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IntroductionUnknown boundary coefficient

Unknown boundaryAbnormal Diffusion

The Mathematical Analysis of

the Diffusion Process and its Applications

Cheng Jin

School of Mathematical Sciences, Fudan University

AIP09, Vienna, Austria

July 24th, 2009

Cheng Jin The Mathematical Analysis of



Outline

1 Introduction

2 Unknown boundary coefficient

Motivation and Modelling

Theoretic Results

Proof of the main results:

3 Unknown boundary

Motivation and Modelling

Forward Problems

Related Inverse Problems

4 Abnormal Diffusion

Related Inverse Problem




Joint research with Masahiro Yamamoto(Tokyo

University, Japan), Shuai Lu (OCIAM, Austria), Wenbin

Chen, Xiaoyi Hu(Fudan University, China)

This research is partly supported by the Programme of

Introducing Talents of Discipline to Universities(No.B08018), the

NSFC (No.10431030), the Shuguang Project and E-Institute of

Shanghai Municipal Education Commission (N.E03004).




We discuss the applications of the Mathematical Models of

Diffusion Process in the Industry and Environments Sciences,

Especially the methods of Inverse Problems.

Mathematical models:

Normal Diffusion: (Temperature Distribution in the steel)

Abnormal Diffusion: (Diffusion in the porous media)




Background

Steel Works:




Math is needed to produce steel safely and effectively, specially

PDE and ”inverse” PDE.




Steel? Heat? Heat equation?

∂u

∂t−∆u = f (1)

It is so easy, just ask junior undergraduate students to code,

everything has been solved, I am happy:)




Wait a moment:

Radiation interface conditions: Nonlinear

To detect the corrosion: Inverse problems

Requirement from the industry: High temperature, safe,

detection, ...

Math can be manufactured?

Minutes Days Weeks Months




The modelling of the dynamics of anomalous processes by

means of differential equations that involve derivatives of fractional

order has provided interesting results in a variety of fields of

science. The most studied and applied model is the fractional

diffusion equations (FDE) which play an important role in

describing anomalous diffusion. A general account of FDEs is given

in R. Metzler and J. Klafter, Phys. Rep. 339, 1 (2000)




Part I

Inverse Problems of Determining the unknownboundary in the heat process:




We want to detect the defects in the thin sheetduring the steel rolling process by the thermalimage:




Sheet and Types of defect

Figure: Sheet with the defects




Mathematical formulation of the problem:We consider the following mathematical model problem in the

heat transfer:

∂u

∂t= ∆u, 0 < t < T , x ∈ Ω (2)

−∂u

∂ν= σ(u4 − u4

A), 0 < t < T , x ∈ ∂Ω (3)

u = a(x), t = 0, x ∈ Ω (4)

where Ω is a simply connected domain in Rn with the C 2 boundary

∂Ω and ν is the outer normal unit with respect to ∂Ω.




Here u represents the temperature distribution and uA is the

air temperature, which is assumed to be a positive constant.

σ(x) ∈ H2(∂Ω) is called the Stefan-Boltzmann radiation

coefficient, which characters the heat transfer between the solid

and air.




Figure: Thermal Image




Some assumptions on the domain:Suppose that the boundary ∂Ω can be divided as two parts Γ0

and Γ1, such that ∂Ω = Γ0 ∪ Γ1. Moreover, there is an one open

set ω ⊂ Γ1.




The Assumptions on the coefficients

We assume that

(A1). uA is a given positive constant.

(A2). the Stenfan-Boltzmann coefficient σ ∈ C 2(∂Ω) and satisfies

σ(x) =

σ0, x ∈ Γ0 ∪ ω

γ(x), x ∈ Γ1 \ ω(5)

where σ0 is a given positive constant and γ is an unknown function.

(A3). The initial value a(x) > c0 > uA, where c0 is a given

constant.




Our inverse problem is

find the unknown functions γ(x), x ∈ Γ1 and the initial value

a(x), x ∈ Ω from u(t, x), (t, x) ∈ (0,T )× Γ1.

