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On a Riesz–Feller space fractional backward diffusion problem with a nonlinear source
Accepted Manuscript On a Riesz–Feller space fractional backward diffusion problem with a nonlinear source Nguyen Huy Tuan, Dinh Nguyen Duy Hai, Le Dinh Long, Van Thinh Nguyen, Mokhtar Kirane PII: DOI: Reference:
S0377-0427(16)00007-8 http://dx.doi.org/10.1016/j.cam.2016.01.003 CAM 10439
To appear in:
Journal of Computational and Applied Mathematics
Received date: 1 September 2015 Revised date: 3 January 2016 Please cite this article as: N.H. Tuan, D.N.D. Hai, L.D. Long, V.T. Nguyen, M. Kirane, On a Riesz–Feller space fractional backward diffusion problem with a nonlinear source, Journal of Computational and Applied Mathematics (2016), http://dx.doi.org/10.1016/j.cam.2016.01.003 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.
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On a Riesz - Feller space fractional backward diffusion problem with a nonlinear source Nguyen Huy Tuana , Dinh Nguyen Duy Haib , Le Dinh Longc , Van Thinh Nguyend , Mokhtar Kiranee,f,∗ a Applied
Analysis Research Group, Faculty of Mathematics and Statistics, Ton Duc Thang University, Ho Chi Minh City, Viet Nam b Department of Mathematics, University of Science, Vietnam National University, 227 Nguyen Van Cu Street, District 5, Ho Chi Minh City, Viet Nam c Institute of Computational Science and Technology in Ho Chi Minh City, Viet Nam d Department of Civil and Environmental Engineering, Seoul National University, Republic of Korea e Laboratoire de Mathmatiques Pˆ ole Sciences et Technologie, Universi´ e de La Rochelle, A´ enue M. Cr´ epeau, 17042 La Rochelle Cedex, France f Nonlinear Analysis and Applied Mathematics (NAAM) Research Group, Department of Mathematics, Faculty of Science, King Abdulaziz University, P.O. Box 80203, Jeddah 21589, Saudi Arabia
Abstract In this paper, a backward diffusion problem for a space-fractional diffusion equation with a nonlinear source in a strip is investigated. This problem is obtained from the classical diffusion equation by replacing the second-order space derivative with a Riesz-Feller derivative of order α ∈ (0, 2]. A nonlinear problem is severely ill-posed, therefore we propose two new modified regularization solutions to solve it. We further show that the approximated problems are well-posed and their solutions converge if the original problem has a classical solution. In addition, the convergence estimates are presented under a priori bounded assumption of the exact solution. For estimating the error of the proposed method, a numerical example has been implemented. Keywords: Space-fractional backward diffusion problem; Ill-posed problem; Regularization; Error estimate. 1. Introduction
5
10
15
In recent decades, fractional operators have been playing more and more important roles in science and engineering [4], for instance, in the theory of viscoelasticity and viscoplasticity (mechanics), in the modeling of polymers and proteins (biochemistry), in the transmission of ultrasound waves (electrical engineering), and in the modeling of human tissue under mechanical loads (medicine), which are referred to [6, 9, 14]. These new fractional-order models are more adequate than the integer-order models, because the fractional order derivatives and integrals enable to describe the memory and hereditary properties of different substance [10]. The space-fractional diffusion equation (SFDE) has been arisen from replacing the standard space partial derivative in the diffusion equation with a space fractional partial derivative. It can be derived from the continuous-time random walk in statistical mechanics, and has a wide range of applications in the theory of probability distribution, especially in the modeling of the high-frequency price dynamics in financial markets [10, 12]. If an initial concentration distribution and boundary conditions are given, a complete recovery of the unknown solution is attainable from solving a well-posed forward problem. The well-posed forward problem, i.e. initial and boundary value problem (IBVP) for the SFDE have been studied extensively in the last few years. P. Agarwal [13, 18, 25, 26] and his coauthors have recently developed some new methods for fractional differential equations. Liu et al. [8] considered stability and convergence of the difference methods for the space-time fractional advection-diffusion equation. Ray et al. [16] investigated application of modified ∗ Corresponding
author Email address: [email protected] (Mokhtar Kirane)
Preprint submitted to Elsevier
20
25
30
decomposition method for the analytical solution of space-fractional diffusion equation. Yang et al. [20] studied several numerical methods for fractional partial differential equations with Riesz space-fractional derivatives. The space fractional diffusion also investigated by some other authors, such as M. Younis et al [17], M.Duman et al [2], F. Liu [19], Z. Q. Chen [1], Y. Povstenko [15]. However, in some practical problems, the boundary data on the whole boundary cannot be obtained. In situation we only may know the noisy data on a part of the boundary or at some interior points of the considered domain, which leads to an inverse problem, namely the space-fractional inverse diffusion problems (SFIDP). To the best of our knowledge, the result from the study of SFIDP is still very limited to some specific cases dealing with homogeneous problems ([[3],[21],[22],[23],[24]]). Motivated by this reason, in this work, we consider a backward problem for the following nonlinear space fractional diffusion equation α ut (x, t) = x Dθ u(x, t) + f (x, t, u(x, t)), (x, t) ∈ R × (0, T ), u(x, t)|x→±∞ = 0, (x, t) ∈ R × (0, T ), (1) u(x, T ) = g(x), x ∈ R, where the space-fractional derivative x Dθα is the Riesz-Feller fractional derivative of order α(0 < α ≤ 2) and skewness θ (|θ| ≤ min{α, 2 − α}, θ 6= ±1) defined in [7], as follows: ( Z∞ (α + θ)π f (x + s) − f (x) Γ(1 + α) α sin ds x Dθ f (x) = π 2 s1+α 0 ) Z∞ (α − θ)π f (x − s) − f (x) + sin ds , 0 < α < 2, 2 s1+α 0 2 D2 f (x) = df (x) , α = 2. x 0 dx2
35
40
45
50
The space-fractional backward diffusion problem is to determine the distribution u(x, t) for 0 < t < T from the final concentration distribution u(x, T ). As it is well-known, the backward diffusion problem is severely ill-posed [22], i.e., solutions do not always exist, and in the case of existence, these do not depend continuously on the given data. In fact, from small noise contaminated physical measurements, the corresponding solutions have large errors. It makes difficult to numerical calculations. Hence, a regularization is in order. In recent years, the homogeneous problem, i.e, f = 0 in Equation (1) has been proposed by some authors. Zheng and Wei [22] used two methods, the spectral regularization and modified equation methods, to solve this problem. In [21], they developed an optimal modified method to solve this problem by an a priori and an a posteriori strategy. In 2014, Zhao et al [24] applied a simplified Tikhonov regularization method to deal with this problem. After then, a new regularization method of iteration type for solving this problem has been introduced by Cheng et al [3]. However, in many practical engineering applications, the diffusion occur in spatially heterogeneous environments, which requires to take account of a nonlinear source term. As mentioned above, up-to-date we have not found any publication dealing with the SFIDP with a nonlinear source term; because it requires a special technique to deal with the fractional terms and nonlinear term for solving this nonlinear problem, this is the most challenging task. Therefore in this study, we try to develop new methods and techniques to overcome this limitation. In comparison with previous studies ([[3],[21],[22],[23],[24]]) on solving the space-fractional backward diffusion problem, our paper shows a significant improvement, because it can deal with the space-fractional inverse diffusion problems with nonlinear source. This paper is organized as follows. In the following section we outline the main results. The proofs of these results is described in Section 3. In Section 4, a numerical example is proposed to show the effectiveness of the regularized methods. Then we end up the manuscript with the conclusion at Section 5.
2
55
2. The main results Let gb(ω) denote the Fourier transform of the integrable function g(x), which defined by 1 gb(ω) := √ 2π
+∞ Z exp(−ixω)g(x)dx,
−∞
i=
√
−1.
In terms of the Fourier transform, we have the following properties for the Riesz-Feller space-fractional derivative (see [11]) α \ x Dθ (g)(ω)
= −ψαθ (ω)b g (ω),
where πθ θπ + isign(ω) sin . ψαθ (ω) = |ω|α cos 2 2 60
(2)
By taking a Fourier transform to (1), we transform Problem (1) into the following differential equation ( u bt (ω, t) = −ψαθ (ω)b u(ω, t) + fb(ω, t, u(ω, t)), u b(ω, T ) = gb(ω).
Solving the latter problem, we have u b(ω, t) = exp
ψαθ (ω)(T
ZT h i − t) gb(ω) − exp ψαθ (ω)(s − T ) fb(ω, s, u(ω, s))ds .
(3)
t
From (3), applying the inverse Fourier transform, we get u(x, t) =
1 √ 2π
+∞ Z exp ψαθ (ω)(T − t)
−∞
h
× gb(ω) −
65
ZT t
i exp ψαθ (ω)(s − T ) fb(ω, s, u(ω, s))ds exp(ixω)dω.
(4)
θπ Note that ψαθ (ω) has the real part |ω|α cos( < s < T ; from (4) that, when |ω| positive 2 ). Since 0 ≤ t θ becomes large, the terms exp ψα (ω)(T − t) and exp ψαθ (ω)(s − t) increases rather quickly: small errors in high-frequency components can blow up and completely destroy the solution for 0 < t < T , therefore recovering the scalar (temperature, pollution) u(x, t) from the measured data g (x) is severely ill-posed. We define a regularization as follows +∞ Z exp ψαθ (ω)(T − t) 1 uβ (x, t) = √ 2π 1 + β exp |ω|α cos πθ T −∞
h × gb (ω) −
2
ZT t
i exp (ψαθ (ω)(s − T ) fb ω, s, uβ (ω, s) ds exp(iωx)dω,
where β is regularization parameter. We introduce the first theorem for main result as following
3
(5)
Theorem 1. Let g ∈ L2 (R) and let f ∈ L∞ (R × [0, T ] × R) satisfy f (x, t, 0) = 0 and |f (x, t, w) − f (x, t, v)| ≤ k|w − v|, 70
for constant k > 0 independent of x,t,w,v. Then there exists a unique solution uβ ∈ C([0, T ]; L2 (R)) to Problem (5). Let g ∈ L2 (R) be measured data such that kg − g kL2 (R) ≤ . Let us choose β such that lim→0 β −1 bounded. a) Suppose that the problem (1) has a unique solution u ∈ C([0, T ], L2 (R)) that satisfies +∞ 2 Z exp |ω|α cos πθ t u dω < ∞, for all t ∈ [0, T ], b (ω, t) 2
−∞
then we have the estimate
75
t
uβ (., t) − u(., t) 2 ≤ β T exp((T − t)2 k 2 )K, L (R)
where K=
√
2β
−1
21 +∞ 2 Z πθ α exp |ω| cos t u b(ω, t) dω . + 2 sup 2
t∈[0,T ] −∞
Moreover, if ut ∈ L2 ((0, T ); L2 (R)) then for all ∈ (0, 1) there exists a constant t > 0 such that 1 − 14
1 1
u(., 0) − uβ (., t ) 2 ≤ E8 4 T 4 ln , L (R) β
where
(6)
21 ZT 2 2 2 E = max kus (., s)kL2 (R) ds , exp((T − t) k )K . 0
b) Suppose that the problem (1) has a unique solution u ∈ C([0, T ], L2 (R)) that satisfies +∞ZT Z
−∞ 0
80
2 exp 2ψαθ (ω)s fb(ω, s, u(ω, s)) dsdω < ∞,
then we have the following estimate
uβ (., t) − u(., t) 2 L (R)
≤
where
D=
√
2β −1 + 4||u(., 0)||2L2 (R) + 4T
+∞ZT Z
−∞ 0
t β T exp (T − t)2 k 2 D,
(7)
(8)
21 2 exp(2ψαθ (ω)s) fb(ω, s, u(ω, s)) dsdω .
Remark 1. 1. If we choose β = , then the results in Theorem 1 hold. The error estimate (8) is no longer valid at t = 0. To approximate u(., 0), we use the similar technique as shown in (6). 2. In Theorem 1, we require the following condition +∞ 2 Z exp |ω|α cos πθ t u dω < ∞. b (ω, t) 2
−∞
4
(9)
85
90
95
If f (x, t, u) = 0 then the left hand side of (9) is equal to ku(., 0)k2L2 (R) . Therefore, the condition (9) is natural and acceptable. t If f (x, t, u) = 0, the error estimate in Theorem 1 is of order β T . It is similar to the homogeneous case in [21, 24]. 3. In Theorem 1, to obtain the error estimate, we require the strong assumptions of u as shown by (7) and (9). This is a limitation of Theorem 1, because there are not many functions u satisfied these conditions, particularly in practice these conditions are more difficult to be satisfied and checked. To improve this limitation, we introduce a new regularization solution and introduce a new technique to estimate the error in the Theorem 2 below. In fact, in this Theorem we only need a weaker assumption for u which is u ∈ C([0, T ]; L2 (R)), and this condition is natural and acceptable. Theorem 2. Let g, f, β be as Theorem 1. Suppose that 0 < kT < 1 and Problem (1) has a unique Then we construct a new regularized solution solution u ∈ C([0, T ]; L2 (R)). Let m ∈ 0, k21T 2 − 1 . Uβ ∈ C([0, T ]; L2 (R)) in such a way that
Uβ (., t) − u(., t) 2 L (R)
≤
β
−1
+ ku(., 0)kL2 (R)
s
1 1+ m t βT , 2 2 1 − (1 + m)T k
where Uβ is the function whose Fourier transform is defined by ZT exp ψαθ (ω)(T − t) exp ψαθ (ω)(s − t) c (ω, t) = gb (ω) − fb(ω, s, Uβ )ds U β α cos πθ T 1 + β exp |ω|α cos πθ T 1 + β exp |ω| 2 2 t Zt β exp |ω|α cos πθ T 2 exp (s − t)ψαθ (ω) fb(ω, s, Uβ )ds. + 1 + β exp |ω|α cos πθ 2 T 0
(10)
Remark 2. 1. Using the same technique as in (6), we obtain the estimate at t = 0 as kUβ (., t )
− u(., 0)kL2 (R)
≤
H2
−1 4
T
1 4
1 ln β
−1 4
,
where 12 21 ZT 1 1 + m H = max kus (., s)k2L2 (R) ds , β −1 + ku(., 0)kL2 (R) . 1 − (1 + m)T 2 k 2 0
100
2. In Theorem 2, our method has a drawback that the condition for k ∈ [0, T1 ) is still too strong which may restrict the class of equations. We are trying to remove this constraint. 3. If k = 0 and f = 0 then Problem (1) becomes a homogeneous problem. The error estimate in Theorem 2 t is of order β T . It is similar to the homogeneous case in [21, 24]. 3. Proof of main results
105
First, we consider the following Lemma
5
Lemma 1. Let t, s ∈ [0, T ]. 1) If s ≥ t, then we have
exp ψαθ (ω)(s − t) t−s ≤ β T . πθ α 1 + β exp |ω| cos 2 T exp ψαθ (ω)(T − t) ≤ β t−T T . 1 + β exp |ω|α cos πθ 2 T
a)
b)
2) If s ≤ t, then we have
β exp ψ θ (ω)(s − t + T ) α t−s ≤ β T . θπ 1 + β exp |ω|α cos 2 T
c)
Proof. First, we prove (a). In fact, we have exp ψαθ (ω)(s − t) exp |ω|α cos( πθ 2 )(s − t − T ) = 1 + β exp |ω|α cos( πθ )T β + exp − |ω|α cos( πθ 2 2 )T exp |ω|α cos( πθ 2 )(s − t − T ) = h i s−t h i T −s+t T T β + exp − |ω|α cos( πθ β + exp − |ω|α cos( πθ 2 )T 2 )T ≤
1 h i s−t T β + exp − |ω|α cos( πθ )T 2
≤ β 110
t−s T
.
Making the change s = T , we obtain (b). Next, we prove (c). In fact from (b), we have exp ψ θ (ω)(T − (t − s)) α πθ α 1 + β exp |ω| cos( 2 )T
≤ β
t−s−T T
,
it follows that
β exp ψ θ (ω)(s − t + T ) α ≤ πθ α 1 + β exp |ω| cos( 2 )T
β
t−s T
.
