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Effects of urban green areas on air temperature in a medium-sized Argentinian city

  • Urban climate is the result of both atmospheric and geographic factors affecting a region, as well as the morphology, structures and human activities in a city. Urban vegetation in particular affects this climate at a local scale and provides many other social, economic and ecological benefits. Thus, it is important to explore the effects of different green areas used for urban and periurban agriculture and forestry activities (UPAF) on daily atmospheric temperature and the required degrees of cooling or refrigerating temperature. Comfort temperatures were defined using a range 18-24°C and analyzed using actual measured as well as forecasted temperatures using a future scenario. Actual temperatures were recorded from September 2013 to August 2014 using digital sensors across eight sites in Rosario, Argentina: three in the central core with no vegetation, one in the central core with street trees, one in an urban agriculture site, one in a public park and two in periurban agricultural areas. Results show that air temperature in the central core with no vegetation were higher than those in other sites with vegetation during day and night, with the exception of the temperatures measured at the central core site with street trees. Findings also show that temperature effects in urban agricultural gardens of approximately 0.2 ha were similar to those of gardens and public parks 2-3 ha in size. Three UPAF types were classified according to cooling degree days, which decreased in order from (1) central core with no trees; (2) central core with street trees and public parks; and (3) urban and periurban agriculture areas. Conversely, the opposite trends for heating degree days were found. Results from this study can be used for integrating UPAF measures into climate change mitigation and urban planning policies in medium-sized cities in the developing world.

    Citation: Alejandra S. Coronel, Susana R. Feldman, Emliano Jozami, Kehoe Facundo, Rubén D. Piacentini, Marielle Dubbeling, Francisco J. Escobedo. Effects of urban green areas on air temperature in a medium-sized Argentinian city[J]. AIMS Environmental Science, 2015, 2(3): 803-826. doi: 10.3934/environsci.2015.3.803

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  • Urban climate is the result of both atmospheric and geographic factors affecting a region, as well as the morphology, structures and human activities in a city. Urban vegetation in particular affects this climate at a local scale and provides many other social, economic and ecological benefits. Thus, it is important to explore the effects of different green areas used for urban and periurban agriculture and forestry activities (UPAF) on daily atmospheric temperature and the required degrees of cooling or refrigerating temperature. Comfort temperatures were defined using a range 18-24°C and analyzed using actual measured as well as forecasted temperatures using a future scenario. Actual temperatures were recorded from September 2013 to August 2014 using digital sensors across eight sites in Rosario, Argentina: three in the central core with no vegetation, one in the central core with street trees, one in an urban agriculture site, one in a public park and two in periurban agricultural areas. Results show that air temperature in the central core with no vegetation were higher than those in other sites with vegetation during day and night, with the exception of the temperatures measured at the central core site with street trees. Findings also show that temperature effects in urban agricultural gardens of approximately 0.2 ha were similar to those of gardens and public parks 2-3 ha in size. Three UPAF types were classified according to cooling degree days, which decreased in order from (1) central core with no trees; (2) central core with street trees and public parks; and (3) urban and periurban agriculture areas. Conversely, the opposite trends for heating degree days were found. Results from this study can be used for integrating UPAF measures into climate change mitigation and urban planning policies in medium-sized cities in the developing world.


    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 $ \beta $ at $ s $ if and only if it is (classically) differentiable at $ s $, and

    $ Cβf(s)sβ=s1βf(s)s, $ (1.1)

    where $ 0 < \beta \leq 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 $ \mathbb{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

    $  Cαtαu(x,t)=Dα2x2u(x,t), $ (1.2)

    where $ 0 < \alpha \le 1, \, x > 0, \, t > 0, $ $ u(x, t) $ is the concentration, and $ D_\alpha $ 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 $ \alpha = 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 $ \Omega \subset\mathbb{R}^N $ ($ N \ge 1 $) be a bounded domain with smooth boundary $ \partial \Omega $, and $ T > 0 $ is a given positive number. In this paper, we investigate the Sobolev equation with a conformable derivative as follows

    $ {Cαtαu+(Δ)βu(x,t)mCαtαΔu=F(u(x,t)),xΩ,t(0,T),u(x,t)=0,xΩ,t(0,T),u(x,0)=u0(x)xΩ $ (1.3)

    where $ \alpha, \beta \in (0, 1] $, $ m > 0 $, the time fractional derivative $ \frac{{ }^{ C}\partial^\alpha}{\partial t^\alpha} $ is the conformable derivative of order $ \alpha, $ defined in Definition 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 $ (-\Delta)^\beta $ 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).

    $ \bullet $ 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

    $ {Cαtαu+(Δ)βu(x,t)=F(u(x,t)),xΩ,t(0,T),u(x,t)=0,xΩ,t(0,T),u(x,0)=u0(x)xΩ. $ (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).

    $ \bullet $ If $ \alpha = 1 $, the main equation of Problem (1.3) becomes the classical equation

    $ ut+(Δ)βu(x,t)mΔut=F(u(x,t)). $ (1.5)

    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

    $ {Dαt(u+kAu)+Aβu=F(t,x,u),in(0,T]×Ω,u(t,x)=0,on(0,T]×Ω,u(0,x)=u0(x),inΩ, $ (1.6)

    where $ 0 < \alpha < 1 $, $ \mathcal A = -\Delta $, $ D^\alpha $ is Caputo fractional derivative operator of order $ \alpha $. 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 \to 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 $ \beta \to 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^{\alpha-1} $ or $ (t^\alpha- s^\alpha)^{- 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 \to 0^+ $. In Section 6, we investigate the convergence of mild solutions when $ \beta \to 1^- $.

    Definition 2.1. Conformable derivative model: Let $ B $ be a Banach space, and the function $ f:[0, \infty) \to B $. Let $ \frac{{ }^{C}\partial^\beta}{\partial t^\beta} $ be the conformable derivative operator of order $ \beta \in (0, 1] $, locally defined by

    $ Cβf(t)tβ:=limh0f(t+ht1β)f(t)hinB $

    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

    $ {Δen(x)=λnen(x),xΩ,en(x)=0,xΩ, $

    admits the eigenvalues $ 0 < \lambda_1 \leq \lambda_2 \leq \dots \leq \lambda_n \leq \dots $ with $ \lambda_n \to \infty $ as $ n \to \infty $. The corresponding eigenfunctions are $ e_n \in H_0^1(\Omega) $.

    Definition 2.2. (Hilbert scale space). We recall the Hilbert scale space, which is given as follows

    $ Hs(Ω)={fL2(Ω)|n=1λ2sn(Ωf(x)en(x)dx)2<} $

    for any $ s \geq 0 $. It is well-known that $ \mathbb H^s (\Omega) $ is a Hilbert space corresponding to the norm

    $ \|f\|_{ \mathbb H^s (\Omega)} = \Bigg( \sum\limits_{n = 1}^\infty \lambda_n^{2s} \Big( \int_\Omega f(x) e_n (x){\rm d}x \Big)^2 \Bigg)^{1/2},\; \; f \in \mathbb H^s (\Omega). $

    Definition 2.3. Let $ {\bf X}_{a, q, \alpha} ((0, T]; B) $ denote the weighted space of all the functions $ \psi \in C ((0, T];X) $ such that

    $ ψXa,q,α((0,T];B):=supt(0,T]taeqtαψ(t,)B<, $

    where $ a, q > 0 $ and $ 0 < \alpha \le 1 $ (see [22]). If $ q = 0 $, we denote $ {\bf X}_{a, q} ((0, T]; B) $ by $ {\mathbb X}_{a} ((0, T]; B) $.

