Application of generalized Convolution Quadrature in Acoustics … · 2016-11-10 · Graz...

Graz University of TechnologyInstitute of Applied Mechanics

Application of generalized Convolution Quadraturein Acoustics and Thermoelasticity

Martin Schanz

joint work with Relindis Rott and Stefan Sauter

Space-Time Methods for PDEsSpecial Semester on Computational Methods in Science and EngineeringRICAM, Linz, Austria, November 10, 2016

> www.mech.tugraz.at

Content

1 Generalized convolution quadrature method (gCQM)Quadrature formulaAlgorithm

2 Acoustics: Absorbing boundary conditionsBoundary element formulationAnalytical solutionNumerical examples

3 Thermoelasticity: Uncoupled formulationBoundary element formulationNumerical example

Martin Schanz gCQM: Acoustics and Thermoelasticity 2 / 39

Content

Convolution integral

Convolution integral with the Laplace transformed function f (s)

y (t) = (f ∗g)(t) =(f (∂t )g

)(t) =

f (t− τ)g (τ)dτ

es(t−τ)g (τ)dτ

︸︷︷︸x (t,s)

Integral is equivalent to solution of ODE

∂tx (t,s) = sx (t,s) + g (t) with x (t = 0,s) = 0

Implicit Euler for ODE , [0,T ] = [0, t1, t2, . . . , tN ], variable time steps∆ti , i = 1,2, . . . ,N

xn (s) =xn−1 (s)

1−∆tns+

∆tn1−∆tns

∑j=1

∆tjgj

∏k=j

11−∆tk s

Convolution integral

Convolution integral with the Laplace transformed function f (s)

y (t) = (f ∗g)(t) =(f (∂t )g

)(t) =

f (t− τ)g (τ)dτ

es(t−τ)g (τ)dτ

︸︷︷︸x (t,s)

Integral is equivalent to solution of ODE

∂tx (t,s) = sx (t,s) + g (t) with x (t = 0,s) = 0

Implicit Euler for ODE , [0,T ] = [0, t1, t2, . . . , tN ], variable time steps∆ti , i = 1,2, . . . ,N

xn (s) =xn−1 (s)

1−∆tns+

∆tn1−∆tns

∑j=1

∆tjgj

∏k=j

11−∆tk s

Time stepping formula

Solution at the discrete time tn

y (tn) =1

f (s)xn (s)ds

f (s)∆tn1−∆tns

gn ds +1

f (s)xn−1 (s)

1−∆tnsds

f (s)xn−1 (s)

1−∆tnsds .

Recursion formula for the implicit Euler

y (tn) =1

f (s)n

∑j=1

∆tjgj

∏k=j

11−∆tk s

∑j=1

∆tjgj1

f (s)n

∏k=j

11−∆tk s

Complex integral is solved with a quadrature formula

y (tn) =1

f (s)xn (s)ds

gn ds +1

f (s)xn−1 (s)

1−∆tnsds

f (s)xn−1 (s)

1−∆tnsds .

y (tn) =1

f (s)n

∑j=1

∆tjgj

∏k=j

11−∆tk s

∑j=1

∆tjgj1

f (s)n

∏k=j

11−∆tk s

y (tn) =1

f (s)xn (s)ds

gn ds +1

f (s)xn−1 (s)

1−∆tnsds

f (s)xn−1 (s)

1−∆tnsds .

y (tn) =1

f (s)n

∑j=1

∆tjgj

∏k=j

11−∆tk s

∑j=1

∆tjgj1

f (s)n

∏k=j

11−∆tk s

Algorithm

First Euler step

y (t1) = f

with implicit assumption of zero initial condition

For all steps n = 2, . . . ,N the algorithm has two steps1 Update the solution vector xn−1 at all integration points s` with an implicit Euler step

xn−1 (s`) =xn−2 (s`)

