Well-posedness and blow-up results for a class of nonlinear fractional Rayleigh-Stokes problem

Jing Na Wang; Ahmed Alsaedi; Bashir Ahmad; Yong Zhou

doi:10.1515/anona-2022-0249

Open Access Published by De Gruyter June 12, 2022

Well-posedness and blow-up results for a class of nonlinear fractional Rayleigh-Stokes problem

Jing Na Wang , Ahmed Alsaedi , Bashir Ahmad and Yong Zhou

From the journal Advances in Nonlinear Analysis

https://doi.org/10.1515/anona-2022-0249

Abstract

In this article, we consider the fractional Rayleigh-Stokes problem with the nonlinearity term satisfies certain critical conditions. The local existence, uniqueness and continuous dependence upon the initial data of ε -regular mild solutions are obtained. Furthermore, a unique continuation result and a blow-up alternative result of ε -regular mild solutions are given in the end.

Keywords: Riemann-Liouville fractional derivative; Rayleigh-Stokes problem; ε-regular mild solutions; well-posedness; blow-up

MSC 2010: 26A33; 35E15; 35R11; 76D03; 35R44

1 Introduction

In recent years, the Rayleigh-Stokes problem has attracted much attention of researchers due to its practical importance in describing the behaviors of non-Newtonian fluids. The problem is used to describe the diffusion of viscous incompressible fluid on rectangular edge when the edge suddenly starts from rest and moves at a constant speed parallel to itself, which is also called as pseudo-parabolic equation or a Sobolev-type equation in the case of first-order derivative. There are many non-Newtonian fluids such as ketchup, drilling mud, blood, shampoo, polymer solutions and many other examples in our daily life. In contrast to Newtonian fluids, non-Newtonian fluids cannot be drowned by a single model, thus researchers studied the Rayleigh-Stokes problem in various non-Newtonian fluids such as heated second grade fluid, Maxwell fluid and Oldroyd-B fluid. Corresponding exact solutions are obtained by researchers in virtue of the simple and double Fourier sine transforms, and we refer readers to [10,11,12,16,19] for more details.

As the development of fractional calculus, the study of Rayleigh-Stokes problem with Riemann-Liouville fractional derivative in generalized non-Newtonian fluids has attracted much attention of researchers because of the flexibility of fractional derivative in describing viscoelastic behaviors. The fractional models are based on replacing the first-order derivative in constitutive equations of integer models by fractional derivative. In [21,24,25], researchers studied the fractional Rayleigh-Stokes problem in heated generalized second grade fluid and generalized Maxwell fluid, and obtained exact solutions by virtue of Fourier sine transform and fractional Laplace transform.

The homogeneous Rayleigh-Stokes problem for a generalized second grade fluid is considered in [8], and the well-posedness and Sobolev regularity results were obtained by using an operator approach. Based on such results, many research results on the inverse problem of fractional Rayleigh-Stokes problem in a generalized second grade fluid were carried out recently, such as the identification of source term and the regularization of solution for fractional Rayleigh-Stokes problem with Gaussian random noise were obtained by a general filter method in [17]. The existence, regularity and regulation estimates of mild solutions were studied in [7,18]. For the nonlinear fractional Rayleigh-Stokes problem in a generalized second grade fluid, the well-posedness results of such problem with initial condition were studied in [28], and the existence and regularization of mild solutions for such problem with finial condition were obtained in [23]. In the most recent work [13], authors discussed the semilinear time fractional Rayleigh-Stokes problem on R N ( N ≥ 1 ) by the Gagliardo-Nirenberg inequality and harmonic analysis method, the global and local well-posedness results were obtained, the continuation and blow-up alternative results, blow-up rate and the integrability in Lebesgue spaces were also presented.

The interpolation–extrapolation scales of Banach spaces are drowned in this article for the reason that the force terms in some nonlinear problems may map spaces into intermediate spaces rather than into themselves, and we refer readers to [1] for the details. The well-posedness of problem with critical nonlinearities in interpolation space has been studied by many researchers, and the continuation and blow-up alternative for mild solutions have been established as well. For example, the local existence and uniqueness of mild solutions for abstract parabolic problems with critical nonlinearities were studied in [6], the local well-posedness, continuation and blow-up alternative for mild solutions of an integro-differential equation were studied in [3] and the application of those results to a strongly damped plate equation with memory was displayed as well. Similar results of the abstract Volterra integro-differential equations were obtained in [4], and the results were applied to some parabolic models with memory. In [5], the semi-linear time-fractional diffusion equation with global Lipschitz condition was discussed and some well-posedness results were obtained. We also refer readers to the references cited therein [9,15,20,22,26,27].

To the best of authors’ knowledge, there are few works on well-posedness of fractional Rayleigh-Stokes problem with critical nonlinearity. Therefore, inspired by the above discussion, we will consider the fractional Rayleigh-Stokes problem with critical nonlinearity and obtain the well-posedness results, continuation and blow-up alternative results in some interpolation spaces. The methods and solution spaces in this article will be different from those of [13]. Let us introduce the fractional Rayleigh-Stokes problem that we will study in this article

(1.1) ∂ t u − ( 1 + γ ∂ t α ) Δ u = f ( t , u ) , x ∈ Ω , t > 0 , u ( t , x ) = 0 , x ∈ ∂ Ω , t > 0 , u ( 0 , x ) = u 0 ( x ) , x ∈ Ω ,

where γ > 0 is a given constant, ∂ t α is the Riemann-Liouville fractional partial derivative of order α ∈ ( 0 , 1 ) , Δ is the Laplace operator, Ω ⊂ R d ( d ≥ 1 ) is a bounded domain with smooth boundary ∂ Ω , u 0 ( x ) is the initial data for u in L 2 ( Ω ) , f : [ 0 , ∞ ] × R → R is an ε -regular map defined in the later section.

The article is organized as follows. In Section 2, the interpolation–extrapolation scales and ε -regular map are briefly introduced, and some concepts and lemmas which will be used in this article are given. Section 3 begins with the definition of ε -regular mild solutions of problem (1.1), and then, the properties of solution operators are discussed and the well-posedness results are obtained for problem (1.1) in the case that the nonlinearity term is an ε -regular map. Furthermore, the continuation and blow-up alternative results are given in Section 4.

2 Preliminaries

In this section, we recall some concepts and lemmas which are useful in the following.

R + is the set of all non-negative real numbers, N + denotes the set of all positive integer numbers, L 2 ( Ω ) denotes the Banach space of all measurable functions on Ω ⊂ R d ( d ≥ 1 ) with the inner product ( ⋅ , ⋅ ) and the norm ‖ v ‖ L 2 ( Ω ) = ∣ ( v , v ) ∣ 1 2 . For any given Banach space Y and a set J ⊂ R + , we denote by C ( J ; Y ) the space of all continuous functions from J into Y equipped with the norm

‖ u ‖ C ( J ; Y ) ≔ sup t ∈ J ‖ u ( t ) ‖ Y .

H 0 1 ( Ω ) is the closure of C 0 ∞ ( Ω ) in H 1 ( Ω ) and equips with the norm

‖ v ‖ H 0 1 ( Ω ) ≔ ‖ ∇ v ‖ L 2 ( Ω ) ,

where C 0 ∞ ( Ω ) is the space of all infinitely differentiable functions with compact support in Ω , H 1 ( Ω ) is a Hilbert space.

We recall some definitions of fractional calculus in the sequel. For more details, we refer the reader to [14].

Definition 2.1

The left Riemann-Liouville fractional integral of order α ∈ ( 0 , 1 ) is defined by

( I t α 0 u ) ( t , x ) = 1 Γ ( α ) ∫ 0 t ( t − s ) α − 1 u ( s , x ) d s , t > 0 ,

provided the right-hand side is pointwise defined on [ 0 , b ] . The left Riemann-Liouville fractional partial derivative of order α ∈ ( 0 , 1 ) is defined by

( ∂ t α u ) ( t , x ) = ∂ t ( I t 1 − α 0 u ( t , x ) ) = 1 Γ ( 1 − α ) ∂ ∂ t ∫ 0 t ( t − s ) − α u ( s , x ) d s , t > 0 .

In the following, we introduce the spectral problem

(2.1) − Δ φ j ( x ) = λ j φ j ( x ) , x ∈ Ω , φ j ( x ) = 0 , x ∈ ∂ Ω ,

where { φ j ( x ) } j = 1 ∞ is an orthogonal basis of H 0 1 ( Ω ) and an orthonormal basis of L 2 ( Ω ) , λ j is the eigenvalue of − Δ corresponding to φ j which satisfies 0 < λ 1 ≤ λ 2 ≤ … λ j ≤ … and has the property that λ j → ∞ as j → ∞ .

In the following, for the convenience of writing we denote A = − Δ , and then, we define the fractional power operator A β for β ≥ 0 as

A β v ≔ ∑ j = 1 ∞ λ j β ( v , φ j ) φ j , v ∈ D ( A β ) .

Obviously, D ( A β ) is a Banach space equipped with the norm

‖ v ‖ D ( A β ) ≔ ∑ j = 1 ∞ λ j 2 β ∣ ( v , φ j ) ∣ 2 1 2 .