    In order to find a precise formulation for solutions, we consider the mild solution in terms of the Fourier series

    $ u(x,t) = \sum\limits_{n = 1}^\infty \langle u(.,t), e_n \rangle e_n (x). $

    Taking the inner product of Problem (1.3) with $ e_n $ gives

    $ {Cαtαu(.,t),en+λβnu(.,t),en+mλnCαtαu(.,t),en=F(.,t),en,t(0,T),u(.,0),en=u0,en, $ (3.1)

    where we repeat that

    $ \left \langle \Delta u(.,t) , e_n \right \rangle = - \lambda_n \langle u(.,t), e_n \rangle. $

    The first equation of (3.1) is a differential equation with a conformable derivative as follows

    $ Cαtαu(.,t),en+λβn1+mλnu(.,t),en=11+mλnF(.,t),en. $

    In view of the result in Theorem 5, [26] and Theorem 3.3, [28] the solution to Problem (1.3) is

    $ u(.,t),en=exp(λβn1+mλntαα)u0,en+11+mλnt0να1exp(λβn1+mλnναtαα)F(.,ν),endν. $

    To simplify the solution formula, we will express the solution in operator equations. Let us set the following operators

    $ Sm,α,β(t)f=nNexp(λβn1+mλntαα)f,enen, $

    and

    $ Pmf=nN(1+mλn)1f,enen $

    for any $ f \in L^2(\Omega) $ in the form $ f = \sum_{n\in \mathbb N} \big\langle f, e_n \big\rangle e_n $. Then the inverse operator of $ {\bf S}_{m, \alpha, \beta}(t) $ is defined by

    $ (Sm,α,β(t))1f=nNexp(λβn1+mλntαα)f,enen $

    The mild solution is given by

    $ u(t)=Sm,α,β(t)u0+t0να1PmSm,α,β(t)(Sm,α,β(ν))1F(ν)dν. $ (3.2)

    For a qualitative analysis of the solution to (3.2), we need the bounded result for the operators in Hilbert scales space.

    Lemma 3.1. (a) Let $ v \in \mathbb H^{s+k-\beta k} (\Omega) $ for any $ k > 0 $. Then we get

    $ Sm,α,β(t)vHs(Ω)CkαkmktαkvHs+kβk(Ω), $ (3.3)

    and for $ 0 \le \nu \le t \le T $,

    $ Sm,α,β(t)(Sm,α,β(ν))1vHs(Ω)Ckαkmk(tανα)kvHs+kβk(Ω). $ (3.4)

    (b) If $ v \in \mathbb H^{s} (\Omega) $, then

    $ Sm,α,β(t)vHs(Ω)vHs(Ω) $ (3.5)

    and

    $ Sm,α,β(t)(Sm,α,β(ν))1vHs(Ω)vHs(Ω) $ (3.6)

    for $ 0 \le \nu \le t \le T $.

    (c) If $ v \in \mathbb H^{s} (\Omega) $, then we have

    $ \Big\| {\bf P}_m v \Big\|_{\mathbb H^s (\Omega)} \le\big\| v \big\|_{\mathbb H^{s} (\Omega)} $

    for any $ s $.

    Proof. For (a), in view of Parseval's equality, we get that

    $ Sm,α,β(t)v2Hs(Ω)=nNλ2snexp(2λβn1+mλntαα)v,en2. $ (3.7)

    Using the inequality $ e^{-y} \le C_k y^{-k} $, we find that

    $ exp(λβn1+mλntαα)Ck(λβn1+mλn)ktαkαk=Ckαk(mk+λ11)λ(1β)kntαk, $ (3.8)

    where we use

    $ (1+mλn)kCk(1+mkλkn)Ck(mk+λ11)λkn. $

    It follows from (3.7) that

    $ Sm,α,β(t)v2Hs(Ω)(Ckαk)2(mk+λ11)2t2αknNλ2s+2k2βknf,en2=(Ckαk)2(mk+λ11)2t2αkv2Hs+kβk(Ω). $

    By a similar explanation, we also get that for $ 0 \le \nu \le t \le T $,

    $ Sm,α,β(t)(Sm,α,β(ν))1v2Hs(Ω)=nNλ2snexp(λβn1+mλn2να2tαα)v,en2(Ckαk)2(mk+λ11)2(tανα)2knNλ2s+2k2βknf,en2=(Ckαk)2(mk+λ11)2(tανα)2kv2Hs+kβk(Ω), $

    where we use the fact that

    $ exp(λβn1+mλnναtαα)Ck(λβn1+mλn)k(tανα)kαk=Ckαk(mk+λ11)λ(1β)kn(tανα)k. $

    Hence, (a) is proved.

    For (b), in view of Parseval's equality, we get that

    $ Sα,β(t)v2Hs(Ω)=nNλ2snexp(2λβn1+mλntαα)v,en2nNλ2snv,en2=v2Hs(Ω), $

    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 $ \left(1+ m \lambda_n \right)^{-1} < 1 $, we can claim it as follows

    $ Pmv2Hs(Ω)=nNλ2sn(1+mλn)2v,en2nNλ2snv,en2=v2Hs(Ω). $

    The proof of Lemma 3.1 is completed.

    Let $ G: \mathbb H^r (\Omega) \to \mathbb H^s (\Omega) $ such that $ G({\bf 0}) = {\bf 0} $ and

    $ G(w1)G(w2)Hs(Ω)Lgw1w2Hr(Ω), $ (4.1)

    for any $ w_1, w_2\in \mathbb H^r (\Omega) $ and $ L_g $ is a postive constant.

    Theorem 4.1. (i) Let $ G: \mathbb H^r (\Omega) \to \mathbb H^s (\Omega) $ such that (4.1) holds. Here $ r, s $ satisfy that $ s \ge r+k-\beta k $ for $ 0 < k < \frac{1}{2}. $ Let the initial datum $ u_0 \in \mathbb H^{r+k-\beta k} (\Omega) $. Then Problem (1.3) has a unique solution $ u_{m, \alpha, \beta } \in L^p(0, T; \mathbb H^r(\Omega)) $, where

    $ 1<p<1b,αkb<ααk. $

    In addition, we get

    $ um,α,β(t)Hr(Ω)2Ckαk(mk+λ11)eμ0TαTbαktbu0Hr+kβk(Ω), $ (4.2)

    where $ C_k $ depends on $ k $.

    (ii) Let us assume that $ b < \frac{ \alpha}{2} $ and $ u_0 \in \mathbb H^{s+k-\beta k+\beta-1} (\Omega) \cap \mathbb H^{r+k-\beta k} (\Omega) $. Then we have

    $ tum,α,β(.,t)Hs(Ω)tααk1u0Hs+kβk+β1(Ω)+(tα1b+t2αbkα1)u0Hr+kβk(Ω). $ (4.3)

    Here the hidden constant depends on $ k, b, \alpha, m, p, \beta, L_g $ ($ L_g $ is defined in (4.1)).

    Proof. Let us define $ { B}: {\bf X}_{b, \mu, \alpha }((0, T]; \mathbb H^r(\Omega)) \to {\bf X}_{b, \mu, \alpha}((0, T]; \mathbb H^r(\Omega)) $, $ \mu > 0 $ by

    $ Bw(t):=Sm,α,β(t)u0+t0να1PmSm,α,β(t)(Sm,α,β(ν))1G(w(ν))dν. $ (4.4)

    Let the zero function $ w_0(t) = 0 $. From the fact $ G(0) = 0 $, we know that

    $ { B} w_0 (t) = {\bf S}_{m, \alpha, \beta}(t) u_0 . $

    In view of (3.4) as in Lemma 3.3, we obtain the following estimate

    $ Bw0(t)Hr(Ω))=Sm,α,β(t)u0Hr(Ω))Ckαk(mk+λ11)tαku0Hr+kβk(Ω). $

    Hence, multiplying both sides of the above expression by $ t^b e^{-\mu t^ \alpha} $, we have that

