Identification of the zeroth-order coefficient in a time fractional diffusion equation

Accepted Manuscript Identification of zero order coefficient in a time fractional diffusion equation

Liangliang Sun, Ting Wei

PII: DOI: Reference:

S0168-9274(16)30172-6 http://dx.doi.org/10.1016/j.apnum.2016.09.005 APNUM 3093

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Applied Numerical Mathematics

Received date: Revised date: Accepted date:

27 January 2016 18 July 2016 9 September 2016

Please cite this article in press as: L. Sun, T. Wei, Identification of zero order coefficient in a time fractional diffusion equation, Appl. Numer. Math. (2016), http://dx.doi.org/10.1016/j.apnum.2016.09.005

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Identification of zero order coefficient in a time fractional diffusion equation Liangliang Suna , Ting Weia,∗ a School

of Mathematics and Statistics, Lanzhou University, Lanzhou 730030, PR China

Abstract This paper is devoted to identify a zero order coefficient in a time-fractional diffusion equation from two boundary measurement data in one-dimensional case. The existence and uniqueness of two kind of weak solutions for the direct problem with Neumann boundary condition are proved. We provide the uniqueness for recovering the zero order coefficient and fractional order simultaneously by the Laplace transformation and Gel’fand-Levitan theory. The identification of the zero order coefficient is formulated into a variational problem by the Tikhonov regularization. The existence, stability and convergence of the solution for the variational problem are provided. We deduce an adjoint problem and then use a conjugate gradient method to solve the variational problem. Two numerical examples are provided to show the effectiveness of the proposed method. Keywords: Fractional diffusion equation; Inverse zero order coefficient; Uniqueness; conjugate gradient method

1. Introduction The fractional diffusion equation has been firstly introduced in physics by Nimatullin [26] to describe diffusions in media with fractal geometry. Adams and Gelhar [1] firstly pointed out that field data in the saturated zone of a ∗ Corresponding

author Email address: [email protected] and [email protected] (Ting Wei)

Preprint submitted to Elsevier

September 12, 2016

5

highly heterogeneous aquifer could not be simulated by the classical advectiondiffusion equation, and the data indicate slower diffusion than the classical one. This anomalous diffusion phenomena attract more attention since it shows many different aspects from the classical diffusion. After this, Zhou and Selim [43] show that the slow diffusion can be characterized by the long-tailed profile in the

10

spatial distribution of densities as the time passes. Hatano and Hatano [12] and Berkowitz et al. [2] applied the continuous-time random walk model to formulate a underground environmental problem, see also Xiong et al. [38]. Metzler and Klafter [22] demonstrated a fractional diffusion equation with respect to a non-Markovian diffusion process with a memory. Raman and Alermany [30] in-

15

vestigated continuous-time random walks on fractals, see also Sokolov et al. [33] and references therein. Ginoa et al. [8] pointed out that the fractional diffusion equation describes relaxation phenomena in complex viscoelastic materials. As we know, the researches for inverse coefficient problems of time fractional diffusion equations are in an initial stage and there are only some results on this

20

topic. Such as Cheng et al. in [4] established the uniqueness in determining fractional order α and space-dependent diffusion coefficient with Dirac delta function as an initial condition based on the Gel’fand-Levitan theory. Li et al. in [18] discussed the uniqueness in determining fractional order α and diffusion coefficient with smooth initial function on the basis of the inverse eigenvalue

25

problem. Zhang in [42] consider an inverse time-dependent diffusion coefficient problem. For other inverse problems, Sakamoto and Yamamoto in [31] considered the stability for the backward problem and an inverse source problem for fractional diffusion equations. Xu et al. in [39] investigated a conditional stability estimate for the Cauchy problem by a Carleman estimate on a special case

30

α = 1/2 in one dimensional case. Zhang et al. [41] obtained the uniqueness in determining source term by analytic continuation and Laplace transform. For the sake of completeness, we also mention some research on theoretical analysis for the direct problem of fractional diffusion equations, refer to [31, 9, 21].

2

In this paper, we consider an inverse zero order coefficient problem ⎧ ⎪ ⎪ ∂ α u(x, t) = uxx (x, t) − p(x)u(x, t) + f (x, t), x ∈ (0, 1), 0 < t ≤ T, ⎪ ⎨ 0+ u(x, 0) = g(x), x ∈ (0, 1), ⎪ ⎪ ⎪ ⎩ u (0, t) = u (1, t) = 0, 0 < t ≤ T, x x

(1.1) α u(x, t) is the Caputo fractional left-sided derivative where 0 < α < 1 and ∂0+

defined by α u(x, t) = ∂0+

1 Γ(1 − α)



t 0

(t − s)−α

∂u (x, s)ds, ∂s

see, e.g., Kilbas et al. [15] and Podlubny [28] for the definition and properties 35

of the Caputo derivative. The inverse zero order coefficient problem is to determine the zero order coefficient p(x) from problem (1.1) and additional boundary data u(0, t) and u(1, t), 0 < t ≤ T . For the inverse zero order coefficient problem, Jin et al. [14] obtained a uniqueness result in determining zero order term coefficient p(x)

40

from the flux measurements in one dimension with the Dirichlet boundary conditions, and some discussions on the inverse potential problem in [13]. In [40], Yamamoto and Zhang gave a conditional stability estimate in determining a zeroth-order coefficient in a half-order fractional diffusion equation by a Carleman estimate. Miller et al. [23] discussed an inverse problem of determining

45

zero order term coefficient p(x) and fractional order α from the internal data. Tuan [34] proved the uniqueness of zero order term coefficient by taking suitable initial distributions only finitely many measurements on the boundary. In this paper, we focus on the uniqueness for determining fractional order α and zero order term coefficient p(x) simultaneously by observation measure-

50

ments u(0, t) and u(1, t), 0 < t ≤ T for a general order 0 < α < 1 and propose a numerical method to find the zero-order coefficient. The remainder of this paper is organized as follows. In Section 2, we present some preliminaries used in Section 3 and Section 4. The existence and uniqueness for the direct problem are then proved in Section 3. In Section 4, we

55

present the uniqueness for inverse zero order coefficient problem. We formulate

3

the inverse problem into a variational problem and show the existence, stability, convergence of solution for the variational problem. Then we deduce the gradient of the regularized functional and present the conjugate gradients algorithm in Section 5. Numerical results for two examples are investigated in Section 6. 60

Finally, we give a conclusion in Section 7 and an appendix for the proof of some theorems.

2. Preliminary Let AC[0, T ] be the space of functions f which are absolutely continuous on [0, T ]. Throughout this paper, we use the following definitions and propositions 65

in [15, 28]. Definition 2.1. The Mittag-Leffler function is ∞ 

Eα,β (z) =

k=0

zk , Γ(αk + β)

z ∈ C,

(2.1)

where α > 0 and β > 0 are arbitrary constants. Definition 2.2. If f (t) ∈ L(0, T ), then for α > 0 the Riemann-Liouville fracα f and right-sided integral ITα− f are defined by tional left-sided integral I0+  t f (s)ds 1 α f (t) = , 0 < t ≤ T, (2.2) I0+ Γ(α) 0 (t − s)1−α

and ITα− f (t) =



1 Γ(α)

T t

f (s)ds , (s − t)1−α

0 ≤ t < T.

(2.3)

Definition 2.3. Let z(t) ∈ AC(0, T ), then for 0 < α < 1 the Caputo fractional α left-sided derivative ∂0+ y(t) and right-sided derivative ∂Tα− y(t) of order α are

defined by α ∂0+ y(t) =

1 Γ(1 − α)



t 0



y (s)ds 1−α  =: (I0+ y )(t), (t − s)α

0 < t ≤ T,

(2.4)

0 ≤ t < T.

(2.5)

and ∂Tα− y(t) = −

1 Γ(1 − α)



T t



y (s)ds  =: −(IT1−α − y )(t), (s − t)α 4

And for 0 < α < 1 the Riemann-Liouville fractional left-sided derivative of order α is defined by α y(t) = D0+

d 1 Γ(1 − α) dt



t

0

d 1−α y(s) ds =: (I0+ y)(t), α (t − s) dt

0 < t ≤ T.

(2.6)

Proposition 2.4. Let 0 < α < 2 and β ∈ R be arbitrary. We suppose that μ is such that πα/2 < μ < min{π, πα}. Then there exists a constant c = c(α, β, μ) > 0 such that | Eα,β (z) |≤

c , 1+ | z |

μ ≤| arg(z) |≤ π.

(2.7)

Proposition 2.5. Let 0 < α < 1 and λ > 0, then we have d Eα,1 (−λtα ) = −λtα−1 Eα,α (−λtα ), dt

t > 0.

(2.8)

Proposition 2.6. Let 0 < α < 1 and λ > 0, then we have ∂tα Eα,1 (−λtα ) = −λEα,1 (−λtα ),

t > 0.

