Numerical Solution of Fractional PDEs

transcript

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

November 2015

William McLeanThe University of New South Wales

Updated November 27, 2015

Part I

Fractional integrals and derivatives

Introduction

This lecture provides some key definitions and results fromfractional calculus, needed for our study of fractional PDEs. Theliterature contains several concepts of fractional differentiation, butwe 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 theRiemann–Liouville fractional integral and derivative.

Outline

Fractional integration

Fractional differentiation

Grunwald–Letnikov definition

Fractional integration

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

I0a+f (x) = f (x)

Ina+f (x) =

aIn−1f (y) dy for n ≥ 1.

We claim

Ina+f (x) =

(x − y)n−1

(n − 1)!f (y) dy for n ≥ 1.

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

I1a+f (x) =

af (y) dy .

Easy proof by induction on n

Let n ≥ 1 and assume

Ina+f (x) =

(x − y)n−1

(n − 1)!f (y) dy .

In+1a+ f (x) =

aIna+f (z) dz =

(z − y)n−1

(n − 1)!f (y) dy dz

(z − y)n−1

(n − 1)!dz f (y) dy

(x − y)n

n!f (y) dy .

Gamma function

Recall that

Γ(α) =

∫ ∞0

e−ttα−1 dt for α > 0,

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

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

Iαa+f (x) =

(x − y)α−1

Γ(α)f (y) dy for x > a.

This definition is consistent with our earlier definition of Ina+

when α = n.

Transposed operator

Putting

〈f , g〉 =

af (x)g(x) dx

we find that ⟨Iαa+f , g

⟩=⟨f , Iαb−g

where the right-sided, Riemann–Liouville fractional integrationoperator of order α is given by

Iαb−g(x) =

(y − x)α−1

Γ(α)g(y) dy for x < b.

Semigroup property

From the recursive definition, we see that

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

Key question: does

Iαa+Iβa+ = Iα+β

a+ for all α > 0 and β > 0?

Consider

Iαa+Iβa+f (x) =

(x − z)α−1

Γ(α)

(z − y)β−1

Γ(β)f (y) dy dz

(x − z)α−1

Γ(α)

(z − y)β−1

Γ(β)dz f (y) dy .

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 + β),

gives∫ x

(x − z)α−1

Γ(α)

(z − y)β−1

Γ(β)dz

=(x − y)(α−1)+(β−1)+1

Γ(α)Γ(β)

0(1− t)α−1tβ−1 dt

=(x − y)α+β−1

Γ(α + β),

implying the desired result.

Gel’fand–Shilov function

Υα(x) =xα−1

Γ(α),

where x+ = max(x , 0), and use the abbreviation

Iα = Iα0+ for α > 0.

The fractional integral is given by the Laplace convolution

Iαf (x) =

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

We easily see that

Υα ∗Υβ = Υα+β for α > 0, β > 0.

In fact, since ∗ is associative,

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

for every continuous f .

It follows thatIαΥβ = Υα+β,

generalising the identity

(m + n)!.

Shifted version

LetΥβ,a(x) = Υβ(x − a).

Using the substitution y = a + t,

Iαa+Υβ,a(x) =

aΥα(x − y)Υβ(y − a) dy

∫ x−a

0Υα(x − a− t)Υβ(t) dt

= (Υα ∗Υβ)(x − a) = Υα+β(x − a),

or in other words,

Iαa+Υβ,a(x) = Υα+β,a(x), x > a.

Fractional differentiationAssume that

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

and write Dn = (d/dx)n.

The Riemann–Liouville fractional derivative is defined by

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.

LemmaFor x > a and β > 0,(

DIβa+ − Iβa+D

)f (x) = f (a)Υβ(x − a).

Note that DΥβ(x) = Υβ−1(x) and Υ1(x) = 1.

By the fundamental theorem of calculus,

Ia+Df (x) =

af ′(y) dy = f (x)− f (a),

sof (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

TheoremIf 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−αa+ f (x) = D

(I2−αa+ 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)

(x − a)1−α

Γ(2− α).

The general case follows in the same way.

Differentiating a shifted Gel’fand–Shilov function

LemmaIf α > 0, β > 0 and x > a, then

Dαa+Υβ,a(x) = Υβ−α,a(x).

Proof.If n − 1 < α < n then

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

Γ(k + 1− α)= Υk+1−α,a(x) = Dαa+Υk+1,a(x)

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+

(f (x)−

n−1∑k=0

Dk f (a)(x − a)k

), x > a,

where n − 1 < α < n, as before.

Alternative representation

LemmaIf 0 < α < 1 and x > a, then

Dαa+f (x) = f (x)Υ1−α(x−a)+

aΥ−α(x−y)

[f (y)−f (x)

Proof.Differentiate the identity

I1−αa+ f (x) = f (x)Υ2−α(x − a) +

aΥ1−α(x − y)

[f (y)− f (x)

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

aΥ−α(x − y)

[f (y)− f (x)

]dx −Υ2−α(x − a)f ′(x).

Representation as a Hadamard finite-part integral

Assume 0 < α < 1 and x > a. Then∫ x

aΥ−α(x − y)

[f (y)− f (x)

x−ε· · · dy

∫ x−ε

aΥ−α(x − y)f (y) dy + f (x)

[Υ1−α(ε)−Υ1−α(x − a)

Dαa+f (x) =f (x)ε−α

Γ(1− α)+

∫ x−ε

aΥ−α(x − y)f (y) dy + O(ε1−α).

and therefore

Dαa+f (x) = “I−αa+ f (x)” = fpε↓0

∫ x−ε

aΥ−α(x − y)f (y) dy .

Grunwald–Letnikov definitionCan we define a fractional derivative (or integral) directly, withoutusing integer-order derivatives and integrals?Denote the backward difference by

∆hf (x) = f (x)− f (x − h).

Can check by induction on k that

∆khf (x) =

k∑j=0

)(−1)j f (x − jh) for k ∈ 0, 1, 2, . . . .

Hence define the fractional backward difference of order α by

∆αh,nf (x) =

n∑j=0

)(−1)j f (x − jh),

with (α

α− 1

2· · · α− j + 1

Γ(α + 1)

Γ(j + 1)Γ(α− j + 1).

An induction on k shows that

∆khf (x) =

0· · ·∫ h

0f (k)(x − t1 − · · · − tk) dt1 · · · dtk ,

and thus

f (k)(x) = limh→0

∆khf (x)

Given x and a, we therefore define

GLDαa+f (x) = lim∆α

h,nf (x)

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

h =x − a

so that nh = x − a is constant.

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)

(x − y)m−α−1

Γ(m − α)f (m)(y) dy ,

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,

GLD−αa+ f (x) = Iαa+f (x), α > 0.

Part II

Useful tools

Introduction

We will make extensive use of the Laplace transform (and someuse of the Fourier transform), first to derive the fractional diffusionequation and then to study properties of the solution. Laplacetransformation also plays a large role in some of the numericalmethods we study, either as part of the method itself or for theerror analysis.

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

Outline

Laplace transforms

Mittag–Leffler function

Wright functions

Laplace transforms

Notation:

f (z) = (Lf )(z) =

∫ ∞0

e−zt f (t) dt.

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 theinversion formula

f (t) =1

∫ a+i∞

a−i∞ezt f (z) dz , a > λ.

Transform of a Gel’fand–Shilov function

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

Υα(z) =1

Γ(α)

∫ ∞0

e−zttα−1 dt =1

Γ(α)

∫ ∞0

e−y(

)α−1 dy

=z−α

Γ(α)

∫ ∞0

e−yyα−1 dy ,

that is,Υα(z) = z−α,

consistent with

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

Laplace transform of an integral

If (t) =

0f (s) ds = (Υ1 ∗ f )(t)

we haveLIf (t) = Υ1(z)f (z) = z−1f (z).

In general, Inf = Υn ∗ f so

LInf (t) = z−n f (z) for n ∈ 0, 1, 2, . . ..

Laplace transform of a derivative

Integration by parts shows

LDf (t) =

∫ ∞0

e−ztDf (t) dt

=[e−zt f (t)

]∞t=0−∫ ∞

0(−z)e−zt f (t) dt

= 0− f (0) + z

∫ ∞0

e−zt f (t) dt,

soLDf (t) = zf (z)− f (0).

Easily verify by induction on n that

LDnf (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),

LCDαf (t) = z−(n−α)g(z)

= zα−n(

zn f (z)−n−1∑k=0

zn−1−kDk f (0)

)and so

LCDαf (t) = zαf (z)−n−1∑k=0

zα−1−kDk f (0).

Laplace transform of a Riemann–Liouville fractionalderivative

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

LDαf (t) = LCDαf (t)+n−1∑k=0

Dk f (0)zα−1−k ,

that is,LDαf (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

Γ(1 + kα).

That is,f (t) = Eα(tα),

where the Mittag–Leffler function is

Eα(z) =∞∑k=0

Γ(1 + kα).

In fact, CDαΥ1 = 0 and for k ≥ 1,

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

CDαf = CDα(Υ1 + Υ1+α + Υ2+α + Υ3+α + · · ·

)= 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, itsuffices to show that as k →∞,∣∣∣∣ zk+1

Γ(1 + (k + 1)α

Γ(1 + kα)

∣∣∣∣ = |z | Γ(1 + kα)

Γ(1 + kα + α)→ 0.

In fact, using Stirling’s approximation,

Γ(x) =

√2π

)x(1 + O(x−1)

)as x →∞,

we find

Γ(1 + kα)

Γ(1 + kα + α)=

1 + kα + α

)α(1 + O(k−1)

Mittag–Leffler function Eα(x) for 0 < α ≤ 1

Mittag–Leffler function Eα(x) for 1 < α ≤ 2

Special choices of α

E0(x) =∞∑k=0

1− z.

E1/2(x) = exp(x2) erfc(−x) =1√2π

∫ ∞−x

exp(x2 − t2) dt.

E1(x) =∞∑k=0

k!= exp(x).

E2(x) =∞∑k=0

(2k)!= cosh(

√x), x ≥ 0.

E2(−x) =∞∑k=0

(−1)kxk

(2k)!= cos(

√x), x ≥ 0.

Fractional relaxation equationProblem: find u satisfying

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

We claim that the solution is

u(t) = Eα(−λtα) =∞∑k=0

(−λ)ktkα

Γ(1 + kα).

In fact,

CDαu(t) = CDα∞∑k=0

(−λ)kΥ1+kα(t)

= 0 +∞∑k=1

(−λ)kΥ1+(k−1)α(t)

= −λ∞∑k=1

(−λ)k−1Υ1+(k−1)α(t) = −λu(t).

Equivalent formulation

The function u(t) = Eα(−λtα) satisfies

I1−αDu(t) + λu(t) = 0.

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

DI1Du(t) + λDIαu(t) = 0.

Therefore, since DI1f (t) = f (t),

Du(t) + λD1−αu(t) = 0.

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

u(z) = LEα(−λtα) =1

z + λz1−α .

Fractional relaxation: u(t) = Eα(−tα)

Integral representation

The reciprocal Gamma function has the representation

Γ(a)=

∫ 0+

−∞eww−a dw ,

where∫ 0+−∞ means integration around a Hankel contour that

encircles the negative real axis and has a counterclockwiseorientation.

TheoremThe Mittag–Leffler function admits the integral representation

Eα(z) =1

∫ 0+

−∞

ewwα−1

wα − zdw

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

Sincezk

Γ(1 + αk)=

∫ 0+

−∞eww−1−αk dw ,

we see that

Eα(z) =∞∑k=0

∫ 0+

−∞eww−1−αk dw

∫ 0+

−∞

∞∑k=0

dw (if |z | < |w |α)

∫ 0+

−∞

1− zw−αdw

∫ 0+

−∞

ewwα−1

wα − zdw .

Positivity

Another integral representation follows from the Laplace inversionformula,

Eα(−λtα) =1

∫ 0+

−∞

ezt dz

z + λz1−α , t > 0.

By collapsing the Hankel contour onto the negative real axis, wefind

Eα(−λtα) =1

∫ ∞0

e−rtλrα sinαπ dr(rα + λ cosαπ

)2+ λ2 sin2 απ

which shows that

Eα(−λtα) > 0 andd

dtEα(−λtα) < 0 for all t > 0.

Asymptotic behaviour

z + z1−α =zα−1

1 + zα=

∞∑n=0

(−1)nz(n+1)α, |z | < 1,

we find that as t →∞,

Eα(−tα) ∼∞∑n=0

(−1)n

∫ 0+

−∞eztz(n+1)α dz

and a substitution gives

∫ 0+

−∞eztz(n+1)α dz

t−(n+1)α

∫ 0+

−∞ezz(n+1)α dz

=t−(n+1)α

Γ(1− (n + 1)α

Eα(−tα) ∼∞∑n=0

(−1)nt−(n+1)α

Γ(1− (n + 1)α

) ,that is,

Eα(−tα) ∼∞∑n=1

(−1)n+1t−nα

Γ(1− nα)=

t−α

Γ(1− α)− t−2α

Γ(1− 2α)+ · · · .

The identity1

Γ(1− z)=

πΓ(z) sinπz

yields an alternative form,

Eα(−tα) ∼ 1

∞∑n=1

(−1)n+1t−nαΓ(nα) sin nπα.

Notice what happens as α→ 1.

Wright functions

The Wright function is defined by

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.

TheoremThe Wright function has the integral representation

Wλ,µ(z) =1

∫ 0+

−∞exp(w + zw−λ

We once again use

Γ(a)=

∫ 0+

−∞eww−a dw ,

and find

Wλ,µ(z) =∞∑k=0

∫ 0+

−∞eww−λk−µ dw

∫ 0+

−∞eww−µ

∞∑k=0

(zw−λ)k

∫ 0+

−∞exp(w + zw−λ)

Wright M-functionOur main concern is with the function

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

Γ(1− z)=

πΓ(z) sinπz

yields an alternative expression

Mα(t) =1

∞∑k=0

(−1)ktk

k!Γ((k + 1)α

)sinπ(k + 1)α.

