Some useful lemmas
Lemma 3.1
([13])
If
\(u,v\)
are any two grid functions in
\(\nu_{h}\), then there exists a linear operator
\(\Lambda_{h}^{\alpha}=h^{-\frac{\alpha}{2}}G^{\frac{1}{2}}\)
such that
$$\bigl(\delta_{h}^{\alpha}u,v\bigr)=\bigl(h^{-\alpha}Gu,v \bigr)=\bigl(\Lambda_{h}^{\alpha }u,\Lambda_{h}^{\alpha}v \bigr), $$
where
$$\begin{aligned} G= \left ( \begin{matrix} g_{0}^{(\alpha)} & g_{-1}^{(\alpha)} & \ldots g_{-M+2}^{(\alpha)}\\ g_{1}^{(\alpha)} & g_{0}^{(\alpha)} & \ldots g_{-M+3}^{(\alpha)}\\ \vdots& \vdots & \ddots \vdots\\ g_{M-2}^{(\alpha)}& g_{M-3}^{(\alpha)} & \cdots g_{0}^{(\alpha)} \end{matrix} \right ); \end{aligned}$$
\(G^{\frac{1}{2}}\)
is the unique positive definite square root of
G, that is, \((G^{\frac{1}{2}})^{2}=G\). Moreover, we have
$$ \bigl(\delta_{h}^{\alpha}u,u\bigr)=\bigl( \Lambda_{h}^{\alpha}u,\Lambda _{h}^{\alpha}u \bigr)= \bigl\Vert \Lambda_{h}^{\alpha}u \bigr\Vert ^{2}, $$
(3.1)
where
\(\delta_{h}^{\alpha}u=(\delta_{h}^{\alpha}u_{1},\delta _{h}^{\alpha}u_{2},\ldots,\delta_{h}^{\alpha}u_{M-1})^{T}\).
Lemma 3.2
If
\(u^{n}\in\nu_{h}\), then we have
$$\begin{aligned} &\bigl(u_{t\bar{t}}^{n},2 u_{\hat{t}}^{n} \bigr)=\bigl( \bigl\Vert u_{t}^{n} \bigr\Vert ^{2}\bigr)_{\bar{t}}, \end{aligned}$$
(3.2)
$$\begin{aligned} & \bigl(\delta_{h}^{\alpha}\bar{u}^{n},2u_{\hat{t}}^{n} \bigr)= \bigl\Vert \Lambda _{h}^{\alpha} u^{n} \bigr\Vert ^{2}_{\hat{t}}. \end{aligned}$$
(3.3)
Proof
Equality (3.2) can be found in [8], so we omit it here. Equality (3.3) can be proved as follows:
$$\begin{aligned} \bigl(\delta_{h}^{\alpha}\bar{u}^{n},2u_{\hat{t}}^{n} \bigr)={}& \frac {1}{2\tau} \bigl(\delta_{h}^{\alpha}u^{n+1}+ \delta_{h}^{\alpha }u^{n-1},u^{n+1}-u^{n-1} \bigr) \\ ={}& \frac{1}{2\tau}\bigl[\bigl(\delta_{h}^{\alpha}u^{n+1},u^{n+1} \bigr)-\bigl(\delta _{h}^{\alpha}u^{n-1},u^{n-1} \bigr) +\bigl(\delta_{h}^{\alpha}u^{n-1},u^{n+1} \bigr)-\bigl(\delta_{h}^{\alpha }u^{n+1},u^{n-1} \bigr)\bigr] \\ ={}&\frac{1}{2\tau}\bigl[\bigl(\delta_{h}^{\alpha}u^{n+1},u^{n+1} \bigr)-\bigl(\delta _{h}^{\alpha}u^{n-1},u^{n-1} \bigr)\bigr] \\ ={}& \bigl\Vert \Lambda_{h}^{\alpha} u^{n} \bigr\Vert ^{2}_{\hat{t}}. \end{aligned}$$
The proof ends. □
Lemma 3.3
([8])
Let
\(\omega(k)\)
and
\(\rho(k)\)
be nonnegative mesh functions. If
\(C > 0\), \(\rho(k)\)
is nondecreasing, and
$$\omega(k)\leq\rho(k)+C \tau\sum_{l=0}^{k-1} \omega(l) $$
for all
k, then
$$\omega(k)\leq\rho(k)e^{Ck\tau} $$
for all
k.
Lemma 3.4
(Discrete Sobolev inequality [24])
For every
\(\frac{1}{2}<\sigma\leq1\), there exists a constant
\(C =C(\sigma) > 0\)
independent of
\(h > 0\)
such that
$$\begin{aligned} \Vert u \Vert _{l_{h}^{\infty}}\leq C \Vert u \Vert _{H^{\sigma}} \end{aligned}$$
for all
\(u\in l_{h}^{2}\).
