fast computation of time convolutions: fractional …...fractional order dynamical systems in...
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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)
Convolutionintegrals
FractionalintegralsDiscretizing Laplaceintegralrepresentation
Numerical results
Solution ofheat equation
Example:modelingcrystal growth
Conclusions
Outline
1 Convolution integrals
2 Fractional integralsDiscretizing Laplace integral representationNumerical results
3 Solution of heat equation
4 Example: modeling crystal growth
5 Conclusions
Convolutionintegrals
FractionalintegralsDiscretizing Laplaceintegralrepresentation
Numerical results
Solution ofheat equation
Example:modelingcrystal growth
Conclusions
Convolution integrals
Convolution in time and space
U(x, t) =
t∫0
∫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.
Convolutionintegrals
FractionalintegralsDiscretizing Laplaceintegralrepresentation
Numerical results
Solution ofheat equation
Example:modelingcrystal growth
Conclusions
Topic I:
Time stepping of fractional integrals
Convolutionintegrals
FractionalintegralsDiscretizing Laplaceintegralrepresentation
Numerical results
Solution ofheat equation
Example:modelingcrystal growth
Conclusions
Time convolution integral
DefinitionFractional order integral (of order α of function f )
Iα[f ](t) =1
Γ(α)
∫ t
0(t − τ)α−1 f (τ) dτ, 0 < α.
DefinitionFractional order derivative
Dα[f ](t) :=1
Γ(dαe − α)
∫ t
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
tk
k !f k0 .
Convolutionintegrals
FractionalintegralsDiscretizing Laplaceintegralrepresentation
Numerical results
Solution ofheat equation
Example:modelingcrystal growth
Conclusions
Time convolution integral
Solve fractional order differential equations
Dα[y ](t) = f(t , y(t)
), yk (t0) = yk
0 , k = 0, 1, · · · bαc.
Convolutionintegrals
FractionalintegralsDiscretizing Laplaceintegralrepresentation
Numerical results
Solution ofheat equation
Example:modelingcrystal growth
Conclusions
Applications
Fractional order dynamical systems in control theory[n∑
k=0
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]
Convolutionintegrals
FractionalintegralsDiscretizing Laplaceintegralrepresentation
Numerical results
Solution ofheat equation
Example:modelingcrystal growth
Conclusions
Convolution with fractional power kernel
Iα[f ](t) =1
Γ(α)
∫ t
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).
Convolutionintegrals
FractionalintegralsDiscretizing Laplaceintegralrepresentation
Numerical results
Solution ofheat equation
Example:modelingcrystal growth
Conclusions
Existing approachesFractional power linear multi-step methods [Lubich, 1986]
Iα[f ](m∆t) ≈ ∆tαm∑
j=0
ωm−j f (j∆t) + ∆tαs∑
j=0
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)
)α
BDF2(
32 − 2ρ + ρ2
2
)−α
Correction weights wmj treat non-smoothness of f at 0,as solution of Vandermonde linear system,solved at each time step.
Convolutionintegrals
FractionalintegralsDiscretizing Laplaceintegralrepresentation
Numerical results
Solution ofheat equation
Example:modelingcrystal growth
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].
Convolutionintegrals
FractionalintegralsDiscretizing Laplaceintegralrepresentation
Numerical results
Solution ofheat equation
Example:modelingcrystal growth
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)
∫ t
0(t − τ)α−1 f (τ) dτ,
=1
Γ(α)
∫ t
0
(1
Γ(1− α)
∫ ∞
0e−ξ (t−τ) ξ−α dξ
)f (τ) dτ,
=1
Γ(α)
1Γ(1− α)
∫ ∞
0
(∫ t
0e−ξ (t−τ) f (τ) dτ
)ξ−α dξ,
=1
Γ(α)
1Γ(1− α)
∫ ∞
0g(ξ, t) ξ−α dξ,
≈Q∑
j=1
g(ξj , t) wj
Convolutionintegrals
FractionalintegralsDiscretizing Laplaceintegralrepresentation
Numerical results
Solution ofheat equation
Example:modelingcrystal growth
Conclusions
Time stepping each mode in transform domain
g(ξj , t) :=
∫ 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
dgdt
(ξ, t) = −ξ g(ξ, t) + f (t), g(ξ, 0) = 0.