    $ tbeμtαBw0(t)Hr(Ω))Ckαk(mk+λ11)tbαku0Hr+kβk(Ω)Ckαk(mk+λ11)Tbαku0Hr+kβk(Ω), $ (4.5)

    where we use $ b \ge \alpha k $. This implies that $ { B} w_0 \in {\bf X}_{b, \mu, \alpha }((0, T]; \mathbb H^r(\Omega)) $. Let any two functions $ w_1, w_2 \in {\bf X}_{b, \mu, \alpha }((0, T]; \mathbb H^r(\Omega)) $. From (4.4) and (3.3), we obtain that

    $ Bw1(t)Bw2(t)Hr(Ω))=t0να1PmSm,α,β(t)(Sm,α,β(ν))1(G(w1(ν))G(w2(ν)))dνHr(Ω))Ckαkmkt0να1(tανα)kG(w1(ν))G(w2(ν))Hr+kβk(Ω))dν. $ (4.6)

    Since the constraint $ s \ge r+k-\beta k $, we know that Sobolev embedding

    $ Hs(Ω)Hr+kβk(Ω). $

    From some above observations and noting (4.1), we get that

    $ tbeμtαBw1(t)Bw2(t)Hr(Ω))Ckαk(mk+λ11)tbeμtαt0να1(tανα)kG(w1(ν))G(w2(ν))Hs(Ω))dνCkLgαk(mk+λ11)tbeμtαt0να1(tανα)kw1(ν)w2(ν)Hr(Ω))dν=CkLgαk(mk+λ11)tbt0να1b(tανα)keμ(tανα)νbeμναw1(ν)w2(ν)Hr(Ω))dν. $ (4.7)

    From the fact that

    $ \Big\| w_1- w_2 \Big\|_{ {\bf X}_{b, \mu, \alpha }((0,T]; \mathbb H^r(\Omega)) } = \sup\limits_{0 \le \nu \le T} \nu^b e^{-\mu \nu^ \alpha } \Big\| w_1(\nu)- w_2(\nu) \Big\|_{\mathbb H^{r}(\Omega)) }, $

    we follows from (4.7) that

    $ sup0tTtbeμtαBw1(t)Bw2(t)Hr(Ω))¯Cw1w2Xb,μ,α((0,T];Hr(Ω))sup0tT[tbt0να1b(tανα)keμ(tανα)dν], $ (4.8)

    where $ \overline C = C_k L_g \alpha^k m^{-k}. $ Let us continue to treat the integral term as follows

    $ J_{1, \mu} (t) = t^b \int_0^t \nu^{ \alpha-1-b} \left( t^ \alpha-\nu^ \alpha \right)^{-k} e^{-\mu(t^ \alpha-\nu^ \alpha)} {\rm d}\nu. $

    In order to control the above integral, we need to change the variable $ \nu = t \xi^{\frac{1}{ \alpha}} $. Then we get the following statement

    $ J1,μ(t)=1αtααk10ξbα(1ξ)keμtα(1ξ)dξ. $

    Next, we provide the following lemma which can be found in [22], Lemma 8, page 9.

    Lemma 4.1. Let $ a_1 > -1 $, $ a_2 > -1 $ such that $ a_1+a_2 \ge -1 $, $ \rho > 0 $ and $ t\in [0, T] $. For $ h > 0 $, the following limit holds

    $ limρ(supt[0,T]th10νa1(1ν)a2eρt(1ν)dν)=0. $

    Since $ 0 < b < \alpha $ and $ 0 < k < \min (\frac{b}{ \alpha}, \; \; 1-\frac{b}{ \alpha}) $, we easily to verify that the following conditions hold

    $ {ααk>0,bα>1, k>1,bαk1. $ (4.9)

    By Lemma 4.1 and (4.9), we have

    $ \lim\limits_{\mu \to +\infty } \sup\limits_{0\le t \le T} J_{1, \mu} (t) = 0. $

    This statement shows that there exists a $ \mu_0 $ such that

    $ ¯Csup0tTJ1,μ0(t)12. $ (4.10)

    Combining (4.8) and (4.10), we obtain

    $ Bw1Bw2Xb,μ0((0,T];Hr(Ω))12w1w2Xb,μ0((0,T];Hr(Ω)) $ (4.11)

    for any $ w_1, w_2 \in {\bf X}_{b, \mu, \alpha }((0, T]; \mathbb H^r(\Omega)) $. This statement tells us that $ B $ is the mapping from $ {\bf X}_{b, \mu, \alpha }((0, T]; \mathbb H^r(\Omega)) $ to itself. By applying Banach fixed point theorem, we deduce that $ B $ has a fixed point $ u_{m, \alpha, \beta} \in {\bf X}_{b, \mu_0, \alpha}((0, T]; \mathbb H^r(\Omega)) $. Hence, we can see that

    $ um,α,β(t)=Sm,α,β(t)u0+t0να1PmSm,α,β(t)(Sm,α,β(ν))1G(um,α,β(ν))dν. $ (4.12)

    Let us show the regularity property of the mild solution $ u_{m, \alpha, \beta} $. Indeed, using the triangle inequality and (4.11) and noting that $ B (v = 0) = {\bf S}_{m, \alpha, \beta}(t) u_0 $, we obtain

    $ um,α,βXb,μ0,α((0,T];Hr(Ω))=Bum,α,βXb,μ0,α((0,T];Hr(Ω))12um,α,βXb,μ0((0,T];Hr(Ω))+B(v=0)Xb,μ0,α((0,T];Hr(Ω))=12um,α,βXb,μ0,α((0,T];Hr(Ω))+Sm,α,β(t)u0Xb,μ0,α((0,T];Hr(Ω)), $

    which combined with (4.5), we get

    $ um,α,βXb,μ0,α((0,T];Hr(Ω))2Ckαk(mk+λ11)Tbαku0Hr+kβk(Ω), $

    which allows us to get that

    $ um,α,β(t)Hr(Ω)˜C1tbu0Hr+kβk(Ω), $ (4.13)

    where $ \widetilde C_1 $ depends on $ k, \mu_0, b, \alpha, m $ and

    $ \widetilde C_1 = 2 C_k \alpha^k \left(m^k+ \lambda_1^{-1} \right) e^{\mu_0 T^ \alpha} T^{b- \alpha k}, $

    and we remind that $ C_k $ depends on $ k $. Note that the improper integral $ \int_0^T t^{-pb} {\rm d}t $ is convergent for $ 1 < p < \frac{1}{b} $, we deduce that

    $ u_{m, \alpha, \beta} \in L^p(0,T; \mathbb H^r(\Omega) ), $

    and the following regularity holds

    $ um,α,βLp(0,T;Hr(Ω))˜Cu0Hr+kβk(Ω), $

    where $ \widetilde C $ depends on $ k, \mu_0, b, \alpha, m, p $. Our next aim is to claim the derivative of the mild solution $ u_{m, \alpha, \beta} $. Applying the following formula

    $ {\rm d} \left( \int_0^t \mathcal K(t,s) {\rm d}s \right) = \int_0^t \partial_t \mathcal K (t,s) {\rm d}s+ \mathcal K(t,t) {\rm d}t, $

    we obtain the following equality

    $ tum,α,β(.,t)=tα1Qm,α,β(t)u0+tα1G(um,α,β(x,t))+tα1t0να1PmQm,α,β(t)(Sm,α,β)1(ν)G(um,α,β(ν))dν, $ (4.14)

    where the operator $ {\bf Q}_{m, \alpha, \beta}(t) $ is defined by

    $ Qm,α,β(t)v=nNλβn1+mλnexp(λβn1+mλntαα)v,enen $

    for any $ v \in L^2(\Omega) $. In view of Parseval's equality, we get that

    $ Qm,α,β(t)v2Hs(Ω)=nNλ2sn(λβn1+mλn)2exp(2λβn1+mλntαα)v,en2. $

    Using (3.8) and noting that $ \frac{ \lambda_n^\beta }{1+ m \lambda_n} \le m^{-1} \lambda_n^{\beta-1} $, we get that

