On Theoretical and Numerical Aspect of Fractional Differential Equations with Purely Integral Conditions
OON THEORETICAL AND NUMERICAL ASPECT OF FRACTIONALDIFFERENTIAL EQUATIONS WITH PURELY INTEGRAL CONDITIONS
SAADOUNE BRAHIMI , AHCENE MERAD AND ADEM KILICMAN Abstract.
In this paper, we are interested in the study of a problem with fractional derivativeshaving boundary conditions of integral types. The problem represents a Caputo type advection-diffusion equation where the fractional order derivative with respect to time with 1 < α <
2. Themethod of the energy inequalities is used to prove the existence and the uniqueness of solutionsof the problem. The finite difference method is also introduced to study the problem numericallyin order to find an approximate solution of the considered problem. Some numerical examplesare presented to show satisfactory results. Introduction
Fractional Partial Differential Equations (FPDE) are considered as generalizations of par-tial differential equations having an arbitrary order and play essential role in engineering, physicsand applied mathematics. Due to the properties of Fractional Differential Equations (FDE), thenon-local relationships in space and time are used to model a complex phenomena, such as inelectroanalytical chemistry, viscoelasticity [10 , , , , , , , , , , , , , , , , , , classical boundary nonlocal conditions , we cancite the works of A. Alikhanov [3 , , , boundary conditions of integralstype (cid:90) v ( x, t ) dx, (cid:90) x n v ( x, t ) dx . For the theoretical study, we use the energy inequalitiesmethod to prove the existence and the uniqueness. However the numerical study is based on thefinite difference method to obtain an approximate numerical solution of the proposed problem. Weuse a uniform discretization of space and time and the fractional operator in the Caputo sensehaving order α (1 < α <
2) is approximated by a scheme called L Date : February 23, 2021.2010
Mathematics Subject Classification.
Key words and phrases.
Fractional derivatives, Caputo derivative, Fractional advection-diffusion equation,Finite difference schemes, Integrals conditions.*Corresponding author: [email protected]. a r X i v : . [ m a t h . NA ] F e b S. BRAHIMI, A. MERAD AND A. KILICMAN differential operators are also approximated by central and advanced numerical schemes. For thestability and convergence of obtained numerical scheme, the conditionally stable method is usedand we prove the convergence. Numerical tests are carried out in order to illustrate satisfactoryresults from the point of view that the values of the approximate solution that is very close to theexact solution. In the process of numerical and graphical results we applied MATLAB software..1.1.
Notions and preleminaries.
In this section we recall some early results that we need, suchas, the definition of Caputo derivative to explain the problem that we shall study in this work:let Γ ( . ) denote the gamma function. For any positive non-integer value 1 < α < , the caputoderivative defined as follows: Definition 1. ( See [12]) . Let us denote by C (0 , the space of continuous fonctions with compactsupport in (0 , , and its bilinear form is given by (( u, w )) = (cid:90) (cid:61) mx u. (cid:61) mx wdx ( m ∈ N ∗ ) , (1)where (cid:61) mx u = x (cid:90) ( x − ξ ) m − ( m − u ( ξ, t ) dξ ( m ∈ N ∗ ) . For m = 1, we have (cid:61) x u = x (cid:90) u ( ξ, t ) dξ and (cid:61) t u = t (cid:90) u ( x, τ ) dτ . The bilinear form (1) isconsidered as scalar product on C (0 ,
1) when is not complete.
Definition 2. ( See [12]) . We denote by B m (0 ,
1) = (cid:26) L (0 , f or m = 0 u/ (cid:61) mx u ∈ L (0 , f or m ∈ N ∗ , the completion of C (0 , for the scalar product defined by (1) .The associated norm to the scalarproduct is given by (cid:107) u (cid:107) B m (0 , = (cid:107)(cid:61) mx u (cid:107) L (0 , = T (cid:90) ( (cid:61) mx u ) dx . Lemma 3. ( See [8]) . For all m ∈ N ∗ , we obtain (cid:107) u (cid:107) B m (0 , ≤ (cid:18) (cid:19) m (cid:107) u (cid:107) L (0 , . (2) Definition 4. ( See [12]) . Let X be a Banach space with the norm (cid:107) u (cid:107) X , and let u : (0 , T ) → X be an abstract functions, by (cid:107) u ( ., t ) (cid:107) X we denote the norm of the element u ( ., t ) ∈ X at a fixed t. We denote by L (0 , T ; X ) the set of all measurable abstract functions u ( ., t ) from (0 , T ) into X such that (cid:107) u (cid:107) L (0 ,T ; X ) = T (cid:90) (cid:107) u ( ., t ) (cid:107) X dt < ∞ . HEORETICAL AND NUMERICAL ASPECT OF FRACTIONAL DIFFERENTIAL EQUATIONS 3
