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[.]

Contents lists available at ScienceDirect

journal homepage: www.elsevier.com/locate/amc

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

b 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 550 0 0 0, Vietnam

a r t i c l e i n f o

Keywords:

Fractional derivative

ill-posed problem

Elliptic equation

Random noise

Regularized solution

a b s t r a c t

Thispaperisconcernedwithfindingthesolutionu(x,t)oftheCauchyproblemfor non-linearfractional ellipticequationwith perturbedinput data Thisstudy shows thatour forwardproblemis severelyill-posedinsense ofHadamard.Forthisill-posedproblem, thetrigonometricofnon-parametricregressionassociatedwiththetruncationmethodis appliedtoconstructaregularizedsolution.Under priorassumptionsfor theexact solu-tion,theconvergencerateisobtainedinbothL2 andH q(forq >0)norm.Moreover,the numericalexampleisalsoinvestigatedtojustifyourresults

© 2019ElsevierInc.Allrightsreserved

1 Introduction

In this work, we focus on finding the solution for the following time fractional elliptic equation

∂α

with the Cauchy condition and initial conditions

 u ( x, t ) = 0 , ( x, t ) ∈ ∂  × ( 0 , T ) ,

u ( x, 0 ) = ρ ( x ) , x ∈  ,

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 ∂α

t u is the Caputo fractional derivative of order α ∈ (1, 2) with respect to t defined in [1–3] as follows

∂α

t u ( x, t ) = ( 2 1 − α )

 t

0 ( t − ω )(1−α) ∂2

∂ω2u ( x, ω ) d ω , ( x, t ) ∈  × ( 0 , T ) , (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

0 096-30 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

do 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 uncon-trollable 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= π2k−1

2n for k = 1 , , n following discrete random models:





random variables Xk, Yk are mutually independent and identically distributed Xk, Yk i.i.d∼ N ( 0 , 1 ) and τk, νk are unknown positive real constants and bounded by constants Mτ, Mν, respectively, i.e., 0 < τk≤ Mτ and 0 < 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 assump-tions on the exact solution, we will proof that the approximate solution will converge to the correct solution both in L2and

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 4 Numerical experiment for various value of α is given in Section 5 We conclude the paper with Section 6

2 Preliminaries

Definition 2.1 ( L2(  ) space) Given that f :  → R is Lebesgue measurable We define

L2() = 

f :



f

2( x ) d x < ∞ 

,

with the inner product and the norm

f1, f2 =



f1( x ) f2( x ) d x, f 2

L2()=



f

2( x ) d x.

For a fixed positive number τ > 0, we define the Sobolev class of function as

Hτ() =



f ∈ L2() :

∞



p=1

λτ

p f, ϕp 2< ∞



,

it is a Hilbert space which endowed with the norm

f 2

H τ ()= ∞

p=1

λτ

p f, ϕp 2.

Given a Banach space B, and p ≥ 2 Let us define the Bochner space Lp( ; B ) = Lp(( , F, P ) ; B ) (see [21] ) as following

Lp( ; B ) = 

u : E u p

B=



 u ( θ ) p

Bd P ( θ ) < ∞ , θ ∈  

,

with the norm

u L p ( ;B )=

E u p B

1

p .

The space of all the L2-value predictable process u (see [22] ) defined as following

VT=

u : sup

≤t≤T u ( ·, t ) L2( ;L2())= sup

≤t≤T

E u ( ·, t ) 2

L2()< ∞ ,

and an equivalent norm on VT,

u V T = sup

0≤t≤T

E u ( ·, t ) 2

L2().

Let us consider the Laplace operator

L u = −  u = − ∂2

∂ x2u.

Since L is positive define, linear densely and self-adjoint operator on  with Dirichlet boundary condition, the eigenvalues

λpand eigenfunctions ϕp( x ) of operator L satisfy (see [23] )

L ϕp( x ) = λpϕp( x ) , x ∈  ,

ϕp( x ) = 0 , x ∈ ∂ 

The spectrum increases to infinity

0 < λ1≤ λ2≤ λ3≤ ≤ λp≤ ,

and the eigenfunctions form an orthonormal basis of L2(  ),



ϕ1=



2

π sin x, ϕ2=



2

π sin 2 x, . . . , ϕp=



2

π sin px, . . .



