First, we introduce a generalized finite difference approximation to discretize the time-fractional derivative. Let \(t_{k}: = k \tau, k=0,1,\ldots,K \), where \(\tau= \frac{T}{K}\) is the time step. To motivate the construction of the scheme, we define the sequence \(\{a_{j}\}|_{j=0}^{K}\) as \(a_{j}=\frac{\tau ^{1-\alpha}}{1-\alpha} ((j+1)^{1-\alpha}-j^{1-\alpha} )\) and introduce the following lemmas [17].
Lemma 2.1
Let
\(g\in C^{2}[0, t_{k}]\)
and
\(0< \alpha< 1\). Then
$$\begin{aligned} & \Biggl\vert \int_{0}^{t_{k}}\frac{g'(t)}{(t_{k}-t)^{\alpha}}\,dt-\dfrac {1}{\tau } \Biggl[a_{0}g(t_{k})- \sum_{j=1}^{k-1}(a_{k-j-1}-a_{k-j})g(t_{j})-a_{k-1}g(t_{0}) \Biggr] \Biggr\vert \\ &\quad \leq \frac{1}{1-\alpha} \biggl(\frac{1-\alpha}{12}+\frac {2^{2-\alpha }}{2-\alpha}- \bigl(1+2^{-\alpha} \bigr) \biggr)\max_{0\leq t \leq t_{k}} \bigl\vert g''(t) \bigr\vert \tau ^{2-\alpha}. \end{aligned}$$
Lemma 2.2
For any
\(G=\{G_{1}, G_{2}, G_{3},\dots\}\)
and
q, we have
$$\begin{aligned} &\sum_{n=1}^{N} \Biggl(a_{0}G_{n}- \sum_{k=1}^{n-1}(a_{n-k-1}-a_{n-k})G_{k}-a_{n-1}q \Biggr)G_{n}\\ &\quad\geq\frac {t_{N}^{-\alpha}}{2}\tau\sum _{n=1}^{N}G_{n}^{2}- \frac {t_{N}^{1-\alpha }}{2(1-\alpha)}q^{2}. \end{aligned}$$
To motivate the construction of the time-discrete scheme, we use the following functions:
$$ \textstyle\begin{cases} v(x,t)=\frac{\partial u(x,t)}{\partial t}, \\ w(x,t)=\frac{1}{\Gamma(1-\alpha)} \int_{0}^{t}\frac{\partial v(x,s)}{\partial s}\frac{ds}{(t-s)^{\alpha}}, \\ z(x,t)=\frac{1}{\Gamma(1-\alpha)} \int_{0}^{t}\frac{\partial u(x,s)}{\partial s}\frac{ds}{(t-s)^{\alpha}}. \end{cases} $$
(2.1)
Introduce the following notation:
$$u^{k-1/2}(x)=\frac{1}{2} \bigl(u^{k}(x)+u^{k-1}(x) \bigr), \qquad \delta _{t}u^{k-1/2}(x)=\frac{1}{\tau} \bigl(u^{k}(x)-u^{k-1}(x) \bigr). $$
Then we have
$$\begin{aligned} & w^{k-1/2}(x)+z^{k-1/2}(x)+\beta u^{k-1/2}(x)- \lambda\frac{\partial^{2} u^{k-1/2}(x)}{\partial x^{2}}=f^{k-1/2}(x,t), \end{aligned}$$
(2.2)
$$\begin{aligned} & v^{k-1/2}(x)=\delta_{t} u^{k-1/2}(x)+r_{1}^{k-1/2}(x), \end{aligned}$$
(2.3)
and there exists a constant \(C_{1}\) such that \(|r_{1}^{k-1/2}|\leq C_{1}\tau^{2}\) for all \(1\leq k\leq K\).