1−∆tn−1s`+

∆tn−1

1−∆tn−1s`gn−1

for ` = 1, . . . ,NQ with the number of integration points NQ .2 Compute the solution of the integral at the actual time step tn

y (tn) = f

∑`=1

ω`f (s`)

1−∆tns`xn−1 (s`)

Essential parameter: NQ = N log(N), integration is dependent on q = ∆tmax∆tmin

Algorithm

First Euler step

y (t1) = f

with implicit assumption of zero initial condition

For all steps n = 2, . . . ,N the algorithm has two steps1 Update the solution vector xn−1 at all integration points s` with an implicit Euler step

xn−1 (s`) =xn−2 (s`)

1−∆tn−1s`+

∆tn−1

1−∆tn−1s`gn−1

for ` = 1, . . . ,NQ with the number of integration points NQ .2 Compute the solution of the integral at the actual time step tn

y (tn) = f

∑`=1

ω`f (s`)

1−∆tns`xn−1 (s`)

Essential parameter: NQ = N log(N), integration is dependent on q = ∆tmax∆tmin

Numerical integration

Integration weights and points

s` = γ(σ`) ω` =4K(k2)

2πiγ′ (σ`)

for N = 25,T = 5, tn =(

)αT ,α = 1.5

Content

Absorbing boundary conditions

Materials with absorbing surfaces

Mechanical modell: Coupling of porousmaterial layer at the boundary

Simpler mechanical model: Impedance boundary condition

v ·nspecific impedance

ρc= α(x) = cosθ

1−√

1−αS (x)

1 +√

1−αS (x)

with density ρ, wave velocity c, and absorption coefficient αS = f (ω)

Absorbing boundary conditions

Materials with absorbing surfaces

Mechanical modell: Coupling of porousmaterial layer at the boundary

Simpler mechanical model: Impedance boundary condition

v ·nspecific impedance

ρc= α(x) = cosθ

1−√

1−αS (x)

1 +√

1−αS (x)

with density ρ, wave velocity c, and absorption coefficient αS = f (ω)

Problem setting

Bounded Lipschitz domain Ω− ⊂ R3 with boundary Γ := ∂ΩΩ+ := R3\Ω− is its unbounded complement.

Linear acoustics for the pressure p

∂tt p− c2∆p = 0 in Ωσ×R>0,

p (x ,0) = ∂t p (x ,0) = 0 in Ωσ,

γσ1 (p)−σ

σ0 (∂t p)= f (x , t) on Γ×R>0

with σ ∈ +,−, wave velocity c, and α absorption coefficient

Integral equation

Single layer ansatz for the density

2−K′ ∗ϕ

)−σ

c(V ∗ ϕ) = f a.e. in Γ×R>0

Retarded potentials

(V ∗ϕ)(x , t) =∫Γ

(y , t− ‖x−y‖

)4π‖x− y‖ dΓy

(K′ ∗ϕ

)(x , t) =

〈n (x) ,y− x〉‖x− y‖2

(y , t− ‖x−y‖

)‖x− y‖ +

(y , t− ‖x−y‖

Single layer potential for the pressure

p (x , t) = (S ∗ϕ)(x , t) :=∫Γ

(y , t− ‖x−y‖

)4π‖x− y‖ dΓy ∀(x , t) ∈ Ωσ×R>0

Integral equation

Single layer ansatz for the density

2−K′ ∗ϕ

)−σ

c(V ∗ ϕ) = f a.e. in Γ×R>0

Retarded potentials

(V ∗ϕ)(x , t) =∫Γ

(y , t− ‖x−y‖

)4π‖x− y‖ dΓy

(K′ ∗ϕ

)(x , t) =

〈n (x) ,y− x〉‖x− y‖2

(y , t− ‖x−y‖

)‖x− y‖ +

(y , t− ‖x−y‖

Single layer potential for the pressure

p (x , t) = (S ∗ϕ)(x , t) :=∫Γ

(y , t− ‖x−y‖

)4π‖x− y‖ dΓy ∀(x , t) ∈ Ωσ×R>0

Solution for the unit ball

Geometry is the unit ball

Right hand side of the impedance boundary condition is

f (x , t) := f (t)Yn,m

Spherical harmonics are eigenfunctions of the boundary integral operators

ZY mn = λ

n for Z ∈V,K,K′,W

It holds

λ(V)n (s) =−sjn (is)h(1)