The admissible set for the unknowns is

Λ =

(σ, a)

∣∣∣∣ ‖σ‖C2(∂Ω) ≤ M, σ > 0; ‖a‖C2(Ω) ≤ M, a(x) > c0




Theorem 1: (Existence, Uniqueness and Bounds)

Suppose that (A1), (A2) and (A3) are satisfied. If the (1) holds,

then, for the nonlinear boundary value problem, there exists a

unique classical solution u(t, x) ∈ C 1,2(Q) such that

uA < u(t, x) ≤ M (6)

and

‖u‖C1,2(Q) ≤ C1 (7)

where C1 is a positive constant depending on M.




Theorem 2: (Uniqueness of IP)

Suppose that (σ1, a1), (σ2, a2) ∈ Λ and (A1), (A2) and (A3) are

satisfied. If it holds that

u1(t, x) = u2(t, x), 0 < t < T , x ∈ Γ1

then we have

σ1(x) = σ2(x), x ∈ ∂Ω

a1(x) = a2(x), x ∈ Ω.




Theorem 3: (Conditional Stability of IP)

Suppose that (σ1, a1), (σ2, a2) ∈ Λ and (A1), (A2) and (A3) are

satisfied. Then, for the inverse problem we discuss, it holds that

‖σ1 − σ2‖C(∂Ω) ≤ C2‖(f1 − f2)esα‖C(Σ1) (8)

where fj(t, x) = uj(t, x), (t, x) ∈ Σ1 = (0,T )× Γ1, j = 1, 2, and

C2 is a positive constant depending on M and

η1 == min(t,x)∈(0,T )×∂Ω(u41 − u4

A).

α is defined in Carleman estimation and s(λ) > 0 is

sufficiently large for λ > λ, where λ > 0 depends on M.




One Remark:

In Theorem 3, the Lipschitz stability estimation is only for the

unknown function σ. For another unknown function a(x), there is

also a conditional stability estimation results. It is a logarithmic

conditional stability estimation, which is too weak from the

numerical point of view.




Definition of the weighted function

Suppose that ω is an open set in Γ1. Then there exits a function

ψ ∈ C 2(Ω) such that

ψ(x) > 0, x ∈ Ω, |∇ψ(x)| > 0, x ∈ Ω

ψ(x) = 0, x ∈ Γ0 ∪ ω,∂ψ

∂ν≤ 0, (t, x) ∈ [0,T ]× (∂Ω \ ω).




Definition of the weighted function

Let

l(t) = t(T − t).

We set

φ(t, x) =eλψ(x)

l(t)(9)

and

α(t, x) =eλψ(x) − e

2λ‖ψ‖C(Ω)

l(t)(10)

where λ is a positive constant.




The equationWe consider a function w ∈ W 1,2

2 (Q), which is a weak solution of

the following problem:

∂w∂t = ∆u + g(t, x), 0 < t < T , x ∈ Ω

∂w∂ν + k(t, x)w = 0, 0 < t < T , x ∈ Γ0

w = 0, < t < T , x ∈ Γ1

w = a, t = 0, x ∈ Ω.

(11)




Carleman estimation: (Imanuvilov, Yamamoto )

Suppose that w is the weak solution of (11) and

‖k‖C((0,T )×Γ1), ‖∂k

∂t‖C((0,T )×Γ1) ≤ M1.

Then there exists a λ, which depends on M1, such that, for

any arbitrary λ ≥ λ, we can choose s0(λ) satisfying: for any

s ≥ s0(λ), the following estimation holds:




∫Q

(sφ)p−1

∣∣∣∣∂w

∂t

∣∣∣∣2 +n∑

i ,j=1

∣∣∣∣ ∂2w

∂xi∂xj

∣∣∣∣2 e2sαdxdt

+

∫Q

((sφ)p+1|∇w |2 + (sφ)p+3w2

)e2sαdxdt

≤ C2

∫Q

(sφ)p|g |2e2sαdxdt + C2

∫(0,T )×ω

(sφ)p(∂w

∂t

)2

e2sαdxdt

+C2

∫(0,T )×ω

((sφ)p+1|∇w |2 + (sφ)p+3w2

)e2sαdxdt

where C2 is a positive constant depending λ.




