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Numerical Solution of Fractional PDEs — Beijing Computational
Science Research Center, November 2015November 2015

Updated November 27, 2015

Introduction

This lecture provides some key definitions and results from fractional calculus, needed for our study of fractional PDEs. The literature contains several concepts of fractional differentiation, but we focus only on the Riemann–Liouville and Caputo definitions, with a brief mention of the Grunwald–Letnikov approach.

In the sequel, we work almost exclusively with the Riemann–Liouville fractional integral and derivative.

Outline

Fractional integration

Motivation: consider the n-fold integration operator Ina+ based at a, defined recursively by

I0 a+f (x) = f (x)

and

We claim

(n − 1)! f (y) dy for n ≥ 1.

The formula holds for n = 1 because (x − y)0/0! = 1 and

I1 a+f (x) =

Let n ≥ 1 and assume

Ina+f (x) =

Then

∫ x

∫ x

a

∫ z

a

=

=

and Γ(n + 1) = n! for any integer n ≥ 0.

For any real α > 0, we define the left-sided, Riemann–Liouville fractional integration operator of order α by

Iαa+f (x) =

This definition is consistent with our earlier definition of Ina+

when α = n.

⟩ = ⟨ f , Iαb−g

Iαb−g(x) =

Semigroup property

From the recursive definition, we see that

Ima+Ina+ = Im+n a+ for all integers m ≥ 0, n ≥ 0.

Key question: does

a+ for all α > 0 and β > 0?

Consider

∫ x

a

=

Putting t = (z − y)/(x − y), we have

z = y + t(x−y), x− z = (1− t)(x−y), z−y = t(x−y),

so the Beta function identity∫ 1

0 (1− t)α−1tβ−1 dt = B(α, β) =

Γ(α)Γ(β)

Γ(α)Γ(β)

∫ 1

= (x − y)α+β−1

Iα = Iα0+ for α > 0.

The fractional integral is given by the Laplace convolution

Iαf (x) =

0 Υα(x − y)f (y) dy = Υα ∗ f (x), x > 0.

We easily see that

In fact, since ∗ is associative,

(Υα ∗Υβ) ∗ f = Υα ∗ (Υβ ∗ f ) = Iα(Iβf ) = Iα+βf = Υα+β ∗ f

for every continuous f .

generalising the identity

Iαa+Υβ,a(x) =

=

= (Υα ∗Υβ)(x − a) = Υα+β(x − a),

or in other words,

Fractional differentiation Assume that

n − 1 < α ≤ n for some n ∈ {1, 2, 3, . . .},

and write Dn = (d/dx)n.

Dαa+f (x) = DnIn−αa+ f (x) for x > a.

whereas the Caputo fractional derivative is defined by

CDαa+f (x) = Iα−na+ Dnf (x) for x > a.

Lemma For x > a and β > 0,(

DIβa+ − I β a+D

) f (x) = f (a)Υβ(x − a).

Proof

By the fundamental theorem of calculus,

Ia+Df (x) =

so f (x) = Ia+Df (x) + f (a)Υ1(x − a),

Thus, Iβa+f (x) = Iβ+1

a+ Df (x) + f (a)Υβ+1(x − a),

and finally

DIβa+f (x) = DIa+ Iβa+Df (x) + f (a)Υβ(x − a).

Relation between Dα and CDα

Theorem If n − 1 < α < n, then

Dαa+f (x) = CDαa+f (x) + n−1∑ k=0

Dk f (a) (x − a)k−α

Γ(k + 1− α) , x > a.

Proof. In the case n = 2, we have 1 < α < 2 and the Lemma gives

Dαa+f (x) = D2I2−α a+ f (x) = D

( I2−α a+ Df (x) + f (a)Υ2−α(x − a)

) = I2−α

a+ D2f (x) +Df (a)Υ2−α(x − a) + f (a)Υ1−α(x − a)

= CDαa+f (x) + f (a) (x − a)−α

Γ(1− α) +Df (a)

Differentiating a shifted Gel’fand–Shilov function

Lemma If α > 0, β > 0 and x > a, then

Dαa+Υβ,a(x) = Υβ−α,a(x).

Dαa+Υβ,a(x) = DnIn−αa+ Υβ,a(x) = DnΥn−α+β,a(x) = Υβ−α,a(x).

In particular, since Υ1,a(x) ≡ 1, if x > a then

Dαa+1(x) = Υ1−α,a(x) = (x − a)−α

Γ(1− α) whereas CDαa+1(x) = 0.

Relation between Dα and CDα restated

Since (x − a)k−α

the relation

Dαa+f (x) = CDαa+f (x) + n−1∑ k=0

Dk f (a) (x − a)k−α

Γ(k + 1− α) , x > a,

may be re-stated in the form

CDαa+f (x) = Dαa+

k!

Alternative representation

Dαa+f (x) = f (x)Υ1−α(x−a)+

∫ x

∫ x

] dy ,

noting that the derivative of the integral on the right is∫ x

a Υ−α(x − y)

[ f (y)− f (x)

Representation as a Hadamard finite-part integral

Assume 0 < α < 1 and x > a. Then∫ x

a Υ−α(x − y)

[ f (y)− f (x)

[ Υ1−α(ε)−Υ1−α(x − a)

] ,

Γ(1− α) +

∫ x−ε

and therefore

Dαa+f (x) = “I−αa+ f (x)” = fp ε↓0

∫ x−ε

Grunwald–Letnikov definition Can we define a fractional derivative (or integral) directly, without using integer-order derivatives and integrals? Denote the backward difference by

hf (x) = f (x)− f (x − h).

Can check by induction on k that

k hf (x) =

k∑ j=0

Hence define the fractional backward difference of order α by

α h,nf (x) =

n∑ j=0

k hf (x) =

and thus

k hf (x)

GLDαa+f (x) = lim α

h,nf (x)

hα ,

where the limit is obtained by sending n→∞ and h→ 0+ keeping

h = x − a

Can show that if m − 1 < α < m, then

GLDαa+f (x) = m−1∑ k=0

f (k)(a) (x − a)k−α

Γ(k + 1− α)

which means that

GLDαa+f (x) = CDαa+f (x) + m−1∑ k=0

f (k)(a) (x − a)k−α

Γ(k + 1− α) = Dαa+f (x).

Furthermore,

Part II

Useful tools

Introduction

We will make extensive use of the Laplace transform (and some use of the Fourier transform), first to derive the fractional diffusion equation and then to study properties of the solution. Laplace transformation also plays a large role in some of the numerical methods we study, either as part of the method itself or for the error analysis.

This lecture also introduces some special functions that are arise in the study of fractional initial-boundary value problems.

Outline

If f is locally integrable on [0,∞), and if

|f (t)| ≤ Ceλt for t > 0,

then f (z) exists and is analytic for <z > λ, and we have the inversion formula

f (t) = 1

Transform of a Gel’fand–Shilov function

For α > 0 and z > 0, the substitution y = tz gives

Υα(z) = 1

Γ(α)

∫ ∞ 0

consistent with

z−α−β = Υα+β(z) = L(Υα ∗Υβ) = Υα(z)Υβ(z) = z−αz−β.

Laplace transform of an integral

Since

we have L{If (t)} = Υ1(z)f (z) = z−1f (z).

In general, Inf = Υn ∗ f so

L{Inf (t)} = z−n f (z) for n ∈ {0, 1, 2, . . .}.

Laplace transform of a derivative

Integration by parts shows

= 0− f (0) + z

so L{Df (t)} = zf (z)− f (0).

Easily verify by induction on n that

L{Dnf (t)} = zn f (z)− n−1∑ k=0

zn−1−kDk f (0).

Laplace transform of a Caputo fractional derivative

If n − 1 < α < n, then

CDαf (t) = In−αg(t) where g(t) = Dnf (t),

so

= zα−n (

zn−1−kDk f (0)

) and so

zα−1−kDk f (0).

Laplace transform of a Riemann–Liouville fractional derivative

If n − 1 < α < n, then

Dαf (t) = CDαf (t) + n−1∑ k=0

Dk f (0)Υk+1−α(t),

and since Υk+1−α(z) = z−(k+1−α) we have

L{Dαf (t)} = L{CDαf (t)}+ n−1∑ k=0

Dk f (0)zα−1−k ,

that is, L{Dαf (t)} = zαf (z).

Mittag–Leffler function

Problem: find f (t) satisfying

CDαf (t) = f (t) for t > 0, with f (0) = 1.

If α = 1 then f (t) = et .

If 0 < α < 1, then we claim

f (t) = ∞∑ k=0

Υ1+kα(t) = ∞∑ k=0

where the Mittag–Leffler function is

Eα(z) = ∞∑ k=0

CDαΥ1+kα = I1−αDΥ1+kα = I1−αΥkα = Υ1+(k−1)α,

so

) = 0 + Υ1 + Υ1+α + Υ2+α + · · · = f .

Also, Υ1(t) ≡ 1 and Υ1+kα(0) = 0 for k ≥ 1,

so f (0) = 1.

Convergence?

We claim Eα(z) is an entire function of z . By the ratio test, it suffices to show that as k →∞, zk+1

Γ ( 1 + (k + 1)α

Γ(x) =

√ 2π

x

( x

e

Special choices of α

∫ ∞ −x

CDαu + λu = 0 for t > 0, with u(0) = 1.

We claim that the solution is

u(t) = Eα(−λtα) = ∞∑ k=0

(−λ)ktkα

Equivalent formulation

I1−αDu(t) + λu(t) = 0.

To obtain an equation involving a Riemann–Liouville fractional derivative, apply DIα and obtain

DI1Du(t) + λDIαu(t) = 0.

Du(t) + λD1−αu(t) = 0.

Taking Laplace transforms, zu(z)− u(0) + λz1−αu(z) = 0, and thus (z + λz1−α)u(z) = 1, showing that

u(z) = L{Eα(−λtα)} = 1

z + λz1−α .

Integral representation

1

Γ(a) =

1

2πi

∫ 0+

encircles the negative real axis and has a counterclockwise orientation.

Theorem The Mittag–Leffler function admits the integral representation

Eα(z) = 1

provided the Hankel contour encloses the disc |w | ≤ |z |1/α.

Proof

= 1

2πi

∫ 0+

Eα(−λtα) = 1

z + λz1−α , t > 0.

By collapsing the Hankel contour onto the negative real axis, we find

Eα(−λtα) = 1

)2 + λ2 sin2 απ

dt Eα(−λtα) < 0 for all t > 0.

Asymptotic behaviour

1 + zα =

Eα(−tα) ∼ ∞∑ n=0

Notice what happens as α→ 1.

Wright functions

Wλ,µ(z) = ∞∑ k=0

k!Γ(λk + µ) , λ > −1, µ ∈ C.

This series converges for all z ∈ C so Wλ,µ is an entire function.

Theorem The Wright function has the integral representation

Wλ,µ(z) = 1

Mα(t) = W−α,1−α(−t) = ∞∑ k=0

(−1)ktk

k!Γ ( 1− (k + 1)α

) , 0 < α < 1, introduced by F. Mainardi in 1994. The identity

1

M1/2(t) = exp(−t2/4)√

Wright M-function Mα(t)

Integral representation and Laplace transform

Putting λ = −α and µ = 1− α in the integral representation of the Wright function, and replacing t by −t, gives

Mα(t) = 1

which allows us to prove the following result.

Theorem For 0 < α < 1, the Laplace transform of Mα is

Mα(z) = Eα(−z).

= 1

2πi

∫ 0+

∫ ∞ −∞

e−iξx f (x) dx .

If f ∈ L1(R) then f is continuous on R and f (ξ)→ 0 as |ξ| → ∞.

Plancherel theorem: Fourier transform extends uniquely to a bounded linear operator F : L2(R)→ L2(R) satisfying

1

2π

Symmetric Wright M-function

Lemma For κ > −1 and 0 < α < 1,∫ ∞

0 tκMα(t) dt =

Proof of the lemma

so∫ ∞ 0

Part III

Anomalous diffusion

The classical diffusion equation,

ut − K∇2u = 0,

describes how the concentration u of a substance evolves over time if, at the microscopic scale, its particles exhibit Brownian motion. The equation can also be derived from a purely macroscopic argument based on conservation of mass and Fick’s law, which states that the mass flux vector equals −K∇u.

In this lecture, we study continuous-time random walks, which provide a generalization of Brownian motion. When the waiting-time distribution obeys a power law, the particles experience trapping and their macroscopic behaviour is said to be subdiffusive. The mean-square displacement of such a particle is proportional to tα for a characteristic exponent in the range 0 < α < 1.

With the help of Fourier and Laplace transformation, we show that the macroscopic concentration obeys a time-fractional PDE,

ut − ∂1−α t Kα∇2u = 0,

where it is convenient to write ∂1−α t = D1−α

0+ for the Riemann–Liouville fractional derivative with respect to t. The constant Kα > 0 is a generalized diffusivity, and the classical diffusion equation then arises as the limiting special case when α→ 1.

Outline

Continuous-time random walks

A walker moves along the x-axis, starting at position x0 at time t0 = 0. At time t1 the walker jumps to x1, then at time t2

jumps to x2, and so on. We assume that the increments

tn = tn − tn−1 and xn = xn − xn−1

are independent, identically distributed random variables with probability density functions ψ(t) and λ(x), respectively. That is,

P(a < tn < b) =

and

a λ(x) dx for −∞ < a < b <∞.

Aim: determine the probability that the particle lies in a given spatial interval at time t.

An example

Suppose that the waiting-time distribution is exponential with parameter τ > 0,

ψ(t) = τ−1e−t/τ for 0 < t <∞,

and that the jump-length distribution is normal with mean 0 and variance σ2,

λ(x) = 1

Thus,

E(tn) = τ, E(t2 n) = τ2, E(xn) = 0, E(x2

n ) = σ2.

The position x(t) of the walker is a step function.

Typical path (τ = 1, σ = 1, x0 = 2)

Random walk in 3D (σx = σy = σz = 1)

Convolutions and probability

We denote the Fourier convolution of f and g by

f ~ g(z) =

f (z − y)g(y) dy , for −∞ < z <∞.

If f (x) = 0 for x < 0 and g(y) = 0 for y < 0 then this integral equals the Laplace convolution

f ∗ g(z) =

0 f (z − y)g(y) dy for z > 0.

Theorem If X and Y are independent random variables with probability density functions f and g, respectively, then the sum Z = X + Y has probability density function f ~ g.

Probability distribution of tn

Let ψn(t) denote the probability density function of the random variable

tn = t1 + t2 + · · ·+ tn,

By the theorem quoted above,

ψn(t) = (ψn−1 ∗ ψ)(t) =

n factors

Survival probability

Let Ψ(t) denote the survival probability, that is, the probability of the walker not jumping within a time t, or equivalently, the probability of remaining stationary for at least a duration t. Then,

Ψ(t) =

∫ ∞ t

0 ψ(s) ds for 0 < t <∞.

It follows that the probability of taking exactly n steps up to time t is

χn(t) =

∫ t

Probability distribution of xn − x0

Let λn(x) denote the probability density function of the random variable

xn − x0 = x1 + x2 + · · ·+ xn,

that is,

Since

n factors

ψ(z) = Lψ(z) =

whereas the characteristic function of λ(x) is its Fourier transform,

λ(ξ) = Fλ(ξ) =

e−iξxλ(x) dx .

Since ψn and λn are n-fold convolutions of ψ and λ, respectively,

ψn(z) = ψ(z)n and λn(ξ) = λ(ξ)n for n ≥ 0.

The characteristic function for the survival probability Ψ = 1− I1ψ is

Ψ(z) = z−1 − z−1ψ(z) = 1− ψ(z)

z .

Probability density

Let p(x , t) denote the probability density function for the position of the particle at time t, that is,

P(a < x(t)− x0 < b) =

a p(x , t) dx .

Since χn(t) is the probability of taking n steps up to time t,

p(x , t) = ∞∑ n=0

z .

ˆp(ξ, z) = LFp(ξ, z) =

ˆp(ξ, z) = ∞∑ n=0

λ(0) =

But if ξ 6= 0 or z > 0, then λ(ξ)ψ(z)

< 1 so

∞∑ n=0

σ √

2ξ2/2,

so

1 + τz ,

Uncertain initial position

Instead of assuming x(0) = x0 is known, we can treat x0 as a random variable with probability density p0(x), so that

P(a < x0 < b) =

a p0(x) dx for −∞ < a < b <∞.

Let λn denote the probability density function for xn (rather than xn − x0, as before), so

P(a < xn < b) =

Since xn = x0 + x1 + · · ·+ xn,

we have λn = p0 ~ λ~ λ~ · · ·~ λ

n factors

Using [λ(ξ)]np0(ξ)

ˆp(ξ, z) = 1− ψ(z)

1− λ(ξ)ψ(z) ,

that is, the only change is to introduce a factor p0(ξ).

Formally, put p0(x) = δ(x − x0) to recover the case when x0 is known with certainty.

Rescaling

Assume now that the probability density functions ψ(t) and λ(x) are normalized to satisfy∫ ∞

0 tψ(t) dt = 1,

x2λ(x) dx = 1.

Let τ > 0 and σ > 0, and let the random variables tn and xn now have the rescaled probability density functions

ψτ (t) = 1

E(tn) = τ, E(xn) = 0, E(x2 n ) = σ2.

We want to investigate what happens as τ and σ tend to zero.

Typical path (σ = 0.1, τ = 0.005)

Detailed view of inset

dk ψ

we have

z

1

ψ(z) = ψ(0) + ψ′(0)z + · · · = 1− z + O(z2) as z → 0,

implies that

1− ψ(τz)

) as τ → 0.

Assume for simplicity that λ(−x) = λ(x). Then λ′′′(0) = 0 and

λ(ξ) = λ(0) + λ′(0)ξ + 1 2 λ ′′(0)ξ2 + · · · = 1− 1

2ξ 2 + O(ξ4).

τz + 1 2σ

2ξ2 × 1 + O(τz)

1 + O(τz + σ2ξ2) .

Limiting probability density Now send σ → 0 and τ → 0 while keeping

σ2

ˆp(ξ, z) = lim τ

p(ξ, t) = 1

p(x , t) = 1√

Notice that

= (z + Kξ2)ˆp(ξ, z)− p(ξ, 0)

= 1− p(ξ, 0) = 0,

where the final step follows because p(x , 0) = δ(x) and so p(ξ, 0) = 1.

Therefore, p(x , t), the probability density for x(t)− x0, the position (relative to x0) of the walker at time t, satisfies the partial differential equation

pt − Kpxx = 0 for 0 < t <∞ and −∞ < x <∞.

Uncertain initial position

τz + 1 2σ

2ξ2 × 1 + O(τz)

1 + O(τz + σ2ξ2) ,

ˆp(ξ, z) = p0(ξ)

= (z + Kξ2)ˆp(ξ, z)− p(ξ, 0)

= p0(ξ)− p(ξ, 0) = 0,

since p(x , 0) = p0(x).

Subdiffusion

Let 0 < α < 1, and suppose now that the waiting time probability density function is a power law:

ψ(t) ∼ A

tψ(t) dt = +∞,

so the preceding analysis of the random walk breaks down.

We make no change to our assumptions on λ(x).

Example

1

σ √

Typical path (α = 0.75, σ = 1)

Behaviour of the characteristic function as z → 0 Assume there exists T > 0 such that

|t1+αψ(t)− A| ≤ Ct−1 for T ≤ t <∞,

and let 0 < z ≤ T−1 (so Tz ≤ 1). Since we know ψ(0) = 1, consider

1− ψ(z) =

where

I1 =

∫ T

I2 =

∫ ∞ T

I3 =

∫ ∞ T

(1− e−zt)At−1−α dt.

Since 0 ≤ 1− e−y ≤ min(1, y) for 0 ≤ y ≤ 1,

we immediately see that

|I2| ≤ ∫ ∞ T

∫ ∞ Tz

≤ Cz1+α

∫ ∞ 1

I3 =

∫ ∞ T

1− e−y

where

∫ Tz

0

AT−αz

1− α ,

we have shown that1− ψ(z)− Bαzα = |I1 + I2 + I4| ≤ CT ,αz for 0 < z ≤ T−1.

Integrating by parts,∫ ∞ 0

) =

which completes the proof of the following result.

Theorem If

then, with Bα = Aα−1Γ(1− α),

ψ(z) = 1− Bαzα + O(z) as z → 0.

Rescaling

xλ(x) dx = 0 and

ψτ (t) = 1

ψτ (t) ∼ Aτα

As before,

z

1

but this time

where Bα = Aα−1Γ(1− α).

Thus,

] .

so

= [ Bατ

Limiting probability density

αzα−1

1 + O ( τ1−αz1−α + ταzα + σξ

) . Once again, send σ → 0 and τ → 0, but now keep

σ2

ˆp(ξ, z) = lim Bατ

zα−1

zα + Kαξ2 .

Notice that we recover the earlier formula by putting α = 1.

Typical path (α = 0.75, σ = 0.1, τα = σ2/2, N = 800)

Detailed view of inset

z + λz1−α

Since

we see that

Kαtα.