n (is) λ(K′)n (s) =

12− is2jn (is)∂h(1)

n (is)

with the spherical Bessel and Hankel functions jn, h(1)n and ∂jn, ∂h(1)

n denoting theirfirst derivatives

Analytical transformation yields time domain solution

f (x , t) := f (t)Yn,m

ZY mn = λ

It holds

n (is) λ(K′)n (s) =

12− is2jn (is)∂h(1)

n (is)

f (x , t) := f (t)Yn,m

ZY mn = λ

It holds

n (is) λ(K′)n (s) =

12− is2jn (is)∂h(1)

n (is)

Solution in time domain

Solution for σ = +1, i.e., outer space, and n = 0Load function f (t) = (ct)υ e−ct

Density function

ϕ+ (t) =− 2

1 + α

bct/2c∑`=0

((ct−2`)υ e−(ct−2`)

− (1 + α)υ

αυ+1 γ

(υ + 1,

1 + α(ct−2`)

ct−2`1+α

)with the incomplete Gamma function γ(a,z) :=

∫ z0 ta−1e−t dt

Pressure solution

p+ (r , t) =− (1 + α)υ

παυ+1γ

(υ + 1,

1 + ατ+

τ1+α

with r > 1, we define τ := ct− (r −1) and (τ)+ := max0,τ

Solution in time domain

Solution for σ = +1, i.e., outer space, and n = 0Load function f (t) = (ct)υ e−ct

Density function

ϕ+ (t) =− 2

1 + α

bct/2c∑`=0

((ct−2`)υ e−(ct−2`)

− (1 + α)υ

αυ+1 γ

(υ + 1,

1 + α(ct−2`)

ct−2`1+α

)with the incomplete Gamma function γ(a,z) :=

∫ z0 ta−1e−t dt

Pressure solution

p+ (r , t) =− (1 + α)υ

παυ+1γ

(υ + 1,

1 + ατ+

τ1+α

with r > 1, we define τ := ct− (r −1) and (τ)+ := max0,τ

Discretization

Spatial discretization: Linear continuous shape functions on linear trianglesTemporal discretization: gCQM with time grading

tn = T( n

, n = 0, . . . ,N with grading exponent χ = 1/υ

Meshes of the unit sphere

h1 = 0.393m h2 = 0.196m h3 = 0.098m h3 = 0.049m

Material data: Air (c = 343.41 m/s)Load function: f (t) = (ct)υ e−ct with υ = 1

2Observation time T = 0.002915905s and β = c∆t/h

Error in time

errrel =

∑n=0

∆tn (u (tn)−uh (tn))2/

∑n=0

∆tn (u (tn))2 eoc = log2

errh+1

)Martin Schanz gCQM: Acoustics and Thermoelasticity 14 / 39

Solution density

0 0.2 0.4 0.6 0.8 1 1.2·10−2

−0.2

time t [s]

+α = 0α = 0.25α = 0.5α = 1analytic α = 0.25analytic α = 0.5analytic α = 1

Solution pressure

0 0.2 0.4 0.6 0.8 1 1.2·10−2

time t [s]

u+[N/m

α = 0α = 0.25α = 0.5α = 1

Relative density error: mesh size

10−1.2 10−1 10−0.8 10−0.6 10−0.4

10−2.5

10−2

10−1.5

mesh size h

err re

∆tvar ,β = 0.125∆tconst ,β = 0.125∆tvar ,β = 0.0625∆tconst ,β = 0.0625eoc = 0.5eoc = 1