3.1. Proof of Theorem 1 115
Proof. We divide the proof into three steps. Step1. Constructing a regularized solution (5). We consider the problem ZT exp ψαθ (ω)(T − t) exp ψαθ (ω)(s − t) c (ω, t) = gb (ω) − fb(ω, s, uβ (ω, s))ds u β α cos πθ T 1 + β exp |ω|α cos πθ T 1 + β exp |ω| 2 2 t 6
or uβ (x, t)
=
exp ψαθ (ω)(T − t) gb (ω) exp(iωx)dω α cos πθ T 1 + β exp |ω| 2 −∞ +∞ZT θ Z exp ψ (ω)(s − t) α 1 fb(ω, s, uβ (ω, s)) exp(iωx)dsdω. −√ α cos πθ T 2π 1 + β exp |ω| 2 −∞ t +∞ Z
1 √ 2π
(11)
First, we prove that Problem (11) has a unique solution uβ that belongs to C([0, T ]; L2 (R)). Denoting 1 1 F (w)(x, t) = √ G(x, t) − √ 2π 2π for all w ∈ C([0, T ]; L2 (R)) and G(x, t) =
+∞ Z
−∞
120
+∞ZT Z
−∞ t
exp ψαθ (ω)(s − t) fb(ω, s, w) exp(iωx)dsdω, 1 + β exp |ω|α cos πθ 2 T
exp ψαθ (ω)(T − t) gb (ω) exp(iωx)dω. 1 + β exp |ω|α cos πθ 2 T
Since f (x, t, 0) = 0, and due to the Lipschitzian property of f (x, t, w) with respect to w, we get G(w) ∈ C([0, T ]; L2 (R)) for every w ∈ C([0, T ]; L2 (R)). We claim that, for every w, v ∈ C([0, T ]; L2 (R)), p ≥ 1, we have !2p k (T − t)p Qp 2 p p 2 kF (w)(., t) − F (v)(., t)kL2 (R) ≤ kw − vk , (12) β p! where Q = max{1, T } and k.k is the sup norm in L2 (R). We shall prove the latter inequality by induction. For p = 1, we have 2 kF (w)(., t) − F (v)(., t)kL2 (R) =
2
b
F (w)(., t) − Fb(v)(., t) 2
L (R)
2 exp ψαθ (ω)(s − t) fb(ω, s, w) − fb(ω, s, v) ds dω = 1 + β exp |ω|α cos πθ 2 T −∞ t +∞ ZT 2 ZT θ Z 2 exp ψ (ω)(s − t) α ds fb(ω, s, w) − fb(ω, s, v) ds ≤ dω 1 + β exp |ω|α cos πθ 2 T −∞ t t 2 +∞ ZT πθ α Z ZT )(s − t) exp |ω| cos( 2 2 ds fb(ω, s, w) − fb(ω, s, v) ds = dω α 1 + β exp |ω| cos πθ 2 T +∞ ZT Z
−∞
≤
≤
t
t
ZT
2 1
b
(T − t)
f (., s, w(., s)) − fb(., s, v(., s)) 2 ds 2 β L (R) t
1 (T − t) β2
ZT t
2
k 2 kw(., s) − v(., s)kL2 (R) ds ≤
7
2 k 2 (T − t)Q kw − vk . β
125
Therefore (12) holds for p = 1. Suppose that (12) holds for p = m (m ≥ 1). We prove that (12) holds for p = m + 1. We have
2
m+1
2
F (w)(., t) − F m+1 (v)(., t) L2 (R) = Fb(F m (w))(., t) − Fb(F m (v))(., t) L2 (R)
2 exp ψαθ (ω)(s − t) fb(ω, s, F m (w)) − fb(ω, s, F m (v)) ds dω = 1 + β exp |ω|α cos πθ 2 T −∞ t 2 +∞ ZT πθ α Z ZT 2 exp |ω| cos( )(s − t) 2 ds fb(ω, s, F m (w)) − fb(ω, s, F m (v)) ds ≤ dω 1 + β exp |ω|α cos πθ 2 T t −∞ t +∞ ZT Z
≤
1 (T − t) β2
=
1 (T − t) β2
+∞ZT Z 2 b f (., s, F m (w)(., s)) − fb(., s, F m (v)(., s)) dsdω
−∞ t
ZT
2
b
f (., s, F m (w)(., s)) − fb(., s, F m (v)(., s))
L2 (R)
ds
t
ZT
≤
1 (T − t)k 2 β2
≤
2m ZT 2(m+1) 1 k k (T − t)m+1 m+1 (T − s)m Qm 2 2 2 (T − t)k kw − vk ds ≤ Q kw − vk . β2 β m! β (m + 1)!
t
2
kF m (w)(., s) − F m (v)(., s)kL2 (R) ds
t
Therefore, by the induction principle, (12) holds for every integer p. From (12), we get kF p (w) − F p (v)kL2 (R) ≤ for every w, v ∈ C([0, T ]; L2 (R)). Since
p p k T2 p √ Q 2 kw − vk , β p!
p p T2 p k √ Q 2 = 0, p→∞ β p! lim
130
135
there exits a positive integer number p0 such that F p0 is a contraction. It follows that F p0 (w) = w has a unique solution uβ ∈ C([0, T ]; L2 (R)). We claim that F (uβ ) = uβ . In fact, one has F (F p0 (uβ )) = F (uβ ). Hence, F p0 (F (uβ )) = F (uβ ). By the uniqueness of the fixed point of F p0 , one has F (uβ ) = uβ , i.e., the equation F (w) = w has a unique solution
uβ ∈ C([0, T ]; L2 (R)). The main purpose of this paper is to estimate the error uβ − u 2 . To this end, we proceed to the next two steps. Step2. Let wβ be the function whose Fourier transform is defined by c (ω, t) w β
=
L (R)
ZT exp ψαθ (ω)(T − t) exp (ψαθ (ω)(s − t) gb(ω) − fb ω, s, wβ (ω, s) ds. 1 + β exp |ω|α cos πθ 1 + β exp |ω|α cos πθ 2 T 2 T t
(13)
We estimate wβ (., t) − uβ (., t)
L2 (R)
.
Using the Parseval’s equality and the inequality (a + b)2 ≤ 2(a2 + b2 ), we get 8
2
wβ (., t) − uβ (., t) 2
L (R)
2 +∞ Z exp ψαθ (ω)(T − t) b (ω) − gb(ω) dω ≤ 2 g 1 + β exp |ω|α cos πθ 2 T −∞ 2 +∞ ZT Z exp ψαθ (ω)(s − t) fb(ω, s, uβ (ω, s)) − fb(ω, s, wβ (ω, s)) ds dω +2 1 + β exp |ω|α cos πθ 2 T −∞ t = A1 + A2 ,
where A1
A2
140
2 +∞ Z exp ψαθ (ω)(T − t) b (ω) − gb(ω) dω, = 2 g 1 + β exp |ω|α cos πθ 2 T −∞ 2 +∞ ZT Z exp ψαθ (ω)(s − t) fb(ω, s, uβ (ω, s)) − fb(ω, s, wβ (ω, s)) ds dω. = 2 1 + β exp |ω|α cos πθ 2 T −∞ t
(14)
(15)
Term (15) can be estimated as follows 2 +∞ θ Z exp ψα (ω)(T − t) gb (ω) − gb(ω) 2 dω A1 ≤ 2 πθ α 1 + β exp |ω| cos( 2 )T −∞ 2β
2t−2T T
≤
2β
2t−2T T
≤
2β
2t−2T T
≤
+∞ Z gb (ω) − gb(ω) 2 dω
−∞
2
kg − gkL2 (R)
(16)
2
and A2
2 +∞ ZT Z exp ψαθ (ω)(s − t) fb(ω, s, wβ (ω, s)) − fb(ω, s, uβ (ω, s)) ds dω = 2 1 + β exp |ω|α cos πθ 2 T −∞ t 2 +∞ZT Z 2 exp ψαθ (ω)(s − t) fb(ω, s, wβ (ω, s)) − fb(ω, s, uβ (ω, s)) dsdω ≤ 2(T − t) 1 + β exp |ω|α cos( πθ 2 )T −∞ t ≤ 2(T − t)
+∞ZT Z
−∞ t 2t
≤ 2(T − t)β T k 2
β
ZT
2t−2s T
β
−2s T
t
2 b f (ω, s, wβ (ω, s)) − fb(ω, s, uβ (ω, s) dsdω
wβ (., s) − uβ (., s) 2 2 ds. L (R)
(17)
Combining (14), (16) and (17), we have
wβ (., t) − uβ (., t) 2 2 L (R)
≤
2β
2t−2T T
2t T
+ 2(T − t)β k 2
9
2
ZT t
β
−2s T
wβ (., s) − uβ (., s) 2 2 ds. L (R)
Hence, β
145
−2t T
wβ (., t) − uβ (., t) 2 2 L (R)
2β
≤
+ 2(T − t)k
−2 2
Using the Gronwall’s inequality, we obtain
2 −2t β T wβ (., t) − uβ (., t) L2 (R)
2
ZT
β
−2s T
t
wβ (., s) − uβ (., s) 2 2 ds. L (R)
≤ 2β −2 exp 2(T − t)2 k 2 2 .
Therefore, we conclude that √ t−T
wβ (., t) − uβ (., t) 2 2β T exp (T − t)2 k 2 . ≤ L (R)
. Step3. We shall estimate the error wβ (., t) − u(., t)
(18)
L2 (R)
Let wβ be the function defined in Step 2. We recall the Fourier transform of u and wβ from (3) and (13) ZT u b(ω, t) = exp ψαθ (ω)(T − t) gb(ω) − exp(ψαθ (ω)(s − T ))fb(ω, s, u(ω, s))ds (19) t
and
c (ω, t) w β
150
=
exp ψαθ (ω)(T − t)
1 + β exp |ω|α cos( πθ 2 )T
gb(ω) −
By direct calculation from (19) and (20) we get c (ω, t) u b(ω, t) − w β exp(ψαθ (ω)(T
=
−
ZT
β exp(|ω| cos( πθ 2 )T ) + β exp(|ω|α cos( πθ 2 )T ) α
=
− t))fb(ω, s, u(ω, s))ds +
1 +
ZT t
ZT t
exp(ψαθ (ω)(s
t
exp(ψαθ (ω)(T − t)) − t)) − 1 + β exp(|ω|α cos( πθ 2 )T )
exp(ψαθ (ω)(s
t
ZT
!
(20)
gb(ω)
exp(ψαθ (ω)(s − t)) fb(ω, s, wβ (ω, s))ds 1 + β exp(|ω|α cos( πθ 2 )T )
exp(ψαθ (ω)(T − t))b g (ω) −
−
T ))fb(ω, s, wβ (ω, s))ds .
ZT t
exp(ψαθ (ω)(s − t))fb(ω, s, u(ω, s))ds
exp(ψαθ (ω)(s − t)) b(ω, s, w (ω, s)) − fb(ω, s, u(ω, s)) ds. f β 1 + β exp(|ω|α cos( πθ 2 )T )
Using the Parseval
2 equality, we obtain
u(., t) − wβ (., t) 2 L (R)
=
+∞ Z 2 c (ω, t) dω u(ω, t) − w b β
−∞
2 +∞ ZT Z β exp(|ω|α cos( πθ )T ) 2 2 g (ω) − exp(ψαθ (ω)(s − t))fb(ω, s, u)ds dω ≤ 2 exp(ψαθ (ω)(T − t))b 1 + β exp(|ω|α cos( πθ 2 )T ) −∞
+2
t
+∞ ZT Z
−∞
= I1 + I2 ,
t
2 exp(ψαθ (ω)(s − t)) fb(ω, s, wβ (ω, s)) − fb(ω, s, u(ω, s)) ds dω 1 + β exp(|ω|α cos( πθ 2 )T ) 10
(21)
where 2 +∞ ZT Z β exp(|ω|α cos( πθ )T ) 2 2 g (ω) − exp(ψαθ (ω)(s − t))fb(ω, s, u)ds dω (22) I1 = 2 exp(ψαθ (ω)(T − t))b 1 + β exp(|ω|α cos( πθ 2 )T ) −∞
t
and
2 +∞ ZT Z θ exp(ψ (ω)(s − t)) α b b I2 = 2 1 + β exp(|ω|α cos( πθ )T ) f (ω, s, wβ (ω, s)) − f (ω, s, u(ω, s)) ds dω, 2 −∞
155
t
where the term I1 is estimated in the following manner I1
=
2 +∞ Z β exp(|ω|α cos( πθ )(T − t)) πθ α 2 2 exp |ω| cos t u b (ω, t) dω 1 + β exp |ω|α cos πθ 2 2 T −∞
≤ ≤
2t
2β T
+∞ 2 Z dω exp |ω|α cos πθ t u b (ω, t) 2
−∞
2β
2t T
+∞ 2 Z πθ α b(ω, t) dω sup exp |ω| cos 2 t u
(23)
t∈[0,T ] −∞
and using the H¨ older inequality, the term I2 is estimated as follows
I2
2 +∞ ZT Z θ exp(ψ (ω)(s − t)) α b b = 2 1 + β exp(|ω|α cos( πθ )T ) f (ω, s, wβ (ω, s)) − f (ω, s, u(ω, s)) ds dω 2 −∞ t 2 +∞ZT Z 2 exp(ψαθ (ω)(s − t)) b b(ω, s, u(ω, s)) dsdω ≤ 2(T − t) f (ω, s, w (ω, s)) − f β πθ 1 + β exp(|ω|α cos( 2 )T ) −∞ t
≤
2(T − t)k
2
ZT t
β
2t−2s T
wβ (., s) − u(., s) 2 2 ds. L (R)
Combining (21), (23) and (24), we have
u(., t) − wβ (., t) 2 2 L (R)
2t
≤ 2β T sup
+∞ 2 Z exp |ω|α cos πθ t u dω b (ω, t) 2
t∈[0,T ] −∞
+ 2(T − t)k 2
ZT
β
2t−2s T
t
Hence, β
−2t T
u(., t) − wβ (., t) 2 2 L (R)
≤
2 sup
+∞ 2 Z exp |ω|α cos πθ t u dω b (ω, t) 2
t∈[0,T ] −∞
+ 2(T − t)k 11
wβ (., s) − u(., s) 2 2 ds. L (R)
2
ZT t
β
−2s T
u(., s) − wβ (., s) 2 2 ds. L (R)
(24)
Using the Gronwall’s inequality, we obtain β
160
−2t T
u(., t) − wβ (., t) 2 2 L (R)
2 exp(2(T − t)2 k 2 ) sup
≤
+∞ 2 Z exp |ω|α cos πθ t u dω. b (ω, t) 2
t∈[0,T ] −∞
Whereupon
12 +∞ 2 Z
t πθ
u(., t) − wβ (., t) 2 exp |ω|α cos ≤ β T exp((T − t)2 k 2 ) 2 sup t u b(ω, t) dω . L (R) 2 t∈[0,T ] −∞
(25)
Combining (18) and (25) with applying the triangle inequality, we have √ t−T
uβ (., t) − u(., t) 2 2β T exp (T − t)2 k 2 ≤ L (R) 21 +∞ 2 Z t πθ 2 2 α exp |ω| cos + β T exp((T − t) k ) 2 sup t u b(ω, t) dω 2 t∈[0,T ] −∞
t
≤ β T exp((T − t)2 k 2 )K,
(26)
where K=
As u(x, t) − u(x, 0) =
√ Rt
21 +∞ 2 Z exp |ω|α cos πθ t u 2β −1 + 2 sup b(ω, t) dω . 2
t∈[0,T ] −∞
us (x, s)ds, it follows that
0
2
ku(., 0) − u(., t)kL2 (R)
≤ t
Due to (26) and (27), we obtain
u(., 0) − uβ (., t) 2 L (R) 165
≤ ≤
Zt 0
2 v u T uZ u kus (., s)k2L2 (R) ds ≤ t t kus (., s)k2L2 (R) ds .
