Notes on Convergence Results for Parabolic Equations with Riemann–Liouville Derivatives

Abstract

Fractional diffusion equations have applications in various fields and in this paper we consider a fractional diffusion equation with a Riemann–Liouville derivative. The main objective is to investigate the convergence of solutions of the problem when the fractional order tends to $1^{-}$. Under some suitable conditions on the Cauchy data, we prove the convergence results in a reasonable sense.

1. Introduction

Fractional differential equations arise in models in many fields such as economics, biology, mechanics, geology, and heat transfer, and usually there are two types of derivatives that are of interest, namely the Caputo and the Riemman–Liouville derivatives. Both of these derivatives are defined by non-local integrals. Note that the Riemann–Liouville derivative of a constant is not zero and if an arbitrary function is constant at the origin, then its fractional derivative has a singularity at the origin, for example, exponential and Mittag–Leffler functions. However, if we study the initial value problem with the Caputo derivative, we often treat the initial values as normal functions, just like the initial value problem with the classical derivative. However, when considering a problem with the Riemman–Liouville derivative its initial value condition usually involves a fractional integral or a fractional derivative (see [1]). In this paper, we examine:

$$\begin{array}{c}\left\{\begin{array}{c}{\mathbf{D}}_{0^{+}}^{\alpha }u-u_{xx}=G(x,t),(x,t)\in (0,\pi )\times (0,T),\hfill \\ u(0,t)=u(\pi ,t)=0,0<t<T,\hfill \\ t^{1-\alpha }{u|}_{t=0}=f\left(x\right),\hfill \end{array}\right.\hfill \end{array}$$

(1)

where ${\mathbf{D}}_{0^{+}}^{\alpha }v$ is called the Riemann–Liouville fractional derivative of v with order $\alpha$, $0<\alpha \le 1$ and it is defined by

$$\begin{array}{c}\hfill {\mathbf{D}}_{0^{+}}^{\alpha }v\left(t\right)=\frac{d}{dt}\left(\frac{1}{\Gamma (1-\alpha )}{\int }_0^t{(t-r)}^{-\alpha }v\left(r\right)dr\right),\end{array}$$

(2)

and ${\mathbf{D}}_{0^{+}}^{\alpha }v\left(t\right)=:{\displaystyle \frac{d}{dt}}v\left(t\right)$ if $\alpha =1$. The diffusion equation arises from many diffusion phenomena that occur in nature (e.g., phase transitions, biochemistry) and equations of time fractional reactions occur in describing “memory” in physics, for example plasma turbulence [2], fractal geometry [3,4], and single-molecular protein dynamics [5] (we also refer the reader to Refs. [6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22]).

In Ref. [23], the authors focused on:

$$\begin{array}{c}\left\{\begin{array}{c}{\mathbf{D}}_{0^{+}}^{\alpha }u=F(t,u\left(t\right)),(x,t)\in (0,\pi )\times (0,T),\hfill \\ t^{1-\alpha }{u|}_{t=0}=\lambda {\int }_0^{T}u\left(t\right)dt+h,h\in \mathbb{R},\hfill \end{array}\right.\hfill \end{array}$$

(3)

where $\lambda \ge 0$ and in [24] the following non-local problem was studied:

$$\begin{array}{c}\left\{\begin{array}{c}{\mathbf{D}}_{0^{+}}^{\alpha }u+F(t,v\left(t\right))=a,t\in (0,T),\hfill \\ {\mathbf{D}}_{0^{+}}^{\beta }v+F(t,u\left(t\right))=b,t\in (0,T),\hfill \\ u\left(0\right)=0,u\left(T\right)={\int }_0^T\varphi \left(t\right)u\left(t\right)dt,\hfill \\ v\left(0\right)=0,v\left(T\right)={\int }_0^T\varphi \left(t\right)v\left(t\right)dt,\hfill \end{array}\right.\hfill \end{array}$$

(4)

where $1<\alpha ,\beta <2$. In Ref. [25], the authors studied the existence and continuation of solutions, and several global existence results were given to a general fractional differential equation (FDE) with a Riemann–Liouville derivative.

Recently, the authors in Ref. [26] investigated the following nonlinear diffusion equation with the Riemann–Liouville derivative

$$\begin{array}{c}\hfill {\mathbf{D}}_{0^{+}}^{\alpha }u-u_{xx}=F(x,t,u),\end{array}$$

and obtained the existence and regularity of mild solutions using the Banach fixed point theorem. In Ref. [27], Luc studied the fractional diffusion equation with the Riemann–Liouville derivative in the form

$${\mathbf{D}}_{0^{+}}^{\alpha }u-u_{xx}=F(x,t),$$

(5)

and under some assumptions on the input data, he obtained the wellposedness of problem (5).

Returning to Problem (1) we see that if $f\in {\mathbb{H}}^s(\Omega )$ for any $s\ge 0$, Problem (1) has a unique solution. The solution to the problem depends on the fractional order $\alpha$ and we denote the solution as $u_{\alpha }$. The main goal of this paper is to see what happens to $u_{\alpha }$ when $\alpha \to 1^{-}$, and this question is inherently difficult.

In this paper, we prove for the first time that the solution $u_{\alpha }$ to our problem converges to the solution of the problem with a classical derivative. Theorem 1 gives a result concerning this convergence in the homogeneous case and Theorem 2 is a convergent result for the linear inhomogeneous case. The main technique of the paper is to use a Lemma in Ref. [28] combined with the evaluation in Hilbert scale spaces.

2. Preliminaries

With $\Omega =(0,\pi )$, note the spectral problem

$$\begin{array}{c}\left\{\begin{array}{cc}\Delta {\mathit{e}}_j\left(x\right)=-{\lambda }_j{\mathit{e}}_j\left(x\right),\hfill & \hfill x\in \Omega ,\\ {\mathit{e}}_j\left(x\right)=0,\hfill & \hfill x\in \partial \Omega ,\end{array}\right.\hfill \end{array}$$

with the eigenvalues

$$\begin{array}{c}\hfill 0<{\lambda }_1\le {\lambda }_2\le \cdots \le {\lambda }_j\le \cdots \mathrm{with}{\lambda }_j\to \infty \mathrm{for}j\to \infty \end{array}$$

and the corresponding eigenfunctions ${\mathit{e}}_j\in H_0^1(\Omega )$.

Definition 1.

The Mittag–Leffler function, which is defined by

$$\begin{array}{c}\hfill E_{\alpha ,\beta }\left(z\right)=\sum _{n=0}^{\infty }\frac{z^n}{\Gamma (n\alpha +\beta )},(z\in \mathbb{C}),\end{array}$$

for $\alpha >0$ and $\beta \in \mathbb{R}$. When $\beta =1$, it is abbreviated as $E_{\alpha }\left(z\right)=E_{\alpha ,1}\left(z\right)$.

We note following lemmas (see for example Ref. [2]).

