fast computation of time convolutions: fractional …...fractional order dynamical systems in...

Convolutionintegrals

FractionalintegralsDiscretizing Laplaceintegralrepresentation

Numerical results

Solution ofheat equation

Example:modelingcrystal growth

Conclusions

Fast computation of time convolutions:fractional integrals and heat equation

Jing-Rebecca Li

Projet POEMSInstitut National de Recherche en Informatique et en Automatique (INRIA)

RocquencourtFRANCE

Groupe de travail "Méthodes numériques", Paris 6

Colloborators:Leslie Greengard (Courant Institute), Donna Calhoun

(CEA), Lucien Brush (Univ. of Washington)

Numerical results

Conclusions

Outline

1 Convolution integrals

2 Fractional integralsDiscretizing Laplace integral representationNumerical results

3 Solution of heat equation

4 Example: modeling crystal growth

5 Conclusions

Numerical results

Conclusions

Convolution integrals

Convolution in time and space

U(x, t) =

∫Ω(τ)

k(x− y, t − τ) f (y, τ) dy dτ, x ∈ Ω(τ).

Naive implementation

Spatial discretization: x1, · · · , xN , y1, · · · , yN ∈ Ω(τ).N unknowns, O(N2) work.Time discretization: t = ∆t , 2∆t , · · · , M∆t .M time steps: O(M2) work, O(M) storage.

Numerical results

Conclusions

Topic I:

Time stepping of fractional integrals

Numerical results

Conclusions

Time convolution integral

DefinitionFractional order integral (of order α of function f )

Iα[f ](t) =1

Γ(α)

0(t − τ)α−1 f (τ) dτ, 0 < α.

DefinitionFractional order derivative

Dα[f ](t) :=1

Γ(dαe − α)

0(t − τ)(dαe−α)−1 f dαe(τ) dτ.

Satisfy

Dα Iα[f ](t) = f (t), Iα Dα[f ](t) = f (t)−bαc∑k=0

k !f k0 .

Numerical results

Conclusions

Time convolution integral

Solve fractional order differential equations

Dα[y ](t) = f(t , y(t)

), yk (t0) = yk

0 , k = 0, 1, · · · bαc.

Numerical results

Conclusions

Applications

Fractional order dynamical systems in control theory[n∑

an−kDαn−k

]y(t) = f (t)

Fractional order laws of deformation in viscoelasticity.

σ(t) = κ−∞Dαε(t), 0 < α < 1.

σ stress, ε strain, κ generalized viscosity of material.Fractional order laws in tracer fluid flows.Fractional order model of neurons.

[Source: Recent Applications of fractional calculus to science andengineering, Lokenath Debnath, IJMMS 2003]

Numerical results

Conclusions

Convolution with fractional power kernel

Iα[f ](t) =1

Γ(α)

0(t − τ)α−1 f (τ) dτ, 0 < α < 1.

Input function f

f (τ) =∑i=1

ai τβi , βi ∈ j + lα, j , l ∈ 0, 1, 2, · · · .

Challenges

Time discretization: t = ∆t , 2∆t , · · · , M∆t .Naive implementation:M time steps: O(M2) work, O(M) storage.f is not smooth.Special quadrature (numerical integration formulae).

Numerical results

Conclusions

Existing approachesFractional power linear multi-step methods [Lubich, 1986]

Iα[f ](m∆t) ≈ ∆tαm∑

ωm−j f (j∆t) + ∆tαs∑

wmj f (j∆t).

Weights: (ω(ρ))α =∑∞

m=0 ωmρm,ω(ρ) generating function of a linear multistep methodfor first order ODEs.

ODE method ω(ρ)α

Backward Euler (1− ρ)−α

Trapezoidal(

1+ρ2(1−ρ)

32 − 2ρ + ρ2

)−α

Correction weights wmj treat non-smoothness of f at 0,as solution of Vandermonde linear system,solved at each time step.

Numerical results

Conclusions

Existing approaches

Direct quadrature [Diethelm 1997, Ford+Simpson 2001].

