Convergence analysis of the time-stepping numerical methods for time-fractional nonlinear subdiffusion equations

Abstract

In 1986, Dixon and McKee (Z Angew Math Mech 66:535–544, 1986) developed a discrete fractional Gronwall inequality, which can be seen as a generalization of the classical discrete Gronwall inequality. However, this generalized discrete Gronwall inequality and its variant (Al-Maskari and Karaa in SIAM J Numer Anal 57:1524–1544, 2019) have not been widely applied in the numerical analysis of the time-stepping methods for the time-fractional evolution equations. The main purpose of this paper is to show how to apply the generalized discrete Gronwall inequality to prove the convergence of a class of time-stepping numerical methods for time-fractional nonlinear subdiffusion equations, including the popular fractional backward difference type methods of order one and two, and the fractional Crank-Nicolson type methods. We obtain the optimal $L^2$ error estimate in space discretization for multi-dimensional problems. The convergence of the fast time-stepping numerical methods is also proved in a simple manner. The present work unifies the convergence analysis of several existing time-stepping schemes. Numerical examples are provided to verify the effectiveness of the present method.

Appendices

Appendix A. Proof of $c_n\ge 0$ for the BN-$\theta$ method

The BN-$\theta$ method reduces to the FBDF-2 method for $\theta =0$ and to the GNGF-2 for $\theta = 1/2$. In this section, we prove $c_n\ge 0$ for the BN-$\theta$ method when $0\le \theta \le 1/2$.

Firstly, we give the proof of the following lemma.

Lemma A.1

For $a^{(-\alpha )}_n=\frac{\varGamma (n+\alpha )}{\varGamma (\alpha )\varGamma (n+1)}$ and $0< \alpha < 1$, we have

$$\begin{aligned} \left( \frac{1+\alpha }{2}\right) ^j a_{n-j}^{(-\alpha )}\le & {} a_{n}^{(-\alpha )},\quad 0\le j \le n-1, \end{aligned}$$

(8.1)

$$\begin{aligned} a_{n}^{(-\alpha -1)}-a_{n}^{(-\alpha )}\le & {} (1+\alpha )\left( \frac{2+\alpha }{2}\right) ^{n-2},\quad n\ge 0. \end{aligned}$$

(8.2)

Proof

From $a^{(-\alpha )}_n=\frac{\varGamma (n+\alpha )}{\varGamma (\alpha )\varGamma (n+1)}$, we obtain $a_{n}^{(-\alpha -1)}-a_{n}^{(-\alpha )} =\frac{n}{\alpha }a_{n}^{(-\alpha )}$ and

$$\begin{aligned} \frac{a^{(-\alpha )}_{n+1}}{a^{(-\alpha )}_n}= & {} \frac{n+\alpha }{n+1}\ge \frac{1+\alpha }{2}, \quad n \ge 1, \end{aligned}$$

(8.3)

$$\begin{aligned} \frac{a_{n+1}^{(-\alpha -1)}-a_{n+1}^{(-\alpha )}}{a_{n}^{(-\alpha -1)}-a_{n}^{(-\alpha )}}= & {} \frac{n+\alpha }{n}\le \frac{2+\alpha }{2},\quad n\ge 2. \end{aligned}$$

(8.4)

Eq. (8.3) implies $\frac{a^{(-\alpha )}_{n}}{a^{(-\alpha )}_{n-j}} =\prod _{k=n-j}^{n-1}\frac{a^{(-\alpha )}_{k+1}}{a^{(-\alpha )}_k} \ge \left( \frac{1+\alpha }{2}\right) ^j$, which completes the proof of (8.1). Obviously, (8.2) holds for $n=0,1$. From (8.4), we obtain $a_{n}^{(-\alpha -1)}-a_{n}^{(-\alpha )} \le \left( \frac{2+\alpha }{2}\right) ^{n-2}(a_{2}^{(-\alpha -1)}-a_{2}^{(-\alpha )}) =(\alpha +1)\left( \frac{2+\alpha }{2}\right) ^{n-2}$. The proof complete. $\square$

For $0\le \theta \le 1/2$, we have the following properties:

$$\begin{aligned} f_1(\theta )= & {} \frac{\theta }{1+\theta \alpha }+ \frac{1-2\theta }{3-2\theta } \le f_1(({2+2\alpha })^{-1})= \frac{1+\alpha }{2+3\alpha }, \end{aligned}$$