Lemma 3.5
(Uniform norm equivalence [25])
For any grid function
\(u=\{u_{j}\}\)
and every
\(1<\alpha\leq2\), we have
$$\begin{aligned} \biggl(\frac{2}{\pi}\biggr)^{\alpha} \vert u \vert _{H^{\alpha/2}}^{2}\leq \bigl(h^{-\alpha }\Delta_{h}^{\alpha}u,u \bigr)\leq \vert u \vert _{H^{\alpha/2}}^{2}. \end{aligned}$$
Lemma 3.6
([26], Lemma 1.4, Ch. 2)
Let
X
be a finite-dimensional Hilbert space with scalar product
\([\cdot,\cdot]\)
and norm
\([\cdot]\), and let
Ϝ
be a continuous mapping from
X
into itself such that
$$\bigl[\digamma(\xi), \xi\bigr] > 0\quad \textit{for any }[\xi] = k > 0. $$
Then there exists
\(\xi\in X\)
with
\([\xi] \leq k \)
such that
Conservation
Theorem 3.1
The scheme (2.9)–(2.11) is conservative in the sense that
$$\varepsilon^{n}=\varepsilon^{n-1}=\cdots= \varepsilon^{0}, $$
where
$$\varepsilon^{n}=\frac{1}{2} \bigl\Vert u_{t}^{n} \bigr\Vert ^{2}+\frac{1}{4} \bigl\Vert \Lambda _{h}^{\alpha} u^{n+1} \bigr\Vert ^{2} + \frac{1}{4} \bigl\Vert \Lambda_{h}^{\alpha}u^{n} \bigr\Vert ^{2}+\frac{h}{2}\sum_{i=0}^{M-1} \bigl[\bigl(1-\cos u_{i}^{n+1}\bigr)+\bigl(1-\cos u_{i}^{n}\bigr)\bigr] $$
is the energy in the discrete sense.
Proof
Making the discrete inner product of (2.9) with \(2 u_{\hat {t}}^{n}\), we get that
$$ \bigl(\bigl(u^{n}\bigr)_{t\bar{t}},2 u_{\hat{t}}^{n}\bigr)+\bigl(\delta_{h}^{\alpha} \bar {u}^{n},2 u_{\hat{t}}^{n}\bigr)=\bigl(\zeta\bigl( \bar{u}^{n}\bigr),2 u_{\hat{t}}^{n}\bigr). $$
(3.4)
Directly computing, we have
$$\begin{aligned} \bigl(\zeta\bigl(\bar{u}^{n}\bigr),2 u_{\hat{t}}^{n}\bigr)={}&\frac{h}{\tau} \sum _{i=0}^{M-1}\bigl(\cos u_{i}^{n+1}- \cos u_{i}^{n-1}\bigr) \\ ={}&-\frac{h}{\tau} \sum_{i=0}^{M-1} \bigl\{ \bigl[\bigl(1-\cos u_{i}^{n+1}\bigr) +\bigl(1-\cos u_{i}^{n}\bigr)\bigr] \\ &{}-\bigl[\bigl(1-\cos u_{i}^{n}\bigr)+\bigl(1-\cos u_{i}^{n-1}\bigr)\bigr]\bigr\} . \end{aligned}$$
(3.5)
Notice that other inner products in (3.4) can be calculated by Lemma 3.2. Substituting (3.2), (3.3), and (3.5) into (3.4), we obtain
$$\begin{aligned} &\bigl( \bigl\Vert u_{t}^{n} \bigr\Vert ^{2} \bigr)_{\bar{t}}+\frac{1}{2\tau} \bigl[\bigl( \bigl\Vert \Lambda_{h}^{\alpha}u^{n+1} \bigr\Vert ^{2}+ \bigl\Vert \Lambda_{h}^{\alpha} u^{n} \bigr\Vert ^{2}\bigr)-\bigl( \bigl\Vert \Lambda_{h}^{\alpha} u^{n} \bigr\Vert ^{2} + \bigl\Vert \Lambda_{h}^{\alpha} u^{n-1} \bigr\Vert ^{2}\bigr)\bigr] \\ &\qquad{}+\frac{h}{\tau} \sum_{i=0}^{M-1} \bigl\{ \bigl[\bigl(1-\cos u_{i}^{n+1}\bigr) +\bigl(1-\cos u_{i}^{n}\bigr)\bigr]-\bigl[\bigl(1-\cos u_{i}^{n}\bigr)+\bigl(1-\cos u_{i}^{n-1} \bigr)\bigr]\bigr\} \\ &\quad=0, \end{aligned}$$
and hence
$$\begin{aligned} \varepsilon^{n}&= \frac{1}{2} \bigl\Vert u_{t}^{n} \bigr\Vert ^{2}+\frac{1}{4} \bigl[\bigl( \bigl\Vert \Lambda_{h}^{\alpha} u^{n+1} \bigr\Vert ^{2}+ \bigl\Vert \Lambda_{h}^{\alpha }u^{n} \bigr\Vert ^{2}\bigr)\bigr] + \frac{h}{2}\sum_{i=0}^{M-1} \bigl[\bigl(1-\cos u_{i}^{n+1}\bigr) +\bigl(1-\cos u_{i}^{n}\bigr)\bigr] \\ &= \frac{1}{2} \bigl\Vert u_{t}^{n-1} \bigr\Vert ^{2}+\frac{1}{4}( \bigl\Vert \Lambda _{h}^{\alpha} u^{n} \bigr\Vert ^{2}+ \bigl\Vert \Lambda_{h}^{\alpha} u^{n-1} \bigr\Vert ^{2} +\frac{h}{2} \sum _{i=0}^{M-1}\bigl[\bigl(1-\cos u_{i}^{n} \bigr)+\bigl(1-\cos u_{i}^{n-1}\bigr)\bigr], \end{aligned}$$
that is,
$$\begin{aligned} \varepsilon^{n}=\varepsilon^{n-1}=\cdots= \varepsilon^{0}. \end{aligned}$$
□
The boundedness, solvability, and convergence
Theorem 3.2
Assume that
\(\{u_{i}^{n}|0 \leq i \leq M, 0 \leq n \leq N\}\)
is a solution of the difference scheme (2.9)–(2.11). Then
$$\begin{aligned} \bigl\Vert u_{t}^{n} \bigr\Vert \leq C,\qquad \bigl\Vert \Lambda_{h}^{\alpha} u^{n} \bigr\Vert \leq C, \qquad \bigl\Vert u^{n} \bigr\Vert \leq C, \qquad \bigl\Vert u^{n} \bigr\Vert _{l_{h}^{\infty}}\leq C. \end{aligned}$$
Here and later, C denotes a generic positive constant; in different places, it may represent different constants.