.

The set SN= span { ϕ1, ϕ2, , ϕN} is called the subspace of L2(  ).

Definition 2.2 (see [1] ) . The Mittag-Leffler function Eα,β( ) are given by the following series

Eα , β( z ) = ∞

k=0

zα

where α > 0 and β ∈ R are arbitrary constants.

Lemma 2.1 (See [4] ) . Given α ∈ [ α1, α2] satisfies 1 < α1< α2< 2 Then there exist positive constants −α, +α such that the follow-ing inequalities hold for any t > 0:

1. −αexp ( λ1/ α

p t ) ≤ Eα ,1( λptα) ≤ c+

αexp ( λ1/ α

p t ) ,

2 t Eα ,2( λptα) ≤ c+

α λ11/ α p

exp ( λ1/ α

p t ) ,

3 tα−1Eα , α( λptα) ≤ c+

αλp1−αexp ( λ1/ α

p t )

Suppose that the solution of (1) is given by the following Fourier series

u ( x, t ) = ∞

p=1

up( t ) ϕp( x ) ,

where we write up( t ) = u ( ·, t ) , ϕp for the usual inner product in L2(  ) The Fourier coefficients up( t ) are solution of the following ordinary differential equation

⎧

⎪

⎪

⎨

⎪

⎪

⎩

∂α

t up− λpup= fp( u ) ,

up( 0 ) = ρp,

∂

∂ t up( 0 ) = ξp.

(7)

where fp( u ) = f ( x , t , u ) , ϕp , ρp= ρ , ϕp and ξp= ξ , ϕp By using the same method as [24–26] , the system (7) has the solution

up( t ) = Eα ,1( λptα) ρp+ t Eα ,2( λptα) ξp+

t

0 ( t − ω )α−1Eα , α( λp( t − ω )α) fp( u ) d ω

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( λptα) ρp+ t Eα ,2( λptα) ξp+

 t

0( t − ω )α−1Eα , α( λp( t − ω )α) fp( u ) d ω  ϕp( x ) (8)

4 N.D Phuong, N.H Tuan and D Baleanu et al / Applied Mathematics and Computation 362 (2019) 124458

Given a sequence of points xk= π2k−1

2n , k = 1 , , n in  ¯ and f ( x ) ∈ C1( ) ¯ We noticed that the p th Fourier coefficients

is fp= f , ϕp , and the approximation of an integral by a Riemann sum

fp≈ π

n

n



k=1

f ( xk) ϕp( xk)

Fortunately, we have the equation of relation between fpand Riemann sum as following

Lemma 2.2. Assume that f ∈ C1( ) ¯ For p = 1 , , n − 1 , we set

fn ;p:= π

n

n



k=1

f ( xk) ϕp( xk) ,

with xk= π2k−1

2n ∈  The Fourier coefficients fp= f , ϕp of the function f can be represented as

fp= fn ;p− Rf

n ;p, where Rn f ;pis called a residual of approximation fp by fn;pand its form is

Rn f ;p= ∞

l=1

( −1 )l

Proof. As for the proof, see Tuan and Erkan [27] , and see also Randall [28, p 144] 

Under the specific assumption, the residual Rn f ;ptends 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 > 1 Then, there exists a constant C ( τ , f ) which depends on f and τ , such that

Proof. The assumption f ∈ Hτ B implies that f 2

H τ ()≥ λτ

p f , ϕp2 Since λp= p2, we deduce that f H τ ()≥ pτ f , ϕp Thus

 Rn f ;p ≤ ∞

l=1

| f , ϕ2ln+p + | f , ϕ2ln −p ≤

∞



l=1



f H τ ()

( 2 ln + p )τ + f H τ ()

( 2 ln − p )τ



.