(8.5)

$$\begin{aligned} f_2(\theta )= & {} 3\theta ^2\left( \frac{1-2\theta }{3-2\theta }\right) + \left( \frac{\theta }{1+\theta }\right) ^3 +\left( \frac{1-2\theta }{3-2\theta }\right) ^3 <\frac{8}{100},\end{aligned}$$

(8.6)

$$\begin{aligned} f_3(\theta )= & {} 4\theta \left( \frac{1-2\theta }{3-2\theta }\right) ^2+ \left( \frac{\theta }{1+\theta }\right) ^3 +\left( \frac{1-2\theta }{3-2\theta }\right) ^3<\frac{8}{100},\end{aligned}$$

(8.7)

$$\begin{aligned} f_4(\theta )= & {} {\theta } \frac{1-2\theta }{3-2\theta }+\frac{3}{8}\bigg [\frac{\theta ^2}{(1+\theta )^2} +\left( \frac{1-2\theta }{3-2\theta }\right) ^2\bigg ]<\frac{9}{100},\end{aligned}$$

(8.8)

$$\begin{aligned} \rho _1= & {} \frac{1-2\theta }{3-2\theta }\le \frac{1}{3},\qquad \rho _2=\frac{\alpha \theta }{1+\alpha \theta }\le \frac{\alpha }{2+\alpha }, \end{aligned}$$

(8.9)

where we used

$$\begin{aligned} \begin{aligned}&\max _{0\le \theta \le 1/2} f_1(\theta )= f_1(({2+2\alpha })^{-1}) ,\qquad \quad \max _{0\le \theta \le 1/2} f_2(\theta )\approx f_2(0.3769)\approx 0.0685, \\&\max _{0\le \theta \le 1/2} f_3(\theta )\approx f_3(0.1681)\approx 0.0602, \quad \max _{0\le \theta \le 1/2} f_4(\theta )\approx f_4(0.2811)\approx 0.0806. \end{aligned} \end{aligned}$$

Proof

From (2.19) and (4.2), we have $c_n= 2b_0{a}^{(-\alpha )}_n -\sum _{j=0}^{n}{b}_j{a}^{(-\alpha )}_{n-j},$ where $b_n$ is given by (4.3).

Step 1) Prove $c_n\ge 0$ for $n\ge 3$. Let

$$\begin{aligned} \rho =\max \{\rho _1,\rho _2\},\quad \rho _3= \rho ({2+\alpha })/{2},\quad \lambda = ({1+\alpha })/{2}. \end{aligned}$$

(8.10)

By (4.3), (8.10), (8.2), and $\sum _{j=1}^{n-1} a_{j}^{(-\alpha )}=a_{n}^{(-\alpha -1)}-1-a_n^{(-\alpha )}$, we have

$$\begin{aligned} b_n/b_0&=\rho _1\rho _2\sum _{j=1}^{n-1} a_{j}^{(-\alpha )} \rho _1^{j-1}\rho _2^{n-j-1} + \rho _2^n + \rho ^n_1a_n^{(-\alpha )} \nonumber \\&\le \rho _1\rho _2\rho ^{n-2}\sum _{j=1}^{n-1} a_{j}^{(-\alpha )}+ \rho _2^n + \rho ^n_1 a_n^{(-\alpha )}\nonumber \\&\le (1+\alpha ) \rho _1\rho _2\rho _3^{n-2} + \rho _2^n+\rho ^n_1a_n^{(-\alpha )}. \end{aligned}$$

(8.11)

From (8.1), we have $\lambda ^{1-n}a_n^{(-\alpha )}/\alpha \ge 1$ for $n\ge 1$. Hence,

$$\begin{aligned} \begin{aligned} b_n/b_0\le&\left[ (1+\alpha ) \rho _1\rho _2\rho _3^{n-2}+\rho _2^n\right] \lambda ^{1-n} {a_n^{(-\alpha )}}/{\alpha } +\rho ^n_1 a_n^{(-\alpha )}\\ =&\left[ \frac{1+\alpha }{\alpha \lambda } \rho _1\rho _2 \left( \frac{\rho _3}{\lambda }\right) ^{n-2} +\frac{\lambda }{\alpha }\left( \frac{\rho _2}{\lambda }\right) ^n+\rho ^n_1\right] a_n^{(-\alpha )}\\ \le&\bigg [\frac{1+\alpha }{\alpha \lambda } \rho _1\rho _2 \left( \frac{\rho _3}{\lambda }\right) +\frac{\lambda }{\alpha }\left( \frac{\rho _2}{\lambda }\right) ^3 +\rho ^3_1 \bigg ]a_n^{(-\alpha )}, \quad n\ge 3, \end{aligned} \end{aligned}$$