(27)
0
ku(., 0) − u(., t)kL2 (R) + u(., t) − uβ (., t) L2 (R) √ t E t+βT ,
where E is defined in Theorem 1. For every ∈ (0, 1), there exists uniquely a positive number t such that √ t β t = β T , i.e., lntt = 2 ln T . Using the inequality ln t > −(1/t) for every t > 0, we get proposition 2
1 1
u(., 0) − uβ (., t ) 2 ≤ E8 4 T 4 L (R)
ln
1 β
− 41
.
Proof (b) of Theorem 1 From (3), we have 170
gb(ω) = exp −ψαθ (ω)T u b(ω, 0) +
ZT 0
exp sψαθ (ω) fb(ω, s, u(ω, s))ds .
12
(28)
Combining (22), (28), using inequality (a + b)2 ≤ 2(a2 + b2 ) and H¨ older inequality, we obtain I1
" # +∞ Zt Z β exp(|ω|α cos( πθ )T ) 2 2 θ θ 2 b b(ω, 0) + exp(ψα (ω)s)f (ω, s, u(ω, s))ds dω = 2 exp(−ψα (ω)t) u 1 + β exp(|ω|α cos( πθ 2 )T ) −∞ 0 2 +∞ Z Zt β exp ψ θ (ω)(s − t + T ) β exp − ψαθ (ω)t α u fb(ω, s, u(ω, s))ds dω = 2 b(ω, 0) + β + exp − |ω|α cos( πθ )T 1 + β exp |ω|α cos( πθ 2 2 )T −∞ 0 2 +∞ Z β exp − ψαθ (ω)t |b ≤ 4 u(ω, 0)|2 dω β + exp − |ω|α cos( πθ )T 2 −∞ 2 +∞ Zt θ Z β exp ψα (ω)(s − t + T ) b +4 f (ω, s, u(ω, s))ds dω 1 + β exp |ω|α cos( πθ )T 2 −∞ 0 2 +∞ Z exp − ψαθ (ω)t 2 |b ≤ 4β u(ω, 0)|2 dω β + exp − |ω|α cos( πθ )T 2 −∞ +∞ T Z Z exp ψ θ (ω)(s − t + T ) 2 2 α fb(ω, s, u(ω, s)) dsdω + 4T β 2 1 + β exp |ω|α cos( πθ )T 2 −∞ 0 ≤ 4β 2
+∞ +∞ZT Z Z 2 2(t−T ) 2(t−T ) u(ω, 0)|2 dω + 4T β 2 β T |b β T exp 2ψαθ (ω)s fb(ω, s, u(ω, s)) dsdω
−∞
≤ β
2t T
4||u(., 0)||2L2 (R)
+ 4T
+∞ZT Z
−∞ 0
exp
−∞ 0
! 2 b f (ω, s, u(ω, s)) dsdω .
2ψαθ (ω)s
(29)
Combining (21), (24) and (29), we have
u(., t) − wβ (., t) 2 2 L (R)
≤ 2(T − t)k +β
2t T
2
ZT
β
2t−2s T
t
u(., s) − wβ (., s) 2 2 ds L (R)
4||u(., 0)||2L2 (R)
+ 4T
+∞ZT Z
exp
−∞ 0
Hence,
2 −2t
β T u(., t) − wβ (., t) 2
! 2 b f (ω, s, u(ω, s)) dsdω .
2ψαθ (ω)s
L (R)
≤ 2(T − t)k 2 +
ZT
β
−2s T
t
4||u(., 0)||2L2 (R)
u(., s) − wβ (., s) 2 2 ds L (R) + 4T
+∞ZT Z
exp
−∞ 0
13
! 2 b f (ω, s, u(ω, s)) dsdω .
2ψαθ (ω)s
175
Using the Gronwall’s inequality, we obtain β
−2t T
u(., t) − wβ (., t) 2 2 L (R)
≤
exp 2(T − t)2 k 2 4||u(., 0)||2L2 (R) + 4T
! 2 exp 2ψαθ (ω)s fb(ω, s, u(ω, s)) dsdω .
+∞ZT Z
−∞ 0
Whereupon
u(., t) − wβ (., t) 2 L (R)
≤
t β T exp (T − t)2 k 2
× 4||u(., 0)||2L2 (R) + 4T
+∞ZT Z
−∞ 0
12 2 exp 2ψαθ (ω)s fb(ω, s, u(ω, s)) dsdω .
(30)
Applying the results of Step 2 and (30), we get the following estimate by using the triangle inequality
uβ (., t) − u(., t) 2 L (R) 21 +∞ZT Z 2 t ≤ β T exp (T − t)2 k 2 4||u(., 0)||2L2 (R) + 4T exp 2ψαθ (ω)s fb(ω, s, u(ω, s)) dsdω +
≤
√
2β
t−T T
−∞ 0
exp (T − t) k β exp (T − t)2 k 2 21 +∞ZT Z 2 √ × 2β −1 + 4||u(., 0)||2L2 (R) + 4T exp 2ψαθ (ω)s fb(ω, s, u(ω, s)) dsdω , t T
2 2
−∞ 0
for all t ∈ [0, T ]. This completes the proof of Theorem 1. 3.2. Proof of Theorem 2 180
We divide it into three steps. Step1. The existence and the uniqueness of a solution of (10). Construction of a regularized solution Uβ . We consider the problem ZT exp ψαθ (ω)(s − t) exp ψαθ (ω)(T − t) c (ω, t) = gb (ω) − fb(ω, s, Uβ )ds U β α cos πθ T 1 + β exp |ω|α cos πθ T 1 + β exp |ω| 2 2 t Zt β exp |ω|α cos πθ T 2 exp (s − t)ψαθ (ω) fb(ω, s, Uβ )ds, + 1 + β exp |ω|α cos πθ 2 T 0
14
(31)
or Uβ (x, t)
exp ψαθ (ω)(T − t) gb (ω) exp(iωx)dω = α cos πθ T 1 + β exp |ω| 2 −∞ +∞ZT θ Z exp ψ (ω)(s − t) α 1 fb(ω, s, Uβ ) exp(iωx)dsdω −√ α cos πθ T 2π 1 + β exp |ω| 2 −∞ t t +∞ πθ Z Z β exp |ω|α cos 2 T 1 exp (s − t)ψαθ (ω) fb(ω, s, Uβ ) exp(iωx)dsdω. + √ 2π 1 + β exp |ω|α cos πθ T 1 √ 2π
+∞ Z
2
−∞ 0
185
Let us define the norm on C([0; T ]; L2 (R)) as follows khk1 = sup β 0≤t≤T
−t T
kh(t)kL2 (R) , for all h ∈ C([0; T ]; L2 (R)).
It is easy to show that k · k1 is a norm of C([0; T ]; L2 (R)). For v ∈ C([0; T ]; L2 (R)), we consider the following function Θ(v)(x, t)
where
=
1 √ A(x, t) 2π +∞ZT Z exp ψαθ (ω)(s − t) 1 fb(ω, s, v) exp(iωx)dsdω −√ α cos πθ T 2π 1 + β exp |ω| 2 −∞ t +∞Zt πθ α Z β exp |ω| cos 2 T 1 exp (s − t)ψαθ (ω) fb(ω, s, v) exp(iωx)dsdω, +√ πθ 2π 1 + β exp |ω|α cos 2 T −∞ 0
A(x, t) =
+∞ Z
−∞
exp ψαθ (ω)(T − t) gb (ω) exp(iωx)dω. 1 + β exp |ω|α cos πθ 2 T
We claim that, for every v1 , v2 ∈ C([0; T ]; L2 (R))
kΘ(v1 ) − Θ(v2 )k1 ≤ kT kv1 − v2 k1 .
15
190
First, we have the two following estimates for all t ∈ [0, T ] 2 +∞ ZT Z exp ψαθ (ω)(s − t) fb(ω, s, v1 ) − fb(ω, s, v2 ) ds dω J1 = α cos πθ T 1 + β exp |ω| 2 −∞ t 2 +∞ZT Z 2 exp ψαθ (ω)(s − t) fb(ω, s, v1 ) − fb(ω, s, v2 ) dsdω ≤ (T − t) 1 + β exp |ω|α cos πθ 2 T −∞ t +∞ZT Z
β
≤ β T k 2 (T − t)
ZT
β
2
sup β
≤ (T − t)
−∞ t
2t
2t T
≤ β k (T − t) 2
2(t−s) T
−2s T
t
2 b f (ω, s, v1 ) − fb(ω, s, v2 ) dsdω
kv1 (., s) − v2 (., s)k2L2 (R) ds −2s T
0≤s≤T
kv1 (., s) − v2 (., s)k2L2 (R)
2t
≤ β T k 2 (T − t)2 kv1 − v2 k21
(32)
and J2
β exp |ω|α cos πθ 2 T exp (s − t)ψαθ (ω) = 1 + β exp |ω|α cos πθ 2 T −∞ 0 2 × fb(ω, s, v1 ) − fb(ω, s, v2 ) ds dω 2 +∞Zt πθ α Z β exp |ω| cos 2 T exp (s − t)ψαθ (ω) ≤ t 1 + β exp |ω|α cos πθ 2 T −∞ 0 2 × fb(ω, s, v1 ) − fb(ω, s, v2 ) dsdω +∞ Zt Z
≤ t
+∞Zt Z
−∞ 0
β
Zt
2t T
2
2t T
2 2
≤ β k t ≤ β k t
2(t−s) T
β
−2s T
0
2 b f (ω, s, v1 ) − fb(ω, s, v2 ) dsdω
kv1 (., s) − v2 (., s)k2L2 (R) ds
sup β
−2s T
0≤s≤T
kv1 (., s) − v2 (., s)k2L2 (R)
2t
(33)
≤ β T k 2 t2 kv1 − v2 k21 .
1 For 0 < t < T , using the inequality (a + b)2 ≤ (1 + m)a2 + (1 + m )b2 for all real numbers a and b and m > 0, we obtain 2
kΘ(v1 )(., t) − Θ(v2 )(., t)kL2 (R)
By choosing m = β
T −t t ,
−2t T
2t
≤ (1 + m)β T k 2 t2 kv1 − v2 k21 2t 1 + 1+ β T k 2 (T − t)2 kv1 − v2 k21 . m
we have
kΘ(v1 )(., t) − Θ(v2 )(., t)k2L2 (R) ≤ k 2 T 2 kv1 − v2 k21 , for all t ∈ (0, T ). 16
(34)
195
On other hand, letting t = 0 in (32), we have kΘ(v1 )(., 0) − Θ(v2 )(., 0)k2L2 (R) ≤ k 2 T 2 kv1 − v2 k21 .
(35)
By letting t = T in (33), we have β −2 kΘ(v1 )(., T ) − Θ(v2 )(., T )k2L2 (R) ≤ k 2 T 2 kv1 − v2 k21 .
(36)
Combining (34), (35) and (36), we obtain β
200
−2t T
kΘ(v1 )(., t) − Θ(v2 )(., t)k2L2 (R) ≤ k 2 T 2 kv1 − v2 k21 , for all t ∈ [0, T ]
which leads to (32). Since kT < 1, Θ is a contraction. It follows that the equation Θ(v) = v has a unique solution Uβ ∈ C([0; T ]; L2 (R)).
. Step 2. Estimate of the error Uβ (., t) − Vβ (., t) 2 L (R)
Let Vβ be the function whose Fourier transform is defined by
exp ψαθ (ω)(T − t) gb(ω) 1 + β exp |ω|α cos πθ 2 T ZT exp ψαθ (ω)(s − t) fb(ω, s, Vβ )ds − α cos πθ T 1 + β exp |ω| 2 t Zt β exp |ω|α cos πθ T 2 exp (s − t)ψαθ (ω) fb(ω, s, Vβ )ds. + 1 + β exp |ω|α cos πθ 2 T 0
Vc β (ω, t) =
(37)
From (31) and (37), we have
θ exp ψ (ω)(T − t) α c gb (ω) − gb(ω) Uβ (ω, t) − Vc β (ω, t) ≤ 1 + β exp |ω|α cos πθ 2 T T Z h i exp ψαθ (ω)(s − t) fb(ω, s, Vβ (ω, s)) − fb(ω, s, Uβ (ω, s)) ds + 1 + β exp |ω|α cos πθ 2 T t Zt β exp |ω|α cos πθ 2 T exp (s − t)ψαθ (ω) + 1 + β exp |ω|α cos πθ T 2 0 h i b b × f (ω, s, Uβ (ω, s)) − f (ω, s, Vβ (ω, s)) ds ≤
β
t−T T
+
Zt 0
≤ β
gb (ω) − gb(ω) +
t−T T
β
t−s T
ZT
β
t−s T
t
b f (ω, s, Vβ (ω, s)) − fb(ω, s, Uβ (ω, s)) ds
b f (ω, s, Vβ (ω, s)) − fb(ω, s, Uβ (ω, s)) ds
gb (ω) − gb(ω) +
ZT 0
17
β
t−s T
b f (ω, s, Vβ (ω, s)) − fb(ω, s, Uβ (ω, s)) ds.
Since m ∈ 0, k21T 2 − 1 , we have that 0 < k <
(a + b)2 ≤
T
√1 . 1+m
1+
1 m
From the inequality
a2 + (1 + m)b2
(38)
for all real number a, b and m > 0, we get
2
c
U (., t) − V (., t) 2 2 (., t) − Vc (., t) = U β
β
β
L (R)
≤
≤ 205
β
L2 (R)
2 2t−2T 1 1+ β T gb − gb L2 (R) m 2 +∞ ZT Z t−s β T fb(ω, s, Vβ (ω, s)) − fb(ω, s, Uβ (ω, s)) ds dω + (1 + m) 1 1+ m
−∞
β
0
2t−2T T
+ (1 + m)T k 2
2
ZT
β
2t−2s T
0
This leads to β
−2t T
Uβ (., t) − Vβ (., t) 2 2 L (R)
≤
1+
1 m
Uβ (., s) − Vβ (., s) 2 2 ds. L (R)
β −2 2
+ (1 + m)T k 2
ZT 0
−2t
β
−2s T
Uβ (., s) − Vβ (., s) 2 2 ds. L (R)
(39)
Set Z(t) = β T kUβ (., t) − Vβ (., t)k2L2 (R) , ∀t ∈ [0, T ]. Since Uβ , Vβ ∈ C([0, T ]; L2 (R)), we see that the function Z is continuous on [0, T ] and attains over there its maximum M at some t0 ∈ [0, T ]. Let M = −2t max β T kUβ (., t) − Vβ (., t)k2L2 (R) . From (39), we obtain t∈[0,T ]
M or
≤
1 1+ β −2 2 + (1 + m)T 2 k 2 M, m
1 − (1 + m)T 2 k 2 M ≤
210
1 1+ β −2 2 . m
This implies that for all t ∈ [0, T ] β Hence
−2t T
Uβ (., t) − Vβ (., t) 2 2 ≤M ≤ L (R)
Uβ (., t) − Vβ (., t) 2 ≤ L (R)
s
−2 2 1 1+ m β . 1 − (1 + m)T 2 k 2
1 1+ m t−T β T . 1 − (1 + m)T 2 k 2
18
Step 3. Estimate u(., t) − Vβ (., t)
L2 (R)
We have
u b(ω, t)
.
= exp ψαθ (ω)(T − t) gb(ω) −
ZT t
exp ψαθ (ω)(s − T ) fb(ω, s, u(ω, s))ds
ZT exp ψαθ (ω)(T − t) gb(ω) − exp ψαθ (ω)(s − T ) fb(ω, s, u(ω, s))ds = 1 + β exp |ω|α cos πθ 2 T t θ α β exp ψα (ω)(T − t) exp |ω| cos πθ 2 T + 1 + β exp |ω|α cos πθ 2 T T Z × gb(ω) − exp ψαθ (ω)(s − T ) fb(ω, s, u(ω, s))ds .
(40)
t
On other hand, we get
u b(ω, T ) = gb(ω) = exp −
215
This implies that
gb(ω) −
ZT t
T ψαθ (ω)
u b(ω, 0) +
ZT 0
exp
sψαθ (ω)
b f (ω, s, u(ω, s))ds .
exp ψαθ (ω)(s − T ) fb(ω, s, u(ω, s))ds
Zt = exp − T ψαθ (ω) u b(ω, 0) + exp (s − T )ψαθ (ω) fb(ω, s, u(ω, s))ds.
