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On a backward problem for nonlinear fractional diffusion equations
Accepted Manuscript On a backward problem for nonlinear fractional diffusion equations Nguyen Huy Tuan, Le Nhat Huynh, Tran Bao Ngoc, Yong Zhou
PII: DOI: Reference:
S0893-9659(18)30388-4 https://doi.org/10.1016/j.aml.2018.11.015 AML 5700
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Applied Mathematics Letters
Received date : 12 October 2018 Revised date : 16 November 2018 Accepted date : 16 November 2018 Please cite this article as: N.H. Tuan, L.N. Huynh, T.B. Ngoc et al., On a backward problem for nonlinear fractional diffusion equations, Applied Mathematics Letters (2018), https://doi.org/10.1016/j.aml.2018.11.015 This is a PDF file of an unedited manuscript that has been accepted for publication. As a service to our customers we are providing this early version of the manuscript. The manuscript will undergo copyediting, typesetting, and review of the resulting proof before it is published in its final form. Please note that during the production process errors may be discovered which could affect the content, and all legal disclaimers that apply to the journal pertain.
Elsevier Editorial System(tm) for Applied Mathematics Letters Manuscript Draft Manuscript Number: AML-D-18-01892R1 Title: On a backward problem for nonlinear fractional diffusion equations Article Type: Regular Article Keywords: Fractional diffusion equation; backward problem; regularization; fixed point theory. Corresponding Author: Professor Tuan Nguyen Huy, Corresponding Author's Institution: Ton Duc Thang University First Author: Tuan Nguyen Huy Order of Authors: Tuan Nguyen Huy; Huynh Le Nhat; Ngoc Tran Bao; Yong Zhou Abstract: In this paper, we consider a backward problem for a time-space fractional diffusion equation with nonlinear source. Under some assumptions, we establish the existence and uniqueness of local mild solutions to the nonlinear problem. We also prove that the backward problem is illposed in the sense of Hadamard. A regularization method is proposed to approximate the solution. Furthermore, the convergence rate for the regularized solution is given.
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On a backward problem for nonlinear fractional diffusion equations Nguyen Huy Tuana , Le Nhat Huynhb , Tran Bao Ngocc , Yong Zhoud,e a Applied
Analysis Research Group, Faculty of Mathematics and Statistics, Ton Duc Thang University, Ho Chi Minh City, Vietnam.
b Faculty of Mathematics and Computer Science, University of Science, Vietnam National University (VNU-HCMC), Ho Chi Minh city, Vietnam. c Institute d Faculty
of Research and Development, Duy Tan University, Da Nang 550000, Vietnam.
of Information Technology, Macau University of Science and Technology, Macau 999078, China of Mathematics and Computational Science, Xiangtan University, Hunan 411105, China
e Faculty
Abstract In this paper, a backward problem for a time-space fractional diffusion with nonlinear source has been considered. Under some assumptions, we establish the existence and uniqueness of mild solutions of a local solution to the nonlinear problem. We also prove that our backward problem is ill-posed in the sense of Hadamard. A regularization method has been proposed to approximate the solution. Furthermore, the convergence rate for the regularized solution can be proved. Keywords: Backward problem, regularization, fractional diffusion equation, fixed point theory
1. Introduction The backward problem of diffusion process is of great importance in engineering and aims at detecting the previous status of physical field from its present information. They have many applications in practice. Indeed, they describe the blurring effect, and the backward problems give the mathematical formulation for the deblurring process in image restoration. In this paper, we consider the backward problem for a time-fractional diffusion equation with nonlinear source α ∂t u(x, t) = Au(x, t) + F(x, t, u(x, t)), (x, t) ∈ Ω × (0, T ), u(x, t) = 0, (x, t) ∈ ∂Ω × (0, T ), u(x, T ) = f (x), x ∈ Ω·
(1)
∂α where Ω is a bounded domain in Rd with sufficiently smooth boundary ∂Ω, α is the Caputo fractional derivative of order ∂t 0 < α < 1 (see [9]), F is a given function-satisfying some hypotheses. The operator A in (1) is a linear densely defined self-adjoint and positive definite elliptic operator on the connected bounded domain Ω with zero Dirichlet boundary condition. Studying fractional partial differential equations involving a final condition is urgent from the applications like stochastic model of living matter in biology; prediction when earthquake occurs; electroencephalography problems of recovering epilepsy points in human brain; determining far electromagnetic waves in universe, etc. In practice, many phenomena cannot be observed at the time t = 0, that is, the initial data may not be known. However, the phenomena can be measured at a backward time t = T in some situations. The problems equipped with these backward in time data are called backward in time problems, which showed their great importance in engineering areas and aimed at detecting the previous status of a physical field from its present information [12, 20]. The initial value problems for the fractional diffusion equation have been studied extensively in recent years, see Diethelm [2], Mainardi [8, 7]. In [6], Luchko established the existence and uniqueness of the solution for the initial boundary (Dirichlet) value problem. In [4], the authors considered the initial boundary value problem for a coupled fractional diffusion system. For homogeneous case of backward problem i.e., F = 0 Yamamoto et al [11] showed that Problem (1) has a unique weak solution in case f ∈ H 2 (Ω). In the case of the fractional order α ∈ (1, 2), T. Wei et al [18] have studied a homogeneus problem and obtained some regularization results. Some results concerning a homogeneous backward problem can be found in [5, 17, 19, 10, 21]. ∗ Corresponding
Author. Email: [email protected] (Nguyen Huy Tuan), [email protected] (Le Nhat Huynh), [email protected] (Tran Bao Ngoc); [email protected] (Yong Zhou) Preprint submitted to Elsevier
