Three landweber iterative methods for solving the initial value problem of time fractional diffusion wave equation on spherically symmetric domain

Inverse Problems in Science and Engineering

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Three Landweber iterative methods for solving the
initial value problem of time-fractional diffusionwave equation on spherically symmetric domain
Fan Yang, Qiao-Xi Sun & Xiao-Xiao Li
To cite this article: Fan Yang, Qiao-Xi Sun & Xiao-Xiao Li (2021): Three Landweber iterative
methods for solving the initial value problem of time-fractional diffusion-wave equation
on spherically symmetric domain, Inverse Problems in Science and Engineering, DOI:
10.1080/17415977.2021.1914603
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INVERSE PROBLEMS IN SCIENCE AND ENGINEERING
/>
Three Landweber iterative methods for solving the initial
value problem of time-fractional diffusion-wave equation on
spherically symmetric domain
Fan Yang, Qiao-Xi Sun and Xiao-Xiao Li
School of Science, Lanzhou University of Technology, Lanzhou, People’s Republic of China
ABSTRACT

ARTICLE HISTORY

In this paper, the inverse problem for identifying the initial value
of time-fractional diffusion wave equation on spherically symmetric
region is considered. The exact solution of this problem is obtained
by using the method of separating variables and the property the
Mittag–Leffler functions. This problem is ill-posed, i.e. the solution(if
exists) does not depend on the measurable data. Three different
kinds landweber iterative methods are used to solve this problem.
Under the priori and the posteriori regularization parameters choice
rules, the error estimates between the exact solution and the regularization solutions are obtained. Several numerical examples are given
to prove the effectiveness of these regularization methods.

Received 6 April 2020
Accepted 28 March 2021
KEYWORDS

Time-fractional diffusion
wave equation; spherically
symmetric; Ill-posed
problem; fractional
Landweber method; inverse
problem
2010 MATHEMATICS
SUBJECT
CLASSIFICATIONS

35R25; 47A52; 35R30

1. Introduction

Nowadays, people find that the fractional derivative has much advantages in solving practical problem, such as in medical engineering [1], chemistry and biochemistry [2], finance
and economics [3–6], inverse scattering [7]. Up to now, a lot of achievements have been
made in solving the direct problems [8–12] of fractional differential equations. However,
when solving practical problems, the initial value, or source term, or diffusion coefficient,
or part of the boundary value [13,14] is unknown, and it is necessary to invert them
through some measurement data, which puts forward the inverse problem of fractional
diffusion equation. The study of the fractional diffusion equation is still on an initial stage,
the direct problem of it has been studied in [15–17].
For the inverse problem of time-fractional diffusion equation as 0 < α < 1, there are a
lot of research results. For identifying the unknown source, one can see [18–26]. About
backward heat conduction problem, one can see [27–34]. About identifying the initial
value problem, one can see [35–37]. For an inverse unknown coefficient problem of a
time-fractional equation, one can see [38,39]. About identifying the source term and initial data simultaneous of time-fractional diffusion equation, one can see [40,41]. About
identifying some unknown parameters in time-fractional diffusion equation, one can see
CONTACT Fan Yang

730050, People’s Republic of China

School of Science, Lanzhou University of Technology, Lanzhou, Gansu

© 2021 Informa UK Limited, trading as Taylor & Francis Group


2

F. YANG ET AL.

[42]. About the inverse problem for the heat equation on a columnar axis-symmetric area,
one can see [43–48]. In [43–45], Landweber regularization method, a simplified Tikhonov
regularization method and a spectral method are used to identify source term on a columnar axis-symmetric area. In [46–48], the authors considered a backward problem on a

columnar axis-symmetric domain. In [46], Yang et al. used the quasi-boundary value
regularization method to solve the inverse problem for determining the initial value of
heat equation with inhomogeneous source on a columnar symmetric domain. The error
estimate between the regular solution and the exact solution under the corresponding regularization parameter selection rules is obtained. Finally, numerical example is given to
verify that the regularization method is very effective for solving this inverse problem.
In [47], Cheng et al. used the modified Tikhonov regularization method to do with the
inverse time problem for an axisymmetric heat equation. Finally, Hölder type error estimate between the approximate solution and exact solution is obtained. In [48], Djerrar
et al. used standard Tikhonov regularization method to deal with an axisymmetric inverse
problem for the heat equation inside the cylinder a ≤ r ≤ b, and numerical examples is
used to show that this method is effective and stable. About the inverse problem for the
time-fractional diffusion equation as 0 < α < 1 on a columnar and spherical symmetric
areas, one can see [49,50]. In [49], Xiong proposed a backward problem model of a timefractional diffusion-heat equation on a columnar axis-symmetric domain. Yang et al. in
[50] used Landweber iterative regularization method to solve identifying the initial value
of time-fractional diffusion equation on spherically symmetric domain. Compare with the
inverse problem of time-fractional diffusion equation as 0 < α < 1, there are little research
result for the inverse problem of time-fractional diffusion wave equation as 1 < α < 2.
Šišková et al. in [51] used the regularization method to solve the inverse source problem of time-fractional diffusion wave equation. Liao et al. in [52] used conjugate gradient
method combined with Morozovs discrepancy principle to identify the unknown source
for the time-fractional diffusion wave equation. Šišková et al. in [53] used the regularization method to deal with an inverse source problem for a time-fractional wave equation.
Gong et al. in [54] used a generalized Tikhonov to identify the time-dependent source
term in a time-fractional diffusion-wave equation. In recent years, in physical oceanography and global meteorology, the inversion of initial boundary value problems has always
been a hot issue. In order to increase the accuracy of numerical weather prediction, usually by the model combined with the observation data of the inversion of initial boundary
value problems, and in numerical weather prediction model, provide a reasonable initial
field. At present, many domestic and foreign ocean circulation model, atmospheric general circulation model, numerical weather prediction model and torrential rain forecasting
model belongs to the inversion of initial boundary value problems during initialization,
so such problems of scientific research application prospect is very broad. Yang et al. in
[55] used the truncated regularization method to solve the inverse initial value problem
of the time-fractional inhomogeneous diffusion wave equation. Yang et al. in [56] used
the Landweber iterative regularization method to solve the inverse problem for identifying
the initial value problem of a space time-fractional diffusion wave equation. Wei et al. in

[57] used the Tikhonov regularization method to solve the inverse initial value problem
of time-fractional diffusion wave equation. Wei et al. in [58] used the conjugate gradient
algorithm combined with Tikhonov regularization method to identify the initial value of
time-fractional diffusion wave equation. Until now, we find that there are few papers for the


INVERSE PROBLEMS IN SCIENCE AND ENGINEERING

3

inverse problem of time-fractional diffusion-wave equation on a columnar axis-symmetric
domain and spherically symmetric domain. In [59], the authors used the Landweber iterative method to solve an inverse source problem of time-fractional diffusion-wave equation
on spherically symmetric domain. It is assumed that the grain is of a spherically symmetric
domain diffusion geometry as illustrated in Figure (a-b), which is actually consistent with
laboratory measurements of helium diffusion from a physical point of view from apatite. As
a consequence of radiogenic production and diffusive loss, u(r, t) which only depends on
the spherically radius r and t denotes the concentration of helium. For the inverse problem
of inversion initial value in spherically symmetric region, there are few research results at
present. Whereupon, in this paper, we consider the inverse problem to identify the initial
value of time-fractional diffusion-wave equation on spherically symmetric region and give
three regularization methods to deal with this inverse problem in order to find a effective
regular method.

