# Fourier Spectral Method for a Class of Nonlinear Schrödinger Models

Authors: Lei Zhang; Weihua Ou Yang; Xuan Liu; Haidong Qu
Journal: Advances in Mathematical Physics
 (2021)

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## Abstract

In this paper, Fourier spectral method combined with modified fourth order exponential time-differencing Runge-Kutta is proposed to solve the nonlinear Schrödinger equation with a source term. The Fourier spectral method is applied to approximate the spatial direction, and fourth order exponential time-differencing Runge-Kutta method is used to discrete temporal direction. The proof of the conservation law of the mass and the energy for the semidiscrete and full-discrete Fourier spectral scheme is given. The error of the semidiscrete Fourier spectral scheme is analyzed in the proper Sobolev space. Finally, several numerical examples are presented to support our analysis.


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## Body

## 1. Introduction

Schrödinger equation, being known as basic assumption of quantum mechanics, is one of the most important equations in quantum mechanics that proposed by Austrian physicist Schrödinger. Known to all, the standard Schrödinger equation is a second-order partial differential equation established by combining the concept of matter wave and wave equation, and it can describe the movement of microparticles. Indeed, each microsystem has a corresponding Schrodinger equation. The specific form of wave function and corresponding energy can be obtained by solving this equation, so as to understand the properties of microsystem. Schrödinger equation shows that in quantum mechanics, particles appear in the form of probability, with uncertainty, and can be ignored at the macroscale.In recent decades, more and more scholars have been interested in studying Schrödinger equations, and the research on Schrödinger equations can be divided into two categories: integral order (or local) and fractional order (or nonlocal) models. Now, we will make a brief literature review.(1)
Schrödinger Equations of Fractional Order. In 2000, firstly, Laskin [1] extended the fractional concept in quantum physics to construct a fractional path integral, and over these paths, the standard nonlinear Schrödinger equation was generalized to the fractional one. Very recently, many scholars studied the numerical solutions of fractional Schrödinger equations. In 2008, Rida et al. [2] applied the Adomian decomposition method to get the analytic and approximate solutions for different types of fractional differential equations. In 2011, the generalized two-dimensional differential transform method was directly applied to solve the coupled Burger’s equations by Liu et al. [3]. In 2012, Wei et al. [4] presented an implicit fully discrete local discontinuous Galerkin finite element method for solving the time-fractional Schrödinger equation with Caputo fractional derivative. In 2013, Secchi et al. [5] obtained the solutions to a class of Schrödinger equations by using the fractional Laplacian. In 2014, the Jacobi spectral collocation method was constructed, and the numerical solutions of the time-fractional Schrödinger equation and the space-fractional Schrödinger equation were got by Bhrawy et al. [6]. In literature [7], Wang et al. proposed a linearly implicit conservative difference scheme for the coupled nonlinear Schrödinger equations with space fractional derivative, and she proved that the scheme can conserve the mass and energy in the discrete level. In 2015, Wang et al. [8] studied the energy conservative Crank-Nicolson difference scheme for nonlinear Riesz space-fractional Schrödinger equations and gave the proof of mass conservation and energy conservation in the discrete sense. In 2016, Ran et al. [9], under some assumptions, an implicit difference scheme was proposed to solve the strongly coupled nonlinear space fractional Schrödinger equations. In the literature [10–12], the semidiscrete scheme was fully constructed by employing finite element method. In 2020, Wang et al. [13] proposed a preconditioning method for the space fractional coupled nonlinear Schrödinger equations. For other literature on the fractional Schrödinger equations, see [14–16](2)
Schrödinger Equations of Integral Order. The study of the standard Schrödinger equations has attracted many scholars. In 1995, Edwards et al. [17] presented a numerical treatment to solve the time-independent nonlinear Schrödinger equation with an external potential. In 1999, Chang et al. [18] studied the finite difference schemes for solving the generalized nonlinear Schrödinger equations, and several schemes, including Crank-Nicolson-type schemes, split step Fourier scheme, and pseudo-spectral scheme, were employed for solving three models of the generalized nonlinear Schrödinger equations. In 2003, Bao et al. [19] studied the performance of time-splitting spectral approximations for general nonlinear Schrödinger equations and proved that the discrete scheme was unconditionally stable and conserved in L1 norm. In 2005, Wang [20] got the numerical solutions to nonlinear Schrödinger equations by using the split-step finite difference method. In 2017, Dehghan et al. [21] presented a numerical scheme based on radial basis function for solving the high-dimensional nonlinear Schrödinger equations using the explicit Runge-Kutta method. In 2019, Ahsan et al. [22] proposed a Haar wavelet collocation method for linear and nonlinear Schrödinger equations, and the stability analysis of this method was given. In 2020, Ban et al. [23] proposed a family of algorithms, which were special implementations of Lubich’s Convolution Quadrature, for the numerical solution to the Schrödinger equation. For other literature on the standard Schrödinger equations, see [24–28]In this paper, we study the numerical solution to the nonlinear Schrödinger equation of the following form:(1)iutx,t+αΔux,t+βux,t2ux,t+γfxux,t=0,with the periodic condition
(2)ux+2π,t=ux,t,and the initial condition
(3)ux,0=u0x,where
(4)Δux,t=∂xxux,t,x,t∈−π,π×0,T.i is the imaginary number of units and i2=−1; u0x is a function with a period of 2π; α, β, and γ are the three real numbers.For this model, two important physical quantities are often considered; these are mass and energy and using the conservation of mass and energy to measure whether the algorithm is effective for this equation. As in [10], we can easily derive the mass and energy conservation
(5)Mt=M0,Et=E0,where
(6)Mt=∫02πux,t2dx,Et=−α∫02πuxx,tu¯xx,tdxβ2∫02πux,t∣4dx+γ∫02πfxux,t2dx.The rest of this paper is organized as follows. In Section 2, the definitions and properties of the related Sobolev space are introduced; based on this space, we establish a semidiscrete Fourier spectral scheme for model (1), whose solution keeps mass and energy conversation for some function fx. Then, the error estimation of this semidiscrete scheme is given in the established Sobolev space. Section 3 gives an introduction of modified fourth order exponential time-differencing Runge-Kutta method that is used to discrete the temporal component, and a proof of mass conservation and energy conservation is given in the discrete sense. In Section 4, we carry out some numerical experiments to verify our conclusions. Finally, Section 5 offers some concluding remarks.


