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(∂˜u∂t+(˜u.∇)˜u)=−∇P+μ∇2˜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˜u−w△˜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˜u−w△˜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 x∈ω 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˜u−wP△˜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 n≥3. 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:
C∞σ(H)=(v∈(C∞(H))n:∇.v=0), | (2.1) |
and the null space of which is given as
v∈(C∞(H))n:v=∇.ϕ,ϕ∈C∞(H). | (2.2) |
For a suitable approach, let Jr = ¯C∞σ(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 v∈Jr, define the following:
A−γv=1Γ(γ)∫∞0υγ−1e−υAudυ. |
Therefore, A−γ is bounded [14], just like the injective operator on Jr. Suppose that A−γ is the inverse of A−γ. For γ>0 we symbolize the space Hγ,r by the extent of A−γ 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(υ−s)γ−1f(s)ds,υ>0. |
The RL fractional derivative for a function v:[0,∞)→Rn of order γ∈Rn is defined by
L0Dγυv(υ)=dndυn(gn−γ∗v)(υ),υ≥0,n−1<γ<n. |
The RL fractional order integral is defined as
Jγυv(υ):=gγ∗v(υ)=1Γ(γ)∫υ0(υ−s)γ−1v(s)ds,υ∈[0,ℑ]. |
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(J1−γυ[v(υ)−v(0)])=ddυ(1Γ(1−γ)∫υ0(υ−s)−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),υ∈C,R(γ)>0. |
Now, let us consider the generalized Mittag-Leffler functions:
Eγ(−υγA)=∫∞0Mγ(s)e−sυγAds |
and
Eγ,γ(−υγA)=∫∞0γsMγ(s)e−sυγAds, |
where
Mγ(υ):=∞∑n=0(−υ)nn!Γ(−γ(n)+1−γ). |
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)=limn→∞fn(x) exists. Assume that there is an integrable g:Rn→[0,∞] with |fn(x)|≤g(x) for each x∈Rn. Then, f is integrable as is fn for each n, and
limn→∞∫Rnfndμ=∫Rnlimn→∞fndμ=∫Rnfdμ. |
(ⅰ) Eγ,γ(−υγA)=12πι∫Eγ,γΓθ(−μυγ)(μI+A)−1dμ;
(ⅱ) AγEγ,γ(−υγA)=12πι∫μγΓθEγ,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 υ≥0;
(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(υ−γ)}(λ)=e−λγ.
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) ∀v∈X,limt→0+Eγ(−υγA)v=v;
(ⅱ) ∀v∈D(A) and υ>0,CDγυEγ(−υγA)v=−AEγ(−υγA)v;
(ⅲ) ∀v∈X and E′γ(−υγA)v=−υγ−1AEγ,γ(−υγA)v;
(ⅳ) forυ>0,Eγ(−υγA)v=I1−γυ{(υγ−1Eγ,γ(−υγA)u)}.
Lemma 2.3. Suppose that 1<r<∞ and γ1≤γ2. Then, there exist a constant C=C(γ1,γ2) such that
|e−υAv|Hγ2,r≤Cυ−(γ2−γ1)|v|Hγ1,rasυ>0forv∈Hγ1,r. |
Moreover, limυ→0υ(γ2−γ1)|e−υAv|γ2,rH=0.
Lemma 2.4. Suppose that 1<r<∞ and γ1≤γ2. For any ℑ>0, there exists a constant C1=C1(γ1,γ2) such that
|Eα(−υγA)|Hγ2,r≤C1υ−α(γ2−γ1)|v|Hγ1,r |
and
|Eγ,γ(−υγA)|Hγ2,r≤C1υ−γ(γ2−γ1)|v|Hγ1,r |
for all v∈Hγ1,r and υ∈(0,T]. Therefore, limυ→0υα(γ2−γ1)|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:I→Rn is continuously differentiable with image g(I)⊂[a,b], where I⊂Rn is some open interval showing that the function
F(s)=−∫g(s)af(υ)dυ |
is continuously differentiable on I.
Theorem 2.4. Let ℑ(υ):υ≥0⊂X be a C0 semi group on X. Then, the following holds
(i) If C:D(G)⊂X→X, then G is said to be dense and close defined by linear operators. Therefore, υ∈[0,∞)→ℑ(υ)x∈X is continuously differentiable for any x∈D(G):
ddυℑ(υ)x=Gℑ(υ)x=ℑ(υ)Gx,forυ>0. |
(ii) Then there exists σ>0 such that Re(λ)>0; given λ∈ρ(C), we have
(λ−C)−1x=∫∞0e−λυℑ(υ)xdυforallx∈X. |
Theorem 2.5. Let γ∈(0,1] and suppose that the positive sectorial operator is A:D(A)⊂X→X. Thus, the operators {Eγ(−υγA):υ≥0} and {Eγ,γ(−υγA):υ≥0} as follows:
Eγ(−υγA)=∫∞0Mγ(s)ℑsυγAds,υ≥0 |
and
Eγ,γ(−υγA)=∫∞0γsMγ(s)ℑsυγds,υ≥0, |
where ℑ(υ):υ≥0} defines the C0 semigroup which is generated by −A.
