1 Introduction

Consider the following nonlinear Schrödinger equation (NLSE):

$$ \textstyle\begin{cases} iu_{t}+\Delta u+\lambda f( \vert u \vert ^{2})u=0, & (X,t)\in \varOmega \times (0,T], \\ u(X,t)=0, & (X,t)\in \partial \varOmega \times (0,T], \\ u(X,0)=u_{0}(X), & X\in \varOmega , \end{cases} $$
(1.1)

where \(X=(x,y)\), \(\varOmega \subset R^{2}\) is a bounded convex domain with Lipschitz boundary ∂Ω. i is the imaginary unit, \(u(X,t)\) is a complex-valued function, \(T\in (0,+\infty )\) is a real parameter, \(f(\vert u\vert ^{2})=\vert u\vert ^{2r}\) (\(r\geq 1\) is an integer) is a smooth real-valued function, and \(u_{0}(X)\) is a known smooth function.

The Schrödinger equation may describe many physical phenomena in optics, mechanics, and plasma physics, and it plays a very important role in various areas of mathematical physics. Numerical methods for this problem have been investigated extensively, e.g., see [1,2,3] for finite difference methods, [4,5,6,7,8] for finite element methods (FEMs), and [9,10,11] for others. Especially, the superconvergence analysis of FEMs for the Schrödinger equation have been studied successfully. For example, [12] used the conforming bilinear element to solve the LSE and obtained the superclose and superconvergence results in \(H^{1}\)-norm for the semi-discrete scheme. [13] derived the same results as [12] for NLSE with the conforming linear triangular element by establishing the relationship between Ritz projection and the linear interpolation. Whereafter, a series of superconvergence results about backward Euler and Crank–Nicolson fully-discrete schemes for NLSE also were studied in [14,15,16,17].

A two-grid method was first introduced by Xu [18, 19] as a discretization technique for nonlinear and nonsymmetric indefinite partial differential equations. The main idea of this method is to use a coarse space (with mesh size H) to produce a rough approximation of the solution, and then use it as the initial guess for one Newton iteration on the fine grid (with mesh size h and \(h\ll H\)). Up to now, the two-grid method was deeply researched for different problems [20,21,22,23,24,25]. Especially, the two-grid method was used to solve the linear Schrödinger equation (LSE) and NLSE in [26,27,28,29,30]. However, this method is rarely considered for nonconforming elements.

As we know, \({\mathit{EQ}_{1}^{\mathrm{rot}}}\) element is an important quadrilateral nonconforming finite element and has been employed to deal with different problems successfully for its good theoretical and numerical behavior [31,32,33,34,35,36,37]. The purpose of this work is to use this element to deal with problem (1.1). By virtue of the special properties of this element, we obtain the superclose and superconvergence results in the broken \(H^{1}\)-norm for the backward Euler fully-discrete scheme. At the same time, in order to reduce the computing cost, we develop a new two-grid algorithm and deduce the corresponding superconvergence results.

The paper is organized as follows. In Sect. 2, \(\mathit{EQ}_{1}^{\mathrm{rot}}\) element and some lemmas are briefly introduced. In Sect. 3, the backward Euler fully-discrete scheme for problem (1.1) is discussed and some important superconvergence results are derived. In Sect. 4, a two-grid scheme of (1.1) is established and the corresponding superconvergence results are obtained. Finally, a numerical experiment is carried out to confirm the theoretical results.

2 Nonconforming \(\mathit{EQ}_{1}^{\mathrm{rot}}\) element and some lemmas

For simplicity, let \(\varOmega \subset {\mathbb{R}}^{2}\) be a convex polygon with edges parallel to the coordinate axes, \({T}_{h}\) be a regular subdivision of Ω. For a given element K with the center point \((x_{K},y_{K})\), its four vertices and edges are denoted as \(a(x_{i}, y_{i})\) (\(i=1,2,3,4\)) and \(F_{i}=\overline{a_{i}a_{i+1}}\) (\(i=1,2,3, 4 \mod 4\)), respectively. We assume that edges \(F_{i}\) (\(i=1,3\)) parallel to x-axis and \(F_{i}\) (\(i=2,4\)) parallel to y-axis, \(h_{x,K}\) and \(h_{y,K}\) denote the half length of element K along x and y-axis, respectively.

The \(\mathit{EQ}_{1}^{\mathrm{rot}}\) finite element \(({K},{P},{\varSigma })\) on K is defined as follows:

$$ {\varSigma }=\{{v}_{1},{v}_{2},{v}_{3},{v}_{4},{v}_{5} \}, \quad\quad {P}=\operatorname{span}\bigl\{ 1,x,y,\varphi (x),\varphi (y)\bigr\} , $$
(2.1)

where

$$ {v}_{i}=\frac{1}{ \vert {F_{i}} \vert } \int_{{F_{i}}}{v}\,ds\quad (i=1,2,3,4),\quad\quad {v}_{5}= \frac{1}{ \vert {K} \vert } \int_{{K}}{v}\,dx\,dy, \varphi (t)=\frac{1}{2} \bigl(3t^{2}-1\bigr), $$

and \(\vert F_{i}\vert \) and \(\vert K\vert \) are the measures of \({F_{i}}\) and K, respectively.

The associated finite element space \({V}_{h}\) can be defined by

$$ {V}_{h}=\biggl\{ v_{h}:{{v_{h}}}\vert _{{K}}\in {P},\forall K\in T_{h}, \int_{F}[v_{h}]\,ds=0, F\subset \partial K \biggr\} , $$

where \([{v}_{h}]\) denotes the jump value of \({v}_{h}\) across the boundary F, and \([{v}_{h}]={v}_{h}\) if \(F\subset \partial \varOmega \).

Obviously, \(\Vert \cdot \Vert _{h}= (\sum_{K\in T_{h}}\vert \cdot \vert _{1,K} ^{2} )^{\frac{1}{2}}\) is a norm over \({V}_{h}\).

Let \(\varPi_{h}\) be the associated interpolation operator over \({V}_{h}\), then we have

$$\begin{aligned} \Vert { u}-{ \varPi }_{h}{ u} \Vert _{0}+h \Vert { u}-{\varPi }_{h}{ u} \Vert _{h}\leq Ch^{2} \vert { u} \vert _{2}, \quad \forall {u}\in {H}^{2}(\varOmega )\cap {H}_{0}^{1}(\varOmega ). \end{aligned}$$
(2.2)

Lemma 2.1

([31, 32])

For \({v}_{h} \in {V}_{h}\), we have

$$\begin{aligned} \biggl\vert \sum_{K\in T_{h}} \int_{\partial K}\frac{\partial {u}}{\partial {n}}{v}_{h} \,ds \biggr\vert \leq \textstyle\begin{cases} Ch \vert {u} \vert _{2} \Vert {v}_{h} \Vert _{h}, & \forall {u}\in {H}^{2}(\varOmega ), \\ Ch^{2} \vert {u} \vert _{3} \Vert {v}_{h} \Vert _{h},& \forall {u}\in {H}^{3}(\varOmega ). \end{cases}\displaystyle \end{aligned}$$
(2.3)

Lemma 2.2

For \({v}_{h} \in {V}_{h}\), we have

$$ \biggl\vert \sum_{K\in T_{h}} \int_{\partial K}\frac{\partial {u}}{\partial {n}}{v}_{h} \,ds \biggr\vert \leq Ch^{2} \vert {u} \vert _{4} \Vert {v}_{h} \Vert _{0}, \quad \forall {u}\in {H}^{4}( \varOmega ). $$
(2.4)

Proof

By introducing two functions

$$ E(x)=\frac{1}{2} \bigl((x-x_{K})^{2}-h_{x,K}^{2} \bigr), \quad\quad F(y)= \frac{1}{2} \bigl((y-y_{K})^{2}-h_{y,K}^{2} \bigr), $$

and notation \({P_{0}}{v_{h}}\vert _{F_{i}}=\frac{1}{\vert F_{i}\vert }\int_{F_{i}}v _{h}\,ds\), which has continuity between elements and vanishes on ∂Ω, and hence the summation

$$ \sum_{K\in T_{h}} \biggl( \int_{F_{2}}- \int_{F_{4}} \biggr)u_{x}{P_{0}} {v _{h}}\,dy=0, \quad\quad \sum_{K\in T_{h}} \biggl( \int_{F_{3}}- \int_{F_{1}} \biggr)u _{y}{P_{0}} {v_{h}}\,dx=0. $$