(2.9)

Proposition 2.7. (See [29])For 0 < α < 1, t > 0, we have 0 < Eα,1 (−t) < 1. Moreover, Eα,1 (−t) is completely monotonic that is (−1)n

dn Eα,1 (−t) ≥ 0, dtn

∀n ∈ N.

(2.10)

Proposition 2.8. ([29])For 0 < α < 1, η > 0, we have 0 ≤ Eα,α (−η) ≤ 70

1 Γ(α) .

Moreover, Eα,α (−η) is a monotonic decreasing function with η > 0. Proposition 2.9. Let y(t) ∈ AC[0, T ]. Then the Caputo fractional derivaα α tive ∂0+ y(t) and the Riemann-Liouville fractional derivative D0+ y(t) exist al-

most everywhere on [0, T ], there is a relationship between the Caputo fractional derivative and the Riemann-Liouville fractional derivative α α y(t) = D0+ y(t) − ∂0+

y(0) −α t , Γ(1 − α)

a.e. t ∈ (0, T ].

(2.11)

Proposition 2.10. (See [37]) For 0 < α < 1 and λ > 0, if q(t) ∈ AC[0, T ], we have 

t

q(τ )(t − τ )α−1 Eα,α (−λ(t − τ )α )dτ  t = q(t) − λ q(τ )(t − τ )α−1 Eα,α (−λ(t − τ )α )dτ, ∂tα

(2.12)

0

0

5

0 < t ≤ T.

In particular, if λ = 0, we have  t ∂tα q(τ )(t − τ )α−1 dτ = Γ(α)q(t), 0

0 < t ≤ T.

(2.13)

Lemma 2.11. (See [35]) For 0 < α < 1, suppose u(t), v(t) ∈ AC[0, T ], then we have  0

T

1−α α D0+ u(t)v(t)dt = (I0+ u)(T )v(T ) +



T 0

u(t)∂Tα− v(t)dt.

(2.14)

3. Existence and uniqueness of a weak solution for the direct problem In this section, we firstly consider the following direct problem for a timefractional diffusion equation in a general bounded domain in the multi-dimensional space. ⎧ ⎪ ⎪ ∂ α u(x, t) = Δu(x, t) − p(x)u(x, t) + f (x, t), ⎪ ⎨ 0+ ⎪ ⎪ ⎪ ⎩ 75

u(x, 0) = g(x), ∂u ∂ν (x, t)

= 0,

x ∈ Ω, 0 < t ≤ T,

x ∈ Ω,

(3.1)

x ∈ ∂Ω, 0 < t ≤ T,

where Ω ⊂ Rd is a bounded domain with sufficiently smooth boundary ∂Ω, and ν is the outward unit normal vector of the boundary ∂Ω. ¯ and p(x) ≥ p > 0 on Ω. In order to define a weak Suppose p(x) ∈ C(Ω) solution to (3.1), we firstly define an operator Ap in H 2 (Ω) by ⎧ ⎨ A ψ = −Δψ + p(x)ψ, x ∈ Ω, p ⎩ ψ ∈ D(A ) = {ψ ∈ H 2 (Ω); dψ = 0, x ∈ ∂Ω}. p dν Let {λk , φk }∞ k=1 be an eigensystem of Ap . We know 0 < λ1 < λ2 ≤ · · · , lim λk = k→∞ ∞  φk 2 ⊂ H (Ω) forms the standard or∞, Ap φk = λk φk , and ϕk = φk  2 L (Ω)

k=1

thogonal basis of L2 (Ω). Especially, for one-dimensional situation, it is well 80

known 0 < λ1 < λ2 < · · · . (See Levitan[16]) Henceforth (·, ·) denotes the scalar product in L2 (Ω). Now we define a fractional power Aγp of Ap with γ ≥ 0 by [27]

 ∞  γ 2 2γ 2 λn |(ψ, ϕn )| < ∞ . D(Ap ) = ψ ∈ L (Ω); n=1

6

Then D(Aγp ) is a Hilbert space with the norm

ψD(Aγp ) =

∞ 

 12 2 λ2γ n |(ψ, ϕn )|

.

n=1

Note that D(Aγp ) ⊂ L2 (Ω), thus D(Aγp ) ⊂ L2 (Ω) ⊂ D(Aγp ) . Here D(Aγp ) γ  −γ denotes the dual space of D(Aγp ). Set D(A−γ p ) = D(Ap ) . For f ∈ D(Ap ) and

ψ ∈ D(Aγp ), we denote the dual scalar product by

−γ f, ψ γ .

And D(A−γ p ) is

also a Hilbert space with the norm

f D(A−γ = p )

∞ 

 12 2 λ−2γ n | f, ϕn |

.

n=1

Now we can define a weak solution to system (3.1) as follows. Definition 3.1. We call that u(x, t) is a weak solution to (3.1) if the following conditions hold u ∈ L2 (0, T ; D(Aγp )) ∩ C([0, T ]; L2 (Ω)),

(3.2)

α ∂0+ u, Ap u ∈ L2 (0, T ; D(Aγ−1 )), p

(3.3)

lim u(·, t) − gL2 (Ω) = 0,

(3.4)

t→0

α γ−1 ∂0+ u, ψ 1−γ

+ (u(·, t), Ap ψ) = (f (·, t), ψ), 0 < t ≤ T, ∀ψ ∈ D(A(3.5) p ).

Lemma 3.2. Assume f ∈ L∞ (0, T ; L2 (Ω)), 0 < α < 1, denote  t (f (·, τ ), ϕn )(t − τ )α−1 Eα,α (−λn (t − τ )α )dτ, t ∈ (0, T ], n = 1, 2, · · · , fn (t) = 0

and fn (0) = 0, then fn (t) ∈ C[0, T ]. Proof. For 0 < t < T , from Proposition 2.5, we have  t |fn (t)| = | (f (·, τ ), ϕn )(t − τ )α−1 Eα,α (−λn (t − τ )α )dτ | 0

≤ f L∞ (0,T ;L2 (Ω)) 1/λn [1 − Eα,1 (−λn tα )] → 0, t → 0+ , 85

hence fn is continuous at t = 0.

7

(3.6)

By Propositions 2.7 and 2.5, it is easy to see sα−1 Eα,α (−λn sα ) is monotonically decreasing in (0, ∞). For any t, t + h ∈ (0, T ], if h > 0 we have |fn (t + h) − fn (t)|  t (f (·, τ ), ϕn ) (t + h − τ )α−1 Eα,α (−λn (t + h − τ )α ) ≤ | 0

−(t − τ )α−1 Eα,α (−λn (t − τ )α ) dτ |    t+h    α−1 α + (f (·, τ ), ϕn )(t + h − τ ) Eα,α (−λn (t + h − τ ) ) dτ   t   t ≤ f L∞ (0,T ;L2 (Ω)) |(t + h − τ )α−1 Eα,α (−λn (t + h − τ )α ) 0

−(t − τ )α−1 Eα,α (−λn (t − τ )α ) |dτ + f L∞ (0,T ;L2 (Ω)) 1/λn [1 − Eα,1 (−λn hα )] =

f L∞ (0,T ;L2 (Ω)) [1 − Eα,1 (−λn tα ) − Eα,1 (−λn hα ) λn +Eα,1 (−λn (t + h)α ) + 1 − Eα,1 (−λn hα )].

It is clear we have lim+ fn (t + h) = fn (t). By the similar deduction, we have h→0

lim− fn (t + h) = fn (t). Therefore fn ∈ C[0, T ]. This completes the proof.

h→0 90

Now we introduce a space AC([0, T ]; L2 (Ω)), which means that u(·, t)L2 (Ω) is absolutely continuous of t on the interval [0, T ] for u ∈ AC([0, T ]; L2 (Ω)). Theorem 3.3. Assume f ∈ AC([0, T ]; L2 (Ω)), g(x) ∈ D(Aγp ) for 0 ≤ γ ≤ ¯ Then there exists a unique weak solution to (3.1) and 1, d < 4 and p ∈ C(Ω). the solution is given by u(x, t) =

∞ 

Eα,1 (−λn tα )(g, ϕn )ϕn (x) +

n=1

where fn (t) =

t 0

∞ 

fn (t)ϕn (x),

(3.7)

n=1

(f (·, τ ), ϕn )(t − τ )α−1 Eα,α (−λn (t − τ )α )dτ . Moreover, we have √

√ 2T gD(Aγp ) + 2λγ−1 f L2 ((0,T )×Ω) , 1  √ uC([0,T ];L2 (Ω)) ≤ 2gL2 (Ω) + C1 f L∞ (0,T ;L2 (Ω)) , √ √ α uL2 (0,T ;D(Aγ−1 2T gD(Aγp ) + 2 2λγ−1 f L2 ((0,T )×Ω) . ∂0+ 1 )) ≤ p

uL2 (0,T ;D(Aγp )) ≤

Proof.