Lemma 5. [ Cauchy inequality with ε ] ( See [13]) . For all ε and arbitrary variables a,b ∈ R , wehave the following inequality: | ab | ≤ ε | a | + 12 ε | b | . (3) Definition 6. ( See [21]) . The left Caputo derivative for < α < can be expressed as c ∂ αt f ( t ) = 1Γ (2 − α ) t (cid:90) f ” ( s )( t − s ) α − ds ; t > . Definition 7. ( See [21]) . The integral of order α of the function f ∈ L [ a, b ] is defined by: I α f ( t ) = 1Γ ( α ) t (cid:90) f ( s )( t − s ) − α ds ; t > . Lemma 8. ( See [1]) . For all real < α < we have the inequality (cid:90) c ∂ αt ( (cid:61) x u ) dx ≤ (cid:90) ( c ∂ αt u ) ( (cid:61) x u ) dx. Lemma 9. ( See [28]) . For all real < α < we have the inequality (cid:90) Q ( c ∂ αt u ) ( (cid:61) x u ) dxdt ≤ (cid:90) Q (cid:16) c ∂ α t (cid:61) x u (cid:17) dxdt. Statement of the problem
In the rectangular domain Q = (cid:8) ( x, t ) ∈ R : 0 < x < , < t < T (cid:9) , where T > , we consider the fractional differential equation: £ v = c ∂ αt v + a ( x, t ) ∂ v∂x + b ( x, t ) ∂v∂x + c ( x, t ) v = g ( x, t ) , where 1 < α < , (4)to the equation (4), we associate the initial conditions: (cid:40) (cid:96)v = v ( x,
0) = Φ( x ) , x ∈ (0 , ,qv = v ( x, ∂t = Ψ( x ) , x ∈ (0 , , (5)and the purely integrals conditions (cid:90) v ( x, t ) dx = µ ( t ) , t ∈ (0 , T ) , (cid:90) xv ( x, t ) dx = E ( t ) , t ∈ (0 , T ) , (6)where Φ , Ψ , µ, E, a, b, c and g are known continuous functions. Assumptions:
1) for all ( x, t ) ∈ Q , we assume that: S. BRAHIMI, A. MERAD AND A. KILICMAN sup Q a ( x, t ) ≤ , sup Q ∂a ( x, t ) ∂x ≥ , inf Q ∂ b ( x, t ) ∂x ≤ , c ( x, t ) ≥ , sup Q ∂c ( x, t ) ∂x ≥ , (7)2) for all ( x, t ) ∈ Q , we assume that:0 < M ≤ ∂ a ( x, t ) ∂x − Q a ( x, t ) −
12 sup Q ∂a ( x, t ) ∂x + 12 inf Q ∂ b ( x, t ) ∂x −
12 sup Q ∂c ( x, t ) ∂x − ∂b ( x, t ) ∂x + 2 c ( x, t ) − ε . (8)3) The functions Φ( x ) and Ψ( x ) satisfy the following compatibility conditions: (cid:90) Φ dx = µ (0) , (cid:90) x Φ dx = E (0) , (cid:90) Ψ dx = µ (cid:48) (0) , (cid:90) x Ψ dx = E (cid:48) (0) . (9)We transform a problem (4) – (6) with nonhomegenous integral conditions to the equivalent prob-lem with homogenous integral conditions, by introducing a new unknown function u defined by v ( x, t ) = (cid:101) u ( x, t ) + U ( x, t ) , (10)where U ( x, t ) = 2(2 − x ) µ ( t ) + 6(2 x − E ( t ) . (11)Now we study a new problem with homegenous integral conditions £ (cid:101) u = c ∂ αt (cid:101) u + a ( x, t ) ∂ (cid:101) u∂x + b ( x, t ) ∂ (cid:101) u∂x + c ( x, t ) (cid:101) u = h ( x, t ) ,(cid:96)v = (cid:101) u ( x,
0) = ϕ ( x ) , x ∈ (0 , ,qv = (cid:101) u ( x, ∂t = ψ ( x ) , x ∈ (0 , , (cid:90) (cid:101) u ( x, t ) dx = 0 , t ∈ (0 , T ) , (cid:90) x (cid:101) u ( x, t ) dx = 0 , t ∈ (0 , T ) , (12)where h ( x, t ) = g ( x, t ) − £ U ( x, t ) ,ϕ ( x ) = Φ( x ) − (cid:96)U,ψ ( x ) = Ψ( x ) − qU and (cid:90) ϕ ( x ) dx = 0 , (cid:90) xϕ ( x ) dx = 0 , (cid:90) ψ ( x ) = 0 , (cid:90) xψ ( x ) = 0 . Again we introduce new function u defined by u ( x, t ) = (cid:101) u ( x, t ) − ψ ( x ) t − ϕ ( x ) , (13)therefore the problem (12) can be given as follow HEORETICAL AND NUMERICAL ASPECT OF FRACTIONAL DIFFERENTIAL EQUATIONS 5 £ u = c ∂ αt u + a ( x, t ) ∂ u∂x + b ( x, t ) ∂u∂x + c ( x, t ) u = f ( x, t ) ,(cid:96)u = u ( x,
0) = 0 , x ∈ (0 , ,qu = u ( x, ∂t = 0 , x ∈ (0 , , (cid:90) u ( x, t ) dx = 0 , t ∈ (0 , T ) , (cid:90) xu ( x, t ) dx = 0 , t ∈ (0 , T ) . (14)Thus, instead of seeking a solution v of the problem (4) − (6), we establish the existence anduniqueness of solution u of the problem (14) and solution v will simply be given by: v ( x, t ) = (cid:101) u ( x, t ) + U ( x, t ) . (15)3. Inequality of energy and its consequences
The solution of the problem (14) can be considered as a solution of the problem in operationalform: Lu = F, where L = ( £ , (cid:96), q ) is considered from B to F , where B is a Banach space of functions u ∈ L ( Q ),whose norm: (cid:107) u (cid:107) B = (cid:18)(cid:90) Q (cid:16) c ∂ α t ( (cid:61) x u ) (cid:17) dxdt + (cid:90) Q ( (cid:61) x u ) dxdt (cid:19) (16)is finite, and F is a Hilbert space consisting of all the elements F = ( f, ,
0) whose norm isgiven by: (cid:107) F (cid:107) F = (cid:18)(cid:90) Q f dxdt (cid:19) . (17)Now we let D ( L ) be the domain of the op´erator L for the set of all functions u such as that: (cid:61) x u, (cid:61) x ( c ∂ αt u ) , (cid:61) x ∂u∂x , (cid:61) x ∂ u∂x ∈ L ( Q ) and u satisfies the integral conditions in problem (14) . Then,
Theorem 10.