Let γ∈(0,1) and consider the A:D(A)⊂X→X {is a positive} sectorial operator. Then for any x∈X, it holds that
L{Eγ(−υγA)x}(λ)=λγ−1(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} x∈X; observe that by Theorem 2.4,
L{Eγ,γ(−υγA)x}(λ)=∫∞0e−λυυγ−1Eγ,γ(−υγA)xdυ=∫∞0e−λυυγ−1(∫∞0γsMγ(s)ℑ(sυγ)xds)dυ. |
Now by using s=ωυ−γ, we conclude the following
L{Eγ,γ(−υγA)x}(λ)=∫∞0e−λυυγ−1(∫∞0γ(ωυ−γ)Mγ(ωυ−γ)T(ω)xυ−γdω)dυ=∫∞0ω(∫∞0γυ−(1+γ)Mγ(ωυ−γ)e−λυdυ)T(ω)xdω. |
Choose
H∗=∫∞0γυ−(1+γ)Mγ(ωυ−γ)e−λυ. |
By taking υ=τω1γ from Lemma 2.4, we have that
H∗=∫∞0γ(τω1γ)−(1+γ)Mγ(ω(τω1γ)−γ)e−λ(τω1γ)ω1γdτ=ω−1∫∞0γτ−(1+γ)Mγ(τ−γ)e−(λω1γ)dτ=ω−1e−λγω. |
Therefore, by Theorem 2.5, we have
L{Eγ,γ(−υγA)x}(λ)=∫∞0e−λγωℑ(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(υ−s)γ−1(A˜u(s)+h(s))dsasυ≥0; |
therefore we get
˜u(υ)=∫υ0Eγ(−(υ−s)γA)ads+∫υ0(υ−s)γ−1Eγ,γ(−(υ−s)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(υ−s)γ−1(−A˜u(s)+F(u(s),w(s))+Pg(s))dsasυ≥0˜u(υ)=aυ+1Γ(γ)∫υ0(υ−s)γ−1(−A˜u(s)+F(˜u(s),w(s))+Pg(s))dswhileυ≥0. |
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)=aλγ−2+F(˜u(λ),w(λ))+Pg(λ)˜u(λ)=aλγ−2(λγ+A)−1+F(˜u(λ),w(λ))(λγ+A)−1+Pg(λ)(λγ+A)−1˜u(λ)=aλ−1λγ−1(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(−(υ−s)γA)ads+∫υ0(υ−s)γ−1Eγ,γ(−(υ−s)γA)F(˜u(s),w(s))ds+∫υ0(υ−s)γ−1Eγ,γ(−(υ−s)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 ˜u∈C([0,∞),Hγ,r) and, for υ∈[0,∞),
˜u(υ)=∫υ0Eγ(−(υ−s)γA)ads+∫υ0(υ−s)γ−1Eγ,γ(−(υ−s)γA)F(˜u(s),w(s))ds+∫υ0(υ−s)γ−1Eγ,γ(−(υ−s)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(−(υ−s)γA)ads˜φ(υ)=∫t0(t−s)γ−1Eγ,γ(−(υ−s)γA)Pg(s)ds˜ϖ(˜u,w)(υ)=∫υ0(υ−s)γ−1Eγ,γ(−(υ−s)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(υ−γ(1−β))for0<β<1andυ→0. | (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(1−β)|Pg(s)|r) |
and
B1=C1maxB(γ(1−β),1−γ(1−β)),B(γ(1−α),1−γ(1−β))L≥MC1maxB(γ(1−β),1−2γ(α−β)),B(γ(1−α),1−2γ(α−β)). |
Theorem 3.1. Let 1<r<∞, 0<γ<1 and (3.1) hold; then, for all at∈Hγ,r. Suppose that
C1|aυ|Hγ,r+B1M∞<14L. |
If a unique function ˜u:[0,∞)→Hγ,r is satisfied and we choose n2r−12<β, then there is α>max(β,12) where M∞=sups∈[0,∞)(sγ(1−β)Pg(s)):
(ⅰ) ˜u(0)=ax and ˜u:[0,∞)→Hγ,r shows continuity;
(ⅱ) ˜u:[0,∞)→Hα,r is continuous and limυ→0υγ(α−β)|˜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, limυ→0υγ(α−β)|˜u(υ)|Hα,r=0 and its genuine form is given by
||˜u||X∞=max(supυ≥0|˜u(υ)|Hγ,r,supt≥0υγ(α−β)|˜u(υ)|Hα,r). |
Now, we know that there exists M such that ˜u,w∈Hα,r such that F:Hα,r∗Hα,r→Jr is bounded and bilinear mapping:
|F(˜u,w)|r≤M|˜u|Hα,r|w|Hα,r|F(˜u,˜u)−F(w,w)|r≤M(|˜u|Hα,r+|w|Hα,r)|˜u−w|Hα,r. |
Step 1:
Consider that ˜u,w∈X∞. The term ˜ϖ(˜u(υ),w(τ)) is a part of C(0,∞),Hγ,r and C([0,ℑ],Hγ,r). By considering ε>0 that is very small and randomly fixing υ0≥0, now again suppose that υ>υ0 (υ<υ0analogously); we have
|˜ϖ(˜u(υ),w(υ))−˜ϖ(˜u(υ0),w(υ0))|Hγ,rds≤∫υυ0(υ−s)γ−1|Eγ,γ(−(υ−s)γA)F(˜u(s),w(s))|Hγ,rds+∫υ00|(υ−s)γ−1−(υ0−s)γ−1Eγ,γ(−(υ−s)γA)F(˜u(s),w(s))|Hγ,rds+∫υ0−ϵ0(υ0−s)γ−1|Eγ,γ(−(υ−s)γA)−Eγ,γ(−(υ0−s)γA)F(˜u(s),w(s))|Hγ,rds+∫υ0υ0−ϵ(υ0−s)γ−1|Eγ,γ(−(υ−s)γA)−Eγ,γ(−(υ0−s)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)≤C1∫υυ0(υ−s)γ(1−β)−1|F(˜u(s),w(s))|rds≤MC1∫υυ0(υ−s)γ(1−β)−1|(˜u(s)|Hα,r,|w(s))|Hα,rds≤MC1∫υυ0(υ−s)γ(1−β)−1s−2γ(α−β)dssups∈(0,υ](s2γ(α−β)|˜u(s)|Hα,r|w(s)|Hα,r) |
=MC1∫1υ0/υ(1−s)γ(1−β)−1s−2γ(α−β)dssups∈(0,υ](s2γ(α−β)|˜u(s)|Hα,r|w(s)|Hα,r). |
By applying the β function properties, ∃δ>0 much smaller as 0<υ−υ0<δ; we have
∫1υ0/υ(1−s)γ(1−β)−1s−2γ(α−β)ds→0 |
which follows that as υ−υ0→0, D11(υ) tends to 0.