## 2. Preliminaries

In this section, based on the semidiscrete Fourier spectral scheme, we prove the mass and energy conversation for this semidiscrete scheme and give the error estimation of the Fourier spectral approximation. First, we introduce the following Sobolev space and some notations.LetΩ=−π,π⊂ℝ, and the space Cp∞ is defined by
(7)Cp∞=u∣u∈C∞Ω,ux−π,t=ux+π,t,x∈ℝ.Define inner product onΩ=−π,π:
(8)ux,t,vx,tΩ=∫−ππux,tv¯x,tdx,and the norm on Ω=−π,π:
(9)∥ux,t∥=∫−ππux,tu¯x,tdx,where v¯x,t is the conjugate of vx,t. We denote the complete space of Cp∞ by Hpn, which can be characterized in the frequency space under the norm
(10)∥u∥n≔∑m≤n∥Dmux,t∥21/2,the seminorm un is defined by
(11)un≔∑0<m≤n∥Dmux,t∥21/2,where Dmux,t=∂mux,t/∂xm.Suppose(12)SN=spaneikx,−N≤k≤N.Then, the orthogonal projection operatorPN:Hpn⟶SN is defined by
(13)PNux=∑k=−NNu^keikx=uN,where u^k, k=−N,⋯,N are the Fourier coefficients of u.According to the definition of inner product,∀vN∈SN, we define the semidiscrete scheme for model (1):
(14)iutN,vNΩ+αΔuN,vNΩ+βuN2uN,vNΩ+γfuN,vNΩ=0,(15)uN0=PNu0x.Usually, when designing a numerical scheme for (14), two important conservation laws often need to be considered: the mass and the total energy of the wave function. In the following lemma, we prove that the semidiscrete Fourier spectral scheme (14) can keep the conservation laws.Lemma 1.
LetuN0,uNx,t∈HP1, then the semidiscrete scheme (14) is conservative in the sense
(16)Mt=M0,0<t<T,Et=E0,0<t<T,where
(17)Mt=∫−ππuNx,t2dx,Et=−α∫−ππuxNx,tu¯xNx,tdx+β2∫−ππuNx,t∣4dx+γ∫−ππfxuNx,t2dx,are the mass and energy of the wave function, respectively.Proof.
LetvN=uN in (14), we have
(18)iutN,uNΩ+αΔuN,uNΩ+βuN2uN,uNΩ+γfuN,uNΩ=0.
Note thatα, β, and γ are three real numbers; thus, taking the imaginary part of the above equation and noting the periodic boundary condition, we get
(19)ImαΔuN,uNΩ=Im∫−ππuxNu¯xN=0,ImβuN2uN,uNΩ=ImβuN2uN,uNΩ=0,ImγfuN,uNΩ=0,which indicates
(20)utN,uNΩ=12ddtuN,uNΩ=12ddtMt=0.
Next, we prove the semidiscrete scheme (14) can keep energy conservation. Let vN=utN in (14), we have
(21)iutN,utNΩ+αΔuN,utNΩ+βuN2uN,utNΩ+γfuN,utNΩ=0.
Taking the real part of the above equation and noting the periodic boundary condition, we have(22)ReiutN,utNΩ=0,(23)ReαΔuN,utNΩ=ReαuxxN,utNΩ=αRe∫−ππuxxNu¯tNdx=−αRe∫−ππuxNu¯x,tNdx=−α2ddt∫−ππuxNu¯xNdx.
Computing the real part ofβuN2uN,utNΩ, we have
(24)ReβuN2uN,utNΩ,=Reβ∫−ππuN2uNu¯tNdx=Reβ∫−ππuNu¯NuNu¯tNdx=β4ddt∫−ππuNu¯NuNu¯Ndx=β4ddt∫−ππuN4dx.
Computing the real part ofγfuN,utNΩ, we have
(25)ReγfuN,utNΩ=γ2ddt∫−ππfuN2dx.