Proof
By Theorem 3.1 we get
$$\begin{aligned} 2\varepsilon^{n}= \bigl\Vert u_{t}^{n} \bigr\Vert ^{2}+\frac{1}{2} \bigl[\bigl( \bigl\Vert \Lambda_{h}^{\alpha} u^{n+1} \bigr\Vert ^{2}+ \bigl\Vert \Lambda_{h}^{\alpha} u^{n} \bigr\Vert ^{2}\bigr)\bigr] + h \sum_{i=0}^{M-1} \bigl[\bigl(1-\cos u_{i}^{n+1}\bigr) +\bigl(1-\cos u_{i}^{n}\bigr)\bigr]=C. \end{aligned}$$
Notice that
$$\begin{aligned} \bigl(1-\cos u_{i}^{n+1}\bigr) +\bigl(1-\cos u_{i}^{n}\bigr) \geq0. \end{aligned}$$
Then
$$\begin{aligned} \bigl\Vert u_{t}^{n} \bigr\Vert \leq C,\qquad \bigl\Vert \Lambda_{h}^{\alpha} u^{n} \bigr\Vert \leq C. \end{aligned}$$
Moreover,
$$\begin{aligned} \frac{ \Vert u^{k+1} \Vert - \Vert u^{k} \Vert }{\tau} \leq \bigl\Vert u_{t}^{k} \bigr\Vert \leq C. \end{aligned}$$
By summing this equality for \(k=0,1,\ldots,n-1\)
\((n\tau\leq T)\), we get
$$\begin{aligned} \bigl\Vert u^{n} \bigr\Vert \leq n\tau C+ \bigl\Vert u^{0} \bigr\Vert \leq CT+ \bigl\Vert u^{0} \bigr\Vert \leq C. \end{aligned}$$
Combining Lemma 3.1 and Lemma 3.5, we derive
$$\begin{aligned} \bigl\vert u^{n} \bigr\vert _{H^{\alpha/2}}^{2}\leq \biggl(\frac{\pi}{2}\biggr)^{\alpha} \bigl(h^{-\alpha} \Delta_{h}^{\alpha}u^{n},u^{n}\bigr)= \biggl( \frac{\pi}{2}\biggr)^{\alpha}\bigl(h^{-\alpha}Gu^{n},u^{n} \bigr)=\biggl(\frac{\pi }{2}\biggr)^{\alpha} \bigl\Vert \Lambda_{h}^{\alpha} u^{n} \bigr\Vert ^{2} \leq C. \end{aligned}$$
Hence
$$\begin{aligned} \bigl\Vert u^{n} \bigr\Vert _{H^{\alpha/2}}^{2}= \bigl\Vert u^{n} \bigr\Vert ^{2}+ \bigl\vert u^{n} \bigr\vert _{H^{\alpha/2}}^{2} \leq C. \end{aligned}$$
According to Lemma (3.4),
$$\begin{aligned} \bigl\Vert u^{n} \bigr\Vert _{l_{h}^{\infty}}\leq C \bigl\Vert u^{n} \bigr\Vert _{H^{\alpha/2}}\leq C. \end{aligned}$$
Thus we get the desired results. □
For the existence and uniqueness of a solution for the difference scheme (2.9)–(2.11), we have the following theorems.
Theorem 3.3
The difference scheme (2.9)–(2.11) has at least one solution.
Proof
We carry out the proof by mathematical induction. Assume that \(u^{0},u^{1},\ldots, u^{n}\) are given solutions. We aim to prove that there exists yet one \(u^{n+1}\) satisfying the difference scheme. Let \(X=\nu_{h}\) with the scalar product \((\cdot,\cdot)\) defined in (2.1). We define the mapping \(\digamma: X\rightarrow X\) such that
$$\begin{aligned} \bigl(\digamma(\omega),\upsilon\bigr)={}&\biggl(\frac{\omega}{\tau^{2}},\upsilon \biggr)-\biggl(\frac{2}{\tau}u_{t}^{n-1},\upsilon\biggr)+ \biggl(\frac{1}{2}\delta _{h}^{\alpha}\omega,\upsilon\biggr) +\bigl(\delta_{h}^{\alpha}u^{n-1},\upsilon\bigr) \\ &{}+\biggl(\frac{\cos u^{n-1}-\cos(\omega+u^{n-1}) }{\omega},\upsilon \biggr), \quad \forall\omega,\upsilon\in X, \end{aligned}$$
where
$$\begin{aligned} &\frac{\cos u^{n-1}-\cos(\omega+u^{n-1}) }{\omega}=(\chi_{1},\chi _{2},\ldots, \chi_{M-1})^{T}, \\ &\chi_{i}=\frac{\cos u_{i}^{n-1}-\cos(\omega_{i}+u_{i}^{n-1}) }{\omega_{i}},\quad 1\leq i\leq M-1. \end{aligned}$$