Now we introduce the construction of abstract interpolation–extrapolation scales, for more information, we refer readers to the monograph [2]. For Banach spaces E 0 and E 1 , if E 1 ↪ E 0 densely, we say the pair ( E 0 , E 1 ) is a densely injected Banach couple. Denote by ( ⋅ , ⋅ ) θ the admissible interpolation functor of exponent θ ∈ ( 0 , 1 ) , we mean that an interpolation functor of exponent θ for the category of densely injected Banach couples such that E 1 ↪ ( E 0 , E 1 ) θ densely. Let E 1 = D ( A ) with the graph norm ‖ ⋅ ‖ E 1 = ‖ A ⋅ ‖ E 0 and A 1 : D ( A 1 ) ⊂ E 1 → E 1 be the realization of A in E 1 , we can define E 2 = ( D ( A 1 ) , ‖ A 1 ⋅ ‖ E 1 ) .

Similarly, for k ∈ N + , we can define E k + 1 = ( D ( A k ) , ‖ A k ⋅ ‖ k ) and A k + 1 = E k + 1 -the realization of A k . Then, for k ∈ N + ∪ { − 1 } and θ ∈ ( 0 , 1 ) , we denote E k + θ = ( E k , E k + 1 ) θ and A k + θ = E k + θ -the realization of A k . We say { ( E β , A β ) ; − 1 ≤ β < ∞ } the interpolation–extrapolation scale over [ − 1 , ∞ ) associated with A and ( ⋅ , ⋅ ) θ .

In this article, for the reason that the scale of fractional power spaces may not be suitable to treat critical problems, we consider the interpolation space X β ≔ E β 2 ( β ≥ 0 ) , thus the norm of X β is given as follows:

‖ v ‖ X β ≔ ∑ j = 1 ∞ λ j β ∣ ( v , φ j ) ∣ 2 1 2 .

It is easy to see that X 0 = L 2 ( Ω ) .

In the following, we introduce a definition given in [6].

Definition 2.2

For ε ≥ 0 , we say a map g is an ε -regular map relative to the pair ( X 0 , X 1 ) if there exist ρ > 1 , δ ( ε ) ∈ [ ρ ε , 1 ) and a positive constant M such that g : X 1 + ε → X δ ( ε ) satisfies

‖ g ( u ) − g ( v ) ‖ X δ ( ε ) ≤ M ‖ u − v ‖ X 1 + ε ( ‖ u ‖ X 1 + ε ρ − 1 + ‖ v ‖ X 1 + ε ρ − 1 + 1 ) , ∀ u , v ∈ X 1 + ε .

Denote by ℱ the family of all ε -regular maps relative to ( X 0 , X 1 ) which satisfy the following conditions:

for u , v ∈ X 1 + ε , ρ > 1 and ρ ε ≤ δ ( ε ) < 1 , f ( t , ⋅ ) satisfies
(2.2) ‖ f ( t , u ) − f ( t , v ) ‖ X δ ( ε ) ≤ C ‖ u − v ‖ X 1 + ε ( ‖ u ‖ X 1 + ε ρ − 1 + ‖ v ‖ X 1 + ε ρ − 1 + w ( t ) t ς ) ,
where C is a positive constant, w ( t ) is a non-decreasing function that satisfies 0 ≤ w ( t ) ≤ a , − δ ( ε ) + ε ≤ ς ≤ 0 , t > 0 ;
for v ∈ X 1 + ε , − δ ( ε ) ≤ ς ˜ ≤ 0 , f ( t , ⋅ ) satisfies
(2.3) ‖ f ( t , v ) ‖ X δ ( ε ) ≤ C ( ‖ v ‖ X 1 + ε ρ + w ( t ) t ς ˜ ) .

It is easy to verify that there exist functions that belong to ℱ , which implies that ℱ is not empty, for example, f ( t , u ) = u ∣ u ∣ ρ − 1 , f ( t , u ) = P ( u ⋅ ∇ ) u , where P is the orthogonal projection, for more details, we refer readers to [4,6] and other references cited therein.

For any b > 0 and η > 0 , we denote by C η ( [ 0 , b ] ; X 1 + ε ) the space of all functions v ∈ C ( ( 0 , b ] ; X 1 + ε ) with the property t η v ( t ) ∈ C ( [ 0 , b ] ; X 1 + ε ) . It is easy to see that C η ( [ 0 , b ] ; X 1 + ε ) is a Banach space under the norm

‖ v ‖ C η ( [ 0 , b ] ; X 1 + ε ) ≔ sup t ∈ [ 0 , b ] t η ‖ v ( t ) ‖ X 1 + ε .

Let

S α , j ( t ) ≔ ℒ − 1 ( ( z + ( 1 + γ z α ) λ j ) − 1 ) ( t ) = 1 2 π i ∫ B e z t ( z + ( 1 + γ z α ) λ j ) − 1 d z ,

where t ≥ 0 , j = 1 , 2 , … , ℒ − 1 ( ⋅ ) denotes the inverse Laplace transform and B = { z : ℜ z = ϖ , ϖ > 0 } is the Brownwich path. From [8, Theorem 2.2], we know that

(2.4) S α , j ( t ) = ∫ 0 ∞ e − ξ t K j ( ξ ) d ξ ,

where

K j ( ξ ) = γ π λ j ξ α sin ( α π ) ( − ξ + λ j γ ξ α cos ( α π ) + λ j ) 2 + ( λ j γ ξ α sin ( α π ) ) 2 .

We introduce some properties of S α , j ( t ) in the following.

Lemma 2.1

[8, Theorem 2.2] The functions S α , j ( t ) , j = 1 , 2 , … have the following properties:

S α , j ( 0 ) = 1 , 0 < S α , j ( t ) ≤ 1 for t ≥ 0 ;
For any j ∈ N + , S α , j ( t ) is completely monotone for t ≥ 0 .

Lemma 2.2

[18, Lemma 2.2] For α ∈ ( 0 , 1 ) and t ≥ 0 , S α , j ( t ) satisfies

S α , j ( t ) ≤ M 1 1 + λ j t 1 − α ,

where j ∈ N + and

M 1 = Γ ( 1 − α ) γ π sin ( α π ) + 1 .

3 Existence of mild solutions

At the beginning, we give the definition of ε -regular mild solutions of problem (1.1). In fact, multiplying both sides of the first equation in (1.1) by φ j ( x ) and taking the Laplace transform, we obtain that

z u j ˆ ( z ) − u j ( 0 ) + ( 1 + γ z α ) λ j u j ˆ ( z ) = f j ˆ ( z ) ,

that is,

u j ˆ ( z ) = ( z + ( 1 + γ z α ) λ j ) − 1 ( u j ( 0 ) + f j ˆ ( z ) ) ,

where u j = ( u , φ j ) , f j = ( f , φ j ) , u j ˆ and f j ˆ stand for the Laplace transform of u j and f j , respectively. And then, from the uniqueness of the inverse Laplace transform, we obtain

u j ( t ) = S α , j ( t ) u 0 , j + ∫ 0 t S α , j ( t − s ) f j ( s , u ( s ) ) d s ,

where u 0 , j = ( u 0 , φ j ) . Denote

S α ( t ) v = ∑ j = 1 ∞ S α , j ( t ) ( v , φ j ) φ j , for v ∈ L 2 ( Ω ) ,

then we have

(3.1) u ( t ) = S α ( t ) u 0 + ∫ 0 t S α ( t − s ) f ( s , u ( s ) ) d s .

Now we can give the definition of ε -regular mild solutions.

Definition 3.1

By an ε -regular mild solution of problem (1.1), we mean a function u ∈ C η ( [ 0 , b ] ; X 1 + ε ) ∩ C ( [ 0 , b ] ; X 1 ) and satisfies (3.1).

In the following, we give some properties of S α ( t ) , which will be useful for us to obtain the main results.

Lemma 3.1

Assume 0 ≤ p < ∞ and 0 ≤ q − p ≤ 2 , for t > 0 and any v ∈ X p , the following inequality holds

(3.2) t ( 1 − α ) ( q − p ) / 2 ‖ S α ( t ) v ‖ X q ≤ M 1 ‖ v ‖ X p ,

where M 1 is a positive constant given in Lemma 2.2.

Proof

From Lemma 2.2, for t > 0 and any v ∈ X p , we have

‖ S α ( t ) v ‖ X q 2 = ∑ j = 1 ∞ λ j q S α , j 2 ( t ) ∣ ( v , φ j ) ∣ 2 ≤ M 1 2 ∑ j = 1 ∞ λ j q ( 1 + λ j t 1 − α ) 2 ∣ ( v , φ j ) ∣ 2 ≤ M 1 2 ∑ j = 1 ∞ λ j q ( 1 + λ j t 1 − α ) q − p ∣ ( v , φ j ) ∣ 2 ≤ M 1 2 t − ( 1 − α ) ( q − p ) ‖ v ‖ X p 2 ,

which implies

t ( 1 − α ) ( q − p ) / 2 ‖ S α ( t ) v ‖ X q ≤ M 1 ‖ v ‖ X p .

This concludes the result.□

Remark 3.1

From the process of proof of Lemma 3.1, we can see that (3.2) holds at t = 0 if and only if p = q . When p ≠ q , the property λ j → ∞ as j → ∞ makes the conclusion (3.2) invalid at t = 0 .