Now for D12(υ),
D12(υ)=C1∫υ00((υ0−s)γ−1−(υ−s)γ−1)(υ−s)−βγ|F(˜u(s),w(s))|rds≤MC1∫υ00((υ0−s)γ−1−(υ−s)γ−1)(υ−s)−βγs−2γ(α−β)dssups∈(0,υ0](s2γ(α−β)|˜u(s)|Hα,r|w(s)|Hα,r). |
It is interesting to note that
∫υ00|(υ0−s)γ−1−(υ−s)γ−1|(υ−s)−βγs−2γ(α−β)ds≤∫υ00(υ−s)γ−1(υ−s)−βγs−2γ(α−β)ds+∫υ00(υ0−s)γ−1(υ−s)−βγs−2γ(α−β)ds≤2∫υ00(υ0−s)γ(1−β)−1(υ−s)−βγs−2γ(α−β)ds=2B(γ(1−β)),1−2γ(α−β). |
We can show this by applying Lebesgue's dominated convergence Theorem 2.1, we get
∫υ00((υ0−s)γ−1−(υ−s)γ−1)(υ−s)−βγs−2γ(α−β)ds→0,asυ→υ0. |
We can say that limυ→υ0D12(υ)=0. For D13(υ), we have
D13(υ)≤∫υ0−ϵ0(υ0−s)γ−1|Eγ,γ(−(υ−s)γA)−Eγ,γ(−(υ0−s)γA)F(˜u(s),w(s))|Hγ,rds≤∫υ0−ϵ0(υ0−s)γ−1((υ−s)−βγ+(υ0−s)−βγ)|F(˜u(s),w(s))|rds≤2MC1∫υ00(υ0−s)γ−1s−2γ(α−β)dssups∈(0,υ0](s2γ(α−β)|˜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
limυ→υ0D13(υ)=∫υ0−ϵ0(υ0−s)γ−1|Eγ,γ(−(υ−s)γA)−Eγ,γ(−(υ0−s)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)≤∫υ0υ0−ϵ(υ0−s)γ−1((υ−s)−βγ+(υ0−s)−βγ)|F(˜u(s),w(s))|rds≤2MC1∫υ00(υ0−s)γ−1s−2γ(α−β)dssups∈[υ0−ϵ,υ0](s2γ(α−β)|˜u(s)|Hα,r|w(s)|Hα,r)→0forε→0. |
It follows that
|˜ϖ(˜u(υ),w(υ))−˜ϖ(˜u(υ0),w(υ0))|Hγ,rds→0whileυ→υ0. |
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:X∞∗X∞→X∞ is a continuous bilinear operator, we consider that, according to Lemma 2.4, we have
|˜ϖ(˜u(υ),w(υ))|Hγ,r=|∫υ0(υ−s)γ−1|Eγ,γ(−(υ−s)γA)F(˜u(s),w(s))|Hγ,rds≤C1∫υ0(υ−s)γ(1−β)−1|F(˜u(s),w(s))|rds≤MC1∫υ0(υ−s)γ(1−β)−1s−2γ(α−β)dssups∈(0,υ](s2γ(α−β)|˜u(s)|Hα,r|w(s)|Hα,r)=MC1B((γ(1−β)),1−2γ(α−β))||u||X∞||w||X∞ |
and
|˜ϖ(˜u(υ),w(υ))|α,rH=|∫υ0(υ−s)γ−1|Eγ,γ(−(υ−s)γA)F(˜u(s),w(s))|Hα,rds≤C1∫υ0(υ−s)γ(1−α)−1|F(˜u(s),w(s))|rds≤MC1∫υ0(υ−s)γ(1−α)−1s−2γ(α−β)dssups∈(0,υ](s2γ(α−β)|˜u(s)|Hα,r|w(s)|Hα,r)=MC1υ−γ(α−β)B((γ(1−α)),1−2γ(α−β))||˜u||X∞||w||X∞. |
As a result of this
supυ∈[0,∞)υγ(α−β)|˜ϖ(˜u(υ),w(υ))|Hα,r≤MC1B(γ(1−α)),1−2γ(α−β)||˜u||X∞||w||X∞. |
To be more specific,
limυ→0υγ(α−β)|˜ϖ(˜u(υ),w(υ))|Hα,r=0. |
Therefore, ˜ϖ(˜u,w)∈X∞ and ||˜ϖ(˜u(υ),w(υ))||X∞≤L||˜u||X∞||w||X∞.
Step 3:
Let 0<υ0<υ. We have that
|˜φ(υ)−˜φ(υ0)|Hγ,r≤∫υυ0(υ−s)γ−1|Eγ,γ(−(υ−s)γA)Pg(s)|Hγ,rds+∫υ00((υ0−s)γ−1−(υ−s)γ−1)|Eγ,γ(−(t−s)γA)Pg(s)|Hγ,rds+∫υ0−ϵ0(υ0−s)γ−1|Eγ,γ(−(υ−s)γA)−Eγ,γ(−(υ0−s)γA)Pg(s)|Hγ,rds+∫υ0υ0−ϵ(υ0−s)γ−1|Eγ,γ(−(υ−s)γA)−Eγ,γ(−(υ0−s)γA)Pg(s)|Hγ,rds≤C1∫υυ0(υ−s)γ(1−β)−1|Pg(s)|rds+C1∫υ00((υ0−s)γ−1−(υ−s)γ−1)(υ−s)−βγ|Pg(s)|rds+C1∫υ0−ϵ0(υ0−s)γ−1|Eγ,γ(−(υ−s)γA)−Eγ,γ(−(υ0−s)γA)Pg(s)|Hγ,rds+2C1∫υ0υ0−ϵ(υ0−s)γ(1−β)−1|Pg(s)|rds≤C1M(υ)∫υυ0(υ−s)γ(1−β)−1s−γ(1−β)ds+C1M(υ)∫υ00((υ−s)γ−1−(υ0−s)γ−1)(υ−s)−βγs−γ(1−β)ds+C1C(υ)∫υ0−ϵ0(υ0−s)γ−1|Eγ,γ(−(υ−s)γA)−Eγ,γ(−(υ0−s)γA)|Hγ,rds+2C1C(υ)∫υ0υ0−ϵ(υ0−s)γ(1−β)−1s−γ(1−β)ds. |
The first two integrals and last integral tend to 0 as υ→υ0; also, ε→0 by using the characteristics of the β function. Now the third integral also tends to 0 as υ→υ0 by using Lemma 2.1. This suggests that
|˜φ(υ)−˜φ(υ0)|Hγ,r→0whileυ→υ0. |
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(υ−s)γ−1|Eγ,γ(−(υ−s)γA)Pg(s)|Hγ,rds≤C1∫υ0(υ−s)γ(1−β)−1|Pg(s)|rds≤C1M(υ)∫υ0(υ−s)γ(1−β)−1s−γ(1−β)ds=C1M(υ)B((γ(1−β)),(1−γ(1−β)))|˜φ(υ)|Hα,r|=∫υ0(υ−s)γ−1|Eγ,γ(−(υ−s)αA)Pg(s)|Hα,rds≤C1∫υ0(υ−s)γ(1−α)−1|Pg(s)|rds≤C1M(υ)∫υ0(υ−s)γ(1−α)−1s−γ(1−β)ds=υ−γ(α−β)C1M(υ)B((γ(1−α)),(1−γ(1−β))). | (3.2) |
