An L q ( L p ) -theory for the time fractional evolution equations with variable coefficients
aa r X i v : . [ m a t h . A P ] M a y AN L q ( L p ) -THEORY FOR THE TIME FRACTIONAL EVOLUTIONEQUATIONS WITH VARIABLE COEFFICIENTS ILDOO KIM, KYEONG-HUN KIM, AND SUNGBIN LIM
Abstract.
We introduce an L q ( L p )-theory for the quasi-linear fractional equa-tions of the type ∂ αt u ( t, x ) = a ij ( t, x ) u x i x j ( t, x ) + f ( t, x, u ) , t > , x ∈ R d . Here, α ∈ (0 , p, q >
1, and ∂ αt is the Caupto fractional derivative of order α .Uniqueness, existence, and L q ( L p )-estimates of solutions are obtained. Theleading coefficients a ij ( t, x ) are assumed to be piecewise continuous in t anduniformly continuous in x . In particular a ij ( t, x ) are allowed to be discontin-uous with respect to the time variable. Our approach is based on classicaltools in PDE theories such as the Marcinkiewicz interpolation theorem, theCalderon-Zygmund theorem, and perturbation arguments. Introduction
Fractional calculus has been used in numerous areas including mathematicalmodeling [24, 44], control engineering [6, 32], electromagnetism [13, 43], polymerscience [3, 29], hydrology [4, 41], biophysics [14, 22], and even finance [36, 40].See also [15, 30, 37, 48] and references therein. The classical heat equation ∂ t u =∆ u describes the heat propagation in homogeneous mediums. The time-fractionaldiffusion equation ∂ αt u = ∆ u , α ∈ (0 , ∂ t u = ∆ u , α ∈ (1 , L q ( L p )-theory for the quasi-linearfractional evolution equation ∂ αt u ( t, x ) = a ij ( t, x ) u x i x j ( t, x ) + b i ( t, x ) u x i ( t, x ) + c ( t, x ) u ( t, x ) + f ( t, x, u ) (1.1)given for t > x ∈ R d . Here α ∈ (0 , , p, q >
1, and ∂ αt denotes the Caputofractional derivative (see (2.3)). The indices i and j move from 1 to d , and thesummation convention with respect to the repeated indices is assumed throughoutthe article. It is assumed that the leading coefficients a ij ( t, x ) are piecewise con-tinuous in t and uniformly continuous in x , and the lower order coefficients b i and c are only bounded measurable functions. We prove that under a mild condition Mathematics Subject Classification.
Key words and phrases.
Fractional diffusion-wave equation, L q ( L p )-theory, L p -theory, Caputofractional derivative, Variable coefficients.This work was supported by Samsung Science and Technology Foundation under Project Num-ber SSTF-BA1401-02. on the nonlinear term f ( t, x, u ) there exists a unique solution u to (1.1) and the L q ( L p )-norms of the derivatives D βx u , | β | ≤
2, are controlled by the L q ( L p )-normof f ( t, x, H α,nq,p, ( T ) (see [39, 31] for the proof).Here is a brief survey of closely related works. In [8, 33] an L q ( L p )-theory forthe parabolic Volterra equations of the type ∂∂t (cid:18) c u + Z t −∞ k ( t − s ) u ( s, x ) ds (cid:19) = ∆ u + f ( t, x, u ) , t ∈ R , x ∈ R d (1.2)is obtained under the conditions k ( t ) ≥ ct − α for small t , c ≥ α ∈ (0 , αq + dp < . (1.3)The results of [8, 33] also cover the case c >
0, however it is obtained only for thecase a ij ( t, x ) = δ ij with the restrictions α ∈ (0 ,
1) and (1.3). If p = q , an L p -theoryof type (1.1) with the variable coefficients a ij ( t, x ) is presented in [46] under thecondition that a ij are uniformly continuous in ( t, x ) and lim | x |→∞ a ij ( t, x ) exists.In [47] an L -theory is obtained for the divergence type equations with generalmeasurable coefficients. Also an eigenfunction expansion method is introducedin [38] to obtain L -estimates of solutions of divergence type equations with C -coefficients.For other approaches to the equations with fractional time derivatives, we referto [21] for the semigroup approach, to [9, 34] for C δ ([0 , T ] , X )-type theory, where X is an appropriate Banach space, and to [7] for BU C − β ([0 , T ] , X )-type estimate,where k u k BUC − β ([0 ,T ] ,X ) = sup t ∈ (0 ,T ] t − β k u ( t ) k X .Our result substantially generalizes above mentioned results in the sense that wedo not impose any algebraic conditions on α , p , and q . The conditions (1.3) and α ∈ (0 ,
1) are used in [8, 33], and the restrictions p = q and α
6∈ { p − , p − − , p , p − } are assumed in [46]. More importantly, in this article the condition on the leadingcoefficients a ij ( t, x ) is considerably weakened. In particular, a ij ( t, x ) depend onboth t and x and can be discontinuous in t . Recall that if p > t , but the condition p = q and the continuity of a ij with respect to ( t, x ) are assumed in [46].Another significance of this article is the method we use. The results of [8, 33, 46]are operator theoretic, [21] is based on H ∞ -functional calculus, and the method of[47, 38] works well only in the Hilbert-space framework. Our approach is purely ana-lytic and is based on the classical tools in PDE theories including the Marcinkiewiczinterpolation theorem and the Calderon-Zygmund theorem. We obtain the meanoscillation (or BMO estimate) of solutions and then apply the Marcinkiewicz in-terpolation theorem to obtain L p -estimates of solutions. To go from L p -theoryto L q ( L p )-theory we show that the kernel appeared in the representation of solu-tions for equations with constant coefficients satisfies the conditions needed for theCalderon-Zygmund theorem. Perturbation and fixed point arguments are used tohandle the variable coefficients and the nonlinear term respectively.The article is organized as follows. Some properties of the fractional derivativesand our main result, Theorem 2.9, are presented in Section 2. The representation of q ( L p )-THEORY FOR FRACTIONAL EQUATIONS 3 solutions to a model equation and an L -estimate of solutions are given in Section3, and BMO and an L q ( L p )-estimate of solutions to a model equation are obtainedin Section 4. The proof of Theorem 2.9 is given in Section 5, and sharp estimatesof kernels related to the representation of solutions are obtained in Section 6.We finish the introduction with some notation used in this article. As usual N = { , , · · · } , R d stands for the Euclidean space of points x = ( x , ..., x d ), B r ( x ) := { y ∈ R d : | x − y | < r } , and B r := B r (0). For multi-indices γ = ( γ , ..., γ d ), γ i ∈ { , , , ... } , x ∈ R d , and functions u ( x ) we set u x i = ∂u∂x i = D i u, D γx u = D γ · ... · D γ d d u,x γ = ( x ) γ ( x ) γ · · · ( x d ) γ d , | γ | = γ + · · · + γ d . We also use D mx to denote a partial derivative of order m with respect to x . For anopen set Ω ⊂ R d by C ∞ c (Ω) we denote the set of infinitely differentiable functionswith compact support in U . For a Banach space F and p > L p ( U, F ) wedenote the set of F -valued Lebesgue-measurable functions u on Ω satisfying k u k L p (Ω ,F ) = (cid:18)Z Ω k u ( x ) k pF dx (cid:19) /p < ∞ . We write f ∈ L p,loc ( U, F ) if ζf ∈ L p ( U, F ) for any real-valued ζ ∈ C ∞ c ( U ). Also L p (Ω) = L p (Ω , R ) and L p = L p ( R d ). We use “:=” to denote a definition. By F and F − we denote the d -dimensional Fourier transform and the inverse Fouriertransform respectively, i.e. F ( f )( ξ ) := 1(2 π ) d/ Z R d e − ix · ξ f ( x ) dx, F − ( f )( x ) := 1(2 π ) d/ Z R d e iξ · x f ( ξ ) dξ. For a Lebesgue set A ⊂ R d , we use | A | to denote its Lebesgue measure and by1 A ( x ) we denote the indicator of A . For a complex number z , ℜ [ z ] is the real partof z . Finally if we write N = N ( a, b, . . . ), this means that the constant N dependsonly on a, b, . . . . 2. Main results
We fix T ∈ (0 , ∞ ) throughout the article. For α > k α ( t ) := t α − Γ( α ) − , t > , where Γ( α ) := R ∞ t α − e − t dt . For functions ϕ ∈ L ((0 , T )) the Riemann-Liouvillefractional integral of the order α > I α ϕ ( t ) = k α ∗ ϕ ( t ) = 1Γ( α ) Z t ( t − s ) α − ϕ ( s ) ds. It is easy to check that I α + β ϕ ( t ) = I α I β ϕ ( t ) ∀ α, β > . Also, by Jensen’s inequality, for any p ∈ [1 , ∞ ], k I α ϕ k L p ((0 ,T )) ≤ N ( T, α ) k ϕ k L p ((0 ,T )) . (2.1)It is also known (see e.g. [39]) that I α : B λ → C λ + α is a bounded operator if λ ≥ λ + α <
1, where B = L ∞ and B λ = C λ if λ > ILDOO KIM, KYEONG-HUN KIM, AND SUNGBIN LIM
Let k ∈ N , k − ≤ α < k , and f ( k − ( t ) be absolutely continuous, where f ( k − ( t )denotes the ( k − f . Then the Caputo fractional derivativeof order α > ∂ αt f ( t ) = 1Γ( k − α ) Z t ( t − s ) k − α − f ( n ) ( s ) ds (2.2)= 1Γ( k − α ) ddt Z t ( t − s ) k − α − h f ( k − ( s ) − f ( k − (0) i ds. (2.3)Note that (2.3) (“Kochubei extension”) is defined for a broader class of functions.Let q ≥
1. For functions f ∈ C k ([0 , T ]), we denote k f k H αq ( T ) = Z T | f | q dt ! /q + Z T | ∂ αt f | q dt ! /q . The following lemma shows that it is irreverent whether one uses (2.2) or (2.3) asthe definition of ∂ αt for functions in H αq ( T ). Lemma 2.1.
The closures H αq ( T ) and ˜ H αq ( T ) of C k ([0 , T ]) in the space L q ((0 , T )) with respect to norms k · k H αq related to (2.2) and (2.3) respectively coincide.Proof. This is obvious because (2.2) and (2.3) are equal for functions f ∈ C k ([0 , T ]). (cid:3) Next we introduce an another fractional derivative. Let D αt denote the Riemann-Liouville fractional derivative of order α which is defined as D αt ϕ ( t ) = ddt ( I − α ϕ )( t ) , α ∈ (0 , . (2.4)It is obvious that ∂ αt ϕ ( t ) = D αt ( ϕ − ϕ (0)) = D αt ϕ ( t ) − ϕ (0) t α Γ(1 − α ) , α ∈ (0 , . (2.5)It is easy to check for any ϕ ∈ L ((0 , T )), D αt I α ϕ = ϕ, α ∈ (0 , . (2.6)Similarly, the equality I α D αt ϕ = ϕ, α ∈ (0 ,
1) (2.7)also holds if I − α ϕ is absolutely continuous and I − α ϕ (0) = 0. Definition 2.2.
Let k − ≤ α < k and f ∈ H αq ( T ) . We write f (0) = 0 if thereexists a sequence f n ∈ C k ([0 , T ]) such that f n (0) = 0 and f n → f in H αq ( T ) .Similarly if α > we write f ′ (0) = 0 if f ′ n (0) = 0 for all n . The following lemma gives sufficient and necessary conditions for f ∈ H αq ( T )and f (0) = 0 (or f ′ (0) = 0). Lemma 2.3. (i) Let α ∈ (0 , and q > . Then f ∈ H αq ( T ) and f (0) = 0 if andonly if f ∈ L q ((0 , T )) , I − α f ∈ H q ( T ) , I − α f ( t ) is continuous, and I − α f (0) = 0 .(ii) Let α ∈ (1 , and q > . Then f ∈ H αq ( T ) and f ′ (0) = 0 if and only if f ∈ H q ( T ) , I − α f ′ ∈ H q ( T ) , I − α f ′ ( t ) is continuous, and I − α f ′ (0) = 0 . q ( L p )-THEORY FOR FRACTIONAL EQUATIONS 5 Proof. (i) Suppose f ∈ H αq ( T ) and f (0) = 0. Take a sequence f n ∈ C ([0 , T ]) sat-isfying f n (0) = 0 and f n → f in H αq ( T ). Then we have f n → f and ddt ( I − α f n ) → ∂ αt f in L q ((0 , T )). It follows that I − α f n converges to I − α f in the space H q ((0 , T )).It also follows that I − α f ( t ) = Z t ∂ αt f ( s ) ds, t ≤ T ( a.e. ) . Since 1 − /q >
0, by the Sobolev embedding theorem I − α f ( t ) is continuous in t and thus the above equality holds for all t , which implies I − α f (0) = 0.Next we prove the “if” part of (i). By the assumption, we can choose a function g ∈ L q ((0 , T )) so that I − α f ( t ) = Z t g ( s ) ds, ∀ t ≤ T. Take a sequence of functions g n ∈ C ([0 , T ]) which converges to g in L q ((0 , T )).Define f n = I α g n . Then f n ∈ C ([0 , T ]), f n (0) = 0, and f n → I α g = f in L q ((0 , T )) . Also ∂ αt f n = ∂ αt I α g n = g n → ∂ αt f in L q ((0 , T )) . (ii) The proof is very similar to (i). We only explain how one can choose asequence to prove the “if” part.By the assumption f ∈ H q ( T ) and q >
1, we may assume f ∈ C ([0 , T ]). Take g ∈ L q ((0 , T )) so that I − α f ′ = R t g ( s ) ds . We choose g n ∈ C ([0 , T ]) which convergesto g in L q ((0 , T )). Define f n ( t ) = I α g n ( t ) + f (0). Then f ′ n (0) = I α − g n (0) = 0, f n → f in L q ((0 , T )), and ∂ αt f n is a Cauchy sequence in L q ((0 , T )). Thus f ∈ H αp ( T ) and f ′ (0) = 0.The lemma is proved. (cid:3) Next we introduce our solution space H α,kq,p ( T ) and related notation. Roughlyspeaking, we write u ∈ H α,kq,p ( T ) if and only if u, ∂ αt u, D kx u ∈ L q ((0 , T ) , L p ) . For p, q > k = 0 , , , · · · , we denote H kp = H kp ( R d ) = { u ∈ L p ( R d ) : D γx u ∈ L p ( R d ) , | γ | ≤ k } , H ,kq,p ( T ) = L q ((0 , T ) , H kp ) , L q,p ( T ) := H , q,p ( T ) , where D γx are derivatives in the distributional sense. Thus u ∈ H ,kq,p ( T ) if and onlyif u ( t, · ) is H kp -valued measurable function satisfying k u k H ,kq,p ( T ) := "Z T k u k qH kp ds /q < ∞ . We extend the real-valued time fractional Sobolev space to L p ( R d )-valued one. Inother words, we consider the completion of C ([0 , T ] × R d ) ∩ L q,p ( T ) with respectto norm k · k L q,p ( T ) + k ∂ αt · k L q,p ( T ) in L q,p ( T ). ILDOO KIM, KYEONG-HUN KIM, AND SUNGBIN LIM
Definition 2.4.