Lemma 3.2

For 0 ≤ p < ∞ and 0 ≤ q − p ≤ 2 , S α ( t ) is strongly continuous in terms of t > 0 on X q . Moreover, for t 2 > t 1 > 0 and any v ∈ X p , the following inequality holds

(3.3) ‖ S α ( t 2 ) v − S α ( t 1 ) v ‖ X q ≤ M 2 λ 1 ( q − p − 2 ) / 2 ( t 1 α − 1 − t 2 α − 1 ) ‖ v ‖ X p ,

where M 2 = Γ ( 1 − α ) π γ sin ( α π ) .

Proof

For any t 2 > t 1 > 0 and v ∈ X p , by virtue of (2.4), we have

(3.4) ‖ S α ( t 2 ) v − S α ( t 1 ) v ‖ X q 2 = ∑ j = 1 ∞ λ j q ∣ S α , j ( t 2 ) − S α , j ( t 1 ) ∣ 2 ∣ ( v , φ j ) ∣ 2 = ∑ j = 1 ∞ λ j q ∫ 0 ∞ ( e − τ t 1 − e − τ t 2 ) K j ( τ ) d τ 2 ∣ ( v , φ j ) ∣ 2 = ∑ j = 1 ∞ λ j q ∫ 0 ∞ ∫ t 1 t 2 e − τ t τ K j ( τ ) d t d τ 2 ∣ ( v , φ j ) ∣ 2 ,

and then, from the expression of K j ( τ ) , we can easily obtain that

0 < K j ( τ ) ≤ τ − α π γ sin ( α π ) λ j ,

which implies

(3.5) ∫ 0 ∞ ∫ t 1 t 2 e − τ t τ K j ( τ ) d t d τ ≤ 1 π γ sin ( α π ) λ j ∫ t 1 t 2 ∫ 0 ∞ e − τ t τ 1 − α d τ d t = Γ ( 2 − α ) π γ sin ( α π ) λ j ∫ t 1 t 2 t α − 2 d t = Γ ( 1 − α ) π γ sin ( α π ) λ j ( t 1 α − 1 − t 2 α − 1 ) .

Substituting (3.5) into (3.4), we obtain that

‖ S α ( t 2 ) v − S α ( t 1 ) v ‖ X q 2 ≤ Γ ( 1 − α ) ( t 1 α − 1 − t 2 α − 1 ) π γ sin ( α π ) 2 λ 1 q − p − 2 ‖ v ‖ X p 2 .

It shows that in the case t 2 → t 1 , ‖ S α ( t 2 ) v − S α ( t 1 ) v ‖ X q → 0 , which implies that S α ( t ) v is strongly continuous for t > 0 . This completes the proof.□

Remark 3.2

In the case p = q , the strong continuity of S α ( t ) still holds at t = 0 on X p . In fact, for any v ∈ X p , taking t 1 = 0 , t 2 > 0 , from the fact that S α , j ( t ) = 1 at t = 0 in Lemma 2.1, we have S α ( t 1 ) v = v . By virtue of Lemma 2.2, we obtain

‖ S α ( t 2 ) v − v ‖ X p 2 = ∑ j = 1 ∞ λ j p ( 1 − S α , j ( t 2 ) ) 2 ∣ ( v , φ j ) ∣ 2 ≤ 2 ( 1 + M 1 2 ) ‖ v ‖ X p 2 .

The Weierstrass discriminance implies that ‖ S α ( t 2 ) v − v ‖ X p converges uniformly. Then we have

lim t 2 → 0 + ‖ S α ( t 2 ) v − v ‖ X p = ∑ j = 1 ∞ lim t 2 → 0 + λ j p ( 1 − S α , j ( t 2 ) ) 2 ∣ ( v , φ j ) ∣ 2 1 2 = 0 ,

which implies that S α ( t ) is strongly continuous for all t ≥ 0 on X p .

In the following, we give some estimations when the source term is ℱ type, which will be useful for verifying the main results. For the convenience in our writing, we use the notation σ as σ = ( 1 − α ) ( 1 + ε − δ ( ε ) ) . Obviously, 0 < σ < 1 .

Lemma 3.3

Assume f ∈ ℱ and 0 < η < 1 ρ , then for t ∈ ( 0 , b ] and any u ∈ C η ( [ 0 , b ] ; X 1 + ε ) , we have

(3.6) ∫ 0 t S α ( t − s ) f ( s , u ( s ) ) d s X 1 + ε ≤ M 1 C B 1 t 1 − σ / 2 ( t − ρ η ‖ u ‖ C η ( [ 0 , b ] ; X 1 + ε ) ρ + w ( t ) t ς ˜ ) ,

where B 1 = max { B ( 1 − σ / 2 , 1 − ρ η ) , B ( 1 − σ / 2 , 1 + ς ˜ ) } .

Proof

For t > 0 and any u ∈ C η ( [ 0 , b ] ; X 1 + ε ) , from Lemma 3.1, we have

∫ 0 t S α ( t − s ) f ( s , u ( s ) ) d s X 1 + ε ≤ ∫ 0 t ‖ S α ( t − s ) f ( s , u ( s ) ) ‖ X 1 + ε d s ≤ M 1 ∫ 0 t ( t − s ) − σ / 2 ‖ f ( s , u ( s ) ) ‖ X δ ( ε ) d s .

Substituting condition (2.3) into above, we have

∫ 0 t S α ( t − s ) f ( s , u ( s ) ) d s X 1 + ε ≤ M 1 ∫ 0 t ( t − s ) − σ / 2 C ( ‖ u ( s ) ‖ X 1 + ε ρ + w ( s ) s ς ˜ ) d s ≤ M 1 C ∫ 0 t ( t − s ) − σ / 2 s − ρ η d s ‖ u ‖ C η ( [ 0 , b ] ; X 1 + ε ) ρ + M 1 C w ( t ) ∫ 0 t ( t − s ) − σ / 2 s ς ˜ d s ≤ M 1 C B ( 1 − σ / 2 , 1 − ρ η ) t 1 − σ / 2 − ρ η ‖ u ‖ C η ( [ 0 , b ] ; X 1 + ε ) ρ + M 1 C w ( t ) B ( 1 − σ / 2 , 1 + ς ˜ ) t 1 − σ / 2 + ς ˜ ,

where B ( ⋅ , ⋅ ) denotes the beta function. From the definition of σ , we know that 1 − σ / 2 > 0 , and then, combined with η < 1 ρ and ς ˜ ≥ − δ ( ε ) > − 1 , we can easily check that the beta functions in the last inequality of above are meaningful. The proof is completed.□

Lemma 3.4

Assume f ∈ ℱ and 0 < η < min 1 ρ , 1 + ς , for any u , v ∈ C η ( [ 0 , b ] ; X 1 + ε ) , denote U = max { ‖ u ‖ C η ( [ 0 , b ] ; X 1 + ε ) , ‖ v ‖ C η ( [ 0 , b ] ; X 1 + ε ) } , then we have

(3.7) ∫ 0 t S α ( t − s ) ( f ( s , u ( s ) ) − f ( s , v ( s ) ) ) d s X 1 + ε ≤ M 1 B 2 C t 1 − σ / 2 ( 2 U ρ − 1 t − η ρ + w ( t ) t − η + ς ) ‖ u − v ‖ C η ( [ 0 , b ] ; X 1 + ε ) ,

where 0 < t ≤ b and B 2 = max { B ( 1 − σ / 2 , 1 − η ρ ) , B ( 1 − σ / 2 , 1 − η + ς ) } .

Proof

For any u , v ∈ C η ( [ 0 , b ] ; X 1 + ε ) , from Lemma 3.1, we have

∫ 0 t S α ( t − s ) ( f ( s , u ( s ) ) − f ( s , v ( s ) ) ) d s X 1 + ε ≤ ∫ 0 t ‖ S α ( t − s ) ( f ( s , u ( s ) ) − f ( s , v ( s ) ) ) ‖ X 1 + ε d s ≤ M 1 ∫ 0 t ( t − s ) − σ / 2 ‖ f ( s , u ( s ) ) − f ( s , v ( s ) ) ‖ X δ ( ε ) d s .