In this paper, we consider the following problem:
⎧
2
⎪
⎪
Dαt u(r, t) − ur (r, t) − urr (r, t) = f (r), 0 < r < r0 , 0 < t < T, 1 < α < 2,
⎪

⎪
r
⎪
⎪
⎪
u(r
,
t)
=
0,
0 ≤ t ≤ T,
⎪
0
⎪
⎪
⎨
u(r, 0) = ϕ(r),
0 ≤ r ≤ r0 ,
⎪
0 ≤ r ≤ r0 ,
ut (r, 0) = ψ(r),
⎪
⎪
⎪
⎪
⎪
0 ≤ t ≤ T,
lim u(r, t) bounded,
⎪
⎪

r→0
⎪
⎪
⎩
u(r, T) = g(r),
0 ≤ r ≤ r0 ,

(1)

where r0 is the radius, Dαt u(r, t) is the Caputo fractional derivative (1 < α < 2), it is defined
as
Dαt u(r, t) =

1
(2 − α)

t
0

∂ 2 u(r, s)
ds
,
∂s2 (t − s)α−1

1 < α < 2.

(2)

The existence and uniqueness of the direct problem solution has been proved in the [60].
The inverse problem is to use the measurement data g(r) and the known function f (r)

to identify the unknown initial data ϕ(r), ψ(r). The inverse initial value problem can be
transformed into two cases:


4

F. YANG ET AL.

(IVP1): Assuming ψ(r) is known, we use the final value data g(r) and the known function
f (r) to invert the initial value ϕ(r).
(IVP2): Assuming ϕ(r) is known, we use the final value data g(r) and the known function
f (r) to invert the initial value ψ(r).
Because the measurements are error-prone, we remark the measurements with error as
f δ and g δ and satisfy
fδ − f

L2 [0,r0 ;r2 ]

g δ (r) − g(r)

≤ δ,

L2 [0,r0 ;r2 ]

(3)
≤ δ,

(4)

where δ > 0. In this paper, L2 [0, r0 ; r2 ] represents the Hilbert space with weight r2 on the

interval [0, r0 ] of the Lebesgue measurable function. (·, ·) and · represent the inner
product and norm of the space of [0, r0 ; r2 ], respectively. · is defined as follows:
· =

r0
0

2

2

r | · | dr

1
2

.

(5)

This paper is organized as follows. In Section 2, we recall and state some preliminary theoretical results. In Section 3, we analyse the ill-posedness of the problem (IVP1) and the
problem (IVP2), and give the conditional stability result. In Section 4, we give the corresponding a priori error estimates and posteriori error estimates for three regularization
methods. In Section 5, we conduct some numerical tests to show the validity of the proposed regularization methods. Since most of the solutions of fractional partial differential
equations contain special functions (Mittag–Leffler functions), and the calculation of these
functions is quite difficult. In this paper, the difficulties are overcome through [61,62].
Finally, we give some concluding remarks.

2. Preliminary results
In this section, we give some important Lemmas.
Lemma 2.1 ([57]): If 0 < α < 2, and β ∈ R be arbitrary. Suppose μ satisfy

min{π , π α}. Then there exists a constant C1 = C(α, β, μ) > 0 such that
|Eα,β (z)| ≤

C1
,
1 + |z|

μ ≤ |arg(z)| ≤ π .

πα
2

<μ<

(6)

Lemma 2.2 ([57]): For 1 < α < 2, β ∈ R and η > 0, we have
Eα,β (−η) =

1
1
+O 2
(β − α)η
η

,

η → ∞.

(7)

2 (−λ T α )
Lemma 2.3: For 12 < γ < 1, a1 and a2 are relaxation factor and satisfy 0 < a1 Eα,1
n
2
2
α
< 1, 0 < a2 T Eα,2 (−λn T ) < 1, m1 ≥ 1, m2 ≥ 1, we have
2
sup [1 − (1 − a1 Eα,1
(−λn T α ))m1 ]γ

λn >0

√
1
≤ a1 m1 ,
α
Eα,1 (−λn T )

(8)


INVERSE PROBLEMS IN SCIENCE AND ENGINEERING

2
sup [1 − (1 − a2 T 2 Eα,2
(−λn T α ))m2 ]γ

λn >0

√
1
≤ a2 m2 .
α
TEα,2 (−λn T )

5

(9)

Proof: Refer to the appendix for the details of the proof.
2
2
α
Lemma 2.4: For 12 < γ < 1, m1 ≥ 1, m2 ≥ 1, λn = ( nπ
r0 ) > 0, 0 < a1 Eα,1 (−λn T ) < 1,
2 (−λ T α ) < 1, we have
0 < a2 T 2 Eα,2
n
p

−

p

2
2
(−λn T α ))m1 Eα,1
(−λn T α ) ≤ c(a1 , p)m1 4 ,

sup (1 − a1 Eα,1

(10)

λn >0

p

p

−

p

2
2
sup (1 − a2 T 2 Eα,2
(−λn T α ))m2 T 2 Eα,2
(−λn T α ) ≤ c(a2 , p)m2 4 ,

λn >0

(11)

p

p

where the constant c(a1 , p) is given by c(a1 , p) = ( 4a1 ) 4 and c(a2 , p) is given by c(a2 , p) =
p

p

( 4a2 ) 4 .
Proof: Refer to the appendix for the details of the proof.
Lemma 2.5: For 1 < α < 2 and any fixed T > 0, there is at most a finite index set I1 =
nπ 2 α
2 α
{n1 , n2 , . . . , nN } such that Eα,1 (−( nπ
r0 ) T ) = 0 for n ∈ I1 and Eα,1 (−( r0 ) T ) = 0 for
n∈
/ I1 . Meanwhile there is at most a finite index set I2 = {m1 , m2 , . . . , mM } such that
nπ 2 α
2 α
Eα,2 (−( nπ
/ I2 .
r0 ) T ) = 0 for n ∈ I2 and Eα,2 (−( r0 ) T ) = 0 for n ∈
Proof: From Lemma 2.2, we know that there exists L0 > 0 such that
Eα,1 −

nπ
r0

2

Tα

≤

1

nπ
2 (1 − α)
r0

2

< 0,
Tα

nπ
r0

2

T α > L0 ,

nπ 2 α
2 α
for 1 < α < 2, thus we know Eα,1 (−( nπ
r0 ) T ) = 0 only if ( r0 ) T ≤ L0 . Since
nπ 2
nπ 2 α
2
limn→+∞ ( nπ
r0 ) = +∞, there are only finite ( r0 ) satisfying ( r0 ) T ≤ L0 . The proof
2 α
for Eα,2 (−( nπ
r0 ) T ) is similar.