It follows from equations (22), (23), (24), and (25); we get the energy conservation law. This completes the proof.Next, we analyze the existence and uniqueness of solution for the semidiscrete scheme (14). We need the following lemma.Lemma 2.
LetuN is the solution for the semidiscrete scheme (14); then, there exists a positive constant C, such that
(26)max∥uN∥1,∥uN∥∞≤C.Proof.
From Lemma1, we have
(27)∥uNt∥02=∥uN0∥02=∥PNu0x∥02≤∥u0∥02=C,(28)∥uNt∥02+β2∫−ππuNx,t4dx=∥uN0∥02+β2∫−ππuNx,04dx≤∥uN0∥02+β2∫−ππmaxuNx,04dx≤∥uN0∥02+βπ∥uNx,0∥∞4≤∥uN0∥02+C∣β∣π∥uNx,0∥14=C.
(27) and (28) lead to
(29)∥uNt∥12=∥uNt∥02+∥uxNt∥02≤C.
Finally, using Sobolev’s embedding theorem, we obtain(30)∥uNt∥∞2≤C,which implies both ∥uNt∥1 and ∥uNt∥∞ are bounded. Therefore
(31)max∥uN∥1,∥uN∥∞≤C.
This completes the proof.The semidiscrete scheme (14) is a first order nonlinear ordinary differential equation with respect to time t on a finite interval; using Lemma 2 and Picard’s theorem, we obtain the following theorem.Theorem 3.
Let the solutionu∈L20,T,HP10,2π, u0∈HP10,2π, then the semi-discrete scheme (14) has unique solution uN in SN.Proof.
The proof is omit.
In the following theorem, we will give the error estimation of the Fourier spectral method to solve problem (1). For convenience, we still use η≤ξ to denote that there exists a positive constant C, which is independent of N, η, and ξ, such that η≤Cξ.Theorem 4.
Letu is the solution of (1); uN is obtained from the semidiscrete scheme (14); then, we have
(32)∥ut−uNt∥2≤N2μ−2.Proof.
LeteNt=uNt−PNut, with uNt in (14) replaced by PNut, for all vN∈SN, we have
(33)ieN,tt,vNΩ+αΔeNt,vNΩ+βuNt2uNt−PNut2PNut,vNΩ+γfeNt,vNΩ=0.
Note thatut−PNut,vNΩ=0,∀vN∈SN, by setting vN=eNt, and taking the imaginary part of (33), we obtain
(34)α2ddteNt,eNtΩ+βImut2ut−uNt2uNt,eNtΩ=0.
Applying the Schwarz inequality to the above equation, we get(35)∣α∣2ddt∥eNt∥2≤∥∣β∣ut2ut−uNt2uNt∥·∥eNt∥≤12∥∣β∣ut2ut−uNt2uNt∥2+12∥eNt∥2.
We note that(36)∣β∣ut2ut−uNt2uNt=βut2ut−uNt+∣β∣uNtu¯tut−utu¯Nt+utu¯Nt−u¯NtuNt=∣β∣ut2ut−uNt+uNtutu¯t−u¯Nt+uNt2ut−uNt.
Then, the factor∥∣β∣ut2ut−uNt2uNt∥2 is bounded by
(37)∥∣β∣ut2ut−uNt2uNt∥≤∥ut−uNt∥≤∥ut−PNut∥+∥eNt∥.
According to equations (35) and (37) and u−PNu,eN=0, we arrive at
(38)ddt∥eNt∥2≤∥ut−PNut∥2+∥eNt∥2,which in accordance with Gronwall inequality, we have
(39)∥eNt∥≤∫0T∥uτ−PNuτ∥2dτ.
Noting that for all0<μ<1, there exists a positive constant C, such that
(40)∥ut−PNut∥μ≤CNμ−1u1.
Finally, a combination of the above estimates leads to(41)∥ut−uNt∥2≤∥ut−PNut∥2+∥PNut−uNt∥2≤N2μ−2∣ut12+∫0T∥uτ−PNuτ∥2dτ≤N2μ−2ut∣12+N2μ−2∫0Tuτ12dτ≤N2μ−2ut∣12+∫0Tuτ12dτ≤N2μ−2.
This completes the proof.


