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A study of the time fractional Navier-Stokes equations for vertical flow

  • Navier-Stokes (NS) equations dealing with gravitational force with time-fractional derivatives are discussed in this paper. These equations can be used to predict fluid velocity and pressure for a given geometry. This paper investigates the local and global existence and uniqueness of mild solutions to NS equations for the time fractional differential operator. We also work on the regularity effects of such types of equations were caused by orthogonal flow.

    Citation: Abdelkader Moumen, Ramsha Shafqat, Azmat Ullah Khan Niazi, Nuttapol Pakkaranang, Mdi Begum Jeelani, Kiran Saleem. A study of the time fractional Navier-Stokes equations for vertical flow[J]. AIMS Mathematics, 2023, 8(4): 8702-8730. doi: 10.3934/math.2023437

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  • Navier-Stokes (NS) equations dealing with gravitational force with time-fractional derivatives are discussed in this paper. These equations can be used to predict fluid velocity and pressure for a given geometry. This paper investigates the local and global existence and uniqueness of mild solutions to NS equations for the time fractional differential operator. We also work on the regularity effects of such types of equations were caused by orthogonal flow.



    To explain the flow of incompressible fluids in fluid mechanics, Navier-Stokes (NS) equations are applied as partial differential equations in order to discuss such fluids. Such equations constitute a modification of the equations composed by the Swiss mathematician Leonhard. In the 18th century, Euler illustrated the movement of frictionless and incompressible fluids. For the most realistic and challenging viscous fluid issue in 1821 Claude-Louis Navier, a French engineer, presented the element of viscosity (friction). Sir George Gabriel Stokes, a British physicist and mathematician, expanded this work in the mid-nineteenth century, however, comprehensive solutions have only been achieved for the case of 2-D flows. The complex vortexes and disturbance, or disruption, that arise in 3-D fluid flows (including gas) as velocities rise are demonstrated to be uncontrolled for every numerical analysis approach except approximation.

    Because they characterize the physics of many scientific and engineering events, NS equations are useful. They might be models of climate, ocean waves, water flow in a pump, or air flow. The NS equations, in both their complete and simplified versions are applied for assistance with aviation and automobile design, blood flow studies, the design of power plants, pollution monitoring and a variety of other activities.

    The NS equation, in modern notation for vertical flow:

    ρ(˜ut+(˜u.)˜u)=P+μ2˜u+ρg (1.1)

    where ˜u is defined as the fluid velocity vector, P is the fluid pressure, ρ is the fluid density, μ is the dynamic viscosity and 2 is the Laplacian operator.

    The NS equations in mathematics demonstrate the conservation of speed and mass for Newtonian fluids. Sometimes the condition applied on equation associated with stress, temperature and body weight. They result from the application of Isaac Newton's second law to water movement, as well as the concept that the fluid stress is the sum of several viscous time (rate and velocity) and pressure respectively which describes the viscous flow. The difference between Euler equations and Navier equations is that the Euler equations model is just an inviscid flow whereas NS equations are viscosity sensitive. For that reason, NS equations are a parabolic with improved analytical properties at the cost of a constrained mathematical framework (for example, they are never fully integrated).

    These equations are the central part of fluid flow modeling. Solving them for a set of specific boundary conditions (entrances, exits and walls) predicts fluid velocity and pressure for a given geometry. In view of the fact that complexity recognizes only a limited number of these equations analytical solutions [1], it is relatively convenient, for instance, to solve these equations for a flow between two parallel plates, or for the flow in a circular pipe. Abbas et al. [2] and Mehmood et al. [3] solved ordinary differential equations. Niazi et al.[4], Iqbal et al. [5], Shafqat et al. [6], Alnahdi et al. [7], Khan et al. [8] and Abuasbeh et al. [9,10,11,12] investigated the existence and uniqueness of the fuzzy fractional evolution equations.

    The Cauchy problem for NS equations in incompressible fluids is given by

    {γυ˜uw˜u+(˜u.)˜u=p+g,υ>0,.˜u=0,˜u|Ω=0,˜u(υ,x)=axsin(γ)+bycos(γ). (1.2)

    If the fluid strikes orthogonally, i.e., if γ=π2 then we have

    ˜u(0)=ax.

    Therefore we have,

    {γυ˜uw˜u+(˜u.)˜u=p+g,υ>0,.˜u=0,˜u|Ω=0,˜u(0,x)=ax, (1.3)

    while γυ denotes the fractional order Caputo derivative. At a point xω and time υ>0 the velocity field vector is represented by ˜u=(˜u1(υ,x),˜u2(υ,x),...,˜un(υ,x)), ρ=ρ(υ,x) denotes the pressure and υ is the viscosity. The gravitational force or body force is represented by g=g(υ,x) while the initial velocity is represented by ax. First of all we assume that Ω has a smooth boundary so in order to remove the pressure term we must apply, Helmholtz-Leray projector P in 1.3, which gives

    {γυ˜uwP˜u+P(˜u.)˜u=Pg,υ>0,.˜u=0,˜u|Ω=0,˜u(0,x)=ax. (1.4)

    In the divergence-free function space under discussion, the operator wP under Dirichlet boundary conditions is mainly the Stokes operator A. So, writing 1.3 in abstract form, we have

    {cDγυ˜u(υ)=A˜u+F(˜u,w)+Pg,υ>0,˜u(0)=ax, (1.5)

    where F(˜u,w) = P(˜u.)w. If the Helmholtz-Leray projection P and the Stokes operator A makes, the solution of (1.5) is also the solution of (1.3). The fundamental goal is to prove the existence and uniqueness of global and local mild solutions of problem (1.5) in Hγ,r. Moreover, we also establish the regularity conclusions which claim that if Pg is Hölder continuous then there is a unique classical solution ˜u(υ) such that A˜u and cDγυ˜u(υ) hold Hölder continuity in Jr.

    In this section, we define the Gamma function, fractional order integral, Riemann-Liouville (RL) fractional derivative, Caputo fractional derivative and some more definitions, lemmas and theorems. For a brief review of fractional calculus definitions and properties, see [13].

    Let the half space in Rn, i.e., Ω=H=(x1,...,xn):xn>0 be the open subset of Rn, where n3. Let 1<r<. Then we have the Hödge projection which is a bounded projection P on (Lr(Ω)n), the range for which is as follows:

    Cσ(H)=(v(C(H))n:.v=0), (2.1)

    and the null space of which is given as

    v(C(H))n:v=.ϕ,ϕC(H). (2.2)

    For a suitable approach, let Jr = ¯Cσ(H)|.|r, which is closed subspace of (Lr(H))n, A=υPΔ is the Stokes operator in Jr containing the domain Dr(A)=Dr(Δ)Jr. The Stokes operator, named George Gabriel Stokes is an unbounded linear operator which is used in the theory of partial differential equations and specifically in the fields of fluid dynamics and electro magnetics;

    Dr(Δ)=v(W2,r(H)n):v|H=0.

    We have to introduce definitions of fractional power spaces that are related to A. For γ>0 and vJr, define the following:

    Aγv=1Γ(γ)0υγ1eυAudυ.

    Therefore, Aγ is bounded [14], just like the injective operator on Jr. Suppose that Aγ is the inverse of Aγ. For γ>0 we symbolize the space Hγ,r by the extent of Aγ with the norm

    |v|Hγ,r=|Aγv|r.

    Definition 2.1. The fractional integration of order γ>0 for a function f is defined as

    Iγ0f(υ)=1Γ(γ)υ0(υs)γ1f(s)ds,υ>0.

    The RL fractional derivative for a function v:[0,)Rn of order γRn is defined by

    L0Dγυv(υ)=dndυn(gnγv)(υ),υ0,n1<γ<n.

    The RL fractional order integral is defined as

    Jγυv(υ):=gγv(υ)=1Γ(γ)υ0(υs)γ1v(s)ds,υ[0,].