For α ∈ (0 , we say u ∈ H α, q,p ( T ) if and only if there exists asequence u n ∈ C ([0 , T ] × R d ) ∩ L q,p ( T ) so that sup n k ∂ αt u n k L q,p ( T ) < ∞ , and k u − u n k L q,p ( T ) → and k ∂ αt u n − ∂ αt u m k L q,p ( T ) → as n and m go to infinity. We call this sequence u n a defining sequence of u . For u ∈ H α, q,p ( T ) , we define ∂ αt u = lim n →∞ ∂ αt u n in L q,p ( T ) , where u n is a defining sequence of u . Obviously H α, q,p ( T ) is a Banach space withthe norm k u k H α, q,p ( T ) = k u k L q,p ( T ) + k ∂ αt u k L q,p ( T ) . Definition 2.5.
For u ∈ H α, q,p ( T ) , we say that u (0 , x ) = 0 if any only if there existsa defining sequence u n such that u n (0 , x ) = 0 ∀ x ∈ R d , ∀ n ∈ N . Similarly we say that u (0 , · ) = 0 and ∂∂t u (0 , · ) = 0 if any only if there exists adefining sequence u n such that u n (0 , x ) = 0 and ∂∂t u n (0 , x ) = 0 ∀ x ∈ R d , ∀ n ∈ N . Let H α,kq,p ( T ) := H α, q,p ( T ) ∩ H ,kq,p ( T )and H α,kq,p, ( T ) be the subspace of H α,kq,p ( T ) such that u (0 , · ) = 0 if α ∈ (0 , u (0 , · ) = 0 and ∂∂t u (0 , · ) = 0 if α ∈ (1 , . Theorem 2.6. (i) The space H α,kq,p ( T ) is a Banach space with the norm k u k H α,kq,p ( T ) := k u k H ,kq,p ( T ) + k u k H α, q,p ( T ) . (ii) The space H α,kq,p, ( T ) is a closed subspace of H α,kq,p ( T ) .(iii) C ∞ c ( R d +1+ ) is dense in H α,kq,p, ( T ) .(iv) For any u ∈ H α, q,p, ( T ) , k u ( t ) k L p ≤ N ( α ) Z t ( t − s ) α − k ∂ αt u ( s ) k L p ds, t ≤ T ( a.e. ) . (2.8) Consequently, k u k q L q,p ( t ) ≤ N ( q, α, T ) Z t Z s ( s − r ) α − k ∂ αt u ( r ) k qL p drds, ∀ t ≤ T. Proof. (i) This is obvious because both H α, q,p ( T ) and H ,kq,p ( T ) are Banach spaces.(ii) Suppose u n ∈ H α,kq,p, ( T ) and u ∈ H α,kq,p ( T ) so that u n → u in H α,kq,p ( T ). Since u n ∈ H α,kq,p, ( T ) for each n , we can find a v n ∈ C ([0 , T ] × R d ) ∩ L q,p ( T ) so that k u n − v n k L q,p ( T ) < n and k ∂ αt u n − ∂ αt v n k L q,p ( T ) < n ,v n (0 , x ) = 0 ∀ x ∈ R d , if α ∈ (0 , , q ( L p )-THEORY FOR FRACTIONAL EQUATIONS 7 and v n (0 , x ) = 0 and ∂∂t v n (0 , x ) = 0 ∀ x ∈ R d , if α ∈ (1 , . Therefore u ∈ H α,kq,p, ( T ) because obviously k v n − u k H α, q,p ( T ) → n → ∞ . This certainly proves (ii).(iii) We take nonnegative smooth functions η ∈ C ∞ c ((1 , η ∈ C ∞ c ( R d ), and η ∈ C ∞ c ( R d ) so that Z ∞ η ( t ) dt = 1 , Z R d η ( x ) dx = 1 , and η ( x ) = 1 if | x | ≤ . For ε , ε , ε >
0, we define η ,ε ( t ) = ε − η ( t/ε ) , η ,ε ( x ) = ε − d η ( x/ε ) ,u ε ( t, x ) = Z ∞ u ( s, x ) η ,ε ( t − s ) ds,u ε ,ε ( t, x ) = Z R d Z ∞ u ( s, y ) η ,ε ( t − s ) η ,ε ( x − y ) dsdy, and u ε ,ε ,ε ( t, x ) = η ( t ) η ( ε x ) Z R d Z ∞ u ( s, y ) η ,ε ( t − s ) η ,ε ( x − y ) dsdy, where η ∈ C ∞ ([0 , ∞ )) such that η ( t ) = 1 for all t ≤ T and vanishes for all large t .Due to the condition η ∈ C ∞ c ((1 , u ε ,ε ,ε ( t, x ) = 0 ∀ t < ε , ∀ x ∈ R d . We can easily check that for any u ∈ H α,kq,p, ( T ) ∂ αt u ε ( t ) = ( ∂ αt u ) ε ( t ) . Hence for any given ε > k u − u ε ,ε ,ε k H α,kq,p ( T ) ≤ k u − u ε k H α,kq,p ( T ) + k u ε − u ε ,ε k H α,kq,p ( T ) + k u ε ,ε − u ε ,ε ,ε k H α,kq,p ( T ) ≤ ε if ε , ε , and ε are small enough. Therefore (iii) is proved.(iv) Due to (iii), it is enough to prove (2.8) only for u ∈ C ∞ c ( R d +1 ). Denote f := ∂ αt u . One can easily check u ( t ) = Z t k α ( t − s ) f ( s ) ds, ∀ t ≤ T in the space L p , which clearly implies (2.8) due to the generalized Minkowski in-equality. The theorem is proved. (cid:3) Assumption 2.7.
Let f ( u ) = f ( t, x, u ) and f = f ( t, x, .(i) There exist T < T < · · · < T ℓ = T and functions a ijk ( t, x ) such that a ij ( t, x ) = ℓ X k =1 a ijk ( t, x ) I ( T k − ,T k ] ( t ) , ( a.e. ) . ILDOO KIM, KYEONG-HUN KIM, AND SUNGBIN LIM (ii) There exist constants δ, K > so that for any k , t , and xδ | ξ | ≤ a ijk ( t, x ) ξ i ξ j ≤ K | ξ | , ∀ ξ ∈ R d , (2.9) and | a ijk ( t, x ) | + | b i ( t, x ) | + | c ( t, x ) | ≤ K. (iii) The coefficients a ijk are uniformly continuous on ( t k − , t k ] × R d for all k =1 , . . . , ℓ and i, j = 1 , . . . , d .(iv) f ∈ L q,p ( T ) and f ( u ) satisfies the following continuity property: for any ε > ,there exists a constant K ε > such that k f ( t, x, u ) − f ( t, x, v ) k L p ≤ ε k u − v k H p + K ε k u − v k L p , (2.10) for any ( t, x ) and u, v ∈ H p . If p = q then we need an additional condition (see the comment below (5.4) forthe reason). Assumption 2.8. If p = q then lim | x |→∞ a ij ( t, x ) exists uniformly in t ∈ (0 , T ) . Here is the main result of this article. The proof will be given in Section 5.
Theorem 2.9.
Let p, q > and Assumptions 2.7 and 2.8 hold. Then the equation ∂ αt u = a ij u x i x j + b i u x i + cu + f ( u ) , t > admits a unique solution u in the class H α, q,p, ( T ) , and for this solution it holds that k u k H α, q,p ( T ) ≤ N k f k L q,p ( T ) , (2.12) where N depends only on d, p, q, δ, K, K ε , T, ℓ , and the modulus of continuity of a ijk .Remark . (i) Due to the definition of our solution space H α, q,p, ( T ), the zeroinitial condition is given to equation (2.11), that is u (0 , x ) = 0 and additionally ∂u∂t (0 , x ) = 0 if α > . (ii) Some examples of f ( u ) satisfying (2.10) can be found e.g. in [19]. Forinstance, let κ := 2 − d/p > h = h ( x ) ∈ L p , and f ( x, u ) := h ( x ) sup x | u | . Then by the Sobolev embedding theorem, k f ( u ) − f ( v ) k L p ≤ k h k L p sup x | u − v | ≤ N k u − v k H κp ≤ ε k u − v k H p + K k u − v k L p . Similarly one can show that f ( t, x, u ) := a ( t, x )( − ∆) δ u also satisfies (2.10) if a ( t, x )is bounded and δ ∈ (0 , q ( L p )-THEORY FOR FRACTIONAL EQUATIONS 9 Some Preliminaries
In this section we introduce some estimates of kernels related to the representa-tion of a solution, and we also present an L -estimate of a solution.The Mittag-Leffler function E α ( z ) is defined as E α ( z ) = ∞ X k =0 z k Γ( αk + 1) , z ∈ C , α > . The series converges for any z ∈ C , and E α ( z ) is an entire function. Using ∂ αt t β = Γ( β + 1)Γ( β + 1 − α ) t β − α , β ≥ α one can easily check that for any constant λ , ϕ ( t ) := E α ( λt α )satisfies ϕ (0) = 1 (also ϕ ′ (0) = 0 if α >
1) and ∂ αt ϕ = λϕ, t > . By taking the Fourier transform to the equation ∂ αt u = ∆ u, t > , u (0) = h, (and u ′ (0) = 0 if α > F ( u )( t, ξ ) = E α ( − t α | ξ | )˜ h . Thus it is naturally needed to findan integrable function p ( t, x ) satisfying F ( p ( t, · ))( ξ ) = E α ( − t α | ξ | ). It is knownthat (see e.g. (12) in [10, Theorem 1.3-4]) E α ( − t ) ∼
11 + | t | , t > . (3.1)Therefore, E α ( − t α | ξ | ) is integrable only if d = 1. If d ≥ F − ( E α ( − t α | ξ | ))might be understood as an improper integral. However, in this article we do notconsider F − ( E α ( − t α | ξ | )). Lemma 3.1. (i) Let d ≥ and α ∈ (0 , . Then there exists a function p ( t, x ) such that p ( t, · ) is integrable in R d and F ( p ( t, · ))( ξ ) = E α ( − t α | ξ | ) . (ii) Let m, n = 0 , , , · · · and denote R = t − α | x | . Then there exist constants C and σ depending only on m, n, d , and α so that if R ≥ | ∂ nt D mx p ( t, x ) | ≤ N t − α ( d + m )2 − n exp {− σt − α − α | x | − α } , (3.2) and if R ≤ | ∂ nt D mx p ( t, x ) | ≤ N | x | − d − m t − n (cid:16) R + R ln R · d =2 ,m =0 + R / · d =1 ,m =0 (cid:17) . (3.3)By (3.2), p ( t, x ) is absolutely continuous on (0 , T ) and lim t → p ( t, x ) = 0 if x = 0.Thus we can define q ( t, x ) := ( I α − p ( t, x ) , α ∈ (1 , D − αt p ( t, x ) , α ∈ (0 , . Since p (0 , x ) = 0 for x = 0, D − αt p ( t, x ) = ∂ − αt p ( t, x ). Lemma 3.2. (i) Let d ≥ , α ∈ (0 , , and m, n = 0 , , , · · · . Denote R = t − α | x | .Then there exist constants N and σ depending only on m , n , d , and α so that if R ≥ | ∂ nt D mx q ( t, x ) | ≤ N t − α ( d + m )2 − n + α − exp {− σt − α − α | x | − α } , (3.4) and if R ≤ | ∂ nt D mx q ( t, x ) | ≤ N | x | − d − m t − n + α − ( R + R ln R · d =2 )+ N | x | − d t − n + α − (cid:16) R / · d =1 + R · d =2 + R ln R · d =4 (cid:17) m =0 . (3.5) (ii) For any t = 0 and x = 0 , ∂ αt p = ∆ p, ∂p∂t = ∆ q. (3.6)One can find similar statements of Lemmas 3.1 and 3.2 in [11, 12, 18, 35]. Forthe sake of completeness, we give an independent and rigorous proof in Section 6. Corollary 3.3.