Thus,

Fractional partial differential equation for p

We have

LF{pt − KαD1−αpxx} = z ˆp(ξ, z)− p(ξ, 0) + Kαz1−αξ2ˆp(ξ, z)

= ( z + Kαz1−αξ2

since

Thus, p satisfies the time-fractional diffusion equation,

pt − KαD1−αpxx = 0 for 0 < t <∞ and −∞ < ξ <∞.

Mean-square displacement

) =

∫ ∞ −∞

Since

E ( (x(t)− x0)2

Introduction

We have shown that in 1D the probability density function for the location of a subdiffusive particle at time t obeys a time-fractional PDE. The concentration u = u(x , t) of a large number of such particles evolves in the same way. Moreover, the 1D analysis can be generalized to higher dimensions to yield the time-fractional diffusion equation

ut − Kα∂ 1−α t ∇2u = 0.

As in the classical case α = 1, the solution u in a bounded spatial domain (and subject to homogeneous boundary conditions) can be constructed by separation of variables to yield a series expansion involving the eigenfunctions of −∇2. We use this representation of u(x , t) to investigate its behaviour.

Outline

Positivity

Initial-boundary value problem

Let denote a bounded, Lipschitz domain in R2 or R3. We seek u = u(x , t) satisfying

ut − Kα∂ 1−α t ∇2u = f (x , t) for x ∈ and t > 0,

u = u0(x) for x ∈ , when t = 0,

and impose homogeneous boundary conditions, either Dirichlet

u = 0 for x ∈ ∂ and t > 0,

or else Neumann,

where n is the outward unit normal to .

Abstract initial-value problem

Let A be a linear operator with dense domain D(A) in a Hilbert space H with inner product u, v and norm u =

√ u, u. Given

u0 ∈ H and f : [0,∞)→ H we seek u : [0,∞)→ H satisfying

u + ∂1−α t Au = f (t) for t > 0,

with u(0) = u0, where u = ut = ∂u/∂t.

Standard example:

H = L2(), Au = −Kα∇2u, D(A) = H2() ∩ H1 0 ().

Integrate to obtain an equivalent Volterra equation in H,

u(t) +

∫ t

∫ t

Eigenfunction expansion Assume that A is self-adjoint and positive-semidefinite, with a complete orthonormal eigensystem, say

Aφm = λmφm for m = 0, 1, 2, . . . ,

with φm, φn = δmn. Number the eigenvalues so that

0 ≤ λ0 ≤ λ1 ≤ λ2 ≤ · · · .

u(t) = ∞∑

um(t)φm where um(t) = u(t), φm.

Likewise, putting fm(t) = f (t), φm and u0m = u0, φm, we have

f (t) = ∞∑

m=0

Dirichlet boundary conditions in 1D

Take = (0, L) and A = −Kαd2/dx2 with homogeneous Dirichlet boundary conditions. Then,

φm(x) =

√ 2

( mπ

L

)2

v(x) = ∞∑

vm = v , φm =

Neumann boundary conditions in 1D

Again take = (0, L) and A = −Kαd2/dx2, but now impose homogeneous Neumann boundary conditions. Then,

φ0(x) = 1√ L

and λ0 = 0,

( mπ

L

)2

Separation of variables

The function u satisfies

u + ∂1−α t Au = f (t) for t > 0, with u(0) = u0.

iff the mth eigenmode satisfies

um + λm∂ 1−α t um = fm(t) for t > 0, with um(0) = u0m,

for m = 0, 1, 2, . . . . Laplace transformation gives

zum(z)− um(0) + λmz1−αum(z) = fm(z)

so

Recall that

E(t)v = ∞∑

m=0

Eα(−λmtα)v , φmφm for t > 0 and v ∈ H,

then the mild solution of the abstract initial-value problem is

u(t) = E(t)u0 +

Caution: E(t + s) 6= E(t)E(s) if 0 < α < 1.

Stability in H Recall that Eα(−λtα) is positive and decreasing for t > 0, and equals 1 at t = 0, so

0 < Eα(−λtα) ≤ 1 for 0 ≤ t <∞ and any λ ≥ 0.

Thus, using Parseval’s identity,

E(t)v2 = ∞∑

m=0

m=0

and therefore

u(t) ≤ u0+

Smoothing property of fractional diffusion

For 0 ≤ r <∞, define the norm

v2 r =

(I + A)r/2v2 = ∞∑

(1 + λm)r v , φm2

and the corresponding closed subspace

Hr = { v ∈ H : vr <∞},

which is a Hilbert space with respect to the inner product that induces · r .

For our standard example H = L2() and A = −Kα∇2, write Hr = Hr

D() or Hr N() to indicate the choice of Dirichlet or

Neumann boundary conditions.

Can prove the following via interpolation and elliptic regularity.

Theorem Suppose that ∂ is C∞. If 0 ≤ r < 1

2 , then Hr D() = H r (),

however if 2j − 3 2 < r < 2j + 1

2 for j ∈ {1, 2, 3, . . .}, then

Hr D() = { v ∈ H r () : v = Av = · · · = Aj−1v = 0 on ∂ }.

In the exceptional case r = 2j − 3 2 , the condition Aj−1v = 0 on ∂

must be replaced by Aj−1v ∈ H1/2().

If is Lipschitz, then the conclusions still hold for r ≤ 1, and in particular H1

D = H1 0 () = { u ∈ H1() : u = 0 on ∂ }.

If is convex or C 1,1, then r ≤ 2 is OK, and in particular H2

D = H2() ∩ H1 0 () = { u ∈ H2() : u = 0 on ∂ }.

Neumann boundary conditions and Sobolev spaces

Theorem Suppose that ∂ is C∞. If 0 ≤ r < 3

2 , then Hr N() = H r (),

however if 2j − 1 2 < r < 2j + 3

2 for j ∈ {1, 2, 3, . . .}, then

Hr N() = { v ∈ H r () :

∂nv = ∂nAv = · · · = ∂nAj−1v = 0 on ∂ }.

In the exceptional case r = 2j − 1 2 , the condition ∂nAj−1v = 0

on ∂ must be replaced by ∂nAj−1v ∈ H1/2().

If is Lipschitz, then the conclusions still hold if r ≤ 1, and in particular H1

N() = H1().

If is convex or C 1,1, then r ≤ 2 is OK, and in particular H2

N() = { u ∈ H2() : ∂nu = 0 on ∂ }.

A 1D example

Consider v(x) = 1 for x ∈ = (0, L). Since∫ L

0 v(x) sin

vr <∞ ⇐⇒ ∞∑ p=0

(1 + 2p)2r−2 <∞,

so v ∈ Hr D() iff 2r − 2 < −1, that is, r < 1

2 . However, v ∈ Hr

0 v(x) cos

Smoothing property of classical diffusion

If α = 1 then Eα(−tα) = e−t so

E(t)v = ∞∑

and thus

)2 .

(1 + λ)µ(e−λt)2 ≤ t−µ(T + λt)µe−2λt ≤ CT ,µt−µ

and so E(t)vr+µ ≤ CT ,µt−µ/2vr for µ ≥ 0.

Weaker smoothing property for subdiffusion

The theorem below shows that if v ∈ Hr then E(t)v ∈ Hr+2 for each t > 0, but E(t)vr+2 may blow up as t → 0+.

Lemma 0 < Eα(−tα) ≤ C min(1, t−α) for 0 < t <∞.

Proof. Follows because

t−α/Γ(1− α) + O(t2−α) as t →∞.

Theorem Let 0 ≤ µ ≤ 2 and 0 ≤ r <∞. If v ∈ Hr , then

E(t)vr+µ ≤ CT t−αµ/2vr for 0 < t ≤ T .

Proof of theorem Put g(t) = Eα(−tα). Since g(λ1/αt) = E (−λtα), we have

E(t)v2 r+µ =

The lemma implies that (assuming 0 ≤ µ ≤ 2)

0 < g(t) ≤ C (1 + tα)−µ/2 for 0 < t <∞,

so, for 0 < t ≤ T ,

g(λ1/αt)2 ≤ C (1 + λtα)−µ = Ct−µα(t−α + λ)−µ

≤ CT t−µα(1 + λ)−µ

and thus

∞∑ m=0

(1 + λm)r v , φm2 = CT t−αµv2 r .

Regularity in time

Let q ∈ {1, 2, 3, . . .}. Similar arguments yield the following estimates.

Lemma The function g(t) = Eα(−tα) satisfies

tq|g (q)(t)| ≤ Cq min(tα, t−α) for 0 < t <∞.

Theorem Let −2 ≤ µ ≤ 2, 0 ≤ r <∞ and q ∈ {1, 2, 3, . . .}. If v ∈ Hr , then

tqE(q)(t)vr+µ ≤ Cq,T t−αµ/2vr for 0 < t ≤ T .

Detailed behaviour as t → 0+

Since

(−1)ptαp

Γ(1 + αp) λp + O(λMtαM) as t → 0+,

and λpmv , φm = Apv , φm, we can show the following.

Theorem Let 0 ≤ r <∞ and M ∈ {1, 2, 3, . . .}. If v ∈ Hr+2M , then

E(t)v = v + M−1∑ p=1

(−1)ptαp

where, given 0 ≤ µ ≤ 2, the remainder operator satisfies

RM(t)vr+µ ≤ CM,T tMα−αµ/2vr for 0 < t ≤ T .

Behaviour of an eigenmode

If the initial data is an eigenfunction of A, say u0 = φm, then the solution of the homogeneous problem is

u(t) = E(t)φm = Eα(−λmtα)φm,

) φm(x) as t → 0+.

This example makes clear the fact that the time derivative u = O(tα−1) is unbounded as t → 0+ no matter how regular the initial data (so long as it is not zero, but in that case u ≡ 0).

Contrast this behaviour with that of the classical diffusion equation (α = 1): if u0 = φm then u(x , t) = e−λmtφm(x) is C∞ for t ≥ 0.

The inhomogeneous problem The function u(t) = E(t)u0 solves the homogeneous problem

u + ∂1−α t Au = 0 for t > 0, with u(0) = u0.

Now consider

which solves the inhomogeneous problem with vanishing initial data:

u + ∂1−α t Au = f (t) for t > 0, with u(0) = 0.

Theorem For 0 ≤ r <∞ and q ∈ {0, 1, 2, . . .}, the function u = E ∗ f satisfies

tqu(q)(t)r ≤ Cq

0 s jf (j)(s)r ds for 0 < t <∞.

Positivity

{ 0, t < 0,

u(t), t ≥ 0, ,

then the Laplace transform of u is related to the Fourier transform of u+ by

u(iy) = u+(y) =

f (t)g(t) dt = 1

u(iy)v(iy) dy .

We can now show that the operator ∂1−α t is positive semidefinite.

Theorem If 0 < β < 1 and if u is real-valued, then∫ ∞

0 (∂βt u)u dt =

Proof. Since L{∂βt u(t)} = zβ u(z),∫ ∞

0 (∂βt u)u dt =

(iy)β|u(iy)|2 dy .

The result follows because u(iy) = u(−iy) and for y > 0,

(±iy)β = (e±iπ/2y)β = yβ ( cos 1

2πβ ± i sin 1 2πβ

) .

Lemma For 0 < α < 1 and suitable u : (0,∞)→ H,∫ ∞

0 ∂1−α

2πα

π

∫ ∞ 0

Proof.

= ∞∑

Digression: fractional derivative at a jump discontinuity

Suppose that

v2(t), t > a,

where v1 : [0, a]→ R and v2 : [a,∞)→ R are C 1 functions. If 0 < t < a, then differentiating the formula

Iαv(t) = Υα ∗ v(t) =

∫ t

= v(0+)Υα(t) +

Iαv(t) =

∫ a

∫ t−a

+

∫ t−a

and therefore, with [v ]a = v2(a)− v1(a) = v(a+)− v(a−),

∂1−αv(t) = v(0+) Υα(t) + [v ]aΥα(t − a)

+

∫ t

Example

Consider the homogeneous equation,

u(t) + ∂1−α t Au(t) = 0.

Take the inner product with u(t) and integrate to obtain∫ T

0 u, u dt +

Letting

Thus, ∫ T

But ∫ T

2u(0)2,

E(t)u0 ≤ u0 for t > 0.

Part V

Introduction

We begin our study of numerical methods for fractional diffusion problems by considering simple explicit and implicit finite difference (and quadrature) schemes in the 1D case:

ut − Kα∂ 1−α t uxx = f (x , t).

These schemes generalize the forward and backward Euler methods for the heat equation. As in the classical setting, the explicit scheme is stable only if the time step is sufficiently small, but the implicit scheme is unconditionally stable.

Unlike the classical Euler methods, the grid stencils extend back through all preceding time levels resulting in a dramatically higher computational cost.

Outline

Implicit Euler method

Consider first the scalar problem (A = λ > 0)

u + λ∂1−α t u = f (t) for 0 < t < T , with u(0) = u0,

where λ > 0. Put t = T/N and define grid points

tn = nt for 0 ≤ n ≤ N.

We want to compute Un ≈ u(tn).

Time-stepping

∫ tn+1

Un+1 − Un + λ

∂1−α t U(t) dt = f n t,

where f n = f (tn) and U is the piecewise-constant function

U(t) = Un for tn ≤ t < tn+1.

Approximation of the fractional derivative

Recalling ∂1−α t v = (Υα ∗ v)t , we have∫ tn+1

tn

=

] U j dt

Γ(α + 1)

tα

] U j dt.

wj = (j + 1)α − 2jα + (j − 1)α for j ≥ 1.

Note that wj < 0 for j ≥ 1, with wj ≈ α(α− 1)jα−2 for large j .

Weights when α = 1/2

Un+1 − Un + λtα

Γ(α + 1)

Thus, starting from U0 = u0 we compute

Un+1 = Un + f n t − λtα

Γ(α + 1)

Classical Euler method

In the limiting case α = 1 we have wj = 0 for j ≥ 1 so

λtα

Γ(α + 1)

and therefore

or equivalently, Un+1 − Un

t + λUn = f n.

Un+1 =

Γ(α + 1)

wn−jU j ,

and we can prove the following discrete analogue of the stability estimate for the continuous problem:

|u(t)| ≤ |u0|+ ∫ t

Theorem If

λtα

|f n|t for 1 ≤ n ≤ N.

Proof

|Un+1| ≤ (1− ρ)|Un|+ |f n|t + ρ

n−1∑ j=0

|wn−j ||U j |

so

|Un+1| ≤ (1− ρ)|Un|+ |f n|t + ρ max 0≤j≤n−1

|U j |

|U j |.

The desired estimate now follows using induction on n.

Example: α = 1/2, λ = 1, f ≡ 0, N = 25, ρ = 0.3192

Example: α = 1/2, λ = 4, f ≡ 0, N = 25, ρ = 1.2766

Explicit Euler method for a PDE

Consider = (0, L) in 1D with boundary ∂ = {0, L}. We seek u = u(x , t) satisfying

ut − Kα∂ 1−α t uxx = f (x , t) for x ∈ and 0 < t < T ,

u = u0(x) for x ∈ , when t = 0,

u = 0 for x ∈ ∂ and 0 < t < T .

Put x = L/P and t = T/N, and define grid points

(xp, tn) = (p x , n t) for 0 ≤ p ≤ P and 0 ≤ n ≤ N.

We want to compute Un p ≈ u(xp, tn).

Second central difference in space Using the approximation

uxx(xp, tn) ≈ Un p+1 − 2Un

p + Un p−1

x2

and letting f n p = f (xp, tn), we discretize in time as before and

arrive at the scheme

Un+1 p − Un

p + U j p−1

x2 = f n

p t

for 0 ≤ n ≤ N − 1 and 1 ≤ p ≤ P − 1, with the initial conditions

U0 p = u0(xp) for 0 ≤ p ≤ P,

and boundary conditions

Stencil

x

t

. . . . . .

. . .

so that

Γ(α + 1)

( mπ

L

)2

satisfy

with φm(0) = 0 = φm(L). Putting

Φm =

For U, V ∈ RP−1 define

U,V = P−1∑ p=1

UpVp x and U = √ U,U.

We find that Φ1, Φ2, . . . . . . , ΦP−1 form an orthogonal basis for RP−1,

Φm,Φm′ = 0 if m 6= m′ and m, m′ ∈ {1, 2, . . . ,P − 1},

and, with θ = mπ/P,

( sin

mπ

2 sin θ

L

2 .

Stability of the discrete Fourier modes Define the discrete Fourier coefficients

Un m =

Un mΦm.

Un+1 m − Un

m t,

|Un m| ≤ |U0

Λm tα

Since

|Um|2Φm2 = L

2

Theorem If

n−1∑ j=0

Problem if α is small

The stability restriction ρ ≤ 1 means that the time step must be chosen so that

tα ≤ Γ(α + 1)

Example

Suppose α = 1/5, Kα = Γ(α + 1) and x = 2× 10−3, then we require

t ≤ (x/2)10 = 10−30.

Implicit Euler method

Again start with the ODE

u + λ∂1−α t u = f (t) for 0 < t < T , with u(0) = u0,

but now integrate over (tn−1, tn) to obtain

u(tn)− u(tn−1) + λ

∫ tn

Un − Un−1 + λ

Weights

=

] U j dt

ωj = (j + 1)α − 2jα + (j − 1)α for j ≥ 1.

Unconditional stability

λtα

Γ(α + 1) .

Thus, starting from U0 = u0, we compute Un for n = 1, 2, . . . , N by solving

(1 + ρ)Un = Un−1 + f n t − ρ n−1∑ j=1

wn−jU j .

Proof

|wj | = − n−1∑ j=1

wj = 1 + (n − 1)α − nα ≤ 1

so (1 + ρ)|Un| ≤ |Un−1|+ |f n|t + ρ max

1≤j≤n−1 |U j |

and therefore

+ ρ

≤ |f n|t + max 1≤j≤n−1

|U j |.

Example: α = 1/2, λ = 1, f ≡ 0, N = 25

Convergence behaviour

|Un − u(tn)|,

|Un − u(tn)|.

Recall that if we compute a sum with M terms,

S = M∑

m=1

Am

in a system of floating-point arithmetic with unit roundoff ε, then

| fl(S)− S | ≤ Mε

|Am|.

Thus, for wj = (j + 1)α − 2jα + (j − 1)α and j large,

| fl(wj)− wj | . εjα whereas wj ≈ α(1− α)jα−2

so the relative rounding error | fl(wj)− wj |/|wj | is of order εj2.

By writing wj = j −j−1 and using expm1 and log1p to evaluate

j ≡ (j + 1)α − jα = jα [ (1 + j−1)α − 1

] = jα

] − 1 ) ,

we can reduce somewhat the rounding error in fl(wj) for large j .

When we compute

(n − j)α|U j | . nα+2ε max 1≤j≤n

|U j |,

which suggests that roundoff might become a problem once nα+2 ≥ ε−1.

Fractional diffusion equation

For = (0, L), we again consider the initial-boundary value problem

ut − Kα∂ 1−α t uxx = f (x , t) for x ∈ and 0 < t < T ,

u = u0(x) for x ∈ , when t = 0,

u = 0 for x ∈ ∂ and 0 < t < T .

The implicit time-stepping scheme leads to

Un p − Un−1

p + U j p−1

x2 = f n

for 1 ≤ n ≤ N and 1 ≤ p ≤ P − 1, with

U0 p = u0(xp) and Un

0 = 0 = Un P .

(I + B)Un = Un−1 + t fn − n−1∑ j=1

wn−jBUj

. . . . . .

. . .

.

Computational cost

At the nth time step, evaluation of the RHS costs O(nP) flops, and the elliptic solve costs O(P) flops. Since

N∑ n=1

Also, the nth time step requires O(nP) active memory locations.

Contrast this with using the implicit Euler method to solve the classical diffusion equation (the case α = 1): each time step requires O(P) flops and O(P) active memory locations, and N time steps requires O(NP) flops.

Conclusion: cost when 0 < α < 1 is N times the cost when α = 1.

Comparison with direct simulation

Let u = u(x , t) be the solution of

ut − Kα∂ 1−α t uxx = f (x , t) for x ∈ = (−L, L) and 0 < t < T ,

u = δ(x) for x ∈ , when t = 0,

u = 0 for x ∈ ∂ = {−L, L} and 0 < t < T .

We can approximate u using the implicit Euler method, or by simulating CTRWs with

ψ(t) = α

1√ 2π

exp(−x2/2).

In this case ψ(t) ∼ A/t1+α as t →∞, with A = α, so Bα = Aα−1Γ(1− α) = Γ(1− α) and we must rescale in such a way that

σ2

Probability densities of CTRWs (15, 000 samples)

Part VI

Introduction

The finite element method provides the simplest approach for discretization of a fractional diffusion problem on a spatial domain of general shape. In the classical method of lines for the heat equation, such a spatial discretization leads to a large, stiff system of first-order ODEs in time, that can be integrated using an appropriate black-box routine. For a time-fractional diffusion problem, we instead obtain a system of fractional-order ODEs.

In this lecture, we seek to estimate the errors from the spatial discretization assuming that the time integration is performed exactly. Suitable approaches for the time discretization include the implicit Euler method described previously and the more accurate schemes addressed in subsequent lectures.

Outline

Method of lines

Spatial domain ⊆ Rd (d = 1, 2 or 3); for simplicity a convex polygon or polyhedron so the elliptic problem is H2-regular.