Convolutionintegrals
FractionalintegralsDiscretizing Laplaceintegralrepresentation
Numerical results
Solution ofheat equation
Example:modelingcrystal growth
Conclusions
Gain in speed and memory
Add them up
Iα[f ](t) ≈Q∑
j=1
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.
Convolutionintegrals
FractionalintegralsDiscretizing Laplaceintegralrepresentation
Numerical results
Solution ofheat equation
Example:modelingcrystal growth
Conclusions
Discrete approximation of kernel
Find quadrature for integral
tα−1 =1
Γ(1− α)
∫ ∞
0e−ξ t ξ−α dξ.
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.
Convolutionintegrals
FractionalintegralsDiscretizing Laplaceintegralrepresentation
Numerical results
Solution ofheat equation
Example:modelingcrystal growth
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 ,∞).
Convolutionintegrals
FractionalintegralsDiscretizing Laplaceintegralrepresentation
Numerical results
Solution ofheat equation
Example:modelingcrystal growth
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(t − τ)α−1 f (τ) dτ.
Convolutionintegrals
FractionalintegralsDiscretizing Laplaceintegralrepresentation
Numerical results
Solution ofheat equation
Example:modelingcrystal growth
Conclusions
Time stepping of history part
Mode
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∑
j=1
g(ξj , t ,∆t) wj .
Convolutionintegrals
FractionalintegralsDiscretizing Laplaceintegralrepresentation
Numerical results
Solution ofheat equation
Example:modelingcrystal growth
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)
.
Convolutionintegrals
FractionalintegralsDiscretizing Laplaceintegralrepresentation
Numerical results
Solution ofheat equation
Example:modelingcrystal growth
Conclusions
Behavior of integrandChange of variable
∫ ∞
0e−ξ tξ−α dξ = γ
∫ ∞
0e−ηγ t dη, γ =
11− α
, η = ξ1γ .
10−1
100
101
102
103
0
0.5
1
1.5
exp(
−ηγ t)
η
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
100
101
102
103
0
0.5
1
1.5
exp(
−ηγ t)
η
integrand as a function of t, α = 0.5
t=0.0001t=0.001t=0.01t=0.1t=1t=10
(b) α = 0.5
10−1
100
101
102
103
0
0.5
1
1.5
exp(
−ηγ t)
η
integrand as a function of t, α = 0.9
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.
Convolutionintegrals
FractionalintegralsDiscretizing Laplaceintegralrepresentation
Numerical results
Solution ofheat equation
Example:modelingcrystal growth
Conclusions
Finite domain of integrationEstimate
γ
∣∣∣∣∣∫ ∞
0e−ηγ t dη −
∫ L
0e−ηγ t dη
∣∣∣∣∣ ≤ e−∆t LγΓ
(1γ
)∆t
1γ
, ∀t ∈ [∆t ,∞).
Choose cutoff
L(∆t , ε) =
− log ε/3 ∆t(1−α)
(1−α)Γ(1−α)
∆t
1−α
= (∆t)α−1(
log3ε
+ (α− 1) log ∆t + log (1− α)Γ(1− α)
)1−α
.
γ
∣∣∣∣∣∫ ∞
0e−ηγ t dη −
∫ L(∆t ,ε)
0e−ηγ t dη
∣∣∣∣∣ ≤ 13
ε, ∀t ∈ [∆t ,∞).