Based on Propositions 2.6 and 2.10, by the separation of variables, we

can obtain a formal solution for direct problem (3.1) as (3.7). 8

95

(1) Verification of u ∈ L2 (0, T ; D(Aγp )) ∩ C([0, T ]; L2 (Ω)). By Proposition 2.5, we have  η 1 tα−1 Eα,α (−λn tα )dt = [1 − Eα,1 (−λn η α )], η > 0. λ n 0 By (3.8) and the Young inequality for the convolution, we obtain   fn (t)2L2 (0,T )

T

≤ ≤

0

1 λ2n

T

2

|(f (·, t), ϕn )| dt 

T 0

0

|t

α−1

(3.8)

2 α

Eα,α (−λn t )|dt

(3.9)



2

|(f (·, t), ϕn )| dt .

Therefore, noting that 0 ≤ γ ≤ 1, one have

≤

u2L2 (0,T ;D(Aγp ))  T  ∞ ∞   α 2 2 2γ−2 2 λ2γ |E (−λ t )| dt|(g, ϕ )| + 2 λ α,1 n n n n n=1

≤

0

2T g2D(Aγp )

n=1

+

T 0

|(f (·, t), ϕn )|2 dt

2λ2γ−2 f 2L2 ((0,T )×Ω) . 1 2

Let us fix t ∈ [0, T ]. From (3.6) and λn ∼ n d (see [5]), combining d < 4, we have u(·, t)2L2 (Ω)

≤ 2

∞ 

|Eα,1 (−λn (t)α )|2 |(g, ϕn )|2 + 2

n=1

∞ 

|fn (t)|2

n=1

∞  1 ≤ 2g2L2 (Ω) + 2f 2L∞ (0,T ;L2 (Ω)) λ2 n=1 n

≤ 2g2L2 (Ω) + C1 f 2L∞ (0,T ;L2 (Ω)) . From the proof of Lemma 3.2 and Proposition 2.7, we have |fn (t + h) − fn (t)| ≤ 6

f L∞ (0,T ;L2 (Ω)) . λn

(3.10)

Combining (3.10), we have u(·, t + h) − u(·, t)2L2 (Ω)

≤

2

∞ 

|Eα,1 (−λn (t + h)α ) − Eα,1 (−λn (t)α )|2 |(g, ϕn )|2

n=1 ∞ 

+2

|fn (t + h) − fn (t)|2

n=1

≤

8g2L2 (Ω) + 72f 2L∞ (0,T ;L2 (Ω))

9

∞  1 < ∞. 2 λ n=1 n

Thus, using the Lebesgue convergence theorem, we have lim u(·, t + h) − u(·, t))L2 (Ω) = 0.

(3.11)

h→0 100

Therefore, u ∈ L2 (0, T ; D(Aγp )) ∩ C([0, T ]; L2 (Ω)). α (2) Verification of ∂0+ u, Ap u ∈ L2 (0, T ; D(Aγ−1 )). p

By Proposition 2.10, we have  t α (f (·, τ ), ϕn )(t − τ )α−1 Eα,α (−λn (t − τ )α )dτ ∂0+ 0  t = (f (·, τ ), ϕn ) − λn (f (·, τ ), ϕn )(t − τ )α−1 Eα,α (−λn (t − τ )α )dτ, 0

(3.12) 0 < t ≤ T.

By (3.8), (3.12) and the Young inequality for the convolution, we have 2   t   α α−1 α  ∂0+ (f (·, τ ), ϕ )(t − τ ) E (−λ (t − τ ) )dτ n α,α n   0

 ≤

2

≤

4



T 0 T 0



2

|(f (·, t), ϕn )| dt + 2

T 0

L2 (0,T )



2

|(f (·, t), ϕn )| dt

T 0

|λn t

α−1

2 α

Eα,α (−λn t )|dt

|(f (·, t), ϕn )|2 dt.

Hence, α u2L2 (0,T ;D(Aγ−1 )) ∂0+ p

≤

2

∞ 

λ2γ−2 λ2n |(g, ϕn )|2 n

n=1 ∞ 

+2

λ2γ−2 n



n=1

105

T 0

2T g2D(Aγp ) + 8λ2γ−2 1

≤

2T g2D(Aγp )

Since Ap u = have

≤

n=1

0

|Eα,1 (−λn τ α )|2 dτ

0

∞  T 

|(f (·, t), ϕn )|2 dt

n=1 0 2γ−2 8λ1 f 2L2 ((0,T )×Ω) .

λn Eα,1 (−λn tα )(g, ϕn )ϕn +

∞  n=1

λn fn (t)ϕn , combining (3.9), one

Ap u2L2 (0,T ;D(Aγ−1 )) p  T  ∞ ∞   α 2 2 2γ−2 2 λ2γ |E (−λ t )| dt|(g, ϕ )| + 2 λ α,1 n n n n n=1

≤

∞ 

T

 2  t   α α−1 α  dt ∂0+ (f (·, τ ), ϕ )(t − τ ) E (−λ (t − τ ) )dτ n α,α n  

≤

+



0

2T g2D(Aγp )

n=1

+

2λ2γ−2 f 2L2 ((0,T )×Ω) . 1 10

T 0

|(f (·, t), ϕn )|2 dt

α Therefore, ∂0+ u, Ap u ∈ L2 (0, T ; D(Aγ−1 )). p

(3) Verification of (3.4) and (3.5). ∞  Since g(x) = (g, ϕn )ϕn (x) ∈ D(Aγp ), by Proposition 2.7 and (3.6), we 110

n=1

have

u(·, t) − g2L2 (Ω) ≤

2

∞ 

|Eα,1 (−λn tα ) − 1|2 |(g, ϕn )|2 + 2

n=1

≤

2g2L2 (Ω) + 2f 2L∞ (0,T ;L2 (Ω))

≤

2g2L2 (Ω)

+

∞ 

∞ 

|fn (t)|2

n=1

λ−2 n

n=1 2 C1 f L∞ (0,T ;L2 (Ω)) .

By the Lebesgue convergence theorem and Lemma 3.2, we have lim u(·, t) − g(·)L2 (Ω) = 0.

t→0

On the other hand, for ∀ ψ ∈ D(Ap ), ψ = ∂tα u(·, t), ψ =

∞ 

∞ 

(ψ, ϕn )ϕn (x), we have

n=1

(−λn )Eα,1 (−λn tα )(g, ϕn )(ψ, ϕn )

n=1 ∞ 

+

α ∂0+ fn (t)(ψ, ϕn ), 0 < t ≤ T.

n=1

(u(·, t), Ap ψ)

=

∞ 

λn Eα,1 (−λn tα )(g, ϕn )(ψ, ϕn )

n=1 ∞ 

+

λn fn (t)(ψ, ϕn ), 0 < t ≤ T.

n=1 α By Proposition 2.10, we know ∂0+ fn (t) = (f (·, t), ϕn ) − λn fn (t), then (3.5) is

satisfied. 115

Now we prove the uniqueness of the weak solution to (3.1). Under the conditions f = 0 and g = 0, we have to prove that the system (3.1) has only a trivial solution. Set vn (t) = (u(·, t), ϕn ), 0 < t ≤ T . From (3.5) and ∂tα u(·, t), ϕn = ∂tα (u(·, t), ϕn )(see e.g. Lemma A.1 in [4]), we have ⎧ ⎨ ∂ α v (t) + λ v (t) = 0, 0 < t ≤ T, n n t n ⎩ v (0) = 0. n

11

The uniqueness of the initial value problem for the fractional ODE (e.g. [15, 28]) implies vn (t) = 0 for 0 ≤ t ≤ T and n ∈ N. Since {ϕn }∞ n=1 is a complete orthogonal basis in L2 (Ω), we have u = 0 in Ω × [0, T ]. Remark 1. The condition for f in Theorem 3.3 can be weaken from AC([0, T ]; L2 (Ω) 120

to L2 (0, T ; L2 (Ω)) such that the weak solution is in 0 H α (0, T ; L2 (Ω))∩L2 (0, T ; H 2 (Ω)). For such, we should extend the definition of ∂0α+ u to a generalized one, see [10]. . Remark 2. In Theorem 3.3, if we strict g(x) ∈ D(Aγp ) with 1

1 2

≤ γ ≤ 1, then

2

one can obtain that ∂t u ∈ L (0, T ; L (Ω)), which means that u is absolutely 125

continuous on t in the sense of L2 norm. Proof. By Proposition 2.5 and the integration by parts, we have ∂t u(x, t)

=

− + +

∞  n=1 ∞ 

(f (·, 0), ϕn )tα−1 Eα,α (−λn tα )ϕn (x)

n=1 ∞  t  n=1



λn tα−1 Eα,α (−λn tα )(g, ϕn )ϕn (x)

0

 (fτ (·, τ ), ϕn )(t − τ )

α−1

Eα,α (−λn (t − τ ) )dτ α

ϕn (x)

I 1 + I2 + I 3 .