Reformulation as a system of differential equations[Staffans 1994, Montseny+Audounet+Matignon 2000,Helie+Matignon 2006, Yuan+Agrawal 2002, Chatterjee2005].

Numerical results

Conclusions

Laplace transform representation of kernel

tα−1 =1

Γ(1− α)

∫ ∞

0e−ξ t ξ−α dξ.

e−ξ t better than tα−1 in time stepping method

Iα[f ](t) =1

Γ(α)

0(t − τ)α−1 f (τ) dτ,

Γ(α)

Γ(1− α)

∫ ∞

0e−ξ (t−τ) ξ−α dξ

)f (τ) dτ,

Γ(α)

1Γ(1− α)

∫ ∞

(∫ t

0e−ξ (t−τ) f (τ) dτ

)ξ−α dξ,

Γ(α)

1Γ(1− α)

∫ ∞

0g(ξ, t) ξ−α dξ,

≈Q∑

g(ξj , t) wj

Numerical results

Conclusions

Time stepping each mode in transform domain

g(ξj , t) :=

0e−ξj (t−τ) f (τ) dτ.

Compute each mode

g(ξj , t) = e−ξj∆tg(ξj , t −∆t) + Ψ(ξj , t ,∆t),Ψ(ξj , t ,∆t) =

∫ tt−∆t e−ξj (t−τ) f (τ) dτ,

Alternatively: differential formulation

(ξ, t) = −ξ g(ξ, t) + f (t), g(ξ, 0) = 0.

Numerical results

Conclusions

Gain in speed and memory

Add them up

Iα[f ](t) ≈Q∑

g(ξj , t) wj

Algorithm complexity

Time discretization: t = ∆t , 2∆t , · · · , M∆t .O(MQ) work, O(Q) memory.In our approach: Q depends on ∆t but not on M.

Numerical results

Conclusions

Discrete approximation of kernel

Find quadrature for integral

tα−1 =1

Γ(1− α)

∫ ∞

Find one discrete approx. valid for wide range of t

tα−1

Γ(α)≈

Q∑j=1

e−ξj t wj ,

Discrete approximation must stay away fromsingularity at t = 0.

Numerical results

Conclusions

In other words

Find quadrature valid for all t ∈ [∆t , Tmax ] or [∆t ,∞).

∣∣∣∣∣∣ tα−1

Γ(α)−

Q∑j=1

e−ξj t wj

∣∣∣∣∣∣ ≤ ε,

∀t ∈ [∆t , Tmax ]

∀t ∈ [∆t ,∞).

Numerical results

Conclusions

Break Iα[f ](t) into history and local parts

Fractional integral in two parts

Iα[f ](t) = Hα[f ](t) + Lα[f ](t).

History part: staying away from singularity

Hα[f ](t) :=1

Γ(α)

∫ t−∆t

0(t − τ)α−1 f (τ) dτ.

Local part: treat singularity in kernel

Lα[f ](t) :=1

Γ(α)

t−∆t(t − τ)α−1 f (τ) dτ.

Numerical results

Conclusions

Time stepping of history part

g(ξ, t ,∆t) :=

∫ t−∆t

0e−ξ (t−τ) f (τ) dτ.

Time stepping of mode

g(ξj , t ,∆t) = e−ξj∆tg(ξj , t −∆t ,∆t) + Ψ(ξj , t ,∆t),

Ψ(ξj , t ,∆t) =

∫ t−∆t

t−2 ∆te−ξj (t−τ) f (τ) dτ,

Quadrature

Hα[f ](t) ≈Q∑

g(ξj , t ,∆t) wj .

Numerical results

Conclusions

Quadrature valid for a family of integrands

Find quadrature valid for all t ∈ [∆t , Tmax ] or [∆t ,∞).

