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

$: Fast computation of time convolutions: fractional …...Fractional order dynamical systems in control theory " Xn k=0 an−kD αn−k # y(t) = f(t) Fractional order laws of deformation$
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) =

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.



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)

∫ 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 .



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∑

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]



Numerical results



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).



Numerical results



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.



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)

∫ 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



Numerical results



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.



Numerical results



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.



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(t − τ)α−1 f (τ) dτ.



Numerical results



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 .



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

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


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


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

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 ,∞).



Numerical results



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


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


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


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


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


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

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.



Numerical results



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.



Numerical results



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.



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

3 .



Numerical results



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.



Numerical results



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


=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


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


=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


=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


ε=1e−006, ∆ t=0.1, Tmax

=∞, Q = 176

(g) Tmax =∞

10−6

10−4

10−2

100

102

L(dt,ε)

η


=∞, 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


=∞, ε=1e−006, γ=2

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

(i) Integrands



Numerical results



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


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


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


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

x

τ

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) =

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

.



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

=

(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.



Numerical results



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.



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) =

(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.



Numerical results



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.



Numerical results



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.



Numerical results



Conclusions

Modeling crystal growth

−1

−0.9

−0.8

−0.7

−0.6

−0.5

−0.4

−0.3

−0.2

−0.1

0 1 2 3 4 5 6 7 8 9 100

1

2

3

4

5

6

7

8

9

10High undercooling (S = 0.5), time = 1.2



Numerical results



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



Numerical results



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.



Numerical results



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.



Numerical results



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.



Numerical results



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.



Numerical results



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.



Numerical results



Conclusions

FRM(2)/SIE(1), S = 0.1, time = 0

−2 −1 0 1 2−2

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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

−2 −1 0 1 2−2

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(b) t=0.25

FRM(2)/SIE(1), S = 0.1, time = 0.5

−2 −1 0 1 2−2

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(c) t=0.5

FRM(2)/SIE(1), S = 0.1, time = 1.5

−2 −1 0 1 2−2

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(d) t=1.5

FRM(2)/SIE(1), S = 0.1, time = 2

−2 −1 0 1 2−2

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(e) t=2

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−3 −2 −1 0 1 2 3−3

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1

2

3FRM(2)/SIE(1), S = 0.1, time = 25


(f) t=25

Figure: Heat diffuses out of the computational domain correctly,simulation can continue until solidification front reaches thecomputational boundary.



Numerical results



Conclusions

Simulation

−1

−0.8

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−1.5 −1 −0.5 0 0.5 1 1.5−1.5

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1

1.5FIT2/FD1, (S = 0.25), time = 0.48




Numerical results



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!

http://www-rocq.inria.fr/~jli

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