Two special cases: (Ai = Airy function)

M1/2(t) =exp(−t2/4)√

πand M1/3(t) = 32/3 Ai(t),

Wright M-function Mα(t)

Integral representation and Laplace transform

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

Mα(t) =1

∫ 0+

−∞exp(w − twα)

w 1−α ,

which allows us to prove the following result.

TheoremFor 0 < α < 1, the Laplace transform of Mα is

Mα(z) = Eα(−z).

Mα(z) =

∫ ∞0

e−ztMα(t) dt

∫ ∞0

e−zt1

∫ 0+

w 1−α dt

∫ 0+

−∞

w 1−α

∫ ∞0

e−(z+wα)t dt dw

∫ 0+

−∞

w 1−α1

z + wαdw

∫ 0+

−∞

ewwα−1

wα − (−z)dw

= Eα(−z).

Fourier transform

Notation:

f (ξ) = Ff (x) =

∫ ∞−∞

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 abounded linear operator F : L2(R)→ L2(R) satisfying

∫ ∞−∞F f (ξ)Fg(ξ) dξ =

∫ ∞−∞

f (x)g(x) dx .

Inversion formula:

f (x) =1

2πlim

R→∞

−Re iξx f (ξ) dξ, −∞ < x <∞.

Symmetric Wright M-function

We will require the Fourier transform of Mα(|x |).

LemmaFor κ > −1 and 0 < α < 1,∫ ∞

0tκMα(t) dt =

Γ(κ+ 1)

Γ(ακ+ 1).

TheoremFor 0 < α < 1,

FMα(|x |) = 2E2α(−ξ2).

Proof of the lemma

∫ ∞0

tκMα(t) dt =

∫ ∞0

∫ 0+

w 1−α dt

∫ 0+

−∞ew∫ ∞

0tκe−tw

w 1−α

and∫ ∞0

tκe−twα

dt = Γ(κ+ 1)

∫ ∞0

e−twα

Υκ+1(t) dt

= Γ(κ+ 1)Υκ+1(wα) = Γ(κ+ 1)(wα)−(κ+1)

so∫ ∞0

tκMα(t) dt =Γ(κ+ 1)

∫ 0+

−∞eww−(ακ+1) dw =

Γ(κ+ 1)

Γ(ακ+ 1).

Proof of the theorem

FMα(|x |) =

∫ ∞−∞

e−iξxMα(|x |) dx

∫ ∞0

Mα(x) cos ξx dx

∫ ∞0

Mα(x)∞∑k=0

(−1)k(ξx)2k

(2k)!dx

= 2∞∑k=0

(−1)kξ2k

∫ ∞0

x2kMα(x) dx

= 2∞∑k=0

(−1)kξ2k

Γ(2k + 1)

Γ(2αk + 1)

= 2∞∑k=0

(−ξ2)k

Γ(1 + 2αk)= 2 E2α(−ξ2).

Part III

Anomalous diffusion

Introduction

The classical diffusion equation,

ut − K∇2u = 0,

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

In this lecture, we study continuous-time random walks, whichprovide a generalization of Brownian motion. When thewaiting-time distribution obeys a power law, the particlesexperience trapping and their macroscopic behaviour is said to besubdiffusive. The mean-square displacement of such a particle isproportional to tα for a characteristic exponent in therange 0 < α < 1.

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

ut − ∂1−αt Kα∇2u = 0,

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

0+ for theRiemann–Liouville fractional derivative with respect to t. Theconstant Kα > 0 is a generalized diffusivity, and the classicaldiffusion equation then arises as the limiting special casewhen α→ 1.

Outline

Continuous-time random walks

Rescaling

Subdiffusion

Continuous-time random walks

A walker moves along the x-axis, starting at position x0 attime 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 withprobability density functions ψ(t) and λ(x), respectively. That is,

P(a < ∆tn < b) =

aψ(t) dt for 0 < a < b <∞,

P(a < ∆xn < b) =

aλ(x) dx for −∞ < a < b <∞.

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

An example

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

ψ(t) = τ−1e−t/τ for 0 < t <∞,

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

λ(x) =1

2πexp

(− x2

)for −∞ < x <∞.

E(∆tn) = τ, E(∆t2n) = 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 integralequals the Laplace convolution

f ∗ g(z) =

0f (z − y)g(y) dy for z > 0.

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

Probability distribution of tn

Let ψn(t) denote the probability density function of the randomvariable

tn = ∆t1 + ∆t2 + · · ·+ ∆tn,

that is,

P(a < tn < b) =

aψn(t) dt for 0 < a < b <∞.

By the theorem quoted above,

ψn(t) = (ψn−1 ∗ ψ)(t) =

0ψn−1(s)ψ(t − s) ds,

soψn = ψ ∗ ψ ∗ · · · ∗ ψ︸︷︷︸

n factors

with ψ0 = δ.

Survival probability

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

Ψ(t) =

∫ ∞t

ψ(s) ds = 1−∫ t

0ψ(s) ds for 0 < t <∞.

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

χn(t) =

0ψn(s)Ψ(t − s) ds for 0 < t <∞.

Probability distribution of xn − x0

Let λn(x) denote the probability density function of the randomvariable

xn − x0 = ∆x1 + ∆x2 + · · ·+ ∆xn,

that is,

P(a < xn − x0 < b) =

aλn(x) dx for −∞ < a < b <∞.

λn(x) = (λn−1 ~ λ)(x) =

∫ ∞−∞

λn−1(y)λ(x − y) dy ,

we haveλn = λ~ λ~ · · ·~ λ︸︷︷︸

n factors

with λ0 = δ.

Characteristic functionsTerminology from probability theory: the characteristic function ofψ(t) is just its Laplace transform,

ψ(z) = Lψ(z) =

∫ ∞0

e−ztψ(t) dt,

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 probabilityΨ = 1− I1ψ is

Ψ(z) = z−1 − z−1ψ(z) =1− ψ(z)

Probability density

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

P(a < x(t)− x0 < b) =

ap(x , t) dx .

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

p(x , t) =∞∑n=0

λn(x)χn(t).

Recalling that χn = ψn ∗Ψ,

χn(z) = ψn(z)Ψ(z) = ψ(z)n1− ψ(z)

Characteristic function

Denote the Fourier–Laplace transform of p by

ˆp(ξ, z) = LFp(ξ, z) =

∫ ∞0

e−zt∫ ∞−∞

e−iξxp(x , t) dx dt.

Using the results derived above,

ˆp(ξ, z) =∞∑n=0

λn(ξ)χn(z)

=∞∑n=0

λ(ξ)nψ(z)n1− ψ(z)

=1− ψ(z)

∞∑n=0

[λ(ξ)ψ(z)

Geometric series

Since λ is a probability density function,

λ(0) =

∫ ∞−∞

λ(x) dx = 1,

and likewise

ψ(0) =

∫ ∞0

ψ(t) dt = 1.

But if ξ 6= 0 or z > 0, then∣∣λ(ξ)ψ(z)

∣∣ < 1 so

∞∑n=0

[λ(ξ)ψ(z)

1− λ(ξ)ψ(z)

and hence

ˆp(ξ, z) =1− ψ(z)

1− λ(ξ)ψ(z).

Earlier example

ψ(t) = τ−1e−t/τ and λ(x) =1

2πexp

(− x2

we find that

ψ(z) =1

1 + τzand λ(ξ) = e−σ

2ξ2/2,

1− ψ(z)

1 + τz,

1− ψ(z)λ(ξ) =1 + τz − exp(−σ2ξ2/2)

1 + τz,

and thusˆp(ξ, z) =

1 + τz − exp(−σ2ξ2/2).

Uncertain initial position

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

P(a < x0 < b) =

ap0(x) dx for −∞ < a < b <∞.

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

P(a < xn < b) =

aλn(x) dx for −∞ < a < b <∞.

Sincexn = x0 + ∆x1 + · · ·+ ∆xn,

we haveλn = p0 ~ λ~ λ~ · · ·~ λ︸︷︷︸

n factors

In particular, λ0 = p0 (rather than δ),

Using[λ(ξ)]np0(ξ)

in the preceding analysis leads to

ˆp(ξ, z) =1− ψ(z)

p0(ξ)

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 isknown with certainty.

Rescaling

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

0tψ(t) dt = 1,

∫ ∞−∞

xλ(x) dx = 0,

∫ ∞−∞

x2λ(x) dx = 1.

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

ψτ (t) =1

)and λσ(x) =

so that

E(∆tn) = τ, E(∆xn) = 0, E(∆x2n ) = 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

MomentsWe saw earlier that ψ(0) = 1 = λ(0). Since

∫ ∞0

e−zt(−t)kψ(t) dt

we have

ψ′(0) = −∫ ∞

0tψ(t) dt = −1.

Similarly,dk λ

∫ ∞−∞

e−iξx(−ix)kλ(x) dx

λ′(0) = −i

∫ ∞−∞

xλ(x) dx = 0,

λ′′(0) = −∫ ∞−∞

x2λ(x) dx = −1.

Characteristic functions

Sinceψτ (z) = ψ(τz) and λσ(ξ) = λ(σξ),

we have

ˆp(ξ, z ;σ, τ) =1− ψ(τz)

1− ψ(τz)λ(σξ).

The Taylor expansion

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

implies that

1− ψ(τz)

z=τz + O(τ2z2)

(1 + O(τz)

)as τ → 0.

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

λ(ξ) = λ(0) + λ′(0)ξ + 12 λ′′(0)ξ2 + · · · = 1− 1

2ξ2 + O(ξ4).

ψ(τz)λ(σξ) =[1− τz + O(τ2z2)

][1− 1

2σ2ξ2 + O(σ4ξ4)

]= 1− τz − 1

2σ2ξ2 + O(τ2z2 + σ4ξ4)

1− ψ(τz)λ(σξ) =(τz + 1

2σ2ξ2)[

1 + O(τz + σ2ξ2)],

ˆp(ξ, z ;σ, τ) =τ

τz + 12σ

2ξ2× 1 + O(τz)

1 + O(τz + σ2ξ2).

Limiting probability densityNow send σ → 0 and τ → 0 while keeping

2τ= K ,

for a fixed K > 0, and obtain

ˆp(ξ, z) = limτ

τz + 12σ

z + Kξ2.

Inverting the Laplace transform, we find

p(ξ, t) =1

∫ a+i∞

a−i∞

ezt dz

z + Kξ2= e−Kξ

and then inverting the Fourier transform,

p(x , t) =1√

4πKtexp

(− x2

Snapshots of p(x , t)

Partial differential equation for p

Notice that

LF(pt − Kpxx

)= z ˆp(ξ, z)− p(ξ, 0) + Kξ2ˆp(ξ, z)

= (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 sop(ξ, 0) = 1.

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

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

Uncertain initial position

When x0 is uncertain,

ˆp(ξ, z ;σ, τ) =τ p0(ξ)

τz + 12σ

2ξ2× 1 + O(τz)

1 + O(τz + σ2ξ2),

so in the scaling limit,

ˆp(ξ, z) =p0(ξ)

z + Kξ2

and thus

LF(pt − Kpxx

)= z ˆp(ξ, z)− p(ξ, 0) + Kξ2ˆp(ξ, z)

= (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 probabilitydensity function is a power law:

ψ(t) ∼ A

t1+αas t →∞,

for some constant A > 0. It follows that∫ ∞0

tψ(t) dt = +∞,

so the preceding analysis of the random walk breaks down.

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

Example

ψ(t) =α

(1 + t)1+αand λ(x) =

2πexp

(− x2

Power law (α = 0.75) vs Exponential (τ = 1) distributions

Typical path (α = 0.75, σ = 1)

Behaviour of the characteristic function as z → 0Assume 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) =

∫ ∞0

(1− e−zt)ψ(t) dt = I1 + I2 + I3,

0(1− e−zt)ψ(t) dt,

∫ ∞T

(1− e−zt)(ψ(t)− At−1−α) dt,

∫ ∞T

(1− e−zt)At−1−α dt.

Since0 ≤ 1− e−y ≤ min(1, y) for 0 ≤ y ≤ 1,

we immediately see that

0 ≤ I1 ≤∫ T

0ztψ(t) dt ≤ zT

∫ ∞0

ψ(t) dt = Tz .

Also, the substitution t = y/z gives

|I2| ≤∫ ∞T

(1− e−zt)Ct−2−α dt ≤ Cz1+α

∫ ∞Tz

(1− e−y )y−2−α dy

≤ Cz1+α

(∫ 1

Tzy−1−α dy +

∫ ∞1

y−2−α dy

)= Cz1+α

((Tz)−α − 1

1 + α

)≤ Cα,T z .

The same substitution t = y/z yields

∫ ∞T

(1− e−zt)At−1−α dt = Azα∫ ∞Tz

1− e−y

y 1+αdy = Bαzα + I4,

Bα = A

∫ ∞0

1− e−y

y 1+αdy and I4 = −Azα

∫ Tz

1− e−y

y 1+αdy .

|I4| ≤ Azα∫ Tz

0y−α dy =

AT−αz

1− α,

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

Integrating by parts,∫ ∞0

1− e−y

y 1+αdy =

∫ ∞0

(1− e−y ) d(−α−1y−α

[− 1− e−y

+ α−1

∫ ∞0

e−yy−α dy

= α−1Γ(1− α) = −Γ(−α),

which completes the proof of the following result.

TheoremIf

ψ(t) = At−1−α + O(t−2−α) as t →∞,

then, with Bα = Aα−1Γ(1− α),

ψ(z) = 1− Bαzα + O(z) as z → 0.

Rescaling

As before, normalize λ(x) so that∫ ∞−∞

xλ(x) dx = 0 and

∫ ∞−∞

x2λ(x) dx = 1,

but now suppose

ψ(t) ∼ A

t1+αas t →∞.

Define the rescaled probability density functions

ψτ (t) =1

)and λσ(x) =

and notice ψτ is again a power law,

ψτ (t) ∼ Aτα

t1+αas t →∞.

Characteristic functions

Hence,

E(∆tn) = +∞, E(∆xn) = 0, E(∆x2n ) = σ2.