Convolutionintegrals
FractionalintegralsDiscretizing Laplaceintegralrepresentation
Numerical results
Solution ofheat equation
Example:modelingcrystal growth
Conclusions
High frequency cut-off
10−1
100
101
102
103
0
0.5
1
1.5
L(dt = 0.01, ε= 0.001)
exp(
−ηγ t)
η
integrand as a function of t, α = 0.1
t=0.01t=0.1t=1t=10
(a) α = 0.1
10−1
100
101
102
103
0
0.5
1
1.5
L(dt = 0.01, ε= 0.001)
exp(
−ηγ t)
η
integrand as a function of t, α = 0.5
t=0.01t=0.1t=1t=10
(b) α = 0.5
10−1
100
101
102
103
0
0.5
1
1.5
L(dt = 0.01, ε= 0.001)
exp(
−ηγ t)
η
integrand as a function of t, α = 0.9
t=0.01t=0.1t=1t=10
(c) α = 0.9
10−4
10−3
10−2
10−1
100
100
101
102
103
L(∆
t, ε)
∆ t
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
100
100
101
102
103
L(∆
t, ε)
∆ t
High frequency cutoff L(∆ t,ε), α = 0.5
ε=0.0001ε=0.001ε=0.01ε=0.1
(e) α = 0.5
10−4
10−3
10−2
10−1
100
100
101
102
103
L(∆
t, ε)
∆ t
High frequency cutoff L(∆ t,ε), α = 0.9
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.
Convolutionintegrals
FractionalintegralsDiscretizing Laplaceintegralrepresentation
Numerical results
Solution ofheat equation
Example:modelingcrystal growth
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
j
∣∣∣∣∣∣ ≤ 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.
Convolutionintegrals
FractionalintegralsDiscretizing Laplaceintegralrepresentation
Numerical results
Solution ofheat equation
Example:modelingcrystal growth
Conclusions
Numerically verify quadrature for t ∈ [tmin, tmax ]
−0.5
0
0.5
1
1.5
η = a η = b
exp(
−ηγ t)
integrand in the dyadic interval [a,b=2a]
τmin
τmax
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.
Convolutionintegrals
FractionalintegralsDiscretizing Laplaceintegralrepresentation
Numerical results
Solution ofheat equation
Example:modelingcrystal growth
Conclusions
Tmax finite
Solve for largest amin = 2jmin : e−aγmin Tmax ≤ q.
e−ηγ Tmax is negligible outside [0, amin].
−0.5
0
0.5
1
1.5
η = 0 η = amin
exp(
−ηγ t)
integrand in the dyadic interval [0,amin
]
τmin
T
max
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.
Convolutionintegrals
FractionalintegralsDiscretizing Laplaceintegralrepresentation
Numerical results
Solution ofheat equation
Example:modelingcrystal growth
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
3 .
Convolutionintegrals
FractionalintegralsDiscretizing Laplaceintegralrepresentation
Numerical results
Solution ofheat equation
Example:modelingcrystal growth
Conclusions
Partition of the interval [0, L] into dyadic intervals
0
0.5
1
1.5
2jmin L
exp(
−ηγ t)
Partition of [0,L] into dyadic intervals
t=∆ t t=T
max
Figure: Partition of the interval [0, L] into dyadic intervals[2j , 2j+1]. On each interval, quadrature is verified numerically.
Convolutionintegrals
FractionalintegralsDiscretizing Laplaceintegralrepresentation
Numerical results
Solution ofheat equation
Example:modelingcrystal growth
Conclusions
Quadrature errors, α = 0.5
10−1
101
103
105
107
10−15
10−10
10−5
100
t = ∆ t t = Tmax
Err
or
t
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
100
102
L(dt,ε)
η
Quad nodes, , Q=64, dt=0.1, Tmax
=100, ε=1e−006, γ=2
(b) 64 Nodes
10−6
10−4
10−2
100
102
0
0.5
1
1.5
L(dt,ε)
exp(
−ηγ t)
η
Quad nodes, , Q=64, dt=0.1, Tmax
=100, ε=1e−006, γ=2
t=0.1t=1t=10t=100
(c) Integrands
10−1
101
103
105
107
10−15
10−10