With 0 < α < 1, let u = u(x , t) be the mild solution of

ut − Kα∂ 1−α t ∇2u = f (x , t) for x ∈ and t > 0,

u = u0(x) for x ∈ , when t = 0,

subject to homogeneous Dirichlet boundary conditions

u = 0 for x ∈ ∂ and t > 0.

We wish (for now) to discretize in space only.

Weak formulation

∫

Kα ∂u

Thus,

ut , v+ a(∂1−α t u, v) = f , v for v ∈ H1

0 (),

and also

ut , v+ ∂1−α t a(u, v) = f , v for v ∈ H1

0 ().

h = max K∈Th

diam(K ).

Let Vh denote the space of real-valued functions on that are continuous piecewise polynomials of degree at most p ≥ 1 with respect to Th, and which vanish on ∂. Hence,

Vh ⊆ H1 0 ().

Seek a finite element solution uh : [0,∞)→ Vh satisfying

uht , χ+ ∂1−α t a(uh, χ) = f , χ for χ ∈ Vh and t > 0,

with uh(0) = u0h ≈ u0 for a suitable u0h ∈ Vh.

Alternative formulation

Ahψ, χ = a(ψ, χ) for ψ, χ ∈ Vh,

and the L2-projector Ph : L2()→ Vh by

Phv , χ = v , χ for v ∈ L2() and χ ∈ Vh.

Since

t Ahuh, χ = ∂1−α t Ahuh, χ,

we see that

uht + ∂1−α t Ahuh, χ = f , χ = Phf , χ for all χ ∈ Vh,

and thus uht + ∂1−α

t Ahuh = Phf for t > 0.

Nodal equations Construct a nodal basis χ1, χ2, . . . , χN for Vh so that

χj(xk) = δjk ,

where x1, x2, . . . , xN are the free (interior) nodes. Thus,

uh(x , t) = N∑

Uk(t)χk(x) where Uk(t) = uh(xk , t).

Define the mass matrix M, stiffness matrix S and load vector f by

Mjk = χk , χj, Sjk = a(χk , χj), fk(t) = f (t), χk

then the nodal vector U = [Uk(t)] satisfies the system of ordinary integro-differential equations

M dU

Discrete eigensystem

The finite dimensional linear operator Ah : Vh → Vh is symmetric and positive-definite, so Vh has an orthonormal basis φh1, φh2, . . . , φhN consisting of eigenfunctions of Ah. Thus,

Ahφ h m = λhmφ

uh(t) = Eh(t)u0h +

0 Eh(t − s)Phf (s) ds,

where the discrete solution operator for the homogeneous problem is defined by

Eh(t)χ = N∑

m=1

Stability

uh(t) ≤ u0h+

Proof. Since 0 < Eα(−s) ≤ 1 for 0 ≤ s <∞,

Eh(t)χ2 = N∑

m=1

χ, φhm2 = χ2,

so Eh(t)χ ≤ χ and the desired estimate follows from the Duhamel formula above.

Error estimates

We wish to estimate the error uh(t)− u(t) in L2() and in H1 0 ().

Our assumption that is convex implies that the elliptic problem,

−∇2u = f in , with u = 0 on ∂,

is H2-regular, that is, the weak solution u ∈ H1 0 (), given by

a(u, v) = f , v for all v ∈ H1 0 (),

necessarily belongs to H2() and

u2 ≤ Cf .

Caution: in this lecture, vr denotes the norm in H r (), so vr <∞ does not guarantee v ∈ Hr

D() unless v vanishes appropriately on ∂.

Error in the elliptic problem

Let uh ∈ Vh be the finite element solution of the elliptic problem above, so that

a(uh, χ) = f , χ for all χ ∈ Vh.

Since the symmetric bilinear form a(u, v) is bounded and coercive on H1

0 (), and using an appropriate quasi-interpolant, we have

uh − u1 ≤ C inf χ∈Vh

χ− u1 ≤ Chr−1ur for 1 ≤ r ≤ p + 1.

The usual duality argument (which relies on H2-regularity) then implies that

uh − u ≤ Ch inf χ∈Vh

χ− u1 ≤ Chrur for 1 ≤ r ≤ p + 1.

Ritz projector

It is convenient to define Rh : H1 0 ()→ Vh by Rhu = uh, or

equivalently,

It follows that R2 h = Rh, with

v − Rhv ≤ Chrvr and v − Rhv1 ≤ Chr−1vr

for 1 ≤ r ≤ p + 1. Also, since

PhAv , χ = Av , χ = a(v , χ) = a(Rhv , χ) = AhRhv , χ

for all v ∈ H1 0 () and χ ∈ Vh, we see that

PhA = AhRh : H1 0 ()→ Vh.

Equation for the error

Returning to the time-dependent case, we split the error into two terms

uh(t)− u(t) = ϑ(t) + %(t), ϑ = uh − Rhu, % = Rhu − u.

Then for all χ ∈ Vh,

ϑt , χ+ ∂1−α t a(ϑ, χ)

= uht , χ+ ∂1−α t a(uh, χ)− Rhut , χ − ∂1−α

t a(Rhu, χ)

= f , χ+ Rhut , χ − ∂1−α t a(u, χ)

= ut , χ+ ∂1−α t a(u, χ)− Rhut , χ − ∂1−α

t a(u, χ)

so ϑ : [0,∞)→ Vh satisfies an equation of the same form as the one for uh (with ut − Rhut playing the role of f )

ϑt , χ+ ∂1−α t a(ϑ, χ) = ut − Rhut , χ.

Estimate for ϑ = uh − Rhu

Lemma For 1 ≤ r ≤ p + 1,

ϑ(t) ≤ u0h − Rhu0+ Chr

ϑ(t) ≤ ϑ(0)+

Ph(ut − Rhut) ≤ ut − Rhut ≤ Chrutr .

Estimate for % = Rhu − u

%(t) ≤ Chr

( u0r +

together with

Quasi-optimal error bound in L2()

Theorem The finite element solution of the time-fractional diffusion equation satisfies

uh(t)− u(t) ≤ u0h − Rhu0+ Chr

( u0r +

) for 1 ≤ r ≤ p + 1.

( · · · )

Realistic smoothness

Consider the homogeneous problem with f ≡ 0 so that u(t) = E(t)u0, and choose p = 1 (piecewise-linears).

For sufficiently small ε > 0, if u0 ∈ H2+ε() and u0 = 0 on ∂, then u0 ∈ H2+ε

D () so by our earlier regularity theorem,

tut2 ≤ Ctεα/2u02+ε,

implying that ut2 = O(tεα/2−1) as t → 0. Choosing u0h = Rhu0

for simplicity, we obtain

uh(t)− u(t) ≤ Ch2u02+ε for 0 ≤ t ≤ T ,

with C = C (T , α, ε,).

Estimate for ϑ(t)1

We saw that

ϑt , χ+ a(∂1−α t ϑ, χ) = −ρt , χ for all χ ∈ Vh.

Choosing χ = Ahϑ(t), we have

ϑt ,Ahϑ = 1

and

−%t ,Ahϑ = −%t ,PhAhϑ = −Ph%t ,Ahϑ = −a(Ph%t , ϑ).

Thus,

1

2

d

Integrating from t = 0 to t = T and using∫ T

0 ∂1−α

Since a(v , v) is equivalent to v2 1,

ϑ(T )1 ≤ C

) .

ϑ(t)1,

≤ Cϑ(0)2 1 + C

∫ t∗

≤ C

) ϑ(t∗)1

1 + C

Quasi-optimal error bound in H1()

Assume now that Th is such that the L2-projector Ph is stable in H1(), that is,

Phv1 ≤ Cv1 for v ∈ H1().

For instance, it suffices to assume that Th is quasi-uniform.

Theorem The finite element solution of the time-fractional diffusion equation satisfies

uh(t)− u(t)1 ≤ Cu0h − Rhu01

+ Chr−1

( u0r +

Proof Recall that uh − u = ϑ+ % and

ϑ(t)1 ≤ C

) .

and since Ph is stable in H1(),

Ph%t(s)1 ≤ Cut(s)− Rhut(s)1 ≤ Chr−1ut(s)r .

The error bound follows because

%(t)1 ≤ %(0)1 +

≤ Chr−1

( u01 +

Define the closed sector

Σψ = { z ∈ C : z 6= 0 and | arg z | ≤ ψ } ∪ {0}.

Our aim now is to prove the following error bound, which does not require any spatial regularity for u0 or f .

Theorem Assume that u0h = Phu0, and fix such that 0 < < π/2. Then,

uh(t)− u(t) ≤ C t−αh2 ( u0+ sup

z∈∂Σπ−

Under Laplace transformation, our problem

ut + ∂1−α t Au = f (t) for t > 0, with u(0) = u0,

becomes zu(z)− u0 + z1−αAu(z) = f (z),

so

Similarly, for uh we have

(zαI + Ah)uh(z) = zα−1gh(z) where gh(z) = u0h + Ph f (z).

Notice that gh(z) = Phg(z) because we assume u0h = Phu0.

Resolvent estimate for A

Theorem Let satisfy 0 < ≤ π/2 and put

M = 1

(zI + A)−1 ≤ M

1 + |z | .

Proof Let z = re iθ with |θ| ≤ π − and r > 0.

If 0 ≤ |θ| ≤ π/2 then |z | ≤ |z + λm| because

0 ≤ <z ≤ <(z + λm) and =z = =(z + λm)

If π/2 ≤ |θ| ≤ π − then |z | ≤ M|z + λm| because

|z | sin ≤ |z | sin |θ| = |=z | = |=(z + λm)| ≤ |z + λm|.

Therefore, since M ≥ 1,

(zI + A)−1v = ∞∑

(zI + A)−1v2 = ∞∑

If |z | ≤ λ1/2, then |z + λm| ≥ λ1 − |z | ≥ λ1/2 so

1 + |z | |z + λm|

= M

( 1 +

2

λ1

Resolvent estimate for Ah

Theorem Let satisfy 0 < ≤ π/2 and put M = 1/ sin. If z ∈ Σπ− then

(zI + Ah)−1 ≤ M

1 + |z | .

Proof. The same proof shows that the conclusion holds with λh1 in place of λ1. But

λh1 = min χ∈Vh

Ahχ, χ χ2

the Laplace inversion formula gives

u(t) = 1

and

where

Error representation

where

Gh(z) = (zαI + Ah)−1Ph − (zαI + A)−1 = G 1 h (z) + G 2

h (z),

G 1 h (z) = (zαI + Ah)−1Ph − Ph(zαI + A)−1,

G 2 h (z) = (Ph − I )(zαI + A)−1.

Lemma For z ∈ Σπ−,

A(zαI + A)−1v ≤ Cv and (zαI + Ah)−1Ahχ ≤ Cχ.

Proof. The identity

A(zαI + A)−1 = (zαI + A− zαI )(zαI + A)−1 = I − zα(zαI + A)−1

implies that

|zα| ≤ C .

Estimate for G 1 h (z)

Lemma G 1

Proof. Recall that PhA = AhRh, so

G 1 h (z) = (zαI + Ah)−1Ph − Ph(zαI + A)−1

= (zαI + Ah)−1 [ Ph(zαI + A)− (zαI + Ah)Ph

] (zαI + A)−1

= (zαI + Ah)−1Ah(Rh − Ph)(zαI + A)−1,

H2-regularity of the elliptic problem gives

G 1 h (z)v ≤ C(Rh − Ph)(zαI + A)−1v ≤ Ch2(zαI + A)−1v2

≤ Ch2A(zαI + A)−1v ≤ Ch2v.

Estimate for G 2 h (z)

Lemma G 2

Proof. As above,

≤ Ch2(zαI + A)−1v2

≤ Ch2A(zαI + A)−1v ≤ Ch2v.

Error estimate

uh(t)− u(t) = 1

with Gh(z)v ≤ Ch2v for z ∈ Σπ−.

Deform the integration contour Γ to ∂Σπ− = Γ+ − Γ−, where

Γ± = { se±i(π−) : 0 < s <∞},

so that uh(t)− u(t) = I+ − I− where

I± = 1

eztzα−1Gh(z)g(z) dz .

The substitution z = se±i(π−) = s(− cos± i sin) gives

I± ≤ Ch2 ( u0+ sup

ds

s .

A second substitution s = w(t cos)−1 shows that the integral on the right equals∫ ∞

0 e−w

Introduction

The Laplace inversion formula yields a contour integral representation of the finite element solution uh(t) to the time-fractional diffusion equation. Applying a quadrature approximation [Lopez-Fernandez+Palencia-2004] leads to a fully discrete solution UN,h(t). The main cost of the method is the computation of uh(zj) at the quadrature point zj for |j | ≤ N, which requires the solution of a system of finite element equations involving the complex parameter zj .

The main advantages of the this approach are high accuracy and easy parallel implementation. A key disadvantage is that the method imposes severe limitations on form of the source term f (t). We outline a modified approach [McLean+Thomee-2010] that sacrifices some accuracy to relax the requirements on f (t).

Outline

Contour integral and quadrature Once again consider

u + ∂1−α t Au = f (t) for t > 0, with u(0) = u0,

and recall that

We now choose for Γ a contour of the form

z = z(ξ) = µ ( 1− sin(δ − iξ)

) for −∞ < ξ <∞,

µ > 0 and 0 < δ < π

2 .

Hyperbola

Since sin(δ − iξ) = sin δ cosh ξ − i cos δ sinh ξ,

we have

) ,

y(ξ) = =z(ξ) = µ cos δ sinh ξ,

so Γ is the left branch of the hyperbola( x − µ µ sin δ

)2

Parameterised integral

We have

u(t) = 1

= 1

2πi

) = µ

) ,

) .

(cosh ξ − sin δ)2 =

cosh ξ + sin δ

1− sin δ .

Double exponential decay

|ez(ξ)t | = ex(ξ)t = exp ( µt(1− sin δ cosh ξ)

) ,

|ez(ξ)t | ≤ exp ( µt − 1

Discretisation error For ξ > 0, let

Q∞(v) = ∞∑

1. v(ζ) is analytic on the strip −r− ≤ =ζ ≤ r+;

2. ∫ r+

3. ∫∞ −∞ |v(ξ ± ir±)| dξ ≤ M±.

Then Q∞(v)− ∫ ∞ −∞

( 1− sin(δ − iζ)

) defines conformal mapping that takes the line <ζ = η to the left branch of the hyperbola(

x − µ µ sin(δ + η)

)2

π

2

Truncation error

so need to estimate

v(ξ) dξ

) .

ea(1−b cosh ξ) dξ ≤ Cea(1−b)L(ab)

where

L(x) =

Proof

Since a(1− b cosh ξ) = a(1− b) + ab(1− cosh ξ) it suffices to estimate

I ≡ ∫ ∞

∫ ∞ 0

=

,

where we used the substitution y = cosh ξ− 1 followed by x = aby . If ab ≥ 1 then I ≤

∫∞ 0 x−1/2e−x dx <∞, whereas if ab ≤ 1, then

I ≤ ∫ 1

0

Quadrature error [Weideman+Trefethen-2007]

Thus, taking a = µt and b = sin(δ ± r±) in the lemma,∫ ∞ −∞ |v(ξ ± ir±)| dξ ≤ M± = Ceµt(1−sin(δ±r±))L

( µt sin(δ ± r±)

≤ CL ( µt sin(δ ± r±)

) − 2πr±/ξ

) ξ.

µt ( 1− sin(δ + r+)

) ,

2 .

The limiting choices r+ = π/2− δ and r− = δ give

− 2π

ξ

( π

) ,

2πδ

ξ .

4δ − π ,

( 2δ

b(δ) ,

|DE+ |+ |DE− |+ |TE | ≤ CL(· · · )e−B(δ)N .

Optimal choice of δ

We find that B(δ) has a unique maximum in the interval π/4 < δ < π/2, namely, at

δ∗ = 1.1721 0423, (near 3π/8 = 1.1780 9724)

and the maximum value is

B(δ∗) = 2.3156 5403.

ξ∗ = 1.0817 9214

N

t ,

e−B(δ∗)N = e−2.3157N = 10.1315−N .

Scalar test problem

L{e−λt} =

We consider

where zj = z(j ξ) and z ′j = z ′(j ξ).

Convergence behaviour

Fixed points with varying t

Given N and τ > 0, suppose we choose ξ∗ and µ∗ as above for t = τ . What if we use the approximation

e−λt ≈ 1

for t near τ?

In the fractional PDE case, we can solve one set of elliptic problems for uh(zj) at the zj optimized for t = τ and use them not just at t = τ but for several values of t in an interval around τ . The zj are then slightly sub-optimal but we avoid having to compute a new set of uh(zj) for each t.

Error for the scalar example with τ = 1

Parallel-in-time algorithm

Combining the above approach to numerical inversion of the Laplace transform with a spatial discretisation by finite elements leads to a fully discrete numerical method that involves no time stepping.

The method is particularly suited to applications in which the solution is required for only a few values of t.

In addition to its high, spectral-order accuracy in time, the method is embarassingly parallel.

Method-of-lines solution

Recall that the semidiscrete finite element solution uh : [0,∞)→ Vh satisfies

uht , χ+ ∂1−α t a(uh, χ) = f (t), χ for χ ∈ Vh and t > 0,

or equivalently,

uht + ∂1−α t Ahuh = Phf (t) for t > 0,

with uh(0) = u0h ≈ u0.

zuh + z1−αAhuh = gh(z)

Fully-discrete solution

If we solve the (complex) finite element equations for uh(z) at each quadrature point z = zj , then we can compute

UN,h(t) = 1

as an approximation to

) z ′(ξ) dξ.

The computational cost is dominated by solving the elliptic problems, and a key advantage of the method is that (unlike in a time-stepping scheme) these elliptic solves can easily be performed in parallel.

Halving the computational cost

Assuming that u0 and f are real-valued, it follows that uh is real-valued and so

uh(z) = uh(z).

Since z(−ξ) = z(ξ) and z ′(−ξ) = −z ′(ξ),

so z−j = zj , uh(z−j) = uh(zj), z ′−j/i = z ′j/i

and therefore

2πi

Since y0 = 0 and x ′0 = 0, it follows that

UN,h(t) = 1

1

π

) ξ.

In particular, it suffices to compute uh(zj) for 0 ≤ j ≤ N.

In matrix terms, we solve the N + 1 linear systems

(zαj M + S)U(zj) = zα−1 j G(zj), 0 ≤ j ≤ N,

where M and S are the mass and stiffness matrices, U(z) is the vector of nodal values of uh(z), and G(zj) is the load vector for gh(z).

Quadrature error

M± = 1

dξ

for η = δ ± r± where 0 < r− < δ < r+ < π/2. We saw earlier that

(zI + Ah)−1v ≤ v |z | sin

for z ∈ Σπ− and v ∈ Vh,

and z ′(ξ + iη)

we have

≤ gh(z) |z | sin

for zα ∈ Σπ−, that is, for z ∈ Σ(π−)/α and hence for z ∈ Σπ−.

Therefore,

dξ

≤ 1

f (z) ≤ Cf , for z ∈ Σπ−,

so that

gh(z) ≤ u0+ Ph f (z) ≤ u0+ Cf , for z ∈ Σπ−,

then we may estimate the quadrature error as before and obtain

UN,h(t)− uh(t) ≤ Ce−B(δ)NL(ct).

UN,h(t)− u(t) ≤ UN,h(t)− uh(t)+ uh(t)− u(t) = O

( L(ct)e−B(δ)N

) + O(t−αh2).

U? N,h(t) =

with εjL∞() ≤ ε, leading to an additional perturbation

U? N,h(t)− UN,h(t) ≤ ε

≤ ε

) ξ

≈ ∫ N ξ

−N ξ eµt(1−sin δ cosh ξ dξ ≤ Ceµ(1−sin δ)L(µt sin δ)

we conclude that

−N≤j≤N g(zj).

lead to eµ∗t(1−sin δ∗) = e1.4224N = 1.422N .

Example

Suppose h = 10−3 and ε = 2−52 ≈ 2.22× 10−16. Then

ε 1.422N ≥ h2 = 10−6

when

So roundoff probably not an issue.

However, “optimal” parameter values might be problematic if a high-accuracy spatial discretisation (say a spectral method) were used. We do not want µ to be too large.

Behaviour of f (z)

f (x , z) = g(x) z + a

(z + a)2 + ω2 ,

Example

If

e−z − e−2z

A more flexible approach [McLean+Thomee-2010]

Now drop the assumption that f (z) is analytic and bounded for z ∈ Σπ−. We will describe a method based on Duhamel’s formula,

u(t) = E(t)u0 +

0 E(t − s)f (s) ds, t > 0.

Recall that E(t), the solution operator for the homogeneous fractional diffusion equation has the series representation

E(t)v = ∞∑

and the integral representation

Noting that∫ t

∫ t

0

1

2πi

∫ Γ

= 1

2πi

we define g(z , t) = eztu0 +

∫ t

and deduce u(t) =

∫ Γ E(z)g(z , t) dz .

Notice that g(z , t) is an entire function of z , with

g(z , t) ≤ u0+

0 f (s) ds for <z ≤ 0 and t ≥ 0.

