A Fractional Diffusion-Telegraph Equation and itsStochastic Solution
.:Federico Polito:.
Department of MathematicsUniversity of Torino
Joint work with .:Mirko D’Ovidio:. (University of Rome)
Outline
1 Prologue
2 Operator
3 Time Change
4 Levy Processes
Part Zero: Prologue
Anomalous Diffusion
Consider the Cauchy problemdα
dtα u(x , t) = λ2 d2
dx2u(x , t), t > 0, x ∈ R,
u(x , 0) = δ(x), 0 < α ≤ 2,ddt u(x , t)
∣∣t=0
= 0, 1 < α ≤ 2,
where dα/dtα represents the so-called Caputo fractionalderivative
dα
dtαu(t) =
1
Γ(m − s)
∫ t
0
dm
dsm u(s)
(t − s)1+α−mds, m − 1 < α < m,
dm
dtmu(t), α = m.
the solution can be written as
u(x , t) =1
2λtα/2Mα
(− |x |λtα/2
), 0 < α ≤ 2,
where the Mainardi–Wright function (M-function)Mα(z) = W−α/2,1−α/2(z) is a specific case of the Wright function
Wa,b(z) =∞∑k=0
zk
k! Γ(ak + b).
The solution reduces to the Gaussian function for α = 1 and to theclassical d’Alambert’s solution to the wave equation for α→ 2.
Mainardi–Wright Function
Part One: Operator
Prabhakar Operator
Prabhakar, TR (1971). A singular integral equation with ageneralized Mittag Leffler function in the kernel. YokohamaMathematical Journal 19, 7–15.
Generalized fractional integral:(Eξα,η,ζ;0+f (·)
)(t) =
∫ t
0(t − y)η−1E ξα,η [ζ(t − y)α] f (y) dy
Mittag–Leffler function:
E ξα,η(x) =∞∑r=0
x r
r !Γ(αr + η)
Γ(r + ξ)
Γ(ξ), α, η, ξ ∈ C, <(α) > 0
Note that E 11,1 = exp(x).
Inverse Operator
Kilbas AA, Saigo, M & Saxena RK (2004). GeneralizedMittag–Leffler function and generalized fractional calculusoperators. Integral Transforms and Special Functions 15(1),31–49.
Generalized fractional derivative:(Dξα,η,ζ;0+f (·)
)(t) =
([Eξα,η,ζ;0+
]−1f (·))
(t)
=dη+θ
dtη+θ
∫ t
0(t − y)θ−1E−ξα,θ [ζ(t − y)α] f (y) dy
where dβ
dtβis a Riemann–Liouville fractional derivative.
∴ D is a Riemann–Liouville-like fractional derivative operator.
Considering that
Im−(η+θ)0+ f (x) =
(E0α,m−(η+θ),ζ,0+f
)(x)
we can write(Dξα,η,ζ,0+f
)(x) =
dη+θ
dxη+θ
(E−ξα,θ,ζ,0+f
)(x)
=dm
dxmIm−(η+θ)
(E−ξα,θ,ζ,0+f
)(x), m = dη + θe
=dm
dxm
(E−ξα,m−η,ζ,0+f
)(x),
where we used the fact that
Eγρ,µ,ω,0+Eσρ,ν,ω,0+f (x) = Eγ+σρ,µ+ν,ω,0+f (x).
Caputo-like Fractional Derivative
Define(Dξα,η,ζ;0+f (·)
)(t) =
(Dξα,η,ζ;0+f (·)
)(t)− f (0+) t−ηE−ξα,1−η(ζtα)
such that ∫ ∞0
e−st(Dξα,η,ζ;0+f (·)
)(t) dt
= sη(1− ζs−α)ξ f (s)− f (0+) sη−1(1− ζs−α)ξ,
If ξ = 0
• D: Riemann–Liouville derivative
• D: Caputo derivative
Part Two: Time Change
Consider the following Cauchy problem.{(Dδν,γ+ν,−1;0+h(x , ·)
)(t) = − ∂
∂x h(x , t), t ≥ 0, x ≥ 0,
h(x , 0+) = δ(x).
• δ = 0 (inverse stable subordinator):
dγ+ν
dtγ+νh(x , t) = − ∂
∂xh(x , t), t ≥ 0, x ≥ 0,
where dκ/dtκ is the Caputo fractional derivative.
• δ = 1 (inverse of sum of two independent stablesubordinators):
dγ+ν
dtγ+νh(x , t) +
dγ
dtγh(x , t) = − ∂
∂xh(x , t), t ≥ 0, x ≥ 0.
• δ = n ∈ N ∪ {0}:
(Dnν,γ+ν,−1;0+h(x , ·)
)(t) = − ∂
∂xh(x , t)
⇔ ∂γ+ν+θ
∂tγ+ν+θ
∫ t
0(t − y)θ−1E−nν,θ [−(t − y)ν ] h(x , y) dy
= − ∂
∂xh(x , t) + δ(x) t−(γ+ν)E−nν,1−(γ+ν)(−t
ν)
⇔n∑
r=0
(n
r
)∂γ−ν(r−1)
∂tγ−ν(r−1)h(x , t)
= − ∂
∂xh(x , t) + δ(x)
n∑r=0
(n
r
)t−(γ−ν(r−1))
Γ (1− (γ − ν (r − 1)))
⇔n∑
r=0
(n
r
)dγ−ν(r−1)
dtγ−ν(r−1)h(x , t) = − ∂
∂xh(x , t)
with 0 < γ − ν(r − 1) < 1 and thus nν < γ + ν < 1.