Lemma 1

(see Ref. [28]). Let ${\displaystyle \frac{3}{4}}\le \alpha <1$ and $\alpha \le \beta \le 1$. Then there exists a constant C which is independent of $\alpha ,\beta ,z$ such that, for any $z<0$,

$$|E_{\alpha ,\beta }\left(z\right)-e^z|\le \frac{C}{1+\left|z\right|}(1-\alpha ).$$

(6)

The proof can be found in Ref. [28].

Lemma 2

(see Ref. [29]). Let $0<\alpha <1$. Then the function $z\mapsto E_{\alpha ,\alpha }\left(z\right)$ has no negative root. Moreover, there exists a constant ${\overline{C}}_{\alpha }$ such that

$$\begin{array}{c}\hfill 0\le E_{\alpha ,\alpha }(-z)\le \frac{{\overline{C}}_{\alpha }}{1+z},z>0.\end{array}$$

(7)

For a positive number $r\ge 0$, we define the Hilbert scale space

$$\begin{array}{c}\hfill H^r(0,\pi )=\left\{f\in L^2(0,\pi ):\sum _{j=1}^{\infty }{j}^{2r}{〈f,\sqrt{\frac{2}{\pi }}sin\left(jx\right)〉}^2<+\infty \right\},\end{array}$$

(8)

with the following norm ${\displaystyle {\parallel u\parallel }_{H^r(0,\pi )}={\left(\sum _{j=1}^{\infty }{j}^{2r}{〈f,\sqrt{\frac{2}{\pi }}sin\left(jx\right)〉}^2\right)}^{\frac{1}{2}}·}$

Theorem 1.

Let the Cauchy data $f\in {\mathbb{H}}^{s-2\beta }(\Omega )\cap {\mathbb{H}}^{s-2\theta }(\Omega )\cap {\mathbb{H}}^{s+2\epsilon }(\Omega )$ any $0<\epsilon <1,s\ge 0$, $0<\beta ,\theta <1$, $\frac{3}{4}\le \alpha <1$, $\beta ,\theta \le {\displaystyle \frac{s}{2}}$ and $\theta <\alpha$. Let $u_{\alpha }$ be the mild solution to Problem (1) and $u^{*}$ be the mild solution to Problem (1) with $\alpha =1$, i.e.,

$$\begin{array}{c}\left\{\begin{array}{c}{u}_t^{*}-u_{xx}^{*}=0,(x,t)\in (0,\pi )\times (0,T),\hfill \\ u^{*}(0,t)=u^{*}(\pi ,t)=0,t\times (0,T),\hfill \\ u^{*}(x,0)=f\left(x\right),\hfill \end{array}\right.\hfill \end{array}$$

(9)

Then we get

$$\begin{array}{cc}\hfill \parallel u_{\alpha }-u^{*}{\parallel }_{L^1(0,T;{\mathbb{H}}^s(\Omega ))}& \le \left(\frac{C\Gamma \left(\alpha \right)T^{\alpha -\alpha \beta }}{\alpha -\alpha \beta }(1-\alpha )\right){\parallel f\parallel }_{{\mathbb{H}}^{s-2\beta }(\Omega )}\hfill \\ \hfill & +C(\theta ,\mu ,T)\left[{(1-\alpha )}^{\mu }+{\left(T^{*}\right)}^{1-\alpha }-1\right]{\parallel f\parallel }_{{\mathbb{H}}^{s-2\theta }(\Omega )}\hfill \\ \hfill & +\frac{T^{1-\beta }}{1-\beta }|\Gamma \left(\alpha \right)-1|{\parallel f\parallel }_{{\mathbb{H}}^{s-2\beta }(\Omega )}\hfill \\ \hfill & +\left[{(1-\alpha )}^{\frac{\alpha \epsilon }{2}}+{\left({\left(T^{*}\right)}^{1-\alpha }-1\right)}^{\epsilon }\right]{\parallel f\parallel }_{{\mathbb{H}}^{s+2\epsilon }(\Omega )},\hfill \end{array}$$

(10)

where $T^{*}=max(T,1)$.

Proof. 

From the paper Ref. [27], we obtain the representation of the mild solution $u_{\alpha }$

$$\begin{array}{c}\hfill {\int }_{\Omega }{u}_{\alpha }(x,t)e_j\left(x\right)dx=\Gamma \left(\alpha \right)t^{\alpha -1}{E}_{\alpha ,\alpha }\left(-j^2{t}^{\alpha }\right)\left({\int }_{\Omega }f\left(x\right)e_j\left(x\right)dx\right).\end{array}$$

(11)

The mild solution of the classical problem is defined by

$$\begin{array}{c}\hfill {\int }_{\Omega }{u}^{*}(x,t)e_j\left(x\right)dx=exp\left(-j^{2}t\right)\left({\int }_{\Omega }f\left(x\right)e_j\left(x\right)dx\right).\end{array}$$

(12)

Subtracting both sides of the two equations above we get

$$\begin{array}{cc}\hfill {\int }_{\Omega }\left(u_{\alpha }(x,t)-u^{*}(x,t)\right)e_j\left(x\right)dx& =\Gamma \left(\alpha \right)t^{\alpha -1}\left(E_{\alpha ,\alpha }\left(-j^2{t}^{\alpha }\right)-exp\left(-j^{2}t\right)\right)\left({\int }_{\Omega }f\left(x\right)e_j\left(x\right)dx\right)\hfill \\ \hfill & +\Gamma \left(\alpha \right)t^{\alpha -1}\left(exp\left(-j^2{t}^{\alpha }\right)-exp\left(-j^{2}t\right)\right)\left({\int }_{\Omega }f\left(x\right)e_j\left(x\right)dx\right)\hfill \\ \hfill & +\Gamma \left(\alpha \right)\left(t^{\alpha -1}-1\right)exp\left(-j^{2}t\right)\left({\int }_{\Omega }f\left(x\right)e_j\left(x\right)dx\right)\hfill \\ \hfill & +\left(\Gamma \left(\alpha \right)-1\right)exp\left(-j^{2}t\right)\left({\int }_{\Omega }f\left(x\right)e_j\left(x\right)dx\right).\hfill \end{array}$$

(13)