Let < ε < T . Then Z R d sup t ∈ [ ε,T ] | D mx q ( t, x ) | dx < ∞ , m = 0 , , . and Z R d sup t ∈ [ ε,T ] | D mt p ( t, x ) | dx < ∞ , m = 0 , . Proof.
By (3.4) and (3.5), it follows that for large | x | sup t ∈ [ ε,T ] | D mx q ( t, x ) | ≤ N e − c | x | − α and for small | x | sup t ∈ [ ε,T ] | D mx q ( t, x ) | ≤ N | x | − d − m (cid:0) | x | + | x | · m =0 (cid:1) , where N and c depend only on d , α , T , and ε . This certainly proves the assertionrelated to q , and p is handled similarly. The corollary is proved. (cid:3) Lemma 3.4.
Let λ > , α ∈ (0 , , and φ be a continuous function on [0 , ∞ ) sothat the Laplace transforms of φ and ∂ αt φ exist, ∂ αt φ + λφ = f ( t ) , t > , (3.7) and φ (0) = 0 ( additionally φ ′ (0) = 0 if α ∈ (1 , . Moreover we assume for each s > , e − st Z t ∂ αt φ ( r ) dr → as t → ∞ . Then φ ( t ) = Z t H α,λ ( t − s ) f ( s ) ds, (3.8) where H α,λ ( t ) = D − αt E α ( − λt α ) if α ∈ (0 , and H α,λ ( t ) = I α − E α ( − λt α ) other-wise. q ( L p )-THEORY FOR FRACTIONAL EQUATIONS 11 Proof.
First, recall that ϕ ( t ) := E α ( − t α ) satisfies ϕ (0) = 1 and ∂ αt ϕ = − ϕ . Let L denote the Laplace transform. Then from L [ ∂ αt ϕ ] = −L [ ϕ ] we get L [ ϕ ]( s ) := Z ∞ e − st ϕ ( t ) dt = s α − s α + 1 . Here, we used the following facts: for β ∈ (0 , L [ h ′ ] = s L [ h ]( s ) , L [ h ∗ g ]( s ) = L [ h ]( s ) · L [ g ]( s ) , L [ k − β ]( s ) = s β − . (3.9)It follows that L [ E α ( − λt α )]( s ) = s α − s α + λ , L [ H α,λ ]( s ) = 1 s α + λ . (3.10)On the other hand, taking the Laplace transform to (3.7) and using (3.9) we get L [ φ ]( s ) = 1 s α + λ · L [ f ]( s ) , s > . This and (3.10) certainly prove the lemma, because to prove equality (3.8) it isenough to show that two functions under consideration have the same Laplacetransform. (cid:3)
Lemma 3.5. (i) Let u ∈ C ∞ c ( R d +1+ ) and denote f := ∂ αt u − ∆ u . Then u ( t, x ) = Z t Z R d q ( t − s, x − y ) f ( s, y ) dyds. (3.11) (ii) Let f ∈ C ∞ c ( R d +1+ ) and define u as in (3.11) . Then u satisfies ∂ αt u = ∆ u + f .Proof. (i) Since q is integrable on (0 , T ) × R d (see Lemma 3.2), it is enough to proveˆ u ( t, ξ ) = Z t ˆ q ( t − s, ξ ) ˆ f ( s, ξ ) ds, (3.12)where ˆ f denotes the Fourier transform of f with respect to x , i.e.ˆ f ( s, ξ ) := F ( f ( s, · ))( ξ ) = 1(2 π ) d/ Z R d e − ix · ξ f ( s, x ) dx. First note that from the definition of f we getˆ f ( t, ξ ) = ∂ αt ˆ u ( t, ξ ) + | ξ | ˆ u ( t, ξ ) , ∀ t > . Therefore by Lemma 3.4ˆ u ( t, ξ ) = Z t H α, | ξ | ( t − s ) ˆ f ( s, ξ ) ds. Hence it is enough to prove ˆ q ( t, ξ ) = H α, | ξ | ( t ) . (3.13)Denote c d = (2 π ) − d/ . If α ∈ (0 ,
1) then by the definitionˆ q ( t, x ) = c d Z R d e − ix · ξ q ( t, x ) dx = c d Z R d e − ix · ξ (cid:20) ddt Z t k α ( t − s ) p ( s, x ) ds (cid:21) dx. Since q ( t, · ) is integrable in R d uniformly in a neighborhood of t > ddt out of the integral. After this using Fubini’stheorem we getˆ q = ddt Z t k α ( t − s ) (cid:20) c d Z R d e − ix · ξ p ( s, x ) dx (cid:21) ds = D − αt E α ( − t α | ξ | ) . Hence (3.13) and (i) are proved. The case α ∈ [1 ,
2) is easier and we skip the proof.(ii) Taking the Fourier transform to (3.11) we getˆ u ( t, ξ ) = Z t H α, | ξ | ( t − s ) ˆ f ( s, ξ ) ds. Note that if one defines φ as in (3.8) then it satisfies (3.7). Consequently, ˆ u satisfies ∂ αt ˆ u ( t, ξ ) + | ξ | ˆ u ( t, ξ ) = ˆ f ( t, ξ ) , and this certainly proves the equality ∂ αt u = ∆ u + f , because ˆ f ( t, · ) = 0 if t is smallenough and thus for each t > F ( ∂ αt u ( t, · )) ( x ) = Z t H α, | ξ | ( t − s ) ∂ αs ˆ f ( s, ξ ) ds = ∂ αt ˆ u ( t, ξ ) . The lemma is proved. (cid:3)
Now we define the operator G by G f ( t, x ) = Z t −∞ Z R d q ( t − s, x − y )∆ f ( s, y ) dyds. Since q is integrable on (0 , T ) × R d , G f is well-defined if f ∈ C ∞ c ( R d +1 ). Alsorecall that D x q ( t, · ) and D x q ( t, · ) are integrable in R d for each t >
0, and thereforeit follows that G f ( t, x ) = lim ε → Z t − ε −∞ (cid:20)Z R d ∆ q ( t − s, x − y ) f ( s, y ) dy (cid:21) ds. (3.14) Lemma 3.6.
Let f ∈ C ∞ c ( R d +1 ) . Then kG f k L ( R d +1 ) ≤ N k f k L ( R d +1 ) (3.15) where N = N ( α, d ) .Proof. Denote q M = q
0, i.e.sup M> ,τ,ξ | I M ( τ, ξ ) | < ∞ . q ( L p )-THEORY FOR FRACTIONAL EQUATIONS 13 Then the claim implies kG M f k L ≤ N k f k L , (3.16)where N is independent of M .By the integration by parts and the change of variables, I M ( τ, ξ ) = i sgn( τ ) Z τM e − sgn( τ ) it E α ( − ( t | τ | ) α || ξ | ) dt + e − iτM E α ( − M α | ξ | ) − . (3.17)Denote for απ < η < min { π, απ } ∆ ∗ η := { z ∈ C : | π − Arg z | < π − η } . Then by [10, (1.3.12)] or [31, Theorem 1.6], | E α ( z ) | = N | z | , z ∈ ∆ ∗ η . Hence the function E α ( z ) is bounded in ∆ ∗ η .Let τ >
0. We take a θ ∈ (0 , π/
3) sufficiently small so that − e − iθ , − e − iαθ ∈ ∆ ∗ η for any θ ∈ [0 , θ ]. Denote C ,τM = { t : t ∈ [0 , τ M ] } , C ,τM = { te − iθ : t ∈ [0 , τ M ] } ,C ,τM = { τ M e − iθ : θ ∈ [0 , θ ] } , and define a contour C τM = C ,τM ∪ C ,τM ∪ C ,τM . Then the contour integralof the function e − iz E α ( − ( zτ ) α | ξ | ) on C τM is zero for any M >
0. Since E α ( z ) isbounded in ∆ ∗ η , (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z C ,τM e − iz E α ( − ( zτ ) α | ξ | ) dz (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ Z θ ( τ M ) e − τM sin θ dθ ≤ N ( θ ) . (3.18)Also, (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z C ,τM e − iz E α ( − ( zτ ) α | ξ | ) dz (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ N Z τM e − t sin θ dt ≤ N ( θ ) . (3.19)Note that θ depends only on α . Hence by (3.17), (3.18), and (3.19), it follows thatif τ > | I M ( τ, ξ ) | is bounded uniformly for M .If τ < C ′ τM = C ,τM ∪ C ′ ,τM ∪ C ′ ,τM where C ′ ,τM = { te iθ : t ∈ [0 , τ M ] } , C ′ ,τM = { τ M e iθ : θ ∈ [0 , θ ] } . Then the same arguments above go through. Thus our claim is proved.To finish the proof, observe that for each ( t, x ) ∈ R d +1 q M ∗ ∆ f ( t, x ) = G M f ( t, x ) → G f ( t, x ) as M → ∞ since ∆ f ∈ C ∞ c ( R d +1 ). Therefore (3.16) and Fatou’s lemma easily yields (3.15). (cid:3) BMO and L q ( L q ) -estimate In this section we obtain BMO and L q ( L p )-estimates for a solution of the modelequation ∂ αt u = ∆ u + f, ( t, x ) ∈ (0 , ∞ ) × R d . Recall that p is integrable with respect to x , F ( p ( t, · )) = E α ( − t α | ξ | ), and q isdefined as q ( t, x ) = ( I α − p ( t, x ) , α ∈ (1 , D − αt p ( t, x ) , α ∈ (0 , . Also recall that for f ∈ C ∞ c ( R d +1+ ), it holds that G f = Z t −∞ (cid:20)Z R d ∆ q ( t − s, x − y ) f ( s, y ) dy (cid:21) ds, (4.1)and by (3.15) the operator G is continuously extended onto L ( R d +1 ). We denotethis extension by the same notation G .For a locally integrable function h on R d +1 , we define the BMO semi-norm of h on R d +1 as k h k BMO ( R d +1 ) = sup Q ∈ Q | Q | Z Q | h ( t, x ) − h Q | dtdx where h Q = | Q | R Q h ( t, x ) dtdx and Q := { Q δ ( t , x ) = ( t − δ /α , t + δ /α ) × B δ ( x ) : δ > , ( t , x ) ∈ R d +1 } . Denote Q δ := Q δ (0 , Lemma 4.1.
Let f ∈ L ( R d +1 ) and vanish on R d +1 \ Q δ . Then − Z Q δ |G f ( t, x ) | dtdx ≤ N k f k L ∞ ( R d +1 ) , where N = N ( d, α ) .Proof. By H¨older’s inequality and Lemma 3.6, Z Q δ |G f ( t, x ) | dtdx ≤ (cid:18)Z Q δ |G f ( t, x ) | dtdx (cid:19) / | Q δ | / ≤ kG f k L ( R d +1 ) | Q δ | / ≤ N k f k L ( R d +1 ) | Q δ | / = N (cid:18)Z Q δ | f ( t, x ) | dtdx (cid:19) / | Q δ | / ≤ N k f k L ∞ ( R d +1 ) | Q δ | . The lemma is proved. (cid:3)
Denote K ( t, x ) = 1 t> ∆ q ( t, x ) . Due to Lemma 3.2(ii), we have K ( t, x ) = 1 t> ∂ t p ( t, x ). Furthermore the followingscaling properties hold (see (6.21) and (6.12) for detail): K ( t, x ) = 1 t> t − − αd K (1 , t − α/ x ) (4.2) ∂ t K ( t, x ) = 1 t> t − − αd ( ∂ t K )(1 , t − α/ x ) , (4.3) q ( L p )-THEORY FOR FRACTIONAL EQUATIONS 15 and ∂∂x i K ( t, x ) = 1 t> t − − α ( d +1)2 ∂∂x i K (1 , t − α/ x ) . (4.4) Lemma 4.2.
There exists a constant N = N ( α, d ) such that(i) for any t > a and η > , Z ta Z | y |≥ η | K ( t − s, y ) | dyds ≤ N ( t − a ) η − /α ; (4.5) (ii) for any t > τ > a , Z a −∞ Z R d | K ( t − s, y ) − K ( τ − s, y ) | dyds ≤ N t − ττ − a ; (4.6) (iii) for any t > a and x ∈ R d , Z a −∞ Z R d | K ( t − s, x + y ) − K ( t − s, y ) | dyds ≤ N | x | ( t − a ) − α/ . (4.7) Proof.
First observe that Z R d +1 (cid:16) | y | /α | K (1 , y ) | + | D x K (1 , y ) | + | ∂ t K (1 , y ) | (cid:17) dy < ∞ , which is an easy consequence of Lemma 3.2. By (4.2), it holds that Z ta Z | y |≥ η | K ( t − s, y ) | dyds = Z ta ( t − s ) − − αd Z | y |≥ η K (1 , ( t − s ) − α/ y ) dyds = Z ta ( t − s ) − Z | y |≥ η ( t − s ) − α/ K (1 , y ) dyds ≤ (cid:18)Z ta η − /α ds (cid:19) (cid:18)Z R d | y | /α K (1 , y ) dy (cid:19) ≤ N ( t − a ) η − /α . Hence (i) is proved.Next we prove (ii) and (iii) on the basis of the scaling property. By (4.3), (4.4),and the mean-value theorem, we have Z a −∞ Z R d | K ( t − s, y ) − K ( τ − s, y ) | dyds ≤ ( t − τ ) Z a −∞ Z R d Z | ∂ s K ( θt + (1 − θ ) τ − s, y ) | dθdyds ≤ ( t − τ ) Z (cid:18)Z a −∞ ( θt + (1 − θ ) τ − s ) − ds (cid:19) (cid:18)Z R d | ∂ s K (1 , y ) | dy (cid:19) dθ ≤ ( t − τ ) (cid:18)Z a −∞ ( τ − s ) − ds (cid:19) ≤ N t − ττ − a . Also, Z a −∞ Z R d | K ( t − s, y + x ) − K ( t − s, y ) | dyds ≤ | x | Z a −∞ Z R d Z | D y K ( t − s, y + θx ) | dθdyds ≤ | x | Z ∞ t − a Z R d | D y K ( s, y ) | dyds ≤ | x | Z ∞ t − a Z R d | t> t − − α D y K (1 , y )) | dyds ≤ N | x | ( t − a ) − α/ . The lemma is proved. (cid:3)
Lemma 4.3.
