The trial function space $U_h$ is chose as the standard linear element space on $T_h$, then $U_h=\left\{u_h\in C\left(\Omega \right):{u_h|}_K\in P_1;{u_h|}_{\partial \Omega }=0\right\}\subset H_0^1\left(\Omega \right)$, where $P_1$ is the set of all linear polynomials on the triangle element $K$.

The test function space $V_h$ is chosed as the piecewise constant space on $T_h^{\mathrm{<mo>\; *\; </mo>}}$, then $V_h=\left\{v_h\in L^2\left(\Omega \right):\forall P_0\in {\Omega }_h^{*},{v_h|}_{K_{P_0}^{*}}=C;\forall P_0\in \partial \Omega ,{v_h|}_{K_{P_0}^{*}}=0\right\}$. Thus, $V_h$ can be viewed as a space spanned by the following basis functions: for $\forall P_0\in {\Omega }_h^{\mathrm{<mo>\; *\; </mo>}}$,

${\varphi }_{P_0}\left(P\right)=\{\begin{array}{ll}\mathrm{1,}\hfill & P\in K_{P_0}^{\mathrm{<mo>\; *\; </mo>}}\mathrm{,}\hfill \\ \mathrm{0,}\hfill & \text{elsewhere}\mathrm{.}\hfill \end{array}$(6)

Firstly, the intermediate function $v\left(x\mathrm{,}y\mathrm{,}t\right)=-\Delta u\left(x\mathrm{,}y\mathrm{,}t\right)$ is introduced, then the original equation can be rewritten as

$\{\begin{array}{ll}{}_0^c{D}_t^{\alpha }u\left(x,y,t\right)-\Delta v\left(x,y,t\right)-\Delta v\left(x,y,t-s\right)\hfill & \hfill \\ =f\left(x,y,t,u\left(x,y,t\right),u\left(x,y,t-s\right)\right),\hfill & \left(x,y,t\right)\in \Omega \times \left(0,T\right],\hfill \\ v\left(x,y,t\right)=-\Delta u\left(x,y,t\right),\hfill & \left(x,y,t\right)\in \Omega \times \left(0,T\right],\hfill \\ u\left(x,y,t\right)=0,\hfill & \left(x,y,t\right)\in \partial \Omega \times \left[0,T\right],\hfill \\ v\left(x,y,t\right)=0,\hfill & \left(x,y,t\right)\in \partial \Omega \times \left[0,T\right],\hfill \\ u\left(x,y,t\right)=\omega \left(x,y,t\right),\hfill & \left(x,y,t\right)\in \Omega \times \left[-s,0\right],\hfill \\ v\left(x,y,t\right)=-\Delta \omega \left(x,y,t\right),\hfill & \left(x,y,t\right)\in \Omega \times \left[-s,0\right].\hfill \end{array}$(7)

Multiplying ${\varphi }_h\in V_h$ and ${\psi }_h\in V_h$ at both sides of the first equation and the second equation of (7), respectively and utilizing Green’s formula to get:

$\{\begin{array}{ll}\left({}_0^c{D}_t^{\alpha }u,{\varphi }_h\right)+a\left(v,{\varphi }_h\right)+a\left(v\left(x,y,t-s\right),{\varphi }_h\right)\hfill & \hfill \\ =\left(f\left(x,y,t,u,u\left(x,y,t-s\right)\right),{\varphi }_h\right),\hfill & \forall {\varphi }_h\in V_h,\hfill \\ \left(v,{\psi }_h\right)=a\left(u,{\psi }_h\right),\hfill & \forall {\psi }_h\in V_h,\hfill \end{array}$(8)

where the bilinear form $a\left(v,{\varphi }_h\right)=-\underset{P_0\in {\Omega }_h}{{\displaystyle {\displaystyle \sum }}}{\varphi }_h\left(P_0\right){\displaystyle {\int }_{\partial K_{P_0}^{*}}}\frac{\partial v}{\partial x}\text{d}y-\frac{\partial v}{\partial y}\text{d}x$.

Choosing the trial function space $U_h$ as described above, and we can obtain the semi-discrete finite volume element format: find $\left(u_h\left(\cdot \mathrm{,}t\right)\mathrm{,}{v}_h\left(\cdot \mathrm{,}t\right)\right)\in U_h\times U_h\left(0\le t\le T\right)$ such that