Thus, we get

$$\begin{array}{cc}\hfill u_{\alpha }(x,t)-u^{*}(x,t)& =\sum _j\Gamma \left(\alpha \right)t^{\alpha -1}\left(E_{\alpha ,\alpha }\left(-j^2{t}^{\alpha }\right)-exp\left(-j^{2}t\right)\right)\left({\int }_{\Omega }f\left(x\right)e_j\left(x\right)dx\right)e_j\left(x\right)\hfill \\ \hfill & +\sum _j\Gamma \left(\alpha \right)\left(t^{\alpha -1}-1\right)exp\left(-j^{2}t\right)\left({\int }_{\Omega }f\left(x\right)e_j\left(x\right)dx\right)e_j\left(x\right)\hfill \\ \hfill & +\sum _j\left(\Gamma \left(\alpha \right)-1\right)exp\left(-j^{2}t\right)\left({\int }_{\Omega }f\left(x\right)e_j\left(x\right)dx\right)e_j\left(x\right)\hfill \\ \hfill & +\sum _j\Gamma \left(\alpha \right)t^{\alpha -1}\left(exp\left(-j^2{t}^{\alpha }\right)-exp\left(-j^{2}t\right)\right)\left({\int }_{\Omega }f\left(x\right)e_j\left(x\right)dx\right)e_j\left(x\right)\hfill \\ \hfill & =J_1(x,t)+J_2(x,t)+J_3(x,t)+J_4(x,t).\hfill \end{array}$$

(14)

Step 1. Estimate of $J_1$. In view of the inequality in Lemma 1, we have for $z>0$ that

$$|E_{\alpha ,\alpha }(-z)-exp(-z)|\le \frac{C}{1+\left|z\right|}(1-\alpha ),$$

and we find that

$$\begin{array}{c}\hfill |E_{\alpha ,\alpha }\left(-j^2{t}^{\alpha }\right)-exp\left(-j^2{t}^{\alpha }\right)|\le \frac{C}{1+j^2{t}^{\alpha }}(1-\alpha ).\end{array}$$

(15)

From $0<\beta <1$, we get

$$\begin{array}{c}\hfill \frac{C}{1+j^2{t}^{\alpha }}\le \frac{C}{{(1+j^2{t}^{\alpha })}^{\beta }}\le C{j}^{-2\beta }{t}^{-\alpha \beta }.\end{array}$$

(16)

Hence, we obtain the following estimate for $0<\beta <1$,

$$\begin{array}{c}\hfill |E_{\alpha ,\alpha }\left(-j^2{t}^{\alpha }\right)-exp\left(-j^2{t}^{\alpha }\right)|\le C{j}^{-2\beta }{t}^{-\alpha \beta }(1-\alpha ).\end{array}$$

(17)

This implies that

$$\begin{array}{cc}\hfill \parallel J_1(.,t){\parallel }_{{\mathbb{H}}^s(\Omega )}^2& =\sum _j{j}^{2s}{|\Gamma \left(\alpha \right)|}^2{t}^{2\alpha -2}{\left(E_{\alpha ,\alpha }\left(-j^2{t}^{\alpha }\right)-exp\left(-j^{2}t\right)\right)}^2{\left({\int }_{\Omega }f\left(x\right)e_j\left(x\right)dx\right)}^2\hfill \\ \hfill & \le C^2{|\Gamma \left(\alpha \right)|}^2{t}^{2\alpha -2-2\alpha \beta }{(1-\alpha )}^2\sum _j{j}^{2s-4\beta }{\left({\int }_{\Omega }f\left(x\right)e_j\left(x\right)dx\right)}^2.\hfill \end{array}$$

(18)

Using Parseval’s equality, we have for any $0<\beta <1$ that

$$\begin{array}{c}\hfill \parallel J_1(.,t){\parallel }_{{\mathbb{H}}^s(\Omega )}\le C\Gamma \left(\alpha \right)t^{\alpha -1-\alpha \beta }(1-\alpha ){\parallel f\parallel }_{{\mathbb{H}}^{s-2\beta }(\Omega )}.\end{array}$$

(19)

Thus

$$\begin{array}{c}\hfill {\parallel J_1\parallel }_{L^1(0,T;{\mathbb{H}}^s(\Omega ))}={\int }_0^T{\parallel J_1(.,t)\parallel }_{{\mathbb{H}}^s(\Omega )}dt\le C\Gamma \left(\alpha \right)(1-\alpha ){\parallel f\parallel }_{{\mathbb{H}}^{s-2\beta }(\Omega )}\left({\int }_0^T{t}^{\alpha -1-\alpha \beta }dt\right).\end{array}$$

(20)

Note the integral ${\displaystyle {\int }_0^T{t}^{\alpha -1-\alpha \beta }dt}$ is convergent and

$${\int }_0^T{t}^{\alpha -1-\alpha \beta }dt=\frac{T^{\alpha -\alpha \beta }}{\alpha -\alpha \beta }·$$

Hence, we get that

$$\begin{array}{c}\hfill {\parallel J_1\parallel }_{L^1(0,T;{\mathbb{H}}^s(\Omega ))}\le \frac{C\Gamma \left(\alpha \right)T^{\alpha -\alpha \beta }}{\alpha -\alpha \beta }(1-\alpha ){\parallel f\parallel }_{{\mathbb{H}}^{s-2\beta }(\Omega )}.\end{array}$$

(21)

Step 2. Estimate of $J_2$. In this step (at the end) since $\theta <\alpha$ we choose $0<\mu <1$ so that $\mu +\theta <\alpha$. Note

$$\begin{array}{c}\hfill {\parallel J_2(.,t)\parallel }_{{\mathbb{H}}^s(\Omega )}^2=\sum _j{j}^{2s}{\left|t^{\alpha -1}-1\right|}^{2}exp\left(-2j^{2}t\right){\left({\int }_{\Omega }f\left(x\right)e_j\left(x\right)dx\right)}^2.\end{array}$$

(22)

We now consider the term $|t^{\alpha -1}-1|$. If $0<t\le 1$, then we use the inequality $1-e^{-z}\le C_{\mu }{z}^{\mu },0<\mu <1$ to obtain

$$\begin{array}{cc}\hfill |t^{\alpha -1}-1|& =t^{\alpha -1}\left(1-t^{1-\alpha }\right)=t^{\alpha -1}\left(1-exp(-(1-\alpha )log(1/t\left)\right)\right)\hfill \\ \hfill & \le t^{\alpha -1}{C}_{\mu }{(1-\alpha )}^{\mu }{\left(log(\frac{1}{t})\right)}^{\mu }.\hfill \end{array}$$

(23)

In view of the inequality $log\left(z\right)\le z$ for any $z\ge 1$, we have that

$${\left(log(\frac{1}{t})\right)}^{\mu }\le t^{-\mu }.$$

Therefore, we obtain for the case $0<t\le 1$ that

$$\begin{array}{c}\hfill \left|t^{\alpha -1}-1\right|\le C_{\mu }{(1-\alpha )}^{\mu }{t}^{\alpha -\mu -1}.\end{array}$$

(24)

For the case $t>1$, we see that

$$\begin{array}{c}\hfill |t^{\alpha -1}-1|=t^{\alpha -1}\left(t^{1-\alpha }-1\right)\le t^{\alpha -1}\left({\left(T^{*}\right)}^{1-\alpha }-1\right),\end{array}$$

(25)

where $T^{*}=max(T,1)$. From (24) and (25), we have that

$$\begin{array}{c}\hfill |t^{\alpha -1}-1|\le t^{\alpha -1}\left[C_{\mu }{(1-\alpha )}^{\mu }{t}^{-\mu }+{\left(T^{*}\right)}^{1-\alpha }-1\right],\end{array}$$