Let f ∈ L ( R d +1 ) and f = 0 on Q δ . Then − Z Q δ − Z Q δ |G f ( t, x ) − G f ( s, y ) | dsdydtdx ≤ N ( d, α ) k f k L ∞ ( R d +1 ) . (4.8) Proof.
First assume f ∈ C ∞ c ( R d +1 ). We claim that − Z Q δ |G f ( t, x ) − G f ( − δ /α , | dtdx ≤ N k f k L ∞ ( R d +1 ) . (4.9)Let ( t, x ) ∈ Q δ . Then |G f ( t, x ) − G f ( − δ /α , |≤ |G f ( t, x ) − G f ( t, | + |G f ( t, − G f ( − δ /α , | =: I + I . We consider I first. I = (cid:12)(cid:12)(cid:12)(cid:12)Z t −∞ Z R d (cid:0) K ( t − s, x − y ) − K ( t − s, − y ) (cid:1) f ( s, y ) dyds (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z t − (2 δ ) /α Z R d · · · dyds + Z − (2 δ ) /α −∞ Z R d · · · dyds (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ Z t − (2 δ ) /α Z R d | K ( t − s, y ) || f ( s, x − y ) | dyds + Z t − (2 δ ) /α Z R d | K ( t − s, y ) || f ( s, − y ) | dyds + Z − (2 δ ) /α −∞ Z R d | K ( t − s, x − y ) − K ( t − s, − y ) | | f ( s, y ) | dyds =: I + I + I . Note that if − (2 δ ) /α < s ≤ t < δ /α and | y | ≤ δ , then f ( s, x − y ) = 0 , f ( s, − y ) = 0 , (4.10) q ( L p )-THEORY FOR FRACTIONAL EQUATIONS 17 because | x − y | ≤ δ , | − y | ≤ δ , and f = 0 on Q δ . Hence by (4.5), I + I is lessthan or equal to N k f k L ∞ ( R d +1 ) Z t − (2 δ ) /α Z | y |≥ δ | K ( t − s, y ) | dyds ≤ N k f k L ∞ ( R d +1 ) ( t + (2 δ ) /α ) δ − /α ≤ N k f k L ∞ ( R d +1 ) . Also, by (4.7) we have I ≤ N k f k L ∞ ( R d +1 ) Z − (2 δ ) /α −∞ Z R d | K ( t − s, x − y ) − K ( t − s, − y ) | dyds ≤ N k f k L ∞ ( R d +1 ) | x | ( t + (2 δ ) /α ) − α/ ≤ N k f k L ∞ ( R d +1 ) . Next, we consider I . Note that I = (cid:12)(cid:12)(cid:12) G f ( t, − G f ( − δ /α , (cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z t −∞ Z R d K ( t − s, y ) f ( s, − y ) dyds − Z − δ /α −∞ Z R d K ( − δ /α − s, − y ) f ( s, − y ) dyds (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z t −∞ Z R d K ( t − s, y ) f ( s, − y ) dsdy − Z − δ /α −∞ Z R d K ( t − s, y ) f ( s, − y ) dyds (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z − δ /α −∞ Z R d h K ( t − s, y ) − K ( − δ /α − s, y ) i f ( s, − y ) dyds (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) =: I + I . Recall (4.10). Thus by (4.5), we have I ≤ Z t − δ /α Z R d | K ( t − s, y ) f ( s, − y ) | dyds ≤ k f k L ∞ ( R d +1 ) Z t − δ /α Z | y |≥ δ | K ( t − s, y ) | dyds ≤ N k f k L ∞ ( R d +1 ) ( t + δ /α ) δ − /α ≤ N k f k L ∞ ( R d +1 ) . Also, I ≤ Z − δ /α − (2 δ ) /α Z R d | K ( t − s, y ) − K ( − δ /α − s, y ) || f ( s, − y ) | dyds + k f k L ∞ ( R d +1 ) Z − (2 δ ) /α −∞ Z R d (cid:12)(cid:12)(cid:12) K ( t − s, y ) − K ( − δ /α − s, y ) (cid:12)(cid:12)(cid:12) dyds =: I + I . By (4.5) again, we have I ≤ k f k L ∞ ( R d +1 ) Z t − (2 δ ) /α Z | t |≥ δ | K ( t − s, t ) | dyds + k f k L ∞ ( R d +1 ) Z − δ /α − (2 δ ) /α Z | y |≥ δ | K ( − δ /α − s, y ) | dyds ≤ N k f k L ∞ ( R d +1 ) . On the other hand, by (4.6) we obtain I ≤ t + δ /α − δ /α + (2 δ ) /α k f k L ∞ ( R d +1 ) ≤ N k f k L ∞ ( R d +1 ) . Hence (4.9) is proved and this obviously implies (4.8) for f ∈ C ∞ c ( R d +1 ).Now we consider the general case, that is f ∈ L ( R d +1 ). We choose a sequenceof functions f n ∈ C ∞ c ( R d +1 ) such that f n = 0 on Q δ , G f n → G f ( a.e. ), and k f n k L ∞ ( R d +1 ) ≤ k f k L ∞ ( R d +1 ) . Then by Fatou’s lemma, − Z Q δ − Z Q δ |G f ( t, x ) − G f ( s, y ) | dtdsdxdy ≤ lim inf n →∞ − Z Q δ − Z Q δ |G f n ( t, x ) − G f n ( s, y ) | dtdsdxdy ≤ N lim inf n →∞ k f n k L ∞ ( R d +1 ) ≤ N k f k L ∞ ( R d +1 ) . The lemma is proved. (cid:3)
Theorem 4.4. (i) For any f ∈ L ( R d +1 ) ∩ L ∞ ( R d +1 ) , kG f k BMO ( R d +1 ) ≤ N ( d, α ) k f k L ∞ ( R d +1 ) . (4.11) (ii) For any p, q ∈ (1 , ∞ ) and f ∈ C ∞ c ( R d +1 ) , kG f k L q ( R ,L p ) ≤ N ( d, p, q, α ) k f k L q ( R ,L p ) . (4.12) Proof. (i) It suffices to prove that for each Q = Q δ ( t , x ) the following holds : − Z Q |G f ( t, x ) − ( G f ) Q | dtdx ≤ N k f k L ∞ ( R d +1 ) . (4.13)Due to the translation invariant property of the operator G , we may assume that( t , x ) = 0. Thus Q = Q δ = ( − δ /α , δ /α ) × B δ . Take ζ ∈ C ∞ c ( R d +1 ) such that ζ = 1 on Q δ and ζ = 0 outside Q δ . Then − Z Q |G f ( t, x ) − ( G f ) Q | dtdx ≤ − Z Q |G ( ζf ) − ( G ( ζf )) Q | dtdx + − Z Q |G ((1 − ζ ) f ) − ( G ((1 − ζ ) f )) Q | dtdx ≤ − Z Q |G ( ζf ) | dtdx + − Z Q − Z Q |G ((1 − ζ ) f )( t, x ) − G ((1 − ζ ) f )( s, y ) | dsdydtdx. Thus (4.13) comes from Lemma 4.1 and Lemma 4.3.(ii)
Step 1 . We prove (4.12) for the case q = p . First we assume q = p ≥
2. Fora measurable function h ( t, x ) on R d +1 , we define the maximal function M h ( t, x ) = sup Q ∈ Q − Z Q | f ( r, z ) | drdz, and the sharp function h ♯ ( t, x ) = sup Q ∈ Q − Z Q | f ( r, z ) − f Q | drdz. q ( L p )-THEORY FOR FRACTIONAL EQUATIONS 19 Then by the Fefferman-Stein theorem and the Hardy-Littlewood maximal theorem k h ♯ k L p ( R d +1 ) ∼ kM h k L p ( R d +1 ) ∼ k h k L p ( R d +1 ) . (4.14)Combining Lemma 3.6 with (4.14), we get for any f ∈ L ( R d +1 ) ∩ L ∞ ( R d +1 ), k ( G f ) ♯ k L ( R d +1 ) ≤ N k f k L ( R d +1 ) . Also, by (4.11) k ( G f ) ♯ k L ∞ ( R d +1 ) ≤ N k f k L ∞ ( R d +1 ) . Note that the map f → ( G f ) ♯ is subadditive since G is a linear operator. Hence bya version of the Marcinkiewicz interpolation theorem (see e.g. [17, Lemma 3.4]),for any p ∈ [2 , ∞ ) there exists a constant N such that k ( G f ) ♯ k L p ( R d +1 ) ≤ N k f k L p ( R d +1 ) , for all f ∈ L ( R d +1 ) ∩ L ∞ ( R d +1 ). Finally by the Fefferman-Stein theorem, we get kG f k L p ( R d +1 ) ≤ N ( d, α, p ) k f k L p ( R d +1 ) . Therefore (4.12) is proved for q = p ∈ [2 , ∞ ).For p ∈ (1 , f, g ∈ C ∞ c ( R d +1 ) and p ′ be theconjugate of p , i.e. p ′ = pp − ∈ (2 , ∞ ). By the integration by parts, the change ofvariable, and Fubini’s theorem, Z R d +1 g ( t, x ) G f ( t, x ) dxdt = Z R d +1 ∆ g ( t, x ) (cid:18)Z R d +1 t>s q ( t − s, x − y ) f ( s, y ) dsdy (cid:19) dxdt = Z R d +1 f ( − s, − y ) (cid:18)Z R d +1 s>t q ( s − t, y − x )∆ g ( − t, − x ) dxdt (cid:19) dsdy = Z R d +1 f ( − s, − y ) G ˜ g ( s, y ) dsdy (4.15)where ˜ g ( t, x ) = g ( − t, − x ). Then by H¨older’s inequality (also recall 2 < p ′ < ∞ ), (cid:12)(cid:12)(cid:12)(cid:12)Z R d +1 g ( t, x ) G f ( t, x ) dxdt (cid:12)(cid:12)(cid:12)(cid:12) ≤ N k f k L p kG ˜ g k L p ′ ≤ N k f k L p k g k L p ′ . Since g ∈ C ∞ c ( R d +1 ) is arbitrary, (4.11) is proved for q = p ∈ (1 ,
2) as well.
Step 2 . Now we prove (4.12) for general p, q ∈ (1 , ∞ ). For each ( t, s ) ∈ R , wedefine the operator K ( t, s ) as follows: K ( t, s ) f ( x ) := Z R d K ( t − s, x − y ) f ( y ) dy, f ∈ C ∞ c . Let p ∈ (1 , ∞ ) and ( t, s ) ∈ R . Then by the mean-value theorem, Lemma 3.2, and(4.2), kK ( t, s ) f k L p = (cid:13)(cid:13)(cid:13)(cid:13)Z R d K ( t − s, x − y ) f ( y ) dy (cid:13)(cid:13)(cid:13)(cid:13) L p ≤ k f k L p Z R d | K ( t − s, y ) | dy ≤ N ( d, α )( t − s ) − k f k L p . Hence the operator K ( t, s ) is uniquely extendible to L p for t = s . Denote Q := [ t , t + δ ) , Q ∗ := [ t − δ, t + 2 δ ) . Note that for t / ∈ Q ∗ and s, r ∈ Q , we have | s − r | ≤ δ, | t − ( t + δ ) | ≥ δ, and recall K ( t − s, x − y ) = 0 if t ≤ s . Thus by Lemma 3.2 and (4.3), it holds that kK ( t, s ) − K ( t, r ) k L ( L p ) = sup k f k Lp =1 (cid:13)(cid:13)(cid:13)(cid:13)Z R d { K ( t − s, x − y ) − K ( t − r, x − y ) } f ( y ) dy (cid:13)(cid:13)(cid:13)(cid:13) L p ≤ sup k f k Lp =1 k f k L p Z R d | K ( t − s, x ) − K ( t − r, x ) | dx = Z R d | K ( t − s, x ) − K ( t − r, x ) | dx ≤ N ( d, α ) | s − r | ( t − ( t + δ )) , where L ( L p ) denotes the bounded linear operator norm on L p . Therefore, Z R \ Q ∗ kK ( t, s ) − K ( t, r ) k L ( L p ) dt ≤ Z R \ Q ∗ | s − r | ( t − ( t + δ )) dt ≤ N | s − r | Z | t − ( t + δ ) |≥ δ t − ( t + δ )) dt ≤ N δ Z ∞ δ t − dt ≤ N. Furthermore, by following the proof of Theorem 1.1 of [20], one can easily checkthat for almost every t outside of the support of f ∈ C ∞ c ( R , L p ), G f ( t, x ) = Z ∞−∞ K ( t, s ) f ( s, x ) ds where G denotes the unique extension on L p ( R d +1 ) which is verified in Step 1.Hence by the Banach space-valued version of the Calder´on-Zygmund theorem [20,Theorem 4.1], our assertion is proved for 1 < q ≤ p .For the remaining case that 1 < p < q < ∞ , we use the duality argument.Define p ′ = p/ ( p −
1) and q ′ = q/ ( q − < q ′ < p ′ , by (4.15) and H¨older’sinequality, Z R d +1 g ( t, x ) G f ( t, x ) dxdt = Z R (cid:18)Z R d f ( − s, − y ) G ˜ g ( s, y ) dy (cid:19) ds ≤ Z R k f ( − s, · ) k L p kG ˜ g ( s, · ) k L p ′ ds ≤ N ( d, α, p ) k f k L q ( R ,L p ) k g k L q ′ ( R ,L p ′ ) for any f, g ∈ C ∞ c ( R d +1 ). Since g is arbitrary, we have kG f k L q ( R ,L p ) ≤ N ( d, α, p ) k f k L q ( R ,L p ) . The theorem is proved. (cid:3) Proof of Theorem 2.9
First we consider the solvability of the model equation.
