to bound the energy for u (0 , t ) with data: g ( t ) < C for some constant C ∈ R

Discrete Boundary procedures CKD : U t + P − 1 QU = 0 U ( · , t ) = ( I − W ) U ( · , t ) + WG ( t ) W = diag ( tanh ) ◮ G is data! ◮ U = ( U 0 , U 1 , . . . U N ) T is the solution. ◮ CKD = Classic K˚ allberg-Davies Relaxation

Discrete Boundary procedures WKD : U t + P − 1 QU = P − 1 W ( G − U ) W = diag ( tanh ) ◮ WKD = Weak K˚ allberg-Davies Relaxation, proven stable

Discrete Boundary procedures SAT : U t + P − 1 QU = P − 1 E 0 ( G − U ) E 0 = diag (1 , 0 , . . . , 0) ◮ SAT = Simultaneous Approximation Term (Carpenter et. al.), proven stable

Discrete Boundary procedures CKD : U t + P − 1 QU = 0 U ( · , t ) = ( I − W ) U ( · , t ) + WG ( t ) W = diag ( tanh ) WKD : U t + P − 1 QU = P − 1 W ( G − U ) W = diag ( tanh ) SAT : U t + P − 1 QU = P − 1 E 0 ( G − U ) E 0 = diag (1 , 0 , . . . , 0) ◮ CKD = Classic K˚ allberg-Davies Relaxation ◮ WKD = Weak K˚ allberg-Davies Relaxation, proven stable ◮ SAT = Simultaneous Approximation Term (Carpenter et. al.), proven stable

Energy estimate for Davies-relaxation Weak Davies relaxation Strong Davies relaxation U t + P − 1 QU = τ P − 1 W tanh ( G − U ) U t + P − 1 QU = 0 U = U + W tanh ( G − U ) ◮ Discrete Energy Method to prove stability ◮ No energy estimate! (For ◮ Multiply with U T P stability(?) proof use GKS ◮ Use summation-by-parts theory DIFFICULT !) property of difference operator ◮ In fact there are counterexamples shoving instability for strong methods. 2 2 ”High Order Difference Methods for Time Dependent PDE”, Bertil Gustafsson ISBN 978-3-540-74992-9, Springer Verlag 2008

Energy estimate for Davies-relaxation cont. d dt � U � 2 P = U 2 1 (1 − 2 w 1 ) + 2 w 1 U 1 G 1 − U 2 N (1 + 2 w N ) + 2 w N U N G N N − 1 � (2 w i U 2 − i − 2 w i U i G i ) (1) i =2 Since w 1 ≥ 1 2 and w i ≥ 0 ∴ Davies Relaxation is proven energy-stable!

Energy estimate for Davies-relaxation cont. d dt � U � 2 P = U 2 1 (1 − 2 w 1 ) + 2 w 1 U 1 G 1 − U 2 N (1 + 2 w N ) + 2 w N U N G N N − 1 � (2 w i U 2 − i − 2 w i U i G i ) (1) i =2 Since w 1 ≥ 1 2 and w i ≥ 0 ∴ Davies Relaxation is proven energy-stable! ◮ Stable, BUT imposing conditions where not needed!

Energy estimate for Davies-relaxation cont. d dt � U � 2 P = U 2 1 (1 − 2 w 1 ) + 2 w 1 U 1 G 1 − U 2 N (1 + 2 w N ) + 2 w N U N G N N − 1 � (2 w i U 2 − i − 2 w i U i G i ) (1) i =2 Since w 1 ≥ 1 2 and w i ≥ 0 ∴ Davies Relaxation is proven energy-stable! ◮ Stable, BUT imposing conditions where not needed! ◮ Solution quality is not affected if � U i − G i � is ”small”.

Energy estimate for Davies-relaxation cont. d dt � U � 2 P = U 2 1 (1 − 2 w 1 ) + 2 w 1 U 1 G 1 − U 2 N (1 + 2 w N ) + 2 w N U N G N N − 1 � (2 w i U 2 − i − 2 w i U i G i ) (1) i =2 Since w 1 ≥ 1 2 and w i ≥ 0 ∴ Davies Relaxation is proven energy-stable! ◮ Stable, BUT imposing conditions where not needed! ◮ Solution quality is not affected if � U i − G i � is ”small”. ◮ Usually we impose time-interpolated G i.e. G i ( t ) = π 6 h G i ( t n )

Spectral Radius of operators part I Spectral radius of operator WITHOUT Boundary conditions Spectral radius of WEAK DAVIES RELAXATION 50 150 40 100 30 20 50 10 Imaginary Imaginary 0 0 −10 −50 −20 −30 −100 −40 −150 −50 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 −140 −120 −100 −80 −60 −40 −20 0 Real −5 Real x 10 (a) Operator without bound- (b) WKD ary conditions Spectral radius of SAT 50 40 30 20 10 Imaginary 0 −10 −20 −30 −40 −50 −6 −5 −4 −3 −2 −1 0 Real (c) SAT