By Lemma 2.1 we have
$$\begin{aligned} &w^{k-1/2}(x) \\ &\quad=\frac{1}{\Gamma(1-\alpha)\tau} \Biggl(a_{0}v^{k-1/2}(x)- \sum _{j=1}^{k-1}(a_{k-j-1}-a_{k-j})v^{j-1/2}(x)-a_{k-1}v^{0}(x) \Biggr)+r_{2}^{k-1/2}, \end{aligned}$$
(2.4)
$$\begin{aligned} &z^{k-1/2}(x) \\ &\quad=\frac{1}{\Gamma(1-\alpha)\tau} \Biggl(a_{0}u^{k-1/2}(x)- \sum _{j=1}^{k-1}(a_{k-j-1}-a_{k-j})u^{j-1/2}(x)-a_{k-1}u^{0}(x) \Biggr)+r_{3}^{k-1/2}, \end{aligned}$$
(2.5)
and there exist constants \(C_{2}\) and\(C_{3}\) such that \(|r_{2}^{k-1/2}|\leq C_{2}\tau^{2-\alpha}\) and \(|r_{3}^{k-1/2}|\leq C_{3}\tau^{2-\alpha}\).
From (2.3)-(2.5) we obtain
$$\begin{aligned} &\frac{1}{\Gamma(1-\alpha)}\frac{1}{\tau} \Biggl(a_{0} \bigl(\delta _{t}u^{k-1/2}(x)+u^{k-1/2}(x) \bigr)- \sum_{j=1}^{k-1}(a_{k-j-1}-a_{k-j}) \bigl(\delta _{t}u^{j-1/2}(x)+u^{j-1/2}(x) \bigr) \\ &\qquad -a_{k-1} \bigl(u_{1}(x)+u_{0}(x) \bigr) \Biggr)+ \beta u^{k-1/2}(x)-\lambda \frac{\partial^{2}u^{k-1/2}(x)}{\partial x^{2}} \\ &\quad =f^{k-1/2}+R^{k-1/2},\quad k=1,2,\ldots,K, \end{aligned}$$
(2.6)
where
$$R^{k-1/2}=- \Biggl\{ \frac{1}{\Gamma(1-\alpha)\tau} \Biggl(a_{0}r_{1}^{k-1/2}- \sum_{j=1}^{k-1}(a_{k-j-1}-a_{k-j})r_{1}^{j-1/2} \Biggr)+r_{2}^{k-1/2}+r_{3}^{k-1/2} \Biggr\} $$
with
$$\begin{aligned} \bigl\vert R^{k-1/2} \bigr\vert &\leq \frac{1}{\Gamma(1-\alpha)\tau} \Biggl(a_{0}+ \sum_{j=1}^{k-1}(a_{k-j-1}-a_{k-j}) \Biggr)C_{1}\tau^{2}+C_{2}\tau ^{2-\alpha }+C_{3}\tau^{2-\alpha} \\ &=\frac{1}{\Gamma(1-\alpha)\tau}(2a_{0}-a_{k-1})C_{1}\tau ^{2}+C_{2}\tau ^{2-\alpha}+C_{3} \tau^{2-\alpha} \\ &\leq \biggl(\frac{2C_{1}}{\Gamma(2-\alpha)}+C_{2}+C_{3} \biggr)\tau ^{2-\alpha}. \end{aligned}$$
(2.7)
Dropping the truncation error \(R^{k-1/2}\) in (2.6), we can easily get the the variation (weak) formulation of (1.1): Find \(u^{k}(x)\in H_{0}^{1}(\Omega)\) such that, for all \(v\in H_{0}^{1}(\Omega)\),
$$\begin{aligned} &\frac{1}{\Gamma(1-\alpha)\tau} \Biggl(a_{0} \bigl( \delta _{t}u^{k-1/2}(x)+u^{k-1/2}(x),v(x) \bigr)-a_{k-1} \bigl(u_{1}(x)+u_{0}(x),v(x) \bigr) \\ &\qquad{}- \sum_{j=1}^{k-1}(a_{k-j-1}-a_{k-j}) \bigl(\delta _{t}u^{j-1/2}(x)+u^{j-1/2}(x),v(x) \bigr) \Biggr) \\ &\qquad{}+\beta \bigl(u^{k-1/2}(x),v(x) \bigr)+\lambda \biggl(\frac{\partial u^{k-1/2}(x)}{\partial x}, \frac{\partial v(x)}{\partial x} \biggr) \\ &\quad= \bigl(f^{k-1/2}(x),v(x) \bigr). \end{aligned}$$
(2.8)
For the semidiscrete problem, we have the following result.