$\{\begin{array}{ll}\left({}_0^c{D}_t^{\alpha }{u}_h,{\varphi }_h\right)+a\left(v_h,{\varphi }_h\right)+a\left(v_h\left(\cdot ,t-s\right),{\varphi }_h\right)\hfill & \hfill \\ =\left(f\left(x,y,t,u_h,u_h\left(x,y,t-s\right)\right),{\varphi }_h\right),\hfill & \forall {\varphi }_h\in V_h,\hfill \\ \left(v_h,{\psi }_h\right)=a\left(u_h,{\psi }_h\right),\hfill & \forall {\psi }_h\in V_h.\hfill \\ u_h\left(x,y,t\right)=0,\hfill & \left(x,y,t\right)\in \partial \Omega \times \left[0,T\right],\hfill \\ v_h\left(x,y,t\right)=0,\hfill & \left(x,y,t\right)\in \partial \Omega \times \left[0,T\right],\hfill \\ u_h\left(x,y,t\right)={\omega }_h\left(x,y,t\right),\hfill & \left(x,y,t\right)\in \Omega \times \left[-s,0\right],\hfill \\ v_h\left(x,y,t\right)={\varpi }_h\left(x,y,t\right),\hfill & \left(x,y,t\right)\in \Omega \times \left[-s,0\right],\hfill \end{array}$(9)

where ${\omega }_h\left(x\mathrm{,}y\mathrm{,}t\right)$ and ${\varpi }_h\left(x\mathrm{,}y\mathrm{,}t\right)$ are the elliptic projections of $\omega \left(x\mathrm{,}y\mathrm{,}t\right)$ and $-\Delta \omega \left(x\mathrm{,}y\mathrm{,}t\right)$ on $U_h$, respectively.

Next, we discrete time fractional order derivatives and deal with nonlinear terms. Let $m$ be a positive integer, and make $\tau =\frac{s}{m}$, $N=\frac{T}{\tau }$, $t_n=n\tau \left(n=0,1,2,\cdots ,N\right)$. For convenience, assume that $N$ is a positive integer.

Lemma 1 [27] Let $0<\alpha <1,a_l^{\alpha }={\left(l+1\right)}^{1-\alpha }-l^{1-\alpha },l=0,1,2,\cdot \cdot \cdot$, then

(I) $1=a_0^{\alpha }>a_1^{\alpha }>a_2^{\alpha }>\cdot \cdot \cdot >a_l^{\alpha }>0;a_l^{\alpha }\to 0,\text{if}l\to \infty ,$(10)

(II) $\left(1-\alpha \right)l^{-\alpha }<a_{l-1}^{\alpha }<\left(1-\alpha \right){\left(l-1\right)}^{-\alpha },l\ge 1.$(11)

Lemma 2 [27] For $\alpha \left(0<\alpha <1\right)$ order Caputo fractional order derivatives, there are

${}_0^c{D}_t^{\alpha }f\left(t_n\right)=\frac{{\tau }^{-\alpha }}{\Gamma \left(2-\alpha \right)}\left[a_0^{\alpha }f\left(t_n\right)-{\displaystyle \underset{k=1}{\overset{n-1}{\sum }}}\left(a_{n-k-1}^{\alpha }-a_{n-k}^{\alpha }\right)f\left(t_k\right)-a_{n-1}^{\alpha }f\left(t_0\right)\right]+R_n,$(12)

where $\left|R_n\right|\le \frac{1}{2\Gamma \left(1-\alpha \right)}\left[\frac{1}{4}+\frac{\alpha }{\left(1-\alpha \right)\left(2-\alpha \right)}\right]\underset{t_0\le t\le t_n}{\max }\left|f_{tt}\right|{\tau }^{2-\alpha }$.

When $t=t_n$, for the right nonlinear term, Taylor’s formula yields:

$\begin{array}{l}f\left(x\mathrm{,}y\mathrm{,}t\mathrm{,}{u}_h\left(x\mathrm{,}y\mathrm{,}{t}_n\right)\mathrm{,}{u}_h\left(x\mathrm{,}y\mathrm{,}{t}_{n-m}\right)\right)\\ =f\left(x\mathrm{,}y\mathrm{,}t\mathrm{,2}{u}_h\left(x\mathrm{,}y\mathrm{,}{t}_{n-1}\right)-u_h\left(x\mathrm{,}y\mathrm{,}{t}_{n-2}\right)\mathrm{,}{u}_h\left(x\mathrm{,}y\mathrm{,}{t}_{n-m}\right)\right)+O\left({\tau }^2\right)\mathrm{.}\end{array}$(13)

Substitute (13) into (9). The applying the Lemma 2 to the fractional derivative term and omitting small quantity terms, we get the fully discrete finite volume element format: find $\left(u_h^n\mathrm{,}{v}_h^n\right)\in U_h\times U_h\left(n=\mathrm{1,2,}\cdot \cdot \cdot \right)$ such that