Outline of the proof of Theorem 3:

Step 1:

By Carleman estimation, we can prove that

‖u1 − u2‖L2((0,T )×Γ1\ω) + ‖∂(u1 − u2)

∂ν‖L2((0,T )×Γ1\ω)

≤ C‖(f1 − f2)e2sα‖

W 1,12 (0,T )×Γ1)

.




Outline of the proof of Theorem 3:

Step 2:

By the nonlinear boundary condition, we have

∂u1

∂ν− ∂u1

∂ν= (σ1 − σ2)(u

41 − u4

A) + σ2(u41 − u4

2).




Some remarks:

For the convenience of proving the results, we assume that

more regularity for the unknown functions. These assumptions can

be weakened by the same method.




Some remarks:

The numerical methods based on our analysis for the inverse

problems have been proposed. The reader can find the related

results in the forthcoming paper from our group.




Future works:

1 The case σ is a piecewise smooth function.

2 Multi-layers cases (reduce the measurements)

3 Other applications




Part II

Inverse Problems of Determining the unknownboundary:




We want to describe the temperature distributioninside the container in the process of continuouscasting:




The governing equations are the heat equation

∂u

∂t= a∆u, inside the layers




The heat transfer on the interfaces:1 The heat transfer between the outer layer and air

2 The heat transfer between the different layers

3 The heat transfer between the inner layer and molten steel




The heat transfer between the air and solid:Stephen-Boltzmann radiation condition:

λ∂u

∂ν|interface = −c(u4 − u4

a)

where ua is the air temperature.

The heat transfer between the molten steel andsolid material:

u = costant




How to model the heat transfer between the different layers?

Previous Formulation:

u+ = u−, λ+∂u+

∂ν= λ−

∂u−

∂ν, on interface

This is not consistent with the experimentresults!

On the interface, the temperatures on the different layers are

different. i.e.

u+ 6= u−, on interface




We consider the case

λ+∂u+

∂ν= −c((u+)4 − u4

∗)

λ−∂u−

∂ν= c((u−)4 − u4

∗)

Let δ → 0Cheng Jin The Mathematical Analysis of



We have the transmission conditions on the interface:

λ+∂u+

∂ν= λ−

∂u−

∂ν

λ+∂u+

∂ν= −c

2((u+)4 − (u−)4)

Nonlinear Transmission conditions!




We consider the one dimensional case

∂tu1(x , t) = α1∂2xu1(x , t), 0 < x < `1, 0 < t < T ,

∂tu2(x , t) = α2∂2xu2(x , t), `1 < x < `2, 0 < t < T ,

∂tu3(x , t) = α3∂2xu3(x , t), `2 < x < `3, 0 < t < T ,

with an initial condition:

u(x , 0) = a(x), 0 < x < `3Cheng Jin The Mathematical Analysis of



and the boundary condition, transmission conditions:

u1(0, t) = uM ,

−λ1∂xu1(`1, t) = σ1(u41(`1, t)− u4

2(`1, t)),

λ1∂xu1(`1, t) = λ2∂xu2(`1, t),

−λ2∂xu2(`2, t) = σ2(u42(`2, t)− u4

3(`2, t)),

λ2∂xu2(`2, t) = λ3∂xu3(`2, t),

−λ3∂xu3(`3, t) = σ3(u43(`3, t)− u4

a),

Here α1, α2, α3 > 0, λ1, λ2, λ3 > 0, σ1, σ2, σ3 > 0, uM > 0, ua ≥ 0

are constants.




We have the result:Theorem: Suppose that a(x) satisfies the compatibility

conditions. Let T be arbitrarily given. Then there exists a unique

solution.

The solutions are within the class:u1 ∈ C1([0,T ];C2[0, `1]),

u2 ∈ C1([0,T ];C2[`1, `2]),

u3 ∈ C1([0,T ];C2[`2, `3]),

(12)




Lemma: Suppose that a(x) satisfies the compatibility conditions.