(26)

for any $0<\mu <1$. In view of the inequality $e^{-z}\le C_{\theta }{z}^{-\theta }$, $0<\theta <1$, one has the following inequality

$$\begin{array}{c}\hfill exp\left(-2j^{2}t\right)\le C_{\theta }^2{t}^{-2\theta }{j}^{-4\theta }.\end{array}$$

(27)

Combining (22), (26), and (27), we deduce that

$$\begin{array}{cc}\hfill \parallel J_2(.,t){\parallel }_{{\mathbb{H}}^s(\Omega )}^2& \le C(\theta ,\mu )t^{2\alpha -2-2\theta }{\left[C_{\mu }{(1-\alpha )}^{\mu }{t}^{-\mu }+{\left(T^{*}\right)}^{1-\alpha }-1\right]}^2\hfill \\ \hfill & \sum _j{j}^{2s-4\theta }{\left({\int }_{\Omega }f\left(x\right)e_j\left(x\right)dx\right)}^2\hfill \\ \hfill & =C(\theta ,\mu )t^{2\alpha -2-2\theta }{\left[C_{\mu }{(1-\alpha )}^{\mu }{t}^{-\mu }+{\left(T^{*}\right)}^{1-\alpha }-1\right]}^2{\parallel f\parallel }_{{\mathbb{H}}^{s-2\theta }(\Omega )}^2.\hfill \end{array}$$

(28)

Thus

$$\begin{array}{c}\hfill {\parallel J}_2(.,t){\parallel }_{{\mathbb{H}}^s(\Omega )}\le C(\theta ,\mu )t^{\alpha -1-\theta }\left[{(1-\alpha )}^{\mu }{t}^{-\mu }+{\left(T^{*}\right)}^{1-\alpha }-1\right]{\parallel f\parallel }_{{\mathbb{H}}^{s-2\theta }(\Omega )},\end{array}$$

(29)

where $C(\theta ,\mu )$ represents a constant that depends on $\mu ,\theta$. Therefore, we have

$$\begin{array}{cc}\hfill {\parallel J_2\parallel }_{L^1(0,T;{\mathbb{H}}^s(\Omega ))}& ={\int }_0^T{\parallel J_2(.,t)\parallel }_{{\mathbb{H}}^s(\Omega )}dt\hfill \\ \hfill & \le C(\theta ,\mu ){(1-\alpha )}^{\mu }{\parallel f\parallel }_{{\mathbb{H}}^{s-2\theta }(\Omega )}\left({\int }_0^T{t}^{\alpha -1-\theta -\mu }dt\right)\hfill \\ \hfill & +C(\theta ,\mu )\left({\left(T^{*}\right)}^{1-\alpha }-1\right){\parallel f\parallel }_{{\mathbb{H}}^{s-2\theta }(\Omega )}\left({\int }_0^T{t}^{\alpha -1-\theta }dt\right).\hfill \end{array}$$

(30)

Note (with $\mu$ chosen so that $\theta +\mu <\alpha$)

$$\begin{array}{c}\hfill {\int }_0^T{t}^{\alpha -1-\theta -\mu }dt=\frac{T^{\alpha -\theta -\mu }}{\alpha -\theta -\mu },\end{array}$$

and

$$\begin{array}{c}\hfill {\int }_0^T{t}^{\alpha -1-\theta }dt=\frac{T^{\alpha -\theta }}{\alpha -\theta }·\end{array}$$

The above imply that

$$\begin{array}{c}\hfill {\parallel J2\parallel }_{L^1(0,T;{\mathbb{H}}^s(\Omega ))}\le C(\theta ,\mu ,T)\left[{(1-\alpha )}^{\mu }+{\left(T^{*}\right)}^{1-\alpha }-1\right]{\parallel f\parallel }_{{\mathbb{H}}^{s-2\theta }(\Omega )}.\end{array}$$

(31)

Step 3. Estimate of $J_3$. Using Parseval’s equality, we have

$$\begin{array}{c}\hfill {\parallel J_3(.,t)\parallel }_{{\mathbb{H}}^s(\Omega )}^2={\left(\Gamma \left(\alpha \right)-1\right)}^2\sum _j{j}^{2s}exp\left(-2j^{2}t\right){\left({\int }_{\Omega }f\left(x\right)e_j\left(x\right)dx\right)}^2.\end{array}$$

(32)

In view of the inequality $e^{-z}\le C_{\beta }{z}^{-\beta }$, we have

$$exp\left(-2j^{2}t\right)\le C_{\beta }^2{j}^{-4\beta }{t}^{-2\beta }.$$

This inequality together with (32) gives

$$\begin{array}{c}\hfill {\parallel J_3(.,t)\parallel }_{{\mathbb{H}}^s(\Omega )}\le |\Gamma \left(\alpha \right)-1|t^{-\beta }{\parallel f\parallel }_{{\mathbb{H}}^{s-2\beta }(\Omega )}.\end{array}$$

(33)

Thus

$$\begin{array}{c}\hfill {\parallel J3\parallel }_{L^1(0,T;{\mathbb{H}}^s(\Omega ))}={\int }_0^T{\parallel J_3(.,t)\parallel }_{{\mathbb{H}}^s(\Omega )}dt\le |\Gamma \left(\alpha \right)-1|{\parallel f\parallel }_{{\mathbb{H}}^{s-2\beta }(\Omega )}\left({\int }_0^T{t}^{-\beta }dt\right).\end{array}$$

(34)

Since $0<\beta <1$, then the proper integral ${\int }_0^T{t}^{-\beta }dt$ is convergent and

$$\begin{array}{c}\hfill {\int }_0^T{t}^{-\beta }dt=\frac{T^{1-\beta }}{1-\beta }·\end{array}$$

(35)

Combining (34) and (35), we obtain

$$\begin{array}{c}\hfill {\parallel J3\parallel }_{L^1(0,T;{\mathbb{H}}^s(\Omega ))}\le \frac{T^{1-\beta }}{1-\beta }|\Gamma \left(\alpha \right)-1|{\parallel f\parallel }_{{\mathbb{H}}^{s-2\beta }(\Omega )}.\end{array}$$

(36)

Step 4. Estimate of $J_4$. We will choose $\mu ={\displaystyle \frac{\alpha }{2}}$ below. In view of Parseval’s equality, we get

$$\begin{array}{c}\hfill {\parallel J_4(.,t)\parallel }_{{\mathbb{H}}^s(\Omega )}^2={\left(\Gamma \left(\alpha \right)\right)}^2{t}^{2\alpha -2}\sum _j{j}^{2s}{\left(exp\left(-j^2{t}^{\alpha }\right)-exp\left(-j^{2}t\right)\right)}^2{\left({\int }_{\Omega }f\left(x\right)e_j\left(x\right)dx\right)}^2.\end{array}$$

