Some inverse problems for time-fractionaldiffusion equations

Masahiro Yamamoto : The University of Tokyo,Honorary Member of Academy of Romanian Scientists

Internationa Workshop on Fractional CalculusGhent Analysis & PDE Center, Ghent University

10 June 2020

Simulation by advection-diffusion model

Field data (Adams & Gelhar, 1992)

t-=0t1

t2t3

From ultraslow dissolution to anomalous diffusion

Anomalous diffusion: not simulated by classical diffusion equation

Security critical value

Security criticalvalue

Anomalous diffusion

dissolution

density

aggregation

monitoring well放出源

location

aggregation –dissolution, then disappear

Optimal control

Mission II: Various

inverse problems

for

fractional partial

differential equations

Grand Research Plan

Parameter identification

Real world problems: e.g., pollution in soil

Mission I: Construct

general theory for fractional

partial differential equations Launch by Gorenflo-Luchko-Yamamoto 2015

Nonlinear theory, Dynamical system, etc.

Classical

theory of PDE

motivations

Sublime to theory

Contents:Three kinds of inverse problems for fractionalpartial differential equations

Messages:We have many interesting and importantinverse problems.

Contents

• §1. Introduction• §2. Determination of fractional orders• §3. Backward problems in time• §4. Determination of coefficients

§1. Introduction

Ω ⊂ Rd with smooth boundary ∂Ω

∂αt

u(x, t) =d∑

i,j=1

∂i(aij(x)∂ju(x, t)) + c(x)u

Forward problem:Solve initial boundary value problem =⇒ predictionInverse problems:How to determine α, aij, initial, boundary value,shape of Ω. =⇒modelling, basement for forward problemsVariety of inverse problems!

Mathematical issues for inverse problems:uniqueness, stability

varieties of inverse problems × varietiesof fractional equations as models=⇒ Varieties 100

Many interesting inverse problems for fractionaldifferential equations!!

§2. Determination of fractionalorders

uα,β

∂α

tu = −(−∆)βu, x ∈ Ω, t > 0,

u|∂Ω = 0, t > 0,u(x, 0) = a(x), x ∈ Ω if 0 < α < 1,u(x, 0) = a(x), ∂tu(x, 0) = 0 if 1 < α < 2.

Inverse problem: determine α ∈ (0, 2) and β ∈ (0, 1)by u(x0, t), 0 < t < T with fixed x0 ∈ Ω.

Remark.• Caputo derivative:

∂αt

v(t) =1

Γ(n − α)

∫ t

0(t−s)n−α−1 dnv

dsn(s)ds, n−1 < α < n.

• We can replace −∆ by∑d

i,j=1∂i(ai j(x)∂ j) + c(x).

Maybe we can treat non-symmetric operators.

About forward problems: a monographKubica-Ryszewska-Yamamoto (Springer, 2020)

Remark. (−∆)β: fractional power of −∆⇐= spatial derivative of order 2βα, β =⇒ important physical parameters

Inverse problem.Let x0 ∈ Ω be fixed. u(x0, t), 0 < t < T =⇒α ∈ ((0, 2) \ 1) and β ∈ (0, 1).

Uniqueness⇐= How much information data have?

Determination of fractional orders at

laboratory based on the micro-model

Comparison of laboratory data

with numerical results by

Monte Carlo method soil particles

Prof. Y. Hatano (Tsukuba University)

tx 2 10

Continuous time random walk

(micro-model) tx 2

Normal diffusion by Fick’s

law

Ranom walk

α=1

Related works on determination oforders

• Hatano, Nakagawa, Wang and Yamamoto 2013• Li and Yamamoto 2015: uniqueness for mutiterm cases• Yu, Jing and Qi 2015• Janno 2016: unique existence• Janno and Kinash 2018• Krasnoschok, Pereverzyev, Siryk and Vasylyeva 2019• Ashurov and Umarov 2020: arXiv:2005.13468v1

Other examples of data:∫Ω

u(x, t)ρ(x)dx, 0 < t < T : ρ: weight

∇u · ν(x0, t), 0 < t < T

ν(x): unit outward normal vector

Preparations0 < λ1 < λ2 < · · · −→ ∞: set of eigenvalues of −∆ withu|∂Ω = 0.