(41)
0
Combining (40) and (41), we obtain ZT exp ψαθ (ω)(T − t) θ b gb(ω) − exp ψα (ω)(s − T ) f (ω, s, u(ω, s))ds u b(ω, t) = 1 + β exp |ω|α cos πθ 2 T t πθ θ α β exp − tψα (ω) exp |ω| cos 2 T + u b(ω, 0) 1 + β exp |ω|α cos πθ 2 T Zt β exp |ω|α cos πθ 2 T exp (s − t)ψαθ (ω) fb(ω, s, u(ω, s))ds. + 1 + β exp |ω|α cos πθ 2 T 0
It follows from (37) and (42) that
u b(ω, t) − Vc β (ω, t) = B1 + B2 + B3 ,
19
(42)
where B1
=
B2
=
B3
=
h i exp ψαθ (ω)(s − t) fb(ω, s, Vβ (ω, s)) − fb(ω, s, u(ω, s)) ds, 1 + β exp |ω|α cos πθ 2 T t β exp − tψαθ (ω) exp |ω|α cos πθ 2 T u b(ω, 0), πθ 1 + β exp |ω|α cos 2 T Zt h i β exp |ω|α cos πθ 2 T exp (s − t)ψαθ (ω) fb(ω, s, u(ω, s)) − fb(ω, s, Vβ (ω, s)) ds. 1 + β exp |ω|α cos πθ 2 T 0
ZT
This leads to
u(ω, t) − Vc b β (ω, t) ≤
|B1 | + |B2 | + |B3 | ZT
≤
β
t−s T
t
t
b f (ω, s, u) − fb(ω, s, Vβ ) ds
+ β T |b u(., 0)| + t
u(., 0)| + β T |b
≤ 220
Using (38), we get
u(., t) − Vβ (., t) 2 2 L (R)
ZT
Zt
β
t−s T
0
β
t−s T
0
b f (ω, s, u) − fb(ω, s, Vβ ) ds.
1 1+ m
−∞
0
2t T
β ku(., 0)k2L2 (R) 2t
+ (1 + m)T k 2 β T
ZT
β
−2s T
0
Thus −2t T
(43)
2
= b u(., t) − Vc β (., t) 2 L (R) 2t 1 β T ku(., 0)k2L2 (R) ≤ 1+ m 2 +∞ ZT Z t−s β T fb(ω, s, u) − fb(ω, s, Vβ ) ds dω + (1 + m)
≤
β
b f (ω, s, u) − fb(ω, s, Vβ ) ds
u(., t) − Vβ (., t) 2 2 L (R)
u(., s) − Vβ (., s) 2 2 ds. L (R)
1 1+ ku(., 0)k2L2 (R) m ZT
2 −2s × + (1 + m)T k 2 β T u(., s) − Vβ (., s) L2 (R) ds.
≤
0
Since u(., t) and Vβ (., t) ∈ C([0, T ]; L2 (R)), the function u(., t) − Vβ (., t) is continuous on [0, T ]. L2 (R)
2 −2t Therefore, there exists a positive N = max β T u(., t) − Vβ (., t) L2 (R) . This implies that N≤
t∈[0,T ]
1+
1 m
ku(., 0)k2L2 (R) + (1 + m)T 2 k 2 N, 20
that is, β
−2t T
1 1+ m ku(., 0)k2L2 (R)
u(., t) − Vβ (., t) 2 2 ≤ N ≤ . L (R) 1 − (1 + m)T 2 k 2
Hence, we obtain the error estimate
u(., t) − Vβ (., t) 2 ≤ L (R)
s
1 1+ m t ku(., 0)kL2 (R) β T . 1 − (1 + m)T 2 k 2
From Step 2 and Step 3, we obtain
Uβ (., t) − u(., t) 2 ≤ Uβ (., t) − Vβ (., t) L2 (R) + u(., t) − Vβ (., t) L2 (R) L (R) s s 1 1 1+ m 1+ m t−T t T + β ku(., 0)kL2 (R) β T ≤ 1 − (1 + m)T 2 k 2 1 − (1 + m)T 2 k 2 s 1 1+ m t β −1 + ku(., 0)kL2 (R) ≤ βT . 1 − (1 + m)T 2 k 2 4. Numerical experiment 225
230
To verify our proposed methods, we carry out the numerical experiment for above regularization methods. The numerical example is implemented for T=1, and with the variation of α ∈ (0, 2]. In order to illustrate the sensitivity of the computational accuracy to the noise of the measurement data, we use the random function to generate the noisy data similar to an observation data. The perturbation was defined as rand(size()), where rand(size()) is a random number, and plays as an amplitude of the errors. The approximation of the regularization solution is computed by the FFT algorithm. In this example, we consider the following Riesz - Feller space fractional backward diffusion problem α ut (x, t) = x Dθ u(x, t) + f (x, t, u), (x, t) ∈ (−π, π) × (0, T ), u(−π, t) = u(π, t) = 0, t ∈ (0, T ), (44) u(x, T ) = g(x), x ∈ (−π, π). Here,
f (x, t, u)
=
−u + h(x, t),
where g(x)
= e sin3 (2x),
h(x, t) = αt
α−1
235
240
( ) h 1 α i h i 1 α α exp(t ) sin (2x) − exp(t ) 3 1 + sin(2x) − 1 + (6) sin(6x) , 4 2 α
3
with x ∈ (−π, π), subjecting to u(−π, t) = u(π, t) = 0, and u(x, 0) = sin3 (2x). The exact solution of Problem (44) is u(x, t) = exp(tα ) sin3 (2x). The regularized solution is defined by (5) with ψαθ (ω) defined by (2). The regularized solution is expected to be closed to the exact solution under a proper discretization as shown below. In general, the whole numerical procedure is proceeded in the following steps: Step 1. Choosing I and J to generate spatial and temporal discretizations as follows xi
=
tj
=
2π , i = 0, I, I 1 j∆t, ∆t = , j = 0, J. J i∆x, ∆x =
21
Of course, the higher value of I and J will provide more stable and accurate numerical calculation, however in the following examples I = J = 201 are satisfied. Step 2. We choose the couple (g , F ) as the observed data including the noise in the manner that g F (., .) 245
rand(size(g)) √ , π (rand(size(F (., .)))) √ . F (., .) + π
= g+ =
Step 3. Set u(., :)β() (xi ) = u(., :)β(),i and u(., :) (xi ) = ui , constructing two following vectors contained all discrete values of u(., :)β and u(., :) denoted by Λβ and Ψ, respectively, Λβ = u(., :)β,0 u(., :)β,1 ... u(., :)β,I−1 u(., :)β,I ∈ RI+1 , Ψ = u(., :)0 u(., :)1 ... u(., :)I−1 u(., :)I ∈ RI+1 . Step 4. Errors between the exact and its regularized solutions are estimated by Absolute error estimation v u I u 1 X E1 = t |u (xi , .) − u(xi , .)|2 . I + 1 i=0 β Relative error estimation
E2 =
250
255
260
265
s
I P
i=0
|uβ (xi , .) − u(xi , .)|2
s
I P
i=0
|u(xi
.
, .)|2
Fig. 1 shows the graphs of the exact solutions of Problem (44) corresponding to α = 0.2, 0.6, 1.4 and 1.8. In this example as shown in Fig. 2, when α is small (close to zero) the concentration distribution u(x, t) does not fluctuate much once t is increased. On the other hand, when t tends to zero the concentration distribution u(x, t) is more fluctuating once the value of α is decreased to zero as shown in Fig. 3. Figs. 4 and 5 show the comparisons between the exact and its regularized solutions at t = 0.5, and for α = 0.2 and 0.8, respectively. Figures 6 and 7 show the comparisons between the exact and its regularized solutions at t = 0.9, and for α = 0.2 and 0.8, respectively. Even for α = 0.2, at the beginning the regularized solution is oscillated around the exact solution, eventually the regularized solution is converged very well to the exact solution once tends to 0. Tables 1 and 2 show the estimate of absolute (E1 ) and relative (E2 ) errors between the exact and its regularized solutions with different values of α (0.2, 0.6, 1.4 and 1.8) at t = 0.5 and 0.9, respectively. It clearly shows the regularized solution converges to the exact solution with different values of α. Tables 3, 4, 5 and 6 show the estimate of absolute (E1 ) and relative (E2 ) errors between the exact and its regularized solutions at the variation of t from 0.1 to 0.8 for a lower value of α = 0.2, and a higher value of α = 1.8, respectively. These tables are not including the values of t at 0.5 and 0.9, because they are already shown in Tables 1 and 2 above. In general, the errors between the exact and its regularized solutions are slightly higher once the order α of the Riesz-Feller fractional derivative is closed to zero. It also confirms again the regularized solution is converged very well to the exact solution in the range of t ∈ (0, 1), and of α ∈ (0, 2]. 5. Conclusion
270
In this study, we introduced two new modified regularization methods as shown in Theorems 1 and 2 for solving a Riesz-Feller space fractional backward diffusion problem with a nonlinear source. In theoretical results, we overcome the strong condition as shown in (7) and (9) by a weaker condition based on Theorem 2, and the error estimates are obtained. In our numerical experiment, it is shown that the above regularized solution converges nicely to the exact solution with different values of α ∈ (0, 2]. 22
(a) α = 0.2
(b) α = 0.6
(c) α = 1.4
(d) α = 1.8 Figure 1: The exact solutions with different α .
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24
(a) α = 0.2
(b) α = 0.6
(c) α = 1.4
(d) α = 1.8
Figure 2: The exact solutions with different α at t=0.2, 0.5, 0.7 and 0.9.
(a) t = 0.2
(b) t = 0.5
(c) t = 0.7
(d) t = 0.9
25
Figure 3: The exact solutions at different t with different values of α (α = 0.2, 0.6, 1.4 and 1.8)
(a) = 10−1
(b) = 10−2
(c) = 10−3
(d) = 10−4
Figure 4: A comparison between the exact and its regularized solutions at t = 0.5 with α = 0.2 .
(a) = 10−1
(b) = 10−2
(c) = 10−3
(d) = 10−4
Figure 5: A comparison between the exact and its regularized solutions at t = 0.5 with α = 1.8.
26
(a) = 10−1
(b) = 10−2
(c) = 10−3
(d) = 10−4
Figure 6: A comparison between the exact and its regularized solutions at t = 0.9 with α = 0.2.
(a) = 10−1
(b) = 10−2
(c) = 10−3
(d) = 10−4
Figure 7: A comparison between the exact and its regularized solutions at t = 0.9 with α = 1.8.
27
1E-01 1E-02 1E-03 1E-04 1E-05
α = 0.2 E1 E2 4.19E-01 6.55E-01 1.69E-01 2.65E-01 6.02E-02 9.40E-02 2.09E-02 3.27E-02 7.30E-03 1.13E-02
α = 0.6 E1 E2 3.62E-01 9.27E-01 1.24E-01 3.17E-01 4.24E-02 1.09E-01 1.47E-02 3.76E-02 5.10E-03 1.30E-02
α = 1.4 E1 E2 8.39E-01 1.55E+00 2.55E-01 4.71E-01 1.04E-01 1.92E-01 3.67E-02 6.77E-02 1.27E-02 2.35E-02
α = 1.8 E1 E2 7.69E-01 9.96E-01 7.43E-02 9.62E-02 4.36E-02 5.65E-02 3.99E-03 5.16E-03 1.24E-04 1.60E-04
Table 1: The absolute (E1 ) and relative (E2 ) error estimates between the exact and its regularized solutions at t = 0.5.
1E-01 1E-02 1E-03 1E-04 1E-05
α = 0.2 E1 E2 3.32E-01 5.86E-01 1.37E-01 2.41E-01 4.85E-02 8.55E-02 1.69E-02 2.97E-02 5.90E-03 1.03E-02
α = 0.6 E1 E2 2.26E-01 4.21E-01 7.50E-02 1.40E-01 2.60E-02 4.84E-02 9.00E-03 1.68E-02 3.10E-03 5.80E-03
α = 1.4 E1 E2 9.17E-01 1.75E+00 2.69E-01 5.13E-01 8.85E-02 1.69E-01 3.06E-02 5.84E-02 1.06E-02 2.02E-02
α = 1.8 E1 E2 5.96E-01 9.97E-01 5.82E-01 9.72E-01 3.00E-02 5.01E-02 3.72E-02 6.22E-02 1.12E-03 1.86E-03
Table 2: The absolute (E1 ) and relative (E2 ) error estimates between the exact and its regularized solutions at t = 0.9.
335
t 1E-01 1E-02 1E-03 1E-04 1E-05
t = 0.1 E1 E2 5.23E-01 7.44E-01 2.09E-01 2.97E-01 7.42E-02 1.05E-01 2.58E-02 3.66E-02 8.90E-03 1.27E-02
t = 0.2 E1 E2 4.94E-01 7.17E-01 1.99E-01 2.88E-01 7.04E-02 1.02E-01 2.45E-02 3.55E-02 8.50E-03 1.23E-02
t = 0.3 E1 E2 4.69E-01 6.95E-01 1.89E-01 2.80E-01 6.69E-02 9.92E-02 2.32E-02 3.45E-02 8.10E-03 1.20E-02
t = 0.4 E1 E2 4.43E-01 6.74E-01 1.79E-01 2.72E-01 6.35E-02 9.65E-02 2.21E-02 3.36E-02 7.70E-03 1.16E-02
Table 3: The absolute (E1 ) and relative (E2 ) error estimates between the exact and its regularized solutions with α = 0.2.
28
t 1E-01 1E-02 1E-03 1E-04 1E-05
t = 0.6 E1 E2 3.96E-01 6.36E-01 1.61E-01 2.58E-01 5.71E-02 9.17E-02 1.99E-03 3.19E-03 6.90E-04 1.11E-03
t = 0.7 E1 E2 3.74E-01 6.19E-01 1.52E-01 2.52E-01 5.41E-02 8.95E-02 1.88E-03 3.11E-03 6.50E-04 1.08E-03
t = 0.8 E1 E2 3.53E-01 6.03E-01 1.44E-01 2.46E-01 5.13E-02 8.75E-02 1.78E-03 3.05E-03 6.20E-04 1.06E-03
Table 4: The absolute (E1 ) and relative (E2 ) error estimates between the exact and its regularized solutions with α = 0.2.
t 1E-01 1E-02 1E-03 1E-04 1E-05
t = 0.1 E1 E2 5.23E+00 9.95E-01 5.08E-01 9.96E-01 9.70E-02 1.84E-02 3.01E-03 5.72E-03 9.19E-04 1.75E-04
t = 0.2 E1 E2 3.56E+00 9.96E-01 3.46E-01 9.67E-01 2.67E-02 7.46E-02 2.01E-03 5.62E-03 6.16E-04 1.72E-01
t = 0.3 E1 E2 2.32E+00 9.98E-01 2.50E-01 9.66E-01 5.98E-02 2.60E-02 2.25E-03 5.47E-03 3.93E-04 1.69E-04
t = 0.4 E1 E2 1.40E+00 9.53E-01 1.36E-01 9.62E-01 5.84E-02 4.14E-02 7.36E-03 5.21E-03 2.30E-04 1.63E-04
Table 5: The absolute (E1 ) and relative (E2 ) error estimates between the exact and its regularized solutions with α = 1.8.
t 1E-01 1E-02 1E-03 1E-04 1E-05
t = 0.6 E1 E2 6.10E-01 9.97E-01 5.97E-01 9.76E-01 4.14E-02 6.77E-02 3.50E-03 5.71E-03 1.22E-04 1.99E-04
t = 0.7 E1 E2 6.54E-01 9.97E-01 6.40E-01 9.75E-01 4.28E-02 6.53E-02 4.06E-03 6.19E-03 1.27E-04 1.94E-04
t = 0.8 E1 E2 6.40E-01 9.97E-01 6.24E-01 9.73E-01 4.06E-02 6.33E-02 3.50E-03 5.46E-03 1.21E-04 1.89E-04
Table 6: The absolute (E1 ) and relative (E2 ) error estimates between the exact and its regularized solutions with α = 1.8.