For the series I1 and I2 , considering Proposition 2.4 and

1 2

≤ γ ≤ 1, we know

that I1 (·, t) + I2 (·, t)2L2 (Ω) ≤

2

∞ 

|λn tα−1 Eα,α (−λn tα )|2 |(g, ϕn )|2 + 2

n=1 ∞ 

∞ 

|(f (·, 0), ϕn )tα−1 Eα,α (−λn tα )|2

n=1

≤

 1−γ α−1 2 ∞   cλn t  c2 2   √ 2 λ2γ |(g, ϕ )| + 2 (f (·, 0), ϕn )2 t2α−2 √ α 2 n n   α 2 λn t (2 λn t ) n=1 n=1

≤

c2 λ1−2γ c2 1 g2D(Aγp ) tα−2 + f 2L∞ (0,T ;L2 (Ω)) tα−2 . 2 2λ1

12

Since 0 < α < 1, thus I1 , I2 ∈ L1 (0, T ; L2 (Ω)). For the series I3 , it hold that

≤ =

I3 L1 (0,T ;L2 (Ω))  ∞  T  t     (fτ (·, τ ), ϕn )(t − τ )α−1 Eα,α (−λn (t − τ )α )dτ  dt  

n=1 ∞ 

0

0



(ft (·, t), ϕn ) ∗ tα−1 Eα,α (−λn tα ) L1 (0,T ) .

n=1 130

Using Young’s inequality for convolutions and (3.8), we have

≤ ≤

I3 L1 (0,T ;L2 (Ω)) ∞  (ft (·, t), ϕn )L1 (0,T ) · tα−1 Eα,α (−λn tα )L1 (0,T ) n=1 ∞  n=1

 ≤ ≤

T 0

λ−1 n (ft (·, t), ϕn )L1 (0,T ) = 

∞ 

 12  |(ft (·, t), ϕn )|

2

n=1



∞ T 

0

∞ 

λ−1 n |(ft (·, t), ϕn )|dt

n=1

 12

λ−2 n

dt

n=1

Cft L1 (0,T ;L2 (Ω)) .

Therefore, ∂t u ∈ L1 (0, T ; L2 (Ω)). 4. The uniqueness of the inverse problem in one-dimensional case Theorem 4.1. Assume p(x), q(x) ∈ C[0, 1], p, q ≥ p > 0 on [0, 1], α, β ∈ (0, 1), g(x) ∈ H 3 (0, 1) such that g  (0) = g  (1) = 0, and f ≡ 0. Let u be the weak solution to (1.1), and v be the weak solution to the following equation β ∂0+ v(x, t) = vxx (x, t) − q(x)v(x, t),

x ∈ (0, 1), 0 < t ≤ T,

with the same initial and boundary conditions in (1.1). Denote φn and ψn as the eigenfunctions for the operators Ap in D(Ap ) and Aq in D(Aq ) satisfying φn (0) = ψn (0) = 1, respectively. Suppose g(x) satisfy g  (0) = 0, g(0) = 0,

(4.1)

(g, φn ) = 0, (g, ψn ) = 0, n = 1, 2, · · · .

(4.2)

and

13

Then u(0, t) = v(0, t) and u(1, t) = v(1, t), 0 < t ≤ T0 ≤ T implies α = β and p(x) = q(x) on [0, 1]. 135

Proof. From Theorem 3.3, the weak solutions u and v are given by u(x, t) =

∞ 

ρn Eα,1 (−λn tα )(g, φn )φn (x),

n=1

v(x, t) =

∞ 

σn Eβ,1 (−μn tβ )(g, ψn )ψn (x),

n=1

where φn and ψn are eigenfunctions of Ap and Aq corresponding to eigenvalues λn and μn satisfying φn (0) = ψn (0) = 1, respectively, and ρn = φn −2 L2 (0,1) , σn = ψn −2 L2 (0,1) , note that ρn = c0 + o(1), σn = c0 + o(1)(see [17]). Next we prove the uniform convergence of series. By the Sobolev embedding theorem, we have φn C[0,1] ≤ C2 φn 

1

H 2 +2ε (0,1)

,

(4.3)

with sufficiently small ε > 0. Moreover, we have φn 

1

1

H 2 +2ε (0,1)

≤ C3 Ap4

+ε

1

φn L2 (0,1) = C3 λn4

+ε

φn L2 (0,1) .

(4.4)

Let t0 > 0 be arbitrarily fixed. By Proposition 2.4 and the Cauchy-Schwarz 140

inequality, for t ≥ t0 , we have ∞ 

≤ ≤ ≤

n=1 ∞ 

max |ρn Eα,1 (−λn tα )(g, φn )φn (x)|

x∈[0,1]

cρn (g, φn )φn C[0,1] 1 + λ n tα n=1

∞ 1 cC2 C3  √ +ε−1 ρn |(g, φn )|λn4 α t0 n=1 ∞ 1  − 3 +2 2 C gL2 (0,1) λn 2 . tα 0 n=1

By λn ∼ n2 (See [5]), the series above is uniform convergent on [0, 1] × [t0 , ∞). By the Weierstrass theorem, the above series is analytic in t > 0. From u(0, t) = v(0, t), u(1, t) = v(1, t), 0 < t ≤ T0 , combining analytic extension method, we

14

have ∞  n=1 ∞ 

ρn Eα,1 (−λn tα )(g, φn ) =

∞ 

σn Eβ,1 (−μn tβ )(g, ψn ), t > 0,

n=1

ρn Eα,1 (−λn tα )(g, φn )φn (1) =

n=1

∞ 

(4.5)

σn Eβ,1 (−μn tβ )(g, ψn )ψn (1), t > (4.6) 0.

n=1

145

From (2.1), we have ∞

Eα,1 (−λn tα )

 (−λn )k tα(k−2) λn t α + t2α Γ(α + 1) Γ(αk + 1)

=

1−

=

λn tα + t2α λ2n Eα,2α+1 (−λn tα ). 1− Γ(α + 1)

k=2

We know that 3   D(A3/2 p ) ⊃ {w ∈ H (0, 1); w (0) = w (1) = 0}.

(4.7)

By the Cauchy-Schwarz inequality, we have |

∞ 

ρn λn (g, φn )|

≤

n=1

n=1

ρn |λn | |(g, φn )|

2

∞ 

 12 ρn λ−1 n

n=1

≤

CgD(A3/2 )

≤

CgH 3 (0,1) .

p

(4.8)

ρn (g, φn ) is convergent uniformly on [0, 1]. Since H 3 (0, 1) →

C 2 [0, 1], thus

∞  n=1

150

 12

3

n=1

∞ 

Similarly,

∞ 

ρn (g, φn ) =

∞ 

σn (g, ψn ) = g(0).

n=1

Thus (4.5) implies − =

∞ ∞   tα ρn λn (g, φn ) + t2α ρn (g, φn )λ2n Eα,2α+1 (−λn tα ) Γ(α + 1) n=1 n=1

∞ ∞   tβ 2β − σn μn (g, ψn ) + t σn (g, ψn )μ2n Eβ,2β+1 (−μn tβ ). (4.9) Γ(β + 1) n=1 n=1

15

It is not hard to prove that  1− 1   λ α tα− 1   n  <∞ sup  α   1 + λ t n n∈N,t≥0 with sufficiently small 1 > 0 such that

1 α

< 1/4. From Proposition 2.4, we

have 2α

|t

∞ 

ρn λ2n (g, φn )Eα,2α+1 (−λn tα )|

∞ 

1−

1

≤

λn α tα− 1 ct (g, φn ) 1 + λn tα n=1

∞  12 21  −1+ α Ctα+ 1 gD(A3/2 ) ρn λ n

≤

CgH 3 (0,1) tα+ 1 .

≤

n=1

α+ 1

1+ ρn λ n

1 α

p

n=1

Thus, we have |t2α

∞ 

ρn λ2n (g, φn )Eα,2α+1 (−λn tα )| = O(tα+ 1 ), t → 0,

n=1

with sufficiently small 1 > 0. Similarly, we have |t2β

∞ 

σn μ2n (g, ψn )Eβ,2β+1 (−μn tβ )| = O(tβ+ 1 ), t → 0,

n=1

From (4.7), g ∈ D(Ap ). Combing (4.8), (4.9) yields −

tα tβ Ap (g)(0) + O(tα+ 1 ) = − Aq (g)(0) + O(tβ+ 1 ), t → 0, Γ(α + 1) Γ(β + 1)

with some 1 > 0. Here Ap (g)(0) = −g  (0) + p(0)g(0) = p(0)g(0) = 0 by g  (0) = 0. We use proof by contradiction. Let α > β. Then dividing by tβ yields −

1 tα−β Ap (g)(0) + O(tα−β+ 1 ) = − Aq (g)(0) + O(t 1 ), t → 0. Γ(α + 1) Γ(β + 1)

Let t → 0, then

1 Γ(β+1) Aq (g)(0)

= q(0)g(0) = 0. This is impossible from (4.1).