Lemma 5.1.
Theorem 2.9 holds for the equation ∂ αt u = ∆ u + f with N = N ( p, q, α, T ) . q ( L p )-THEORY FOR FRACTIONAL EQUATIONS 21 Proof.
First we prove the uniqueness. Suppose that u ∈ H α, q,p, ( T ) and ∂ αt u = ∆ u .Due to Lemma 2.6 (iii), there exists a u n ∈ C ∞ c ( R d +1 ) such that u n → u in H α, q,p, ( T ). Due to Lemma 3.5 (i), u n ( t, x ) = Z t Z R d q ( t − s, x − y ) f n ( s, y ) dyds, where f n := ∂ αt u n − ∆ u n . Since f n → L q,p ( T ), we obtain the uniqueness fromLemma 2.6 (iv) and Theorem 4.4.For the existence and (2.12), first assume f ∈ C ∞ c ( R d +1+ ) and define u = G f .Then all the claims follows from Lemma 3.5(ii), Theorem 4.4, and (2.8). For general f one can consider an approximation f n → f , and above arguments show that G f n is a Cauchy sequence in H α, q,p, ( T ) and the limit becomes a solution of ∂ αt u = ∆ u + f . (cid:3) The following two results will be used later when we extend results proved forsmall T to the case when T is arbitrary. Lemma 5.2.
Let u ∈ H α, q,p, ( ˜ T ) and ˜ T ≤ T . Then there exists ˜ u ∈ H α, q,p, ( T ) suchthat ˜ u ( t ) = u ( t ) for all t ≤ ˜ T , and k ˜ u k H α, q,p ( T ) ≤ N k u k H α, q,p ( ˜ T ) , (5.1) where N is from Lemma 5.1 and is independent of ˜ T .Proof. Denote f = ∂ αt u , and let ˜ u ∈ H α, q,p, ( T ) be the solution of˜ u αt = ∆˜ u + ( f − ∆ u )1 t ≤ ˜ T , t ≤ T. Then by Lemma 5.1, k ˜ u k H α, q,p ( T ) ≤ N k ( f − ∆ u )1 t ≤ ˜ T k L q,p ( T ) ≤ N k u k H α, q,p ( ˜ T ) . Next observe that for t ≤ ˜ T , ∂ αt (˜ u − u ) = ∆˜ u + f − ∆ u − f = ∆(˜ u − u ) . It follows from Lemma 5.1 that ˜ u = u for t ≤ ˜ T . The lemma is proved. (cid:3) Lemma 5.3.
Let < ˜ T < T and u, ˜ u ∈ H α, q,p, ( T ) . Assume that u ( t ) = ˜ u ( t ) t ≤ ˜ T ( a.e. ) . (5.2) Then ¯ u ( t ) := u ( ˜ T + t ) − ˜ u ( ˜ T + t ) ∈ H α, q,p, ( T − ˜ T ) .Proof. We take u ε ,ε ,ε from the proof of Lemma 2.6 (iii) which is defined as u ε ,ε ,ε ( t, x ) = η ( t ) η ( ε x ) Z R d Z ∞ u ( s, y ) η ,ε ( t − s ) η ,ε ( x − y ) dsdy. As shown before, for any ε > k u − u ε ,ε ,ε k H α, q,p ( T ) ≤ ε if ε , ε , and ε are small enough. Hence we can take a sequence ( a n , b n , c n ) so that k u − u a n ,b n ,c n k H α, q,p ( T ) → n → ∞ and k ˜ u − ˜ u a n ,b n ,c n k H α, q,p ( T ) → n → ∞ . Observe that u a n ,b n ,c n ( t, x ) = ˜ u a n ,b n ,c n ( t, x ) ∀ t ≤ ( ˜ T + a n ) ∧ T due to (5.2) and the fact that η ,ε ∈ C ∞ c (( ε , ε )). Thus¯ u n ( t, x ) := u a n ,b n ,c n ( ˜ T + t, x ) − ˜ u a n ,b n ,c n ( ˜ T + t, x )is a defining sequence of ¯ u such that¯ u n (0 , x ) = 0 and ∂∂t ¯ u n (0 , x ) = 0 ∀ x ∈ R d , ∀ n ∈ N . Therefore the lemma is proved. (cid:3)
Proof of Theorem 2.9 . Step 1 . Denote ¯ f = b i u x i + cu + f ( u ). Then k ¯ f ( u ) − ¯ f ( v ) k L p ≤ N k u − v k H p + k f ( u ) − f ( v ) k L p ≤ ε k u − v k H p + K k u − v k L p + k f ( u ) − f ( v ) k L p . Thus considering ¯ f in place of f we may assume b i = c = 0. Step 2 . Let f = f be independent of u . Assume that a ij are independent of( t, x ). In this case obviously we may assume a ij = δ ij , and therefore the resultsfollow from Lemma 5.1. Step 3 . Let f = f . Suppose that Theorem 2.9 holds with some matrix ¯ a =(¯ a ij ( t, x )) in place ( a ij ( t, x )). We prove that there exists ε = ε ( N ) > ( t,x ) | a ij ( t, x ) − ¯ a ij ( t, x ) | ≤ ε , then Theorem 2.9 also holds for ( a ij ( t, x )) with 2 N in place of N . To prove this,due to the method of continuity, we only need to prove that (2.12) holds given thata solution u of (2.11) already exists. Note that u satisfies ∂ αt u = ¯ a ij u x i x j + ¯ f , ¯ f = f + ( a ij − ¯ a ij ) u x i x j . Hence by the assumption, k u k H α, q,p ( T ) ≤ N k ( a ij − ¯ a ij ) u x i x j k L q,p ( T ) + N k f k L q,p ( T ) ≤ N sup t,x | a ij − ¯ a ij | k u k H α, q,p ( T ) + N k f k L q,p ( T ) . Thus it is enough to take ε = (2 N ) − . Step 4 . Let a ij = a ij ( x ) depend only on x and f = f . In this case one canrepeat the classical perturbation arguments to prove the claims. Below we givea detail for the sake of the completeness. As before we only need to prove thatthere exists a constant N independent of u such that (2.12) holds given that u isa solution. By Steps 2 and 3, there exists ε > p, q, α, K , and T such that the theorem holds true if there exists any point x ∈ R d such thatsup x | a ij ( x ) − a ij ( x ) | ≤ ε . (5.3) q ( L p )-THEORY FOR FRACTIONAL EQUATIONS 23 Recall that a ij is uniformly continuous. Let δ < ε such that | a ij ( x ) − a ij ( y ) | ≤ ε / , if | x − y | < δ . Choose a partition of unity φ n , n = 1 , , · · · so that φ n = φ ( x − x n δ ) for some x n ∈ R d and φ ∈ C ∞ c ( B (0)) satisfying 0 ≤ φ ≤ φ = 1 for | x | ≤
1. Denote¯ φ n = φ ( x − x n δ ). Then ¯ φ n = 1 on the support of φ n , and u n = φ n u satisfies ∂ αt u n = a ijn ( u n ) x i x j + f n where a ijn = ¯ φ n a ij ( x ) + (1 − ¯ φ n ) a ij ( x n ) ,f n = f φ n − a ij u x i ( φ n ) x j − a ij u ( φ n ) x i x j . It is easy to check ( a ijn ) satisfies (2.9) and (5.3) with x n in place of x . By [19,Lemma 6.7] and Step 3, if p = q then for any t ≤ T , k u k q H , q,p ( t ) ∼ ∞ X n =1 k uφ n k q H , q,p ( t ) ≤ N ∞ X n =1 k f n k q L q,p ( t ) (5.4) ≤ N ∞ X n =1 h k u x ( φ n ) x k q L q,p ( t ) + k u ( φ n ) xx k q L q,p ( t ) + k f φ n k q L q,p ( t ) i ≤ N k u k q H , q,p ( t ) + N k f k q L q,p ( t ) . (5.5)The last inequality above is also from Lemma 6.7 of [19]. We emphasize thatequivalence relation in (5.4) and inequality (5.5) hold in general only if p = q oronly finite φ n are non-zero functions. Hence, if q = p then we take sufficientlylarge R, M > P Mn =1 φ n ( x ) = 1 on B R and vanishes for | x | ≥ R , andthe oscillation of a ij on the complement of B R/ is less then ε /
2. Denote φ =1 − P Mn =1 φ n . Then one can repeat the above calculations and use the relation k u k H p ≤ ε k u k H p + N k u k L p to conclude that for all t ≤ T k u k q H , q,p ( t ) ≤ N k u k q L q,p ( t ) + N k f k q L q,p ( t ) ≤ N Z t (cid:20)Z s ( s − r ) − α ( k ∂ αt u ( r ) k H p + k f ( r ) k L p ) dr (cid:21) q ds + N k f k q L q,p ( t ) ≤ N Z t ( t − s ) − α k u k q H , q,p ( s ) ds + N k f k q L q,p ( T ) , where Theorem 2.6 (iv) is used in the second inequality. Hence by a version ofGronwall’s lemma (see e.g. [45, Corollary 2]) we get k u k q H , q,p ( T ) ≤ N k f k q L q,p ( T ) . (5.6)From equation (2.11), we easily conclude that k ∂ αt u k L q,p ( T ) ≤ N k u k H , q,p ( T ) + k f k L q,p ( T ) , and therefore (5.6) certainly leads to the a priori estimate. Step 5 . Let f = f , and a ij be uniformly continuous in ( t, x ). As before, weonly need to prove the a priori estimate (2.12).Let N be the constant from Step 4 so that a priori estimate (2.12) holds when-ever a ij are independent of t . Take ε from Step 3 corresponding to this N . Wewill apply Step 3 with ¯ a ij = a ij ( t , x ) for some t . Take κ > | a ij ( t, x ) − a ij ( s, x ) | ≤ ε / , if | t − s | ≤ κ. Also take an integer N so that T /N ≤ κ , and denote ˜ T i = iT /N . By Steps 3 and4 applied with ¯ a ij ( t, x ) = a ij (0 , x ), (2.12) holds with T and 2 N in place of T and N respectively. Now we use use the induction. Suppose that the a priori estimate(2.12) holds for ˜ T k < T with N independent of u . This constant N may dependon k . Take ˜ u from Lemma 5.2 corresponding to ˜ T = ˜ T k . Denote¯ u ( t, x ) = ( u − ˜ u )( ˜ T + t, x ) , ¯ f ( t, x ) = f ( ˜ T + t, x ) . Then one can easily check that ¯ u satisfies ∂ αt ¯ u = a ij ( ˜ T + t, x )¯ u x i x j + ¯ f + ( a ij ( ˜ T + t, x ) − δ ij )˜ u x i x j ( ˜ T + t, x ) , t ≤ T − ˜ T .