29
S0377-0427(16)00007-8 http://dx.doi.org/10.1016/j.cam.2016.01.003 CAM 10439
To appear in:
Journal of Computational and Applied Mathematics
Received date: 1 September 2015 Revised date: 3 January 2016 Please cite this article as: N.H. Tuan, D.N.D. Hai, L.D. Long, V.T. Nguyen, M. Kirane, On a Riesz–Feller space fractional backward diffusion problem with a nonlinear source, Journal of Computational and Applied Mathematics (2016), http://dx.doi.org/10.1016/j.cam.2016.01.003 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.
Manuscript Click here to view linked References
On a Riesz - Feller space fractional backward diffusion problem with a nonlinear source Nguyen Huy Tuana , Dinh Nguyen Duy Haib , Le Dinh Longc , Van Thinh Nguyend , Mokhtar Kiranee,f,∗ a Applied
Analysis Research Group, Faculty of Mathematics and Statistics, Ton Duc Thang University, Ho Chi Minh City, Viet Nam b Department of Mathematics, University of Science, Vietnam National University, 227 Nguyen Van Cu Street, District 5, Ho Chi Minh City, Viet Nam c Institute of Computational Science and Technology in Ho Chi Minh City, Viet Nam d Department of Civil and Environmental Engineering, Seoul National University, Republic of Korea e Laboratoire de Mathmatiques Pˆ ole Sciences et Technologie, Universi´ e de La Rochelle, A´ enue M. Cr´ epeau, 17042 La Rochelle Cedex, France f Nonlinear Analysis and Applied Mathematics (NAAM) Research Group, Department of Mathematics, Faculty of Science, King Abdulaziz University, P.O. Box 80203, Jeddah 21589, Saudi Arabia
Abstract In this paper, a backward diffusion problem for a space-fractional diffusion equation with a nonlinear source in a strip is investigated. This problem is obtained from the classical diffusion equation by replacing the second-order space derivative with a Riesz-Feller derivative of order α ∈ (0, 2]. A nonlinear problem is severely ill-posed, therefore we propose two new modified regularization solutions to solve it. We further show that the approximated problems are well-posed and their solutions converge if the original problem has a classical solution. In addition, the convergence estimates are presented under a priori bounded assumption of the exact solution. For estimating the error of the proposed method, a numerical example has been implemented. Keywords: Space-fractional backward diffusion problem; Ill-posed problem; Regularization; Error estimate. 1. Introduction
5
10
15
In recent decades, fractional operators have been playing more and more important roles in science and engineering [4], for instance, in the theory of viscoelasticity and viscoplasticity (mechanics), in the modeling of polymers and proteins (biochemistry), in the transmission of ultrasound waves (electrical engineering), and in the modeling of human tissue under mechanical loads (medicine), which are referred to [6, 9, 14]. These new fractional-order models are more adequate than the integer-order models, because the fractional order derivatives and integrals enable to describe the memory and hereditary properties of different substance [10]. The space-fractional diffusion equation (SFDE) has been arisen from replacing the standard space partial derivative in the diffusion equation with a space fractional partial derivative. It can be derived from the continuous-time random walk in statistical mechanics, and has a wide range of applications in the theory of probability distribution, especially in the modeling of the high-frequency price dynamics in financial markets [10, 12]. If an initial concentration distribution and boundary conditions are given, a complete recovery of the unknown solution is attainable from solving a well-posed forward problem. The well-posed forward problem, i.e. initial and boundary value problem (IBVP) for the SFDE have been studied extensively in the last few years. P. Agarwal [13, 18, 25, 26] and his coauthors have recently developed some new methods for fractional differential equations. Liu et al. [8] considered stability and convergence of the difference methods for the space-time fractional advection-diffusion equation. Ray et al. [16] investigated application of modified ∗ Corresponding
author Email address: [email protected] (Mokhtar Kirane)
Preprint submitted to Elsevier
20
25
30
decomposition method for the analytical solution of space-fractional diffusion equation. Yang et al. [20] studied several numerical methods for fractional partial differential equations with Riesz space-fractional derivatives. The space fractional diffusion also investigated by some other authors, such as M. Younis et al [17], M.Duman et al [2], F. Liu [19], Z. Q. Chen [1], Y. Povstenko [15]. However, in some practical problems, the boundary data on the whole boundary cannot be obtained. In situation we only may know the noisy data on a part of the boundary or at some interior points of the considered domain, which leads to an inverse problem, namely the space-fractional inverse diffusion problems (SFIDP). To the best of our knowledge, the result from the study of SFIDP is still very limited to some specific cases dealing with homogeneous problems ([[3],[21],[22],[23],[24]]). Motivated by this reason, in this work, we consider a backward problem for the following nonlinear space fractional diffusion equation α ut (x, t) = x Dθ u(x, t) + f (x, t, u(x, t)), (x, t) ∈ R × (0, T ), u(x, t)|x→±∞ = 0, (x, t) ∈ R × (0, T ), (1) u(x, T ) = g(x), x ∈ R, where the space-fractional derivative x Dθα is the Riesz-Feller fractional derivative of order α(0 < α ≤ 2) and skewness θ (|θ| ≤ min{α, 2 − α}, θ 6= ±1) defined in [7], as follows: ( Z∞ (α + θ)π f (x + s) − f (x) Γ(1 + α) α sin ds x Dθ f (x) = π 2 s1+α 0 ) Z∞ (α − θ)π f (x − s) − f (x) + sin ds , 0 < α < 2, 2 s1+α 0 2 D2 f (x) = df (x) , α = 2. x 0 dx2
35
40
45
50
The space-fractional backward diffusion problem is to determine the distribution u(x, t) for 0 < t < T from the final concentration distribution u(x, T ). As it is well-known, the backward diffusion problem is severely ill-posed [22], i.e., solutions do not always exist, and in the case of existence, these do not depend continuously on the given data. In fact, from small noise contaminated physical measurements, the corresponding solutions have large errors. It makes difficult to numerical calculations. Hence, a regularization is in order. In recent years, the homogeneous problem, i.e, f = 0 in Equation (1) has been proposed by some authors. Zheng and Wei [22] used two methods, the spectral regularization and modified equation methods, to solve this problem. In [21], they developed an optimal modified method to solve this problem by an a priori and an a posteriori strategy. In 2014, Zhao et al [24] applied a simplified Tikhonov regularization method to deal with this problem. After then, a new regularization method of iteration type for solving this problem has been introduced by Cheng et al [3]. However, in many practical engineering applications, the diffusion occur in spatially heterogeneous environments, which requires to take account of a nonlinear source term. As mentioned above, up-to-date we have not found any publication dealing with the SFIDP with a nonlinear source term; because it requires a special technique to deal with the fractional terms and nonlinear term for solving this nonlinear problem, this is the most challenging task. Therefore in this study, we try to develop new methods and techniques to overcome this limitation. In comparison with previous studies ([[3],[21],[22],[23],[24]]) on solving the space-fractional backward diffusion problem, our paper shows a significant improvement, because it can deal with the space-fractional inverse diffusion problems with nonlinear source. This paper is organized as follows. In the following section we outline the main results. The proofs of these results is described in Section 3. In Section 4, a numerical example is proposed to show the effectiveness of the regularized methods. Then we end up the manuscript with the conclusion at Section 5.
2
55
2. The main results Let gb(ω) denote the Fourier transform of the integrable function g(x), which defined by 1 gb(ω) := √ 2π
+∞ Z exp(−ixω)g(x)dx,
−∞
i=
√
−1.
In terms of the Fourier transform, we have the following properties for the Riesz-Feller space-fractional derivative (see [11]) α \ x Dθ (g)(ω)
= −ψαθ (ω)b g (ω),
where πθ θπ + isign(ω) sin . ψαθ (ω) = |ω|α cos 2 2 60
(2)
By taking a Fourier transform to (1), we transform Problem (1) into the following differential equation ( u bt (ω, t) = −ψαθ (ω)b u(ω, t) + fb(ω, t, u(ω, t)), u b(ω, T ) = gb(ω).
Solving the latter problem, we have u b(ω, t) = exp
ψαθ (ω)(T
ZT h i − t) gb(ω) − exp ψαθ (ω)(s − T ) fb(ω, s, u(ω, s))ds .
(3)
t
From (3), applying the inverse Fourier transform, we get u(x, t) =
1 √ 2π
+∞ Z exp ψαθ (ω)(T − t)
−∞
h
× gb(ω) −
65
ZT t
i exp ψαθ (ω)(s − T ) fb(ω, s, u(ω, s))ds exp(ixω)dω.
(4)
θπ Note that ψαθ (ω) has the real part |ω|α cos( < s < T ; from (4) that, when |ω| positive 2 ). Since 0 ≤ t θ becomes large, the terms exp ψα (ω)(T − t) and exp ψαθ (ω)(s − t) increases rather quickly: small errors in high-frequency components can blow up and completely destroy the solution for 0 < t < T , therefore recovering the scalar (temperature, pollution) u(x, t) from the measured data g (x) is severely ill-posed. We define a regularization as follows +∞ Z exp ψαθ (ω)(T − t) 1 uβ (x, t) = √ 2π 1 + β exp |ω|α cos πθ T −∞
h × gb (ω) −
2
ZT t
i exp (ψαθ (ω)(s − T ) fb ω, s, uβ (ω, s) ds exp(iωx)dω,
where β is regularization parameter. We introduce the first theorem for main result as following
3
(5)
Theorem 1. Let g ∈ L2 (R) and let f ∈ L∞ (R × [0, T ] × R) satisfy f (x, t, 0) = 0 and |f (x, t, w) − f (x, t, v)| ≤ k|w − v|, 70
for constant k > 0 independent of x,t,w,v. Then there exists a unique solution uβ ∈ C([0, T ]; L2 (R)) to Problem (5). Let g ∈ L2 (R) be measured data such that kg − g kL2 (R) ≤ . Let us choose β such that lim→0 β −1 bounded. a) Suppose that the problem (1) has a unique solution u ∈ C([0, T ], L2 (R)) that satisfies +∞ 2 Z exp |ω|α cos πθ t u dω < ∞, for all t ∈ [0, T ], b (ω, t) 2
−∞
then we have the estimate
75
t
uβ (., t) − u(., t) 2 ≤ β T exp((T − t)2 k 2 )K, L (R)
where K=
√
2β
−1
21 +∞ 2 Z πθ α exp |ω| cos t u b(ω, t) dω . + 2 sup 2
t∈[0,T ] −∞
Moreover, if ut ∈ L2 ((0, T ); L2 (R)) then for all ∈ (0, 1) there exists a constant t > 0 such that 1 − 14
1 1
u(., 0) − uβ (., t ) 2 ≤ E8 4 T 4 ln , L (R) β
where
(6)
21 ZT 2 2 2 E = max kus (., s)kL2 (R) ds , exp((T − t) k )K . 0
b) Suppose that the problem (1) has a unique solution u ∈ C([0, T ], L2 (R)) that satisfies +∞ZT Z
−∞ 0
80
2 exp 2ψαθ (ω)s fb(ω, s, u(ω, s)) dsdω < ∞,
then we have the following estimate
uβ (., t) − u(., t) 2 L (R)
≤
where
D=
√
2β −1 + 4||u(., 0)||2L2 (R) + 4T
+∞ZT Z
−∞ 0
t β T exp (T − t)2 k 2 D,
(7)
(8)
21 2 exp(2ψαθ (ω)s) fb(ω, s, u(ω, s)) dsdω .
Remark 1. 1. If we choose β = , then the results in Theorem 1 hold. The error estimate (8) is no longer valid at t = 0. To approximate u(., 0), we use the similar technique as shown in (6). 2. In Theorem 1, we require the following condition +∞ 2 Z exp |ω|α cos πθ t u dω < ∞. b (ω, t) 2
−∞
4
(9)
85
90
95
If f (x, t, u) = 0 then the left hand side of (9) is equal to ku(., 0)k2L2 (R) . Therefore, the condition (9) is natural and acceptable. t If f (x, t, u) = 0, the error estimate in Theorem 1 is of order β T . It is similar to the homogeneous case in [21, 24]. 3. In Theorem 1, to obtain the error estimate, we require the strong assumptions of u as shown by (7) and (9). This is a limitation of Theorem 1, because there are not many functions u satisfied these conditions, particularly in practice these conditions are more difficult to be satisfied and checked. To improve this limitation, we introduce a new regularization solution and introduce a new technique to estimate the error in the Theorem 2 below. In fact, in this Theorem we only need a weaker assumption for u which is u ∈ C([0, T ]; L2 (R)), and this condition is natural and acceptable. Theorem 2. Let g, f, β be as Theorem 1. Suppose that 0 < kT < 1 and Problem (1) has a unique Then we construct a new regularized solution solution u ∈ C([0, T ]; L2 (R)). Let m ∈ 0, k21T 2 − 1 . Uβ ∈ C([0, T ]; L2 (R)) in such a way that
Uβ (., t) − u(., t) 2 L (R)
≤
β
−1
+ ku(., 0)kL2 (R)
s
1 1+ m t βT , 2 2 1 − (1 + m)T k
where Uβ is the function whose Fourier transform is defined by ZT exp ψαθ (ω)(T − t) exp ψαθ (ω)(s − t) c (ω, t) = gb (ω) − fb(ω, s, Uβ )ds U β α cos πθ T 1 + β exp |ω|α cos πθ T 1 + β exp |ω| 2 2 t Zt β exp |ω|α cos πθ T 2 exp (s − t)ψαθ (ω) fb(ω, s, Uβ )ds. + 1 + β exp |ω|α cos πθ 2 T 0
(10)
Remark 2. 1. Using the same technique as in (6), we obtain the estimate at t = 0 as kUβ (., t )
− u(., 0)kL2 (R)
≤
H2
−1 4
T
1 4
1 ln β
−1 4
,
where 12 21 ZT 1 1 + m H = max kus (., s)k2L2 (R) ds , β −1 + ku(., 0)kL2 (R) . 1 − (1 + m)T 2 k 2 0
100
2. In Theorem 2, our method has a drawback that the condition for k ∈ [0, T1 ) is still too strong which may restrict the class of equations. We are trying to remove this constraint. 3. If k = 0 and f = 0 then Problem (1) becomes a homogeneous problem. The error estimate in Theorem 2 t is of order β T . It is similar to the homogeneous case in [21, 24]. 3. Proof of main results
105
First, we consider the following Lemma
5
Lemma 1. Let t, s ∈ [0, T ]. 1) If s ≥ t, then we have
exp ψαθ (ω)(s − t) t−s ≤ β T . πθ α 1 + β exp |ω| cos 2 T exp ψαθ (ω)(T − t) ≤ β t−T T . 1 + β exp |ω|α cos πθ 2 T
a)
b)
2) If s ≤ t, then we have
β exp ψ θ (ω)(s − t + T ) α t−s ≤ β T . θπ 1 + β exp |ω|α cos 2 T
c)
Proof. First, we prove (a). In fact, we have exp ψαθ (ω)(s − t) exp |ω|α cos( πθ 2 )(s − t − T ) = 1 + β exp |ω|α cos( πθ )T β + exp − |ω|α cos( πθ 2 2 )T exp |ω|α cos( πθ 2 )(s − t − T ) = h i s−t h i T −s+t T T β + exp − |ω|α cos( πθ β + exp − |ω|α cos( πθ 2 )T 2 )T ≤
1 h i s−t T β + exp − |ω|α cos( πθ )T 2
≤ β 110
t−s T
.
Making the change s = T , we obtain (b). Next, we prove (c). In fact from (b), we have exp ψ θ (ω)(T − (t − s)) α πθ α 1 + β exp |ω| cos( 2 )T
≤ β
t−s−T T
,
it follows that
β exp ψ θ (ω)(s − t + T ) α ≤ πθ α 1 + β exp |ω| cos( 2 )T
β
t−s T
.