(37)

Using the inequality $|e^{-a}-e^{-b}|\le C_{\epsilon }{|a-b|}^{\epsilon }$, see Refs. [30,31] page 12–13, for any $a,b>0$, $0<\epsilon <1$, we get

$$\begin{array}{c}\hfill |exp\left(-j^2{t}^{\alpha }\right)-exp\left(-j^{2}t\right)|\le C_{\epsilon }{j}^{2\epsilon }{|t^{\alpha }-t|}^{\epsilon }.\end{array}$$

(38)

From (26), we obtain the following inequality

$$\begin{array}{cc}\hfill |t^{\alpha }-t|=t|t^{\alpha -1}-1|& \le t^{\alpha }\left[C_{\mu }{(1-\alpha )}^{\mu }{t}^{-\mu }+{\left(T^{*}\right)}^{1-\alpha }-1\right]\hfill \\ \hfill & \le C_{\mu }{(1-\alpha )}^{\mu }{t}^{\alpha -\mu }+\left({\left(T^{*}\right)}^{1-\alpha }-1\right)t^{\alpha },\hfill \end{array}$$

(39)

for any $0<\mu <1$. Thus, we obtain that

$$\begin{array}{c}\hfill {|t}^{\alpha }{-t|}^{2\epsilon }\le C(\mu ,\epsilon )\left({(1-\alpha )}^{2\epsilon \mu }{t}^{2\epsilon \alpha -2\epsilon \mu }+{\left({\left(T^{*}\right)}^{1-\alpha }-1\right)}^{2\epsilon }{t}^{2\epsilon \alpha }\right).\end{array}$$

(40)

From some previous observations, we have

$$\begin{array}{cc}\hfill & t^{2\alpha -2}\sum _j{j}^{2s}{\left(exp\left(-j^2{t}^{\alpha }\right)-exp\left(-j^{2}t\right)\right)}^2{\left({\int }_{\Omega }f\left(x\right)e_j\left(x\right)dx\right)}^2\hfill \\ \hfill & \le C(\mu ,\epsilon ){(1-\alpha )}^{2\epsilon \mu }{t}^{2\epsilon \alpha -2\epsilon \mu +2\alpha -2}\sum _j{j}^{2s+4\epsilon }{\left({\int }_{\Omega }f\left(x\right)e_j\left(x\right)dx\right)}^2\hfill \\ \hfill & +C(\mu ,\epsilon ){\left({\left(T^{*}\right)}^{1-\alpha }-1\right)}^{2\epsilon }{t}^{2\epsilon \alpha +2\alpha -2}\sum _j{j}^{2s+4\epsilon }{\left({\int }_{\Omega }f\left(x\right)e_j\left(x\right)dx\right)}^2.\hfill \end{array}$$

(41)

We now choose $\mu ={\displaystyle \frac{\alpha }{2}}$. Using the inequality $\sqrt{a+b}\le \sqrt{a}+\sqrt{b}$ for any $a,b\ge 0$, we have

$$\begin{array}{cc}\hfill \parallel J_4(.,t){\parallel }_{{\mathbb{H}}^s(\Omega )}& \le \Gamma \left(\alpha \right)C(\alpha ,\epsilon ){(1-\alpha )}^{\frac{\alpha \epsilon }{2}}{t}^{\frac{\epsilon \alpha }{2}+\alpha -1}{\parallel f\parallel }_{{\mathbb{H}}^{s+2\epsilon }(\Omega )}\hfill \\ \hfill & +\Gamma \left(\alpha \right)C(\alpha ,\epsilon ){\left({\left(T^{*}\right)}^{1-\alpha }-1\right)}^{\epsilon }{t}^{\epsilon \alpha +\alpha -1}{\parallel f\parallel }_{{\mathbb{H}}^{s+2\epsilon }(\Omega )}.\hfill \end{array}$$

(42)

The latter estimate implies that

$$\begin{array}{cc}\hfill {\parallel J_4\parallel }_{L^1(0,T;{\mathbb{H}}^s(\Omega ))}& ={\int }_0^T{\parallel J_4(.,t)\parallel }_{{\mathbb{H}}^s(\Omega )}dt\hfill \\ \hfill & \le \Gamma \left(\alpha \right)C(\alpha ,\epsilon ){(1-\alpha )}^{\frac{\alpha \epsilon }{2}}{\parallel f\parallel }_{{\mathbb{H}}^{s+2\epsilon }(\Omega )}\left({\int }_0^T{t}^{\frac{\epsilon \alpha }{2}+\alpha -1}dt\right)\hfill \\ \hfill & +\Gamma \left(\alpha \right)C(\alpha ,\epsilon ){\left({\left(T^{*}\right)}^{1-\alpha }-1\right)}^{\epsilon }{\parallel f\parallel }_{{\mathbb{H}}^{s+2\epsilon }(\Omega )}\left({\int }_0^T{t}^{\epsilon \alpha +\alpha -1}dt\right).\hfill \end{array}$$

(43)

We note that the two integrals ${\int }_0^T{t}^{\frac{\epsilon \alpha }{2}+\alpha -1}dt$ and ${\int }_0^T{t}^{\epsilon \alpha +\alpha -1}dt$ are convergent. Thus, it follows from (43) that

$$\begin{array}{c}\hfill {\parallel J4\parallel }_{L^1(0,T;{\mathbb{H}}^s(\Omega ))}\lesssim \left[{(1-\alpha )}^{\frac{\alpha \epsilon }{2}}+{\left({\left(T^{*}\right)}^{1-\alpha }-1\right)}^{\epsilon }\right]{\parallel f\parallel }_{{\mathbb{H}}^{s+2\epsilon }(\Omega )},\end{array}$$

(44)

where the hidden constant depends on $\alpha ,\epsilon$. □

Remark 1.

Note that the right hand side of (10) goes to 0 when $\alpha \to 1^{-}$.

3. Inhomogeneous Case

In this section, we consider the following diffusion equation with the inhomogeneous term

$$\begin{array}{c}\left\{\begin{array}{c}{\mathbf{D}}_{0^{+}}^{\alpha }u-u_{xx}=G(x,t),(x,t)\in (0,\pi )\times (0,T),\hfill \\ u=0,(x,t)\in \partial \Omega \times (0,T),\hfill \\ t^{1-\alpha }{u|}_{t=0}=f\left(x\right).\hfill \end{array}\right.\hfill \end{array}$$

(45)

In order to reduce the calculation duplication process, we take the function $f=0$. The goal of this section is to prove the convergence of the mild solution to problem (45) when $\alpha \to 1^{-}$.

Theorem 2.