So we can obtain that

$$\begin{aligned}& \sum_{K\in T_{h}} \int_{\partial K}\frac{\partial {u}}{\partial {n}}{v}_{h} \,ds \\ & \quad = \sum_{K\in T_{h}} \biggl( \int_{F_{2}}- \int_{F_{4}} \biggr)u_{x}v_{h}\,dy + \sum _{K\in T_{h}} \biggl( \int_{F_{3}}- \int_{F_{1}} \biggr)u_{y}v_{h}\,dx \\ & \quad = \sum_{K\in T_{h}} \biggl( \int_{F_{2}}- \int_{F_{4}} \biggr)u_{x}(v_{h}- {P_{0}} {v_{h}})\,dy+\sum_{K\in T_{h}} \biggl( \int_{F_{3}}- \int_{F_{1}} \biggr)u_{y}(v_{h}-{P_{0}} {v_{h}})\,dx \\ & \quad = \sum_{K\in T_{h}} \int_{K} \biggl[u_{xx} \biggl((y-y_{K})v_{hy}+ \biggl((y-y _{K})^{2}-\frac{h_{y,K}^{2}}{3} \biggr) \frac{v_{hyy}}{2} \biggr) \biggr]\,dx\,dy \\ & \quad \quad {} +\sum_{K\in T_{h}} \int_{K} \biggl[u_{yy} \biggl((x-x_{K})v_{hx}+ \biggl((x-x _{K})^{2}-\frac{h_{x,K}^{2}}{3} \biggr) \frac{v_{hxx}}{2} \biggr) \biggr]\,dx\,dy \\ & \quad \triangleq A_{1}+A_{2}, \end{aligned}$$
(2.5)

where we use the expressions

$$\begin{aligned}& (v_{h}-{P_{0}} {v_{h}})\vert _{F_{i}}=(y-y_{K})v_{hy}+ \biggl((y-y_{K})^{2}- \frac{h _{y,K}^{2}}{3} \biggr)\frac{v_{hyy}}{2}, \quad i=2,4, \\& (v_{h}-{P_{0}} {v_{h}})\vert _{F_{i}}=(x-x_{K})v_{hx}+ \biggl((x-x_{K})^{2}- \frac{h _{x,K}^{2}}{3} \biggr)\frac{v_{hxx}}{2}, \quad i=1,3. \end{aligned}$$

Now we begin to estimate \(A_{1}\), which can be rewritten as

$$\begin{aligned} A_{1} =&\sum_{K\in T_{h}} \int_{K} u_{xx}(y-y_{K})v_{hy} \,dx\,dy + \frac{1}{2}\sum_{K\in T_{h}} \int_{K}u_{xx}(y-y_{K})^{2}v_{hyy} \,dx\,dy \\ & {} - \sum_{K\in T_{h}}\frac{h_{y,K}^{2}}{6} \int_{K}u_{xx}v_{hyy}\,dx\,dy \triangleq B_{1}+B_{2}+B_{3}. \end{aligned}$$
(2.6)

Firstly, for all \(v_{hy}\in \operatorname{span}\{1,y\}\), we know that

$$ v_{hy}(x,y)=v_{hy}(x,y_{K})+(y-y_{K})v_{hyy}. $$

Noting that \(F(y)\vert _{F_{1},F_{3}}=0\), \(F'(y)=y-y_{K}\), \(F(y)= \frac{1}{6} (F^{2}(y) )''-\frac{h_{y,K}^{2}}{3}\), and by Green’s formula, we have

$$\begin{aligned} B_{1} =&\sum_{K\in T_{h}} \int_{K} u_{xx}F'(y)v_{hy} \,dx\,dy \\ =& -\sum_{K\in T_{h}} \int_{K} u_{xxy}F(y)v_{hy}\,dx\,dy \\ =&-\sum_{K\in T_{h}} \int_{K} u_{xxy} \biggl[\frac{1}{6} \bigl(F^{2}(y) \bigr)''-\frac{h_{y,K}^{2}}{3} \biggr]v_{hy}\,dx\,dy \\ =&\sum_{K\in T_{h}} \biggl[\frac{1}{6} \int_{K} u_{xxyy} \bigl(F^{2}(y) \bigr)'v_{hy}\,dx\,dy + \int_{K} \frac{h_{y,K}^{2}}{3}v_{hy}\,dx\,dy \biggr] \\ =&\sum_{K\in T_{h}}\frac{1}{6} \int_{K} u_{xxyy} \bigl(F^{2}(y) \bigr)'v _{hy}\,dx\,dy + \sum_{K\in T_{h}} \frac{h_{y,K}^{2}}{3} \int_{K}u_{xxy}v _{hy}\,dx\,dy \\ & {} -\sum_{K\in T_{h}}\frac{h_{y,K}^{2}}{3} \int_{K}u_{xxy}(y-y_{K})v _{hyy} \,dx\,dy \\ \triangleq & B_{11}+B_{12}+B_{13}. \end{aligned}$$
(2.7)

By the inverse inequality, term \(B_{11}\) can be estimated as

$$\begin{aligned} B_{11} =&\sum_{K\in T_{h}} \frac{1}{6} \int_{K} u_{xxyy} \bigl(F^{2}(y) \bigr)'v_{hy}\,dx\,dy \\ {\le }&\sum_{K\in T_{h}}Ch^{3}_{y,K} \Vert u_{xxyy} \Vert _{0,K} \Vert v_{hy} \Vert _{0,K} \le Ch^{2} \vert u \vert _{4} \Vert v_{h} \Vert _{0}. \end{aligned}$$
(2.8)

For the term \(B_{12}\), it can be written as

$$\begin{aligned} B_{12} =& \sum_{K\in T_{h}} \frac{h_{y,K}^{2}}{3} \int_{K}u_{xxy}v_{hy}\,dx\,dy \\ =& \sum_{K\in T_{h}}\frac{h_{y,K}^{2}}{3} \biggl[ \biggl( \int_{F_{3}}- \int_{F_{1}} \biggr)u_{xxy}v_{h}\,dx - \int_{K}u_{xxyy}v_{h}\,dx\,dy \biggr]. \end{aligned}$$
(2.9)

Noting that

$$\begin{aligned}& \sum_{K\in T_{h}} \biggl( \int_{F_{3}}- \int_{F_{1}} \biggr)u_{xxy}v_{h}\,dx= \sum _{K\in T_{h}} \biggl( \int_{F_{3}}- \int_{F_{1}} \biggr)u_{xxy}(v_{h}- {P_{0}} {v_{h}})\,dx \\& \quad =\sum_{K\in T_{h}} \int_{K} \biggl[u_{xxyy} \biggl((x-x_{K})v_{hx}+ \biggl((x-x _{K})^{2}+\frac{h_{x,K}^{2}}{3} \biggr) \frac{v_{hxx}}{2} \biggr) \biggr]\,dx\,dy \\& \quad \leq \sum_{K\in T_{h}}\bigl(Ch_{x,K} \vert u \vert _{4} \Vert v_{hx} \Vert _{0,K} + Ch_{x,K} ^{2} \vert u \vert _{4} \Vert v_{hxx} \Vert _{0,K}\bigr)\leq C \vert u \vert _{4} \Vert v_{h} \Vert _{0}, \end{aligned}$$
(2.10)

and substituting (2.10) into (2.9), we can derive

$$\begin{aligned} B_{12}\leq Ch^{2} \vert u \vert _{4} \Vert v_{h} \Vert _{0}. \end{aligned}$$
(2.11)

As to the term \(B_{13}\), noting that \(F(y)\vert _{F_{1},F_{3}}=0\) and \(F'(y)=y-y_{K}\), we have

$$\begin{aligned} B_{13} =&-\sum_{K\in T_{h}} \frac{h_{y,K}^{2}}{3} \int_{K}u_{xxy}F'(y)v _{hyy} \,dx\,dy \\ =&\sum_{K\in T_{h}}\frac{h_{y,K}^{2}}{3} \int_{K}u_{xxyy}F(y)v_{hyy}\,dx\,dy \leq Ch^{2} \vert u \vert _{4} \Vert v_{h} \Vert _{0}. \end{aligned}$$
(2.12)

Combining with (2.7)–(2.9) and (2.12), we can obtain

$$\begin{aligned} B_{1}\leq Ch^{2} \vert u \vert _{4} \Vert v_{h} \Vert _{0}. \end{aligned}$$
(2.13)

Secondly, noting that \((y-y_{K})^{2}=\frac{1}{3} [ (F^{2}(y) )''+h_{y,K}^{2} ]\), we have

$$\begin{aligned}& B_{2}+B_{3} \\& \quad = \frac{1}{6}\sum_{K\in T_{h}} \int_{K}u_{xx} \bigl[ \bigl(F^{2}(y) \bigr)''+h_{y,K}^{2} \bigr]v_{hyy}\,dx\,dy - \sum_{K\in T_{h}} \frac{h_{y,K} ^{2}}{6} \int_{K}u_{xx}v_{hyy}\,dx\,dy \\& \quad = \frac{1}{6}\sum_{K\in T_{h}} \int_{K}u_{xx} \bigl(F^{2}(y) \bigr)''v _{hyy}\,dx\,dy \frac{1}{6}\sum _{K\in T_{h}} \int_{K}u_{xxyy}F^{2}(y)v_{hyy}\,dx \,dy \\& \quad \leq Ch^{4} \vert u \vert _{4} \Vert v_{hyy} \Vert _{0}\leq Ch^{2} \vert u \vert _{4} \Vert v_{h} \Vert _{0}. \end{aligned}$$
(2.14)