(8.12)

where we used $\rho _1\le 1/3$, $\rho _2/\lambda <1$, and $\rho _3/\lambda <1$. Direct calculation yields

$$\begin{aligned} \frac{1+\alpha }{\alpha } \frac{\rho _1\rho _2\rho _3}{\lambda ^2} =\left\{ \begin{aligned}&\frac{2\alpha (2+\alpha )}{ 1+\alpha } \frac{1-2\theta }{3-2\theta }\frac{\theta ^2}{(1+\alpha \theta )^2} \le {3\theta ^2} \frac{1-2\theta }{3-2\theta } ,&\rho _1\le \rho _2 ,\\&\frac{2(2+\alpha )}{1+\alpha } \left( \frac{1-2\theta }{3-2\theta }\right) ^2 \frac{ \theta }{1+\alpha \theta } \le 4\theta \left( \frac{1-2\theta }{3-2\theta }\right) ^2,&\rho _1> \rho _2, \end{aligned} \right. \nonumber \\ \end{aligned}$$

(8.13)

$$\begin{aligned} \frac{1}{\alpha } \frac{\rho _2^3}{\lambda ^2} +\rho ^3_1 =\frac{4\alpha ^2}{ (1+\alpha )^2}\frac{\theta ^3}{(1+\theta \alpha )^3} +\left( \frac{1-2\theta }{3-2\theta }\right) ^3 \le \left( \frac{\theta }{1+\theta }\right) ^3 +\left( \frac{1-2\theta }{3-2\theta }\right) ^3.\nonumber \\ \end{aligned}$$

(8.14)

Combining (8.6), (8.7), (8.12), (8.13), and (8.14), yields

$$\begin{aligned} \begin{aligned} b_n/b_0\le&\max \{f_2(\theta ),f_3(\theta )\}a_n^{(-\alpha )} \le \frac{2}{25} a_n^{(-\alpha )},\quad n\ge 3. \end{aligned} \end{aligned}$$

(8.15)

From (8.5), we have

$$\begin{aligned} {b_1}/{b_0}=\rho _2 + a_1^{(-\alpha )}\rho _1 =\alpha \left( \frac{\theta }{1+\theta \alpha }+ \frac{1-2\theta }{3-2\theta }\right) \le \frac{\alpha (1+\alpha )}{2+3\alpha }. \end{aligned}$$

(8.16)

Combining (8.15) and (8.16) yields

$$\begin{aligned} \begin{aligned} \frac{b_0a_n^{(-\alpha )}+b_1a_{n-1}^{(-\alpha )}+ b_n}{b_0a_n^{(-\alpha )}} \le&\frac{27}{25}+\frac{3\alpha (1+\alpha )}{(2+\alpha )(2+3\alpha )} \le \frac{27}{25}+ \frac{2}{5}=\frac{37}{25}\, , \end{aligned} \end{aligned}$$

(8.17)

where we used ${a_{n-1}^{(-\alpha )}} \le \frac{n}{n-1+\alpha }{a_{n}^{(-\alpha )}} \le \frac{3}{2+\alpha }{a_{n}^{(-\alpha )}}$ for $n\ge 3$.

Using (8.9), we obtain

$$\begin{aligned}&1+\alpha -2\rho _1 \ge 1+\alpha -2/3=({1+3\alpha })/{3},\\&1+\alpha -2\rho _2 \ge 1+\alpha - {2\alpha }/({2+\alpha }) = ({\alpha ^2+\alpha +2})/({2+\alpha }),\\&1+\alpha -2\rho _3 \ge 1+\alpha -\rho ({2+\alpha }) \ge 1+\alpha -({2+\alpha })/{3} =({1+2\alpha })/{3}, \end{aligned}$$