November 16, 2018
If F(u(x, t)) = F(x, t), Tuan et al [12] gave a quasi-boundary value regularization method to approximate the solution. In [13], the authors applied the Tikhonov method to regularizing the linear backward problem. Very recent, Yang et al [20] considered the problem (1) by using the method of quasi-reversibility. To the best of our knowledge, we could not find any result on backward problems for nonlinear time-fractional diffusion equations. The backward problems were treated for a long time, e.g. see [14, 15] which are corresponding to Problem (1) with derivatives of integer-orders i.e, α = 1. The difficulties of a backward problem can be briefly discribed as follows. Firstly, since the fractional derivative ∂αt u is defined on the time interval (0, t), it is impossible to transform a final value problem for fractional diffusion equation to an initial-value problem by using some substitution methods. Secondly, the formula of the mild solution of a backward problem is more complex than the initial problem. In fact, we can show the solution operators of an initial problem for fractional diffusion equation is bounded for all time in the interval [0, T ] (see [4]). However, the formula of the mild solution of a backward problem for fractional diffusion equation shows that the solution operators are unbounded at the time t = 0. Our main goal in this paper is to investigate the existence and regularization of local solutions of Problem (1). This paper is organized as follows. In section 2, we give a formula of the mild solution to Problem (1). Then we show the existence and uniqueness of the local solution. Moreover, we also give the instability of the mild solution. In section 3, we give a regularization method. 2. Existence and uniqueness of the mild solution Since A is introduced as above, the eigenvalues of A satisfy 0 < λ1 ≤ λ2 ≤ λ3 ≤ · · · ≤ λ j ≤ · · · with λ j → ∞ as j → ∞. The corresponding eigenfunctions are denoted respectively by ϕ j ∈ H01 (Ω). The functions ϕ j are normalized so that {ϕ j }∞j=1 is an orthonormal basis of L2 (Ω). Defining ( ) ∞ X D E 2k k 2 2 λ j | v, ϕ j | < +∞ , H (Ω) = v ∈ L (Ω) : j=1
2
k
where h·, ·i is the inner product in L (Ω), then H (Ω) is a Hilbert space equipped with norm kvk
Hk (Ω)
=
∞ X j=1
D E 1/2 2 λ2k . j | v, ϕ j |
Now, we consider the following definition and lemmas which are useful for our main results. Definition 1. (see [9]) The Mittag-Leffler function is Eα,β (z) =
∞ X k=0
zk , z∈C Γ(αk + β)
where α > 0 and β ∈ R are arbitrary constants. Lemma 1. (see [11]) For λ > 0 and 0 < α < 1, we have d Eα,1 (−λtα ) = −λtα−1 Eα,α (−λtα ), t > 0· dt Lemma 2. (see [11]) The following equality holds for λ > 0, α > 0 and m ∈ N dm Eα,1 (−λtα ) = −λtα−m Eα,α−m+1 (−λtα ), t > 0· dtm
(2)
Lemma 3. see [9] Let α ∈ (0, 1) then Eα,1 (−z) > 0 for any z > 0. Moreover, there exist three positive constants M−1,α , M+1,α , M+2,α such that M−1,α 1+z
≤ Eα,1 (−z) ≤
M+1,α 1+z
, Eα,α (−z) ≤
M+2,α 1+z
(3)
·
If α ∈ [α0 , α1 ] for any 0 < α0 < α1 < 1 then by [16], the constant M−1,α , M+1,α , M+1,α can be chosen which depends only α0 , α1 . 2
Definition 2. The function u is called a mild solution of Problem (1) if u ∈ L∞ (0, T ; L2 (Ω)) and satisfies that ∞ X E Eα,1 (−λ j tα ) D f, ϕ ϕ j (x) j Eα,1 (−λ j T α ) j=1 Z ∞ X Eα,1 (−λ j tα )Eα,α − λ j (T − s)α D E T α−1 − F(u(., s)), ϕ j ds ϕ j (x) α) 0 (T − s) E (−λ T α,1 j j=1 ∞ Z t X D E ! α−1 α (t − s) Eα,α − λ j (t − s) F(., s, u(., s)), ϕ j ds ϕ j (x). +
u(x, t) =
(4)
0
j=1
Theorem 1. Let T > 0. F satisfy kF(v(., t)) − F(w(., t))kL2 (Ω) ≤ K(t)kv(., t) − w(., t)kL2 (Ω) for any t ∈ [0, T ] and K is bounded. Assume that the integral L(α, T ) = T
2α+1 2
M−1,α
RT 0
(5)
(T − t)−2 K 2 (t)dt converges to |A(T )|2 which satisfies
A(T ) M+2,α
+ B sup |K(t)| < 1.
Let f ∈ H2 (Ω). Then Problem (1) has a unique mild solution in L∞ (0, T ; L2 (Ω)).
Remark 1. Assume that K(t) ≤ (T − t)m for m > 1. Then we see that the integral Let Ω = (0, π). It is easy to check that λ j = j2 and |A(T )|2 ≤ P 2 B ≤ ∞j=1 j12 = π6 . Once choice for T is
(6)
0≤t≤T
m−1
T m−1
RT 0
(T − t)−2 K 2 (t)dt is convergent.
and sup0≤t≤T |K(t)| ≤ T m . Moreover, we also have
1 1 3 1 (m − 1)M+ α+m− 2 2,α m T < min 2 , − M π 1,α
which implies that (6) holds.
Remark 2. For considering the direct problem, we can use the assumption (5) and the initial condition belongs to L2 (Ω). However, for considering the inverse problem, we have to give the final conditon which belongs to H 2 (Ω). The studying Problem (1) is more difficult if F has a general form, such as F = F(x; t; u; ∇u). We think the solution does not belong to L∞ (0, T ; L2 (Ω)). Our study is considering the solution which is in L∞ (0, T ; H 2 (Ω)) and this topic is a future work.
Remark 3. If α ∈ (1, 2), we need to one addition conditon ut at t = 0. Assume that ut (x, 0) = 0 then we can show the existence of the mild solution of Problem (1) by the method in the case α ∈ (0, 1). However, the technique for the case α ∈ (1, 2) is more complex. Proof. We divide the proof into two parts. Part A The existence and uniqueness of the mild solution. Assume that u satisfies problem (1) then by using the method in [3] (p. 230), we have ∞ ∞ Z t X X D E D E ! α α−1 α (t − s) Eα,α − λ j (t − s) F(u(., s)), ϕ j ds ϕ j (x). u(x, t) = Eα,1 (−λ j t ) u(., 0), ϕ j ϕ j (x) + j=1
0
j=1
Substitute t by T into the above equation and express u(x, 0) in term of u(x, T ), we obtain (4). Set Z ∞ X Eα,1 (−λ j tα )Eα,α − λ j (T − s)α D E T α−1 G1 v(x, t) = F(v(., s)), ϕ j ds ϕ j (x), α) 0 (T − s) E (−λ T α,1 j j=1 G2 v(x, t) =
∞ Z X j=1
0
t
(t − s)
α−1
α
Eα,α − λ j (t − s)
D
E ! F(., s, u(., s)), ϕ j ds ϕ j (x).
3
Since Lemma 2, we get kG1 v(., t) − G1 w(., t)k2L2 (Ω) Z 2 ∞ X Eα,1 (−λ j tα )Eα,α − λ j (T − s)α D E T α−1 = F(v(., s)) − F(w(., s)), ϕ j ds α) 0 (T − s) E (−λ T α,1 j j=1 2 2α ∞ Z T M− 2 X D E2 λjT 1,α 2α−2 (T − s) ≤ T + F(v(., s)) − F(w(., s)), ϕ j ds M2,α j=1 0 λ2j (T − s)2α M− 2 Z T 1,α 2α+1 ≤T (T − s)−2 kF(v(., s)) − F(w(., s))k2L2 (Ω) ds. + M2,α 0
(7)
In view of (6), we get Z T Z T −2 2 (T − s) kF(v(., s)) − F(w(., s))kL2 (Ω) ds ≤ (T − s)−2 K 2 (s)kv(., s) − w(., s)k2L2 (Ω) ds 0 0 ! Z T −2 2 ≤ (T − s) K (s)ds kv(., s) − w(., s)k2L∞ (0,T ;L2 (Ω)) .