Due to Lemma 5.3, ¯ u is contained in H α,nq,p, ( T ). Thus by the result of Steps 3 and4 with ¯ a ij = a ij ( ˜ T k , x ), we have k ¯ u k q H α, q,p ( T ) ≤ (2 N ) q Z ˜ T k +1 ˜ T k k f + ( a ij − δ ij )˜ u x i x j k qL p dt ≤ N k ˜ u k q H , q,p ( T ) + N k f k q L q,p ( T ) ≤ N k u k q H α, q,p ( ˜ T k ) + k f k q L q,p ( T ) ≤ N k f k q L q,p ( T ) , where the third inequality is due to (5.1) and the last inequality is from the as-sumption. Hence k u k q H α, q,p ( ˜ T k +1 ) ≤ k ˜ u k q H α, q,p ( ˜ T k +1 ) + k ¯ u k q H α, q,p ( ˜ T ) ≤ N k f k q L q,p ( T ) . As the induction goes through, the a priori estimate (2.12) is proved. We emphasizethat for each k the constant N varies, however for each ˜ T k we use the result inStep 4 and therefore the choice of ˜ T k +1 (or the difference | ˜ T k +1 − ˜ T k | ) does notdepend on k , and therefore we can reach up to T by finite such steps. Step 6 . Let f = f and a ij = P ℓk =1 a ijk ( t, x ) I ( T k − ,T k ] ( t ). If ℓ = 1, (2.12) comesdirectly from Step 5. If ℓ >
1, we use the induction argument used in Step 5. Theonly difference is that to estimate the solution on [ T k , T k +1 ) we use the result ofStep 5, in place of the result in Step 4. Step 7 . The general non-linear case. We modify the proof of Theorem 5.1 of[19]. For each u ∈ H α, q,p, ( T ) consider the equation ∂ αt v = a ij + f ( u ) , t ≤ T. By the above results, this equation has a unique solution v ∈ H α, q,p, ( T ). By denoting v = R u we can define an operator R : H α, q,p, ( T ) → H α, q,p, ( T ). By the results forthe linear case, for each t ≤ T , kR u − R v k q H α, q,p ( t ) ≤ N k f ( u ) − f ( v ) k q L q,p ( t ) ≤ N ε q k u − v k q H , q,p ( t ) + N K qε k u − v k q L q,p ( t ) ≤ N ε q k u − v k q H α, q,p ( t ) + N Z t ( t − s ) − α k u − v k q H α, q,p ( s ) ds, q ( L p )-THEORY FOR FRACTIONAL EQUATIONS 25 where N depends also on ε , and Theorem 2.6 (iv) is used in the last inequality.Next, we fix ε so that θ := N ε q < /
4. Then repeating the above inequality andusing the identity Z t ( t − s ) − α Z s ( s − s ) − α · · · Z s n − ( s n − − s n ) − α ds n · · · ds = { Γ( α ) } n Γ( nα + 1) t nα , we get kR m u − R m v k q H α, q,p ( T ) ≤ m X k =0 (cid:18) mk (cid:19) θ m − k ( T α N ) k { Γ( α ) } k Γ( kα + 1) k u − v k q H α, q,p ( T ) ≤ m θ m (cid:20) max k (cid:18) ( θ − T α N Γ( α )) k Γ( kα + 1) (cid:19)(cid:21) k u − v k q H α, q,p ( T ) ≤ m N k u − v k q H α, q,p ( T ) . For the second inequality above we use P mk =0 (cid:18) mk (cid:19) = 2 m . It follows that if m is sufficiently large then R m is a contraction in H α, q,p, ( T ), and this yields all theclaims. The theorem is proved. (cid:3) Kernels p and q The kernel p ( t, x ) . In this subsection, we prove Lemma 3.1(i). In other words,we introduce a kernel p ( t, x ) which is integrable with respect to x and satisfies F{ p ( t, · ) } ( ξ ) = E α ( − t α | ξ | ) . (6.1)Let Γ( z ) denote the gamma function which can be defined (see [2, Section 1.1])for z ∈ C \ { , − , − , . . . } asΓ( z ) = lim n →∞ n ! n z z ( z + 1) · · · ( z + n ) . Note that Γ( z ) is a meromorphic function with simple poles at the nonpositiveintegers. From the definition, for z ∈ C \ { , − , − , . . . } , z Γ( z ) = Γ( z + 1) . (6.2)By (6.2), one can easily check that for k = 0 , , , . . . ,Res z = − k Γ( z ) = lim z →− k ( z + k )Γ( z ) = ( − k k ! , (6.3)where Res z = − k Γ( z ) denotes the residue of Γ( z ) at z = − k . It is also well-known(see. e.g. [2, Theorem 1.1.4]) that if ℜ [ z ] > ℜ [ ω ] > z ) = Z ∞ t z − e − t dt, Z t z − (1 − t ) ω − dt = Γ( z )Γ( ω )Γ( z + ω ) . (6.4)Using Stirling’s approximation (see [2, Corollary 1.4.3] or [16, (1.2.3)])Γ( z ) ∼ √ πe ( z − ) log z e − z , | z | → ∞ , (6.5) one can easily show that for fixed a ∈ R , ([16, (1.2.2)]) | Γ( a + i b ) | ∼ √ π | b | a − e − a − π | b | , | b | → ∞ . (6.6)Let m, n, µ, ν be fixed integers satisfying 0 ≤ m ≤ µ , 0 ≤ n ≤ ν . Assume that thecomplex parameters c , . . . , c ν , d , . . . , d µ and positive real parameters γ , . . . , γ ν , δ , . . . , δ µ are given so that P ∩ P = ∅ where P := (cid:26) − d j + kδ j ∈ C : j ∈ { , . . . , m } , k = 0 , , , . . . (cid:27) P := (cid:26) − c j + kγ j ∈ C : j ∈ { , . . . , n } , k = 0 , , , . . . (cid:27) . If either m = 0 or n = 0, then by the definition P ∩ P = ∅ . For the aboveparameters, the Fox H-function H ( r ) ( r >
0) is defined as H ( r ) := H mnνµ (cid:20) r (cid:12)(cid:12)(cid:12) ( c , γ ) · · · ( c n , γ n ) ( c n +1 , γ n +1 ) · · · ( c ν , γ ν )( d , δ ) · · · ( d m , δ m ) ( d m +1 , δ m +1 ) · · · ( d µ , δ µ ) . (cid:21) := 12 π i Z L Q mj =1 Γ( d j + δ j z ) Q nj =1 Γ(1 − c j − γ j z ) Q µj = m +1 Γ(1 − d j − δ j z ) Q νj = n +1 Γ( c j + γ j z ) r − z dz, (6.7)where the contour L is chosen appropriately depending on the parameters. Somespecial cases needed in our setting are specified below. In this article, we addition-ally assume that parameters c , . . . , c ν and d , . . . , d µ are real and µ X j =1 δ j − ν X i =1 γ i > , n X i =1 γ i − ν X i = n +1 γ i + m X j =1 δ j − µ X j = m +1 δ j > . (6.8)Under (6.8), we can choose the contour L of two different types. Hankel contour L h is a loop starting at the point −∞ + i ρ and ending at the point −∞ + i ρ where ρ < < ρ , which encircles all the poles of P once in the positive directionbut none of the poles of P . Bromwich contour L v is a vertical contour, which gofrom γ − i ∞ to γ + i ∞ where γ ∈ R and leaves all the poles of P to the rightand all poles of P to the left. Braaksma [5] showed that the contour integral (6.7)makes sense along L h and L v for r ∈ (0 , ∞ ), and two integrals along L h and L v coincide (see [16, Section 1.2]). Furthermore, if P µj =1 δ j − P νi =1 γ i > H ( r )is an analytic function on (0 , ∞ ) and can be represented (see [16, Theorem 1.2]) as H ( r ) = m X i =1 ∞ X k =0 Res z = ˆ d ik " Q mj =1 Γ( d j + δ j z ) Q nj =1 Γ(1 − c j − γ j z ) Q µj = m +1 Γ(1 − d j − δ j z ) Q νj = n +1 Γ( c j + γ j z ) r − z , (6.9)where ˆ d ik := − ( d i + k ) /δ i ∈ P are poles of the integrand in the contour integral, i ∈ { , . . . , m } , and k = 0 , , , · · · . Note that on the negative real half-axis theMittag-Leffler function E α ( z ) = ∞ X k =0 z k Γ( αk + 1) , α > , can be written as E α ( − x ) = H (cid:20) x (cid:12)(cid:12)(cid:12) (0 , ,
1) (0 , α ) (cid:21) ( x > . (6.10) q ( L p )-THEORY FOR FRACTIONAL EQUATIONS 27 Indeed, by (6.9) and (6.3)H (cid:20) x (cid:12)(cid:12)(cid:12) (0 , ,
1) (0 , α ) (cid:21) = ∞ X k =0 Res z = − k (cid:20) Γ( z )Γ(1 − z )Γ(1 − αz ) x − z (cid:21) = ∞ X k =0 ( − k Γ(1 + k ) k !Γ( αk + 1) x k = ∞ X k =0 ( − x ) k Γ( αk + 1) = E α ( − x ) . We introduce some notion to introduce the asymptotic behavior of H ( r ). Let j s (1 ≤ j s ≤ m ) be the number such that ρ s := d j s δ j s = min (cid:20) d j δ j (cid:21) (min ∅ := ∞ )where the minimum are taken over all d j δ j so that ˆ d j is a simple pole. Similarly let j c (1 ≤ j c ≤ m ) be the number such that ρ c := d j c δ j c = min (cid:20) d j δ j (cid:21) where the minimum are taken over all d j δ j so that ˆ d j is a pole with order n c ≥ n c denotes the smallest number of the orders for non-simple poles.The following result can be proved on the basis of (6.9). See [16, Corollary1.12.1] for the proof. Theorem 6.1. (i) If ρ s < ρ c , then for r ≤ | H ( r ) | ≤ N r ρ s ; (ii) if ρ s ≥ ρ c , then for r ≤ | H ( r ) | ≤ N r ρ c | ln r | n c − . An upper bound of H ( r ) on [1 , ∞ ) is also well-known if n = 0 and m = µ in(6.7). See [16, Corollary 1.10.2] and [16, (2.2.2)]. Theorem 6.2.
Suppose that n = 0 and m = µ in (6.7) . Then for r ≥ , | H ( r ) | ≤ N r (Λ+1 / ω − exp {− ωη − /ω r /ω } , (6.11) where Λ := µ X j =1 d j − ν X i =1 c i + ν − µ , ω := µ X j =1 δ j − ν X i =1 γ i , η := µ Y j =1 δ δ j j ν Y i =1 γ − γ i i . Theorem 6.3.
Suppose that n = 0 amd m = µ in (6.7) . Then ddr H ( r ) = − r − H µ +1 0 ν +1 µ +1 (cid:20) r (cid:12)(cid:12)(cid:12) ( c , γ ) · · · ( c ν , γ ν ) (0 , d , δ ) · · · ( d µ , δ µ ) (1 , (cid:21) . (6.12)Now we define p ( t, x ) so that (6.1) holds. Fix α ∈ (0 ,
2) and let p ( t, x ) := π − d | x | − d H (cid:20) t − α | x | (cid:12)(cid:12)(cid:12) (1 , α )( d ,
1) (1 , (cid:21) = π − d | x | − d π i Z L v Γ( d + z )Γ(1 + z )Γ(1 + αz ) (cid:18) t − α | x | (cid:19) − z dz. (6.13) One can easily check that (6.8) is satisfied. Hence we can take the Bromwichcontour L = L v , which runs along from − γ − i ∞ to − γ + i ∞ , i.e. L v := { z ∈ C : ℜ [ z ] = − γ } where0 < γ < min { , d − , α } if d ≥ < γ < min { , α } if d = 1 . Note that min { ρ s , ρ c } = ρ s = 1 if d ≥ ρ c = 1 , n c = 2 if d = 2 ρ s = if d = 1.By Theorem 6.1 and Theorem 6.2, p ( t, · ) ∈ L ( R d ) , ∀ t > . For the asymptotic behavior of the integrand in (6.13) we use Stirling’s approxima-tion. Write z = − γ + i τ , τ ∈ ( −∞ , ∞ ). Then by (6.6), for ρ ∈ (0 , ∞ ), t ∈ (0 , ∞ ),and large | τ | , (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Γ( d + z )Γ(1 + z )Γ(1 + αz ) (cid:18) t − α ρ (cid:19) − z (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ N (cid:18) t − α ρ (cid:19) γ | τ | c e − c | τ | , (6.14)where c := − γ (2 − α ) + d − , c := π − α ) . Therefore, sup z ∈ L (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Γ( d + z )Γ(1 + z )Γ(1 + αz ) (cid:18) t − α ρ (cid:19) − z (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ N (cid:18) t − α ρ (cid:19) γ , (6.15)because for fixed a / ∈ { , − , − , − , . . . } the mapping b Γ( a + i b ) is a continuousfunction and does not vanish on R . Hence (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z L Γ( d + z )Γ(1 + z )Γ(1 + αz ) (cid:18) t − α ρ (cid:19) − z dz (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ Z ∞−∞ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Γ( d − γ + i τ )Γ(1 − γ + i τ ))Γ(1 − αγ + i ατ ) (cid:18) t − α ρ (cid:19) γ − i τ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) dτ ≤ N t − αγ ρ γ ( N + Z | τ |≥ | τ | c e − c | τ | dτ ) ≤ N t − αγ ρ γ . (6.16)We remark that (6.14), (6.15), and (6.16) hold for any 0 < γ < min { , d , α } .Now we prove (6.1). First assume d ≥
2. By the formula for the Fourier trans-form of a radial function (see [42, Theorem IV.3.3]), we have F{ p ( t, · ) } ( ξ ) = 2 d/ | ξ | d − Z ∞ ρ − d H (cid:20) t − α ρ (cid:12)(cid:12)(cid:12) (1 , α )( d ,
1) (1 , (cid:21) J d − ( | ξ | ρ ) dρ, where J d − is the Bessel function of the first kind of order d −
1, i.e. for r ∈ [0 , ∞ ), J d − ( r ) = ∞ X k =0 ( − k k !Γ( k + d ) (cid:16) r (cid:17) k − d/ . q ( L p )-THEORY FOR FRACTIONAL EQUATIONS 29 It is well-known (e.g. [16, (2.6.3)]) that if m > − J m ( t ) = ( O ( t m ) , t → O ( t − / ) , t → ∞ . Thus Z ∞ | ρ − d +2 γ J d − ( | ξ | ρ ) | dρ ≤ Z ρ γ − dρ + Z ∞ ρ − d +2 γ − dρ < ∞ (6.17)since 0 < γ < min { , d − , α } . By the definition of the Fox H-function, Z ∞ ρ − d H (cid:20) t − α ρ (cid:12)(cid:12)(cid:12) (1 , α )( d ,
1) (1 , (cid:21) J d − ( | ξ | ρ ) dρ = 12 π i Z ∞ ρ − d "Z L Γ( d + z )Γ(1 + z )Γ(1 + αz ) (cid:18) t − α ρ (cid:19) − z dz J d − ( | ξ | ρ ) dρ. (6.18)Combining (6.16) and (6.17), we have Z ∞ Z L (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ρ − d Γ( d + z )Γ(1 + z )Γ(1 + αz ) (cid:18) t − α ρ (cid:19) − z J d − ( | ξ | ρ ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) | dz | dρ < ∞ . Thus we can apply Fubini’s theorem to (6.18). Furthermore, since − d < ℜ [ − d − z ] < − ∀ z ∈ L, by using the formula [16, (2.6.4)] Z ∞ ρ − d − z J d − ( | ξ | ρ ) dρ = 2 − d − z | ξ | d − z Γ( − z )Γ( d + z ) , we get Z ∞ ρ − d H (cid:20) t − α ρ (cid:12)(cid:12)(cid:12) (1 , α )( d ,
1) (1 , (cid:21) J d − ( | ξ | ρ ) dρ = 12 π i Z L (cid:20)Z ∞ ρ − d − z J d − ( | ξ | ρ ) dρ (cid:21) Γ( d + z )Γ(1 + z )Γ(1 + αz ) (cid:18) t − α (cid:19) − z dz = | ξ | d − d/ π i Z L Γ( d + z )Γ(1 + z )Γ( − z )Γ(1 + αz )Γ( d + z ) ( t − α | ξ | − ) − z dz = | ξ | d − d/ π i Z − L Γ(1 − z )Γ( z )Γ(1 − αz ) ( t α | ξ | ) − z dz = | ξ | d − d/ H (cid:20) t α | ξ | (cid:12)(cid:12)(cid:12) (0 , ,
1) (0 , α ) (cid:21) = | ξ | d − d/ E α ( − t α | ξ | )where the last equality is due to (6.10). Therefore (6.1) is proved for d ≥ d = 1. By (3.10), L (cid:2) E α ( − t α | ξ | ) (cid:3) )( s ) = Z ∞ e − st E α ( − t α | ξ | ) dt = s α − s α + | ξ | . Thus it suffices to prove L [ F { p ( t, · ) } ] ( s ) = Z ∞ e − st F { p ( t, · ) } ( ξ ) dt = s α − s α + | ξ | . (6.19) By Theorems 6.1 and 6.2 it holds that for each s > Z ∞ Z R e − st | p ( t, x ) | dxdt < ∞ . Hence by Fubini’s theorem, L [ F { p ( t, · ) } ] ( s ) = F {L [ p ( · , x )] ( s ) } . Due to (6.4) and(6.16), L [ p ( · , x )] ( s ) = Z ∞ e − st p ( t, x ) dt = π − | x | − Z ∞ e − st " π i Z L Γ( + z )Γ(1 + z )Γ(1 + αz ) (cid:18) t − α | x | (cid:19) − z dz dt = π − | x | − π i Z L Γ( + z )Γ(1 + z )Γ(1 + αz ) (cid:18) | x | (cid:19) − z (cid:20)Z ∞ e − st t αz dt (cid:21) dz = π − | x | − s − π i Z L Γ( 12 + z )Γ(1 + z ) (cid:18) s α | x | (cid:19) − z dz. Furthermore by [16, (2.9.19)],12 π i Z L Γ( 12 + z )Γ(1 + z ) (cid:18) s α | x | (cid:19) − z dz = H (cid:20) s α | x | (cid:12)(cid:12)(cid:12) ( ,
1) (1 , (cid:21) = 2 (cid:18) s α/ | x | (cid:19) / K / ( s α/ | x | ) , where K η ( z ) is called the modified Bessel function of the second kind or the Mac-donald function and it satisfies (see [1, 9.7.2]) K / ( z ) = r π z − / e − z , | arg z | < π . Hence we obtain L [ p ( · , x )] ( s ) = s α/ − {− s α/ | x |} , which obviously implies that F {L [ p ( · , x )] ( s ) } ( ξ ) = s α/ − Z ∞−∞ e − i xξ exp {− s α/ | x |} dx = s α/ − · s α/ s α + | ξ | = s α − s α + | ξ | . Therefore (6.19) is proved, and (6.1) holds.6.2.