And then, the condition (2.2) implies that

∫ 0 t S α ( t − s ) ( f ( s , u ( s ) ) − f ( s , v ( s ) ) ) d s X 1 + ε ≤ M 1 ∫ 0 t ( t − s ) − σ / 2 C ‖ u ( s ) − v ( s ) ‖ X 1 + ε ( ‖ u ( s ) ‖ X 1 + ε ρ − 1 + ‖ v ( s ) ‖ X 1 + ε ρ − 1 + w ( s ) s ς ) d s ≤ 2 M 1 C U ρ − 1 ∫ 0 t ( t − s ) − σ / 2 s − η ρ d s ‖ u − v ‖ C η ( [ 0 , b ] ; X 1 + ε ) + M 1 C w ( t ) ∫ 0 t ( t − s ) − σ / 2 s − η + ς d s ‖ u − v ‖ C η ( [ 0 , b ] ; X 1 + ε ) = 2 M 1 C U ρ − 1 B ( 1 − σ / 2 , 1 − η ρ ) t 1 − σ / 2 − η ρ ‖ u − v ‖ C η ( [ 0 , b ] ; X 1 + ε ) + M 1 C w ( t ) B ( 1 − σ / 2 , 1 − η + ς ) t 1 − σ / 2 − η + ς ‖ u − v ‖ C η ( [ 0 , b ] ; X 1 + ε ) ,

where B ( ⋅ , ⋅ ) denotes the beta function. Similarly, combined with the facts σ ∈ ( 0 , 1 ) and η < min 1 ρ , 1 + ς , we know that 1 − σ / 2 > 0 , ρ η < 1 and 1 − η + ς > 0 , which imply that the beta functions in the last inequality of above are meaningful. The proof is completed.□

In the following, we present an existence result and investigate the behaviors of ε -regular mild solutions of problem (1.1) at t = 0 . Furthermore, the dependence of ε -regular mild solutions on initial conditions is also obtained. Before giving our main results, let us verify the relationships between the parameters at first. From the constraint on ς and the relation that δ ( ε ) < 1 , we have that

1 + ς − ( 1 − α ) ε / 2 ≥ 1 − δ ( ε ) + ( 1 + α ) ε / 2 > 0

and

1 − σ / 2 + ς ≥ 1 − σ / 2 − δ ( ε ) + ε = ( 1 − δ ( ε ) + ε ) ( 1 + α ) / 2 > 0 .

By virtue of the fact that δ ( ε ) ≥ ρ ε , we obtain

1 − ( 1 − α ) ( 1 − δ ( ε ) ) / 2 ρ − ( 1 − α ) ε / 2 = 1 + α 2 ρ + 1 − α 2 ρ ( δ ( ε ) − ρ ε ) > 0 .

Theorem 3.1

Assume f ∈ ℱ and u 0 ∈ X 1 , let ( 1 − α ) ε / 2 < η < min 1 + ς , 1 − ( 1 − α ) ( 1 − δ ( ε ) ) / 2 ρ , then there exists b > 0 such that problem (1.1) has a unique ε -regular mild solution in [ 0 , b ] , and the solution u ( t , x ) satisfies

t η ‖ u ( t ) ‖ X 1 + ε → 0 as t → 0 + .

Moreover, if u ( t , x ) and z ( t , x ) are two ε -regular mild solutions of problem (1.1) corresponding to the initial value conditions u 0 , z 0 ∈ X 1 , respectively, then for any t ∈ [ 0 , b ] , we have

t η ‖ u ( t ) − z ( t ) ‖ X 1 + ε ≤ C 1 ‖ u 0 − z 0 ‖ X 1 ,

where C 1 = 4 M 1 b η − ( 1 − α ) ε / 2 .

Proof

From the condition η < 1 − ( 1 − α ) ( 1 − δ ( ε ) ) / 2 ρ , we have η < 1 ρ , which implies that B 1 and B 2 exist and are well-defined, we denote B max = max { B 1 , B 2 } . For u 0 ∈ X 1 , we take 0 < r ≤ 1 and choose b > 0 such that for any t ∈ [ 0 , b ] , the following inequalities hold:

(3.8) M 1 t η − ( 1 − α ) ε / 2 ‖ u 0 ‖ X 1 ≤ r 2 , M 1 C B max t 1 + η − σ / 2 − ρ η r ρ − 1 ≤ 1 4

and

(3.9) max { M 1 C B 1 t 1 + η − σ / 2 + ς ˜ w ( t ) , M 1 C B 2 t 1 − σ / 2 + ς w ( t ) } ≤ r 4 .

For any u ∈ C η ( [ 0 , b ] ; X 1 + ε ) , we define an operator T as follows:

(3.10) ( T u ) ( t ) = S α ( t ) u 0 + ∫ 0 t S α ( t − s ) f ( s , u ( s ) ) d s , t ≥ 0 .

The main idea of our proof here is to prove that T exists a unique fixed point in C η ( [ 0 , b ] ; X 1 + ε ) , and then to check that this fixed point belongs to C ( [ 0 , b ] ; X 1 ) .

Let

B ( r , b ) = { u ∈ C η ( [ 0 , b ] ; X 1 + ε ) : sup t ∈ [ 0 , b ] t η ‖ u ( t ) ‖ X 1 + ε ≤ r } .

Now we shall prove that T maps B ( r , b ) into itself and T is a contraction.

Step I. T maps B ( r , b ) into B ( r , b ) .

We start by showing that for any u ∈ B ( r , b ) , T u ∈ C ( ( 0 , b ] ; X 1 + ε ) . In fact, for 0 < t 1 < t 2 ≤ b , we have

‖ ( T u ) ( t 2 ) − ( T u ) ( t 1 ) ‖ X 1 + ε ≤ ‖ S α ( t 2 ) u 0 − S α ( t 1 ) u 0 ‖ X 1 + ε + ∫ t 1 t 2 S α ( t 2 − s ) f ( s , u ( s ) ) d s X 1 + ε + ∫ 0 t 1 ( S α ( t 2 − s ) − S α ( t 1 − s ) ) f ( s , u ( s ) ) d s X 1 + ε ≕ ℐ 1 + ℐ 2 + ℐ 3 .

Since t 1 > 0 , according to the strong continuity of S α ( t ) v for v ∈ X 1 and combining with the inequality (3.3), we have

ℐ 1 ≤ M 2 λ 1 ( ε − 2 ) / 2 ( t 1 α − 1 − t 2 α − 1 ) ‖ u 0 ‖ X 1 ,

which implies that ℐ 1 goes to zero as t 2 → t 1 .

And then, from the process of proof of Lemma 3.3, we obtain that

(3.11) ℐ 2 ≤ M 1 C ∫ t 1 t 2 ( t 2 − s ) − σ / 2 ( ‖ u ( s ) ‖ X 1 + ε ρ + w ( s ) s ς ˜ ) d s ≤ M 1 C r ρ ∫ t 1 t 2 ( t 2 − s ) − σ / 2 s − ρ η d s + M 1 C w ( t 2 ) ∫ t 1 t 2 ( t 2 − s ) − σ / 2 s ς ˜ d s = M 1 C r ρ t 2 1 − σ / 2 − ρ η ∫ t 1 t 2 1 ( 1 − s ) − σ / 2 s − ρ η d s + M 1 C w ( t 2 ) t 2 1 − σ / 2 + ς ˜ ∫ t 1 t 2 1 ( 1 − s ) − σ / 2 s ς ˜ d s .

The ranges of σ , η , ρ and ς ˜ show that − σ / 2 > − 1 , − ρ η > − 1 and ς ˜ ≥ − δ ( ε ) > − 1 , which imply that ( 1 − s ) − σ / 2 s − ρ η and ( 1 − s ) − σ / 2 s ς ˜ are integrable for s ∈ ( t 1 t 2 , 1 ) , thus the last inequality of (3.11) converges to zero as t 2 → t 1 , which implies that ℐ 2 converges to zero as t 2 → t 1 .

For any 0 < ξ < t 1 , we can estimate ℐ 3 as follows:

ℐ 3 ≤ ∫ 0 t 1 ‖ ( S α ( t 2 − s ) − S α ( t 1 − s ) ) f ( s , u ( s ) ) ‖ X 1 + ε d s ≤ ∫ 0 t 1 − ξ ‖ ( S α ( t 2 − s ) − S α ( t 1 − s ) ) f ( s , u ( s ) ) ‖ X 1 + ε d s + ∫ t 1 − ξ t 1 ‖ ( S α ( t 2 − s ) − S α ( t 1 − s ) ) f ( s , u ( s ) ) ‖ X 1 + ε d s ≕ ℐ 31 + ℐ 32 .

From Lemma 3.1 and (2.3), we have

‖ ( S α ( t 2 − s ) − S α ( t 1 − s ) ) f ( s , u ( s ) ) ‖ X 1 + ε ≤ 2 M 1 ( t 1 − s ) − σ / 2 ‖ f ( s , u ( s ) ) ‖ X δ ( ε ) ≤ 2 M 1 C ( t 1 − s ) − σ / 2 ( s − η ρ r ρ + w ( s ) s ς ˜ ) ,

and then, from − σ / 2 > − 1 , − ρ η > − 1 and ς ˜ ≥ − δ ( ε ) > − 1 , we know that 2 M 1 C ( t 1 − s ) − σ / 2 ( s − η ρ r ρ + w ( s ) s ς ˜ ) is integrable for s ∈ ( 0 , t 1 ) . Consequently, in the case s ∈ ( 0 , t 1 − ξ ) , by virtue of Lebesgue’s dominated convergence theorem and Lemma 3.2, we obtain ℐ 31 → 0 as t 2 → t 1 . When s ∈ ( t 1 − ξ , t 1 ) , the above inequality implies that

ℐ 32 ≤ 2 M 1 C r ρ ∫ t 1 − ξ t 1 ( t 1 − s ) − σ / 2 s − ρ η d s + 2 M 1 C w ( t 1 ) ∫ t 1 − ξ t 1 ( t 1 − s ) − σ / 2 s ς ˜ d s ,

in light of the integrability of the integrand terms, we obtain that ℐ 32 → 0 as ξ → 0 + . Thus, we obtain that ℐ 3 → 0 as t 2 → t 1 .