Under assumptions (7) - (8) , the condition satisfied then we have the estimate (cid:107) u (cid:107) B ≤ C (cid:107) Lu (cid:107) F , (18) where C is a positive constant and independent of u where u ∈ D ( L ) .Proof. Multiplying the fractional differential equation in the problem (14) by
M u = − (cid:61) x u andintegrating it on Q we obtain − (cid:90) Q ( c ∂ αt u ) (cid:61) x udxdt − (cid:90) Q a ( x, t ) ∂ u∂x (cid:61) x udxdt − (cid:90) Q b ( x, t ) ∂u∂x (cid:61) x udxdt − (cid:90) Q c ( x, t ) u (cid:61) x udxdt = − (cid:90) Q f (cid:61) x udxdt. (19) S. BRAHIMI, A. MERAD AND A. KILICMAN
Integrating by parts of four integrals in the left side of (19), we get − (cid:90) Q ( c ∂ αt u ) (cid:61) x udxdt = 2 (cid:90) Q ( c ∂ αt (cid:61) x u ) ( (cid:61) x u ) dxdt, (20) − (cid:90) Q a ( x, t ) ∂ u∂x (cid:61) x udxdt = 4 (cid:90) Q ∂ a∂x ( (cid:61) x u ) dx − (cid:90) Q au dxdt − (cid:90) Q ∂a ∂x (cid:0) (cid:61) x u (cid:1) dx, (21) − (cid:90) Q b ( x, t ) ∂u∂x (cid:61) x udx = (cid:90) Q ∂ b∂x (cid:0) (cid:61) x u (cid:1) dx − (cid:90) Q ∂b∂x ( (cid:61) x u ) dx, (22) − (cid:90) Q c ( x, t ) u (cid:61) x udx = − (cid:90) Q ∂ c∂x (cid:0) (cid:61) x u (cid:1) dx + 2 (cid:90) Q c ( (cid:61) x u ) dx (23)Substituting (20) − (23) in (19), we have2 (cid:90) Q ( c ∂ αt (cid:61) x u ) ( (cid:61) x u ) dx + 4 (cid:90) Q ∂ a∂x ( (cid:61) x u ) dx − (cid:90) Q au dx − (cid:90) Q ∂a ∂x (cid:0) (cid:61) x u (cid:1) dx + (cid:90) Q ∂ b∂x (cid:0) (cid:61) x u (cid:1) dx − (cid:90) Q ∂b∂x ( (cid:61) x u ) dx − (cid:90) Q ∂ c∂x (cid:0) (cid:61) x u (cid:1) dx + 2 (cid:90) Q c ( (cid:61) x u ) dx = − (cid:90) Q f (cid:61) x udx. (24)By the elementary inequalities in lemmas (8), (9) respectively and assumptions (7) − (8) give2 (cid:90) Q (cid:16) c ∂ α t ( (cid:61) x u ) (cid:17) dxdt + (cid:90) Q (4 ∂ a∂x − au − ∂a ∂x + 12 inf ∂ b∂x − ∂b∂x −
12 sup ∂ c∂x + 2 c ) ( (cid:61) x u ) dxdt ≤ − (cid:90) Q f (cid:61) x udxdt. (25)The estimate of the right side of (25) gives: (cid:90) Q (cid:16) c ∂ α t ( (cid:61) x u ) (cid:17) dxdt + (cid:90) Q (4 ∂ a∂x − au − ∂a ∂x dxdt + 12 inf ∂ b∂x − ∂b∂x − ∂ c∂x + 2 c − ε ) ( (cid:61) x u ) dxdt ≤ ε (cid:90) Q f dxdt. (26) HEORETICAL AND NUMERICAL ASPECT OF FRACTIONAL DIFFERENTIAL EQUATIONS 7
So, by using the assumptions (7) − (8) we find2 (cid:90) Q (cid:16) c ∂ α t ( (cid:61) x u ) (cid:17) dxdt + M (cid:90) Q ( (cid:61) x u ) dxdt ≤ ε (cid:90) Q f dxdt (27)Finally, we obtain a priori estimate (cid:107) u (cid:107) B ≤ C (cid:107) Lu (cid:107) F , (28)where C = (cid:18) ε min (2 , M ) (cid:19) . (cid:3) Corollary 11.
A strong solution of problem (14) is unique if it exists, and depends continuouslyon F = ( f, , . Corollary 12.
The range of the operator L is closed in F and R ( L ) = R ( L ) . Existence of solutions
In thei section, we prove the uniqueness of solution, if there is a solution. However, we havenot demonstrated it yet. To do it, we will just prove that R ( L ) is dense in F. Theorem 13.
Let us suppose that the assumptions (7) − (8) and integral conditions (6) are filled,and for ω ∈ L ( Q ) and for all u ∈ D ( L ) , we have (cid:90) Q £ u.ωdxdt = 0 , (29) then ω almost everywhere in Q. Proof.
We can rewrite the equation (29) as follows (cid:90) Q ( c ∂ αt uω ) dxdt = − (cid:90) Q a ( x, t ) ∂ u∂x ωdxdt − (cid:90) Q b ( x, t ) ∂u∂x ωdxdt − (cid:90) Q c ( x, t ) uωdxdt, (30)Further, we express the function ω in terms of u as follows : ω = − (cid:61) x u (31)Substituting ω by its representation (31) in (30) , integrating by parts, and taking into account theconditions (6), we obtain:2 (cid:90) Q ( c ∂ αt (cid:61) x u ) (cid:61) x udxdt = − (cid:90) Q ∂ a∂x ( (cid:61) x u ) dxdt + 2 (cid:90) Q au dxdt + (cid:90) Q ∂ a∂x ( (cid:61) x u ) dxdt − (cid:90) Q ∂ b∂x ( (cid:61) x u ) dxdt + 3 (cid:90) Q ∂b∂x ( (cid:61) x u ) dxdt + (cid:90) Q ∂ c∂x ( (cid:61) x u ) dxdt − (cid:90) Q c ( (cid:61) x u ) dxdt, (cid:3) S. BRAHIMI, A. MERAD AND A. KILICMAN on using under assumptions (7) − (8) and conditions (9), we obtain2 (cid:90) Q ( c ∂ αt (cid:61) x u ) (cid:61) x udxdt = − (cid:90) Q (4 ∂ a∂x + 4 sup au + 12 ∂ a∂x −
12 inf ∂ b∂x + 3 ∂b∂x + 2 sup ∂ c∂x − c ) ( (cid:61) x u ) dxdt, and this leads that 2 (cid:90) Q ( c ∂ αt (cid:61) x u ) (cid:61) x udxdt ≤ − (cid:18) ε + M (cid:19) (cid:90) Q ( (cid:61) x u ) dxdt. By lemmas (2) , (3) and (4) we obtain2 (cid:90) Q (cid:16) c ∂ α t ( (cid:61) x u ) (cid:17) dxdt ≤ − (cid:18) ε + M (cid:19) (cid:90) Q ( (cid:61) x u ) dxdt. Then ( (cid:61) x u ) = 0 (32)and we obtain u = 0 . So u = 0 in Ω wich gives ω = 0 in L ( Q ) . Finite Difference Method
Discretization of the problem.