Finally, substituting estimates (2.13) and (2.14) into (2.6), we obtain

$$\begin{aligned} A_{1}\leq Ch^{2} \vert u \vert _{4} \Vert v_{h} \Vert _{0}. \end{aligned}$$
(2.15)

And similarly, we can derive the result

$$\begin{aligned} A_{2}\leq Ch^{2} \vert u \vert _{4} \Vert v_{h} \Vert _{0}. \end{aligned}$$
(2.16)

Combining with (2.15) and (2.16), the desired result is obtained. □

Remark 2.1

In [38], the authors derived the following result:

$$ \biggl\vert \sum_{K\in T_{h}} \int_{\partial K}\frac{\partial {u}}{\partial {n}}{v}_{h} \,ds \biggr\vert \leq Ch^{2} \vert {u} \vert _{5} \Vert {v}_{h} \Vert _{0}, \quad \forall {u}\in {H}^{5}(\varOmega ), {v}_{h} \in {V}_{h}. $$
(2.17)

Obviously, the regularity requirement of u is stronger than our result.

3 Backward Euler fully-discrete scheme and superconvergence results

The variational form of (1.1) is to find \(u\in H_{0}^{1}(\varOmega )\) such that

$$ \textstyle\begin{cases} i(u_{t},v)-(\nabla u, \nabla v)+\lambda (f( \vert u \vert ^{2})u,v)=0, & v\in H _{0}^{1}(\varOmega ), \\ u(X,0)=u_{0}(X), & X\in \varOmega , \end{cases} $$
(3.1)

where \((u,v)=\int_{\varOmega }u\overline{v}\,dx\,dy\) denotes the inner product, v̅ is the conjugate of v.

Given a time step \(\tau =T/N\), where N is a positive integer, we shall approximate the solution at times \(t_{n}=n\tau \), \(n=0,1,\ldots,N\). For a given smooth function \(\phi^{n}\) on \([0,T]\), define \(\phi^{n}=\phi (X,t^{n})\), \({{\partial }_{t}}\phi^{n}=\frac{\phi^{n}- \phi^{n-1}}{\tau }\), and \({{\partial }_{t}}\nabla \phi^{n}=\frac{ \nabla \phi^{n}-\nabla \phi^{n-1}}{\tau }\).

Equation (3.1) has the following equivalent formulation:

$$ \textstyle\begin{cases} i({{\partial }_{t}}u^{n},v)-(\nabla u^{n}, \nabla v)+\lambda (f( \vert u ^{n} \vert ^{2})u^{n},v)=i(R_{1}^{n},v), & v\in H_{0}^{1}(\varOmega ), \\ u(X,0)=u_{0}(X), & X\in \varOmega , \end{cases} $$
(3.2)

where \(R_{1}^{n}={{\partial }_{t}} u^{n}-u_{t}^{n}\). Furthermore, we have

$$\begin{aligned} \bigl\Vert R_{1}^{n} \bigr\Vert _{0}^{2} =& \biggl\Vert \frac{1}{\tau } \int_{t_{n-1}}^{t_{n}}(t _{n}-t)u_{tt} \,dt \biggr\Vert _{0}^{2}\leq \frac{1}{\tau^{2}} \int_{\varOmega } \biggl( \int_{t_{n-1}}^{t_{n}}(t_{n}-t)u_{tt} \,ds \biggr)^{2}\,dx\,dy \\ \leq &\frac{C}{\tau^{2}} \int_{\varOmega } \biggl( \int_{t_{n-1}}^{t_{n}}(t _{n}-t)\,dt \biggr)^{2} \biggl( \int_{t_{n-1}}^{t_{n}}u_{tt}\,ds \biggr)^{2}\,dx\,dy \leq C\tau \int_{t_{n-1}}^{t_{n}} \Vert u_{tt} \Vert _{0}^{2}\,dt. \end{aligned}$$
(3.3)

The backward Euler fully-discrete scheme of (3.1) is to find \(U^{n}\in V_{h}\) such that

$$ \textstyle\begin{cases} i({{\partial }_{t}} U^{n},v_{h})-(\nabla U^{n}, \nabla v_{h})_{h}+ \lambda (f( \vert U^{n} \vert ^{2})U^{n},v_{h})=0, & v_{h}\in V_{h}, \\ U^{0}=\varPi_{h}u_{0}(X), & X\in \varOmega , \end{cases} $$
(3.4)

where \((\cdot ,\cdot )_{h}= \sum_{K\in {T_{h}}}(\cdot ,\cdot )_{K}\).

In order to carry out the error estimate and superclose analysis, we introduce the following assumption.

Assumption 3.1

Let \(u^{n}\) and \(U^{n}\) be the solutions of (1.1) and (3.4), respectively, for \(n = 1,2,\ldots,N\), then there exists \(0< h_{0}<1\) such that, for \(0 < h< h_{0}\), \(n=1,2,\ldots,N\), it holds

$$\begin{aligned} \bigl\Vert u^{n}-U^{n} \bigr\Vert _{0,\infty } < 1, \end{aligned}$$
(3.5)

which means \(\Vert U^{n}\Vert _{0,\infty } < C\).

Regarding the proof of Assumption 3.1, one can refer to [15] for details.

For simplicity, we write

$$ u^{n}-U^{n}=\bigl(u^{n}-\varPi_{h}u^{n} \bigr)+\bigl(\varPi_{h}u^{n}-U^{n}\bigr)\triangleq \rho ^{n} + \theta^{n}. $$

Then we have the following results.

Theorem 3.1

Assume that \(u^{n}\) and \(U^{n}\) are the solutions of (1.1) and (3.4), respectively. If \(u\in H^{4}(\varOmega ) \cap H^{1}_{0}(\varOmega )\), \(u_{t}\in H^{4}(\varOmega )\), \(u_{tt}\in H ^{2}(\varOmega )\), \(u_{ttt}\in L^{2}(\varOmega )\), we have

$$\begin{aligned} \bigl\Vert \theta^{n} \bigr\Vert _{0} + \bigl\Vert \theta^{n} \bigr\Vert _{h}= O \bigl(h^{2} + \tau \bigr). \end{aligned}$$
(3.6)

Proof

From (1.1) and (3.4), we have the result

$$\begin{aligned}& i \bigl({{\partial }_{t}}\bigl(u^{n}-U^{n} \bigr),v_{h} \bigr)- \bigl(\nabla \bigl(u^{n}-U ^{n}\bigr),v_{h} \bigr)_{h}+\lambda \bigl(f \bigl( \bigl\vert u^{n} \bigr\vert ^{2} \bigr)u^{n}-f\bigl( \bigl\vert U^{n} \bigr\vert ^{2}\bigr)U ^{n},v_{h} \bigr) \\& \quad =i\bigl(R_{1}^{n},v_{h}\bigr) - \sum _{K\in T_{h}} \int_{\partial K}\frac{\partial {u}^{n}}{\partial {n}}\cdot \overline{v_{h}} \,ds, \end{aligned}$$
(3.7)

which can be rewritten as

$$\begin{aligned} i\bigl({{\partial }_{t}}\theta^{n},v_{h} \bigr)-\bigl(\nabla \theta^{n},\nabla v _{h} \bigr)_{h} =&-\lambda \bigl(f\bigl( \bigl\vert u^{n} \bigr\vert ^{2}\bigr)u^{n}-f\bigl( \bigl\vert U^{n} \bigr\vert ^{2}\bigr)U^{n},v_{h} \bigr) \\ & {} -i\bigl({{\partial }_{t}}\rho^{n},v_{h} \bigr)+i\bigl(R_{1}^{n},v_{h}\bigr) - \sum _{K \in T_{h}} \int_{\partial K}\frac{\partial {u}^{n}}{\partial {n}} \cdot \overline{v_{h}} \,ds. \end{aligned}$$
(3.8)

Taking \(v_{h}=\theta^{n}\) in (3.8), we have

$$\begin{aligned} i\bigl({{\partial }_{t}}\theta^{n}, \theta^{n}\bigr)-\bigl(\nabla \theta^{n},\nabla \theta^{n}\bigr)_{h} =&-\lambda \bigl(f\bigl( \bigl\vert u^{n} \bigr\vert ^{2}\bigr)u^{n}-f\bigl( \bigl\vert U^{n} \bigr\vert ^{2}\bigr)U^{n}, \theta^{n} \bigr) \\ & {} -i\bigl({{\partial }_{t}}\rho^{n}, \theta^{n}\bigr)+i\bigl(R_{1}^{n}, \theta^{n}\bigr) - \sum_{K\in T_{h}} \int_{\partial K}\frac{\partial {u}^{n}}{\partial {n}}\cdot \overline{ \theta^{n}} \,ds. \end{aligned}$$
(3.9)