Similarly, β > α is also impossible. Therefore, α = β. (4.5) and (4.6) turn into ∞  n=1 ∞  n=1

α

ρn Eα,1 (−λn t )(g, φn ) =

∞ 

σn Eα,1 (−μn tα )(g, ψn ), t > 0,

n=1

ρn Eα,1 (−λn tα )(g, φn )φn (1) =

∞  n=1

16

(4.10)

σn Eα,1 (−μn tα )(g, ψn )ψn (1), t(4.11) > 0.

155

Next we take the Laplace transform for Eα,1 (−λn tα ):  ∞ z α−1 e−zt Eα,1 (−λn tα )dt = α , Re z > 0. z + λn 0

(4.12)

By Proposition 2.4, ρn = o(1), and λn ∼ n2 , combining the Cauchy-Schwarz inequality, one have |e−tRe

z

∞  n=1

ρn Eα,1 (−λn tα )(g, φn )|

≤ ce−tRe z gL2 (0,1) {

n=1

≤ C4 e−tRe and e−tRe

z 1 tα

∞ 

z 1 tα ,

1 ρn 2 (1+λn tα )2 }

t > 0,

is integrable in t ∈ (0, ∞) for fixed z satisfying Re z > 0. By the

Lebesgue convergence theorem, we have 

∞

e−zt

0

∞ 

ρn Eα,1 (−λn tα )(g, φn )dt =

n=1

∞ 

z α−1 , Re z > 0. + λn

ρn (g, φn )

zα

σn (g, ψn )

zα

n=1

Similarly, 

∞ 0

e−zt

∞ 

σn Eα,1 (−μn tα )(g, ψn )dt =

n=1

∞  n=1

z α−1 , Re z > 0. + μn

Hence (4.10) yields ∞ ∞  ρn (g, φn )  σn (g, ψn ) = , Re z > 0. z α + λn z α + μn n=1 n=1

That is

∞ ∞  ρn (g, φn )  σn (g, ψn ) = , Re ξ > 0. ξ + λn ξ + μn n=1 n=1

(4.13)

From g ∈ L2 (0, 1) and the Cauchy-Schwarz inequality, one obtain ∞  ρn (g, φn ) ≤ gL2 (0,1) ξ + λn n=1



∞ 

ρn (ξ + λn )2 n=1

 12 .

As λn ∼ n2 and ρn = o(1), we can see that both sides of (4.13) are internally closed uniform convergence in ξ ∈ C\({−λn }n≥1 ∪ {−μn }n≥1 ). Therefore, employing the Weierstrass theorem, we can analytically continue both sides of (4.13) in ξ and (4.13) holds for ξ ∈ C\({−λn }n≥1 ∪ {−μn }n≥1 ).

17

160

Now we prove λ1 = μ1 . Using proof by contradiction, let λ1 < μ1 . Then we can take a suitable disk which only include −λ1 and does not include others. Integrating (4.13) in the disk, we have 2πiρ1 (g, φ1 ) = 0. This is impossible since ρ1 = 0 and (g, φ1 ) = 0. Thus λ1 = μ1 . Repeating this argument, we have λn = μn , n = 1, 2, · · · , then (4.5) and (4.6) yield ∞  n=1 ∞ 

α

ρn Eα,1 (−λn t )(g, φn ) =

∞  n=1

ρn Eα,1 (−λn tα )(g, φn )φn (1) =

n=1 165

σn Eα,1 (−λn tα )(g, ψn ), t > 0, ∞ 

(4.14)

σn Eα,1 (−λn tα )(g, ψn )ψn (1), t(4.15) > 0.

n=1

Next we will prove ρn = σn . Similarly the arguments for (4.13), first taking the Laplace transform for (4.14) and (4.15), then integrating in a suitable disk, we have ρn (g, φn ) = σn (g, ψn ),

(4.16)

ρn (g, φn )φn (1) = σn (g, ψn )ψn (1).

(4.17)

By (4.2), (4.16) and (4.17), we have φn (1) = ψn (1).

(4.18)

Therefore, ρn = σn , n ≥ 1 (See e.g. Theorem 1 of Murayama [25] pp. 318). By 170

means of Gel’fand-Levitan theory [7], we obtain p(x) = q(x) on x ∈ [0, 1].

5. Variational method and the conjugate gradient method In this section, we solve numerically the zero-order coefficient p(x) by problem (1.1) and the additional boundary conditions u(0, t) = h0 (t), u(1, t) = h1 (t). The inverse coefficient problem is formulated into a variational problem by using the Tikhhonov regularization. Then the existence, stability and convergence of minimizer for the variational problem are provided. We deduce the

18

175

gradient of functional by means of a sensitivity problem and an adjoint problem, and then using a conjugate gradient method to solve the variational problem. In the following, we assume the zero-order coefficient p(x) ∈ L∞ (0, 1) and the initial data g(x) ∈ H 2 (0, 1) and the source function f (x, t) ∈ L2 , we give a weak formulation for direct problem (1.1) and prove that its solution exists

180

uniquely. Let u(x, t) be the smooth solution of (1.1). Denote v(x, t) = u(x, t) − g(x), then v satisfy the following problem ⎧ α ⎪ ¯ ⎪ ⎪ ∂0+ v(x, t) − vxx (x, t) + p(x)v(x, t) = f (x, t), ⎨ v(x, 0) = 0, x ∈ (0, 1), ⎪ ⎪ ⎪ ⎩ v (0, t) = a(t), v (1, t) = b(t), 0 < t ≤ T, x

x ∈ (0, 1), 0 < t ≤ T,

x

(5.1) where f¯(x, t) = f (x, t) + g  (x) − p(x)g(x), a(t) = −g  (0) and b(t) = −g  (1). Denote Q = (0, 1) × (0, T ), I = (0, T ), Λ = (0, 1) and for any s > 0 define a space B s (Q) := 0 H s (0, T ; L2 (0, 1)) ∩ L2 (0, T ; H 1 (0, 1))

(5.2)

equipped with the norm 1/2  . uB s (Q) = u2H s (0,T ;L2 (0,1)) + u2L2 (0,T ;H 1 (0,1))

(5.3)

Refer to [19], we know that B s (Q) is a Hilbert space. Based on [19, 36], we can deduce a weak formulation for problem (5.1) as follow. α

Find v ∈ B 2 (Q) such that A(v, w) = F (w),

α

∀w ∈ B 2 (Q),

(5.4)

where α/2

α/2

A(v, w) := (D0+ v, DT − w)Q + (vx , wx )Q + (pv, w)Q , F (w) := (f¯, w)Q − (a(t), w(0, t))I + (b(t), w(1, t))I , 185

where (·, ·)Q and (·, ·)I are the inner products in L2 (Q) and L2 (I), respectively. Similar to the proofs in [19, 36], we could obtain the following theorem. 19

Theorem 5.1. Assume that f ∈ L2 (Q), g ∈ H 2 (0, 1), p ∈ L∞ (0, 1) and p(x) ≥ p > 0. Then there exists a unique solution v(x, t) to (5.4) and the solution satisfies vB α2 (Q) ≤ C(α, T, p)(f L2 (Q) + gH 2 (0,1) + pL∞ (0,1) gL2 (0,1) ),

(5.5)

where C(α, T, p) > 0 is a constant independent of v. Define a forward operator F : p(x) ∈ D(F) → (up (0, t), up (1, t)) ∈ L2 (0, T ) × L2 (0, T ),

(5.6)

where D(F) = {p(x) ∈ L∞ (0, 1)| 0 < p ≤ p(x) ≤ p} and up (x, t) = vp (x, t) + α

g(x) ∈ B 2 (Q) where vp (x, t) is the solution for weak formulation (5.4) under f ∈ L2 (Q) and g ∈ H 2 (0, 1). Thus the inverse problem is formulated into solving the following abstract operator equation F(p) = h(t)  (h0 (t), h1 (t)).