Obviously, the mapping Ϝ is continuous. Moreover, we have
$$\begin{aligned} \bigl(\digamma(\omega),\omega\bigr)={}&\frac{ \Vert \omega \Vert ^{2}}{\tau^{2}}-\frac {2}{\tau} \bigl(u_{t}^{n-1},\omega\bigr)+\frac{1}{2}\bigl( \delta_{h}^{\alpha }\omega,\omega\bigr) +\bigl( \delta_{h}^{\alpha}u^{n-1},\omega\bigr) \\ &{}+\biggl(\frac{\cos u^{n-1}-\cos(\omega+u^{n-1}) }{\omega},\omega\biggr). \end{aligned}$$
Applying Young’s inequality and Theorem 3.2, we have
$$\begin{aligned} \frac{2}{\tau}\bigl(u_{t}^{n-1},\omega\bigr)&=\biggl(2 \sqrt{2}u_{t}^{n-1},\frac {1}{\sqrt{2}\tau}\omega\biggr) \\ &\leq 4 \bigl\Vert u_{t}^{n-1} \bigr\Vert ^{2}+ \frac{1}{4}\frac{ \Vert \omega \Vert ^{2}}{\tau ^{2}} \\ &\leq 8\varepsilon^{0}+ \frac{1}{4}\frac{ \Vert \omega \Vert ^{2}}{\tau^{2}} \end{aligned}$$
and
$$\begin{aligned} \bigl(\delta_{h}^{\alpha}u^{n-1},\omega\bigr)&\leq \frac{\tau^{2}}{2} \bigl\Vert \delta_{h}^{\alpha}u^{n-1} \bigr\Vert ^{2}+ \frac{1}{2}\frac{ \Vert \omega \Vert ^{2}}{\tau^{2}} \\ &\leq 4\tau^{2} h^{-\alpha}g_{0}^{(\alpha)} \varepsilon^{0}+\frac {1}{2}\frac{ \Vert \omega \Vert ^{2}}{\tau^{2}}, \end{aligned}$$
where we used the fact that
$$\begin{aligned} \bigl\Vert \delta_{h}^{\alpha}u^{n-1} \bigr\Vert ^{2}&=\bigl(h^{-\alpha}Gu^{n-1},h^{-\alpha }Gu^{n-1} \bigr) \\ &\leq h^{-\alpha} \lambda_{\max}(G) \bigl(h^{-\alpha}G u^{n-1},u^{n-1}\bigr) \\ &\leq 2h^{-\alpha}g_{0}^{(\alpha)} \bigl\Vert \Lambda_{h}^{\alpha}u^{n-1} \bigr\Vert ^{2} \\ &\leq 8h^{-\alpha}g_{0}^{(\alpha)}\varepsilon^{0}. \end{aligned}$$
Noticing that
$$\begin{aligned} \bigl(\delta_{h}^{\alpha}\omega,\omega\bigr)=\bigl( \Lambda_{h}^{\alpha}\omega ,\Lambda_{h}^{\alpha} \omega\bigr)\geq0 \end{aligned}$$
and
$$\begin{aligned} \biggl(\frac{\cos u^{n-1}-\cos(\omega+u^{n-1}) }{\omega},\omega\biggr)=h\sum_{i=1}^{M-1} \bigl(\cos u_{i}^{n-1}-\cos\bigl(\omega_{i}+u_{i}^{n-1} \bigr)\bigr)\leq2L, \end{aligned}$$
where \(L=X_{R}-X_{L}\), we have
$$\begin{aligned} \bigl(\digamma(\omega),\omega\bigr)\geq\frac{1}{4}\frac{ \Vert \omega \Vert ^{2}}{\tau^{2}}- \bigl[\bigl(8+4\tau^{2} h^{-\alpha}g_{0}^{(\alpha )} \bigr)\varepsilon^{0}+2L\bigr]. \end{aligned}$$
Let
$$\begin{aligned} K=2\tau\sqrt{\bigl(8+4\tau^{2} h^{-\alpha}g_{0}^{(\alpha)} \bigr)\varepsilon^{0}+2L}. \end{aligned}$$
Then, for any \(\|\omega\|>K\), we have \((\digamma(\omega),\omega)>0\). By Lemma 3.6 there exists \(\omega\in X\) such that \(\digamma (\omega)=0\). Setting \(\omega=u^{n+1}-u^{n-1}\), we get the desired result. □
Theorem 3.4
The solution of the difference scheme (2.9)–(2.11) is unique.
Proof
Suppose that two sequences \(u^{0},u^{1}, \ldots,u^{n},v\) and \(u^{0},u^{1}, \ldots,u^{n}, v^{\prime}\) both satisfy the difference scheme (2.9)–(2.11). Let \(v^{*}=v-v^{\prime}\). Then we obtain that
$$ \frac{1}{\tau^{2}}v_{i}^{*}+ \delta_{h}^{\alpha}v_{i}^{*}-p_{i}=0,\quad 1\leq i\leq M-1, $$
(3.6)
where \(\{p_{i}\}\) are given by (3.8).