$\{\begin{array}{ll}\left(D_{\tau }^{\alpha }{u}_h^n,{\varphi }_h\right)+a\left(v_h^n,{\varphi }_h\right)+a\left(v_h^{n-m},{\varphi }_h\right)\hfill & \hfill \\ =\left(f\left(x,y,t_n,2u_h^{n-1}-u_h^{n-2},u_h^{n-m}\right),{\varphi }_h\right),\hfill & \forall {\varphi }_h\in V_h,\hfill \\ \left(v_h^n,{\psi }_h\right)=a\left(u_h^n,{\psi }_h\right),\hfill & \forall {\psi }_h\in V_h,\hfill \\ u_h^n=0,\hfill & \left(x,y\right)\in \partial \Omega ,\hfill \\ v_h^n=0,\hfill & \left(x,y\right)\in \partial \Omega ,\hfill \\ u_h^l={\omega }_h^l,\hfill & l=-m,\cdot \cdot \cdot ,-1,0,\hfill \\ v_h^l={\varpi }_h^l,\hfill & l=-m,\cdot \cdot \cdot ,-1,0,\hfill \end{array}$(14)

where $D_{\tau }^{\alpha }{u}_h^n=\frac{{\tau }^{-\alpha }}{\Gamma \left(2-\alpha \right)}\left[a_0^{\alpha }{u}_h^n-{\displaystyle {\sum }_{k=1}^{n-1}}\left(a_{n-k-1}^{\alpha }-a_{n-k}^{\alpha }\right)u_h^k-a_{n-1}^{\alpha }{u}_h^0\right]$.

For the purpose of subsequent research, the following definitions and lemmas are presented.

Definition 1. [28] The interpolation operator ${\Pi }^{\mathrm{<mo>\; *\; </mo>}}$ is defined as follows:

${\Pi }^{*}:H_0^1\to V_h,{\Pi }^{*}u={\displaystyle \underset{P_0\in {\Omega }_h^{*}}{\sum }}u\left(P_0\right){\varphi }_{P_0},\forall u\in H_0^1.$(15)

Definition 2. [28] Introduce the elliptic projection operator $P_h\mathrm{<mo>\; :\; </mo>}{H}^2\left(\Omega \right)\cap H_0^1\left(\Omega \right)\to U_h$, defined by the following format:

$a\left(P_{h}u\mathrm{,}{v}_h\right)=a\left(u\mathrm{,}{v}_h\right)\mathrm{,}\forall v_h\in V_h\mathrm{.}$(16)

Lemma 3. [28] Let $u\in {\text{W}}^{\mathrm{3,}p}\left(\Omega \right)\cap H_0^1\left(\Omega \right)$, and $P_{h}u$ is the elliptic projection of $u$, then

$\Vert u-P_{h}u\Vert \le C{h}^2{\Vert u\Vert }_{\mathrm{3,}p}\mathrm{.}$(17)

Lemma 4. [18] [28] Let $\left|\Vert u_h\Vert \right|={\left(u_h\mathrm{,}{\Pi }^{\mathrm{<mo>\; *\; </mo>}}{u}_h\right)}^{\frac{1}{2}}$, we have

(I) $\left(u_h\mathrm{,}{\Pi }^{\mathrm{<mo>\; *\; </mo>}}{w}_h\right)=\left(w_h\mathrm{,}{\Pi }^{\mathrm{<mo>\; *\; </mo>}}{u}_h\right)\mathrm{,}\forall u_h\mathrm{,}{w}_h\in U_h\mathrm{,}$(18)

(II) $\left(u_h\mathrm{,}{\Pi }^{\mathrm{<mo>\; *\; </mo>}}{u}_h\right)\ge \mathrm{0,}\forall u_h\in U_h\mathrm{,}$(19)

(III) $\left|\left(u_h\mathrm{,}{\Pi }^{\mathrm{<mo>\; *\; </mo>}}{w}_h\right)\right|\le \left|\Vert u_h\Vert \right|\cdot \left|\Vert w_h\Vert \right|\mathrm{,}\forall u_h\mathrm{,}{w}_h\in U_h\mathrm{.}$(20)

Lemma 5. [18] [28] $\left|\Vert \cdot \Vert \right|$ and $\Vert \cdot \Vert$ are equivalent on $U_h$, i.e., there exist constants $C_1,C_2>0$ such that

$C_1\Vert u_h\Vert \le \left|\Vert u_h\Vert \right|\le C_2\Vert u_h\Vert \mathrm{,}\forall u_h\in U_h\mathrm{.}$(21)

Lemma 6. [28] There exist positive constants $\alpha$ and $M$ such that the following equations hold

$a\left(u_h\mathrm{,}{\Pi }^{\mathrm{<mo>\; *\; </mo>}}{w}_h\right)=a\left(w_h\mathrm{,}{\Pi }^{\mathrm{<mo>\; *\; </mo>}}{u}_h\right)\mathrm{,}\forall u_h\mathrm{,}{w}_h\in U_h\mathrm{,}$(22)

$a\left(u_h\mathrm{,}{\Pi }^{\mathrm{<mo>\; *\; </mo>}}{u}_h\right)\ge \alpha {\Vert u_h\Vert }_1^2\mathrm{,}\forall u_h\in U_h\mathrm{,}$(23)

$\left|a\left(u_h\mathrm{,}{\Pi }^{\mathrm{<mo>\; *\; </mo>}}{w}_h\right)\right|\le M{\Vert u_h\Vert }_1{\Vert w_h\Vert }_1\mathrm{,}\forall u_h\mathrm{,}{w}_h\in U_h\mathrm{.}$(24)

Lemma 7. [29] Let $\left\{z_k\right\}\mathrm{,}\left\{g_k\right\}$ be two non-negative sequences. If

$z^k\le K+{\displaystyle \underset{i=0}{\overset{k}{\sum }}}{g}_i{z}_i\mathrm{,}k\ge \mathrm{0,}$(25)

then

$z^k\le K\cdot \exp \left({\displaystyle \underset{i=0}{\overset{k}{\sum }}}{g}_i\right),k\ge 0,$(26)

where $K$ is a positive constant.

