Numerical Methods and Comparison for the Dirac Equation in the Nonrelativistic Limit Regime

Abstract

We analyze rigorously error estimates and compare numerically spatial/temporal resolution of various numerical methods for the discretization of the Dirac equation in the nonrelativistic limit regime, involving a small dimensionless parameter $0<\varepsilon \ll 1$ which is inversely proportional to the speed of light. In this limit regime, the solution is highly oscillatory in time, i.e. there are propagating waves with wavelength $O(\varepsilon ^2)$ and O(1) in time and space, respectively. We begin with several frequently used finite difference time domain (FDTD) methods and obtain rigorously their error estimates in the nonrelativistic limit regime by paying particular attention to how error bounds depend explicitly on mesh size h and time step $\tau$ as well as the small parameter $\varepsilon$. Based on the error bounds, in order to obtain ‘correct’ numerical solutions in the nonrelativistic limit regime, i.e. $0<\varepsilon \ll 1$, the FDTD methods share the same $\varepsilon$-scalability on time step and mesh size as: $\tau =O(\varepsilon ^3)$ and $h=O(\sqrt{\varepsilon })$. Then we propose and analyze two numerical methods for the discretization of the Dirac equation by using the Fourier spectral discretization for spatial derivatives combined with the symmetric exponential wave integrator and time-splitting technique for temporal derivatives, respectively. Rigorous error bounds for the two numerical methods show that their $\varepsilon$-scalability is improved to $\tau =O(\varepsilon ^2)$ and $h=O(1)$ when $0<\varepsilon \ll 1$. Extensive numerical results are reported to support our error estimates.

Appendices

Appendix 1

Proof of Theorem 2.1 for the LFFD method

Define the local truncation error $\tilde{\xi }^{n}=(\tilde{\xi }_0^n,\tilde{\xi }_1^n,\ldots ,\tilde{\xi }_M^n)^T\in X_{M}$ of the LFFD (2.6) with (2.10) and (2.11) as follows, for $0\le j\le M-1$ and $n\ge 1$,

$$\begin{aligned} \tilde{\xi }_{j}^{n}:&=\left[ i\delta _{t}\varPhi + \frac{i}{\varepsilon }\sigma _1\delta _{x}\varPhi -\frac{1}{\varepsilon ^2}\sigma _3\varPhi + \left( A_{1,j}^n\sigma _1-V_j^nI_2\right) \varPhi \right] _{t=t_n,x=x_j}, \end{aligned}$$

(6.1)

$$\begin{aligned} \tilde{\xi }_{j}^{0}:&=i\delta _{t}^+\varPhi (0,x_{j})+\frac{i}{\varepsilon }\sigma _1\delta _x\varPhi _0(x_j)-\left( \frac{1}{\varepsilon ^2}\sigma _3 +V_j^0I_2-A_{1,j}^0\sigma _1\right) \varPhi _0(x_j). \end{aligned}$$

(6.2)

Applying the Taylor expansion in (6.1) and (6.2) we obtain for $j=0,1,\ldots , M-1$ and $n\ge 1$,

$$\begin{aligned} \tilde{\xi }_{j}^0=i\tau \partial _{tt}\varPhi (\tau ^{\prime },x_j)+ \frac{i}{\varepsilon }h^2\partial _{xxx}\varPhi _0(x^\prime ),\quad \tilde{\xi }_{j}^n=i\tau ^2\partial _{ttt}\varPhi (t^\prime ,x_j) +\frac{i}{\varepsilon }h^2\partial _{xxx}\varPhi (t_n,x^{\prime \prime }), \end{aligned}$$

where $t^\prime \in (0,\tau )$, $t^{\prime \prime }\in (t_{n-1},t_{n+1})$, $x^\prime ,x^{\prime \prime }\in (x_{j-1},x_{j+1})$. Noticing (2.1) and the assumptions (A) and (B), we have

$$\begin{aligned} |\tilde{\xi }_{j}^0|\lesssim \frac{\tau }{\varepsilon ^4}+\frac{h^2}{\varepsilon }, \quad |\tilde{\xi }_{j}^n|\lesssim \frac{\tau ^2}{\varepsilon ^6}+\frac{h^2}{\varepsilon }, \quad j=0,1,\ldots , M-1, \quad n\ge 1, \end{aligned}$$

(6.3)

which immediately implies

$$\begin{aligned} \Vert \tilde{\xi }^n\Vert _{l^\infty }=\max _{0\le j\le M-1}|\tilde{\xi }_j^n| \lesssim \frac{\tau ^2}{\varepsilon ^6}+\frac{h^2}{\varepsilon }, \ \Vert \tilde{\xi }^n\Vert _{l^2}\lesssim \Vert \tilde{\xi }^n\Vert _{l^\infty } \lesssim \frac{\tau ^2}{\varepsilon ^6}+\frac{h^2}{\varepsilon }, \ n\ge 1. \end{aligned}$$

(6.4)

Subtracting (2.6) from (6.1), noticing (2.32), we get for $0\le j\le M-1$ and $n\ge 1$,

$$\begin{aligned} i\delta _t\mathbf{e} _{j}^{n}=-\frac{i}{\varepsilon }\sigma _1\delta _x\mathbf{e} _{j}^{n}+\frac{1}{\varepsilon ^2}\sigma _3 \mathbf{e} _{j}^{n}+\left( V_j^{n}I_2-A_{1,j}^{n}\sigma _1 \right) \mathbf{e} _{j}^{n}+\tilde{\xi }_j^n, \end{aligned}$$