## 3. The Full-Discrete Fourier Spectral Scheme and Its Conservation

In this section, we first introduce the full-discrete Fourier spectral scheme for (1) and then give the proof of its conservation properties.


### 3.1. The Full-Discrete Fourier Spectral Scheme

Due to the periodicity of the solution, it suffices to compute the model in the intervalΩ=−π,π. We choose N interpolation points xN=x1,x2,⋯,xN, and xj=2πj/N, −N/2≤j≤N/2−1. Then, the semidiscrete scheme (14) can be discretized in space by seeking the approximate solution
(42)uxj=∑k∞≤N/2u~keikxj,t>0,where
(43)u~k=CkN∑j=1nuxjeikxj,is the kth discrete Fourier coefficient, Ck=1 for ∣k∣<N/2, and Ck=2 for ∣k∣=N/2. Thus, for each frequency k, we have
(44)i∂u~k∂t+αk2u~k+βFF−1u~k2F−1u~k+γFfF−1u~k=0,u~x,0=Fu0.Leth1,h2,⋯,hQ be the time steps, satisfying
(45)∑q=1Qhq=T,tq=∑j=1qhj,L=iαk2,Fu~k,t=iβFF−1u~k2F−1u~k+iγFfF−1u~k,then (44) is equivalent to
(46)u~ktq+hq+1=eLhq+1u~ktq+eLhq+1∫0hq+1e−LsFu~ktq+s,tq+sds.More detail about this equation can be find in [29, 30]. Now, let u~k,q=u~ktq and using modified fourth order exponential time-differencing Runge-Kutta (ETDRK4) method to approximate the integral equation (46), we have
(47)λq=eLhq+12u~k,q+L−1eLhq+12−IFu~k,q,tq,χq=eLhq+12u~k,q+L−1eLhq+12−IFλq,tq+hq+12,νq=eLhq+12λq+L−1eLhq+12−I·2Fχq,tq+hq+12−Fu~k,q,tq,u~k,q+1=eLhq+1u~k,q+hq+1θ1Fu~k,q,tq+2θ2Fλq,tq+hq+12+Fχq,tq+hq+12+θ3Fνq,tq+hq+1,where
(48)θ1=4ϕ3Lhq+1−3ϕ2Lhq+1+ϕ1Lhq+1,θ2=ϕ2Lhq+1−2ϕ3Lhq+1,θ3=4ϕ3Lhq+1−ϕ2Lhq+1,with the functions ϕ1, ϕ2, and ϕ3 being defined by
(49)ϕ1=ez−1z,ϕ2=ez−1−zz2,ϕ3=ez−1−z−z2/2z3.Finally, by using inverse Fourier transform, we obtainuj,q=F−1u~k,q is the approximate solution of uxj,tq.For simplicity, the equal step-size is used to discrete time directions. The method presented in this paper allows one to choose different step sizes which are determined by adaptive strategy given in [31]. Next, we proof that the full-discrete scheme (47) satisfies two conservation laws given by Lemma 1.