As before,

ˆp(ξ, z ;σ, τ) =1− ψ(τz)

1− ψ(τz)λ(σξ),

withλ(ξ) = 1− 1

2ξ2 + O(ξ4) as ξ → 0,

but this time

ψ(z) = 1− Bαzα + O(z) as z → 0,

where Bα = Aα−1Γ(1− α).

1− ψ(τz)

Bαταzα + O(τz)

z= Bατ

αzα−1[1 + O(τ1−αz1−α)

Likewise,

ψ(τz)λ(σξ) =[1− Bατ

αzα + O(τz)][

1− 12σ

2ξ2 + O(σ4ξ4)]

= 1− Bαταzα − 1

2σ2ξ2 + O(τz + τ2αz2α + σ4ξ4)

1− ψ(τz)λ(σξ) = Bαταzα + 1

2σ2ξ2 + O(τz + τ2αz2α + σ4ξ4)

=[Bατ

αzα + 12σ

2ξ2][

1 + O(τ1−αz1−α + ταzα + σ2ξ2)].

Limiting probability density

Therefore,

ˆp(ξ, z ;σ, τ) =Bατ

αzα−1

Bαταzα + 12σ

1 + O(τ1−αz1−α)

1 + O(τ1−αz1−α + ταzα + σξ

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

2Bατα= Kα,

for a fixed Kα > 0, and obtain

ˆp(ξ, z) = limBατ

αzα−1

Bαταzα + 12σ

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

Recall that

Lt→zEα(−λtα) =1

z + λz1−α

andFx→ξMα/2(|x/µ|) = 2µEα(−µ2ξ2).

ˆp(ξ, z) =zα−1

zα + Kαξ2=

z + λz1−α if λ = Kαξ2,

we see that

p(ξ, t) = Eα(−Kαtαξ2) = Eα(−µ2ξ2) if µ =√

Kαtα.

p(x , t) =1

KαtαMα/2

(|x |√Kαtα

Snapshots of p(x , t) when α = 2/3.

Fractional partial differential equation for p

We have

LFpt − KαD1−αpxx = z ˆp(ξ, z)− p(ξ, 0) + Kαz1−αξ2ˆp(ξ, z)

=(z + Kαz1−αξ2

)ˆp(ξ, z)− p(ξ, 0) = 0

ˆp(ξ, z) =zα−1

zα + Kαξ2=

z + Kαz1−αξ2and p(ξ, 0) = δ(ξ) = 1.

Thus, p satisfies the time-fractional diffusion equation,

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

Mean-square displacement

V (t) = E((x(t)− x0)2

∫ ∞−∞

x2p(x , t) dx for t > 0.

V (z) =

∫ ∞−∞

x2p(x , z) dx = − d2

dξ2ˆp(ξ, z)

∣∣∣∣ξ=0

= − d2

(z + Kαz1−αξ2

)−1∣∣∣∣ξ=0

= 2Kαz−1−α,

it follows that V (t) = 2KαΥ1+α(t), that is,

E((x(t)− x0)2

2Kαtα

Γ(1 + α)∝ tα.

Part IV

The time-fractional diffusion equation

Introduction

We have shown that in 1D the probability density function for thelocation of a subdiffusive particle at time t obeys a time-fractionalPDE. The concentration u = u(x , t) of a large number of suchparticles evolves in the same way. Moreover, the 1D analysis canbe generalized to higher dimensions to yield the time-fractionaldiffusion equation

ut − Kα∂1−αt ∇2u = 0.

As in the classical case α = 1, the solution u in a bounded spatialdomain (and subject to homogeneous boundary conditions) can beconstructed by separation of variables to yield a series expansioninvolving the eigenfunctions of −∇2. We use this representationof u(x , t) to investigate its behaviour.

Outline

Initial-boundary value problem

Smoothing property of fractional diffusion

Positivity

Initial-boundary value problem

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

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

u = u0(x) for x ∈ Ω, when t = 0,

and impose homogeneous boundary conditions, either Dirichlet

u = 0 for x ∈ ∂Ω and t > 0,

or else Neumann,

∂n= 0 for x ∈ ∂Ω and t > 0,

where n is the outward unit normal to Ω.

Abstract initial-value problem

Let A be a linear operator with dense domain D(A) in a Hilbertspace 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−αt Au = f (t) for t > 0,

with u(0) = u0, where u = ut = ∂u/∂t.

Standard example:

H = L2(Ω), Au = −Kα∇2u, D(A) = H2(Ω) ∩ H10 (Ω).

Integrate to obtain an equivalent Volterra equation in H,

u(t) +

0Υα(t − s)Au(s) ds = u0 +

0f (s) ds, t > 0.

Eigenfunction expansionAssume that A is self-adjoint and positive-semidefinite, with acomplete orthonormal eigensystem, say

Aφm = λmφm for m = 0, 1, 2, . . . ,

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

0 ≤ λ0 ≤ λ1 ≤ λ2 ≤ · · · .

(These assumptions hold for our standard example.) Thus,

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) =∞∑

fm(t)φm and u0 =∞∑

u0mφm.

Dirichlet boundary conditions in 1D

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

φm(x) =

Lx and λm = Kα

for m ∈ 1, 2, 3, . . ., so

v(x) =∞∑

vmφm(x)

is just the sine series expansion of v , where

vm = 〈v , φm〉 =

0v(x) sin

Lx dx .

Neumann boundary conditions in 1D

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

φ0(x) =1√L

and λ0 = 0,

φm(x) =

Lx and λm = Kα

for m ≥ 1,

v(x) =∞∑

vmφm(x)

is just the cosine series expansion of v .

Separation of variables

The function u satisfies

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

iff the mth eigenmode satisfies

um + λm∂1−αt 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)

um(z) =u0m + fm(z)

z + λmz1−α .

Duhamel formula and the mild solution

Recall that

LEα(−λtα) =1

z + λz1−α ,

um(t) = Eα(−λmtα)u0m +

0Eα(−λm(t − s)α

)fm(s) ds.

Define the solution operator for the homogeneous problem,

E(t)v =∞∑

Eα(−λmtα)〈v , φm〉φm for t > 0 and v ∈ H,

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

u(t) = E(t)u0 +

0E(t − s)f (s) ds, t > 0.

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

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

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

Thus, using Parseval’s identity,

‖E(t)v‖2 =∞∑

Eα(−λmtα)2〈v , φm〉2 ≤∞∑

〈v , φm〉2 = ‖v‖2,

and therefore

‖E(t)v‖ ≤ ‖v‖ for t ≥ 0 and v ∈ H.

Hence, the mild solution satisfies the stability estimate

‖u(t)‖ ≤ ‖u0‖+

0‖f (s)‖ ds for t ≥ 0.

Smoothing property of fractional diffusion

For 0 ≤ r <∞, define the norm

‖v‖2r =

∥∥(I + A)r/2v‖2 =∞∑

(1 + λm)r 〈v , φm〉2

and the corresponding closed subspace

Hr = v ∈ H : ‖v‖r <∞,

which is a Hilbert space with respect to the inner product thatinduces ‖ · ‖r .

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

D(Ω) or HrN(Ω) to indicate the choice of Dirichlet or

Neumann boundary conditions.

Dirichlet boundary conditions and Sobolev spaces

Can prove the following via interpolation and elliptic regularity.

TheoremSuppose that ∂Ω is C∞. If 0 ≤ r < 1

2 , then HrD(Ω) = H r (Ω),

however if 2j − 32 < r < 2j + 1

2 for j ∈ 1, 2, 3, . . ., then

HrD(Ω) = v ∈ H r (Ω) : v = Av = · · · = Aj−1v = 0 on ∂Ω .

In the exceptional case r = 2j − 32 , the condition Aj−1v = 0 on ∂Ω

must be replaced by Aj−1v ∈ H1/2(Ω).

If Ω is Lipschitz, then the conclusions still hold for r ≤ 1, and inparticular H1

D = H10 (Ω) = u ∈ H1(Ω) : u = 0 on ∂Ω .

If Ω is convex or C 1,1, then r ≤ 2 is OK, and in particularH2

D = H2(Ω) ∩ H10 (Ω) = u ∈ H2(Ω) : u = 0 on ∂Ω .

Neumann boundary conditions and Sobolev spaces

TheoremSuppose that ∂Ω is C∞. If 0 ≤ r < 3

2 , then HrN(Ω) = H r (Ω),

however if 2j − 12 < r < 2j + 3

2 for j ∈ 1, 2, 3, . . ., then

HrN(Ω) = v ∈ H r (Ω) :

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

In the exceptional case r = 2j − 12 , the condition ∂nAj−1v = 0

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

If Ω is Lipschitz, then the conclusions still hold if r ≤ 1, and inparticular H1

N(Ω) = H1(Ω).

If Ω is convex or C 1,1, then r ≤ 2 is OK, and in particularH2

N(Ω) = u ∈ H2(Ω) : ∂nu = 0 on ∂Ω .

A 1D example

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

0v(x) sin

Lx dx = L

1− (−1)m

and (1 + λm)r ∼ (1 + m2)r , we see that

‖v‖r <∞ ⇐⇒∞∑p=0

(1 + 2p)2r−2 <∞,

so v ∈ HrD(Ω) iff 2r − 2 < −1, that is, r < 1

2 . However,v ∈ Hr

N(Ω) for all r ≥ 0 because∫ L

0v(x) cos

Lx dx = 0 for all m ≥ 1.

Smoothing property of classical diffusion

If α = 1 then Eα(−tα) = e−t so

E(t)v =∞∑

e−λmt〈v , φm〉φm

and thus

‖E(t)v‖2r+µ =

∞∑m=0

(1 + λm)r+µ(e−λmt〈v , φm〉

If 0 < t ≤ T and λ ≥ 0, then

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

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

Weaker smoothing property for subdiffusion

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

Lemma0 < Eα(−tα) ≤ C min(1, t−α) for 0 < t <∞.

Proof.Follows because

Eα(−tα) =

1 + O(tα) as t → 0+,

t−α/Γ(1− α) + O(t2−α) as t →∞.

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

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

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

‖E(t)v‖2r+µ =

∞∑m=0

(1 + λm)r+µg(λ1/αm t)2〈v , φm〉2.

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

‖E(t)v‖2r+µ ≤ CT

∞∑m=0

(1 + λm)r 〈v , φm〉2 = CT t−αµ‖v‖2r .

Regularity in time

Let q ∈ 1, 2, 3, . . .. Similar arguments yield the followingestimates.

LemmaThe function g(t) = Eα(−tα) satisfies

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

TheoremLet −2 ≤ µ ≤ 2, 0 ≤ r <∞ and q ∈ 1, 2, 3, . . .. If v ∈ Hr , then

tq‖E(q)(t)v‖r+µ ≤ Cq,T t−αµ/2‖v‖r for 0 < t ≤ T .

Detailed behaviour as t → 0+

Eα(−λtα) =M−1∑p=0

(−1)ptαp

Γ(1 + αp)λp + O(λMtαM) as t → 0+,

and λpm〈v , φm〉 = 〈Apv , φm〉, we can show the following.

TheoremLet 0 ≤ r <∞ and M ∈ 1, 2, 3, . . .. If v ∈ Hr+2M , then

E(t)v = v +M−1∑p=1

(−1)ptαp

Γ(1 + αp)Apv + RM(t)AMv ,

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

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

Behaviour of an eigenmode

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

u(t) = E(t)φm = Eα(−λmtα)φm,

u(x , t) =

(1− λmtα

Γ(1 + α)+ O(t2α)

)φm(x) as t → 0+.

This example makes clear the fact that the time derivativeu = O(tα−1) is unbounded as t → 0+ no matter how regular theinitial 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 problemThe function u(t) = E(t)u0 solves the homogeneous problem

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

Now consider

u(t) = E ∗ f (t) =

0E(t − s)f (s) ds,

which solves the inhomogeneous problem with vanishing initialdata:

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

TheoremFor 0 ≤ r <∞ and q ∈ 0, 1, 2, . . ., the function u = E ∗ fsatisfies

tq‖u(q)(t)‖r ≤ Cq

q∑j=0

0s j‖f (j)(s)‖r ds for 0 < t <∞.

Positivity

If we write

u+(t) =

0, t < 0,

u(t), t ≥ 0,,

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

u(iy) = u+(y) =

∫ ∞0

e−iytu(t) dt.

Thus, the Parseval–Plancherel identity,∫ ∞−∞

f (t)g(t) dt =1

∫ ∞−∞

f (ξ)g(ξ) dξ,

implies that ∫ ∞0

u(t)v(t) dt =1

∫ ∞−∞

u(iy)v(iy) dy .

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

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

0(∂βt u)u dt =

cos 12πβ

∫ ∞0

yβ|u(iy)|2 dy ≥ 0.

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 12πβ

H-valued case

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

0〈∂1−α

t Au, u〉 dt =sin 1

∫ ∞0

y 1−α‖A1/2u(iy)‖2 dy ≥ 0.

Proof.

∫ ∞0〈∂1−α

t Au, u〉 dt =

∫ ∞0

∞∑m=0

λm(∂1−αt um)um dt

=∞∑

λmcos 1

2π(1− α)

∫ ∞0

y 1−α|um(iy)|2 dy .

Digression: fractional derivative at a jump discontinuity

Suppose that

v(t) =

v1(t), 0 ≤ t < a,

v2(t), t > a,

where v1 : [0, a]→ R and v2 : [a,∞)→ R are C 1 functions. If0 < t < a, then differentiating the formula

Iαv(t) = Υα ∗ v(t) =

0Υα(s)v(t − s) ds

∂1−αt v(t) = DIαv1(t) = v1(0)Υα(t) +

0Υα(s)v ′1(t − s) ds

= v(0+)Υα(t) +

0Υα(t − s)v ′(s) ds.

However, if t > a then

Iαv(t) =

0Υα(t − s)v1(s) ds +

aΥα(t − s)v2(s) ds

t−aΥα(s)v1(t − s) ds +

∫ t−a

0Υα(s)v2(t − s) ds

∂1−αv(t) = v1(0)Υα(t)− v1(a)Υα(t − a) + v2(a)Υα(t − a)

t−aΥα(s)v ′1(t − s) ds +

∫ t−a

0Υα(s)v ′2(t − s) ds

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

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

0Υα(t − s)v ′(s) ds +

aΥα(t − s)v ′(s) ds.