Theorem 1. The fully discrete finite volume element format (14) has a unique solution.

Proof. The homogeneous equations corresponding to (14) are

$\{\begin{array}{ll}\left(u_h^n,{\varphi }_h\right)+\mu a\left(v_h^n,{\varphi }_h\right)=0,\hfill & \forall {\varphi }_h\in V_h,\hfill \\ \left(v_h^n,{\psi }_h\right)=a\left(u_h^n,{\psi }_h\right),\hfill & \forall {\psi }_h\in V_h,\hfill \end{array}$(27)

where $\mu ={\tau }^{\alpha }\Gamma \left(2-\alpha \right)$.

To prove that (14) is uniquely solvable, it is sufficient to prove that its corresponding homogeneous equations (27) has only zero solutions. Let ${\varphi }_h={\Pi }^{\mathrm{<mo>\; *\; </mo>}}{v}_h^n\mathrm{,}{\psi }_h={\Pi }^{\mathrm{<mo>\; *\; </mo>}}{u}_h^n$ in (27), then

$\left(u_h^n\mathrm{,}{\Pi }^{\mathrm{<mo>\; *\; </mo>}}{v}_h^n\right)+\mu a\left(v_h^n\mathrm{,}{\Pi }^{\mathrm{<mo>\; *\; </mo>}}{v}_h^n\right)=\mathrm{0,}$(28)

$\left(v_h^n\mathrm{,}{\Pi }^{\mathrm{<mo>\; *\; </mo>}}{u}_h^n\right)=a\left(u_h^n\mathrm{,}{\Pi }^{\mathrm{<mo>\; *\; </mo>}}{u}_h^n\right)\mathrm{.}$(29)

According to Lemma 4

$\left(v_h^n\mathrm{,}{\Pi }^{\mathrm{<mo>\; *\; </mo>}}{u}_h^n\right)=\left(u_h^n\mathrm{,}{\Pi }^{\mathrm{<mo>\; *\; </mo>}}{v}_h^n\right)\mathrm{.}$(30)

Substituting (30) into (29) gives

$\left(u_h^n\mathrm{,}{\Pi }^{\mathrm{<mo>\; *\; </mo>}}{v}_h^n\right)=a\left(u_h^n\mathrm{,}{\Pi }^{\mathrm{<mo>\; *\; </mo>}}{u}_h^n\right)\mathrm{.}$(31)

Substituting (31) into (28) gives

$a\left(u_h^n\mathrm{,}{\Pi }^{\mathrm{<mo>\; *\; </mo>}}{u}_h^n\right)+\mu a\left(v_h^n\mathrm{,}{\Pi }^{\mathrm{<mo>\; *\; </mo>}}{v}_h^n\right)=0.$(32)

According to Lemma 6

$a\left(u_h^n\mathrm{,}{\Pi }^{\mathrm{<mo>\; *\; </mo>}}{u}_h^n\right)\ge \alpha {\Vert u_h^n\Vert }_1^2\ge \mathrm{0,}a\left(v_h^n\mathrm{,}{\Pi }^{\mathrm{<mo>\; *\; </mo>}}{v}_h^n\right)\ge \alpha {\Vert v_h^n\Vert }_1^2\ge 0.$(33)

We get

${\Vert u_h^n\Vert }_1={\Vert v_h^n\Vert }_1=0.$(34)

So $u_h^n=0,v_h^n=0$. According to the induction principle, the homogeneous equations has only zero solutions, so the format (14) is uniquely solvable.

Theorem 2. Assuming that $\left\{u\mathrm{,}v\right\}$ and $\left\{u_h^n\mathrm{,}{v}_h^n\right\}$ are solutions of the original Equation (7) and the finite volume element format (14), respectively, and that the requirement of sufficient regularity is satisfied, then there exists a constant $C$ such that

$\Vert u^n-u_h^n\Vert \le C\left(h^2+{\tau }^{2-\alpha }\right)\mathrm{,}$(35)

$\Vert v^n-v_h^n\Vert \le C\left(h^2+{\tau }^{2-\alpha }\right)\mathrm{.}$(36)

Proof. Let

${\eta }^n=u^n-P_h{u}^n,{\epsilon }^n=P_h{u}^n-u_h^n,$(37)

${\theta }^n=v^n-P_h{v}^n,{\rho }^n=P_h{v}^n-v_h^n,$(38)

then

$u^n-u_h^n={\eta }^n+{\epsilon }^n\mathrm{,}$(39)