(6.5)

where the boundary and initial conditions are given as

$$\begin{aligned} \mathbf{e} _0^{n}=\mathbf{e} _{M}^{n}, \quad \mathbf{e} _{-1}^{n}=\mathbf{e} _{M-1}^n, \quad n\ge 0,\quad \mathbf{e} _j^0=\mathbf{0}, \quad j=0,1,\ldots ,M. \end{aligned}$$

(6.6)

For the first step, we have

$$\begin{aligned} \Vert \mathbf{e} ^1\Vert _{l^2}=\tau \Vert \tilde{\xi }^0\Vert _{l^2}\lesssim \frac{\tau ^2}{\varepsilon ^4}+\frac{\tau h^2}{\varepsilon }\lesssim \frac{h^2}{\varepsilon }+\frac{\tau ^2}{\varepsilon ^6}. \end{aligned}$$

(6.7)

Denote ${\mathcal {E}}^{n+1}$ for $n=0, 1, \ldots$ as

$$\begin{aligned} {\mathcal {E}}^{n+1}=&\Vert \mathbf{e} ^{n+1}\Vert ^2_{l^2}+\Vert \mathbf{e} ^{n}\Vert _{l^2}^2+2\,\text {Re}\left( \tau h\sum \limits _{j=0}^{M-1}(\mathbf{e} _j^{n+1})^*\sigma _1\delta _x\mathbf{e} ^n_j\right) \nonumber \\&-2\,\text {Im}\left( \frac{\tau h}{\varepsilon ^2}\sum \limits _{j=0}^{M-1}(\mathbf{e} _j^{n+1})^*\sigma _3\mathbf{e} _j^n\right) ; \end{aligned}$$

(6.8)

and under the stability condition (2.33), e.g., $\tau \le \frac{\varepsilon ^2\tau _1 h}{\varepsilon ^2h V_{\mathrm{max}}+\sqrt{h^2+\varepsilon ^2(1+\varepsilon h A_{1,\mathrm{max}})^2}}$ with $\tau _1=\frac{1}{4}$, which implies $\frac{\tau }{h}\le \frac{1}{4}$ and $\frac{\tau }{\varepsilon ^2}\le \frac{1}{4}$, using Cauchy inequality, we can get that

$$\begin{aligned} \frac{1}{2}\left( \Vert \mathbf{e} ^{n+1}\Vert ^2_{l^2}+\Vert \mathbf{e} ^{n}\Vert _{l^2}^2\right) \le {\mathcal {E}}^{n+1}\le \frac{3}{2}\left( \Vert \mathbf{e} ^{n+1}\Vert ^2_{l^2}+\Vert \mathbf{e} ^{n}\Vert _{l^2}^2\right) , \quad n\ge 0. \end{aligned}$$

(6.9)

It follows from (6.7) that

$$\begin{aligned} {\mathcal {E}}^1\lesssim \left( \frac{h^2}{\varepsilon }+\frac{\tau ^2}{\varepsilon ^6}\right) ^2. \end{aligned}$$

(6.10)

Multiplying (6.5) from the left by $2h\tau (\mathbf{e} _j^{n+1}+\mathbf{e} _j^{n-1})^*$, taking the imaginary part, then summing for $j=0,1,\ldots ,M-1$, using Cauchy inequality, noting (6.4) and (6.9), we get for $n\ge 1$,

$$\begin{aligned} {\mathcal {E}}^{n+1}-{\mathcal {E}}^{n}\lesssim&h\tau \sum _{j=0}^{M-1}\left( (A_{1,\mathrm max}+V_\mathrm{max})|\mathbf{e} _j^n|+|\tilde{\xi }_j^n| \right) (|\mathbf{e} _j^{n+1}|+|\mathbf{e} _j^{n-1}|)\\ \lesssim&\tau ({\mathcal {E}}^{n+1}+{\mathcal {E}}^{n})+\tau \left( \frac{h^2}{\varepsilon }+\frac{\tau ^2}{\varepsilon ^6}\right) ^2, \quad n\ge 0. \end{aligned}$$

Summing the above inequality for $n=1, 2, \ldots , m-1$, we get

$$\begin{aligned} {\mathcal {E}}^m-{\mathcal {E}}^1\lesssim \tau \sum _{k=1}^m{\mathcal {E}}^k+m\tau \left( \frac{h^2}{\varepsilon }+\frac{\tau ^2}{\varepsilon ^6}\right) ^2, \quad 1\le m\le \frac{T}{\tau }. \end{aligned}$$

(6.11)

Taking $\tau _0$ sufficiently small, using the discrete Gronwall’s inequality and noticing (6.10), we obtain from the above equation that

$$\begin{aligned} {\mathcal {E}}^m \lesssim \left( \frac{h^2}{\varepsilon }+\frac{\tau ^2}{\varepsilon ^6}\right) ^2, \quad 1\le m\le \frac{T}{\tau }, \end{aligned}$$

(6.12)

which immediately implies the error bound (2.34) in view of (6.9). $\square$

Appendix 2

Proof of Theorem 3.1 for the sEWI-FP method

Define the error function $\mathbf{e} ^n(x)$ for $n=0,1,\ldots$ as

$$\begin{aligned} \mathbf{e} ^n(x)= \begin{pmatrix} e_1^n(x)\\ e_2^n(x) \\ \end{pmatrix} :=P_M\varPhi (t_n,x)-\varPhi _M^n(x)=\sum _{l=-M/2}^{M/2-1}\widehat{\mathbf{e} }_l^n e^{i\mu _l(x-a)},\ a\le x\le b.\quad \end{aligned}$$