Since E(z) ∼ z−1I as |z | → ∞ with z ∈ Σπ−, we expect

E0(z) = E(z)− z−1I

to decay more rapidly than E(z), which is needed to compensate for the disappearance of the factor ezt in the integrand.

Theorem If 0 ≤ σ ≤ 1 and v ∈ D(Aσ), then

E0(z)v ≤ C,σ Aσv |z |1+ασ

for z ∈ Σπ−,

= [ zα−1I − z−1(zαI + A)

] (zαI + A)−1

]σ Aσ

] (zαI + A)−1

(zαI + A)−1 ≤ 1

σ ≤ ( 1

]1−σ ≤ (1 + 1

1

2πi

∫ Γ

g(z , t)

w0(z , t) = E0(z)g(z , t) = w(z , t)− z−1g(z , t).

and w(z , t) = E(z)g(z , t) denotes the solution of the (complex) elliptic problem

(zαI + A)w(z , t) = zα−1g(z , t).

Similarly, if we define the spatially discrete operators

Eh(z) = zα−1(zI + Ah)−1 and E0h(z) = Eh(z)− z−1I

and put

then

w0h(z , t) dz

where w0h(z , t) = E0h(z)gh(z , t) = wh(z , t)− z−1gh(z , t), and wh(z , t) is computed by solving the (complex) finite element equations

(zαI + Ah)wh(z , t) = zα−1gh(z , t).

Fully-discrete scheme The equal-weight quadrature approximation∫

Γ w0h(z , t) dz ≈

w0h(zj , t)z ′j ξ.

Lemma If z = z(ξ + iη) and z ′ = z ′(ξ + iη), then for 0 < σ1 + α−1 = σ ≤ 1,

wh(z , t)z ′ ≤ 2C,σ,σ1

wh(z , t) = E0h(z) ( eztu0h

= C,σ |z ′| |z | Aσhu0h |z |ασ

and, since 2 + ασ1 = 1 + α(σ1 + α−1) = 1 + ασ,

z−1E0h(z)Phf (0)z ′ ≤ C,σ1 |z ′| Aσ1

h u0h |z |2+ασ1

= C,σ1

Quadrature error

) ,

= (1− e−|ξ| sin δ)2 + e−2|ξ| cos2 δ

≥ (1− sin δ)2,

it follows that

µασ .

Set r = r± such that 0 < δ − r < δ + r < π/2, and estimate∫ ∞ −∞

w ( z(ξ ± ir)

) z ′(ξ ± ir)

µασ

∫ ∞ 0

µασ e−2πr/ξ.

At the same time,

eµt(1−sin δ)

µασ e−ασNξ.

Setting 2πr/ξ = ασNξ, and choosing µ > 0 to minimise eµt(1−sin δ)/µασ, we arrive at the following estimate.

Theorem For the flexible scheme described above, if

ξ =

√ 2πr

( − √

2πrασN ) .

The error bound suggests choosing δ = π/4 and r slightly less than π/4.

Example

Taking

2 , σ = 1

gives 2πrασ = π2/4 so the decay factor in the error bound is of order

e− 1 2π √ N = e−1.5708

√ N ,

for our earlier method.

0 K (tn − s)f (s) ds ≈

n∑ j=0

wn−j f (tj), tj = j t,

where the convolution weights wn = wn(t) are computed directly from the Laplace transform K (z) rather than the kernel K (t). This approach can be advantageous if K (z) is simpler than K (t).

Convolution quadrature can be used to approximate any function of the form K ∗ f , and in particular the fractional integral Iαf = Υα ∗ f .

Outline

Assume that K (z) is analytic and satisfies

|K (z)| ≤ C |z |−µ for z ∈ Σπ−, where µ > 0.

Then, as before,

K (t) = 1

Since∫ t

∫ t

0

( 1

2πi

∫ Γ

K ∗ f (t) = 1

The associated ODE

so that

so y is the solution of the initial-value problem

dy

dt − zy = f (t) for t > 0, with y(0) = 0.

Implicit Euler method

Y n = Y n(z) ≈ y(tn; z) where tn = n t,

by solving

Y n − Y n−1

t − zY n = F n for n ≥ 1, with Y 0 = 0,

where F n = f (tn). Then,

K ∗ f (tn) = 1

Y n(z)ζn.

∞∑ n=0

Y nζn − ∞∑

n=−1

so the finite difference equation implies that

1− ζ t

Thus,

δ(ζ)t−1 − z where δ(ζ) = 1− ζ.

Recall

K (z)Y n(z) dz .

For t and |ζ| sufficiently small, Γ passes to the left of the pole at z = δ(ζ)t−1, and

∞∑ n=0

= − 1

2πi

∫ Γ

z − δ(ζ)t−1 F (ζ).

Here, the integrand is O(|z |−1−µ) so Cauchy’s theorem gives

− 1

2πi

∫ Γ

( δ(ζ)t−1

( δ(ζ)t−1

K ( δ(ζ)t−1

Example

Suppose K (t) = Υα(t) and so K (z) = z−α. Then

K ( δ(ζ)t−1

) = ( (1− ζ)t−1

)−α = tα(1− ζ)−α

) (−1)n.

Note that wn > 0 because w0 = tα and, for n ≥ 1,( −α n

) (−1)n =

Higher-order methods

To improve on the implicit Euler method, recall that the polynomial

Y n + Y n

+ 1

p!

tp (t − tn) · · · (t − tn−p+1)

of degree p takes the value Y j at t = tj for n − p ≤ j ≤ n, where Y n = Y n − Y n−1 denotes the backward difference. Differentiating with respect to t and setting t = tn leads to the backward differentiation formula (BDF)

y ′(tn) ≈ Y n

1

t

` − zY n = F n for n ≥ 1,

with starting values Y 0 = Y−1 = · · · = Y−p = 0. Generating function still satisfies

δ(ζ)

t Y (ζ)− zY (ζ) = F (ζ) but now δ(ζ) =

p∑ `=1

(1− ζ)`

| arg δ(ζ)| ≤ π − α for |ζ| < 1,

for the following values of α.

p α

1 90

2 90

3 86

4 73

5 51

6 17

Operational calculus

∫ t

In this way, since L{K ∗ f } = K (z)f (z),

we have

eztK (z)f (z) dz .

Explanation: if K (t) = 1 then K (z) = z−1 so

∂−1f (t) =

0 f (s) ds and L{∂−1f } = z−1f (z).

Example

For K (t) = Υα(t) = tα−1/Γ(α) we have K (z) = z−α and so

∂−αf (t) = Υα ∗ f (t) = Iαf (t)

is the fractional integral of order α > 0, with

L{∂−αf } = z−αf (z).

Example

If K (t) = eat then K (z) = (z − a)−1 so

(∂ − a)−1f (t) =

with L{(∂ − a)−1f } = (z − a)−1f (z).

Theorem

Proof.

∫ t

=

= 1

2πi

∫ Γ

= 1

2πi

∫ Γ

Discrete operational calculus

0≤tj≤t wj(t)f (t − tj) for t > 0.

In particular, at t = tn,

K (∂t)f (tn) = n∑

j=0

If K (t) = 1 then K (z) = z−1 so

K ( δ(ζ)t−1

∂−1 t f (t) =

∫ t

Example

is the solution of the initial-value problem

y − ay = f (t) for t > 0, with y(0) = 0.

The BDF solution Y n satisfies( δ(ζ)t−1 − a

) Y (ζ) = F (ζ), F n = f (tn),

so if ( δ(ζ)t−1 − a

)−1 = ∞∑ n=0

Example

y ′(tn) ≈ Y n

2 (1− ζ)2 = 3 2 (1− ζ)(1− 1

3ζ).

( α + n − 1

wn =

( 2

3

Since

wn = 1

We can use this representation to show the following result.

Theorem

Proof

so

)−1 = ∞∑ n=0

= ∑

= 1

2πi

∫ Γ

0≤tj≤t w? j (t; z)f (t − tj) dz

= 1

2πi

∫ Γ

Summary

1

2πi

∫ Γ

2πi

∫ Γ

w(ζ) = ∞∑ n=0

wnζ n = K

( δ(ζ)t−1

2πi

∫ Γ

Accuracy of convolution quadrature

wj f (t − tj)χ(t − tj), χ(t) =

{ 1, t > 0,

0, t < 0,

L{K (∂t)f } =

= ∞∑ j=0

= ∞∑ j=0

wjζ j

= K ( δ(e−z t)t−1

) f (z).

δ(e−h)h−1 = 1 + O(hp) as h→ 0,

so

δ(e−zt)t−1 = zδ(e−h)h−1, h = zt,

= z + O(zp+1tp)

) f (z).

Notice also

= 1

2πi

∫ Γ

] f (t) dz .

(A1) There exist 0 < < π/2 and −∞ < µ <∞ such that the function G (z) is analytic with |G (z)| ≤ C |z |−µ for | arg z | < π−.

(A2) The linear multistep method is strongly A-stable of order p ≥ 1, that is,

I δ(ζ) is analytic in a neighbourhood of the closed unit disk |ζ| ≤ 1,

I for ζ in this neighbourhood, δ(ζ) = 0 iff ζ = 1,

I there exists 1 > such that | arg δ(ζ)| ≤ π − 1 for |ζ| < 1,

I h−1δ(e−h) = 1 + O(hp) as h→ 0.

Theorem ([Lubich-2004])

If assumptions (A1) and (A2) hold, then for 0 < t <∞,

G (∂t)tβ−1 − G (∂)tβ−1 ≤ {Ctµ−1+β−ptp, p ≤ β,

Ctµ−1tβ, 0 < β ≤ p.

Multiplication property

Notice that since

K1 ∗ (K2 ∗ f ) = (K1 ∗ K2) ∗ f and L{K1 ∗ K2} = K1(z)K2(z),

we have K1(∂)K2(∂) = (K1K2)(∂),

The analogous identity holds in the discrete case.

Theorem

Proof

= ∑

∑ 0≤tk≤t−tj

f (t − tj − tk)

= ∑

n−j wn

) f (t − tn).

k ζ j+k

n−j

K ∗ (f ∗ g) = (K ∗ f ) ∗ g ,

or equivalently,

) ∗ g .

K (∂t)(f ∗ g) = ( K (∂t)f

) ∗ g ,

because

L {

) f (z)g(z).

=

so

) ∗ f (p)(t),

and the same formula holds with ∂ replaced by ∂t . Hence,

[G (∂t)− G (∂) ] f (t)

≤ Ctµ−1 p−1∑ k=0

|f (k)(0)|tk+1

Correction terms

p−2∑ j=0

w∼nj f (tj)

and choose the extra weights w∼nj so that the modified quadrature rule is exact for polynomials up to degree p − 2:

p−1∑ j=1

w∼njΥk(tj) = G (∂)Υk(tn)− G (∂t)Υk(tn)

for 1 ≤ k ≤ p − 1. Unfortunately, the matrix [Υk(tj)] is badly conditioned.

An alternative approach works if tn is bounded away from 0.

Application to fractional diffusion

For simplicity, we suppose f ≡ 0 so that, after integrating in time, our initial-value problem takes the form

u + ∂−αAu = u0,

Thus, u = u0 − (I + ∂−αA)−1∂−αAu0,

which suggests seeking U(t) ≈ u(t) such that

U = u0 − (I + ∂−αt A)−1∂−αAu0.

Thus,

which leads to the implicit scheme

W n + n∑

wn−jAW j = −Υ1+α(tn)Au0 for n ≥ 1,

with W 0 = 0, where the weights wn = wn(t) are given by

[ δ(ζ)t−1

]−α = ∞∑ n=0

wnζ n.

Fully-discrete version: Uh = u0h + Wh where W 0 h = 0 and

W n h +

n∑ j=0

Error bound for nonsmooth initial data Since

u − u0 = −(I + ∂−αA)∂−αAu0,

U − u0 = −(I + ∂−αt A)∂−αAu0,

the error from the time discretization is

U − u = [ G (∂)− G (∂t)

] ∂−αu0,

Theorem ([Cuesta+Lubich+Palencia-2006])

Ct−1−αt1+αu0, p ≥ 2.

Proof

Since

= zα(zαI + A)−1 [ (zαI + A)− zαI

] = zα

] and (zαI + A)−1 ≤ C |z |−α, we have

G (z) ≤ C |z |α for z ∈ Σπ−.

Noting that ∂−αu0 = Υ1+α(t)u0 = tαu0/Γ(1 + α), we apply the theorem with µ = −α and β = 1 + α to conclude

U(t)− u(t) = [G (∂)tβ − G (∂t)tβ

] u0

t−α−1t1+α, p ≥ 1 + α.

Error bound for smooth initial data Recall that

u(t) = E(t)u0 = u0 − tα

and observe that

= Υ1+2α(t)A2u0,

We therefore consider

U(t) = u0 −Υ1+α(t)Au0 + (I + ∂−αt A)−1Υ1+2α(t)A2u0.

The error is

U(t)− u(t) = [ (I + ∂−αt A)−1 − (I + ∂−αA)−1

] Υ1+2α(t)A2u0

] Υ1+2α(t)Au0,

where, once again, G (z) = (I + z−αA)−1A. Applying the theorem with µ = −α and β = 1 + 2α we have

U(t)− u(t) ≤ CAu0 ×

t−1−αt1+2α, p ≥ 1 + 2α.

For instance, if p = 2 and 1/2 ≤ α < 1, then

U(t)− u(t) ≤ Ctα−2t2Au0.

Part IX

Introduction

We consider a class of time-stepping methods in which u(t) is approximated by a piecewise polynomial U in t. Continuity across the time levels is enforced only weakly. These fully implicit methods are flexible and robust, and can achieve high accuracy, but are rather complicated to implement in general and have a somewhat higher computational cost than many simpler time-stepping schemes. Crucially, they allow the use of highly non-uniform grids.

Once again, we largely ignore the spatial discretization.

Outline

Stability

Convergence

Discontinuous piecewise-polynomial approximation in time

Let t = (tn)Nn=0 be a vector of time levels satisfying

0 = t0 < t1 < t2 < · · · < tN = T ,

and denote the nth open subinterval and its length by

In = (tn−1, tn) and tn = tn − tn−1 for 1 ≤ n ≤ N.

For each n, choose a closed subspace Sn ⊆ H1 0 () and an

integer pn ≥ 0, and write S = (Sn)Nn=0 and p = (pn)Nn=1.

We define our trial space W =W(t,S,p) to consist of those functions U : (0,T )→ H1

0 () such that U|In is a polynomial of degree at most pn in t with coefficients in Sn for 1 ≤ n ≤ N.

Example

Choose Sn = H1 0 () and pn = p independent of n.

Example

Choose Sn = Vh to be the usual continuous piecewise-linear finite element space with respect to a triangulation Th of and enforcing a homogeneous Dirichlet boundary condition.

Thus, our methods are conforming in space but non-conforming in time.

For U ∈ W, we denote the one-sided limits and the jumps at tn by

Un ± = U(t±n ) = lim

−

for 0 ≤ n ≤ N, with the convention that U0 − ∈ S0 (even though I0

is undefined).

Weak formulation

Recall that the mild solution u of our initial-boundary value problem for the fractional diffusion equation satisfies

u(t), v+ a ( ∂1−α t u(t), v

) = f (t), v for all v ∈ H1

0 (),

where u = ut = ∂u/∂t. Hence, for suitable v : In → H1 0 (),∫

In

)] dt =

∫ In

f (t), v(t) dt.

In the discontinuous Galerkin (DG) method we seek U ∈ W satisfying, for all X ∈ W and for 1 ≤ n ≤ N,

Un−1 + ,X n−1

+ +

)] dt

+ +

In addition, we require that U0 − = U0 for a suitable

approximation U0 ∈ S0 to the given initial data u0.

Example

Take pn = 0 and Sn = Vh for all n, so that U and X are piecewise constant in time. Writing Un = Un

− and χ = X n −, we have

U(t) = Un = Un−1 + and X (t) = χ = X n−1

+ for all t ∈ In,

Un − Un−1, χ+

) dt

f (t), χ dt for all χ ∈ Sn,

which is essentially the implicit Euler scheme we considered earlier, but using finite elements instead of finite differences in space.

Fully implicit time stepping

If ψn 0 , ψn

1 , . . . , ψn pn is a basis for the space of polynomials of degree

at most pn, and if χ1, χ2, . . . , χMn is a basis for Sn (so Mn = dim Sn), then we can write

U(x , t) =

Starting from the known (approximate) initial data

U0 −(x) = U0(x) =

M0∑ l=1

U0 l χ

we compute the unknown Unr l by solving the

[(pn + 1)Mn]× [(pn + 1)Mn] linear system determined by the DG equations on In for successive n = 1, 2, . . . , N.

Stability

We now prove a series of technical lemmas that will establish unconditional stability of DG time stepping. This robustness is a key benefit of the method, and helps justify its relatively high computational cost. The stability estimate below also shows that most of the jumps [U]j must be small for large N.

Theorem If U0

( (0,T );H

DG solution U ∈ W, and for 1 ≤ n ≤ N,

Un −2 +

Global bilinear form

[U]n−1,X n−1 + +

∫ In

f ,X dt,

and summing over n, we see that U ∈ W is the DG solution iff

GN(U,X ) = U0,X 0 ++

where the bilinear form GN is defined by

GN(U,X ) = U0 −,X

+ N∑

)] dt.

Lemma

−2−Un−1 − 2

2U 2, twice the LHS equals

2[U]n−1,Un−1 + + Un

−2 − Un−1 + 2

= [U]n−1, [U]n−1 + Un−1 + + Un−1

− + Un −2 − Un−1

− 2 + Un −2 − Un−1

+ 2

− 2.

Lemma

2

0 −2

(∂βt u)v dt =

(∂βt u)u dt = cos 1

2πβ

π

∫ ∞ 0

Lemma For 0 < β < 1, and real-valued u and v,∫ ∞

0 (∂βt u)v dt

Proof

Noting that u(−iy) = u(iy) and v(−iy) = v(iy), and using the Cauchy–Schwarz inequality, we have∫ ∞

0 (∂βt u)v dt

= 1

π

∫ ∞ 0

2

∫ T

≤ ∫ T

1

0 ∂βt v , v dt.

Proof. Extend u and v by zero. For any µ > 0, the mth Fourier coefficients satisfy

2

∫ T

∫ ∞ 0

) ,

and the result follows by summing over m, using Parseval’s identity and choosing µ = cos 1

2πβ.

GN(U,X ) = U0,X 0 ++

1 2U

0 +2 + 1

2

2U 0 +2 and cancel the

term 1 2U

∫ tN

g(t) = I1−αA−1/2f (t) and v(t) = A1/2U(t),

so that

Piecewise-constant case

If pn = 0 then U(t) ≤ Un∗ − = max0≤n≤N Un for 0 ≤ t ≤ tN .

Since

[U]n2

∫ tn∗

Un∗ − UN

Piecewise-linear case

{ Un−1

∫ t1

− + [U]n−1,

≤ (2 + 2)

( 2U0,U0

) ,

and by choosing n∗ such that UIn∗ = max1≤n≤N UIn we see

UIn ≤ 8

0 f (t) dt

) for 1 ≤ n ≤ N.

However, for p ≥ 2 we have not been able to prove such an L∞(L2) stability bound that mimics the one for the continuous problem:

u(t) ≤ u0+

Convergence

For simplicity, assume now that Sn = H1 0 () for all n (so no spatial

discretization). We decompose the DG error as

U − u = ϑ+ %, ϑ = U − Πu, % = Πu − u,

where the quasi-interpolant Πu ∈ W(t,S,p) is defined by the conditions

(Πu)n− = u(t−n ) and

(u − Πu)tq−1 dt = 0

for 1 ≤ q ≤ pn and 1 ≤ n ≤ N, with (Πu)0 − = u(0).

Example

Example

where avgIn(u) = t−1 n

∫ In

(Πu)(t)− u(t) =

tn − t

t2 n

=

tn − t

t2 n

u − ΠuIn ≤ 2

u − ΠuIn ≤ 3tn

∫ In

In the general case we have the following estimate.

Theorem ([Schoetzau+Schwab-2000])

(u − Πu)′(t)2 dt ≤ Cε(pn, q)

( tn

2

)] dt.

GN(U,X ) = UN − ,X

)] dt.

GN(U,X ) = U0,X 0 ++

and, since [u]n = 0, the exact solution satisfies

GN(u,X ) = u0,X 0 ++

Furthermore, the construction of Π ensures

%n− = 0 and

so

GN(%,X ) =

∫ tN

0 +,

Therefore,

t %,X dt

for all X ∈ W(t,p,S), showing that ϑ is the DG solution with initial data U0 − u0 and source term −∂1−α

t A%. By applying the stability result to ϑ it is possible to prove the following error estimate.

h-Version accuracy

Theorem ([Mustapha-2015])

tn = (n/N)γT and p = (1, p, p, . . . , p)

and

u(j)(t)1 ≤ Ctσ−j for 0 < t ≤ T and 1 ≤ j ≤ p + 1,

then

U(t)− u(t) ≤ C trN ≤ CN−r for 0 ≤ t ≤ T ,

where

r =

min{γ(σ + 1 2α−

2α− 1 2} −

Theorem ([McLean+Mustapha-2015])

Suppose f ≡ 0, tn = n t and pn = 0 for

Updated November 27, 2015

Introduction

This lecture provides some key definitions and results from fractional calculus, needed for our study of fractional PDEs. The literature contains several concepts of fractional differentiation, but we focus only on the Riemann–Liouville and Caputo definitions, with a brief mention of the Grunwald–Letnikov approach.