To be more precise,
υγ(α−β)|˜φ(υ)|Hα,r≤C1M(υ)B((γ(1−α)),(1−γ(1−β)))→0whileυ→υ0. |
Because ˜M(t)→0 is the same as υ→0 owing to assumption, this result confirms that ||˜φ(υ)||∞≤˜B1˜M∞ as ˜φ(t)∈X∞. Now for aυ∈Hγ,r, according to Lemma 2.1 obvious that
Eγ(−υγA)a∈C([0,∞),Hγ,r) |
and
Eγ(−υγA)a∈C([0,∞),Hα,r). |
Therefore by Theorem 2.3, we can say that
∫υ0Eγ(−(υ−s)γA)ads∈C([0,∞),Hγ,r)and∫υ0Eγ(−(υ−s)γA)ads∈C([0,∞),Hα,r). |
This implies that for all υ∈(0,ℑ], and together with Theorem 2.2,
∫υ0Eγ(−(υ−s)γA)ads∈X∞. |
By using the above condition, we get
υγ(α−β)∫υ0Eγ(−(υ−s)γA)ads∈C([0,∞),Hα,r) |
||∫υ0Eγ(−(υ−s)γA)ads||X∞≤∫υ0||Eγ(−(υ−s)γA)a||X∞ds≤C1∫υ0|a|Hγ,rds≤C1|a|(t)Hγ,r. |
According to Theorem 3.1, the inequality
||∫υ0Eγ(−υγA)ads+˜φ(υ)||X∞≤||∫υ0Eγ(−υγA)ads||X∞+||˜φ(υ)||X∞≤∫υ0||Eγ(−υγA)a||X∞ds+||˜φ(υ)||X∞≤14L |
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 υ→0, we must show that
limυ→0∫υ0Eγ(−(υ−s)γA)ads=0limυ→0∫υ0(υ−s)γ−1Eγ,γ(−(υ−s)γA)Pg(s)ds=0limυ→0∫υ0(υ−s)γ−1Eγ,γ(−(υ−s)γA)F(˜u(s),w(s))ds=0 |
in Hγ,r. In fact limυ→0˜ψ(υ)=0 and limυ→0˜φ(υ)=0(limυ→0˜M(υ)=0) due to (3.2). So,
∫υ0(υ−s)γ−1|Eγ,γ(−(υ−s)γA)F(˜u(s),˜u(s))|Hγ,rds≤C∫υ0(υ−s)γ(1−β)−1|F(˜u(s),˜u(s))|rds≤MC∫υ0(υ−s)γ(1−β)−1|˜u(s)|2Hα,rds≤MC∫υ0(υ−s)γ(1−β)−1s−2γ(α−β)dssups∈(0,υ]{s2γ(α−β)|˜u(s)|2Hα,r}=MCB((γ(1−β)),1−2γ(α−β))sups∈(0,υ]{s2γ(α−β)|˜u(s)|2Hα,r}→0asυ→υ0. |
Theorem 3.2. Let 1<r<∞, 0<γ<1 and (3.1) holds; then, suppose that
n2r−12<β. |
Then there is a unique function ˜u:[0,∞)→Hγ,r and α>max(β,12) for all at∈Hγ,r there exists ℑ∗>0 satisfying the following
(ⅰ) ˜u(0)=aυ and ˜u:[0,ℑ∗]→Hγ,r is continuous,
(ⅱ) ˜u:[0,ℑ∗]→Hα,r is continuous, and limυ→0υγ(α−β)|˜u(υ)|Hα,r=0,
(ⅲ) for υ∈[0,ℑ∗], ˜u satisfies Definition 2.4.
Proof. Consider the space of all of the curves X=X[T] and ˜u:(0,ℑ]→Hγ,r; now, suppose that α=1+β2 so we have the following
(ⅰ) ˜u:[0,ℑ]→Hγ,r is continuous.
(ⅱ) ˜u:(0,ℑ]→Hα,r is continuous and limυ→0υγ(α−β)|˜u(υ)|Hα,r=0,
its original form is given by
||˜u||X=supυ∈[0,ℑ](υγ(α−β)|˜u(υ)|Hα,r). |
Similar to the conclusion of Theorem 3.1, it is worth noting that the function g:X∗X→X is linearly mappable and continuous. By Lemma 2.1 the function φ(υ)∈X for all υ∈(0,ℑ]. It is simple to assert that
Eγ(−υγA)a∈C([0,ℑ]),Hγ,r)Eγ(−υγA)a∈C([0,ℑ],Hα,r). |
Therefore by Theorem 2.3, we can say that
∫υ0Eγ(−(υ−s)γA)ads∈C([0,ℑ],Hγ,r)∫υ0Eγ(−(υ−s)γA)ads∈C([0,ℑ],Hα,r). |
For all υ∈(0,ℑ], using Theorem 2.2 implies that
∫υ0Eγ(−(υ−s)γA)ads∈X∞. |
We have
||˜u||X=supυ∈[0,ℑ](υγ(α−β)|˜u(υ)|Hα,r). |
By using the above condition, we get
υγ(α−β)∫υ0Eγ(−(υ−s)γA)ads∈C([0,ℑ],Hα,r). |
Now, consider a sufficiently small ℑ∗>0, such that
||∫υ0Eγ(−υγA)ads+˜φ(υ)||X[ℑ∗]≤||∫υ0Eγ(−υγA)ads||X[ℑ∗]+||˜φ(υ)||X[ℑ∗]<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
aυ∈Hγ,rwithn2r−12<γ. |
For aυ∈Hγ,r, Problem (1.5) has a unique mild solution ˜u in Jr, Aα˜u shows continuity on (0,ℑ]. Furthermore, υγ(α−β)Aα˜u(υ) is bounded while υ→0. Additionally, ˜u also shows continuity in [0,ℑ].