$v^n-v_h^n={\theta }^n+{\rho }^n\mathrm{.}$(40)

Let $t=t_n$ in (8) and subtract from (14). According to the elliptic projection in Definition 2, we get

$\begin{array}{l}\left(D_{\tau }^{\alpha }{\eta }^n+D_{\tau }^{\alpha }{\epsilon }^n\mathrm{,}{\varphi }_h\right)+a\left({\rho }^n\mathrm{,}{\varphi }_h\right)+a\left({\rho }^{n-m}\mathrm{,}{\varphi }_h\right)\\ =\left(f\left(x\mathrm{,}y\mathrm{,}{t}_n\mathrm{,2}{u}^{n-1}-u^{n-2}\mathrm{,}{u}^{n-m}\right)-f\left(x\mathrm{,}y\mathrm{,}{t}_n\mathrm{,2}{u}_h^{n-1}-u_h^{n-2}\mathrm{,}{u}_h^{n-m}\right)\mathrm{,}{\varphi }_h\right)\\ +\left(R_1\mathrm{,}{\varphi }_h\right)\mathrm{,}\forall {\varphi }_h\in V_h\mathrm{,}\end{array}$(41)

$\left({\theta }^n+{\rho }^n\mathrm{,}{\psi }_h\right)=a\left({\epsilon }^n\mathrm{,}{\psi }_h\right)\mathrm{,}\forall {\psi }_h\in V_h\mathrm{,}$(42)

where $\Vert R_1\Vert \le c_0{\tau }^{2-\alpha }$ and $c_0$ is a positive constant that does not depend on $\tau \mathrm{,}h$.

Let ${\varphi }_h={\Pi }^{\mathrm{<mo>\; *\; </mo>}}{\epsilon }^n$ and ${\psi }_h={\Pi }^{\mathrm{<mo>\; *\; </mo>}}{\rho }^n$ in (41) and (42), respectively, and use Lemma 6 to get

$\begin{array}{l}\left(D_{\tau }^{\alpha }{\eta }^n+D_{\tau }^{\alpha }{\epsilon }^n\mathrm{,}{\Pi }^{\mathrm{<mo>\; *\; </mo>}}{\epsilon }^n\right)+a\left({\rho }^n\mathrm{,}{\Pi }^{\mathrm{<mo>\; *\; </mo>}}{\epsilon }^n\right)+a\left({\rho }^{n-m}\mathrm{,}{\Pi }^{\mathrm{<mo>\; *\; </mo>}}{\epsilon }^n\right)\\ =\left(f\left(x\mathrm{,}y\mathrm{,}{t}_n\mathrm{,2}{u}^{n-1}-u^{n-2}\mathrm{,}{u}^{n-m}\right)-f\left(x\mathrm{,}y\mathrm{,}{t}_n\mathrm{,2}{u}_h^{n-1}-u_h^{n-2}\mathrm{,}{u}_h^{n-m}\right)\mathrm{,}{\Pi }^{\mathrm{<mo>\; *\; </mo>}}{\epsilon }^n\right)\\ +\left(R_1\mathrm{,}{\Pi }^{\mathrm{<mo>\; *\; </mo>}}{\epsilon }^n\right)\mathrm{,}\end{array}$(43)

$a\left({\rho }^n\mathrm{,}{\Pi }^{\mathrm{<mo>\; *\; </mo>}}{\epsilon }^n\right)=a\left({\epsilon }^n\mathrm{,}{\Pi }^{\mathrm{<mo>\; *\; </mo>}}{\rho }^n\right)=\left({\theta }^n\mathrm{,}{\Pi }^{\mathrm{<mo>\; *\; </mo>}}{\rho }^n\right)+\left({\rho }^n\mathrm{,}{\Pi }^{\mathrm{<mo>\; *\; </mo>}}{\rho }^n\right)\mathrm{.}$(44)

Let ${\psi }_h={\Pi }^{\mathrm{<mo>\; *\; </mo>}}{\rho }^{n-m}$ in (42), and use Lemma 6 to get

$a\left({\rho }^{n-m}\mathrm{,}{\Pi }^{\mathrm{<mo>\; *\; </mo>}}{\epsilon }^n\right)=a\left({\epsilon }^n\mathrm{,}{\Pi }^{\mathrm{<mo>\; *\; </mo>}}{\rho }^{n-m}\right)=\left({\theta }^n\mathrm{,}{\Pi }^{\mathrm{<mo>\; *\; </mo>}}{\rho }^{n-m}\right)+\left({\rho }^n\mathrm{,}{\Pi }^{\mathrm{<mo>\; *\; </mo>}}{\rho }^{n-m}\right)\mathrm{.}$(45)