    $ Qm,α,β(t)v2Hs(Ω)(Ckαkmk1)2t2αknNλ2s+2k2βk+2β2nv,en2, $

    which implies that

    $ Qm,α,β(t)vHs(Ω)Ckαkmk1tαkvHs+kβk+β1(Ω) $ (4.15)

    for any $ v \in \mathbb H^{s+k-\beta k+\beta-1} (\Omega) $. In a similar technique as above, we also get that

    $ Qm,α,β(t)(Sm,α,β)1(ν)vHs(Ω)Ckαkmk1(tανα)kvHs+kβk+β1(Ω), $ (4.16)

    for any $ 0 \le \nu \le 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

    $ tα1Qm,α,β(t)u0Hs(Ω)Ckαkmk1tααk1u0Hs+kβk+β1(Ω). $ (4.17)

    Using global Lipschitz of $ G $ as in (4.1) and the fact that $ G({\bf 0}) = {\bf 0} $, the second term on the right hand side of (4.14) is estimated as follows

    $ tα1G(um,α,β(x,t))Hs(Ω)˜C1tα1Lgum,α,βHr(Ω)˜C1Lgtα1bu0Hr+kβk(Ω), $ (4.18)

    where we use (4.13). Let us now to treat the third integral term in (4.14). By (4.16), we obtain that

    $ t0να1PmQm,α,β(t)(Sm,α,β)1(ν)G(um,α,β(ν))dνHs(Ω)Ckαkmk1t0να1(tανα)kG(um,α,β(ν))Hs+kβk+β1(Ω)dν. $ (4.19)

    Since $ \beta \le 1 $ and $ 0 < k < 1 $, we can easily verify that $ s+k-\beta k+\beta-1 \le s $, which implies the Sobolev embedding $ \mathbb H^{s}(\Omega) \hookrightarrow \mathbb H^{s+k-\beta k+\beta-1 }(\Omega) $ is true. From these above observations and using (4.13), we derive that

    $ t0να1(tανα)kG(um,α,β(ν))Hs+kβk+β1(Ω)dνC(s,k,β)t0να1(tανα)kG(um,α,β(ν))Hs(Ω)dνLgC(s,k,β)t0να1(tανα)kum,α,β(ν)Hr(Ω)dνLgC(s,k,β)˜C1u0Hr+kβk(Ω)(t0να1b(tανα)kdν). $ (4.20)

    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

    $ (t0να1b(tανα)kdν)2=(t0να12bνα12(tανα)kdν)2(t0να12bdν)(t0να1(tανα)2kdν)=tα2bα2b(t0να1(tανα)2kdν), $ (4.21)

    where $ \alpha > 2b $. By changing to a new variable $ z = \nu^ \alpha $, we derive that $ {\rm d}z = \alpha \nu^{ \alpha-1} {\rm d}\nu $. Hence, we infer that

    $ t0να1(tανα)2kdν=1αtα0(tαz)2kdz=tα(12k)α(12k), $ (4.22)

    where we note that $ 0 < k < \frac{1}{2} $. Combining (4.21) and (4.22), we obtain that the following inequality

    $ t0να1b(tανα)kdνtαbkαα(12k)(α2b). $ (4.23)

    By (4.23) and following from (4.19) and (4.20), we have

    $ t0να1Qm,α,β(t)(Sm,α,β)1(ν)G(um,α,β(ν))dνHs(Ω)Ckαkmk1t0να1(tανα)kG(um,α,β(ν))Hs+kβk+β1(Ω)dνCkαkmk1LgC(s,k,β)˜C1α(12k)(α2b)tαbkαu0Hr+kβk(Ω). $ (4.24)

    Summarizing the above results (4.14), (4.17), (4.18), (4.24) and using the triangle inequality, we obtain the following assertion

    $ tum,α,β(.,t)Hs(Ω)tα1Qm,α,β(t)u0Hs(Ω)+tα1G(um,α,β(x,t))Hs(Ω)+tα1t0να1PmQm,α,β(t)(Sm,α,β)1(ν)G(um,α,β(ν))dνHs(Ω)Ckαkmk1tααk1u0Hs+kβk+β1(Ω)+˜C1Lgtα1bu0Hr+kβk(Ω)+Ckαkmk1LgC(s,k,β)˜C1α(12k)(α2b)t2αbkα1u0Hr+kβk(Ω) $

    which shows (4.3). The proof is completed.

    The main purpose of this section is to investigate the convergence of mild solutions to Problem (1.3) when $ m \to 0^+ $. Our result gives us an interesting connection between the solution of the Sobolev equation and the parabolic equation.

    Theorem 5.1. Let $ G: \mathbb H^r (\Omega) \to \mathbb H^s (\Omega) $ such that $ G({\bf 0}) = {\bf 0} $ and (4.1) holds. Here $ r, s $ satisfy that $ s \ge r+k-\beta k $ for $ 0 < k < \frac{1}{2}. $ Let the initial datum $ u_0 \in \mathbb H^{r+k-\beta k} (\Omega) $. Let $ u_{m, \alpha, \beta} $ and $ u^*_{ \alpha, \beta} $ be the mild solutions to Problem (1.3) with $ m > 0 $ and $ m = 0 $ respectively. Then we get the following estimate

    $ um,α,β(t)uα,β(t)Hr(Ω)[m2γεγ+εl2+mk]T(α,γ,l,b,k)E1,α(Lgt), $

    where $ \alpha k \le b < \frac{ \alpha}{2} $, $ \gamma = \frac{k(1-\beta)+l\left(\beta-\frac{\varepsilon}{2} \right) }{\beta+1-\frac{\varepsilon}{2}} $, $ 0 < l < k (1-\beta) $ and $ 1 < \varepsilon < \min \Big(2, \frac{2k(1-\beta)+2l \beta}{l} \Big) $.

    Remark 5.1. Note that $ m^{ \frac{2\gamma- \varepsilon \gamma+\varepsilon l }{2} } + m^{k} $ tends to zero when $ m \to 0^+ $. Hence, we can deduce that $ \Big\| u_{m, \alpha, \beta}(t)- u^*_{ \alpha, \beta}(t) \Big\|_{\mathbb H^{r} (\Omega)} \to 0 $ when $ m \to 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

    $ uα,β(t)=Sα,β(t)u0+t0να1Sα,β(t)(Sα,β(ν))1G(uα,β(ν))dν, $ (5.1)

    where we provide two operators that have the following Fourier series representation as follows

    $ Sα,β(t)f=nNexp(λβntαα)f,enen $

    and

    $ (Sα,β(ν))1f=nNexp(λβnναα)f,enen. $

    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

    $ um,α,β(t)uα,β(t)=(Sm,α,β(t)Sα,β(t))u0+t0να1Sα,β(t)(Sα,β(ν))1(G(um,α,β(ν))G(uα,β(ν)))dν+t0να1[PmSm,α,β(t)(Sm,α,β(ν))1PmSα,β(t)(Sα,β(ν))1]G(um,α,β(ν))dν+t0να1(PmI)Sα,β(t)(Sα,β(ν))1G(um,α,β(ν))dν=M1+M2+M3+M4. $ (5.2)

    Next, we estimate the four terms on the right hand of (5.2) in $ \mathbb H^r(\Omega) $ space. We divide this process into four steps as below.

    Step 1. Estimate of $ M_1 $. In view of the inequality $ \Big| e^{-c}- e^{-d} \Big| \le C_\gamma \max (e^{-c}, e^{-d}) |c-d|^\gamma \ \text{with}\ \gamma > 0, $ let $ c = \frac{ \lambda_n^\beta }{1+ m \lambda_n} \frac{ t^ \alpha}{ \alpha} $ and $ d = \lambda_n^\beta \frac{ t^ \alpha}{ \alpha} $, we get the following inequality

    $ |exp(λβn1+mλntαα)exp(λβntαα)|Cγexp(λβn1+mλntαα)(λβnλβn1+mλn)γtαγαγCγCl(λβn1+mλn)l(tαα)l(λβnλβn1+mλn)γtαγαγCγClαlγλβln(mλ1+βn)γ(11+mλn)γltαγαl $ (5.3)

    by $ e^{-c} > e^{-d} $ and (3.8).