Proof. Step 1:
Let
K(υ)=sups∈(0,υ]sγ(α−β)|Aα˜u(s)|r |
and
ζ(υ)=˜ϖ(˜u,˜u)(υ)=∫υ0(υ−s)γ−1Eγ,γ(−(υ−s)γA)F(˜u(s),˜u(s))ds. |
According to the summary of Step 2 in Theorem 3.1, ζ(υ) shows continuity on [0,ℑ] and Aαζ(υ) is continuous on (0,ℑ] and also exists, then,
Aα|˜ϖ(˜u(υ),˜u(υ))|Hα,r=|∫υ0(υ−s)γ−1|Eγ,γ(−(υ−s)γA)AαF(˜u(s),˜u(s))|Hα,rds≤C1∫υ0(υ−s)γ(1−α)−1|AαF(˜u(s),˜u(s))|rds≤MC1∫υ0(υ−s)γ(1−α)−1s−2γ(α−β)dssups∈[0,υ]{s2γ(α−β)|Aαu(s)2|Hα,r}. |
We have
K2(υ)=sups∈(0,υ]s2γ(α−β)|Aα˜u(s)2|r. |
Using the above equation, it gives the final result such that
|Aαζ(υ)|r≤MC1B(γ(1−α),1−2γ(α−β))K2(υ)υ−γ(α−β). | (4.1) |
Take into account the integral ˜φ(υ). As the (3.1) holds then the inequality
|Pg(s)|r≤M(υ)sγ(1−β) |
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,ℑ]. Before discussing continuity, we signify
M(υ)=sups∈(0,υ]{sγ(1−β)|AαPg(s)|r}|Aα˜φ(υ)|Hα,r|=∫υ0(υ−s)γ−1|Eγ,γ(−(t−s)αA)AαPg(s)|Hα,rds≤C1∫υ0(υ−s)γ(1−α)−1|AαPg(s)|rds≤C1M(υ)∫υ0(υ−s)γ(1−α)−1s−γ(1−β)ds|Aα˜φ(υ)|r|=υ−γ(α−β)C1M(υ)B((γ(1−α)),(1−γ(1−β))). | (4.2) |
As υ→0, we get that M(υ)=0 and |Pg(υ)|r=0(υ−γ(α−β)). As υ→0, we get that |Aαζ(υ)|r=0(υ−γ(α−β)) according to (4.2). We show that the function ˜φ is continuous in Jr. In reality, considering 0≤υ0<υ<ℑ, we have
|˜φ(υ)−˜φ(υ0)|r≤C3∫υυ0(υ−s)γ(1−β)−1|Pg(s)|rds+C3∫υ00((υ0−s)γ−1−(υ−s)γ−1)(υ−s)−βγ|Pg(s)|rds+C3∫υ0−ϵ0(υ0−s)γ−1|Eγ,γ(−(υ−s)γA)−Eγ,γ(−(υ0−s)γA)Pg(s)|Hγ,rds+2C3∫υ0υ0−ϵ(υ0−s)γ(1−β)−1|Pg(s)|rds≤C3M(υ)∫υυ0(υ−s)γ(1−β)−1s−γ(1−β)ds+C3M(υ)∫υ00((υ−s)γ−1−(υ0−s)γ−1)(υ−s)−βγs−γ(1−β)ds+C3M(υ)∫υ0−ϵ0(υ0−s)γ−1|Eγ,γ(−(υ−s)γA)−Eγ,γ(−(υ0−s)γA)|rs−γ(1−β)ds+2C3M(υ)∫υ0υ0−ϵ(υ0−s)γ(1−β)−1s−γ(1−β)ds. |
Further, we assume the function ∫υ0Eγ(−(υ−s)γA)ads. From Theorem 2.2, it is clear that
|Aα∫υ0Eγ(−(υ−s)γA)ads|r=∫υ0|AαEγ(−(υ−s)γA)a|Hγ,rds. |
From Lemma 2.4,
≤C1∫υ0υ−γ(α−β)|Aβa|r=C1υ−γ(α−β)|aυ|Hα,r |
and
limυ→0υγ(α−β)|Aα∫υ0Eγ(−(υ−s)γA)ads|r=limυ→0υγ(α−β)∫υ0|Eγ(−(υ−s)γA)a|Hγ,rds=0. |
Step 2:
We now build the result through successive approximation:
˜u0(υ)=∫υ0Eγ(−(υ−s)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,ℑ]. Moreover, Kn(0)=0. However, according to (4.1) and (4.3), Kn(υ) satisfies the inequality
Kn+1(υ)≤K0(υ)+MC1B((γ(1−α)),1−2γ(α−β))Kn2(υ). | (4.4) |
Select ℑ>0 so that
4MC1B((γ(1−α)),1−2γ(α−β))K0(υ)<1. | (4.5) |
Then, the sequence Kn(ℑ) is bounded; following a simple description of (4.4), so we get
Kn(ℑ)≤ρ(ℑ),forn=0,1,2,..., | (4.6) |
while (4.4) is just like the quadratic equation; we have
MC1B((γ(1−α)),1−2γ(α−β))Kn2(υ)−Kn+1(υ)+K0(υ)≥0. |
Here, a=MC1B((γ(1−α)),1−2γ(α−β))Kn2(υ), b=−1 and c=K0(υ). After applying the quadratic formula, we obtain
Kn(υ)=1−√1−4MC1B((γ(1−α)),1−2γ(α−β))K0(υ)2MC1B((γ(1−α)),1−2γ(α−β)). |
Considering (4.6),
ρ(υ)=1−√1−4MC1B((γ(1−α)),1−2γ(α−β))K0(υ)2MC1B((γ(1−α)),1−2γ(α−β)). |
Conversely, Kn(υ)≤ρ(υ) holds for any υ∈(0,ℑ]. Correspondingly, ρ(υ)≤2K0(υ). Suppose the following concept of equality:
yn+1(υ)=∫υ0(υ−s)γ−1Eγ,γ(−(υ−s)γA)F(˜un+1(s),˜un+1(s))−F(˜un(s),˜un(s))ds. |
For υ∈(0,ℑ] and n=0,1,2,..., we denote yn=˜un+1−˜un:
Yn(υ)=sups∈(0,υ]sγ(α−β)|Aαyn(s)|r. |
Using Lemma 2.3, we get
|F(˜un+1(s),˜un+1(s))−F(˜un(s),˜un(s))|r≤M(|˜un+1|+|˜un|)|˜un+1−˜un|≤M(|˜un+1|+|˜un|)Aαynsups∈(0,υ]s−γ(α−β)sγ(α−β)≤M(|Aα˜un+1|+|Aα˜un|)ynsups∈(0,υ]s−γ(α−β)sγ(α−β)≤M(sups∈(0,υ]sγ(α−β)|Aα˜un+1|+sups∈(0,υ]sγ(α−β)|Aα˜un|)yns−γ(α−β)≤M(Kn+1+Kn)yns−γ(α−β). |
Then, we set
≤M(Kn+1+Kn)yns−γ(α−β)Aαsups∈(0,υ]s−γ(α−β)sγ(α−β) |
≤M(Kn+1+Kn){sups∈(0,υ]sγ(α−β)Aαyn}s−2γ(α−β) |
|F(˜un+1(s),˜un+1(s))−F(˜un(s),˜un(s))|r≤M(Kn+1(s)+Kn(υ))Yn(s)s−2γ(α−β). |
We get the results from Step 2 in Theorem 3.1:
υγ(α−β)|Aαyn+1(υ)|r≤2MC1B((γ(1−α)),1−γ(1−β))ρ(υ)Yn(υ). |
It provides
Yn+1(ℑ)≤2MC1B((γ(1−α)),1−2γ(α−β))ρ(ℑ)Yn(ℑ)≤4MC1B(γ(1−α),1−2γ(α−β))K0(ℑ)Yn(ℑ). | (4.7) |
In accordance with (4.5) and (4.7), we get
limn→0Yn+1(ℑ)Yn(ℑ)≤4MC1B((γ(1−α)),1−2γ(α−β))K0(ℑ)≤1. |