(5.7)

α

From Theorem 5.1, we know that up ∈ B 2 (Q). By the trace theorem in [20], we know up (0, t), up (1, t) ∈ H 4 (0, T ) → L2 (0, T ) compactly, thus the operator α

F : D(F) → L2 (0, T ) × L2 (0, T ) is compact. The inverse zero-order coefficient is ill-posed. Take p∗ ∈ D(F). In order to ensure the stability of numerical solution, we introduce the Tikhonov regularization functional J(p)

μ 1 F(p) − hδ 2L2 (0,T )×L2 (0,T ) + p − p∗ 2L2 (0,1) 2 T 2 

2

2 1 1 T μ 1 δ up (0, t) − h0 (t) dt + up (1, t) − hδ1 (t) dt + = (p(x) − p∗ (x))2 dx, 2 0 2 0 2 0 (5.8)

=

where μ > 0 is a regularization parameter, and hδ0 , hδ1 are the noisy functions of 190

h0 , h1 . Therefore the zero order coefficient identification problem is transformed into solving a variational problem J(pδμ ) = min J(p). p∈D(F )

20

(5.9)

By using the similar proof methods in [6, 32], we can obtain the existence of minimizer pδμ and its sequence stability on hδ , and the sequence convergence is also obtained. See the following theorems and their proofs in Appendix. Theorem 5.2. There exists at least one minimizer pδμ ∈ D(F) for the varia195

tional problem (5.9). Theorem 5.3. Let μ > 0 and let {(h0k , h1k )} and {pk } be sequences where (h0k , h1k ) → (hδ0 , hδ1 ) in L2 (0, T ) × L2 (0, T ) and pk is a minimizer of (5.9) with (hδ0 , hδ1 ) replaced by (h0k , h1k ). Then there exists a convergent subsequence of {pk } and the limit of every convergent subsequence is a minimizer of (5.9).

200

Definition 5.4. p† is called p∗ -minimum-norm solution (p∗ -MNS) of (5.7), if F(p† ) = h and p† − p∗ L2 (0,1) = min{p − p∗ L2 (0,1) | F(p) = h}. Theorem 5.5. Let p† be the p∗ -minimum-norm solution of F(p) = (h0 , h1 ) and (hδ0 , hδ1 ) ∈ L2 (0, T ) × L2 (0, T ) with h0 − hδ0  + h1 − hδ1  ≤ δ and let μ(δ) satisfy μ(δ) → 0 and δ 2 /μ(δ) → 0 as δ → 0. Then every sequence pδμkk , where δk → 0, μk = μ(δk ) and pδμkk is a minimizer of (5.9) with (hδ0 , hδ1 ) replaced by (hδ0k , hδ1k ) and μ = μk , has a convergent subsequence. The limit of every convergent subsequence is a p∗ -minimum-norm solution. In addition, if p† is unique, then lim pδμ(δ) = p† .

δ→0

In the following, we focus on the numerical method for solving problem (5.9). The key task is to deduce the gradient of J. We suppose the solution up is smooth enough and we give a formal deduction. 205

Let the coefficient p(x) be perturbed by a small amount δp(x), then the forward 



solution has a small change up+δp − up = up δp + o(δp) where up is the Fr´ echet 

derivative of the forward map up . Denote ω = up δp, then we know ω satisfy Sensitive problem: ⎧ ⎪ ⎪ ∂ α ω(x, t) = ωxx − p(x)ω − δp(x)up (x, t), ⎪ ⎨ 0+ ω(x, 0) = 0, x ∈ (0, 1), ⎪ ⎪ ⎪ ⎩ ω (0, t) = ω (1, t) = 0, 0 < t ≤ T. x

x

21

(x, t) ∈ (0, 1) × (0, T ], (5.10)

From (5.8), we have δJ(p)

= J(p + δp) − J(p)  T  T (up (0, t) − hδ0 (t))ω(0, t)dt + (up (1, t) − hδ1 (t))ω(1, t)dt = 0  0 1 +μ (p − p∗ ) · δpdx + o(ω(0, t)L2 (0,T ) + ω(1, t)L2 (0,T ) + δpL∞ (0,1) ). 0

(5.11) α ω(x, t) ∂0+

Since ω(x, 0) = 0, by Proposition 2.9, we have

=

α D0+ ω(x, t).

From

(5.10) and Lemma 2.11, we have  T 1 − {p(x)ω + δp(x)up (x, t)}v(x, t)dxdt  T0 10 α = (∂0+ ω(x, t) − ωxx )v(x, t)dxdt 0 0  T 1  1 1−α I0+ ω(x, T )v(x, T )dx + ω(x, t)(∂Tα− v − vxx )(x, t)dxdt = 0 0 0 T + (ω(1, t)vx (1, t) − ω(0, t)vx (0, t)dt. 0

(5.12)

Let v(x, t) satisfy the following problem. Adjoint problem: ⎧ α ⎪ ⎪ ⎪ ∂T − v(x, t) = vxx − p(x)v, (x, t) ∈ (0, 1) × (0, T ], ⎪ ⎪ ⎪ ⎨ v(x, T ) = 0, x ∈ (0, 1), ⎪ ⎪ vx (1, t) = up (1, t) − hδ1 (t), t ∈ (0, T ], ⎪ ⎪ ⎪ ⎪ ⎩ −v (0, t) = u (0, t) − hδ (t), t ∈ (0, T ]. x p 0 From (5.13) and (5.10), (5.12) yields  T  δ ω(0, t)(up (0, t) − h0 (t))dt + 0



=−

T



0

1 0

T 0

(5.13)

ω(1, t)(up (1, t) − hδ1 (t))dt (5.14)

δp(x)up (x, t)v(x, t)dxdt.

Thus (5.11) yields  δJ(p)

=−

T 0





1 0

δp(x)up (x, t)v(x, t)dxdt + μ

1 0

(p − p∗ ) · δpdx

+ o(ω(0, t)L2 (0,T ) + ω(1, t)L2 (0,T ) + δpL∞ (0,1) ). For the sensitivity problem, using estimate (5.5), we know w(0, t)L2 (0,T ) , w(1, t)L2 (0,T ) ≤ CwB α2 (Q) ≤ CδpL∞ (0,1) up (x, t)L2 (Q) , 22

thus they are the same order infinitesimal of δp, then we know the gradient of the functional J(p) is Jp = −



T 0

up (x, t)v(x, t)dt + μ(p(x) − p∗ (x)).

(5.15)

We use the conjugate gradient method (CGM) to search the minimizer of functional J(p). Let pk be the kth approximate solution to p(x), then the iteration formula is pk+1 = pk + βk dk , k = 0, 1, 2, · · · ,

(5.16)

where βk is the step size and dk is a descent direction in the k-th iteration. The conjugate gradient method use the following iteration formula to compute the descent direction dk = −Jp k + γk dk−1 ,

(5.17)

where γk is conjugate coefficient calculated by  (Jp )2 dx  γk = Ω  k 2 , γ0 = 0. (J ) dx Ω pk−1 210

(5.18)

The step size βk can be obtained approximately in the following deduction. From (5.8), we have

≈

(5.19) J(pk + βk dk )  T 1 (upk (0, t) + βk ωk (0, t) − hδ0 (t))2 dt 2 0   μ 1 1 T (upk (1, t) + βk ωk (1, t) − hδ1 (t))2 dt + (pk + βk dk − p∗ )2 dx, + 2 0 2 0

where ωk is the solution of (5.10) with δp = dk . Let  T dJ ≈ (upk (0, t) + βk ωk (0, t) − hδ0 (t))ωk (0, t)dt dβk 0  1  T δ (upk (1, t) + βk ωk (1, t) − h1 (t))ωk (1, t)dt + μ (pk + βk dk − p∗ )dk dx + 0

=

0

0,

we can get a step size βk by T T 1 (upk (0, t) − hδ0 (t))ωk (0, t)dt + 0 (upk (1, t) − hδ1 (t))ωk (1, t)dt + μ 0 (pk − p∗ )dk dx . βk = − 0 T 2 T 2 1 2 ω (0, t)dt + ω (1, t)dt + μ d dx k k k 0 0 0 (5.20) 23

Therefore, we have the following CGM to solve the variational problem (5.9). (a) Initialize p0 = p, and set k = 0, d0 = −Jp 0 ; 215

(b) Solve the direct problem (1.1) with p = pk , and determine the residual upk (0, t) − hδ0 (t) and upk (1, t) − hδ1 (t); (c) Solve the adjoint problem (5.13) and determine the gradient Jp k by (5.15); (d) Calculate the conjugate coefficient γk by (5.18) and the descent direction

220

dk by (5.17); (e) Solve the sensitivity problem (5.10) for ωk with δp = dk ; (f) Calculate the step length βk by (5.20); (g) Update the zero order term pk by (5.16). (h) Calculate the project of pk onto [p, p], i.e. taking pk = min(p, max(p, pk ));

225

(i) Increase k by one and go to Step (b), repeat the above procedure until a stopping criterion is satisfied.

6. Numerical experiments In this section, we present the numerical results for two examples in onedimensional case to show the effectiveness of the the conjugate gradient algo230

rithm. The noisy data are generated by adding random perturbations, i.e., hδi = hi + εhi · (2 rand(size(hi )) − 1), i = 0, 1, where ε is a relative noise level and rand(·) generate random numbers uniformly distributed on [0, 1]. The corresponding noise level is calculated by δ = hδ0 − h0 L2 (0,T ) + hδ1 − h1 L2 (0,T ) . To show the accuracy of numerical solution, we compute the approximate L2 error denoted by rek =

pk (x) − p(x)L2 (0,1) p(x)L2 (0,1)

,

(6.1)

where pk (x) is the coefficient term reconstructed at the kth iteration, and p(x) 235

is the exact solution. 24

The residual Ek at the kth iteration is given by Ek = upk (0, t) − hδ0 (t)L2 (0,T ) + upk (1, t) − hδ1 (t)L2 (0,T ) .