Let $G\in L^{\infty }(0,T;{\mathbb{H}}^{s+2\epsilon }(\Omega ))\cap L^{\infty }(0,T;{\mathbb{H}}^{s-2\beta }(\Omega ))\cap L^{\infty }(0,T;{\mathbb{H}}^{s-\theta }(\Omega ))$, for any $\epsilon >0$, $s\ge 0$, $0<\beta ,\theta <1$, $\frac{3}{4}\le \alpha <1$, $\beta ,\theta \le {\displaystyle \frac{s}{2}}$, $\beta <{\displaystyle \frac{1}{2}}$ and $\theta <\alpha$. Let $u_{\alpha }$ be the mild solution of Problem (45) and let $u^{*}$ be the mild solution to the classical problem, i.e.,

$$\begin{array}{c}\left\{\begin{array}{c}{u}_t^{*}-u_{xx}^{*}=G(x,t),(x,t)\in (0,\pi )\times (0,T),\hfill \\ u^{*}=0,(x,t)\in \partial \Omega \times (0,T),\hfill \\ u^{*}(x,0)=f\left(x\right),x\in \Omega .\hfill \end{array}\right.\hfill \end{array}$$

(46)

Then we have the following estimate

$$\begin{array}{cc}\hfill \parallel u_{\alpha }(.,t)-u^{*}(.,t){\parallel }_{{\mathbb{H}}^s(\Omega )}& \le (1-\alpha ){\parallel G\parallel }_{L^{\infty }(0,T;{\mathbb{H}}^{s-2\beta }(\Omega ))}\hfill \\ \hfill & +\left({(1-\alpha )}^{\frac{\alpha \epsilon }{2}}+{\left({\left(T^{*}\right)}^{1-\alpha }-1\right)}^{\epsilon }\right){\parallel G\parallel }_{L^{\infty }(0,T;{\mathbb{H}}^{s+2\epsilon }(\Omega ))}\hfill \\ \hfill & +\sqrt{C_{\alpha }{T}^{\frac{\alpha }{2}}{(1-\alpha )}^{\frac{\alpha }{2}}+\frac{T^{\alpha }}{\alpha }\left({\left(T^{*}\right)}^{1-\alpha }-1\right)}{\parallel G\parallel }_{L^{\infty }(0,T;{\mathbb{H}}^{s-\theta }(\Omega ))}.\hfill \end{array}$$

(47)

Proof. 

The mild solution $u_{\alpha }$ of Problem (45) is defined by (see Ref. [26])

$$\begin{array}{c}\hfill {\int }_{\Omega }{u}_{\alpha }(x,t)e_j\left(x\right)dx={\int }_0^t{(t-r)}^{\alpha -1}{E}_{\alpha ,\alpha }\left(-j^2{(t-r)}^{\alpha }\right)\left({\int }_{\Omega }G(x,r)e_j\left(x\right)dx\right)dr.\end{array}$$

(48)

The mild solution to the classical problem is given by

$$\begin{array}{c}\hfill {\int }_{\Omega }{u}^{*}(x,t)e_j\left(x\right)dx={\int }_0^{t}exp(-j^2(t-r))\left({\int }_{\Omega }G(x,r)e_j\left(x\right)dx\right)dr.\end{array}$$

(49)

From two above equalities, we get that

$$\begin{array}{cc}\hfill & u_{\alpha }(x,t)-u^{*}(x,t)\hfill \\ \hfill & =\sum _{j=1}^{\infty }\left[{\int }_0^t{(t-r)}^{\alpha -1}\left[E_{\alpha ,\alpha }\left(-j^2{(t-r)}^{\alpha }\right)-exp(-j^2(t-r))\right]\left({\int }_{\Omega }G(x,r)e_j\left(x\right)dx\right)dr\right]e_j\left(x\right)\hfill \\ \hfill & +\sum _{j=1}^{\infty }\left[{\int }_0^t{(t-r)}^{\alpha -1}\left[exp(-j^2{(t-r)}^{\alpha })-exp(-j^2(t-r))\right]\left({\int }_{\Omega }G(x,r)e_j\left(x\right)dx\right)dr\right]e_j\left(x\right)\hfill \\ \hfill & +\sum _{j=1}^{\infty }\left[{\int }_0^t\left[{(t-r)}^{\alpha -1}-1\right]exp(-j^2(t-r))\left({\int }_{\Omega }G(x,r)e_j\left(x\right)dx\right)dr\right]e_j\left(x\right)\hfill \\ \hfill & ={\mathbb{H}}_1(x,t)+{\mathbb{H}}_2(x,t)+{\mathbb{H}}_3(x,t).\hfill \end{array}$$

(50)

Step 1. Estimate of ${\mathbb{H}}_1$. In order to estimate ${\mathbb{H}}_1$, we use the bound (17) and obtain

$$\begin{array}{c}\hfill |E_{\alpha ,\alpha }\left(-j^2{(t-r)}^{\alpha }\right)-exp(-j^2{(t-r)}^{\alpha })|\le C{j}^{-2\beta }{(t-r)}^{-\alpha \beta }(1-\alpha ).\end{array}$$

(51)

The term ${\mathbb{H}}_1$ is bounded by

$$\begin{array}{cc}\hfill {\parallel {\mathbb{H}}_1\parallel }_{{\mathbb{H}}^s(\Omega )}^2& =\sum _{j=1}^{\infty }{j}^{2s}{\left[{\int }_0^t{(t-r)}^{\alpha -1}\left[E_{\alpha ,\alpha }\left(-j^2{(t-r)}^{\alpha }\right)-exp(-j^2(t-r))\right]\left({\int }_{\Omega }G(x,r)e_j\left(x\right)dx\right)dr\right]}^2\hfill \\ \hfill & \le \sum _{j=1}^{\infty }{j}^{2s}\left({\int }_0^t{(t-r)}^{\alpha -1}dr\right)\hfill \\ \hfill & [{\int }_0^t{(t-r)}^{\alpha -1}{\left[E_{\alpha ,\alpha }\left(-j^2{(t-r)}^{\alpha }\right)-exp(-j^2(t-r))\right]}^2{\left({\int }_{\Omega }G(x,r)e_j\left(x\right)dx\right)}^{2}dr,\hfill \end{array}$$

(52)

where we have used the Hölder inequality. In view of (51), we obtain

$$\begin{array}{cc}\hfill {\parallel {\mathbb{H}}_1\parallel }_{{\mathbb{H}}^s(\Omega )}^2& \lesssim {(1-\alpha )}^2\sum _{j=1}^{\infty }{j}^{2s-4\beta }\left({\int }_0^t{(t-r)}^{\alpha -1-2\alpha \beta }{\left({\int }_{\Omega }G(x,r)e_j\left(x\right)dx\right)}^{2}dr\right)\hfill \\ \hfill & ={(1-\alpha )}^2{\int }_0^t{(t-r)}^{\alpha -1-2\alpha \beta }\parallel G(.,r){|}_{{\mathbb{H}}^{s-2\beta }(\Omega )}^{2}dr\hfill \\ \hfill & \le {(1-\alpha )}^2{\parallel G\parallel }_{L^{\infty }(0,T;{\mathbb{H}}^{s-2\beta }(\Omega ))}^2\left({\int }_0^t{(t-r)}^{\alpha -1-2\alpha \beta }dr\right).\hfill \end{array}$$

(53)

Note

$${\int }_0^t{(t-r)}^{\alpha -1-2\alpha \beta }dr=\frac{t^{\alpha -2\alpha \beta }}{\alpha -2\alpha \beta },$$

and we recall that $\beta <\frac{1}{2}$. Thus, we can deduce that

$$\begin{array}{c}\hfill {\parallel {\mathbb{H}}_1\parallel }_{{\mathbb{H}}^s(\Omega )}\lesssim (1-\alpha ){\parallel G\parallel }_{L^{\infty }(0,T;{\mathbb{H}}^{s-2\beta }(\Omega )).}\end{array}$$

(54)

Step 2. Estimate of ${\mathbb{H}}_2$.