10−5
100
t = ∆ t t = Tmax
Err
or
t
Quadrature error of ∫0∞ exp(−ηγ t), α = 0.5
ε=1e−006, ∆ t=0.1, Tmax
=1000, Q = 80
(d) Tmax = 1000
10−6
10−4
10−2
100
102
L(dt,ε)
η
Quad nodes, , Q=80, dt=0.1, Tmax
=1000, ε=1e−006, γ=2
(e) 80 Nodes
10−6
10−4
10−2
100
102
0
0.5
1
1.5
L(dt,ε)
exp(
−ηγ t)
η
Quad nodes, , Q=80, dt=0.1, Tmax
=1000, ε=1e−006, γ=2
t=0.1t=1t=10t=100
(f) Integrands
10−1
101
103
105
107
10−7
10−6
10−5
Err
or
t
Quadrature error of ∫0∞ exp(−ηγ t), α = 0.5
ε=1e−006, ∆ t=0.1, Tmax
=∞, Q = 176
(g) Tmax =∞
10−6
10−4
10−2
100
102
L(dt,ε)
η
Quad nodes, , Q=176, dt=0.1, Tmax
=∞, ε=1e−006, γ=2
(h) 176 Nodes
10−6
10−4
10−2
100
102
0
0.5
1
1.5
L(dt,ε)
exp(
−ηγ t)
η
Quad nodes, , Q=176, dt=0.1, Tmax
=∞, ε=1e−006, γ=2
t=0.1t=1t=10t=100
(i) Integrands
Convolutionintegrals
FractionalintegralsDiscretizing Laplaceintegralrepresentation
Numerical results
Solution ofheat equation
Example:modelingcrystal growth
Conclusions
Quadrature errors, α = 0.1
10−1
101
103
105
107
10−15
10−10
10−5
100
t = ∆ t t = Tmax
Err
or
t
Quadrature error of ∫0∞ exp(−ηγ t), α = 0.1
ε=1e−006, ∆ t=0.1, Tmax
=100, Q = 128
(j) Tmax = 100, Q = 128
10−1
101
103
105
107
10−15
10−10
10−5
100
t = ∆ t t = Tmax
Err
or
t
Quadrature error of ∫0∞ exp(−ηγ t), α = 0.1
ε=1e−006, ∆ t=0.1, Tmax
=1000, Q = 152
(k) Tmax = 1000, Q = 152
10−1
101
103
105
107
10−7
10−6
10−5
Err
or
t
Quadrature error of ∫0∞ exp(−ηγ t), α = 0.1
ε=1e−006, ∆ t=0.1, Tmax
=∞, Q = 212
(l) Tmax =∞, Q = 212
Convolutionintegrals
FractionalintegralsDiscretizing Laplaceintegralrepresentation
Numerical results
Solution ofheat equation
Example:modelingcrystal growth
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.
Convolutionintegrals
FractionalintegralsDiscretizing Laplaceintegralrepresentation
Numerical results
Solution ofheat equation
Example:modelingcrystal growth
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.
Convolutionintegrals
FractionalintegralsDiscretizing Laplaceintegralrepresentation
Numerical results
Solution ofheat equation
Example:modelingcrystal growth
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 ,∞].
Convolutionintegrals
FractionalintegralsDiscretizing Laplaceintegralrepresentation
Numerical results
Solution ofheat equation
Example:modelingcrystal growth
Conclusions
Topic II
Fast algorithm to solve heat equation
Convolutionintegrals
FractionalintegralsDiscretizing Laplaceintegralrepresentation
Numerical results
Solution ofheat equation
Example:modelingcrystal growth
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.
Convolutionintegrals
FractionalintegralsDiscretizing Laplaceintegralrepresentation
Numerical results
Solution ofheat equation
Example:modelingcrystal growth
Conclusions
Moving boundary
Γ(τ) y
x
τ
Figure: Boundary motion is prescribed
Convolutionintegrals
FractionalintegralsDiscretizing Laplaceintegralrepresentation
Numerical results
Solution ofheat equation
Example:modelingcrystal growth
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...
Convolutionintegrals
FractionalintegralsDiscretizing Laplaceintegralrepresentation
Numerical results
Solution ofheat equation
Example:modelingcrystal growth
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) =
t∫0
∫Ω(τ)
k (x− y, t − τ) f (y, τ) dy dτ,
DL(µ)(x, t) =
t∫0
∫Γ(τ)
∂
∂ny
(k(x− y, t − τ)
)µ(y, τ) dσy dτ.