which leads to

$$\begin{aligned}&\frac{\rho _{1}\rho _{2}}{1+\alpha -2\rho _{3}}+\frac{1}{1+\alpha } \frac{\rho ^{2_{2}}}{1+\alpha -2\rho _{2}} + \frac{\alpha }{2}\frac{\rho ^{2_{1}}}{1+\alpha -2\rho _{1}} \nonumber \\&\le \frac{3\alpha }{1+2\alpha }\frac{\theta }{1+\theta \alpha }\frac{1-2\theta }{3-2\theta } +\frac{\alpha (2+\alpha )}{(1+\alpha )(2+\alpha +\alpha ^2)} \frac{\alpha \theta ^{2}}{(1+\alpha \theta )^2}\nonumber \\&\qquad +\,\frac{3\alpha }{2(1+3\alpha )}\left( \frac{1-2\theta }{3-2\theta }\right) ^{2} \nonumber \\&\le {\theta } \frac{1-2\theta }{3-2\theta }+ \frac{3}{8}\left[ \frac{\alpha \theta ^{2}}{(1+\alpha \theta )^{2}} +\left( \frac{1-2\theta }{3-2\theta }\right) ^2\right] \nonumber \\&\le {\theta } \frac{1-2\theta }{3-2\theta }+\frac{3}{8}\left[ \frac{\theta ^2}{(1+\theta )^2} +\left( \frac{1-2\theta }{3-2\theta }\right) ^{2}\right] < \frac{9}{100}. \qquad \text {(By (8.8))} \end{aligned}$$

(8.18)

From (8.11), $a_j^{(-\alpha )}\le a_2^{(-\alpha )}=\alpha (1+\alpha )/2$, and the following inequality,

$$\begin{aligned} \sum _{j=2}^{n-1}\rho _k^j a_{n-j}^{(-\alpha )}= & {} \sum _{j=2}^{n-1}\frac{\rho _k^j}{\lambda ^j} \left( \lambda ^ja_{n-j}^{(-\alpha )}\right) \le a_{n}^{(-\alpha )} \sum _{j=2}^{n-1}\frac{\rho _k^j}{\lambda ^j} \le a_{n}^{(-\alpha )} \frac{\rho _k^2/\lambda ^2}{1-\rho _k/\lambda } \\= & {} \frac{\rho ^2_k}{\lambda (\lambda -\rho _k)} a_{n}^{(-\alpha )} =\frac{4\rho ^2_k}{(1+\alpha )(1+\alpha -2\rho _k)} a_{n}^{(-\alpha )}, \quad k=1,2,3, \end{aligned}$$

we obtain

$$\begin{aligned} \sum _{j=2}^{n-1}\frac{{b}_j}{b_0}{a}^{(-\alpha )}_{n-j}\le & {} \sum _{j=2}^{n-1} \left[ (1+\alpha ) \rho _1\rho _2\rho _3^{j-2} + \rho _2^j + \frac{\alpha (1+\alpha )}{2}\rho ^j_1\right] a_{n-j}^{(-\alpha )}\nonumber \\\le & {} {4} \left( \frac{\rho _1\rho _2}{1+\alpha -2\rho _3} +\frac{1}{1+\alpha }\frac{\rho ^2_2}{1+\alpha -2\rho _2} +\frac{\alpha }{2}\frac{\rho ^2_1}{1+\alpha -2\rho _1}\right) a_{n}^{(-\alpha )}\nonumber \\\le & {} \frac{9}{25}a_{n}^{(-\alpha )}. \qquad \text {(By 8.8)} \end{aligned}$$

(8.19)

Combining (8.17) and (8.19) yields

$$\begin{aligned} b_0^{-1}\sum _{j=0}^{n}{b}_j{a}^{(-\alpha )}_{n-j}=\sum _{j=2}^{n-1}({b}_j/b_0){a}^{(-\alpha )}_{n-j} +\left( b_0a_n^{(-\alpha )}+b_1a_{n-1}^{(-\alpha )}+ b_n\right) /b_0 \le \frac{46}{25}a_n^{(-\alpha )}, \end{aligned}$$

which leads to

$$\begin{aligned} c_n= 2b_0{a}^{(-\alpha )}_n -b_0^{-1}\sum _{j=0}^{n}{b}_j{a}^{(-\alpha )}_{n-j} \ge b_0\left( 2 - \frac{46}{25}\right) a_n^{(-\alpha )} =\frac{4}{25}b_0{a}^{(-\alpha )}_n \ge 0,\,n\ge 3. \end{aligned}$$