Comparing the imaginary parts of (3.9), we get

$$\begin{aligned}& \frac{1}{\tau }\operatorname{Re}\bigl(\theta^{n}- \theta^{n-1},\theta^{n}\bigr) \\& \quad =\frac{1}{2\tau } \bigl( \bigl\Vert \theta^{n} \bigr\Vert _{0}^{2}- \bigl\Vert \theta^{n-1} \bigr\Vert _{0} ^{2}+ \bigl\Vert \theta^{n}- \theta^{n-1} \bigr\Vert _{0}^{2} \bigr) \\& \quad =-\operatorname{Re}\bigl({{\partial }_{t}}\rho^{n}, \theta^{n}\bigr)-\operatorname{Im}\lambda \bigl(f\bigl( \bigl\vert u^{n} \bigr\vert ^{2}\bigr)u ^{n}-f\bigl( \bigl\vert U^{n} \bigr\vert ^{2}\bigr)U^{n}, \theta^{n} \bigr) \\& \quad\quad {} +\operatorname{Re}\bigl(R_{1}^{n},\theta^{n}\bigr) - \operatorname{Im} \sum_{K\in T_{h}} \int_{\partial K}\frac{ \partial {u^{n}}}{\partial {n}}\cdot \overline{ \theta^{n}} \,ds, \end{aligned}$$
(3.10)

which implies

$$\begin{aligned}& \bigl\Vert \theta^{n} \bigr\Vert _{0}^{2}- \bigl\Vert \theta^{n-1} \bigr\Vert _{0}^{2} \\& \quad \leq -2\tau \operatorname{Re}\bigl({{\partial }_{t}}\rho^{n}, \theta^{n}\bigr)-2\tau \operatorname{Im} \lambda \bigl(f\bigl( \bigl\vert u^{n} \bigr\vert ^{2}\bigr)u^{n}-f\bigl( \bigl\vert U^{n} \bigr\vert ^{2}\bigr)U^{n}, \theta^{n} \bigr) \\& \quad\quad {} +2\tau \operatorname{Re}\bigl(R_{1}^{n},\theta^{n}\bigr) - 2\tau \operatorname{Im}\sum_{K\in T_{h}} \int_{\partial K}\frac{\partial {u^{n}}}{\partial {n}}\cdot \overline{ \theta^{n}} \,ds\triangleq \sum_{i=1}^{4}D_{i}. \end{aligned}$$
(3.11)

Now, we start to estimate each term \(D_{i}\) (\(i=1,2,3,4\)) one by one.

Applying ε-Young’s inequality, we obtain

$$\begin{aligned} \vert D_{1} \vert \leq \bigl\vert 2\tau \bigl({{ \partial }_{t}}\rho^{n},\theta^{n}\bigr) \bigr\vert \leq Ch^{4} \int_{t_{n-1}}^{t_{n}} \vert u_{t} \vert _{2}^{2}\,dt + C\tau \bigl\Vert \theta^{n} \bigr\Vert _{0} ^{2}. \end{aligned}$$
(3.12)

By using the continuity of \(f(s)\) and Assumption 3.1, we have

$$\begin{aligned} \vert D_{2} \vert \leq & 2\tau \vert \lambda \vert \bigl\vert \bigl(f\bigl( \bigl\vert u^{n} \bigr\vert ^{2}\bigr)u^{n}-f\bigl( \bigl\vert U^{n} \bigr\vert ^{2}\bigr)U ^{n},\theta^{n} \bigr) \bigr\vert \\ =& 2\tau \vert \lambda \vert \bigl\vert \bigl(f\bigl( \bigl\vert u^{n} \bigr\vert ^{2}\bigr)u^{n}-f\bigl( \bigl\vert u^{n} \bigr\vert ^{2}\bigr)U^{n} +f \bigl( \bigl\vert u^{n} \bigr\vert ^{2} \bigr)U^{n}-f\bigl( \bigl\vert U^{n} \bigr\vert ^{2}\bigr)U^{n},\theta^{n} \bigr) \bigr\vert \\ =& C\tau \bigl\Vert f\bigl( \bigl\vert u^{n} \bigr\vert ^{2}\bigr) \bigr\Vert _{0,\infty } \bigl\vert \bigl( \rho^{n}+\theta^{n},\theta ^{n}\bigr) \bigr\vert \\ & {} +C\tau \bigl\Vert U^{n} \bigr\Vert _{0,\infty }\Vert f'({\xi_{1}})\Vert _{0,\infty }\bigl( \bigl\Vert u^{n} \bigr\Vert _{0,\infty } + \bigl\Vert U^{n} \bigr\Vert _{0,\infty }\bigr) \bigl\vert \bigl(\rho^{n}+ \theta^{n},\theta ^{n}\bigr) \bigr\vert \\ \leq & C\tau \bigl\vert \bigl(\rho^{n}+\theta^{n}, \theta^{n}\bigr) \bigr\vert \leq C\tau h^{4} \bigl\vert u ^{n} \bigr\vert _{2}^{2} + C\tau \bigl\Vert \theta^{n} \bigr\Vert _{0}^{2}, \end{aligned}$$
(3.13)

where \(\xi_{1}\) lies between \(\vert u^{n}\vert ^{2}\) and \(\vert U^{n}\vert ^{2}\).

With the help of the result (3.3) and Lemma 2.2, terms \(D_{3}\) and \(D_{4}\) can be estimated as

$$\begin{aligned}& \vert D_{3} \vert \leq 2\tau \bigl\vert \bigl(R_{1}^{n},\theta^{n}\bigr) \bigr\vert \leq 2\tau \bigl\Vert R_{1}^{n} \bigr\Vert _{0} \bigl\Vert \theta^{n} \bigr\Vert _{0}\leq C \tau^{2} \int_{t_{n-1}}^{t_{n}} \Vert u_{tt} \Vert _{0}^{2}\,dt + C\tau \bigl\Vert \theta^{n} \bigr\Vert _{0}^{2}, \end{aligned}$$
(3.14)
$$\begin{aligned}& \vert D_{4} \vert \leq C\tau h^{4} \bigl\vert u^{n} \bigr\vert _{4} \bigl\Vert \theta^{n} \bigr\Vert _{0}\leq C\tau h ^{4} \bigl\vert u^{n} \bigr\vert _{4}^{2} + C\tau \bigl\Vert \theta^{n} \bigr\Vert _{0}^{2}. \end{aligned}$$
(3.15)

Combining the above estimates (3.11)–(3.15) yields

$$\begin{aligned} \bigl\Vert \theta^{n} \bigr\Vert _{0}^{2}- \bigl\Vert \theta^{n-1} \bigr\Vert _{0}^{2} \leq & Ch^{4} \biggl( \tau \bigl\Vert u^{n} \bigr\Vert _{4}^{2} + \int_{t_{n-1}}^{t_{n}} \vert u_{t} \vert _{2}^{2}\,dt \biggr) \\ & {} + C\tau^{2} \int_{t_{n-1}}^{t_{n}} \Vert u_{tt} \Vert _{0}^{2}\,dt + C\tau \bigl\Vert \theta^{n} \bigr\Vert _{0}^{2}. \end{aligned}$$
(3.16)

Summing (3.16) up with respect to n and noting \(\theta^{0}=0\), we have

$$\begin{aligned} \bigl\Vert \theta^{n} \bigr\Vert _{0}^{2} \leq & Ch^{4} \int_{0}^{T} \vert u_{t} \vert _{2}^{2}\,dt + C \tau^{2} \int_{0}^{T} \Vert u_{tt} \Vert _{0}^{2}\,dt \\ & {} + C\tau h^{4}\sum_{i=1}^{n} \bigl\Vert u^{i} \bigr\Vert _{4}^{2} + C\tau \sum_{i=1}^{n} \bigl\Vert \theta^{i} \bigr\Vert _{0}^{2}. \end{aligned}$$
(3.17)

By Gronwall’s lemma, we can derive

$$\begin{aligned} \bigl\Vert \theta^{n} \bigr\Vert _{0}^{2}\leq Ch^{4} \int_{0}^{T} \vert u_{t} \vert _{2}^{2}\,dt + C\tau ^{2} \int_{0}^{T} \Vert u_{tt} \Vert _{0}^{2}\,dt + C\tau h^{4}\sum _{i=1}^{n} \bigl\Vert u ^{i} \bigr\Vert _{4}^{2}, \end{aligned}$$
(3.18)

which implies

$$\begin{aligned} \bigl\Vert \theta^{n} \bigr\Vert _{0} = O\bigl(h^{2} + \tau \bigr). \end{aligned}$$
(3.19)