Theorem 2.1
The semidiscrete problem (2.8) is unconditionally stable in the sense that, for all
\(\tau \geq 0\),
$$\bigl\Vert u^{n} \bigr\Vert _{1}+2\tau\sum _{k=1}^{n} \bigl\Vert u^{k-1/2} \bigr\Vert _{1}^{2}\leq C \Bigl( \Vert u_{1} \Vert ^{2}+ \Vert u_{0} \Vert ^{2}+ \max _{1\leq k\leq K} \bigl\Vert f^{k-1/2} \bigr\Vert ^{2} \Bigr), $$
where
\(n=1,2,\ldots,K\), and
\(C=\frac{\max\{\beta, \gamma\}}{\min\{ \beta , \gamma\}}+\frac{2T^{1-\alpha}}{\min\{\beta, \gamma\}\Gamma (2-\alpha )}+ \frac{\Gamma(1-\alpha)T^{1+\alpha}}{\min\{\beta, \gamma\}}\)
is a constant.
Proof
Taking \(v = \delta_{t} u^{k-1/2}+u^{k-1/2}\) in (2.8), we have
$$\begin{aligned} &\frac{1}{\Gamma(1-\alpha)\tau} \Biggl(a_{0} \bigl( \delta _{t}u^{k-1/2}+u^{k-1/2}, \delta _{t}u^{k-1/2}+u^{k-1/2} \bigr)-a_{k-1} \bigl(u_{1}+u_{0},\delta _{t}u^{k-1/2}+u^{k-1/2} \bigr) \\ &\qquad{}- \sum_{j=1}^{k-1}(a_{k-j-1}-a_{k-j}) \bigl(\delta _{t}u^{j-1/2}+u^{j-1/2}, \delta_{t}u^{k-1/2}+u^{k-1/2} \bigr) \Biggr) \\ &\qquad{}+\beta \bigl(u^{k-1/2},\delta_{t}u^{k-1/2}+u^{k-1/2} \bigr)+\lambda \biggl(\frac{\partial u^{k-1/2}}{\partial x },\frac{\partial\delta _{t}u^{k-1/2}+u^{k-1/2}}{\partial x} \biggr) \\ &\quad = \bigl(f^{k-1/2},\delta_{t}u^{k-1/2}+u^{k-1/2} \bigr). \end{aligned}$$
(2.9)
We first sum both sides of (2.9) for k from 1 to n, and then, using Lemma 2.2 and the Cauchy-Schwarz inequality for the first term of the left-hand side of (2.9), we obtain
$$\begin{aligned} &\sum_{k=1}^{n} \frac{1}{\Gamma(1-\alpha)\tau} \Biggl(a_{0} \bigl\Vert \delta _{t}u^{k-1/2}+u^{k-1/2} \bigr\Vert ^{2}-a_{k-1} \bigl(u_{1}+u_{0},\delta _{t}u^{k-1/2}+u^{k-1/2} \bigr) \\ &\qquad{}-\sum_{j=1}^{k-1}(a_{k-j-1}-a_{k-j}) \bigl(\delta _{t}u^{j-1/2}+u^{j-1/2}, \delta_{t}u^{k-1/2}+u^{k-1/2} \bigr) \Biggr) \\ &\quad \geq \frac{1}{\Gamma(1-\alpha)\tau}\sum_{k=1}^{n} \Biggl(a_{0} \bigl\Vert \delta _{t}u^{k-1/2}+u^{k-1/2} \bigr\Vert -\sum_{j=1}^{k-1}(a_{k-j-1}-a_{k-j}) \bigl\Vert \delta _{t}u^{j-1/2}+u^{j-1/2} \bigr\Vert \\ &\qquad{}-a_{k-1} \Vert u_{1}+u_{0} \Vert \Biggr) \bigl\Vert \delta_{t}u^{k-1/2}+u^{k-1/2} \bigr\Vert \\ &\quad \geq\frac{t_{n}^{-\alpha}}{2\Gamma(1-\alpha)}\sum_{k=1}^{n} \bigl\Vert \delta _{t}u^{k-1/2}+u^{k-1/2} \bigr\Vert ^{2}-\frac{t_{n}^{1-\alpha}}{2\tau\Gamma (2-\alpha)} \Vert u_{1}+u_{0} \Vert ^{2}. \end{aligned}$$