3.1. Proof of Theorem 1 115
Proof. We divide the proof into three steps. Step1. Constructing a regularized solution (5). We consider the problem ZT exp ψαθ (ω)(T − t) exp ψαθ (ω)(s − t) c (ω, t) = gb (ω) − fb(ω, s, uβ (ω, s))ds u β α cos πθ T 1 + β exp |ω|α cos πθ T 1 + β exp |ω| 2 2 t 6
or uβ (x, t)
=
exp ψαθ (ω)(T − t) gb (ω) exp(iωx)dω α cos πθ T 1 + β exp |ω| 2 −∞ +∞ZT θ Z exp ψ (ω)(s − t) α 1 fb(ω, s, uβ (ω, s)) exp(iωx)dsdω. −√ α cos πθ T 2π 1 + β exp |ω| 2 −∞ t +∞ Z
1 √ 2π
(11)
First, we prove that Problem (11) has a unique solution uβ that belongs to C([0, T ]; L2 (R)). Denoting 1 1 F (w)(x, t) = √ G(x, t) − √ 2π 2π for all w ∈ C([0, T ]; L2 (R)) and G(x, t) =
+∞ Z
−∞
120
+∞ZT Z
−∞ t
exp ψαθ (ω)(s − t) fb(ω, s, w) exp(iωx)dsdω, 1 + β exp |ω|α cos πθ 2 T
exp ψαθ (ω)(T − t) gb (ω) exp(iωx)dω. 1 + β exp |ω|α cos πθ 2 T
Since f (x, t, 0) = 0, and due to the Lipschitzian property of f (x, t, w) with respect to w, we get G(w) ∈ C([0, T ]; L2 (R)) for every w ∈ C([0, T ]; L2 (R)). We claim that, for every w, v ∈ C([0, T ]; L2 (R)), p ≥ 1, we have !2p k (T − t)p Qp 2 p p 2 kF (w)(., t) − F (v)(., t)kL2 (R) ≤ kw − vk , (12) β p! where Q = max{1, T } and k.k is the sup norm in L2 (R). We shall prove the latter inequality by induction. For p = 1, we have 2 kF (w)(., t) − F (v)(., t)kL2 (R) =
2
b
F (w)(., t) − Fb(v)(., t) 2
L (R)
2 exp ψαθ (ω)(s − t) fb(ω, s, w) − fb(ω, s, v) ds dω = 1 + β exp |ω|α cos πθ 2 T −∞ t +∞ ZT 2 ZT θ Z 2 exp ψ (ω)(s − t) α ds fb(ω, s, w) − fb(ω, s, v) ds ≤ dω 1 + β exp |ω|α cos πθ 2 T −∞ t t 2 +∞ ZT πθ α Z ZT )(s − t) exp |ω| cos( 2 2 ds fb(ω, s, w) − fb(ω, s, v) ds = dω α 1 + β exp |ω| cos πθ 2 T +∞ ZT Z
−∞
≤
≤
t
t
ZT
2 1
b
(T − t)
f (., s, w(., s)) − fb(., s, v(., s)) 2 ds 2 β L (R) t
1 (T − t) β2
ZT t
2
k 2 kw(., s) − v(., s)kL2 (R) ds ≤
7
2 k 2 (T − t)Q kw − vk . β
125
Therefore (12) holds for p = 1. Suppose that (12) holds for p = m (m ≥ 1). We prove that (12) holds for p = m + 1. We have
2
m+1
2
F (w)(., t) − F m+1 (v)(., t) L2 (R) = Fb(F m (w))(., t) − Fb(F m (v))(., t) L2 (R)
2 exp ψαθ (ω)(s − t) fb(ω, s, F m (w)) − fb(ω, s, F m (v)) ds dω = 1 + β exp |ω|α cos πθ 2 T −∞ t 2 +∞ ZT πθ α Z ZT 2 exp |ω| cos( )(s − t) 2 ds fb(ω, s, F m (w)) − fb(ω, s, F m (v)) ds ≤ dω 1 + β exp |ω|α cos πθ 2 T t −∞ t +∞ ZT Z
≤
1 (T − t) β2
=
1 (T − t) β2
+∞ZT Z 2 b f (., s, F m (w)(., s)) − fb(., s, F m (v)(., s)) dsdω
−∞ t
ZT
2
b
f (., s, F m (w)(., s)) − fb(., s, F m (v)(., s))
L2 (R)
ds
t
ZT
≤
1 (T − t)k 2 β2
≤
2m ZT 2(m+1) 1 k k (T − t)m+1 m+1 (T − s)m Qm 2 2 2 (T − t)k kw − vk ds ≤ Q kw − vk . β2 β m! β (m + 1)!
t
2
kF m (w)(., s) − F m (v)(., s)kL2 (R) ds
t
Therefore, by the induction principle, (12) holds for every integer p. From (12), we get kF p (w) − F p (v)kL2 (R) ≤ for every w, v ∈ C([0, T ]; L2 (R)). Since
p p k T2 p √ Q 2 kw − vk , β p!
p p T2 p k √ Q 2 = 0, p→∞ β p! lim
130
135
there exits a positive integer number p0 such that F p0 is a contraction. It follows that F p0 (w) = w has a unique solution uβ ∈ C([0, T ]; L2 (R)). We claim that F (uβ ) = uβ . In fact, one has F (F p0 (uβ )) = F (uβ ). Hence, F p0 (F (uβ )) = F (uβ ). By the uniqueness of the fixed point of F p0 , one has F (uβ ) = uβ , i.e., the equation F (w) = w has a unique solution
uβ ∈ C([0, T ]; L2 (R)). The main purpose of this paper is to estimate the error uβ − u 2 . To this end, we proceed to the next two steps. Step2. Let wβ be the function whose Fourier transform is defined by c (ω, t) w β
=
L (R)
ZT exp ψαθ (ω)(T − t) exp (ψαθ (ω)(s − t) gb(ω) − fb ω, s, wβ (ω, s) ds. 1 + β exp |ω|α cos πθ 1 + β exp |ω|α cos πθ 2 T 2 T t
(13)
We estimate wβ (., t) − uβ (., t)
L2 (R)
.
Using the Parseval’s equality and the inequality (a + b)2 ≤ 2(a2 + b2 ), we get 8
2
wβ (., t) − uβ (., t) 2
L (R)
2 +∞ Z exp ψαθ (ω)(T − t) b (ω) − gb(ω) dω ≤ 2 g 1 + β exp |ω|α cos πθ 2 T −∞ 2 +∞ ZT Z exp ψαθ (ω)(s − t) fb(ω, s, uβ (ω, s)) − fb(ω, s, wβ (ω, s)) ds dω +2 1 + β exp |ω|α cos πθ 2 T −∞ t = A1 + A2 ,
where A1
A2
140
2 +∞ Z exp ψαθ (ω)(T − t) b (ω) − gb(ω) dω, = 2 g 1 + β exp |ω|α cos πθ 2 T −∞ 2 +∞ ZT Z exp ψαθ (ω)(s − t) fb(ω, s, uβ (ω, s)) − fb(ω, s, wβ (ω, s)) ds dω. = 2 1 + β exp |ω|α cos πθ 2 T −∞ t
(14)
(15)
Term (15) can be estimated as follows 2 +∞ θ Z exp ψα (ω)(T − t) gb (ω) − gb(ω) 2 dω A1 ≤ 2 πθ α 1 + β exp |ω| cos( 2 )T −∞ 2β
2t−2T T
≤
2β
2t−2T T
≤
2β
2t−2T T
≤
+∞ Z gb (ω) − gb(ω) 2 dω
−∞
2
kg − gkL2 (R)
(16)
2
and A2
2 +∞ ZT Z exp ψαθ (ω)(s − t) fb(ω, s, wβ (ω, s)) − fb(ω, s, uβ (ω, s)) ds dω = 2 1 + β exp |ω|α cos πθ 2 T −∞ t 2 +∞ZT Z 2 exp ψαθ (ω)(s − t) fb(ω, s, wβ (ω, s)) − fb(ω, s, uβ (ω, s)) dsdω ≤ 2(T − t) 1 + β exp |ω|α cos( πθ 2 )T −∞ t ≤ 2(T − t)
+∞ZT Z
−∞ t 2t
≤ 2(T − t)β T k 2
β
ZT
2t−2s T
β
−2s T
t
2 b f (ω, s, wβ (ω, s)) − fb(ω, s, uβ (ω, s) dsdω
wβ (., s) − uβ (., s) 2 2 ds. L (R)
(17)
Combining (14), (16) and (17), we have
wβ (., t) − uβ (., t) 2 2 L (R)
≤
2β
2t−2T T
2t T
+ 2(T − t)β k 2
9
2
ZT t
β
−2s T
wβ (., s) − uβ (., s) 2 2 ds. L (R)
Hence, β
145
−2t T
wβ (., t) − uβ (., t) 2 2 L (R)
2β
≤
+ 2(T − t)k
−2 2
Using the Gronwall’s inequality, we obtain
2 −2t β T wβ (., t) − uβ (., t) L2 (R)
2
ZT
β
−2s T
t
wβ (., s) − uβ (., s) 2 2 ds. L (R)
≤ 2β −2 exp 2(T − t)2 k 2 2 .
Therefore, we conclude that √ t−T
wβ (., t) − uβ (., t) 2 2β T exp (T − t)2 k 2 . ≤ L (R)
. Step3. We shall estimate the error wβ (., t) − u(., t)
(18)
L2 (R)
Let wβ be the function defined in Step 2. We recall the Fourier transform of u and wβ from (3) and (13) ZT u b(ω, t) = exp ψαθ (ω)(T − t) gb(ω) − exp(ψαθ (ω)(s − T ))fb(ω, s, u(ω, s))ds (19) t
and
c (ω, t) w β
150
=
exp ψαθ (ω)(T − t)
1 + β exp |ω|α cos( πθ 2 )T
gb(ω) −
By direct calculation from (19) and (20) we get c (ω, t) u b(ω, t) − w β exp(ψαθ (ω)(T
=
−
ZT
β exp(|ω| cos( πθ 2 )T ) + β exp(|ω|α cos( πθ 2 )T ) α
=
− t))fb(ω, s, u(ω, s))ds +
1 +
ZT t
ZT t
exp(ψαθ (ω)(s
t
exp(ψαθ (ω)(T − t)) − t)) − 1 + β exp(|ω|α cos( πθ 2 )T )
exp(ψαθ (ω)(s
t
ZT
!
(20)
gb(ω)
exp(ψαθ (ω)(s − t)) fb(ω, s, wβ (ω, s))ds 1 + β exp(|ω|α cos( πθ 2 )T )
exp(ψαθ (ω)(T − t))b g (ω) −
−
T ))fb(ω, s, wβ (ω, s))ds .
ZT t
exp(ψαθ (ω)(s − t))fb(ω, s, u(ω, s))ds
exp(ψαθ (ω)(s − t)) b(ω, s, w (ω, s)) − fb(ω, s, u(ω, s)) ds. f β 1 + β exp(|ω|α cos( πθ 2 )T )
Using the Parseval
2 equality, we obtain
u(., t) − wβ (., t) 2 L (R)
=
+∞ Z 2 c (ω, t) dω u(ω, t) − w b β
−∞
2 +∞ ZT Z β exp(|ω|α cos( πθ )T ) 2 2 g (ω) − exp(ψαθ (ω)(s − t))fb(ω, s, u)ds dω ≤ 2 exp(ψαθ (ω)(T − t))b 1 + β exp(|ω|α cos( πθ 2 )T ) −∞
+2
t
+∞ ZT Z
−∞
= I1 + I2 ,
t
2 exp(ψαθ (ω)(s − t)) fb(ω, s, wβ (ω, s)) − fb(ω, s, u(ω, s)) ds dω 1 + β exp(|ω|α cos( πθ 2 )T ) 10
(21)
where 2 +∞ ZT Z β exp(|ω|α cos( πθ )T ) 2 2 g (ω) − exp(ψαθ (ω)(s − t))fb(ω, s, u)ds dω (22) I1 = 2 exp(ψαθ (ω)(T − t))b 1 + β exp(|ω|α cos( πθ 2 )T ) −∞
t
and
2 +∞ ZT Z θ exp(ψ (ω)(s − t)) α b b I2 = 2 1 + β exp(|ω|α cos( πθ )T ) f (ω, s, wβ (ω, s)) − f (ω, s, u(ω, s)) ds dω, 2 −∞
155
t
where the term I1 is estimated in the following manner I1
=
2 +∞ Z β exp(|ω|α cos( πθ )(T − t)) πθ α 2 2 exp |ω| cos t u b (ω, t) dω 1 + β exp |ω|α cos πθ 2 2 T −∞
≤ ≤
2t
2β T
+∞ 2 Z dω exp |ω|α cos πθ t u b (ω, t) 2
−∞
2β
2t T
+∞ 2 Z πθ α b(ω, t) dω sup exp |ω| cos 2 t u
(23)
t∈[0,T ] −∞
and using the H¨ older inequality, the term I2 is estimated as follows
I2
2 +∞ ZT Z θ exp(ψ (ω)(s − t)) α b b = 2 1 + β exp(|ω|α cos( πθ )T ) f (ω, s, wβ (ω, s)) − f (ω, s, u(ω, s)) ds dω 2 −∞ t 2 +∞ZT Z 2 exp(ψαθ (ω)(s − t)) b b(ω, s, u(ω, s)) dsdω ≤ 2(T − t) f (ω, s, w (ω, s)) − f β πθ 1 + β exp(|ω|α cos( 2 )T ) −∞ t
≤
2(T − t)k
2
ZT t
β
2t−2s T
wβ (., s) − u(., s) 2 2 ds. L (R)
Combining (21), (23) and (24), we have
u(., t) − wβ (., t) 2 2 L (R)
2t
≤ 2β T sup
+∞ 2 Z exp |ω|α cos πθ t u dω b (ω, t) 2
t∈[0,T ] −∞
+ 2(T − t)k 2
ZT
β
2t−2s T
t
Hence, β
−2t T
u(., t) − wβ (., t) 2 2 L (R)
≤
2 sup
+∞ 2 Z exp |ω|α cos πθ t u dω b (ω, t) 2
t∈[0,T ] −∞
+ 2(T − t)k 11
wβ (., s) − u(., s) 2 2 ds. L (R)
2
ZT t
β
−2s T
u(., s) − wβ (., s) 2 2 ds. L (R)
(24)
Using the Gronwall’s inequality, we obtain β
160
−2t T
u(., t) − wβ (., t) 2 2 L (R)
2 exp(2(T − t)2 k 2 ) sup
≤
+∞ 2 Z exp |ω|α cos πθ t u dω. b (ω, t) 2
t∈[0,T ] −∞
Whereupon
12 +∞ 2 Z
t πθ
u(., t) − wβ (., t) 2 exp |ω|α cos ≤ β T exp((T − t)2 k 2 ) 2 sup t u b(ω, t) dω . L (R) 2 t∈[0,T ] −∞
(25)
Combining (18) and (25) with applying the triangle inequality, we have √ t−T
uβ (., t) − u(., t) 2 2β T exp (T − t)2 k 2 ≤ L (R) 21 +∞ 2 Z t πθ 2 2 α exp |ω| cos + β T exp((T − t) k ) 2 sup t u b(ω, t) dω 2 t∈[0,T ] −∞
t
≤ β T exp((T − t)2 k 2 )K,
(26)
where K=
As u(x, t) − u(x, 0) =
√ Rt
21 +∞ 2 Z exp |ω|α cos πθ t u 2β −1 + 2 sup b(ω, t) dω . 2
t∈[0,T ] −∞
us (x, s)ds, it follows that
0
2
ku(., 0) − u(., t)kL2 (R)
≤ t
Due to (26) and (27), we obtain
u(., 0) − uβ (., t) 2 L (R) 165
≤ ≤
Zt 0
2 v u T uZ u kus (., s)k2L2 (R) ds ≤ t t kus (., s)k2L2 (R) ds .