Remark 2.1: The index sets I1 and I2 may be empty, that means the singular values for the

operators K1 and K2 are not zeros. Here and below, all the results for I1 = ∅ and I2 = ∅
are regarded as the special cases.
2
Lemma 2.6 ([57]): For 1 < α < 2 and λn = ( nπ
r0 ) , there exists positive constants C, C
depending on α, T such that

C
C
≤ |Eα,1 (−λn T α )| ≤
,
λn
λn

n ∈ I1 ,

(12)

C
C
≤ |Eα,2 (−λn T α )| ≤
,
λn
λn

n ∈ I2 .

(13)

6

F. YANG ET AL.

nπ 2 α
2
2
Lemma 2.7: For 12 < γ < 1, m3 ≥ 1, m4 ≥ 1, λn = ( nπ
r0 ) > 0, 0 < a1 Eα,1 (−( r0 ) T ) <
2 (−( nπ )2 T α ) < 1, we have
1, 0 < a2 T 2 Eα,2
r0

a1 C2 r04 − p
) 4
pπ 4

T

and C4 = (

α

p

p

(1 + n2 )− 2 ≤ C3 (m3 + 1)− 4 ,
m4

2

nπ
−
r0

2
1 − a2 T 2 Eα,2

where C3 = (

m3

2

nπ
−
r0

2
1 − a1 Eα,1

T

α

p

(14)
p

(1 + n2 )− 2 ≤ C4 (m4 + 1)− 4 ,

(15)

a2 T 2 C2 r04 − p
) 4.
pπ 4

Proof: Refer to the appendix for the details of the proof.
2 (−( nπ )2 T α ) < 1, 0 <
Lemma 2.8: For 12 < γ < 1, m5 ≥ 1, m6 ≥ 1, λn > 0, 0 < a1 Eα,1
r0
nπ 2 α
2
2
a2 T Eα,2 (−( r0 ) T ) < 1, we have
γ +1

1 − a1 Eα,1

m5

Tα

p

− 2(γ +1)

p

where C5 = ( π 20 )− 2 ( 2a1 (γp +1) )

m6

2

T

α

p

− 2(γ +1)

p

(1 + n2 )− 2 ≤ C5 m5

nπ
−
r0

γ +1
1 − a2 T γ +1 Eα,2
Cr2

2

nπ

r0

−

,
p

− 2(γ +1)

p

(1 + n2 )− 2 ≤ C6 m6

and C6 = (

(16)

,

(17)

CTr02 − p 2a2 (γ +1) − 2(γp+1)
) 2( p )
.
π2

Proof: Refer to the appendix for the details of the proof.
γ +1

2 α

γ +1
Lemma 2.9: For 0 < γ ≤ 1, m5 ≥ 1, m6 ≥ 1, 0 < a1 Eα,1 (−( nπ
r0 ) T ) < 1, 0 < a2 T
γ +1

2 α
Eα,2 (−( nπ
r0 ) T ) < 1, we have

1−

γ +1
1 − a1 Eα,1

Eα,1
1−

nπ
−
r0

γ +1
1 − a2 T γ +1 Eα,2

TEα,2

2

nπ
−

r0
2

m5

Tα
1

≤ (a1 m5 ) γ +1 ,

Tα

nπ
−
r0

nπ
−
r0

(18)

2

2

m6

Tα
1

≤ (a2 m6 ) γ +1 .

(19)

Tα

Proof: Refer to the appendix for the details of the proof.
Lemma 2.10: For a1 > 0, a2 > 0, p > 0, m1 > 0, m2 > 0, we have
Cr2
F(x) = 20
π

a1 C 2 r 4
1 − 2 40
x π

m1 −1

p

(x)− 2 −1 ≤ C7 (m1 + 1)−

p+2
4

,

(20)

INVERSE PROBLEMS IN SCIENCE AND ENGINEERING

G(x) =
where C7 =

TCr02
π2

1−

Cr02 a1 Cr04 − p+2
(
) 4
π 2 (p+2)π 4

m2 −1

a2 T 2 C2 r04
x2 π 4

and C8 =

p

(x)− 2 −1 ≤ C8 (m2 + 1)−

p+2
4

,

7

(21)

TCr02 a2 TCr04 − p+2
(
) 4 .
π 2 (p+2)π 4

Proof: Refer to the appendix for the details of the proof.
Lemma 2.11: For a1 > 0, a2 > 0, p > 0, m5 ≥ 1, m6 ≥ 1, we have
Cr2
F(x) = 20
π
TCr02
G(x) =
π2
where C9 =

γ +1 m5 −1

Cr02
xπ 2

1 − a1

1 − a2 T

γ +1 m6 −1

Cr02
xπ 2

γ +1

p

p+2
Cr02 Cr02 − p −1 a1 − 2(γ
+1)
( ) 2 ( p+2 )
π2 π2

and C10 =

p+2

− 2(γ +1)

(x)− 2 −1 ≤ C9 (m5 + 1)
p

,

(22)
p+2

− 2(γ +1)

(x)− 2 −1 ≤ C10 (m6 + 1)

,

(23)

p+2
TCr02 TCr02 − p −1 a1 − 2(γ
+1) .
( π 2 ) 2 ( p+2 )
π2

Proof: Refer to the appendix for the details of the proof.

3. The ill-posedness and the conditional stability
Define
∞
p

2

p

2

(1 + n2 ) 2 (fn , Rn (r)) ,

H = f ∈ L [0, r0 ; r ];

(24)

n=1

where (·, ·) is the inner product in L2 [0, r0 ; r2 ], then H p is a Hilbert space with the norm
∞

ϕ(·)

Hp

p

(1 + n2 ) 2 (ϕn , Rn (r)) ,

:=
n=1

and
∞

ψ(·)

Hp

p

(1 + n2 ) 2 (ψn , Rn (r)) .

:=

n=1

Theorem 3.1: Let ϕ(r), ψ(r) ∈ L2 [0, r0 ; r2 ], then there exists a unique weak solution and
the weak solution for (1) is given by
∞

u(r, t) =

nπ
r0

t α−1 Eα,α −

n=1

+ Eα,1 −
+ tEα,2 −

nπ
r0
nπ
r0

2

2

t α (f (r), Rn (r))

t α (ϕ(r), Rn (r))

2

t α (ψ(r), Rn (r)) Rn (r),

(25)


8

F. YANG ET AL.

where (ϕ(r), Rn (r)) and (ψ(r), Rn (r)) are the Fourier coefficients.
Let t = T in (25), we have
∞

u(r, T) =

nπ
r0

T α−1 Eα,α −

n=1
2

nπ
r0

+ Eα,1 −

T α (f (r), Rn (r))

T α (ϕ(r), Rn (r))
2

nπ
r0

+ TEα,2 −

2

T α (ψ(r), Rn (r)) Rn (r) = g(r).

Denote
∞

T α−1 Eα,α −

g1 (r) := g(r) −
n=1

2

nπ
r0

nπ
r0

T α fn + TEα,2 −

2

T α ψn Rn (r),

and
∞

T α−1 Eα,α −

g2 (r) := g(r) −
n=1

nπ
r0

2

T α fn + Eα,1 −

nπ
r0

2

T α ϕn Rn (r).

