On an initial boundary value problem for fractional pseudo-parabolic equation with conformable derivative

: In this paper, we study the initial boundary value problem of the pseudo-parabolic equation with a conformable derivative. We focus on investigating the existence of the global solution and examining the derivative’s regularity. In addition, we contributed two interesting results. Firstly, we proved the convergence of the mild solution of the pseudo-parabolic equation to the solution of the parabolic equation. Secondly, we examine the convergence of solution when the order of the derivative of the fractional operator approaches 1 − . Our main techniques used in this paper are Banach fixed point theorem and Sobolev embedding. We also apply di ff erent techniques to evaluate the convergence of generalized integrals encountered.


Introduction
Fractional calculus is one of today's most popular mathematical tools to model real-world problems.More specifically, it has been applied to model evolutionary systems involving memory effects on dynamical systems.Partial differential equations (PDEs) with fractional operators are important in describing phenomena in many fields such as physics, biology and chemistry [1,2].Based on the generalizations of fractional derivatives by famous mathematicians such as Euler, Lagrange, Laplace, Fourier, Abel and Liouville, today's mathematicians have explored and introduced many more types of fractional derivatives such as Riemann-Liouville, Caputo, Liouville, Weyl, Riesz and Hifler [3][4][5][6].
PDEs with conformable derivatives attract interested mathematicians using different approaches because of their wide range of applications, such as electrical circuits [7] and chaotic systems in dynamics [8].We recognize that the conformable and classical derivatives have a close relationship.There is an interesting observation that: If f is a real function and s > 0, then f has a conformable fractional derivative of order β at s if and only if it is (classically) differentiable at s, and where 0 < β ≤ 1.Another surprising observation is that Eq (1.1) will not hold if f is defined in a general Banach space.We can better understand why the ODEs with the conformable derivative on R have been studied so much.In addition, the relevant research in infinite-dimensional spaces, such as Banach or Hilbert space, is still limited, which motivates us to investigate some types of PDEs with conformable derivatives in Hilbert or Sobolev spaces.Besides, in some phenomena, the conformable derivative is better simulated than the classical derivative.In [9], the authors considered the conformable diffusion equation where 0 < α ≤ 1, x > 0, t > 0, u(x, t) is the concentration, and D α represents the generalized diffusion coefficient, which was applied in the description of a subdiffusion process.In particular, the conformable diffusion equation (1.2) reduces to the normal diffusion equation if α = 1.A natural and fundamental question is, "Does the conformable diffusion model predict better than the normal diffusion model?".The results in [9] show that the conformable derivative model agrees better with the experimental data than the normal diffusion equation.
Let Ω ⊂ R N (N ≥ 1) be a bounded domain with smooth boundary ∂Ω, and T > 0 is a given positive number.In this paper, we investigate the Sobolev equation with a conformable derivative as follows x ∈ Ω, t ∈ (0, T ), u(x, t) = 0, x ∈ ∂Ω, t ∈ (0, T ), where α, β ∈ (0, 1], m > 0, the time fractional derivative C ∂ α ∂t α is the conformable derivative of order α, defined in Definition 2.1.The function F represents the external forces or the advection term of a diffusion phenomenon, etc., and the function u 0 is the initial condition specified later.The operator (−∆) β is the fractional Laplacian operator, which is well-defined in [10] (see page 3).In the sense of distribution, the study of weak solutions to Problem (1.3) is still limited compared to the classical problem.Thus, the fundamental knowledge in the distributive sense for Problem (1.3) is still open and challenging, which is the main reason and motivation for us to study the problem from the perspective of the semigroup.
Next, we mention some results related to Problem (1.3).There are two interesting observations regarding Problem (1.3).
• If we take m = 0 in Problem (1.3), then we obtain an initial boundary value problem of the parabolic equation with a conformable operator as follows (1.4) The latest results on the well-posedness of solutions to Problem (1.4) are shown in more detail in [11], and the authors used the Hilbert scales space technique to prove the local existence of the mild solution to Problem (1.4).
• If α = 1, the main equation of Problem (1.3) becomes the classical equation Equation (1.5) is familiar to mathematicians about PDEs, called the pseudo-parabolic equation, also known as the Sobolev equation.The pseudo-parabolic equation describes a series of important physical processes, such as the permeation of a homogeneous liquid through fractured rock, population aggregation and one-way propagation of the nonlinear dispersion length wave [12,13].Equation (1.5) has been studied extensively; for details, see [12][13][14] and references given there.