    Thus by using the definitions of the RL fractional-order integral, we construct the Caputo fractional order differential operator.

    Definition 2.2. [15] The Caputo fractional order derivative is defined as follows

    c0Dγυv(υ)=ddυ(J1γυ[v(υ)v(0)])=ddυ(1Γ(1γ)υ0(υs)γ[v(s)v(0)]ds),υ>0.

    Definition 2.3. [16] The Mittag-Leffler function was introduced by Magnus Gustaf (Gösta) Mittag-Leffler (Swedish mathematician) in 1902. It is a simple conclusion of the exponential function. Recently, some researchers have been focusing on the Mittag-Leffler function because of its application in the analysis of fractional differential equations (FDEs). It occurs often in the solutions of FDEs and fractional integral equations. The Mittag-Leffler function with the one-parameter Eγ(t) is defined as

    Eγ(υ)=k=0υkΓ(γk+1),υC,R(γ)>0.

    Now, let us consider the generalized Mittag-Leffler functions:

    Eγ(υγA)=0Mγ(s)esυγAds

    and

    Eγ,γ(υγA)=0γsMγ(s)esυγAds,

    where

    Mγ(υ):=n=0(υ)nn!Γ(γ(n)+1γ).

    The function Mγ is also known as the Mainardi function. In order to distinguish between the fundamental solutions for some standard boundary value problems, Mainardi introduced such a type of functions that is a special type of Wright function. It is impressive that the Mainardi function plays a role as a bridge between the classical abstract theories and fractional theories.

    Theorem 2.1. Suppose that fn:Rn[,] denotes (Lebesgue) measurable functions such that the pointwise limit f(x)=limnfn(x) exists. Assume that there is an integrable g:Rn[0,] with |fn(x)|g(x) for each xRn. Then, f is integrable as is fn for each n, and

    limnRnfndμ=Rnlimnfndμ=Rnfdμ.

    (ⅰ) Eγ,γ(υγA)=12πιEγ,γΓθ(μυγ)(μI+A)1dμ;

    (ⅱ) AγEγ,γ(υγA)=12πιμγΓθEγ,γ(μυγ)(μI+A)1dμ.

    Proof. See the results in [17].

    Let γ(0,1) and 1<r<, λ>0. Then Mainardi function possess the following properties:

    (ⅰ) Mγ(υ)0 for all υ0;

    (ⅱ) 0υrMγ(υ)dυ=Γ(r+1)Γ(γr+1);

    (ⅲ) L{γυMγ(υ)}(z)=Eγ,γ(z);

    (ⅳ) L{Mγ(υ)}(z)=Eγ(z);

    (ⅴ) L{γυ(1+γ)Mγ(υγ)}(λ)=eλγ.

    Proof. The proof of this proposition can be found in [18].

    Lemma 2.1. For υ>0, the operators Eγ(υγA) and Eγ,γ(υγA) in the uniform operator topology are continuous and well defined from X to X. Then continuity is uniform on [r,) for every r>0.

    Lemma 2.2. [19] Let 0<γ<1. Then

    (ⅰ) vX,limt0+Eγ(υγA)v=v;

    (ⅱ) vD(A) and υ>0,CDγυEγ(υγA)v=AEγ(υγA)v;

    (ⅲ) vX and Eγ(υγA)v=υγ1AEγ,γ(υγA)v;

    (ⅳ) forυ>0,Eγ(υγA)v=I1γυ{(υγ1Eγ,γ(υγA)u)}.

    Lemma 2.3. Suppose that 1<r< and γ1γ2. Then, there exist a constant C=C(γ1,γ2) such that

    |eυAv|Hγ2,rCυ(γ2γ1)|v|Hγ1,rasυ>0forvHγ1,r.

    Moreover, limυ0υ(γ2γ1)|eυAv|γ2,rH=0.

    Lemma 2.4. Suppose that 1<r< and γ1γ2. For any >0, there exists a constant C1=C1(γ1,γ2) such that

    |Eα(υγA)|Hγ2,rC1υα(γ2γ1)|v|Hγ1,r

    and

    |Eγ,γ(υγA)|Hγ2,rC1υγ(γ2γ1)|v|Hγ1,r

    for all vHγ1,r and υ(0,T]. Therefore, limυ0υα(γ2γ1)|Eγ(υγA)v|Hγ2,r=0.

    Proof. The proof of this lemma can be studied in [17].

    Theorem 2.2. If f(υ) is defined on the interval [c,d] is Riemann-integrable, then |f(υ)| is also Riemann integrable defined on the interval [c,d] and

    |dcf(υ)dυ|dc|f(υ)|dυ.

    Theorem 2.3. Suppose that f:[a,b]Rn is continuous and g:IRn is continuously differentiable with image g(I)[a,b], where IRn is some open interval showing that the function

    F(s)=g(s)af(υ)dυ

    is continuously differentiable on I.

    Theorem 2.4. Let (υ):υ0X be a C0 semi group on X. Then, the following holds

    (i) If C:D(G)XX, then G is said to be dense and close defined by linear operators. Therefore, υ[0,)(υ)xX is continuously differentiable for any xD(G):

    ddυ(υ)x=G(υ)x=(υ)Gx,forυ>0.

    (ii) Then there exists σ>0 such that Re(λ)>0; given λρ(C), we have

    (λC)1x=0eλυ(υ)xdυforallxX.

    Theorem 2.5. Let γ(0,1] and suppose that the positive sectorial operator is A:D(A)XX. Thus, the operators {Eγ(υγA):υ0} and {Eγ,γ(υγA):υ0} as follows:

    Eγ(υγA)=0Mγ(s)sυγAds,υ0

    and

    Eγ,γ(υγA)=0γsMγ(s)sυγds,υ0,

    where (υ):υ0} defines the C0 semigroup which is generated by A.

    Let γ(0,1) and consider the A:D(A)XX {is a positive} sectorial operator. Then for any xX, it holds that

    L{Eγ(υγA)x}(λ)=λγ1(λγ+A)1x,L{Eγ,γ(υγA)x}(λ)=(λγ+A)1x.

    Proof. The first equality can be proved {analogously the second equality is for any} xX; observe that by Theorem 2.4,

    L{Eγ,γ(υγA)x}(λ)=0eλυυγ1Eγ,γ(υγA)xdυ=0eλυυγ1(0γsMγ(s)(sυγ)xds)dυ.

    Now by using s=ωυγ, we conclude the following

    L{Eγ,γ(υγA)x}(λ)=0eλυυγ1(0γ(ωυγ)Mγ(ωυγ)T(ω)xυγdω)dυ=0ω(0γυ(1+γ)Mγ(ωυγ)eλυdυ)T(ω)xdω.

    Choose

    H=0γυ(1+γ)Mγ(ωυγ)eλυ.

    By taking υ=τω1γ from Lemma 2.4, we have that

    H=0γ(τω1γ)(1+γ)Mγ(ω(τω1γ)γ)eλ(τω1γ)ω1γdτ=ω10γτ(1+γ)Mγ(τγ)e(λω1γ)dτ=ω1eλγω.

    Therefore, by Theorem 2.5, we have

    L{Eγ,γ(υγA)x}(λ)=0eλγω(ω)xdω=(λγ+A)1x.

    Lemma 2.5. If ˜u(υ) is the solution of (1.5) for ˜u(0)=ax, then ˜u(υ) is given as

    ˜u(υ)=aυ+1Γ(γ)υ0(υs)γ1(A˜u(s)+h(s))dsasυ0;

    therefore we get

    ˜u(υ)=υ0Eγ((υs)γA)ads+υ0(υs)γ1Eγ,γ((υs)γA)h(s)ds,

    where

    Eγ(υ)=0q(θ)Mγ(θ)Q(υγθ)dθ.