We will choose $\mu ={\displaystyle \frac{\alpha }{2}}$ below. Using Hölder’s inequality, the term ${\mathbb{H}}_2$ is bounded by

$$\begin{array}{cc}\hfill {\parallel {\mathbb{H}}_2\parallel }_{{\mathbb{H}}^s(\Omega )}^2& =\sum _{j=1}^{\infty }{j}^{2s}{\left[{\int }_0^t{(t-r)}^{\alpha -1}\left[exp\left(-j^2{(t-r)}^{\alpha }\right)-exp(-j^2(t-r))\right]\left({\int }_{\Omega }G(x,r)e_j\left(x\right)dx\right)dr\right]}^2\hfill \\ \hfill & \le \sum _{j=1}^{\infty }{j}^{2s}\left({\int }_0^t{(t-r)}^{\alpha -1}dr\right)\hfill \\ \hfill & [{\int }_0^t{(t-r)}^{\alpha -1}{\left[exp\left(-j^2{(t-r)}^{\alpha }\right)-exp(-j^2(t-r))\right]}^2{\left({\int }_{\Omega }G(x,r)e_j\left(x\right)dx\right)}^{2}dr.\hfill \end{array}$$

(55)

Using (38) and (40) and the inequality ${(a+b)}^{\epsilon }\le C_{\epsilon }(a^{\epsilon }+b^{\epsilon })$ for any $a,b\ge 0$, $ϵ>0$, we find for any $0\le r\le t$ that

$$\begin{array}{cc}\hfill & |exp\left(-j^2{(t-r)}^{\alpha }\right)-exp\left(-j^2(t-r)\right)|\le C_{\epsilon }{j}^{2\epsilon }{|{(t-r)}^{\alpha }-(t-r)|}^{\epsilon }\hfill \\ \hfill & \lesssim j^{2\epsilon }\left[{(1-\alpha )}^{\mu \epsilon }{(t-r)}^{(\alpha -\mu )\epsilon }+{\left({\left(T^{*}\right)}^{1-\alpha }-1\right)}^{\epsilon }{(t-r)}^{\alpha \epsilon }\right].\hfill \end{array}$$

(56)

Now choose $\mu ={\displaystyle \frac{\alpha }{2}}$. Thus, we see

$$\begin{array}{cc}\hfill & |exp\left(-j^2{(t-r)}^{\alpha }\right)-exp\left(-j^2(t-r)\right){|}^2\hfill \\ \hfill & \lesssim j^{4\epsilon }{(1-\alpha )}^{2\mu \epsilon }{(t-r)}^{\alpha \epsilon }+j^{4\epsilon }{\left({\left(T^{*}\right)}^{1-\alpha }-1\right)}^{2\epsilon }{(t-r)}^{2\alpha \epsilon }.\hfill \end{array}$$

(57)

This implies that

$$\begin{array}{cc}\hfill & \mathit{The}\mathit{right}\mathit{hand}\mathit{side}\mathit{of}\left(55\right)\hfill \\ \hfill & \lesssim {(1-\alpha )}^{\alpha \epsilon }\sum _{j=1}^{\infty }{j}^{2s+4\epsilon }{\int }_0^t{(t-r)}^{\alpha -1+\alpha \epsilon }{\left({\int }_{\Omega }G(x,r)e_j\left(x\right)dx\right)}^{2}dr\hfill \\ \hfill & +{\left({\left(T^{*}\right)}^{1-\alpha }-1\right)}^{2\epsilon }\sum _{j=1}^{\infty }{j}^{2s+4\epsilon }{\int }_0^t{(t-r)}^{\alpha -1+2\alpha \epsilon }{\left({\int }_{\Omega }G(x,r)e_j\left(x\right)dx\right)}^{2}dr.\hfill \end{array}$$

(58)

Hence, we obtain that

$$\begin{array}{cc}\hfill {\parallel {\mathbb{H}}_2\parallel }_{{\mathbb{H}}^s(\Omega )}^2& \lesssim {(1-\alpha )}^{\alpha \epsilon }{\int }_0^t{(t-r)}^{\alpha -1+\alpha \epsilon }{\parallel G(.,r)\parallel }_{{\mathbb{H}}^{s+2\epsilon }(\Omega )}^{2}dr\hfill \\ \hfill & +{\left({\left(T^{*}\right)}^{1-\alpha }-1\right)}^{2\epsilon }{\int }_0^t{(t-r)}^{\alpha -1+2\alpha \epsilon }{\parallel G(.,r)\parallel }_{{\mathbb{H}}^{s+2\epsilon }(\Omega )}^{2}dr.\hfill \end{array}$$

(59)

Note the convergence of the two proper integrals ${\int }_0^t{(t-r)}^{\alpha -1+\alpha \epsilon }dr$ and ${\int }_0^t{(t-r)}^{\alpha -1+2\alpha \epsilon }dr$, and we have the following estimate

$$\begin{array}{cc}\hfill {\parallel {\mathbb{H}}_2\parallel }_{{\mathbb{H}}^s(\Omega )}^2& \lesssim \left({(1-\alpha )}^{\alpha \epsilon }+{\left({\left(T^{*}\right)}^{1-\alpha }-1\right)}^{2\epsilon }\right){\parallel G\parallel }_{L^{\infty }(0,T;{\mathbb{H}}^{s+2\epsilon }(\Omega ))}^2.\hfill \end{array}$$

(60)

Thus, we deduce that

$$\begin{array}{c}\hfill {\parallel \mathbb{H}2\parallel }_{{\mathbb{H}}^s(\Omega )}\lesssim \left({(1-\alpha )}^{\frac{\alpha \epsilon }{2}}+{\left({\left(T^{*}\right)}^{1-\alpha }-1\right)}^{\epsilon }\right){\parallel G\parallel }_{L^{\infty }(0,T;{\mathbb{H}}^{s+2\epsilon }(\Omega ))}.\end{array}$$

(61)

Step 3. Estimate of ${\mathbb{H}}_3$.