(8)
0
Combining (7) and (8), we get kG1 v − G1 wkL∞ (0,T ;L2 (Ω)) ≤ T
2α+1 2
M−1,α
M+2,α
A(T )kv(., s) − w(., s)kL∞ (0,T ;L2 (Ω))
(9)
Firstly, we note that for 0 ≤ t ≤ T ∞ D D E 2 X E 2 2 F(v(., t)) − F(w(., t)), ϕ j ≤ F(v(., t)) − F(w(., t)), ϕ j ≤ kF(v) − F(w)kL∞ (0,T ;L2 (Ω)) .
(10)
j=1
Moreover, it follows from Lemma 2 which together with Eα,1 (0) = 1 that Z t 1 1 − Eα,1 (−λ j tα ) . (t − s)α−1 Eα,α (−λ j (t − s)α )ds = λj 0
On the other hand, by Lemma 3, we know that Eα,1 (−λ j tα ) ≥ 0, which leads to (t − s)α−1 Eα,α (−λ j (t − s)α ) ≥ 0 for all t ≥ s ≥ 0. Hence Z t 0≤ (t − s)α−1 Eα,α −λ j (t − s)α ds, t ≥ s ≥ 0. (11) 0
Combining (10) and (11) yields
kG2 v(., t) − G2 w(., t)k2L2 (Ω) =
∞ Z X j=1
t 0
D E !2 (t − s)α−1 Eα,α (−λ j (t − s)α ) F(v(., s)) − F(w(., s)), ϕ j ds
≤ kF(v) −
F(w)k2L∞ (0,T ;L2 (Ω))
∞ Z X j=1
t
0
(t − s)
α−1
Eα,α (−λ j (t − s) )ds
∞ X 1 ≤ kKk2L∞ (0,T ) kv − wk2L∞ (0,T ;L2 (Ω)) λ2 j=1
α
!2
(12)
j
where kKkL∞ (0,T ) = sup0≤t≤T |K(t)|. It is known that λ j ≥ C j2/d for j ∈ N, where C is a positive constant independent of j; see e.g., Courant and Hilbert [1]. Hence for 0 < d < 4, we see that B2 =
∞ ∞ X 1X 1 1 ≤ < ∞. λ2 C j=1 n d4 j=1 j
Therefore kG2 v − G2 wkL∞ (0,T ;L2 (Ω)) ≤ BkKkL∞ (0,T ) kv − wkL∞ (0,T ;L2 (Ω)) .
(13) 4
Consider the following operator Pv =
∞ X E Eα,1 (−λ j tα ) D f, ϕ j ϕ j (x) + G1 v(x, t) + G2 v(x, t). α Eα,1 (−λ j T ) j=1
Using (9), (12) and (13), we deduce that kPv − PwkL∞ (0,T ;L2 (Ω)) ≤ kG1 v − G1 wkL∞ (0,T ;L2 (Ω)) + kG2 v − G2 wkL∞ (0,T ;L2 (Ω)) ≤ L(α, T )kv − wkL∞ (0,T ;L2 (Ω)) .
(14)
If w = 0 then since f ∈ H2 (Ω), we know that kPwk2L2 (Ω) =
∞ ∞ X E!2 D E2 Eα,1 (−λ j tα ) D 1 X f, ϕ ≤ (1 + λ j )2 f, ϕ j < ∞ j − α 2 Eα,1 (−λ j T ) |M1,α | j=1 j=1
which leads to Pw ∈ L2 (Ω). Hence if v ∈ L∞ (0, T ; L2 (Ω)) then Pv ∈ L∞ (0, T ; L2 (Ω)). In view of (14) and noting that L(α, T ) < 1 , we conclude that P is a contraction mapping. Hence, P has a unique fixed point u ∈ L∞ (0, T ; L2 (Ω)). Part B The instability of the mild solution. In order to illustrate the ill-posedness of the backward problem through an example, let f ∈ H2 (Ω) and and an explicitly defned function f ∈ L2 (Ω) given by ϕm f (x) = f (x) + √ . λm where = (m) = √1λ (for some positive m) represents the possible measurement noise. The corresponding solution to the m backward problem with such a noisy fnal data can be represented as v (x, t) =
∞ X Eα,1 (−λ j tα ) D E f , ϕ j ϕ j (x) + G1 v (x, t) + G2 v (x, t). α) E (−λ T α,1 j j=1
(15)
This implies that ∞
X
E Eα,1 (−λ j tα ) D kv (., t) − u(., t)kL2 (Ω) ≥
f − f, ϕ j ϕ j (x)
2 α L (Ω) Eα,1 (−λ j T ) j=1 − kG1 v (., t) − G1 u(., t)kL2 (Ω) + kG2 v (., t) − G2 u(., t)kL2 (Ω) . E
(16)
(−λ tα )
m The first term of (16) is √λ α,1 α . The next step is to estimate the second norm on the right-hand side of (16). Indeed, m E α,1 (−λm T ) from (9), (13), and using the triangle inequality, we get
kG1 v (., t) − G1 u(., t)kL2 (Ω) + kG2 v (., t) − G2 u(., t)kL2 (Ω) ≤ T
2α+1 2
M−1,α M+2,α
A(T )kv − ukL∞ (0,T ;L2 (Ω)) + BkKkL∞ (0,T ) kv − ukL∞ (0,T ;L2 (Ω))
= L(α, T )kv − ukL∞ (0,T ;L2 (Ω)) .