    Proof. [20] By using the above lemma rewriting (1.5) and applying the RL derivative, we get

    ˜u(υ)=˜u(0)+1Γ(γ)υ0(υs)γ1(A˜u(s)+F(u(s),w(s))+Pg(s))dsasυ0˜u(υ)=aυ+1Γ(γ)υ0(υs)γ1(A˜u(s)+F(˜u(s),w(s))+Pg(s))dswhileυ0.

    By applying the Laplace transform, we have the following

    ˜u(λ)=aλ2+1λγ{A˜u(λ)}+1λγ{F˜u(λ),w(λ)}+1λγ{Pg(λ)}.

    Simplification yields

    (λγ+A)˜u(λ)=aλγ2+F(˜u(λ),w(λ))+Pg(λ)˜u(λ)=aλγ2(λγ+A)1+F(˜u(λ),w(λ))(λγ+A)1+Pg(λ)(λγ+A)1˜u(λ)=aλ1λγ1(λγ+A)1+F(˜u(λ),w(λ))(λγ+A)1+Pg(λ)(λγ+A)1.

    By taking the inverse Laplace transform and applying the convolution theorem, we get

    ˜u(υ)=υ0Eγ((υs)γA)ads+υ0(υs)γ1Eγ,γ((υs)γA)F(˜u(s),w(s))ds+υ0(υs)γ1Eγ,γ((υs)γA)Pg(s)ds.

    Definition 2.4 A Problem (1.5) has global mild solution of a function ˜u:[0,)Hγ,r in Hγ,r, if ˜uC([0,),Hγ,r) and, for υ[0,),

    ˜u(υ)=υ0Eγ((υs)γA)ads+υ0(υs)γ1Eγ,γ((υs)γA)F(˜u(s),w(s))ds+υ0(υs)γ1Eγ,γ((υs)γA)Pg(s)ds.

    Definition 2.5 Let 0<T<. A Problem (1.5) has local mild solution of a function ˜u:[0,T]Hγ,r in Hγ,r if ˜u([0,T],Hγ,r) and for υ[0,T], ˜u satisfies the Definition [4]. Conveniently, we respectively define three operators ˜ψ(υ), ˜φ(υ) and ˜ϖ(˜u,w)(υ):

    ˜ψ(υ)=υ0Eγ((υs)γA)ads˜φ(υ)=t0(ts)γ1Eγ,γ((υs)γA)Pg(s)ds˜ϖ(˜u,w)(υ)=υ0(υs)γ1Eγ,γ((υs)γA)F(˜u(s),w(s))ds.

    To provide suitable circumstances for the existence and uniqueness of mild solutions [21] to the Problem (1.5) in Hγ,r, we consider this section. Suppose that we have (K). For υ>0, Pg is continuous and

    |pg(υ)|r=0(υγ(1β))for0<β<1andυ0. (3.1)

    To deal with the mild solution of Problem (1.5) with global existence in Hγ,r, let

    M(υ)=sups(0,υ](sγ(1β)|Pg(s)|r)

    and

    B1=C1maxB(γ(1β),1γ(1β)),B(γ(1α),1γ(1β))LMC1maxB(γ(1β),12γ(αβ)),B(γ(1α),12γ(αβ)).

    Theorem 3.1. Let 1<r<, 0<γ<1 and (3.1) hold; then, for all atHγ,r. Suppose that

    C1|aυ|Hγ,r+B1M<14L.

    If a unique function ˜u:[0,)Hγ,r is satisfied and we choose n2r12<β, then there is α>max(β,12) where M=sups[0,)(sγ(1β)Pg(s)):

    (ⅰ) ˜u(0)=ax and ˜u:[0,)Hγ,r shows continuity;

    (ⅱ) ˜u:[0,)Hα,r is continuous and limυ0υγ(αβ)|˜u(υ)|Hα,r=0;

    (ⅲ) For υ[0,), ˜u satisfies Definition 2.4.

    Proof. Here we explain X which is subspace of all of the curves and X=X[], ˜u:(0,)Hγ,r. Now suppose that α=1+β2 such that the following is true:

    (ⅰ) ˜u:[0,)Hγ,r is continuous and bounded.

    (ⅱ) ˜u:(0,)Hα,r is continuous and bounded; furthermore, limυ0υγ(αβ)|˜u(υ)|Hα,r=0 and its genuine form is given by

    ||˜u||X=max(supυ0|˜u(υ)|Hγ,r,supt0υγ(αβ)|˜u(υ)|Hα,r).

    Now, we know that there exists M such that ˜u,wHα,r such that F:Hα,rHα,rJr is bounded and bilinear mapping:

    |F(˜u,w)|rM|˜u|Hα,r|w|Hα,r|F(˜u,˜u)F(w,w)|rM(|˜u|Hα,r+|w|Hα,r)|˜uw|Hα,r.

    Step 1:

    Consider that ˜u,wX. The term ˜ϖ(˜u(υ),w(τ)) is a part of C(0,),Hγ,r and C([0,],Hγ,r). By considering ε>0 that is very small and randomly fixing υ00, now again suppose that υ>υ0 (υ<υ0analogously); we have

    |˜ϖ(˜u(υ),w(υ))˜ϖ(˜u(υ0),w(υ0))|Hγ,rdsυυ0(υs)γ1|Eγ,γ((υs)γA)F(˜u(s),w(s))|Hγ,rds+υ00|(υs)γ1(υ0s)γ1Eγ,γ((υs)γA)F(˜u(s),w(s))|Hγ,rds+υ0ϵ0(υ0s)γ1|Eγ,γ((υs)γA)Eγ,γ((υ0s)γA)F(˜u(s),w(s))|Hγ,rds+υ0υ0ϵ(υ0s)γ1|Eγ,γ((υs)γA)Eγ,γ((υ0s)γA)F(˜u(s),w(s))|Hγ,rds=D11(υ)+D12(υ)+D13(υ)+D14(υ).

    In the view of Lemma 2.4, we consider every term separately for D11(υ) and get the following

    D11(υ)C1υυ0(υs)γ(1β)1|F(˜u(s),w(s))|rdsMC1υυ0(υs)γ(1β)1|(˜u(s)|Hα,r,|w(s))|Hα,rdsMC1υυ0(υs)γ(1β)1s2γ(αβ)dssups(0,υ](s2γ(αβ)|˜u(s)|Hα,r|w(s)|Hα,r)
    =MC11υ0/υ(1s)γ(1β)1s2γ(αβ)dssups(0,υ](s2γ(αβ)|˜u(s)|Hα,r|w(s)|Hα,r).

    By applying the β function properties, δ>0 much smaller as 0<υυ0<δ; we have

    1υ0/υ(1s)γ(1β)1s2γ(αβ)ds0

    which follows that as υυ00, D11(υ) tends to 0.

    Now for D12(υ),

    D12(υ)=C1υ00((υ0s)γ1(υs)γ1)(υs)βγ|F(˜u(s),w(s))|rdsMC1υ00((υ0s)γ1(υs)γ1)(υs)βγs2γ(αβ)dssups(0,υ0](s2γ(αβ)|˜u(s)|Hα,r|w(s)|Hα,r).

    It is interesting to note that

    υ00|(υ0s)γ1(υs)γ1|(υs)βγs2γ(αβ)dsυ00(υs)γ1(υs)βγs2γ(αβ)ds+υ00(υ0s)γ1(υs)βγs2γ(αβ)ds2υ00(υ0s)γ(1β)1(υs)βγs2γ(αβ)ds=2B(γ(1β)),12γ(αβ).

    We can show this by applying Lebesgue's dominated convergence Theorem 2.1, we get

    υ00((υ0s)γ1(υs)γ1)(υs)βγs2γ(αβ)ds0,asυυ0.