In the sequel, we work almost exclusively with the Riemann–Liouville fractional integral and derivative.

Outline

Fractional integration

Motivation: consider the n-fold integration operator Ina+ based at a, defined recursively by

I0 a+f (x) = f (x)

and

We claim

(n − 1)! f (y) dy for n ≥ 1.

The formula holds for n = 1 because (x − y)0/0! = 1 and

I1 a+f (x) =

Let n ≥ 1 and assume

Ina+f (x) =

Then

∫ x

∫ x

a

∫ z

a

=

=

and Γ(n + 1) = n! for any integer n ≥ 0.

For any real α > 0, we define the left-sided, Riemann–Liouville fractional integration operator of order α by

Iαa+f (x) =

This definition is consistent with our earlier definition of Ina+

when α = n.

⟩ = ⟨ f , Iαb−g

Iαb−g(x) =

Semigroup property

From the recursive definition, we see that

Ima+Ina+ = Im+n a+ for all integers m ≥ 0, n ≥ 0.

Key question: does

a+ for all α > 0 and β > 0?

Consider

∫ x

a

=

Putting t = (z − y)/(x − y), we have

z = y + t(x−y), x− z = (1− t)(x−y), z−y = t(x−y),

so the Beta function identity∫ 1

0 (1− t)α−1tβ−1 dt = B(α, β) =

Γ(α)Γ(β)

Γ(α)Γ(β)

∫ 1

= (x − y)α+β−1

Iα = Iα0+ for α > 0.

The fractional integral is given by the Laplace convolution

Iαf (x) =

0 Υα(x − y)f (y) dy = Υα ∗ f (x), x > 0.

We easily see that

In fact, since ∗ is associative,

(Υα ∗Υβ) ∗ f = Υα ∗ (Υβ ∗ f ) = Iα(Iβf ) = Iα+βf = Υα+β ∗ f

for every continuous f .

generalising the identity

Iαa+Υβ,a(x) =

=

= (Υα ∗Υβ)(x − a) = Υα+β(x − a),

or in other words,

Fractional differentiation Assume that

n − 1 < α ≤ n for some n ∈ {1, 2, 3, . . .},

and write Dn = (d/dx)n.

Dαa+f (x) = DnIn−αa+ f (x) for x > a.

whereas the Caputo fractional derivative is defined by

CDαa+f (x) = Iα−na+ Dnf (x) for x > a.

Lemma For x > a and β > 0,(

DIβa+ − I β a+D

) f (x) = f (a)Υβ(x − a).

Proof

By the fundamental theorem of calculus,

Ia+Df (x) =

so f (x) = Ia+Df (x) + f (a)Υ1(x − a),

Thus, Iβa+f (x) = Iβ+1

a+ Df (x) + f (a)Υβ+1(x − a),

and finally

DIβa+f (x) = DIa+ Iβa+Df (x) + f (a)Υβ(x − a).

Relation between Dα and CDα

Theorem If n − 1 < α < n, then

Dαa+f (x) = CDαa+f (x) + n−1∑ k=0

Dk f (a) (x − a)k−α

Γ(k + 1− α) , x > a.

Proof. In the case n = 2, we have 1 < α < 2 and the Lemma gives

Dαa+f (x) = D2I2−α a+ f (x) = D

( I2−α a+ Df (x) + f (a)Υ2−α(x − a)

) = I2−α

a+ D2f (x) +Df (a)Υ2−α(x − a) + f (a)Υ1−α(x − a)

= CDαa+f (x) + f (a) (x − a)−α

Γ(1− α) +Df (a)

Differentiating a shifted Gel’fand–Shilov function

Lemma If α > 0, β > 0 and x > a, then

Dαa+Υβ,a(x) = Υβ−α,a(x).

Dαa+Υβ,a(x) = DnIn−αa+ Υβ,a(x) = DnΥn−α+β,a(x) = Υβ−α,a(x).

In particular, since Υ1,a(x) ≡ 1, if x > a then

Dαa+1(x) = Υ1−α,a(x) = (x − a)−α

Γ(1− α) whereas CDαa+1(x) = 0.

Relation between Dα and CDα restated

Since (x − a)k−α

the relation

Dαa+f (x) = CDαa+f (x) + n−1∑ k=0

Dk f (a) (x − a)k−α

Γ(k + 1− α) , x > a,

may be re-stated in the form

CDαa+f (x) = Dαa+

k!

Alternative representation

Dαa+f (x) = f (x)Υ1−α(x−a)+

∫ x

∫ x

] dy ,

noting that the derivative of the integral on the right is∫ x

a Υ−α(x − y)

[ f (y)− f (x)

Representation as a Hadamard finite-part integral

Assume 0 < α < 1 and x > a. Then∫ x

a Υ−α(x − y)

[ f (y)− f (x)

[ Υ1−α(ε)−Υ1−α(x − a)

] ,

Γ(1− α) +

∫ x−ε

and therefore

Dαa+f (x) = “I−αa+ f (x)” = fp ε↓0

∫ x−ε

Grunwald–Letnikov definition Can we define a fractional derivative (or integral) directly, without using integer-order derivatives and integrals? Denote the backward difference by

hf (x) = f (x)− f (x − h).

Can check by induction on k that

k hf (x) =

k∑ j=0

Hence define the fractional backward difference of order α by

α h,nf (x) =

n∑ j=0

k hf (x) =

and thus

k hf (x)

GLDαa+f (x) = lim α

h,nf (x)

hα ,

where the limit is obtained by sending n→∞ and h→ 0+ keeping

h = x − a

Can show that if m − 1 < α < m, then

GLDαa+f (x) = m−1∑ k=0

f (k)(a) (x − a)k−α

Γ(k + 1− α)

which means that

GLDαa+f (x) = CDαa+f (x) + m−1∑ k=0

f (k)(a) (x − a)k−α

Γ(k + 1− α) = Dαa+f (x).

Furthermore,

Part II

Useful tools

Introduction

We will make extensive use of the Laplace transform (and some use of the Fourier transform), first to derive the fractional diffusion equation and then to study properties of the solution. Laplace transformation also plays a large role in some of the numerical methods we study, either as part of the method itself or for the error analysis.

This lecture also introduces some special functions that are arise in the study of fractional initial-boundary value problems.

Outline

If f is locally integrable on [0,∞), and if

|f (t)| ≤ Ceλt for t > 0,

then f (z) exists and is analytic for <z > λ, and we have the inversion formula

f (t) = 1

Transform of a Gel’fand–Shilov function

For α > 0 and z > 0, the substitution y = tz gives

Υα(z) = 1

Γ(α)

∫ ∞ 0

consistent with

z−α−β = Υα+β(z) = L(Υα ∗Υβ) = Υα(z)Υβ(z) = z−αz−β.

Laplace transform of an integral

Since

we have L{If (t)} = Υ1(z)f (z) = z−1f (z).

In general, Inf = Υn ∗ f so

L{Inf (t)} = z−n f (z) for n ∈ {0, 1, 2, . . .}.

Laplace transform of a derivative

Integration by parts shows

= 0− f (0) + z

so L{Df (t)} = zf (z)− f (0).

Easily verify by induction on n that

L{Dnf (t)} = zn f (z)− n−1∑ k=0

zn−1−kDk f (0).

Laplace transform of a Caputo fractional derivative

If n − 1 < α < n, then

CDαf (t) = In−αg(t) where g(t) = Dnf (t),

so

= zα−n (

zn−1−kDk f (0)

) and so

zα−1−kDk f (0).

Laplace transform of a Riemann–Liouville fractional derivative

If n − 1 < α < n, then

Dαf (t) = CDαf (t) + n−1∑ k=0

Dk f (0)Υk+1−α(t),

and since Υk+1−α(z) = z−(k+1−α) we have

L{Dαf (t)} = L{CDαf (t)}+ n−1∑ k=0

Dk f (0)zα−1−k ,

that is, L{Dαf (t)} = zαf (z).

Mittag–Leffler function

Problem: find f (t) satisfying

CDαf (t) = f (t) for t > 0, with f (0) = 1.

If α = 1 then f (t) = et .

If 0 < α < 1, then we claim

f (t) = ∞∑ k=0

Υ1+kα(t) = ∞∑ k=0

where the Mittag–Leffler function is

Eα(z) = ∞∑ k=0

CDαΥ1+kα = I1−αDΥ1+kα = I1−αΥkα = Υ1+(k−1)α,

so

) = 0 + Υ1 + Υ1+α + Υ2+α + · · · = f .

Also, Υ1(t) ≡ 1 and Υ1+kα(0) = 0 for k ≥ 1,

so f (0) = 1.

Convergence?

We claim Eα(z) is an entire function of z . By the ratio test, it suffices to show that as k →∞, zk+1

Γ ( 1 + (k + 1)α

Γ(x) =

√ 2π

x

( x

e

Special choices of α

∫ ∞ −x

CDαu + λu = 0 for t > 0, with u(0) = 1.

We claim that the solution is

u(t) = Eα(−λtα) = ∞∑ k=0

(−λ)ktkα

Equivalent formulation

I1−αDu(t) + λu(t) = 0.

To obtain an equation involving a Riemann–Liouville fractional derivative, apply DIα and obtain

DI1Du(t) + λDIαu(t) = 0.

Du(t) + λD1−αu(t) = 0.

Taking Laplace transforms, zu(z)− u(0) + λz1−αu(z) = 0, and thus (z + λz1−α)u(z) = 1, showing that

u(z) = L{Eα(−λtα)} = 1

z + λz1−α .

Integral representation

1

Γ(a) =

1

2πi

∫ 0+

encircles the negative real axis and has a counterclockwise orientation.

Theorem The Mittag–Leffler function admits the integral representation

Eα(z) = 1

provided the Hankel contour encloses the disc |w | ≤ |z |1/α.

Proof

= 1

2πi

∫ 0+

Eα(−λtα) = 1

z + λz1−α , t > 0.

By collapsing the Hankel contour onto the negative real axis, we find

Eα(−λtα) = 1

)2 + λ2 sin2 απ

dt Eα(−λtα) < 0 for all t > 0.

Asymptotic behaviour

1 + zα =

Eα(−tα) ∼ ∞∑ n=0

Notice what happens as α→ 1.

Wright functions

Wλ,µ(z) = ∞∑ k=0

k!Γ(λk + µ) , λ > −1, µ ∈ C.

This series converges for all z ∈ C so Wλ,µ is an entire function.

Theorem The Wright function has the integral representation

Wλ,µ(z) = 1

Mα(t) = W−α,1−α(−t) = ∞∑ k=0

(−1)ktk

k!Γ ( 1− (k + 1)α

) , 0 < α < 1, introduced by F. Mainardi in 1994. The identity

1

M1/2(t) = exp(−t2/4)√

Wright M-function Mα(t)

Integral representation and Laplace transform

Putting λ = −α and µ = 1− α in the integral representation of the Wright function, and replacing t by −t, gives

Mα(t) = 1

which allows us to prove the following result.

Theorem For 0 < α < 1, the Laplace transform of Mα is

Mα(z) = Eα(−z).

= 1

2πi

∫ 0+

∫ ∞ −∞

e−iξx f (x) dx .

If f ∈ L1(R) then f is continuous on R and f (ξ)→ 0 as |ξ| → ∞.

Plancherel theorem: Fourier transform extends uniquely to a bounded linear operator F : L2(R)→ L2(R) satisfying

1

2π

Symmetric Wright M-function

Lemma For κ > −1 and 0 < α < 1,∫ ∞

0 tκMα(t) dt =

Proof of the lemma

so∫ ∞ 0

Part III

Anomalous diffusion

The classical diffusion equation,

ut − K∇2u = 0,

describes how the concentration u of a substance evolves over time if, at the microscopic scale, its particles exhibit Brownian motion. The equation can also be derived from a purely macroscopic argument based on conservation of mass and Fick’s law, which states that the mass flux vector equals −K∇u.

In this lecture, we study continuous-time random walks, which provide a generalization of Brownian motion. When the waiting-time distribution obeys a power law, the particles experience trapping and their macroscopic behaviour is said to be subdiffusive. The mean-square displacement of such a particle is proportional to tα for a characteristic exponent in the range 0 < α < 1.

With the help of Fourier and Laplace transformation, we show that the macroscopic concentration obeys a time-fractional PDE,

ut − ∂1−α t Kα∇2u = 0,

where it is convenient to write ∂1−α t = D1−α

0+ for the Riemann–Liouville fractional derivative with respect to t. The constant Kα > 0 is a generalized diffusivity, and the classical diffusion equation then arises as the limiting special case when α→ 1.

Outline

Continuous-time random walks

A walker moves along the x-axis, starting at position x0 at time t0 = 0. At time t1 the walker jumps to x1, then at time t2

jumps to x2, and so on. We assume that the increments

tn = tn − tn−1 and xn = xn − xn−1

are independent, identically distributed random variables with probability density functions ψ(t) and λ(x), respectively. That is,

P(a < tn < b) =

and

a λ(x) dx for −∞ < a < b <∞.

Aim: determine the probability that the particle lies in a given spatial interval at time t.

An example

Suppose that the waiting-time distribution is exponential with parameter τ > 0,

ψ(t) = τ−1e−t/τ for 0 < t <∞,

and that the jump-length distribution is normal with mean 0 and variance σ2,

λ(x) = 1

Thus,

E(tn) = τ, E(t2 n) = τ2, E(xn) = 0, E(x2

n ) = σ2.

The position x(t) of the walker is a step function.

Typical path (τ = 1, σ = 1, x0 = 2)

Random walk in 3D (σx = σy = σz = 1)

Convolutions and probability

We denote the Fourier convolution of f and g by

f ~ g(z) =

f (z − y)g(y) dy , for −∞ < z <∞.

If f (x) = 0 for x < 0 and g(y) = 0 for y < 0 then this integral equals the Laplace convolution

f ∗ g(z) =

0 f (z − y)g(y) dy for z > 0.

Theorem If X and Y are independent random variables with probability density functions f and g, respectively, then the sum Z = X + Y has probability density function f ~ g.

Probability distribution of tn

Let ψn(t) denote the probability density function of the random variable

tn = t1 + t2 + · · ·+ tn,

By the theorem quoted above,

ψn(t) = (ψn−1 ∗ ψ)(t) =

n factors

Survival probability

Let Ψ(t) denote the survival probability, that is, the probability of the walker not jumping within a time t, or equivalently, the probability of remaining stationary for at least a duration t. Then,

Ψ(t) =

∫ ∞ t

0 ψ(s) ds for 0 < t <∞.

It follows that the probability of taking exactly n steps up to time t is

χn(t) =

∫ t

Probability distribution of xn − x0

Let λn(x) denote the probability density function of the random variable

xn − x0 = x1 + x2 + · · ·+ xn,

that is,

Since

n factors

ψ(z) = Lψ(z) =

whereas the characteristic function of λ(x) is its Fourier transform,

λ(ξ) = Fλ(ξ) =

e−iξxλ(x) dx .

Since ψn and λn are n-fold convolutions of ψ and λ, respectively,

ψn(z) = ψ(z)n and λn(ξ) = λ(ξ)n for n ≥ 0.

The characteristic function for the survival probability Ψ = 1− I1ψ is

Ψ(z) = z−1 − z−1ψ(z) = 1− ψ(z)

z .

Probability density

Let p(x , t) denote the probability density function for the position of the particle at time t, that is,

P(a < x(t)− x0 < b) =

a p(x , t) dx .

Since χn(t) is the probability of taking n steps up to time t,

p(x , t) = ∞∑ n=0

z .

ˆp(ξ, z) = LFp(ξ, z) =

ˆp(ξ, z) = ∞∑ n=0

λ(0) =

But if ξ 6= 0 or z > 0, then λ(ξ)ψ(z)

< 1 so

∞∑ n=0

σ √

2ξ2/2,

so

1 + τz ,

Uncertain initial position

Instead of assuming x(0) = x0 is known, we can treat x0 as a random variable with probability density p0(x), so that

P(a < x0 < b) =

a p0(x) dx for −∞ < a < b <∞.

Let λn denote the probability density function for xn (rather than xn − x0, as before), so

P(a < xn < b) =

Since xn = x0 + x1 + · · ·+ xn,

we have λn = p0 ~ λ~ λ~ · · ·~ λ

n factors

Using [λ(ξ)]np0(ξ)

ˆp(ξ, z) = 1− ψ(z)

1− λ(ξ)ψ(z) ,

that is, the only change is to introduce a factor p0(ξ).

Formally, put p0(x) = δ(x − x0) to recover the case when x0 is known with certainty.

Rescaling

Assume now that the probability density functions ψ(t) and λ(x) are normalized to satisfy∫ ∞

0 tψ(t) dt = 1,

x2λ(x) dx = 1.

Let τ > 0 and σ > 0, and let the random variables tn and xn now have the rescaled probability density functions

ψτ (t) = 1

E(tn) = τ, E(xn) = 0, E(x2 n ) = σ2.

We want to investigate what happens as τ and σ tend to zero.

Typical path (σ = 0.1, τ = 0.005)

Detailed view of inset

dk ψ

we have

z

1

ψ(z) = ψ(0) + ψ′(0)z + · · · = 1− z + O(z2) as z → 0,

implies that

1− ψ(τz)

) as τ → 0.

Assume for simplicity that λ(−x) = λ(x). Then λ′′′(0) = 0 and

λ(ξ) = λ(0) + λ′(0)ξ + 1 2 λ ′′(0)ξ2 + · · · = 1− 1

2ξ 2 + O(ξ4).

τz + 1 2σ

2ξ2 × 1 + O(τz)

1 + O(τz + σ2ξ2) .

Limiting probability density Now send σ → 0 and τ → 0 while keeping

σ2

ˆp(ξ, z) = lim τ

p(ξ, t) = 1

p(x , t) = 1√

Notice that

= (z + Kξ2)ˆp(ξ, z)− p(ξ, 0)

= 1− p(ξ, 0) = 0,

where the final step follows because p(x , 0) = δ(x) and so p(ξ, 0) = 1.

Therefore, p(x , t), the probability density for x(t)− x0, the position (relative to x0) of the walker at time t, satisfies the partial differential equation

pt − Kpxx = 0 for 0 < t <∞ and −∞ < x <∞.

Uncertain initial position

τz + 1 2σ

2ξ2 × 1 + O(τz)

1 + O(τz + σ2ξ2) ,

ˆp(ξ, z) = p0(ξ)

= (z + Kξ2)ˆp(ξ, z)− p(ξ, 0)

= p0(ξ)− p(ξ, 0) = 0,

since p(x , 0) = p0(x).

Subdiffusion

Let 0 < α < 1, and suppose now that the waiting time probability density function is a power law:

ψ(t) ∼ A

tψ(t) dt = +∞,

so the preceding analysis of the random walk breaks down.

We make no change to our assumptions on λ(x).

Example

1

σ √

Typical path (α = 0.75, σ = 1)

Behaviour of the characteristic function as z → 0 Assume there exists T > 0 such that

|t1+αψ(t)− A| ≤ Ct−1 for T ≤ t <∞,

and let 0 < z ≤ T−1 (so Tz ≤ 1). Since we know ψ(0) = 1, consider

1− ψ(z) =

where

I1 =

∫ T

I2 =

∫ ∞ T

I3 =

∫ ∞ T

(1− e−zt)At−1−α dt.

Since 0 ≤ 1− e−y ≤ min(1, y) for 0 ≤ y ≤ 1,

we immediately see that

|I2| ≤ ∫ ∞ T

∫ ∞ Tz

≤ Cz1+α

∫ ∞ 1

I3 =

∫ ∞ T

1− e−y

where

∫ Tz

0

AT−αz

1− α ,

we have shown that1− ψ(z)− Bαzα = |I1 + I2 + I4| ≤ CT ,αz for 0 < z ≤ T−1.

Integrating by parts,∫ ∞ 0

) =

which completes the proof of the following result.

Theorem If

then, with Bα = Aα−1Γ(1− α),

ψ(z) = 1− Bαzα + O(z) as z → 0.

Rescaling

xλ(x) dx = 0 and

ψτ (t) = 1

ψτ (t) ∼ Aτα

As before,

z

1

but this time

where Bα = Aα−1Γ(1− α).

Thus,

] .

so

= [ Bατ

Limiting probability density

αzα−1

1 + O ( τ1−αz1−α + ταzα + σξ

) . Once again, send σ → 0 and τ → 0, but now keep

σ2

ˆp(ξ, z) = lim Bατ

zα−1

zα + Kαξ2 .

Notice that we recover the earlier formula by putting α = 1.

Typical path (α = 0.75, σ = 0.1, τα = σ2/2, N = 800)

Detailed view of inset

z + λz1−α

Since

we see that

Kαtα.