Substitute (44) and (45) into (43) to obtain

$\begin{array}{l}\left(D_{\tau }^{\alpha }{\epsilon }^n\mathrm{,}{\Pi }^{\mathrm{<mo>\; *\; </mo>}}{\epsilon }^n\right)+\left({\rho }^n\mathrm{,}{\Pi }^{\mathrm{<mo>\; *\; </mo>}}{\rho }^n\right)\\ =-\left(D_{\tau }^{\alpha }{\eta }^n\mathrm{,}{\Pi }^{\mathrm{<mo>\; *\; </mo>}}{\epsilon }^n\right)-\left({\theta }^n\mathrm{,}{\Pi }^{\mathrm{<mo>\; *\; </mo>}}{\rho }^n\right)-\left({\theta }^n\mathrm{,}{\Pi }^{\mathrm{<mo>\; *\; </mo>}}{\rho }^{n-m}\right)-\left({\rho }^n\mathrm{,}{\Pi }^{\mathrm{<mo>\; *\; </mo>}}{\rho }^{n-m}\right)\\ +\left(f\left(x\mathrm{,}y\mathrm{,}{t}_n\mathrm{,2}{u}^{n-1}-u^{n-2}\mathrm{,}{u}^{n-m}\right)-f\left(x\mathrm{,}y\mathrm{,}{t}_n\mathrm{,2}{u}_h^{n-1}-u_h^{n-2}\mathrm{,}{u}_h^{n-m}\right)\mathrm{,}{\Pi }^{\mathrm{<mo>\; *\; </mo>}}{\epsilon }^n\right)\\ +\left(R_1\mathrm{,}{\Pi }^{\mathrm{<mo>\; *\; </mo>}}{\epsilon }^n\right)\mathrm{.}\end{array}$(46)

From the definition of $D_{\tau }^{\alpha }{\epsilon }^n$, multiplying $\mu ={\tau }^{\alpha }\Gamma \left(2-\alpha \right)$ on both sides of the (46) to get

$\begin{array}{l}\left({\epsilon }^n,{\Pi }^{*}{\epsilon }^n\right)+\mu \left({\rho }^n,{\Pi }^{*}{\rho }^n\right)\\ =-\mu \left(D_{\tau }^{\alpha }{\eta }^n,{\Pi }^{*}{\epsilon }^n\right)-\mu \left({\theta }^n,{\Pi }^{*}{\rho }^n\right)-\mu \left({\theta }^n,{\Pi }^{*}{\rho }^{n-m}\right)-\mu \left({\rho }^n,{\Pi }^{*}{\rho }^{n-m}\right)\\ +\mu \left(f\left(x,y,t_n,2u^{n-1}-u^{n-2},u^{n-m}\right)-f\left(x,y,t_n,2u_h^{n-1}-u_h^{n-2},u_h^{n-m}\right),{\Pi }^{*}{\epsilon }^n\right)\\ +\mu \left(R_1,{\Pi }^{*}{\epsilon }^n\right)+{\displaystyle \underset{k=1}{\overset{n-1}{\sum }}}\left(a_{n-k-1}^{\alpha }-a_{n-k}^{\alpha }\right)\left({\epsilon }^k,{\Pi }^{*}{\epsilon }^n\right).\end{array}$(47)

In order to estimate the order of convergence of the solution $u_h^n$, we analyze the terms at the right side of the equation (47). According to the properties of the function $f$, it follows that

$\begin{array}{l}\left|f\left(x\mathrm{,}y\mathrm{,}{t}_n\mathrm{,2}{u}^{n-1}-u^{n-2}\mathrm{,}{u}^{n-m}\right)-f\left(x\mathrm{,}y\mathrm{,}{t}_n\mathrm{,2}{u}_h^{n-1}-u_h^{n-2}\mathrm{,}{u}_h^{n-m}\right)\right|\\ \le c_1\left|\left(2u^{n-1}-u^{n-2}\right)-\left(2u_h^{n-1}-u_h^{n-2}\right)\right|+c_2\left|u^{n-m}-u_h^{n-m}\right|\\ \le c_1\left|2\left({\eta }^{n-1}+{\epsilon }^{n-1}\right)-\left({\eta }^{n-2}+{\epsilon }^{n-2}\right)\right|+c_2\left|{\eta }^{n-m}+{\epsilon }^{n-m}\right|\\ \le 2c_3\left|{\eta }^{n-1}\right|+2c_3\left|{\epsilon }^{n-1}\right|+c_3\left|{\eta }^{n-2}\right|+c_3\left|{\epsilon }^{n-2}\right|+c_3\left|{\eta }^{n-m}\right|+c_3\left|{\epsilon }^{n-m}\right|\mathrm{,}\end{array}$(48)

where $c_3=\max \left\{c_1,c_2\right\}$.