    We can say that limυυ0D12(υ)=0. For D13(υ), we have

    D13(υ)υ0ϵ0(υ0s)γ1|Eγ,γ((υs)γA)Eγ,γ((υ0s)γA)F(˜u(s),w(s))|Hγ,rdsυ0ϵ0(υ0s)γ1((υs)βγ+(υ0s)βγ)|F(˜u(s),w(s))|rds2MC1υ00(υ0s)γ1s2γ(αβ)dssups(0,υ0](s2γ(αβ)|˜u(s)|Hα,r|w(s)|Hα,r).

    Considering the uniform continuity, and by using Lebesgue's dominated convergence theorem, we get D13(υ) as follows

    limυυ0D13(υ)=υ0ϵ0(υ0s)γ1|Eγ,γ((υs)γA)Eγ,γ((υ0s)γA)F(˜u(s),w(s))|Hγ,rds=0.

    For D14(υ), using calculations, we estimate that, according to the β function properties

    D14(υ)υ0υ0ϵ(υ0s)γ1((υs)βγ+(υ0s)βγ)|F(˜u(s),w(s))|rds2MC1υ00(υ0s)γ1s2γ(αβ)dssups[υ0ϵ,υ0](s2γ(αβ)|˜u(s)|Hα,r|w(s)|Hα,r)0forε0.

    It follows that

    |˜ϖ(˜u(υ),w(υ))˜ϖ(˜u(υ0),w(υ0))|Hγ,rds0whileυυ0.

    The operator's continuity ˜ϖ(˜u,w) can be demonstrated in C((0,),Hα,r) following the preceding debate.

    Step 2:

    To show that ˜ϖ:XXX is a continuous bilinear operator, we consider that, according to Lemma 2.4, we have

    |˜ϖ(˜u(υ),w(υ))|Hγ,r=|υ0(υs)γ1|Eγ,γ((υs)γA)F(˜u(s),w(s))|Hγ,rdsC1υ0(υs)γ(1β)1|F(˜u(s),w(s))|rdsMC1υ0(υs)γ(1β)1s2γ(αβ)dssups(0,υ](s2γ(αβ)|˜u(s)|Hα,r|w(s)|Hα,r)=MC1B((γ(1β)),12γ(αβ))||u||X||w||X

    and

    |˜ϖ(˜u(υ),w(υ))|α,rH=|υ0(υs)γ1|Eγ,γ((υs)γA)F(˜u(s),w(s))|Hα,rdsC1υ0(υs)γ(1α)1|F(˜u(s),w(s))|rdsMC1υ0(υs)γ(1α)1s2γ(αβ)dssups(0,υ](s2γ(αβ)|˜u(s)|Hα,r|w(s)|Hα,r)=MC1υγ(αβ)B((γ(1α)),12γ(αβ))||˜u||X||w||X.

    As a result of this

    supυ[0,)υγ(αβ)|˜ϖ(˜u(υ),w(υ))|Hα,rMC1B(γ(1α)),12γ(αβ)||˜u||X||w||X.

    To be more specific,

    limυ0υγ(αβ)|˜ϖ(˜u(υ),w(υ))|Hα,r=0.

    Therefore, ˜ϖ(˜u,w)X and ||˜ϖ(˜u(υ),w(υ))||XL||˜u||X||w||X.

    Step 3:

    Let 0<υ0<υ. We have that

    |˜φ(υ)˜φ(υ0)|Hγ,rυυ0(υs)γ1|Eγ,γ((υs)γA)Pg(s)|Hγ,rds+υ00((υ0s)γ1(υs)γ1)|Eγ,γ((ts)γA)Pg(s)|Hγ,rds+υ0ϵ0(υ0s)γ1|Eγ,γ((υs)γA)Eγ,γ((υ0s)γA)Pg(s)|Hγ,rds+υ0υ0ϵ(υ0s)γ1|Eγ,γ((υs)γA)Eγ,γ((υ0s)γA)Pg(s)|Hγ,rdsC1υυ0(υs)γ(1β)1|Pg(s)|rds+C1υ00((υ0s)γ1(υs)γ1)(υs)βγ|Pg(s)|rds+C1υ0ϵ0(υ0s)γ1|Eγ,γ((υs)γA)Eγ,γ((υ0s)γA)Pg(s)|Hγ,rds+2C1υ0υ0ϵ(υ0s)γ(1β)1|Pg(s)|rdsC1M(υ)υυ0(υs)γ(1β)1sγ(1β)ds+C1M(υ)υ00((υs)γ1(υ0s)γ1)(υs)βγsγ(1β)ds+C1C(υ)υ0ϵ0(υ0s)γ1|Eγ,γ((υs)γA)Eγ,γ((υ0s)γA)|Hγ,rds+2C1C(υ)υ0υ0ϵ(υ0s)γ(1β)1sγ(1β)ds.

    The first two integrals and last integral tend to 0 as υυ0; also, ε0 by using the characteristics of the β function. Now the third integral also tends to 0 as υυ0 by using Lemma 2.1. This suggests that

    |˜φ(υ)˜φ(υ0)|Hγ,r0whileυυ0.

    In order to evaluate the continuity of ~φ(υ) in Hα,r, we must go along with the same pattern as in Hγ,r. On the contrary,

    |˜φ(υ)|Hγ,r|=υ0(υs)γ1|Eγ,γ((υs)γA)Pg(s)|Hγ,rdsC1υ0(υs)γ(1β)1|Pg(s)|rdsC1M(υ)υ0(υs)γ(1β)1sγ(1β)ds=C1M(υ)B((γ(1β)),(1γ(1β)))|˜φ(υ)|Hα,r|=υ0(υs)γ1|Eγ,γ((υs)αA)Pg(s)|Hα,rdsC1υ0(υs)γ(1α)1|Pg(s)|rdsC1M(υ)υ0(υs)γ(1α)1sγ(1β)ds=υγ(αβ)C1M(υ)B((γ(1α)),(1γ(1β))). (3.2)

    To be more precise,

    υγ(αβ)|˜φ(υ)|Hα,rC1M(υ)B((γ(1α)),(1γ(1β)))0whileυυ0.

    Because ˜M(t)0 is the same as υ0 owing to assumption, this result confirms that ||˜φ(υ)||˜B1˜M as ˜φ(t)X. Now for aυHγ,r, according to Lemma 2.1 obvious that

    Eγ(υγA)aC([0,),Hγ,r)

    and

    Eγ(υγA)aC([0,),Hα,r).

    Therefore by Theorem 2.3, we can say that

    υ0Eγ((υs)γA)adsC([0,),Hγ,r)andυ0Eγ((υs)γA)adsC([0,),Hα,r).

    This implies that for all υ(0,], and together with Theorem 2.2,

    υ0Eγ((υs)γA)adsX.

    By using the above condition, we get

    υγ(αβ)υ0Eγ((υs)γA)adsC([0,),Hα,r)
    ||υ0Eγ((υs)γA)ads||Xυ0||Eγ((υs)γA)a||XdsC1υ0|a|Hγ,rdsC1|a|(t)Hγ,r.

    According to Theorem 3.1, the inequality

    ||υ0Eγ(υγA)ads+˜φ(υ)||X||υ0Eγ(υγA)ads||X+||˜φ(υ)||Xυ0||Eγ(υγA)a||Xds+||˜φ(υ)||X14L

    continues to hold, resulting in F having a unique fixed point.

