A conservative sine pseudo-spectral-difference method for multi-dimensional coupled Gross–Pitaevskii equations

Abstract

In this paper, a sine pseudo-spectral-difference scheme that preserves the discrete mass and energy is presented and analyzed for the coupled Gross–Pitaevskii equations with Dirichlet boundary conditions in several spatial dimensions. The Crank–Nicolson finite difference method is employed for approximating the time derivative, and the second-order sine spectral differentiation matrix is deduced and applied in spatial discretization. Without any restrictions on the grid ratios, optimal error estimates are established by utilizing the discrete energy method and the equivalence of (semi-)norms. An accelerated algorithm is developed to speed up the numerical implementation with the help of fast sine transform. Numerical examples are tested to confirm the effectiveness and high accuracy of the method.

Appendices

Appendix A: Proof of Lemma 2.1

To obtain the second-order SSDM, the following lemma needs to be verified at first.

Lemma A.1 1

Denote 𝜃_x := μ_x(x − x_j), 𝜃_y := μ_y(y − y_k), $\theta _{x}^{\prime }:=\mu _{x}(x+x_{j}-2a)$, $\theta _{y}^{\prime }:=\mu _{y}(y+y_{k}-2c)$. The interpolation basis functions can be reformulated as:

$$ \begin{array}{@{}rcl@{}} \mathrm{X}_{j}(x)=\mathfrak{S}(N_{x},\mu_{x},\theta_{x})-\mathfrak{S}(N_{x},\mu_{x},\theta_{x}^{\prime}),\ \ \mathrm{Y}_{k}(y)=\mathfrak{S}(N_{y},\mu_{y},\theta_{y})-\mathfrak{S}(N_{y},\mu_{y},\theta_{y}^{\prime}), \end{array} $$

with $\mathfrak {S}(N,\mu ,\theta ):=\frac {1}{2N}\big [\sin \limits (N\theta )\cot \frac {\theta }{2}-\cos \limits (N\theta )\big ]$.

Proof

We just prove the assertion of X_j(x). The other one can be deduced correspondingly. Applying the prosthaphaeresis formula and the definitions of 𝜃_x and $\theta _{x}^{\prime }$, (??) implies that

$$ \begin{array}{@{}rcl@{}} \mathrm{X}_{j}(x)&=&\frac{1}{2N_{x}}\sum\limits_{l=1}^{N_{x}-1}\bigg[1+2\cos(\theta_{x}l)-\big(1+2\cos(\theta_{x}^{\prime}l)\big)\bigg]\\ &=&\frac{1}{2N_{x}}\bigg[\sin\big((N_{x}-\frac{1}{2})\theta_{x}\big)\csc\frac{\theta_{x}}{2} -\sin\big((N_{x}-\frac{1}{2})\theta_{x}^{\prime}\big)\csc\frac{\theta_{x}^{\prime}}{2}\bigg]. \end{array} $$

Invoking to the identity

$$ \sin\big((N-\frac{1}{2})\theta\big)\csc\frac{\theta}{2}=\frac{\sin(N\theta)\cos\frac{\theta}{2} -\cos(N\theta)\sin\frac{\theta}{2}}{\sin\frac{\theta}{2}} =\sin(N\theta)\cot\frac{\theta}{2}-\cos(N\theta), $$

we obtain the desired results and complete the proof of Lemma A.1. □

Proof Proof of Lemma 2.1

For simplicity, we also verify the assertion in the x-direction and the assertion of y-direction is equally acceptable. By means of the above lemma, the second-order derivative of X_j(x) can be given by:

$$ \begin{array}{@{}rcl@{}} \mathrm{X}_{j}^{\prime\prime}(x)= \mathfrak{\ddot{S}}(N_{x},\mu_{x},\theta_{x})-\mathfrak{\ddot{S}}(N_{x},\mu_{x},\theta_{x}^{\prime}), \end{array} $$

where the second-order derivative $\mathfrak {\ddot {S}}(N,\mu ,\theta )$ is obtained directly by:

$$ \begin{array}{@{}rcl@{}} &\mathfrak{\ddot{S}}(N,\mu,\theta)=-\frac{N\mu^{2}}{2}\sin(N\theta)\cot\frac{\theta}{2}-\frac{\mu^{2}}{2}\cos(N\theta)\csc^{2}\frac{\theta}{2}+ \frac{\mu^{2}}{4N}\sin(N\theta)\frac{\cos\frac{\theta}{2}}{\sin^{3}\frac{\theta}{2}}+\frac{N\mu^{2}}{2}\cos(N\theta). \end{array} $$