On the other hand, taking \(v_{h}={{\partial }_{t}}\theta^{n}\) in (3.8), we obtain

$$\begin{aligned}& i\bigl({{\partial }_{t}}\theta^{n},{{ \partial }_{t}}\theta^{n}\bigr)-\bigl(\nabla \theta^{n},\nabla {{\partial }_{t}}\theta^{n} \bigr)_{h} \\& \quad =-\lambda \bigl(f\bigl( \bigl\vert u ^{n} \bigr\vert ^{2}\bigr)u^{n}-f\bigl( \bigl\vert U^{n} \bigr\vert ^{2}\bigr)U^{n},{{\partial }_{t}} \theta^{n} \bigr) \\& \quad\quad {} -i\bigl({{\partial }_{t}}\rho^{n},{{\partial }_{t}}\theta^{n}\bigr)+i\bigl(R_{1}^{n}, {{\partial }_{t}}\theta^{n}\bigr) - \sum _{K\in T_{h}} \int_{\partial K}\frac{ \partial {u}}{\partial {n}}\cdot \overline{{{\partial }_{t}}\theta ^{n}} \,ds. \end{aligned}$$
(3.20)

Comparing the real parts of (3.20), we get

$$\begin{aligned}& \frac{1}{\tau }\operatorname{Re}\bigl(\nabla \theta^{n},\nabla \theta^{n}-\nabla \theta^{n-1}\bigr)_{h} \\& \quad =\frac{1}{2\tau } \bigl( \bigl\Vert \nabla \theta^{n} \bigr\Vert _{0}^{2}- \bigl\Vert \nabla \theta^{n-1} \bigr\Vert _{0}^{2}+ \bigl\Vert \nabla \theta^{n}-\nabla \theta^{n-1} \bigr\Vert _{0} ^{2} \bigr) \\& \quad =-\operatorname{Im}\bigl({{\partial }_{t}}\rho^{n},{{\partial }_{t}}\theta^{n}\bigr) + \operatorname{Re} \lambda \bigl(f\bigl( \bigl\vert u^{n} \bigr\vert ^{2}\bigr)u^{n}-f\bigl( \bigl\vert U^{n} \bigr\vert ^{2}\bigr)U^{n},{{ \partial }_{t}} \theta^{n} \bigr) \\& \quad\quad {} + \operatorname{Im}\bigl(R_{1}^{n},{{\partial }_{t}} \theta^{n}\bigr) + \operatorname{Re}\sum_{K\in T_{h}} \int_{\partial K}\frac{\partial {u^{n}}}{\partial {n}}\cdot \overline{ {{\partial }_{t}}\theta^{n}} \,ds, \end{aligned}$$
(3.21)

which implies

$$\begin{aligned}& \bigl\Vert \theta^{n} \bigr\Vert _{h}^{2}- \bigl\Vert \theta^{n-1} \bigr\Vert _{h}^{2} \\& \quad \leq -2\tau \operatorname{Im}\bigl({{\partial }_{t}}\rho^{n},{{\partial }_{t}}\theta ^{n}\bigr) + 2\tau \operatorname{Re}\lambda \bigl(f\bigl( \bigl\vert u^{n} \bigr\vert ^{2}\bigr)u^{n}-f \bigl( \bigl\vert U^{n} \bigr\vert ^{2} \bigr)U^{n}, {{\partial }_{t}}\theta^{n} \bigr) \\& \quad\quad {} +2\tau \operatorname{Im}\bigl(R_{1}^{n},{{\partial }_{t}} \theta^{n}\bigr) + 2\tau \operatorname{Re} \sum_{K\in T_{h}} \int_{\partial K}\frac{\partial {u^{n}}}{\partial {n}}\cdot \overline{{{\partial }_{t}}\theta^{n}} \,ds \\& \quad \leq C\tau h^{4} \biggl( \bigl\Vert u^{n} \bigr\Vert _{4}^{2} + \int_{t_{n-1}}^{t_{n}} \vert u _{t} \vert _{2}^{2}\,dt \biggr)+ C\tau^{2} \int_{t_{n-1}}^{t_{n}} \Vert u_{tt} \Vert _{0} ^{2}\,dt + C\tau \bigl\Vert {{\partial }_{t}}\theta^{n} \bigr\Vert _{0}^{2}. \end{aligned}$$
(3.22)

Now we estimate the term \(\Vert {{\partial }_{t}}\theta^{n}\Vert _{0}^{2}\). To do this, take difference between two time levels n and \(n-1\) of (3.8) and multiply by \(\frac{1}{\tau }\) on both sides, then set \(v_{h}= {{\partial }_{t}}\theta^{n}\) to get

$$\begin{aligned}& i \bigl({{\partial }_{t}}\bigl({{\partial }_{t}}\theta^{n}\bigr),{{\partial } _{t}} \theta^{n} \bigr)-\bigl(\nabla {{\partial }_{t}} \theta^{n},\nabla {{\partial } _{t}}\theta^{n} \bigr) \\& \quad =-\lambda \bigl({{\partial }_{t}}M^{n},{{\partial } _{t}}\theta^{n} \bigr)-i\bigl({{\partial }_{t}} \bigl({{\partial }_{t}}\rho^{n}\bigr), {{\partial }_{t}}\theta^{n}\bigr) \\& \quad\quad {} +i\bigl({{\partial }_{t}}R_{1}^{n},{{ \partial }_{t}}\theta^{n}\bigr) - \sum _{K\in T_{h}} \int_{\partial K}\frac{\partial ({{\partial }_{t}}u ^{n})}{\partial {n}}\cdot \overline{{{\partial }_{t}}\theta^{n}} \,ds, \end{aligned}$$
(3.23)

where \(M^{n}=f(\vert u^{n}\vert ^{2})u^{n}-f(\vert U^{n}\vert ^{2})U^{n}\).

From [15], we know that

$$\begin{aligned}& \bigl\Vert {{\partial }_{t}}M^{n} \bigr\Vert _{0}^{2} \leq C \bigl( \bigl\Vert {{\partial }_{t}} \theta^{n} \bigr\Vert _{0}^{2} + \bigl\Vert {{\partial }_{t}}\eta^{n} \bigr\Vert _{0} ^{2} + \bigl\Vert \eta^{n-1} \bigr\Vert _{0}^{2} + \bigl\Vert \theta^{n-1} \bigr\Vert _{0}^{2} \bigr), \end{aligned}$$
(3.24)
$$\begin{aligned}& \bigl\Vert {{\partial }_{t}}\bigl({{\partial }_{t}}\rho^{n}\bigr)\bigr\Vert _{0}^{2} \leq \frac{Ch ^{4}}{\tau } \int^{t_{n}}_{t_{n-2}} \vert u_{tt} \vert _{2}^{2}\,dt, \end{aligned}$$
(3.25)
$$\begin{aligned}& \bigl\Vert {{\partial }_{t}}R_{1}^{n} \bigr\Vert _{0}^{2} \leq C\tau \int^{t_{n}}_{t _{n-2}} \vert u_{ttt} \vert _{0}^{2}\,dt. \end{aligned}$$
(3.26)

Further, we have by Lemma 2.2 that

$$\begin{aligned} \biggl\vert \sum_{K\in T_{h}} \int_{\partial K}\frac{\partial ({{\partial } _{t}}u^{n})}{\partial {n}}\cdot \overline{{{\partial }_{t}}\theta^{n}} \,ds \biggr\vert \leq& Ch^{2} \bigl\vert {{\partial }_{t}}u^{n} \bigr\vert _{4} \bigl\Vert {{\partial }_{t}}\theta^{n} \bigr\Vert _{0} \\ \leq & Ch^{4}\bigl( \bigl\vert {{\partial }_{t}}u^{n} \bigr\vert _{4}^{2} + \bigl\Vert {{\partial }_{t}} \theta^{n} \bigr\Vert _{0}^{2} \bigr) \\ \leq& Ch^{4} \biggl(\frac{1}{\tau } \int^{t_{n}}_{t _{n-1}} \Vert u_{t} \Vert _{4}^{2}\,dt + \bigl\Vert {{\partial }_{t}} \theta^{n} \bigr\Vert _{0}^{2} \biggr). \end{aligned}$$
(3.27)