Therefore, combined with above discussion, we obtain that ‖ ( T u ) ( t 2 ) − ( T u ) ( t 1 ) ‖ X 1 + ε → 0 as t 2 → t 1 for 0 < t 1 < t 2 ≤ b , which implies that ( T u ) ( t ) is continuous for t ∈ ( 0 , b ] .

Now we show that t η ( T u ) ( t ) ∈ C ( [ 0 , b ] ; X 1 + ε ) .

From the above discussion, we can easily obtain that t η ( T u ) ( t ) ∈ C ( ( 0 , b ] ; X 1 + ε ) , there only left to prove the continuity of t η ( T u ) ( t ) at t = 0 . For any u ∈ B ( r , b ) and t ∈ ( 0 , b ] , from Lemmas 3.1 and 3.3, we can see that

(3.12) t η ‖ ( T u ) ( t ) ‖ X 1 + ε ≤ t η ‖ S α ( t ) u 0 ‖ X 1 + ε + t η ∫ 0 t S α ( t − s ) f ( s , u ( s ) ) d s X 1 + ε ≤ M 1 t η − ( 1 − α ) ε / 2 ‖ u 0 ‖ X 1 + M 1 C B 1 t 1 + η − σ / 2 − ρ η r ρ + M 1 C B 1 w ( t ) t 1 + η − σ / 2 + ς ˜ ,

where B 1 is given in Lemma 3.3. According to the facts that ( 1 − α ) ε / 2 < η < 1 − ( 1 − α ) ( 1 − δ ( ε ) ) / 2 ρ and ς ˜ ≥ − δ ( ε ) , we obtain 1 + η − σ / 2 − ρ η > 1 + η − σ / 2 − ( 1 − ( 1 − α ) ( 1 − δ ( ε ) ) / 2 ) = η − ( 1 − α ) ε / 2 > 0 and 1 + η − σ / 2 + ς ˜ ≥ 1 + η − σ / 2 − δ ( ε ) = ( 1 − δ ( ε ) ) ( 1 + α ) / 2 + η − ( 1 − α ) ε / 2 > 0 , which imply that t η ‖ ( T u ) ( t ) ‖ X 1 + ε → 0 as t → 0 + , thus we obtain that t η ( T u ) ( t ) ∈ C ( [ 0 , b ] ; X 1 + ε ) .

Now we show that for any u ∈ B ( r , b ) and all t ∈ [ 0 , b ] , t η ‖ ( T u ) ( t ) ‖ X 1 + ε ≤ r .

In fact, substituting (3.8) and (3.9) into (3.12), we can obtain that

t η ‖ ( T u ) ( t ) ‖ X 1 + ε ≤ M 1 t η − ( 1 − α ) ε / 2 ‖ u 0 ‖ X 1 + M 1 C B 1 t 1 + η − σ / 2 ( t − ρ η r ρ + w ( t ) t ς ˜ ) ≤ r 2 + r 4 + r 4 = r .

Therefore, we obtain that T maps B ( r , b ) into itself.

Step II. T is a contraction.

For any u , v ∈ B ( r , b ) , from Lemma 3.4, (3.8) and (3.9), we have

t η ‖ ( T u ) ( t ) − ( T v ) ( t ) ‖ X 1 + ε = t η ∫ 0 t S α ( t − s ) ( f ( s , u ( s ) ) − f ( s , v ( s ) ) ) d s X 1 + ε ≤ M 1 B 2 C ( 2 r ρ − 1 t 1 + η − σ / 2 − η ρ + w ( t ) t 1 − σ / 2 + ς ) ‖ u − v ‖ C η ( [ 0 , b ] ; X 1 + ε ) ≤ 1 2 + r 4 ‖ u − v ‖ C η ( [ 0 , b ] ; X 1 + ε ) ≤ 3 4 ‖ u − v ‖ C η ( [ 0 , b ] ; X 1 + ε ) .

Thus, we obtain that T is a contraction.

The Banach fixed point theorem implies that T exists a fixed point u ( t , x ) in B ( r , b ) .

Step III. The fixed point u ( t , x ) ∈ C ( [ 0 , b ] ; X 1 ) .

From Step I , we know that the fixed point u ∈ C ( ( 0 , b ] ; X 1 + ε ) , and then, combined with the fact that X 1 + ε ↪ X 1 , analogous to the process in proving the continuity of ( T u ) ( t ) in Step I , we can obtain that u ∈ C ( ( 0 , b ] ; X 1 ) , here we omit the proof process. There only left to prove the continuity of u ( t ) at t = 0 in X 1 .

Actually, for t > 0 , from the fact that u is a fixed point of T in B ( r , b ) , we have

‖ u ( t ) − u 0 ‖ X 1 ≤ ‖ S α ( t ) u 0 − u 0 ‖ X 1 + ∫ 0 t S α ( t − s ) f ( s , u ( s ) ) d s X 1 ≕ O 1 + O 2 .

From Remark 3.2, we can easily obtain that O 1 → 0 as t → 0 + .

Now we estimate O 2 .

In fact, by using an analogous argument presented in the process of proof of Lemma 3.3, we have

O 2 ≤ ∫ 0 t ‖ S α ( t − s ) f ( s , u ( s ) ) ‖ X 1 d s ≤ M 1 C ∫ 0 t ( t − s ) − ( 1 − α ) ( 1 − δ ( ε ) ) / 2 ( ‖ u ( s ) ‖ X 1 + ε ρ + w ( s ) s ς ˜ ) d s ≤ M 1 C ∫ 0 t ( t − s ) − ( 1 − α ) ( 1 − δ ( ε ) ) / 2 s − ρ η d s ( sup s ∈ [ 0 , t ] s η ‖ u ( s ) ‖ X 1 + ε ) ρ + M 1 C ∫ 0 t ( t − s ) − ( 1 − α ) ( 1 − δ ( ε ) ) / 2 w ( s ) s ς ˜ d s ≤ M 1 C B ( 1 − ( 1 − α ) ( 1 − δ ( ε ) ) / 2 , 1 − η ρ ) t 1 − ( 1 − α ) ( 1 − δ ( ε ) ) / 2 − η ρ ( sup s ∈ [ 0 , t ] ( s η ‖ u ( s ) ‖ X 1 + ε ) ) ρ + M 1 C w ( t ) B ( 1 − ( 1 − α ) ( 1 − δ ( ε ) ) / 2 , 1 + ς ˜ ) t 1 − ( 1 − α ) ( 1 − δ ( ε ) ) / 2 + ς ˜ ,

where the facts 0 < α < 1 and − ς ˜ ≤ δ ( ε ) < 1 are used, which show that the beta functions in the last inequality of above are well-defined, and 1 − ( 1 − α ) ( 1 − δ ( ε ) ) / 2 + ς ˜ ≥ ( 1 − δ ( ε ) ) ( 1 + α ) / 2 > 0 can be checked as well. According to the constraint of η , we can obtain that the exponents of t in the last inequality of above are positive. Thus, the right-hand side of above inequalities converges to zero as t → 0 + , which implies u ( t , x ) ∈ C ( [ 0 , b ] ; X 1 ) .

Based on the discussion of above steps, we conclude that u ( t , x ) is an ε -regular mild solution of problem (1.1).

Now we show that t η ‖ u ( t ) ‖ X 1 + ε → 0 as t → 0 + .

From Lemmas 3.1, 3.3 and (3.8), we have

t η ‖ u ( t ) ‖ X 1 + ε ≤ M 1 t η − ( 1 − α ) ε / 2 ‖ u 0 ‖ X 1 + M 1 C B ( 1 − σ / 2 , 1 − ρ η ) t 1 + η − σ / 2 − ρ η sup s ∈ [ 0 , t ] ( s η ‖ u ( s ) ‖ X 1 + ε ) ρ + M 1 C w ( t ) B ( 1 − σ / 2 , 1 + ς ˜ ) t 1 + η − σ / 2 + ς ˜ ≤ M 1 t η − ( 1 − α ) ε / 2 ‖ u 0 ‖ X 1 + M 1 C B 1 t 1 + η − σ / 2 − ρ η r ρ − 1 sup s ∈ [ 0 , t ] ( s η ‖ u ( s ) ‖ X 1 + ε ) + M 1 C w ( t ) B 1 t 1 + η − σ / 2 + ς ˜ ≤ M 1 t η − ( 1 − α ) ε / 2 ‖ u 0 ‖ X 1 + 1 4 sup s ∈ [ 0 , t ] ( s η ‖ u ( s ) ‖ X 1 + ε ) + M 1 C w ( t ) B 1 t 1 + η − σ / 2 + ς ˜ ,

which implies

sup s ∈ [ 0 , t ] ( s η ‖ u ( s ) ‖ X 1 + ε ) ≤ 4 3 ( M 1 t η − ( 1 − α ) ε / 2 ‖ u 0 ‖ X 1 + M 1 C w ( t ) B 1 t 1 + η − σ / 2 + ς ˜ ) → 0 as t → 0 + .

Step IV. The continuous dependence of ε -regular mild solutions on the initial value conditions.