Example

Stability via an energy argument

Consider the homogeneous equation,

u(t) + ∂1−αt Au(t) = 0.

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

0〈u, u〉 dt +

0〈∂1−α

t Au, u〉 dt = 0.

Letting

u∗(t) =

u(t), 0 < t < T ,

0, t > T ,

we have ∫ T

0〈∂1−α

t Au, u〉 dt =

∫ ∞0〈∂1−α

t Au∗, u∗〉 dt ≥ 0.

Thus, ∫ T

0〈u, u〉 dt ≤ 0.

But ∫ T

0〈u, u〉 dt =

[12〈u, u〉

= 12‖u(T )‖2 − 1

2‖u(0)‖2,

so ‖u(T )‖ ≤ ‖u(0)‖, which again shows that

‖E(t)u0‖ ≤ ‖u0‖ for t > 0.

Part V

Simple finite difference schemes

Introduction

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

ut − Kα∂1−αt uxx = f (x , t).

These schemes generalize the forward and backward Euler methodsfor the heat equation. As in the classical setting, the explicitscheme is stable only if the time step is sufficiently small, but theimplicit scheme is unconditionally stable.

Unlike the classical Euler methods, the grid stencils extend backthrough all preceding time levels resulting in a dramatically highercomputational cost.

Outline

Explicit Euler method for a fractional ODE

Explicit Euler method for a fractional PDE

Implicit Euler method

Explicit Euler method for a fractional ODE

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

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

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

tn = n∆t for 0 ≤ n ≤ N.

We want to computeUn ≈ u(tn).

Time-stepping

Integrating the ODE gives

u(tn+1)− u(tn) + λ

∫ tn+1

∂1−αt u(t) dt =

∫ tn+1

f (t) dt,

which suggests the time-stepping scheme

Un+1 − Un + λ

∫ tn+1

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

∂1−αt U(t) dt = (Υα ∗ U)(tn+1)− (Υα ∗ U)(tn)

∫ tn+1

Υα(tn+1 − t)Un dt

+n−1∑j=0

∫ tj+1

[Υα(tn+1 − t)−Υα(tn − t)

]U j dt

=∆tα

Γ(α + 1)

n∑j=0

wn−jUj .

Weights

Here, ∫ tn+1

Υα(tn+1 − t) dt = Υα+1(∆t) =∆tα

Γ(α + 1)

so w0 = 1, and for 0 ≤ j ≤ n − 1,

∆tα

Γ(α + 1)wn−j =

∫ tj+1

[Υα(tn+1 − t)−Υα(tn − t)

]U j dt.

Find that

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

Implementation

In this way, we arrive at

Un+1 − Un +λ∆tα

Γ(α + 1)

n∑j=0

wn−jUj = f n ∆t.

Thus, starting from U0 = u0 we compute

Un+1 = Un + f n ∆t − λ∆tα

Γ(α + 1)

n∑j=0

wn−jUj

for n = 0, 1, 2, . . . , N.

Classical Euler method

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

λ∆tα

Γ(α + 1)

n∑j=0

wn−jUj = λ∆t Un

and therefore

Un+1 − Un + λ∆t Un = f n ∆t,

or equivalently,Un+1 − Un

∆t+ λUn = f n.

Conditional stabilitySince w0 = 1 we have

Un+1 =

(1− λ∆tα

Γ(α + 1)

)Un + f n ∆t − λ∆tα

Γ(α + 1)

n−1∑j=0

wn−jUj ,

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

|u(t)| ≤ |u0|+∫ t

0|f (s)| ds for t > 0.

TheoremIf

λ∆tα

Γ(α + 1)≤ 1

|Un| ≤ |U0|+n−1∑j=0

|f n|∆t for 1 ≤ n ≤ N.

Put ρ = λ∆tα/Γ(α + 1) ≤ 1, then

|Un+1| ≤ (1− ρ)|Un|+ |f n|∆t + ρ

n−1∑j=0

|wn−j ||U j |

and we find

n−1∑j=0

|wn−j | = −n∑

wj = 1 + nα − (n + 1)α ≤ 1,

|Un+1| ≤ (1− ρ)|Un|+ |f n|∆t + ρ max0≤j≤n−1

|U j |

≤ |f n|∆t + max1≤j≤n

|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 Ω = (0, L) in 1D with boundary ∂Ω = 0, L. We seeku = u(x , t) satisfying

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

u = 0 for x ∈ ∂Ω 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 computeUnp ≈ u(xp, tn).

Second central difference in spaceUsing the approximation

uxx(xp, tn) ≈Unp+1 − 2Un

p + Unp−1

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

arrive at the scheme

Un+1p − Un

p −Kα ∆tα

Γ(α + 1)

n∑j=0

wn−jU jp+1 − 2U j

p + U jp−1

∆x2= f n

p ∆t

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

U0p = u0(xp) for 0 ≤ p ≤ P,

and boundary conditions

Un0 = 0 = Un

P for 1 ≤ n ≤ N.

Stencil

Matrix–vector formulationLet

Un2...

UnP−2

UnP−1

and A =Kα

2 −1−1 2 −1

. . .. . .

−1 2 −1−1 2

so that

Un+1 −Un +∆tα

Γ(α + 1)

n∑j=0

wn−jAUj = ∆t fn

and hence

Un+1 = Un + ∆t fn − ∆tα

Γ(α + 1)

n∑j=0

wn−jAUj .

Eigenvectors

Recall that

φm(x) = sinmπ

Lx and λm = Kα

satisfy

−Kαd2

dx2φm = λmφm for x ∈ Ω = (0, L),

with φm(0) = 0 = φm(L). Putting

φm(x1)φm(x2)

...φm(xP−1)

and Λm =Kα

(2 sin

2L∆x

we find thatAΦm = ΛmΦm.

Discrete L2-inner product and -norm

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 basisfor RP−1,

〈Φm,Φm′〉 = 0 if m 6= m′ and m, m′ ∈ 1, 2, . . . ,P − 1,

and, with θ = mπ/P,

‖Φm‖2 =P−1∑p=1

∆x =L

P∑p=1

sin2 pθ

2− cos(P + 1)θ sin Pθ

2 sin θ

Stability of the discrete Fourier modesDefine the discrete Fourier coefficients

〈Un,Φm〉‖Φm‖2

, 1 ≤ m ≤ P − 1,

so that

Un =P−1∑m=1

UnmΦm.

For 1 ≤ m ≤ P − 1,

Un+1m − Un

m +Λm ∆tα

Γ(α + 1)

n−1∑j=0

wn−j Ujm = f n

m ∆t,

with U0m = u0m, so our earlier analysis gives

|Unm| ≤ |U0

m|+n−1∑j=0

|f nm|∆t provided

Λm ∆tα

Γ(α + 1)≤ 1.

Stability of the full solution

‖U‖2 =P−1∑m=1

|Um|2‖Φm‖2 =L

P−1∑m=1

|Um|2,

and Λm ≤ 4Kα/∆x2, we can show the following.

TheoremIf

ρ ≡ 4Kα ∆tα

Γ(α + 1) ∆x2≤ 1

‖Un‖2 ≤ 2‖U0‖2 + 2tn

n−1∑j=0

‖f j‖2 ∆t for 1 ≤ n ≤ N.

Problem if α is small

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

∆tα ≤ Γ(α + 1)

4Kα∆x2.

This is a severe restriction if α is small.

Example

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

∆t ≤ (∆x/2)10 = 10−30.

Therefore natural to consider implicit methods.

Again start with the ODE

u + λ∂1−αt 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

tn−1

∂1−αt u(t) dt =

∫ tn

tn−1

f (t) dt,

and compute Un ≈ u(tn) via

Un − Un−1 + λ

∫ tn

tn−1

∂1−αt U(t) dt = f n ∆t,

whereU(t) = Un for tn−1 < t ≤ tn.

Weights

We find that∫ tn

tn−1

∂1−αt U(t) dt = (Υα ∗ U)(tn)− (Υα ∗ U)(tn−1)

∫ tn

tn−1

Υα(tn − t)Un dt

+n−1∑j=1

∫ tj

tj−1

[Υα(tn − t)−Υα(tn−1 − t)

]U j dt

=∆tα

Γ(α + 1)

n∑j=1

ωn−jUj ,

where, as before ω0 = 1 and

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

Unconditional stability

In this way,

Un − Un−1 + ρ

n∑j=1

wn−jUj = f j ∆t, ρ =

λ∆tα

Γ(α + 1).

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

(1 + ρ)Un = Un−1 + f n ∆t − ρn−1∑j=1

wn−jUj .

Theorem

|Un| ≤ |U0|+n∑

|f j |∆t for 1 ≤ n ≤ N.

We have

n−1∑j=1

|wn−j | =n−1∑j=1

|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

|Un| ≤ |f n|∆t +|Un−1|1 + ρ

1 + ρmax

1≤j≤n−1|U j |

≤ |f n|∆t + max1≤j≤n−1

|U j |.

The desired estimate follows using induction on n.

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

Convergence behaviour

E1(N) = max1≤n≤N

|Un − u(tn)|,

E2(N) = max0.5≤tn≤T

|Un − u(tn)|.

N E1(N) E2(N)

80 1.344e-02 3.951e-03

160 8.303e-03 0.6944 2.023e-03 0.9654

320 5.005e-03 0.7303 1.028e-03 0.9771

640 2.911e-03 0.7821 5.196e-04 0.9844

1280 1.650e-03 0.8186 2.617e-04 0.9892

2560 9.159e-04 0.8495 1.316e-04 0.9925

5120 4.998e-04 0.8739 6.602e-05 0.9947

10240 2.688e-04 0.8949 3.310e-05 0.9963

Remark on computing the weights

Recall that if we compute a sum with M terms,

S =M∑

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

| fl(S)− S | ≤ Mε

1−Mε

M∑m=1

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 toevaluate

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

]= jα

(exp[α log(1 + j−1)

]− 1),

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

When we compute

S = ρ

n∑j=1

wn−jUj

the estimates above yield

| fl(S)− S | . nεn∑

(n − j)α|U j | . nα+2ε max1≤j≤n

|U j |,

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

Fractional diffusion equation

For Ω = (0, L), we again consider the initial-boundary valueproblem

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

u = 0 for x ∈ ∂Ω and 0 < t < T .

The implicit time-stepping scheme leads to

Unp − Un−1

p − Kα ∆tα

Γ(α + 1)

n∑j=1

wn−jU jp+1 − 2U j

p + U jp−1

∆x2= f n

p ∆t

for 1 ≤ n ≤ N and 1 ≤ p ≤ P − 1, with

U0p = u0(xp) and Un

0 = 0 = UnP .

Stencil

Matrix–vector formulation

We have

Un −Un−1 +∆tα

Γ(α + 1)

n∑j=1

wn−jAUj = ∆t fn

(I + B)Un = Un−1 + ∆t fn −n−1∑j=1

wn−jBUj

B =∆tα

Γ(α + 1)A =

Kα ∆tα

Γ(α + 1)∆x2

2 −1−1 2 −1

. . .. . .

−1 2 −1−1 2

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

n ≈ N2

the overall cost of N time steps is O(N2P).

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

Contrast this with using the implicit Euler method to solve theclassical diffusion equation (the case α = 1): each time steprequires O(P) flops and O(P) active memory locations, and Ntime 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−αt uxx = f (x , t) for x ∈ Ω = (−L, L) and 0 < t < T ,

u = δ(x) for x ∈ Ω, when t = 0,

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

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

ψ(t) =α

(1 + t)αand λ(x) =

1√2π

exp(−x2/2).

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

2Γ(1− α)τα= Kα

Implicit Euler solutions (α = 2/3)

Probability densities of CTRWs (15, 000 samples)

Part VI

Spatial discretization via finite elements

Introduction

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

In this lecture, we seek to estimate the errors from the spatialdiscretization assuming that the time integration is performedexactly. Suitable approaches for the time discretization include theimplicit Euler method described previously and the more accurateschemes addressed in subsequent lectures.

Outline

Method of lines

Error estimates

Non-smooth data

Method of lines

Spatial domain Ω ⊆ Rd (d = 1, 2 or 3); for simplicity a convexpolygon 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−αt ∇2u = f (x , t) for x ∈ Ω and t > 0,

subject to homogeneous Dirichlet boundary conditions

u = 0 for x ∈ ∂Ω and t > 0.

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

Weak formulation

Au = −Kα∇2u and a(u, v) =

Kα∇u · ∇v dx .

First Green identity:∫Ω

Au v dx = a(u, v)−∫∂Ω

Kα∂u

∂nv ds.

〈ut , v〉+ a(∂1−αt u, v) = 〈f , v〉 for v ∈ H1

0 (Ω),

and also

〈ut , v〉+ ∂1−αt a(u, v) = 〈f , v〉 for v ∈ H1

0 (Ω).

Finite element space

Family of triangulations Th of Ω, where as usual

h = maxK∈Th

diam(K ).

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

Vh ⊆ H10 (Ω).

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

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

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

Alternative formulation

Define the discrete elliptic operator Ah : Vh → Vh by

〈Ahψ, χ〉 = a(ψ, χ) for ψ, χ ∈ Vh,

and the L2-projector Ph : L2(Ω)→ Vh by

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

∂1−αt a(uh, χ) = ∂1−α

t 〈Ahuh, χ〉 = 〈∂1−αt Ahuh, χ〉,

we see that

〈uht + ∂1−αt Ahuh, χ〉 = 〈f , χ〉 = 〈Phf , χ〉 for all χ ∈ Vh,

and thusuht + ∂1−α

t Ahuh = Phf for t > 0.

Nodal equationsConstruct 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 ordinaryintegro-differential equations

dt+ ∂1−α

t SU = f(t) for t > 0.

