THE FUNDAMENTAL SOLUTIONS FOR FRACTIONAL EVOLUTION EQUATIONS OF PARABOLIC TYPE

where 0 < α≤ 1, Γ(α) is the gamma function, {A(t) : t ∈ [0,T]} is a family of linear closed operators defined on dense set D(A) in a Banach space E into E, u is the unknown Evalued function, u0 ∈ D(A), and f is a given E-valued function defined on [0,T]. It is assumed that D(A) is independent of t. Let B(E) denote the Banach space of all linear bounded operators in E endowed with the topology defined by the operator norm. We need the following conditions. (A1) The operator [A(t) + λI]−1 exists in B(E) for any λ with Reλ≥ 0 and ∥∥∥[A(t) + λI]−1∥∥∥≤ C |λ|+ 1 , (1.2)


Introduction
In this paper, we consider the fractional integral evolution equation  (1.4) for all t 1 ,t 2 ∈ [0,T], where C and β are positive constants and 0 < β ≤ 1 (the constants C and β are independent of t 1 and t 2 ). Under condition (A 1 ), each operator −A(s), s ∈ [0,T], generates an analytic semigroup exp(−tA(s)), t > 0, and there exists a positive constant C independent of both t and s such that where n = 0,1, t > 0, s ∈ [0,T] [9,10]. In Section 2, we will construct the fundamental solution of the homogeneous fractional differential equation We will prove the existence and uniqueness of the solution of (1.6), with the initial condition v(0) = u 0 ∈ D(A). (1.7) The continuous dependence of the solutions of (1.1) on the elements u 0 and the function f is proved.
In Section 3, we give an application to a mixed problem of a parabolic partial differential equation of fractional order.

The fundamental solution
We say that u is a strong solution of the fractional integral equation Let h be an E-valued function defined on [0, T]. If dh(t)/dt and the integral t τ (t − θ) −α (dh(θ)/dθ)dθ exist in the sense of Bochner, then we use the following definition of the fractional derivative τ D α t h(t): (see [7,13,16]). If u is a strong solution of (1.1), then the fractional derivative Mahmoud M. El-Borai 199 exists and is continuous in t ∈ [0,T]. In this case, we notice that Using (1.1), (2.1), and (2.3), we get The converse is also true. In other words, if d α u(t)/dt α is continuous in t ∈ [0,T] and u represents a solution of the Cauchy problem (2.5), (2.6), then u represents a strong solution of (1.1) (this means that the integral equation (1.1) is equivalent to the Cauchy problem (2.5), (2.6)). We will consider integrals of operator-valued functions. We denote by ψ(t,s) the integral where ζ α is a probability density function defined on [0,∞) such that its Laplace transform is given by . More details about this probability density function can be found in [8]. Proof. The existence of the considered improper integral is clear for η > 0, t,s ∈ [0,T]. If (2.9) 200 The fundamental solutions

It can be proved under conditions (A 1 ) and (A 2 ) that
where the positive constant C is independent of t, s, η, and τ. We estimate the norm of the first term on the right-hand side of (2.10) by using condition (A 2 ) and (1.5). We estimate also the norms of the second term and the last term of the right-hand side of (2.10) by using (2.12), (1.5), and (2.11), respectively. We thus find that the norm of the left-hand side of (2.10) is bounded by This completes the proof.
where C is a positive constant independent of t, η.
Proof. Let { f n } be a sequence of functions defined by We consider the integrals (2.20) Using (2.19) and (2.20), we get for all t,η ∈ [0,T], t − η ≥ . Clearly, whereas for x ∈ D(A), Using (2.23) and noticing that f satisfies condition (A 3 ), we deduce that the sequence { f n } uniformly converges to f with respect to t ∈ [0,T]. Using (2.22), we get, for any positive number , the inequality for a sufficiently large n. Consequently, This completes the proof. Let Using condition (A 2 ), we get Using Lemma 2.1, we conclude that ϕ 1 is uniformly continuous in t, τ in the uniform topology provided that t − τ ≥ > 0. Now one verifies, by induction, that all the functions ϕ k , k = 1,2,..., are uniformly continuous in t, τ in the uniform topology for t − τ ≥ , t,τ ∈ [0,T], and It is easy to see that The function ϕ is uniformly continuous in the uniform topology in t, τ provided that 0 ≤ τ ≤ t − , ≤ t ≤ T for any > 0. Using Fubini's theorem, we deduce that ϕ is the unique solution of the integral equation where the positive constant C does not depend on t 1 , t 2 , or τ.
Using (2.34), we get We estimate the norm of the first term on the right-hand side of (2.48) by using (2.47) and the norm of the second term by using (2.28) and (2.33). After simple calculations, the required result follows.