It is clear evident that 2 ln + p ≥ ln and 2 ln − p ≥ ln (because of 1 ≤ p ≤ n − 1 ), hence

 Rn f ;p ≤ ∞

l=1



f H τ ()

( ln )τ + f H τ ()

( ln )τ



≤ 2 f H τ ()

nτ

∞



l=1

1

By putting

C ( τ , f ) = 2 f H τ ()

the proof of Lemma is complete 

Assume g ( x ) is a unknown function from  into R Its observed value at xkis  gkwhich drawn from random model  gk=

g ( xk) + εk Here εkare mutually independent and identically distributed εk

i i.d∼ N ( 0 , σ2

k) Minh and his partner [29] used the non-parametric least square method to generate estimators of g ( x ) in SN,

 gN n( x ) =

N



p=1



π

n

n



k=1



gkϕp( xk)



Especially, if σ2

k ≤ M2, using the properties of mutually independent and identically distributed of normal variables, Nane and Tuan [30] also have

E



π

n

n



k=1

εkϕp( xk)

2

= π2

n2

n



k=1

E ε2

kϕ2

p( xk) ≤ π

3 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 ) = 0 and the source

f ( x, t , u ) = 1

2 T c+α

∞



p=1

λα−1 α

p exp

− λ1/ α

Problem (1) – (2) has a unique solution u ( x , t ) = 0 Indeed, from (8) we have

u ( x, t ) = ∞

p=1

 t

0 ( t − ω )α−1Eα , α( λp( t − ω )α) fp( u )( ω ) d ω



ϕp( x ) ,

the Parseval identity leads to

u ( ·, t ) 2

L2()=

∞



p=1

 t

0 ( t − ω )α−1Eα , α( λp( t − ω )α) fp( u )( ω ) d ω

2

,

by the Hölder inequality, we have

u ( ·, t ) 2

L2()≤ ∞

p=1

t

0

d ω t

0



( t − ω )α−1Eα , α( λp( t − ω )α) fp( u )( ω ) 

2d ω ,

As the source function is set by (14) and using Lemma 2.1 , we have either

u ( ·, t ) 2

L2()≤ 1

4 T2

∞



p=1

t

0

d ω t

0 u, ϕp 2d ω ≤ 1

4 T

t

0 u ( ·, ω ) 2

L2()d ω ,

by using the Gronwall inequality, we get u ( ·, t ) 2

L2()= 0 or u ( x , t ) = 0

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 − 1 From the function estimation formula (12) , the estimator of ρ , ξ in SN are



ρn−1

n ( x ) =

n−1



p=1



π

n

n



k=1

1

√

n Xkϕp( xk)



ϕp( x ) ,



ξn−1

n ( x ) =

n−1



p=1



π

n

n



k=1

1

√

n Ykϕp( xk)



ϕp( 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 ) ,

 u ( x, 0 ) = ρ n−1

n ( x ) , x ∈  ,

 ut( x, 0 ) = ξ n−1

n ( x ) , x ∈ 

(15)

From the result (8) leads to

 u ( x, t ) =

n−1



p=1



Eα ,1( λptα) ρ n−1

n ;p + tEα ,2( λptα) ξ n−1

n ;p +

t

0 ( t − ω )α−1Eα , α( λp( t − ω )α) fp(  u ) d ω  ϕp( x ) , (16)

where ρn−1

n ;p =  ρn−1

n , ϕp

, ξn−1

n ;p =  ξn−1

n , ϕp

We need to show that it has a unique solution  u ( x , t ) ∈ VT We start by setting

J (  u ) ( x, t ) =

n−1



p=1



Eα ,1( λptα) ρ n−1

n ;p + tEα ,2( λptα) ξ n−1

n ;p +

t

0 ( t − ω )α−1Eα , α( λp( t − ω )α) fp(  u ) d ω  ϕp( x )

Noting that  u is a trigonometric polynomial with order less than n with respect to the variable x Using the Parseval identity,

we get

J (  u ) ( t ) − J (  v ) ( t ) 2

L2()=

n−1



p=1

 t

0( t − ω )α−1Eα , α( λp( t − ω )α) ( fp(  u ) − fp(  v ) ) d ω

2

.