We will choose $\mu =\frac{\alpha }{2}$ below and recall $\theta <\alpha$. By Parseval’s equality, we obtain that

$$\begin{array}{cc}\hfill {\parallel {\mathbb{H}}_3\parallel }_{{\mathbb{H}}^s(\Omega )}^2& =\sum _{j=1}^{\infty }{j}^{2s}{\left[{\int }_0^t\left[{(t-r)}^{\alpha -1}-1\right]exp(-j^2(t-r))\left({\int }_{\Omega }G(x,r)e_j\left(x\right)dx\right)dr\right]}^2\hfill \\ \hfill & \le \left({\int }_0^t|{(t-r)}^{\alpha -1}-1|dr\right)\hfill \\ \hfill & \sum _{j=1}^{\infty }{j}^{2s}\left[{\int }_0^t|{(t-r)}^{\alpha -1}-1|exp(-2j^2(t-r)){\left({\int }_{\Omega }G(x,r)e_j\left(x\right)dx\right)}^{2}dr\right].\hfill \end{array}$$

In view of (26) (and choosing $\mu ={\displaystyle \frac{\alpha }{2}}$) we have for $0\le r\le t$ that

$$\begin{array}{c}\hfill |{(t-r)}^{\alpha -1}-1|\le C_{\alpha }{(t-r)}^{\frac{\alpha }{2}-1}{(1-\alpha )}^{\frac{\alpha }{2}}+\left({\left(T^{*}\right)}^{1-\alpha }-1\right){(t-r)}^{\alpha -1},\end{array}$$

(62)

where $C_{\alpha }$ indicates a constant which depends on $\alpha$. The inequality (62) implies that

$$\begin{array}{cc}\hfill {\int }_0^t|{(t-r)}^{\alpha -1}-1|dr& \le C_{\alpha }{(1-\alpha )}^{\frac{\alpha }{2}}\left({\int }_0^t{(t-r)}^{\frac{\alpha }{2}-1}dr\right)\hfill \\ \hfill & +\left({\left(T^{*}\right)}^{1-\alpha }-1\right)\left({\int }_0^t{(t-r)}^{\alpha -1}dr\right).\hfill \end{array}$$

(63)

Note

$$\begin{array}{c}\hfill {\int }_0^t{(t-r)}^{\frac{\alpha }{2}-1}dr=\frac{2}{\alpha }{t}^{\frac{\alpha }{2}},{\int }_0^t{(t-r)}^{\alpha -1}dr=\frac{1}{\alpha }{t}^{\alpha }.\end{array}$$

Thus, it follows from (63) that

$$\begin{array}{c}\hfill {\int }_0^t|{(t-r)}^{\alpha -1}-1|dr\le C_{\alpha }{T}^{\frac{\alpha }{2}}{(1-\alpha )}^{\frac{\alpha }{2}}+\frac{T^{\alpha }}{\alpha }\left({\left(T^{*}\right)}^{1-\alpha }-1\right).\end{array}$$

(64)

By applying the inequality $e^{-j^2(t-r)}\le C_{\theta }{j}^{-\theta }{(t-r)}^{-\theta }$ for any $\theta >0$, we get the following estimate:

$$\begin{array}{cc}\hfill & \sum _{j=1}^{\infty }{j}^{2s}\left[{\int }_0^t\right|{(t-r)}^{\alpha -1}-1|exp(-2j^2(t-r)){\left({\int }_{\Omega }G(x,r)e_j\left(x\right)dx\right)}^{2}dr\hfill \\ \hfill & \le {\int }_0^t|{(t-r)}^{\alpha -1}-1|{(t-r)}^{-\theta }\left(\sum _{j=1}^{\infty }{j}^{2s-2\theta }{\left({\int }_{\Omega }G(x,r)e_j\left(x\right)dx\right)}^2\right)dr\hfill \\ \hfill & \le {\parallel G\parallel }_{L^{\infty }(0,T;{\mathbb{H}}^{s-\theta }(\Omega ))}^2\left({\int }_0^t{(t-r)}^{\alpha -1-\theta }dr+{\int }_0^t{(t-r)}^{-\theta }dr\right),\hfill \end{array}$$

(65)

where we have used the triangle inequality. Then (recall $\theta <\alpha$) from the convergence of the two proper integrals above, we deduce

$$\begin{array}{c}\hfill {\parallel {\mathbb{H}}_3\parallel }_{{\mathbb{H}}^s(\Omega )}^2\le \left[C_{\alpha }{T}^{\frac{\alpha }{2}}{(1-\alpha )}^{\frac{\alpha }{2}}+\frac{T^{\alpha }}{\alpha }\left({\left(T^{*}\right)}^{1-\alpha }-1\right)\right]{\parallel G\parallel }_{L^{\infty }(0,T;{\mathbb{H}}^{s-\theta }(\Omega ))}^2.\end{array}$$

(66)

Combining (54), (61), and (66), we get

$$\begin{array}{cc}\hfill {\parallel u_{\alpha }(.,t)-u^{*}(.,t)\parallel }_{{\mathbb{H}}^s(\Omega )}& \le {\parallel {\mathbb{H}}_1\parallel }_{{\mathbb{H}}^s(\Omega )}+{\parallel {\mathbb{H}}_2\parallel }_{{\mathbb{H}}^s(\Omega )}+{\parallel {\mathbb{H}}_3\parallel }_{{\mathbb{H}}^s(\Omega )}\hfill \\ \hfill & \le (1-\alpha ){\parallel G\parallel }_{L^{\infty }(0,T;{\mathbb{H}}^{s-2\beta }(\Omega ))}\hfill \\ \hfill & +\left({(1-\alpha )}^{\frac{\alpha \epsilon }{2}}+{\left({\left(T^{*}\right)}^{1-\alpha }-1\right)}^{\epsilon }\right){\parallel G\parallel }_{L^{\infty }(0,T;{\mathbb{H}}^{s+2\epsilon }(\Omega ))}\hfill \\ \hfill & +\sqrt{C_{\alpha }{T}^{\frac{\alpha }{2}}{(1-\alpha )}^{\frac{\alpha }{2}}+\frac{T^{\alpha }}{\alpha }\left({\left(T^{*}\right)}^{1-\alpha }-1\right)}{\parallel G\parallel }_{L^{\infty }(0,T;{\mathbb{H}}^{s-\theta }(\Omega ))}.\hfill \end{array}$$

(67)

□

Remark 2.

Note the right hand side of (47) tends to zero when $\alpha \to 1^{-}$.

4. Conclusions

In this work, the main objective was to investigate the convergence of solutions to the problem when the fractional order tends to $1^{-}$ with the Riemann–Liouville derivative. In the future, we hope to investigate the convergence towards $1^{-}$ with the Caputo derivative, the Atangana Baleanu Caputo derivative, and some other non-integer order derivatives.