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A new four stages symmetric two-step method with vanished phase-lag and its first derivative for the numerical integration of the Schrödinger equation

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Abstract

In this paper a four stages symmetric two-step method with vanished phase-lag and its first derivative with high algebraic order is developed for the first time in the literature. More pricesly in this paper the following are presented: (1) the phase-lag analysis for the construction of the new high algebraic order method, (2) the construction of the new method, (3) the error analysis based on the radial Schrödinger equation, (4) the interval of periodicity analysis and the stability analysis, (5) a new error estimation procedure based on the phase-lag (6) the numerical tests for the study of the efficiency of the new obtained method which are based on the numerical solution of the Schrödinger equation.

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Notes

  1. Where S is a set of distinct points.

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T. E. Simos is a Highly Cited Researcher (http://isihighlycited.com/), Active Member of the European Academy of Sciences and Arts. Active Member of the European Academy of Sciences. Corresponding Member of European Academy of Arts, Sciences and Humanities.

Appendices

Appendix 1: Formulae of the derivatives of \(q_{n}\)

Formulae of the derivatives which presented in the formulae of the Local Truncation Errors:

$$\begin{aligned} q_{n}^{(2)}= & {} \left( V(x)-V_{c} + G\right) \, q(x) \\ q_{n}^{(3)}= & {} \left( {\frac{d}{dx}}g \left( x \right) \right) q \left( x \right) \\&+ \left( g \left( x \right) +G \right) {\frac{d}{dx}}q \left( x \right) \\ q_{n}^{(4)}= & {} \left( {\frac{d^{2}}{d{x}^{2}}}g \left( x \right) \right) q \left( x \right) \\&+2\, \left( {\frac{d}{dx}}g \left( x \right) \right) {\frac{d}{dx}}q \left( x \right) \\&+ \left( g \left( x \right) +G \right) ^{2}q \left( x \right) \\ q_{n}^{(5)}= & {} \left( {\frac{d^{3}}{d{x}^{3}}}g \left( x \right) \right) q \left( x \right) \\&+3\, \left( {\frac{d^{2}}{d{x}^{2}}}g \left( x \right) \right) {\frac{d}{dx}}q \left( x \right) \\&+4\, \left( g \left( x \right) +G \right) q \left( x \right) {\frac{d}{dx}}g \left( x \right) \\&+ \left( g \left( x \right) +G \right) ^{2}{\frac{d}{dx}}q \left( x \right) \\ q_{n}^{(6)}= & {} \left( {\frac{d^{4}}{d{x}^{4}}}g \left( x \right) \right) q \left( x \right) \\&+4\, \left( {\frac{d^{3}}{d{x}^{3}}}g \left( x \right) \right) {\frac{d}{dx}}q \left( x \right) \\&+7\, \left( g \left( x \right) +G \right) q \left( x \right) {\frac{d^{2}}{d{x}^{2}}}g \left( x \right) \\&+4\, \left( {\frac{d}{dx}}g \left( x \right) \right) ^{2}q \left( x \right) \\&+6\, \left( g \left( x \right) +G \right) \left( {\frac{d}{dx}}q \left( x \right) \right) {\frac{d}{dx}}g \left( x \right) \\&+ \left( g \left( x \right) +G \right) ^{3}q \left( x \right) \\ q_{n}^{(7)}= & {} \left( {\frac{d^{5}}{d{x}^{5}}}g \left( x \right) \right) q \left( x \right) \\&+5\, \left( {\frac{d^{4}}{d{x}^{4}}}g \left( x \right) \right) {\frac{d}{dx}}q \left( x \right) \end{aligned}$$
$$\begin{aligned}&+11\, \left( g \left( x \right) +G \right) q \left( x \right) {\frac{d^{3}}{d{x}^{3}}}g \left( x \right) \\&+15\, \left( {\frac{d}{dx}}g \left( x \right) \right) q \left( x \right) \\&{\frac{d^{2}}{d{x}^{2}}}g \left( x \right) +13\, \left( g \left( x \right) +G \right) \\&\left( {\frac{d}{dx}}q \left( x \right) \right) {\frac{d^{2}}{d{x}^{2}}}g \left( x \right) \\&+10\, \left( {\frac{d}{dx}}g \left( x \right) \right) ^{2}{\frac{d}{dx}}q \left( x \right) \\&+9\, \left( g \left( x \right) +G \right) ^{2}q \left( x \right) \\&{\frac{d}{dx}}g \left( x \right) + \left( g \left( x \right) +G \right) ^{3}{\frac{d}{dx}}q \left( x \right) \\ q_{n}^{(8)}= & {} \left( {\frac{d^{6}}{d{x}^{6}}}g \left( x \right) \right) q \left( x \right) \\&+6\, \left( {\frac{d^{5}}{d{x}^{5}}}g \left( x \right) \right) {\frac{d}{dx}}q \left( x \right) \\&+16\, \left( g \left( x \right) +G \right) q \left( x \right) {\frac{d^{4}}{d{x}^{4}}}g \left( x \right) \\&+26\, \left( {\frac{d}{dx}}g \left( x \right) \right) q \left( x \right) \\&{\frac{d^{3}}{d{x}^{3}}}g \left( x \right) +24\, \left( g \left( x \right) +G \right) \\&\left( {\frac{d}{dx}}q \left( x \right) \right) {\frac{d^{3}}{d{x}^{3}}}g \left( x \right) \\&+15\, \left( {\frac{d^{2}}{d{x}^{2}}}g \left( x \right) \right) ^{2}q \left( x \right) \end{aligned}$$
$$\begin{aligned}&+48\, \left( {\frac{d}{dx}}g \left( x \right) \right) \\&\left( {\frac{d}{dx}}q \left( x \right) \right) {\frac{d^{2}}{d{x}^{2}}}g \left( x \right) \\&+22\, \left( g \left( x \right) +G \right) ^{2}q \left( x \right) \\&{\frac{d^{2}}{d{x}^{2}}}g \left( x \right) +28\, \left( g \left( x \right) +G \right) \\&q \left( x \right) \left( {\frac{d}{dx}}g \left( x \right) \right) ^{2}\\&+12\, \left( g \left( x \right) +G \right) ^{2} \\&\left( {\frac{d}{dx}}q \left( x \right) \right) {\frac{d}{dx}}g \left( x \right) \\&+ \left( g \left( x \right) +G \right) ^{4}q \left( x \right) \\&\ldots \end{aligned}$$