Concerning the study of the existence and blowup of solutions to pseudo-parabolic equations, we refer the reader to [15][16][17].For the convenience of readers, we next list some interesting results related to pseudo-parabolic with fractional derivative.Luc et al. in [18] considered fractional pseudo-parabolic equation with Caputo derivative where 0 < α < 1, A = −∆, D α is Caputo fractional derivative operator of order α.They studied the local and global existence of solutions to Problem (1.6) when the nonlinear term F is the global Lipschitz.Based on the work in [18] , there are many related results in the spirit of a semigroup representation of the form of the Fourier series.In [12], the authors studied nonlinear time-fractional pseudo-parabolic equations with Caputo derivative on both bounded and unbounded domains by different methods and techniques from [18].In [19], the authors studied the nonlocal in time problem for a pseudo-parabolic equation with fractional time and space in the linear case.Tuan et al. [20] derived the nonlinear pseudo-parabolic equation with a nonlocal type of integral condition.In [21], the authors considered the time-space pseudo-parabolic equation with the Riemann-Liouville time-fractional derivative, and they applied the Galerkin method to show the global and local existence of solutions.
As far as we know, there has not been any work that considers the initial boundary value problem (1.3) with a conformable derivative.The main results and methods of the present paper are described in detail as follows • Firstly, we prove the existence of the global solution to Problem (1.3).The main idea is to use Banach fixed point theorem with the new weighted norm used in [22].In order to prove the regularity and the derivative of the mild solution, we need to apply some complicated techniques on Hilbert scales for nonlinearity terms.Compared with [11], our method has very different characteristics.It is important to emphasize that proving the existence of the global solution is difficult, which is demonstrated in our current paper, but not in the paper [11].• Secondly, we investigate the convergence of solution to Problem (1.3) when m → 0 + , which does not appear in the works related to fractional pseudo-parabolic equation.This result allows us to get the relationship between the solution of Sobolev equation and parabolic diffusion equation.To overcome the difficulty, we need to control the improper integrals and control the parameters.This pioneering work can open up some new research directions for finding the relationship between the solutions of the pseudo-parabolic equation and the parabolic equation.• Finally, we prove the convergence of the solution when the order of derivative β → 1 − .This direction of research was motivated by the recent paper [23].Since the current model has a nonlinear source function, the processing technique for the proof in this paper seems to be more complicated than that of [23].
The greatest difficulty in solving this problem is the study of many integrals containing singular terms, such as s α−1 or (t α − s α ) −m .To overcome these difficulties, we need to use ingenious calculations and techniques to control the convergence of several generalized integrals.This paper is organized as follows.Section 2 provides some definitions.In Section 3, we give the definition of the mild solution and some important lemmas for the proof of the main results.Section 4 shows the global existence of the solution to Problem (1.3).In addition, we present the regularity result for the derivative of the mild solution.Section 5 shows the convergence of the solution to Problem (1.3) when m → 0 + .In Section 6, we investigate the convergence of mild solutions when β → 1 − .

Preliminary results
for each t > 0. (For more details on the above definition, we refer the reader to [24][25][26][27].) In this section, we introduce the notation and the functional setting used in our paper.Recall the spectral problem The corresponding eigenfunctions are e n ∈ H 1 0 (Ω).Definition 2.2.(Hilbert scale space).We recall the Hilbert scale space, which is given as follows for any s ≥ 0. It is well-known that H s (Ω) is a Hilbert space corresponding to the norm Definition 2.3.Let X a,q,α ((0, T ]; B) denote the weighted space of all the functions ψ ∈ C((0, T ]; X) such that ∥ψ∥ X a,q,α ((0,T ];B) := sup where a, q > 0 and 0 < α ≤ 1 (see [22]).If q = 0, we denote X a,q ((0, T ]; B) by X a ((0, T ]; B).

The mild solution and some lemmas
In order to find a precise formulation for solutions, we consider the mild solution in terms of the Fourier series u(x, t) = ∞ n=1 ⟨u(., t), e n ⟩e n (x).