Now, we consider a uniform subdivision of intervals [0 , , T ] as follows x i = ih ; i = 0 , ..., N and t k = kh t ; k = 0 , ..., M. Then, denote by v ki the approximate solution of v ( x i , t k ) at points ( x i , t k ), and the operator L is defined by L = a ∂ ∂x + b ∂∂x + c, L ( . ) ki = a ki ∂ ( . ) ∂x + b ki ∂ ( . ) ∂x + c ki (33)where a ki = a ( x i , t k ) , b ki = b ( x i , t k ) , c ki = c ( x i , t k ) . From the Taylor devlopment of function v at the point ( x i , t k ) we have (cid:18) ∂ v∂x (cid:19) ki = 1 h (cid:0) v ki − − v ki + v ki +1 (cid:1) + O (cid:0) h (cid:1) , (cid:18) ∂v∂x (cid:19) ki = v ki +1 − v ki h + O ( h ) . (34)Substituting (34) in the operateur L ki expressed in(33) gives Lv k +1 i = (cid:32) a k +1 i h + b k +1 i h (cid:33) v k +1 i +1 + (cid:32) c k +1 i − a k +1 i h − b k +1 i h (cid:33) v k +1 i + a k +1 i h v k +1 i − . (35)The discretization of Caputo derivative fractional operator c ∂ αt v [17] with 1 < α < c ∂ αt v ) k +1 i (cid:39) γ k (cid:88) j =0 (cid:0) v k − j − i − v k − j i + v k − j +1 i (cid:1) d j . (36)where (cid:26) d j = ( j + 1) − α − j − α d = 1; k = 1 , ..., M , γ = h − αt Γ (3 − α ) . HEORETICAL AND NUMERICAL ASPECT OF FRACTIONAL DIFFERENTIAL EQUATIONS 9
Writing fractional differential equation (4) in points ( ih, ( k + 1) h t ), we find γ k (cid:88) j = 0 (cid:0) v k − j − i − v k − j i + v k − j +1 i (cid:1) d j + Lv k +1 i = g k +1 i , i = 1 , N − F k +1 i v k +1 i − + A k +1 i v k +1 i + B k +1 i v k +1 i +1 − γd k v k i + γd k v k − i + γ k − (cid:88) j =1 S j d j + γ (cid:0) v − i − v i + v i (cid:1) d k = g k +1 i (38)where A k +1 i = γ + c k +1 i − a k +1 i h − b k +1 i h , B k +1 i = a k +1 i h + b k +1 i h ,F k +1 i = a k +1 i h , S j = v k − j − i − v k − j i + v k − j +1 i . In order to eliminate v − i , we use initial condition (5), and we find (cid:18) ∂v∂t (cid:19) ni (cid:39) v ni − v n − i h t therefore v − i (cid:39) Φ i − h t Ψ i = v i − h t Ψ i , i = 1 , N − . (39)Substituting (39) in (38) , we obtain F k +1 i v k +1 i − + A k +1 i v k +1 i + B k +1 i v k +1 i +1 − γd k v ki + γd k v k − i + γ k − (cid:88) j =1 S j d j = d k γv i + d k γh t Ψ i − d k γv i + g k +1 i . (40)For k = 0, the relation (40) gives F i v i − + A i v i + B i v i +1 = γv i + γh t Ψ i + g i with i = 1 , N − . (41)By conditions (6) , and trapezoid method we obtain, v = 2 µ ( h t ) − E ( h t ) h + 2 N − (cid:88) j =1 ( jh − v j , v N = 2 E ( h t ) h − N − (cid:88) j =1 jhv j . For i = 1, (cid:0) A + 2 F ( h − (cid:1) v + (cid:0) B + 2 F (2 h − (cid:1) v + 2 F N − (cid:88) j =3 ( jh − v j = γv + γh t Ψ + g + 2 F h ( E ( h t ) − µ ( h t )) . (42)For i = N − − B N − N − (cid:88) j =1 jhv j + (cid:0) F N − − B N − ( N − h (cid:1) v N − + (cid:0) A N − − B N − ( N − h (cid:1) v N − = γv N − + γh t Ψ N − + g N − − B N − h E ( h t ) . (43) Matrix’s form
We denote by w i = γv i + γh t Ψ i + g i , y = 2 F h ( E ( h t ) − µ ( h t )) , z N − = − B N − h E ( h t ) ,P = ( l i,j ) N − ,N − is square matrix and defined by l , = A + 2 F ( h − , l , = B + 2 F (2 h − ,l N − ,N − = F N − − B N − ( N − h, l N − ,N − = A N − − B N − ( N − h , l i,j = F ( jh −
1) when i = 1 , j = 3 , N −
10 when | i − j | ≥ i = 2 , N − A i when i = j, i = 2 , N − B i when i = j − , i = 1 , N − F i when i = j + 1 , i = 2 , N − − B N − jh when i = N − , j = 1 , N − . Taking account (41) , (42) , and (43) , we obtain the matrix system P .V = H (44)where H = W + R , W = (cid:0) w , w , ..., w N − (cid:1) T , R = (cid:0) y , , , ..., , z N − (cid:1) T . To solve the system (44) we can apply one of direct methods.5.2.
General case.