6 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 J (  u ) ( t ) − J (  v ) ( t ) 2

L2()≤ 1

4 T E

 t

0  u ( ·, ω ) − v ( ·, ω ) 2

L2()d ω 

≤ sup

0≤ω ≤T

1

4 E  u ( ·, ω ) − v ( ·, ω ) 2

L2(),

this leads to

J (  u ) − J (  v ) V T ≤ 1

2  u − v V T,

it means J is the contraction with respect to the norm · V T By the Banach fixed point theorem, there is a unique function



u ∈ VT such that J (  u ) =  u almost surely The rest of the example will show that the mean square error between exact Cauchy data ( ρ , ξ ) and their estimators ( ρ n−1

n ;  ξn−1

n ) are small, but the mean square error between the solution u and  u is very large.

Small change in data. By the Parseval identity, we have

ρ n−1

n 2

L2()=

n−1



p=1

π

n

n



k=1

1

√

n Xkϕp( xk)

2

.

Using (13) , we have

E ρ n−1− ρ2

L2()=

n−1



p=1

π

n2 = π ( n − 1 )

By a similar argument as above, we also have

E ξ n−1− ξ2

L2()= π ( n − 1 )

Large change in the solution. From (16) , we have

E  u ( ·, t ) 2

L2()≥ E 

Eα ,1( λn−1tα) ρ n−1

n ;n−1+ tEα ,2( λn−1tα) ξ n−1

n ;n−1

+

t

0( t − ω )α−1Eα , α( λn−1( t − ω )α) fn−1(  u ) d ω 2=: E ( I1+ I2+ I3)2.

The inequality 2 ( a + b + )2≥ a2− 4 b2− 42 for all a , b , ∈ R leads to

2 E  u ( ·, t ) 2

From Lemma 2.1 and (13) , we have

E I2

1= ( Eα ,1( λn−1tα) )2

E



π

n

n



k=1

1

√

n Xkϕn−1( xk)

2

≥ ( c−α)2π

2 λ1/ α

Similarly, we have

E I2

2= ( t Eα ,2( λn−1tα) )2E



π

n

n



k=1

1

√

n Ykϕn−1( xk)

2

≤ ( c+

α)2π

n2

1

λ1/ α

n−1

exp

2 λ1/ α

Using Lemma 2.1 again, we have

E I23≤ E  t

0

c+αλ1 −α

n−1exp

λ1/ α

n−1( t − ω ) fn−1(  u ) d ω 2,

and the Hölder inequality leads to

E I23≤ ( c+α)2λ21 −α

n−1 E

  t

0

d ω  t

0

exp

2 λ1/ α

n−1( t − ω ) f2

n−1(  u ) d ω



≤ 1

4 T E

t

0  u , ϕn−1 2d ω ≤ 1

40sup≤t≤TE  u ( ·, t ) 2

Taking (17)–(20) together, we have

2 E  u ( ·, t ) 2

L2()≥ π exp

2 λ1/ α

n−1t ( c−α)2

n2 − 4 ( c+)2

n2

1

λ1/ α

n−1

− sup

0≤t≤TE  u ( ·, t ) 2

L2(),

this leads to

3 sup

0≤t≤TE  u ( ·, t ) 2

L2()≥ π exp

2 λ1/ α

n−1t ( c−α)2

n2 − 4 ( c+)2

n2

1

λ1/ α

n−1

.

Thus

lim

n→∞

E ρ n−1− ρ2

L2()+ E  ξn−1− ξ2

L2() = 0 ,

however

lim

n→∞  u − u 2

V T = ∞ ,

we can conclude that Problem (1) – (2) is ill-posed.

4 Regularized solution and convergence estimate

4.1 Regularized solution

Given observation values ( ρ k, ξ k) of Cauchy data ( ρ , ξ ) at xk= π2k−1

2n for k = 1 , , n Given an integer number N such that 1 < 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,



ρN

n( x ) =

N



i=1



π

n

n



k=1

ρ ( xk) + τkXk!