Let T be arbitrarily given. Then there exits a t0 > 0 such that the

problem has a unique solution within the classu1 ∈ C1([0, t0];C2[0, `1]),

u2 ∈ C1([0, t0];C2[`1, `2]),

u3 ∈ C1([0, t0];C2[`2, `3]).




Outline of the proof:1 By Green functions, the solution can be represented by the

integral equations:

u1(x , t) = u1(x , t) + α1

∫ t

0G1(t − τ, x , `1)f1(τ)dτ (13)

u2(x , t) = u2(x , t)− α2

∫ t

0G2(t − τ, x − `1, 0)

λ1

λ2f1(τ)dτ(14)

+α2

∫ t

0G2(t − τ, x − `1, `2 − `1)f2(τ)dτ

u3(x , t) = u3(x , t)− α3

∫ t

0G3(t − τ, x − `2, 0)

λ2

λ3f2(τ)dτ(15)

+α3

∫ t

0G3(t − τ, x − `2, `3 − `2)f3(τ)dτ.




where

u1(x , t) =

∫ `1

0a(ξ)G1(t, x , ξ)dξ +

∫ t

0uM∂xG1(t − τ, x , 0)dτ

u2(x , t) =

∫ `2−`1

0a(ξ + `1)G2(t, x , ξ)dξ,

u3(x , t) =

∫ `3−`2

0a(ξ + `2)G3(t, x , ξ)dξ.

2 By the nonlinear interface transmission conditions, we have

f1 = K1(f1, f2)

f2 = K2(f1, f2, f3)

f3 = K3(f2, f3)

3 K = (K1,K2,K3) is a contractive operator when t0 is small

enough.Cheng Jin The Mathematical Analysis of



Lemma: Suppose that there exists the solution u1, u2, u3 and

a > 0, uM > ua > 0 . Then we have

u1(`1, t) > 0, u2(`1, t) > 0, u2(`2, t) > 0,

u3(`2, t) > 0, u3(`3, t) > 0, for t > 0.

Lemma: Suppose that there exists the solution u1, u2, u3 and let

T > 0 be arbitrarily fixed. Then it holds that

max‖u1‖C([0,`1]×[0,T ]), ‖u2‖C([`1,`2]×[0,T ]), ‖u3‖C([`2,`3]×[0,T ])

≤ max‖a‖C [0,`3], |uM |, |ua|.




Numerical Methods: we propose a numerical scheme on the

basis of the finite difference method.

The difference scheme is

un+1m,i − un

m,i

∆t=

αi

∆x2

(un+1m,i+1 − 2un+1

m,i + un+1m,i−1

),

where unm,i means the temperature ui in the i-th layer at time

tn = (n − 1)∆t and position xm = (m − 1)∆x .




Treat the nonlinear interface condition. Setting a fictitious point

un+10,1 on the right hand side of u1(`1, t)

−λ1

un+10,1 − un+1

m1−1,1

2∆x= σ1

((un

m1,1)3un+1

m1,1− (un

m1,2)3un+1

m1,2

).

Moreover

un+1m1,1

− unm1,1

∆t=

α1

(∆x)2

(un+1m1−1,1 − 2un+1

m1,1+ un+1

0,1

).

Then eliminating un+10,1 , we have

(1+2τ1+τ1τ2(unm1,1)

3)un+1m1,1

−2τ1un+1m1−1,1−τ1τ2(u

nm1,2)

3un+1m1,2

= unm1,1,

where

τ1 = α1∆t/(∆x)2, τ2 = 2(∆x)σ1/λ1.




Numerical Simulations:Setting uM = 1873, a(x) ≡ 303 and the parameters are

chosen as:α1 = 4.02× 10−2, λ1 = 40, σ1 = 4.88× 10−4, `1 = 0.06,

α2 = 1.7× 10−3, λ2 = 1.2, σ2 = 4.88× 10−4, `2 − `1 = 0.1,

α3 = 9.3× 10−3, λ3 = 7.3, σ3 = 4.88× 10−4, `3 − `2 = 0.04.




