The Rothe method for multi-term time fractional integral diffusion equations

In this paper we study a class of multi-term time fractional integral diffusion equations. Results on existence, uniqueness and regularity of a strong solution are provided through the Rothe method. Several examples are given to illustrate the applicability of main results.


1.
Introduction. The fractional calculus as a natural generalization of the classical integer order calculus was introduced firstly in the 17th century by L'Hospital to inquire Leibniz what meaning could be ascribed to d n u dt n , if n is a fraction. Since then, it attracted attention of many famous mathematicians such as Abel, Euler, Fourier, Laplace, Liouville, Riemann and so forth. However, its development was slow, only a century later, in 1819, Lacroix managed to answer the question concerning the expression d 0.5 u(t) dt 0.5 for the function u(t) = t. Until the twentieth century, many researchers found that some physical phenomena for viscoelastic matertials lead to their description in terms of noninteger order differential equations, for example, fractional Kelvin-Voigt constitutive laws and fractional Maxwell model. After that, more and more researchers have paid their attention to fractional calculus, which is not only a generalization of the classical integer order where α ∈ (0, 1). Note that the results in [19] are important contributions for the development of fractional diffusion equations. However, to obtain the existence and uniqueness for (1), the condition is required in [19]. This means that [19,Lemma 3.4] provides only a local (in time) unique strong solution of (1). For example, the following fractional integral diffusion equation problem cannot be solved by using main results of [19]          ∂u(t, x) ∂t − ∂ 2 u(t, x) ∂x 2 = 1 Γ(0.5) t 0 (t − s) −0.5 u(s, x) ds in (0, 10) × (0, π), u(0, x) = u 0 (x) for all x ∈ [0, π], u(t, 0) = u(t, π) = 0 for all t ∈ [0, 10], since 10 1.5 Γ(1.5) > 1 and the smallness condition (2) is not satisfied. On the other hand, in [19], the order of the derivative α is constrained to the interval (0, 1). In some situations, the fractional integral diffusion equations can require that the order α is not constrained to (0, 1) and the integral term is not a single one but it is a multi-term. Being motivated by the above analysis, in this paper, we will remove these drawbacks and investigate the following multi-term time fractional integral diffusion equation, (FIDE) for short, in a Hilbert space where the constants a i , α i , i = 1, . . . , k are such that a i ≥ 0, α i > 0, −A : D(A) ⊂ H → H is an infinitesimal generator of a C 0 -semigroup of contractions in H, the function f : [0, T ] → H is Lipschitz continuous, 0 I αi t u denotes the fractional integral of order α i > 0 of u, and u 0 ∈ D(A). It is obvious that if a 1 = 1, a i = 0 for i = 2, . . . , k and α 1 ∈ (0, 1), then (3) reduces to (1).
We now give the definition of strong solution for (FIDE). In this paper, we will study a class of multi-term time fractional integral diffusion equations (FIDE) in (3). The main results are delivered on existence, uniqueness and regularity of a strong solution to (FIDE). We apply the Rothe method, see [9,10,28] for details, combined with a surjectivity result for m-accretive operator, see [2,21,26].
The outline of the paper is as follows. In Section 2, we recall the basic material we use in next sections. In Section 3, we investigate the multi-term time fractional integral diffusion equation (3) and provide our main results. Several examples which illustrate the applicability of main results are given in Section 4.

2.
Notation and preliminaries. In this section we recall the basic notation and various results which are needed in the sequel, see [2,3,4,11,14,18,23,24,26]. We first present definitions from the fractional calculus theory.
Definition 2.1. (Riemann-Liouville fractional order integral) Let X be a Banach space, y ∈ L 1 (0, T ; X) and α > 0. The fractional integral of order α > 0 of y is defined by where Γ is the gamma function given by In what follows, we denote by 0 I 0 t the identity operator, i.e., we write 0 I 0 t y(t) = y(t) for a.e. t ∈ [0, T ].
Let X be a Banach space. We denote the space of all absolutely continuous functions from [0, T ] to X by AC(0, T ; X). Definition 2.2. (Caputo fractional order derivative with 0 < α ≤ 1) Let X be a Banach space, y ∈ AC(0, T ; X) and α ∈ (0, 1]. The Caputo fractional derivative of order α ∈ (0, 1] of the function y is defined by It is clear that the Caputo derivative of order α = 1 reduces to the classical first-order derivative, that is, we have C 0 D 1 t y(t) = y (t) for a.e. t ∈ [0, T ]. Let X be a real normed space with its dual X * , and the norms of X and X * be denoted by · and · X * , respectively. The value of a functional x ∈ X * at a point x ∈ X will be conveniently denoted by x * , x . The symbols "→" and " " stand for the strong convergence in X and weak convergence in X. The duality mapping J : X → 2 X * is defined by It is well-known, see [2,Theorem 1.2], that if X * is strictly convex, then the duality mapping J is single valued and demicontinuous. In particular, if X is a Hilbert space, then the duality mapping J becomes the identity operator I. Definition 2.3. Let X be a Banach space and A : D(A) ⊂ X → 2 X be a multivalued mapping. Then A is called (i) accretive, if for all x 1 , x 2 ∈ D(A), y 1 ∈ Ax 1 , y 2 ∈ Ax 2 and z * ∈ J(x 1 − x 2 ), we have for all λ > 0, x i ∈ D(A) and y i ∈ Ax i , i = 1, 2.
Let A : D(A) ⊂ X → X be an accretive operator. Let us define the operators J λ : R(I + λA) → D(A) and A λ : R(I + λA) → X by where R(I + λA) denotes the range of the operator I + λA and the operator A λ is called the Yosida approximation of A (for details, see [20, p.151] or [2, p.101]). We also recall the following important properties of m-accretive operator, which can be found in [2, Proposition 3.4, p.103].
Proposition 1. Let A : X → 2 X be an m-accretive operator. Then A is closed and if λ n ∈ R and x n ∈ X are such that then y ∈ Ax. If X * is uniformly convex, then A is demiclosed, and if Recall that an operator A : X → 2 X is said to be closed, if x n → x, y n → y and y n ∈ Ax n , then y ∈ Ax. Also, A is said to be demiclosed, if x n → x, y n y and y n ∈ Ax n yield y ∈ Ax.

