Graz University of Technology Institute of Applied Mechanics Application of generalized Convolution Quadrature in Acoustics and Thermoelasticity Martin Schanz joint work with Relindis Rott and Stefan Sauter Space-Time Methods for PDEs Special Semester on Computational Methods in Science and Engineering RICAM, Linz, Austria, November 10, 2016 > www.mech.tugraz.at
Content Generalized convolution quadrature method (gCQM) 1 Quadrature formula Algorithm 2 Acoustics: Absorbing boundary conditions Boundary element formulation Analytical solution Numerical examples 3 Thermoelasticity: Uncoupled formulation Boundary element formulation Numerical example Martin Schanz gCQM: Acoustics and Thermoelasticity 2 / 39
Content Generalized convolution quadrature method (gCQM) 1 Quadrature formula Algorithm 2 Acoustics: Absorbing boundary conditions Boundary element formulation Analytical solution Numerical examples 3 Thermoelasticity: Uncoupled formulation Boundary element formulation Numerical example Martin Schanz gCQM: Acoustics and Thermoelasticity 2 / 39
Content Generalized convolution quadrature method (gCQM) 1 Quadrature formula Algorithm 2 Acoustics: Absorbing boundary conditions Boundary element formulation Analytical solution Numerical examples 3 Thermoelasticity: Uncoupled formulation Boundary element formulation Numerical example Martin Schanz gCQM: Acoustics and Thermoelasticity 2 / 39
Content Generalized convolution quadrature method (gCQM) 1 Quadrature formula Algorithm 2 Acoustics: Absorbing boundary conditions Boundary element formulation Analytical solution Numerical examples 3 Thermoelasticity: Uncoupled formulation Boundary element formulation Numerical example Martin Schanz gCQM: Acoustics and Thermoelasticity 3 / 39
Convolution integral Convolution integral with the Laplace transformed function ˆ f ( s ) t � ˆ � � y ( t ) = ( f ∗ g )( t ) = f ( ∂ t ) g ( t ) = f ( t − τ ) g ( τ ) d τ 0 t 1 � � ˆ e s ( t − τ ) g ( τ ) d τ = f ( s ) d s 2 π i C 0 � �� � x ( t , s ) Integral is equivalent to solution of ODE ∂ ∂ t x ( t , s ) = sx ( t , s )+ g ( t ) x ( t = 0 , s ) = 0 with Implicit Euler for ODE , [ 0 , T ] = [ 0 , t 1 , t 2 ,..., t N ] , variable time steps ∆ t i , i = 1 , 2 ,..., N n n x n ( s ) = x n − 1 ( s ) ∆ t n 1 ∑ ∏ 1 − ∆ t n s + 1 − ∆ t n s g n = ∆ t j g j 1 − ∆ t k s j = 1 k = j Martin Schanz gCQM: Acoustics and Thermoelasticity 4 / 39
Convolution integral Convolution integral with the Laplace transformed function ˆ f ( s ) t � ˆ � � y ( t ) = ( f ∗ g )( t ) = f ( ∂ t ) g ( t ) = f ( t − τ ) g ( τ ) d τ 0 t 1 � � ˆ e s ( t − τ ) g ( τ ) d τ = f ( s ) d s 2 π i C 0 � �� � x ( t , s ) Integral is equivalent to solution of ODE ∂ ∂ t x ( t , s ) = sx ( t , s )+ g ( t ) x ( t = 0 , s ) = 0 with Implicit Euler for ODE , [ 0 , T ] = [ 0 , t 1 , t 2 ,..., t N ] , variable time steps ∆ t i , i = 1 , 2 ,..., N n n x n ( s ) = x n − 1 ( s ) ∆ t n 1 ∑ ∏ 1 − ∆ t n s + 1 − ∆ t n s g n = ∆ t j g j 1 − ∆ t k s j = 1 k = j Martin Schanz gCQM: Acoustics and Thermoelasticity 4 / 39
Time stepping formula Solution at the discrete time t n y ( t n ) = 1 � ˆ f ( s ) x n ( s ) d s 2 π i C ˆ f ( s )∆ t n f ( s ) x n − 1 ( s ) = 1 1 − ∆ t n s g n d s + 1 � � ˆ 1 − ∆ t n s d s 2 π i 2 π i C C � � f ( s ) x n − 1 ( s ) 1 g n + 1 � =ˆ ˆ 1 − ∆ t n s d s . f ∆ t n 2 π i C Recursion formula for the implicit Euler n n 1 1 � ˆ ∑ ∏ y ( t n ) = f ( s ) ∆ t j g j 1 − ∆ t k s d s 2 π i j = 1 k = j C � � n − 1 n 1 1 1 � = ˆ ˆ ∑ ∏ g n + ∆ t j g j f ( s ) f 1 − ∆ t k s d s ∆ t n 2 π i k = j j = 1 C Complex integral is solved with a quadrature formula Martin Schanz gCQM: Acoustics and Thermoelasticity 5 / 39
Time stepping formula Solution at the discrete time t n y ( t n ) = 1 � ˆ f ( s ) x n ( s ) d s 2 π i C ˆ f ( s )∆ t n f ( s ) x n − 1 ( s ) = 1 1 − ∆ t n s g n d s + 1 � � ˆ 1 − ∆ t n s d s 2 π i 2 π i C C � � f ( s ) x n − 1 ( s ) 1 g n + 1 � =ˆ ˆ 1 − ∆ t n s d s . f ∆ t n 2 π i C Recursion formula for the implicit Euler n n 1 1 � ˆ ∑ ∏ y ( t n ) = f ( s ) ∆ t j g j 1 − ∆ t k s d s 2 π i j = 1 k = j C � � n − 1 n 1 1 1 � = ˆ ˆ ∑ ∏ g n + ∆ t j g j f ( s ) f 1 − ∆ t k s d s ∆ t n 2 π i k = j j = 1 C Complex integral is solved with a quadrature formula Martin Schanz gCQM: Acoustics and Thermoelasticity 5 / 39