The numerical results under the grid: ∆t = 0.001,∆x = 0.001 are

shown in Figure 1 and Figure 2 for T = 0.02 hour and T = 0.1

hour:




The problem is a nonlinear initial/boundary value problem for

the heat equation. It is believed that the solution will convergence

to the solution of the stationary problem. Our numerical results

show the asymptotic behavior of the solution




Purpose: Detect the corrosion inside from the thermal image

outside.

Based on the forward model!




The part which contains some corrosion:




The temperature distribution outside




Problem:

Is it possible to detect the corrosion from the thermal image

outside?

Is it possible to have some stable and fast algorithms to

reconstruct the corrosion, especially the depth of the

corrosion?

the numerical results should be consistent with the experiment

data.

Remark: This method is a cheap way.




The mathematical formulation:

The governing equation:

∂u

∂t= a∆u, (0,T )× Ω

where u is the temperature distribution.

It is obvious that this problem can be treated as an inverse

problem for the heat equation!




Notations:Ω is bounded domain in Rn and Γ = Γ1 ∪ Γ2.

Γ1 is the part of the boundary which is fixed

Γ2 is the part of boundary which the corrosion may happen.

T is a fixed constant

T can not be too big!Cheng Jin The Mathematical Analysis of



The mathematical formulation:

∂u

∂t= α∆u, (0,T )× Ω

u = c0, (0,T )× Γ2

∂u

∂ν= c(u4 − u4

a), (0,T )× Γ1

The additional data is given

u = f , (0,T )× Γ1




The picture of the section which contains corrosion:




We will study:

1 The additional information is enough to determine the

unknown boundary? (Uniqueness)

2 If so, is it possible to give a fast and stable reconstruction

algorithm?




Difficulties:

1 ill-posedness: Cauchy problem for heat equation

2 no initial data

3 T is not so big




Previous work on this topic:

Transform to an optimization problem!

not so good!




Theoretic analysis: (one dimensional model)

Ω = (0, l), Γ1 = 0, Γ2 = l

∂u

∂t=∂2u

∂x2, 0 < x < l , 0 < t < T

u(t, 0) = f (t), 0 < t < T

∂u

∂x(t, 0) = g(t), 0 < t < T

u(t, l) = c0, 0 < t < T




Uniqueness Results:

Suppose that T = ∞, ‖f ‖C2 ≤ M and ‖g‖C2 ≤ M. Assume

that there exists a constant γ0 > 0 such that

‖f − c0‖ ≥ γ0

If, for two l , l , the solutions u and u satisfy the previous equations

and boundary conditions, then we have l = l .




Outline of the proof:

Assume that l < l .

1 By the uniqueness of the Cauchy problem for heat equation,

we can conclude

u(x , t) = u(x , t), 0 < x < l , 0 < t < T .

2 We have that

∂u

∂t=∂2u

∂x2, l < x < l , 0 < t < T

u(t, l) = c0, 0 < t < T

u(t, l) = c0, 0 < t < TCheng Jin The Mathematical Analysis of



3 By the result for the heat equation, it holds that

‖u − c0‖C(l ,l) −→ 0, t →∞.

4 By the unique continuation for the heat equation, we have

that

|u(t, 0)− c0| → 0, t →∞

This is the contradiction to our assumption!




Remarks:

The contribution of the initial data is not important

This result is true for multidimensional cases

The conditional stability can be obtained.

(The several conditional stability results of the inverse problem

for the parabolic equations with the given initial condition

have obtained by the Italy group and other researchers.)




Numerical Algorithm:We consider

∂u∂t = ∂2u

∂x2 , 0 < x < l , 0 < t < T

u(t, 0) = f1(t), 0 < t < T

u(t, l) = f2(t), 0 < t < T

∂u∂x (t, 0) = h(t), 0 < t < T

∂u∂x (t, l) = g(t), 0 < t < T

u(0, x) = u0(x)