It is readily checked that, for k ≥ k − (cid:88) j =1 S j d j = ( d − d ) v k − i + d v ki + d k − v i + ( d k − − d k − ) v i + k − (cid:88) m =2 σ m v k − mi (45)where σ m = d m − − d m + d m +1 , m = 2 , k − . Lemma 14. If k ≥ the numerical scheme (40) is equivalent to F k +1 i v k +1 i − + A k +1 i v k +1 i + B k +1 i v k +1 i +1 = − γ k − (cid:88) m =1 σ m v k − mi + γ (2 − d ) v k i + γ ( d k − d k − ) v i + γd k h t Ψ i + g k +1 i , for i = 1 , . . . , N − Proof.
From the scheme (40), we have F k +1 i v k +1 i − + A k +1 i v k +1 i + B k +1 i v k +1 i +1 − γd k v ki + γd k v k − i + γ k − (cid:88) j =1 S j d j = d k γv i + d k γh t Ψ i − d k γv i + g k +1 i (cid:3) so F k +1 i v k +1 i − + A k +1 i v k +1 i + B k +1 i v k +1 i +1 + γ k − (cid:88) j =2 S j d i + γ ( v k +1 i − v ki + v k − i ) d + γ ( v i − v i + v − i ) d k = g k +1 i (47)using (45) we obtain F k +1 i v k +1 i − + A k +1 i v k +1 i + B k +1 i v k +1 i +1 = − γ k − (cid:88) m =1 σ m v k − mi + γ (2 − d ) v ki + γ ( d k − d k − ) v i HEORETICAL AND NUMERICAL ASPECT OF FRACTIONAL DIFFERENTIAL EQUATIONS 11 + γd k h t Ψ i + g k +1 i , for i = 1 , . . . , N − , and by trapezoid method we obtain: For i = 1, (cid:0) A k +11 + 2 F k +11 ( h − (cid:1) v k +11 + (cid:0) B k +11 + 2 F k +11 (2 h − (cid:1) v k +12 + 2 F k +11 N − (cid:88) j =3 ( jh − v k +1 j = 2 F k +11 h ( E (( k + 1) h t ) − µ (( k + 1) h t )) − γ k − (cid:88) m =1 σ m v k − m + γ ( d k − d k − ) v + γd k h t Ψ + g k +11 . For i = N − − B k +1 N − N − (cid:88) j =1 jhv k +1 j + (cid:0) F k +1 N − − B k +1 N − ( N − h (cid:1) v k +1 N − + (cid:0) A k +1 N − − B k +1 N − ( N − h (cid:1) v k +1 N − = − B k +1 N − h E (( k + 1) h t ) − γ k − (cid:88) m =1 σ m v k − mN − + γ (2 − d ) v kN − + γ ( d k − d k − ) v N − + γd k h t Ψ N − + g k +1 N − . (49) Matrix’s form
We take expression (48) for i = 2 , N − ?? ), (49) to formulate the matrixsystems: P k +1 V k +1 = H k +1 ; k ≥ V , V are known (50)where P k +1 = (cid:0) l k +1 i,j (cid:1) N − ,N − is square matrix defined by l k +11 , = A k +11 + 2 F k +11 ( h − , l k +11 , = B k +11 + 2 F k +11 (2 h − ,l k +1 N − ,N − = F k +1 N − − B k +1 N − ( N − h, l k +1 N − ,N − = A k +1 N − − B k +1 N − ( N − h , l k +1 i,j = F k +11 ( jh −
1) when i = 1 , j = 3 , N −
10 when | i − j | ≥ i = 2 , N − A k +1 i when i = j, i = 2 , N − B k +1 i when i = j − , i = 1 , N − F k +1 i when i = j + 1 , i = 2 , N − − B k +1 N − jh when i = N − , j = 1 , N − V k +1 = (cid:0) v k +11 , ..., v k +1 N − (cid:1) T ; H k +1 = − γ k − (cid:88) m =1 σ m V k − m + W k +1 + R k +1 ; k ≥ W k +1 = (cid:0) w k +11 , w k +12 , ..., w k +1 N − (cid:1) T , R k +1 = (cid:0) y k +11 , , , ..., , z k +1 N − (cid:1) T ,w k +1 i = γ (2 − d ) v ki + γ ( d k − d k − ) v i + γd k h t Ψ i + g k +1 i ,y k +11 = 2 F k +11 h ( E (( k + 1) h t ) − µ (( k + 1) h t )) ; z k +1 N − = − B k +1 N − h E (( k + 1) h t ) . In order to prove system (50) has a unique solution we denote ρ as an eigenvalue of the matrix P k , and X = ( x , x , ..., x N − ) T is an nonzero eigenvector corresponding to ρ . Then, we choose i such as | x i | = max {| x j | : j = 1; ... ; N − } then N − (cid:88) j =1 l i,j x j = ρx i ; i = 1; N − ρ = l i,i + N − (cid:88) j =1 j (cid:54) = i l i, j x j x i . (51)Substituting the values of l i,j into (51) , and taking into account that F ki , a ki are negative and (cid:12)(cid:12)(cid:12) x j x i (cid:12)(cid:12)(cid:12) ≤ i = 1, ρ = (cid:0) A k +11 + 2 F k +11 ( h − (cid:1) + (cid:0) B k +11 + 2 F k +11 (2 h − (cid:1) x x + 2 F k +11 N − (cid:88) j =3 ( jh − x j x = γ + c k +1 i − F k +11 − B k +11 (cid:18) − x x (cid:19) + 2 F k +11 N − (cid:88) j =2 ( jh − x j x . For i = N − ρ = l i,i + N − (cid:88) j =1 j (cid:54) = i l i, j x j x i = A k +1 N − − B k +1 N − ( N − h + (cid:0) F k +1 N − − B k +1 N − ( N − h (cid:1) (cid:18) x N − x N − (cid:19) − B k +1 N − N − (cid:88) j =1 jh x j x N − = γ + c k +1 N − − B k +1 N − + F k +1 N − (cid:18) x N − x N − − (cid:19) − B k +1 N − ( N − h − B k +1 N − N − (cid:88) j =1 jh x j x N − . For i = 2 , N − ρ = l i,i + N − (cid:88) j =1 j (cid:54) = i l i,j x j x i = A k +1 i + F k +1 i x i − x i + B k +1 i x i +1 x i = γ + c k +1 i − B k +1 i − F k +1 i + F k +1 i x i − x i + B k +1 i x i +1 x i = γ + c k +1 i + F k +1 i (cid:18) x i − x i − (cid:19) + a k +1 i + hb k +1 i h (cid:18) x i +1 x i − (cid:19) . (52)From the above we conclude for i = 1 , N − , if b k +1 i < , B k +1 N − < ρ > . If b k +1 i > h ≤ min ≤ i ≤ N − (cid:16) − a k +1 i b k +1 i (cid:17) , ρ > , then all eigenvalues of matrix P k +1 are strictly positive,therefore P k +1 is invertible. HEORETICAL AND NUMERICAL ASPECT OF FRACTIONAL DIFFERENTIAL EQUATIONS 13
Stability.