    Here $ \gamma $ and $ l $ are two positive constants that are later chosen. For $ 1 < \varepsilon < 2 $ and any $ z > 0 $, we easily verify that $ (1+z)^{\frac{2}{\varepsilon}} \ge 1+ z > z, $ which implies

    $ (1+z)γl=(1+z)2εε(γl)2>zε(γl)2 $ (5.4)

    for any $ \gamma > l > 0 $.

    In (5.4), by choosing $ z = 1+ m \lambda_n $ and after some simple calculation, we have $ \left(1+ m \lambda_n\right)^{\gamma-l} > (m \lambda_n)^{\frac{\varepsilon (\gamma-l)}{2} }, $ which leads to the following inequality

    $ (11+mλn)γl(mλn)ε(lγ)2. $ (5.5)

    Combining (5.3) and (5.5), we find that

    $ |exp(λβn1+mλntαα)exp(λβntαα)|C(α,γ,l)m2γεγ+εl2tαγαlλβl+βγ+γ+ε(lγ)2n. $ (5.6)

    Therefore, we have the following estimate

    $ (Sm,α,β(t)Sα,β(t))u02Hr(Ω))=nNλ2rn|exp(λβn1+mλntαα)exp(λβntαα)|2u0,en2|C(α,γ,l)|2m2γεγ+εlt2αγ2αlnNλ2r2βl+2βγ+2γ+ε(lγ)nu0,en2, $

    which implies that

    $ (Sm,α,β(t)Sα,β(t))u0Hr(Ω))C(α,γ,l)m2γεγ+εl2tαγαlu0Hrβl+βγ+γ+ε(lγ)2(Ω). $ (5.7)

    Next, we explain how to choose the parameters $ l, \varepsilon, \gamma $. Since $ \beta < 1 $, we can choose $ l $ such that

    $ 0<l<min(k(1β),1ββk)=k(1β), $

    which implies that $ 2k(1-\beta)+2l \beta > l $, that is $ \frac{2k(1-\beta)+2l \beta}{l} > 1 $. Then we can choose $ \varepsilon $ such that

    $ 1<ε<min(2,2k(1β)+2lβl). $

    Let us choose $ \gamma $ such that

    $ γ=k(1β)+l(βε2)β+1ε2. $

    It is easy to verify that $ \gamma > \frac{l+l\left(\beta-\frac{\varepsilon}{2} \right) }{\beta+1-\frac{\varepsilon}{2}} = l $ and the following equality

    $ r-\beta l + \beta \gamma+ \gamma + \frac{\varepsilon (l-\gamma)}{2} = r+k-\beta k. $

    Then the estimate (5.7) becomes

    $ M1Hr(Ω)=(Sm,α,β(t)Sα,β(t))u0Hr(Ω))C(α,γ,l)m2γεγ+εl2tαγαlu0Hr+kβk(Ω). $ (5.8)

    Step 2. Estimate of $ M_2 $. Let $ \psi \in \mathbb H^r (\Omega) $. For $ 0 \le \nu \le t \le T $, we can also get

    $ Sα,β(t)(Sα,β(ν))1ψ2Hr(Ω)=nNλ2rnexp(2λβnναtαα)ψ,en2ψ2Hr(Ω), $ (5.9)

    where we use that

    $ exp(λβn1+mλnναtαα)1. $

    By (5.9) and Sobolev embedding $ \mathbb H^{s}(\Omega)) \hookrightarrow \mathbb H^{r}(\Omega) $, we find that

    $ t0να1Sα,β(t)(Sα,β(ν))1(G(um,α,β(ν))G(uα,β(ν)))dνHr(Ω)t0να1G(um,α,β(ν))G(uα,β(ν))Hr(Ω)dνC(r,s)t0να1G(um,α,β(ν))G(uα,β(ν))Hs(Ω)dν. $ (5.10)

    By the global Lipschitz property of $ G $ as in (4.1) and noting (5.10), it follows that

    $ M2Hr(Ω)LgC(r,s)t0να1um,α,β(ν)uα,β(ν)Hr(Ω)dν. $ (5.11)

    Step 3. Estimate of $ M_3 $. Let $ f \in \mathbb H^{r+k-\beta k } (\Omega) $. Then using Parseval' s equality, we have the following identity

    $ [PmSm,α,β(t)(Sm,α,β(ν))1PmSα,β(t)(Sα,β(ν))1]f2Hr(Ω))=nNλ2rn(1+mλn)2|exp(λβn1+mλntαναα)exp(λβntαναα)|2f,en2. $ (5.12)

    By a similar explanation as in (5.6), we find that

    $ |exp(λβn1+mλntαναα)exp(λβntαναα)|C(α,γ,l)m2γεγ+εl2(tανα)γlλβl+βγ+γ+ε(lγ)2n. $ (5.13)

    By (5.12) and (5.13), we get that

    $ [PmSm,α,β(t)(Sm,α,β(ν))1PmSα,β(t)(Sα,β(ν))1]fHr(Ω))C(α,γ,l)m2γεγ+εl2(tανα)γlfHr+kβk(Ω). $ (5.14)

    In view of (5.14) and $ s \ge r+k -\beta k $, we derive that

    $ [PmSm,α,β(t)(Sm,α,β(ν))1PmSα,β(t)(Sα,β(ν))1]G(uα,β(ν))Hr(Ω))C(α,γ,l)m2γεγ+εl2(tανα)γlG(um,α,β(ν))Hr+kβk(Ω)C(α,γ,l,s)m2γεγ+εl2(tανα)γlG(um,α,β(ν))Hs(Ω). $ (5.15)

    By using global Lipschitz property of $ G $ as in (4.1) and noting (4.2), we infer that

    $ G(um,α,β(ν))Hs(Ω)Lgum,α,β(ν)Hr(Ω)2CkLgαk(mk+λ11)eμ0TαTbαkνbu0Hr+kβk(Ω)(mk+λ11)νbu0Hr+kβk(Ω), $ (5.16)

    where the hidden constant depends on $ k, \alpha, L_g, \mu_0, b $. Combining (5.15) and (5.16), we get the following estimate

    $ M3Hr(Ω)m2γεγ+εl2(mk+λ11)u0Hr+kβk(Ω)t0να1b(tανα)γldν. $ (5.17)

    Since $ \gamma > l $ and noting that $ b < \alpha $, we infer that

    $ t0να1b(tανα)γldνtα(γl)t0να1bdν=tα(γl)+αbαb. $ (5.18)

    By (5.17) and (5.18), we obtain

    $ M3Hr(Ω)m2γεγ+εl2(mk+λ11)tα(γl)+αbαbu0Hr+kβk(Ω). $ (5.19)

    Step 4. Estimate of $ M_4 $. By Parseval's equality, we know that

    $ (PmI)Sα,β(t)(Sα,β(ν))1f2Hr(Ω))=nNλ2rn(mλn1+mλn)2exp(2λβntαναα)f,en2. $ (5.20)

    Using the inequality $ e^{-z} \le C(\varepsilon_0) z^{-\varepsilon_0} $, we get the following inequality

    $ exp(2λβntαναα)C(ε0,α)λ2βε0n(tανα)2ε0. $ (5.21)

    By the inequality $ (1+z)^2 > z^{\varepsilon_1} $ for $ 1 < \varepsilon_1 < 2 $, we find that the following inequality

    $ (mλn1+mλn)2m2ε1λ2ε1n. $ (5.22)

    By (5.20)–(5.22), we derive that

    $ (PmI)Sα,β(t)(Sα,β(ν))1f2Hr(Ω))C(ε0,α)m2ε1(tανα)2ε0nNλ2r+2ε12βε0nf,en2. $

    Hence, by using Parseval's equality, we obtain the following estimate

    $ (PmI)Sα,β(t)(Sα,β(ν))1fHr(Ω))C(ε0,α)m1ε12(tανα)ε0fHrβε0+1ε12(Ω)). $ (5.23)