Thus,

Fractional partial differential equation for p

We have

LF{pt − KαD1−αpxx} = z ˆp(ξ, z)− p(ξ, 0) + Kαz1−αξ2ˆp(ξ, z)

= ( z + Kαz1−αξ2

since

Thus, p satisfies the time-fractional diffusion equation,

pt − KαD1−αpxx = 0 for 0 < t <∞ and −∞ < ξ <∞.

Mean-square displacement

) =

∫ ∞ −∞

Since

E ( (x(t)− x0)2

Introduction

We have shown that in 1D the probability density function for the location of a subdiffusive particle at time t obeys a time-fractional PDE. The concentration u = u(x , t) of a large number of such particles evolves in the same way. Moreover, the 1D analysis can be generalized to higher dimensions to yield the time-fractional diffusion equation

ut − Kα∂ 1−α t ∇2u = 0.

As in the classical case α = 1, the solution u in a bounded spatial domain (and subject to homogeneous boundary conditions) can be constructed by separation of variables to yield a series expansion involving the eigenfunctions of −∇2. We use this representation of u(x , t) to investigate its behaviour.

Outline

Positivity

Initial-boundary value problem

Let denote a bounded, Lipschitz domain in R2 or R3. We seek u = u(x , t) satisfying

ut − Kα∂ 1−α t ∇2u = f (x , t) for x ∈ and t > 0,

u = u0(x) for x ∈ , when t = 0,

and impose homogeneous boundary conditions, either Dirichlet

u = 0 for x ∈ ∂ and t > 0,

or else Neumann,

where n is the outward unit normal to .

Abstract initial-value problem

Let A be a linear operator with dense domain D(A) in a Hilbert space H with inner product u, v and norm u =

√ u, u. Given

u0 ∈ H and f : [0,∞)→ H we seek u : [0,∞)→ H satisfying

u + ∂1−α t Au = f (t) for t > 0,

with u(0) = u0, where u = ut = ∂u/∂t.

Standard example:

H = L2(), Au = −Kα∇2u, D(A) = H2() ∩ H1 0 ().

Integrate to obtain an equivalent Volterra equation in H,

u(t) +

∫ t

∫ t

Eigenfunction expansion Assume that A is self-adjoint and positive-semidefinite, with a complete orthonormal eigensystem, say

Aφm = λmφm for m = 0, 1, 2, . . . ,

with φm, φn = δmn. Number the eigenvalues so that

0 ≤ λ0 ≤ λ1 ≤ λ2 ≤ · · · .

u(t) = ∞∑

um(t)φm where um(t) = u(t), φm.

Likewise, putting fm(t) = f (t), φm and u0m = u0, φm, we have

f (t) = ∞∑

m=0

Dirichlet boundary conditions in 1D

Take = (0, L) and A = −Kαd2/dx2 with homogeneous Dirichlet boundary conditions. Then,

φm(x) =

√ 2

( mπ

L

)2

v(x) = ∞∑

vm = v , φm =

Neumann boundary conditions in 1D

Again take = (0, L) and A = −Kαd2/dx2, but now impose homogeneous Neumann boundary conditions. Then,

φ0(x) = 1√ L

and λ0 = 0,

( mπ

L

)2

Separation of variables

The function u satisfies

u + ∂1−α t Au = f (t) for t > 0, with u(0) = u0.

iff the mth eigenmode satisfies

um + λm∂ 1−α t um = fm(t) for t > 0, with um(0) = u0m,

for m = 0, 1, 2, . . . . Laplace transformation gives

zum(z)− um(0) + λmz1−αum(z) = fm(z)

so

Recall that

E(t)v = ∞∑

m=0

Eα(−λmtα)v , φmφm for t > 0 and v ∈ H,

then the mild solution of the abstract initial-value problem is

u(t) = E(t)u0 +

Caution: E(t + s) 6= E(t)E(s) if 0 < α < 1.

Stability in H Recall that Eα(−λtα) is positive and decreasing for t > 0, and equals 1 at t = 0, so

0 < Eα(−λtα) ≤ 1 for 0 ≤ t <∞ and any λ ≥ 0.

Thus, using Parseval’s identity,

E(t)v2 = ∞∑

m=0

m=0

and therefore

u(t) ≤ u0+

Smoothing property of fractional diffusion

For 0 ≤ r <∞, define the norm

v2 r =

(I + A)r/2v2 = ∞∑

(1 + λm)r v , φm2

and the corresponding closed subspace

Hr = { v ∈ H : vr <∞},

which is a Hilbert space with respect to the inner product that induces · r .

For our standard example H = L2() and A = −Kα∇2, write Hr = Hr

D() or Hr N() to indicate the choice of Dirichlet or

Neumann boundary conditions.

Can prove the following via interpolation and elliptic regularity.

Theorem Suppose that ∂ is C∞. If 0 ≤ r < 1

2 , then Hr D() = H r (),

however if 2j − 3 2 < r < 2j + 1

2 for j ∈ {1, 2, 3, . . .}, then

Hr D() = { v ∈ H r () : v = Av = · · · = Aj−1v = 0 on ∂ }.

In the exceptional case r = 2j − 3 2 , the condition Aj−1v = 0 on ∂

must be replaced by Aj−1v ∈ H1/2().

If is Lipschitz, then the conclusions still hold for r ≤ 1, and in particular H1

D = H1 0 () = { u ∈ H1() : u = 0 on ∂ }.

If is convex or C 1,1, then r ≤ 2 is OK, and in particular H2

D = H2() ∩ H1 0 () = { u ∈ H2() : u = 0 on ∂ }.

Neumann boundary conditions and Sobolev spaces

Theorem Suppose that ∂ is C∞. If 0 ≤ r < 3

2 , then Hr N() = H r (),

however if 2j − 1 2 < r < 2j + 3

2 for j ∈ {1, 2, 3, . . .}, then

Hr N() = { v ∈ H r () :

∂nv = ∂nAv = · · · = ∂nAj−1v = 0 on ∂ }.

In the exceptional case r = 2j − 1 2 , the condition ∂nAj−1v = 0

on ∂ must be replaced by ∂nAj−1v ∈ H1/2().

If is Lipschitz, then the conclusions still hold if r ≤ 1, and in particular H1

N() = H1().

If is convex or C 1,1, then r ≤ 2 is OK, and in particular H2

N() = { u ∈ H2() : ∂nu = 0 on ∂ }.

A 1D example

Consider v(x) = 1 for x ∈ = (0, L). Since∫ L

0 v(x) sin

vr <∞ ⇐⇒ ∞∑ p=0

(1 + 2p)2r−2 <∞,

so v ∈ Hr D() iff 2r − 2 < −1, that is, r < 1

2 . However, v ∈ Hr

0 v(x) cos

Smoothing property of classical diffusion

If α = 1 then Eα(−tα) = e−t so

E(t)v = ∞∑

and thus

)2 .

(1 + λ)µ(e−λt)2 ≤ t−µ(T + λt)µe−2λt ≤ CT ,µt−µ

and so E(t)vr+µ ≤ CT ,µt−µ/2vr for µ ≥ 0.

Weaker smoothing property for subdiffusion

The theorem below shows that if v ∈ Hr then E(t)v ∈ Hr+2 for each t > 0, but E(t)vr+2 may blow up as t → 0+.

Lemma 0 < Eα(−tα) ≤ C min(1, t−α) for 0 < t <∞.

Proof. Follows because

t−α/Γ(1− α) + O(t2−α) as t →∞.

Theorem Let 0 ≤ µ ≤ 2 and 0 ≤ r <∞. If v ∈ Hr , then

E(t)vr+µ ≤ CT t−αµ/2vr for 0 < t ≤ T .

Proof of theorem Put g(t) = Eα(−tα). Since g(λ1/αt) = E (−λtα), we have

E(t)v2 r+µ =

The lemma implies that (assuming 0 ≤ µ ≤ 2)

0 < g(t) ≤ C (1 + tα)−µ/2 for 0 < t <∞,

so, for 0 < t ≤ T ,

g(λ1/αt)2 ≤ C (1 + λtα)−µ = Ct−µα(t−α + λ)−µ

≤ CT t−µα(1 + λ)−µ

and thus

∞∑ m=0

(1 + λm)r v , φm2 = CT t−αµv2 r .

Regularity in time

Let q ∈ {1, 2, 3, . . .}. Similar arguments yield the following estimates.

Lemma The function g(t) = Eα(−tα) satisfies

tq|g (q)(t)| ≤ Cq min(tα, t−α) for 0 < t <∞.

Theorem Let −2 ≤ µ ≤ 2, 0 ≤ r <∞ and q ∈ {1, 2, 3, . . .}. If v ∈ Hr , then

tqE(q)(t)vr+µ ≤ Cq,T t−αµ/2vr for 0 < t ≤ T .

Detailed behaviour as t → 0+

Since

(−1)ptαp

Γ(1 + αp) λp + O(λMtαM) as t → 0+,

and λpmv , φm = Apv , φm, we can show the following.

Theorem Let 0 ≤ r <∞ and M ∈ {1, 2, 3, . . .}. If v ∈ Hr+2M , then

E(t)v = v + M−1∑ p=1

(−1)ptαp

where, given 0 ≤ µ ≤ 2, the remainder operator satisfies

RM(t)vr+µ ≤ CM,T tMα−αµ/2vr for 0 < t ≤ T .

Behaviour of an eigenmode

If the initial data is an eigenfunction of A, say u0 = φm, then the solution of the homogeneous problem is

u(t) = E(t)φm = Eα(−λmtα)φm,

) φm(x) as t → 0+.

This example makes clear the fact that the time derivative u = O(tα−1) is unbounded as t → 0+ no matter how regular the initial data (so long as it is not zero, but in that case u ≡ 0).

Contrast this behaviour with that of the classical diffusion equation (α = 1): if u0 = φm then u(x , t) = e−λmtφm(x) is C∞ for t ≥ 0.

The inhomogeneous problem The function u(t) = E(t)u0 solves the homogeneous problem

u + ∂1−α t Au = 0 for t > 0, with u(0) = u0.

Now consider

which solves the inhomogeneous problem with vanishing initial data:

u + ∂1−α t Au = f (t) for t > 0, with u(0) = 0.

Theorem For 0 ≤ r <∞ and q ∈ {0, 1, 2, . . .}, the function u = E ∗ f satisfies

tqu(q)(t)r ≤ Cq

0 s jf (j)(s)r ds for 0 < t <∞.

Positivity

{ 0, t < 0,

u(t), t ≥ 0, ,

then the Laplace transform of u is related to the Fourier transform of u+ by

u(iy) = u+(y) =

f (t)g(t) dt = 1

u(iy)v(iy) dy .

We can now show that the operator ∂1−α t is positive semidefinite.

Theorem If 0 < β < 1 and if u is real-valued, then∫ ∞

0 (∂βt u)u dt =

Proof. Since L{∂βt u(t)} = zβ u(z),∫ ∞

0 (∂βt u)u dt =

(iy)β|u(iy)|2 dy .

The result follows because u(iy) = u(−iy) and for y > 0,

(±iy)β = (e±iπ/2y)β = yβ ( cos 1

2πβ ± i sin 1 2πβ

) .

Lemma For 0 < α < 1 and suitable u : (0,∞)→ H,∫ ∞

0 ∂1−α

2πα

π

∫ ∞ 0

Proof.

= ∞∑

Digression: fractional derivative at a jump discontinuity

Suppose that

v2(t), t > a,

where v1 : [0, a]→ R and v2 : [a,∞)→ R are C 1 functions. If 0 < t < a, then differentiating the formula

Iαv(t) = Υα ∗ v(t) =

∫ t

= v(0+)Υα(t) +

Iαv(t) =

∫ a

∫ t−a

+

∫ t−a

and therefore, with [v ]a = v2(a)− v1(a) = v(a+)− v(a−),

∂1−αv(t) = v(0+) Υα(t) + [v ]aΥα(t − a)

+

∫ t

Example

Consider the homogeneous equation,

u(t) + ∂1−α t Au(t) = 0.

Take the inner product with u(t) and integrate to obtain∫ T

0 u, u dt +

Letting

Thus, ∫ T

But ∫ T

2u(0)2,

E(t)u0 ≤ u0 for t > 0.

Part V

Introduction

We begin our study of numerical methods for fractional diffusion problems by considering simple explicit and implicit finite difference (and quadrature) schemes in the 1D case:

ut − Kα∂ 1−α t uxx = f (x , t).

These schemes generalize the forward and backward Euler methods for the heat equation. As in the classical setting, the explicit scheme is stable only if the time step is sufficiently small, but the implicit scheme is unconditionally stable.

Unlike the classical Euler methods, the grid stencils extend back through all preceding time levels resulting in a dramatically higher computational cost.

Outline

Implicit Euler method

Consider first the scalar problem (A = λ > 0)

u + λ∂1−α t u = f (t) for 0 < t < T , with u(0) = u0,

where λ > 0. Put t = T/N and define grid points

tn = nt for 0 ≤ n ≤ N.

We want to compute Un ≈ u(tn).

Time-stepping

∫ tn+1

Un+1 − Un + λ

∂1−α t U(t) dt = f n t,

where f n = f (tn) and U is the piecewise-constant function

U(t) = Un for tn ≤ t < tn+1.

Approximation of the fractional derivative

Recalling ∂1−α t v = (Υα ∗ v)t , we have∫ tn+1

tn

=

] U j dt

Γ(α + 1)

tα

] U j dt.

wj = (j + 1)α − 2jα + (j − 1)α for j ≥ 1.

Note that wj < 0 for j ≥ 1, with wj ≈ α(α− 1)jα−2 for large j .

Weights when α = 1/2

Un+1 − Un + λtα

Γ(α + 1)

Thus, starting from U0 = u0 we compute

Un+1 = Un + f n t − λtα

Γ(α + 1)

Classical Euler method

In the limiting case α = 1 we have wj = 0 for j ≥ 1 so

λtα

Γ(α + 1)

and therefore

or equivalently, Un+1 − Un

t + λUn = f n.

Un+1 =

Γ(α + 1)

wn−jU j ,

and we can prove the following discrete analogue of the stability estimate for the continuous problem:

|u(t)| ≤ |u0|+ ∫ t

Theorem If

λtα

|f n|t for 1 ≤ n ≤ N.

Proof

|Un+1| ≤ (1− ρ)|Un|+ |f n|t + ρ

n−1∑ j=0

|wn−j ||U j |

so

|Un+1| ≤ (1− ρ)|Un|+ |f n|t + ρ max 0≤j≤n−1

|U j |

|U j |.

The desired estimate now follows using induction on n.

Example: α = 1/2, λ = 1, f ≡ 0, N = 25, ρ = 0.3192

Example: α = 1/2, λ = 4, f ≡ 0, N = 25, ρ = 1.2766

Explicit Euler method for a PDE

Consider = (0, L) in 1D with boundary ∂ = {0, L}. We seek u = u(x , t) satisfying

ut − Kα∂ 1−α t uxx = f (x , t) for x ∈ and 0 < t < T ,

u = u0(x) for x ∈ , when t = 0,

u = 0 for x ∈ ∂ and 0 < t < T .

Put x = L/P and t = T/N, and define grid points

(xp, tn) = (p x , n t) for 0 ≤ p ≤ P and 0 ≤ n ≤ N.

We want to compute Un p ≈ u(xp, tn).

Second central difference in space Using the approximation

uxx(xp, tn) ≈ Un p+1 − 2Un

p + Un p−1

x2

and letting f n p = f (xp, tn), we discretize in time as before and

arrive at the scheme

Un+1 p − Un

p + U j p−1

x2 = f n

p t

for 0 ≤ n ≤ N − 1 and 1 ≤ p ≤ P − 1, with the initial conditions

U0 p = u0(xp) for 0 ≤ p ≤ P,

and boundary conditions

Stencil

x

t

. . . . . .

. . .

so that

Γ(α + 1)

( mπ

L

)2

satisfy

with φm(0) = 0 = φm(L). Putting

Φm =

For U, V ∈ RP−1 define

U,V = P−1∑ p=1

UpVp x and U = √ U,U.

We find that Φ1, Φ2, . . . . . . , ΦP−1 form an orthogonal basis for RP−1,

Φm,Φm′ = 0 if m 6= m′ and m, m′ ∈ {1, 2, . . . ,P − 1},

and, with θ = mπ/P,

( sin

mπ

2 sin θ

L

2 .

Stability of the discrete Fourier modes Define the discrete Fourier coefficients

Un m =

Un mΦm.

Un+1 m − Un

m t,

|Un m| ≤ |U0

Λm tα

Since

|Um|2Φm2 = L

2

Theorem If

n−1∑ j=0

Problem if α is small

The stability restriction ρ ≤ 1 means that the time step must be chosen so that

tα ≤ Γ(α + 1)

Example

Suppose α = 1/5, Kα = Γ(α + 1) and x = 2× 10−3, then we require

t ≤ (x/2)10 = 10−30.

Implicit Euler method

Again start with the ODE

u + λ∂1−α t u = f (t) for 0 < t < T , with u(0) = u0,

but now integrate over (tn−1, tn) to obtain

u(tn)− u(tn−1) + λ

∫ tn

Un − Un−1 + λ

Weights

=

] U j dt

ωj = (j + 1)α − 2jα + (j − 1)α for j ≥ 1.

Unconditional stability

λtα

Γ(α + 1) .

Thus, starting from U0 = u0, we compute Un for n = 1, 2, . . . , N by solving

(1 + ρ)Un = Un−1 + f n t − ρ n−1∑ j=1

wn−jU j .

Proof

|wj | = − n−1∑ j=1

wj = 1 + (n − 1)α − nα ≤ 1

so (1 + ρ)|Un| ≤ |Un−1|+ |f n|t + ρ max

1≤j≤n−1 |U j |

and therefore

+ ρ

≤ |f n|t + max 1≤j≤n−1

|U j |.

Example: α = 1/2, λ = 1, f ≡ 0, N = 25

Convergence behaviour

|Un − u(tn)|,

|Un − u(tn)|.

Recall that if we compute a sum with M terms,

S = M∑

m=1

Am

in a system of floating-point arithmetic with unit roundoff ε, then

| fl(S)− S | ≤ Mε

|Am|.

Thus, for wj = (j + 1)α − 2jα + (j − 1)α and j large,

| fl(wj)− wj | . εjα whereas wj ≈ α(1− α)jα−2

so the relative rounding error | fl(wj)− wj |/|wj | is of order εj2.

By writing wj = j −j−1 and using expm1 and log1p to evaluate

j ≡ (j + 1)α − jα = jα [ (1 + j−1)α − 1

] = jα

] − 1 ) ,

we can reduce somewhat the rounding error in fl(wj) for large j .

When we compute

(n − j)α|U j | . nα+2ε max 1≤j≤n

|U j |,

which suggests that roundoff might become a problem once nα+2 ≥ ε−1.

Fractional diffusion equation

For = (0, L), we again consider the initial-boundary value problem

ut − Kα∂ 1−α t uxx = f (x , t) for x ∈ and 0 < t < T ,

u = u0(x) for x ∈ , when t = 0,

u = 0 for x ∈ ∂ and 0 < t < T .

The implicit time-stepping scheme leads to

Un p − Un−1

p + U j p−1

x2 = f n

for 1 ≤ n ≤ N and 1 ≤ p ≤ P − 1, with

U0 p = u0(xp) and Un

0 = 0 = Un P .

(I + B)Un = Un−1 + t fn − n−1∑ j=1

wn−jBUj

. . . . . .

. . .

.

Computational cost

At the nth time step, evaluation of the RHS costs O(nP) flops, and the elliptic solve costs O(P) flops. Since

N∑ n=1

Also, the nth time step requires O(nP) active memory locations.

Contrast this with using the implicit Euler method to solve the classical diffusion equation (the case α = 1): each time step requires O(P) flops and O(P) active memory locations, and N time steps requires O(NP) flops.

Conclusion: cost when 0 < α < 1 is N times the cost when α = 1.

Comparison with direct simulation

Let u = u(x , t) be the solution of

ut − Kα∂ 1−α t uxx = f (x , t) for x ∈ = (−L, L) and 0 < t < T ,

u = δ(x) for x ∈ , when t = 0,

u = 0 for x ∈ ∂ = {−L, L} and 0 < t < T .

We can approximate u using the implicit Euler method, or by simulating CTRWs with

ψ(t) = α

1√ 2π

exp(−x2/2).

In this case ψ(t) ∼ A/t1+α as t →∞, with A = α, so Bα = Aα−1Γ(1− α) = Γ(1− α) and we must rescale in such a way that

σ2

Probability densities of CTRWs (15, 000 samples)

Part VI

Introduction

The finite element method provides the simplest approach for discretization of a fractional diffusion problem on a spatial domain of general shape. In the classical method of lines for the heat equation, such a spatial discretization leads to a large, stiff system of first-order ODEs in time, that can be integrated using an appropriate black-box routine. For a time-fractional diffusion problem, we instead obtain a system of fractional-order ODEs.

In this lecture, we seek to estimate the errors from the spatial discretization assuming that the time integration is performed exactly. Suitable approaches for the time discretization include the implicit Euler method described previously and the more accurate schemes addressed in subsequent lectures.

Outline

Method of lines

Spatial domain ⊆ Rd (d = 1, 2 or 3); for simplicity a convex polygon or polyhedron so the elliptic problem is H2-regular.