∣∣∣∣∣∣∫ ∞

0C e−ξ t ξ−α dξ −

Q∑j=1

e−ξj t wj

∣∣∣∣∣∣ ≤ ε,

∀t ∈ [∆t , Tmax ]

∀t ∈ [∆t ,∞).

tα−1

Γ(α)=

∫ ∞

0C e−ξ t ξ−α dξ, C =

1Γ(α)Γ(1− α)

Numerical results

Conclusions

Behavior of integrandChange of variable

∫ ∞

0e−ξ tξ−α dξ = γ

∫ ∞

0e−ηγ t dη, γ =

11− α

, η = ξ1γ .

10−1

−ηγ t)

integrand as a function of t, α = 0.1

t=0.0001t=0.001t=0.01t=0.1t=1t=10

(a) α = 0.1

10−1

−ηγ t)

t=0.0001t=0.001t=0.01t=0.1t=1t=10

(b) α = 0.5

10−1

−ηγ t)

t=0.0001t=0.001t=0.01t=0.1t=1t=10

(c) α = 0.9

Figure: The behavior of e−ηγ t for several values of α and t . Thedecay of the integrand is fast when t is large and it is slow when tis small.

Numerical results

Conclusions

Finite domain of integrationEstimate

∣∣∣∣∣∫ ∞

0e−ηγ t dη −

0e−ηγ t dη

∣∣∣∣∣ ≤ e−∆t LγΓ

, ∀t ∈ [∆t ,∞).

Choose cutoff

L(∆t , ε) =

− log ε/3 ∆t(1−α)

(1−α)Γ(1−α)

1−α

= (∆t)α−1(

log3ε

+ (α− 1) log ∆t + log (1− α)Γ(1− α)

)1−α

∣∣∣∣∣∫ ∞

0e−ηγ t dη −

∫ L(∆t ,ε)

0e−ηγ t dη

∣∣∣∣∣ ≤ 13

ε, ∀t ∈ [∆t ,∞).

Numerical results

Conclusions

High frequency cut-off

10−1

L(dt = 0.01, ε= 0.001)

−ηγ t)

t=0.01t=0.1t=1t=10

(a) α = 0.1

10−1

L(dt = 0.01, ε= 0.001)

−ηγ t)

t=0.01t=0.1t=1t=10

(b) α = 0.5

10−1

L(dt = 0.01, ε= 0.001)

−ηγ t)

t=0.01t=0.1t=1t=10

(c) α = 0.9

10−4

10−3

10−2

10−1

t, ε)

High frequency cutoff L(∆ t,ε), α = 0.1

ε=0.0001ε=0.001ε=0.01ε=0.1

(d) α = 0.1

10−4

10−3

10−2

10−1

t, ε)

ε=0.0001ε=0.001ε=0.01ε=0.1

(e) α = 0.5

10−4

10−3

10−2

10−1

t, ε)

ε=0.0001ε=0.001ε=0.01ε=0.1

(f) α = 0.9

Figure: Top: The domain of integration is cut off at the frequencycutoff L(∆t , ε). Bottom: The frequency cutoff L(∆t , ε) as afunction of ∆t for various values of α and ε.

Numerical results

Conclusions

Approximation on dyadic intervals

Divide [0, L] into disjoint dyadic intervals [a, b := 2a]

Choose Gauss-Legendre quadrature of lowest order,Qa, ηa

j and weights vaj , to numerically satisfy

∣∣∣∣∣∣∫ b

ae−ηγ tdη −

Qa∑j=1

e−(ηaj )γ t va

∣∣∣∣∣∣ ≤ 1/3 ε (b − a)

γ L(∆t , ε), t ∈ [tmin, tmax ],

e−aγ tmax = q, e−bγ tmin = 1− q,

q is a small factor, picked to be 10−4.On [a, b], e−ηγ t is almost ≡ 0 if t > tmax .On [a, b], e−ηγ t is almost ≡ 1 if t < tmin.

Numerical results

Conclusions

Numerically verify quadrature for t ∈ [tmin, tmax ]

−0.5

η = a η = b

−ηγ t)

integrand in the dyadic interval [a,b=2a]

Figure: On the dyadic interval [a, b], e−ηγ t is almost identically 0if t > tmax , and it is almost identically 1 if t < tmin.