(17)
This gives that Eα,1 (−λm tα ) . kv (., t) − u(., t)kL2 (Ω) + L(α, T )kv − ukL∞ (0,T ;L2 (Ω)) ≥ √ λm Eα,1 (−λm T α )
(18)
Since the function Eα,1 (−λm tα ) is an decreasing function for 0 ≤ t ≤ T , we know that (1 + L(α, T )) kv − ukL∞ (0,T ;L2 (Ω)) ≥ kv (., t) − u(., t)kL2 (Ω) + L(α, T )kv − ukL∞ (0,T ;L2 (Ω)) √ Eα,1 (−λm tα ) 1 λm (1 + λm T α ) ≥ sup √ = √ ≥ . M−1,α λm Eα,1 (−λm T α ) λm Eα,1 (−λm T α ) 0≤t≤T
(19)
Therefore, we conclude that
λm (1 + λm T α ) = ∞. m→+∞ (1 + L(α, T )) M− 1,α
lim kv − ukL∞ (0,T ;L2 (Ω)) ≥ lim
m→+∞
√
(20)
So, the latter inequality shows that that even if the noise level = √1λ goes to zero, as m → +∞, the instability always happens m backwards in time. Hence, the need for a regularization method has been ascertained. 5
3. Regularization and error estimate The main idea of a general filter regularization method is to replace the term R(, λ j )
Eα,1 (−λ j tα ) Eα,1 (−λ j T α )
by the following quantity
Eα,1 (−λ j tα ) Eα,1 (−λ j T α )
in order to obtain the following regularized solution U (x, t) =
∞ X
R(, λ j )
j=1
Theorem 2. Let R(, λ j ) satisfy the estimates R(, λ j )
Eα,1 (−λ j tα ) D E f , ϕ j ϕ j (x) + G1 U (x, t) + G2 U (x, t). Eα,1 (−λ j T α )
Eα,1 (−λ j tα ) ≤ C , Eα,1 (−λ j T α )
for p > 0 and where C , C are such that
(21)
1 − R(, λ j ) ≤ C λ pj
(22)
lim C−1 = lim C = lim C = 0.
→0
→0
(23)
→0
Assume that f ∈ H p (Ω). Then we have kU − ukL∞ (0,T ;L2 (Ω)) ≤
C + C k f kH p (Ω) . 1 − L(α, T )
(24)
Proof. From the definition of G1 and G2 , we have
∞ ∞
X
α D α D X E E E (−λ t ) E (−λ t ) α,1 j α,1 j kU (., t) − u(., t)kL2 (Ω) ≤
R(, λ j ) f , ϕ j ϕ j (x) − f, ϕ j ϕ j (x)
α α
Eα,1 (−λ j T ) Eα,1 (−λ j T ) j=1 j=1 L2 (Ω) | {z }
(25)
Next, we estimate I1 as follows v v u u tX tX ∞ ∞ 2 Eα,1 (−λ j tα ) 2 D E2 E Eα,1 (−λ j tα ) 2 D f, ϕ j 2 1 − R(, λ ) I1 ≤ f − f, ϕ + R(, λ j ) j j Eα,1 (−λ j T α ) Eα,1 (−λ j T α ) j=1 j=1
(26)
I1
+ kG1 U − G1 ukL2 (Ω) + kG2 U − G2 ukL2 (Ω) .
≤ C k f − f |L2 (Ω) + C k f kH p (Ω) ≤ C + C k f kH p (Ω) .
Combining (14), (25), (26), we deduce that kU (., t) − u(., t)kL2 (Ω) ≤ C + C k f kH p (Ω) + L(α, T )kU − ukL∞ (0,T ;L2 (Ω)) .
(27)
The right hand side of (27) is independent of t which implies that kU − ukL∞ (0,T ;L2 (Ω)) ≤ C + C k f kH p (Ω) + L(α, T )kU − ukL∞ (0,T ;L2 (Ω)) .
(28)
Since L(α, T ) < 1, we obtain (24). Corollary 1. Let us choose R(, λ j ) as below 1, R(, λ j ) = 0,
if if
1 λ j ≤ ln( ), 1 λ j > ln( ).
(29)
Firstly, we check the condition (22). Indeed, we have R(, λ j )
! Eα,1 (−λ j tα ) 1 + λj 1 1 ≤ R(, λ ) ≤ 1 + ln( ) . j Eα,1 (−λ j T α ) M−α M−α
This implies that R(, λ j ) fulfills the condition (22) with C = 1 + ln( 1 ). Next, we check the condition (22). It is also see that −p 1 −p 1 1 − R(, λ j ) λ−p ≤ ln( ) . Hence, R(, λ ) fulfills the condition (22) with C = ln( ) . Then kU − ukL∞ (0,T ;L2 (Ω)) is of j j −p order max + ln( 1 ), ln( 1 ) . 6
E
(−λ T α )
Corollary 2. Let us choose R(, λ j ) := r +Eα,1α,1 (−λj j T α ) , for any 0 < r < 1. Firstly, using the inequality Eα,1 (−z) ≤ 1 for z ≥ 0, we get Eα,1 (−λ j tα ) Eα,1 (−λ j tα ) 1 = ≤ r ≤ −r . (30) R(, λ j ) α r α Eα,1 (−λ j T ) + Eα,1 (−λ j T ) + Eα,1 (−λ j T α ) Hence, R(, λ j ) fulfills the condition (22) with C = −r and using the fact that Eα,1 (−z) ≥
1 − R(, λ j ) λ−1 j =
r λ−1 j r + Eα,1 (−λ j T α )
≤
r λ−1 j Eα,1 (−λ j T α )
≤
M−1,α z ,
α r (λ−1 j + T )
M−1,α Eα,1 (−λ j T α )
≤
we get α λ−1 j +T
M−1,α
r.
(31)
These above imply that R(, λ j ) fulfills the condition (22) with C = r and p = 1. Then kU − ukL∞ (0,T ;L2 (Ω)) is of observations order max C , C = max r , 1−r . References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20] [21]
R. Courant, D. Hilbert, Methods of Mathematical Physics Interscience. New York; (1953). K. Diethelm, The analysis of fractional differential equationst, Springer, Berlin, 2010. A. Kilbas, H.M. Srivastava, J.J. Trujillo, Theory and Applications of Fractional Differential Equations, Elsevier, Amsterdam, 2006. L. Li, L. Jin, S. Fang, Existence and uniqueness of the solution to a coupled fractional diffusion system, Adv. Difference Equ. 2015, 2015:370. J.J. Liu and M. Yamamoto, A backward problem for the time-fractional diffusion equation, Appl. Anal. 89 (2010), 1769-1788. Y. Luchko, Some uniqueness and existence results for the initial-boundary-value problems for the generalized time-fractional diffusion equation, Comput. Math. Appl., 59 (2010), 1766-1772. J.A. Machado, V. Kiryakova and F. Mainardi, A poster about the recent history of fractional calculus, Frac. Calc. Appl. Anal., 13 (2010), 329-334. F. Mainardi, Fractional calculus and wave in linear viscoelasticity, Imperial College Press, London, 2010. I. Podlubny, Fractional differential equations, Academic Press, London, 1999. C. Ren, X. Xu, S. Lu, Regularization by projection for a backward problem of the time-fractional diffusion equation, J. Inverse Ill-Posed Probl., 22 (2014), 121-139. K. Sakamoto, M. Yamamoto, Initial value/boundary value problems for fractional diffusion-wave equations and applications to some inverse problems. J. Math. Anal. Appl., 382 (2011), 426-447. N.H. Tuan, L.D. Long, N.V. Thinh, T. Tran, On a final value problem for the time-fractional diffusion equation with inhomogeneous source, Inverse Problem in Science and Engineering, 25 (9)(2017), 1367-1395. N.H. Tuan, L.D. Long, S. Tatar, Tikhonov regularization method for a backward problem for the inhomogeneous time-fractional diffusion equation, Appl. Anal., 97 (2018), 842-863. N.H. Tuan, D.D. Trong, On a backward parabolic problem with local Lipschitz source, J. Math. Anal. Appl. 414 (2014), 678-692. V.A. Vo, N.H. Tuan, Identification of the initial condition in backward problem with nonlinear diffusion and reaction, J. Math. Anal. Appl., 452 (2017), 176-187. D.D. Trong; E. Nane, N.D. Minh; N.H. Tuan, Continuity of solutions of a class of fractional equations, Potential Anal., 49 (2018), 423-478. G.J. Wang, T. Wei, B.Y. Zhou, Optimal error bound and simplified Tikhonov regularization method for a backward problem for the time-fractional diffusion equation, J. Comput. Appl. Math., 279 (2015), 277-292. T. Wei, Y. Zhang, The backward problem for a time-fractional diffusion-wave equation in a bounded domain, Comput. Math. Appl., 75 (2018), 3632-3648. T. Wei, G.J. Wang, A modified quasi-boundary value method for the backward time-fractional diffusion problem, ESAIM Math. Model. Numer. Anal., 48 (2014), 603-621. F. Yang, Y.P. Ren, X.X. Li, The quasi-reversibility method for a final value problem of the time-fractional diffusion equation with inhomogeneous source, Math. Methods Appl. Sci., 41 (2018), 1774-1795. M. Yang, J. Liu Solving a final value fractional diffusion problem by boundary condition regularization, Applied Numerical Mathematics, 66 (2013), 45-58.