Remark 2. It is known, see [17, Theorem 1.4.3(b)] and [2, Proposition 3.3], that if
We conclude this section by recalling the Gronwall inequalities which proofs can be found in [ with a positive constant c independent of N (or τ ). Then, there exists a positive constant c, independent of N (or τ ) such that x n ≤ c y n + τ n−1 j=1 y j for n = 1, . . . , N.
Therefore, for a constant C > 0 independent of N (or τ ), we have If, instead of (4), we assume x n ≤ c y n + c τ n j=1 x j for n = 1, . . . , N, and τ is sufficient small, then x n ≤ c y n + τ n j=1 y j for n = 1, . . . , N, where c > 0 is a constant. Then Moreover, if g is nondecreasing, then

Time fractional integral diffusion equation.
In this section, we pass to the formulation of the Rothe method to prove the existence and uniqueness of a strong solution to (FIDE) in (3). It consists in the time discretization in which we define an approximate sequence by using an implicit (backward) Euler formula. Then, at each time step, we will solve a stationary equation. Finally, we construct the piecewise constant and piecewise affine interpolants, and prove a convergence result.
In the rest of the paper, C > 0 denotes a constant whose value may change from line to line. Let N ∈ N be fixed and denote τ = T N . Let f 0 τ = f (0) and for n = 1, . . . , N , where t n τ = nτ . Now, we apply the Rothe method to (FIDE) in (3) and consider the following discretized problem called the Rothe problem for (FIDE).
The following result provides conditions for the solvability of Problem 1.
Proof. We first prove the estimate (6). For n = 1, we have

Multiplying the above equation by u 1
τ and then applying the m-accretivity of A, cf. Remark 2, we obtain Since f is Lipschitz continuous on [0, T ], then there exists a constant m f > 0 such that f n τ ≤ m f for all τ > 0 and n ≥ 0. This combined with (8) implies Next, for n = 2, . . . , N , we multiply the equation (5) by u n τ to obtain It follows from m-accretivity of the operator A that Summing up the above inequalities from 2 to l, where 2 ≤ l ≤ N , we obtain Combining this estimate with the fact that aiT α i Γ(αi+1) and then applying the discrete version of the Gronwall inequality in Lemma 2.4, for all τ ∈ (0, τ 0 ), we have u l τ ≤ C for all 2 ≤ l ≤ N . The latter combined with (9) implies that the estimate (6) holds. Now we shall verify the estimate (7). For n = 1, it is obvious that Consider the n-th equation in (5) and the (n − 1)-th equation in (5), and then subtract these two equalities to get Multiplying the above equality by u n τ − u n−1 τ and then applying the m-accretivity of the operator A, we obtain The boundedness of {u n τ }, see (6), guarantees that u n τ − u n−1 Since f is a Lipschitz continuous function, then there exists a constant L f > 0 such that We use this inequality and the fact that Summing up the above inequalities from 2 to l, where 2 ≤ l ≤ N , and then using (10), we have for all l = 2, . . . , N . This means that the estimate (7) holds, which completes the proof of the lemma.
Subsequently, for a given τ > 0, we define the piecewise affine function u τ , and the piecewise constant interpolant functions u τ and f τ as follows The main result of this paper is obtained in the following theorem.
According to the estimate (7), we easily calculate that for all t ∈ (t n−1 , t n ]. On the other hand, by the estimate (6), we can easily obtain, for t ∈ (t n−1 , t n ], that By the definitions of u τ , u τ , v τ and f τ , it is clear that (5) is equivalent to the following problem Analogously Subtracting the equalities (14) and (15), and then multiplying the result by u τ (t) − u h (t), we have Thus, we deduce the following estimate From the estimates (6), (7) and inequality (12), we get, for all t ∈ (t n−1 , t n ] Hence, we obtain By using the m-accretivity of the operator A, the estimates (6), (7), the inequality (13), from (16), we get