(6.13)

Using the triangular inequality and standard interpolation result, we get

$$\begin{aligned} \Vert \varPhi ( t_n,x)-\varPhi _M^n(x)\Vert _{L^2}\le & {} \Vert \varPhi ( t_n,x)-P_M\varPhi (t_n,x)\Vert _{L^2}+\Vert \mathbf{e} ^n(x)\Vert _{L^2}\nonumber \\\le & {} h^{m_0}+\Vert \mathbf{e} ^n(x)\Vert _{L^2}, \end{aligned}$$

(6.14)

where $0\le n\le \frac{T}{\tau }$, and the above result means that we only need estimate $\Vert \mathbf{e} ^n(x)\Vert _{L^2}$.

Define the local truncation error $\xi ^n(x)=\sum _{l=-M/2}^{M/2-1}\widehat{\xi }_l^ne^{i\mu _l(x-a)}\in Y_M$ of the sEWI-FP (3.17) for $n\ge 1$ as

$$\begin{aligned} \widehat{\xi }_l^n= \widehat{(\varPhi (t_{n+1}))}_l +2i\sin (\frac{\tau \varGamma _l}{\varepsilon ^2})\,\widehat{(\varPhi (t_n))}_l-\widehat{(\varPhi (t_{n-1})}_l +2i\frac{\varepsilon ^2}{\delta _l}\sin (\frac{\tau \delta _l}{\varepsilon ^2})\widehat{(G(t_n)\varPhi (t_n))}_l, \end{aligned}$$

(6.15)

and for $n=0$ as

$$\begin{aligned} \widehat{\xi }_l^0=\widehat{(\varPhi (\tau ))}_l -e^{-i\tau \varGamma _l/\varepsilon ^2}\widehat{(\varPhi (0))}_l +\varepsilon ^2\varGamma _l^{-1}\left[ I_2-e^{-\frac{i\tau }{\varepsilon ^2}\varGamma _l}\right] \widehat{(G(0)\varPhi (0))}_l, \end{aligned}$$

(6.16)

where we write $\varPhi (t)$ and G(t) in short for $\varPhi (t,x)$ and G(t, x), respectively.

Firstly, we estimate the local truncation error $\xi ^n(x)$. Multiplying both sides of the Dirac equation (2.1) by $e^{i\mu _l(x-a)}$ and integrating over the interval (a, b), we easily recover the equations for $(\widehat{\varPhi (t)})_l$, which are exactly the same as (3.6) with $\varPhi _M$ being replaced by $\varPhi (t,x)$. Replacing $\varPhi _M$ with $\varPhi (t,x)$, we use the same notations $\widehat{F}_l^n(s)$ as in (3.7) and the time derivatives of $\widehat{F}_l^n(s)$ enjoy the same properties of time derivatives of $\varPhi (t,x)$. Thus, the same representation (3.12) holds for $(\widehat{\varPhi (t_n)})_l$ with $n\ge 1$. From the derivation of the EWI method, it is clear that the error $\xi ^n(x)$ comes from the approximations for the integrals in (3.13) and (3.14), and we have

$$\begin{aligned} \widehat{\xi }^0_l=-i\int _0^{\tau }e^{\frac{i(s-\tau )}{\varepsilon ^2}\varGamma _l}(\widehat{F}^0_l(s) -\widehat{F}^0_l(0))ds=-i\int _0^{\tau }\int _0^s e^{\frac{i(s-\tau )}{\varepsilon ^2}\varGamma _l} \partial _{s_1}\widehat{F}^0_l(s_1)\,ds_1ds, \end{aligned}$$

and for $n\ge 1$

$$\begin{aligned} \widehat{\xi }^n_l=&-i\int _0^{\tau }\cos ((s-\tau )\delta _l/\varepsilon ^2) \int _0^s\int _{-s_1}^{s_1}\partial _{s_2s_2} \widehat{F}^{n}_l(s_2)\,ds_2ds_1ds \nonumber \\&+\int _0^\tau \sin \left( \frac{(s-\tau )\varGamma _l}{\varepsilon ^2}\right) \int _{-s}^s\partial _{s_1}\widehat{F}_l^n(s_1) \,ds_1\,ds. \end{aligned}$$

(6.17)

For $n=0$, the above equalities imply $|\widehat{\xi }^0_l|\lesssim \int _0^{\tau }\int _0^s|\partial _{s_1}\widehat{F}^0_l(s_1)|ds_1ds$ and by the Bessel inequality and assumptions (C) and (D), we find

$$\begin{aligned} \Vert \xi ^0(x)\Vert _{L^2}^2=&(b-a)\sum \limits _{l=-M/2}^{M/2-1}|\widehat{\xi }_l^0|^2 \lesssim (b-a)\tau ^2\int _0^\tau \int _0^s\sum \limits _{l=-M/2}^{M/2-1}|\partial _{s_1} \widehat{F}^0_l(s_1)|^2\,ds_1ds\\ \lesssim&\tau ^2\int _0^\tau \int _0^s\Vert \partial _{s_1}(G(s_1)\varPhi (s_1))\Vert _{L^2}^2\,ds_1ds \lesssim \frac{\tau ^4}{\varepsilon ^4}. \end{aligned}$$