7
PII: DOI: Reference:
S0893-9659(18)30388-4 https://doi.org/10.1016/j.aml.2018.11.015 AML 5700
To appear in:
Applied Mathematics Letters
Received date : 12 October 2018 Revised date : 16 November 2018 Accepted date : 16 November 2018 Please cite this article as: N.H. Tuan, L.N. Huynh, T.B. Ngoc et al., On a backward problem for nonlinear fractional diffusion equations, Applied Mathematics Letters (2018), https://doi.org/10.1016/j.aml.2018.11.015 This is a PDF file of an unedited manuscript that has been accepted for publication. As a service to our customers we are providing this early version of the manuscript. The manuscript will undergo copyediting, typesetting, and review of the resulting proof before it is published in its final form. Please note that during the production process errors may be discovered which could affect the content, and all legal disclaimers that apply to the journal pertain.
Elsevier Editorial System(tm) for Applied Mathematics Letters Manuscript Draft Manuscript Number: AML-D-18-01892R1 Title: On a backward problem for nonlinear fractional diffusion equations Article Type: Regular Article Keywords: Fractional diffusion equation; backward problem; regularization; fixed point theory. Corresponding Author: Professor Tuan Nguyen Huy, Corresponding Author's Institution: Ton Duc Thang University First Author: Tuan Nguyen Huy Order of Authors: Tuan Nguyen Huy; Huynh Le Nhat; Ngoc Tran Bao; Yong Zhou Abstract: In this paper, we consider a backward problem for a time-space fractional diffusion equation with nonlinear source. Under some assumptions, we establish the existence and uniqueness of local mild solutions to the nonlinear problem. We also prove that the backward problem is illposed in the sense of Hadamard. A regularization method is proposed to approximate the solution. Furthermore, the convergence rate for the regularized solution is given.
*Manuscript Click here to view linked References
On a backward problem for nonlinear fractional diffusion equations Nguyen Huy Tuana , Le Nhat Huynhb , Tran Bao Ngocc , Yong Zhoud,e a Applied
Analysis Research Group, Faculty of Mathematics and Statistics, Ton Duc Thang University, Ho Chi Minh City, Vietnam.
b Faculty of Mathematics and Computer Science, University of Science, Vietnam National University (VNU-HCMC), Ho Chi Minh city, Vietnam. c Institute d Faculty
of Research and Development, Duy Tan University, Da Nang 550000, Vietnam.
of Information Technology, Macau University of Science and Technology, Macau 999078, China of Mathematics and Computational Science, Xiangtan University, Hunan 411105, China
e Faculty
Abstract In this paper, a backward problem for a time-space fractional diffusion with nonlinear source has been considered. Under some assumptions, we establish the existence and uniqueness of mild solutions of a local solution to the nonlinear problem. We also prove that our backward problem is ill-posed in the sense of Hadamard. A regularization method has been proposed to approximate the solution. Furthermore, the convergence rate for the regularized solution can be proved. Keywords: Backward problem, regularization, fractional diffusion equation, fixed point theory
1. Introduction The backward problem of diffusion process is of great importance in engineering and aims at detecting the previous status of physical field from its present information. They have many applications in practice. Indeed, they describe the blurring effect, and the backward problems give the mathematical formulation for the deblurring process in image restoration. In this paper, we consider the backward problem for a time-fractional diffusion equation with nonlinear source α ∂t u(x, t) = Au(x, t) + F(x, t, u(x, t)), (x, t) ∈ Ω × (0, T ), u(x, t) = 0, (x, t) ∈ ∂Ω × (0, T ), u(x, T ) = f (x), x ∈ Ω·
(1)
∂α where Ω is a bounded domain in Rd with sufficiently smooth boundary ∂Ω, α is the Caputo fractional derivative of order ∂t 0 < α < 1 (see [9]), F is a given function-satisfying some hypotheses. The operator A in (1) is a linear densely defined self-adjoint and positive definite elliptic operator on the connected bounded domain Ω with zero Dirichlet boundary condition. Studying fractional partial differential equations involving a final condition is urgent from the applications like stochastic model of living matter in biology; prediction when earthquake occurs; electroencephalography problems of recovering epilepsy points in human brain; determining far electromagnetic waves in universe, etc. In practice, many phenomena cannot be observed at the time t = 0, that is, the initial data may not be known. However, the phenomena can be measured at a backward time t = T in some situations. The problems equipped with these backward in time data are called backward in time problems, which showed their great importance in engineering areas and aimed at detecting the previous status of a physical field from its present information [12, 20]. The initial value problems for the fractional diffusion equation have been studied extensively in recent years, see Diethelm [2], Mainardi [8, 7]. In [6], Luchko established the existence and uniqueness of the solution for the initial boundary (Dirichlet) value problem. In [4], the authors considered the initial boundary value problem for a coupled fractional diffusion system. For homogeneous case of backward problem i.e., F = 0 Yamamoto et al [11] showed that Problem (1) has a unique weak solution in case f ∈ H 2 (Ω). In the case of the fractional order α ∈ (1, 2), T. Wei et al [18] have studied a homogeneus problem and obtained some regularization results. Some results concerning a homogeneous backward problem can be found in [5, 17, 19, 10, 21]. ∗ Corresponding