Mahmoud M. El-Borai 205
We will make use of the inequality for any u 0 ∈ E, Q(t)u 0 satisfies the fractional differential equation Proof. We set We will determine the operator-valued function U(t) such that Q(t)u 0 satisfies (2.50).
Using formally Lemma 2.3, we get [11,14,15]). The operator-valued function U(t) can be obtained by successive approximations, that is, we put Using the properties of ϕ k and Fubini's theorem, one easily shows, by induction, that It is clear that U(t) is given by Using (2.33), we get It is easy to see that (2.58) Using condition (A 2 ) and Lemma 2.4, we find that where t 2 > t 1 , t 1 ,t 2 ∈ [0,T], and C is a positive constant independent of t 1 , t 2 .
Recalling that ψ(t − η,η) is uniformly continuous in t, η, provided that t − η ≥ > 0, and using (2.14), (2.59), one can verify without difficulty that t 0 ψ(t − η,η)U(η)dη is uniformly continuous (in the norm of B(E)) in t ∈ [0,T]. Using (2.57), we get Q(t) ≤ C for all t ∈ [0,T], where C is a positive constant independent of t. It is also obvious that Q(0) = A −1 (0) and Q(t)u 0 is contiguous in t ∈ [0,T] for every u 0 ∈ E. We prove now that the range of Q(t) is included in D(A) for 0 < t ≤ T.

The operator-valued function A(t)ψ(t − η,η)U(η) can be written in the form
(2.60) By using (2.11) and (2.57), we find that the norm of the first term on the right-hand side of (2.60) is bounded by C(t − η) γ−1 . By using (1.5) and (2.59), we find that the norm of the second term on the right-hand side of (2.60) is bounded by C(t − η) γ−δ−1 (where C is a generic positive constant independent of both t and η). Using these estimations and Mahmoud M. El-Borai 207 It can be proved that there are two positive constants C and δ such that  Proof. We set (2.65) Then we determine the function V such that u satisfies (2.5). The proof is carried out similar to that of Theorem 2.5.

Proof. We introduce the bounded operators A n (t) = A(t)[I + (1/n)A(t)] −1 . It is known that
where s,t,τ ∈ [0,T] and C is a positive constant independent of t, τ, s, and n. Consider the following Cauchy problem: The solution of the Cauchy problem (2.69) is unique. To prove this fact, suppose g n (t) = 0. Then w n (t) satisfies (2.70) for every n, where C n is a positive constant. It follows that w n (t) = 0 for all t ∈ [0,T]. Noticing that g n is continuous in t ∈ [0,T] for every n = 1,2,... and A n (t) is a bounded operator that varies continuously in t ∈ [0,T] (in the uniform topology), then it is easy to see, with the help of (2.5), that the unique solution of the Cauchy problem (2.69) is given by  It can be shown that the sequence {g n } uniformly converges to zero in E with respect to t ∈ [0,T]. Consequently, by using (2.14), (2.33), (2.51), (2.67), and (2.71), we get v(t) = lim n→∞ v n (t) uniformly with respect to t ∈ [0,T]. Since v n (t) is uniquely defined as a solution of the Cauchy problem (2.68), v(t) is also unique.
The continuous dependence of the solution of the Cauchy problem (2.5), (2.6) on f and u 0 is established from formula (2.64) (cf. [1]).
It must be noticed that the fractional differential equations have many important applications in different areas of applied mathematics (see [5,6,12]).

Application
Let Ω be a bounded domain in the real n-dimensional Euclidean space R n . For any 0 < T < ∞, denote by Q T the cylinder {(x, t) : x ∈ Ω, 0 < t < T} and by ∂Ω the boundary of Ω.
We consider the differential operator where A * (x,t,D) is said to be uniformly elliptic in Q T if the coefficients a q (x,t) are bounded in Q T and (−1) m Re |q|=2m a q (x,t)ξ q ≥ C|ξ| 2m , for all (x,t) ∈ Q T and for all real ξ, where C is a positive constant independent of x, t, ξ, and 1 ··· D qn n , D j = ∂ ∂x j , |q| = q 1 + ··· + q n , q = q 1 ,..., q n is a multi-index .
Theorem 3.1. Assume that A * (x,t,D) is uniformly elliptic in Q T , that (I) and (II) hold, and that ∂Ω is of class C 2m . Then there exists a unique strong solution of problem (3.3), (3.4).
Proof. Writing (3.3) in the form d α u dt α + A * (t) + kI u = f (t) + ku, (3.6) we see that for some constant k, the operator A * (t) + kI satisfies conditions (A 1 ) and (A 2 ). Using formula (2.64), we get A(t) = A * (t) + kI. (3.7) It can be proved that the last integral equation has the unique required solution u(t). This completes the proof.