On cauchy problem for nonlinear fractional differential equation with random discrete data

On Cauchy problem for nonlinear fractional differential equation with random discrete data Applied Mathematics and Computation 362 (2019) 124458 Contents lists available at ScienceDirect Applied Mathe[.]

Applied Mathematics and Computation 362 (2019) 124458 Contents lists available at ScienceDirect Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc On Cauchy problem for nonlinear fractional differential equation with random discrete data Nguyen Duc Phuong a,b, Nguyen Huy Tuan c,∗, Dumitru Baleanu d,e, Tran Bao Ngoc f a Faculty of Fundamental Science, Industrial University of Ho Chi Minh City, Vietnam Department of Mathematics and Computer Science VNUHCM - University of Science, 227 Nguyen Van Cu Str., Dist 5, HoChiMinh City, Vietnam c Applied Analysis Research Group Faculty of Mathematics and Statistics, Ton Duc Thang University, Ho Chi Minh City, Vietnam d Department of Mathematics, Cankaya University, Ankara, Turkey e Institute of Space Sciences, Magurele, Bucharest, Romania f Institute of Research and Development, Duy Tan University, Da Nang 550000, Vietnam b a r t i c l e i n f o a b s t r a c t Keywords: Fractional derivative ill-posed problem Elliptic equation Random noise Regularized solution This paper is concerned with finding the solution u(x, t) of the Cauchy problem for nonlinear fractional elliptic equation with perturbed input data This study shows that our forward problem is severely ill-posed in sense of Hadamard For this ill-posed problem, the trigonometric of non-parametric regression associated with the truncation method is applied to construct a regularized solution Under prior assumptions for the exact solution, the convergence rate is obtained in both L2 and Hq (for q > 0) norm Moreover, the numerical example is also investigated to justify our results © 2019 Elsevier Inc All rights reserved Introduction In this work, we focus on finding the solution for the following time fractional elliptic equation ∂tα u + u = f (x, t, u ), (x, t ) ∈  × (0, T ), (1) with the Cauchy condition and initial conditions  u(x, t ) = 0, (x, t ) ∈ ∂  × (0, T ), u(x, ) = ρ (x ), x ∈ , ut (x, ) = ξ (x ), x ∈  (2) where  = (0, π ) ⊂ R is a bounded connected domain with a smooth boundary ∂ , and T is a given positive real number The time fractional derivative ∂tα u is the Caputo fractional derivative of order α ∈ (1, 2) with respect to t defined in [1–3] as follows ∂tα u(x, t ) = (2 − α )  t (t − ω )(1−α ) ∗ ∂2 u(x, ω )dω, (x, t ) ∈  × (0, T ), ∂ω2 (3) Corresponding author E-mail addresses: nguyenducphuong@iuh.edu.vn (N.D Phuong), nguyenhuytuan@tdtu.edu.vn (N.H Tuan), dumitru@cankaya.edu.tr (D Baleanu), tranbaongoc@hcmuaf.edu.vn (T.B Ngoc) https://doi.org/10.1016/j.amc.2019.05.029 096-30 03/© 2019 Elsevier Inc All rights reserved 2 N.D Phuong, N.H Tuan and D Baleanu et al / Applied Mathematics and Computation 362 (2019) 124458 where (·) is the Gamma function Recall that, our purpose is to seek the solution u(x, t) from the given data ρ , ξ ∈ L2 () Unfortunately, in practice, we not have these functions Instead, we have only their measurable value at finite points, and the measurement always contains the errors These errors could come from a controllable or uncontrollable source One may find many studies the problem with the first type of error which is often called deterministic [4–7] Otherwise, if the errors come from an uncontrollable source such as wind, humidity, etc., then the model is random Therefore, in this work we assume that the value of ρ (x), ξ (x) are measured at fixed design points xk = π 2k2−1 n for k = 1, , n following discrete random models: ρk = ρ (xk ) + τk Xk , (4) ξk = ξ (xk ) + νkYk , (5) i.i.d random variables Xk , Yk are mutually independent and identically