Appendix 2: The new four stages twelfth algebraic order symmetric two-step method with vanished phase-lag (phase-fitted)

We consider the family of four stages symmetric two-step methods (12) with

$$\begin{aligned} a_{0} = -\frac{59}{113400}, \, a_{1} = \frac{15}{112}, \, a_{2}=-2, \, b_{2} = -\frac{7}{50}, \, b_{{1}}\, = \, {\frac{71}{600}} \end{aligned}$$

Applying the method (12) with coefficients \(a_{j}, \, j=0(1)2, \, \, b_{i}, \, i=1,2\) given above to the scalar test equation (4), we obtain the difference equation given by (5) with:

$$\begin{aligned} A_{1} (v)= & {} 1+{v}^{2} \left[ {\frac{229}{7788}}-{\frac{140\,{v}^{2}}{649} \left( -{\frac{127\,{v}^{2}}{16934400}}-{\frac{11}{5600}} \right) }+{\frac{127\,{v}^{4}}{78503040}} \right] \\ A_{0} (v)= & {} -2+{v}^{2} \left[ b_{{0}}-{\frac{140}{649}}-{\frac{280\,{v}^{2}}{649} \left( {\frac{711}{5600}}-{\frac{2923\,{v}^{2}}{1693440}} \right) } \right] \end{aligned}$$