Similarly, for $n\ge 1$, we obtain

$$\begin{aligned}&\Vert \xi ^n(x)\Vert _{L^2}^2=(b-a)\sum \limits _{l=-M/2}^{M/2-1}|\widehat{\xi }_l^n|^2\\&\lesssim \tau ^3\int _0^{\tau }\int _0^s\int _{-s_1}^{s_1} \sum \limits _{l=-\frac{M}{2}}^{\frac{M}{2}-1}|\partial _{s_2s_2}\widehat{F}_l^n(s_2)|^2\,ds_2ds_1ds\\&\quad +\tau ^2\int _0^\tau \int _{-s}^{s}\frac{(\tau -s)^2}{\varepsilon ^4}\sum \limits _{l=-\frac{M}{2}}^{\frac{M}{2}-1} |\partial _{\theta _1}\widehat{F}_l^{n-1}(\theta _1)|^2\,d\theta _1\,d\theta \,ds\\&\lesssim \tau ^6\Vert \partial _{tt}(G(t)\varPhi (t))\Vert _{L^\infty ([0,T]; (L^2)^2)}^2+\frac{\tau ^6}{\varepsilon ^4}\Vert \partial _{t}(G(t)\varPhi (t))\Vert _{L^\infty ([0,T]; (L^2)^2)}^2 \lesssim \frac{\tau ^6}{\varepsilon ^8}, \end{aligned}$$

where we have used the assumptions (C) and (D). Hence, we derive that

$$\begin{aligned} \Vert \xi ^0(x)\Vert _{L^2}\lesssim \frac{\tau ^2}{\varepsilon ^2},\quad \Vert \xi ^n(x)\Vert _{L^2}\lesssim \frac{\tau ^3}{\varepsilon ^4},\quad n\ge 1. \end{aligned}$$

(6.18)

Now, we look at the error equations. For each fixed $l=-M/2,\ldots ,M/2-1$, subtracting (3.17) from (6.15), we obtain the equation for the error vector function as

$$\begin{aligned} \widehat{\mathbf{e} }^0_l=\mathbf{0},\quad \widehat{\mathbf{e} }^1_l=\widehat{\xi }^0_l; \quad \widehat{\mathbf{e} }^{n+1}_l-\widehat{\mathbf{e} }^{n-1}_l= -2i\sin (\tau \varGamma _l/\varepsilon ^2)\widehat{\mathbf{e} }^n_l +\widehat{R}^n_l +\widehat{\xi }^n_l, \end{aligned}$$

(6.19)

where $1\le n\le \frac{T}{\tau }-1,$ and $R^n(x)=\sum \limits _{l=-M/2}^{M/2-1}\widehat{R}_l^ne^{i\mu _l(x-a)}\in Y_M$ for $n\ge 1$ is given by

$$\begin{aligned} \widehat{R}^n_l=2i\varepsilon ^2\delta _l^{-1}\sin (\tau \delta _l/\varepsilon ^2)\left( \widehat{(G(t_n)\varPhi (t_n))}_l-\widehat{(G(t_n)\varPhi ^n_M)}_l\right) . \end{aligned}$$

(6.20)

Since $|\varepsilon ^2\delta _l^{-1}\sin (\tau \delta _l/\varepsilon ^2)|\le \tau$, from (6.20) and the assumption (D), we get

$$\begin{aligned} \Vert R^n(x)\Vert _{L^2}^2=&(b-a)\sum \limits _{l=-M/2}^{M/2-1}|\widehat{R}_l^n|^2\nonumber \\ \lesssim&(b-a)\tau ^2\sum \limits _{l=-M/2}^{M/2-1}\left| \widehat{(G(t_n)\varPhi (t_n))}_l-(\widehat{G(t_n)\varPhi _M^n})_l\right| ^2 \nonumber \\ \lesssim&\tau ^2\Vert G(t_n)\varPhi ( t_n,x)-G(t_n)\varPhi _M^n(x)\Vert _{L^2}^2\lesssim \tau ^2\Vert \varPhi ( t_n,x)-\varPhi _M^n(x)\Vert _{L^2}^2\nonumber \\ \lesssim&\tau ^2 h^{2m_0} +\tau ^2\Vert \mathbf{e} ^{n}(x)\Vert _{L^2}^2. \end{aligned}$$

(6.21)

Multiplying both sides of (6.19) by $\left( \widehat{\mathbf{e} }^n_l\right) ^*$ from left, taking the real parts, we obtain

$$\begin{aligned} \text {Re}\left( (\widehat{\mathbf{e} }^n_l)^*\widehat{\mathbf{e} }^{n+1}_l\right) -\text {Re}\left( (\widehat{\mathbf{e} }^n_l)^*\widehat{\mathbf{e} }^{n-1}_l\right) =\text {Re}\left( (\widehat{\mathbf{e} }^n_l)^*(\widehat{R}^n_l +\widehat{\xi }^n_l)\right) , \end{aligned}$$

which implies

$$\begin{aligned} |\widehat{\mathbf{e} }^{n+1}_l|^2+|\widehat{\mathbf{e} }^{n}_l|^2 -|\widehat{\mathbf{e} }^{n+1}_l-\widehat{\mathbf{e} }^{n}_l|^2= & {} |\widehat{\mathbf{e} }^{n}_l|^2+|\widehat{\mathbf{e} }^{n-1}_l|^2 -|\widehat{\mathbf{e} }^{n}_l-\widehat{\mathbf{e} }^{n-1}_l|^2\nonumber \\&+2\,\text {Re} \left( (\widehat{\mathbf{e} }^n_l)^*(\widehat{R}^n_l +\widehat{\xi }^n_l)\right) . \end{aligned}$$