Hence the series ∑∞n=0Yn(ℑ) is convergent. For υ∈(0,ℑ] it shows that the series ∑∞n=0υγ(α−β)Aαyn(υ) is uniformly convergent. Thus, the sequence {υγ(α−β)Aα˜un(υ)} is uniformly convergent in (0,ℑ]. This signifies that limn→∞˜un(υ)=˜u(υ)∈D(Aα) and limn→∞υγ(α−β)Aα˜un(υ)=υγ(α−β)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 A−α is bounded. As a result, the function K(υ)=sups∈(0,υ]sβ(α−γ)|Aα˜un(s)|r satisfies
K(υ)≤ρ(υ)≤2K0(υ),forυ∈(0,υ] | (4.8) |
and
ηn=sups∈(0,ℑ]s2γ(α−β)|˜F(˜un(s),˜un(s))−˜F(˜u(s),˜u(s))|r≤M(Kn(ℑ)+K(ℑ))sups∈(0,ℑ]sυγ(α−β)|Aα(˜un(s)−˜u(s))|r→0asn→∞. |
Lastly, it is necessary to confirm that ˜u is a mild solution of (1.5) in [0,ℑ]. Because
|ϖ(˜un,˜un)(υ)−ϖ(˜u,˜u)(υ)|r≤∫υ0(υ−s)γ−1ηns−2γ(α−β)ds=υβγηn→0,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,ℑ], we can say that (4.9) holds and ˜u∈C([0,ℑ],Jr). Furthermore, the uniform convergence of υγ(α−β)Aαun(υ) to tγ(α−β)Aα˜u(υ) shows continuity of Aα˜u(υ). By (4.8) and K0(0)=0, we obtain that |Aαu(υ)|r=0υ(−γ(α−β)) 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(υ−s)γ−1Eγ,γ(−(υ−s)γA)(F(˜u(s),˜u(s))−F(˜w(s),˜w(s)))ds. |
Now, we defining the following functions:
K=maxsups∈(0,υ]sγ(α−β)|Aαu(s)|r,sups∈(0,υ]sγ(α−β)|Aαv(s)|r. |
According to Theorem 3.1 and Lemma 2.4, we get
|Aαy(υ)|r≤MC1K∫υ0(υ−s)γ(1−α)−1s−γ(α−β)|Aαy(s)|rds. |
For υ∈(0,ℑ], the Gronwall inequality demonstrates that Aαy(υ)=0. It shows that for υ∈[0,ℑ], 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(1−α)), which is given as
|Pg(υ)−Pg(s)|r≤L|υ−s|θforall0<υ,s≤ℑ. | (5.1) |
Definition 5.1. The problem (1.5) has a solution that is a classical solution of a function ˜u:[0,ℑ]→Jr, and ˜u∈C([0,ℑ],Jr) with cDυυ∈C([0,ℑ],Jr), which satisfies (1.5) for all υ∈(0,ℑ] and accepts values in D(A).
Lemma 5.1. Let Definition 5.1 be satisfied if
φ1(υ)=∫υ0(υ−s)γ−1Eγ,γ(−(υ−s)γA)(Pg(s),Pg(υ))ds,forυ∈(0,ℑ] |
so Aφ1(υ)∈Cθ([0,ℑ],Jr) and φ1(υ)∈D(A).
Proof. From Lemma 2.4 and (5.1), we fixed υ∈(0,ℑ],
(υ−s)γ−1|AEγ,γ(−(υ−s)γA)(Pg(s),Pg(υ))|r≤C1(υ−s)−1|Pg(s)−Pg(υ)|r≤C1L(υ−s)θ−1∈L1([0,ℑ],Jr). |
Then
|Aφ(υ)|r≤(υ−s)γ−1|AEγ,γ(−(υ−s)γA)(Pg(s),Pg(υ))|rds≤C1L∫υ0(υ−s)θ−1ds≤C1Lθυθ<∞; |
we get φ1(υ)∈D(A) by the closeness property of A. We must demonstrate that Aφ1(υ) has Hölder continuity. Because
ddυ(υγ−1Eγ,γ(−μυγ))=(υγ−2Eγ,γ−1(−μυγ)) |
it follows that
ddυ(υγ−1AEγ,γ(−υγA))=12πι∫Γθ(υγ−2Eγ,γ−1(−μυγ))A(μI+A)−1dμ=12πι∫Γθ(γ−2Eγ,γ−1(−μυγ))dμ−12πι∫Γθ(υγ−2Eγ,γ−1(−μυγ))A(μI+A)−1dμ. |
Substituting −μυγ=ξ implies that −υγdμ=dξ. So we have that dμ=−1υγdξ:
=12πι∫Γθ(−υγ−2Eγ,γ−1(ξ))1υγdξ−12πι∫Γθ(υγ−2Eγ,γ−1(ξ))ξυγA(−ξυγI+A)−11υγdξ. |
Since
||(μI+A)−1||≤C|μ| |
and
||ddυ(υγ−1AEγ,γ(−υγA))||≤Cγυ−2for0<υ≤ℑ |
for every 0<s<υ≤ℑ, we apply the mean value theorem
||(υγ−1AEγ,γ(−υγA))−(sγ−1AEγ,γ(−sγA))||=||∫υsddτ(τγ−1AEγ,γ(−τγA))dτ||≤∫υs||ddτ(τγ−1AEγ,γ(−τγA))||dτ≤∫υsτ−2dτ=Cγ(s−1−υ−1). | (5.2) |
Now for 0<υ<υ+ℏ≤ℑ, we suppose that ℏ>0, so
Aφ1(υ+ℏ)−Aφ1(υ)=∫υ0(υ+ℏ−s)γ−1AEγ,γ(−(υ+ℏ−s)γA)(Pg(s)−Pg(υ))ds−(υ−s)γ−1AEγ,γ(−(υ−s)γA)(Pg(s)−Pg(υ))ds+∫υ0(υ+ℏ−s)γ−1AEγ,γ(−(υ+ℏ−s)γA)(Pg(υ)−Pg(υ+ℏ))ds+∫υ+ℏυ(υ+ℏ−s)γ−1AEγ,γ(−(υ+ℏ−s)γA)(Pg(s)−Pg(υ+ℏ))ds=ℏ1(υ)+ℏ2(υ)+ℏ3(υ). | (5.3) |
For convenience, we solve each term separately. For ℏ1(υ), from (5.2) and (5.1) we have
|ℏ1(υ)|r≤∫υ0||(υ+ℏ−s)γ−1AEγ,γ(−(υ+ℏ−s)γA)−(υ−s)γ−1AEγ,γ(−(υ−s)γA)||r(Pg(s)−Pg(υ))ds≤CγLℏ∫υ0(υ+ℏ−s)−1(υ−s)θ−1ds≤CγLℏ∫υ0(s+ℏ)−1(υ−s)θ−1ds≤CγL∫h0ℏℏ+ssθ−1ds+CγLℏ∫∞ℏsℏ+ssθ−1ds≤CγLℏθ. | (5.4) |
For ℏ2(t), using Lemma 2.4 and (5.1), we get
|ℏ2(υ)|r≤∫υ0(υ+ℏ−s)γ−1|AEγ,γ(−(υ+ℏ−s)γA)(Pg(υ)−Pg(υ+ℏ))|rds≤C1∫υ0(υ+ℏ−s)−1|Pg(υ)−Pg(υ+ℏ)|rds≤C1ℏθ∫υ0(υ+ℏ−s)−1ds=C1L[lnℏ−ln(υ+ℏ)]ℏθ. | (5.5) |
Now for ℏ3(υ), using Lemma 2.4 and (5.1), we have
|ℏ3(υ)|r≤∫υ+ℏυ(υ+ℏ−s)γ−1|AEγ,γ(−(υ+ℏ−s)γA)(Pg(s)−Pg(υ+ℏ))|rds≤C1∫υ+ℏυ(υ+ℏ−s)−1|Pg(s)−Pg(υ+ℏ)|rds≤C1L∫υ+ℏυ(υ+ℏ−s)θ−1ds=C1Lℏθθ. | (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 aυ∈D(A).