distributed Xk , Yk ∼ N (0, ) and τ k , ν k are unknown positive real constants and bounded by constants Mτ , Mν , respectively, i.e., < τk ≤ Mτ and < νk ≤ Mν Fractional derivatives have been found to be more flexible than the traditional integer derivative in describing practical phenomena Hence, in the last decades, fractional calculus and derivatives have much success in many field of science, for instance, physics [8–11], signal and image processing [12,13], environmental science [14,15], control theory [16,17] and so on One of the most important reason of using fractional differential equations in these and other applications is its non-local behavior property which is different from the integer order differential operator which is so called local behavior [18–20] Due to the difficulty of the fractional derivative and the ill-posedness, at least up to our knowledge, the results on problems for time-fractional elliptic equation are very few Recently, Tuan and his partner [4] had considered Problem (1) in the deterministic case However, there are no publications on the Cauchy problem for nonlinear fractional elliptic equation (1) with discrete random noise data (4)–(5), and that is also the main motivation for us to conduct this article The aim of this paper is to find the approximate solution for the exact solution by the trigonometric of non-parametric regression associated with the truncation method is applied to construct a regularized solution Then under some assumptions on the exact solution, we will proof that the approximate solution will converge to the correct solution both in L2 and Hq (for q > 0) norm This paper is organized as follows In Section 2, we briefly review the main definitions and present the discretization scheme for Fourier coefficients In Section 3, we give an example to demonstrate the ill-poseness of the problem The main results are discussed in Section Numerical experiment for various value of α is given in Section We conclude the paper with Section Preliminaries Definition 2.1 (L2 () space) Given that f :  → R is Lebesgue measurable We define L2 () =    f :  f ( x )d x < ∞ , with the inner product and the norm f1 , f2 =   f 2L2 () = f1 (x ) f2 (x )dx,   f (x )dx For a fixed positive number τ > 0, we define the Sobolev class of function as  Hτ () = f ∈ L () : ∞   λτp f, ϕ p < ∞ , p=1 it is a Hilbert space which endowed with the norm f 2Hτ () = ∞  λτp f, ϕ p p=1 Given a Banach space B, and p ≥ Let us define the Bochner space L p (; B ) = L p ((, F, P ); B ) (see [21]) as following  L p (; B ) = u : EuBp =    u(θ )Bp dP(θ ) < ∞, θ ∈  , with the norm uL p (;B ) = EuBp p The space of all the L2 -value predictable process u (see [22]) defined as following VT = u : sup 0≤t≤T u(·, t )L2 (;L2 ()) = sup 0≤t≤T Eu(·, t ) L2 () and β ∈ R are arbitrary constants − + Lemma 2.1 (See [4]) Given α ∈ [α , α ] satisfies < α < α < Then there exist positive constants cα , cα such that the following inequalities hold for any t > 0: − + cα exp(λ1p/α t ) ≤ Eα ,1 (λ pt α ) ≤ cα exp(λ1p/α t ), + t Eα ,2 (λ pt α ) ≤ cα λ1p/α exp(λ1p/α t ), −α + t α −1 Eα ,α (λ pt α ) ≤ cα λ p α exp(λ1p/α t ) Suppose that the solution of (1) is given by the following Fourier series u(x, t ) = ∞  u p (t )ϕ p (x ), p=1 where we write u p (t ) = u(·, t ), ϕ p for the usual inner product in L2 () The Fourier coefficients up (t) are solution of the following ordinary differential equation ⎧ α ∂ u − λ p u p = f p ( u ), ⎪ ⎪ t p ⎨ u p (0 ) = ρ p , ⎪ ⎪ ⎩ ∂ u (0 ) = ξ p ∂t p (7) where f p (u ) = f (x, t, u ), ϕ p , ρ p = ρ , ϕ p and ξ p = ξ , ϕ p By using the same method as [24–26], the system (7) has the solution u p (t ) = Eα ,1 (λ pt α )ρ p + t Eα ,2 (λ pt α )ξ p +  t (t − ω )α−1 Eα,α (λ p (t − ω )α ) f p (u )dω Definition 2.3 The function u(x, t) is called a mild solution of the system (1)–(2) if