Let

G (t, x , y) =2

l

∞∑n=0

e−λnt sin(n +1

2)π

lx sin(n +

1

2)π

ly

Then the solution can be expressed by

u(t, x) =∞∑

n=0

Ane−λnt sin(n +

1

2)π

lx +

∫ t

0

∂G

∂x(t − s, x , 0)f1(s)ds

+

∫ t

0G (t − s, x , l)g(s)ds, 0 < x < l , 0 < t < T




Evaluate:Let

w(t, x) =

∫ tj

ti

∂G

∂x(t − s, x , 0)ds

It satisfies

∂w∂t = ∂2w

∂x2 , 0 < x < l , 0 < t < T

w(0, x) = 0

∂w∂x (t, l) = 0

w(t, 0) =

1, t ∈ [ti , tj ]

0, otherwiae




Numerical Algorithm:

1 First, we fix a suitable initial data and choose a l0 By the

formula, we can solve the forward problem and obtain∂u0∂x (t, 0).

2 Compare u0∂x (t, 0) with h(t) for t > T1 and choose another l1.

Goto Step 1

3 Stopping criteria:

‖∂u0

∂x(t, 0)− h(t)‖ ≤ ε, for t > T1




Key Point:

At each step, we just compare the numerical results with the

data for t > T1, i.e.

‖∂u0

∂x(t, 0)− h(t)‖ ≤ ε, for t > T1




Numerical results:




Future works:

1 Multi-dimensional case (fast algorithm)

2 Multi-layers cases (reduce the measurements)

3 Other applications




Part III

One Inverse Problem Example in FractionalDiffusion:




.

Consider a periodically kicked rotator whose dynamical

equation is generated by the following Hamiltonian:

H =p2

2− K cos(x)

∞∑m=−∞

δ(t −m)

Assume that m ∈ N, then it can be rewritten as:

pn+1 = pn + K sin(xn)

xn+1 = xn + pn+1

There is a critical value Kc ≈ 0.97 for this kicked rotator, and

if K ≤ Kc the phase space is regular otherwise the chaos will

appear.Cheng Jin The Mathematical Analysis of



Greene, J.M., A method for determining a stochastic transition .

1979. J. Math. Phys. 20, 1183-1201.




The following figure called stochastic sea shows the phase

space (p, x) with different initial conditions with K = 1.1.

Figure: Strange Kinetics, Nature, May 6, 1993.




This new topological object in phase space (p, x) called

cantorus has some properties

A closed curve except for an infinite number of gaps

The measure of cantorus is zero.

Particles can pass through the gaps

The particle gets trapped inside the area between the island

border and nearest cantorus.

This is why one can observe high densities of dots in the stochastic

sea.




Once the particle was trapped between the border of islands

and the nearest cantorus, the (Mean Squared Displacement) will

behavior as ⟨x2(t)− 〈x(t)〉2

⟩= Dt2µ

where µ < 1/2.

Sometimes µ > 1/2, which is corresponding to strong

enhancement of transport (ballistic or jet motion). This

phenomenon can be observed in plasma fusion and spread of

environmental pollution and etc.

To Describe these phenomenons we need turn to

fractional diffusion equation!




In order to describe the abnormal diffusion in inhomogeneous

media a new time fractional diffusion equation whose order varies

with space proposed by A.V. Chechkin using CTRW.

∂p(x , t)

∂t=

∂2

∂x2

(K (x)D

1−β(x)t p(x , t)

),

where

D1−β(x)t p(x , t) :=

1

Γ(β(x))

∂

∂t

∫ t

0

p(x , τ)

(t − τ)1−β(x)dτ




For simplicity we assume that the media are composite of two

kinds of different materials, i.e.,

K (x) =

K+, x > 0

K−, x < 0.β(x) =

β+, x > 0

β−, x < 0.

p(±∞, t) = 0 and p(x , 0) = f (x). Performing the Laplace

transformation of and considering the initial/boundary condition

gives:

p±(x , s) =

∫ +∞

−∞

exp(− |x−ξ|sv±√

K±

)s(√

K+s−v+ +√

K−s−v−) f (ξ)dξ

where v± =β±2

.