(6.2)

In an iteration algorithm, the most important work is to find a suitable stopping rule. In this study we use the well-known discrepancy principle [24], i.e. we choose k satisfying the following inequality Ek ≤ τ δ < Ek−1 ,

(6.3)

where τ > 1 is a constant and can be taken heuristically to be 1.01, as suggested by Hanke and Hansen [11]. If the noise level is 0, then we take k = 10 for Example 1 and k = 20 for Example 2. Without loss of generality, we set T = 1 and take p = 0.02, p = 10 and p∗ ≡ 240

1, which are far away from the exact solutions. In fact, in our computations, the approximate solution at each iteration is located in [p, p] such that the project step is not used really. The grid points on [0, 1] and [0, T ] are always both 101 when solving the direct problem, sensitive problem and adjoint problem by using a finite element method given in [36]. We consider the following two

245

examples. Example 1. Suppose p(x) = 1 + x2 and the boundary data u(0, t) and u(1, t) are obtained by solving the direct problem (1.1) with f (x, t) = 0, u(x, 0) = x5/2 (1 − x)3/2 by using the finite element method. The numerical results for Example 1 by using the discrepancy principle for

250

various noise levels in the case of α = 0.3, 0.7 are shown in Figures 1(a) and 1(b), respectively by taking μ = 0, and in Figures 2(a) and 2(b), respectively by taking μ = 0.01δ 3/2 . We can see that the numerical results of the zero order coefficient for Example 1 match the exact ones quite well even up to 5% noise added in the ”exact” Dirichlet data u(0, t) and u(1, t). We note that for small

255

μ, the numerical results are very similar to the results by using μ = 0. If μ is a little big, the numerical results become worse. Thus we suggest μ = 0 in our proposed algorithm. Example 2. Suppose p(x) = 1 + sin(πx) and take u(x, t) = t1+α cos(πx) + 1 + 25

2

2 exact ε=0, k=10 ε=0.01,k=6 ε=0.03,k=6 ε=0.05, k=6

1.8

1.6 k

p(x) and p (x)

1.6 p(x) and pk(x)

exact ε=0, k=10 ε=0.01,k=7 ε=0.03,k=5 ε=0.05, k=5

1.8

1.4

1.4

1.2

1.2

1

1

0.8

0

0.2

0.4

0.6

0.8

0.8

1

0

0.2

0.4

x

0.6

0.8

1

x

(a) α = 0.3

(b) α = 0.7

Figure 1: The numerical results for Example 1 for various noise levels with μ = 0.

2

2 exact ε=0, k=10 ε=0.01,k=6 ε=0.03,k=6 ε=0.05, k=6

1.8

1.6 k

p(x) and p (x)

1.6 p(x) and pk(x)

exact ε=0, k=10 ε=0.01,k=7 ε=0.03,k=5 ε=0.05, k=5

1.8

1.4

1.4

1.2

1.2

1

1

0.8

0

0.2

0.4

0.6

0.8

0.8

1

x

0

0.2

0.4

0.6

0.8

1

x

(a) α = 0.3

(b) α = 0.7

Figure 2: The numerical results for Example 1 for various noise levels with μ = 0.01δ 3/2 .

26

x7/2 (1 − x)5/2 , and the initial function and source function can be calculated 260

by the expression of u(x, t). For this example, equation (1.1) is inhomogeneous. The boundary data are u(0, t) = t1+α + 1 and u(1, t) = −t1+α + 1. The numerical results for Example 2 by using the discrepancy principle for various noise levels in the case of α = 0.3, 0.7 are shown in Figures 3(a) and 3(b), respectively by taking μ = 0, in Figures 4(a) and 4(b), respectively by

265

taking μ = 0.01δ 3/2 . We can also see that the numerical results of the zero order coefficient for Example 2 are accurate even up to 5% noise except a small interval near the end x = 1. The numerical results by small parameters μ are almost same with the results by using μ = 0. If μ are chosen little big, the numerical results become worse. We do not show them. It is observed that the numerical results for two examples are stable to noise

270

and the stopping steps are quite small which means the convergence is quick. 2

2.2

2

1.8

1.8 p(x) and p (x)

k

p(x) and pk(x)

1.6 1.6

1.4

1.4

1.2 1.2

exact ε=0, k=10 ε=0.01,k=9 ε=0.03,k=9 ε=0.05, k=9

1

0.8

0

0.2

0.4

0.6

exact ε=0, k=10 ε=0.01,k=10 ε=0.03,k=9 ε=0.05, k=9

1

0.8

0.8

1

x

0

0.2

0.4

0.6

0.8

1

x

(a) α = 0.3

(b) α = 0.7

Figure 3: The numerical results for Example 2 for various noise levels with μ = 0.

In Table 1, we show the relative numerical errors rek and the stop steps in parentheses for Example 1 with respective to different α and ε. It can be seen that the numerical results become a little worse when the relative noise levels 275

increase and are not sensitive to the fractional order α. All the stop steps are less than 10 which means the proposed method has a good convergence.

27

2

2.2

2

1.8

1.8 p(x) and p (x)

k

p(x) and pk(x)

1.6 1.6

1.4

1.4

1.2 1.2

exact ε=0, k=10 ε=0.01,k=9 ε=0.03,k=9 ε=0.05, k=9

1

0.8

0

0.2

0.4

0.6

exact ε=0, k=10 ε=0.01,k=10 ε=0.03,k=9 ε=0.05, k=9

1

0.8

0.8

1

0

0.2

x

0.4

0.6

0.8

1

x

(a) α = 0.3

(b) α = 0.7

Figure 4: The numerical results for Example 2 for various noise levels with μ = 0.01δ 3/2 .

Table 1: Numerical results for Example 1 with various α and ε.

α\

0.0025

0.0050

0.0100

0.0500

0.100

0.1500

0.1

0.0199(6)

0.0199(6)

0.0200(6)

0.0219 (6)

0.0260 (5)

0.0349(5)

0.3

0.0203(8)

0.0204(8)

0.0209(6)

0.0240 (6)

0.0257 (5)

0.0311 (5)

0.5

0.0199 (9)

0.0199 (8)

0.0208(7)

0.0227 (5)

0.0265(5)

0.0318(5)

0.7

0.0193 (8)

0.0193 (8)

0.0200 (7)

0.0246 (5)

0.0311(5)

0.0388(5)

0.9

0.0189 (7)

0.0192 (5)

0.0195(5)

0.0245 (5)

0.0346 (5)

0.0463(5)

28

7. Conclusions In this paper, we prove the existence and uniqueness of a weak solution to the fractional diffusion equation with Neumann boundary conditions. The 280

uniqueness for the inverse fractional order α and zero-order term coefficient from boundary measurements is obtained by using Laplace transform and analytic continuation as well as the Gel’fand-Levitan theory in one-dimensional situation. We propose a variational method to solve the inverse zero-order coefficient problem and use the conjugate gradient method to find the approximation of

285

the regularized solution. The existence, stability and convergence of the solution of the variational problem are provided. The numerical experiments for two numerical examples show that our proposed method is effective.

Acknowledgments This work is supported by the NSF of China (11371181) and the Fundamen290

tal Research Funds for the Central Universities (lzujbky-2013-k02).

Appendix ∗

Lemma 7.1. For any pn ∈ D(F), if pn  p in L∞ (0, 1) and F(pn )  q = (q0 (t), q1 (t)) in L2 (0, T ) × L2 (0, T ), then p ∈ D(F) and q = F(p). ∗

Proof. From pn  p in L∞ (0, 1), we know p ∈ L∞ (0, 1) and pn  p in L2 (0, 1). Since D(F) is a convex set, using the Mazur Theorem (e.g., Corollary 3.8 (pp 61) in [3]), we know that 1 pn = pk → p in L2 (0, 1). n n

(7.1)

k=1

Thus there exists a subsequence, still denoted by pn , such that pn → p, a.e. x ∈ 295

(0, 1). Note that pn ∈ D(F), we have p ∈ D(F). Denote vpn (x, t) as the solution of (5.4) with p = pn , then F(pn ) = (vpn (0, t)+ g(0), vpn (1, t) + g(1)). We only need to prove that q = (vp (0, t) + g(0), vp (1, t) +

29

g(1)) where vp (x, t) is the solution of (5.4). From Theorem 5.1, we have α/2

α/2

(D0+ vpn , DT − w)Q + ((vpn )x , wx )Q + (pn vpn , w)Q

(7.2)

= (f, w)Q + (g  , w)Q − (pn g, w)Q − (a(t), w(0, t))I + (b(t), w(1, t))I ,

α

∀w ∈ B 2 (Q)

and vpn B α2 (Q) ≤ C(α, T, p, p)(f L2 (Q) + gH 2 (0,1) ),

(7.3)

where C(α, T, p, p) > 0. Since B 2 (Q) → L2 (Q) compactly and the trace of any function in B 2 (Q) α

α

is in H α/4 (0, T ) and H α/4 (0, T ) → L2 (0, T ) compactly, see [20], there exists a α

subsequence, still denoted by vpn , such that vpn (x, t)  z(x, t) in B 2 (Q) and vpn (x, t) → z(x, t) in L2 (Q), and vpn (0, t) → z(0, t), vpn (1, t)) → z(1, t), 300

in L2 (0, T ).