    Step 4:

    In order to verify that ˜u(υ)aυ in Hγ,r as υ0, we must show that

    limυ0υ0Eγ((υs)γA)ads=0limυ0υ0(υs)γ1Eγ,γ((υs)γA)Pg(s)ds=0limυ0υ0(υs)γ1Eγ,γ((υs)γA)F(˜u(s),w(s))ds=0

    in Hγ,r. In fact limυ0˜ψ(υ)=0 and limυ0˜φ(υ)=0(limυ0˜M(υ)=0) due to (3.2). So,

    υ0(υs)γ1|Eγ,γ((υs)γA)F(˜u(s),˜u(s))|Hγ,rdsCυ0(υs)γ(1β)1|F(˜u(s),˜u(s))|rdsMCυ0(υs)γ(1β)1|˜u(s)|2Hα,rdsMCυ0(υs)γ(1β)1s2γ(αβ)dssups(0,υ]{s2γ(αβ)|˜u(s)|2Hα,r}=MCB((γ(1β)),12γ(αβ))sups(0,υ]{s2γ(αβ)|˜u(s)|2Hα,r}0asυυ0.

    Theorem 3.2. Let 1<r<, 0<γ<1 and (3.1) holds; then, suppose that

    n2r12<β.

    Then there is a unique function ˜u:[0,)Hγ,r and α>max(β,12) for all atHγ,r there exists >0 satisfying the following

    (ⅰ) ˜u(0)=aυ and ˜u:[0,]Hγ,r is continuous,

    (ⅱ) ˜u:[0,]Hα,r is continuous, and limυ0υγ(αβ)|˜u(υ)|Hα,r=0,

    (ⅲ) for υ[0,], ˜u satisfies Definition 2.4.

    Proof. Consider the space of all of the curves X=X[T] and ˜u:(0,]Hγ,r; now, suppose that α=1+β2 so we have the following

    (ⅰ) ˜u:[0,]Hγ,r is continuous.

    (ⅱ) ˜u:(0,]Hα,r is continuous and limυ0υγ(αβ)|˜u(υ)|Hα,r=0,

    its original form is given by

    ||˜u||X=supυ[0,](υγ(αβ)|˜u(υ)|Hα,r).

    Similar to the conclusion of Theorem 3.1, it is worth noting that the function g:XXX is linearly mappable and continuous. By Lemma 2.1 the function φ(υ)X for all υ(0,]. It is simple to assert that

    Eγ(υγA)aC([0,]),Hγ,r)Eγ(υγA)aC([0,],Hα,r).

    Therefore by Theorem 2.3, we can say that

    υ0Eγ((υs)γA)adsC([0,],Hγ,r)υ0Eγ((υs)γA)adsC([0,],Hα,r).

    For all υ(0,], using Theorem 2.2 implies that

    υ0Eγ((υs)γA)adsX.

    We have

    ||˜u||X=supυ[0,](υγ(αβ)|˜u(υ)|Hα,r).

    By using the above condition, we get

    υγ(αβ)υ0Eγ((υs)γA)adsC([0,],Hα,r).

    Now, consider a sufficiently small >0, such that

    ||υ0Eγ(υγA)ads+˜φ(υ)||X[]||υ0Eγ(υγA)ads||X[]+||˜φ(υ)||X[]<14L

    holds, resulting in F having a unique fixed point.

    This part is devoted to the iteration method's evaluation of a local mild solution to Problem (1.5) in Jr. Consider that α=1+β2.

    Theorem 4.1. Suppose that 1<r<, 0<γ<1 and (3.1) holds. Suppose that

    aυHγ,rwithn2r12<γ.

    For aυHγ,r, Problem (1.5) has a unique mild solution ˜u in Jr, Aα˜u shows continuity on (0,]. Furthermore, υγ(αβ)Aα˜u(υ) is bounded while υ0. Additionally, ˜u also shows continuity in [0,].

    Proof. Step 1:

    Let

    K(υ)=sups(0,υ]sγ(αβ)|Aα˜u(s)|r

    and

    ζ(υ)=˜ϖ(˜u,˜u)(υ)=υ0(υs)γ1Eγ,γ((υs)γA)F(˜u(s),˜u(s))ds.

    According to the summary of Step 2 in Theorem 3.1, ζ(υ) shows continuity on [0,] and Aαζ(υ) is continuous on (0,] and also exists, then,

    Aα|˜ϖ(˜u(υ),˜u(υ))|Hα,r=|υ0(υs)γ1|Eγ,γ((υs)γA)AαF(˜u(s),˜u(s))|Hα,rdsC1υ0(υs)γ(1α)1|AαF(˜u(s),˜u(s))|rdsMC1υ0(υs)γ(1α)1s2γ(αβ)dssups[0,υ]{s2γ(αβ)|Aαu(s)2|Hα,r}.

    We have

    K2(υ)=sups(0,υ]s2γ(αβ)|Aα˜u(s)2|r.

    Using the above equation, it gives the final result such that

    |Aαζ(υ)|rMC1B(γ(1α),12γ(αβ))K2(υ)υγ(αβ). (4.1)

    Take into account the integral ˜φ(υ). As the (3.1) holds then the inequality

    |Pg(s)|rM(υ)sγ(1β)

    is accomplished by a continuous function M(υ). Now considering the Step 3 of Theorem 3.1, we show that Aα˜φ(υ) shows continuity on (0,]. Before discussing continuity, we signify

    M(υ)=sups(0,υ]{sγ(1β)|AαPg(s)|r}|Aα˜φ(υ)|Hα,r|=υ0(υs)γ1|Eγ,γ((ts)αA)AαPg(s)|Hα,rdsC1υ0(υs)γ(1α)1|AαPg(s)|rdsC1M(υ)υ0(υs)γ(1α)1sγ(1β)ds|Aα˜φ(υ)|r|=υγ(αβ)C1M(υ)B((γ(1α)),(1γ(1β))). (4.2)

    As υ0, we get that M(υ)=0 and |Pg(υ)|r=0(υγ(αβ)). As υ0, we get that |Aαζ(υ)|r=0(υγ(αβ)) according to (4.2). We show that the function ˜φ is continuous in Jr. In reality, considering 0υ0<υ<, we have

    |˜φ(υ)˜φ(υ0)|rC3υυ0(υs)γ(1β)1|Pg(s)|rds+C3υ00((υ0s)γ1(υs)γ1)(υs)βγ|Pg(s)|rds+C3υ0ϵ0(υ0s)γ1|Eγ,γ((υs)γA)Eγ,γ((υ0s)γA)Pg(s)|Hγ,rds+2C3υ0υ0ϵ(υ0s)γ(1β)1|Pg(s)|rdsC3M(υ)υυ0(υs)γ(1β)1sγ(1β)ds+C3M(υ)υ00((υs)γ1(υ0s)γ1)(υs)βγsγ(1β)ds+C3M(υ)υ0ϵ0(υ0s)γ1|Eγ,γ((υs)γA)Eγ,γ((υ0s)γA)|rsγ(1β)ds+2C3M(υ)υ0υ0ϵ(υ0s)γ(1β)1sγ(1β)ds.

    Further, we assume the function υ0Eγ((υs)γA)ads. From Theorem 2.2, it is clear that

    |Aαυ0Eγ((υs)γA)ads|r=υ0|AαEγ((υs)γA)a|Hγ,rds.

    From Lemma 2.4,

    C1υ0υγ(αβ)|Aβa|r=C1υγ(αβ)|aυ|Hα,r

    and

    limυ0υγ(αβ)|Aαυ0Eγ((υs)γA)ads|r=limυ0υγ(αβ)υ0|Eγ((υs)γA)a|Hγ,rds=0.

    Step 2:

    We now build the result through successive approximation:

    ˜u0(υ)=υ0Eγ((υs)γA)ads+˜φ(υ)˜un+1(υ)=˜u0(υ)+ϖ(˜un,˜un)(υ),forn=0,1,2.... (4.3)

    We know that denotes increasing and continuous functions on [0,]. Moreover, Kn(0)=0. However, according to (4.1) and (4.3), Kn(υ) satisfies the inequality

    Kn+1(υ)K0(υ)+MC1B((γ(1α)),12γ(αβ))Kn2(υ). (4.4)

    Select >0 so that

    4MC1B((γ(1α)),12γ(αβ))K0(υ)<1. (4.5)

    Then, the sequence Kn() is bounded; following a simple description of (4.4), so we get

    Kn()ρ(),forn=0,1,2,..., (4.6)

    while (4.4) is just like the quadratic equation; we have

    MC1B((γ(1α)),12γ(αβ))Kn2(υ)Kn+1(υ)+K0(υ)0.