### 3.2. The Conservation Laws for ETDRK4

In this subsection, we prove that the scheme given by (47) can keep mass and energy conservation. First, see the following two explanatory notes about ETDRK4.Note 1.
Equation (46) is obtained by integrating
(50)e−Ltu~k·,t=e−LtFu~k,t,from t=tq to t=tq+1. Then, from the reasoning process of the method ETDRK4 [29, 30], we get the following form
(51)e−Ltq+1u^N,k,q+1=e−Ltqu^N,k,q+hq+1Φtq,u^N,k,q+Ohq+15,where
(52)Φtq,u^N,k,q=∑j=14hj−1j!dj−1dtj−1e−LtFu^N,k,q,tt=tq,u^N,k,q=FuNx,tq,here, uNx,tq is obtained from the semidiscrete scheme (14). Then, if ignoring the error term Ohq+15, using the method of ETDRK4 to solve (46) is equivalent to get the solution to (50) by the scheme (51).Note 2.
Assume that the functionϕ mentioned above satisfies the Lipschitz condition with parameter λ>0, that is
(53)∣Φt,u1−Φt,u2∣≤λu1−u2.
Actually,ϕ should be third-order continuous differentiable with respect to u due to using the method of ETDRK4, which ensures that ϕ satisfies this assumption easily.In Lemma1, we have proved the mass conservation for the semidiscrete scheme (14). Then, if it holds ∣uNxj,tq−uj,q∣≤Ohmaxn, the full-discrete scheme ETDRK4 is mass conservative, where uNxj,tq is the solution to (14) and uj,q is obtained from (47) and inverse Fourier transform. Next, we give the proof in the following theorem.Theorem 5 (mass conservation).
LetuNxj,tq be obtained from the semidiscrete scheme (14), uj,q=F−1u~k,q, and u~k,q is computed by (47). Define the discrete mass Mtq=∑jΩjuj,q, Ω=⋃jΩj. Then, ignoring the error comes from machine calculation, we have
(54)uNxj,tq−uj,q≤Ohmax4,where hmax=max1≤i≤Qhi, q=0,1,2,⋯,Q. Furthermore,
(55)Mt0~Mt1~⋯~MtQ.Proof.
Without loss of generality, we only need to proveuNxj,tQ−uj,Q≤Ohmax4. From Note 1, we have
(56)e−Ltq+1u~k,q+1=e−Ltqu~k,q+hq+1Φtq,u~k,q.
Taking the module of equation (51) minus equation (56), and let e~q=u^N,k,q−u~k,q, it follows
(57)e−Ltq+1∣e~q+1∣≤e−Ltq∣e~q∣+hq+1∣Φtq,u^N,k,q−Φtq,u~k,q∣+Ohq+15.
Using Note2, we get
(58)e−Ltq+1∣e~q+1∣≤e−Ltq∣e~q∣+hq+1λ∣e~q∣+Ohq+15,
q=0,1,⋯,Q−1. Summing the above inequalities from q=0 to q=Q−1, we obtain
(59)e−LT∣e~Q∣≤∣e~0∣+λ∑q=0Q−1hq+1∣e~q∣+Ohmax4.
LetC1=eLT using Gronwall inequality leads to
(60)∣e~M∣≤eλC1∑q=0Q−1hq+1C1∣e~0∣+Ohmax4.
Note thate~0=0 and denote eλC1T by C, the desired result follows,
(61)∣uNxj,tQ−uj,Q∣=∣F−1e~Q∣=∣e~Q∣≤C·Ohmax4=Ohmax4.
Moreover, we have(62)uj,0−uj,Q≤uj,0−uNxj,t0+uNxj,tQ−uj,Q≤∣uj,0−uNxj,t0∣+uNxj,tQ−uj,Q≤2Ohmax4=Ohmax4.
The above inequalities imply(63)M0~Mt1~⋯~MtQ,as N is big enough and hmax⟶0.Theorem 6 (energy conservation).
Let(64)uj,q=F−1u~k,q,and u~k,q is obtained from (47). Define the discrete local-energy
(65)Ejtq=Ωj−αu¯j,qδα/2uj,q+β2∣uj,q4,and the discrete energy
(66)Etq=∑jEjtq,here, δα/2uj,q is a numerical approximation of the fractional Laplacian operator −Δα/2, Ω=⋃jΩj. Then, we have
(67)∣Ej0−Ejtq∣≤Ohmax4,q=1,2,⋯,Q.
Furthermore,Et0~Et1~⋯~EtQ.Proof.
Without loss of generality, we only need to prove(68)∣Ej0−EjtQ∣≤Ohmax4.
Define bounded linear operator(69)A∣uj,q∣≔Ejtq,then we have
(70)∣Ej0−EjtQ∣=∥A∣uj,0∣−∣uj,Q∣∥≤C∣uj,0∣−∣uj,Q∣,where C>0 is a positive constant. By Theorem 5,
(71)∣Ej0−EjtQ∣≤Ohmax4,which implies
(72)E0~E1~⋯~EQ,as N is big enough and hmax⟶0.


## 3.1. The Full-Discrete Fourier Spectral Scheme

Due to the periodicity of the solution, it suffices to compute the model in the intervalΩ=−π,π. We choose N interpolation points xN=x1,x2,⋯,xN, and xj=2πj/N, −N/2≤j≤N/2−1. Then, the semidiscrete scheme (14) can be discretized in space by seeking the approximate solution
(42)uxj=∑k∞≤N/2u~keikxj,t>0,where
(43)u~k=CkN∑j=1nuxjeikxj,is the kth discrete Fourier coefficient, Ck=1 for ∣k∣<N/2, and Ck=2 for ∣k∣=N/2. Thus, for each frequency k, we have
(44)i∂u~k∂t+αk2u~k+βFF−1u~k2F−1u~k+γFfF−1u~k=0,u~x,0=Fu0.Leth1,h2,⋯,hQ be the time steps, satisfying
(45)∑q=1Qhq=T,tq=∑j=1qhj,L=iαk2,Fu~k,t=iβFF−1u~k2F−1u~k+iγFfF−1u~k,then (44) is equivalent to
(46)u~ktq+hq+1=eLhq+1u~ktq+eLhq+1∫0hq+1e−LsFu~ktq+s,tq+sds.More detail about this equation can be find in [29, 30]. Now, let u~k,q=u~ktq and using modified fourth order exponential time-differencing Runge-Kutta (ETDRK4) method to approximate the integral equation (46), we have
(47)λq=eLhq+12u~k,q+L−1eLhq+12−IFu~k,q,tq,χq=eLhq+12u~k,q+L−1eLhq+12−IFλq,tq+hq+12,νq=eLhq+12λq+L−1eLhq+12−I·2Fχq,tq+hq+12−Fu~k,q,tq,u~k,q+1=eLhq+1u~k,q+hq+1θ1Fu~k,q,tq+2θ2Fλq,tq+hq+12+Fχq,tq+hq+12+θ3Fνq,tq+hq+1,where
(48)θ1=4ϕ3Lhq+1−3ϕ2Lhq+1+ϕ1Lhq+1,θ2=ϕ2Lhq+1−2ϕ3Lhq+1,θ3=4ϕ3Lhq+1−ϕ2Lhq+1,with the functions ϕ1, ϕ2, and ϕ3 being defined by
(49)ϕ1=ez−1z,ϕ2=ez−1−zz2,ϕ3=ez−1−z−z2/2z3.Finally, by using inverse Fourier transform, we obtainuj,q=F−1u~k,q is the approximate solution of uxj,tq.For simplicity, the equal step-size is used to discrete time directions. The method presented in this paper allows one to choose different step sizes which are determined by adaptive strategy given in [31]. Next, we proof that the full-discrete scheme (47) satisfies two conservation laws given by Lemma 1.