According to the Cauchy-Schwarz inequality and (48), we get

$\begin{array}{l}\left|\mu \left(f\left(x\mathrm{,}y\mathrm{,}{t}_n\mathrm{,2}{u}^{n-1}-u^{n-2}\mathrm{,}{u}^{n-m}\right)-f\left(x\mathrm{,}y\mathrm{,}{t}_n\mathrm{,2}{u}_h^{n-1}-u_h^{n-2}\mathrm{,}{u}_h^{n-m}\right)\mathrm{,}{\Pi }^{\mathrm{<mo>\; *\; </mo>}}{\epsilon }^n\right)\right|\\ \le \frac{1}{48c_3^2}\mu {\left|\Vert f\left(x\mathrm{,}y\mathrm{,}{t}_n\mathrm{,2}{u}^{n-1}-u^{n-2}\mathrm{,}{u}^{n-m}\right)-f\left(x\mathrm{,}y\mathrm{,}{t}_n\mathrm{,2}{u}_h^{n-1}-u_h^{n-2}\mathrm{,}{u}_h^{n-m}\right)\Vert \right|}^2\\ +12c_3^2\mu {\left|\Vert {\epsilon }^n\Vert \right|}^2\\ \le \frac{1}{48c_3^2}(4c_3^2\mu {\left|\Vert {\eta }^{n-1}\Vert \right|}^2+4c_3^2\mu {\left|\Vert {\epsilon }^{n-1}\Vert \right|}^2+c_3^2\mu {\left|\Vert {\eta }^{n-2}\Vert \right|}^2\\ +c_3^2\mu {\left|\Vert {\epsilon }^{n-2}\Vert \right|}^2+c_3^2\mu {\left|\Vert {\eta }^{n-m}\Vert \right|}^2+c_3^2\mu {\left|\Vert {\epsilon }^{n-m}\Vert \right|}^2)+12c_3^2\mu {\left|\Vert {\epsilon }^n\Vert \right|}^2\\ \le \frac{1}{12}\mu {\left|\Vert {\eta }^{n-1}\Vert \right|}^2+\frac{1}{12}\mu {\left|\Vert {\epsilon }^{n-1}\Vert \right|}^2+\frac{1}{48}\mu {\left|\Vert {\eta }^{n-2}\Vert \right|}^2+\frac{1}{48}\mu {\left|\Vert {\epsilon }^{n-2}\Vert \right|}^2\\ +\frac{1}{48}\mu {\left|\Vert {\eta }^{n-m}\Vert \right|}^2+\frac{1}{48}\mu {\left|\Vert {\epsilon }^{n-m}\Vert \right|}^2+12c_3^2\mu {\left|\Vert {\epsilon }^n\Vert \right|}^2\mathrm{.}\end{array}$(49)

For the remaining terms at the right side, using the Cauchy-Schwartz inequality and substituting (49) into (47) to obtain

$\begin{array}{l}\left({\epsilon }^n,{\Pi }^{*}{\epsilon }^n\right)+\mu \left({\rho }^n,{\Pi }^{*}{\rho }^n\right)\\ \le \frac{1}{2}\mu {\left|\Vert D_{\tau }^{\alpha }{\eta }^n\Vert \right|}^2+\frac{3}{4}\mu {\left|\Vert {\epsilon }^n\Vert \right|}^2+3\mu {\left|\Vert {\theta }^n\Vert \right|}^2+\mu {\left|\Vert {\rho }^n\Vert \right|}^2+\frac{11}{24}\mu {\left|\Vert {\rho }^{n-m}\Vert \right|}^2\\ +\frac{1}{12}\mu {\left|\Vert {\eta }^{n-1}\Vert \right|}^2+\frac{1}{12}\mu {\left|\Vert {\epsilon }^{n-1}\Vert \right|}^2+\frac{1}{48}\mu {\left|\Vert {\eta }^{n-2}\Vert \right|}^2\\ +\frac{1}{48}\mu {\left|\Vert {\epsilon }^{n-2}\Vert \right|}^2+\frac{1}{48}\mu {\left|\Vert {\eta }^{n-m}\Vert \right|}^2+\frac{1}{48}\mu {\left|\Vert {\epsilon }^{n-m}\Vert \right|}^2+12c_3^2\mu {\left|\Vert {\epsilon }^n\Vert \right|}^2\\ +\mu {\left|\Vert R_1\Vert \right|}^2+{\displaystyle \underset{k=1}{\overset{n-1}{\sum }}}2\left(a_{n-k-1}^{\alpha }-a_{n-k}^{\alpha }\right){\left|\Vert {\epsilon }^k\Vert \right|}^2+\frac{1}{8}{\left|\Vert {\epsilon }^n\Vert \right|}^2.\end{array}$(50)