Taking the inner product of Problem (1.3) with e n gives In view of the result in Theorem 5, [26] and Theorem 3.3, [28] the solution to Problem (1.3) is u(., t), e n = exp To simplify the solution formula, we will express the solution in operator equations.Let us set the following operators for any f ∈ L 2 (Ω) in the form f = n∈N f, e n e n .Then the inverse operator of S m,α,β (t) is defined by The mild solution is given by For a qualitative analysis of the solution to (3.2), we need the bounded result for the operators in Hilbert scales space. ) and for any s.
Proof.For (a), in view of Parseval's equality, we get that (3.7) Using the inequality e −y ≤ C k y −k , we find that exp where we use It follows from (3.7) that . By a similar explanation, we also get that for 0 , where we use the fact that exp Hence, (a) is proved.For (b), in view of Parseval's equality, we get that which allows us to conclude the proof of (3.5).The proof of (3.6) is similar to (3.5), and we omit it here.For (c), noting that (1 + mλ n ) −1 < 1, we can claim it as follows The proof of Lemma 3.1 is completed.□

Well-posedness of Problem (1.3)
for any w 1 , w 2 ∈ H r (Ω) and L g is a postive constant.
3) has a unique solution u m,α,β ∈ L p (0, T ; H r (Ω)), where In addition, we get where C k depends on k.
(ii) Let us assume that b < α 2 and u 0 ∈ Here the hidden constant depends on k, b, α, m, p, β, L g (L g is defined in (4.1)).
Proof.Let us define B : X b,µ,α ((0, T ]; H r (Ω)) → X b,µ,α ((0, T ]; H r (Ω)), µ > 0 by Let the zero function w 0 (t) = 0. From the fact G(0) = 0, we know that In view of (3.4) as in Lemma 3.3, we obtain the following estimate Hence, multiplying both sides of the above expression by t b e −µt α , we have that where we use b ≥ αk.This implies that Bw 0 ∈ X b,µ,α ((0, T ]; H r (Ω)).Let any two functions w 1 , w 2 ∈ X b,µ,α ((0, T ]; H r (Ω)).From (4.4) and (3.3), we obtain that Since the constraint s ≥ r + k − βk, we know that Sobolev embedding From some above observations and noting (4.1), we get that From the fact that , we follows from (4.7) that sup 0≤t≤T where Let us continue to treat the integral term as follows In order to control the above integral, we need to change the variable ν = tξ 1 α .Then we get the following statement Next, we provide the following lemma which can be found in [22], Lemma 8, page 9.
For h > 0, the following limit holds , we easily to verify that the following conditions hold (4.9)By Lemma 4.1 and (4.9), we have lim This statement shows that there exists a µ 0 such that Combining (4.8) and (4.10), we obtain for any w 1 , w 2 ∈ X b,µ,α ((0, T ]; H r (Ω)).This statement tells us that B is the mapping from X b,µ,α ((0, T ]; H r (Ω)) to itself.By applying Banach fixed point theorem, we deduce that B has a fixed point u m,α,β ∈ X b,µ 0 ,α ((0, T ]; H r (Ω)).Hence, we can see that Let us show the regularity property of the mild solution u m,α,β .Indeed, using the triangle inequality and (4.11) and noting that B(v = 0) = S m,α,β (t)u 0 , we obtain , which combined with (4.5), we get which allows us to get that where C 1 depends on k, µ 0 , b, α, m and and we remind that C k depends on k.Note that the improper integral and the following regularity holds where C depends on k, µ 0 , b, α, m, p.Our next aim is to claim the derivative of the mild solution u m,α,β .
Applying the following formula we obtain the following equality where the operator Q m,α,β (t) is defined by for any v ∈ L 2 (Ω).In view of Parseval's equality, we get that Using (3.8) and noting that which implies that for any v ∈ H s+k−βk+β−1 (Ω).In a similar technique as above, we also get that for any 0 ≤ ν ≤ t.Let us go back to the right hand side of (4.14).By (4.15), we evaluate the first term on the right hand side of (4.14) as follows Using global Lipschitz of G as in (4.1) and the fact that G(0) = 0, the second term on the right hand side of (4.14) is estimated as follows where we use (4.13).Let us now to treat the third integral term in (4.14).By (4.16), we obtain that Since β ≤ 1 and 0 < k < 1, we can easily verify that s + k − βk + β − 1 ≤ s, which implies the Sobolev embedding H s (Ω) → H s+k−βk+β−1 (Ω) is true.From these above observations and using (4.13), we derive that ) Let us now treat the integral term on the right hand side of (4.20).Controlling it is really not that simple task.By applying Hölder inequality, we find that where α > 2b.By changing to a new variable z = ν α , we derive that dz = αν α−1 dν.Hence, we infer that where we note that 0 < k < 1 2 .Combining (4.21) and (4.22), we obtain that the following inequality Summarizing the above results (4.14), (4.17), (4.18), (4.24) and using the triangle inequality, we obtain the following assertion which shows (4.3).The proof is completed.□

The convergence of the mild solution when m → 0 +
The main purpose of this section is to investigate the convergence of mild solutions to Problem (1.3) when m → 0 + .Our result gives us an interesting connection between the solution of the Sobolev equation and the parabolic equation.+m k tends to zero when m → 0 + .Hence, we can deduce that u m,α,β (t)− → 0 when m → 0 + .