Since, we have F k +1 i + A k +1 i + B k +1 i = γ + c k +1 i , F k +1 i ≤ , A k +1 i + B k +1 i ≥ , then we let u k +1 i be the approximate solution of (48) , and e k +1 i , the error at point ( x i , t k +1 ) definedby v k +1 i − u k +1 i = e k +1 i , and (cid:13)(cid:13) E k (cid:13)(cid:13) = Max ≤ i ≤ N − | e ki | , E k = (cid:0) e k , ..., e kN − (cid:1) T , for k = 0 we apply (41) we get (cid:13)(cid:13) E (cid:13)(cid:13) ≤ (cid:0) γ + c i (cid:1) (cid:13)(cid:13) E (cid:13)(cid:13) = (cid:0) F i + A i + B i (cid:1) (cid:13)(cid:13) E (cid:13)(cid:13) = (cid:0) F i (cid:13)(cid:13) E (cid:13)(cid:13) + (cid:0) A i + B i (cid:1) (cid:13)(cid:13) E (cid:13)(cid:13)(cid:1) ≤ (cid:0)(cid:0) A i + B i (cid:1) (cid:13)(cid:13) E (cid:13)(cid:13) + F i (cid:12)(cid:12) e i − (cid:12)(cid:12)(cid:1) ≤ Max ≤ i ≤ N − (cid:12)(cid:12) F i e i − + A i e i + B i e i +1 (cid:12)(cid:12) = γ (cid:13)(cid:13) E (cid:13)(cid:13) so (cid:13)(cid:13) E (cid:13)(cid:13) (cid:22) γγ + c i (cid:13)(cid:13) E (cid:13)(cid:13) (cid:22) (cid:13)(cid:13) E (cid:13)(cid:13) . (53)Therefore the method is stable. Lemma 15.
For k ≥ the scheme (47) is stable and we have (cid:13)(cid:13) E k +1 (cid:13)(cid:13) ≤ C (cid:13)(cid:13) E (cid:13)(cid:13) , C > , for all k ≥ Proof.
We use the mathematical induction. (cid:3)
We assume (cid:13)(cid:13) E j (cid:13)(cid:13) ≤ c j (cid:13)(cid:13) E (cid:13)(cid:13) , and C max = max c j ; where c j (cid:31) , j = 1 , k from (48) we get F k +1 i e k +1 i − + A k +1 i e k +1 i + B k +1 i e k +1 i +1 = − γ k − (cid:88) m =1 σ m e k − mi + γ (2 − d ) e k i + γ ( d k − d k − ) e i , i = 1 , N − , so (cid:0) γ + c k +1 i (cid:1) (cid:13)(cid:13) E k +1 (cid:13)(cid:13) ≤ (cid:0)(cid:0) A k +1 i + B k +1 i (cid:1) (cid:13)(cid:13) E k +1 (cid:13)(cid:13) + F k +1 i (cid:12)(cid:12) e k +1 i − (cid:12)(cid:12)(cid:1) ≤ (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) − γ k − (cid:88) m =1 σ m e k − mi + γ (2 − d ) e k i + γ ( d k − d k − ) e i (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ≤ γ (cid:32) k − (cid:88) m =1 | σ m | (cid:13)(cid:13) E k − mi (cid:13)(cid:13) + (2 − d ) (cid:13)(cid:13) e ki (cid:13)(cid:13) + ( d k − − d k ) (cid:13)(cid:13) e i (cid:13)(cid:13)(cid:33) ≤ γC max (cid:32) k − (cid:88) m =1 | σ m | + 2 − d + d k − − d k (cid:33) (cid:13)(cid:13) E (cid:13)(cid:13) ≤ γC max (cid:0) − − α (cid:1) (cid:13)(cid:13) E (cid:13)(cid:13) , where k − (cid:88) m =1 | σ m | + 2 − d + d k − − d k = 5 − − α , < σ m < , − < d k − d k − < , < − d < (cid:13)(cid:13) E k +1 (cid:13)(cid:13) ≤ C (cid:13)(cid:13) E (cid:13)(cid:13) ; C = C max (cid:0) − − α (cid:1) . (54)Therefore the method is stable. Convergence.