    Next, we need to choose the appropriate parameters $ \varepsilon_0, \varepsilon_1 $. Let $ \varepsilon_0 = k \in (0, \frac{1}{2}) $ and $ \varepsilon_1 = 2-2k $, we can verify that $ 1 < \varepsilon_1 < 2 $. Hence, it follows from (5.23) that

    $ (PmI)Sα,β(t)(Sα,β(ν))1fHr(Ω))C(k,α)mk(tανα)kfHrβk+k(Ω)). $

    By (5.16) and Sobolev embedding $ \mathbb H^{s}(\Omega)) \hookrightarrow \mathbb H^{r-\beta k +k}(\Omega), $ we derive that

    $ (PmI)Sα,β(t)(Sα,β(ν))1G(um,α,β(ν))Hr(Ω))C(k,α)mk(tανα)kG(um,α,β(ν))Hrβk+k(Ω))C(k,α,s)mk(tανα)kG(um,α,β(ν))Hs(Ω))mk(tανα)k(mk+λ11)νbu0Hr+kβk(Ω), $

    which implies that

    $ M4Hr(Ω)mk(mk+λ11)u0Hr+kβk(Ω)t0να1b(tανα)kdνmk(mk+λ11)tαbkαα(12k)(α2b)u0Hr+kβk(Ω), $ (5.24)

    where we use (4.23). Combining (5.8), (5.11), (5.19) and (5.24), we obtain that

    $ um,α,β(t)uα,β(t)Hr(Ω)M1Hr(Ω)+M2Hr(Ω)+M3Hr(Ω)+M4Hr(Ω)C(α,γ,l)m2γεγ+εl2tαγαlu0Hr+kβk(Ω)+m2γεγ+εl2(mk+λ11)tα(γl)+αbαbu0Hr+kβk(Ω)+mk(mk+λ11)tαbkαα(12k)(α2b)u0Hr+kβk(Ω)+LgC(r,s)t0να1um,α,β(ν)uα,β(ν)Hr(Ω)dν. $ (5.25)

    Here we note that $ 1-2k > 0 $ and $ \alpha-2b > 0 $. Since the fact that $ \gamma > l $ and $ b < \min (\alpha, (1-k) \alpha) $, it is obvious to see that

    $ t^{ \alpha \gamma- \alpha l} \le T^{ \alpha \gamma- \alpha l}, $
    $ \frac{t^{ \alpha(\gamma-l)+ \alpha-b}}{ \alpha-b} \le \frac{T^{ \alpha(\gamma-l)+ \alpha-b}}{ \alpha-b}, $
    $ \frac{t^{ \alpha-b-k \alpha}}{ \sqrt{ \alpha (1-2k) ( \alpha-2b)} } \le \frac{T^{ \alpha-b-k \alpha}}{ \sqrt{ \alpha (1-2k) ( \alpha-2b)} }, $

    which motivate us to put

    $ \mathbb T^*( \alpha, \gamma, l, b, k) = \max \Big( T^{ \alpha \gamma- \alpha l},\; \frac{T^{ \alpha(\gamma-l)+ \alpha-b}}{ \alpha-b}, \frac{T^{ \alpha-b-k \alpha}}{ \sqrt{ \alpha (1-2k) ( \alpha-2b)} } \Big) . $

    It follows from (5.25) that

    $ um,α,β(t)uα,β(t)Hr(Ω)[m2γεγ+εl2+mk]T(α,γ,l,b,k)+Lgt0να1um,α,β(ν)uα,β(ν)Hr(Ω)dν. $ (5.26)

    To continue to go further in the proof, we now need to recall the following Lemma introduced in [29].

    Lemma 5.1. Let $ v \in L^1[0, T] $. Consider some postive constant $ A, B, \beta', \gamma' $ such that $ \beta'+\gamma' > 1 $ and

    $ v(t)A+Bt0(tr)β1rγ1v(r)dr. $

    Then for $ 0 < t \le T $, we get

    $ v(t)AEβ,γ(B(Γ(β))1β+γ1t) $

    Looking Lemma 5.1 and (5.26), we set

    $ v(t) = \Big\| u_{m, \alpha, \beta}(t)- u^*_{ \alpha, \beta}(t) \Big\|_{\mathbb H^{r} (\Omega)},\; \; A = \Big[ m^{ \frac{2\gamma- \varepsilon \gamma+\varepsilon l }{2} } + m^{ \frac{2\gamma- \varepsilon \gamma+\varepsilon l }{2} } + m^{k} \Big] \mathbb T^*( \alpha, \gamma, l, b, k), $

    $ B = L_g, \; \; \beta' = 1 $ and $ \gamma' = \alpha. $ Then we deduce that

    $ um,α,β(t)uα,β(t)Hr(Ω)[m2γεγ+εl2+mk]T(α,γ,l,b,k)E1,α(Lgt). $

    The proof of Theorem 5.1 is completed.

    Theorem 6.1. Let $ G: \mathbb H^r (\Omega) \to \mathbb H^s (\Omega) $ such that (4.1) holds. Here $ r, s $ satisfy that $ s > r+k-\beta k $ for $ 0 < k < \frac{1}{2}. $ Let the initial datum $ u_0 \in \mathbb H^{r+k-\beta k+ \varepsilon} (\Omega) $ for any $ \varepsilon > 0 $. Then we get the following estimate

    $ um,α,βum,αXαk(0,T;Hr(Ω))Dβ(ε,k)u0Hr+kβk+ε(Ω)+|Eβ(r,s,k)|u0Hr+kβk(Ω), $

    where the hidden constants depends on $ \alpha, k, T, b $. Here

    $ {\bf D}_\beta (\varepsilon,k) = \left| 1- \lambda_1^{1-\beta} \right|^{\frac{ 2k-\beta k+ \varepsilon}{\beta}} + (1-\beta)^{ \varepsilon } , \quad \quad {\bf E}_\beta (r,s,k) = \left| 1- \lambda_1^{1-\beta} \right|^{\frac{ k+s-r}{\beta}} + (1-\beta)^{ s- r-k +\beta k} $

    for any $ \varepsilon > 0 $.

    Proof. For $ m > 0 $, let $ u_{m, \alpha, \beta} $ and $ u^{**}_{m, \alpha} $ be the mild solutions to Problem (1.3) with $ 0 < \beta < 1 $ and $ \beta = 1 $ respectively. Let us recall the formula of these two solutions

    $ um,α,β(t):=Sm,α,β(t)u0+t0να1PmSm,α,β(t)(Sm,α,β(ν))1G(um,α,β(ν))dν. $ (6.1)

    and

    $ um,α(t):=Sm,α,1(t)u0+t0να1PmSm,α,1(t)(Sm,α,1(ν))1G(um,α(ν))dν. $ (6.2)

    Subtracting (6.1) from (6.2) on each side, we derive that

    $ um,α,β(t)um,α(t)=(Sm,α,β(t)u0Sm,α,1(t)u0)+t0να1[PmSm,α,β(t)(Sm,α,β(ν))1PmSm,α,1(t)(Sm,α,1(ν))1]G(um,α(ν))dν+t0να1PmSm,α,β(t)(Sm,α,β(ν))1[G(um,α,β(ν))G(um,α(ν))]dν=N1+N2+N3. $ (6.3)

    Step 1. Estimate of $ N_1 $. By Parseval' s equality, we have that the following equality

    $ N12Hr(Ω)=(Sm,α,β(t)u0Sm,α,1(t)u0)2Hr(Ω)=nNλ2rn|exp(λβn1+mλntαα)exp(λn1+mλntαα)|2u0,en2=λn>1λ2rn|exp(λβn1+mλntαα)exp(λn1+mλntαα)|2u0,en2+λn1λ2rn|exp(λβn1+mλntαα)exp(λn1+mλntαα)|2u0,en2=N1,1+N1,2. $ (6.4)