Assume that u ( t , x ) , z ( t , x ) ∈ C η ( [ 0 , b ] ; X 1 + ε ) ∩ C ( [ 0 , b ] ; X 1 ) are two ε -regular mild solutions of problem (1.1) obtained from above under the given initial value conditions u 0 and z 0 , respectively, thus they have the following forms:

u ( t , x ) = S α ( t ) u 0 + ∫ 0 t S α ( t − s ) f ( s , u ( s , x ) ) d s , z ( t , x ) = S α ( t ) z 0 + ∫ 0 t S α ( t − s ) f ( s , z ( s , x ) ) d s .

For u 0 , z 0 ∈ X 1 , from Lemmas 3.1, 3.4, (3.8) and (3.9), we can estimate t η ‖ u ( t ) − z ( t ) ‖ X 1 + ε as follows:

t η ‖ u ( t ) − z ( t ) ‖ X 1 + ε ≤ t η ‖ S α ( t ) ( u 0 − z 0 ) ‖ X 1 + ε + t η ∫ 0 t S α ( t − s ) ( f ( s , u ( s ) ) − f ( s , z ( s ) ) ) d s X 1 + ε ≤ M 1 t η − ( 1 − α ) ε / 2 ‖ u 0 − z 0 ‖ X 1 + 2 M 1 B 2 C t η + 1 − σ / 2 − η ρ r ρ − 1 sup s ∈ [ 0 , t ] ( s η ‖ u ( s ) − z ( s ) ‖ X 1 + ε ) + M 1 B 2 C t 1 − σ / 2 + ς w ( t ) sup s ∈ [ 0 , t ] ( s η ‖ u ( s ) − z ( s ) ‖ X 1 + ε ) ≤ M 1 t η − ( 1 − α ) ε / 2 ‖ u 0 − z 0 ‖ X 1 + 3 4 sup s ∈ [ 0 , t ] ( s η ‖ u ( s ) − z ( s ) ‖ X 1 + ε ) ,

which implies

t η ‖ u ( t ) − z ( t ) ‖ X 1 + ε ≤ 4 M 1 t η − ( 1 − α ) ε / 2 ‖ u 0 − z 0 ‖ X 1 ≤ 4 M 1 b η − ( 1 − α ) ε / 2 ‖ u 0 − z 0 ‖ X 1 .

Thus, the continuous dependence of ε -regular mild solutions on the initial value conditions is proved. The proof is completed.□

4 Continuation and blow-up alternative

In this section, we consider the continuation and blow-up alternative for the ε -regular mild solution obtained by Theorem 3.1. We give the definition of continuation of ε -regular mild solutions at the beginning.

Definition 4.1

For the ε -regular mild solution u ∈ C η ( [ 0 , b ] ; X 1 + ε ) ∩ C ( [ 0 , b ] ; X 1 ) of problem (1.1), we say that v is a continuation of u in [ 0 , b ˜ ] if v ∈ C η ( [ 0 , b ˜ ] ; X 1 + ε ) ∩ C ( [ 0 , b ˜ ] ; X 1 ) is an ε -regular mild solution for b ˜ > b and v ( t ) = u ( t ) whenever t ∈ [ 0 , b ] .

Theorem 4.1

Assume the conditions in Theorem 3.1 hold and let u be an ε -regular mild solution of problem (1.1) in [ 0 , b ] , then there exists a unique continuation u ˜ of u in [ 0 , b ˜ ] for some b ˜ > b .

Proof

Take b ˜ > b and choose r ˜ > 0 such that for u 0 ∈ X 1 and any t ∈ [ b , b ˜ ] , the following inequalities hold:

(4.1) M 2 λ 1 ( ε − 2 ) / 2 t η ( b α − 1 − t α − 1 ) ‖ u 0 ‖ X 1 ≤ r ˜ 3 , M 1 C w ( t ) t 1 + η − σ / 2 + ς ˜ ∫ b t 1 ( 1 − s ) − σ / 2 s ς ˜ d s ≤ r ˜ 6 , M 2 λ 1 ( ε − δ ( ε ) − 1 ) / 2 C m ρ t η B ( α , 1 − ρ η ) ( b α − ρ η − t α − ρ η ) + t α − ρ η ∫ b t 1 ( 1 − s ) α − 1 s − ρ η d s ≤ r ˜ 6 , M 2 λ 1 ( ε − δ ( ε ) − 1 ) / 2 C w ( b ) t η B ( α , 1 + ς ˜ ) ( b α + ς ˜ − t α + ς ˜ ) + t α + ς ˜ ∫ b t 1 ( 1 − s ) α − 1 s ς ˜ d s ≤ r ˜ 6

and

(4.2) M 1 C m ρ t 1 + η − σ / 2 − η ρ ∫ b t 1 ( 1 − s ) − σ / 2 s − η ρ d s ≤ r ˜ 6 , M 1 C w ( t ) t 1 − σ / 2 + ς ∫ b t 1 ( 1 − s ) − σ / 2 s − η + ς d s ≤ 1 3 ,

where m = max { r , r ˜ + b ˜ η ‖ u ( b ) ‖ X 1 + ε } . From the condition ( 1 − α ) ε / 2 < η < min 1 + ς , 1 − ( 1 − α ) ( 1 − δ ( ε ) ) / 2 ρ , we can easily check that 1 + η − σ / 2 − ρ η > 0 , 1 + η − σ / 2 + ς ˜ ≥ 1 + η − σ / 2 − δ ( ε ) > 0 . The facts ρ η < 1 and 1 + ς ˜ ≥ 1 − δ ( ε ) > 0 imply that the beta functions in (4.1) are well-defined. We can also verify that functions ( 1 − s ) − σ / 2 s ς ˜ , ( 1 − s ) − σ / 2 s − η ρ and ( 1 − s ) − σ / 2 s − η + ς are integrable for s ∈ b t , 1 and t ∈ ( b , b ˜ ] .

We define the set B ( r ˜ , b ˜ ) of all functions v ∈ C η ( [ 0 , b ˜ ] ; X 1 + ε ) ∩ C ( [ 0 , b ˜ ] ; X 1 ) such that v ( t ) = u ( t ) for t ∈ [ 0 , b ] and sup s ∈ [ b , t ] s η ‖ v ( s ) − u ( b ) ‖ X 1 + ε ≤ r ˜ for t ∈ [ b , b ˜ ] . For any v ∈ B ( r ˜ , b ˜ ) and t ∈ [ 0 , b ˜ ] , we still use the definition of T in (3.10).

Now we prove that T maps B ( r ˜ , b ˜ ) into B ( r ˜ , b ˜ ) .

For any v ∈ B ( r ˜ , b ˜ ) , in the case t ∈ [ 0 , b ] , we know that v ( t ) = u ( t ) , then from Theorem 3.1, we obtain that T maps B ( r ˜ , b ˜ ) into itself and ( T v ) ( t ) = ( T u ) ( t ) = u ( t ) . Now we prove that ( T v ) ( t ) ∈ C η ( [ 0 , b ˜ ] ; X 1 + ε ) ∩ C ( [ 0 , b ˜ ] ; X 1 ) . In fact, using a similar process of proof provided in Theorem 3.1, we can show that for any v ∈ B ( r ˜ , b ˜ ) , ( T v ) ( t ) ∈ C η ( [ 0 , b ˜ ] ; X 1 + ε ) ∩ C ( [ 0 , b ˜ ] ; X 1 ) , here we omit the details.

There left to prove that sup s ∈ [ b , t ] s η ‖ ( T v ) ( s ) − u ( b ) ‖ X 1 + ε ≤ r ˜ holds for t ∈ [ b , b ˜ ] .

In fact, for any v ∈ B ( r ˜ , b ˜ ) , from Theorem 3.1 and the definition of B ( r ˜ , b ˜ ) , we have u ( b ) = S α ( b ) u 0 + ∫ 0 b S α ( b − s ) f ( s , u ( s ) ) d s = S α ( b ) u 0 + ∫ 0 b S α ( b − s ) f ( s , v ( s ) ) d s , then for any t ∈ [ b , b ˜ ] , we obtain

‖ ( T v ) ( t ) − u ( b ) ‖ X 1 + ε ≤ ‖ ( S α ( t ) − S α ( b ) ) u 0 ‖ X 1 + ε + ∫ b t S α ( t − s ) f ( s , v ( s ) ) d s X 1 + ε + ∫ 0 b ( S α ( t − s ) − S α ( b − s ) ) f ( s , v ( s ) ) d s X 1 + ε ≕ Q 1 + Q 2 + Q 3 .

From Lemma 3.2, we obtain that

Q 1 ≤ M 2 λ 1 ( ε − 2 ) / 2 ( b α − 1 − t α − 1 ) ‖ u 0 ‖ X 1 .

And then, similar to (3.11), we use Lemma 3.1 to estimate the second term, that is,

Q 2 ≤ ∫ b t ‖ S α ( t − s ) f ( s , v ( s ) ) ‖ X 1 + ε d s ≤ M 1 C ∫ b t ( t − s ) − σ / 2 s − ρ η d s ( sup s ∈ [ b , t ] s η ‖ v ( s ) ‖ X 1 + ε ) ρ + M 1 C w ( t ) ∫ b t ( t − s ) − σ / 2 s ς ˜ d s ≤ M 1 C t 1 − σ / 2 − ρ η ∫ b t 1 ( 1 − s ) − σ / 2 s − ρ η d s ( sup s ∈ [ b , t ] s η ‖ v ( s ) ‖ X 1 + ε ) ρ + M 1 C w ( t ) t 1 − σ / 2 + ς ˜ ∫ b t 1 ( 1 − s ) − σ / 2 s ς ˜ d s .