Representation of q ( t, x ) and K ( t, x ) . Let α ∈ (0 ,
2) and recall p ( t, x ) := π − d | x | − d H (cid:20) t − α | x | (cid:12)(cid:12)(cid:12) (1 , α )( d ,
1) (1 , (cid:21) . By Theorems 6.2 and (6.12), if t = 0 and x = 0 then p ( t, x ) is differentiable in t and lim t → p ( t, x ) = 0. Thus we can define q ( t, x ) := ( I α − t p ( t, x ) , α ∈ (1 , D − αt p ( t, x ) , α ∈ (0 , , K ( t, x ) := ∂∂t p ( t, x ) . It has also been called the modified Bessel function of the third kind. q ( L p )-THEORY FOR FRACTIONAL EQUATIONS 31 In this subsection, we derive the following representations: q ( t, x ) = π − d | x | − d t α − H (cid:20) t − α | x | (cid:12)(cid:12)(cid:12) ( α, α )( d ,
1) (1 , (cid:21) (6.20)and K ( t, x ) = π − d | x | − d t − H (cid:20) t − α | x | (0 , α )( d ,
1) (1 , (cid:21) . (6.21)We consider the Bromwich contour L v := { z ∈ C : ℜ [ z ] = − γ } , < γ < min { , d , α } . Let β >
0. By (6.16), Z t ( t − s ) β − Z L (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Γ( d + z )Γ(1 + z )Γ(1 + αz ) (cid:18) s − α | x | (cid:19) − z (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) | dz | ! ds ≤ N | x | γ Z t ( t − s ) β − ( t − s ) − αγ ds < ∞ . Thus by Fubini’s theorem and (6.4), I βt p ( t, x ) = 1Γ( β ) Z t ( t − s ) β − p ( s, x ) ds = π − d | x | − d Γ( β ) Z t ( t − s ) β − " π i Z L Γ( d + z )Γ(1 + z )Γ(1 + αz ) (cid:18) s − α | x | (cid:19) − z dz ds = π − d | x | − d π i Z L Γ( d + z )Γ(1 + z )Γ(1 + αz ) (cid:20) β ) Z t ( t − s ) β − s αz ds (cid:21) (cid:18) | x | (cid:19) − z dz = π − d | x | − d t β π i Z L Γ( d + z )Γ(1 + z )Γ(1 + β + αz ) (cid:18) t − α | x | (cid:19) − z dz (6.22)= π − d | x | − d t β H (cid:20) t − α | x | (cid:12)(cid:12)(cid:12) (1 + β, α )( d ,
1) (1 , (cid:21) . Therefore, (6.20) is proved for α ∈ (1 , t . Write z = − γ + i τ . Following themethod used to prove (6.14) and (6.15), for any α ∈ (0 ,
2) and β ∈ [0 ,
2) we have (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Γ( d + z )Γ(1 + z )Γ(1 + β + αz ) (cid:18) | x | (cid:19) − z (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12)(cid:12) ddt t αz (cid:12)(cid:12)(cid:12)(cid:12) ≤ N ( d, γ, β ) t − (cid:18) t − α | x | (cid:19) γ h ( γ + | τ | ) / (1 | τ |≥ | τ | c − β e − c | τ | + 1 | τ |≤ ) i . (6.23) Hence the time derivative of the integrand in (6.22) is integrable in z uniformly ina neighborhood of t >
0, and thus by the dominated convergence theorem, ddt H (cid:20) t − α | x | (cid:12)(cid:12)(cid:12) (1 + α, α )( d ,
1) (1 , (cid:21) = 12 π i ddt (Z L Γ( d + z )Γ(1 + z )Γ(1 + α + αz ) (cid:18) t − α | x | (cid:19) − z dz ) = t − π i Z L Γ( d + z )Γ(1 + z )Γ(1 + α + αz ) αz (cid:18) t − α | x | (cid:19) − z dz. Using the relation αz Γ( α + αz + 1) = 1Γ( α + αz ) − α Γ( α + αz + 1) , we obtain ddt H (cid:20) t − α | x | (cid:12)(cid:12)(cid:12) (1 + α, α )( d ,
1) (1 , (cid:21) = t − π i Z L ( Γ( d + z )Γ(1 + z )Γ( α + αz ) − α Γ( d + z )Γ(1 + z )Γ( α + αz + 1) ) (cid:18) t − α | x | (cid:19) − z dz = t − (cid:26) H (cid:20) t − α | x | (cid:12)(cid:12)(cid:12) ( α, α )( d ,
1) (1 , (cid:21) − α H (cid:20) t − α | x | (cid:12)(cid:12)(cid:12) (1 + α, α )( d ,
1) (1 , (cid:21)(cid:27) . Thus, D − αt p ( t, x ) = ddt I αt p ( t, x )= ddt (cid:26) π − d | x | − d t α H (cid:20) t − α | x | (cid:12)(cid:12)(cid:12) (1 + α, α )( d ,
1) (1 , (cid:21)(cid:27) = π − d/ | x | − d (cid:26) αt α − H (cid:20) t − α | x | (cid:12)(cid:12)(cid:12) (1 + α, α )( d ,
1) (1 , (cid:21) + t α ddt H (cid:20) t − α | x | (cid:12)(cid:12)(cid:12) (1 + α, α )( d ,
1) (1 , (cid:21) (cid:27) = π − d/ | x | − d t α − H (cid:20) t − α | x | (cid:12)(cid:12)(cid:12) ( α, α )( d ,
1) (1 , (cid:21) . Similarly, using the relation αz Γ( αz + 1) = 1Γ( αz ) , we have K ( t, x ) = ddt H (cid:20) t − α | x | (cid:12)(cid:12)(cid:12) (1 , α )( d ,
1) (1 , (cid:21) = π − d | x | − d t − H (cid:20) t − α | x | (0 , α )( d ,
1) (1 , (cid:21) . Therefore (6.20) and (6.21) are proved. q ( L p )-THEORY FOR FRACTIONAL EQUATIONS 33 Estimates of p ( t, x ) and q ( t, x ) . In this subsection we prove Lemma 3.1(ii)and Lemma 3.2. Since the case α = 1 is easier, we assume α = 1.By (6.7) and (6.9) with n = 0 and m = µ ,H µ νµ (cid:20) r (cid:12)(cid:12)(cid:12) ( c , γ ) · · · ( c ν , γ ν )( d , δ ) · · · ( d µ , δ µ ) (cid:21) = m X i =1 ∞ X k =0 Res z = ˆ d ik " Q µj =1 Γ( d j + δ j z ) Q νj =1 Γ( c j + γ j z ) r − z = m X i =1 ∞ X k =0 lim z → ˆ d ik ((cid:18) ddz (cid:19) n ik − ( z − ˆ d ik ) n ik ( n ik − Q µj =1 Γ( d j + δ j z ) Q νj =1 Γ( c j + γ j z ) r − z !) , where ˆ d ik = − ( d i + k ) /δ i ∈ P is a pole of the integrand in the contour integraland n ik is its order for i = 1 , . . . , µ and k = 0 , , , · · · .Let R := t − α | x | , and denoteH pk,l ( R ) := H k + l k + l k + l (cid:20) R (cid:12)(cid:12)(cid:12) (1 , α ) (0 , · · · (0 , d ,
1) (1 , · · · (1 ,
1) (1 , (cid:21) and H qk,l ( R ) := H k + l k + l k + l (cid:20) R (cid:12)(cid:12)(cid:12) ( α, α ) (0 , · · · (0 , d ,
1) (1 , · · · (1 ,
1) (1 , (cid:21) , where k, l = 0 , , , · · · . Then by (6.12), (cid:12)(cid:12)(cid:12) D x i (cid:16) | x | − d H pk, ( R ) (cid:17)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12) − dx i | x | − d − H pk, ( R ) − x i | x | − d − H pk, ( R ) (cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12) − x i | x | − d − [ d H pk, ( R ) + 2H pk, ( R )] (cid:12)(cid:12)(cid:12) ≤ N | x | − d − (cid:12)(cid:12)(cid:12) d H pk, ( R ) + 2H pk, ( R ) (cid:12)(cid:12)(cid:12) , and (cid:12)(cid:12)(cid:12) D x j D x i (cid:16) | x | − d H pk, ( R ) (cid:17)(cid:12)(cid:12)(cid:12) = | D x j (cid:16) − x i | x | − d − [ d H pk, ( R ) + 2H pk, ( R )] (cid:17) | = | ( − δ ij | x | − d − + ( d + 2) x i x j | x | − d − )[ d H pk, ( R ) + 2H pk, ( R )]+ 2 x i x j | x | − d − [ d H pk, ( R ) + 2H pk, ( R )] |≤ N | x | − d − X l =1 | d H pk,l − ( R ) + 2H pk,l ( R ) | . Inductively, for any m = 3 , , · · · and k = 0 , , · · · , we have (cid:12)(cid:12)(cid:12) D mx (cid:16) | x | − d H pk, ( R ) (cid:17)(cid:12)(cid:12)(cid:12) ≤ N | x | − d − m m X l =1 (cid:12)(cid:12)(cid:12) d H pk,l − ( R ) + 2H pk,l ( R ) (cid:12)(cid:12)(cid:12) . Hence | ∂ nt D mx p ( t, x ) | = (cid:12)(cid:12)(cid:12)(cid:12) D mx (cid:18) | x | − d ∂ nt H (cid:20) R (cid:12)(cid:12)(cid:12) (1 , α )( d ,
1) (1 , (cid:21)(cid:19)(cid:12)(cid:12)(cid:12)(cid:12) ≤ N t − n n X k =1 (cid:12)(cid:12)(cid:12) D mx (cid:16) | x | − d H pk, ( R ) (cid:17)(cid:12)(cid:12)(cid:12) ≤ N | x | − d − m t − n n X k =1 m X l =1 (cid:12)(cid:12)(cid:12) d H pk,l − ( R ) + 2H pk,l ( R ) (cid:12)(cid:12)(cid:12) . (6.24)Similarly, | D mx (cid:16) | x | − d H qk, ( R ) (cid:17) | ≤ N | x | − d − m m X l =1 (cid:12)(cid:12)(cid:12) d H qk,l − ( R ) + 2H qk,l ( R ) (cid:12)(cid:12)(cid:12) and | ∂ nt D mx q ( t, x ) | ≤ N t − n + α − n X k =1 (cid:12)(cid:12)(cid:12) D mx (cid:16) | x | − d H qk, ( R ) (cid:17)(cid:12)(cid:12)(cid:12) ≤ N | x | − d − m t − n + α − n X k =0 m X l =1 (cid:12)(cid:12)(cid:12) d H qk,l − ( R ) + 2H qk,l ( R ) (cid:12)(cid:12)(cid:12) . (6.25)Now we prove (3.2) and (3.4). If R ≥
1, by Theorem 6.2H pk,l ( R ) ≤ N R ( d + k + l ) / (2 − α ) exp {− σR − α } and H qk,l ( R ) ≤ N R ( d + k + l +1 − α ) / (2 − α ) exp {− σR − α } , where σ = (2 − α ) α α/ (2 − α ) . Hence by (6.24) | ∂ nt D mx p ( t, x ) | ≤ N | x | − d − m t − n n X k =1 m X l =0 R ( d + k + l ) / (2 − α ) exp {− σR − α }≤ N ( t − α/ ) d + m t − n exp {− ( σ/ t − α − α | x | − α }≤ N t − α ( d + m )2 − n exp {− ( σ/ t − α − α | x | − α } . Similarly, by (6.25) | ∂ nt D mx q ( t, x ) | ≤ N t − α ( d + m )2 − n + α − exp {− ( σ/ t − α − α | x | − α } . Thus (3.2) and (3.4) are proved.To prove (3.3), we recall (6.2). For k, l = 0 , , . . . , denoteΘ pk,l ( z ) = Γ( d + z ) { Γ(1 + z ) } k + l +1 Γ(1 + αz ) { Γ( z ) } k + l = Γ( d + z )Γ(1 + z )Γ(1 + αz ) z k + l . q ( L p )-THEORY FOR FRACTIONAL EQUATIONS 35 First we assume that d is an odd number. Then Θ pk,l has simple poles at d j = − − j and d j = − d − j for j = 0 , , , · · · . Due to (6.3) and (6.9), for R ≤ pk,l ( R ) = X i =1 ∞ X j =0 Res z = d ij h Θ pk,l R − z i = X i =1 ∞ X j =0 lim z → d ij (cid:16) ( z − d ij )Θ pk,l ( z ) R − z (cid:17) = ∞ X j =0 ( − − j ) k + l · ( − j j ! · Γ( d − − j )Γ(1 − α − jα ) R j + ∞ X j =0 (cid:18) − d − j (cid:19) k + l · ( − j j ! · Γ(1 − d − j )Γ(1 − αd − αj ) R d + j = ( − k + l Γ( d − − α ) R + (cid:18) − d (cid:19) k + l Γ(1 − d )Γ(1 − αd ) R d + O ( R )( R d +1 + R ) , where O ( R ) is bounded in (0 ,