(6.22)

Multiplying both sides of (6.19) by $\left( \widehat{\mathbf{e} }_l^{n+1}-2\widehat{\mathbf{e} }_l^{n}+\widehat{\mathbf{e} }_l^{n-1}\right) ^*$ from left, taking the real parts, we have

$$\begin{aligned}&\left| \widehat{\mathbf{e} }^{n+1}_l- \widehat{\mathbf{e} }^n_l\right| ^2-\left| \widehat{\mathbf{e} }^{n}_l- \widehat{\mathbf{e} }^{n-1}_l\right| ^2 \nonumber \\ =&2\,\text {Im}\left( (\widehat{\mathbf{e} }^{n+1}_l)^*\sin (\tau \varGamma _l/\varepsilon ^2)\widehat{\mathbf{e} }^{n}_l\right) - 2\,\text {Im}\left( (\widehat{\mathbf{e} }^{n}_l)^*\sin (\tau \varGamma _l/\varepsilon ^2)\widehat{\mathbf{e} }^{n-1}_l\right) \nonumber \\&+\text {Re}\left( (\widehat{\mathbf{e} }^{n+1}_l- 2\widehat{\mathbf{e} }^{n}_l+\widehat{\mathbf{e} }^{n-1}_l)^*(\widehat{R}^n_l+\widehat{\xi }_l^n)\right) . \end{aligned}$$

(6.23)

Summing (6.22) and (6.23), then applying Cauchy inequality and triangle inequality, we get

$$\begin{aligned}&|\widehat{\mathbf{e} }^{n+1}_l|^2+|\widehat{\mathbf{e} }^{n}_l|^2-2\, \text {Im}\left( (\widehat{\mathbf{e} }^{n+1}_l)^*\sin (\tau \varGamma _l/\varepsilon ^2) \widehat{\mathbf{e} }^{n}_l\right) \nonumber \\ \le&|\widehat{\mathbf{e} }^{n}_l|^2+|\widehat{\mathbf{e} }^{n-1}_l|^2 -2\text {Im}\left( (\widehat{\mathbf{e} }^{n}_l)^*\sin (\tau \varGamma _l/\varepsilon ^2) \widehat{\mathbf{e} }^{n-1}_l\right) \nonumber \\&+\tau (|\widehat{\mathbf{e} }^{n+1}_l|^2+|\widehat{\mathbf{e} }^{n-1}_l|^2) +\frac{1}{\tau }(|\widehat{R}^n_l|^2+|\widehat{\xi }^n_l|^2). \end{aligned}$$

(6.24)

Denote

$$\begin{aligned} \mathcal{E}^n=\left\| \mathbf{e} ^{n+1}(x)\right\| _{L^2}^2+\left\| \mathbf{e} ^{n}(x)\right\| _{L^2}^2-2(b-a)\sum \limits _{l=-M/2}^{M/2-1} \text {Im}\left( (\widehat{\mathbf{e} }^{n+1}_l)^*\sin (\tau \varGamma _l/\varepsilon ^2)\widehat{\mathbf{e} }^{n}_l\right) , \end{aligned}$$

(6.25)

and it follows from the stability constraint (3.25) that the matrix $l^2$ norm satisfies $\Vert \sin (\frac{\tau \varGamma _l}{\varepsilon ^2})\Vert _{l^2}\le \sin (\tau \delta _l/\varepsilon ^2)\le \sin (\pi /3)=\sqrt{3}/2$, which yield the following conclusion

$$\begin{aligned} \mathcal{E}^n\ge&\sum _{k=n}^{n+1}\left\| \mathbf{e} ^{k}(x)\right\| _{L^2}^2- \frac{\sqrt{3}}{2}(b-a)\sum \limits _{l=-M/2}^{M/2-1} \left( \left| \widehat{\mathbf{e} }^{n+1}_l\right| ^2+\left| \widehat{\mathbf{e} }^{n}_l\right| ^2\right) \nonumber \\ =&\frac{2-\sqrt{3}}{2}(\left\| \mathbf{e} ^{n+1}(x)\right\| _{L^2}^2+ \left\| \mathbf{e} ^{n}(x)\right\| _{L^2}^2). \end{aligned}$$

(6.26)

Multiplying (6.24) by $b-a$ and summing together for $l=-M/2, \ldots , M/2-1$, in view of the Bessel inequality, we obtain

$$\begin{aligned} \mathcal{E}^{n}-\mathcal{E}^{n-1}\lesssim&\tau (\left\| \mathbf{e} ^{n+1}(x)\right\| _{L^2}^2+\left\| \mathbf{e} ^{n}(x)\right\| _{L^2}^2+\left\| \mathbf{e} ^{n-1}(x)\right\| _{L^2}^2)\nonumber \\&+\frac{1}{\tau }\Vert R^n(x)\Vert ^2_{L^2}+\frac{1}{\tau }\Vert \xi ^n(x)\Vert ^2_{L^2},\quad n\ge 1. \end{aligned}$$

(6.27)