Comparing the imaginary part of (3.23) with the above estimations, we obtain

$$\begin{aligned}& \frac{ \Vert {{\partial }_{t}}\theta^{n} \Vert _{0}^{2}- \Vert {{\partial }_{t}} \theta^{n-1} \Vert _{0}^{2}}{2\tau } \\ & \quad \leq \frac{Ch^{4}}{\tau } \int^{t_{n}}_{t_{n-2}} \vert u_{tt} \vert _{2}^{2}\,dt + C\tau \int^{t_{n}}_{t_{n-2}} \vert u_{ttt} \vert _{0}^{2}\,dt + \frac{Ch^{4}}{ \tau } \int^{t_{n}}_{t_{n-1}} \Vert u_{t} \Vert _{4}^{2}\,dt \\ & \quad\quad {} + \bigl\Vert \eta^{n-1} \bigr\Vert _{0}^{2} + \bigl\Vert \theta^{n-1} \bigr\Vert _{0}^{2} + C \bigl\Vert {{\partial }_{t}}\theta^{n} \bigr\Vert _{0}^{2}. \end{aligned}$$
(3.28)

Summing (3.28) up with respect to n leads to

$$\begin{aligned} \bigl\Vert {{\partial }_{t}}\theta^{n} \bigr\Vert _{0}^{2} \leq& Ch^{4} \int^{T}_{0} \bigl( \vert u _{tt} \vert _{2}^{2} + \Vert u_{t} \Vert _{4}^{2} \bigr)\,dt + C\tau^{2} \int^{T}_{0} \vert u _{ttt} \vert _{0}^{2}\,dt \\ &{}+C\tau \sum_{i=2}^{n} \bigl( \bigl\Vert \eta^{i-1} \bigr\Vert _{0}^{2} + \bigl\Vert \theta^{i-1} \bigr\Vert _{0}^{2} \bigr) + C\sum _{i=2}^{n} \bigl\Vert {{\partial }_{t}}\theta^{i} \bigr\Vert _{0}^{2} + \bigl\Vert {{\partial }_{t}}\theta^{1} \bigr\Vert _{0}^{2}. \end{aligned}$$
(3.29)

Setting \(n=1\) and taking \(v_{h}={{\partial }_{t}}\theta^{1}\) in (3.8), with an argument similar to (3.19), we can derive that

$$\begin{aligned} \bigl\Vert {{\partial }_{t}}\theta^{1} \bigr\Vert _{0}\leq C\bigl(h^{2}+\tau \bigr), \end{aligned}$$
(3.30)

which together with (3.19), (3.29), and (3.30) gives

$$\begin{aligned} \bigl\Vert {{\partial }_{t}}\theta^{n} \bigr\Vert _{0}\leq C\bigl(h^{2}+\tau \bigr). \end{aligned}$$
(3.31)

Substituting (3.31) into (3.22) and summing up from 1 to n yields

$$\begin{aligned} \bigl\Vert \theta^{n} \bigr\Vert _{h} = O\bigl(h^{2} + \tau \bigr). \end{aligned}$$
(3.32)

Thus the proof is complete. □

Now we will introduce a proper interpolation postprocessing operator to get the global superconvergence result. For this purpose, we further assume that \(T_{h}\) has been obtained from \(T_{2h}\) by dividing each element into four congruent rectangles. Let \(\mathcal{T}=\bigcup_{i=1} ^{4} K_{i}\), \(L_{1}\), \(L_{2}\), \(L_{3}\), and \(L_{4}\) be four edges. As in [31, 38], we define the interpolation operator \(\varPi_{2h}\) on the partition \(T_{2h}\):

$$ \textstyle\begin{cases} \varPi_{2h}u\vert _{\mathcal{T}}\in P_{2}(\mathcal{T}), \quad\quad\ \forall \mathcal{T}\in T_{2h}, \\ \int_{L_{i}}(\varPi_{2h}u-u)\,ds=0, \quad i=1,2,3,4, \\ \int_{K_{1}\cup K_{3}}(\varPi_{2h}u-u)\,dx\,dy=0, \quad\quad \int_{K_{2}\cup K_{4}}(\varPi_{2h}u-u)\,dx\,dy=0, \end{cases} $$

where \(P_{2}( \mathcal{T})\) denotes the set of polynomials of degree 2.

It has been shown in [38] that the interpolation operator \(\varPi_{2h}\) defined above satisfies the following properties:

$$\begin{aligned}& \varPi_{2h}\varPi_{h}u=\varPi_{2h}u, \quad \quad \Vert u-\varPi_{2h}u \Vert _{h}\leq Ch^{r} \vert u \vert _{r+1,\varOmega }, \quad 0\leq r\leq 2, \end{aligned}$$
(3.33)
$$\begin{aligned}& \Vert \varPi_{2h}v \Vert _{h}\leq C \Vert v \Vert _{h}, \quad \forall v\in V_{h}. \end{aligned}$$
(3.34)

Theorem 3.2

Under the same assumptions of Theorem 3.1, we have

$$\begin{aligned} \bigl\Vert {u^{n}}-{\varPi }_{2h}{U^{n}} \bigr\Vert _{h} = O\bigl(h^{2}+ \tau \bigr). \end{aligned}$$
(3.35)

Proof

Noticing that

$$ {u^{n}}-{\varPi }_{2h}{U^{n}}={u^{n}}-{ \varPi }_{2h}{\varPi }_{h}{u^{n}}+ {\varPi }_{2h}{\varPi }_{h}{u^{n}}-{\varPi }_{2h}{U^{n}}, $$
(3.36)

by (3.33) and interpolation error estimates, we have

$$\begin{aligned} \bigl\Vert {u^{n}}-{\varPi }_{2h}{\varPi }_{h}{U^{n}} \bigr\Vert _{h}= \bigl\Vert {u^{n}}-{\varPi }_{2h} {u^{n}} \bigr\Vert _{h}\leq Ch^{2} \bigl\vert {u^{n}} \bigr\vert _{3}. \end{aligned}$$
(3.37)

Consequently, it follows from (3.34) and Theorem 3.1 that

$$\begin{aligned} \bigl\Vert {\varPi }_{2h}{\varPi }_{h}{u^{n}}-{ \varPi }_{2h}{U^{n}} \bigr\Vert _{h} =&\Vert {\varPi } _{2h}\bigl({\varPi }_{h}{u^{n}}-{U^{n}} \bigr)\Vert _{h} \\ \leq &\bigl\Vert {\varPi }_{h}{u^{n}}-{U^{n}} \bigr\Vert _{h} = O\bigl(h^{2} + \tau \bigr). \end{aligned}$$
(3.38)

From (3.36)–(3.38), we can derive the result (3.35) directly. □

Remark 3.1

Theorems 3.1–3.2 are also valid to the \(Q_{1}^{\mathrm{rot}}\) element [39] on square meshes.

4 The two-grid finite element scheme and error analysis

In this section, we design a two-grid finite element algorithm (see Algorithm 4.1 below) for problem (1.1) to reduce the computing cost. The idea of the two-grid method is to reduce the nonlinear problem on a fine grid into a linear system through solving a nonlinear problem on a coarse grid. \(T_{H}\) and \(T_{h}\) are two regular subdivisions of Ω with two different mesh sizes H and h (\(h\ll H\)), and the corresponding \(\mathit{EQ}_{1}^{\mathrm{rot}}\) finite element spaces \(V_{H}\) and \(V_{h}\) (which will be called the coarse-grid space and the fine-grid space), respectively.

Algorithm 4.1

  • Step 1. Find \(u^{n}_{H}\in V_{H}\) (\(n=1,2,\ldots,N\)) such that

    $$ \textstyle\begin{cases} i({{\partial }_{t}} u_{H}^{n},v_{H})-(\nabla u_{H}^{n}, \nabla v_{H})+ \lambda (f( \vert u_{H}^{n} \vert ^{2})u_{H}^{n},v_{H})=0, & v_{H}\in V_{H}, \\ u_{H}^{0}=\varPi_{h}u_{0}(X)\in V_{H}, & X\in \varOmega . \end{cases} $$
    (4.1)

  • Step 2. Find \(u^{n}_{h}\in V_{h}\) (\(n=1,2,\ldots,N\)) such that

    $$ \textstyle\begin{cases} i({{\partial }_{t}} u_{h}^{n},v_{h})-(\nabla u_{h}^{n}, \nabla v_{h})+ \lambda (\widetilde{f( \vert u_{H}^{n} \vert ^{2}})u_{h}^{n},v_{h})=0, & v_{h} \in V_{h}, \\ u_{h}^{0}=\varPi_{h}u_{0}(X)\in V_{h}, & X\in \varOmega , \end{cases} $$
    (4.2)

    where \(\widetilde{f(\vert u_{H}^{n}\vert ^{2})}=f(\vert u_{H}^{n}\vert ^{2})+f'(\vert u_{H} ^{n}\vert ^{2})(\vert u_{h}^{n-1}\vert ^{2}-\vert u_{H}^{n}\vert ^{2})\).

Now we consider the error estimates in the broken \(H^{1}\)-norm for Algorithm 4.1.