u ∈ C([0, T]; L2 ()) and satisfies that u(x, t ) = ∞   p=1 Eα ,1 ( λ p t α ) ρ p + t Eα ,2 ( λ p t α ) ξ p +  t  (t − ω )α−1 Eα,α (λ p (t − ω )α ) f p (u )dω ϕ p (x ) (8) N.D Phuong, N.H Tuan and D Baleanu et al / Applied Mathematics and Computation 362 (2019) 124458 ¯ ¯ Given a sequence of points xk = π 2k2−1 n , k = 1, , n in  and f (x ) ∈ C () We noticed that the pth Fourier coefficients is f p = f, ϕ p , and the approximation of an integral by a Riemann sum fp ≈ n π n f ( xk )ϕ p ( xk ) k=1 Fortunately, we have the equation of relation between fp and Riemann sum as following ¯ For p = 1, , n − 1, we set Lemma 2.2 Assume that f ∈ C () fn;p := n π n f ( xk )ϕ p ( xk ), k=1 with xk = π 2k2−1 n ∈  The Fourier coefficients f p = f, ϕ p of the function f can be represented as f p = fn;p − Rnf ;p , f where Rn;p is called a residual of approximation fp by fn; p and its form is Rnf ;p = ∞  (−1 )l f2ln+ p − f2ln−p (9) l=1 Proof As for the proof, see Tuan and Erkan [27], and see also Randall [28, p 144]  f Under the specific assumption, the residual Rn;p tends to zero as the number of points n tends to infinity This is reflected in the following Lemma Lemma 2.3 Assume that f ∈ Hτ () with τ > Then, there exists a constant C (τ , f ) which depends on f and τ , such that  f  Rn;p  ≤ C (τ , f ) , τ for all ≤ p ≤ n − 1, n (10) Proof The assumption f ∈ HBτ implies that f 2Hτ () ≥ λτp f, ϕ p Since λ p = p2 , we deduce that f Hτ () ≥ pτ f, ϕ p Thus ∞ ∞  f    Rn;p  ≤ | f, ϕ2ln+ p | + | f, ϕ2ln−p | ≤ l=1  l=1  f Hτ () f Hτ () + (2ln + p)τ (2ln − p)τ It is clear evident that 2ln + p ≥ ln and 2ln − p ≥ ln (because of ≤ p ≤ n − 1), hence   ∞ ∞  f   f Hτ () f Hτ () f Hτ ()  Rn;p  ≤ + ≤ τ τ τ (ln ) (ln ) n lτ l=1 l=1 (11) By putting C (τ , f ) = f Hτ () , nτ the proof of Lemma is complete  Assume g(x) is a unknown function from  into R Its observed value at xk is  gk which drawn from random model  gk = i.i.d g(xk ) + εk Here ε k are mutually independent and identically distributed εk ∼ N (0, σk2 ) Minh and his partner [29] used the non-parametric least square method to generate estimators of g(x) in SN ,  gNn (x ) = N  p=1   n π  gk ϕ p ( xk ) ϕ p ( x ) n (12) k=1 Especially, if σk2 ≤ M2 , using the properties of mutually independent and identically distributed of normal variables, Nane and Tuan [30] also have  E n π n k=1 2 εk ϕ p ( xk ) = n π2  n2 k=1 Eεk2 ϕ p2 (xk ) ≤ π n M2 (13) N.D Phuong, N.H Tuan and D Baleanu et al / Applied Mathematics and Computation 362 (2019) 124458 ill-posedness of the problem In this section, we give an example in order to demonstrate the ill-posedness of the problem in the sense of Hadamard (the solution is not stable) Proposed the problem We consider the simplest special case of choosing ϕ (x ) = ξ (x ) = and the source f (x, t, u ) = ∞  αα−1 λ p exp −λ1p/α t u, ϕ p ϕ p (x ) + T cα (14) p=1 Problem (1)–(2) has a unique solution u(x, t ) = Indeed, from (8) we have u(x, t ) =  ∞  t p=1 (t − ω )α−1 Eα,α (λ p (t − ω )α ) f p (u )(ω )dω  ϕ p ( x ), the Parseval identity leads to u(·, t ) L2 ()  ∞  = t p=1 2 (t − ω )α−1 Eα,α (λ p (t − ω )α ) f p (u )(ω )dω , by the Hölder inequality, we have u(·, t )2L2 () ≤ ∞   p=1 t dω  t   (t − ω )α−1 Eα,α (λ p (t − ω )α ) f p (u )(ω ) dω, As the source function is set by (14) and using Lemma 2.1, we have either u(·, t )2L2 () ≤  t  t ∞   t dω u, ϕ p dω ≤ u(·, ω )2L2 () dω, 4T 4T 0 p=1 by using the Gronwall inequality, we get u(·, t )L2 () = or u(x, t ) = The problem with perturbed data Let us consider the problem (1)–(2) with random models (4)–(5) Here, we choose τk = νk = √1n and set N = n − From the function estimation formula (12), the estimator of ρ , ξ in SN are ρ nn−1 (x ) = n−1  p=1 ξnn−1 (x ) = n−1   n  p=1  n π k=1 √ Xk ϕ p (xk ) n  n π n k=1 ϕ p ( x ), √ Yk ϕ p (xk ) n ϕ p ( x ), respectively Now, let us define the problem