Author. Email: [email protected] (Nguyen Huy Tuan), [email protected] (Le Nhat Huynh), [email protected] (Tran Bao Ngoc); [email protected] (Yong Zhou) Preprint submitted to Elsevier
November 16, 2018
If F(u(x, t)) = F(x, t), Tuan et al [12] gave a quasi-boundary value regularization method to approximate the solution. In [13], the authors applied the Tikhonov method to regularizing the linear backward problem. Very recent, Yang et al [20] considered the problem (1) by using the method of quasi-reversibility. To the best of our knowledge, we could not find any result on backward problems for nonlinear time-fractional diffusion equations. The backward problems were treated for a long time, e.g. see [14, 15] which are corresponding to Problem (1) with derivatives of integer-orders i.e, α = 1. The difficulties of a backward problem can be briefly discribed as follows. Firstly, since the fractional derivative ∂αt u is defined on the time interval (0, t), it is impossible to transform a final value problem for fractional diffusion equation to an initial-value problem by using some substitution methods. Secondly, the formula of the mild solution of a backward problem is more complex than the initial problem. In fact, we can show the solution operators of an initial problem for fractional diffusion equation is bounded for all time in the interval [0, T ] (see [4]). However, the formula of the mild solution of a backward problem for fractional diffusion equation shows that the solution operators are unbounded at the time t = 0. Our main goal in this paper is to investigate the existence and regularization of local solutions of Problem (1). This paper is organized as follows. In section 2, we give a formula of the mild solution to Problem (1). Then we show the existence and uniqueness of the local solution. Moreover, we also give the instability of the mild solution. In section 3, we give a regularization method. 2. Existence and uniqueness of the mild solution Since A is introduced as above, the eigenvalues of A satisfy 0 < λ1 ≤ λ2 ≤ λ3 ≤ · · · ≤ λ j ≤ · · · with λ j → ∞ as j → ∞. The corresponding eigenfunctions are denoted respectively by ϕ j ∈ H01 (Ω). The functions ϕ j are normalized so that {ϕ j }∞j=1 is an orthonormal basis of L2 (Ω). Defining ( ) ∞ X D E 2k k 2 2 λ j | v, ϕ j | < +∞ , H (Ω) = v ∈ L (Ω) : j=1
2
k
where h·, ·i is the inner product in L (Ω), then H (Ω) is a Hilbert space equipped with norm kvk
Hk (Ω)
=
∞ X j=1
D E 1/2 2 λ2k . j | v, ϕ j |
Now, we consider the following definition and lemmas which are useful for our main results. Definition 1. (see [9]) The Mittag-Leffler function is Eα,β (z) =
∞ X k=0
zk , z∈C Γ(αk + β)
where α > 0 and β ∈ R are arbitrary constants. Lemma 1. (see [11]) For λ > 0 and 0 < α < 1, we have d Eα,1 (−λtα ) = −λtα−1 Eα,α (−λtα ), t > 0· dt Lemma 2. (see [11]) The following equality holds for λ > 0, α > 0 and m ∈ N dm Eα,1 (−λtα ) = −λtα−m Eα,α−m+1 (−λtα ), t > 0· dtm
(2)
Lemma 3. see [9] Let α ∈ (0, 1) then Eα,1 (−z) > 0 for any z > 0. Moreover, there exist three positive constants M−1,α , M+1,α , M+2,α such that M−1,α 1+z
≤ Eα,1 (−z) ≤
M+1,α 1+z
, Eα,α (−z) ≤
M+2,α 1+z
(3)
·
If α ∈ [α0 , α1 ] for any 0 < α0 < α1 < 1 then by [16], the constant M−1,α , M+1,α , M+1,α can be chosen which depends only α0 , α1 . 2
Definition 2. The function u is called a mild solution of Problem (1) if u ∈ L∞ (0, T ; L2 (Ω)) and satisfies that ∞ X E Eα,1 (−λ j tα ) D f, ϕ ϕ j (x) j Eα,1 (−λ j T α ) j=1 Z ∞ X Eα,1 (−λ j tα )Eα,α − λ j (T − s)α D E T α−1 − F(u(., s)), ϕ j ds ϕ j (x) α) 0 (T − s) E (−λ T α,1 j j=1 ∞ Z t X D E ! α−1 α (t − s) Eα,α − λ j (t − s) F(., s, u(., s)), ϕ j ds ϕ j (x). +
u(x, t) =
(4)
0
j=1
Theorem 1. Let T > 0. F satisfy kF(v(., t)) − F(w(., t))kL2 (Ω) ≤ K(t)kv(., t) − w(., t)kL2 (Ω) for any t ∈ [0, T ] and K is bounded. Assume that the integral L(α, T ) = T
2α+1 2
M−1,α
RT 0
(5)
(T − t)−2 K 2 (t)dt converges to |A(T )|2 which satisfies
A(T ) M+2,α
+ B sup |K(t)| < 1.
Let f ∈ H2 (Ω). Then Problem (1) has a unique mild solution in L∞ (0, T ; L2 (Ω)).
Remark 1. Assume that K(t) ≤ (T − t)m for m > 1. Then we see that the integral Let Ω = (0, π). It is easy to check that λ j = j2 and |A(T )|2 ≤ P 2 B ≤ ∞j=1 j12 = π6 . Once choice for T is
(6)
0≤t≤T
m−1
T m−1
RT 0
(T − t)−2 K 2 (t)dt is convergent.
and sup0≤t≤T |K(t)| ≤ T m . Moreover, we also have
1 1 3 1 (m − 1)M+ α+m− 2 2,α m T < min 2 , − M π 1,α
which implies that (6) holds.
Remark 2. For considering the direct problem, we can use the assumption (5) and the initial condition belongs to L2 (Ω). However, for considering the inverse problem, we have to give the final conditon which belongs to H 2 (Ω). The studying Problem (1) is more difficult if F has a general form, such as F = F(x; t; u; ∇u). We think the solution does not belong to L∞ (0, T ; L2 (Ω)). Our study is considering the solution which is in L∞ (0, T ; H 2 (Ω)) and this topic is a future work.