In order the above four stages method (12) to have vanished the phase-lag (phase-fitted) we have the following equation:

$$\begin{aligned} \mathrm{Phase-Lag(PL)}= & {} \frac{1}{2}\,\frac{T_{6}}{1+{v}^{2} \left( {\frac{229}{7788}}-{\frac{140\,{v}^{2}}{649} \left( -{\frac{127\,{v}^{2}}{16934400}}-{\frac{11}{5600}} \right) }+{\frac{127\,{v}^{4}}{78503040}} \right) } = 0 \end{aligned}$$

where

$$\begin{aligned} T_{6}= & {} 2\, \Bigl ( 1+{v}^{2} \Bigl ( {\frac{229}{7788}}-{\frac{140\,{v}^{2}}{649} \left( -{\frac{127\,{v}^{2}}{16934400}}-{\frac{11}{5600}} \right) }\\&+{\frac{127\,{v}^{4}}{78503040}} \Bigr ) \Bigr ) \cos \left( v \right) -2+{v}^{2} \left( b_{{0}}-{\frac{140}{649}}-{\frac{280\,{v}^{2}}{649} \left( {\frac{711}{5600}}-{\frac{2923\,{v}^{2}}{1693440}} \right) } \right) \end{aligned}$$

If we solve the above equation, we obtain the coefficient \(b_{0}\) of the new developed four stages symmetric two-step method:

$$\begin{aligned} b_{{0}}\, = \,-\frac{T_{7}}{19625760\,{v}^{2}} \end{aligned}$$

where

$$\begin{aligned} T_{7}= & {} 127\,\cos \left( v \right) {v}^{6}+14615\,{v}^{6}+16632\,\cos \left( v \right) {v}^{4}\\&-1075032\,{v}^{4}+1154160\,\cos \left( v \right) {v}^{2}\\&-4233600\,{v}^{2}+39251520\,\cos \left( v \right) -39251520 \end{aligned}$$

For cancellations purposes we give also the Taylor series expansion :

$$\begin{aligned} b_{{0}}= & {} {\frac{4505}{3894}}-{\frac{45469\,{v}^{12}}{1697361329664000}}+{\frac{12079\,{v}^{14}}{20368335955968000}} \\&-{\frac{26123\,{v}^{16}}{4155140535017472000}}+{\frac{697769\,{v}^{18}}{16579010734719713280000}} + \ldots \end{aligned}$$

The behavior of the coefficient of the new four stages two-step method is presented in Fig. 3.

Fig. 3
figure 3

Behavior of the coefficient of the new obtained twelfth algebraic order four stages two-step method for several values of \(v = \phi \, h\)

Based on the above coefficients, we can find the local truncation error of the new developed two stage two-step method (12) (mentioned as FourStageTwoStep100D) which is given by:

$$\begin{aligned}&LTE _{FourStageTwoStep120D}= \\&\quad - \frac{45469}{1697361329664000}\, h^{14} \, \Biggl ( q_{n}^{(14)} - \phi ^{12} \, q_{n}^{(2)} \Biggr ) + O \left( h^{14} \right) \end{aligned}$$

The asymptotic form of the Local Truncation Error (after application to the test problem (20)) is given by:

$$\begin{aligned}&LTE _{FourStageTwoStep120D}= \\&\quad - \frac{45469}{282893554944000}\, h^{14} \, \Biggl [ \Biggl ( g \left( x \right) \, q \left( x \right) \Biggr )\, G^{6} + \cdots \Biggr ] + O \left( h^{14} \right) \end{aligned}$$

The \(s-v\) plane for the new obtained four stages symmetric two-step method is shown in Fig. 4.

Fig. 4
figure 4

\(s-v\) plane of the new produced four stages two-step twelfth algebraic order method with vanished phase-lag (phase-fitted)

Remark 11

For the case where \(s=v\) (i.e. see the surroundings of the first diagonal of the \(s-v\) plane), the interval of periodicity for this four stages symmetric two-step method is equal to: \(\left( 0, 480 \right) \).

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Liang, M., Simos, T.E. A new four stages symmetric two-step method with vanished phase-lag and its first derivative for the numerical integration of the Schrödinger equation. J Math Chem 54, 1187–1211 (2016). https://doi.org/10.1007/s10910-016-0615-x

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