    Here, a=MC1B((γ(1α)),12γ(αβ))Kn2(υ), b=1 and c=K0(υ). After applying the quadratic formula, we obtain

    Kn(υ)=114MC1B((γ(1α)),12γ(αβ))K0(υ)2MC1B((γ(1α)),12γ(αβ)).

    Considering (4.6),

    ρ(υ)=114MC1B((γ(1α)),12γ(αβ))K0(υ)2MC1B((γ(1α)),12γ(αβ)).

    Conversely, Kn(υ)ρ(υ) holds for any υ(0,]. Correspondingly, ρ(υ)2K0(υ). Suppose the following concept of equality:

    yn+1(υ)=υ0(υs)γ1Eγ,γ((υs)γA)F(˜un+1(s),˜un+1(s))F(˜un(s),˜un(s))ds.

    For υ(0,] and n=0,1,2,..., we denote yn=˜un+1˜un:

    Yn(υ)=sups(0,υ]sγ(αβ)|Aαyn(s)|r.

    Using Lemma 2.3, we get

    |F(˜un+1(s),˜un+1(s))F(˜un(s),˜un(s))|rM(|˜un+1|+|˜un|)|˜un+1˜un|M(|˜un+1|+|˜un|)Aαynsups(0,υ]sγ(αβ)sγ(αβ)M(|Aα˜un+1|+|Aα˜un|)ynsups(0,υ]sγ(αβ)sγ(αβ)M(sups(0,υ]sγ(αβ)|Aα˜un+1|+sups(0,υ]sγ(αβ)|Aα˜un|)ynsγ(αβ)M(Kn+1+Kn)ynsγ(αβ).

    Then, we set

    M(Kn+1+Kn)ynsγ(αβ)Aαsups(0,υ]sγ(αβ)sγ(αβ)
    M(Kn+1+Kn){sups(0,υ]sγ(αβ)Aαyn}s2γ(αβ)
    |F(˜un+1(s),˜un+1(s))F(˜un(s),˜un(s))|rM(Kn+1(s)+Kn(υ))Yn(s)s2γ(αβ).

    We get the results from Step 2 in Theorem 3.1:

    υγ(αβ)|Aαyn+1(υ)|r2MC1B((γ(1α)),1γ(1β))ρ(υ)Yn(υ).

    It provides

    Yn+1()2MC1B((γ(1α)),12γ(αβ))ρ()Yn()4MC1B(γ(1α),12γ(αβ))K0()Yn(). (4.7)

    In accordance with (4.5) and (4.7), we get

    limn0Yn+1()Yn()4MC1B((γ(1α)),12γ(αβ))K0()1.

    Hence the series n=0Yn() is convergent. For υ(0,] it shows that the series n=0υγ(αβ)Aαyn(υ) is uniformly convergent. Thus, the sequence {υγ(αβ)Aα˜un(υ)} is uniformly convergent in (0,]. This signifies that limn˜un(υ)=˜u(υ)D(Aα) and limnυγ(αβ)Aα˜un(υ)=υγ(αβ)Aα˜u(υ) uniformly. From the boundedness theorem, a function f continuous on a bounded and closed interval is necessarily a bounded function. Therefore, Aα is closed if Aα is bounded. As a result, the function K(υ)=sups(0,υ]sβ(αγ)|Aα˜un(s)|r satisfies

    K(υ)ρ(υ)2K0(υ),forυ(0,υ] (4.8)

    and

    ηn=sups(0,]s2γ(αβ)|˜F(˜un(s),˜un(s))˜F(˜u(s),˜u(s))|rM(Kn()+K())sups(0,]sυγ(αβ)|Aα(˜un(s)˜u(s))|r0asn.

    Lastly, it is necessary to confirm that ˜u is a mild solution of (1.5) in [0,]. Because

    |ϖ(˜un,˜un)(υ)ϖ(˜u,˜u)(υ)|rυ0(υs)γ1ηns2γ(αβ)ds=υβγηn0,asn

    we get ϖ(˜un,˜un)(υ)ϖ(˜u,˜u)(υ). We obtain it by taking the limits of (4.2) on both sides;

    ˜u(υ)=˜u0(υ)+ϖ(˜u,˜u)(υ). (4.9)

    Suppose that ˜u(0)=aυ. For υ[0,], we can say that (4.9) holds and ˜uC([0,],Jr). Furthermore, the uniform convergence of υγ(αβ)Aαun(υ) to tγ(αβ)Aα˜u(υ) shows continuity of Aα˜u(υ). By (4.8) and K0(0)=0, we obtain that |Aαu(υ)|r=0υ(γ(αβ)) is clear.

    Step 3:

    Now, we have to suppose that ˜u and ˜w are mild solutions of (1.5) because to prove that the mild solutions are unique. Let y=˜u˜w; we consider the equality

    y(υ)=υ0(υs)γ1Eγ,γ((υs)γA)(F(˜u(s),˜u(s))F(˜w(s),˜w(s)))ds.

    Now, we defining the following functions:

    K=maxsups(0,υ]sγ(αβ)|Aαu(s)|r,sups(0,υ]sγ(αβ)|Aαv(s)|r.

    According to Theorem 3.1 and Lemma 2.4, we get

    |Aαy(υ)|rMC1Kυ0(υs)γ(1α)1sγ(αβ)|Aαy(s)|rds.

    For υ(0,], the Gronwall inequality demonstrates that Aαy(υ)=0. It shows that for υ[0,], y(υ)=˜u(υ)˜w(υ)0. Hence, the mild solutions are unique.

    In this part, we evaluate the regularity's behavior [22] of a solution ˜u which satisfies (1.5). We will assume throughout this section that (k1) : Pg(υ) has Hölder's continuity with an exponent θ(0,γ(1α)), which is given as

    |Pg(υ)Pg(s)|rL|υs|θforall0<υ,s. (5.1)

    Definition 5.1. The problem (1.5) has a solution that is a classical solution of a function ˜u:[0,]Jr, and ˜uC([0,],Jr) with cDυυC([0,],Jr), which satisfies (1.5) for all υ(0,] and accepts values in D(A).

    Lemma 5.1. Let Definition 5.1 be satisfied if

    φ1(υ)=υ0(υs)γ1Eγ,γ((υs)γA)(Pg(s),Pg(υ))ds,forυ(0,]

    so Aφ1(υ)Cθ([0,],Jr) and φ1(υ)D(A).

    Proof. From Lemma 2.4 and (5.1), we fixed υ(0,],

    (υs)γ1|AEγ,γ((υs)γA)(Pg(s),Pg(υ))|rC1(υs)1|Pg(s)Pg(υ)|rC1L(υs)θ1L1([0,],Jr).

    Then

    |Aφ(υ)|r(υs)γ1|AEγ,γ((υs)γA)(Pg(s),Pg(υ))|rdsC1Lυ0(υs)θ1dsC1Lθυθ<;

    we get φ1(υ)D(A) by the closeness property of A. We must demonstrate that Aφ1(υ) has Hölder continuity. Because

    ddυ(υγ1Eγ,γ(μυγ))=(υγ2Eγ,γ1(μυγ))

    it follows that

    ddυ(υγ1AEγ,γ(υγA))=12πιΓθ(υγ2Eγ,γ1(μυγ))A(μI+A)1dμ=12πιΓθ(γ2Eγ,γ1(μυγ))dμ12πιΓθ(υγ2Eγ,γ1(μυγ))A(μI+A)1dμ.