ϕp( xk)



ϕp( x ) ,



ξN

n (x ) =

N



i=1



π

n

n



k=1

ξ ( xk) + νkYk

!

ϕp( xk)



ϕp( x )

We are considering a problem

⎧

⎪

⎪

⎨

⎪

⎪

⎩

∂α

t U N+  U N= f

x, t , U N , ( x, t ) ∈  × ( 0 , T ) ,



UN( x , t ) = 0 , ( x , t ) ∈ ∂  × ( 0 , T ) ,



UN( x, 0 ) = ρ N

n( x ) , x ∈  ,



UN

t ( x, 0 ) =  ξN

n( x ) , x ∈ 

(21)

Similar results have been obtained (8) , it leads to the solution of (21) as following



UN( x, t ) =

N



p=1



Eα ,1( λptα) ρ N

n ;p+ t Eα ,2( λptα)  ξN

n ;p+

t

0( t − ω )α−1Eα , α( λp( t − ω )α) fp( U N) d ω  ϕp( x ) (22)

where ρN

n ;p= 

ρN

n, ϕp

, ξN

n ;p= 

ξN

n, ϕp

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 > 0 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)

Theorem 4.1. The nonlinear integral equation (22) has a unique solution U N∈ VT.

Proof. We first put

J ( u )( x, t ) =

N



p=1



Eα ,1( λptα) ρ N

n ;p+ t Eα ,2( λptα)  ξN

n ;p+

t

0( t − ω )α−1Eα , α( λp( t − ω )α) fp( u ) d ω  ϕp( 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 ) V T ≤

" 

T2K2( c+α)2λ21 −α

1 exp

2 T λ1/ α

8 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 = 1

E J ( u )( t ) − J ( v )( t ) 2

L2()= E

N



n=1

 t

0 ( t − ω )α−1Eα , α( λp( t − ω )α) ( fp( u ) − fp( v ) ) d ω

2

≤ T K2K2

αλ21 −α

1 exp

2 T λ1/ α

t

0 u ( ·, ω ) − v ( ·, ω ) 2

L2()d ω ,

this leads to

J ( u ) − J ( v ) V T ≤

"

T2K2( c+α)2λ21 −α

1 exp

2 T λ1/ α N

1! u − v V T.

Then, we assume (25) to hold for m , we will prove it for m + 1 This can be done as follows

E ## Jm+1( u ) − Jm+1( v ) ##2

L2()= E ## J

Jm+1( u ) − J

Jm+1( v ) 2

L2()

≤ T 

K2( c+α)2λ21 −α

1 exp

2 T λ1/ α N

m+1

E

 t

0

( T t )m

m ! u ( ·, ω ) − v ( ·, ω ) 2

L2()d ω

Therefore, by the induction principle, we say that (25) holds Since

lim

m→∞

" 

T2K2( c+α)2λ21 −α

1 exp

2 T λ1/ α

there exists a positive integer m0such that Jm0 is a contraction with respect to the norm · V T It follows that the equation

Jm0( u ) = u has a unique solution U N∈ VT, it means Jm0( U N) = U Nalmost surely Moreover, since J ( Jm0( U N)) = J ( U N) almost surely, then Jm0( J ( 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

of Jm0, we conclude that J ( U N) = U N has a unique solution U N∈ VT 

4.2.1 The error in L2(  ) norm

Theorem 4.2. Let ρ ∈ C1[0 , π ] ∩ Hτ() , ξ ∈ C1[0 , π ] ∩ Hν() , where τ , ν > 1, and 0 < N < n Furthermore, assume that Prob-lem (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

λ2μ

p exp

2 λ1/ α

for some constant  , then we have

E UN( ·, t ) − u ( ·, t ) 2

L2()≤  N

n exp

2 λ1/ α

N t  + 4 λ−2μ

N exp

−2 λ1/ α

N ( T − t )  

exp 4 t T K2

c+α 2λ21 −α 1

!