STANIS LAW MIGÓRSKI AND SHENGDA ZENG
We combine this inequality with (12), (17) and (18) to deduce Now, we consider the last term in the inequality (19). It follows from the Cauchy inequality with ε > 0, see [26], that for all ε > 0. Now, we choose the constant ε in the following way .

This choice implies that
Taking (20) into account in (19) and then integrating it over [0, t], we obtain Thus, applying the integral version of the Gronwall inequality, see Lemma 2.5, we conclude for all t ∈ [0, T ], where C > 0 is independent of τ , h and t. This means that {u τ } ⊂ C(0, T ; H) is a Cauchy sequence, and therefore, there exists u ∈ C(0, T ; H) such that u τ → u in C(0, T ; H), as τ → 0. Further, by the fact that the function u τ is uniformly Lipschitz continuous for all τ > 0, we conclude that u is Lipschitz continuous as well.
Next, we will show that u is a strong solution of (FIDE) in (3). From the convergence u τ → u in C(0, T ; H), as τ → 0, we know that u τ (t) → u(t) in H for all t ∈ [0, T ], as τ → 0. However, the inequality (12) ensures that u τ (t) → u(t) in H for all t ∈ [0, T ], as τ → 0. According to the m-accretivity of the operator A, the uniform convexity of H, from Proposition 1, we know that A is demiclosed, thus, Au τ (t) Au(t) in H for all t ∈ [0, T ], as τ → 0. In addition, it is clear that 0 I αi Let w ∈ H. We multiply (14) by w and integrate the result on [0, t]. We have Letting τ → 0 and applying the Lebesgue dominated convergence theorem, see [16,Theorem 1.65], we infer Since w ∈ H is arbitrary, we have for all t ∈ [0, T ]. This implies that for a.e. t ∈ (0, T ), because u is a Lipschitz continuous function. From the convergence u 0 = u τ (0) → u(0) in H, as τ → 0, we have u(0) = u 0 . This means that u is a strong solution to (FIDE) in (3). It remains to prove uniqueness of strong solution to (FIDE) in (3). To that end, assume that u 1 and u 2 are two solutions to (3). Then, we have where j = 1, 2. Subtracting the above equalities, multiplying the result by u 1 (t) − u 2 (t), and using the m-accretivity of the operator A, we have We again apply the Cauchy inequality with ε > 0 to obtain where ε > 0 is chosen such that Integrating the inequality (21) on [0, t], we get Hence Applying the integral Gronwall inequality, see Lemma 2.5, we conclude u 1 = u 2 , which completes the proof of the theorem. For this problem we choose k = 1, a 1 = 1, α 1 = 1 and T = 10. We denote by u : [0, T ] → L 2 (0, π; R) the function u(t)(x) = u(t, x) for all t ∈ [0, T ], x ∈ [0, π], and put H = L 2 (0, π; R), f = 0 and Au = − ∂ 2 u ∂x 2 with D(A) = H 2 (0, π; R) = u ∈ L 2 (0, π; R) | u ∈ L 2 (0, π; R) ⊂ H.
It is obvious that all hypotheses of Theorem 3.3 are satisfied. Therefore, by this theorem, we know that the problem has a unique strong solution in u ∈ C(0, T ; H) and u is Lipschitz continuous. For this problem, we have k = 1, a 1 = 1, α 1 = 0.5 and T = 10. We denote by u : [0, T ] → L 2 (0, π; R) the function u(t)(x) = u(t, x) for all t ∈ [0, T ], x ∈ [0, π], H = L 2 (0, π; R), f (t) = tx 2 and the operator A is defined as in Example 1. It is clear that all assumptions of Theorem 3.3 hold. Therefore, by this theorem, we deduce that the problem has a unique strong solution in C(0, T ; H) and it is a Lipschitz continuous function.
Example 3. Consider the following multi-term fractional integral diffusion equation of the form As before, we denote by u : [0, T ] → L 2 (0, π; R) the function u(t)(x) = u(t, x) for all t ∈ [0, T ], x ∈ [0, π], H = L 2 (0, π; R) and the operator A is defined as in Example 1. It is obvious that all hypotheses of Theorem 3.3 are satisfied, and so by this theorem we deduce that the problem has a unique strong solution in C(0, T ; H) and the solution is a Lipschitz continuous function.