Remark 3. If α ∈ (1, 2), we need to one addition conditon ut at t = 0. Assume that ut (x, 0) = 0 then we can show the existence of the mild solution of Problem (1) by the method in the case α ∈ (0, 1). However, the technique for the case α ∈ (1, 2) is more complex. Proof. We divide the proof into two parts. Part A The existence and uniqueness of the mild solution. Assume that u satisfies problem (1) then by using the method in [3] (p. 230), we have ∞ ∞ Z t X X D E D E ! α α−1 α (t − s) Eα,α − λ j (t − s) F(u(., s)), ϕ j ds ϕ j (x). u(x, t) = Eα,1 (−λ j t ) u(., 0), ϕ j ϕ j (x) + j=1
0
j=1
Substitute t by T into the above equation and express u(x, 0) in term of u(x, T ), we obtain (4). Set Z ∞ X Eα,1 (−λ j tα )Eα,α − λ j (T − s)α D E T α−1 G1 v(x, t) = F(v(., s)), ϕ j ds ϕ j (x), α) 0 (T − s) E (−λ T α,1 j j=1 G2 v(x, t) =
∞ Z X j=1
0
t
(t − s)
α−1
α
Eα,α − λ j (t − s)
D
E ! F(., s, u(., s)), ϕ j ds ϕ j (x).
3
Since Lemma 2, we get kG1 v(., t) − G1 w(., t)k2L2 (Ω) Z 2 ∞ X Eα,1 (−λ j tα )Eα,α − λ j (T − s)α D E T α−1 = F(v(., s)) − F(w(., s)), ϕ j ds α) 0 (T − s) E (−λ T α,1 j j=1 2 2α ∞ Z T M− 2 X D E2 λjT 1,α 2α−2 (T − s) ≤ T + F(v(., s)) − F(w(., s)), ϕ j ds M2,α j=1 0 λ2j (T − s)2α M− 2 Z T 1,α 2α+1 ≤T (T − s)−2 kF(v(., s)) − F(w(., s))k2L2 (Ω) ds. + M2,α 0
(7)
In view of (6), we get Z T Z T −2 2 (T − s) kF(v(., s)) − F(w(., s))kL2 (Ω) ds ≤ (T − s)−2 K 2 (s)kv(., s) − w(., s)k2L2 (Ω) ds 0 0 ! Z T −2 2 ≤ (T − s) K (s)ds kv(., s) − w(., s)k2L∞ (0,T ;L2 (Ω)) .
(8)
0
Combining (7) and (8), we get kG1 v − G1 wkL∞ (0,T ;L2 (Ω)) ≤ T
2α+1 2
M−1,α
M+2,α
A(T )kv(., s) − w(., s)kL∞ (0,T ;L2 (Ω))
(9)
Firstly, we note that for 0 ≤ t ≤ T ∞ D D E 2 X E 2 2 F(v(., t)) − F(w(., t)), ϕ j ≤ F(v(., t)) − F(w(., t)), ϕ j ≤ kF(v) − F(w)kL∞ (0,T ;L2 (Ω)) .
(10)
j=1
Moreover, it follows from Lemma 2 which together with Eα,1 (0) = 1 that Z t 1 1 − Eα,1 (−λ j tα ) . (t − s)α−1 Eα,α (−λ j (t − s)α )ds = λj 0
On the other hand, by Lemma 3, we know that Eα,1 (−λ j tα ) ≥ 0, which leads to (t − s)α−1 Eα,α (−λ j (t − s)α ) ≥ 0 for all t ≥ s ≥ 0. Hence Z t 0≤ (t − s)α−1 Eα,α −λ j (t − s)α ds, t ≥ s ≥ 0. (11) 0
Combining (10) and (11) yields
kG2 v(., t) − G2 w(., t)k2L2 (Ω) =
∞ Z X j=1
t 0
D E !2 (t − s)α−1 Eα,α (−λ j (t − s)α ) F(v(., s)) − F(w(., s)), ϕ j ds
≤ kF(v) −
F(w)k2L∞ (0,T ;L2 (Ω))
∞ Z X j=1
t
0
(t − s)
α−1
Eα,α (−λ j (t − s) )ds
∞ X 1 ≤ kKk2L∞ (0,T ) kv − wk2L∞ (0,T ;L2 (Ω)) λ2 j=1
α
!2
(12)
j
where kKkL∞ (0,T ) = sup0≤t≤T |K(t)|. It is known that λ j ≥ C j2/d for j ∈ N, where C is a positive constant independent of j; see e.g., Courant and Hilbert [1]. Hence for 0 < d < 4, we see that B2 =
∞ ∞ X 1X 1 1 ≤ < ∞. λ2 C j=1 n d4 j=1 j
Therefore kG2 v − G2 wkL∞ (0,T ;L2 (Ω)) ≤ BkKkL∞ (0,T ) kv − wkL∞ (0,T ;L2 (Ω)) .
(13) 4
Consider the following operator Pv =
∞ X E Eα,1 (−λ j tα ) D f, ϕ j ϕ j (x) + G1 v(x, t) + G2 v(x, t). α Eα,1 (−λ j T ) j=1
Using (9), (12) and (13), we deduce that kPv − PwkL∞ (0,T ;L2 (Ω)) ≤ kG1 v − G1 wkL∞ (0,T ;L2 (Ω)) + kG2 v − G2 wkL∞ (0,T ;L2 (Ω)) ≤ L(α, T )kv − wkL∞ (0,T ;L2 (Ω)) .
(14)
If w = 0 then since f ∈ H2 (Ω), we know that kPwk2L2 (Ω) =
∞ ∞ X E!2 D E2 Eα,1 (−λ j tα ) D 1 X f, ϕ ≤ (1 + λ j )2 f, ϕ j < ∞ j − α 2 Eα,1 (−λ j T ) |M1,α | j=1 j=1
which leads to Pw ∈ L2 (Ω). Hence if v ∈ L∞ (0, T ; L2 (Ω)) then Pv ∈ L∞ (0, T ; L2 (Ω)). In view of (14) and noting that L(α, T ) < 1 , we conclude that P is a contraction mapping. Hence, P has a unique fixed point u ∈ L∞ (0, T ; L2 (Ω)). Part B The instability of the mild solution. In order to illustrate the ill-posedness of the backward problem through an example, let f ∈ H2 (Ω) and and an explicitly defned function f ∈ L2 (Ω) given by ϕm f (x) = f (x) + √ . λm where = (m) = √1λ (for some positive m) represents the possible measurement noise. The corresponding solution to the m backward problem with such a noisy fnal data can be represented as v (x, t) =
∞ X Eα,1 (−λ j tα ) D E f , ϕ j ϕ j (x) + G1 v (x, t) + G2 v (x, t). α) E (−λ T α,1 j j=1
(15)
This implies that ∞
X
E Eα,1 (−λ j tα ) D kv (., t) − u(., t)kL2 (Ω) ≥
f − f, ϕ j ϕ j (x)
2 α L (Ω) Eα,1 (−λ j T ) j=1 − kG1 v (., t) − G1 u(., t)kL2 (Ω) + kG2 v (., t) − G2 u(., t)kL2 (Ω) . E
(16)
(−λ tα )
m The first term of (16) is √λ α,1 α . The next step is to estimate the second norm on the right-hand side of (16). Indeed, m E α,1 (−λm T ) from (9), (13), and using the triangle inequality, we get
kG1 v (., t) − G1 u(., t)kL2 (Ω) + kG2 v (., t) − G2 u(., t)kL2 (Ω) ≤ T
2α+1 2
M−1,α M+2,α
A(T )kv − ukL∞ (0,T ;L2 (Ω)) + BkKkL∞ (0,T ) kv − ukL∞ (0,T ;L2 (Ω))
= L(α, T )kv − ukL∞ (0,T ;L2 (Ω)) .