Summing (6.27) for $n=1,\ldots ,m-1$, using (6.21) and (6.18), we derive

$$\begin{aligned} \mathcal{E}^{m-1}-\mathcal{E}^0\lesssim \tau \sum \limits _{k=1}^{m}\left\| \mathbf{e} ^{k}(x)\right\| _{L^2}^2+\frac{m\tau ^5}{\varepsilon ^8}+m\tau h^{2m_0},\quad 1\le m\le \frac{T}{\tau }. \end{aligned}$$

(6.28)

Since $\mathbf{e} ^0(x)=\mathbf{0}$ and $\mathcal{E}^{m-1}$ is bounded from below (6.26), we have for $1\le m\le \frac{T}{\tau }$,

$$\begin{aligned} \frac{2-\sqrt{3}}{2}(\Vert \mathbf{e} ^{m}(x)\Vert _{L^2}^2+\Vert \mathbf{e} ^{m-1}(x)\Vert _{L^2}^2)-\Vert \mathbf{e} ^{1}(x)\Vert _{L^2}^2\lesssim \tau \sum \limits _{k=1}^{m}\Vert \mathbf{e} ^{k}(x)\Vert _{L^2}^2+\frac{m\tau ^5}{\varepsilon ^8}+m\tau h^{2m_0}. \end{aligned}$$

(6.29)

Noticing $\Vert \mathbf{e} ^1(x)\Vert _{L^2}\lesssim \frac{\tau ^2}{\varepsilon ^2}\lesssim \frac{\tau ^2}{\varepsilon ^4}$, the discrete Gronwall’s inequality will imply that for sufficiently small $\tau$,

$$\begin{aligned} \left\| \mathbf{e} ^{m}(x)\right\| _{L^2}^2\lesssim h^{2m_0}+\frac{\tau ^4}{\varepsilon ^8},\quad 1\le m\le \frac{T}{\tau }. \end{aligned}$$

(6.30)

Combining (6.14) and (6.30), we draw the conclusion (3.26). $\square$

Appendix 3

Extensions of the sEWI-FS (3.16–3.17) and TSFP (4.4) in 2D and 3D

The sEWI-FS (3.16–3.17), sEWI-FP (3.18–3.19) and TSFP (4.4) can be easily extended to 2D and 3D with tensor grids by modifying the matrices $\varGamma _l$ in (3.8) and G(t, x) in (4.5) in the TSFP case. For the reader’s convenience, we present the modifications of $\varGamma _l$ in (3.8) and G(t, x) in (4.5) in 2D and 3D as follows.

For the Dirac equation (1.21) in 2D, i.e. we take $d=2$ in (1.21). The problem is truncated on $\varOmega =(a_1, b_1)\times (a_2, b_2)$ with mesh sizes $h_1=(b_1-a_1)/M_1$ and $h_2=(b_2-a_2)/M_2$ ($M_1,M_2$ two even positive integers) in the x- and y-direction, respectively. The wave function $\varPhi$ is a two-component vector, and the matrix $\varGamma _l$ in (3.8) will be replaced by

$$\begin{aligned} \varGamma _{jk}=\begin{pmatrix} 1 &{} \varepsilon \mu _j^{(1)}-i\varepsilon \mu _k^{(2)} \\ \varepsilon \mu _j^{(1)}+i\varepsilon \mu _k^{(2)} &{} -1\\ \end{pmatrix},\ \mu _j^{(1)}=\frac{2j\pi }{b_1-a_1},\ \mu _k^{(2)}=\frac{2k\pi }{b_2-a_2}, \end{aligned}$$

(6.31)

where $-\frac{M_1}{2}\le j\le \frac{M_1}{2}-1$, $-\frac{M_2}{2}\le k\le \frac{M_2}{2}-1$, and the Schur decomposition $\varGamma _{jk}=Q_{jk}D_{jk}Q_{jk}^*$ is given as

$$\begin{aligned} Q_{jk}=\begin{pmatrix} \frac{1+\delta _{jk}}{\sqrt{2\delta _{jk}(1+\delta _{jk})}} &{}\frac{-\varepsilon \mu _{j}^{(1)}+i\varepsilon \mu _k^{(2)}}{\sqrt{2\delta _{jk}(1+\delta _{jk})}}\\ \frac{\varepsilon \mu _{j}^{(1)}+i\varepsilon \mu _k^{(2)}}{\sqrt{2\delta _{jk}(1+\delta _{jk})}} &{}\frac{1+\delta _{jk}}{\sqrt{2\delta _{jk}(1+\delta _{jk})}} \end{pmatrix}, \quad D_{jk}=\begin{pmatrix} \delta _{jk} &{}0\\ 0 &{}-\delta _{jk}\\ \end{pmatrix}, \end{aligned}$$

(6.32)

where

$$\begin{aligned} \delta _{jk}=\sqrt{1+\varepsilon ^2(\mu _{j}^{(1)})^2+\varepsilon ^2(\mu _k^{(2)})^2}. \end{aligned}$$

(6.33)

