A Linear Finite Difference Scheme for the Two-Dimensional Nonlinear Schrödinger Equation with Fractional Laplacian

Abstract

In this paper, we propose a conservative three-layer linearized difference scheme for the two-dimensional nonlinear Schrödinger equation with fractional Laplacian. The difference scheme can be strictly proved to be uniquely solvable, conservation of mass and energy in the discrete sense. Furthermore, it is shown that the difference scheme is unconditionally convergent and stable under $l^{\infty }$-norm by discrete energy method. The convergence order is $\mathcal {O}(\tau ^2+h^2)$ with time step $\tau$ and mesh size h. Numerical examples are given to demonstrate the theoretical results.

Appendix A

For the coefficients of the discrete fractional Laplacian,

$$\begin{aligned} a_{j,k}^{(\alpha )} =\frac{1}{(2\pi )^2}\int _{[-\pi ,\pi ]^2}{\left( 4\sin (\frac{\eta _1}{2})^2+4\sin (\frac{\eta _2}{2})^2\right) ^\frac{\alpha }{2}\mathrm {e}^{-{\mathbf{i}}(j\eta _1+k\eta _2)}\mathrm {d}\eta }, \end{aligned}$$

we denote

$$\begin{aligned} \varPsi (\eta ) = \left( 4\sin (\frac{\eta _1}{2})^2+4\sin (\frac{\eta _2}{2})^2\right) ^\frac{\alpha }{2}. \end{aligned}$$

And we denote the $s'$th Sobolev space

$$\begin{aligned} W^{s}( {\mathcal {T} })=\left\{ u: \forall |\beta | \leqslant s, D^{\beta } u \in L^{2}({\mathcal {T} })\right\} , \end{aligned}$$

where $\mathcal {T}=[-\pi , \pi ]^2$ . Since $\varPsi (\eta )\in W^s({\mathcal {T} })$, here $s= 1+\alpha + \epsilon$ for any small $\epsilon >0$. It can be seen from book [14] on page 72, there is

$$\begin{aligned} \sum _{j,k}|a_{j,k}^{(\alpha )}|^2(1+(j^2+k^2))^{{s(\alpha )}}<\infty . \end{aligned}$$

For bigger j, k, we have

$$\begin{aligned} |a_{j,k}^{(\alpha )}|\le \frac{1}{(1+(j^2+k^2))^{{\frac{s(\alpha )}{2}}}}. \end{aligned}$$

Assume $u\in \mathcal {B}^{2+\alpha }$. Denote

$$\begin{aligned} u^L(x,y)=\left\{ \begin{array}{rcl} u(x,y), &{} &{} (x,y)\in (-L,L)^2\\ 0, &{} &{} (x,y)\in \mathbb {R}^2\backslash (-L,L)^2 \end{array}. \right. \end{aligned}$$

We have

$$\begin{aligned} (-\varDelta )^{\frac{\alpha }{2}}u-Au^L=(-\varDelta )^{\frac{\alpha }{2}}u -(-\varDelta _h)^{\frac{\alpha }{2}}u+(-\varDelta _h)^{\frac{\alpha }{2}}u-Au^L. \end{aligned}$$

From Lemma 2.1, we have

$$\begin{aligned} |(-\varDelta )^{\frac{\alpha }{2}}u-(-\varDelta _h)^{\frac{\alpha }{2}}u| \le C_0h^2\int _{\mathbb {R}^2}\left( 1+|\eta |\right) ^{\alpha +2}|u^{*}(\eta _1,\eta _2)|\mathrm {d}\eta _1\mathrm {d}\eta _2. \end{aligned}$$

And

$$\begin{aligned}&(-\varDelta _h)^{\frac{\alpha }{2}}u(x_{i_1},y_{i_2})-Au^L(x_{i_1},y_{i_2})\nonumber \\&\quad =h^{-\alpha }\sum _{j,k\in \mathbb {Z}^2}a_{j,k}^{(\alpha )}u(x_{i_1}+jh,y_{i_2}+kh)-h^{-\alpha }\sum _{j,k\in \mathbb {Z}^2}a_{j,k}^{(\alpha )}u^{L}(x_{i_1}+jh,y_{i_2}+kh)\nonumber \\&\quad =h^{-\alpha }\sum _{{j,k}\in \mathbb {Z}^2}a_{j,k}^{(\alpha )}(u(x_{i_1}+jh,y_{i_2}+kh) -u^L(x_{i_1}+jh,y_{i_2}+kh))\nonumber \\&\quad =h^{-\alpha }\sum _{{j}=-\infty }^{+{\infty }}\sum _{{k} =-\infty }^{-{i_2}}a_{j,k}^{(\alpha )}u(x_{i_1}+{j}h,y_{i_2}+kh) +h^{-\alpha }\sum _{{j}=-\infty }^{+\infty }\sum _{{k} =M-{i_2}}^{+{\infty }}a_{j,k}^{(\alpha )}u(x_{i_1}+jh,y_{i_2}+kh)\nonumber \\&\qquad +h^{-\alpha }\sum _{{j}=-\infty }^{-{i_1}}\sum _{{k} =-{i_2}}^{M-{i_2}}a_{j,k}^{(\alpha )}u(x_{i_1}+{j}h,y_{i_2}+kh)+h^{-\alpha }\sum _{{j} =M-{i_1}}^{+\infty }\sum _{{k}=-{i_2}}^{M-{i_2}}a_{j,k}^{(\alpha )}u(x_{i_1} +jh,y_{i_2}+kh)\nonumber \\&\quad =: {i}+{ii} +{iii} +{iv} \end{aligned}$$