Then we have
∞

g1 (r) =

2

nπ
r0

ϕn Eα,1 −
n=1

T α Rn (r),

(26)

and
∞

g2 (r) =

ψn TEα,2 −
n=1

nπ
r0

2

T α Rn (r).

(27)

Now we put the definitions of ϕn and ψn into (26) and (27), then the problem (IVP1) and
the problem (IVP2) become the following integral equations
(K1 ϕ)(ξ ) =

r0
0

κ1 (r, ξ )ϕ(ξ ) dξ = g1 (r),

(28)

where the integral kernel is
∞

κ1 (r, ξ ) =

Eα,1 −
n=1

nπ
r0

2

T α Rn (r)Rn (ξ ).

(29)

And
(K2 ψ)(ξ ) =

r0
0

κ2 (r, ξ )ψ(ξ ) dξ = g2 (r),

(30)

where the integral kernel is
∞

κ2 (r, ξ ) =

TEα,2 −
n=1

nπ
r0

2

T α Rn (r)Rn (ξ ),

(31)


INVERSE PROBLEMS IN SCIENCE AND ENGINEERING

9

and due to [59], we know the linear operators K1 and K2 are compact from L2 [0, r0 ; r2 ] to
L2 [0, r0 ; r2 ]. The problem (IVP1) and the problem (IVP2) are ill-posed. √
Let K1∗ be the adjoint of K1 and K2∗ be the adjoint of K2 . Since Rn (r) =
standard

orthogonal system with weight r2

in the L2 [0, r
2

nπ
r0

2
K1∗ K1 Rn (ξ ) = Eα,1
−

0

r
2nπ sin( nπ
r )

√ 3 nπ r 0 is a

r0
2
; r ], it is easy to verity

r0

T α Rn (ξ ),

and
2

nπ
r0

2
K2∗ K2 Rn (ξ ) = T 2 Eα,2
−

T α Rn (ξ ).

(1)

2 α
Hence, the singular values of K1 are σ1n = |Eα,1 (−( nπ
r0 ) T )|. Define

ψn(1) (r) =

⎧
⎪
⎪
⎪
⎪

⎨Rn (r),

⎪
⎪
⎪
⎪
⎩−Rn (r),

nπ
r0

2

nπ
−
r0

2

Eα,1 −
Eα,1

Tα

≥ 0,
(32)

Tα

< 0.

(1)

2
2
It is clear that {ψn }∞
n=1 are orthonormal in L [0, r0 ; r ], we can verity
(1)

K1 Rn (ξ ) = σ1n ψn(1) (r) = Eα,1 −
(1)

K1∗ ψn(1) (r) = σ1n Rn (ξ ) = Eα,1 −

nπ
r0

2

nπ
r0

2

T α Rn (r),

(33)

T α ψn(1) (ξ ).

(34)

(1)
Therefore, the singular system of K1 is (σ1n
; Rn , ψn(1) ).
(2)
(2)
By the similar verification, we know the singular system of K2 is (σ2n ; Rn , ψn ), where
(2)
2 α
σ2n
= |TEα,2 (−( nπ
r0 ) T )| and

ψn(2) (r)

=

⎧
⎪
⎪
⎪
⎪
⎨Rn (r),

⎪
⎪
⎪
⎪
⎩−Rn (r),

nπ
r0

2

nπ
−
r0

2

Eα,2 −
Eα,2

Tα

≥ 0,
(35)

Tα

< 0.

In the following, the integral kernels given in (29) and (31) are rewritten as
∞

κ1 (r, ξ ) =

Eα,1 −

n=1,n∈I
/1
∞

κ2 (r, ξ ) =

TEα,2 −
n=1,n∈I
/2

nπ
r0
nπ
r0

2

T α Rn (r)Rn (ξ ),
2

T α Rn (r)Rn (ξ ).

(36)

(37)


10

F. YANG ET AL.

It is not hard to prove that the kernel spaces of the operators K1 and K2 are
N(K1 ) = span {Rn ; n ∈ I1 } for I1 = ∅; N(K1 ) = {0} for I1 = ∅,
N(K2 ) = span {Rn ; n ∈ I2 } for I2 = ∅; N(K2 ) = {0} for I2 = ∅,
and the ranges of the operators K1 and K2 are

R(K1 ) =

⎧
⎪
⎪
⎪
⎪
⎨

g1 ∈ L2 [0, r0 ; r2 ] | (g1 , Rn ) = 0, n ∈
/ I1 ;

⎪
⎪
⎪
⎪
⎩

⎛
⎜
⎜
⎜
⎜
n=1,n∈I

/ 1⎝
∞

R(K2 ) =

⎫
⎪
⎪
⎪
⎪
⎬

⎞2

⎟
⎟
⎟ < +∞ ,
⎟
⎪
⎪
⎠
⎪
⎪
⎭

(g1 , Rn )
Eα,1 −

⎧
⎪

⎪
⎪
⎪
⎨

2

nπ
r0

Tα

g2 ∈ L2 [0, r0 ; r2 ] | (g2 , Rn ) = 0, n ∈
/ I2 ;

⎪
⎪
⎪
⎪
⎩

⎛
⎜
⎜
⎜
⎜
n=1,n∈I
/ 2⎝

⎞2

∞

(g2 , Rn )
2

nπ
r0

TEα,2 −

Tα

⎫
⎪
⎪
⎪
⎪
⎬

⎟
⎟
⎟ < +∞ .
⎟
⎪
⎪
⎠
⎪
⎪
⎭

Therefore, we have the following existence of the solutions for the integral equations:
Theorem 3.2: If I1 = ∅, for any g1 ∈ R(K1 ), there exists a unique solution in L2 [0, r0 ; r2 ]
for the integral Equation (28) given by
∞

g1n

ϕ(r) =
n=1

Eα,1

nπ
−
r0

2

Rn (r).

(38)

Tα

If I1 = ∅, for any g1 ∈ R(K1 ), there exists infinitely many solutions for the integral
equation (28), but exists only one best approximate solution in L2 [0, r0 ; r2 ] as
∞

g1n

ϕ(r) =
n=1,n∈I
/1

Eα,1

nπ
−
r0

2

Rn (r).
Tα

(39)


INVERSE PROBLEMS IN SCIENCE AND ENGINEERING

11

∞
Proof: Suppose ϕ(ξ ) = ∞
n=1 ϕn Rn (ξ ), put g1 =
n=1,n∈I
/ 1 g1n Rn (r) into (28), according
to the orthonormality of {Rn }, it is not hard to obtain the results.

Theorem 3.3: If I2 = ∅, for any g2 ∈ R(K2 ), there exists a unique solution in L2 [0, r0 ; r2 ]
for the integral equation (30) given by
∞

g2n

ψ(r) =
n=1

TEα,2

Rn (r).

2

nπ
−
r0

(40)

Tα

If I2 = ∅, for any g2 ∈ R(K2 ), there exists infinitely many solutions for the integral
equation (30), but exists only one best approximate solution as
∞

g2n

ψ(r) =

n=1,n∈I
/2

2

nπ
−
r0

TEα,2

Rn (r).