(2.10)
The third term can be written as
$$\begin{aligned} &\lambda\sum_{k=1}^{n} \biggl(\frac{\partial u^{k-1/2}}{\partial x}, \frac{\partial\delta_{t}u^{k-1/2}+u^{k-1/2}}{\partial x} \biggr) \\ &\quad =\frac{\lambda}{2\tau}\sum _{k=1}^{n} \bigl( \bigl\vert u^{k} \bigr\vert _{1}^{2}- \bigl\vert u^{k-1} \bigr\vert _{1}^{2} \bigr)+\lambda\sum_{k=1}^{n} \bigl\vert u^{k-1/2} \bigr\vert _{1}^{2} \\ &\quad =\frac{\lambda}{2\tau} \bigl( \bigl\vert u^{n} \bigr\vert _{1}^{2}- \Vert u_{1} \Vert ^{2} \bigr)+\lambda\sum_{k=1}^{n} \bigl\vert u^{k-1/2} \bigr\vert _{1}^{2}. \end{aligned}$$
(2.11)
Similarly, the second term can be written as
$$\begin{aligned} &\beta \bigl(u^{k-1/2}(x),\delta_{t}u^{k-1/2}+u^{k-1/2} \bigr) \\ &\quad=\frac {\beta }{2\tau} \bigl( \bigl\Vert u^{n} \bigr\Vert ^{2}- \Vert u_{0} \Vert ^{2} \bigr)+\beta \sum_{k=1}^{n} \bigl\Vert u^{k-1/2} \bigr\Vert ^{2}. \end{aligned}$$
(2.12)
Using the Young inequality for the right-hand side of (2.9), we have
$$\begin{aligned} & \sum_{k=1}^{n} \bigl(f^{k-1/2},\delta_{t}u^{k-1/2}+u^{k-1/2} \bigr) \\ &\quad\leq\sum_{k=1}^{n} \biggl( \frac{\Gamma(1-\alpha)}{2 t_{n}^{-\alpha }} \bigl\Vert f^{k-1/2} \bigr\Vert ^{2}+ \frac{t_{n}^{-\alpha}}{2\Gamma(1-\alpha )} \bigl\Vert \delta _{t}u^{k-1/2}+u^{k-1/2} \bigr\Vert ^{2} \biggr). \end{aligned}$$
(2.13)
From (2.10)-(2.13) we get the following relation:
$$\begin{aligned} &\frac{t_{n}^{-\alpha}}{2\Gamma(1-\alpha)}\sum_{k=1}^{n} \bigl\Vert \delta _{t}u^{k-1/2}+u^{k-1/2} \bigr\Vert ^{2}-\frac{t_{n}^{1-\alpha}}{2\tau\Gamma (2-\alpha)} \Vert u_{1}+u_{0} \Vert ^{2} \\ &\qquad{}+\frac{\beta}{2\tau} \bigl( \bigl\Vert u^{n} \bigr\Vert ^{2}- \Vert u_{0} \Vert ^{2} \bigr)+\beta \sum_{k=1}^{n} \bigl\Vert u^{k-1/2} \bigr\Vert ^{2}+\frac{\lambda}{2\tau } \bigl( \bigl\vert