Let v ( x i ; t k +1 ) as the exact solution and v k +1 i is the approximate solution ofscheme (37) , we put v ( x i ; t k +1 ) − v k +1 i = (cid:15) k +1 i ; for i = 1 , N − , k = 1 , M −
1. The scheme L defined on (36)verified ([26]) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∂ α v∂t α − (cid:18) ∂ α v∂t α (cid:19) L (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ O ( h t ) (55)substitution into (37) and using (34) , (55) leads to γ k (cid:88) j = 0 (cid:16) v ( x i ; t k − j − ) − (cid:15) k − j − i − (cid:16) v ( x i ; t k − j ) − (cid:15) k − ji (cid:17) + (cid:16) v ( x i ; t k − j +1 ) − (cid:15) k − j +1 i (cid:17)(cid:17) d j + L (cid:0) v ( x i ; t k +1 ) − (cid:15) k +1 i (cid:1) = g k +1 i . then γ k (cid:88) j = 0 ( v ( x i ; t k − j − ) − v ( x i ; t k − j ) + ( v ( x i ; t k − j +1 ))) d j + Lv ( x i ; t k +1 ) − γ k (cid:88) j = 0 (cid:16) (cid:15) k − j − i − (cid:15) k − ji + (cid:15) k − j +1 i (cid:17) d j − L(cid:15) k +1 i = g k +1 i so ∂ α v ( x, t ) ∂t α + O ( h t ) + Lv ( x ; t ) + O ( h ) − γ k (cid:88) j = 0 (cid:16) (cid:15) k − j − i − (cid:15) k − ji + (cid:15) k − j +1 i (cid:17) d j − L(cid:15) k +1 i = g k +1 i . hence γ k (cid:88) j = 0 (cid:16) (cid:15) k − j − i − (cid:15) k − ji + (cid:15) k − j +1 i (cid:17) d j + L(cid:15) k +1 i = O ( h + h t ) . (56)Taking (cid:12)(cid:12) (cid:15) kl (cid:12)(cid:12) = (cid:13)(cid:13) (cid:15) k (cid:13)(cid:13) = Max ≤ i ≤ N − (cid:12)(cid:12) (cid:15) ki (cid:12)(cid:12) ; (cid:15) k = (cid:0) (cid:15) k , , ..., (cid:15) kN − (cid:1) T ; (cid:13)(cid:13) (cid:15) i (cid:13)(cid:13) = 0for k = 0 we get F i (cid:15) i − + A i (cid:15) i + B i (cid:15) i +1 = γ(cid:15) i + O ( h + h t ) with i = 1 , N − , (57)we have (cid:13)(cid:13) (cid:15) (cid:13)(cid:13) = (cid:12)(cid:12) (cid:15) l (cid:12)(cid:12) ≤ (cid:0) F i + A i + B i (cid:1) (cid:12)(cid:12) (cid:15) l (cid:12)(cid:12) ≤ (cid:0)(cid:0) A i + B i (cid:1) (cid:12)(cid:12) (cid:15) l (cid:12)(cid:12) + F i (cid:12)(cid:12) (cid:15) l (cid:12)(cid:12)(cid:1) ≤ Max ≤ i ≤ N − (cid:12)(cid:12) F i (cid:15) i − + A i (cid:15) l + B i (cid:15) l (cid:12)(cid:12) = O ( h + h t ) . hence (cid:13)(cid:13) (cid:15) (cid:13)(cid:13) ≤ O ( h + h t ) . (58)We assume : (cid:12)(cid:12)(cid:12) (cid:15) jl (cid:12)(cid:12)(cid:12) ≤ O ( h + h t ); j = 1 , k from (56) we get F k +1 i (cid:15) k +1 i − + A k +1 i (cid:15) k +1 i + B k +1 i (cid:15) k +1 i +1 = − γ k − (cid:88) m =1 σ m (cid:15) k − mi + γ (2 − d ) (cid:15) ki + O ( h + h t ) (59) HEORETICAL AND NUMERICAL ASPECT OF FRACTIONAL DIFFERENTIAL EQUATIONS 15 we have (cid:13)(cid:13) (cid:15) k +1 (cid:13)(cid:13) ≤ (cid:0) γ + c k +1 i (cid:1) (cid:12)(cid:12) (cid:15) k +1 l (cid:12)(cid:12) = (cid:0) F k +1 i + A k +1 i + B k +1 i (cid:1) (cid:12)(cid:12) (cid:15) k +1 l (cid:12)(cid:12) ≤ (cid:0) F k +1 i (cid:12)(cid:12) (cid:15) k +1 i − (cid:12)(cid:12) + (cid:0) A k +1 i + B k +1 i (cid:1) (cid:12)(cid:12) (cid:15) k +1 l (cid:12)(cid:12)(cid:1) ≤ Max ≤ i ≤ N − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − γ k − (cid:88) m =1 σ m (cid:15) k − mi + γ (2 − d ) (cid:15) ki + O ( h + h t ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ γ k − (cid:88) m =1 σ m (cid:13)(cid:13) (cid:15) k − m (cid:13)(cid:13) + γ (2 − d ) (cid:13)(cid:13) (cid:15) k (cid:13)(cid:13) + O ( h + h t ) ≤ γ (cid:32) k − (cid:88) m =1 σ m + (2 − d ) (cid:33) O ( h + h t ) + O ( h + h t )hence (cid:13)(cid:13) (cid:15) k +1 (cid:13)(cid:13) ≤ γγ + c k +1 i O ( h + h t ) + 1 γ + c k +1 i O ( h + h t ) ≤ O ( h + h t ) . (60)Therefore, the method is convergent. 6. Applications
In this section, we give some numerical investigation tests.
Example 16.