Performing the inverse Laplace transform on the will give the

analytical solution of equation:

p±(x , t) =

∫ t

0dτ

∫ +∞

−∞

1

ζ1/v±lv( τ

ζ1/v±

)(t − τ)−v−√K−

Ev+−v−,1−v−

(−

√K+

K−(t − τ)v+−v−

)f (ξ)dξ

where ζ = (√

K±/|x − ξ|)1

v± , Eα,β(z) =∞∑

k=0

zk

Γ(αk + β)is

generalized Mittag-Leffler function, and lv (x) is asymmetric Levy

stable PDF with the Levy index ν.




This solution has some properties when f (x) = δ(x):

〈x(t)〉 =( √K+tv+

Γ(1 + v+)−√

K−tv−

Γ(1 + v−)

)and MSD is⟨x2(t)−〈x(t)〉2

⟩= A+K+tβ+ + A

√K−K+t(β++β−)/2 + A−K−tβ−

where

A± =2

Γ(1 + β±)− 1

[Γ(1 + β±/2)]2

A =2

Γ(1 + β−/2)Γ(1 + β−/2)− 2

Γ(1 + (β− + β+)/2)




We use Grunwald-Letnikow formula to approximate the

fractional derivatives, that is,

D1−β(x)t p = lim

∆t→0

1

∆t1−β(x)

[ t∆t

]∑k=0

(−1)k

(1− β(x)

k

)p(x , t − k∆t)

In fact we replace p(x , t − k∆t) by p(x , t − (k − 1)∆t) to avoid

unstability in computation. And the interface condition is:

p(0+, t) = p(0−, t)

K (0+)d

dxD

1−β(0+)t p(0+, t) = K (0−)

d

dxD

1−β(0−)t p(0−, t)




Figure 1. The former one shows numerical solution of normal diffusion.

The latter one displays the numerical solution of fractional diffusion with

β+ = 1, β− = 0.8, K+ = 1.2 and K− = 1.




Proposition

The parameters can be uniquely determined by measurement at

one fixed point.

The proof is quite simple and it also indicates the numerical

method to get the parameters. For simplicity, assume that

f (x) = δ(x) and it gives

p±(x , s) =exp

(− |x |sv±√

K±

)s(√

K+s−v+ +√

K−s−v−)

where 0 < v− ≤ v+ ≤ 1/2. Without losing generality, let x0 ≥ 0

p+(x0, s) =1

s1−v+

1√K+ +

√K−sv+−v−

exp(− x0s

v+√K+

).




Then

lims→0

sµp+(x0, s) =

0, µ > 1− v+

1√K+

, µ = 1− v+.

∞, µ < 1− v+.

So by choosing different µ, we can determine v+ and K+. Then,

let s = 1, then

p+(x0, 1) =1√

K+ +√

K−exp

(− x0√

K+

)and by algebraic computation we can get

K− =

(1

p+(x0, 1)exp

(− x0√

K+

)−√

K+

)2

.




Finally we try to get the last parameter by algebra

computation

v− = 1− logs

(1√

K−p+(x0, s)exp

(− x0s

v+√K+

)−√

K+√K−

s1−v+

)

where s 6= 1. According to the above analysis, the existence and

uniqueness of parameter estimation were obtained.




In practice we usually turn to Tikhonov regularization method

to recover unknown parameters, i.e.,

β(x),K (x) = argmin ‖p(x0, s)−pε(x0, s)‖+α1‖β(x)‖+α2‖K (x)‖

where pε(x0, s) = L[pε(x0, t)], and pε(x0, t) is the data with error.

α1 and α2 are regularization parameters.




Consider the simple case of K and β, we choose

α1 = α2 = 0, and pε given by the numerical solution which

obtained before where ε = O(∆x + ∆t). The final result is

β+ = 1.0000, β− = 0.9539, K+ = 0.9702 and K− = 0.9343.

(exact value β± = K± = 1)

β+ = 0.9949, β− = 0.8425, K+ = 1.1764 and K− = 0.9144.

(exact value β+ = 1, β− = 0.8,K+ = 1.2,K− = 1.)




Thank You!!!




International Conference on Inverse Problems

April 26th – 29th, 2010, Wuhan, China

http://www.math.cuhk.edu.hk/special/icip2010/


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