Proof. Step 1:
We have that aυ∈D(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(υ−s)γ−1Eγ,γ(−(υ−s)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) |
˜φ∈C([0,ℑ],Jr) follows from Theorem 4.1. We can write ˜φ(υ)=˜φ1(υ)+˜φ2(υ), while
˜φ1(υ)=∫υ0(υ−s)γ−1Eγ,γ(−(υ−s)γA)(Pg(s),Pg(υ))ds˜φ2(υ)=∫υ0(υ−s)γ−1Eγ,γ(−(υ−s)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,ℑ] we can say that ˜φ2(υ)∈D(A) and ˜φ2(υ)∈Cν((0,ℑ],Jr). Now we have to show that cDγυ(˜φ)∈C((0,ℑ],Jr). Taking ˜φ(0)=0 and in view of Lemma 2.2(ⅳ),
cDγυ˜φ(υ)=ddυ(I1−γυ˜φ(υ))=ddυ(Eγ(−υγA)∗Pg). |
It is still necessary to demonstrate that (Eγ(−υγA)∗Pg) is differentiability continuous in Jr. Suppose that 0<h≤ℑ−υ; therefore, we have
1ℏ[(Eγ(−(υ+ℏ)γA)∗Pg)−(Eγ(−υγA)∗Pg)]=∫υ01ℏ[(Eγ(−(υ+ℏ−s)γA)Pg(s))−(Eγ(−(υ−s)γA)Pg(s))]ds+1ℏ∫υ+ℏυ(Eγ(−(υ+ℏ−s)γA)Pg(s))ds. |
Notice that
∫υ01ℏ|(Eγ(−(υ+ℏ−s)γA)Pg(s))−(Eγ(−(υ−s)γA)Pg(s))|rds |
≤1ℏ∫υ0|(Eγ(−(υ+h−s)γA)Pg(s))|rds+1ℏ∫υ0|(Eγ(−(υ−s)γA)Pg(s))|rds. |
In view of Lemma 2.4,
≤C1M(υ)1ℏ∫υ0(υ+ℏ−s)−γs−γ(1−β)ds+C1M(υ)1ℏ∫υ0(υ−s)−γs−γ(1−β)ds≤C1M(υ)1ℏ((υ+ℏ)1−γ+υ1−γ)B(1−γ,1−γ(1−β)), |
the dominated convergence theorem applies; then, we get
limℏ→0∫υ01ℏ[(Eγ(−(υ+ℏ−s)γA)Pg(s))−(Eγ(−(υ−s)γA)Pg(s))]ds=E′γ(−(υ−s)γA)Pg(s). |
From Lemma 2.2(ⅲ), we have
E′γ(−υγA)u=−υγ−1AEγ,γ(−υγA)u∫E′γ(−υγA)u=−∫υγ−1AEγ,γ(−υγA)uEγ(−υγA)u=−∫υγ−1AEγ,γ(−υγA)u. |
Therefore,
E′γ(−(υ−s)γA)Pg(s)=−∫υ0(υ−s)γ−1AEγ,γ(−(υ−s)γA)Pg(s)ds=A˜φ(υ). |
Conversely,
1ℏ∫υ+hυEγ(−(υ+ℏ−s)γA)Pg(s)ds. |
Let s∗=υ+ℏ−s so ds∗=−ds and after setting the conditions [s=υ implies s∗=ℏ] and [s=υ+ℏ implies s∗=0], we have
1ℏ∫0ℏEγ(−(s∗)γA)Pg(υ+ℏ−s∗)(−ds∗). |
By replacing s∗→s, we get
1ℏ∫h0Eγ(−sγA)Pg(υ+ℏ−s)(ds)=1ℏ∫ℏ0Eγ(−sγA)[Pg(υ+ℏ−s)−Pg(υ−s)+Pg(υ−s)−Pg(υ)+Pg(υ)]ds=1ℏ∫ℏ0Eγ(−sγA)(Pg(υ+ℏ−s)−Pg(υ−s))ds+1ℏ∫ℏ0Eγ(−sγA)(Pg(υ−s)−Pg(υ))ds+1ℏ∫ℏ0Eγ(−sγA)Pg(υ)ds. |
From Lemmas 2.1–2.4 and (5.1), we have
|1ℏ∫ℏ0Eγ(−sγA)(Pg(υ+ℏ−s)−Pg(υ−s))|rds≤C1Lℏθ|1ℏ∫ℏ0Eγ(−sγA)(Pg(υ−s)−Pg(υ))|rds≤C1Lℏθθ+1. |
From Lemma 2.1(ⅰ),
limh→01h∫h0Eγ(−sγA)Pg(υ)ds=Pg(υ)limh→01h∫υ+hυEγ((υ+h−s)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˜φ2∈C((0,ℑ],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˜φ∈C((0,ℑ],Jr).