The matrix $\int _{t_n}^{t_{n+1}}G(t,\mathbf{x} )dt$ in (4.5) becomes $\int _{t_n}^{t_{n+1}}G(t,\mathbf{x} )dt$ and the Schur decomposition $\int _{t_n}^{t_{n+1}}G(t,\mathbf{x} )dt=P_\mathbf{x} \varLambda _\mathbf{x} P_{\mathbf{x} }^*$ with $V_\mathbf{x} ^{(1)}=\int _{t_n}^{t_{n+1}}V(t,\mathbf{x} )dt$, $A_{l,\mathbf{x} }^{(1)}=\int _{t_n}^{t_{n+1}}A_l(t,\mathbf{x} )dt$ for $l=1,2$, $\lambda _\mathbf{x} ^{(1)}=\sqrt{|A_{1,\mathbf{x} }^{(1)}|^2+|A_{2,\mathbf{x} }^{(1)}|^2}$, $\varLambda _\mathbf{x} =\mathrm{diag}(\varLambda _{\mathbf{x} ,+},\varLambda _{\mathbf{x} ,-})$, $\varLambda _{\mathbf{x} ,\pm }=V_\mathbf{x} ^{(1)}\pm \lambda _\mathbf{x} ^{(1)}$, and $P_\mathbf{x} =I_2$ if $\lambda _\mathbf{x} ^{(1)}=0$ and otherwise

$$\begin{aligned} P_\mathbf{x} =\begin{pmatrix}\frac{1}{\sqrt{2}}&{}\frac{A_{1,\mathbf{x} }^{(1)}-iA_{2,\mathbf{x} }^{(1)}}{\sqrt{2}\lambda _\mathbf{x} ^{(1)}}\\ -\frac{A_{1,\mathbf{x} }^{(1)}+iA_{2,\mathbf{x} }^{(1)}}{\sqrt{2}\lambda _\mathbf{x} ^{(1)}}&{}\frac{1}{\sqrt{2}} \end{pmatrix}. \end{aligned}$$

(6.34)

For the Dirac equation (1.9) in 3D, i.e. we take $d=3$ in (1.9). The problem is truncated on $\varOmega =(a_1, b_1)\times (a_2, b_2)\times (a_3, b_3)$ with mesh sizes $h_1=(b_1-a_1)/M_1$, $h_2=(b_2-a_2)/M_2$ and $h_3=(b_3-a_3)/M_3$ ($M_1,M_2,M_3$ three even positive integers) in x-, y- and z-direction, respectively. The wave function $\varPsi$ is a four-component vector, and the matrix $\varGamma _l$ in (3.8) will be replaced by $\varGamma _{jkl}$ as:

$$\begin{aligned} \varGamma _{jkl}=\begin{pmatrix} 1 &{} 0 &{} \varepsilon \mu _l^{(3)} &{} \varepsilon \mu _j^{(1)}-i\varepsilon \mu _k^{(2)} \\ 0 &{} 1 &{} \varepsilon \mu _j^{(1)}+i\varepsilon \mu _k^{(2)} &{} -\varepsilon \mu _l^{(3)} \\ \varepsilon \mu _l^{(3)} &{} \varepsilon \mu _j^{(1)}-i\varepsilon \mu _k^{(2)} &{} -1 &{} 0 \\ \varepsilon \mu _j^{(1)}+i\varepsilon \mu _k^{(2)} &{} -\varepsilon \mu _l^{(3)} &{} 0 &{} -1 \\ \end{pmatrix}, \end{aligned}$$

(6.35)

where $-\frac{M_1}{2}\le j\le \frac{M_1}{2}-1,-\frac{M_2}{2}\le k\le \frac{M_2}{2}-1,-\frac{M_3}{2}\le l\le \frac{M_3}{2}-1$ and

$$\begin{aligned} \mu _j^{(1)}=\frac{2j\pi }{b_1-a_1},\quad \mu _k^{(2)}=\frac{2k\pi }{b_2-a_2},\quad \mu _l^{(3)}=\frac{2l\pi }{b_3-a_3}. \end{aligned}$$

(6.36)

The eigenvalues of $\varGamma _{jkl}$ are

$$\begin{aligned} \delta _{jkl}, \delta _{jkl}, -\delta _{jkl}, -\delta _{jkl},\quad \text {with}\quad \delta _{jkl}=\sqrt{1+\varepsilon ^2\left| \mu _j^{(1)}\right| ^2+\varepsilon ^2\left| \mu _{k}^{(2)}\right| ^2+ \varepsilon ^2\left| \mu _l^{(3)}\right| ^2}. \end{aligned}$$

The corresponding eigenvectors are

$$\begin{aligned} \mathbf {v}^{(1)}_{jkl}=\begin{pmatrix} 1+\delta _{jkl}\\ 0\\ \varepsilon \mu _l^{(3)}\\ \varepsilon \mu _j^{(1)}+i\varepsilon \mu _{k}^{(2)}\end{pmatrix},\quad \mathbf {v}^{(2)}_{jkl}=\begin{pmatrix} 0\\ 1+\delta _{jkl}\\ \varepsilon \mu _j^{(1)}-i\varepsilon \mu _{k}^{(2)}\\ -\varepsilon \mu _l^{(3)}\end{pmatrix}, \end{aligned}$$

and

$$\begin{aligned} \mathbf {v}^{(3)}_{jkl}=\begin{pmatrix} -\varepsilon \mu _l^{(3)}\\ -\varepsilon \mu _j^{(1)}-i\varepsilon \mu _{k}^{(2)}\\ 1+\delta _{jkl}\\ 0\end{pmatrix},\quad \mathbf {v}^{(4)}_{jkl}=\begin{pmatrix} -\varepsilon \mu _j^{(1)}+i\varepsilon \mu _{k}^{(2)}\\ \varepsilon \mu _l^{(3)}\\ 0\\ 1+\delta _{jkl}\end{pmatrix}. \end{aligned}$$