, where

 = 6 ( c+α)2

π M2

τ+ C2( τ , ρ ) + π M2

τ+ C2( ν , ξ )

λ1/ α

1

.

Remark 4.1. By choosing N : = N ( n ) such that

λN ( n )≤

δ

2 T ln n

α

, 0 < δ < 1 .

If t = T , the mean square error E ## UN( x , t ) − u ( x , t ) ##2

L2()is of order

max



nδ−1;

δ

2 T ln n

−2μα

.

Otherwise, it is of order

max



nδ−1;

δ

2 T ln n

−2μα

n−δ T ( T −t )



.

Proof. We define the nonlinear integral equation as following

uN( x, t ) =

N



p=1



Eα ,1( λptα) ρp+ t Eα ,2( λptα) ξp+

 t

0( t − ω )α−1Eα , α( λp( t − ω )α) fp( uN) d ω  ϕp( x ) (28)

Obviously, we have

E U N( ·, t ) − u ( ·, t ) 2

L2()≤ 2 E U N( ·, t ) − uN( ·, t ) 2

L2()+ 2 E uN( ·, t ) − u ( ·, t ) 2

L2().

Therefore, the proof of this theorem is divided into three steps as following

Step 1. Estimate E U N( ·, t ) − uN( ·, t ) 2

L2()

From (22) and (28) , we have



UN( x, t ) − uN( x, t ) =

N



p=1



Eα ,1( λptα)



ρN

n ;p− ρp ϕp( x )

+

N



p=1



t Eα ,2( λptα) 

ξN

n ;p− ξp ϕp( x )

+

N



p=1

 t

0( t − ω )α−1Eα , α( λp( t − ω )α)

fp( U N) − fp( uN) d ω  ϕp( x )

= : M1+ M2+ M3,

it follows that

E U N( ·, t ) − uN( ·, t ) 2

L2()≤ 3 E M12

L2()+ 3 E M22

L2()+ 3 E M32

Sub-step 1.1. Estimate E M12

L2()

Recall that Lemma 2.2 give us the discrete form of the Fourier coefficient ρp= ρn ;p− Rρ n ;p. In addition to that



ρN

n ;p=

n



k=1



ρkϕp( xk) , ρn ;p=

n



k=1

hence,

M1=

N



p=1

Eα ,1( λptα)



π

n

n



k=1

τkXkϕp( xk) − Rρ n ;p



ϕp( x )

Using the fact that ( a + b )2≤ 2 a2+ 2 b2for all a , b ∈ R and the Parseval’s identity, we obtain

E M12

L2()≤ 2

N



p=1



Eα ,1( λptα) 

2

⎡

⎣ E

π

n

n



k=1

τkXkϕp( xk)

2

+

Rρ

n ;p

2

⎤

⎦

≤ 2

N



p=1

cα+ 2exp ( 2 λ1/ α

p t )



π M2

τ

n + C2( τ , ρ ) n 12τ



≤ 2 N

c+α 2exp ( 2 λ1/ α

N t )



π M2

τ+ C2( τ , ρ )

n



Sub-step 1.2. Estimate E M22

L2()

Do the same as step before, then we have

E M22

L2()≤ 2

N



p=1



t Eα ,2( λptα) 

2

⎡

⎣ E

π

n

n



k=1

νkYkϕp( xk)

2

+ Rξ

n ;p

!2

⎤

⎦

≤ 2

N



p=1

cα+ 2λ11/ α

1

exp ( 2 λ1/ α

p t )



π M2

ν

n + C2( ν , ρ ) n 12ν



≤ 2 N

c+α

2 1

λ1/ α

1

exp ( 2 λ1/ α

N t )



π M2

ν+ C2( ν , ρ )

n



Sub-step 1.3. Estimate M32

L2()