that we are going to compare with above problem ⎧ α ∂t u +  u = f (x, t,  u ), (x, t ) ∈  × (0, T ), ⎪ ⎪ ⎨ u(x, t ) = 0, (x, t ) ∈ ∂  × (0, T ),  nn−1 (x ), u ( x, ) = ρ x ∈ , ⎪ ⎪ ⎩ n −1   ut (x, ) = ξn (x ), x ∈  (15) From the result (8) leads to  u(x, t ) =   t  n−1   −1 −1 nn;p Eα ,1 ( λ p t α ) ρ + tEα ,2 (λ pt α )ξnn;p + (t − ω )α−1 Eα,α (λ p (t − ω )α ) f p ( u )d ω ϕ p ( x ) , p=1 (16)    −1 −1 where ρnn;p = ρnn−1 , ϕ p , ξnn;p = ξnn−1 , ϕ p We need to show that it has a unique solution  u(x, t ) ∈ VT We start by setting J ( u )(x, t ) =  t  n−1   −1 −1 nn;p Eα ,1 ( λ p t α ) ρ + tEα ,2 (λ pt α )ξnn;p + (t − ω )α−1 Eα,α (λ p (t − ω )α ) f p ( u )d ω ϕ p ( x ) p=1 Noting that  u is a trigonometric polynomial with order less than n with respect to the variable x Using the Parseval identity, we get u )(t ) − J( v )(t )2L2 () = J (   n−1   p=1 t 2 (t − ω )α−1 Eα,α (λ p (t − ω )α )( f p ( u ) − f p ( v ) )dω N.D Phuong, N.H Tuan and D Baleanu et al / Applied Mathematics and Computation 362 (2019) 124458 By f(x, t, u) defined in (14), Lemma 2.1 and the Hölder inequality, we have E 4T E J (  u )(t ) − J( v )(t )2L2 () ≤   t u(·, ω ) −  v(·, ω )2L2 () dω  E u(·, ω ) −  v(·, ω )2L2 () , ≤ sup 0≤ω≤T this leads to u ) − J( v )VT ≤  u − vVT , J (  it means J is the contraction with respect to the norm ·V By the Banach fixed point theorem, there is a unique function T  u ∈ VT such that J( u) =  u almost surely The rest of the example will show that the mean square error between exact Cauchy nn−1 ; ξnn−1 ) are small, but the mean square error between the solution u and  data (ρ , ξ ) and their estimators (ρ u is very large Small change in data By the Parseval identity, we have ρnn−1 2L2 () = n−1   n p=1 2 n π √ Xk ϕ p (xk ) n k=1 Using (13), we have n−1 − ρ2L2 () = Eρ n−1  π = n2 p=1 π (n − ) n2 By a similar argument as above, we also have Eξn−1 − ξ 2L2 () = π (n − ) n2 Large change in the solution From (16), we have  −1 α n−1 nn;n E u(·, t )L2 () ≥ E Eα ,1 (λn−1 t α )ρ −1 + tEα ,2 (λn−1 t )ξn;n−1  + t (t − ω )α−1 Eα,α (λn−1 (t − ω )α ) fn−1 ( u )d ω  =: E(I1 + I2 + I3 )2 The inequality 2(a + b + c )2 ≥ a2 − 4b2 − 4c2 for all a, b, c ∈ R leads to 2E u(·, t )L2 () ≥ EI21 − 4EI22 − 4EI23 (17) From Lemma 2.1 and (13), we have  EI21 = (Eα ,1 (λn−1 t α ) ) E n Similarly, we have EI22 2 n π  = (t Eα ,2 (λn−1 t α ) ) E k=1 √ Xk ϕn−1 (xk ) n 2 n π n k=1 ≥ √ Yk ϕn−1 (xk ) n ≤ (cα− )2 π n2 α exp 2λ1n/−1 t (cα+ )2 π α exp 2λ1n/−1 t n2 λ1/α n−1 (18) (19) Using Lemma 2.1 again, we have EI23 ≤ E  t 1−α α cα+ λn−1 exp  α (t − ω ) fn−1 ( u )d ω , λ1n/−1 and the Hölder inequality leads to EI23 ≤ ( cα ) + ≤ E 4T 1−αα n−1 λ  t  E t dω  t u, ϕn−1 dω ≤  exp 2λ (t − ω ) fn2−1 ( u )d ω  sup E u(·, t )L2 () 0≤t≤T Taking (17)–(20) together, we have /α n−1 α 2E u(·, t )L2 () ≥ π exp 2λ1n/−1 t  (cα− )2 n2 ( c + )2 − n2 (20) α λ1n/−1  − sup E u(·, t )L2 () , 0≤t≤T N.D Phuong, N.H Tuan and D Baleanu et al / Applied Mathematics and Computation 362 (2019) 124458 this leads to α sup E u(·, t )L2 () ≥ π exp 2λ1n/−1 t  (cα− )2 n2 0≤t≤T Thus − ( c + )2 n2 α λ1n/−1  n−1 − ρ2L2 () + Eξn−1 − ξ 2L2 () = 0, lim Eρ n→∞ however lim  u − uVT = ∞, n→∞ we can conclude that Problem (1)–(2) is ill-posed Regularized solution and convergence estimate 4.1 Regularized solution k , ξk ) of Cauchy data (ρ , ξ ) at xk = π 2k2−1 Given observation values (ρ n for k = 1, , n Given an integer number N such that < N < n, which later on will play a role of the regularization parameter The formula (12) provides the estimates of ρ , ξ in subspace SN , ρnN (x ) = N  i=1 N n ξ (x ) = N   n π n  ρ (xk ) + τk Xk ϕ p (xk ) ϕ p (x ), k=1 n π n i=1  !  ! ξ (xk ) + νkYk ϕ p (xk ) ϕ p (x ) k=1 We are considering a problem ⎧ α N ∂t U + UN = f x, t, UN , ⎪ ⎪ ⎨UN (x, t ) = 0, N (x, ) = ρ nN (x ), U ⎪ ⎪ ⎩ N (x, ) = ξN (x ), U n t (x, t ) ∈  × (0, T ), (x, t ) ∈ ∂  × (0, T ), x ∈ , (21) x ∈  Similar results have been obtained (8), it leads to the solution of (21) as following N (x, t ) = U   t  N   nN;p + t Eα ,2 (λ pt α )ξnN;p + Eα ,1 ( λ p t α ) ρ (t − ω )α−1 Eα,α (λ p (t − ω )α ) f p (UN )dω ϕ p (x ) p=1 (22)    where ρnN;p = ρnN , ϕ p , ξnN;p = ξnN , ϕ p We propose a new regularized solution as (22) and the regularization parameter N will be chosen depending on n 4.2 Convergence estimate In this work, the nonlinear reaction term f(x, t, u) is assumed to have the globally Lipschitz property with respect to the third variable, i.e., there exists a constant K > independent of x, t, u, v such that f (·, t, u(·, t )) − f (·, t, v(·, t ))L2 () ≤ K u(·, t ) − v(·, t )L2 () , for all u, v ∈ L2 () (23) N ∈ VT Theorem 4.1 The nonlinear integral equation (22) has a unique solution U Proof We first put J(u )(x, t ) = N   nN;p + t Eα ,2 (λ pt α )ξnN;p + Eα ,1 ( λ p t α ) ρ  p=1 t  (t − ω )α−1 Eα,α (λ p (t − ω )α ) f p (u )dω ϕ p (x ) (24) for u ∈ VT We would like to verify that J is a contraction mapping By induction, we can prove that for every u, v ∈ VT , " Jm (u ) − Jm (v )VT ≤ 1−αα T K (cα+ )2 λ1 exp 2T λ1N/α m!  m u − vVT (25) N.D Phuong, N.H Tuan and D Baleanu et al / Applied Mathematics and Computation 362 (2019) 124458 We first consider the base case, m = EJ(u )(t ) − J(v )(t ) L2 () =E  N  2 t (t − ω )α−1 Eα,α (λ p (t − ω )α )( f p (u ) − f p (v ) )dω n=1 1−αα ≤ T K Kα2 λ1 this leads to " 1−αα exp 2T λ1N/α T K (cα+ )2 λ1 J(u ) − J(v )VT ≤  exp 2T λ1N/α E t u(·, ω ) − v(·, ω )2L2 () dω, 1! u − vVT Then, we assume (25) to hold for m, we will prove it for m + This can be done as follows # #2  # #2 = E#J Jm+1 (u ) − J Jm+1 (v ) # L () m+1  t (T t )m exp 2T λ1N/α E u(·, ω ) − v(·, ω )2L2 () dω m! E#Jm+1 (u ) − Jm+1 (v )# 1−αα ≤ T K (cα+ )2 λ1 L2 () Therefore, by the induction principle, we say that (25) holds Since " 1−αα T K (cα+ )2 λ1 lim exp 2T λ1N/α  m = 0, m! m→∞ (26) there exists a positive integer m0 such that Jm0 is a contraction with respect to the norm ·V It follows that the equation T N ∈ VT , it means Jm0 (U N ) = U N almost surely Moreover, since J(Jm0 (U N )) = J(U N ) almost Jm0 (u ) = u has a unique solution U N )) = J(U N ) almost surely Hence J(U N ) is also a fixed point of Jm0 By the uniqueness of the fixed point surely, then Jm0 (J(U N ) = U N has a unique solution U N ∈ VT  of Jm0 , we conclude that J(U 4.2.1 The error in L2 () norm Theorem 4.2 Let ρ ∈ C [0, π ] ∩ Hτ (), ξ ∈ C [0, π ] ∩ Hν (), where τ , ν > 1, and < N < n Furthermore, assume that Problem (1)–(2) has unique solution u ∈ C([0, T]; L2 ()) If there exists a positive number μ such that sup ∞  0≤t≤T p=1 λ2pμ exp 2λ1p/α (T − t ) u, ϕ p ≤ , t ∈ [0, T ], for some constant , then we have EU N (·, t ) − u(·, t )2L2 () ≤ where N n −2μ exp 2λ1N/α t  + 4λN (27)  exp −2λ1N/α (T − t )  exp 4t T K cα+ λ21 1−α α ! ,   π M2τ + C (ν, ξ )  = 6(cα+ )2 π M2τ + C (τ , ρ ) + λ11/α Remark 4.1 By choosing N := N(n) such that  λN (n) ≤ α δ 2T ln n , < δ < # #2 If t = T , the mean square error E#U N (x, t ) − u(x, t )#  max  nδ −1 ; δ 2T max  nδ −1 ; δ 2T  −2μα ln n is of order ln n Otherwise, it is of order  L2 () −2μα  n − Tδ (T −t ) Proof We define the nonlinear integral equation as following uN (x, t ) =  t  N   Eα ,1 (λ pt α )ρ p +t Eα ,2 (λ pt α )ξ p + (t − ω )α−1 Eα,α (λ p (t − ω )α ) f p (uN )dω ϕ p (x ) p=1 (28) N.D Phuong, N.H Tuan and D Baleanu et al / Applied Mathematics and Computation 362 (2019) 124458 Obviously, we have N N N (·, t ) − u(·, t )22 N EU L () ≤ 2EU (·, t ) − u (·, t )L2 () + 2Eu (·, t ) − u (·, t )L2 () Therefore, the proof of this theorem is divided into three steps as following N (·, t ) − uN (·, t )2 Step Estimate EU L2 () From (22) and (28), we have N (x, t ) − uN (x, t ) = U N   N  n;p − ρ p ϕ p (x ) Eα ,1 ( λ p t α ) ρ p=1 + N    t Eα ,2 (λ pt α ) ξnN;p − ξ p ϕ p (x ) p=1 + N   t p=1  (t − ω )α−1 Eα,α (λ p (t − ω )α ) f p (UN ) − f p (uN ) dω ϕ p (x ) = : M1 + M2 + M3 , it follows that 2 N (·, t ) − uN (·, t )22 EU L () ≤ 3EM1 L2 () + 3EM2 L2 () + 3EM3 L2 () Sub-step 1.1 Estimate EM1 2L2 () (29) ρ Recall that Lemma 2.2 give us the discrete form of the Fourier coefficient ρ p = ρn;p − Rn;p In addition to that ρnN;p = n  ρk ϕ p (xk ), ρn;p = k=1 M1 = ρ ( xk )ϕ p ( xk ), (30) k=1 hence, N  n   Eα ,1 ( λ p t α ) n π n p=1  ρ τk Xk ϕ p (xk ) − Rn;p ϕ p (x ) k=1 Using the fact that (a + b)2 ≤ 2a2 + 2b2 for all a, b ∈ R and the Parseval’s identity, we obtain ⎡   N   EM1 2L2 () ≤ Eα ,1 (λ pt α ) ⎣E N  cα+ exp(2λ1p/α t ) p=1 ≤ 2N cα+ exp(2λ1N/α t )  π n p=1 ≤2 n   2 τk Xk ϕ p (xk ) + C (τ , ρ ) n π M2τ + C (τ , ρ ) n  ρ + Rn;p k=1 π M2τ n 2τ ⎤ ⎦  (31) Sub-step 1.2 Estimate EM2 2L2 () Do the same as step before, then we have ⎡   N   EM2 2L2 () ≤ t E α , ( λ p t α ) ⎣E p=1 π n n  2 νkYk ϕ p (xk ) ξ + Rn;p !2 ⎤ ⎦ k=1   1 π M2ν /α exp ( λ t ) + C ( ν , ρ ) p n n 2ν λ11/α p=1   π M2ν + C (ν, ρ ) /α ≤ 2N cα+ exp ( λ t ) N n λ11/α ≤2 N  cα+ Sub-step 1.3 Estimate M3 2L2 () Now, using again Parseval’s identity, we obtain M3 2L2 () = N   p=1 t  (t − ω )α−1 Eα,α (λ p (t − ω )α ) f p (UN ) − f p (uN ) dω , (32) 10 N.D Phuong, N.H Tuan and D Baleanu et al / Applied Mathematics and Computation 362 (2019) 124458 by the Hölder inequality, we have M3 2L2 () = N   t p=1 dω  t  t   (t − ω )α−1 Eα,α (λ p (t − ω )α ) f p (UN ) − f p (uN )) dω By Lemma 2.1, we have M3 2L2 () = N   t p=1 ≤ T cα+ dω λ21  1−α cα+ λ p α exp N   1−α α t  exp  λ1N/α (t − ω ) f p (UN ) − f p (uN ) dω p=1  λ1p/α (t − ω ) f p (UN ) − f p (uN ) dω Since the Lipschitz property of f(u) (23), we have 2 1−α  t N  (·, ω ) − uN (·, ω )22 dω exp 2λ1N/α (t − ω ) U M3 2L2 () ≤ T K cα+ λ1 α L () (33) Substituting (31)–(33) into (29), we have # #2 N (·, t ) − uN (·, t )# E#U + 3T K + cα L2 1−αα λ ()  t N exp(2λ1N/α t )  n ≤ N (·, ω ) − uN (·, ω )22 dω, exp 2λ1N/α (t − ω ) EU L () (34) ! multiplying both sides of (34) with exp −2λ1N/α t , we have /α N (·, t ) − uN (·, t )22 EU L () exp −2λN t  2 1−α t N N (·, ω ) − uN (·, ω )22 dω ≤  + 3T K cα+ λ1 α exp −2λ1N/α ω EU L () n Since N n is independent of variable t, thanks to Gronwall’s inequality EU N (·, t ) − uN (·, t )2L2 () exp −2λ1N/α t ≤ So finally we have EU N (·, t ) − uN (·, t )2L2 () ≤ N n Step Estimate EuN (·, t ) − u(·, t )22 N  u, ϕ p ϕ p = p=1  λ21 1−α α ! (35) Let us consider the truncation version of solution u in (8) as follows N   Eα ,1 (λ pt α )ρ p + tEα ,2 (λ pt α )ξ p p=1  + n exp 2λ1N/α t  exp 3t T K cα+ L () PN u(x, t ) = N  2 1−α !  exp 3t T K cα+ λ1 α t  (t − ω )α−1 Eα,α (λ p (t − ω )α ) f p (u )dω ϕ p (x ), (36) then we have EuN (·, t ) − u2L2 () ≤ 2EuN (·, t ) − Pn u(·, t )2L2 () + 2EPn u(·, t ) − u(·, t )2L2 () Firstly, from (28) and (36), we have uN (x, t ) − Pn u(x, t ) =  N  p=1 t The Parseval identity leads to EuN (·, t ) − Pn u(·, t )2L2 () =  (t − ω )α−1 Eα,α (λ p (t − ω )α ) f p (uN ) − f p (u ) dω ϕ p (x )  N  p=1 t 2 (t − ω )α−1 Eα,α (λ p (t − ω )α ) f p (uN ) − f p (u ) dω Similar arguments apply to the case estimate M3 2L2 () in Part 1, we have EuN (·, t ) − Pn u(·, t )2L2 () ≤ 2T K cα+ λ21 1−α  t α N u (·, ω ) − u(·, ω )2L2 () dω exp 2λ1N/α (t − ω ) (37) N.D Phuong, N.H Tuan and D Baleanu et al / Applied Mathematics and Computation 362 (2019) 124458 11 Secondly, from the assumption (27) on the solution u, we have ∞  EPn u(·, t ) − u(·, t )2L2 () = μ exp −2λ1N/α (T − t ) 1 u(·, t ), ϕ p ≤ λ−2 N (38) p=N+1 To this end, taking (33) and (38) together, we have 1−α  t uN (·, t ) − u(·, t )2L2 () ≤ 2T K cα+ λ21 α exp 2λ1N/α (t − ω ) uN (·, ω ) − u(·, ω )2L2 () (ω )dω −2μ + 2λN exp −2λ1N/α (T − t ) 1 ! Multiplying both sides with exp −2λ1N/α t , it becomes uN (·, t ) − u(·, t )2L2 () exp 2λ1N/α t 2 