Relative pressure error: mesh size

10−1.2 10−1 10−0.8 10−0.6 10−0.4

10−2

10−1

mesh size h

err re

∆tvar ,β = 0.25∆tconst ,β = 0.25∆tvar ,β = 0.125∆tconst ,β = 0.125∆tvar ,β = 0.0625∆tconst ,β = 0.0625eoc = 1

Relative density error: time step

10−5 10−4

10−2.5

10−2

10−1.5

time step size ∆t

err re

mesh 2, ∆tconst

mesh 2, ∆tvar

mesh 3, ∆tconst

mesh 3, ∆tvar

mesh 4, ∆tconst

mesh 4, ∆tvareoc = 0.5eoc = 1

Relative pressure error: time step

10−5 10−4

10−2

10−1

time step size ∆t

err re

mesh 2, ∆tconst

mesh 2, ∆tvar

mesh 3, ∆tconst

mesh 3, ∆tvar

mesh 4, ∆tconst

mesh 4, ∆tvareoc = 2eoc = 1

Problem description

Atrium at University of Zurich (Irchel campus)

Mesh: 7100 elementsTime interval [0,T = 0.15s] with grading

(n−1)2

)∆tconst with ∆tconst = 0.00037s ⇒ N = 405, Ngraded = 248

Sound pressure field

t ≈ 0.028 s

α = 0.1 α = 0.5 α = 1

t ≈ 0.064 s

Sound pressure level

0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16−30

time t [s]

]α = 0.1α = 0.5α = 1

Content

Uncoupled thermoelasticity

Governing equations for temperature θ(x, t) and displacement u(x, t)

κ θ,jj (x, t)− θ(x, t) = 0

µ ui,jj (x, t) + (λ + µ)uj,ij (x, t)− (3λ + 2µ)α θ,i (x, t) = 0

κ thermal diffusivity, α thermal expansion coefficient, λ,µ Lamé constantsBoundary integral formulation

c (y)θ(y, t) =∫Γ

[Θ∗q](x,y, t)− [Q ∗θ](x,y, t)dΓ

cij (y)uj (y, t) =∫Γ

Uij (x,y)tj (x, t)−Tij (x,y) uj (x, t)

+ [Gi ∗q](x,y, t)− [Fi ∗θ](x,y, t)dΓ

withΘ(x,y, t) and Q(x,y, t) kernels of the heat equationUij (x,y) and Tij (x,y) kernels from elastostaticsGi (x,y, t) and Fi (x,y, t) kernels for the one sided coupling

Uncoupled thermoelasticity

Governing equations for temperature θ(x, t) and displacement u(x, t)

κ θ,jj (x, t)− θ(x, t) = 0

µ ui,jj (x, t) + (λ + µ)uj,ij (x, t)− (3λ + 2µ)α θ,i (x, t) = 0

κ thermal diffusivity, α thermal expansion coefficient, λ,µ Lamé constantsBoundary integral formulation

c (y)θ(y, t) =∫Γ

[Θ∗q](x,y, t)− [Q ∗θ](x,y, t)dΓ

cij (y)uj (y, t) =∫Γ

Uij (x,y)tj (x, t)−Tij (x,y) uj (x, t)

+ [Gi ∗q](x,y, t)− [Fi ∗θ](x,y, t)dΓ

withΘ(x,y, t) and Q(x,y, t) kernels of the heat equationUij (x,y) and Tij (x,y) kernels from elastostaticsGi (x,y, t) and Fi (x,y, t) kernels for the one sided coupling

Boundary element formulation

Spatial discretisation on some mesh

θ(x, t) =ND

∑k=1

ψk (x) θ

k (t) q(x, t) =NN

∑k=1

χk (x) qk (t)

uj (x, t) =ND

∑k=1

ψk (x) uk

i (t) tj (x, t) =NN

∑k=1

χk (x) tk

Semi-discrete BEM

Cθθθ(t) = [ΘΘΘ∗q](t)− [Q∗θθθ](t)