−∆φkj = λkφkj, 1 ≤ j ≤ mk, mk: multiplicity of λk

φkj, 1 ≤ j ≤ mk: orthonormal basis(φkj, φi j) :=

∫Ωφkj(x)φki(x)dx = 1 if i = j, = 0 if i , j

λk, φkj1≤ j≤mk : eigensystem of −∆ with u|∂Ω = 0

Fractional power of −∆:

(−∆)βv =∞∑

k=1

λβ

k

mk∑j=1

(v, φkj)φkj,

v ∈ D((−∆)β) ⊂ H2β(Ω) : Sobolev-Slobodeckij space

(−∆)−βa =sin πβ

π

∫ ∞

0η−β(−∆ + η)−1 adη

in L2(Ω) with 0 < β < 1 (e.g., Pazy).

Eα,1(z) =∞∑

k=0

zk

Γ(αk + 1): Mittag-Leffler function

Theorem 1 (uniqueness) .Let a ∈ H2

0(Ω) if d = 1, 2, 3 (a ∈ H2γ

0(Ω) with γ > d

4 ),

a ≥ 0,. 0 or ≤ 0,. 0 in Ω,

and

∃k0 ∈ N,mk0∑j=1

(a, φk0 j)φk0 j(x0) , 0, λk0 , 1.

Then uα,β(x0, t), 0 < t < T⇐⇒ (α, β) ∈ (0, 2) \ 1 × (0, 1) 1 to 1

Simplified case .Assume mk = 1: λk is simple for all k, −∆φk = λkφk.Corollary (uniqueness) .(i) a ∈ H2

0(Ω) if d = 1, 2, 3,

a ≥ 0,. 0 or ≤ 0,. 0 in Ω

(ii) ∃k0 ∈ N, (a, φk0 )φk0 (x0) , 0 and λk0 , 1.Then uα,β(x0, t), 0 < t < T ⇐⇒(α, β) ∈ ((0, 2) \ 1) × (0, 1) 1 to 1

Remark.• (ii) is essential for uniqueness of β.

Let λ1 = 1 and −∆φ1 = φ1. Thenuα,β(x, t) = Eα,1(−tα)φ1(x) for x ∈ Ω, t > 0.No information of β!

• (i) =⇒ uniqueness for α• Tatar-Ulusoy (2013):

(a, φk) > 0 for all k =⇒ uniqueness.Our proof producces the same conclusion.

• Let λk , 1 for all k.Then a(x0) , 0 =⇒ uniqueness for β.

Main ingredients for proof.(1) Eigenfunction expansion(2) Asymptotics of Mittag-Lefller function

Eα,1(−t) =N∑ℓ=1

(−1)ℓ+1

Γ(1 − αℓ)1

tℓ+ O

(1

t1+N

), t > 0,→ ∞

(3) Strong maximum principle for (−∆)β:

Lemma (strong maximum principle for (−∆)β):a ≥ 0, . 0, a|∂Ω = 0 =⇒ (−∆)βa(x) > 0, x ∈ ΩProof:β = 1 =⇒ classical strong maximum principle0 < β < 1: strong maximum principle =⇒(−∆ + η)−1a > 0 in Ω =⇒

((−∆)−β)a(x) =sin πβ

π

∫ ∞

0η−β(−∆ + η)−1a dη > 0

Proof for simple eigenvalues

uα,β (x0 , t) =∞∑

k=1Eα,1(−λβ

ktα)(a, φk )φk(x0)

=1

Γ(1 − α)1

tα

∞∑k=1

(a, φk )φk (x0)