Making the inner product of (3.6) with \(v^{*}\), we obtain that
$$ \biggl(\frac{1}{\tau^{2}}v^{*},v^{*} \biggr)+\bigl(\delta_{h}^{\alpha}v^{*},v^{*} \bigr)-\bigl(p,v^{*}\bigr)=0. $$
(3.7)
Noticing that
$$\begin{aligned} p_{i}&=\frac{\cos v_{i}- \cos u_{i}^{n-1}}{v_{i}- u_{i}^{n-1}}- \frac{\cos v_{i}^{\prime}- \cos u_{i}^{n-1}}{v_{i}^{\prime}- u_{i}^{n-1}} \\ &= \int_{0}^{1}\sin\bigl[\lambda v_{i}^{\prime} +(1- \lambda) u_{i}^{n-1}\bigr]\,d \lambda- \int_{0}^{1}\sin\bigl[\lambda v_{i} +(1- \lambda) u_{i}^{n-1}\bigr]\,d \lambda \\ &= -2 \int_{0}^{1}\cos\biggl[\lambda\frac{v_{i}+v_{i}^{\prime}}{2}+(1- \lambda)u_{i}^{n-1}\biggr] \sin\biggl(\lambda \frac{v_{i}^{*}}{2}\biggr)\,d \lambda, \end{aligned}$$
(3.8)
we have
$$\begin{aligned} \bigl\vert p_{i}v_{i}^{*} \bigr\vert \leq \frac{1}{2}\bigl(v_{i}^{*}\bigr)^{2}, \end{aligned}$$
and therefore
$$\begin{aligned} \bigl\vert \bigl(p,v^{*}\bigr) \bigr\vert \leq\frac{1}{2} \bigl\Vert v^{*} \bigr\Vert ^{2}. \end{aligned}$$
Thus for the right-hand side of (3.7), we get
$$ \biggl(\frac{1}{\tau^{2}}v^{*},v^{*} \biggr)+\bigl(\delta_{h}^{\alpha }v^{*},v^{*} \bigr)-\bigl(p,v^{*}\bigr)\geq\biggl(\frac{1}{\tau^{2}}- \frac{1}{2}\biggr) \bigl\Vert v^{*} \bigr\Vert ^{2}. $$
(3.9)
Then supposing that \(\tau< \sqrt{2}\) and combining (3.7) and (3.9), we derive
$$\begin{aligned} \bigl\Vert v^{*} \bigr\Vert =0,\quad \mbox{i.e., } v=v^{\prime}. \end{aligned}$$
The proof ends. □
Denote
$$\begin{aligned} e_{i}^{n}=U_{i}^{n}-u_{i}^{n},\quad 0\leq i \leq M, 0\leq n \leq N. \end{aligned}$$
Theorem 3.5
Assume that problem (1.2)–(1.4) has a smooth solution and
\(\{u_{i}^{n}| 0\leq i \leq M, 0\leq n \leq N\} \)
is the solution of the finite difference scheme (2.9)–(2.11). If
\(\frac{\tau^{2}}{h^{\alpha}}\leq S\) (\(0< S<+\infty\)), then there exists a positive constant C such that
$$\begin{aligned} \bigl\Vert e^{n} \bigr\Vert _{l_{h}^{\infty}} \leq C \bigl(h^{2}+\tau^{2}\bigr). \end{aligned}$$
Proof
Subtracting (2.9)–(2.11) from (2.6)–(2.8), respectively, we get the following error equations:
$$\begin{aligned} &\bigl(e_{i}^{n}\bigr)_{t\bar{t}}+ \delta_{h}^{\alpha}\bar{e}_{i}^{n}=V \bigl( \bar {e}_{i}^{n}\bigr)+r_{i}^{n},\quad 1\leq i \leq M-1,1\leq n \leq N, \end{aligned}$$
(3.10)
$$\begin{aligned} &e_{0}^{n}=0,\qquad e_{M}^{n} = 0, \quad 0\leq n \leq N, \end{aligned}$$
(3.11)
$$\begin{aligned} &e_{i}^{0}=0,\quad 1\leq i \leq M-1, \end{aligned}$$
(3.12)
where
$$\begin{aligned} V \bigl(\bar{e}_{i}^{n}\bigr)={}& \zeta\bigl(\bar{U}_{i}^{n}\bigr)-\zeta\bigl(\bar {u}_{i}^{n}\bigr) \\ ={}& {-}2 \int_{0}^{1}\cos\biggl[\lambda\frac{U_{i}^{n+1}+u_{i}^{n+1}}{2}+(1- \lambda)\frac{U_{i}^{n-1}+u_{i}^{n-1}}{2}\biggr] \\ & {}\times\sin\biggl[\lambda\frac{U_{i}^{n+1}-u_{i}^{n+1}}{2}+(1- \lambda)\frac{U_{i}^{n-1}-u_{i}^{n-1}}{2} \biggr]\,d \lambda. \end{aligned}$$
(3.13)
Similarly to the proof of Theorem 3.1, making the discrete inner product of (3.10) with \(2 e_{\hat{t}}^{n}\), we get that