(41)

Tα

Proof: The proof is similar to the Theorem 3.2.
We have the following theorem on conditional stability:
Theorem 3.4: When ϕ(r) satisfies the a-priori bound condition
ϕ(r)

Hp

≤ E1 ,

p > 0,

(42)

where E1 and p are positive constants, we have
2
p+2

ϕ(r) ≤ C11 E1

p
p+2

g1

,

(43)

p

π
p+2 is a constant.
where C11 = ( Cr
2)
2

0

Proof: Due to (39) and Hăolder inequality, we have


(r)

2



2
g1n

n2 =

=
n=1,n∈I
/1

n=1,n∈I
/1

4
p+2

∞
2
Eα,1
−

nπ
r0

2

Tα

Tα
⎞

2
g1n
p+2
Eα,1

2

g1n

⎛
⎜ ∞
⎜
≤⎜
⎜
⎝n=1,n∈I
/1

nπ
r0
2p
p+2

g1n

=
n=1,n∈I

/1

2
Eα,1
−

nπ
−
r0

2

Tα

⎟
⎟
⎟
⎟
⎠

2
p+2

⎛
⎝

∞

n=1,n∈I
/1

⎞
2 ⎠
g1n

p
p+2

.

(44)


12

F. YANG ET AL.

Applying Lemma 2.6 and (39), we obtain
∞
n=1,n∈I
/1

∞

2
g1n
p+2

−

Eα,1

2

nπ
r0

2
g1n

≤
n=1,n∈I
/1

Tα

2
Eα,1
−

∞
n=1,n∈I
/1
∞

≤
n=1,n∈I
/1

2

Tα

p

p

n2 π 2
Cr02

ϕn2

=

nπ
r0

λn
C

p

π2
Cr02

ϕ

2
Hp .

(45)

Combining (44) and (45), we can get
2
p+2
Hp

ϕ(r) ≤ C11 ϕ

p
p+2

g1

.

Theorem 3.5: As ψ(r) satisfies a-priori bound condition
ψ(r)

Hp

≤ E2 ,

p > 0,

(46)

where E2 and p are positive constants, then
2
p+2

ψ(r) ≤ C12 E2

p
p+2

g2

,

(47)

p

π
p+2 is a constant.
where C12 = ( TCr
2)
2

0

Proof: The proof is similar to the Theorem 3.4, so it is omitted.

4. Regularization method and error estimation
By referring to [59,64,65], we find a classical Landweber regularization method and two
fractional Landweber iterative regularization methods, but it has not been explained which
of these methods is better. So, in this section, we mainly make use of two kinds of fractional
Landweber regularization methods and the classical Landweber regularization method to
solve the problem (IVP1) and the problem (IVP2). The error estimates between the exact

solution and the corresponding regular solution are given, respectively. By using three
regularization methods to solve the same problem, an optimal regularization method is
obtained.
By [64], the fractional Landweber regularization solution is given as follows:

∞

ϕ

m1 ,δ

1−

2
1 − a1 Eα,1

=
n=1,n∈I
/1

Eα,1

nπ
−
r0

nπ
−
r0

2

2

m1 γ

Tα
δ
Rn (r),
g1n

Tα

(48)


INVERSE PROBLEMS IN SCIENCE AND ENGINEERING

13

and

1−

∞

ψ

m2 ,δ

2
1 − a2 T 2 Eα,2

=
n=1,n∈I
/2

TEα,2

2

nπ
−
r0
2

nπ
−
r0

m2 γ

Tα
δ
Rn (r).
g2n

(49)

Tα

By [59], the Landweber regularization solution is given as follows:

∞

ϕ

m3 ,δ

=
n=1,n∈I
/1

Eα,1

2

nπ
r0

2
1 − 1 − a1 Eα,1
−

2

nπ
−
r0

m3

Tα
δ
Rn (r),
g1n

(50)

Tα

and
1−

∞

ψ

m4 ,δ

=
n=1,n∈I
/2

TEα,2

2

nπ
−

r0

2
1 − a2 T 2 Eα,2

2

nπ
−
r0

m4

Tα
δ
Rn (r).
g2n

(51)

Tα

By [65], the modified iterative regularization solution is given as follows:

γ +1

1 − 1 − a1 Eα,1

∞

ϕ

m5 ,δ

=
n=1,n∈I
/1

Eα,1

2

nπ
r0

−

2

nπ
−
r0

m5

Tα
δ
Rn (r),
g1n

(52)

Tα

and

∞

ψ

m6 ,δ

1−

γ +1
1 − a2 T γ +1 Eα,2

=
n=1,n∈I
/2

TEα,2

nπ
−
r0

nπ
−
r0

2

2

m6

Tα
δ
Rn (r),
g2n

Tα

where a1 , a2 are relaxation factors and satisfy a1 , a2 > 0,

1
2

< γ ≤ 1.

(53)


14

F. YANG ET AL.

4.1. The priori error estimate
Lemma 4.1: Suppose f (x) and g(x) be ∈ L2 (0, r0 ; r2 ), there is a constant M1 to make:

∞

T α−1 Eα,α −

Q(r) = g(r) −
n=1

≤

2[ g

2
L2 [0,r0 ;r2 ]

+ M21 f

2

nπ
r0

T α fn Rn (r)

2
],
L2 [0,r0 ;r2 ]

C1 r02
.
n2 π 2 T

where M1 =

Proof: From Lemma 2.1, we have
Eα,α −

2

nπ
r0

Tα

C1

≤

nπ
r0

2

=
Tα

C1 r02
.
n2 π 2 T α

Thus,

∞

Q(r)

2

n=1
∞

=

gn − T

α−1

nπ
−
r0

Eα,α

n=1
∞

2
gn2 + 2T 2α−2 Eα,α
−

≤2
n=1

≤ 2[ g
where M1 =

2
L2 [0,r0 ;r2 ]

C1 r02
. We complete
n2 π 2 T

+ M21

f

2

nπ
r0

T α−1 Eα,α −

= g(r) −

T α fn Rn (r)

2

2

2

T

nπ
r0

2

α

fn
∞

Tα

fn2
n=1

2
L2 [0,r0 ;r2 ] ],

the proof of Lemma 4.1.

Remark 4.1: Suppose
δ

δ

∞

Q (r) = g (r) −
n=1

T α−1 Eα,α −

nπ
r0

2

T α fnδ Rn (r) .

From Lemma 4.1, (3) and (4), we easily know Q(r) − Qδ (r) ≤

2(M21 + 1)δ.

Define an orthogonal projection operator Q1 : L2 [0, r0 ; r2 ] → R(K1 ), combining Equations (3) and (4), we have:

Q1 g1δ − Q1 g1 ≤ g1δ − g1 = g δ − g ≤

2(M21 + 1)δ.

Define an orthogonal projection operator Q2 : L2 [0, r0 ; r2 ] → R(K2 ), combining Equations (3) and (4), we have:

Q2 g2δ − Q2 g2 ≤ g2δ − g2 = g δ − g ≤

2(M21 + 1)δ.