To obtain the second-order SSDM, we divide the proof into two cases:

• (i) If x = x_p≠x_j, the definitions of 𝜃_x and $\theta _{x}^{\prime }$ lead to:

  $$ \begin{array}{@{}rcl@{}} \sin(N_{x}\theta_{x})=\sin(N_{x}\theta_{x}^{\prime})=0,\ \ \cos(N_{x}\theta_{x})=\cos(N_{x}\theta_{x}^{\prime})=(-1)^{j+p}. \end{array} $$

  Substituting the results into $\mathrm {X}_{j}^{\prime \prime }(x)$ and using the definition of μ_x, for j≠p, one has:

  $$ \begin{array}{@{}rcl@{}} \mathrm{X}_{j}^{\prime\prime}(x_{p})&=&(-1)^{j+p+1}\frac{{\mu_{x}^{2}}}{2}(\csc^{2}\frac{\theta_{x}}{2}-\csc^{2}\frac{\theta_{x}^{\prime}}{2})\\ &=&(-1)^{j+p+1}\frac{{\mu_{x}^{2}}}{2}\bigg[\csc^{2}\big(\frac{\mu_{x}}{2}(j-p)h_{x}\big)- \csc^{2}\big(\frac{\mu_{x}}{2}(j+p)h_{x}\big)\bigg]. \end{array} $$

  This completes the proof of the first case of (??).

• (ii) If x = x_p = x_j, the first term of $\mathrm {X}_{j}^{\prime \prime }(x)$, i.e. $\mathfrak {\ddot {S}}(N_{x},\mu _{x},\theta _{x})$ equals to the following:

  $$ \begin{array}{@{}rcl@{}} &\frac{\frac{{\mu_{x}^{2}}}{4N_{x}}\sin(N_{x}\theta_{x})\cos\frac{\theta_{x}}{2}-\frac{N_{x}{\mu_{x}^{2}}}{2}\sin(N_{x}\theta_{x})\cos\frac{\theta_{x}}{2}\sin^{2} \frac{\theta_{x}}{2}-\frac{{\mu_{x}^{2}}}{2}\cos(N_{x}\theta_{x})\sin\frac{\theta_{x}}{2} }{\sin^{3}\frac{\theta_{x}}{2}}+\frac{N_{x}{\mu_{x}^{2}}}{2}\cos(N_{x}\theta_{x}). \end{array} $$

  We apply Taylor’s expansion for each term of the numerator and find:

  $$ \begin{array}{@{}rcl@{}} &\sin(N_{x}\theta_{x})\cos\frac{\theta_{x}}{2}=\Big(N_{x}\theta_{x} -\frac{(N_{x}\theta_{x})^{3}}{3!}+\cdots\Big)\Big(1-\frac{(\frac{\theta_{x}}{2})^{2}}{2!}+\cdots\Big)\\ &\ \ \ \ \ \ \ \ \ \ \ \quad \ \ \ \ \ \ \ \ =N_{x}\theta_{x}-N_{x}\big(\frac{\theta_{x}}{2}\big)^{3}-\frac{4{N_{x}^{3}}}{3}\big(\frac{\theta_{x}}{2}\big)^{3}+O\big({\theta_{x}^{5}}\big),\\ &\sin(N_{x}\theta_{x})\cos\frac{\theta_{x}}{2}\sin^{2}\frac{\theta_{x}}{2}=\Big(N_{x}\theta_{x} -\frac{(N_{x}\theta_{x})^{3}}{3!}+\cdots\Big)\Big(1-\frac{(\frac{\theta_{x}}{2})^{2}}{2!}+\cdots\Big)\Big(\frac{\theta_{x}}{2}-\frac{(\frac{\theta_{x}}{2})^{3}}{3!}+\cdots\Big)^{2}\\ &\ \ \ \ \ \ \ \ \ \ \ \quad \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =2N_{x}\big(\frac{\theta_{x}}{2}\big)^{3}+O\big({\theta_{x}^{5}}\big),\\ &\cos(N_{x}\theta_{x})\sin\frac{\theta_{x}}{2}=\Big(1-\frac{(N_{x}\theta_{x})^{2}}{2!}+\cdots\Big)\Big(\frac{\theta_{x}}{2}-\frac{(\frac{\theta_{x}}{2})^{3}}{3!}+\cdots\Big)\\ &\ \ \ \ \ \ \ \ \ \ \ \quad \ \ \ \ \ \ \ \ =\frac{\theta_{x}}{2}-\frac{1}{6}\big(\frac{\theta_{x}}{2}\big)^{3}-2{N_{x}^{2}}\big(\frac{\theta_{x}}{2}\big)^{3}+O\big({\theta_{x}^{5}}\big). \end{array} $$