$$\begin{aligned} &\bigl( \bigl\Vert e_{t}^{n} \bigr\Vert ^{2}\bigr)_{\bar{t}}+\frac{1}{2\tau} \bigl[\bigl( \bigl\Vert \Lambda_{h}^{\alpha} e^{n+1} \bigr\Vert ^{2}+ \bigl\Vert \Lambda_{h}^{\alpha} e^{n} \bigr\Vert ^{2}\bigr)-\bigl( \bigl\Vert \Lambda_{h}^{\alpha} e^{n} \bigr\Vert ^{2}+ \bigl\Vert \Lambda_{h}^{\alpha} e^{n-1} \bigr\Vert ^{2}\bigr)\bigr] \\ &\quad =\bigl(V\bigl(\bar{e}^{n}\bigr)+r^{n},2 e_{\hat{t}}^{n}\bigr). \end{aligned}$$
(3.14)
By summing this equality for \(n=1,2,\ldots,k\)
\((k\tau\leq T)\), we derive
$$\begin{aligned} & \bigl\Vert e_{t}^{k} \bigr\Vert ^{2}+\frac{1}{2} \bigl\Vert \Lambda_{h}^{\alpha} e^{k+1} \bigr\Vert ^{2}+\frac{1}{2} \bigl\Vert \Lambda _{h}^{\alpha} e^{k} \bigr\Vert ^{2} \\ &\quad = \bigl\Vert e_{t}^{0} \bigr\Vert ^{2}+ \frac{1}{2} \bigl\Vert \Lambda_{h}^{\alpha} e^{1} \bigr\Vert ^{2}+\frac{1}{2} \bigl\Vert \Lambda_{h}^{\alpha } e^{0} \bigr\Vert ^{2}+ \tau\sum_{n=1}^{k}\bigl(V\bigl( \bar{e}^{n}\bigr)+r^{n},2 e_{\hat{t}}^{n} \bigr). \end{aligned}$$
(3.15)
From (3.13) we get
$$\begin{aligned} \bigl\vert V \bigl(\bar{e}_{i}^{n}\bigr) \bigr\vert &\leq \biggl\vert 2 \int_{0}^{1}\sin\biggl[\lambda\frac {e_{i}^{n+1}}{2}+(1- \lambda)\frac{e_{i}^{n-1}}{2}\biggr]\,d \lambda \biggr\vert \\ &\leq \int_{0}^{1} \bigl\vert \lambda e_{i}^{n+1}+(1- \lambda)e_{i}^{n-1} \bigr\vert \,d \lambda \\ &\leq \int_{0}^{1} \bigl\vert \tau\lambda \bigl(e_{i}^{n}\bigr)_{t}+\lambda e_{i}^{n}+(1- \lambda)e_{i}^{n-1} \bigr\vert \,d \lambda \\ &=\frac{1}{2}\tau \bigl\vert \bigl(e_{i}^{n} \bigr)_{t} \bigr\vert +\frac{1}{2} \bigl\vert e_{i}^{n} \bigr\vert +\frac {1}{2} \bigl\vert e_{i}^{n-1} \bigr\vert , \end{aligned}$$
and hence
$$ \bigl\Vert V \bigl(\bar{e}^{n}\bigr) \bigr\Vert ^{2} \leq\frac{3}{4} \bigl(\tau^{2} \bigl\Vert e_{t}^{n} \bigr\Vert ^{2}+ \bigl\Vert e^{n} \bigr\Vert ^{2}+ \bigl\Vert e^{n-1} \bigr\Vert ^{2}\bigr). $$
(3.16)
Noticing that \(e_{i}^{n}=e_{i}^{0}+\tau\sum_{j=0}^{n-1}(e_{i}^{j})_{t}\) and using the Cauchy–Schwarz inequality, we obtain
$$\begin{aligned} \bigl\Vert e^{n} \bigr\Vert ^{2}&\leq 2 \bigl\Vert e^{0} \bigr\Vert ^{2}+2\tau^{2} \Biggl\Vert \sum_{j=0}^{n-1}e_{t}^{j} \Biggr\Vert ^{2} \\ & \leq 2 \bigl\Vert e^{0} \bigr\Vert ^{2}+2T\tau\sum_{j=0}^{n-1} \bigl\Vert e_{t}^{j} \bigr\Vert ^{2}. \end{aligned}$$
(3.17)
According to (3.16)–(3.17) and noticing that \(e_{\hat{t}}^{n}=\frac{1}{2} e_{t}^{n}+\frac{1}{2} e_{t}^{n-1}\), we get
$$\begin{aligned} \Biggl\vert \tau\sum_{n=1}^{k} \bigl(V\bigl(\bar{e}^{n}\bigr),2 e_{\hat{t}}^{n}\bigr) \Biggr\vert \leq{}& \tau \sum_{n=1}^{k} \biggl( \bigl\Vert V \bigl(\bar{e}^{n}\bigr) \bigr\Vert ^{2}+\frac{1}{2} \bigl\Vert e_{t}^{n} \bigr\Vert ^{2}+\frac{1}{2} \bigl\Vert e_{t}^{n-1} \bigr\Vert ^{2}\biggr) \\ \leq{}& 3T \bigl\Vert e^{0} \bigr\Vert ^{2}+ \frac{3\tau^{3}+2\tau}{4} \bigl\Vert e_{t}^{k} \bigr\Vert ^{2} \\ &{}+\biggl(\frac{3}{4}\tau^{3}+3T^{2}\tau+\tau\biggr) \sum_{n=0}^{k-1} \bigl\Vert e_{t}^{n} \bigr\Vert ^{2}. \end{aligned}$$
(3.18)
On the other hand,
$$\begin{aligned} \Biggl\vert \tau\sum_{n=1}^{k} \bigl(r^{n},2 e_{\hat{t}}^{n}\bigr) \Biggr\vert \leq{}& \tau\sum_{n=1}^{k}\bigl( \bigl\Vert r^{n} \bigr\Vert ^{2}+ \bigl\Vert e_{\hat{t}}^{n} \bigr\Vert ^{2}\bigr) \\ \leq{}& \tau \sum_{n=1}^{k} \bigl\Vert r^{n} \bigr\Vert ^{2} +\frac{1}{2}\tau \sum_{n=0}^{k} \bigl\Vert e_{t}^{n} \bigr\Vert ^{2} +\frac{1}{2}\tau \sum_{n=0}^{k-1} \bigl\Vert e_{t}^{n} \bigr\Vert ^{2}. \end{aligned}$$