Time stepping formula Solution at the discrete time t n y ( t n ) = 1 � ˆ f ( s ) x n ( s ) d s 2 π i C ˆ f ( s )∆ t n f ( s ) x n − 1 ( s ) = 1 1 − ∆ t n s g n d s + 1 � � ˆ 1 − ∆ t n s d s 2 π i 2 π i C C � � f ( s ) x n − 1 ( s ) 1 g n + 1 � =ˆ ˆ 1 − ∆ t n s d s . f ∆ t n 2 π i C Recursion formula for the implicit Euler n n 1 1 � ˆ ∑ ∏ y ( t n ) = f ( s ) ∆ t j g j 1 − ∆ t k s d s 2 π i j = 1 k = j C � � n − 1 n 1 1 1 � = ˆ ˆ ∑ ∏ g n + ∆ t j g j f ( s ) f 1 − ∆ t k s d s ∆ t n 2 π i k = j j = 1 C Complex integral is solved with a quadrature formula Martin Schanz gCQM: Acoustics and Thermoelasticity 5 / 39
Algorithm First Euler step � � 1 y ( t 1 ) = ˆ f g 1 ∆ t 1 with implicit assumption of zero initial condition For all steps n = 2 ,..., N the algorithm has two steps Update the solution vector x n − 1 at all integration points s ℓ with an implicit Euler step 1 x n − 2 ( s ℓ ) ∆ t n − 1 x n − 1 ( s ℓ ) = + g n − 1 1 − ∆ t n − 1 s ℓ 1 − ∆ t n − 1 s ℓ for ℓ = 1 ,..., N Q with the number of integration points N Q . Compute the solution of the integral at the actual time step t n 2 � � N Q ˆ 1 f ( s ℓ ) y ( t n ) = ˆ ∑ f g n + ω ℓ x n − 1 ( s ℓ ) ∆ t n 1 − ∆ t n s ℓ ℓ = 1 Essential parameter: N Q = N log ( N ) , integration is dependent on q = ∆ t max ∆ t min Martin Schanz gCQM: Acoustics and Thermoelasticity 6 / 39
Algorithm First Euler step � � 1 y ( t 1 ) = ˆ f g 1 ∆ t 1 with implicit assumption of zero initial condition For all steps n = 2 ,..., N the algorithm has two steps Update the solution vector x n − 1 at all integration points s ℓ with an implicit Euler step 1 x n − 2 ( s ℓ ) ∆ t n − 1 x n − 1 ( s ℓ ) = + g n − 1 1 − ∆ t n − 1 s ℓ 1 − ∆ t n − 1 s ℓ for ℓ = 1 ,..., N Q with the number of integration points N Q . Compute the solution of the integral at the actual time step t n 2 � � N Q ˆ 1 f ( s ℓ ) y ( t n ) = ˆ ∑ f g n + ω ℓ x n − 1 ( s ℓ ) ∆ t n 1 − ∆ t n s ℓ ℓ = 1 Essential parameter: N Q = N log ( N ) , integration is dependent on q = ∆ t max ∆ t min Martin Schanz gCQM: Acoustics and Thermoelasticity 6 / 39
Numerical integration Integration weights and points � k 2 � 4 K γ ′ ( σ ℓ ) s ℓ = γ ( σ ℓ ) ω ℓ = 2 π i � n � α T , α = 1 . 5 for N = 25 , T = 5 , t n = N Martin Schanz gCQM: Acoustics and Thermoelasticity 7 / 39
Content Generalized convolution quadrature method (gCQM) 1 Quadrature formula Algorithm 2 Acoustics: Absorbing boundary conditions Boundary element formulation Analytical solution Numerical examples 3 Thermoelasticity: Uncoupled formulation Boundary element formulation Numerical example Martin Schanz gCQM: Acoustics and Thermoelasticity 8 / 39
Absorbing boundary conditions Materials with absorbing surfaces Mechanical modell: Coupling of porous material layer at the boundary Simpler mechanical model: Impedance boundary condition p Z = v · n specific impedance � Z ( x ) = α ( x ) = cos θ 1 − 1 − α S ( x ) � ρ c 1 + 1 − α S ( x ) with density ρ , wave velocity c , and absorption coefficient α S = f ( ω ) Martin Schanz gCQM: Acoustics and Thermoelasticity 9 / 39
Absorbing boundary conditions Materials with absorbing surfaces Mechanical modell: Coupling of porous material layer at the boundary Simpler mechanical model: Impedance boundary condition p Z = v · n specific impedance � Z ( x ) = α ( x ) = cos θ 1 − 1 − α S ( x ) � ρ c 1 + 1 − α S ( x ) with density ρ , wave velocity c , and absorption coefficient α S = f ( ω ) Martin Schanz gCQM: Acoustics and Thermoelasticity 9 / 39
Problem setting Bounded Lipschitz domain Ω − ⊂ R 3 with boundary Γ := ∂ Ω Ω + := R 3 \ Ω − is its unbounded complement. Linear acoustics for the pressure p in Ω σ × R > 0 , ∂ tt p − c 2 ∆ p = 0 in Ω σ , p ( x , 0 ) = ∂ t p ( x , 0 ) = 0 1 ( p ) − σα γ σ c γ σ 0 ( ∂ t p )= f ( x , t ) on Γ × R > 0 with σ ∈ { + , −} , wave velocity c , and α absorption coefficient Martin Schanz gCQM: Acoustics and Thermoelasticity 10 / 39
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