## 3.2. The Conservation Laws for ETDRK4

In this subsection, we prove that the scheme given by (47) can keep mass and energy conservation. First, see the following two explanatory notes about ETDRK4.Note 1.
Equation (46) is obtained by integrating
(50)e−Ltu~k·,t=e−LtFu~k,t,from t=tq to t=tq+1. Then, from the reasoning process of the method ETDRK4 [29, 30], we get the following form
(51)e−Ltq+1u^N,k,q+1=e−Ltqu^N,k,q+hq+1Φtq,u^N,k,q+Ohq+15,where
(52)Φtq,u^N,k,q=∑j=14hj−1j!dj−1dtj−1e−LtFu^N,k,q,tt=tq,u^N,k,q=FuNx,tq,here, uNx,tq is obtained from the semidiscrete scheme (14). Then, if ignoring the error term Ohq+15, using the method of ETDRK4 to solve (46) is equivalent to get the solution to (50) by the scheme (51).Note 2.
Assume that the functionϕ mentioned above satisfies the Lipschitz condition with parameter λ>0, that is
(53)∣Φt,u1−Φt,u2∣≤λu1−u2.
Actually,ϕ should be third-order continuous differentiable with respect to u due to using the method of ETDRK4, which ensures that ϕ satisfies this assumption easily.In Lemma1, we have proved the mass conservation for the semidiscrete scheme (14). Then, if it holds ∣uNxj,tq−uj,q∣≤Ohmaxn, the full-discrete scheme ETDRK4 is mass conservative, where uNxj,tq is the solution to (14) and uj,q is obtained from (47) and inverse Fourier transform. Next, we give the proof in the following theorem.Theorem 5 (mass conservation).
LetuNxj,tq be obtained from the semidiscrete scheme (14), uj,q=F−1u~k,q, and u~k,q is computed by (47). Define the discrete mass Mtq=∑jΩjuj,q, Ω=⋃jΩj. Then, ignoring the error comes from machine calculation, we have
(54)uNxj,tq−uj,q≤Ohmax4,where hmax=max1≤i≤Qhi, q=0,1,2,⋯,Q. Furthermore,
(55)Mt0~Mt1~⋯~MtQ.Proof.
Without loss of generality, we only need to proveuNxj,tQ−uj,Q≤Ohmax4. From Note 1, we have
(56)e−Ltq+1u~k,q+1=e−Ltqu~k,q+hq+1Φtq,u~k,q.
Taking the module of equation (51) minus equation (56), and let e~q=u^N,k,q−u~k,q, it follows
(57)e−Ltq+1∣e~q+1∣≤e−Ltq∣e~q∣+hq+1∣Φtq,u^N,k,q−Φtq,u~k,q∣+Ohq+15.
Using Note2, we get
(58)e−Ltq+1∣e~q+1∣≤e−Ltq∣e~q∣+hq+1λ∣e~q∣+Ohq+15,
q=0,1,⋯,Q−1. Summing the above inequalities from q=0 to q=Q−1, we obtain
(59)e−LT∣e~Q∣≤∣e~0∣+λ∑q=0Q−1hq+1∣e~q∣+Ohmax4.
LetC1=eLT using Gronwall inequality leads to
(60)∣e~M∣≤eλC1∑q=0Q−1hq+1C1∣e~0∣+Ohmax4.
Note thate~0=0 and denote eλC1T by C, the desired result follows,
(61)∣uNxj,tQ−uj,Q∣=∣F−1e~Q∣=∣e~Q∣≤C·Ohmax4=Ohmax4.
Moreover, we have(62)uj,0−uj,Q≤uj,0−uNxj,t0+uNxj,tQ−uj,Q≤∣uj,0−uNxj,t0∣+uNxj,tQ−uj,Q≤2Ohmax4=Ohmax4.
The above inequalities imply(63)M0~Mt1~⋯~MtQ,as N is big enough and hmax⟶0.Theorem 6 (energy conservation).
Let(64)uj,q=F−1u~k,q,and u~k,q is obtained from (47). Define the discrete local-energy
(65)Ejtq=Ωj−αu¯j,qδα/2uj,q+β2∣uj,q4,and the discrete energy
(66)Etq=∑jEjtq,here, δα/2uj,q is a numerical approximation of the fractional Laplacian operator −Δα/2, Ω=⋃jΩj. Then, we have
(67)∣Ej0−Ejtq∣≤Ohmax4,q=1,2,⋯,Q.
Furthermore,Et0~Et1~⋯~EtQ.Proof.
Without loss of generality, we only need to prove(68)∣Ej0−EjtQ∣≤Ohmax4.
Define bounded linear operator(69)A∣uj,q∣≔Ejtq,then we have
(70)∣Ej0−EjtQ∣=∥A∣uj,0∣−∣uj,Q∣∥≤C∣uj,0∣−∣uj,Q∣,where C>0 is a positive constant. By Theorem 5,
(71)∣Ej0−EjtQ∣≤Ohmax4,which implies
(72)E0~E1~⋯~EQ,as N is big enough and hmax⟶0.