1−α  t ≤ 2T K cα+ λ1 α exp −2λ1N/α (ω ) uN (·, ω ) − u(·, ω )2L2 () dω −2μ + 2λN exp −2λ1N/α T 1 (39) Using Gronwall’s inequality, we have ( μ ) 2 1−α ! uN (·, t ) − u(·, t )2L2 () ≤ λ−2 exp −2λ1N/α (T − t ) 1 exp 3tT K cα+ λ1 α N (40) Step Taking (29), (35) and (40) into account, we have N (·, t ) − u(·, t )22 EU L () ≤ N n −2μ exp 2λ1N/α t  + 2λN  exp −2λ1N/α (T − t ) 1 exp 3tT K cα+ λ21 1−α α ! (41)  This completes poof of the theorem 4.2.2 The error in Hq () norm One more important and interesting result is the estimation on a Hilbert scale, for instance, Hq () Although this is more difficult than in L2 () space, it plays an important role in physical modeling and engineering Of course, we need stronger assumption on the solution u(x, t) in order to estimate the error in Hq norm Theorem 4.3 Suppose that Problem (1)–(2) has a unique solution u(x, t) such that sup ∞  0≤t≤T p=1 exp 2λ1p/α (T − t +  ) u, ϕ p ≤ 2 , t ∈ [0, T ], (42) where  > and 2 are real constant The error estimate in Hq () norm will be   2 1−α q N N (·, t ) − u(·, t )2 q EU exp 2λ1N/α t  exp 3tT K cα+ λ1 α H () ≤ 2λN n  !  + 2λqN exp −2λ1N/α (T − t +  ) 2 exp 2λqN T K cα+ λ21 1−α α ! Proof With uN was defined in (28), we have N N N (·, t ) − u(·, t )2 q N EU H () ≤ 2EU (·, t ) − u (·, t )Hq () + 2Eu (·, t ) − u (·, t )Hq () (43) Step We have q N N (·, t ) − uN (·, t )2 q N EU H () ≤ λN EU (·, t ) − u (·, t )L2 () , and because of the result (35), we have   ! 2 1−α q N N (·, t ) − uN (·, t )2 q EU exp 2λ1N/α t  exp 3tT K cα+ λ1 α H () ≤ λN n (44) Step Estimate uN (·, t ) − u(·, t )2Hq () Analysis similar to that in the proof of Theorem 4.2 shows that EuN (·, t ) − u(·, t )2Hq () ≤ 2EuN (·, t ) − PN u(·, t )2Hq () + 2EPN u(·, t ) − u(·, t )2Hq () (45) 12 N.D Phuong, N.H Tuan and D Baleanu et al / Applied Mathematics and Computation 362 (2019) 124458 We first estimate u (·, t ) − PN u(·, t ) N Hq ()  N  = 2λ q N p=1 t + λqN T K cα λ21 ≤ 2 (t − ω )α−1 Eα,α (λ p (t − ω )α ) f p (uN ) − f p (u ) dω 1−α  t α N u (·, ω ) − u(·, ω )2L2 () dω exp 2λ1N/α (t − ω ) (46) Our next goal is to estimate ∞  PN u(·, t ) − u(·, t )2Hq () = λqp u, ϕ p (47) p=N+1 It follows from ∞  λqp u, ϕ p = p=N+1 ∞  λqp exp −2λ1p/α (T − t +  ) exp 2λ1p/α (T − t +  ) u, ϕ p p=N+1 In order to that, let us consider the function g(x ) = xa exp(−bx1/α ) which is strictly decreasing when a ≤ αb x1/α Since λp tends to infinity as p tends to infinity, there exists a index p = N large enough such that q≤ (T − t +  ) α λ1N/α If we set a = p and b = 2(T − t +  ) for all p > N, we have g(λ p ) = λqp exp −2λ1p/α (T − t +  ) ≤ g(λN ) Hence, ∞  PN u(·, t ) − u(·, t )2Hq () = g(λ p ) exp 2λ1p/α (T − t +  ) u, ϕ p p=N+1 ∞  ≤ g(λN ) exp 2λ1p/α (T − t +  ) u, ϕ p p=N+1 ≤ λ exp −2λ1N/α (T − t +  ) 2 q N (48) Finally, from (45), (46) and (48) we can assert that 1−α  t uN (·, t ) − u(·, t )2Hq () ≤ 2λqN T K cα+ λ21 α exp 2λ1N/α (t − ω ) uN (·, ω ) − u(·, ω )2L2 () dω + 2λqN exp −2λ1N/α (T − t +  ) 2 ! Multiplying both sides with exp −2λ1N/α t , it becomes /α uN (·, t ) − u(·, t )2Hq () e−2λN t ≤ 2λqN T K cα+ λ21 1−α  α t /α e−2λN ω uN (·, ω ) − u(·, ω )2L2 () dω + 2λqN exp −2λ1N/α (T +  ) 2 Applying the Gronwall inequality, we obtain /α uN (·, t ) − u(·, t )2Hq () e−2λN t   ≤ 2λqN exp −2λ1N/α (T +  ) 2 exp 2λqN T K cα+ λ21 1−α α ! , this leads to   1−α ! uN (·, t ) − u(·, t )2Hq () ≤ 2λqN exp −2λ1N/α (T − t +  ) 2 exp 2λqN T K cα+ λ21 α Step Taking (44), (45) and (49) together, we have q N (·, t ) − u(·, t )2 q EU H () ≤ 2λN N   exp 2λ1N/α t  exp 3tT K cα+ n  λ21 1−α α ! + 2λqN exp −2λ1N/α (T − t +  ) 2 exp 2λqN T K cα+ This completes poof of the theorem  (49) λ21 1−α α ! N.D Phuong, N.H Tuan and D Baleanu et al / Applied Mathematics and Computation 362 (2019) 124458 13 Fig Compare of the exact solution and the regularization solution for α = 1.2 Numerical illustration