    Substituting μυγ=ξ implies that υγdμ=dξ. So we have that dμ=1υγdξ:

    =12πιΓθ(υγ2Eγ,γ1(ξ))1υγdξ12πιΓθ(υγ2Eγ,γ1(ξ))ξυγA(ξυγI+A)11υγdξ.

    Since

    ||(μI+A)1||C|μ|

    and

    ||ddυ(υγ1AEγ,γ(υγA))||Cγυ2for0<υ

    for every 0<s<υ, we apply the mean value theorem

    ||(υγ1AEγ,γ(υγA))(sγ1AEγ,γ(sγA))||=||υsddτ(τγ1AEγ,γ(τγA))dτ||υs||ddτ(τγ1AEγ,γ(τγA))||dτυsτ2dτ=Cγ(s1υ1). (5.2)

    Now for 0<υ<υ+, we suppose that >0, so

    Aφ1(υ+)Aφ1(υ)=υ0(υ+s)γ1AEγ,γ((υ+s)γA)(Pg(s)Pg(υ))ds(υs)γ1AEγ,γ((υs)γA)(Pg(s)Pg(υ))ds+υ0(υ+s)γ1AEγ,γ((υ+s)γA)(Pg(υ)Pg(υ+))ds+υ+υ(υ+s)γ1AEγ,γ((υ+s)γA)(Pg(s)Pg(υ+))ds=1(υ)+2(υ)+3(υ). (5.3)

    For convenience, we solve each term separately. For 1(υ), from (5.2) and (5.1) we have

    |1(υ)|rυ0||(υ+s)γ1AEγ,γ((υ+s)γA)(υs)γ1AEγ,γ((υs)γA)||r(Pg(s)Pg(υ))dsCγLυ0(υ+s)1(υs)θ1dsCγLυ0(s+)1(υs)θ1dsCγLh0+ssθ1ds+CγLs+ssθ1dsCγLθ. (5.4)

    For 2(t), using Lemma 2.4 and (5.1), we get

    |2(υ)|rυ0(υ+s)γ1|AEγ,γ((υ+s)γA)(Pg(υ)Pg(υ+))|rdsC1υ0(υ+s)1|Pg(υ)Pg(υ+)|rdsC1θυ0(υ+s)1ds=C1L[lnln(υ+)]θ. (5.5)

    Now for 3(υ), using Lemma 2.4 and (5.1), we have

    |3(υ)|rυ+υ(υ+s)γ1|AEγ,γ((υ+s)γA)(Pg(s)Pg(υ+))|rdsC1υ+υ(υ+s)1|Pg(s)Pg(υ+)|rdsC1Lυ+υ(υ+s)θ1ds=C1Lθθ. (5.6)

    Hence we can say that Aφ1(υ) is Hölder continuous by combining all of the above results.

    Theorem 5.1. Consider that the supposition of Theorem 4.1 is satisfied. If (5.1) holds, then the classical [23] mild solutions of (1.5) are obtained for every aυD(A).

    Proof. Step 1:

    We have that aυD(A). So, υ0Eγ(υγA)ads is said to be classical solution in regard to the following problem:

    {cDγυ˜u(υ)=A˜u,υ>0,˜u(0)=ax. (5.7)

    We also verify that

    ˜φ(υ)=υ0(υs)γ1Eγ,γ((υs)γA)Pg(s)ds

    is a classical approach to the following problem:

    {cDγυ˜u(υ)=A˜u+Pg(υ),υ>0,˜u(0)=0. (5.8)

    ˜φC([0,],Jr) follows from Theorem 4.1. We can write ˜φ(υ)=˜φ1(υ)+˜φ2(υ), while

    ˜φ1(υ)=υ0(υs)γ1Eγ,γ((υs)γA)(Pg(s),Pg(υ))ds˜φ2(υ)=υ0(υs)γ1Eγ,γ((υs)γA)Pg(υ)ds.

    We can say that ˜φ1(υ)D(A) according to Lemma 5.1. In order to prove the same conclusion for ˜φ2(υ), according to Lemma 2.2(ⅲ), we realized that

    A˜φ2(υ)=Pg(υ)Eγ(υγA)Pg(υ).

    It follows that (5.1):

    |A˜φ2(υ)|(1+C1)|Pg(υ)|r.

    So, for υ(0,] we can say that ˜φ2(υ)D(A) and ˜φ2(υ)Cν((0,],Jr). Now we have to show that cDγυ(˜φ)C((0,],Jr). Taking ˜φ(0)=0 and in view of Lemma 2.2(ⅳ),

    cDγυ˜φ(υ)=ddυ(I1γυ˜φ(υ))=ddυ(Eγ(υγA)Pg).

    It is still necessary to demonstrate that (Eγ(υγA)Pg) is differentiability continuous in Jr. Suppose that 0<hυ; therefore, we have

    1[(Eγ((υ+)γA)Pg)(Eγ(υγA)Pg)]=υ01[(Eγ((υ+s)γA)Pg(s))(Eγ((υs)γA)Pg(s))]ds+1υ+υ(Eγ((υ+s)γA)Pg(s))ds.

    Notice that

    υ01|(Eγ((υ+s)γA)Pg(s))(Eγ((υs)γA)Pg(s))|rds
    1υ0|(Eγ((υ+hs)γA)Pg(s))|rds+1υ0|(Eγ((υs)γA)Pg(s))|rds.

    In view of Lemma 2.4,

    C1M(υ)1υ0(υ+s)γsγ(1β)ds+C1M(υ)1υ0(υs)γsγ(1β)dsC1M(υ)1((υ+)1γ+υ1γ)B(1γ,1γ(1β)),

    the dominated convergence theorem applies; then, we get

    lim0υ01[(Eγ((υ+s)γA)Pg(s))(Eγ((υs)γA)Pg(s))]ds=Eγ((υs)γA)Pg(s).

    From Lemma 2.2(ⅲ), we have

    Eγ(υγA)u=υγ1AEγ,γ(υγA)uEγ(υγA)u=υγ1AEγ,γ(υγA)uEγ(υγA)u=υγ1AEγ,γ(υγA)u.

    Therefore,

    Eγ((υs)γA)Pg(s)=υ0(υs)γ1AEγ,γ((υs)γA)Pg(s)ds=A˜φ(υ).

    Conversely,

    1υ+hυEγ((υ+s)γA)Pg(s)ds.

    Let s=υ+s so ds=ds and after setting the conditions [s=υ implies s=] and [s=υ+ implies s=0], we have

    10Eγ((s)γA)Pg(υ+s)(ds).

    By replacing ss, we get

    1h0Eγ(sγA)Pg(υ+s)(ds)=10Eγ(sγA)[Pg(υ+s)Pg(υs)+Pg(υs)Pg(υ)+Pg(υ)]ds=10Eγ(sγA)(Pg(υ+s)Pg(υs))ds+10Eγ(sγA)(Pg(υs)Pg(υ))ds+10Eγ(sγA)Pg(υ)ds.

    From Lemmas 2.1–2.4 and (5.1), we have

    |10Eγ(sγA)(Pg(υ+s)Pg(υs))|rdsC1Lθ|10Eγ(sγA)(Pg(υs)Pg(υ))|rdsC1Lθθ+1.

    From Lemma 2.1(ⅰ),

    limh01hh0Eγ(sγA)Pg(υ)ds=Pg(υ)limh01hυ+hυEγ((υ+hs)γA)Pg(υ)ds=Pg(υ);

    we deduce that Eγ(υγA)Pg holds differentiability aυ and ddυ(Eγ(υγA)Pg)+=A˜φ(υ)+Pg(υ). Similarly Eγ(υγA)Pg holds differentiability at υ and ddυ(Eγ(υγA)Pg)=A˜φ(υ)+Pg(υ). We can prove that A˜φ=A˜φ1+A˜φ2C((0,],Jr). Clearly, we can say that according to Lemma 2.1, ˜φ2(υ) shows continuity and ˜φ2(υ)=Pg(υ)Eγ(υγA)Pg(υ) because of Lemma 2.2(ⅲ). Consequently, A˜φ1(υ) is continuous according to Lemma 5.1. Thus, cDγυ˜φC((0,],Jr).