(A.1)

Assume u(x, y) is the algebraic decay function, i.e. $u(x,y) = \frac{1}{(x^2+y^2)^{\delta }}, (\delta >0)$. For the first term ${i}$ of Eq. (A.1), we have

$$\begin{aligned} {i}=&h^{-\alpha }\sum _{{j} =-\infty }^{+{\infty }}\sum _{{k}=-\infty }^{-{i_2}}a_{j,k}^{(\alpha )}u(x_{i_1} +{j}h,y_{i_2}+kh)\nonumber \\ \le&h^{-\alpha }\sum _{{j}=-\infty }^{+{\infty }}\sum _{{k}=-\infty }^{-{i_2}} \frac{1}{(1+({j}^{2}+{k}^2))^{{\frac{s(\alpha )}{2}}}}u(x_{i_1} +{j}h,y_{i_2}+kh)\nonumber \\ \le&h^{-\alpha }\frac{1}{L^{2\delta }}\sum _{{j}=-\infty }^{+{\infty }}\sum _{{k} =-\infty }^{-{i_2}}\frac{1}{(1+({j}^{2}+{k}^2))^{{\frac{s(\alpha )}{2}}}}. \end{aligned}$$

(A.2)

There is a constant C, which satisfies

$$\begin{aligned} \sum _{{j}=-\infty }^{+{\infty }}\sum _{{k}=-\infty }^{-{i_2}} \frac{1}{(1+({j}^{2}+{k}^2))^{{\frac{s(\alpha )}{2}}}}\le&C \sum _{{j}=1}^{+{\infty }}\sum _{{k}=1}^{+{\infty }} \frac{1}{(1+({j}^{2}+{k}^2))^{{\frac{s(\alpha )}{2}}}}\\ \le&C\int _1^{+\infty }\int _1^{+\infty } \frac{1}{(1+({x}^{2}+{y}^2))^{{\frac{s(\alpha )}{2}}}}dxdy\\ =&C\int _0^{2\pi }\int _1^{+\infty }\frac{1}{(1+|r|{{^2}})^{{\frac{s(\alpha )}{2}}}}rdrd\theta \\ =&2\pi C \int _1^{+\infty }\frac{1}{(1+|r|{{^2}})^{{\frac{s(\alpha )}{2}}}}rdr\\ \le&2\pi C\int _1^{+\infty }\frac{1}{|r|^{{s(\alpha )}-1}}dr\\ =&{\frac{2\pi C}{{s(\alpha )}-2}}. \end{aligned}$$

Since ${s(\alpha )}>2$, for Eq. (A.2), we have

$$\begin{aligned} {i}\le {\frac{2\pi C}{{s(\alpha )}-2}}h^{-\alpha }L^{-2\delta }. \end{aligned}$$

For the remaining items in Eq. (A.1), the same inequality can be obtained. Now denote $L=h^{-\beta }$. Therefore, we have

$$\begin{aligned} |(-\varDelta _h)^{\frac{\alpha }{2}}u(x_{i_1},y_{i_2})-Au^L(x_{i_1},y_{i_2})|&\le {\frac{8\pi C}{{s(\alpha )}-2}}h^{-\alpha }L^{-2\delta }\\&\le {\frac{8\pi C}{{s(\alpha )}-2}}h^{-\alpha +2\delta \beta }. \end{aligned}$$

When $\delta \beta >1+\alpha /2$, we have

$$\begin{aligned} |(-\varDelta _h)^{\frac{\alpha }{2}}u-Au^L | \le {\frac{8\pi C}{{s(\alpha )}-2}}h^{2}. \end{aligned}$$

Therefore, when $\delta \beta \ge 1+\alpha /2$, i.e, the solution u decay enough fast, we can have

$$\begin{aligned} | (-\varDelta )^{\frac{\alpha }{2}} u-Au^L | \le h^2 \left( C_0\int _{\mathbb {R}^2}\left( 1+|\eta |\right) ^{\alpha +2}|u^{*}(\eta _1,\eta _2)|\mathrm {d}\eta _1\mathrm {d}\eta _2+{\frac{8\pi C}{{s(\alpha )}-2}} \right) .\end{aligned}$$