In this section, a numerical illustration of varying fractional order α = 1.2, 1.5, 1.8 is proposed to show the efficiency of the regularization method that we considered in this paper For t ∈ [0, 1], we consider the forward problem (1)–(2) according to take the sought for initial functions, ρ (x ) = 2(π x − x2 ), ξ (x ) = (π x − x2 ) and the source function as following 2ut (2−α ) − ( + t + t ), α ∈ ( 1, ) (2 − α ) (2 − α )(2 + t + t ) For the equation ∂tα u + u = f (x, t, u ), we should also mention that a lot of theory and numerical techniques for Caputo f (x, t, u ) = time fractional derivative have been addressed [31–33], and the method of approximation for the Laplacian operator can be found in many research papers, see [34,35] However, our simple numerical example has the exact solution u(x, t ) = (π x − x2 )(2 + t + t ), so we no need to present the discretization scheme for it Suppose a sample of n observations of ρ (x), ξ (x) at xk following random models ρk = ρ (xk ) + 0.5 · Xk , ξk = ξ (xk ) + 0.5 · Yk , k , ξk are generated by the where xk = π 2k2−1 for k = 1, , n and random errors Xk , Yk ∼ N (0, ) The random values ρ n pseudo random number algorithm in the MATLAB program as following: i.i.d ρk = ρ (xk ) + 0.5 · normrnd (0, ), ξk = ξ (xk ) + 0.5 · normrnd (0, ) In (22), the regularized solution N (x, t ) = U  t  N   nN;p + t Eα ,2 (λ pt α )ξnN;p + Eα ,1 ( λ p t α ) ρ (t − ω )α−1 Eα,α (λ p (t − ω )α ) f p (UN )dω ϕ p (x ), (50) p=1 with the regularization parameter N is chosen as following * N := α + δ 2T ln n , < δ < 1, where a is integer number which is a truncation of a real number a We discrete Equation (50) by using a finite scheme on a uniform mesh on interval [0, T], T = 1, defined by t0 = < t1 < t2 < · · · < ti < · · · < tN , ti = i · t, t = 1/N, N (x, ) = ρ nN (x ) (see Result (12)), then we start by using U N (x, ti ) = U  ti  N   nN;p + ti Eα ,2 (λ ptiα )ξnN;p + Eα ,1 (λ ptiα )ρ (ti − ω )α−1 Eα,α (λ p (ti − ω )α ) f p (UN )dω ϕ p (x ), p=1 14 N.D Phuong, N.H Tuan and D Baleanu et al / Applied Mathematics and Computation 362 (2019) 124458 Fig Compare of the exact solution and the regularization solution for α = 1.5 Fig Compare of the exact solution and the regularization solution for α = 1.8 where the nonlinear term can be approximated by an iterative scheme i   j=1 tj t j−1 (ti − ω )α−1 Eα,α (λ p (ti − ω )α ) f p (UN (x, t j−1 ))dω, (51) with the Mittag-Leffler function is calculated by the Matlab function mlf which was designed by Professor Igor Podlubny To compare the difference between the exact solution and the regularized solution at ti ∈ [0, 1], we use the mean squared error (MSE) as follows: MSE (ti ) = n ) ( N (xk , ti ) u ( xk , ti ) − U n (52) k=1 Figs 1–3 shows results, the mean squared error between exact solution and the regularized solution for varying fractional order α = 1.2, 1.5, 1.8 at t = 0.15, t = 0.30 and t = 0.75 respectively At a fixed time t, we see that the mean squared error N.D Phuong, N.H Tuan and D Baleanu et al / Applied Mathematics and Computation 362 (2019) 124458 15 are better when the number of observations n larger Moreover, the larger the fractional order α is, the better the rate of convergence we get This implies that the main results correctly reflect the behavior of convergence rate Conclusion We have proposed the example to show that Problem (1),(2) with initial data perturbed by random noise The trigonometric of non-parametric regression associated with the truncation method is used to offer a regularized solution Error estimates are provided in both L2 () and Hq () spaces The study is ended up by the numerical example to justify our main result Besides the results that the article achieves, this research has thrown up two disadvantages of the method in 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