Step 2:
Consider ˜u as the mild solution of (1.5). In order to draw the conclusion that F(˜u,˜u)∈Cθ((0,ℑ],Jr), we have to show that Aα˜u holds Hölder continuity in Jr according to Theorem 3.1. Now for 0<υ<υ+ℏ, let ℏ>0. We know that ψ(υ)=∫υ0Eγ(−(υ−s)γ)ads. Denoting ϕ(υ):=∫υ0Eγ(−υγA)ads, and by Lemma 2.2(ⅲ) we have
ddυEγ(−υγA)a=−υγ−1AEγ,γ(−υγA)a. |
Integrating on both sides, we get
∫υ0ddsEγ(−(υ−s)γA)ads=−∫υ0sγ−1AEγ,γ(−sγA)ads. |
After applying the limits, we have
Eγ(−υγA)a=∫υ0sγ−1AEγ,γ(−sγA)ads. |
Using the above results and Lemma 2.4, then
|Aαϕ(υ+ℏ)−Aγϕ(υ)|r=|∫υ+ℏυsγ−1AαEγ,γ(−sγA)ads|r≤∫υ+ℏυsγ−1|Aα−βEγ,γ(−sγA)Aβa|rds≤C1∫υ+ℏυsγ(1+β−α)−1ds|Aβa|r=C1|a|Hγ,rγ(1+β−α)((υ+ℏ)γ(1+β−α)−υγ(1+β−α))=C1|a|Hγ,rγ(1+β−α)ℏγ(1+β−α). |
Thus Aαϕ∈Cθ((0,ℑ],Jr). For every small ε>0 we take ℏ as ε≤υ<υ+ℏ≤ℑ. We have
|Aαϕ(υ+ℏ)−Aγϕ(υ)|r≤|∫υ+ℏυ(υ+ℏ−s)γ−1AαEγ,γ(−(υ+ℏ−s)γA)Pg(s)ds|r+|∫υ0Aα((υ+ℏ−s)γ−1Eγ,γ(−(υ+ℏ−s)γA)−(υ−s)γ−1Eγ,γ(−(υ−s)γA))Pg(s)ds|r.=ϕ1(υ)+ϕ2(υ). |
Using Lemma 2.4 and (5.1), we have
ϕ1(υ)≤C1∫υ+ℏυ(υ+ℏ−s)γ(1−α)−1|Pg(s)|rds≤C1M(υ)∫υ+ℏυ(υ+ℏ−s)γ(1−α)−1s−γ(1−α)−1ds≤M(υ)C1γ(1−α)ℏγ(1−α)υ−γ(1−α)−1≤M(υ)C1γ(1−α)ℏγ(1−α)ε−γ(1−α)−1. |
To estimate ϕ2, we have
ddυ(υγ−1AαEγ,γ(−υγA))=12πı∫Γμα(υγ−2Eγ,γ−1(−μυγ))A(μI+A)−1dμ=12πı∫Γ′−(−ξυγ)α(υγ−2Eγ,γ−1(ξ))(−ξυγI+A)−11υγdξ, |
which yields that
ddυ(υγ−1AαEγ,γ(−υγA))≤Cγυγ(1−α)−2. |
Now, we apply the mean value theorem:
||(υγ−1AαEγ,γ(−υγA))−(sγ−1AαEγ,γ(−sγA))||=||∫υsddτ(τγ−1AαEγ,γ(−τγA))dτ||≤∫υs||ddτ(τγ−1AαEγ,γ(−τγA))||dτ≤∫υsτγ(1−α)−2dτ=Cγ(sγ(1−α)−1−υγ(1−α)−1). |
Thus,
ϕ2(t)≤|∫υ0Aα((υ+ℏ−s)γ−1Eγ,γ(−(υ+ℏ−s)γA)−(υ−s)γ−1Eγ,γ(−(υ−s)γA))Pg(s)ds|r≤∫υ0((υ−s)γ(1−α)−1−(υ+ℏ−s)γ(1−α)−1)|Pg(s)|rds≤CγM(υ)(∫υ0(υ−s)γ(1−α)−1s−γ(1−β)ds−∫υ+ℏ0(υ+ℏ−s)γ(1−α)−1s−γ(1−β)ds)+CγM(υ)∫υ+ℏυ(υ+ℏ−s)γ(1−α)−1s−γ(1−β)ds≤CγM(υ)(υγ(β−α)−(υ+ℏ)β−α)B(γ(1−α),1−γ(1−β))+CγM(υ)ℏγ(1−α)υ−γ(1−β)≤CγM(υ)ℏγ(α−β)[ε(ε+h)]γ(β−α)+CγM(υ)ℏγ(1−α)ε−γ(1−β). |
This shows that Aαϕ∈Cθ([ε,ℑ],Jr). Therefore, Aαϕ∈Cθ((0,ℑ],Jr) due to an arbitrary ε. Try to remember that ζ(υ)=∫υ0(υ−s)γ−1Eγ,γ(−(υ−s)γA)F(˜u(s),˜u(s))ds. We know that |F(˜u(s),˜u(s)|r≤MK2(υ)s−2γ(α−β), while K(υ)=sups∈(0,υ]sγ(α−β)|Aαu(s)|r is bounded and continuous in (0,ℑ]. A simple reasoning allows us to take the Hölder continuity of Aζ∈Cθ((0,ℑ],Jr). Therefore we have that Aα˜u(υ) = Aαϕ(υ)+Aαφ(υ)+Aαζ(υ)∈Cθ((0,ℑ],Jr). We know that F(˜u,˜u)∈Cθ((0,ℑ],Jr) is verified; similarly following Step 2, this yields that cDγυζ∈Cθ((0,ℑ],Jr), Aζ∈Cθ((0,ℑ],Jr) and cDγυζ=−Aζ+F(˜u,˜u). So we get that cDγυζ∈Cθ((0,ℑ],Jr), A˜u∈Cθ((0,ℑ],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, cDγυ∈Cν((0,ℑ],Jr) and A˜u∈Cν((0,ℑ],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˜φ∈Cγ(1−β)((0,ℑ],Jr). It is also sufficient to demonstrate that, for every ε>0 there is A˜φ∈Cγ(1−β)((ε,ℑ],Jr). Now by using Lemma 2.2(ⅲ) we take ℏ as ε≤υ<υ+ℏ≤ℑ;
|A˜φ(υ+ℏ)−A˜φ(υ)|r=|∫υ+ℏυ−sγ−1A2Eγ,γ(−sγA)ads|r≤C1∫υ+ℏυs−γ(1−β)−1ds|a|Hγ,r=C1|a|Hγ,rγ(υ−γ(1−β)−(υ+ℏ)−γ(1−β))=C1|a|Hγ,rγℏγ(1−β)[˜ε(˜ε+ℏ)]γ(1−β). |
Using (6) and rewriting ˜φ(υ) in the form of
˜φ(υ)=˜φ1(υ)+˜φ2(υ)=∫υ0(υ−s)γ−1Eγ,γ(−(υ−s)γA)(Pg(s)−Pg(υ))ds+∫υυ0(υ−s)γ−1Eγ,γ(−(υ−s)γA)Pg(υ)ds |
as υ∈(0,ℑ], we can say that A˜φ1(υ)∈Cν)((0,ℑ],Jr) and A˜φ2(υ)∈Cθ((0,ℑ],Jr). These results come from Lemma 5.1 and (5.8). As we know that F(˜u,˜u)∈Cθ((0,ℑ],Jr) and ζ(υ) is verified analogously, so we can conclude that Aζ(υ)∈Cν((0,ℑ],Jr). Consequently, cDγυ˜u=A˜u+F(˜u,˜u)+Pg∈Cν((0,ℑ],Jr) as A˜u∈Cν((0,ℑ],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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