Discrete eigensystem

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

Ahφhm = λhmφ

hm for 1 ≤ m ≤ N,

with 〈φhm, φhn〉 = δmn, and consequently

uh(t) = Eh(t)u0h +

0Eh(t − s)Phf (s) ds,

where the discrete solution operator for the homogeneous problemis defined by

Eh(t)χ =N∑

Eα(−λhmtα)〈χ, φhm〉φhm for χ ∈ Vh.

Stability

TheoremThe finite element solution is stable in L2(Ω):

‖uh(t)‖ ≤ ‖u0h‖+

0‖Phf (s)‖ ds for t ≥ 0.

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

‖Eh(t)χ‖2 =N∑

∣∣Eα(−λhmtα)〈χ, φhm〉∣∣2 ≤ N∑

∣∣〈χ, φhm〉∣∣2 = ‖χ‖2,

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

Error estimates

We wish to estimate the error uh(t)− u(t) in L2(Ω) and in H10 (Ω).

Our assumption that Ω is convex implies that the elliptic problem,

−∇2u = f in Ω, with u = 0 on ∂Ω,

is H2-regular, that is, the weak solution u ∈ H10 (Ω), given by

a(u, v) = 〈f , v〉 for all v ∈ H10 (Ω),

necessarily belongs to H2(Ω) and

‖u‖2 ≤ C‖f ‖.

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

D(Ω) unless v vanishesappropriately on ∂Ω.

Error in the elliptic problem

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

a(uh, χ) = 〈f , χ〉 for all χ ∈ Vh.

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

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

‖uh − u‖1 ≤ C infχ∈Vh

‖χ− u‖1 ≤ Chr−1‖u‖r for 1 ≤ r ≤ p + 1.

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

‖uh − u‖ ≤ Ch infχ∈Vh

‖χ− u‖1 ≤ Chr‖u‖r for 1 ≤ r ≤ p + 1.

Ritz projector

It is convenient to define Rh : H10 (Ω)→ Vh by Rhu = uh, or

equivalently,

a(Rhv , χ) = a(v , χ) for all χ ∈ Vh.

It follows that R2h = Rh, with

‖v − Rhv‖ ≤ Chr‖v‖r and ‖v − Rhv‖1 ≤ Chr−1‖v‖r

for 1 ≤ r ≤ p + 1. Also, since

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

for all v ∈ H10 (Ω) and χ ∈ Vh, we see that

PhA = AhRh : H10 (Ω)→ Vh.

Equation for the error

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

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

Then for all χ ∈ Vh,

〈ϑt , χ〉+ ∂1−αt a(ϑ, χ)

= 〈uht , χ〉+ ∂1−αt a(uh, χ)− 〈Rhut , χ〉 − ∂1−α

t a(Rhu, χ)

= 〈f , χ〉+ 〈Rhut , χ〉 − ∂1−αt a(u, χ)

= 〈ut , χ〉+ ∂1−αt a(u, χ)− 〈Rhut , χ〉 − ∂1−α

t a(u, χ)

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

〈ϑt , χ〉+ ∂1−αt a(ϑ, χ) = 〈ut − Rhut , χ〉.

Estimate for ϑ = uh − Rhu

LemmaFor 1 ≤ r ≤ p + 1,

‖ϑ(t)‖ ≤ ‖u0h − Rhu0‖+ Chr

0‖ut‖r ds

Proof.Stability in L2(Ω) gives

‖ϑ(t)‖ ≤ ‖ϑ(0)‖+

0‖Ph(ut − Rhut)‖ ds,

and here ϑ(0) = uh0 − Rhu0 with

‖Ph(ut − Rhut)‖ ≤ ‖ut − Rhut‖ ≤ Chr‖ut‖r .

Estimate for % = Rhu − u

LemmaFor 1 ≤ r ≤ p + 1,

‖%(t)‖ ≤ Chr

(‖u0‖r +

0‖ut‖r ds

Proof.Write

%(t) = %(0) +

0%t(s) ds

and use‖%t‖ = ‖ut − Rhut‖ ≤ Chr‖ut‖r

together with

‖%(0)‖ = ‖Rhu0 − u0‖ ≤ Chr‖u0‖r .

Quasi-optimal error bound in L2(Ω)

TheoremThe finite element solution of the time-fractional diffusion equationsatisfies

‖uh(t)− u(t)‖ ≤ ‖u0h − Rhu0‖+ Chr

(‖u0‖r +

0‖ut(s)‖r ds

)for 1 ≤ r ≤ p + 1.

In practice, this bound is unlikely to be useful for r > 2 because uwill not be sufficiently smooth for

(· · ·)

on the RHS to be finite.

Realistic smoothness

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

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

D (Ω) so by our earlier regularity theorem,

‖tut‖2 ≤ Ctεα/2‖u0‖2+ε,

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

for simplicity, we obtain

‖uh(t)− u(t)‖ ≤ Ch2‖u0‖2+ε for 0 ≤ t ≤ T ,

with C = C (T , α, ε,Ω).

Estimate for ‖ϑ(t)‖1

We saw that

〈ϑt , χ〉+ a(∂1−αt ϑ, χ) = 〈−ρt , χ〉 for all χ ∈ Vh.

Choosing χ = Ahϑ(t), we have

〈ϑt ,Ahϑ〉 =1

dt〈ϑ,Ahϑ〉 =

dta(ϑ, ϑ)

witha(∂1−α

t ϑ,Ahϑ) = 〈∂1−αt Ahϑ,Ahϑ〉

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

dta(ϑ, ϑ) + 〈∂1−α

t Ahϑ,Ahϑ〉 = −a(Ph%t , ϑ).

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

0〈∂1−α

t Ahϑ,Ahϑ〉 dt ≥ 0,

we have

a(ϑ(T ), ϑ(T )

)− a(ϑ(0), ϑ(0)

)≤ −2

0a(Ph%t , ϑ) dt

Since a(v , v) is equivalent to ‖v‖21,

‖ϑ(T )‖1 ≤ C

(‖ϑ(0)‖1 +

0‖Ph%t‖1 dt

Choose t∗ such that

‖ϑ(t∗)‖1 = max0≤t≤T

‖ϑ(t)‖1,

‖ϑ(T )‖1‖ϑ(t∗)‖1 ≤ ‖ϑ(t∗)‖21

≤ C‖ϑ(0)‖21 + C

∫ t∗

0‖Ph%t‖1‖ϑ‖1 dt

(‖ϑ(0)‖1 +

0‖Ph%t‖1 dt

)‖ϑ(t∗)‖1

‖ϑ(T )‖21 ≤ C‖ϑ(0)‖2

0‖Ph%t(t)‖1‖ϑ(t)‖1 dt

Quasi-optimal error bound in H1(Ω)

Assume now that Th is such that the L2-projector Ph is stablein H1(Ω), that is,

‖Phv‖1 ≤ C‖v‖1 for v ∈ H1(Ω).

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

TheoremThe finite element solution of the time-fractional diffusion equationsatisfies

‖uh(t)− u(t)‖1 ≤ C‖u0h − Rhu0‖1

+ Chr−1

(‖u0‖r +

0‖ut(s)‖r ds

)for 2 ≤ r ≤ p + 1.

ProofRecall that uh − u = ϑ+ % and

‖ϑ(t)‖1 ≤ C

(‖ϑ(0)‖1 +

0‖Ph%t(s)‖1 dt

Here,‖ϑ(0)‖1 = ‖u0h − Rhu0‖1,

and since Ph is stable in H1(Ω),

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

The error bound follows because

‖%(t)‖1 ≤ ‖%(0)‖1 +

0‖%t(s)‖1 ds

= ‖u0 − Rhu0‖1 +

0‖ut(s)− Rhut(s)‖1 ds

≤ Chr−1

(‖u0‖1 +

0‖ut(s)‖1 ds

Non-smooth data

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 notrequire any spatial regularity for u0 or f .

TheoremAssume that u0h = Phu0, and fix ϕ such that 0 < ϕ < π/2. Then,

‖uh(t)− u(t)‖ ≤ C t−αh2(‖u0‖+ sup

z∈∂Σπ−ϕ

‖f (z)‖)

for 0 < t ≤ T .

Under Laplace transformation, our problem

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

becomeszu(z)− u0 + z1−αAu(z) = f (z),

(zαI + A)u(z) = zα−1g(z) where g(z) = u0 + f (z).

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

Recall that Aφm = λmφm with 0 ≤ λ1 < λ2 < · · · .

TheoremLet ϕ satisfy 0 < ϕ ≤ π/2 and put

sinϕ.

If z ∈ Σπ−ϕ then

‖(zI + A)−1‖ ≤ M

|z |≤(

1 + |z |.

ProofLet z = re iθ with |θ| ≤ π − ϕ 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 − ϕ then |z | ≤ M|z + λm| because

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

Therefore, since M ≥ 1,

|z + λm|≤ M

and so (zI + A)−1 exists with

(zI + A)−1v =∞∑

〈v , φm〉z + λm

φm for v ∈ L2(Ω).

Hence, Parseval’s identity yields the first estimate:

‖(zI + A)−1v‖2 =∞∑

∣∣∣∣〈v , φm〉z + λm

∣∣∣∣2≤(

)2 ∞∑m=1

|〈v , φm〉|2 =

|z |‖v‖)2

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

1 + |z ||z + λm|

≤ 1 + λ1/2

λ1/2= 1 +

If |z | ≥ λ1/2, then

1 + |z ||z + λm|

≤ 2|z |/λ1 + |z ||z | sinϕ

and the second estimate follows at once.

Resolvent estimate for Ah

TheoremLet ϕ satisfy 0 < ϕ ≤ π/2 and put M = 1/ sinϕ. If z ∈ Σπ−ϕ then

‖(zI + Ah)−1‖ ≤ M

|z |≤(

1 + |z |.

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

λh1 = minχ∈Vh

〈Ahχ, χ〉‖χ‖2

= minχ∈Vh

〈Aχ, χ〉‖χ‖2

≥ minv∈H1

0 (Ω)

〈Av , v〉‖v‖2

= λ1,

so 2/λh1 ≤ 2/λ1.

Integral representations

Sinceu(z) = zα−1(zαI + A)−1g(z)

anduh(z) = zα−1(zαI + Ah)−1Phg(z)

the Laplace inversion formula gives

u(t) =1

eztzα−1(zαI + A)−1g(z) dz

uh(t) =1

eztzα−1(zαI + Ah)−1Phg(z) dz ,

Γ = a + iy : −∞ < y <∞ for any a > 0.

Error representation

uh(t)− u(t) =1

eztzα−1Gh(z)g(z) dz

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

h (z),

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

G 2h (z) = (Ph − I )(zαI + A)−1.

LemmaFor z ∈ Σπ−ϕ,

‖A(zαI + A)−1v‖ ≤ C‖v‖ and ‖(zαI + Ah)−1Ahχ‖ ≤ C‖χ‖.

Proof.The identity

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

implies that

‖A(zαI + A)−1‖ ≤ 1 + |zα| M

|zα|≤ C .

Similarly,

(zαI + Ah)−1Ah = I − zα(zαI + Ah)−1.

Estimate for G 1h (z)

Lemma‖G 1

h (z)v‖ ≤ Ch2‖v‖ for z ∈ Σπ−ϕ.

Proof.Recall that PhA = AhRh, so

G 1h (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 1h (z)v‖ ≤ C‖(Rh − Ph)(zαI + A)−1v‖ ≤ Ch2‖(zαI + A)−1v‖2

≤ Ch2‖A(zαI + A)−1v‖ ≤ Ch2‖v‖.

Estimate for G 2h (z)

Lemma‖G 2

h (z)v‖ ≤ Ch2‖v‖ for z ∈ Σπ−ϕ.

Proof.As above,

‖G 2h (z)v‖ = ‖(Ph − I )(zαI + A)−1v‖

≤ Ch2‖(zαI + A)−1v‖2

≤ Ch2‖A(zαI + A)−1v‖≤ Ch2‖v‖.

Error estimate

Thus, we have shown that

uh(t)− u(t) =1

eztzα−1Gh(z)g(z) dz

with ‖Gh(z)v‖ ≤ Ch2‖v‖ for z ∈ Σπ−ϕ.

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

Γ± = se±i(π−ϕ) : 0 < s <∞,

so that uh(t)− u(t) = I+ − I− where

I± =1

∫Γ±

eztzα−1Gh(z)g(z) dz .

The substitution z = se±i(π−ϕ) = s(− cosϕ± i sinϕ) gives

‖I±‖ ≤ Ch2(‖u0‖+ sup

z∈Γ±

‖f (z)‖)∫ ∞

0e−st cosϕsα

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

0e−w

t cosϕ

)α dw

w= Cα,ϕt−α.

Part VII

Methods based on the Laplace transform

Introduction

The Laplace inversion formula yields a contour integralrepresentation of the finite element solution uh(t) to thetime-fractional diffusion equation. Applying a quadratureapproximation [Lopez-Fernandez+Palencia-2004] leads to a fullydiscrete solution UN,h(t). The main cost of the method is thecomputation of uh(zj) at the quadrature point zj for |j | ≤ N,which requires the solution of a system of finite element equationsinvolving the complex parameter zj .

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

Outline

Contour integral and quadrature

Parallel-in-time algorithm

A more flexible approach

Contour integral and quadratureOnce again consider

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

and recall that

u(t) =1

eztzα−1(zαI + A)−1g(z) dz ,

whereg(z) = u0 + f (z).

We now choose for Γ a contour of the form

z = z(ξ) = µ(1− sin(δ − iξ)

)for −∞ < ξ <∞,

where the parameters µ and δ satisfy

µ > 0 and 0 < δ <π

Hyperbola

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

we have

x(ξ) = <z(ξ) = µ(1− sin δ cosh ξ

y(ξ) = =z(ξ) = µ cos δ sinh ξ,

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

µ cos δ

Notice that the asymptotes are

y = ±(x − µ) cot δ.