λβk

+ O(

1

t2α

)

=1

Γ(1 − α)1

tα(−∆)βa(x0) + O

(1

t2α

)as t → ∞

uα,β(x0 , t) = uα1 ,β1(x0 , t), 0 < t < T =⇒

1Γ(1 − α)

1tα

(−∆)βa(x0) + O(

1

t2α

)=

1Γ(1 − α1)

1

tα1(−∆)β1 a(x0) + O

1

t2α1

as t → ∞ =⇒

(−∆)βa(x0) , 0, (−∆)β1 a(x0) , 0 by x0 ∈ Ω =⇒ α = α1

Proof of β = β1.Set ak = (a, φk)φk (x0).

N∑ℓ=1

(−1)ℓ+1

Γ(1 − αℓ)tαℓ

∞∑k=1

ak

λβℓk

=

N∑ℓ=1

(−1)ℓ+1

Γ(1 − αℓ)tαℓ

∞∑k=1

ak

λβ1ℓk

+ O(

1

tα(1+N)

)

as t → ∞ =⇒ ∞∑k=1

ak

λβℓk

=

∞∑k=1

ak

λβ1ℓk

, ℓ ∈ N \ m/αm∈N

Let k0 ∈ N such that a1 = · · · = ak0−1 = 0 and ak0, 0, λk0

, 1.

Assume λk0> 1 (similar for other case): Let β1 > β =⇒

ak0

(1 − (λ

β−β1k0

)ℓ)+

∞∑k=k0+1

ak

((λβk0/λβk

)ℓ − (λβk0/λβ1k

)ℓ)= 0

for ℓ ∈ N \ m/αm∈N . ∣∣∣∣∣∣λβ−β1k0

∣∣∣∣∣∣ ,∣∣∣∣∣∣λβk0

/λβk

∣∣∣∣∣∣ ,∣∣∣∣∣∣λβk0

/λβ1k

∣∣∣∣∣∣ < 1

for k ≥ k0 + 1.

Choose ℓn ∈ N \ m/αm∈N such that limn→∞ ℓn = ∞

=⇒ ak0 = 0: contradiction

§3. Fractional diffusion equationsbackward in time

with Profs G. Floridia (UniversitaMediterranea di Reggio Calabria) and Z. Li(Shandong University of Technology)−Au :=

∑ni,j=1

∂i(ai j(x)∂ ju) + c(x)u,

where c ≤ 0,D(A) = H2(Ω) ∩ H10(Ω).

∂αt

u = −Au in Ω,u|∂Ω = 0, u(·, T) = b.

unique existenceu ∈ C([0, T]; L2(Ω)) ∩ C((0, T]; H2(Ω) ∩ H1

0(Ω))

to ∂αt

u = −Au with u|∂Ω = 0 and u(·, T) = b(Sakamoto-Yamamoto 2011)

• memory effect or weak smoothing• Different from α = 1.

Preliminaries(I) 0 < λ1 ≤ λ2 ≤ ....: eignvalues of A with multiplicities,φk: eigenfunction for λk, (φ j, φk) = δjk. ∥ · ∥ = ∥ · ∥L2(Ω)

(II) Eα,β(z) =∑∞

k=0zk

Γ(αk+β) (Mittag-Leffler function).

(III) S(t)a =∑∞

k=1(a, φk)Eα,1(−λktα)φk,

K(t)a =∑∞

k=1tα−1(a, φk)Eα,α(−λktα)φk, a ∈ L2(Ω)

=⇒∥S(t)a∥H2(Ω) ≤ Ct−α∥a∥, ∥S(t)a∥ ≤ C∥a∥

∥AγK(t)a∥ ≤ Ctα(1−γ)−1∥a∥, 0 ≤ γ ≤ 1, t > 0

(IV) ∂α

tu = −Au + F in Ω,

u|∂Ω = 0, u(·, 0) = a.