Numerical results

Conclusions

Tmax finite

Solve for largest amin = 2jmin : e−aγmin Tmax ≤ q.

e−ηγ Tmax is negligible outside [0, amin].

−0.5

η = 0 η = amin

−ηγ t)

integrand in the dyadic interval [0,amin

Figure: On the dyadic interval [0, amin], e−ηγ t is almost identically0 if t = Tmax , and it is almost identically 1 if t < tmin.

Numerical results

Conclusions

Quadrature valid for t ∈ [∆t ,∞)

No special interval [0, amin].

The smallest dyadic interval is [2jmin , 2jmin+1],jmin largest integer satisfying 2jmin ≤ ε

Numerical results

Conclusions

Partition of the interval [0, L] into dyadic intervals

2jmin L

−ηγ t)

Partition of [0,L] into dyadic intervals

t=∆ t t=T

Figure: Partition of the interval [0, L] into dyadic intervals[2j , 2j+1]. On each interval, quadrature is verified numerically.

Numerical results

Conclusions

Quadrature errors, α = 0.5

10−1

10−15

10−10

10−5

t = ∆ t t = Tmax

Quadrature error of ∫0∞ exp(−ηγ t), α = 0.5

ε=1e−006, ∆ t=0.1, Tmax

=100, Q = 64

(a) Tmax = 100

10−6

10−4

10−2

L(dt,ε)

Quad nodes, , Q=64, dt=0.1, Tmax

=100, ε=1e−006, γ=2

(b) 64 Nodes

10−6

10−4

10−2

L(dt,ε)

−ηγ t)

=100, ε=1e−006, γ=2

t=0.1t=1t=10t=100

(c) Integrands

10−1

10−15

10−10

10−5

t = ∆ t t = Tmax

ε=1e−006, ∆ t=0.1, Tmax

=1000, Q = 80

(d) Tmax = 1000

10−6

10−4

10−2

L(dt,ε)

=1000, ε=1e−006, γ=2

(e) 80 Nodes

10−6

10−4

10−2

L(dt,ε)

−ηγ t)

=1000, ε=1e−006, γ=2

t=0.1t=1t=10t=100

(f) Integrands

10−1

10−7

10−6

10−5

ε=1e−006, ∆ t=0.1, Tmax

=∞, Q = 176

(g) Tmax =∞

10−6

10−4

10−2

L(dt,ε)

=∞, ε=1e−006, γ=2

(h) 176 Nodes

10−6

10−4

10−2

L(dt,ε)

−ηγ t)

=∞, ε=1e−006, γ=2

t=0.1t=1t=10t=100

(i) Integrands

Numerical results

Conclusions

Quadrature errors, α = 0.1

10−1

10−15

10−10

10−5

t = ∆ t t = Tmax

ε=1e−006, ∆ t=0.1, Tmax

=100, Q = 128

(j) Tmax = 100, Q = 128

10−1

10−15

10−10

10−5

t = ∆ t t = Tmax

ε=1e−006, ∆ t=0.1, Tmax

=1000, Q = 152

(k) Tmax = 1000, Q = 152

10−1

10−7

10−6

10−5

ε=1e−006, ∆ t=0.1, Tmax

=∞, Q = 212

(l) Tmax =∞, Q = 212

Numerical results

Conclusions

Number of modes as function of ∆t and ε

ε = 10−3

α = 0.5 α = 0.1Tmax 100 1000 ∞ ∞

∆t = 0.1 30 38 54 64∆t = 0.05 32 40 56 68∆t = 0.025 34 42 58 72∆t = 0.0125 36 44 60 76∆t = 0.00625 38 46 62 80∆t = 0.003125 40 48 64 84

Table: Number of quadrature nodes Q(∆t , ε), α = 0.5 andα = 0.1.