Theorem 4.1

Let u and \(u_{h}^{n}\) be the solutions of problem (1.1) and the two-grid Algorithm 4.1, respectively. If \(u\in H^{4}(\varOmega )\cap W^{2,\infty }(\varOmega )\cap H^{1}_{0}( \varOmega )\), \(u_{t}\in H^{4}(\varOmega )\), \(u_{tt}\in H^{2}(\varOmega )\), and \(u_{ttt}\in L^{2}(\varOmega )\), there holds

$$\begin{aligned} \bigl\Vert \theta^{n} \bigr\Vert _{0} + \bigl\Vert \theta^{n} \bigr\Vert _{h}= O \bigl[h^{2} + \tau + H^{4} \vert {\ln }H \vert ^{ \frac{1}{2}}\bigr]. \end{aligned}$$
(4.3)

Proof

From (1.1) and (4.2), similar to (3.11), we have the result

$$\begin{aligned}& \bigl\Vert \theta^{n} \bigr\Vert _{0}^{2}- \bigl\Vert \theta^{n-1} \bigr\Vert _{0}^{2} \\& \quad =-2\tau \operatorname{Re}\bigl({{\partial }_{t}}\rho^{n}, \theta^{n}\bigr)-\operatorname{Im}\lambda \bigl(f\bigl( \bigl\vert u ^{n} \bigr\vert ^{2}\bigr)u_{h}^{n}-f\bigl( \bigl\vert u_{H}^{n} \bigr\vert ^{2} \bigr)u_{h}^{n},\theta^{n} \bigr) \\& \quad\quad {} +2\tau \operatorname{Re}\bigl(R_{1}^{n},\theta^{n}\bigr) - 2\tau \operatorname{Im}\sum_{K\in T_{h}} \int_{\partial K}\frac{\partial {u^{n}}}{\partial {n}}\cdot \overline{ \theta^{n}} \,ds\triangleq \sum_{i=1}^{4}M_{i}. \end{aligned}$$
(4.4)

We only need to estimate the term \(M_{2}\). In fact, by using the continuity of \(f(s)\) and Taylor expansions, we have

$$\begin{aligned} f\bigl( \bigl\vert u^{n} \bigr\vert ^{2} \bigr)=f\bigl( \bigl\vert u_{H}^{n} \bigr\vert ^{2}\bigr)+f'\bigl( \bigl\vert u_{H}^{n} \bigr\vert ^{2}\bigr) \bigl( \bigl\vert u^{n} \bigr\vert ^{2}- \bigl\vert u _{H}^{n} \bigr\vert ^{2}\bigr)+\frac{f''(\xi_{2})}{2!}\bigl( \bigl\vert u^{n} \bigr\vert ^{2}- \bigl\vert u_{H}^{n} \bigr\vert ^{2}\bigr)^{2}, \end{aligned}$$
(4.5)

where \(\xi_{1}\) lies between \(\vert u^{n}\vert ^{2}\) and \(\vert u_{H}^{n}\vert ^{2}\).

Then \(M_{2}\) can be expressed as

$$\begin{aligned} \vert M_{2} \vert \leq & 2\tau \vert \lambda \vert \bigl\vert \bigl(f\bigl( \bigl\vert u^{n} \bigr\vert ^{2}\bigr)u^{n}- \widetilde{f\bigl( \bigl\vert u_{H}^{n} \bigr\vert ^{2}}\bigr)u_{h}^{n}, \theta^{n} \bigr) \bigr\vert \\ =& 2\tau \vert \lambda \vert \bigl\vert \bigl(f\bigl( \bigl\vert u^{n} \bigr\vert ^{2}\bigr)u^{n}- \widetilde{f \bigl( \bigl\vert u_{H}^{n} \bigr\vert ^{2}} \bigr)u^{n} + \widetilde{f\bigl( \bigl\vert u_{H}^{n} \bigr\vert ^{2}}\bigr)u ^{n} - \widetilde{f\bigl( \bigl\vert u_{H}^{n} \bigr\vert ^{2}} \bigr)u_{h}^{n},\theta^{n} \bigr) \bigr\vert \\ =& 2\tau \vert \lambda \vert \bigl\vert \bigl(f\bigl( \bigl\vert u^{n} \bigr\vert ^{2}\bigr)u^{n}- \widetilde{f \bigl( \bigl\vert u_{H}^{n} \bigr\vert ^{2}} \bigr)u^{n},\theta^{n} \bigr) \bigr\vert \\ & {} + 2\tau \vert \lambda \vert \bigl\vert \bigl(\widetilde{f\bigl( \bigl\vert u_{H}^{n} \bigr\vert ^{2}}\bigr) \bigl(u^{n} - u _{h}^{n}\bigr), \theta^{n} \bigr) \bigr\vert \triangleq E_{1}+E_{2}. \end{aligned}$$
(4.6)

Firstly, for the term \(E_{1}\), we have

$$\begin{aligned} E_{1} =& 2\tau \vert \lambda \vert \biggl\vert \biggl( \biggl[f'\bigl( \bigl\vert u_{H}^{n} \bigr\vert ^{2}\bigr) \bigl( \bigl\vert u^{n} \bigr\vert ^{2}- \bigl\vert u _{h}^{n-1} \bigr\vert ^{2}\bigr) + \frac{f''( \vert \xi_{2} \vert ^{2})}{2}\bigl( \bigl\vert u^{n} \bigr\vert ^{2}- \bigl\vert u_{H} ^{n} \bigr\vert ^{2}\bigr)^{2}\biggr]u^{n}, \theta^{n} \biggr) \biggr\vert \\ =& 2\tau \vert \lambda \vert \bigl\vert \bigl( f'\bigl( \bigl\vert u_{H}^{n} \bigr\vert ^{2}\bigr) \bigl( \bigl\vert u^{n} \bigr\vert ^{2}- \bigl\vert u^{n-1} \bigr\vert ^{2}\bigr)u ^{n}, \theta^{n} \bigr) \bigr\vert \\ & {} + 2\tau \vert \lambda \vert \bigl\vert \bigl( f'\bigl( \bigl\vert u_{H}^{n} \bigr\vert ^{2}\bigr) \bigl( \bigl\vert u^{n-1} \bigr\vert ^{2}- \bigl\vert u _{h}^{n-1} \bigr\vert ^{2}\bigr)u^{n}, \theta^{n} \bigr) \bigr\vert \\ & {} + \tau \vert \lambda \vert \bigl\vert \bigl( f'' \bigl( \vert \xi_{2} \vert ^{2}\bigr) \bigl( \bigl\vert u^{n} \bigr\vert ^{2}- \bigl\vert u_{H} ^{n} \bigr\vert ^{2}\bigr)^{2}u^{n}, \theta^{n} \bigr) \bigr\vert \\ \triangleq & E_{11}+E_{12}+E_{13}. \end{aligned}$$
(4.7)

Applying the boundedness of u, \(f(s)\), and Theorem 3.1, we can derive that

$$\begin{aligned} E_{11}+E_{12}\leq C\tau \bigl(h^{4}+ \tau^{2}\bigr). \end{aligned}$$
(4.8)

Similar to (3.13), \(E_{13}\) can be estimated as

$$\begin{aligned} E_{13}\leq C\tau \bigl( \bigl\Vert \bigl( \bigl\vert u^{n} \bigr\vert ^{2}- \bigl\vert u_{H}^{n} \bigr\vert ^{2}\bigr)^{2} \bigr\Vert _{0}^{2} + \bigl\Vert \theta^{n} \bigr\Vert _{0}^{2} \bigr). \end{aligned}$$
(4.9)

Notice that the term \(\Vert (\vert u^{n}\vert ^{2}-\vert u_{H}^{n}\vert ^{2})^{2}\Vert _{0}^{2}\) can be rewritten as

$$\begin{aligned} \bigl\Vert \bigl( \bigl\vert u^{n} \bigr\vert ^{2}- \bigl\vert u_{H}^{n} \bigr\vert ^{2}\bigr)^{2} \bigr\Vert _{0}^{2} \leq &C \bigl\Vert \bigl( \bigl\vert u^{n} \bigr\vert ^{2}- \bigl\vert u _{H}^{n} \bigr\vert ^{2}\bigr)^{2} \bigr\Vert _{0}^{2} \leq C \bigl\Vert u^{n}-u_{H}^{n} \bigr\Vert _{0,\infty } ^{2} \bigl\Vert u^{n}-u_{H}^{n} \bigr\Vert _{0}^{2} \\ \leq & \bigl( \bigl\Vert u^{n}-\varPi_{H}u^{n} \bigr\Vert _{0,\infty }^{2} + \bigl\Vert \varPi_{H}u^{n} - u _{H}^{n} \bigr\Vert _{0,\infty }^{2} \bigr) \bigl\Vert u^{n}-u_{H}^{n} \bigr\Vert _{0}^{2}. \end{aligned}$$
(4.10)