=⇒u(t) = S(t)a +

∫ t

0K(t − s)F(s)ds, t > 0.

(V) S(t) : L2(Ω) −→ H2(Ω) ∩ H10(Ω) is isomorphism for t > 0.

Sakamoto-Yamamoto (2011):

§3.2. Non self-adjoint A

−Lu :=n∑

i, j=1

∂i(aij(x)∂ ju) +n∑

j=1

bj(x)∂ ju(x) + c(x)u

=: −Au + Bu∂α

tu = −Lu in Ω,

u|∂Ω = 0, u(T) = b.

Theorem 3.For b ∈ H2(Ω) ∩ H1

0(Ω), there exists unique solution u with same

regularity as in Theorem 2.

Key to ProofFirst Step.∂α

tu = −Au + Bu in Ω, u|∂Ω = 0, u(·, 0) = a =⇒

ua(t) = S(t)a +∫ t

0 K(t − s)Bua(s)ds, where S(t), K(t) are

constructed for −A =⇒ b = S(T)a +∫ T

0 K(T − s)Bua(s)ds⇐⇒

a = S(T)−1b −

∫ T

0K(T − s)Bua(s)ds

= S(T)−1b − S(T)−1

∫ T

0K(T − s)Bua (s)ds = S(T)−1b −Ma

Here

Ma := S(T)−1∫ T

0K(T − s)Bua (s)ds

Unique solvability for a?

Second Step: compactness of M : L2(Ω) −→ L2(Ω).We can prove ∥Au(t)∥ ≤ Ct−α∥a∥ (Sakamoto-Yamamoto for B = 0)

Let 0 < δ < 12 .

∥∥∥∥∥∥∥A1+δ∫ T

0K(T − s)Bua (s)ds

∥∥∥∥∥∥∥ =∥∥∥∥∥∥∥∫ T

0A

12 +δK(T − s)A

12 Bua(s)ds

∥∥∥∥∥∥∥≤ C

∫ T

0(T − s)

α( 12 −δ)−1∥Aua (s)∥ds ≤ C

∫ T

0(T − s)

α( 12 −δ)−1

s−αds∥a∥ ≤ C∥a∥.

=⇒

∥∥∥∥∥∥∥AδS(T)−1∫ T

0K(T − s)Bua (s)ds

∥∥∥∥∥∥∥ ≤ C

∥∥∥∥∥∥∥A1+δ∫ T

0K(T − s)Bua (s)ds

∥∥∥∥∥∥∥ ≤ C∥a∥.

=⇒M : L2(Ω) −→ H2δ (Ω) is bounded

M : L2(Ω) −→ L2(Ω) is compact

Third Step:Fredholm equation of second kind:a = S(T)−1b −Ma

Well-posedness⇐= ”b = 0 =⇒ a = 0”

⇐⇒ Backward uniqueness∂α

tu = −Lu in Ω,

u|∂Ω = 0, u(·, T) = 0 =⇒ u(·, 0) = 0.

Fourth Step: Backward uniqueness.Pm, m ∈ N: eigenprojection for eigenvalue λm of L

Closed subspace spanned by all the generalized eigenfunctionsis L2(Ω). (Agmon)

=⇒ Pma = 0, m ∈ N imply a = 0.

Setum (t) := Pmu(t), am := Pma

=⇒

∂αt

um = (−λm + Dm )um =: Jum (Jordan form) where Dℓm = 0 with large ℓ.

=⇒ um(t) = Eα,1(Jtα )am =∞∑

k=0

(Jtα)k

Γ(αk + 1)am

Set um = (u1m , ..., u

Nm )T and am = (a1

m , ..., aNm )T where T : tranpose

We can represent

u1m (t) = Eα,1(−λmtα )a1

m + · · ·

u2m (t) = Eα,1(−λmtα )a2

m + · · ·· · · · · ·uN

m (t) = Eα,1(−λmtα )aNm

u(T) = 0 =⇒ um(T) = 0

By Eα,1(−λmTα) , 0 =⇒aN

m = 0, thenaN−1

m = 0, then .....a1

m = 0.=⇒ am = 0 for m ∈ N.