(17)
This gives that Eα,1 (−λm tα ) . kv (., t) − u(., t)kL2 (Ω) + L(α, T )kv − ukL∞ (0,T ;L2 (Ω)) ≥ √ λm Eα,1 (−λm T α )
(18)
Since the function Eα,1 (−λm tα ) is an decreasing function for 0 ≤ t ≤ T , we know that (1 + L(α, T )) kv − ukL∞ (0,T ;L2 (Ω)) ≥ kv (., t) − u(., t)kL2 (Ω) + L(α, T )kv − ukL∞ (0,T ;L2 (Ω)) √ Eα,1 (−λm tα ) 1 λm (1 + λm T α ) ≥ sup √ = √ ≥ . M−1,α λm Eα,1 (−λm T α ) λm Eα,1 (−λm T α ) 0≤t≤T
(19)
Therefore, we conclude that
λm (1 + λm T α ) = ∞. m→+∞ (1 + L(α, T )) M− 1,α
lim kv − ukL∞ (0,T ;L2 (Ω)) ≥ lim
m→+∞
√
(20)
So, the latter inequality shows that that even if the noise level = √1λ goes to zero, as m → +∞, the instability always happens m backwards in time. Hence, the need for a regularization method has been ascertained. 5
3. Regularization and error estimate The main idea of a general filter regularization method is to replace the term R(, λ j )
Eα,1 (−λ j tα ) Eα,1 (−λ j T α )
by the following quantity
Eα,1 (−λ j tα ) Eα,1 (−λ j T α )
in order to obtain the following regularized solution U (x, t) =
∞ X
R(, λ j )
j=1
Theorem 2. Let R(, λ j ) satisfy the estimates R(, λ j )
Eα,1 (−λ j tα ) D E f , ϕ j ϕ j (x) + G1 U (x, t) + G2 U (x, t). Eα,1 (−λ j T α )
Eα,1 (−λ j tα ) ≤ C , Eα,1 (−λ j T α )
for p > 0 and where C , C are such that
(21)
1 − R(, λ j ) ≤ C λ pj
(22)
lim C−1 = lim C = lim C = 0.
→0
→0
(23)
→0
Assume that f ∈ H p (Ω). Then we have kU − ukL∞ (0,T ;L2 (Ω)) ≤
C + C k f kH p (Ω) . 1 − L(α, T )
(24)
Proof. From the definition of G1 and G2 , we have
∞ ∞
X
α D α D X E E E (−λ t ) E (−λ t ) α,1 j α,1 j kU (., t) − u(., t)kL2 (Ω) ≤
R(, λ j ) f , ϕ j ϕ j (x) − f, ϕ j ϕ j (x)
α α
Eα,1 (−λ j T ) Eα,1 (−λ j T ) j=1 j=1 L2 (Ω) | {z }
(25)
Next, we estimate I1 as follows v v u u tX tX ∞ ∞ 2 Eα,1 (−λ j tα ) 2 D E2 E Eα,1 (−λ j tα ) 2 D f, ϕ j 2 1 − R(, λ ) I1 ≤ f − f, ϕ + R(, λ j ) j j Eα,1 (−λ j T α ) Eα,1 (−λ j T α ) j=1 j=1
(26)
I1
+ kG1 U − G1 ukL2 (Ω) + kG2 U − G2 ukL2 (Ω) .
≤ C k f − f |L2 (Ω) + C k f kH p (Ω) ≤ C + C k f kH p (Ω) .
Combining (14), (25), (26), we deduce that kU (., t) − u(., t)kL2 (Ω) ≤ C + C k f kH p (Ω) + L(α, T )kU − ukL∞ (0,T ;L2 (Ω)) .
(27)
The right hand side of (27) is independent of t which implies that kU − ukL∞ (0,T ;L2 (Ω)) ≤ C + C k f kH p (Ω) + L(α, T )kU − ukL∞ (0,T ;L2 (Ω)) .
(28)
Since L(α, T ) < 1, we obtain (24). Corollary 1. Let us choose R(, λ j ) as below 1, R(, λ j ) = 0,
if if
1 λ j ≤ ln( ), 1 λ j > ln( ).
(29)
Firstly, we check the condition (22). Indeed, we have R(, λ j )
! Eα,1 (−λ j tα ) 1 + λj 1 1 ≤ R(, λ ) ≤ 1 + ln( ) . j Eα,1 (−λ j T α ) M−α M−α
This implies that R(, λ j ) fulfills the condition (22) with C = 1 + ln( 1 ). Next, we check the condition (22). It is also see that −p 1 −p 1 1 − R(, λ j ) λ−p ≤ ln( ) . Hence, R(, λ ) fulfills the condition (22) with C = ln( ) . Then kU − ukL∞ (0,T ;L2 (Ω)) is of j j −p order max + ln( 1 ), ln( 1 ) . 6
E
(−λ T α )
Corollary 2. Let us choose R(, λ j ) := r +Eα,1α,1 (−λj j T α ) , for any 0 < r < 1. Firstly, using the inequality Eα,1 (−z) ≤ 1 for z ≥ 0, we get Eα,1 (−λ j tα ) Eα,1 (−λ j tα ) 1 = ≤ r ≤ −r . (30) R(, λ j ) α r α Eα,1 (−λ j T ) + Eα,1 (−λ j T ) + Eα,1 (−λ j T α ) Hence, R(, λ j ) fulfills the condition (22) with C = −r and using the fact that Eα,1 (−z) ≥
1 − R(, λ j ) λ−1 j =
r λ−1 j r + Eα,1 (−λ j T α )
≤
r λ−1 j Eα,1 (−λ j T α )
≤
M−1,α z ,
α r (λ−1 j + T )
M−1,α Eα,1 (−λ j T α )
≤
we get α λ−1 j +T
M−1,α
r.
(31)
These above imply that R(, λ j ) fulfills the condition (22) with C = r and p = 1. Then kU − ukL∞ (0,T ;L2 (Ω)) is of observations order max C , C = max r , 1−r . References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20] [21]
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