Numerical results

Conclusions

Number of modes as function of ∆t and ε

ε = 10−6

α = 0.5 α = 0.1Tmax 100 1000 ∞ ∞

∆t = 0.1 64 80 176 212∆t = 0.05 68 84 180 220∆t = 0.025 72 88 184 228∆t = 0.0125 76 92 188 236∆t = 0.00625 80 96 192 244∆t = 0.003125 84 100 200 252

Table: Number of quadrature nodes Q(∆t , ε), α = 0.5 andα = 0.1.

Numerical results

Conclusions

Our work

A different discrete integral representation of tα−1

Limit use of integral representation to t ≥ ∆t .Adaptively obtain quadrature which is accurate forlarger and larger values of t .Quadrature obtained is accepted either when Tmax , ana priori longest simulation time, is reached, or when thequadrature is sufficiently accurate for all t ∈ [t ,∞].

Numerical results

Conclusions

Topic II

Fast algorithm to solve heat equation

Numerical results

Conclusions

Heat equation

Domain ΩT :=∏T

τ=0 Ω(t), boundary ΓT :=∏T

τ=0 Γ(t),

Solution U ∈ C2,1(ΩT )⋂

C(ΩT ).

∂tU(x, t)− a ∇2U(x, t) = f (x, t), (x, t) ∈ ΩT , (source)

U(x, 0) = u0(x), x ∈ Ω(0), (init data)U(x, t) = g(x, t), (x, t) ∈ ΓT . (bdy data)

For simplicity of notation: a = 1.

Numerical results

Conclusions

Moving boundary

Γ(τ) y

Figure: Boundary motion is prescribed

Numerical results

Conclusions

Applications: heat equation

Diffusion of heat: crystal growth.

Brownian motion: photon diffusion in tissue.

Incorporate heat equation solverin larger simulation effort.Part of inner loop, splitting...

Numerical results

Conclusions

Our approach

Solution as integral transformsU(x, t) = U0(x, t) + Us(x, t) + DL(µ)(x, t), (x, t) ∈ ΩT .

U0(x, t) =

∫Ω(0)

k (x− y, t) u0(y) dy

Us(x, t) =

∫Ω(τ)

k (x− y, t − τ) f (y, τ) dy dτ,

DL(µ)(x, t) =

∫Γ(τ)

(k(x− y, t − τ)

)µ(y, τ) dσy dτ.

Free space heat kernel: k (x, t) = e−‖x‖2/4t

(4πt)d2

Numerical results

Conclusions

Differences from Topic I

Laplace transform representation

tα−1 =1

Γ(1− α)

∫ ∞

Fourier transform representation

e−‖x‖2/4t

(4πt)d2

)d ∫s∈Rd

e−is·x e−‖s‖2t ds.

DefinitionFourier and inverse Fourier transforms

f (s) =

∫x∈Rd

eis·x f (x) dx, f (x) =

)d ∫s∈Rd

e−is·x f (s) ds.

Numerical results

Conclusions

Heat equation solver

To compute: Us(x, t) =t∫

∫Ω(τ)

e−‖(x−y)‖2/4(t−τ)

(4π(t−τ))d2

f (y, τ) dy dτ

Compute local part

U locs (x, t ,∆t) =

t∫t−∆t

∫Ω(τ)

e−‖(x−y)‖2/4(t−τ)

(4π(t − τ))d2

f (y, τ) dy dτ.

History part

UHs (s, t ,∆t) =

t−∆t∫0

e−‖s‖2(t−τ)

∫Ω(τ)

ei s · y f (y, τ) dy dτ.

computed via

UsH(s, m) = e−‖s‖2∆t Us

H(s, m − 1) +

∫ t−∆t

t−2∆t· · ·dτ.

Numerical results

Conclusions

Differences from Topic IHistory update

e−‖s‖2∆t UsH(s, m − 1) + Θ(s, m) −→ Us

H(s, m)

Discretize inverse Fourier transform in s

UsH(·, m) =

)d ∫s∈Rd

e−is·x UsH(·, m) ds.

fast computation of time convolutions: fractional …...fractional order dynamical systems in...

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