## 4. Applying

In this section, we present two examples to illustrate our results. As in [10], we define the convergence orders in (l∞) norm sense as follows
(73)orderh=logEh1,N/Eh2,Nlogh1/h2,orderN=logEh,N1/Eh,N2logN2/N1,where Eh,N is given by
(74)Eh,N=maxj,puxj,tp−uj,p.All the experiments were done on a computer with the following configuration: Intel(R) Core (TM) i5-8250 CPU @ 1.60 GHz 1.80 GHz and involved using MATLAB 2015b.


### 4.1. Example 1

We first consider the periodic initial Schrödinger equation (1) with the following conditions: 0≤t≤10, α=1,β=−1,γ=−1, fx=sin2x−cosx−expcosx+12+1, and u0x=expcosx+1. The exact solution is
(75)ux,t=expcosx+1exp−it.By FFT and ETDRK4, we compute numerical solution of this example. The point-wise errors (inl∞ sense) against various N are shown in Table 1; then, we plot in semilog scale in Figure 1, which clearly indicates the exponential convergence in space direction, and the fourth order accuracy in time direction is shown in Table 2. The exact solution is shown in Figure 2, and the solution surface is shown in Figure 3, from which we can see that the two pictures are fit very well. Finally, we compute the mass and the energy of the numerical solution at different times. As one can see in Table 3, the scheme has an excellent conservation property and is in well agreement with the results in Theorem 5.Table 1
Errors in spatial direction for Example 1.
hNErrorl∞OrderN1e–469.6321e–2—1e–486.5321e–39.35391e–4103.4651e–413.16011e–4121.7032e–516.52461e–4147.0652e–720.64541e–4164.1237e–821.27601e–4184.3511e–919.09361e–4205.3511e–1019.8909The errors inl∞ sense vs. various N for Example 1 with h=10−4.Figure 1
The log-errors (inl∞ sense) vs. N for Example 1.Table 2
Errors in temporal direction for Example 1.
NhErrorl∞Orderh321/1001.2268e–5—321/2008.2231e–73.8991321/4006.4230e–83.6784321/8004.9085e–93.7099321/16004.6652e–103.3953321/32003.1237e–113.9006321/64002.3511e–123.7318The errors inl∞ sense vs. different time step size h for Example 1 with N=32.Figure 2
The exact solution on the region0,10×−π,π for Example 1.Figure 3
The approximate solution on the region0,10×−π,π for Example 1.Table 3
The conservation of mass and energy for Example 1.
Conservationt=0t=2t=6t=8t=10Mass (∗102)1.0583387071.0583387081.0583387101.0583387121.058338713Energy (∗103)1.83272995521.83272995511.83272995491.83272995481.8327299547The values of mass and energy at timet=0, t=2, t=6, t=8, and t=10, for Example 1 with N=16 and h=10−2.


### 4.2. Example 2

We then consider the periodic initial Schrödinger equation (1) with the following conditions: 0≤t≤50, α=1,β=−1,γ=−1, fx=cos2x, and u0x=sinx. The exact solution to the equation is
(76)ux,t=sinxexpi−1.5t.We compute the numerical solution for this example until the timet=50 with time step h=0.01 on the uniform grid with N=32. We display the errors that derived from temporal direction in Table 4. By Theorem 5, the scheme can keep the mass and energy conservation. Table 5 validates this result very well. It can be seen from Figures 4 and 5 that the approximate solution is in good agreement with the exact solution.Table 4
Errors in temporal direction for Example 2.
NhErrorl∞Orderh321/508.6535e–07—321/1005.4832e–083.9802321/2004.1220e–093.7336321/4003.1075e–103.7295321/8002.2452e–113.7908321/16001.5247e–123.8802321/32001.0521e–133.8572321/64001.0012e–143.3935The errors inl∞ sense vs. different time step size h for Example 2 with N=32.Table 5
The conservation of mass and energy for Example 2.
Conservationt=0t=10t=20t=30t=50Mass3.14159265353.14159265363.14159265363.14159265363.1415926537Energy1.1780972451.1780972451.1780972451.1780972451.178097245The values of mass and energy at timet=0, t=10, t=20, t=30, and t=50, for Example 2 with N=16 and h=10−2.Figure 4
The exact solution on the region0,60×−π,π for Example 2.Figure 5
The approximate solution on the region0,60×−π,π for Example 2.