Ceu(t) = Ut(t)−Tu(t) + [G∗q](t)− [F∗θθθ](t)

Temporal discretisation with gCQMto solve the thermal equationto perform the convolution of known data for the coupling terms

[G∗q](t) [F∗θθθ](t)

Boundary element formulation

Spatial discretisation on some mesh

θ(x, t) =ND

∑k=1

ψk (x) θ

k (t) q(x, t) =NN

∑k=1

χk (x) qk (t)

uj (x, t) =ND

∑k=1

ψk (x) uk

i (t) tj (x, t) =NN

∑k=1

χk (x) tk

Semi-discrete BEM

Cθθθ(t) = [ΘΘΘ∗q](t)− [Q∗θθθ](t)

Ceu(t) = Ut(t)−Tu(t) + [G∗q](t)− [F∗θθθ](t)

Temporal discretisation with gCQMto solve the thermal equationto perform the convolution of known data for the coupling terms

[G∗q](t) [F∗θθθ](t)

Problem setting

Cube under restrictive boundary conditions to enforce a 1-d solution

θ(t > 0) = 1

y or z

• •

Material data:α = 1 κ = 1λ = 0 µ = 0.5

Time discretisation:constant tn = n∆t

increasing tn =

(n−1)2

graded tn = N∆t( n

Spatial discretisations

Mesh 1 Mesh 2 Mesh 3

Temperature solution: gCQM, graded

0 1 2 3 4

time t [s]

mesh 1mesh 2mesh 3analytic

Displacement solution: gCQM, graded

0 1 2 3 4

time t [s]

mesh 1mesh 2mesh 3analytic

Temperature solution: error, mesh 2

0 1 2 3 4

·10−2

time t [s]

mesh 2, constantmesh 2, gradedmesh 2, increasing

Temperature solution: error, mesh 3

0 1 2 3 4

·10−2

time t [s]

Displacement solution: error, mesh 2

0 1 2 3 4

·10−2

time t [s]

Displacement solution: error, mesh 3

0 1 2 3 4

4·10−3

time t [s]

Temperature error L2 mesh 3

10−1.8 10−1.6 10−1.4 10−1.2 10−1

10−5

10−4

time step size ∆t

err re

lmesh 3, constmesh 3, gradedmesh 3, increasingeoc = 2eoc = 1

Temperature error Lmax mesh 3

10−1.8 10−1.6 10−1.4 10−1.2 10−1

10−2.5

10−2

10−1.5

time step size ∆t

bsmesh 3, constmesh 3, gradedmesh 3, increasingeoc = 0.7eoc = 1

Displacement error L2 mesh 3

10−1.8 10−1.6 10−1.4 10−1.2 10−1

10−3

10−2

time step size ∆t

err re

mesh 3, constmesh 3, gradedmesh 3, increasingeoc = 1.2

Displacement error Lmax mesh 3

10−1.8 10−1.6 10−1.4 10−1.2 10−1

10−3

10−2

time step size ∆t

mesh 3, constmesh 3, gradedmesh 3, increasingeoc = 0.5eoc = 1.3

Conclusions

Indirect BE formulation in time domain for absorbing BC in acoustics

Direct BE formulation for uncoupled thermoelasticity

Time discretisation with generalized Convolution Quadrature Method

Expected rate of convergence in time

Application to real world problems possible

Fast methods to compress matrices is to be done

Fast method only for matrix-vector products

Possible extension to variable space-time formulation

Conclusions

Graz University of TechnologyInstitute of Applied Mechanics

Application of generalized Convolution Quadraturein Acoustics and Thermoelasticity

Martin Schanz

joint work with Relindis Rott and Stefan Sauter

Space-Time Methods for PDEsSpecial Semester on Computational Methods in Science and EngineeringRICAM, Linz, Austria, November 10, 2016

> www.mech.tugraz.at

Application of generalized Convolution Quadrature in Acoustics … · 2016-11-10 · Graz...

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