  Since $\theta _{x} \rightarrow 0$ as $x_{p} \rightarrow x_{j}$. We combine these four equalities and obtain:

  $$ \begin{array}{@{}rcl@{}} \mathfrak{\ddot{S}}(N_{x},\mu_{x},\theta_{x}) \rightarrow -\frac{{\mu_{x}^{2}}}{6}+\frac{N_{x}{\mu_{x}^{2}}}{2}-\frac{{N_{x}^{2}}{\mu_{x}^{2}}}{3},\quad\text{as} x_{p} \rightarrow x_{j}. \end{array} $$

  For the second term of $\mathrm {X}_{j}^{\prime \prime }(x)$, i.e., $\mathfrak {\ddot {S}}(N_{x},\mu _{x},\theta _{x}^{\prime })$, since $\sin \limits \frac {\theta _{x}^{\prime }}{2}\neq 0$, it is not difficult to check:

  $$ \begin{array}{@{}rcl@{}} \mathfrak{\ddot{S}}(N_{x},\mu_{x},\theta_{x}^{\prime})=-\frac{{\mu_{x}^{2}}}{2}(-1)^{j+p}\csc^{2}\frac{\theta_{x}^{\prime}}{2}+\frac{N_{x}{\mu_{x}^{2}}}{2}(-1)^{j+p} =\frac{N_{x}{\mu_{x}^{2}}}{2}-\frac{{\mu_{x}^{2}}}{2}\csc^{2}(j\mu h_{x}). \end{array} $$

  Substituting $\mathfrak {\ddot {S}}(N_{x},\mu _{x},\theta _{x})$ and $\mathfrak {\ddot {S}}(N_{x},\mu _{x},\theta _{x}^{\prime })$ into $\mathrm {X}_{j}^{\prime \prime }(x)$, for j = p, one has:

  $$ \begin{array}{@{}rcl@{}} \mathrm{X}_{j}^{\prime\prime}(x_{p})=-\frac{{\mu_{x}^{2}}}{6}-\frac{{N_{x}^{2}}{\mu_{x}^{2}}}{3}+\frac{{\mu_{x}^{2}}}{2}\csc^{2}(j\mu h_{x}). \end{array} $$

  This completes the proof of (??) and obtains the claimed results in Lemma 2.1.

□

Appendix B: Proof of Lemma 3.1

Always, we prove the assertion in the x-direction and get the other one identically. Differentiating (??) two times and taking x = x_p yield:

$$ \begin{array}{@{}rcl@{}} ({S_{2}^{x}})_{j,p}=\frac{2}{N_{x}}\sum\limits_{l=1}^{N_{x}-1}\bigg[-(\mu_{x}l)^{2} \sin\big(\mu_{x}l(x_{j}-a)\big)\sin\big(\mu_{x}l(x_{p}-a)\big)\bigg]. \end{array} $$

Denote e_l := μ_xl, the spatial discretization gives:

$$ \begin{array}{@{}rcl@{}} ({S_{2}^{x}})_{j,p}=\frac{2}{N_{x}}\sum\limits_{l=1}^{N_{x}-1}\bigg[-(\mu_{x}l)^{2} \sin(\mu_{x}l\cdot \text{jh}_{x}) \sin(\mu_{x}l\cdot \text{ph}_{x})\bigg]=\frac{2}{N_{x}}\sum\limits_{l=1}^{N_{x}-1}\big(-{e_{l}^{2}} \sin \frac{\pi \text{jl}}{N_{x}} \sin \frac{\pi \text{lp}}{N_{x}}\big). \end{array} $$

Let

$$ (S_{N_{x}})_{j,p}=(S_{N_{x}}^{T})_{p,j}=\sqrt{\frac{2}{N_{x}}}\sin \frac{\pi \text{jp}}{N_{x}},\quad {\Lambda}_{x}=\text{diag}(-{e_{1}^{2}},-{e_{2}^{2}},\cdots,-e_{N_{x}-1}^{2}), $$

it is easy to find:

$$ \begin{array}{@{}rcl@{}} ({\Lambda}_{x}S_{N_{x}})_{j,p}&=\sum\limits_{l=1}^{N_{x}-1}({\Lambda}_{x})_{j,l}(S_{N_{x}})_{l,p}=({\Lambda}_{x})_{j,j}(S_{N_{x}})_{j,p}=-{e_{j}^{2}}\sqrt{\frac{2}{N_{x}}}\sin \frac{\pi jp}{N_{x}}. \end{array} $$

and

$$ \begin{array}{@{}rcl@{}} (S_{N_{x}}^{T}{\Lambda}_{x}S_{N_{x}})_{j,p}=\sum\limits_{l=1}^{N_{x}-1}(S_{N_{x}}^{T})_{j,l}({\Lambda}_{x}S_{N_{x}})_{l,p} =\sum\limits_{l=1}^{N_{x}-1}\sqrt{\frac{2}{N_{x}}}\sin \frac{\pi jl}{N_{x}}(-{e_{l}^{2}})\sqrt{\frac{2}{N_{x}}}\sin \frac{\pi \text{lp}}{N_{x}}=({S_{2}^{x}})_{j,p}. \end{array} $$