## 4.1. Example 1

We first consider the periodic initial Schrödinger equation (1) with the following conditions: 0≤t≤10, α=1,β=−1,γ=−1, fx=sin2x−cosx−expcosx+12+1, and u0x=expcosx+1. The exact solution is
(75)ux,t=expcosx+1exp−it.By FFT and ETDRK4, we compute numerical solution of this example. The point-wise errors (inl∞ sense) against various N are shown in Table 1; then, we plot in semilog scale in Figure 1, which clearly indicates the exponential convergence in space direction, and the fourth order accuracy in time direction is shown in Table 2. The exact solution is shown in Figure 2, and the solution surface is shown in Figure 3, from which we can see that the two pictures are fit very well. Finally, we compute the mass and the energy of the numerical solution at different times. As one can see in Table 3, the scheme has an excellent conservation property and is in well agreement with the results in Theorem 5.Table 1
Errors in spatial direction for Example 1.
hNErrorl∞OrderN1e–469.6321e–2—1e–486.5321e–39.35391e–4103.4651e–413.16011e–4121.7032e–516.52461e–4147.0652e–720.64541e–4164.1237e–821.27601e–4184.3511e–919.09361e–4205.3511e–1019.8909The errors inl∞ sense vs. various N for Example 1 with h=10−4.Figure 1
The log-errors (inl∞ sense) vs. N for Example 1.Table 2
Errors in temporal direction for Example 1.
NhErrorl∞Orderh321/1001.2268e–5—321/2008.2231e–73.8991321/4006.4230e–83.6784321/8004.9085e–93.7099321/16004.6652e–103.3953321/32003.1237e–113.9006321/64002.3511e–123.7318The errors inl∞ sense vs. different time step size h for Example 1 with N=32.Figure 2
The exact solution on the region0,10×−π,π for Example 1.Figure 3
The approximate solution on the region0,10×−π,π for Example 1.Table 3
The conservation of mass and energy for Example 1.
Conservationt=0t=2t=6t=8t=10Mass (∗102)1.0583387071.0583387081.0583387101.0583387121.058338713Energy (∗103)1.83272995521.83272995511.83272995491.83272995481.8327299547The values of mass and energy at timet=0, t=2, t=6, t=8, and t=10, for Example 1 with N=16 and h=10−2.


## 4.2. Example 2

We then consider the periodic initial Schrödinger equation (1) with the following conditions: 0≤t≤50, α=1,β=−1,γ=−1, fx=cos2x, and u0x=sinx. The exact solution to the equation is
(76)ux,t=sinxexpi−1.5t.We compute the numerical solution for this example until the timet=50 with time step h=0.01 on the uniform grid with N=32. We display the errors that derived from temporal direction in Table 4. By Theorem 5, the scheme can keep the mass and energy conservation. Table 5 validates this result very well. It can be seen from Figures 4 and 5 that the approximate solution is in good agreement with the exact solution.Table 4
Errors in temporal direction for Example 2.
NhErrorl∞Orderh321/508.6535e–07—321/1005.4832e–083.9802321/2004.1220e–093.7336321/4003.1075e–103.7295321/8002.2452e–113.7908321/16001.5247e–123.8802321/32001.0521e–133.8572321/64001.0012e–143.3935The errors inl∞ sense vs. different time step size h for Example 2 with N=32.Table 5
The conservation of mass and energy for Example 2.
Conservationt=0t=10t=20t=30t=50Mass3.14159265353.14159265363.14159265363.14159265363.1415926537Energy1.1780972451.1780972451.1780972451.1780972451.178097245The values of mass and energy at timet=0, t=10, t=20, t=30, and t=50, for Example 2 with N=16 and h=10−2.Figure 4
The exact solution on the region0,60×−π,π for Example 2.Figure 5
The approximate solution on the region0,60×−π,π for Example 2.


## 5. Conclusion

In this paper, we propose a method to solve the Schrödinger equations by using the method of Fourier spectral and modified fourth order exponential time-differencing Runge-Kutta, and we give the semidiscrete and full-discrete schemes for the equations. We prove that the schemes can keep the conservation law of mass and energy for full-discrete schemes. Numerical examples verify that our conclusions are correct.



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