Then the Schur decomposition $\varGamma _{jkl}=Q_{jkl}D_{jkl}Q^*_{jkl}$ is given as

$$\begin{aligned} D_{jkl}=\mathrm {diag}(\delta _{jkl},\delta _{jkl},-\delta _{jkl},-\delta _{jkl}),\quad Q_{jkl}=\frac{1}{\sqrt{2\delta _{jkl}(1+\delta _{jkl})}}\left( \mathbf {v}^{(1)}_{jkl}, \mathbf {v}^{(2)}_{jkl},\mathbf {v}^{(3)}_{jkl},\mathbf {v}^{(4)}_{jkl}\right) . \end{aligned}$$

The matrix $\int _{t_n}^{t_{n+1}}G(t,\mathbf{x} )dt$ in (4.5) becomes $\int _{t_n}^{t_{n+1}}G(t,\mathbf{x} )dt$ and the Schur decomposition $\int _{t_n}^{t_{n+1}}G(t,\mathbf{x} )dt=P_\mathbf{x} \varLambda _\mathbf{x} P_{\mathbf{x} }^*$ with $V_\mathbf{x} ^{(1)}=\int _{t_n}^{t_{n+1}}V(t,\mathbf{x} )dt$, $A_{l,\mathbf{x} }^{(1)}=\int _{t_n}^{t_{n+1}}A_l(t,\mathbf{x} )dt$ for $l=1,2,3$, $\lambda _\mathbf{x} ^{(1)}=\sqrt{|A_{1,\mathbf{x} }^{(1)}|^2+|A_{2,\mathbf{x} }^{(1)}|^2+|A_{3,\mathbf{x} }^{(1)}|^2}$, $\varLambda _\mathbf{x} =\mathrm{diag}(\varLambda _{\mathbf{x} ,+},\varLambda _{\mathbf{x} ,+},\varLambda _{\mathbf{x} ,-},\varLambda _{\mathbf{x} ,-})$, $\varLambda _{\mathbf{x} ,\pm }=V_\mathbf{x} ^{(1)}\pm \lambda _\mathbf{x} ^{(1)}$, and $P_\mathbf{x} =I_4$ if $\lambda _\mathbf{x} ^{(1)}=0$ and otherwise $P_{\mathbf{x} }=(\mathbf{u}_\mathbf{x} ^{(1)},\mathbf{u}_\mathbf{x} ^{(2)},\mathbf{u}_\mathbf{x} ^{(3)},\mathbf{u}_\mathbf{x} ^{(4)})$,

$$\begin{aligned} \mathbf{u}_\mathbf{x} ^{(1)}=\begin{pmatrix}\frac{-A_{1,\mathbf{x} }^{(1)}+iA_{2,\mathbf{x} }^{(1)}}{\sqrt{2}\lambda _\mathbf{x} ^{(1)}}\\ \frac{A_{3,\mathbf{x} }^{(1)}}{\sqrt{2}\lambda _\mathbf{x} ^{(1)}}\\ 0\\ \frac{1}{\sqrt{2}}\end{pmatrix},\quad \mathbf{u}_\mathbf{x} ^{(2)}=\begin{pmatrix}\frac{-A_{3,\mathbf{x} }^{(1)}}{\sqrt{2}\lambda _\mathbf{x} ^{(1)}}\\ \frac{-A_{1,\mathbf{x} }^{(1)}-iA_{2,\mathbf{x} }^{(1)}}{\sqrt{2}\lambda _\mathbf{x} ^{(1)}}\\ \frac{1}{\sqrt{2}}\\ 0\end{pmatrix}, \end{aligned}$$

and

$$\begin{aligned} \mathbf{u}_\mathbf{x} ^{(3)}=\begin{pmatrix}0\\ \frac{1}{\sqrt{2}}\\ \frac{A_{1,\mathbf{x} }^{(1)}-iA_{2,\mathbf{x} }^{(1)}}{\sqrt{2}\lambda _\mathbf{x} ^{(1)}}\\ \frac{-A_{3,\mathbf{x} }}{\sqrt{2}\lambda _\mathbf{x} ^{(1)}}\end{pmatrix},\quad \mathbf{u}_\mathbf{x} ^{(4)}=\begin{pmatrix}\frac{1}{\sqrt{2}}\\ 0\\ \frac{A_{3,\mathbf{x} }^{(1)}}{\sqrt{2}\lambda _\mathbf{x} ^{(1)}}\\ \frac{A_{1,\mathbf{x} }^{(1)}+iA_{2,\mathbf{x} }^{(1)}}{\sqrt{2}\lambda _\mathbf{x} ^{(1)}}\end{pmatrix} . \end{aligned}$$

For the Dirac equation (1.9) in 2D, we simply let $\mu _{l}^{(3)}=0$, $A_3(t,\mathbf{x} )\equiv 0$ in the above 3D case; and for the Dirac equation (1.9) in 1D, we let $\mu _{k}^{(2)}=\mu _{l}^{(3)}=0$, $A_2(t,\mathbf{x} )=A_3(t,\mathbf{x} )\equiv 0$ in the above 3D case. Then the sEWI-FP (3.18–3.19) and TSFP (4.4) can be designed accordingly for the Dirac equation (1.9) in 2D and 1D.