From Lemma 3.2 and (2.3), we can estimate the third term as follows:

Q 3 ≤ ∫ 0 b ‖ ( S α ( t − s ) − S α ( b − s ) ) f ( s , v ( s ) ) ‖ X 1 + ε d s ≤ M 2 λ 1 ( ε − δ ( ε ) − 1 ) / 2 ∫ 0 b ( ( b − s ) α − 1 − ( t − s ) α − 1 ) ‖ f ( s , v ( s ) ) ‖ X δ ( ε ) d s ≤ M 2 λ 1 ( ε − δ ( ε ) − 1 ) / 2 C ∫ 0 b ( ( b − s ) α − 1 − ( t − s ) α − 1 ) ( ‖ v ( s ) ‖ X 1 + ε ρ + w ( s ) s ς ˜ ) d s ≤ M 2 λ 1 ( ε − δ ( ε ) − 1 ) / 2 C ( sup s ∈ [ 0 , b ] s η ‖ v ( s ) ‖ X 1 + ε ) ρ ∫ 0 b ( ( b − s ) α − 1 − ( t − s ) α − 1 ) s − ρ η d s + M 2 λ 1 ( ε − δ ( ε ) − 1 ) / 2 C w ( b ) ∫ 0 b ( ( b − s ) α − 1 − ( t − s ) α − 1 ) s ς ˜ d s = M 2 λ 1 ( ε − δ ( ε ) − 1 ) / 2 C ( sup s ∈ [ 0 , b ] s η ‖ v ( s ) ‖ X 1 + ε ) ρ ∫ 0 b ( b − s ) α − 1 s − ρ η d s − ∫ 0 t ( t − s ) α − 1 s − ρ η d s + M 2 λ 1 ( ε − δ ( ε ) − 1 ) / 2 C ( sup s ∈ [ 0 , b ] s η ‖ v ( s ) ‖ X 1 + ε ) ρ ∫ b t ( t − s ) α − 1 s − ρ η d s + M 2 λ 1 ( ε − δ ( ε ) − 1 ) / 2 C w ( b ) ∫ 0 b ( b − s ) α − 1 s ς ˜ d s − ∫ 0 t ( t − s ) α − 1 s ς ˜ d s + M 2 λ 1 ( ε − δ ( ε ) − 1 ) / 2 C w ( b ) ∫ b t ( t − s ) α − 1 s ς ˜ d s .

From the definition of beta function, and by virtue of the facts ρ η < 1 , ς ˜ ≥ − δ ( ε ) > − 1 , we can compute the integrals of above inequality as follows:

∫ 0 b ( b − s ) α − 1 s − ρ η d s = b α − ρ η B ( α , 1 − ρ η ) , ∫ 0 b ( b − s ) α − 1 s ς ˜ d s = b α + ς ˜ B ( α , 1 + ς ˜ ) ,

similarly,

∫ 0 t ( t − s ) α − 1 s − ρ η d s = t α − ρ η B ( α , 1 − ρ η ) , ∫ 0 t ( t − s ) α − 1 s ς ˜ d s = t α + ς ˜ B ( α , 1 + ς ˜ ) .

Substituting the above results into the original inequality, we obtain that

Q 3 ≤ M 2 λ 1 ( ε − δ ( ε ) − 1 ) / 2 C B ( α , 1 − ρ η ) ( sup s ∈ [ 0 , b ] s η ‖ v ( s ) ‖ X 1 + ε ) ρ ( b α − ρ η − t α − ρ η ) + M 2 λ 1 ( ε − δ ( ε ) − 1 ) / 2 C ( sup s ∈ [ 0 , b ] s η ‖ v ( s ) ‖ X 1 + ε ) ρ t α − ρ η ∫ b t 1 ( 1 − s ) α − 1 s − ρ η d s + M 2 λ 1 ( ε − δ ( ε ) − 1 ) / 2 C w ( b ) B ( α , 1 + ς ˜ ) ( b α + ς ˜ − t α + ς ˜ ) + t α + ς ˜ ∫ b t 1 ( 1 − s ) α − 1 s ς ˜ d s .

Observing that

sup s ∈ [ 0 , t ] s η ‖ v ( s ) ‖ X 1 + ε ≤ sup s ∈ [ 0 , t ] s η ‖ v ( s ) − u ( b ) ‖ X 1 + ε + t η ‖ u ( b ) ‖ X 1 + ε ≤ m ,

and in virtue of conditions (4.1) and (4.2), we obtain that

t η ‖ T v ( t ) − u ( b ) ‖ X 1 + ε ≤ t η M 2 λ 1 ( ε − 2 ) / 2 ( b α − 1 − t α − 1 ) ‖ u 0 ‖ X 1 + M 1 C t 1 + η − σ / 2 − ρ η m ρ ∫ b t 1 ( 1 − s ) − σ / 2 s − ρ η d s + M 1 C w ( t ) t 1 + η − σ / 2 + ς ˜ ∫ b t 1 ( 1 − s ) − σ / 2 s ς ˜ d s + M 2 λ 1 ( ε − δ ( ε ) − 1 ) / 2 C m ρ t η B ( α , 1 − ρ η ) ( b α − ρ η − t α − ρ η ) + t α − ρ η ∫ b t 1 ( 1 − s ) α − 1 s − ρ η d s + M 2 λ 1 ( ε − δ ( ε ) − 1 ) / 2 C w ( b ) t η B ( α , 1 + ς ˜ ) ( b α + ς ˜ − t α + ς ˜ ) + t α + ς ˜ ∫ b t 1 ( 1 − s ) α − 1 s ς ˜ d s ≤ r ˜ 3 + r ˜ 6 + r ˜ 6 + r ˜ 6 + r ˜ 6 = r ˜ .

Thus, we conclude that T maps B ( r ˜ , b ˜ ) into B ( r ˜ , b ˜ ) .

In the following, we prove that T is a contraction on B ( r ˜ , b ˜ ) .

For any v 1 , v 2 ∈ B ( r ˜ , b ˜ ) , it is easy to see that ( T v 1 ) ( t ) − ( T v 2 ) ( t ) = 0 for t ∈ [ 0 , b ] , we just have to discuss the case when t ∈ [ b , b ˜ ] , in this case, we have

( T v 1 ) ( t ) − ( T v 2 ) ( t ) = ∫ b t S α ( t − s ) ( f ( s , v 1 ( s ) ) − f ( s , v 2 ( s ) ) ) d s .

It follows from the process of proof of Lemma 3.4 that

t η ‖ ( T v 1 ) ( t ) − ( T v 2 ) ( t ) ‖ X 1 + ε ≤ 2 M 1 C ( V max ) ρ − 1 t η ∫ b t ( t − s ) − σ / 2 s − η ρ d s ‖ v 1 − v 2 ‖ C η ( [ 0 , b ˜ ] ; X 1 + ε ) + M 1 C w ( t ) t η ∫ b t ( t − s ) − σ / 2 s − η + ς d s ‖ v 1 − v 2 ‖ C η ( [ 0 , b ˜ ] ; X 1 + ε )

≤ 2 M 1 C m ρ − 1 t 1 + η − σ / 2 − η ρ ∫ b t 1 ( 1 − s ) − σ / 2 s − η ρ d s ‖ v 1 − v 2 ‖ C η ( [ 0 , b ˜ ] ; X 1 + ε ) + M 1 C w ( t ) t 1 − σ / 2 + ς ∫ b t 1 ( 1 − s ) − σ / 2 s − η + ς d s ‖ v 1 − v 2 ‖ C η ( [ 0 , b ˜ ] ; X 1 + ε ) ,

where V max = max { ‖ v 1 ‖ C η ( [ 0 , b ˜ ] ; X 1 + ε ) , ‖ v 2 ‖ C η ( [ 0 , b ˜ ] ; X 1 + ε ) } .

Applying the fact r ˜ m ≤ 1 into the first condition of (4.2), we obtain that

(4.3) M 1 C m ρ − 1 t 1 + η − σ / 2 − η ρ ∫ b t 1 ( 1 − s ) − σ / 2 s − η ρ d s ≤ 1 6 r ˜ m ≤ 1 6 .

Combined with the above discussion, by virtue of (4.3) and the second condition of (4.2), we obtain that

t η ‖ ( T v 1 ) ( t ) − ( T v 2 ) ( t ) ‖ X 1 + ε ≤ 2 3 ‖ v 1 − v 2 ‖ C η ( [ 0 , b ˜ ] ; X 1 + ε ) ,

which implies that T is a strict contraction. By virtue of Banach fixed point theorem, we obtain that there exists a unique fixed point u ˜ ∈ B ( r ˜ , b ˜ ) , which implies that u ˜ is a continuation of u in [ 0 , b ˜ ] .

Now we prove the continuation u ˜ is unique in [ 0 , b ˜ ] .