INVERSE PROBLEMS IN SCIENCE AND ENGINEERING

15

4

Theorem 4.1: Let m1 = [( Eδ1 ) p+2 ]. If the a-priori condition (42) and Lemma 4.1 hold, we
have the following convergence estimate
2
p+2

p

ϕ m1 ,δ (r) − ϕ(r) ≤ C13 E1 δ p+2 ,
Cr2

p

where C13 = c(a1 , p)( π 20 )− 2 +
than or equal to x.

(54)

2a1 (M21 + 1), and [x] denotes the largest integer smaller

Proof: Using the triangle inequality, we have
ϕ m1 ,δ (r) − ϕ(r) ≤ ϕ m1 ,δ (r) − ϕ m1 (r) + ϕ(r) − ϕ m1 (r) = I1 + I2 .
From (8) and Remark 4.1, we have
I12 = ϕ m1 ,δ (r) − ϕ m1 (r)
1−

∞

2

2
1 − a1 Eα,1

=
n=1,n∈I
/1

1−
≤

sup

n≥1,n∈I
/1

2
Eα,1

2

nπ
r0

Eα,1 −

2
1 − a1 Eα,1

2

nπ
−
r0

2

nπ
−
r0

2

δ
(g1n
− g1n )Rn (r)

Tα
2

nπ
−
r0

m1 γ

Tα

m1 2γ

Tα

∞

δ
(g1n
− g1n )2

n=1,n∈I
/1

Tα

≤ 2a1 m1 (M21 + 1)δ 2 .
Then we can get
2a1 m1 (M21 + 1)δ.

ϕ m1 ,δ (r) − ϕ m1 (r) ≤

(55)

For the second part,
I22 = ϕ(r) − ϕ m1 (r)
∞

2

1− 1−

2
1 − a1 Eα,1

=
n=1,n∈I
/1

∞

Eα,1

2
1 − a1 Eα,1

≤
n=1,n∈I
/1

Eα,1

nπ
−
r0

nπ
−
r0

2

nπ
−
r0
nπ
−
r0

2

2

2

m1 γ

g1n Rn (r)

Tα
2

m1

Tα
g1n Rn (r)

Tα

2

Tα


16

F. YANG ET AL.

∞

2
1 − a1 Eα,1

=
n=1,n∈I
/1

=
n=1,n∈I
/1

n=1,n∈I
/1

2m1

Tα
2
g1n

Tα

2
1 − a1 Eα,1

nπ
−
r0

2

2
1 − a1 Eα,1

nπ
−
r0

2

∞

=

2

nπ
r0

2

−
Eα,1

∞

2

nπ
−
r0

2m1

T

α

ϕn2 (r)
2m1

T

α

(1 + n2 )−p (1 + n2 )p ϕn2 (r)

≤ E12 sup(B(n))2 ,

(56)

n≥1

p

2 (−( nπ )2 T α ))m1 (1 + n2 )− 2 .
where B(n) := (1 − a1 Eα,1
r0
From (10), we have

nπ
−
r0

2

B(n) ≤

2
1 − a1 Eα,1

nπ
−
r0

2

≤

2
1 − a1 Eα,1

≤

Cr02
π2

m1

T

α

m1

T

α

p

(n2 )− 2
Cr02
π2

p

−2

p

2
−
Eα,1

nπ
r0

2

Tα

p

−2

−

p

c(a1 , p)m1 4 .

(57)

Combining (55), (56) and (57), Theorem 4.1 is proved.
4

Theorem 4.2: Let m2 = [( Eδ2 ) p+2 ]. If the a-priori condition (46) and Lemma 4.1 hold, we
have the following convergence estimate
2
p+2

p

ψ m2 ,δ (r) − ψ(r) ≤ C14 E2 δ p+2 ,
where C14 = c(a2 , p)(
than or equal to x.

Cr02 T − p
) 2
π2

+

(58)

2a2 (M21 + 1), and [x] denotes the largest integer smaller

Proof: The proof process is similar to Theorem 4.1, so it is omitted.
4

Theorem 4.3: Let m3 = [( Eδ1 ) p+2 ]. If the a-priori condition (42) and Lemma 4.1 hold, we
have the following convergence estimate
2
p+2

p

ϕ m3 ,δ (r) − ϕ(r) ≤ C15 E1 δ p+2 ,
where C15 =
to x.

(59)

2a1 (M21 + 1) + C3 , and [x] denotes the largest integer smaller than or equal


INVERSE PROBLEMS IN SCIENCE AND ENGINEERING

17

Proof: By the triangle inequality, we have
ϕ m3 ,δ (r) − ϕ(r) ≤ ϕ m3 ,δ (r) − ϕ m3 (r) + ϕ m3 (r) − ϕ(r) = I1 + I2 .
On the one hand, from Lemma 4.1, we have
I12 = ϕ m3 ,δ (r) − ϕ m3 (r)
1−

∞

2

=
n=1,n∈I
/1

2

nπ
−
r0

2
1 − a1 Eα,1

nπ
r0

2
−
Eα,1

2

2

m3

Tα
δ
− g1n )Rn (r)
(g1n

Tα

≤ 2(M21 + 1) sup(A(n))2 δ 2 ,
n≥1

where A(n) :=

2 (−( nπ )2 T α ))m3
1−(1−a1 Eα,1

r
0

2 α
Eα,1 (−( nπ
r ) T )

.

0

From Bernoulli inequality, we can deduce that
1−

2
1 − a1 Eα,1

nπ
−
r0

m3

2

T

α

≤

√
nπ
a1 m3 Eα,1 −
r0

2

Tα .

Thus
ϕ m3 ,δ (r) − ϕ m3 (r) ≤

2a1 m3 (M21 + 1)δ.

(60)

On the other hand, using the a-priori bound condition, we can deduce that
I22 = ϕ m3 (r) − ϕ(r)
1−

∞

2

=
n=1,n∈I
/1

∞

=
n=1,n∈I
/1

Eα,1

2
(1 − a1 Eα,1
−
2
Eα,1

n=1,n∈I
/1

nπ
−
r0
nπ 2 α
) T
r0

nπ
−
r0

∞

=

2
(1 − a1 Eα,1
−

2

nπ
−
r0

2
1 − a1 Eα,1

2

2

−1
g1n Rn (r)

Tα
2m3
2
g1n

Tα

nπ 2 α
) T

r0

2

m3

Tα

2m3

ϕn2 (r)


18

F. YANG ET AL.

∞

=
n=1,n∈I
/1

2m3

nπ 2 α
) T
r0

2

(1 − a1 Eα,1
−

(1 + n2 )−p (1 + n2 )p ϕn2 (r)

≤ sup(B(n))2 E12 ,

(61)

n≥1

p

2 (−( nπ )2 T α ))m3 (1 + n2 )− 2 .
where B(n) := (1 − a1 Eα,1
r0
From (16), we have
p

B(n) ≤ C3 (m3 + 1)− 4 .

(62)

Combining (60), (61) and (62), Theorem 4.3 is proved.
4

Theorem 4.4: Let m4 = [( Eδ2 ) p+2 ]. If the a-priori condition (46) and Lemma 4.1 hold, we
have the following convergence estimate
2
p+2

p

ψ m4 ,δ (r) − ψ(r) ≤ C16 E2 δ p+2 ,
where C16 =
to x.