In order to construct a time-change, our aim is to find the solutionto the Cauchy problem{(
Dδν,γ+ν,−1;0+h(x , ·))
(t) = − ∂∂x h(x , t), t ≥ 0, x ≥ 0,
h(x , 0+) = δ(x).
First result:
The stochastic solution to the above problem, δ > 0, is given bythe law of the hitting time Eδt = inf{s : Vδ
s /∈ (0, t)}, t ≥ 0, of
Vδt =
n∑r=0
rV(γ+ν)n/δ−rν(nr)V
δ/nt
, t ≥ 0,
where rV(γ+ν)n/δ−rνt , r = 1, . . . n, and V
δ/nt are independent
stable subordinators, n = dδe is the ceiling of δ andνδ < γ + ν < 1.
Remark:The Laplace–Laplace transform of the stochastic solution h(x , t) is
˜h(z , s) =sγ+ν−1 (1 + s−ν)
δ
sγ+ν (1 + s−ν)δ + z, z > 0, s > 0.
• δ = 0 (inverse stable subordinator):
˜h(z , s) =sν+γ−1
sν+γ + z
• δ = 1 (inverse of sum of two indep stable subordinators):
˜h(z , s) =sν+γ−1 + sγ−1
sν+γ + sγ + z
Part Three: Levy Processes
Let Ξxt , t ≥ 0, be a Levy process starting from x ∈ Rd , with
characteristics (a,Q,Π). We introduce the semigroup
Tt f (x) = Ef (Ξxt ) =
∫Rd
f (y)P(Ξxt ∈ dy)
with infinitesimal generator
limt→0
1
t(Tt f − f ) = Af ,
The Fourier symbol of the process Ξ0t , is
Ψ(ξ) = i〈a, ξ〉+1
2〈ξ,Qξ〉+
∫Rd\{0}
(1− e i〈z,ξ〉 + i〈z , ξ〉I|z|<1
)Π(dz).
The Borel measure Π(·) is the so-called Levy measure satisfying∫Rd\{0}(1 ∧ |z |2) Π(dz) <∞, where, as usual, |z |2 = 〈z , z〉.
P(Ξxt ∈ dy)/dy is the density of the Levy process Ξx
t . This meansthat
E exp(iξΞ0
t
)=
∫Rd
e iξyP(Ξ0t ∈ dy) = exp (−tΨ(ξ))
and that Ψ(·) completely determines the density of the Levyprocess.
Examples:
• if Ψ(ξ) = |ξ|2α, α ∈ (0, 1], then A = −(−4)α is thefractional Laplacian. The process Ξ0
t is isotropic stable andbecomes a Brownian motion for α = 1.
• if d = 1 and Ψ(ξ) = λ(1− e iξ), then Ξt is a Poisson processon Z+ with rate λ > 0 and
Af (x) = λ{f (x)− f (x − 1)}1Z+(x)
is the governing operator written as λ times the discretegradient on Z+. Also, we usually write Af = λ(I − B)f .
• if d = 1 and Ψ(ξ) = λ(1 + iξ − e iξ), the correspondingprocess is the compensated Poisson on R with rate λ > 0.The generator takes the form
Af = λ(I − B)f − λf ′.
We introduce the time-change operator
Hγ,ν,δt =
∫ ∞0
h(dy , t)Ty
where h(y , t) is the law of Eδt , with δ ∈ (0,∞), and Tt is thesemigroup associated to the generator A.Let Rd+1 := Rd × (0,∞) and define the function space Ak as
Ak(Rd+1) =
{u : Rd+1 7→ R+ s.t.
k−1∑j=0
aj∂α+j
∂tα+ju ∈ L1(Rd+1), α ∈ (0, 1], aj > 0 ∀ j ,
|u(x , t)| ≤ g(x)tβ−1, β > 0, g ∈ L∞(Rd)
}, k = 1, 2, . . . .
Main result:
Let δ > 0, n = dδe, and δν < γ + ν ≤ 1, γ, ν ∈ (0, 1). Theunique solution to
g ∈ Adγ+νe(Rd+1),(Dδν,γ+ν,−1;0+g(x , ·)
)(t) = Ag(x , t), x ∈ Rd , t > 0,
g(x , 0) = f (x),
with f ∈ D(A), is written as
g(x , t) = Hγ,ν,δt f (x) = Ef
(ΞxEδt
).
One-dimensional case with 0 < γ + ν ≤ 2
In this general case we do not have any relation with processes.The solution to(Dδν,γ+ν,−λ;0+g(x , ·)
)(t) = c ∂2
∂x2g(x , t), x ∈ R, t > 0,
g(x , 0+) = δ(x),∂∂t g(x , t)
∣∣t↓0 = 0.
is not known explicitely. Anyway we have that the Fourier–Laplacetransform can be written as
ˆg(β, s) =sγ+ν−1 (1 + λs−ν)
δ
sγ+ν (1 + λs−ν)δ + cβ2.
Inverting the Fourier–Laplace transform we get
g(β, t) =∞∑r=0
(−cβ2tγ+ν)rE rδν,r(γ+ν)+1 (−λtν)
and
g(x , s) =1
2sc1/2s(γ+ν)/2
(1 + λs−ν
)δ/2e− |x|
c1/2s(γ+ν)/2(1+λs−ν)
δ/2
.
If you wish, you can download the paper:
M D’Ovidio & F Polito, arXiv:1307.1696, 2013.