(27)
0
ku(., 0) − u(., t)kL2 (R) + u(., t) − uβ (., t) L2 (R) √ t E t+βT ,
where E is defined in Theorem 1. For every ∈ (0, 1), there exists uniquely a positive number t such that √ t β t = β T , i.e., lntt = 2 ln T . Using the inequality ln t > −(1/t) for every t > 0, we get proposition 2
1 1
u(., 0) − uβ (., t ) 2 ≤ E8 4 T 4 L (R)
ln
1 β
− 41
.
Proof (b) of Theorem 1 From (3), we have 170
gb(ω) = exp −ψαθ (ω)T u b(ω, 0) +
ZT 0
exp sψαθ (ω) fb(ω, s, u(ω, s))ds .
12
(28)
Combining (22), (28), using inequality (a + b)2 ≤ 2(a2 + b2 ) and H¨ older inequality, we obtain I1
" # +∞ Zt Z β exp(|ω|α cos( πθ )T ) 2 2 θ θ 2 b b(ω, 0) + exp(ψα (ω)s)f (ω, s, u(ω, s))ds dω = 2 exp(−ψα (ω)t) u 1 + β exp(|ω|α cos( πθ 2 )T ) −∞ 0 2 +∞ Z Zt β exp ψ θ (ω)(s − t + T ) β exp − ψαθ (ω)t α u fb(ω, s, u(ω, s))ds dω = 2 b(ω, 0) + β + exp − |ω|α cos( πθ )T 1 + β exp |ω|α cos( πθ 2 2 )T −∞ 0 2 +∞ Z β exp − ψαθ (ω)t |b ≤ 4 u(ω, 0)|2 dω β + exp − |ω|α cos( πθ )T 2 −∞ 2 +∞ Zt θ Z β exp ψα (ω)(s − t + T ) b +4 f (ω, s, u(ω, s))ds dω 1 + β exp |ω|α cos( πθ )T 2 −∞ 0 2 +∞ Z exp − ψαθ (ω)t 2 |b ≤ 4β u(ω, 0)|2 dω β + exp − |ω|α cos( πθ )T 2 −∞ +∞ T Z Z exp ψ θ (ω)(s − t + T ) 2 2 α fb(ω, s, u(ω, s)) dsdω + 4T β 2 1 + β exp |ω|α cos( πθ )T 2 −∞ 0 ≤ 4β 2
+∞ +∞ZT Z Z 2 2(t−T ) 2(t−T ) u(ω, 0)|2 dω + 4T β 2 β T |b β T exp 2ψαθ (ω)s fb(ω, s, u(ω, s)) dsdω
−∞
≤ β
2t T
4||u(., 0)||2L2 (R)
+ 4T
+∞ZT Z
−∞ 0
exp
−∞ 0
! 2 b f (ω, s, u(ω, s)) dsdω .
2ψαθ (ω)s
(29)
Combining (21), (24) and (29), we have
u(., t) − wβ (., t) 2 2 L (R)
≤ 2(T − t)k +β
2t T
2
ZT
β
2t−2s T
t
u(., s) − wβ (., s) 2 2 ds L (R)
4||u(., 0)||2L2 (R)
+ 4T
+∞ZT Z
exp
−∞ 0
Hence,
2 −2t
β T u(., t) − wβ (., t) 2
! 2 b f (ω, s, u(ω, s)) dsdω .
2ψαθ (ω)s
L (R)
≤ 2(T − t)k 2 +
ZT
β
−2s T
t
4||u(., 0)||2L2 (R)
u(., s) − wβ (., s) 2 2 ds L (R) + 4T
+∞ZT Z
exp
−∞ 0
13
! 2 b f (ω, s, u(ω, s)) dsdω .
2ψαθ (ω)s
175
Using the Gronwall’s inequality, we obtain β
−2t T
u(., t) − wβ (., t) 2 2 L (R)
≤
exp 2(T − t)2 k 2 4||u(., 0)||2L2 (R) + 4T
! 2 exp 2ψαθ (ω)s fb(ω, s, u(ω, s)) dsdω .
+∞ZT Z
−∞ 0
Whereupon
u(., t) − wβ (., t) 2 L (R)
≤
t β T exp (T − t)2 k 2
× 4||u(., 0)||2L2 (R) + 4T
+∞ZT Z
−∞ 0
12 2 exp 2ψαθ (ω)s fb(ω, s, u(ω, s)) dsdω .
(30)
Applying the results of Step 2 and (30), we get the following estimate by using the triangle inequality
uβ (., t) − u(., t) 2 L (R) 21 +∞ZT Z 2 t ≤ β T exp (T − t)2 k 2 4||u(., 0)||2L2 (R) + 4T exp 2ψαθ (ω)s fb(ω, s, u(ω, s)) dsdω +
≤
√
2β
t−T T
−∞ 0
exp (T − t) k β exp (T − t)2 k 2 21 +∞ZT Z 2 √ × 2β −1 + 4||u(., 0)||2L2 (R) + 4T exp 2ψαθ (ω)s fb(ω, s, u(ω, s)) dsdω , t T
2 2
−∞ 0
for all t ∈ [0, T ]. This completes the proof of Theorem 1. 3.2. Proof of Theorem 2 180
We divide it into three steps. Step1. The existence and the uniqueness of a solution of (10). Construction of a regularized solution Uβ . We consider the problem ZT exp ψαθ (ω)(s − t) exp ψαθ (ω)(T − t) c (ω, t) = gb (ω) − fb(ω, s, Uβ )ds U β α cos πθ T 1 + β exp |ω|α cos πθ T 1 + β exp |ω| 2 2 t Zt β exp |ω|α cos πθ T 2 exp (s − t)ψαθ (ω) fb(ω, s, Uβ )ds, + 1 + β exp |ω|α cos πθ 2 T 0
14
(31)
or Uβ (x, t)
exp ψαθ (ω)(T − t) gb (ω) exp(iωx)dω = α cos πθ T 1 + β exp |ω| 2 −∞ +∞ZT θ Z exp ψ (ω)(s − t) α 1 fb(ω, s, Uβ ) exp(iωx)dsdω −√ α cos πθ T 2π 1 + β exp |ω| 2 −∞ t t +∞ πθ Z Z β exp |ω|α cos 2 T 1 exp (s − t)ψαθ (ω) fb(ω, s, Uβ ) exp(iωx)dsdω. + √ 2π 1 + β exp |ω|α cos πθ T 1 √ 2π
+∞ Z
2
−∞ 0
185
Let us define the norm on C([0; T ]; L2 (R)) as follows khk1 = sup β 0≤t≤T
−t T
kh(t)kL2 (R) , for all h ∈ C([0; T ]; L2 (R)).
It is easy to show that k · k1 is a norm of C([0; T ]; L2 (R)). For v ∈ C([0; T ]; L2 (R)), we consider the following function Θ(v)(x, t)
where
=
1 √ A(x, t) 2π +∞ZT Z exp ψαθ (ω)(s − t) 1 fb(ω, s, v) exp(iωx)dsdω −√ α cos πθ T 2π 1 + β exp |ω| 2 −∞ t +∞Zt πθ α Z β exp |ω| cos 2 T 1 exp (s − t)ψαθ (ω) fb(ω, s, v) exp(iωx)dsdω, +√ πθ 2π 1 + β exp |ω|α cos 2 T −∞ 0
A(x, t) =
+∞ Z
−∞
exp ψαθ (ω)(T − t) gb (ω) exp(iωx)dω. 1 + β exp |ω|α cos πθ 2 T
We claim that, for every v1 , v2 ∈ C([0; T ]; L2 (R))
kΘ(v1 ) − Θ(v2 )k1 ≤ kT kv1 − v2 k1 .
15
190
First, we have the two following estimates for all t ∈ [0, T ] 2 +∞ ZT Z exp ψαθ (ω)(s − t) fb(ω, s, v1 ) − fb(ω, s, v2 ) ds dω J1 = α cos πθ T 1 + β exp |ω| 2 −∞ t 2 +∞ZT Z 2 exp ψαθ (ω)(s − t) fb(ω, s, v1 ) − fb(ω, s, v2 ) dsdω ≤ (T − t) 1 + β exp |ω|α cos πθ 2 T −∞ t +∞ZT Z
β
≤ β T k 2 (T − t)
ZT
β
2
sup β
≤ (T − t)
−∞ t
2t
2t T
≤ β k (T − t) 2
2(t−s) T
−2s T
t
2 b f (ω, s, v1 ) − fb(ω, s, v2 ) dsdω
kv1 (., s) − v2 (., s)k2L2 (R) ds −2s T
0≤s≤T
kv1 (., s) − v2 (., s)k2L2 (R)
2t
≤ β T k 2 (T − t)2 kv1 − v2 k21
(32)
and J2
β exp |ω|α cos πθ 2 T exp (s − t)ψαθ (ω) = 1 + β exp |ω|α cos πθ 2 T −∞ 0 2 × fb(ω, s, v1 ) − fb(ω, s, v2 ) ds dω 2 +∞Zt πθ α Z β exp |ω| cos 2 T exp (s − t)ψαθ (ω) ≤ t 1 + β exp |ω|α cos πθ 2 T −∞ 0 2 × fb(ω, s, v1 ) − fb(ω, s, v2 ) dsdω +∞ Zt Z
≤ t
+∞Zt Z
−∞ 0
β
Zt
2t T
2
2t T
2 2
≤ β k t ≤ β k t
2(t−s) T
β
−2s T
0
2 b f (ω, s, v1 ) − fb(ω, s, v2 ) dsdω
kv1 (., s) − v2 (., s)k2L2 (R) ds
sup β
−2s T
0≤s≤T
kv1 (., s) − v2 (., s)k2L2 (R)
2t
(33)
≤ β T k 2 t2 kv1 − v2 k21 .
1 For 0 < t < T , using the inequality (a + b)2 ≤ (1 + m)a2 + (1 + m )b2 for all real numbers a and b and m > 0, we obtain 2
kΘ(v1 )(., t) − Θ(v2 )(., t)kL2 (R)
By choosing m = β
T −t t ,
−2t T
2t
≤ (1 + m)β T k 2 t2 kv1 − v2 k21 2t 1 + 1+ β T k 2 (T − t)2 kv1 − v2 k21 . m
we have
kΘ(v1 )(., t) − Θ(v2 )(., t)k2L2 (R) ≤ k 2 T 2 kv1 − v2 k21 , for all t ∈ (0, T ). 16
(34)
195
On other hand, letting t = 0 in (32), we have kΘ(v1 )(., 0) − Θ(v2 )(., 0)k2L2 (R) ≤ k 2 T 2 kv1 − v2 k21 .
(35)
By letting t = T in (33), we have β −2 kΘ(v1 )(., T ) − Θ(v2 )(., T )k2L2 (R) ≤ k 2 T 2 kv1 − v2 k21 .
(36)
Combining (34), (35) and (36), we obtain β
200
−2t T
kΘ(v1 )(., t) − Θ(v2 )(., t)k2L2 (R) ≤ k 2 T 2 kv1 − v2 k21 , for all t ∈ [0, T ]
which leads to (32). Since kT < 1, Θ is a contraction. It follows that the equation Θ(v) = v has a unique solution Uβ ∈ C([0; T ]; L2 (R)).
. Step 2. Estimate of the error Uβ (., t) − Vβ (., t) 2 L (R)
Let Vβ be the function whose Fourier transform is defined by
exp ψαθ (ω)(T − t) gb(ω) 1 + β exp |ω|α cos πθ 2 T ZT exp ψαθ (ω)(s − t) fb(ω, s, Vβ )ds − α cos πθ T 1 + β exp |ω| 2 t Zt β exp |ω|α cos πθ T 2 exp (s − t)ψαθ (ω) fb(ω, s, Vβ )ds. + 1 + β exp |ω|α cos πθ 2 T 0
Vc β (ω, t) =
(37)
From (31) and (37), we have
θ exp ψ (ω)(T − t) α c gb (ω) − gb(ω) Uβ (ω, t) − Vc β (ω, t) ≤ 1 + β exp |ω|α cos πθ 2 T T Z h i exp ψαθ (ω)(s − t) fb(ω, s, Vβ (ω, s)) − fb(ω, s, Uβ (ω, s)) ds + 1 + β exp |ω|α cos πθ 2 T t Zt β exp |ω|α cos πθ 2 T exp (s − t)ψαθ (ω) + 1 + β exp |ω|α cos πθ T 2 0 h i b b × f (ω, s, Uβ (ω, s)) − f (ω, s, Vβ (ω, s)) ds ≤
β
t−T T
+
Zt 0
≤ β
gb (ω) − gb(ω) +
t−T T
β
t−s T
ZT
β
t−s T
t
b f (ω, s, Vβ (ω, s)) − fb(ω, s, Uβ (ω, s)) ds
b f (ω, s, Vβ (ω, s)) − fb(ω, s, Uβ (ω, s)) ds
gb (ω) − gb(ω) +
ZT 0
17
β
t−s T
b f (ω, s, Vβ (ω, s)) − fb(ω, s, Uβ (ω, s)) ds.
Since m ∈ 0, k21T 2 − 1 , we have that 0 < k <
(a + b)2 ≤
T
√1 . 1+m
1+
1 m
From the inequality
a2 + (1 + m)b2
(38)
for all real number a, b and m > 0, we get
2
c
U (., t) − V (., t) 2 2 (., t) − Vc (., t) = U β
β
β
L (R)
≤
≤ 205
β
L2 (R)
2 2t−2T 1 1+ β T gb − gb L2 (R) m 2 +∞ ZT Z t−s β T fb(ω, s, Vβ (ω, s)) − fb(ω, s, Uβ (ω, s)) ds dω + (1 + m) 1 1+ m
−∞
β
0
2t−2T T
+ (1 + m)T k 2
2
ZT
β
2t−2s T
0
This leads to β
−2t T
Uβ (., t) − Vβ (., t) 2 2 L (R)
≤
1+
1 m
Uβ (., s) − Vβ (., s) 2 2 ds. L (R)
β −2 2
+ (1 + m)T k 2
ZT 0
−2t
β
−2s T
Uβ (., s) − Vβ (., s) 2 2 ds. L (R)
(39)
Set Z(t) = β T kUβ (., t) − Vβ (., t)k2L2 (R) , ∀t ∈ [0, T ]. Since Uβ , Vβ ∈ C([0, T ]; L2 (R)), we see that the function Z is continuous on [0, T ] and attains over there its maximum M at some t0 ∈ [0, T ]. Let M = −2t max β T kUβ (., t) − Vβ (., t)k2L2 (R) . From (39), we obtain t∈[0,T ]
M or
≤
1 1+ β −2 2 + (1 + m)T 2 k 2 M, m
1 − (1 + m)T 2 k 2 M ≤
210
1 1+ β −2 2 . m
This implies that for all t ∈ [0, T ] β Hence
−2t T
Uβ (., t) − Vβ (., t) 2 2 ≤M ≤ L (R)
Uβ (., t) − Vβ (., t) 2 ≤ L (R)
s
−2 2 1 1+ m β . 1 − (1 + m)T 2 k 2
1 1+ m t−T β T . 1 − (1 + m)T 2 k 2
18
Step 3. Estimate u(., t) − Vβ (., t)
L2 (R)
We have
u b(ω, t)
.
= exp ψαθ (ω)(T − t) gb(ω) −
ZT t
exp ψαθ (ω)(s − T ) fb(ω, s, u(ω, s))ds
ZT exp ψαθ (ω)(T − t) gb(ω) − exp ψαθ (ω)(s − T ) fb(ω, s, u(ω, s))ds = 1 + β exp |ω|α cos πθ 2 T t θ α β exp ψα (ω)(T − t) exp |ω| cos πθ 2 T + 1 + β exp |ω|α cos πθ 2 T T Z × gb(ω) − exp ψαθ (ω)(s − T ) fb(ω, s, u(ω, s))ds .
(40)
t
On other hand, we get
u b(ω, T ) = gb(ω) = exp −
215
This implies that
gb(ω) −
ZT t
T ψαθ (ω)
u b(ω, 0) +
ZT 0
exp
sψαθ (ω)
b f (ω, s, u(ω, s))ds .
exp ψαθ (ω)(s − T ) fb(ω, s, u(ω, s))ds
Zt = exp − T ψαθ (ω) u b(ω, 0) + exp (s − T )ψαθ (ω) fb(ω, s, u(ω, s))ds.