This completes the proof of ${S_{2}^{x}}$. For A₁, multiplying $-{h_{x}^{2}}$ on both sides of the decomposition formula in Lemma 3.1 and following the same procedure, one has:

$$ \begin{array}{@{}rcl@{}} (S_{N_{x}}^{T}{\Lambda}_{1}S_{N_{x}})_{j,p}=\sum\limits_{l=1}^{N_{x}-1}(S_{N_{x}}^{T})_{j,l}({\Lambda}_{1}S_{N_{x}})_{l,p} =\frac{2}{N_{x}}\sum\limits_{l=1}^{N_{x}-1}\bigg[\sin \frac{\pi jl}{N_{x}}\sin \frac{\pi \text{lp}}{N_{x}}(2-2\cos\frac{l\pi}{N_{x}})\bigg]. \end{array} $$

Applying the prosthaphaeresis formula on the right-hand side of the equality, it follows that

$$ \begin{array}{@{}rcl@{}} (S_{N_{x}}^{T}{\Lambda}_{1}S_{N_{x}})_{j,p}=\vartheta_{2}-\vartheta_{1}, \end{array} $$

where

$$ \begin{array}{@{}rcl@{}} \vartheta_{1}:=\frac{2}{N_{x}}\sum\limits_{l=1}^{N_{x}-1}\bigg[\cos \frac{\pi (j+p)l}{N_{x}}-\cos \frac{\pi (j-p)l}{N_{x}}\bigg],\quad \vartheta_{2}:=\vartheta_{1}\cos\frac{l\pi}{N_{x}}. \end{array} $$

Substituting 𝜗₁ into 𝜗₂ and utilizing the prosthaphaeresis formula again, the equality mentioned above can be reformulated as follows:

$$ \begin{array}{@{}rcl@{}} (S_{N_{x}}^{T}{\Lambda}_{1}S_{N_{x}})_{j,p}=\vartheta_{21}-\vartheta_{22}-\vartheta_{1}, \end{array} $$

where

$$ \begin{array}{@{}rcl@{}} &\vartheta_{21}:=\frac{1}{N_{x}}\sum\limits_{l=1}^{N_{x}-1}\bigg[\cos \frac{\pi \big(j+(p+1)\big)l}{N_{x}}-\cos \frac{\pi \big(j-(p+1)\big)l}{N_{x}}\bigg],\\ &\vartheta_{22}:=\frac{1}{N_{x}}\sum\limits_{l=1}^{N_{x}-1}\bigg[\cos \frac{\pi \big(j+(p-1)\big)l}{N_{x}}-\cos \frac{\pi \big(j-(p-1)\big)l}{N_{x}}\bigg]. \end{array} $$

All these three items on the right-hand side have the same structures. Resorting to the identity:

$$ \sum\limits_{l=1}^{N_{x}-1}\cos \frac{\pi ml}{N_{x}}=\left\{ \begin{array}{lll} N_{x}&-1, \ \ \ \ \ \ m=0\\ -&1,\ \ \ \ \ \ \ \ \ \ m \in (0, 2N_{x}-1)\ \textrm{and is even}\\ &0,\ \ \ \ \ \ \ \ \ \ m \in (0, 2N_{x}-1)\ \textrm{and is odd} \end{array} \right., $$

the following observations can be easily obtained:

$$ \vartheta_{1}=\left\{ \begin{array}{llll} -2, \quad &j=p\\ 0,\quad &j\neq p \end{array} \right.,\quad \vartheta_{21}=\left\{ \begin{array}{llll} -1, \ \ j=p+1\\ 0,\ \ \ j \neq p+1 \end{array} \right.,\quad \vartheta_{22}=\left\{ \begin{array}{lll} -1, \ \ j=p-1\\ 0,\ \ \ j \neq p-1 \end{array} \right.. $$

Substituting these results into the equality, we obtain the assertion of A₁ directly. This completes the proof of Lemma 3.1.