(3.19)
Substituting (3.18) and (3.19) into (3.15) and noticing that \(e^{0}=0\), we get
$$\begin{aligned} &\biggl(1-\biggl(\frac{3\tau^{3}}{4}+\tau\biggr)\biggr) \bigl\Vert e_{t}^{k} \bigr\Vert ^{2}+ \frac{1}{2} \bigl\Vert \Lambda_{h}^{\alpha} e^{k+1} \bigr\Vert ^{2}+\frac{1}{2} \bigl\Vert \Lambda _{h}^{\alpha} e^{k} \bigr\Vert ^{2} \\ &\quad \leq \bigl\Vert e_{t}^{0} \bigr\Vert ^{2}+\frac{1}{2} \bigl\Vert \Lambda_{h}^{\alpha} e^{1} \bigr\Vert ^{2}+\tau \sum _{n=1}^{k} \bigl\Vert r^{n} \bigr\Vert ^{2} \\ &\qquad{}+\biggl(\frac{3}{4}\tau^{3}+3T^{2}\tau+2\tau \biggr) \sum_{n=0}^{k-1} \bigl\Vert e_{t}^{n} \bigr\Vert ^{2}. \end{aligned}$$
(3.20)
In addition,
$$\begin{aligned} &\biggl(1-\biggl(\frac{3\tau^{3}}{4}+\tau\biggr)\biggr) \bigl\Vert e_{t}^{k} \bigr\Vert ^{2}+\frac{1}{2} \bigl\Vert \Lambda_{h}^{\alpha} e^{k+1} \bigr\Vert ^{2}+\frac{1}{2} \bigl\Vert \Lambda _{h}^{\alpha} e^{k} \bigr\Vert ^{2} \\ &\quad\geq\biggl(1-\biggl(\frac{3\tau^{3}}{4}+\tau\biggr)\biggr) \bigl\Vert e_{t}^{k} \bigr\Vert ^{2}+\frac {1}{2} \tau^{2} \bigl\Vert \Lambda_{h}^{\alpha} e_{t}^{k} \bigr\Vert ^{2} \\ &\quad\geq\biggl(1-\biggl(\frac{3\tau^{3}}{4}+\tau\biggr)+\frac{1}{2} \tau^{2}h^{-\alpha }\lambda_{\min}(G)\biggr) \bigl\Vert e_{t}^{k} \bigr\Vert ^{2}. \end{aligned}$$
Supposing that \(\sigma=1-(\frac{3\tau^{3}}{4}+\tau) +\frac{1}{2}\tau^{2}h^{-\alpha}\lambda_{\min}(G)>0\) (it is easily to verify that when \(\tau\leq0.72\), the condition is satisfied), then inequality (3.20) can be rewritten as
$$\begin{aligned} \bigl\Vert e_{t}^{k} \bigr\Vert ^{2}+\frac{1}{2\sigma} \bigl\Vert \Lambda_{h}^{\alpha} e^{k} \bigr\Vert ^{2} \leq{}&\frac{1}{\sigma} \bigl\Vert e_{t}^{0} \bigr\Vert ^{2}+\frac{1}{2\sigma} \bigl\Vert \Lambda_{h}^{\alpha} e^{1} \bigr\Vert ^{2}+\frac{\tau}{\sigma} \sum_{n=1}^{k} \bigl\Vert r^{n} \bigr\Vert ^{2} \\ &{}+\frac{3\tau^{2}+12T^{2}+8}{4\sigma}\tau\sum_{n=0}^{k-1} \biggl( \bigl\Vert e_{t}^{n} \bigr\Vert ^{2}+ \frac{1}{2\sigma} \bigl\Vert \Lambda_{h}^{\alpha} e^{n} \bigr\Vert ^{2}\biggr). \end{aligned}$$
(3.21)
Let
$$\begin{aligned} \begin{aligned} &\rho(k)=\frac{1}{\sigma} \bigl\Vert e_{t}^{0} \bigr\Vert ^{2}+\frac{1}{2\sigma} \bigl\Vert \Lambda_{h}^{\alpha} e^{1} \bigr\Vert ^{2}+\frac{\tau}{\sigma} \sum _{n=1}^{k} \bigl\Vert r^{n} \bigr\Vert ^{2}, \\ &H(k)= \bigl\Vert e_{t}^{k} \bigr\Vert ^{2}+\frac{1}{2\sigma} \bigl\Vert \Lambda_{h}^{\alpha} e^{k} \bigr\Vert ^{2}. \end{aligned} \end{aligned}$$
(3.22)
Substituting (3.22) into (3.21), we derive
$$H(k)\leq\rho(k)+\frac{3\tau^{2}+12T^{2}+8}{4\sigma}\tau\sum_{n=0}^{k-1} H(n). $$
Applying Lemma 3.3 to \(\rho(k)\) and \(H(k)\), we arrive at
$$ H(k) \leq\rho(k) e^{\frac{3\tau^{2}+12T^{2}+8}{4\sigma}k\tau}\leq \rho(k) e^{\frac{(3\tau^{2}+12T^{2}+8)T}{4\sigma}}. $$
(3.23)
Noticing that if τ and h are sufficiently small, then we have
$$e_{i}^{0}=0,\qquad r_{i}^{n}=O \bigl(h^{2}+\tau^{2}\bigr), \quad 1\leq i\leq M-1, 1\leq n\leq N-1. $$
Applying the Taylor expansion, we get