Since

$$\begin{aligned}& \bigl\Vert u^{n}-\varPi_{H}u^{n} \bigr\Vert _{0,\infty }^{2}\leq CH^{4} \bigl\Vert u^{n} \bigr\Vert _{2, \infty }^{2}, \end{aligned}$$
(4.11)
$$\begin{aligned}& \bigl\Vert u^{n}-u_{H}^{n} \bigr\Vert _{0}^{2}\leq C\bigl(H^{4}+ \tau^{2}\bigr), \end{aligned}$$
(4.12)

and \(\Vert v_{h}\Vert _{0,\infty }\leq C{\vert \ln h\vert ^{\frac{1}{2}}}\Vert v_{h}\Vert _{h}\) [40], we have

$$\begin{aligned} \bigl\Vert \varPi_{H}u^{n} - u_{H}^{n} \bigr\Vert _{0,\infty }^{2}\leq C \vert {\ln }H \vert \bigl\Vert \varPi_{H}u ^{n} - u_{H}^{n} \bigr\Vert _{h}^{2}\leq C \vert {\ln }H \vert \bigl(H^{4}+\tau^{2}\bigr). \end{aligned}$$
(4.13)

Then from (4.10)–(4.13), we know that

$$\begin{aligned} \bigl\Vert \bigl( \bigl\vert u^{n} \bigr\vert ^{2}- \bigl\vert u_{H}^{n} \bigr\vert ^{2}\bigr)^{2} \bigr\Vert _{0}^{2} \leq C\bigl[H^{4} + \tau^{2} + \vert {\ln }H \vert \bigl(H^{4}+\tau^{2}\bigr)\bigr]\bigl(H^{4}+ \tau^{2}\bigr). \end{aligned}$$
(4.14)

Further, when τ is small enough, there holds

$$\begin{aligned} \bigl\Vert \bigl( \bigl\vert u^{n} \bigr\vert ^{2}- \bigl\vert u_{H}^{n} \bigr\vert ^{2}\bigr)^{2} \bigr\Vert _{0}^{2} \leq CH^{8} \vert {\ln }H \vert , \end{aligned}$$
(4.15)

which implies

$$\begin{aligned} E_{1}\leq C\tau \bigl(h^{4} + \tau^{2} + H^{8} \vert {\ln }H \vert \bigr). \end{aligned}$$
(4.16)

Secondly, for the term \(E_{2}\), we have

$$\begin{aligned} E_{2}\leq C\tau \bigl( \bigl\Vert u^{n}-u_{h}^{n} \bigr\Vert _{0}^{2} + \bigl\Vert \theta^{n} \bigr\Vert _{0}^{2} \bigr) \leq C\tau \bigl(h^{4}+ \tau^{2}\bigr). \end{aligned}$$
(4.17)

Finally, substituting (4.16) and (4.17) into (4.6), we obtain

$$\begin{aligned} \vert M_{2} \vert \leq C\tau \bigl(h^{4} + \tau^{2} + H^{8} \vert {\ln }H \vert \bigr), \end{aligned}$$
(4.18)

and substituting (3.12), (3.14), (3.15), and (4.18) into (4.4) yields

$$\begin{aligned} \bigl\Vert \theta^{n} \bigr\Vert _{0}^{2}- \bigl\Vert \theta^{n-1} \bigr\Vert _{0}^{2}\leq C\tau \bigl(h^{4} + \tau^{2} + H^{8} \vert {\ln }H \vert \bigr) + C\tau \bigl\Vert \theta^{n} \bigr\Vert _{0}^{2} . \end{aligned}$$
(4.19)

Then summing (4.19) up with respect to n and noting \(\theta^{0}=0\), we have

$$\begin{aligned} \bigl\Vert \theta^{n} \bigr\Vert _{0}^{2}\leq C\bigl(h^{4} + \tau^{2} + H^{8} \vert {\ln }H \vert \bigr) + C \tau \sum _{i=1}^{n} \bigl\Vert \theta^{i} \bigr\Vert _{0}^{2}. \end{aligned}$$
(4.20)

An application of Gronwall’s lemma yields

$$\begin{aligned} \bigl\Vert \theta^{n} \bigr\Vert _{0}^{2}\leq C\bigl(h^{4} + \tau^{2} + H^{8} \vert {\ln }H \vert \bigr), \end{aligned}$$
(4.21)

which implies that

$$\begin{aligned} \bigl\Vert \theta^{n} \bigr\Vert _{0} = O\bigl(h^{2} + \tau + H^{4} \vert {\ln }H \vert ^{\frac{1}{2}}\bigr). \end{aligned}$$
(4.22)

Thus with the similar arguments to the estimates of (3.32) and (4.22), we can also derive

$$\begin{aligned} \bigl\Vert \theta^{n} \bigr\Vert _{h} = O\bigl(h^{2} + \tau + H^{4} \vert {\ln }H \vert ^{\frac{1}{2}}\bigr), \end{aligned}$$
(4.23)

which is the desired result. □

Similar to the proof of Theorem 3.2, we can derive the following superconvergence results.

Theorem 4.2

Under the same assumptions of Theorem 4.1, and setting \(h=H^{2}(\vert {\ln }H\vert )^{\frac{1}{4}}\), we can derive

$$\begin{aligned} \bigl\Vert {u^{n}}-{\varPi }_{2h}{u^{n}_{h}} \bigr\Vert _{h} = O\bigl(h^{2}+ \tau \bigr). \end{aligned}$$
(4.24)

5 Numerical experiment

In this section, we present the following numerical example to confirm the theoretical analysis, which comes from [8, 16, 41].

Consider the cubic-quintic Schrödinger equation (\(f(s)=-s+s^{2}\))

$$ \textstyle\begin{cases} iu_{t}+\Delta u- \vert u \vert ^{2}u + \vert u \vert ^{4}u=g, & (X,t)\in \varOmega \times (0,T], \\ u(X,t)=0, & (X,t)\in \partial \varOmega \times (0,T], \\ u(X,0)=u_{0}(X), & X\in \varOmega , \end{cases} $$
(5.1)

on \(\varOmega = [0, 1]^{2}\) with the exact solution

$$\begin{aligned} {u}=e^{it+(x+y)/2}\bigl( 1+3t^{2} \bigr)x(1-x)y(1-y), \end{aligned}$$

where g is given corresponding to the exact solution u.

The domain Ω is divided into families \(T_{H}\) and \(T_{h}\) of rectangular meshes, and \(V_{H}\), \(V_{h}\) are \(\mathit{EQ}_{1}^{\mathrm{rot}}\) finite element spaces defined on \(T_{H}\), \(T_{h}\), respectively. In such a way, to obtain enough accuracy, it suffices to take \(h = O(H^{2}\vert {\ln }H\vert ^{\frac{1}{4}})\) in both the broken \(H^{1}\)-norm and \(L^{2}\)-norm. Under the same condition of computing environment and control strategy, the numerical results and CPU times are shown in Tables 1–2, respectively. The exact solution u at time \(t=1\) and FEM solution \({u}_{h}\) at time \(t=1\) with mesh size \(h=1/32\) are pictured in Figs. 1 and 2, respectively.

Figure 1
figure 1

The exact solutions \(u_{1}\) (Real part) and \(u_{2}\) (Imaginary part) at \(t=1\)

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Figure 2
figure 2

The FEM solutions \(u_{h1}\) (Real part) and \(u_{h2}\) (Imaginary part) at \(t=1\)

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Table 1 Numerical results of two-grid Algorithm 4.1 at \(t = 1\) (\(\tau =h^{2}\), \(h\approx H^{2}\))
Full size table
Table 2 Numerical results of the usual Galerkin FEM at \(t = 1\) (\(\tau =h^{2}\))
Full size table

It can be seen from Tables 1–2 that \(\Vert {u}-{u}_{h}\Vert _{h}\) is convergence at rate of \(O(h)\), \(\Vert {u}-{u}_{h}\Vert _{0}\), \(\Vert {\varPi }_{h} {u}-{u}_{h}\Vert _{h}\) and \(\Vert {u}-{\varPi }_{2h}{u}\Vert _{h}\) are convergence at rate of \(O(h^{2})\), which coincide with the theoretical analysis. Thus the two-grid FEM is more efficient in solving NLSE than the usual Galerkin FEM.

6 Conclusions

In this paper, we applied low order nonconforming \(\mathit{EQ}_{1}^{\mathrm{rot}}\) finite element to solve the nonlinear Schrödinger equation, and derived the global superconvergence results for the backward Euler fully-discrete scheme and a type of two-grid scheme, respectively. A numerical example is presented to demonstrate the theoretical results. The method presented in this paper is suitable for the standard Galerkin finite element and can be extended to dealing with other nonlinear problems.