Proof.In the case m = 0, thanks for the results on [11], the mild solution to Problem (1.3) is given by the following operator equation where we provide two operators that have the following Fourier series representation as follows It's worth emphasizing that the existence of the solution to Equation (5.1) has been demonstrated in [11].Subtracting (5.1) from (4.12), we get the following equality by some simple calculations Next, we estimate the four terms on the right hand of (5.2) in H r (Ω) space.We divide this process into four steps as below.
Step  Here γ and l are two positive constants that are later chosen.For 1 < ε < 2 and any z > 0, we easily verify that (1 + z) (5.4) for any γ > l > 0.
Proof.For m > 0, let u m,α,β and u * * m,α be the mild solutions to Problem (1.3) with 0 < β < 1 and β = 1 respectively.Let us recall the formula of these two solutions and Subtracting (6.1) from (6.2) on each side, we derive that Step 1. Estimate of N 1 .By Parseval' s equality, we have that the following equality (6.4) For the term N 1,1 , since λ n > 1 and 0 < β < 1, we note that λ β n 1+mλ n < λ n 1+mλ n .Hence, we have the following inequality exp In view of the above inequality Since the fact that λ n > 1, we know that . Using the inequality 1 − e −y ≤ C(µ)y µ for any µ > 0, we find that where we note that 0 < log(y) ≤ y for any y > 1, which implies that Combining (6.5) and (6.6), we derive that exp If we make the assumption ε ′ ≤ γ ′ , then (1 + mλ n ) ε ′ −γ ′ ≤ 1, which allows us to obtain that
Let us choose µ = ε k and γ ′ = ε ′ = k for any ε > 0. Then we get the following estimate Before mention to N 1,2 , we provide a set N = n ∈ N : λ n ≤ 1 .Let us give the observation that if N is an empty set, then N 1,2 = 0.If N is a non-empty set, then λ 1 ≤ 1.For the term N 1,2 , we note that λ β n 1+mλ n > λ n 1+mλ n since λ n ≤ 1 and 0 < β < 1.Hence, we have that exp By using the fact that e Since λ 1 ≤ λ n ≤ 1, it is obvious to see that Combining (6.8) and (6.9), we find that exp Hence, it follows from (6.10) that Let Hence, in view of Parseval's equality, we get the following estimate Combining (6.4), (6.7) and (6.11), we obtain the following estimate By taking the square root of both sides of the above expression and using the inequality Here we denote where we observe that D Step 2. Estimate of N 2 .We confirm the following result for any ε 0 > 0 using the method similar to Step 1, . Since s > r + k − βk, we know that ε 0 = s − r − k + βk > 0, and using global Lipschitz property of G, we derive that where C 1 = C(k, r, s, β, α), and we have the following observation In view of (4.2), we obtain that the following upper bound where C 2 = C(k, α, m, µ 0 , T, b).From two latter estimations as above, we infer that .
From the two above observations, we confirm the following statement H r (Ω)) dν, (6.16)where C 3 = C k α k m −k K g .Combining (6.3), (6.12), (6.15) and (6.16), we deduce the following estimate dν, (6.17)where C = C(k, r, ε, α).By the Hölder inequality and in combination with (4.22), we get the estimate of the third term on the right hand side of (6.17 Combining (6.17), (6.18) and the inequality (a + b + c)   In view of Lemma 3.1 as in [30], we obtain the following upper bound where C α is a positive constant that depends on α.By (6.21) and ( 6 ≤ AC α .
The proof is completed.□