u^{n} \bigr\vert _{1}^{2}- \Vert u_{1} \Vert ^{2} \bigr)+\lambda\sum _{k=1}^{n} \bigl\vert u^{k-1/2} \bigr\vert _{1}^{2} \\ &\quad \leq\sum_{k=1}^{n} \biggl( \frac{\Gamma(1-\alpha)}{2 t_{n}^{-\alpha }} \bigl\Vert f^{k-1/2} \bigr\Vert ^{2}+ \frac{t_{n}^{-\alpha}}{2\Gamma(1-\alpha )} \bigl\Vert \delta _{t}u^{k-1/2}+u^{k-1/2} \bigr\Vert ^{2} \biggr). \end{aligned}$$
(2.14)
Denoting \(A:=\max\{\beta, \gamma\}\) and \(B:=\min\{\beta,\gamma\} \), we have
$$\begin{aligned} &\frac{B}{2\tau} \bigl\Vert u^{n} \bigr\Vert _{1}^{2}+B\sum_{k=1}^{n} \bigl\Vert u^{k-1/2} \bigr\Vert _{1}^{2} \\ &\quad \leq\frac{A}{2\tau} \bigl( \Vert u_{1} \Vert ^{2}+ \Vert u_{0} \Vert ^{2} \bigr)+\frac {t_{n}^{1-\alpha }}{2\tau\Gamma(2-\alpha)} \Vert u_{1}+u_{0} \Vert ^{2}+\frac{\Gamma(1-\alpha)}{2 t_{n}^{-\alpha}}\sum _{k=1}^{n} \bigl\Vert f^{k-1/2} \bigr\Vert ^{2}. \end{aligned}$$
(2.15)
Multiplying both sides of this inequality at \(\frac{2\tau}{B}\), we obtain
$$\begin{aligned} & \bigl\Vert u^{n} \bigr\Vert _{1}^{2}+2\tau\sum_{k=1}^{n} \bigl\Vert u^{k-1/2} \bigr\Vert _{1}^{2} \\ &\quad \leq \frac{t_{n}^{1-\alpha}}{B \Gamma(2-\alpha )} \Vert u_{1}+u_{0} \Vert ^{2}+\frac{A}{B} \bigl( \Vert u_{1} \Vert ^{2}+ \Vert u_{0} \Vert ^{2} \bigr)+ \frac {\tau\Gamma(1-\alpha)t_{n}^{\alpha}}{B}\sum_{k=1}^{n} \bigl\Vert f^{k-1/2} \bigr\Vert ^{2} \\ &\quad \leq \biggl(\frac{A}{B}+\frac{2t_{n}^{1-\alpha}}{B\Gamma(2-\alpha)}+ \frac{T\Gamma(1-\alpha)t_{n}^{\alpha}}{B} \biggr) \Bigl( \Vert u_{1} \Vert ^{2}+ \Vert u_{0} \Vert ^{2}+ \max_{1\leq k\leq n} \bigl\Vert f^{k-1/2} \bigr\Vert ^{2} \Bigr) \\ &\quad \leq \biggl(\frac{A}{B}+\frac{2T^{1-\alpha}}{B\Gamma(2-\alpha)}+ \frac {\Gamma(1-\alpha)T^{1+\alpha}}{B} \biggr) \Bigl( \Vert u_{1} \Vert ^{2}+ \Vert u_{0} \Vert ^{2}+ \max_{1\leq k\leq K} \bigl\Vert f^{k-1/2} \bigr\Vert ^{2} \Bigr). \end{aligned}$$
(2.16)
The proof is completed. □
Theorem 2.2
Let
\(u(x,t)\) (\(\{u^{k}=u(t_{k})\}_{k=0}^{K}\)) be the exact solution of (1.1), and
\(\{u_{\varsigma}^{k}\}_{k=0}^{K}\)