Parameterised integral

We have

u(t) =1

eztzα−1(zαI + A)−1g(z) dz

∫ ∞−∞

ez(ξ)tz(ξ)α(z(ξ)αI + A

)−1g(z(ξ)

)z ′(ξ)

z(ξ)dξ

z(ξ) = µ(1− sin(δ − iξ)

(1− sin δ cosh ξ + i cos δ sinh ξ

z ′(ξ) = iµ cos(δ − iξ) = µ(− sin δ sinh ξ + i cos δ cosh ξ

Find that∣∣∣∣z ′(ξ)

∣∣∣∣2 =cosh2 ξ − sin2 δ

(cosh ξ − sin δ)2=

cosh ξ + sin δ

cosh ξ − sin δ≤ 1 + sin δ

1− sin δ.

Double exponential decay

Key factor in the behaviour of the integrand is

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

and since cosh ξ ≥ 12 exp(|ξ|),

|ez(ξ)t | ≤ exp(µt − 1

2µt sin δ exp(|ξ|)).

Discretisation errorFor ∆ξ > 0, let

Q∞(v) =∞∑

j=−∞v(j ∆ξ) ∆ξ.

Theorem ([Trefethen+Weideman-2014])

Let r± > 0 and assume

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

2.∫ r+

−r− |v(ξ + iη)| dη → 0 as |ξ| → ∞;

3.∫∞−∞ |v(ξ ± ir±)| dξ ≤ M±.

Then ∣∣∣∣Q∞(v)−∫ ∞−∞

v(ξ) dξ

∣∣∣∣ ≤ DE+ + DE−

DE± =M±

exp(2πr±/∆ξ)− 1.

Conformal mapping

The formulaz = Φ(ζ) = µ

(1− sin(δ − iζ)

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

x − µµ sin(δ + η)

µ cos(δ + η)

Need to ensure0 < δ + η <

so that the hyperbola crosses into the left half-plane.

Truncation error

In practice, compute

QN(v) =N∑

j=−Nv(j ∆ξ) ∆ξ,

so need to estimate

TE =∑|j |>N

v(j ∆ξ) ∆ξ

then the triangle inequality gives∣∣∣∣QN(v)−∫ ∞−∞

v(ξ) dξ

∣∣∣∣ ≤ |DE+ |+ |DE− |+ |TE |.

In our application, if ζ = ξ + iη then

|v(ζ)| ≤ C |eΦ(ζ)t | ≤ C exp(µt(1− sin(δ + η) cosh ξ

Lemma ([Lopez-Fernandez+Palencia-2004])

For a > 0 and 0 < b ≤ 1,∫ ∞0

ea(1−b cosh ξ) dξ ≤ Cea(1−b)L(ab)

L(x) =

1, x ≥ 1,

x, x ≤ 1.

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

I ≡∫ ∞

0eab(1−cosh ξ) dξ =

∫ ∞0

e−aby dy√y(y + 2)

∫ ∞0

e−x dx√x(x + 2ab)

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

dx√x(x + 2ab)

∫ ∞1

e−x dx

and the substitution x = abt/e gives∫ 1

dx√x(x + 2ab)

∫ e/(ab)

dt√t(t + 2e)

≤ C + loge

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±)

|DE± | ≤M±

e2πr±/∆ξ − 1≤ M±e−2πr±/∆ξ

≤ CL(µt sin(δ ± r±)

)exp[µt(1− sin(δ ± r±)

)− 2πr±/∆ξ

At the same time,

|TE | ≤∑|j |>N

|v(j ∆ξ)|∆ξ . 2|v(N ∆ξ)|∆ξ

≤ C exp(µt(1− sin δ cosh(N ∆ξ)

)∆ξ.

Choice of parameters

To balance TE+, TE− and DE, we want

µt(1− sin(δ + r+)

)− 2πr+

∆ξ= µt

(1− sin(δ − r−)

)− 2πr−

= µt(1− sin δ cosh(N ∆ξ)

while satisfying

0 < δ − r− < δ + r+ <π

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

− 2π

2− δ)

= µt − 2πδ

∆ξ= µt

(1− sin δ cosh(N ∆ξ)

µt =π

∆ξ(4δ − π) and µt sin δ cosh(N ∆ξ) =

∆ξ.

Eliminating µt,

sin δ cosh(N ∆ξ) =2δ

4δ − π,

so we defineb(δ) = cosh−1

(4δ − π) sin δ

)and obtain

∆ξ =b(δ)

Nand µ =

π(4δ − π)

In this case,

− 2π

2− δ)

= −B(δ)N where B(δ) =π(π − 2δ)

b(δ),

and the error bound for the quadrature rule is

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

Optimal choice of δ

We find that B(δ) has a unique maximum in theinterval 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.

Given N and t, the corresponding optimal parameters are

∆ξ∗ =1.0817 9214

Nand µ∗ = 4.4920 7528

leading to an error bound with the decay factor

e−B(δ∗)N = e−2.3157N = 10.1315−N .

Scalar test problem

For λ > 0, recall that

Le−λt =

∫ ∞0

e−(z+λ)t dt = (z + λ)−1.

We consider

e−λt =1

ezt(z + λ)−1 dz

∫ ∞−∞

ez(ξ)t(z(ξ) + λ

)−1z ′(ξ) dξ

≈ QN =1

N∑j=−N

ezj t(zj + λ)−1z ′j ∆ξ,

wherezj = z(j ∆ξ) and z ′j = z ′(j ∆ξ).

Convergence behaviour

Quadrature points for different N (with t = 1)

Quadrature points for different t (with N = 15)

Fixed points with varying t

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

e−λt ≈ 1

N∑j=−N

ezj t(zj + λ)−1z ′j ∆ξ

for t near τ?

In the fractional PDE case, we can solve one set of ellipticproblems for uh(zj) at the zj optimized for t = τ and use them notjust 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 tocompute 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 theLaplace transform with a spatial discretisation by finite elementsleads to a fully discrete numerical method that involves no timestepping.

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

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

Method-of-lines solution

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

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

or equivalently,

uht + ∂1−αt Ahuh = Phf (t) for t > 0,

with uh(0) = u0h ≈ u0.

Under Laplace transformation in time, we obtain

zuh + z1−αAhuh = gh(z)

where gh(z) = u0h + Ph f (z).

Fully-discrete solution

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

UN,h(t) =1

N∑j=−N

ezj t uh(zj)z ′j ∆ξ

as an approximation to

uh(t) =1

ezt uh(z) dz =1

∫ ∞−∞

ez(ξ)t uh

(z(ξ)

)z ′(ξ) dξ.

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

Halving the computational cost

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

uh(z) = uh(z).

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

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

and therefore

−1∑j=−N

ezj t uh(zj)z ′j ∆ξ =1

N∑j=1

ezj t uh(zj)z ′j ∆ξ.

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

UN,h(t) =1

2πex0t uh(x0)y ′0 ∆ξ +

N∑j=1

<(ezj t uh(zj)z ′j/i

)∆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−1j G(zj), 0 ≤ j ≤ N,

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

Quadrature error

To estimate DE± we must bound

M± =1

∫ ∞−∞

∥∥ez(ξ+iη)t uh

(z(ξ + iη)

)z ′(ξ + iη)

∥∥ dξ

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

‖(zI + Ah)−1v‖ ≤ ‖v‖|z | sinϕ

for z ∈ Σπ−ϕ and v ∈ Vh,

and ∣∣∣∣z ′(ξ + iη)

z(ξ + iη)

∣∣∣∣ ≤√

1 + sin(δ + η)

1− sin(δ + η)=

cos(δ + η).

Since(zαI + Ah)uh(z) = zα−1gh(z),

we have

‖uh(z)‖ =∥∥zα−1(zαI + Ah)−1gh(z)

∥∥ ≤ ‖gh(z)‖|z | sinϕ

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

Therefore,

M± ≤1

2π sinϕ

∫ ∞−∞

∣∣ez(ξ±ir±)t∣∣∥∥gh

(z(ξ ± ir±)

)∥∥ |z ′(ξ ± ir±)||z(ξ ± ir±)|

π sinϕ cos(δ ± r±)

∫ ∞−∞

∣∣ez(ξ±ir±)t∣∣∥∥gh

(z(ξ ± ir±)

)∥∥ dξ,

and similarly for TE.

Conclusion: if f (z) is analytic with

‖f (z)‖ ≤ Cf ,ϕ for z ∈ Σπ−ϕ,

so that

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

then we may estimate the quadrature error as before and obtain

‖UN,h(t)− uh(t)‖ ≤ Ce−B(δ)NL(ct).

Thus, by the triangle inequality,

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

(L(ct)e−B(δ)N

)+ O(t−αh2).

Roundoff

In practice, roundoff means that we compute

U?N,h(t) =

N∑j=−N

ezj t uh(zj)z ′j (1 + εj) ∆ξ

with ‖εj‖L∞(Ω) ≤ ε, leading to an additional perturbation

‖U?N,h(t)− UN,h(t)‖ ≤ ε

N∑j=−N

|ezj t |‖uh(zj)‖|z ′j |∆ξ

≤ ε

π sinϕ cos δ

N∑j=−N

|ezj t |‖gh(zj)‖∆ξ

N∑j=−N

|ezj t |∆ξ =N∑

j=−Nexp(µt(1− sin δ cosh(j ∆ξ)

)∆ξ

≈∫ N ∆ξ

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

we conclude that

‖U?N,h(t)− UN,h(t)‖ ≤ Cεeµt(1−sin δ)L(µt sin δ) max

−N≤j≤N‖g(zj)‖.

The optimal choices

δ∗ = 1.1721 and µ∗ = 4.4921N/t

lead toeµ∗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

N ≥ log(10−6/ε)

log 1.422= 63.14.

So roundoff probably not an issue.

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

Behaviour of f (z)

Example

If f (x , t) = g(x)e−at cosωt, then

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

(z + a)2 + ω2,

which has simple poles at z = −a± iω.

Example

f (x , t) =

g(x), 1 < t < 2,

0, otherwise,

f (x , z) =

1e−ztg(x) dt = g(x)

e−z − e−2z

and ∣∣e−2z(ξ)∣∣ = eµ(sin δ cosh ξ−1).

A more flexible approach [McLean+Thomee-2010]

Now drop the assumption that f (z) is analytic and boundedfor z ∈ Σπ−ϕ. We will describe a method based on Duhamel’sformula,

u(t) = E(t)u0 +

0E(t − s)f (s) ds, t > 0.

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

E(t)v =∞∑

Eα(−λmtα)〈v , φm〉φm, Aφm = λmφm,

and the integral representation

E(t)v =1

ezt E(z)v dz , E(z) = zα−1(zαI + A)−1.

Noting that∫ t

0E(t − s)f (s) ds =

ez(t−s)E(z)f (s) dz ds

∫ΓE(z)

0ez(t−s)f (s) ds dz ,

we defineg(z , t) = eztu0 +

0ez(t−s)f (s) ds

and deduceu(t) =

∫ΓE(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−ϕ, we expect

E0(z) = E(z)− z−1I

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

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

‖E0(z)v‖ ≤ Cϕ,σ‖Aσv‖|z |1+ασ

for z ∈ Σπ−ϕ,

Cϕ,σ =

)1−σ( 1

We have

E0(z) = E(z)− zI

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

=[zα−1I − z−1(zαI + A)

](zαI + A)−1

= −z−1A(zαI + A)−1

= −z−1[A(zαI + A)−1

]1−σ[(zαI + A)−1

]σAσ

A(zαI + A)−1 =[(zαI + A)− zα

](zαI + A)−1

= I − zα(zαI + A)−1.

For z ∈ Σπ−ϕ, our resolvent estimate gives

‖(zαI + A)−1‖ ≤ 1

|z |α sinϕ,

so ∥∥[(zαI + A)−1]σ∥∥ ≤ ∥∥(zαI + A)−1

∥∥σ ≤ ( 1

|z |α sinϕ

)σand ∥∥[A(zαI + A)−1

]1−σ∥∥ ≤ (1 +1

)1−σ.

‖E0(z)v‖ ≤ 1

)1−σ( 1

|z |α sinϕ

)σ‖Aσv‖.

Modified representation

Since E(z) = z−1I + E0(z) and

z−1g(z , t) dt = resz=0

g(z , t)

z= g(0, t) = u0 +

0f (s) ds.

we have

u(t) = g(0, t) +1

w0(z , t) dz

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

gh(z , t) = eztu0h +

0ez(t−s)Phf (s) ds,

uh(t) = gh(0, t) +1

∫ΓE0h(z)gh(z , t) dz

= u0h +

0Phf (s) ds +

w0h(z , t) dz

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

(zαI + Ah)wh(z , t) = zα−1gh(z , t).

Fully-discrete schemeThe equal-weight quadrature approximation∫

Γw0h(z , t) dz ≈

N∑j=−N

w0h(zj , t)z ′j ∆ξ

leads to

UN,h(t) = gh(0, t) +1

N∑j=−N

w0h(zj , t)z ′j ∆ξ.

LemmaIf z = z(ξ + iη) and z ′ = z ′(ξ + iη), then for0 < σ1 + α−1 = σ ≤ 1,

‖wh(z , t)z ′‖ ≤ 2Cϕ,σ,σ1

cos(δ + η)

eµt(1−sin δ)

|z |ασ

(‖Aσhu0h‖

+ ‖Aσ1h Phf (0)‖+

0‖Aσ1

h Phf ′(s)‖ ds

Outline of Proof

Integrating by parts, we find that

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

)+ E0h(z)

(ezt − 1

zPhf (0) +

ez(t−s) − 1

zPhf ′(s) ds

‖E0h(z)u0hz ′‖ ≤ Cϕ,σ|z ′|‖Aσhu0h‖|z |1+ασ

= Cϕ,σ|z ′||z |‖Aσhu0h‖|z |ασ

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

‖z−1E0h(z)Phf (0)z ′‖ ≤ Cϕ,σ1 |z ′|‖Aσ1

h u0h‖|z |2+ασ1

= Cϕ,σ1

|z ′||z |‖Aσ1

h u0h‖|z |ασ

Now recall that |z ′/z | ≤ 2/ cos(δ + η).

Quadrature error

We saw previously that

|z(ξ + iη)| = µ(cosh ξ − sin(δ + η)

and since

cosh ξ − sin δ12 e |ξ|

= 1 + e−2|ξ| − 2e−|ξ| sin δ

= (1− e−|ξ| sin δ)2 + e−2|ξ| cos2 δ

≥ (1− sin δ)2,

it follows that

|z(ξ + iη)|ασ≤(

1− sin(δ + η)

)ασ e−ασ|ξ|

µασ.