With 0 < α < 1, let u = u(x , t) be the mild solution of

ut − Kα∂ 1−α t ∇2u = f (x , t) for x ∈ and t > 0,

u = u0(x) for x ∈ , when t = 0,

subject to homogeneous Dirichlet boundary conditions

u = 0 for x ∈ ∂ and t > 0.

We wish (for now) to discretize in space only.

Weak formulation

∫

Kα ∂u

Thus,

ut , v+ a(∂1−α t u, v) = f , v for v ∈ H1

0 (),

and also

ut , v+ ∂1−α t a(u, v) = f , v for v ∈ H1

0 ().

h = max K∈Th

diam(K ).

Let Vh denote the space of real-valued functions on that are continuous piecewise polynomials of degree at most p ≥ 1 with respect to Th, and which vanish on ∂. Hence,

Vh ⊆ H1 0 ().

Seek a finite element solution uh : [0,∞)→ Vh satisfying

uht , χ+ ∂1−α t a(uh, χ) = f , χ for χ ∈ Vh and t > 0,

with uh(0) = u0h ≈ u0 for a suitable u0h ∈ Vh.

Alternative formulation

Ahψ, χ = a(ψ, χ) for ψ, χ ∈ Vh,

and the L2-projector Ph : L2()→ Vh by

Phv , χ = v , χ for v ∈ L2() and χ ∈ Vh.

Since

t Ahuh, χ = ∂1−α t Ahuh, χ,

we see that

uht + ∂1−α t Ahuh, χ = f , χ = Phf , χ for all χ ∈ Vh,

and thus uht + ∂1−α

t Ahuh = Phf for t > 0.

Nodal equations Construct a nodal basis χ1, χ2, . . . , χN for Vh so that

χj(xk) = δjk ,

where x1, x2, . . . , xN are the free (interior) nodes. Thus,

uh(x , t) = N∑

Uk(t)χk(x) where Uk(t) = uh(xk , t).

Define the mass matrix M, stiffness matrix S and load vector f by

Mjk = χk , χj, Sjk = a(χk , χj), fk(t) = f (t), χk

then the nodal vector U = [Uk(t)] satisfies the system of ordinary integro-differential equations

M dU

Discrete eigensystem

The finite dimensional linear operator Ah : Vh → Vh is symmetric and positive-definite, so Vh has an orthonormal basis φh1, φh2, . . . , φhN consisting of eigenfunctions of Ah. Thus,

Ahφ h m = λhmφ

uh(t) = Eh(t)u0h +

0 Eh(t − s)Phf (s) ds,

where the discrete solution operator for the homogeneous problem is defined by

Eh(t)χ = N∑

m=1

Stability

uh(t) ≤ u0h+

Proof. Since 0 < Eα(−s) ≤ 1 for 0 ≤ s <∞,

Eh(t)χ2 = N∑

m=1

χ, φhm2 = χ2,

so Eh(t)χ ≤ χ and the desired estimate follows from the Duhamel formula above.

Error estimates

We wish to estimate the error uh(t)− u(t) in L2() and in H1 0 ().

Our assumption that is convex implies that the elliptic problem,

−∇2u = f in , with u = 0 on ∂,

is H2-regular, that is, the weak solution u ∈ H1 0 (), given by

a(u, v) = f , v for all v ∈ H1 0 (),

necessarily belongs to H2() and

u2 ≤ Cf .

Caution: in this lecture, vr denotes the norm in H r (), so vr <∞ does not guarantee v ∈ Hr

D() unless v vanishes appropriately on ∂.

Error in the elliptic problem

Let uh ∈ Vh be the finite element solution of the elliptic problem above, so that

a(uh, χ) = f , χ for all χ ∈ Vh.

Since the symmetric bilinear form a(u, v) is bounded and coercive on H1

0 (), and using an appropriate quasi-interpolant, we have

uh − u1 ≤ C inf χ∈Vh

χ− u1 ≤ Chr−1ur for 1 ≤ r ≤ p + 1.

The usual duality argument (which relies on H2-regularity) then implies that

uh − u ≤ Ch inf χ∈Vh

χ− u1 ≤ Chrur for 1 ≤ r ≤ p + 1.

Ritz projector

It is convenient to define Rh : H1 0 ()→ Vh by Rhu = uh, or

equivalently,

It follows that R2 h = Rh, with

v − Rhv ≤ Chrvr and v − Rhv1 ≤ Chr−1vr

for 1 ≤ r ≤ p + 1. Also, since

PhAv , χ = Av , χ = a(v , χ) = a(Rhv , χ) = AhRhv , χ

for all v ∈ H1 0 () and χ ∈ Vh, we see that

PhA = AhRh : H1 0 ()→ Vh.

Equation for the error

Returning to the time-dependent case, we split the error into two terms

uh(t)− u(t) = ϑ(t) + %(t), ϑ = uh − Rhu, % = Rhu − u.

Then for all χ ∈ Vh,

ϑt , χ+ ∂1−α t a(ϑ, χ)

= uht , χ+ ∂1−α t a(uh, χ)− Rhut , χ − ∂1−α

t a(Rhu, χ)

= f , χ+ Rhut , χ − ∂1−α t a(u, χ)

= ut , χ+ ∂1−α t a(u, χ)− Rhut , χ − ∂1−α

t a(u, χ)

so ϑ : [0,∞)→ Vh satisfies an equation of the same form as the one for uh (with ut − Rhut playing the role of f )

ϑt , χ+ ∂1−α t a(ϑ, χ) = ut − Rhut , χ.

Estimate for ϑ = uh − Rhu

Lemma For 1 ≤ r ≤ p + 1,

ϑ(t) ≤ u0h − Rhu0+ Chr

ϑ(t) ≤ ϑ(0)+

Ph(ut − Rhut) ≤ ut − Rhut ≤ Chrutr .

Estimate for % = Rhu − u

%(t) ≤ Chr

( u0r +

together with

Quasi-optimal error bound in L2()

Theorem The finite element solution of the time-fractional diffusion equation satisfies

uh(t)− u(t) ≤ u0h − Rhu0+ Chr

( u0r +

) for 1 ≤ r ≤ p + 1.

( · · · )

Realistic smoothness

Consider the homogeneous problem with f ≡ 0 so that u(t) = E(t)u0, and choose p = 1 (piecewise-linears).

For sufficiently small ε > 0, if u0 ∈ H2+ε() and u0 = 0 on ∂, then u0 ∈ H2+ε

D () so by our earlier regularity theorem,

tut2 ≤ Ctεα/2u02+ε,

implying that ut2 = O(tεα/2−1) as t → 0. Choosing u0h = Rhu0

for simplicity, we obtain

uh(t)− u(t) ≤ Ch2u02+ε for 0 ≤ t ≤ T ,

with C = C (T , α, ε,).

Estimate for ϑ(t)1

We saw that

ϑt , χ+ a(∂1−α t ϑ, χ) = −ρt , χ for all χ ∈ Vh.

Choosing χ = Ahϑ(t), we have

ϑt ,Ahϑ = 1

and

−%t ,Ahϑ = −%t ,PhAhϑ = −Ph%t ,Ahϑ = −a(Ph%t , ϑ).

Thus,

1

2

d

Integrating from t = 0 to t = T and using∫ T

0 ∂1−α

Since a(v , v) is equivalent to v2 1,

ϑ(T )1 ≤ C

) .

ϑ(t)1,

≤ Cϑ(0)2 1 + C

∫ t∗

≤ C

) ϑ(t∗)1

1 + C

Quasi-optimal error bound in H1()

Assume now that Th is such that the L2-projector Ph is stable in H1(), that is,

Phv1 ≤ Cv1 for v ∈ H1().

For instance, it suffices to assume that Th is quasi-uniform.

Theorem The finite element solution of the time-fractional diffusion equation satisfies

uh(t)− u(t)1 ≤ Cu0h − Rhu01

+ Chr−1

( u0r +

Proof Recall that uh − u = ϑ+ % and

ϑ(t)1 ≤ C

) .

and since Ph is stable in H1(),

Ph%t(s)1 ≤ Cut(s)− Rhut(s)1 ≤ Chr−1ut(s)r .

The error bound follows because

%(t)1 ≤ %(0)1 +

≤ Chr−1

( u01 +

Define the closed sector

Σψ = { z ∈ C : z 6= 0 and | arg z | ≤ ψ } ∪ {0}.

Our aim now is to prove the following error bound, which does not require any spatial regularity for u0 or f .

Theorem Assume that u0h = Phu0, and fix such that 0 < < π/2. Then,

uh(t)− u(t) ≤ C t−αh2 ( u0+ sup

z∈∂Σπ−

Under Laplace transformation, our problem

ut + ∂1−α t Au = f (t) for t > 0, with u(0) = u0,

becomes zu(z)− u0 + z1−αAu(z) = f (z),

so

Similarly, for uh we have

(zαI + Ah)uh(z) = zα−1gh(z) where gh(z) = u0h + Ph f (z).

Notice that gh(z) = Phg(z) because we assume u0h = Phu0.

Resolvent estimate for A

Theorem Let satisfy 0 < ≤ π/2 and put

M = 1

(zI + A)−1 ≤ M

1 + |z | .

Proof Let z = re iθ with |θ| ≤ π − and r > 0.

If 0 ≤ |θ| ≤ π/2 then |z | ≤ |z + λm| because

0 ≤ <z ≤ <(z + λm) and =z = =(z + λm)

If π/2 ≤ |θ| ≤ π − then |z | ≤ M|z + λm| because

|z | sin ≤ |z | sin |θ| = |=z | = |=(z + λm)| ≤ |z + λm|.

Therefore, since M ≥ 1,

(zI + A)−1v = ∞∑

(zI + A)−1v2 = ∞∑

If |z | ≤ λ1/2, then |z + λm| ≥ λ1 − |z | ≥ λ1/2 so

1 + |z | |z + λm|

= M

( 1 +

2

λ1

Resolvent estimate for Ah

Theorem Let satisfy 0 < ≤ π/2 and put M = 1/ sin. If z ∈ Σπ− then

(zI + Ah)−1 ≤ M

1 + |z | .

Proof. The same proof shows that the conclusion holds with λh1 in place of λ1. But

λh1 = min χ∈Vh

Ahχ, χ χ2

the Laplace inversion formula gives

u(t) = 1

and

where

Error representation

where

Gh(z) = (zαI + Ah)−1Ph − (zαI + A)−1 = G 1 h (z) + G 2

h (z),

G 1 h (z) = (zαI + Ah)−1Ph − Ph(zαI + A)−1,

G 2 h (z) = (Ph − I )(zαI + A)−1.

Lemma For z ∈ Σπ−,

A(zαI + A)−1v ≤ Cv and (zαI + Ah)−1Ahχ ≤ Cχ.

Proof. The identity

A(zαI + A)−1 = (zαI + A− zαI )(zαI + A)−1 = I − zα(zαI + A)−1

implies that

|zα| ≤ C .

Estimate for G 1 h (z)

Lemma G 1

Proof. Recall that PhA = AhRh, so

G 1 h (z) = (zαI + Ah)−1Ph − Ph(zαI + A)−1

= (zαI + Ah)−1 [ Ph(zαI + A)− (zαI + Ah)Ph

] (zαI + A)−1

= (zαI + Ah)−1Ah(Rh − Ph)(zαI + A)−1,

H2-regularity of the elliptic problem gives

G 1 h (z)v ≤ C(Rh − Ph)(zαI + A)−1v ≤ Ch2(zαI + A)−1v2

≤ Ch2A(zαI + A)−1v ≤ Ch2v.

Estimate for G 2 h (z)

Lemma G 2

Proof. As above,

≤ Ch2(zαI + A)−1v2

≤ Ch2A(zαI + A)−1v ≤ Ch2v.

Error estimate

uh(t)− u(t) = 1

with Gh(z)v ≤ Ch2v for z ∈ Σπ−.

Deform the integration contour Γ to ∂Σπ− = Γ+ − Γ−, where

Γ± = { se±i(π−) : 0 < s <∞},

so that uh(t)− u(t) = I+ − I− where

I± = 1

eztzα−1Gh(z)g(z) dz .

The substitution z = se±i(π−) = s(− cos± i sin) gives

I± ≤ Ch2 ( u0+ sup

ds

s .

A second substitution s = w(t cos)−1 shows that the integral on the right equals∫ ∞

0 e−w

Introduction

The Laplace inversion formula yields a contour integral representation of the finite element solution uh(t) to the time-fractional diffusion equation. Applying a quadrature approximation [Lopez-Fernandez+Palencia-2004] leads to a fully discrete solution UN,h(t). The main cost of the method is the computation of uh(zj) at the quadrature point zj for |j | ≤ N, which requires the solution of a system of finite element equations involving the complex parameter zj .

The main advantages of the this approach are high accuracy and easy parallel implementation. A key disadvantage is that the method imposes severe limitations on form of the source term f (t). We outline a modified approach [McLean+Thomee-2010] that sacrifices some accuracy to relax the requirements on f (t).

Outline

Contour integral and quadrature Once again consider

u + ∂1−α t Au = f (t) for t > 0, with u(0) = u0,

and recall that

We now choose for Γ a contour of the form

z = z(ξ) = µ ( 1− sin(δ − iξ)

) for −∞ < ξ <∞,

µ > 0 and 0 < δ < π

2 .

Hyperbola

Since sin(δ − iξ) = sin δ cosh ξ − i cos δ sinh ξ,

we have

) ,

y(ξ) = =z(ξ) = µ cos δ sinh ξ,

so Γ is the left branch of the hyperbola( x − µ µ sin δ

)2

Parameterised integral

We have

u(t) = 1

= 1

2πi

) = µ

) ,

) .

(cosh ξ − sin δ)2 =

cosh ξ + sin δ

1− sin δ .

Double exponential decay

|ez(ξ)t | = ex(ξ)t = exp ( µt(1− sin δ cosh ξ)

) ,

|ez(ξ)t | ≤ exp ( µt − 1

Discretisation error For ξ > 0, let

Q∞(v) = ∞∑

1. v(ζ) is analytic on the strip −r− ≤ =ζ ≤ r+;

2. ∫ r+

3. ∫∞ −∞ |v(ξ ± ir±)| dξ ≤ M±.

Then Q∞(v)− ∫ ∞ −∞

( 1− sin(δ − iζ)

) defines conformal mapping that takes the line <ζ = η to the left branch of the hyperbola(

x − µ µ sin(δ + η)

)2

π

2

Truncation error

so need to estimate

v(ξ) dξ

) .

ea(1−b cosh ξ) dξ ≤ Cea(1−b)L(ab)

where

L(x) =

Proof

Since a(1− b cosh ξ) = a(1− b) + ab(1− cosh ξ) it suffices to estimate

I ≡ ∫ ∞

∫ ∞ 0

=

,

where we used the substitution y = cosh ξ− 1 followed by x = aby . If ab ≥ 1 then I ≤

∫∞ 0 x−1/2e−x dx <∞, whereas if ab ≤ 1, then

I ≤ ∫ 1

0

Quadrature error [Weideman+Trefethen-2007]

Thus, taking a = µt and b = sin(δ ± r±) in the lemma,∫ ∞ −∞ |v(ξ ± ir±)| dξ ≤ M± = Ceµt(1−sin(δ±r±))L

( µt sin(δ ± r±)

≤ CL ( µt sin(δ ± r±)

) − 2πr±/ξ

) ξ.

µt ( 1− sin(δ + r+)

) ,

2 .

The limiting choices r+ = π/2− δ and r− = δ give

− 2π

ξ

( π

) ,

2πδ

ξ .

4δ − π ,

( 2δ

b(δ) ,

|DE+ |+ |DE− |+ |TE | ≤ CL(· · · )e−B(δ)N .

Optimal choice of δ

We find that B(δ) has a unique maximum in the interval π/4 < δ < π/2, namely, at

δ∗ = 1.1721 0423, (near 3π/8 = 1.1780 9724)

and the maximum value is

B(δ∗) = 2.3156 5403.

ξ∗ = 1.0817 9214

N

t ,

e−B(δ∗)N = e−2.3157N = 10.1315−N .

Scalar test problem

L{e−λt} =

We consider

where zj = z(j ξ) and z ′j = z ′(j ξ).

Convergence behaviour

Fixed points with varying t

Given N and τ > 0, suppose we choose ξ∗ and µ∗ as above for t = τ . What if we use the approximation

e−λt ≈ 1

for t near τ?

In the fractional PDE case, we can solve one set of elliptic problems for uh(zj) at the zj optimized for t = τ and use them not just at t = τ but for several values of t in an interval around τ . The zj are then slightly sub-optimal but we avoid having to compute a new set of uh(zj) for each t.

Error for the scalar example with τ = 1

Parallel-in-time algorithm

Combining the above approach to numerical inversion of the Laplace transform with a spatial discretisation by finite elements leads to a fully discrete numerical method that involves no time stepping.

The method is particularly suited to applications in which the solution is required for only a few values of t.

In addition to its high, spectral-order accuracy in time, the method is embarassingly parallel.

Method-of-lines solution

Recall that the semidiscrete finite element solution uh : [0,∞)→ Vh satisfies

uht , χ+ ∂1−α t a(uh, χ) = f (t), χ for χ ∈ Vh and t > 0,

or equivalently,

uht + ∂1−α t Ahuh = Phf (t) for t > 0,

with uh(0) = u0h ≈ u0.

zuh + z1−αAhuh = gh(z)

Fully-discrete solution

If we solve the (complex) finite element equations for uh(z) at each quadrature point z = zj , then we can compute

UN,h(t) = 1

as an approximation to

) z ′(ξ) dξ.

The computational cost is dominated by solving the elliptic problems, and a key advantage of the method is that (unlike in a time-stepping scheme) these elliptic solves can easily be performed in parallel.

Halving the computational cost

Assuming that u0 and f are real-valued, it follows that uh is real-valued and so

uh(z) = uh(z).

Since z(−ξ) = z(ξ) and z ′(−ξ) = −z ′(ξ),

so z−j = zj , uh(z−j) = uh(zj), z ′−j/i = z ′j/i

and therefore

2πi

Since y0 = 0 and x ′0 = 0, it follows that

UN,h(t) = 1

1

π

) ξ.

In particular, it suffices to compute uh(zj) for 0 ≤ j ≤ N.

In matrix terms, we solve the N + 1 linear systems

(zαj M + S)U(zj) = zα−1 j G(zj), 0 ≤ j ≤ N,

where M and S are the mass and stiffness matrices, U(z) is the vector of nodal values of uh(z), and G(zj) is the load vector for gh(z).

Quadrature error

M± = 1

dξ

for η = δ ± r± where 0 < r− < δ < r+ < π/2. We saw earlier that

(zI + Ah)−1v ≤ v |z | sin

for z ∈ Σπ− and v ∈ Vh,

and z ′(ξ + iη)

we have

≤ gh(z) |z | sin

for zα ∈ Σπ−, that is, for z ∈ Σ(π−)/α and hence for z ∈ Σπ−.

Therefore,

dξ

≤ 1

f (z) ≤ Cf , for z ∈ Σπ−,

so that

gh(z) ≤ u0+ Ph f (z) ≤ u0+ Cf , for z ∈ Σπ−,

then we may estimate the quadrature error as before and obtain

UN,h(t)− uh(t) ≤ Ce−B(δ)NL(ct).

UN,h(t)− u(t) ≤ UN,h(t)− uh(t)+ uh(t)− u(t) = O

( L(ct)e−B(δ)N

) + O(t−αh2).

U? N,h(t) =

with εjL∞() ≤ ε, leading to an additional perturbation

U? N,h(t)− UN,h(t) ≤ ε

≤ ε

) ξ

≈ ∫ N ξ

−N ξ eµt(1−sin δ cosh ξ dξ ≤ Ceµ(1−sin δ)L(µt sin δ)

we conclude that

−N≤j≤N g(zj).

lead to eµ∗t(1−sin δ∗) = e1.4224N = 1.422N .

Example

Suppose h = 10−3 and ε = 2−52 ≈ 2.22× 10−16. Then

ε 1.422N ≥ h2 = 10−6

when

So roundoff probably not an issue.

However, “optimal” parameter values might be problematic if a high-accuracy spatial discretisation (say a spectral method) were used. We do not want µ to be too large.

Behaviour of f (z)

f (x , z) = g(x) z + a

(z + a)2 + ω2 ,

Example

If

e−z − e−2z

A more flexible approach [McLean+Thomee-2010]

Now drop the assumption that f (z) is analytic and bounded for z ∈ Σπ−. We will describe a method based on Duhamel’s formula,

u(t) = E(t)u0 +

0 E(t − s)f (s) ds, t > 0.

Recall that E(t), the solution operator for the homogeneous fractional diffusion equation has the series representation

E(t)v = ∞∑

and the integral representation

Noting that∫ t

∫ t

0

1

2πi

∫ Γ

= 1

2πi

we define g(z , t) = eztu0 +

∫ t

and deduce u(t) =

∫ Γ E(z)g(z , t) dz .

Notice that g(z , t) is an entire function of z , with

g(z , t) ≤ u0+

0 f (s) ds for <z ≤ 0 and t ≥ 0.