Step 2) Prove $c_n> 0$ for $n=0,1,2$. Obviously, $c_0= 2b_0 - {b}_0\ge b_0 >0$ and

$$\begin{aligned} c_1= (2-\alpha ) b_0- b_1 \ge \left( 2 -\alpha -\frac{\alpha +\alpha ^2}{2+3\alpha }\right) b_0 =\frac{4(1-\alpha ^2) + 3\alpha }{2+3\alpha }b_0>0, \end{aligned}$$

where we used (8.16). By (8.5) and (8.9), we obtain

$$\begin{aligned} \begin{aligned} b_2/b_0 = \frac{\alpha ^2\theta }{1+\theta \alpha }f_1(\theta ) +\frac{\alpha (1+\alpha )}{2}\left( \frac{1-2\theta }{3-2\theta }\right) ^2 \le \frac{\alpha ^2}{2+\alpha } \frac{1+\alpha }{2+3\alpha } +\frac{\alpha (1+\alpha )}{2}\frac{1}{9}. \end{aligned} \end{aligned}$$

From the above inequality and (8.16), we have

$$\begin{aligned} c_2/b_0 =&{a}^{(-\alpha )}_2 -\left( ({b}_1/b_0)\alpha +{b}_2/b_0\right) \\ \ge&\frac{\alpha (1+\alpha )}{2}-\left( \frac{\alpha ^2(1+\alpha )}{2+3\alpha } +\frac{\alpha ^2}{2+\alpha } \frac{1+\alpha }{2+3\alpha } + \frac{\alpha (1+\alpha )}{2} \frac{1}{9}\right) \\ =&{\alpha (1+\alpha )}\left( \frac{4}{9} - \frac{\alpha }{2+3\alpha } -\frac{\alpha }{(2+\alpha )(2+3\alpha )}\right) \\ \ge&{\alpha (1+\alpha )}\left( \frac{4}{9} - \frac{1}{5} -\frac{1}{15}\right) =\frac{8\alpha (1+\alpha )}{45}>0. \end{aligned}$$

The proof is complete. $\square$

Appendix B. Proofs of Lemmas 3 and 7

Proof of Lemma 3

For $\sigma \ge 0$, (3.4) follows from

$$\begin{aligned} \sum _{j=1}^{n-1}(n-j)^{-\alpha -1}j^{\sigma } \le n^{\sigma }\sum _{j=1}^{n-1}(n-j)^{-\alpha -1} \lesssim n^{\sigma } \sum _{j=1}^{\infty }j^{-\alpha -1} \lesssim n^{\sigma }. \end{aligned}$$

Next, we prove (3.4) for $\sigma <0$.

For $n\ge 2$, there exists $j_n=\lceil n/2 \rceil$ and $x_0=j_n/n \in (0,1)$ such that

$$\begin{aligned} \sum _{j=1}^{n-1}(n-j)^{-\alpha -1}j^{\sigma } =&\sum _{j=1}^{j_n}(n-j)^{-\alpha -1}j^{\sigma } + \sum _{j=j_n+1}^{n-1}(n-j)^{-\alpha -1}j^{\sigma }\\ \le&\sum _{j=1}^{j_n}(n-j_n)^{-\alpha -1}j^{\sigma } + \sum _{j=j_n+1}^{n-1}(n-j)^{-\alpha -1}j_n^{\sigma }\\ \lesssim&n^{-\alpha -1}\sum _{j=1}^{n-1}j^{\sigma } + n^{\sigma } \sum _{j=1}^{n-1}j^{-\alpha -1}. \end{aligned}$$

Using $\sum _{j=1}^{n-1}j^{-\alpha -1}\lesssim 1$ and $\sum _{j=1}^{n-1}j^{\sigma } \lesssim n^{\sigma +1}\log (n)$ completes the proof of (3.4).

By $0\le a_{n}^{(-\alpha )}\lesssim n^{\alpha -1}$, one has

$$\begin{aligned} \sum _{j=1}^{n}a_{n-j}^{(-\alpha )}j^{\sigma }\lesssim n^{\sigma }+ \sum _{j=1}^{n-1} {(n-j)}^{\alpha -1}j^{\sigma }. \end{aligned}$$

Repeating the proof of (3.4) finishes the proof of (3.5). The proof is completed. $\square$