    For the term $ N_{1, 1} $, since $ \lambda_n > 1 $ and $ 0 < \beta < 1 $, we note that $ \frac{ \lambda_n^\beta }{1+ m \lambda_n} < \frac{ \lambda_n }{1+ m \lambda_n} $. Hence, we have the following inequality

    $ \exp\left( - \frac{ \lambda_n^\beta }{1+ m \lambda_n} \frac{ t^ \alpha}{ \alpha} \right) > \exp\left( - \frac{ \lambda_n }{1+ m \lambda_n} \frac{ t^ \alpha}{ \alpha} \right) . $

    In view of the above inequality

    $ \Big| e^{-c}- e^{-d} \Big| \le C_{\gamma'} \max (e^{-c}, e^{-d}) |c-d|^{\gamma'},\; \; \; \gamma' > 0 $

    with $ c = \exp\left(- \frac{ \lambda_n^\beta }{1+ m \lambda_n} \frac{ t^ \alpha}{ \alpha} \right) $ and $ d = \exp\left(- \frac{ \lambda_n }{1+ m \lambda_n} \frac{ t^ \alpha}{ \alpha} \right) $, we derive that

    $ |exp(λβn1+mλntαα)exp(λn1+mλntαα)|Cγexp(λβn1+mλntαα)|λnλβn1+mλn|γC(γ,ε)(λβn1+mλntαα)ε|λnλβn1+mλn|γ=C(γ,ε)αεtαε(1+mλn)εγλβεn|λnλβn|γ. $ (6.5)

    Since the fact that $ \lambda_n > 1 $, we know that $ \Big| \lambda_n- \lambda_n^\beta \Big|^{\gamma'} = \lambda_n^{ \gamma'} \left(1- \lambda_n^{\beta-1} \right)^{\gamma'}. $ Using the inequality $ 1- e^{-y} \le C(\mu) y^\mu $ for any $ \mu > 0 $, we find that

    $ 1λβ1n=1exp[(1β)log(λn)]C(μ)(1β)μlogμ(λn)C(μ)(1β)μλμn, $

    where we note that $ 0 < \log(y) \le y $ for any $ y > 1 $, which implies that

    $ |λnλβn|γ=λγn(1λβ1n)γC(μ,γ)(1β)μγλμγ+γn. $ (6.6)

    Combining (6.5) and (6.6), we derive that

    $ |exp(λβn1+mλntαα)exp(λn1+mλntαα)|C(μ,γ,ε)αεtαε(1+mλn)εγ(1β)μγλβεnλμγ+γn. $

    If we make the assumption $ \varepsilon' \le \gamma' $, then $ \left(1+ m \lambda_n \right)^{\varepsilon'-\gamma'} \le 1 $, which allows us to obtain that

    $ N1,1|C(μ,γ,ε,α)|2t2αε(1β)2μγλn>1λ2r+2μγ+2γ2βεnu0,en2|C(μ,γ,ε,α)|2t2αε(1β)2μγu02Hr+μγ+γβε(Ω). $

    Let us choose $ \mu = \frac{\varepsilon}{k} $ and $ \gamma' = \varepsilon' = k $ for any $ \varepsilon > 0 $. Then we get the following estimate

    $ N1,1|C(ε,k,α)|2t2αk(1β)2εu02Hr+kβk+ε(Ω). $ (6.7)

    Before mention to $ N_{1, 2} $, we provide a set $ \overline N = \Big\{ n \in \mathbb N: \lambda_n \le 1 \Big\}. $ Let us give the observation that if $ \overline N $ is an empty set, then $ N_{1, 2} = 0 $. If $ \overline N $ is a non-empty set, then $ \lambda_1 \le 1 $. For the term $ N_{1, 2} $, we note that $ \frac{ \lambda_n^\beta }{1+ m \lambda_n} > \frac{ \lambda_n }{1+ m \lambda_n} $ since $ \lambda_n \le 1 $ and $ 0 < \beta < 1 $. Hence, we have that

    $ \exp\left( - \frac{ \lambda_n^\beta }{1+ m \lambda_n} \frac{ t^ \alpha}{ \alpha} \right) < \exp\left( - \frac{ \lambda_n }{1+ m \lambda_n} \frac{ t^ \alpha}{ \alpha} \right) . $

    By using the fact that

    $ \Big| e^{-c}- e^{-d} \Big| \le C_{\gamma_1} \max (e^{-c}, e^{-d}) |c-d|^{\gamma_1},\; \; \; \gamma_1 > 0 $

    with $ c = \exp\left(- \frac{ \lambda_n^\beta }{1+ m \lambda_n} \frac{ t^ \alpha}{ \alpha} \right) $ and $ d = \exp\left(- \frac{ \lambda_n }{1+ m \lambda_n} \frac{ t^ \alpha}{ \alpha} \right) $, we derive that

    $ |exp(λβn1+mλntαα)exp(λn1+mλntαα)|C(γ1)exp(λn1+mλntαα)|λnλβn1+mλn|γ1C(γ1,ε1)(λn1+mλntαα)ε1|λnλβn1+mλn|γ1=C(γ1,ε1)αε1tαε1(1+mλn)ε1γ1λε1n|λnλβn|γ1. $ (6.8)

    Since $ \lambda_1 \le \lambda_n \le 1 $, it is obvious to see that

    $ |λnλβn|γ1=(λβnλn)γ1=λβγ1n|1λ1βn|γ1λβγ1n|1λ1β1|γ1. $ (6.9)

    Combining (6.8) and (6.9), we find that

    $ |exp(λβn1+mλntαα)exp(λn1+mλntαα)|C(γ1,ε1,α)|1λ1β1|γ1tαε1(1+mλn)ε1γ1λε1+βγ1n. $ (6.10)

    Hence, it follows from (6.10) that

    $ N1,2=λn1λ2rn|exp(λβn1+mλntαα)exp(λn1+mλntαα)|2u0,en2|C(γ1,ε1,α)|2|1λ1β1|2γ1t2αε1n=1(1+mλn)2ε12γ1λ2r2ε1+2βγ1nu0,en2. $

    Let $ \varepsilon_1 = k $ and $ \gamma_1 = \frac{2k+\varepsilon }{\beta}-k $. Since $ 0 < \beta < 1 $, we know that

    $ \varepsilon_1 < \gamma_1 , \quad 2r-2\varepsilon_1+ 2\beta \gamma_1 = 2r+2k-2\beta k +2\varepsilon. $

    Hence, in view of Parseval's equality, we get the following estimate

    $ N1,2C(k,r,ε,α)|1λ1β1|4k2βk+2εβt2αku02Hr+kβk+ε(Ω). $ (6.11)

    Combining (6.4), (6.7) and (6.11), we obtain the following estimate

    $ N12Hr(Ω)=N1,1+N1,2C(k,r,ε,α)[|1λ1β1|4k2βk+2εβ+(1β)2ε]t2αku02Hr+kβk+ε(Ω). $

    By taking the square root of both sides of the above expression and using the inequality $ \sqrt{a+b} \le \sqrt{a}+ \sqrt{b} $ for any $ a, b \ge 0 $, we get

    $ N1Hr(Ω)C(k,r,ε,α)Dβ(ε,k)tαku0Hr+kβk+ε(Ω). $ (6.12)

    Here we denote

    $ {\bf D}_\beta (\varepsilon,k) = \left| 1- \lambda_1^{1-\beta} \right|^{\frac{ 2k-\beta k+ \varepsilon}{\beta}} + (1-\beta)^{ \varepsilon } , \quad \varepsilon > 0, $

    where we observe that $ {\bf D}_\beta (\varepsilon, k) \to 0, \ \beta \to 1^{-}. $

    Step 2. Estimate of $ N_2 $. We confirm the following result for any $ \varepsilon_0 > 0 $ using the method similar to Step 1,