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

∥∥w(z(ξ ± ir)

)z ′(ξ ± ir)

∥∥ dξ ≤ Ceµt(1−sin δ)

µασ

∫ ∞0

e−ασξ dξ

|DE± | ≤ Ceµt(1−sin δ)

µασe−2πr/∆ξ.

At the same time,

∑|j |>N

∥∥wh(zj , t)z ′j∥∥∆ξ ≤ C

eµt(1−sin δ)

µασ

∞∑j=N+1

e−ασj ∆ξ ∆ξ

|TE | ≤ Ceµt(1−sin δ)

µασe−ασN∆ξ.

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

TheoremFor the flexible scheme described above, if

∆ξ =

√2πr

ασNand µ =

t(1− sin δ),

then‖UN,h(t)− uh(t)‖ ≤ Ctασ exp

(−√

2πrασN).

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

Example

Taking

4, α =

2, σ = 1

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

e−12π√N = e−1.5708

√N ,

compared to

e−B(δ∗)N = e−2.3157N = 10.1315−N

for our earlier method.

Part VIII

Convolution Quadrature

Introduction

Convolution quadrature [Lubich-1988, Lubich-2004] refers to anapproximation of the form∫ tn

0K (tn − s)f (s) ds ≈

n∑j=0

wn−j f (tj), tj = j ∆t,

where the convolution weights wn = wn(∆t) are computed directlyfrom the Laplace transform K (z) rather than the kernel K (t). Thisapproach can be advantageous if K (z) is simpler than K (t).

Convolution quadrature can be used to approximate any functionof the form K ∗ f , and in particular the fractionalintegral Iαf = Υα ∗ f .

Outline

Contour integrals again

Operational calculus

Application to fractional diffusion

Contour integrals again

Assume that K (z) is analytic and satisfies

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

Then, as before,

K (t) =1

eztK (z) dz , t > 0.

Since∫ t

0K (t − s)f (s) ds =

ez(t−s)K (z) dz

)f (s) ds,

by reversing the order of integration we obtain

K ∗ f (t) =1

0ez(t−s)f (s) ds dz

The associated ODE

y(t; z) =

0ez(t−s)f (s) ds

so that

K ∗ f (t) =1

K (z)y(t; z) dz .

Notice thatdy

dt= f (t) + z

0ez(t−s)f (s) ds

so y is the solution of the initial-value problem

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

Compute

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

K (z)y(tn; z) dz

K (z)Y n(z) dz .

Generating functionLet

Y (ζ) = Y (ζ; z) =∞∑n=0

Y n(z)ζn.

Putting Y−1 = 0, we have

∞∑n=0

(Y n − Y n−1)ζn =∞∑n=0

Y nζn −∞∑

n=−1

Y nζn+1 = (1− ζ)Y (ζ),

so the finite difference equation implies that

1− ζ∆t

Y (ζ)− zY (ζ) = F (ζ).

Y (ζ; z) =F (ζ)

δ(ζ)∆t−1 − zwhere δ(ζ) = 1− ζ.

Recall

(K ∗ f )(tn) ≈ 1

K (z)Y n(z) dz .

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

∞∑n=0

(K ∗ f )(tn)ζn ≈ 1

K (z)Y (ζ; z) dz

= − 1

K (z) dz

z − δ(ζ)∆t−1F (ζ).

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

K (z) dz

z − δ(ζ)∆t−1= K

(δ(ζ)∆t−1

Thus,K ∗ f (ζ) ≈ K

(δ(ζ)∆t−1

)F (ζ).

WeightsDefine wn = wn(∆t) by

K(δ(ζ)∆t−1

)= w(ζ) =

∞∑n=0

wnζn,

then because

K(δ(ζ)∆t−1

)F (ζ) =

( ∞∑n=0

)( ∞∑m=0

Fmζm)

=∞∑n=0

( n∑j=0

wn−jFj

we conclude that

(K ∗ f )(tn) ≈n∑

wn−jFj =

n∑j=0

wjFn−j .

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

= ∆tα∞∑n=0

(−αn

)(−1)nζn,

showing that

wn(∆t) = ∆tα(−αn

)(−1)n.

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

)(−1)n =

−α1× −α− 1

2× · · · × −α− n + 1

n(−1)n

1× α + 1

2× · · · × α + n − 1

(α + n − 1

Weights when α = 1/2 and ∆t = 1

Higher-order methods

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

Y n +∆Y n

∆t(t − tn) +

∆2Y n

∆t2(t − tn)(t − tn−1) + · · ·

∆pY n

∆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 thebackward differentiation formula (BDF)

y ′(tn) ≈ ∆Y n

∆2Y n

∆t2∆t + · · ·

∆pY n

∆tp(∆t)(2∆t) . . .

((p − 1)∆t

Simplifying:

y ′(tn) ≈ ∆Y n

∆2Y n

∆t+ · · ·+ 1

∆pY n

p∑`=1

∆`Y n

Compute Y n(z) ≈ y(tn; z) by solving

p∑`=1

∆`Y n

`− zY n = F n for n ≥ 1,

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

δ(ζ)

∆tY (ζ)− zY (ζ) = F (ζ) but now δ(ζ) =

p∑`=1

(1− ζ)`

A(α)-Stability

The BDF of order p has the property that

| arg δ(ζ)| ≤ π − α for |ζ| < 1,

for the following values of α.

Operational calculus

Notation:

K (∂)f (t) = K ∗ f (t) =

0K (t − s)f (s) ds for t > 0.

In this way, sinceLK ∗ f = K (z)f (z),

we have

K (∂)f (t) =1

eztK (z)f (z) dz .

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

∂−1f (t) =

0f (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) =

0ea(t−s)f (s) ds

withL(∂ − a)−1f = (z − a)−1f (z).

Theorem

K (∂) =1

K (z)(∂ − z)−1 dz .

Proof.

K (∂)f (t) = (K ∗ f )(t) =

0K (t − s)f (s) ds

ez(t−s)K (z) dz f (s) ds

0ez(t−s)f (s) ds dz

K (z)(∂ − z)−1f (t) dz .

Discrete operational calculus

Notation:

K (∂∆t)f (t) =∑

0≤tj≤twj(∆t)f (t − tj) for t > 0.

In particular, at t = tn,

K (∂∆t)f (tn) =n∑

wj f (tn − tj) =n∑

wj f (tn−j),

or equivalently,

K (∂∆t)F n =n∑

wjFn−j =

n∑j=0

wn−jFj .

Example

If K (t) = 1 then K (z) = z−1 so

K(δ(ζ)∆t−1

(1− ζ

1− ζ= ∆t

∞∑n=0

for |ζ| < 1, showing that wj = ∆t and hence

∂−1∆t f (t) =

∑0≤tj≤t

f (t − tj) ∆t

≈∫ t

0f (t − s) ds =

0f (s) ds = ∂−1f (t).

Example

The function

y(t) = (∂ − a)−1f (t) =

0ea(t−s)f (s) ds

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

wnζn,

thenY n =

n∑j=0

wn−jFj = (∂∆t − a)−1f (tn).

Example

For the BDF of order p = 2,

y ′(tn) ≈ ∆Y n

∆2Y n

andδ(ζ) = (1− ζ) + 1

2 (1− ζ)2 = 32 (1− ζ)(1− 1

Recalling that

(1− ζ)−α =∞∑n=0

(α + n − 1

we see that the weights for ∂−α∆t are

)α n∑j=0

(α + n − j − 1

n − j

)(α + j − 1

)3−j .

Integral representation of the weights

K(δ(ζ)∆t−1

)=∞∑n=0

wnζn,

for any sufficiently small ε > 0 we have

∮|ζ|=ε

K(δ(ζ)∆t−1

)ζn+1

We can use this representation to show the following result.

Theorem

K (∂∆t) =1

K (z)(∂∆t − z)−1 dz .

We have

∫|ζ|=ε

K(δ(ζ)∆t−1

)ζn+1

K(δ(ζ)∆t−1

K (z) dz

δ(ζ)∆t−1 − zdζ

K (z)1

∫|ζ|=ε

(δ(ζ)∆t−1 − z

ζn+1dζ︸︷︷︸

w?n (∆t; z)

Since (δ(ζ)∆t−1 − z

)−1=∞∑n=0

w?n (∆t; z)ζn,

it follows that

K (∂∆t)f (t) =∑

0≤tj≤twj f (t − tj)

0≤tj≤t

K (z)w?j (∆t; z) dzf (t − tj)

K (z)∑

0≤tj≤tw?j (∆t; z)f (t − tj) dz

K (z)(∂∆t − z)−1f (t) dz .

Summary

K (∂)f (t) =

0K (t − s)f (s) ds =

eztK (z)f (z) dz

K (z)y(t; z) dz =1

K (z)(∂ − z)−1f (t) dz .

w(ζ) =∞∑n=0

wnζn = K

(δ(ζ)∆t−1

K (∂∆t)F n =n∑

wn−jFj =

∮|ζ|=ε

w(ζ)F (ζ)

ζn+1dζ

K (z)Y n(z) dz =1

K (z)(∂∆t − z)−1f (t) dz .

Accuracy of convolution quadrature

K (∂∆t)f (t) =∞∑j=0

wj f (t − tj)χ(t − tj), χ(t) =

1, t > 0,

0, t < 0,

we have, with ζ = e−z ∆t ,

LK (∂∆t)f =

∫ ∞0

e−zt∞∑j=0

wj f (t − tj)χ(t − tj) dt

=∞∑j=0

∫ ∞tj

e−zt f (t − tj) dt

=∞∑j=0

∫ ∞0

e−z(t+tj )f (t) dt =∞∑j=0

∫ ∞0

e−zt f (t) dt

= K(δ(e−z ∆t)∆t−1

)f (z).

The BDF of order p satisfies

δ(e−h)h−1 = 1 + O(hp) as h→ 0,

δ(e−z∆t)∆t−1 = zδ(e−h)h−1, h = z∆t,

= z + O(zp+1∆tp)

and thus

LK (∂∆t)f = K(z + O(zp+1∆tp)

)f (z).

Notice also

K (∂∆t)f (t)− K (∂)f (t)

K (z)[(∂∆t − z)−1 − (∂ − z)−1

]f (t) dz .

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

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

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

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

I there exists ϕ1 > ϕ such that | arg δ(ζ)| ≤ π − ϕ1 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−p∆tp, p ≤ β,

Ctµ−1∆tβ, 0 < β ≤ p.

Multiplication property

Notice that since

K1 ∗ (K2 ∗ f ) = (K1 ∗ K2) ∗ f and LK1 ∗ K2 = K1(z)K2(z),

we haveK1(∂)K2(∂) = (K1K2)(∂),

so in particular K1(∂) commutes with K2(∂).

The analogous identity holds in the discrete case.

Theorem

K1(∂∆t)K2(∂∆t) = (K1K2)(∂∆t).

On the one hand,

K1(∂∆t)K2(∂∆t)f (t) =∑

0≤tj≤tw 1j K2(∂∆t)f (t − tj)

0≤tj≤tw 1j

∑0≤tk≤t−tj

f (t − tj − tk)

0≤tn≤t

∑j+k=n

w 1j w 2

k f (t − tj+k)

0≤tn≤t

( n∑j=0

w 1j w 2

n−j︸︷︷︸wn

)f (t − tn).

On the other hand,

(δ(ζ)∆t−1

)=∞∑j=0

w 1j ζ

j∞∑k=0

w 2k ζ

=∞∑n=0

∑j+k=n

w 1j w 2

k ζj+k

=∞∑n=0

( n∑j=0

w 1j w 2

=∞∑n=0

wnζn.

Example

∂−α∆t ∂−β∆t = ∂

−(α+β)∆t .

Associativity

Associativity of the Laplace convolution means that

K ∗ (f ∗ g) = (K ∗ f ) ∗ g ,

or equivalently,

K (∂)(f ∗ g) =(K (∂)f

)∗ g .

The analogous identity holds in the discrete case,

K (∂∆t)(f ∗ g) =(K (∂∆t)f

)∗ g ,

because

K (∂∆t)(f ∗ g)

= K(δ(e−z∆t)∆t−1

)f (z)g(z).

RemarkTaylor expansion gives

f (t) =

p−1∑k=0

f (k)(0)

k!tk +

(p − 1)!

0(t − s)p−1f (p)(s) ds

p−1∑k=0

f (k)(0)Υk+1(t) + (Υp ∗ f (p))(t),

G (∂)f (t) =

p−1∑k=0

f (k)(0)G (∂)tk +(G (∂)Υp

)∗ f (p)(t),

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

∣∣[G (∂∆t)− G (∂)]f (t)

∣∣ ≤ Ctµ−1p−1∑k=0

|f (k)(0)|∆tk+1

+ C ∆tp∫ t

0(t − s)µ−1|f (p)(s)| ds.

Correction terms

If p ≥ 2, put

G (∂∆t)∼f (tn) = G (∂∆t)f (tn) +

p−2∑j=0

w∼nj f (tj)

and choose the extra weights w∼nj so that the modified quadraturerule 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 badlyconditioned.

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,

or equivalently,

(I + ∂−αA)u = (I + ∂−αA)u0 − ∂−αAu0.

Thus,u = u0 − (I + ∂−αA)−1∂−αAu0,

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

U = u0 − (I + ∂−α∆t A)−1∂−αAu0.

U = u0 + W where (I + ∂−α∆t A)W = −∂−αAu0,

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

wnζn.

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

W nh +

n∑j=0

wn−jAhW jh = −Υ1+α(tn)Ahu0h for n ≥ 1.

Error bound for nonsmooth initial dataSince

u − u0 = −(I + ∂−αA)∂−αAu0,

U − u0 = −(I + ∂−α∆t A)∂−αAu0,

the error from the time discretization is

U − u =[G (∂)− G (∂∆t)

]∂−αu0,

whereG (z) = (I + z−αA)−1A.