Proof of Lemma 7

The condition (2.5) and Lemma 1 yield the following linear system

$$\begin{aligned} \begin{aligned} \sum _{j=1}^mw^{(m)}_{n,j}j^{\sigma _k}&= \frac{\varGamma (\sigma _k+1)}{\varGamma (\sigma _k+1-\alpha )}n^{\sigma _k-\alpha } -\sum _{j=1}^n\omega _{n-j}^{(\alpha )}j^{\sigma _k}\\&=O(n^{-\alpha -1}) + O(n^{\sigma _k-\alpha -p}), \qquad 1\le k \le m, \end{aligned}\end{aligned}$$

which leads to

$$\begin{aligned} |w^{(m)}_{n,k}|\lesssim n^{-\alpha -1} + n^{\sigma _m-p-\alpha },\quad 1\le k \le m. \end{aligned}$$

(8.20)

Combining (3.8), Lemma 3, $\omega _n^{(\alpha )}=O(n^{-\alpha -1})$, (2.20), and (8.20) leads to

$$\begin{aligned} |W^{(m)}_{n,k}| \lesssim n^{\max \{-\alpha -1,\sigma _m-p-\alpha \}},\quad 1\le k \le m. \end{aligned}$$

(8.21)

Combining (3.2), (3.5), and (8.21) yields (3.15), which ends the proof. $\square$

Appendix C. Proof of Theorem 3

Proof

We show a sketch of the proof. Let $\theta ^n={}_Fu_{h}^n-u_h^n$. By (5.6), (2.15), and $\theta ^n=\varepsilon _{n}=0$ for $0\le n \le n_0-1$, we obtain

$$\begin{aligned} \begin{aligned} \frac{1}{\tau ^{\alpha }}\sum _{j=n_0}^n\omega _{n-j}^{(\alpha )}(\theta ^j,v) +(\nabla \theta ^n,\nabla v)&=\left( P_{h}\left( f({}_Fu^{n}_h)-f(u^{n}_h)\right) ,v\right) \\&\quad -\frac{1}{\tau ^{\alpha }}\sum _{j=1}^{n-n_0}\varepsilon _{n-j}\omega _{n-j}^{(\alpha )}(\theta ^j + u_h^j-u_h^0,v). \end{aligned}\nonumber \\ \end{aligned}$$

(8.22)

Similar to (3.19), we can obtain the equivalent form of (8.22) as

$$\begin{aligned} \begin{aligned}&({\mathcal {A}}^{\alpha ,n_0-1}_{\tau } \theta ^n,v) + ({\mathcal {B}}^{\alpha ,n_0-1}\nabla \theta ^j,\nabla v) \\&\quad = \big ( {\mathcal {B}}^{\alpha ,n_0-1}{\widetilde{F}}^n ,v\big ) -\sum _{j=n_0}^{n} {\widetilde{b}}_{n-j}(\theta ^j,v)-(H^n,v), \end{aligned} \end{aligned}$$

(8.23)

where ${\widetilde{F}}^n=f({}_Fu^{n}_h)-f(u^{n}_h))$, and

$$\begin{aligned} \begin{aligned} {\widetilde{b}}_n= \frac{1}{\tau ^{\alpha }}\sum _{j=n_0}^nb_{n-j}\varepsilon _{j}\omega _{j}^{(\alpha )},\quad H^n=\sum _{k=n_0}^{n} {b}_{n-k}\sum _{j=1}^{k-n_0} \varepsilon _{k-j}\omega _{k-j}^{(\alpha )}(u_h^j-u_h^0). \end{aligned} \end{aligned}$$

By $\omega _{n}^{(\alpha )}=O(n^{-\alpha -1})$ and (3.4), we can easily obtain

$$\begin{aligned} |{\widetilde{b}}_n | \lesssim \varepsilon n^{-\alpha -1}. \end{aligned}$$

By the boundedness of $\Vert u_h^n\Vert$, $b_n=O(n^{-\alpha -1})$, and $\omega _n^{(\alpha )}=O(n^{-\alpha -1})$, we derive

$$\begin{aligned} \Vert H^n\Vert \lesssim \sum _{k=n_0}^{n} |{b}_{n-k}|\sum _{j=1}^{k} |\varepsilon _{k-j}\omega _{k-j}^{(\alpha )} | \lesssim \varepsilon \sum _{j=n_0}^{n} |{b}_{n-k}|\lesssim \varepsilon . \end{aligned}$$

Following the proof of Theorem 2, we can easily arrive at (5.8), the details are omitted. The proof is complete. $\square$