be the solution of variation (weak) formulation (2.8). Then we have the following error estimate:
$$\bigl\Vert u^{n}-u_{\varsigma}^{n} \bigr\Vert _{1}^{2}+2\tau\sum_{k=1}^{n} \bigl\Vert u^{n-1/2}-u_{\varsigma}^{n-1/2} \bigr\Vert _{1}^{2}\leq C \bigl(\tau ^{2-\alpha} \bigr)^{2}. $$
Proof
Denoting \(\rho^{k}=u^{k}-u_{\varsigma}^{k}\), we obtain
$$\begin{aligned} &\frac{1}{\Gamma(1-\alpha)\tau} \Biggl(a_{0} \bigl( \delta_{t}\rho ^{k-1/2}+ \rho^{k-1/2},v \bigr)- \sum _{j=1}^{k-1}(a_{k-j-1}-a_{k-j}) \bigl(\delta_{t}\rho^{j-1/2}+\rho^{j-1/2},v \bigr) \Biggr) \\ &\qquad{}+\beta \bigl(u^{k-1/2},v \bigr)+\lambda \biggl(\frac{\partial\rho ^{k-1/2}}{\partial x}, \frac{\partial v}{\partial x} \biggr) \\ &\quad = \bigl(R^{k-1/2},v \bigr). \end{aligned}$$
(2.17)
Taking \(v=\delta_{t}\rho^{k-1/2}+\rho^{k-1/2}\) in (2.17) yields
$$\begin{aligned} &\frac{1}{\Gamma(1-\alpha)\tau} \Biggl(a_{0} \bigl( \delta_{t}\rho ^{k-1/2}+ \rho^{k-1/2}, \delta_{t}\rho^{k-1/2}+\rho^{k-1/2} \bigr) \\ &\qquad{}- \sum_{j=1}^{k-1}(a_{k-j-1}-a_{k-j}) \bigl(\delta_{t}\rho ^{j-1/2}+\rho ^{j-1/2}, \delta_{t}\rho^{k-1/2}+\rho^{k-1/2} \bigr) \Biggr) \\ &\qquad{}+\beta \bigl(u^{k-1/2},\delta_{t}\rho^{k-1/2}+ \rho^{k-1/2} \bigr)+\lambda \biggl(\frac{\partial\rho^{k-1/2}}{\partial x},\frac{\partial\delta _{t}\rho ^{k-1/2}+\rho^{k-1/2}}{\partial x} \biggr) \\ &\quad = \bigl(R^{k-1/2},\delta_{t}\rho^{k-1/2}+ \rho^{k-1/2} \bigr). \end{aligned}$$
(2.18)
Summing up for k from 1 to n and using Lemma 2.2, we obtain
$$\begin{aligned} & \sum_{k=1}^{n} \frac{1}{\tau\Gamma(1-\alpha)} \Biggl\{ a_{0} \bigl(\delta _{t} \rho^{k-1/2}+\rho^{k-1/2},\delta_{t}\rho^{k-1/2}+ \rho ^{k-1/2} \bigr) \\ &\qquad{}- \sum_{j=1}^{k-1}(a_{k-j-1}-a_{k-j}) \bigl(\delta_{t}\rho^{j-1/2}+\rho ^{j-1/2}, \delta_{t}\rho^{k-1/2}+\rho^{k-1/2} \bigr) \Biggr\} \\ &\quad \geq \frac{t_{n}^{-\alpha}}{2\Gamma(1-\alpha)}\sum_{k=1}^{n} \bigl\Vert \delta _{t}\rho^{k-1/2}+\rho^{k-1/2} \bigr\Vert ^{2}. \end{aligned}$$
(2.19)
In addition, similarly to the proof of Theorem 2.1, we can write the following relations:
$$\begin{aligned} &\sum_{k=1}^{n} \bigl( \rho^{k-1/2},\delta_{t}\rho^{k-1/2}+\rho ^{k-1/2} \bigr)=\frac {1}{2\tau} \bigl\Vert \rho^{n} \bigr\Vert ^{2}+\sum_{k=1}^{n} \bigl\Vert \rho^{k-1/2} \bigr\Vert ^{2}, \end{aligned}$$
(2.20)
$$\begin{aligned} & \sum_{k=1}^{n} \biggl( \frac{\partial\rho^{k-1/2}}{\partial x},\frac {\partial \delta_{t}\rho^{k-1/2}+\rho^{k-1/2}}{\partial x} \biggr)=\frac{1}{2\tau } \bigl\vert \rho ^{n} \bigr\vert _{1}^{2}+\sum _{k=1}^{n} \bigl\vert \rho^{k-1/2} \bigr\vert _{1}^{2} \end{aligned}$$
(2.21)
and
$$\begin{aligned} \sum_{k=1}^{n} \bigl(R^{k-1/2},\delta_{t}\rho^{k-1/2}+ \rho^{k-1/2} \bigr)\leq{} &\frac{t_{n}^{-\alpha}}{2\Gamma(1-\alpha)}\sum _{k=1}^{n} \bigl\vert \delta _{t}\rho ^{k-1/2}+\rho^{k-1/2} \bigr\vert _{0}^{2} \\ &{}+\frac{t_{n}^{\alpha}\Gamma(1-\alpha)}{2}\sum_{k=1}^{n} \bigl\Vert R^{k-1/2} \bigr\Vert _{0}^{2}. \end{aligned}$$
(2.22)
From (2.19)-(2.22) we obtain the following relation:
$$\begin{aligned} &\frac{\beta}{2\tau} \bigl\Vert \rho^{n} \bigr\Vert ^{2}+\beta\sum_{k=1}^{n} \bigl\Vert \rho ^{k-1/2} \bigr\Vert ^{2}+\frac{\lambda}{2\tau} \bigl\vert \rho^{n} \bigr\vert _{1}^{2}+\lambda\sum _{k=1}^{n} \bigl\vert \rho^{k-1/2} \bigr\vert _{1}^{2} \\ &\quad\leq\frac{t_{n}^{\alpha}\Gamma (1-\alpha )}{2}\sum _{k=1}^{n} \bigl\Vert R^{k-1/2} \bigr\Vert _{0}^{2}. \end{aligned}$$
(2.23)
From (2.7) we can find a constant \(C':=\frac{2C_{1}}{\Gamma (2-\alpha)}+C_{2}+C_{3}\) such that \(\vert R^{k-1/2} \vert \leq C'\tau^{2-\alpha}\). Similarly to (2.16), we obtain
$$\begin{aligned} \bigl\Vert \rho^{n} \bigr\Vert _{1}^{2}+2\tau\sum_{k=1}^{n} \bigl\Vert \rho ^{k-1/2} \bigr\Vert _{1}^{2}&\leq \frac{t_{n}^{\alpha}\Gamma(1-\alpha)}{B}\tau\sum_{k=1}^{n} \bigl\Vert R^{k-1/2} \bigr\Vert _{0}^{2} \\ &\leq\frac{T^{1+\alpha}\Gamma(1-\alpha)}{B} \max_{1\leq k\leq n} \bigl\Vert R^{k-1/2} \bigr\Vert _{0}^{2} \\ &\leq\frac{T^{1+\alpha}\Gamma(1-\alpha)}{B}{C^{\prime}}^{2} \bigl(\tau^{2-\alpha} \bigr)^{2}. \end{aligned}$$
(2.24)
Letting \(C=\frac{T^{1+\alpha}\Gamma(1-\alpha)}{B}{C^{\prime}}^{2}\), the theorem is proved. □