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

For t > 0,

‖U(t)− u(t)‖ ≤

Ct−1∆t‖u0‖, p = 1,

Ct−1−α∆t1+α‖u0‖, p ≥ 2.

G (z) = (I + z−αA)−1A = zα(zαI + A)−1A

= zα(zαI + A)−1[(zαI + A)− zαI

]= zα

[I − zα(zαI + A)−1

]and ‖(zαI + A)−1‖ ≤ C |z |−α, we have

‖G (z)‖ ≤ C |z |α for z ∈ Σπ−ϕ.

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

‖U(t)− u(t)‖ =∥∥[G (∂)tβ − G (∂∆t)tβ

∥∥/Γ(1 + α)

≤ C‖u0‖ ×

t−p∆tp, p ≤ 1 + α,

t−α−1∆t1+α, p ≥ 1 + α.

Error bound for smooth initial dataRecall that

u(t) = E(t)u0 = u0 −tα

Γ(1 + α)Au0 + · · · as t → 0,

and observe that

(I + ∂−αA)[u(t)− u0 + Υ1+α(t)Au0

]= u0 − (I + ∂−αA)u0 + (I + ∂−αA)Υ1+α(t)Au0

= Υ1+2α(t)A2u0,

u(t) = u0 −Υ1+α(t)Au0 + (I + ∂−αA)−1Υ1+2α(t)A2u0.

We therefore consider

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

The error is

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

]Υ1+2α(t)A2u0

=[G (∂∆t)− G (∂)

]Υ1+2α(t)Au0,

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

‖U(t)− u(t)‖ ≤ C‖Au0‖ ×

tα−p∆tp, p ≤ 1 + 2α,

t−1−α∆t1+2α, p ≥ 1 + 2α.

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

‖U(t)− u(t)‖ ≤ Ctα−2∆t2‖Au0‖.

Part IX

Discontinuous Galerkin methods for

time stepping

Introduction

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

Once again, we largely ignore the spatial discretization.

Outline

Discontinuous piecewise-polynomial approximation in time

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 ⊆ H10 (Ω) 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 thosefunctions U : (0,T )→ H1

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

Example

Choose Sn = H10 (Ω) and pn = p independent of n.

Example

Choose Sn = Vh to be the usual continuous piecewise-linear finiteelement space with respect to a triangulation Th of Ω andenforcing a homogeneous Dirichlet boundary condition.

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

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

Un± = U(t±n ) = lim

t→t±n

U(t) and [U]n = Un+ − Un

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 valueproblem for the fractional diffusion equation satisfies

〈u(t), v〉+ a(∂1−αt u(t), v

)= 〈f (t), v〉 for all v ∈ H1

0 (Ω),

where u = ut = ∂u/∂t. Hence, for suitable v : In → H10 (Ω),∫

[〈u(t), v(t)〉+ a

(∂1−αt u(t), v(t)

)]dt =

〈f (t), v(t)〉 dt.

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

〈Un−1+ ,X n−1

+ 〉+

[〈U(t),X (t)〉+ a

(∂1−αt U(t),X (t)

= 〈Un−1− ,X n−1

+ 〉+

〈f (t),X (t)〉 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 piecewiseconstant 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,

with U = 0 on In, so

〈Un − Un−1, χ〉+

a(∂1−αt U(t), χ

〈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 ψn0 , ψn

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

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

U(x , t) =

pn∑r=0

Mn∑l=1

Unrl ψ

nr (t)χl(x) for x ∈ Ω and t ∈ In.

Starting from the known (approximate) initial data

U0−(x) = U0(x) =

M0∑l=1

U0l χ

0l (x) for x ∈ Ω,

we compute the unknown Unrl by solving the

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

Stability

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

TheoremIf U0

− = U0 ∈ H and f ∈ L2

((0,T );H

)then there exists a unique

DG solution U ∈ W, and for 1 ≤ n ≤ N,

‖Un−‖2 +

n−1∑j=1

‖[U]j‖2 +

∫ tn

0a(∂1−α

t U,U) dt

≤ ‖U0‖2 +

∫ tn

0〈A−1f , I1−αf 〉 dt.

Global bilinear form

By rewriting the nth DG equation as

〈[U]n−1,X n−1+ 〉+

[〈U,X 〉+ a

(∂1−αt U,X

)]dt =

〈f ,X 〉 dt,

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

GN(U,X ) = 〈U0,X 0+〉+

∫ tN

0〈f (t),X (t)〉 dt for all X ∈ W,

where the bilinear form GN is defined by

GN(U,X ) = 〈U0−,X

0+〉+

N∑n=1

〈[U]n−1,X n−1+ 〉

[〈U(t),X (t)〉+ a

(∂1−αt U(t),X (t)

〈[U]n−1,Un−1+ 〉+

〈U,U〉 dt =1

(‖[U]n−1‖2 +‖Un

−‖2−‖Un−1− ‖2

Proof.Since 〈U,U〉 = (d/dt) 1

2‖U‖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

= ‖[U]n−1‖2 + ‖Un−1+ ‖2 − ‖Un−1

− ‖2 + ‖Un−‖2 − ‖Un−1

+ ‖2

= ‖[U]n−1‖2 + ‖Un−‖2 − ‖Un−1

− ‖2.

GN(U,U) = 12‖U

0+‖2 + 1

2‖UN−‖2 + 1

N−1∑n=1

‖[U]n‖2

∫ tN

0a(∂1−α

t U,U) dt.

Proof.

〈U0−,U

0+〉+

N∑n=1

(〈[U]n−1,Un−1

+ 〉+

〈U,U〉 dt

= 〈U0−,U

0+〉+ 1

2 [U]0 − 12‖U

0−‖2︸︷︷︸

12‖U

0+‖2

+ 12‖U

N−‖2 + 1

N−1∑n=1

‖[U]n‖2.

To prove the next lemma, recall the identities∫ ∞0

(∂βt u)v dt =

∫ ∞−∞

(iy)β u(iy)v(iy) dy

and, when u is real-valued,∫ ∞0

(∂βt u)u dt =cos 1

∫ ∞0

yβ|u(iy)|2 dy ≥ 0.

LemmaFor 0 < β < 1, and real-valued u and v,∣∣∣∣∫ ∞

0(∂βt u)v dt

∣∣∣∣≤ 1

cos 12πβ

(∫ ∞0

(∂βt u)u dt

)1/2(∫ ∞0

(∂βt v)v dt

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

0(∂βt u)v dt

∣∣∣∣ ≤ 1

∫ ∞−∞|y |β|u(iy)||v(iy)| dy

∫ ∞0

(yβ/2|u(iy)|)(yβ/2|v(iy)|) dy

(∫ ∞0

yβ|u(iy)|2 dy

)1/2(∫ ∞0

yβ|v(iy)|2 dy

cos 12πβ

(∫ ∞0

(∂βt u)u dt

)1/2(∫ ∞0

(∂βt v)v dt

LemmaFor 0 < β < 1,

∣∣∣∣∫ T

0〈∂βt u, v〉 dt

∣∣∣∣ ≤ ∫ T

0〈∂βt u, u〉 dt +

cos2 12πβ

0〈∂βt v , v〉 dt.

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

0(∂βt um)vm dt = 2

∫ ∞0

(∂βt um)vm dt

cos 12πβ

0(∂βt um)um dt +

0(∂βt vm)vm dt

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

2πβ.

Proof of the stability theorem

The DG solution U satisfies

GN(U,X ) = 〈U0,X 0+〉+

∫ tN

0〈f (t),X (t)〉 dt for all X ∈ W,

and by choosing X = U the second lemma gives

12‖U

0+‖2 + 1

2‖UN−‖2 + 1

N−1∑n=1

‖[U]n‖2

∫ tN

0a(∂1−α

t U,U) dt = 〈U0,U0+〉+

∫ tN

0〈f (t),U(t)〉 dt.

Now use 〈U0,U0+〉 ≤ 1

2‖U0‖2 + 1

2‖U0+‖2 and cancel the

term 12‖U

0+‖2.

‖UN−‖2 +

N−1∑n=1

‖[U]n‖2

∫ tN

0a(∂1−α

t U,U) dt = ‖U0‖2 + 2

∫ tN

0〈f (t),U(t)〉 dt.

Write 〈f (t),U(t)〉 = 〈∂1−αt g(t), v(t)〉 where

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

so that

∫ tN

0〈f (t),U(t)〉 dt ≤

∫ tN

0a(∂1−α

t U,U) dt

cos2 12πβ

∫ tN

0〈A−1f (t), I1−αf (t)〉 dt.

Piecewise-constant case

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

‖Un∗− ‖2 ≤ ‖Un∗

− ‖2 +n∗−1∑n=1

‖[U]n‖2

∫ tn∗

0a(∂1−α

t U,U) dt = ‖U0‖2 + 2

∫ tn∗

0〈f (t),U(t)〉 dt

and U0 = U0−, we have

‖Un∗− ‖‖UN

−‖ ≤ ‖Un∗− ‖2 ≤

(‖U0‖+

∫ tn∗

0‖f (t)‖ dt

)‖Un∗− ‖,

‖Un−‖ ≤ ‖U0‖+ 2

∫ tn

0‖f (t)‖ dt, 0 ≤ n ≤ N.

Piecewise-linear case

If pn = 1 for all n, then

‖U‖In ≡ supt∈In‖U(t)‖ = max

‖Un−1

+ , ‖Un−‖.

For the first subinterval,

‖U‖2I1 ≤ ‖U

0+‖2 + ‖U1

−‖2 ≤ 2〈U0,U0+〉+ 2

∫ t1

0〈f (t),U(t)〉 dt

‖U‖I1 ≤ 2‖U0‖+ 2

∫ t1

0‖f (t)‖ dt.

For n ≥ 2, since Un−1+ = Un−1

− + [U]n−1,

‖Un−1+ ‖2 ≤ 2‖Un−1

− ‖2 + 2‖[U]n−1‖2

≤ (2 + 2)

(2〈U0,U0

+〉+ 2

∫ tn

0〈f (t),U(t)〉 dt

and by choosing n∗ such that ‖U‖In∗ = max1≤n≤N ‖U‖In we see

‖U‖In ≤ 8

(‖U0‖+

∫ tn

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‖+

0‖f (s)‖ ds, 0 ≤ t ≤ T .

Convergence

For simplicity, assume now that Sn = H10 (Ω) 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 theconditions

(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

If pn = 0 then (Πu)(t) = u(tn) for t ∈ In.

Example

If pn = 1 then

(Πu)(t) = u(tn) +u(tn)− avgIn(u)

∆tn/2(t − tn) for t ∈ In,

where avgIn(u) = ∆t−1n

u dt. Can show that

(Πu)(t)− u(t) =

∫ tn

tu′(s) ds − 2

tn − t

∆t2n

(s − tn−1)u′(s) ds

∫ tn

t(t − s)u′′(s) ds +

tn − t

∆t2n

(s − tn−1)2u′′(s) ds,

for t ∈ In, and hence

‖u − Πu‖In ≤ 2

‖u′(t)‖ dt ≤ 2∆tn‖u′‖In ,

‖u − Πu‖In ≤ 3∆tn

‖u′′(t)‖ dt ≤ 3∆t2n‖u′′‖In .

In the general case we have the following estimate.

Theorem ([Schoetzau+Schwab-2000])

For 0 ≤ q ≤ pn,∫In

‖(u − Πu)′(t)‖2 dt ≤ Cε(pn, q)

(∆tn

)2q ∫In

‖u(q+1)(t)‖2 dt,

ε(p, q) =(p − q)!

(p + q)!.

Notice that

ε(p, p) =1

(2p)!.

Recall

GN(U,X ) = 〈U0+,X

0+〉+

N−1∑n=1

〈[U]n,X n+〉

[〈U(t),X (t)〉+ a

(∂1−αt U(t),X (t)

Integration by parts yields a dual representation

GN(U,X ) = 〈UN− ,X

N− 〉 −

N−1∑n=1

〈Un−, [X ]n〉

[−〈U(t), X (t)〉+ a

(∂1−αt U(t),X (t)

For all X ∈ W(t,p,S), the DG solution satisfies

GN(U,X ) = 〈U0,X 0+〉+

0〈f (t),X (t)〉 dt,

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

GN(u,X ) = 〈u0,X0+〉+

0〈f (t),X (t)〉 dt.

Furthermore, the construction of Π ensures

%n− = 0 and

〈%, X 〉 dt = 0 for 1 ≤ n ≤ N,

GN(%,X ) =

∫ tN

0a(∂1−α

t %,X ) dt.

Thus,GN(U − u,X ) = 〈U0 − u0,X

0+〉,

and since U − u = ϑ+ % we have

GN(ϑ,X ) = GN(U − u − %,X ) = GN(U − u,X )− GN(%,X ).

Therefore,

GN(ϑ,X ) = 〈U0 − u0,X0+〉+

∫ tN

0〈−A∂1−α

t %,X 〉 dt

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

t A%. By applying thestability result to ϑ it is possible to prove the following errorestimate.

h-Version accuracy

Theorem ([Mustapha-2015])

Let γ ≥ 1 and σ > 0. If

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

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

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

minγ(σ + 1

2α−12 ), 2 + 1

2α−12, p = 1,

minγ(σ + 12α−

12 ), p + 1 + 1

2α−12 −

12 , p ≥ 2.

Piecewise-constants on a uniform mesh

Theorem ([McLean+Mustapha-2015])

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

1. For 0 < α < 1, if u0 ∈ H then

‖Un − u(tn)‖ ≤ Ct−1n ∆t‖u0‖.

2. for 0 < α ≤ 1/2, if A2u0 ∈ H then

‖Un − u(tn)‖ ≤ Ct2α−1n ∆t‖A2u0‖.

3. for 1/2 ≤ α < 1, if A1/αu0 ∈ H then

‖Un − u(tn)‖ ≤ C ∆t‖A1/αu0‖.

Corollary

‖Un − u(tn)‖ ≤ Ctrα−1n ∆t‖Aru0‖ for 0 ≤ r ≤ min2, 1/α.

Part X

Numerical Solution of Fractional PDEs

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