    $ [PmSm,α,β(t)(Sm,α,β(ν))1PmSm,α,1(t)(Sm,α,1(ν))1]ψHr(Ω)C(k,r,ε0,α)[|1λ1β1|2kβk+ε0β+(1β)ε0](tανα)kψHr+kβk+ε0(Ω). $

    Since $ s > r+k-\beta k $, we know that $ \varepsilon_0 = s- r-k +\beta k > 0 $, and using global Lipschitz property of $ G $, we derive that

    $ [PmSm,α,β(t)(Sm,α,β(ν))1PmSm,α,1(t)(Sm,α,1(ν))1]G(um,α(ν))Hr(Ω)¯C1Eβ(r,s,k)(tανα)kG(um,α(ν))Hs(Ω)Kg¯C1Eβ(r,s,k)(tανα)kum,α(ν)Hr(Ω), $

    where $ \overline C_1 = C (k, r, s, \beta, \alpha) $, and we have the following observation

    $ \quad {\bf E}_\beta (r,s,k) = \left| 1- \lambda_1^{1-\beta} \right|^{\frac{ k+s-r}{\beta}} + (1-\beta)^{ s- r-k +\beta k} \to 0, \quad \beta \to 1^{-}. $

    In view of (4.2), we obtain that the following upper bound

    $ um,α(ν)Hr(Ω)¯C2νbu0Hr+kβk(Ω), $

    where $ \overline C_2 = C(k, \alpha, m, \mu_0, T, b). $ From two latter estimations as above, we infer that

    $ N2Hr(Ω)t0να1[PmSm,α,β(t)(Sm,α,β(ν))1PmSm,α,1(t)(Sm,α,1(ν))1]G(um,α(ν))Hr(Ω)dν¯C3Eβ(r,s,k)u0Hr+kβk(Ω)t0να1b(tανα)kdν, $ (6.13)

    where $ \overline C_3 = K_g \overline C_1 \overline C_2 $. Using (4.23), we obtain that the following inequality

    $ t0να1b(tανα)kdνtαbkαα(12k)(α2b). $ (6.14)

    Combining (6.13) and (6.14), we obtain that

    $ N2Hr(Ω)¯C3tαbkαα(12k)(α2b)Eβ(r,s,k)u0Hr+kβk(Ω). $ (6.15)

    Step 3. Estimate of $ N_3 $. By a similar argument as in (4.6), we find that

    $ t0να1PmSm,α,β(t)(Sm,α,β(ν))1[G(um,α,β(ν))G(um,α(ν))]dνHr(Ω))Ckαkmkt0να1(tανα)kG(um,α,β(ν))G(um,α(ν))Hr+kβk(Ω))dν. $

    Since $ s > r+k-\beta k $ and using global Lipschitz property of $ G $, we obtain that

    $ G(um,α,β(ν))G(um,α(ν))Hr+kβk(Ω))G(um,α,β(ν))G(um,α(ν))Hs(Ω))Kgum,α,β(ν)um,α(ν)Hr(Ω)). $

    From the two above observations, we confirm the following statement

    $ N3Hr(Ω)¯C3t0να1(tανα)kum,α,β(ν)um,α(ν)Hr(Ω))dν, $ (6.16)

    where $ \overline C_3 = C_k \alpha^k m^{-k}K_g. $ Combining (6.3), (6.12), (6.15) and (6.16), we deduce the following estimate

    $ um,α,β(t)um,α(t)Hr(Ω)N1Hr(Ω)+N2Hr(Ω)+N3Hr(Ω)˜CtαkDβ(ε,k)u0Hr+kβk+ε(Ω)+¯C3tαbkαα(12k)(α2b)Eβ(r,s,k)u0Hr+kβk(Ω)+¯C3t0να1(tανα)kum,α,β(ν)um,α(ν)Hr(Ω))dν, $ (6.17)

    where $ \widetilde C = C (k, r, \varepsilon, \alpha) $. 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) as follows

    $ (t0να1(tανα)kum,α,β(ν)um,α(ν)Hr(Ω))dν)2(t0να1(tανα)2kdν)(t0να1um,α,β(ν)um,α(ν)2Hr(Ω))dν)tα(12k)α(12k)(t0να1um,α,β(ν)um,α(ν)2Hr(Ω))dν). $ (6.18)

    Combining (6.17), (6.18) and the inequality $ (a+b+c)^2 \le 3 a^2+ 3 b^2 + 3 c^2 $, we derive that

    $ um,α,β(t)um,α(t)2Hr(Ω)3|˜C|2t2αk|Dβ(ε,k)|2u02Hr+kβk+ε(Ω)+3|¯C3|2t2α2b2αkα(12k)(α2b)|Eβ(r,s,k)|2u02Hr+kβk(Ω)+|¯C3|2tα(12k)α(12k)t0να1um,α,β(ν)um,α(ν)2Hr(Ω))dν. $ (6.19)

    Multiplying both sides of (6.19) by $ t^{2 \alpha k} $, we get the following estimate

    $ t2αkum,α,β(t)um,α(t)2Hr(Ω)3|˜C|2|Dβ(ε,k)|2u02Hr+kβk+ε(Ω)+3|¯C3|2t2α2bα(12k)(α2b)|Eβ(r,s,k)|2u02Hr+kβk(Ω)+|¯C3|2tαα(12k)t0να1um,α,β(ν)um,α(ν)2Hr(Ω))dν, $

    which implies that

    $ t2αkum,α,β(t)um,α(t)2Hr(Ω)3|˜C|2|Dβ(ε,k)|2u02Hr+kβk+ε(Ω)+3|¯C3|2T2α2bα(12k)(α2b)|Eβ(r,s,k)|2u02Hr+kβk(Ω)+|¯C3|2Tαα(12k)t0να12αkν2αkum,α,β(ν)um,α(ν)2Hr(Ω))dν. $ (6.20)

    Looking Lemma 5.1 and (6.20), we set $ v(t) = t^{2 \alpha k} \Big\| u_{m, \alpha, \beta} (t)- u^{**}_{m, \alpha} (t) \Big\|_{\mathbb H^r(\Omega)}^2, $

    $ \overline A = 3 |\widetilde C|^2 \Big| {\bf D}_\beta (\varepsilon,k) \Big|^2 \big\| u_0 \big\|_{\mathbb H^{r+k-\beta k +\varepsilon} (\Omega)}^2+ \frac{ 3 |\overline C_3|^2 T^{2 \alpha-2b }}{ { \alpha (1-2k) ( \alpha-2b)} } \Big| {\bf E}_\beta (r,s,k)\Big|^2 \big\| u_0 \big\|_{\mathbb H^{r+k-\beta k} (\Omega)}^2 $

    and

    $ \overline B = \frac{|\overline C_3 |^2 T^{ \alpha}}{ \alpha(1-2k)},\; \; \beta' = 1,\; \; \gamma' = \alpha-2 \alpha k. $

    By applying Lemma 5.1, we obtain that

    $ t2αkum,α,β(t)um,α(t)2Hr(Ω)¯AE1,α2αk(¯B(Γ(β))1β+γ1t)=¯AE1,α2αk(¯Bt). $ (6.21)

    In view of Lemma 3.1 as in [30], we obtain the following upper bound

    $ E1,α2αk(¯Bt)Cα, $ (6.22)

    where $ \mathbb C_ \alpha $ is a positive constant that depends on $ \alpha. $ By (6.21) and (6.22), we get

    $ tαkum,α,β(t)um,α(t)Hr(Ω)¯ACα. $

    From Deinition 2.3, we find

    $ um,α,βum,αXαk(0,T;Hr(Ω))¯ACα. $

    The proof is completed.

    Huy Tuan Nguyen is supported by the Van Lang University. Chao Yang is supported by the Ph.D. Student Research and Innovation Fund of the Fundamental Research Funds for the Central Universities (3072022GIP2403).

    The authors declare there is no conflict of interest.

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