Free space heat kernel: k (x, t) = e−‖x‖2/4t
(4πt)d2
.
Convolutionintegrals
FractionalintegralsDiscretizing Laplaceintegralrepresentation
Numerical results
Solution ofheat equation
Example:modelingcrystal growth
Conclusions
Differences from Topic I
Laplace transform representation
tα−1 =1
Γ(1− α)
∫ ∞
0e−ξ t ξ−α dξ.
Fourier transform representation
e−‖x‖2/4t
(4πt)d2
=
(1
2π
)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) =
(1
2π
)d ∫s∈Rd
e−is·x f (s) ds.
Convolutionintegrals
FractionalintegralsDiscretizing Laplaceintegralrepresentation
Numerical results
Solution ofheat equation
Example:modelingcrystal growth
Conclusions
Heat equation solver
To compute: Us(x, t) =t∫
0
∫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.
Convolutionintegrals
FractionalintegralsDiscretizing Laplaceintegralrepresentation
Numerical results
Solution ofheat equation
Example:modelingcrystal growth
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) =
(1
2π
)d ∫s∈Rd
e−is·x UsH(·, m) ds.
Similar to Topic I ∫ ∞
0e−ξ t ξ−α dξ
but now depends on x∫s∈Rd
e−is·x e−‖s‖2t ds.
Convolutionintegrals
FractionalintegralsDiscretizing Laplaceintegralrepresentation
Numerical results
Solution ofheat equation
Example:modelingcrystal growth
Conclusions
Heat equation solver
Suitable for linear, piecewise constant heat equation.Geometry can be complex. Domain can be unbounded.
No need for artificial boundary conditions.Far field behavior is automatically satisfied.
Treat moving boundary Γ(T ) naturally,part of spatial quadrature.
Accuracy of method is the accuracy of the time andspace quadratures.Can be high order accuracy. Automatically stable.
Convolutionintegrals
FractionalintegralsDiscretizing Laplaceintegralrepresentation
Numerical results
Solution ofheat equation
Example:modelingcrystal growth
Conclusions
CitationsBreaking heat kernel into history and local parts.Bounded domains. Greengard and Strain, 1990.
Unbounded domains: quadrature of Fourier integral.Greengard and Lin, 2000.
Non-equispaced FFT. Dutt and Rokhlin, 1993, Liu andNguygen, 1998, Greengard and Lee, 2004 ( provided code).
2D code. Application to modeling of crystal growth. J. L.
Improved Fourier quadrature. Extension to otherfractional kernels. J. L.
Local layer potential computations. Analysis andnumerics. J. L.
Convolutionintegrals
FractionalintegralsDiscretizing Laplaceintegralrepresentation
Numerical results
Solution ofheat equation
Example:modelingcrystal growth
Conclusions
Modeling crystal growth
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0 1 2 3 4 5 6 7 8 9 100
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9
10High undercooling (S = 0.5), time = 1.2
Convolutionintegrals
FractionalintegralsDiscretizing Laplaceintegralrepresentation
Numerical results
Solution ofheat equation
Example:modelingcrystal growth
Conclusions
Phase field modelphase φ = 0: bulk solid, φ = 1: bulk liquid.
Diffuse interface of mean thickness ε.temp u = T−TM
∆Tu = 0 : melting temp,
u = −1 : initial undercooled temp.
Liquid, T = Tm− Delta T
Solid, T = Tm
interface of width eps
Figure: Phase field model
Convolutionintegrals
FractionalintegralsDiscretizing Laplaceintegralrepresentation
Numerical results
Solution ofheat equation
Example:modelingcrystal growth
Conclusions
Phase field model
Ω = Rd .
ut −∇2u = −30φ2(1− φ)2
Sφt ,
ε2
mφt = φ(1− φ)(φ− 1
2[1 + AnN(x , y , t)])
+ 30φ2(1− φ)2εαSu
+ ε2[− ∂
∂x
(η(θ)η′(θ)
∂φ
∂y
)+
∂
∂y
(η(θ)η′(θ)
∂φ
∂x
)+∇ ·
(η2(θ)∇φ
)],
η(θ) = 1 + γ cos kθ.