We consider a problem (4 − with α = , a ( x, t ) = − x − t, b ( x, t ) = x + t, c ( x ) = 2 , g ( x, t ) = (cid:18) √ π + 2 t √ t (cid:19) e x , φ ( x ) = ψ ( x ) = 0 , µ ( t ) = ( e − t , E ( t ) = t . The analytical solution is given by v ( x, t ) = t e x . The approximate solution u ( x, t ) with A. E is the absolute error.Table 1. h = 0 . h t = 0 . h v ( x, t )0 . . e − . . e − . . e − . . e − . . e − . . e − . . e − . . e − . . e − u ( x, t )1 . e − . e − . e − . e − . e − . e − . e − . e − . e − A . E . e − . e − . e − . e − . e − . e − . e − . e − . e − h Figure 1. α = 1 . h t = 0 . h = 0 . h t =0 . h v ( x, t )0 . . e − . . e − . . e − . . e − . . e − . . e − . . e − . . e − . . e − u ( x, t )4 . e − . e − . e − . e − . e − . e − . e − . e − . e − A . E . e − . e − . e − . e − . e − . e − . e − . e − . e − h = 0 . h t = . h v ( x, t )0 . . e − . . e − . . e − . . e − . . e − . . e − . . e − . . e − . . e − u ( x, t )1 . e − . e − . e − . e − . e − . e − . e − . e − . e − A . E e − e − e − e − e − e − e − e − e − HEORETICAL AND NUMERICAL ASPECT OF FRACTIONAL DIFFERENTIAL EQUATIONS 17
Figure 2. α = 1 . h t = 0 . Figure 3. α = 1 . h t = 0 . h t takes smallvalues very close to zero. that is, for h t = 0 . h t = 0 . h t = 0 . O ( h + h t ) . (a) FIGURE 4. h t = − (b) FIGURE 5. h t = − (c) FIGURE 6. h t = − For k = 1 (second iteration)Table 4 shows the absolute error for space step h = 0 . h h t = − . . e − . . e − . . e − . . e − . . e − . . e − . . e − . . e − . . e − h t = − . e − . e − . e − . e − . e − . e − . e − . e − . e − h t = − . e − . e − . e − . e − . e − . e − . e − . e − . e − u after two steps 2 h t tends to the exact solution when h t close to zero, with convergenceorder O ( h + h t ) . Table 5. The absolute error for h = . ; h t = − i = 1 , ,
18 19 ,
27 28 ,
36 37 ,
45 46 ,
54 55 ,
63 64 ,
72 73 ,
81 82 ,
89 90 , − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − FIGURE 7, α = 1 . h t = 10 − Table 6. The absolute error for h = 0 . h t = − i = 1 , ,
18 19 ,
27 28 ,
36 37 ,
45 46 ,
54 55 ,
63 64 ,
72 73 ,
81 82 ,
89 90 , − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − − From tables 5, 6 and Fig 7, 8 with space step h = . , we see that the approximate solution u tends to the exact solution v when the step h t ( h t = 10 − , h t = 10 − ) takes values close tozero,with convergence order O ( h + h t ) . Example 17.
We take: α = , a ( x, t ) = − x − t, b ( x, t ) = x − t , c ( x ) = x + 2 t, g ( x, t ) =(4 √ t + ( t + 1) ( x + 2 t ) e x , Φ( x ) = e x ; , ψ ( x ) = 2 e x , µ ( t ) = ( t + 1) , E ( t ) = ( t + 1) . The exact analytical solution of this problem is given by v ( x, t ) = ( t + 1) e x . FIGURE 8 α = 1 . h t = 10 − The tables 7, 8 and 9 show the values of the absolute error.Table 7. h = . , h t = − h A . E . . e − . . e − . . e − . . e − . . e − . . e − . . e − . . e − . . e − h = . , h t = − h A . E . . e − . . e − . . e − . . e − . . e − . . e − . . e − . . e − . . e − HEORETICAL AND NUMERICAL ASPECT OF FRACTIONAL DIFFERENTIAL EQUATIONS 21
FIGURE 9. α = 1 . h t = 10 − FIGURE 10 α = 1 . h t = 10 − Table 9. h = 0 . , h t = − h A . E . . e − . . e − . . e − . . e − . . e − . . e − . . e − . . e − . . e − FIGURE 11 α = 1 . h t = 10 − (a) FIGURE 12. h t = − (b) FIGURE 13. h t = − (c) FIGURE 14. h t = − Fig. 12 h t = 0 .
001 Fig. 13 h t = 0 . h t = 0 . h = 0 . h t ( − , − , − ) takes a value close to zero, with convergence order O ( h + h t ) . For h = . , α = 1 . h = 0 .
01 and the time step h t decreases towards zero ( h t = 0 . h t = 0 . , h t = 0 . , the approximate solution u tendsto the exact solution v , in the case where h t = 0 . u and v are almost identical.Table 10 shows the error norm (cid:13)(cid:13) E k (cid:13)(cid:13) ∞ for defferent value of α defined by (cid:13)(cid:13) E k (cid:13)(cid:13) ∞ = Max ≤ i ≤ N − N − (cid:88) i =1 | e i | , where E k = V k − U k = (cid:0) e k , ..., e kN − (cid:1) T HEORETICAL AND NUMERICAL ASPECT OF FRACTIONAL DIFFERENTIAL EQUATIONS 23
Table 10 , h = 0 . h t − − − (cid:13)(cid:13) E (cid:13)(cid:13) ∞ for α = 1 . α = 1 . α = 1 . α = 1 . α = 1 . . e ∗ − . ∗ − . ∗ − . ∗ − . ∗ − . ∗ − . ∗ − . ∗ − . ∗ − . ∗ − . ∗ − . ∗ − . ∗ − . ∗ − . ∗ − We see in the table 10, for the space step h = 0 . , and for the defferent values of α, the errornorm tends to zeros when the time step h t takes values close to zeros, with convergence order O ( h + h t ) . Conclusion
In this paper, we study a problem with fractional derivatives with boundary conditions ofintegral types. The study concerns a Caputo-type advection-diffusion equation where the fractionalorder derivative α with respect to time with 1 < α <
2. The existence and uniqueness areproven by the method of energy inequalities. The numerical study of this problem based on thefinite difference method. Applications on certain examples clearly show that the numerical resultsobtained are very satisfactory, where we see the approximate solution u tends to the exact solution v for the defferent value of α. References [1] A Akilandeeswariy, K Ba achandran, N Annapoorani; Solvability of hyperbolic fractional partial DifferantialEequations, Journal of Applied Analysisand Computation, 7(4), (2017), 1570–1585.[2] A. Anguraj, P. Karthikeyan; Existence of solutions for fractional semilinear evolution boundary value problem,Commun. Appl. Anal. 14 (2010), 505–514.[3] A. A. Alikhanov, On the stability and convergence of nonlocal difference schemes, Differ. Equ. 46(7), (2010),949–961.[4] A. A. Kilbas, H. M. Srivastava, J. J. Trujillo; Theory and Applications of Fractional Differential Equations,Elsevier, Amsterdam, 2006.[5] A. A. 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Day, Parabolic equations and thermodynamics, Quart. Appl. Math., Vol. 50, No. 3 (1992), 523–533. , Department of Mathematics, Faculty of Exact Sciences, Larbi Ben M’hidi University, Oum ElBouaghi, Algeria
Email address : [email protected], merad [email protected], [email protected] Department of Mathematics and Institute for Mathematical Research, Universiti Putra Malaysia,Serdang 43400 UPM, Selangor, Malaysia
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