(63)

2a2 (M21 + 1) + C4 , and [x] denotes the largest integer smaller than or equal

Proof: The proof process is similar to Theorem 4.3, so it is omitted.
2(γ +1)

Theorem 4.5: Let m5 = [( Eδ1 ) p+2 ]. If the a-priori condition (42) and Lemma 4.1 hold, we
have the following convergence estimate
2
p+2

p

ϕ m5 ,δ (r) − ϕ(r) ≤ C17 E1 δ p+2 ,

(64)

1

where C17 = a1γ +1 2(M21 + 1) + C5 , and [x] denotes the largest integer smaller than or
equal to x.
Proof: By the triangle inequality, we have

ϕ m5 ,δ (r) − ϕ(r) ≤ ϕ m5 ,δ (r) − ϕ m5 (r) + ϕ m5 (r) − ϕ(r) = I1 + I2 .

(65)

On the one hand, we have
∞

γ +1

1 − 1 − a1 Eα,1

I1 =
n=1,n∈I
/1

Eα,1

−

nπ
−
r0

nπ
r0
2

2

m5

Tα
δ
− g1n )Rn (r) .
(g1n

Tα

Due to (18) and Lemma 4.1, we have
1

I1 ≤ (a1 m5 ) γ +1 2(M21 + 1)δ.

(66)


INVERSE PROBLEMS IN SCIENCE AND ENGINEERING

19

On the other hand, we have
γ +1
1 − a1 Eα,1

∞

I2 =
n=1,n∈I
/1

∞

=
n=1,n∈I
/1
∞

=
n=1,n∈I
/1

Eα,1

nπ
−
r0

2

g1n Rn (r)

2

γ +1
1 − a1 Eα,1

nπ
−
r0

2

nπ
r0

−

2

Tα

Tα

nπ
−
r0

γ +1

m5

2

γ +1
1 − a1 Eα,1

≤ E1 sup 1 − a1 Eα,1
n≥1

nπ

−
r0

m5

T

α

ϕn Rn (r)
m5

T

α

p

p

(n2 + 1)− 2 (n2 + 1) 2 ϕn Rn (r)
m5

Tα

p

(n2 + 1)− 2 .

From (16), we have

p

− 2(γ +1)

I2 ≤ C5 m5

E1 .

(67)

Combining (65), (66) and (67), Theorem 4.5. is proved.
2(γ +1)

Theorem 4.6: Let m6 = [( Eδ2 ) p+2 ]. If the a-priori condition (46) and Lemma 4.1 hold, we
have the following convergence estimate
2
p+2

p

ψ m6 ,δ (r) − ψ(r) ≤ C18 E2 δ p+2 ,

(68)

1

where C18 = a2γ +1 2(M21 + 1) + C6 , and [x] denotes the largest integer smaller than or
equal to x.
Proof: The proof process is similar to Theorem 4.5, so it is omitted.
4.2. The posteriori error estimate

Assume τ1 >
m1 satisfying

2(M21 + 1) be the constant, the selection of m1 = m1 (δ) ∈ N0 is that when
K1 ϕ m1 ,δ (r) − g1δ (r) ≤ τ1 δ

appears for the first time, the iteration stops, where g1δ (r) ≥ τ1 δ.
Lemma 4.2: Assume ρ(m1 ) = K1 ϕ m1 ,δ (r) − g1δ (r) , then
(a) ρ(m1 ) is a continuous function;
(b) limm1 →0 ρ(m1 ) = g1δ (r) ;

(69)


20

F. YANG ET AL.

(c) limm1 →∞ ρ(m1 ) = 0;
(d) ρ(m1 ) is strictly decreasing for any m1 ∈ (0, +∞).
Theorem 4.7: If formula (3) and (4) is true, then the regularization parameter
m1 = m1 (δ) ∈ N0 satisfies
E1
δ

m1 ≤ C19
where C19 = (

τ1 −

4
p+2

,

(70)

4

√C7

2(M21 +1)

) p+2 .

Proof: From (48), we have

∞

1−

R m 1 g1 =
n=1,n∈I
/1

2

nπ
−
r0

Eα,1

2

nπ
−
r0

2
1 − a1 Eα,1

m1 γ

Tα
g1n Rn (r).

Tα

For ∀g1 ∈ H 2 [0, r0 ; r2 ], we have
K1 Rm1 g1 − g1

2

∞

=

1− 1−
n=1,n∈I

/1
∞

≤
n=1,n∈I
/1

2
1 − a1 Eα,1

2

nπ
−
r0

T

α

g1n Rn (r)
2

m1

Tα

2

m1 γ

2

nπ
−
r0

2
1 − a1 Eα,1

g1n Rn (r)

.

2 (−( nπ )2 T α )| < 1, we have K R
Since |1 − a1 Eα,1
1 m1 − I ≤ 1.
r0
From the Morozov’s discrepancy principle, we obtain

K1 Rm1 −1 g1 − g1 ≥ K1 Rm1 −1 g1δ − g1δ − (K1 Rm1 −1 − I)(g1 − g1δ )
≥ (τ1 −

2(M21 + 1))δ.

Using the a-prior bound condition, we have
K1 Rm1 −1 g1 − g1
⎛⎡
∞

=
n=1,n∈I
/1

2
⎝⎣1 − 1 − a1 Eα,1

∞

≤
n=1,n∈I
/1

2
1 − a1 Eα,1

nπ
−
r0

nπ
−
r0

2

m1 −1

Tα

⎤γ

⎦ − 1⎠ g1n Rn (r)

m1 −1

2

T

α

⎞

Eα,1 −

nπ
r0

2

Tα


INVERSE PROBLEMS IN SCIENCE AND ENGINEERING
p

21

p

· (1 + n2 )− 2 ϕn (1 + n2 ) 2
≤ sup H(n)E1 ,
n≥1

p

2 (−( nπ )2 T α ))m1 −1 |E (−( nπ )2 T α )|(1 + n2 )− 2 .
where H(n) := (1 − a1 Eα,1
α,1
r0
r0

2(M21 + 1))δ. From (20), we obtain

So we have H(n)E1 ≥ (τ1 −

H(n) ≤ C7 (m1 + 1)−

p+2
4

.

So we have
p+2
4

C7 (m1 + 1)−
Then we obtain

E1 ≥ (τ1 −
⎞

⎛
m1 ≤ ⎝

C7
τ1 −

2(M21 + 1))δ.
4
p+2

4
p+2

E1
δ

⎠

2(M21 + 1)

.

The convergence result is given in the following theorem.
Theorem 4.8: Assuming that Lemma 4.1 and (39) are valid, and the regularization parameter are given by Equation (70), then
2
p+2

p

ϕ m1 ,δ (r) − ϕ(r) ≤ C20 E1 δ p+2 ,
where C20 = C11 (τ1 +

p

2(M21 + 1)) p+2 +

(71)

2a1 C19 (M21 + 1).

Proof: Using the triangle inequality, we have
ϕ m1 ,δ (r) − ϕ(r) ≤ ϕ m1 ,δ (r) − ϕ m1 (r) + ϕ m1 (r) − ϕ(r) .