Now, using again Parseval’s identity, we obtain

M32

L2()=

N



p=1

  t

0( t − ω )α−1Eα , α( λp( t − ω )α)

fp( U N) − fp( uN) d ω 2,

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

M32

L2()=

N



p=1

 t

0

d ω  t

0



( t − ω )α−1Eα , α( λp( t − ω )α)

fp( U N) − fp( uN)) 2d ω

By Lemma 2.1 , we have

M32

L2()=

N



p=1

 t

0

d ω  t

0



c+αλ1 −α

p exp

λ1/ α

p ( t − ω ) fp( U N) − fp( uN) 2d ω

≤ T

c+α

2

λ21 −α 1

N



p=1

t

0



exp

λ1/ α

N ( t − ω ) fp( U N) − fp( uN) 2d ω

Since the Lipschitz property of f ( u ) (23) , we have

M32

L2()≤ T K2

c+α 2λ21 −α 1

t

0

exp

2 λ1/ α

N ( t − ω ) U N( ·, ω ) − uN( ·, ω ) 2

Substituting (31) – (33) into (29) , we have

E ##  UN( ·, t ) − uN( ·, t ) ##2

L2()≤ N exp ( 2 λ1/ α

N t )

+ 3 T K2

c+α 2λ21 −α 1

 t

0

exp

2 λ1/ α

N ( t − ω ) E U N( ·, ω ) − uN( ·, ω ) 2

multiplying both sides of (34) with exp −2 λ1/ α

N t !

E U N( ·, t ) − uN( ·, t ) 2

L2()exp

−2 λ1/ α

N t

≤ N

n  + 3 T K2

c+α

2

λ21 −α 1

 t

0

exp

−2 λ1/ α

N ω E U N( ·, ω ) − uN( ·, ω ) 2

L2()d ω

Since N n is independent of variable t , thanks to Gronwall’s inequality

E UN( ·, t ) − uN( ·, t ) 2

L2()exp

−2 λ1/ α

N t ≤  N

n  

exp 3 t T K2

c+α

2

λ21 −α 1

!

.

So finally we have

E UN( ·, t ) − uN( ·, t ) 2

L2()≤  N

n exp

2 λ1/ α

N t  

exp 3 t T K2

c+α 2λ21 −α 1

!

Step 2. Estimate E uN( ·, t ) − u ( ·, t ) 2

L2(). Let us consider the truncation version of solution u in (8) as follows

PNu ( x, t ) =

N



p=1

u, ϕp ϕp=

N



p=1



Eα ,1( λptα) ρp+ tEα ,2( λptα) ξp

+

t

0( t − ω )α−1Eα , α( λp( t − ω )α) fp( u ) d ω  ϕp( x ) , (36)

then we have

E uN( ·, t ) − u 2

L2()≤ 2 E uN( ·, t ) − Pnu ( ·, t ) 2

L2()+ 2 E Pnu ( ·, t ) − u ( ·, t ) 2

Firstly, from (28) and (36) , we have

uN( x, t ) − Pnu ( x, t ) =

N



p=1

 t

0( t − ω )α−1Eα , α( λp( t − ω )α)

fp( uN) − fp( u ) d ω



ϕp( x )

The Parseval identity leads to

E uN( ·, t ) − Pnu ( ·, t ) 2

L2()=

N



p=1

 t

0 ( t − ω )α−1Eα , α( λp( t − ω )α)

fp( uN) − fp( u ) d ω

2

.

Similar arguments apply to the case estimate M32

L2()in Part 1, we have

E uN( ·, t ) − Pnu ( ·, t ) 2

L2()≤ 2 T K2

c+α 2λ21 −α 1

t exp

2 λ1/ α

N ( t − ω ) uN( ·, ω ) − u ( ·, ω ) 2

L2()d ω

3 ill-posedness of the problem< /b>

In this section, we give an example in order to demonstrate the ill-posedness of the problem in... solution as (22) and the regularization parameter N

will be chosen depending on n

4.2 Convergence estimate

In this work, the nonlinear. .. L2(), for all u , v ∈ L2() (23)

Theorem 4.1. The nonlinear integral equation (22) has a unique solution U