    Step 2:

    Consider ˜u as the mild solution of (1.5). In order to draw the conclusion that F(˜u,˜u)Cθ((0,],Jr), we have to show that Aα˜u holds Hölder continuity in Jr according to Theorem 3.1. Now for 0<υ<υ+, let >0. We know that ψ(υ)=υ0Eγ((υs)γ)ads. Denoting ϕ(υ):=υ0Eγ(υγA)ads, and by Lemma 2.2(ⅲ) we have

    ddυEγ(υγA)a=υγ1AEγ,γ(υγA)a.

    Integrating on both sides, we get

    υ0ddsEγ((υs)γA)ads=υ0sγ1AEγ,γ(sγA)ads.

    After applying the limits, we have

    Eγ(υγA)a=υ0sγ1AEγ,γ(sγA)ads.

    Using the above results and Lemma 2.4, then

    |Aαϕ(υ+)Aγϕ(υ)|r=|υ+υsγ1AαEγ,γ(sγA)ads|rυ+υsγ1|AαβEγ,γ(sγA)Aβa|rdsC1υ+υsγ(1+βα)1ds|Aβa|r=C1|a|Hγ,rγ(1+βα)((υ+)γ(1+βα)υγ(1+βα))=C1|a|Hγ,rγ(1+βα)γ(1+βα).

    Thus AαϕCθ((0,],Jr). For every small ε>0 we take as ευ<υ+. We have

    |Aαϕ(υ+)Aγϕ(υ)|r|υ+υ(υ+s)γ1AαEγ,γ((υ+s)γA)Pg(s)ds|r+|υ0Aα((υ+s)γ1Eγ,γ((υ+s)γA)(υs)γ1Eγ,γ((υs)γA))Pg(s)ds|r.=ϕ1(υ)+ϕ2(υ).

    Using Lemma 2.4 and (5.1), we have

    ϕ1(υ)C1υ+υ(υ+s)γ(1α)1|Pg(s)|rdsC1M(υ)υ+υ(υ+s)γ(1α)1sγ(1α)1dsM(υ)C1γ(1α)γ(1α)υγ(1α)1M(υ)C1γ(1α)γ(1α)εγ(1α)1.

    To estimate ϕ2, we have

    ddυ(υγ1AαEγ,γ(υγA))=12πıΓμα(υγ2Eγ,γ1(μυγ))A(μI+A)1dμ=12πıΓ(ξυγ)α(υγ2Eγ,γ1(ξ))(ξυγI+A)11υγdξ,

    which yields that

    ddυ(υγ1AαEγ,γ(υγA))Cγυγ(1α)2.

    Now, we apply the mean value theorem:

    ||(υγ1AαEγ,γ(υγA))(sγ1AαEγ,γ(sγA))||=||υsddτ(τγ1AαEγ,γ(τγA))dτ||υs||ddτ(τγ1AαEγ,γ(τγA))||dτυsτγ(1α)2dτ=Cγ(sγ(1α)1υγ(1α)1).

    Thus,

    ϕ2(t)|υ0Aα((υ+s)γ1Eγ,γ((υ+s)γA)(υs)γ1Eγ,γ((υs)γA))Pg(s)ds|rυ0((υs)γ(1α)1(υ+s)γ(1α)1)|Pg(s)|rdsCγM(υ)(υ0(υs)γ(1α)1sγ(1β)dsυ+0(υ+s)γ(1α)1sγ(1β)ds)+CγM(υ)υ+υ(υ+s)γ(1α)1sγ(1β)dsCγM(υ)(υγ(βα)(υ+)βα)B(γ(1α),1γ(1β))+CγM(υ)γ(1α)υγ(1β)CγM(υ)γ(αβ)[ε(ε+h)]γ(βα)+CγM(υ)γ(1α)εγ(1β).

    This shows that AαϕCθ([ε,],Jr). Therefore, AαϕCθ((0,],Jr) due to an arbitrary ε. Try to remember that ζ(υ)=υ0(υs)γ1Eγ,γ((υs)γA)F(˜u(s),˜u(s))ds. We know that |F(˜u(s),˜u(s)|rMK2(υ)s2γ(αβ), while K(υ)=sups(0,υ]sγ(αβ)|Aαu(s)|r is bounded and continuous in (0,]. A simple reasoning allows us to take the Hölder continuity of AζCθ((0,],Jr). Therefore we have that Aα˜u(υ) = Aαϕ(υ)+Aαφ(υ)+Aαζ(υ)Cθ((0,],Jr). We know that F(˜u,˜u)Cθ((0,],Jr) is verified; similarly following Step 2, this yields that cDγυζCθ((0,],Jr), AζCθ((0,],Jr) and cDγυζ=Aζ+F(˜u,˜u). So we get that cDγυζCθ((0,],Jr), A˜uCθ((0,],Jr) and cDγυ˜u=A˜u+F(˜u,˜u)+Pg. Hence, we can say that ˜u has a classical solution.

    Theorem 5.2. Suppose (5.1) holds if ˜u is a classical solution of (1.5); thus, cDγυCν((0,],Jr) and A˜uCν((0,],Jr).

    Proof. Let ˜u be the classical solution of Problem (1.5) and also suppose that(5.1) holds; so, ˜u(υ)=˜φ(υ)+ϕ(υ)+ζ(υ) shows that A˜φCγ(1β)((0,],Jr). It is also sufficient to demonstrate that, for every ε>0 there is A˜φCγ(1β)((ε,],Jr). Now by using Lemma 2.2(ⅲ) we take as ευ<υ+;

    |A˜φ(υ+)A˜φ(υ)|r=|υ+υsγ1A2Eγ,γ(sγA)ads|rC1υ+υsγ(1β)1ds|a|Hγ,r=C1|a|Hγ,rγ(υγ(1β)(υ+)γ(1β))=C1|a|Hγ,rγγ(1β)[˜ε(˜ε+)]γ(1β).

    Using (6) and rewriting ˜φ(υ) in the form of

    ˜φ(υ)=˜φ1(υ)+˜φ2(υ)=υ0(υs)γ1Eγ,γ((υs)γA)(Pg(s)Pg(υ))ds+υυ0(υs)γ1Eγ,γ((υs)γA)Pg(υ)ds

    as υ(0,], we can say that A˜φ1(υ)Cν)((0,],Jr) and A˜φ2(υ)Cθ((0,],Jr). These results come from Lemma 5.1 and (5.8). As we know that F(˜u,˜u)Cθ((0,],Jr) and ζ(υ) is verified analogously, so we can conclude that Aζ(υ)Cν((0,],Jr). Consequently, cDγυ˜u=A˜u+F(˜u,˜u)+PgCν((0,],Jr) as A˜uCν((0,],Jr). This proof is now complete.

    This study demonstrates the existence-uniqueness of local and global mild solutions. Meanwhile, we offer a local reasonable solution in S. The NS equations with time-fractional derivatives of order γ(0,1) were used to simulate anomaly diffusion in fractal media. We also demonstrate the existence of regular, classical solutions to these equations in S. The concept could be expanded upon by including MHD effects, the concept put forth in this mwork could be developed further, observability could be added and other activities could be generalized in future work. This is an interesting area with a lot of study going on that could lead to a lot of different applications and theories.

    The authors declare that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this paper.

    All authors contributed equally to the writing of this paper. All authors have read and agreed to the published version of the manuscript.

    The authors would like to thank the anonymous referees for their valuable comments and suggestions.



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