(7.4)

Then we have α/2

α/2

α/2

α

α/2

(D0+ vpn , DT − w)Q → (D0+ z, DT − w)Q , ∀ w ∈ B 2 (Q), α 2

((vpn )x , wx )Q → (zx , wx )Q , ∀ w ∈ B (Q), α 2

(vpn , w)Q → (z, w)Q , ∀ w ∈ B (Q).

(7.5) (7.6) (7.7)

∗

By vpn (x, t) → z(x, t) in L2 (Q) and pn  p in L∞ (0, 1), we have (pn vpn , w)Q = (pn (vpn − z), w)Q + (pn z, w)Q → (pz, w)Q , n → ∞.

(7.8)

Combining (7.5)-(7.8) and taking the limit n → ∞ in (7.2), we obtain α/2

α/2

(D0+ z, DT − w)Q + (zx , wx )Q + (pz, w)Q

(7.9)

= (f, w)Q + (g  , w)Q − (pg, w)Q − (a(t), w(0, t))I + (b(t), w(1, t))I . Therefore z(x, t) = vp (x, t) and F(p) = (z(0, t) + g(0), z(1, t) + g(1)). By (7.4), we have F(pn ) → (z(0, t) + g(0), z(1, t) + g(1)) in L2 (0, T ) × L2 (0, T ). Note that the condition F(pn )  q = (q0 (t), q1 (t)) in L2 (0, T ) × L2 (0, T ), we know 305

q = (z(0, t) + g(0), z(1, t) + g(1)) = F(p).

30

Proof of Theorem 5.2: Since the functional J is nonnegative and D(F) is nonempty, there exists d = such that J(pn ) → d.

inf

p∈D(F )

J(p). Thus, there exist a sequence pn ∈ D(F)

Since {pn ∈ D(F)} is bounded in L∞ (0, 1) and F(pn ) is bounded in L2 (0, T )× ∗

L2 (0, T ), there exists a subsequence, still denoted by pn , such that pn  pδμ in L∞ (0, 1) and F(pn )  q in L2 × L2 . By Lemma 7.1, we know pδμ ∈ D(F), q = F(pδμ ). Moreover we have pn  pδμ in L2 (0, 1). Based on the weak lower semicontinuity of L2 -norm, we have μ δ μ pμ − p∗ 2L2 (0,1) ≤ lim inf pn − p∗ 2L2 (0,1) . n→∞ 2 2

(7.10)

and upδμ (0, t) − hδ0 (t)2L2 (0,T ) ≤ lim inf upn (0, t) − hδ0 (t)2L2 (0,T ) (7.11) n→∞

upδμ (1, t) − hδ1 (t)2L2 (0,T ) ≤ lim inf upn (1, t) − hδ1 (t)2L2 (0,T ) . (7.12) n→∞

Therefore, we have d ≤ J(pδμ ) ≤ lim inf J(pn ) = d. n→∞

310

Then

pδμ

is a minimizer.

Proof of Theorem 5.3: By the definition of pk we have F(pk ) − (h0k , h1k )2L2 (0,T )×L2 (0,T ) + μpk − p∗ 2L2 (0,1)

(7.13)

≤ F(p) − (h0k , h1k )2L2 (0,T )×L2 (0,T ) + μp − p∗ 2L2 (0,1) , ∀p ∈ D(F). Note that pk L∞ (0,1) and F(pk )L2 (0,T )×L2 (0,T ) are bounded, then there exist a subsequence {pm } of {pk } and p¯ such that ∗

pm  p¯ in L∞ (0, 1), and F(pm )  q, in L2 (0, T ) × L2 (0, T ).

(7.14)

From Lemma 7.1, we know p¯ ∈ D(F) and q = F(¯ p). By the weak lower semicontinuity of the norm we have ¯ p − p∗ L2 (0,1) ≤ lim inf pm − p∗ L2 (0,1) , F(¯ p) − (hδ0 , hδ1 )L2 (0,T )×L2 (0,T ) ≤ lim inf F(pm ) − (h0m , h1m )L2 (0,T )×L2 (0,T ) . 31

(7.15)

Moreover, by (7.13) we have F(¯ p) − (hδ0 , hδ1 )2L2 (0,T )×L2 (0,T ) + μ¯ p − p∗ 2L2 (0,1) ≤

lim inf(F(pm ) − (h0m , h1m )2L2 (0,T )×L2 (0,T ) + μpm − p∗ 2L2 (0,1) )

≤

lim sup(F(pm ) − (h0m , h1m )2L2 (0,T )×L2 (0,T ) + μpm − p∗ 2L2 (0,1) )

≤ = 315

lim (F(p) − (h0m , h1m )2L2 (0,T )×L2 (0,T ) + μp − p∗ 2L2 (0,1) )

m→∞

F(p) − (hδ0 , hδ1 )2L2 (0,T )×L2 (0,T ) + μp − p∗ 2L2 (0,1)

for all p ∈ D(F). This implies that p¯ is a minimizer of (5.8) and lim (F(pm ) − (h0m , h1m )2L2 (0,T )×L2 (0,T ) + μpm − p∗ 2L2 (0,1) )

m→∞

=

F(¯ p) − (hδ0 , hδ1 )2L2 (0,T )×L2 (0,T ) + μ¯ p − p∗ 2L2 (0,1) .

(7.16)

Now we prove that pm → p¯ in L2 (0, 1). Assume that pm  p¯. Then p − p∗  and there exists a subsequence pmk of pm c := lim sup pm − p∗  > ¯ such that pmk − p∗  → c and pmk satisfies (7.16). By (7.16), we have lim F(pmk ) − (h0mk , h1mk )2L2 (0,T )×L2 (0,T )

n→∞

=

F(¯ p) − (hδ0 , hδ1 )2L2 (0,T )×L2 (0,T ) + μ(¯ p − p∗ 2L2 (0,1) − c2 )

<

F(¯ p) − (hδ0 , hδ1 )2L2 (0,T )×L2 (0,T ) .

This is a contradiction from (7.15). 320

Proof of Theorem 5.5: By the definition of pδμkk , we arrive at F(pδμkk ) − (hδ0k , hδ1k )2 + μk pδμkk − p∗ 2 ≤

F (p† ) − (hδ0k , hδ1k )2 + μk p† − p∗ 2

= δk2 + μk p† − p∗ 2 . Therefore, let k → ∞ on both sides of above inequality, one have lim F(pδμkk ) = h = (h0 , h1 ).

k→∞

(7.17)

By pδμkk − p∗ 2 ≤ δk2 /μk + p† − p∗ 2 , we have lim sup pδμkk − p∗ 2L2 (0,1) ≤ p† − p∗ 2L2 (0,1) . k→∞

32

(7.18)

Note that {pδμkk } ⊂ D(F) is bounded, thus there exists a subsequence, still ∗

denoted by {pδμkk } and p0 ∈ L∞ (0, 1) such that pδμkk  p0 as k → ∞ in L∞ (0, 1). Using (7.17) and Theorem 7.1, we obtain p0 ∈ D(F) and F(p0 ) = (h0 , h1 ). By the definition of p† and (7.18), we have p0 − p∗ L2 (0,1)

325

≤

lim inf pδμkk − p∗ L2 (0,1) ≤ lim sup pδμkk − p∗ L2 (0,1)

≤

p† − p∗ L2 (0,1) ≤ p0 − p∗ L2 (0,1) .

k→∞

k→∞

(7.19)

This implies that p0 −p∗ L2 (0,1) = p† −p∗ L2 (0,1) . Thus p0 is also a p∗ −M N S. Now we proof that pδμkk → p0 in L2 (0, 1). From (7.19), we know limk→∞ pδμkk − p∗ L2 (0,1) = p0 − p∗ L2 (0,1) , by pδμkk − p∗  p0 − p∗ in L2 (0, 1), we have pδμkk − p∗ → p0 − p∗ in L2 (0, 1). Thus pδμkk → p0 in L2 (0, 1). If p† is unique, then, we obtain lim pδμ(δ) = p† .

δ→0

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