Assume that u 1 and u 2 are two continuations of the ε -regular mild solution u obtained by Theorem 3.1 in [ 0 , b ˜ ] . From the definition of continuation, it is obvious that u 1 ( t ) = u 2 ( t ) = u ( t ) for t ∈ [ 0 , b ] . Thus, we only need to check that u 1 ( t ) = u 2 ( t ) for t ∈ [ b , b ˜ ] in the sequel. From the proof process of Lemma 3.4, we have

‖ u 1 ( t ) − u 2 ( t ) ‖ X 1 + ε = ∫ 0 t S α ( t − s ) ( f ( s , u 1 ( s ) ) − f ( s , u 2 ( s ) ) ) d s X 1 + ε = ∫ b t S α ( t − s ) ( f ( s , u 1 ( s ) ) − f ( s , u 2 ( s ) ) ) d s X 1 + ε ≤ M 1 ∫ b t ( t − s ) − σ / 2 C ‖ u 1 ( s ) − u 2 ( s ) ‖ X 1 + ε ( ‖ u 1 ( s ) ‖ X 1 + ε ρ − 1 + ‖ u 2 ( s ) ‖ X 1 + ε ρ − 1 + w ( s ) s ς ) d s ≤ M 1 C M 3 ∫ b t ( t − s ) − σ / 2 ‖ u 1 ( s ) − u 2 ( s ) ‖ X 1 + ε d s ,

where M 3 = ( sup s ∈ [ b , b ˜ ] ‖ u 1 ( s ) ‖ X 1 + ε ) ρ − 1 + ( sup s ∈ [ b , b ˜ ] ‖ u 2 ( s ) ‖ X 1 + ε ) ρ − 1 + w ( b ˜ ) b ς . For the reason that ( t − s ) − σ / 2 is continuous and nonnegative for s ∈ ( b , t ) , the singular Grönwall’s inequality implies that the relation u 1 ( t ) = u 2 ( t ) also holds for t ∈ [ b , b ˜ ] . Thus, we conclude that u ˜ is the unique continuation of u in [ 0 , b ˜ ] . This completes the proof.□

Remark 4.1

We note that the conclusion in Theorem 4.1 still holds if we change the existence time b of u ( t ) into the maximal time we can obtain from Theorem 3.1.

Theorem 4.2

Assume the conditions in Theorem 3.1 hold and u is an ε -regular mild solution of problem (1.1) with maximal time of existence b max . Then either b max = ∞ or limsup t → b max − ‖ u ( t ) ‖ X 1 + ε = ∞ .

Proof

Assume that the maximal time of existence b max < ∞ and there exists a positive constant R such that the solution u ( t ) satisfies sup t ∈ ( 0 , b max ) t η ‖ u ( t ) ‖ X 1 + ε ≤ R . In the following, we shall prove that this assumption is a contradiction.

Taking a sequence { t n } n ∈ N + ( t n < b max ) which satisfies t n → b max − as n → ∞ . Now we show that { u ( t n ) } n ∈ N + is a Cauchy sequence in X 1 + ε .

In fact, for 0 < t m < t n < b max , we have

u ( t n ) − u ( t m ) = ( S α ( t n ) − S α ( t m ) ) u 0 + ∫ 0 t n S α ( t n − s ) f ( s , u ( s ) ) d s − ∫ 0 t m S α ( t m − s ) f ( s , u ( s ) ) d s = ( S α ( t n ) − S α ( t m ) ) u 0 + ∫ t m t n S α ( t n − s ) f ( s , u ( s ) ) d s + ∫ 0 t m ( S α ( t n − s ) − S α ( t m − s ) ) f ( s , u ( s ) ) d s .

By using an analogous argument presented in the process of proof of Theorem 3.1 (Step I ) and Theorem 4.1 ( Q 3 ), we can easily obtain that

‖ u ( t n ) − u ( t m ) ‖ X 1 + ε ≤ ‖ ( S α ( t n ) − S α ( t m ) ) u 0 ‖ X 1 + ε + ∫ t m t n S α ( t n − s ) f ( s , u ( s ) ) d s X 1 + ε + ∫ 0 t m ( S α ( t n − s ) − S α ( t m − s ) ) f ( s , u ( s ) ) d s X 1 + ε ≤ M 2 λ 1 ( ε − 2 ) / 2 ( t m α − 1 − t n α − 1 ) ‖ u 0 ‖ X 1 + M 1 C R ρ t n 1 − σ / 2 − ρ η ∫ t m t n 1 ( 1 − s ) − σ / 2 s − ρ η d s + M 1 C w ( t n ) t n 1 − σ / 2 + ς ˜ ∫ t m t n 1 ( 1 − s ) − σ / 2 s ς ˜ d s + M 2 λ 1 ( ε − δ ( ε ) − 1 ) / 2 C ∫ 0 t m ( ( t m − s ) α − 1 − ( t n − s ) α − 1 ) ( R ρ s − ρ η + w ( t m ) s ς ˜ ) d s ,

where ( 1 − s ) − σ / 2 s − ρ η , ( 1 − s ) − σ / 2 s ς ˜ and ( ( t m − s ) α − 1 − ( t n − s ) α − 1 ) ( R ρ s − ρ η + w ( t m ) s ς ˜ ) are integrable according to our earlier discussion. Moreover, for given θ > 0 , since { t n } n ∈ N + ( t n < b max ) is a sequence which converges to b max − , then there exists an N ∈ N + such that for any n , m ≥ N , ∣ t n − t m ∣ can be small enough so that

M 2 λ 1 ( ε − 2 ) / 2 ( t m α − 1 − t n α − 1 ) ‖ u 0 ‖ X 1 < θ 3 , M 1 C R ρ t n 1 − σ / 2 − ρ η ∫ t m t n 1 ( 1 − s ) − σ / 2 s − ρ η d s + M 1 C w ( t n ) t n 1 − σ / 2 + ς ˜ ∫ t m t n 1 ( 1 − s ) − σ / 2 s ς ˜ d s < θ 3 , M 2 λ 1 ( ε − δ ( ε ) − 1 ) / 2 C ∫ 0 t m ( ( t m − s ) α − 1 − ( t n − s ) α − 1 ) ( R ρ s − ρ η + w ( t m ) s ς ˜ ) d s < θ 3 .

Thus, for n , m ≥ N , we have

‖ u ( t n ) − u ( t m ) ‖ X 1 + ε < θ ,

which implies that { u ( t n ) } n ∈ N + is a Cauchy sequence in X 1 + ε , it follows that lim t → b max − u ( t ) exists and is finite, thus we can extend u ( t ) from ( 0 , b max ) to ( 0 , b max ] in X 1 + ε . Furthermore, Theorem 4.1 implies that u ( t ) can be extended to a bigger interval than ( 0 , b max ] , which contradicts the assumption that b max is the maximal time of existence. Therefore, we conclude that either lim sup t → b max − ‖ u ( t ) ‖ X 1 + ε = ∞ or b max = ∞ . The proof is completed.□

5 Conclusion

In this article, we investigated the fractional Rayleigh-Stokes problem in a heated second grade fluid with critical conditions. The interpolation–extrapolation scale and ε -regular map were introduced at first. And then, we gave the definition of ε -regular mild solutions and some properties of solution operator. The existence and uniqueness of ε -regular mild solutions of problem (1.1) were obtained by virtue of the fixed point theorem. We also established some properties of the ε -regular mild solution when t → 0 + in a weighted continuous function space and the continuous dependence on initial conditions. Moreover, we presented the unique continuation and blow-up alternative results for the ε -regular mild solution. Note that if we redefine the weighted continuous function space as the space of all continuous functions v ∈ C ( ( 0 , b ] , X 1 + ε ) with lim t → 0 + t η v ( t ) exists and is finite, the well-posedness results in Theorem 3.1, the unique continuation result in Theorem 4.1 and the blow-up alternative result in Theorem 4.2 still hold. Particularly, under the above definition of weighted continuous function space, we can extend our results into the case η = ( 1 − α ) ε / 2 , the difference is that in this situation we can only obtain that lim t → 0 + t η ‖ v ( t ) ‖ X 1 + ε exists and is finite rather than goes to zero obtained in Theorem 3.1.

Funding information: The Deanship of Scientific Research (DSR) at King Abdulaziz University, Jeddah, Saudi Arabia has funded this project, under grant no. (FP-036-43).
Conflict of interest: The authors state no conflict of interest.

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Received: 2021-10-11

Revised: 2022-03-01

Accepted: 2022-03-02

Published Online: 2022-06-12

This work is licensed under the Creative Commons Attribution 4.0 International License.

Well-posedness and blow-up results for a class of nonlinear fractional Rayleigh-Stokes problem

Abstract

1 Introduction

2 Preliminaries

Definition 2.1

Definition 2.2

Lemma 2.1

Lemma 2.2

3 Existence of mild solutions

Definition 3.1

Lemma 3.1

Proof

Remark 3.1

Lemma 3.2

Proof

Remark 3.2

Lemma 3.3

Proof

Lemma 3.4

Proof

Theorem 3.1

Proof

4 Continuation and blow-up alternative

Definition 4.1

Theorem 4.1

Proof

Remark 4.1

Theorem 4.2

Proof

5 Conclusion

References

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