Thus a = 0!

§4. Inverse coefficient problems

Let 0 < α < 1, Ω ⊂ Rd: bounded,ω ⊂ Ω: arbitrarily fixed subdomain.

up(x, t)

∂αt

u = ∆u + p(x)u(x, t), x ∈ Ω ⊂ Rd,∂νu|∂Ω = 0, u(x, 0) = a(x), x ∈ Ω

Inverse coefficient problem.up|ω×(0,T) =⇒ p(x), x ∈ Ω.

Similar result with data ∂νu|Γ×(0,T) with Γ ⊂ ∂Ω.

Limited references:(1) Dirichlet-to-Neumann map:Imanuvilov-Li-Yamamoto (2016),Kian-Oksanen-Soccorsi-Yamamoto (2018), etc.(2) ∂1/2

tu − ∂2

x: 1DYamamoto-Zhang (2012)(3) Integral transform to wave equation:Miller-Yamamoto (2013)(4) Spectral approach for 1D case:Cheng-Nakagawa-Yamamoto-Yamazaki (2009),Li-Zhang-Jia-Yamamoto (2013), etc.

Li-Zhang-Jia-Yamamoto (2013):φ′′

k(x) + p(x)φk(x) = −λkφk, φ′

k(0) = φ′

k(1) = 0

ψ′′k

(x) + q(x)ψk(x) = −µkψk, ψ′k(0) = ψ′

k(1) = 0 for k ∈ N

=⇒Let a ∈ H4

0(0, 1) and (a, φk) , 0 for all k ∈ N. If

up(i, t) = uq(i, t) for 0 < t < T and i = 0, 1, thenp(x) = q(x) for 0 < x < 1.

Remark. We can determine α uniquely.

Scenario of proof:Data of u = [eigenfunction expansion] =⇒ Spectral data= [Gel’fand-Levitan theory] =⇒ UniquenessStep 1:(1) 1D case: well by λn = c0n + O(1/n)(2) General dimensions: bottleneckbecause only λn = c0n2/d + o(n2/d).Step 2:Multidimensional Gel’fand-Levitan theory(e.g., Nachman-Sylvester-Uhlmann)

Class of initial values for uniqueness.

A: k ∈ N; (a, φk) , 0 = N.Uniqueness for 1D case, ??? for multidimensions

B: #k ∈ N; (a, φk) , 0 = ∞.Uniqueness ????

C: #k ∈ N; (a, φk) , 0 < ∞.Uniqueness holds. ⇐=We discuss here!

λk: eigenvalue of −∆ − p(x) with ∂νu|∂Ω = 0Pk: orthogonal projection onto Ker (−λk − (∆ + p))µk: eigenvalue of −∆ − q(x) with ∂νu|∂Ω = 0Qk: orthogonal projection onto Ker (−µk − (∆ + q))

up

∂α

tu = ∆u + p(x)u, x ∈ Ω ⊂ Rd, t > 0,

∂νu|∂Ω = 0, u(x, 0) = a(x).

Inverse problem:ω ⊂ Ω: arbitraily given subdomainDetermine p in Ω by up|ω×(0,T).

Theorem 4 (Jing-Yamamoto: 2020).Let p, q, a ∈ H∞(Ω),

#k ∈ N; Pka . 0 in Ω < ∞

and |a(x)| > 0, x ∈ Ω.Then up = uq in ω × (0, T) implies p = q in Ω.

Remark.(i) The set of a for the uniqueness is dense (not very specialchoice).(ii) #k ∈ N; Pka . 0 in Ω = N =⇒ Uniqueness for 1D case

Thank you very much.