$$u_{i}^{1}= u_{i}^{0}+\tau \psi(x_{i})+ \frac{\tau^{2}}{2}\bigl[-(-\Delta)^{\frac{\alpha}{2}}u_{i}^{0}- \sin \bigl(u_{i}^{0}\bigr)\bigr]+O\bigl(\tau^{3} \bigr), \quad 1\leq i \leq M-1. $$
According to Lemma 2.1, we obtain
$$-(-\Delta)^{\frac{\alpha}{2}}u_{i}^{0}=-\delta_{h}^{\alpha }u_{i}^{0}+O \bigl(h^{2}\bigr). $$
Therefore, we have the following equality:
$$ e_{i}^{1}=O\bigl(\tau^{3}+h^{2} \tau^{2}\bigr),\quad 1\leq i\leq M-1. $$
(3.24)
Due to \(e_{i}^{0}=0, \frac{\tau^{2}}{h^{\alpha}}< S\), and (3.24), we derive
$$\begin{aligned} & \bigl\Vert e_{t}^{0} \bigr\Vert ^{2} \leq L \max_{1\leq i\leq M-1}\biggl( \frac{e_{i}^{1}}{\tau}\biggr)^{2} \leq C \bigl(\tau^{2}+h^{2} \tau\bigr)^{2}\leq C \bigl(\tau^{2}+h^{2} \bigr)^{2}=O\bigl(\tau ^{2}+h^{2} \bigr)^{2}, \end{aligned}$$
(3.25)
$$\begin{aligned} & \bigl\Vert \Lambda_{h}^{\alpha} e^{1} \bigr\Vert ^{2}\leq h^{-\alpha}\lambda_{\max }(G) \bigl\Vert e^{1} \bigr\Vert ^{2} \\ &\phantom{\bigl\Vert \Lambda_{h}^{\alpha} e^{1} \bigr\Vert ^{2}}\leq 2h^{-\alpha}g_{0}^{(\alpha)}L \max _{1\leq i\leq M-1}\bigl(e_{i}^{1}\bigr)^{2} \\ &\phantom{\bigl\Vert \Lambda_{h}^{\alpha} e^{1} \bigr\Vert ^{2}}\leq C \frac{\tau^{2}}{h^{\alpha}}\bigl(\tau^{2}+h^{2} \tau\bigr)^{2} \\ &\phantom{\bigl\Vert \Lambda_{h}^{\alpha} e^{1} \bigr\Vert ^{2}}\leq C\cdot S \bigl(\tau^{2}+h^{2} \bigr)^{2} \\ &\phantom{\bigl\Vert \Lambda_{h}^{\alpha} e^{1} \bigr\Vert ^{2}} = O\bigl(\tau^{2}+h^{2} \bigr)^{2}. \end{aligned}$$
(3.26)
On the other hand,
$$ \tau\sum_{n=1}^{k} \bigl\Vert r^{n} \bigr\Vert ^{2}\leq T \max _{1\leq n\leq k} \bigl\Vert r^{n} \bigr\Vert ^{2} \leq T\cdot L \max_{1\leq n\leq k, 1\leq i\leq M-1}\bigl(r_{i}^{n} \bigr)^{2}=O\bigl(\tau^{2}+h^{2} \bigr)^{2}. $$
(3.27)
Combining (3.25), (3.26), and (3.27), we derive
$$ \rho(k)=O \bigl(h^{2}+\tau^{2} \bigr)^{2}. $$
(3.28)
Furthermore, from (3.23) and (3.28) we immediately obtain the following results:
$$\bigl\Vert e_{t}^{k} \bigr\Vert \leq C \bigl(h^{2}+\tau^{2}\bigr), \qquad\bigl\Vert \Lambda_{h}^{\alpha} e^{k} \bigr\Vert \leq C\bigl(h^{2}+\tau^{2} \bigr). $$
Hence
$$\begin{aligned} \frac{ \Vert e^{k+1} \Vert - \Vert e^{k} \Vert }{\tau} \leq \bigl\Vert e_{t}^{k} \bigr\Vert \leq C \bigl(h^{2}+\tau^{2}\bigr). \end{aligned}$$
By summing this equality for \(k=0,1,\ldots,n-1\)
\((n\tau\leq T)\), we get
$$\begin{aligned} \bigl\Vert e^{n} \bigr\Vert \leq n\tau C\bigl(h^{2}+ \tau^{2}\bigr)+ \bigl\Vert e^{0} \bigr\Vert =O \bigl(h^{2}+\tau^{2}\bigr). \end{aligned}$$
It follows from Lemma 3.1 and Lemma 3.5 that
$$\begin{aligned} \bigl\vert e^{n} \bigr\vert _{H^{\alpha/2}}^{2}\leq \biggl(\frac{\pi}{2}\biggr)^{\alpha} \bigl\Vert \Lambda_{h}^{\alpha} e^{n} \bigr\Vert ^{2}=O\bigl(h^{2}+ \tau^{2}\bigr)^{2}, \end{aligned}$$
and hence
$$\begin{aligned} \bigl\Vert e^{n} \bigr\Vert _{H^{\alpha/2}}^{2}= \bigl\Vert e^{n} \bigr\Vert ^{2}+ \bigl\vert e^{n} \bigr\vert _{H^{\alpha /2}}^{2}=O\bigl(h^{2}+ \tau^{2}\bigr)^{2}. \end{aligned}$$
According to Lemma 3.4,
$$\begin{aligned} \bigl\Vert e^{n} \bigr\Vert _{l_{h}^{\infty}}\leq C \bigl\Vert e^{n} \bigr\Vert _{H^{\alpha /2}}=O\bigl(h^{2}+ \tau^{2}\bigr). \end{aligned}$$
Thus we get the desired results. □