Convolutionintegrals
FractionalintegralsDiscretizing Laplaceintegralrepresentation
Numerical results
Solution ofheat equation
Example:modelingcrystal growth
Conclusions
Undercooling
Dimensionless undercooling parameter
S =c ∆T
L,
c: specific heat per unit volume,L: latent heat per unit volume.
Typical numerical simulations reported in literature:0.05 ≤ S ≤ 1.Experiments: 0.001 ≤ S ≤ 0.1.
Convolutionintegrals
FractionalintegralsDiscretizing Laplaceintegralrepresentation
Numerical results
Solution ofheat equation
Example:modelingcrystal growth
Conclusions
Low undercooling
Crystals grow slowly. Extent of thermal field exceedsize of growing crystal by several orders of magnitude.
Nanometers (10−9) for the solid-liquid interfacemicrons (10−6) for the details of growth featuresmillimeters (10−3) or centimeters for thermal field.
Long simulation times.
Computationally challenging.Resolve details.Account for large thermal field.
Convolutionintegrals
FractionalintegralsDiscretizing Laplaceintegralrepresentation
Numerical results
Solution ofheat equation
Example:modelingcrystal growth
Conclusions
Standard approaches
Finite difference/elements discretization ofboth φ and u.
Finite computational domain Σ.
Simple artificial boundary conditions on ∂Σfor both φ and u.
Use Σ large enough to contain support of both φ and u.
Spatial adaptivity required to reduce the degrees offreedom in large Σ.
Convolutionintegrals
FractionalintegralsDiscretizing Laplaceintegralrepresentation
Numerical results
Solution ofheat equation
Example:modelingcrystal growth
Conclusions
Our approach
Standard finite difference discretization of φ.
New heat equation solver for u.
Computational domain Σ contains the support of φbut not u. Support of u is allowed to exit Σ.
Σ can be enlarged as solid-liquid interface grows,even after support of u exits.
All Fourier modes of u are stored.
Difficult to do with finite difference/elements methods.
Convolutionintegrals
FractionalintegralsDiscretizing Laplaceintegralrepresentation
Numerical results
Solution ofheat equation
Example:modelingcrystal growth
Conclusions
FRM(2)/SIE(1), S = 0.1, time = 0
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2
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0
u: colorbarφ: contours at 0.01, 0.1, 0.5, 0.99
(a) t=0
FRM(2)/SIE(1), S = 0.1, time = 0.25
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u: colorbarφ: contours at 0.01, 0.1, 0.5, 0.99
(b) t=0.25
FRM(2)/SIE(1), S = 0.1, time = 0.5
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u: colorbarφ: contours at 0.01, 0.1, 0.5, 0.99
(c) t=0.5
FRM(2)/SIE(1), S = 0.1, time = 1.5
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1
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u: colorbarφ: contours at 0.01, 0.1, 0.5, 0.99
(d) t=1.5
FRM(2)/SIE(1), S = 0.1, time = 2
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u: colorbarφ: contours at 0.01, 0.1, 0.5, 0.99
(e) t=2
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0
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0
1
2
3FRM(2)/SIE(1), S = 0.1, time = 25
u: colorbarφ: contours at 0.8, 0.85, 0.9, 0.99
(f) t=25
Figure: Heat diffuses out of the computational domain correctly,simulation can continue until solidification front reaches thecomputational boundary.
Convolutionintegrals
FractionalintegralsDiscretizing Laplaceintegralrepresentation
Numerical results
Solution ofheat equation
Example:modelingcrystal growth
Conclusions
Simulation
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0
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1
1.5FIT2/FD1, (S = 0.25), time = 0.48
u: colorbarφ: contours at 0.01, 0.1, 0.5, 0.99
Convolutionintegrals
FractionalintegralsDiscretizing Laplaceintegralrepresentation
Numerical results
Solution ofheat equation
Example:modelingcrystal growth
Conclusions
Conclusions
M: num of time steps, O(MQ) work, O(Q) storage.Q: number of modes in transform variable.
Schrodinger kernel?
References at http://www-rocq.inria.fr/~jli
Merci!