(72)

From Theorem 4.1, we have
ϕ

m1 ,δ

(r) − ϕ

m1

(r) ≤

2a1 m1 (M21

+ 1)δ ≤

2a1 C19 (M21

E1
+ 1)
δ

2
p+2

δ.

(73)

2 (−( nπ )2 T α ))m1 ] < 1, we
For the second part, combining (33) and 0 < 1 − (1 − a1 Eα,1
r0
have

K1 (ϕ m1 (r) − ϕ(r))
∞

=

1− 1−
n=1,n∈I
/1

∞

≤
n=1,n∈I
/1

2
1 − a1 Eα,1

2
1 − a1 Eα,1

nπ
−
r0

nπ
−
r0

m1 γ

2

T

α

m1

2

T

α

g1n Rn (r)

g1n Rn (r)


22

F. YANG ET AL.

∞

≤
n=1,n∈I
/1
∞

+
n=1,n∈I
/1

m1

2

nπ
−
r0

2
1 − a1 Eα,1

T

δ
(g1n − g1n
)Rn (r)
m1

2

nπ
−
r0

2
1 − a1 Eα,1

α

T

α

δ

g1n
Rn (r) .

From (3), (4) and the Morozov’s discrepancy principle, we have
K1 ϕ m1 (r) − g1 (r) ≤ τ1 +

2(M21 + 1) δ.

Since
ϕ m1 (r) − ϕ(r)

Hp

1− 1−

∞

≤
n=1,n∈I
/1

Eα,1
2
1 − a1 Eα,1
−

∞

≤
n=1,n∈I

/1

Eα,1

nπ
−
r0
2

nπ
r0

nπ
−
r0

2

2

g1n Rn (r)
Hp

m1

Tα
g1n Rn (r)

Tα
Hp

⎤1
2

⎢ ∞
⎢
≤⎢
⎢
⎣n=1,n∈I
/1

≤⎣

m1 γ

Tα

Tα

⎡

⎡

2

nπ
−
r0

2

1 − a1 Eα,1

2
g1n
2
Eα,1
−

2

nπ
r0

Tα

⎥
⎥
(1 + n2 )p ⎥
⎥
⎦

⎤1

∞

2

(1 + n2 )p ϕn2 ⎦

≤ E1 .

n=1,n∈I
/1

Combining Theorem 3.4 and (70), we have
ϕ

m1

(r) − ϕ(r) ≤ C11 τ1 +

2(M21

+ 1)

p
p+2

2
p+2

p

E1 δ p+2 .

(74)

Combining (72), (73) and (74), we obtain the convergence estimate.
Assuming that τ2 > 2(M21 + 1) is the given constant, the selection of m2 = m2 (δ) ∈
N0 is that when m2 satisfying

K2 ψ m2 ,δ (r) − g2δ (r) ≤ τ2 δ
appears for the first time, the iteration stops, where g2δ (r) ≥ τ2 δ.

(75)


INVERSE PROBLEMS IN SCIENCE AND ENGINEERING

23

Lemma 4.3: Assume ρ(m2 ) = K2 ψ m2 ,δ (r) − g2δ (r) , then
(a)
(b)
(c)
(d)

ρ(m2 ) is a continuous function;
limm2 →0 ρ(m2 ) = g2δ ;
limm2 →∞ ρ(m2 ) = 0;
ρ(m2 ) is strictly decreasing for any m2 ∈ (0, +∞).

Theorem 4.9: If formula (3) and (4) is true, then the regularization parameter
m2 = m2 (δ) ∈ N0 satisfies:
m2 ≤ C21
where C21 = (

τ2 −

√C8

2(M21 +1)

E2
δ

4
p+2

,

(76)

4

) p+2 .

Proof: The proof process is similar to Theorem 4.7, so it is omitted.
Theorem 4.10: Assuming that (3), (4) and (41) are valid, and the regularization parameters
are given by (76), then we have
2
p+2

p

ψ m2 ,δ (r) − ψ(r) ≤ C22 E2 δ p+2 ,
where C22 =

2a2 C21 (M21 + 1) + C12 (τ2 +

(77)

p

2(M21 + 1)) p+2 .

Proof: The proof process is similar to Theorem 4.8, so it is omitted.
Assuming that τ3 > 2(M21 + 1) is the given constant, the selection of m3 = m3 (δ) ∈
N0 is that when m3 satisfying
K1 ϕ m3 ,δ (r) − g1δ (r) ≤ τ3 δ

(78)

appears for the first time, the iteration stops, where g1δ (r) ≥ τ3 δ.
Lemma 4.4: Assuming ρ(m3 ) = K1 ϕ m3 ,δ (r) − g1δ (r) , then
(a)
(b)
(c)
(d)

ρ(m3 ) is a continuous function;
limm3 →0 ρ(m3 ) = g1δ ;
limm3 →∞ ρ(m3 ) = 0;
ρ(m3 ) is strictly decreasing for any m3 ∈ (0, +∞).

Theorem 4.11: If formula (3) and (4) is true, then the regularization parameter
m3 = m3 (δ) ∈ N0 satisfies:
m3 ≤ C23
where C23 = (

τ3 −

√C7

2(M21 +1)

4

) p+2 .

E1
δ

4
p+2

,

(79)


24

F. YANG ET AL.

Proof: From (50), we have
nπ
r0

2
−

1 − 1 − a1 Eα,1

∞

Rm3 g1 =
n=1,n∈I
/1

2

nπ
−
r0

Eα,1

Tα

m3

g1n Rn (r).

Tα

For ∀g1 ∈ H 2 [0, r0 ; r2 ], we have
K1 Rm3 g1 − g1

2

∞

=
n=1,n∈I
/1

2
1 − 1 − a1 Eα,1

∞
2
1 − a1 Eα,1

=
n=1,n∈I
/1

nπ
−
r0

nπ
−
r0

2

m3

Tα

2

∞

g1n Rn (r) −

g1n Rn (r)
n=1,n∈I
/1

2m3

2

T

α

2
g1n
.

2 (−( nπ )2 T α )| < 1, we have K R
Since |1 − a1 Eα,1
1 m3 − I ≤ 1.
r0
On the one hand, we have

K1 Rm3 −1 g1 − g1 ≥ K1 Rm3 −1 g1δ − g1δ − (K1 Rm3 −1 − I)(g1 − g1δ )
≥ (τ3 −

2(M21 + 1))δ.

On the other hand, from the a-priori bound condition, we know
K1 Rm3 −1 g1 − g1
∞

=
n=1,n∈I
/1

2
1 − a1 Eα,1

nπ
−
r0

2

2
1 − a1 Eα,1

nπ
−
r0

2

∞

=
n=1,n∈I
/1

p

m3 −1

T

α

g1n Rn (r)
m3 −1

T

α

Eα,1 −

2

nπ
r0

p

· (1 + n2 )− 2 ϕn (1 + n2 ) 2

≤ sup H(n)E1 ,
n≥1

p

2 (−( nπ )2 T α ))m3 −1 |E (−( nπ )2 T α )|(1 + n2 )− 2 .
where H(n) := (1 − a1 Eα,1
α,1
r0
r0

Tα