Since E(z) ∼ z−1I as |z | → ∞ with z ∈ Σπ−, we expect

E0(z) = E(z)− z−1I

to decay more rapidly than E(z), which is needed to compensate for the disappearance of the factor ezt in the integrand.

Theorem If 0 ≤ σ ≤ 1 and v ∈ D(Aσ), then

E0(z)v ≤ C,σ Aσv |z |1+ασ

for z ∈ Σπ−,

= [ zα−1I − z−1(zαI + A)

] (zαI + A)−1

]σ Aσ

] (zαI + A)−1

(zαI + A)−1 ≤ 1

σ ≤ ( 1

]1−σ ≤ (1 + 1

1

2πi

∫ Γ

g(z , t)

w0(z , t) = E0(z)g(z , t) = w(z , t)− z−1g(z , t).

and w(z , t) = E(z)g(z , t) denotes the solution of the (complex) elliptic problem

(zαI + A)w(z , t) = zα−1g(z , t).

Similarly, if we define the spatially discrete operators

Eh(z) = zα−1(zI + Ah)−1 and E0h(z) = Eh(z)− z−1I

and put

then

w0h(z , t) dz

where w0h(z , t) = E0h(z)gh(z , t) = wh(z , t)− z−1gh(z , t), and wh(z , t) is computed by solving the (complex) finite element equations

(zαI + Ah)wh(z , t) = zα−1gh(z , t).

Fully-discrete scheme The equal-weight quadrature approximation∫

Γ w0h(z , t) dz ≈

w0h(zj , t)z ′j ξ.

Lemma If z = z(ξ + iη) and z ′ = z ′(ξ + iη), then for 0 < σ1 + α−1 = σ ≤ 1,

wh(z , t)z ′ ≤ 2C,σ,σ1

wh(z , t) = E0h(z) ( eztu0h

= C,σ |z ′| |z | Aσhu0h |z |ασ

and, since 2 + ασ1 = 1 + α(σ1 + α−1) = 1 + ασ,

z−1E0h(z)Phf (0)z ′ ≤ C,σ1 |z ′| Aσ1

h u0h |z |2+ασ1

= C,σ1

Quadrature error

) ,

= (1− e−|ξ| sin δ)2 + e−2|ξ| cos2 δ

≥ (1− sin δ)2,

it follows that

µασ .

Set r = r± such that 0 < δ − r < δ + r < π/2, and estimate∫ ∞ −∞

w ( z(ξ ± ir)

) z ′(ξ ± ir)

µασ

∫ ∞ 0

µασ e−2πr/ξ.

At the same time,

eµt(1−sin δ)

µασ e−ασNξ.

Setting 2πr/ξ = ασNξ, and choosing µ > 0 to minimise eµt(1−sin δ)/µασ, we arrive at the following estimate.

Theorem For the flexible scheme described above, if

ξ =

√ 2πr

( − √

2πrασN ) .

The error bound suggests choosing δ = π/4 and r slightly less than π/4.

Example

Taking

2 , σ = 1

gives 2πrασ = π2/4 so the decay factor in the error bound is of order

e− 1 2π √ N = e−1.5708

√ N ,

for our earlier method.

0 K (tn − s)f (s) ds ≈

n∑ j=0

wn−j f (tj), tj = j t,

where the convolution weights wn = wn(t) are computed directly from the Laplace transform K (z) rather than the kernel K (t). This approach can be advantageous if K (z) is simpler than K (t).

Convolution quadrature can be used to approximate any function of the form K ∗ f , and in particular the fractional integral Iαf = Υα ∗ f .

Outline

Assume that K (z) is analytic and satisfies

|K (z)| ≤ C |z |−µ for z ∈ Σπ−, where µ > 0.

Then, as before,

K (t) = 1

Since∫ t

∫ t

0

( 1

2πi

∫ Γ

K ∗ f (t) = 1

The associated ODE

so that

so y is the solution of the initial-value problem

dy

dt − zy = f (t) for t > 0, with y(0) = 0.

Implicit Euler method

Y n = Y n(z) ≈ y(tn; z) where tn = n t,

by solving

Y n − Y n−1

t − zY n = F n for n ≥ 1, with Y 0 = 0,

where F n = f (tn). Then,

K ∗ f (tn) = 1

Y n(z)ζn.

∞∑ n=0

Y nζn − ∞∑

n=−1

so the finite difference equation implies that

1− ζ t

Thus,

δ(ζ)t−1 − z where δ(ζ) = 1− ζ.

Recall

K (z)Y n(z) dz .

For t and |ζ| sufficiently small, Γ passes to the left of the pole at z = δ(ζ)t−1, and

∞∑ n=0

= − 1

2πi

∫ Γ

z − δ(ζ)t−1 F (ζ).

Here, the integrand is O(|z |−1−µ) so Cauchy’s theorem gives

− 1

2πi

∫ Γ

( δ(ζ)t−1

( δ(ζ)t−1

K ( δ(ζ)t−1

Example

Suppose K (t) = Υα(t) and so K (z) = z−α. Then

K ( δ(ζ)t−1

) = ( (1− ζ)t−1

)−α = tα(1− ζ)−α

) (−1)n.

Note that wn > 0 because w0 = tα and, for n ≥ 1,( −α n

) (−1)n =

Higher-order methods

To improve on the implicit Euler method, recall that the polynomial

Y n + Y n

+ 1

p!

tp (t − tn) · · · (t − tn−p+1)

of degree p takes the value Y j at t = tj for n − p ≤ j ≤ n, where Y n = Y n − Y n−1 denotes the backward difference. Differentiating with respect to t and setting t = tn leads to the backward differentiation formula (BDF)

y ′(tn) ≈ Y n

1

t

` − zY n = F n for n ≥ 1,

with starting values Y 0 = Y−1 = · · · = Y−p = 0. Generating function still satisfies

δ(ζ)

t Y (ζ)− zY (ζ) = F (ζ) but now δ(ζ) =

p∑ `=1

(1− ζ)`

| arg δ(ζ)| ≤ π − α for |ζ| < 1,

for the following values of α.

p α

1 90

2 90

3 86

4 73

5 51

6 17

Operational calculus

∫ t

In this way, since L{K ∗ f } = K (z)f (z),

we have

eztK (z)f (z) dz .

Explanation: if K (t) = 1 then K (z) = z−1 so

∂−1f (t) =

0 f (s) ds and L{∂−1f } = z−1f (z).

Example

For K (t) = Υα(t) = tα−1/Γ(α) we have K (z) = z−α and so

∂−αf (t) = Υα ∗ f (t) = Iαf (t)

is the fractional integral of order α > 0, with

L{∂−αf } = z−αf (z).

Example

If K (t) = eat then K (z) = (z − a)−1 so

(∂ − a)−1f (t) =

with L{(∂ − a)−1f } = (z − a)−1f (z).

Theorem

Proof.

∫ t

=

= 1

2πi

∫ Γ

= 1

2πi

∫ Γ

Discrete operational calculus

0≤tj≤t wj(t)f (t − tj) for t > 0.

In particular, at t = tn,

K (∂t)f (tn) = n∑

j=0

If K (t) = 1 then K (z) = z−1 so

K ( δ(ζ)t−1

∂−1 t f (t) =

∫ t

Example

is the solution of the initial-value problem

y − ay = f (t) for t > 0, with y(0) = 0.

The BDF solution Y n satisfies( δ(ζ)t−1 − a

) Y (ζ) = F (ζ), F n = f (tn),

so if ( δ(ζ)t−1 − a

)−1 = ∞∑ n=0

Example

y ′(tn) ≈ Y n

2 (1− ζ)2 = 3 2 (1− ζ)(1− 1

3ζ).

( α + n − 1

wn =

( 2

3

Since

wn = 1

We can use this representation to show the following result.

Theorem

Proof

so

)−1 = ∞∑ n=0

= ∑

= 1

2πi

∫ Γ

0≤tj≤t w? j (t; z)f (t − tj) dz

= 1

2πi

∫ Γ

Summary

1

2πi

∫ Γ

2πi

∫ Γ

w(ζ) = ∞∑ n=0

wnζ n = K

( δ(ζ)t−1

2πi

∫ Γ

Accuracy of convolution quadrature

wj f (t − tj)χ(t − tj), χ(t) =

{ 1, t > 0,

0, t < 0,

L{K (∂t)f } =

= ∞∑ j=0

= ∞∑ j=0

wjζ j

= K ( δ(e−z t)t−1

) f (z).

δ(e−h)h−1 = 1 + O(hp) as h→ 0,

so

δ(e−zt)t−1 = zδ(e−h)h−1, h = zt,

= z + O(zp+1tp)

) f (z).

Notice also

= 1

2πi

∫ Γ

] f (t) dz .

(A1) There exist 0 < < π/2 and −∞ < µ <∞ such that the function G (z) is analytic with |G (z)| ≤ C |z |−µ for | arg z | < π−.

(A2) The linear multistep method is strongly A-stable of order p ≥ 1, that is,

I δ(ζ) is analytic in a neighbourhood of the closed unit disk |ζ| ≤ 1,

I for ζ in this neighbourhood, δ(ζ) = 0 iff ζ = 1,

I there exists 1 > such that | arg δ(ζ)| ≤ π − 1 for |ζ| < 1,

I h−1δ(e−h) = 1 + O(hp) as h→ 0.

Theorem ([Lubich-2004])

If assumptions (A1) and (A2) hold, then for 0 < t <∞,

G (∂t)tβ−1 − G (∂)tβ−1 ≤ {Ctµ−1+β−ptp, p ≤ β,

Ctµ−1tβ, 0 < β ≤ p.

Multiplication property

Notice that since

K1 ∗ (K2 ∗ f ) = (K1 ∗ K2) ∗ f and L{K1 ∗ K2} = K1(z)K2(z),

we have K1(∂)K2(∂) = (K1K2)(∂),

The analogous identity holds in the discrete case.

Theorem

Proof

= ∑

∑ 0≤tk≤t−tj

f (t − tj − tk)

= ∑

n−j wn

) f (t − tn).

k ζ j+k

n−j

K ∗ (f ∗ g) = (K ∗ f ) ∗ g ,

or equivalently,

) ∗ g .

K (∂t)(f ∗ g) = ( K (∂t)f

) ∗ g ,

because

L {

) f (z)g(z).

=

so

) ∗ f (p)(t),

and the same formula holds with ∂ replaced by ∂t . Hence,

[G (∂t)− G (∂) ] f (t)

≤ Ctµ−1 p−1∑ k=0

|f (k)(0)|tk+1

Correction terms

p−2∑ j=0

w∼nj f (tj)

and choose the extra weights w∼nj so that the modified quadrature rule is exact for polynomials up to degree p − 2:

p−1∑ j=1

w∼njΥk(tj) = G (∂)Υk(tn)− G (∂t)Υk(tn)

for 1 ≤ k ≤ p − 1. Unfortunately, the matrix [Υk(tj)] is badly conditioned.

An alternative approach works if tn is bounded away from 0.

Application to fractional diffusion

For simplicity, we suppose f ≡ 0 so that, after integrating in time, our initial-value problem takes the form

u + ∂−αAu = u0,

Thus, u = u0 − (I + ∂−αA)−1∂−αAu0,

which suggests seeking U(t) ≈ u(t) such that

U = u0 − (I + ∂−αt A)−1∂−αAu0.

Thus,

which leads to the implicit scheme

W n + n∑

wn−jAW j = −Υ1+α(tn)Au0 for n ≥ 1,

with W 0 = 0, where the weights wn = wn(t) are given by

[ δ(ζ)t−1

]−α = ∞∑ n=0

wnζ n.

Fully-discrete version: Uh = u0h + Wh where W 0 h = 0 and

W n h +

n∑ j=0

Error bound for nonsmooth initial data Since

u − u0 = −(I + ∂−αA)∂−αAu0,

U − u0 = −(I + ∂−αt A)∂−αAu0,

the error from the time discretization is

U − u = [ G (∂)− G (∂t)

] ∂−αu0,

Theorem ([Cuesta+Lubich+Palencia-2006])

Ct−1−αt1+αu0, p ≥ 2.

Proof

Since

= zα(zαI + A)−1 [ (zαI + A)− zαI

] = zα

] and (zαI + A)−1 ≤ C |z |−α, we have

G (z) ≤ C |z |α for z ∈ Σπ−.

Noting that ∂−αu0 = Υ1+α(t)u0 = tαu0/Γ(1 + α), we apply the theorem with µ = −α and β = 1 + α to conclude

U(t)− u(t) = [G (∂)tβ − G (∂t)tβ

] u0

t−α−1t1+α, p ≥ 1 + α.

Error bound for smooth initial data Recall that

u(t) = E(t)u0 = u0 − tα

and observe that

= Υ1+2α(t)A2u0,

We therefore consider

U(t) = u0 −Υ1+α(t)Au0 + (I + ∂−αt A)−1Υ1+2α(t)A2u0.

The error is

U(t)− u(t) = [ (I + ∂−αt A)−1 − (I + ∂−αA)−1

] Υ1+2α(t)A2u0

] Υ1+2α(t)Au0,

where, once again, G (z) = (I + z−αA)−1A. Applying the theorem with µ = −α and β = 1 + 2α we have

U(t)− u(t) ≤ CAu0 ×

t−1−αt1+2α, p ≥ 1 + 2α.

For instance, if p = 2 and 1/2 ≤ α < 1, then

U(t)− u(t) ≤ Ctα−2t2Au0.

Part IX

Introduction

We consider a class of time-stepping methods in which u(t) is approximated by a piecewise polynomial U in t. Continuity across the time levels is enforced only weakly. These fully implicit methods are flexible and robust, and can achieve high accuracy, but are rather complicated to implement in general and have a somewhat higher computational cost than many simpler time-stepping schemes. Crucially, they allow the use of highly non-uniform grids.

Once again, we largely ignore the spatial discretization.

Outline

Stability

Convergence

Discontinuous piecewise-polynomial approximation in time

Let t = (tn)Nn=0 be a vector of time levels satisfying

0 = t0 < t1 < t2 < · · · < tN = T ,

and denote the nth open subinterval and its length by

In = (tn−1, tn) and tn = tn − tn−1 for 1 ≤ n ≤ N.

For each n, choose a closed subspace Sn ⊆ H1 0 () and an

integer pn ≥ 0, and write S = (Sn)Nn=0 and p = (pn)Nn=1.

We define our trial space W =W(t,S,p) to consist of those functions U : (0,T )→ H1

0 () such that U|In is a polynomial of degree at most pn in t with coefficients in Sn for 1 ≤ n ≤ N.

Example

Choose Sn = H1 0 () and pn = p independent of n.

Example

Choose Sn = Vh to be the usual continuous piecewise-linear finite element space with respect to a triangulation Th of and enforcing a homogeneous Dirichlet boundary condition.

Thus, our methods are conforming in space but non-conforming in time.

For U ∈ W, we denote the one-sided limits and the jumps at tn by

Un ± = U(t±n ) = lim

−

for 0 ≤ n ≤ N, with the convention that U0 − ∈ S0 (even though I0

is undefined).

Weak formulation

Recall that the mild solution u of our initial-boundary value problem for the fractional diffusion equation satisfies

u(t), v+ a ( ∂1−α t u(t), v

) = f (t), v for all v ∈ H1

0 (),

where u = ut = ∂u/∂t. Hence, for suitable v : In → H1 0 (),∫

In

)] dt =

∫ In

f (t), v(t) dt.

In the discontinuous Galerkin (DG) method we seek U ∈ W satisfying, for all X ∈ W and for 1 ≤ n ≤ N,

Un−1 + ,X n−1

+ +

)] dt

+ +

In addition, we require that U0 − = U0 for a suitable

approximation U0 ∈ S0 to the given initial data u0.

Example

Take pn = 0 and Sn = Vh for all n, so that U and X are piecewise constant in time. Writing Un = Un

− and χ = X n −, we have

U(t) = Un = Un−1 + and X (t) = χ = X n−1

+ for all t ∈ In,

Un − Un−1, χ+

) dt

f (t), χ dt for all χ ∈ Sn,

which is essentially the implicit Euler scheme we considered earlier, but using finite elements instead of finite differences in space.

Fully implicit time stepping

If ψn 0 , ψn

1 , . . . , ψn pn is a basis for the space of polynomials of degree

at most pn, and if χ1, χ2, . . . , χMn is a basis for Sn (so Mn = dim Sn), then we can write

U(x , t) =

Starting from the known (approximate) initial data

U0 −(x) = U0(x) =

M0∑ l=1

U0 l χ

we compute the unknown Unr l by solving the

[(pn + 1)Mn]× [(pn + 1)Mn] linear system determined by the DG equations on In for successive n = 1, 2, . . . , N.

Stability

We now prove a series of technical lemmas that will establish unconditional stability of DG time stepping. This robustness is a key benefit of the method, and helps justify its relatively high computational cost. The stability estimate below also shows that most of the jumps [U]j must be small for large N.

Theorem If U0

( (0,T );H

DG solution U ∈ W, and for 1 ≤ n ≤ N,

Un −2 +

Global bilinear form

[U]n−1,X n−1 + +

∫ In

f ,X dt,

and summing over n, we see that U ∈ W is the DG solution iff

GN(U,X ) = U0,X 0 ++

where the bilinear form GN is defined by

GN(U,X ) = U0 −,X

+ N∑

)] dt.

Lemma

−2−Un−1 − 2

2U 2, twice the LHS equals

2[U]n−1,Un−1 + + Un

−2 − Un−1 + 2

= [U]n−1, [U]n−1 + Un−1 + + Un−1

− + Un −2 − Un−1

− 2 + Un −2 − Un−1

+ 2

− 2.

Lemma

2

0 −2

(∂βt u)v dt =

(∂βt u)u dt = cos 1

2πβ

π

∫ ∞ 0

Lemma For 0 < β < 1, and real-valued u and v,∫ ∞

0 (∂βt u)v dt

Proof

Noting that u(−iy) = u(iy) and v(−iy) = v(iy), and using the Cauchy–Schwarz inequality, we have∫ ∞

0 (∂βt u)v dt

= 1

π

∫ ∞ 0

2

∫ T

≤ ∫ T

1

0 ∂βt v , v dt.

Proof. Extend u and v by zero. For any µ > 0, the mth Fourier coefficients satisfy

2

∫ T

∫ ∞ 0

) ,

and the result follows by summing over m, using Parseval’s identity and choosing µ = cos 1

2πβ.

GN(U,X ) = U0,X 0 ++

1 2U

0 +2 + 1

2

2U 0 +2 and cancel the

term 1 2U

∫ tN

g(t) = I1−αA−1/2f (t) and v(t) = A1/2U(t),

so that

Piecewise-constant case

If pn = 0 then U(t) ≤ Un∗ − = max0≤n≤N Un for 0 ≤ t ≤ tN .

Since

[U]n2

∫ tn∗

Un∗ − UN

Piecewise-linear case

{ Un−1

∫ t1

− + [U]n−1,

≤ (2 + 2)

( 2U0,U0

) ,

and by choosing n∗ such that UIn∗ = max1≤n≤N UIn we see

UIn ≤ 8

0 f (t) dt

) for 1 ≤ n ≤ N.

However, for p ≥ 2 we have not been able to prove such an L∞(L2) stability bound that mimics the one for the continuous problem:

u(t) ≤ u0+

Convergence

For simplicity, assume now that Sn = H1 0 () for all n (so no spatial

discretization). We decompose the DG error as

U − u = ϑ+ %, ϑ = U − Πu, % = Πu − u,

where the quasi-interpolant Πu ∈ W(t,S,p) is defined by the conditions

(Πu)n− = u(t−n ) and

(u − Πu)tq−1 dt = 0

for 1 ≤ q ≤ pn and 1 ≤ n ≤ N, with (Πu)0 − = u(0).

Example

Example

where avgIn(u) = t−1 n

∫ In

(Πu)(t)− u(t) =

tn − t

t2 n

=

tn − t

t2 n

u − ΠuIn ≤ 2

u − ΠuIn ≤ 3tn

∫ In

In the general case we have the following estimate.

Theorem ([Schoetzau+Schwab-2000])

(u − Πu)′(t)2 dt ≤ Cε(pn, q)

( tn

2

)] dt.

GN(U,X ) = UN − ,X

)] dt.

GN(U,X ) = U0,X 0 ++

and, since [u]n = 0, the exact solution satisfies

GN(u,X ) = u0,X 0 ++

Furthermore, the construction of Π ensures

%n− = 0 and

so

GN(%,X ) =

∫ tN

0 +,

Therefore,

t %,X dt

for all X ∈ W(t,p,S), showing that ϑ is the DG solution with initial data U0 − u0 and source term −∂1−α

t A%. By applying the stability result to ϑ it is possible to prove the following error estimate.

h-Version accuracy

Theorem ([Mustapha-2015])

tn = (n/N)γT and p = (1, p, p, . . . , p)

and

u(j)(t)1 ≤ Ctσ−j for 0 < t ≤ T and 1 ≤ j ≤ p + 1,

then

U(t)− u(t) ≤ C trN ≤ CN−r for 0 ≤ t ≤ T ,

where

r =

min{γ(σ + 1 2α−

2α− 1 2} −

Theorem ([McLean+Mustapha-2015])

Suppose f ≡ 0, tn = n t and pn = 0 for

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