(41)
0
Combining (40) and (41), we obtain ZT exp ψαθ (ω)(T − t) θ b gb(ω) − exp ψα (ω)(s − T ) f (ω, s, u(ω, s))ds u b(ω, t) = 1 + β exp |ω|α cos πθ 2 T t πθ θ α β exp − tψα (ω) exp |ω| cos 2 T + u b(ω, 0) 1 + β exp |ω|α cos πθ 2 T Zt β exp |ω|α cos πθ 2 T exp (s − t)ψαθ (ω) fb(ω, s, u(ω, s))ds. + 1 + β exp |ω|α cos πθ 2 T 0
It follows from (37) and (42) that
u b(ω, t) − Vc β (ω, t) = B1 + B2 + B3 ,
19
(42)
where B1
=
B2
=
B3
=
h i exp ψαθ (ω)(s − t) fb(ω, s, Vβ (ω, s)) − fb(ω, s, u(ω, s)) ds, 1 + β exp |ω|α cos πθ 2 T t β exp − tψαθ (ω) exp |ω|α cos πθ 2 T u b(ω, 0), πθ 1 + β exp |ω|α cos 2 T Zt h i β exp |ω|α cos πθ 2 T exp (s − t)ψαθ (ω) fb(ω, s, u(ω, s)) − fb(ω, s, Vβ (ω, s)) ds. 1 + β exp |ω|α cos πθ 2 T 0
ZT
This leads to
u(ω, t) − Vc b β (ω, t) ≤
|B1 | + |B2 | + |B3 | ZT
≤
β
t−s T
t
t
b f (ω, s, u) − fb(ω, s, Vβ ) ds
+ β T |b u(., 0)| + t
u(., 0)| + β T |b
≤ 220
Using (38), we get
u(., t) − Vβ (., t) 2 2 L (R)
ZT
Zt
β
t−s T
0
β
t−s T
0
b f (ω, s, u) − fb(ω, s, Vβ ) ds.
1 1+ m
−∞
0
2t T
β ku(., 0)k2L2 (R) 2t
+ (1 + m)T k 2 β T
ZT
β
−2s T
0
Thus −2t T
(43)
2
= b u(., t) − Vc β (., t) 2 L (R) 2t 1 β T ku(., 0)k2L2 (R) ≤ 1+ m 2 +∞ ZT Z t−s β T fb(ω, s, u) − fb(ω, s, Vβ ) ds dω + (1 + m)
≤
β
b f (ω, s, u) − fb(ω, s, Vβ ) ds
u(., t) − Vβ (., t) 2 2 L (R)
u(., s) − Vβ (., s) 2 2 ds. L (R)
1 1+ ku(., 0)k2L2 (R) m ZT
2 −2s × + (1 + m)T k 2 β T u(., s) − Vβ (., s) L2 (R) ds.
≤
0
Since u(., t) and Vβ (., t) ∈ C([0, T ]; L2 (R)), the function u(., t) − Vβ (., t) is continuous on [0, T ]. L2 (R)
2 −2t Therefore, there exists a positive N = max β T u(., t) − Vβ (., t) L2 (R) . This implies that N≤
t∈[0,T ]
1+
1 m
ku(., 0)k2L2 (R) + (1 + m)T 2 k 2 N, 20
that is, β
−2t T
1 1+ m ku(., 0)k2L2 (R)
u(., t) − Vβ (., t) 2 2 ≤ N ≤ . L (R) 1 − (1 + m)T 2 k 2
Hence, we obtain the error estimate
u(., t) − Vβ (., t) 2 ≤ L (R)
s
1 1+ m t ku(., 0)kL2 (R) β T . 1 − (1 + m)T 2 k 2
From Step 2 and Step 3, we obtain
Uβ (., t) − u(., t) 2 ≤ Uβ (., t) − Vβ (., t) L2 (R) + u(., t) − Vβ (., t) L2 (R) L (R) s s 1 1 1+ m 1+ m t−T t T + β ku(., 0)kL2 (R) β T ≤ 1 − (1 + m)T 2 k 2 1 − (1 + m)T 2 k 2 s 1 1+ m t β −1 + ku(., 0)kL2 (R) ≤ βT . 1 − (1 + m)T 2 k 2 4. Numerical experiment 225
230
To verify our proposed methods, we carry out the numerical experiment for above regularization methods. The numerical example is implemented for T=1, and with the variation of α ∈ (0, 2]. In order to illustrate the sensitivity of the computational accuracy to the noise of the measurement data, we use the random function to generate the noisy data similar to an observation data. The perturbation was defined as rand(size()), where rand(size()) is a random number, and plays as an amplitude of the errors. The approximation of the regularization solution is computed by the FFT algorithm. In this example, we consider the following Riesz - Feller space fractional backward diffusion problem α ut (x, t) = x Dθ u(x, t) + f (x, t, u), (x, t) ∈ (−π, π) × (0, T ), u(−π, t) = u(π, t) = 0, t ∈ (0, T ), (44) u(x, T ) = g(x), x ∈ (−π, π). Here,
f (x, t, u)
=
−u + h(x, t),
where g(x)
= e sin3 (2x),
h(x, t) = αt
α−1
235
240
( ) h 1 α i h i 1 α α exp(t ) sin (2x) − exp(t ) 3 1 + sin(2x) − 1 + (6) sin(6x) , 4 2 α
3
with x ∈ (−π, π), subjecting to u(−π, t) = u(π, t) = 0, and u(x, 0) = sin3 (2x). The exact solution of Problem (44) is u(x, t) = exp(tα ) sin3 (2x). The regularized solution is defined by (5) with ψαθ (ω) defined by (2). The regularized solution is expected to be closed to the exact solution under a proper discretization as shown below. In general, the whole numerical procedure is proceeded in the following steps: Step 1. Choosing I and J to generate spatial and temporal discretizations as follows xi
=
tj
=
2π , i = 0, I, I 1 j∆t, ∆t = , j = 0, J. J i∆x, ∆x =
21
Of course, the higher value of I and J will provide more stable and accurate numerical calculation, however in the following examples I = J = 201 are satisfied. Step 2. We choose the couple (g , F ) as the observed data including the noise in the manner that g F (., .) 245
rand(size(g)) √ , π (rand(size(F (., .)))) √ . F (., .) + π
= g+ =
Step 3. Set u(., :)β() (xi ) = u(., :)β(),i and u(., :) (xi ) = ui , constructing two following vectors contained all discrete values of u(., :)β and u(., :) denoted by Λβ and Ψ, respectively, Λβ = u(., :)β,0 u(., :)β,1 ... u(., :)β,I−1 u(., :)β,I ∈ RI+1 , Ψ = u(., :)0 u(., :)1 ... u(., :)I−1 u(., :)I ∈ RI+1 . Step 4. Errors between the exact and its regularized solutions are estimated by Absolute error estimation v u I u 1 X E1 = t |u (xi , .) − u(xi , .)|2 . I + 1 i=0 β Relative error estimation
E2 =
250
255
260
265
s
I P
i=0
|uβ (xi , .) − u(xi , .)|2
s
I P
i=0
|u(xi
.
, .)|2
Fig. 1 shows the graphs of the exact solutions of Problem (44) corresponding to α = 0.2, 0.6, 1.4 and 1.8. In this example as shown in Fig. 2, when α is small (close to zero) the concentration distribution u(x, t) does not fluctuate much once t is increased. On the other hand, when t tends to zero the concentration distribution u(x, t) is more fluctuating once the value of α is decreased to zero as shown in Fig. 3. Figs. 4 and 5 show the comparisons between the exact and its regularized solutions at t = 0.5, and for α = 0.2 and 0.8, respectively. Figures 6 and 7 show the comparisons between the exact and its regularized solutions at t = 0.9, and for α = 0.2 and 0.8, respectively. Even for α = 0.2, at the beginning the regularized solution is oscillated around the exact solution, eventually the regularized solution is converged very well to the exact solution once tends to 0. Tables 1 and 2 show the estimate of absolute (E1 ) and relative (E2 ) errors between the exact and its regularized solutions with different values of α (0.2, 0.6, 1.4 and 1.8) at t = 0.5 and 0.9, respectively. It clearly shows the regularized solution converges to the exact solution with different values of α. Tables 3, 4, 5 and 6 show the estimate of absolute (E1 ) and relative (E2 ) errors between the exact and its regularized solutions at the variation of t from 0.1 to 0.8 for a lower value of α = 0.2, and a higher value of α = 1.8, respectively. These tables are not including the values of t at 0.5 and 0.9, because they are already shown in Tables 1 and 2 above. In general, the errors between the exact and its regularized solutions are slightly higher once the order α of the Riesz-Feller fractional derivative is closed to zero. It also confirms again the regularized solution is converged very well to the exact solution in the range of t ∈ (0, 1), and of α ∈ (0, 2]. 5. Conclusion
270
In this study, we introduced two new modified regularization methods as shown in Theorems 1 and 2 for solving a Riesz-Feller space fractional backward diffusion problem with a nonlinear source. In theoretical results, we overcome the strong condition as shown in (7) and (9) by a weaker condition based on Theorem 2, and the error estimates are obtained. In our numerical experiment, it is shown that the above regularized solution converges nicely to the exact solution with different values of α ∈ (0, 2]. 22
(a) α = 0.2
(b) α = 0.6
(c) α = 1.4
(d) α = 1.8 Figure 1: The exact solutions with different α .
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(a) α = 0.2
(b) α = 0.6
(c) α = 1.4
(d) α = 1.8
Figure 2: The exact solutions with different α at t=0.2, 0.5, 0.7 and 0.9.
(a) t = 0.2
(b) t = 0.5
(c) t = 0.7
(d) t = 0.9
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Figure 3: The exact solutions at different t with different values of α (α = 0.2, 0.6, 1.4 and 1.8)
(a) = 10−1
(b) = 10−2
(c) = 10−3
(d) = 10−4
Figure 4: A comparison between the exact and its regularized solutions at t = 0.5 with α = 0.2 .
(a) = 10−1
(b) = 10−2
(c) = 10−3
(d) = 10−4
Figure 5: A comparison between the exact and its regularized solutions at t = 0.5 with α = 1.8.
26
(a) = 10−1
(b) = 10−2
(c) = 10−3
(d) = 10−4
Figure 6: A comparison between the exact and its regularized solutions at t = 0.9 with α = 0.2.
(a) = 10−1
(b) = 10−2
(c) = 10−3
(d) = 10−4
Figure 7: A comparison between the exact and its regularized solutions at t = 0.9 with α = 1.8.
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1E-01 1E-02 1E-03 1E-04 1E-05
α = 0.2 E1 E2 4.19E-01 6.55E-01 1.69E-01 2.65E-01 6.02E-02 9.40E-02 2.09E-02 3.27E-02 7.30E-03 1.13E-02
α = 0.6 E1 E2 3.62E-01 9.27E-01 1.24E-01 3.17E-01 4.24E-02 1.09E-01 1.47E-02 3.76E-02 5.10E-03 1.30E-02
α = 1.4 E1 E2 8.39E-01 1.55E+00 2.55E-01 4.71E-01 1.04E-01 1.92E-01 3.67E-02 6.77E-02 1.27E-02 2.35E-02
α = 1.8 E1 E2 7.69E-01 9.96E-01 7.43E-02 9.62E-02 4.36E-02 5.65E-02 3.99E-03 5.16E-03 1.24E-04 1.60E-04
Table 1: The absolute (E1 ) and relative (E2 ) error estimates between the exact and its regularized solutions at t = 0.5.
1E-01 1E-02 1E-03 1E-04 1E-05
α = 0.2 E1 E2 3.32E-01 5.86E-01 1.37E-01 2.41E-01 4.85E-02 8.55E-02 1.69E-02 2.97E-02 5.90E-03 1.03E-02
α = 0.6 E1 E2 2.26E-01 4.21E-01 7.50E-02 1.40E-01 2.60E-02 4.84E-02 9.00E-03 1.68E-02 3.10E-03 5.80E-03
α = 1.4 E1 E2 9.17E-01 1.75E+00 2.69E-01 5.13E-01 8.85E-02 1.69E-01 3.06E-02 5.84E-02 1.06E-02 2.02E-02
α = 1.8 E1 E2 5.96E-01 9.97E-01 5.82E-01 9.72E-01 3.00E-02 5.01E-02 3.72E-02 6.22E-02 1.12E-03 1.86E-03
Table 2: The absolute (E1 ) and relative (E2 ) error estimates between the exact and its regularized solutions at t = 0.9.
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t 1E-01 1E-02 1E-03 1E-04 1E-05
t = 0.1 E1 E2 5.23E-01 7.44E-01 2.09E-01 2.97E-01 7.42E-02 1.05E-01 2.58E-02 3.66E-02 8.90E-03 1.27E-02
t = 0.2 E1 E2 4.94E-01 7.17E-01 1.99E-01 2.88E-01 7.04E-02 1.02E-01 2.45E-02 3.55E-02 8.50E-03 1.23E-02
t = 0.3 E1 E2 4.69E-01 6.95E-01 1.89E-01 2.80E-01 6.69E-02 9.92E-02 2.32E-02 3.45E-02 8.10E-03 1.20E-02
t = 0.4 E1 E2 4.43E-01 6.74E-01 1.79E-01 2.72E-01 6.35E-02 9.65E-02 2.21E-02 3.36E-02 7.70E-03 1.16E-02
Table 3: The absolute (E1 ) and relative (E2 ) error estimates between the exact and its regularized solutions with α = 0.2.
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t 1E-01 1E-02 1E-03 1E-04 1E-05
t = 0.6 E1 E2 3.96E-01 6.36E-01 1.61E-01 2.58E-01 5.71E-02 9.17E-02 1.99E-03 3.19E-03 6.90E-04 1.11E-03
t = 0.7 E1 E2 3.74E-01 6.19E-01 1.52E-01 2.52E-01 5.41E-02 8.95E-02 1.88E-03 3.11E-03 6.50E-04 1.08E-03
t = 0.8 E1 E2 3.53E-01 6.03E-01 1.44E-01 2.46E-01 5.13E-02 8.75E-02 1.78E-03 3.05E-03 6.20E-04 1.06E-03
Table 4: The absolute (E1 ) and relative (E2 ) error estimates between the exact and its regularized solutions with α = 0.2.
t 1E-01 1E-02 1E-03 1E-04 1E-05
t = 0.1 E1 E2 5.23E+00 9.95E-01 5.08E-01 9.96E-01 9.70E-02 1.84E-02 3.01E-03 5.72E-03 9.19E-04 1.75E-04
t = 0.2 E1 E2 3.56E+00 9.96E-01 3.46E-01 9.67E-01 2.67E-02 7.46E-02 2.01E-03 5.62E-03 6.16E-04 1.72E-01
t = 0.3 E1 E2 2.32E+00 9.98E-01 2.50E-01 9.66E-01 5.98E-02 2.60E-02 2.25E-03 5.47E-03 3.93E-04 1.69E-04
t = 0.4 E1 E2 1.40E+00 9.53E-01 1.36E-01 9.62E-01 5.84E-02 4.14E-02 7.36E-03 5.21E-03 2.30E-04 1.63E-04
Table 5: The absolute (E1 ) and relative (E2 ) error estimates between the exact and its regularized solutions with α = 1.8.
t 1E-01 1E-02 1E-03 1E-04 1E-05
t = 0.6 E1 E2 6.10E-01 9.97E-01 5.97E-01 9.76E-01 4.14E-02 6.77E-02 3.50E-03 5.71E-03 1.22E-04 1.99E-04
t = 0.7 E1 E2 6.54E-01 9.97E-01 6.40E-01 9.75E-01 4.28E-02 6.53E-02 4.06E-03 6.19E-03 1.27E-04 1.94E-04
t = 0.8 E1 E2 6.40E-01 9.97E-01 6.24E-01 9.73E-01 4.06E-02 6.33E-02 3.50E-03 5.46E-03 1.21E-04 1.89E-04
Table 6: The absolute (E1 ) and relative (E2 ) error estimates between the exact and its regularized solutions with α = 1.8.
29