Then, we can get

$\begin{array}{l}\left(1-\frac{3}{4}\mu -12c_3^2\mu -\frac{1}{8}\right){\left|\Vert {\epsilon }^n\Vert \right|}^2\\ \le \frac{1}{2}\mu {\left|\Vert D_{\tau }^{\alpha }{\eta }^n\Vert \right|}^2+3\mu {\left|\Vert {\theta }^n\Vert \right|}^2+\frac{11}{24}\mu {\left|\Vert {\rho }^{n-m}\Vert \right|}^2+\frac{1}{12}\mu {\left|\Vert {\eta }^{n-1}\Vert \right|}^2+\frac{1}{12}\mu {\left|\Vert {\epsilon }^{n-1}\Vert \right|}^2\\ +\frac{1}{48}\mu {\left|\Vert {\eta }^{n-2}\Vert \right|}^2+\frac{1}{48}\mu {\left|\Vert {\epsilon }^{n-2}\Vert \right|}^2+\frac{1}{48}\mu {\left|\Vert {\eta }^{n-m}\Vert \right|}^2+\frac{1}{48}\mu {\left|\Vert {\epsilon }^{n-m}\Vert \right|}^2\\ +\mu {\left|\Vert R_1\Vert \right|}^2+{\displaystyle \underset{k=1}{\overset{n-1}{\sum }}}2\left(a_{n-k-1}^{\alpha }-a_{n-k}^{\alpha }\right){\left|\Vert {\epsilon }^k\Vert \right|}^2.\end{array}$(51)

According to the definition of $\mu$, and take a sufficiently small $\tau$ to get

$\begin{array}{c}{\left|\Vert {\epsilon }^n\Vert \right|}^2\le 4\mu \left({\left|\Vert D_{\tau }^{\alpha }{\eta }^n\Vert \right|}^2+{\left|\Vert {\theta }^n\Vert \right|}^2+{\left|\Vert {\eta }^{n-1}\Vert \right|}^2+{\left|\Vert {\eta }^{n-2}\Vert \right|}^2+{\left|\Vert {\eta }^{n-m}\Vert \right|}^2+{\left|\Vert R_1\Vert \right|}^2\right)\\ +\frac{11}{18}\mu {\left|\Vert {\rho }^{n-m}\Vert \right|}^2+\frac{1}{9}\mu {\left|\Vert {\epsilon }^{n-1}\Vert \right|}^2+\frac{1}{36}\mu {\left|\Vert {\epsilon }^{n-2}\Vert \right|}^2+\frac{1}{36}\mu {\left|\Vert {\epsilon }^{n-m}\Vert \right|}^2\\ +{\displaystyle \underset{k=1}{\overset{n-1}{\sum }}}\frac{8}{3}\left(a_{n-k-1}^{\alpha }-a_{n-k}^{\alpha }\right){\left|\Vert {\epsilon }^k\Vert \right|}^2.\end{array}$(52)

(52) will be used in the subsequent proof to estimate the convergence order of the solution $u_h^n$.

In order to estimate the order of convergence of the solution $v_h^n$, the terms at the right side of (47) are analyzed. Similar to the derivation of (52), we get

$\begin{array}{c}\mu {\left|\Vert {\rho }^n\Vert \right|}^2\le 6\mu \left({\left|\Vert D_{\tau }^{\alpha }{\eta }^n\Vert \right|}^2+{\left|\Vert {\theta }^n\Vert \right|}^2+{\left|\Vert {\eta }^{n-1}\Vert \right|}^2+{\left|\Vert {\eta }^{n-2}\Vert \right|}^2+{\left|\Vert {\eta }^{n-m}\Vert \right|}^2+{\left|\Vert R_1\Vert \right|}^2\right)\\ +\frac{9}{4}\mu {\left|\Vert {\rho }^{n-m}\Vert \right|}^2+\frac{1}{6}\mu {\left|\Vert {\epsilon }^{n-1}\Vert \right|}^2+\frac{1}{24}\mu {\left|\Vert {\epsilon }^{n-2}\Vert \right|}^2+\frac{1}{24}\mu {\left|\Vert {\epsilon }^{n-m}\Vert \right|}^2\\ +{\displaystyle \underset{k=1}{\overset{n-1}{\sum }}}\left(a_{n-k-1}^{\alpha }-a_{n-k}^{\alpha }\right){\left|\Vert {\epsilon }^k\Vert \right|}^2.\end{array}$(53)

(53) will be used in the subsequent proof to estimate the convergence order of the solution $v_h^n$.

Let $d=\frac{T}{s}$. If $d$ is an integer, then the time period $T$ is divided into $d$ segments, where the length of each segment is $s$. If $d$ is not an integer, then the time period $T$ is divided into $\left[d\right]+1$ segments, where the length of the first $\left[d\right]$ segments are $s$ and the length of the last segment is $T-\left[d\right]s$. Since $m=\frac{s}{\tau }$, then $n=\mathrm{1,2,}\cdot \cdot \cdot \mathrm{,}m$ on the first segment, $n=m+\mathrm{1,}m+\mathrm{2,}\cdot \cdot \cdot \mathrm{,2}m$ on the second segment, and so on. Now let’s first prove that (35)-(36) are true on the first segment.

Starting from (52) we estimate the order of convergence of the solution $u_h^n$. The function is known at $\left[-s\mathrm{,0}\right]$ and $n=\mathrm{1,2,}\cdot \cdot \cdot \mathrm{,}m$, so ${\rho }^{n-m}=0$, ${\epsilon }^{n-m}=0$ on the first segment. Substitute ${\rho }^{n-m}=0$, ${\epsilon }^{n-m}=0$ into (52) to get