1] by (6.5) and (6.2). Hence | H pk, ( R ) | ≤ N ( R + R / · d =1 ) . (6.26)Moreover since d (cid:18) − d (cid:19) k + l − + 2 (cid:18) − d (cid:19) k + l = 0 l = 1 , , . . . , (6.27)it holds that for all l ≥ | d H pk,l − ( R ) + 2H pk,l ( R ) | ≤ N R. (6.28)If d is an even number, Θ pk,l has a simple pole at d j = − − j for 0 ≤ j ≤ d − d j = − d − j for j = 0 , , , · · · . Hence by (6.3) and (6.9),for R ≤ pk,l ( R ) = d − X j =0 Res z = d j h Θ pk,l R − z i d =2 + ∞ X j =0 Res z = d j h Θ pk,l R − z i = d − X j =0 lim z → d j (cid:16) ( z − d j )Θ pk,l ( z ) R − z (cid:17) d =2 + ∞ X j =0 lim z → d j ddz (cid:16) ( z − d j ) Θ pk,l ( z ) R − z (cid:17) = d − X j =0 ( − − j ) k + l · ( − j j ! · Γ( d − − j )Γ(1 − α (1 + j )) R j · d =2 − ∞ X j =0 (cid:18) − d − j (cid:19) k + l · ( − d +2 j − j ! (cid:0) d + j − (cid:1) ! · R d + j ln R Γ(1 − αd − jα )+ ∞ X j =0 Res z = d j h Θ pk,l ( z ) i R d + j = ( − k + l Γ( d − − α ) R · d =2 − (cid:18) − d (cid:19) k + l · ( − d − (cid:0) d − (cid:1) ! · R d ln R Γ(1 − αd ) + O ( R )( R d + R ) , where O ( R ) is bounded in (0 ,
1] due to (6.5) and (6.2). Hence | H pk, ( R ) | ≤ ( R + R ln R · d =2 ) . (6.29)Moreover by (6.27) again, if l ≥ | d H pk,l − ( R ) + 2H pk,l ( R ) | ≤ N R. (6.30)Therefore due to (6.24), (6.26), (6.28), (6.29), and (6.30), we obtain for any R ≤ | ∂ nt p ( t, x ) | ≤ N | x | − d t − n ( R + R / · d =1 + R ln R · d =2 ) , and for any R ≤ m ∈ N | ∂ nt D mx p ( t, x ) | ≤ N | x | − d − m t − n R, where N depends only on d , m , n , and α . Therefore Lemma 3.1(ii) is proved.Next we prove (3.5). For k, l = 0 , , . . . , denoteΘ qk,l ( z ) = Γ( d + z ) { Γ(1 + z ) } k + l +1 Γ( α + αz ) { Γ( z ) } k + l = Γ( d + z )Γ(1 + z )Γ( α (1 + z )) z k + l . We remark that Θ qk,l ( z ) does not have a pole at z = − d = 2. First weassume that d is an odd number. Then Θ qk,l has a simple pole at d j = − − j and d j = − d − j for j = 0 , , , · · · . By (6.3) and (6.9), for R ≤ qk,l ( R ) = X i =1 ∞ X j =0 Res z = d ij h Θ qk,l R − z i = X i =1 ∞ X j =0 lim z → d ij (cid:16) ( z − d ij )Θ qk,l ( z ) R − z (cid:17) = ∞ X j =0 ( − − j ) k + l · ( − j (1 + j )! · Γ( d − − j )Γ( − α (1 + j )) R j + ∞ X j =0 (cid:18) − d − j (cid:19) k + l · ( − j j ! · Γ(1 − d − j )Γ( α − αd − jα ) R d + j = − ( − k + l Γ( d − − α ) R + (cid:18) − d (cid:19) k + l Γ(1 − d )Γ( α − αd ) R d + O ( R )( R d +1 + R ) , (6.31)where O ( R ) is bounded in (0 ,
1] as before. Hence | H qk, ( R ) | ≤ N ( R + R d/ · d =1 , ) . (6.32)Furthermore, by (6.27) and (6.31) if l ≥ | d H qk,l − ( R ) + 2H qk,l ( R ) | ≤ N ( R + R / · d =1 ) . (6.33)On the other hand, if d is even and greater than 2 then Θ qk,l has a simple poleat d j = − − j for 1 ≤ j ≤ d − d j = − d − j for j = 0 , , , · · · . If d = 2, then Θ qk,l has a simple pole at − q ( L p )-THEORY FOR FRACTIONAL EQUATIONS 37 d j = − − j for j = 1 , , · · · . Thus by (6.3) and (6.9), for R ≤ qk,l ( R ) = d − X j =1 Res z = d j h Θ qk,l R − z i d ≥ + Res z = − h Θ qk,l R − z i d =2 + Res z = d h Θ qk,l R − z i · d ≥ + ∞ X j =1 Res z = d j h Θ qk,l R − z i = d − X j =1 lim z → d j (cid:16) ( z − d j )Θ qk,l ( z ) R − z (cid:17) d ≥ + ( − k + l R · d =2 + lim z → d ddz (cid:16) ( z − d ) Θ qk,l ( z ) R − z (cid:17) d ≥ + ∞ X j =1 lim z → d j ddz (cid:16) ( z − d j ) Θ qk,l ( z ) R − z (cid:17) . By the product rule of the differentiation, the above term equals d − X j =1 ( − − j ) k + l · ( − j j ! · Γ( d − − j )Γ( − αj ) R j · d ≥ + ( − k + l R · d =2 + Res z = d h Θ qk,l ( z ) i R d · d ≥ − (cid:18) − d (cid:19) k + l · ( − d − ( d − · R d ln R Γ( α − αd ) 1 d ≥ + ∞ X j =1 Res z = d j h Θ qk,l ( z ) i R d + j − ∞ X j =1 (cid:18) − d − j (cid:19) k + l · ( − d +2 j − j !( d + j − · R d + j ln R Γ( α − αd − αj )= − ( − k + l · Γ( d − − α ) R · d ≥ + ( − k + l R · d =2 + Res z = d h Θ qk,l ( z ) i R d · d ≥ − (cid:18) − d (cid:19) k + l · ( − d − ( d − · R d ln R Γ( α − αd ) 1 d ≥ + O ( R ) (cid:0) R + R d +1 (1 + | ln R | ) (cid:1) , where O ( R ) is again bounded in (0 , | H qk, ( R ) | ≤ N (cid:0) R + R · d =2 + R ln R · d =4 (cid:1) (6.34)and by (6.27) | d H qk,l − ( R ) + 2H qk,l ( R ) | ≤ N ( R + R ln R · d =2 ) . (6.35)Combining (6.25), (6.32), (6.33), (6.34), and (6.35), for any R ≤ m we have | ∂ nt D mx q ( t, x ) |≤ N | x | − d − m t − n + α − ( R + R ln R · d =2 )+ N | x | − d t − n + α − h R / · d =1 + R · d =2 + R ln R · d =4 i m =0 , where N depends only on d , m , n , and α . Therefore Lemma 3.2(i) is proved. Finally we prove Lemma 3.2(ii), that is for any t = 0 an x = 0, ∂ αt p = ∆ p, ∂p∂t = ∆ q. (6.36)To prove the first assertion above one may try to show that their Fourier transformscoincide. But due to the singularity of ∆ p ( t, · ) near zero, we instead prove Z T Z R d ∂ αt p ( t, x ) h ( t ) φ ( x ) dxdt = Z T Z R d ∆ p ( t, x ) h ( t ) φ ( x ) dxdt (6.37)for any φ ∈ C ∞ c ( R d \ { } ) and h ( t ) ∈ C ∞ c ((0 , T )).Note that t − α | x | ≥ c > hφ , and thus (3.2) implies that ∂p∂t ( t, x ) , ∂ αt p ( t, x ), and ∆ p ( t, x ) are bounded on the support of hφ . Hence bothsides of (6.37) make sense. By the integration by parts, Z T ∂ αt p ( t, x ) hdt = Z T p ( T − t, x ) D αt H ( t ) dt, where H ( t ) := h ( T − t ). Recall that by Parseval’s identity Z R d f ¯ gdx = Z R d ( F f )( ¯ F g ) dξ, ∀ f, g ∈ L ( R d ) . Considering an approximation of f by functions in L ( R d ) one can easily provethat Parseval’s identity holds if f ∈ L ( R d ) and g is in the Schwartz class. Hencethe left side of (6.37) equals Z T (cid:20)Z R d p ( T − t, x ) φ ( x ) dx (cid:21) ∂ αt H ( t ) dt = Z T (cid:20)Z R d E α ( − ( T − t ) α | ξ | ) F ( φ )( ξ ) dξ (cid:21) ∂ αt H ( t ) dt. Observe that F ( φ )( ξ ) → | ξ | → ∞ and ∂ αt H ( t ) = 0 if t issufficiently small. Hence by (3.1) we can apply Fubini’s theorem and show that thelast term above is equal to Z R d "Z T E α ( − ( T − t ) α | ξ | ) ∂ αt H ( t ) dt F ( φ )( ξ ) dξ = Z R d Z T E α ( − t α | ξ | ) h ( t ) F (∆ φ )( ξ ) dtdξ = Z T Z R d p ( t, x ) h ( t )∆ φ ( x ) dtdx = Z T Z R d ∆ p ( t, x ) h ( t ) φ ( x ) dxdt. For the first equality above we use the integration by parts in time variable andthe fact ∂ αt E α ( − t α | ξ | ) = −| ξ | E α ( − t α | ξ | ), and Parseval’s identity is used for thesecond equality. Therefore the first assertion of (6.37) is proved.Next we prove ∆ q = ∂p∂t . Note that due to (3.2), if x = 0 then D x p ( · , x )and D x p ( · , x ) are bounded on (0 , T ) uniformly in a neighborhood of x . Hence, q ( L p )-THEORY FOR FRACTIONAL EQUATIONS 39 if α ∈ (0 , q = ∆ ddt Z t k α ( t − s ) p ( s, x ) ds = ddt ∆ Z t k α ( t − s ) p ( s, x ) ds = ddt Z t k α ( t − s )∆ p ( s, x ) ds = D − α ∂ αt p = ∂p∂t . Similarly, if α ∈ (1 , q = ∆( I α − t p ) = I α − t (∆ p ) = I α − t ∂ αt p = ∂p∂t . Thus the second assertion of (6.36) is proved.
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Department of Mathematics, Korea University, 1 Anam-dong, Sungbuk-gu, Seoul,136-701, Republic of Korea
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E-mail address : [email protected] Department of Mathematics, Korea University, 1 Anam-dong, Sungbuk-gu, Seoul,136-701, Republic of Korea
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