Development
Since the main purpose of this paper is to solve numerically (1) using symmetric multiderivative multistep methods, we consider an eightstep fourthderivative method of the form
$$ \begin{aligned} y_{n\pm 4}+a_{3}y_{n\pm 3}+a_{2}y_{n\pm 2}+a_{1}y_{n\pm 1}+a_{0}y_{n} &=h^{2} \bigl(b_{3}y''_{n\pm 3}+b_{2}y''_{n\pm 2}+b_{1}y''_{n\pm 1}+b_{0}y''_{n} \bigr) \\ &\quad {}+h^{4} \bigl(c_{3}y^{(4)}_{n\pm 3}+c_{2}y^{(4)}_{n\pm 2}+c_{1}y^{(4)}_{n \pm 1}+c_{0}y^{(4)}_{n} \bigr), \end{aligned} $$
(7)
where
$$\begin{aligned}& c_{3}=\frac{1}{500},\qquad c_{2}= \frac{1}{500},\qquad c_{1}= \frac{1}{100},\qquad c_{0}=\frac{1}{250}, \\& y_{n\pm i}=y_{n+i}+y_{ni},\quad i=1,2,3,4, \end{aligned}$$
and \(a_{j}\), \(b_{j}\), \(j=0,1,2,3\), are real constant coefficients that should be computed. It is natural that all the properties and characteristics of any numerical method are deduced from its coefficients. However, the coefficients must be carefully generated in order to achieve superior properties. In this paper, several goals are pursued simultaneously so that we can produce a better and more accurate method. Not only do we want to control the interval of periodicity so that \((0,+\infty )\) is produced, but also we intend to have the Schrödinger error coefficient reduced. Applying the new method (7) to the scalar equation (3), CE of the method (7) will be as follows:
$$ A_{4}(v) (y_{n\pm 4})+A_{3}(v) (y_{n\pm 3})+A_{2}(v) (y_{n\pm 2})+A_{1}(v) (y_{n \pm 1})+A_{0}(v)y_{n}=0, $$
(8)
where
$$\begin{aligned}& A_{0}(v) =a_{0}+b_{0}v^{2}+ \frac{1}{250}v^{4}, \\& A_{1}(v) =a_{1}+b_{1}v^{2} \frac{1}{100}v^{4}, \\& A_{2}(v) =a_{2}+b_{2}v^{2}+ \frac{1}{500}v^{4}, \\& A_{3}(v) =a_{3}+b_{3}v^{2} \frac{1}{500}v^{4}, \\& A_{4}(v) =1. \end{aligned}$$
Now, to control the interval of periodicity, we will generate the coefficients so that CE of the scheme is
$$ \lambda ^{8}2\cos (4v)\lambda ^{4}+1=0. $$
(9)
To this end, the polynomials \(A_{1}(v)\), \(A_{2}(v)\), and \(A_{3}(v)\) of the characteristic function must be equal to zero, which produces three equations. On the other hand, because the new method has eight free parameters, we need five more independent linear equations to obtain unique coefficients. Now, to control the Schrödinger error coefficient, suppose that PL and its derivatives up to four are zero. This leads to 8 equations with 8 unknowns. According to Theorem 1.3 with \(k=4\), PL is
$$ \mathrm{PL}= \frac{2A_{1}(v)\cos (v)+2A_{2}(v)\cos (2v)+2A_{3}(v)\cos (3v)+2A_{4}(v)\cos (4v)+A_{0}}{2A_{1}(v)+8A_{2}(v)+18A_{3}(v)+32A_{4}(v)}. $$
Using Maple 18, we have solved this system, and the obtained coefficients are given in Appendix A. The Taylor series expansions of the coefficients are
$$\begin{aligned}& a_{0}= 2+\frac{794\text{,}137}{47\text{,}250}v^{2} \frac{1\text{,}190\text{,}687}{70\text{,}875}v^{4}+ \frac{109\text{,}930\text{,}193}{15\text{,}592\text{,}500}v^{6} \cdots , \\& a_{1}= \frac{424\text{,}453}{63\text{,}000}v^{2}+ \frac{108\text{,}841}{13\text{,}500}v^{4} \frac{56\text{,}802\text{,}077}{15\text{,}592\text{,}500}v^{6}+ \cdots , \\& a_{2}= \frac{13\text{,}103}{157\text{,}500}v^{2}\frac{13\text{,}087}{236\text{,}250}v^{4} + \frac{595\text{,}357}{4\text{,}455\text{,}000}v^{6}\cdots , \\& a_{3}= \frac{1\text{,}495\text{,}957}{945\text{,}000}v^{2}+ \frac{557\text{,}087}{1\text{,}417\text{,}500}v^{4}  \frac{246\text{,}769}{15\text{,}592\text{,}500}v^{6}+ \cdots , \\& b_{0}= \frac{38\text{,}137}{47\text{,}250}\frac{643\text{,}193}{141\text{,}750}v^{2} + \frac{67\text{,}477\text{,}807}{15\text{,}592\text{,}500}v^{4} \frac{811\text{,}862\text{,}971}{521\text{,}235\text{,}000}v^{6}+ \cdots , \\& b_{1}= \frac{424\text{,}453}{63\text{,}000}\frac{54\text{,}353}{6750}v^{2} + \frac{56\text{,}802\text{,}077}{15\text{,}592\text{,}500}v^{4} \frac{77\text{,}581\text{,}573}{90\text{,}090\text{,}000}v^{6}+ \cdots , \\& b_{2}= \frac{13\text{,}103}{157\text{,}500}+\frac{25\text{,}229}{472\text{,}500}v^{2}  \frac{595\text{,}357}{4\text{,}455\text{,}000}v^{4}+\frac{21\text{,}123\text{,}139}{1\text{,}351\text{,}350\text{,}000}v^{6} \cdots , \\& b_{3}= \frac{1\text{,}495\text{,}957}{945\text{,}000}\frac{138\text{,}563}{354\text{,}375}v^{2}+ \frac{246\text{,}769}{15\text{,}592\text{,}500}v^{4}\frac{39\text{,}543\text{,}703}{36\text{,}486\text{,}450\text{,}000}v^{6} \cdots . \end{aligned}$$
The local truncation error of (7), namely \(\mathrm{LTE} _{\mathrm{New}}\), is obtained using the usual Taylor series expansion and is given by
$$ \begin{aligned} \mathrm{LTE}_{\mathrm{New}}&=\frac{326\text{,}687}{5\text{,}670\text{,}000}h^{10} \bigl[\omega ^{10}y_{n}+5 \omega ^{8}y^{(2)}_{n}+10 \omega ^{6}y^{(4)}_{n}+10\omega ^{4}y^{(6)}_{n} +5\omega ^{2}y^{(8)}_{n}+y^{(10)}_{n} \bigr] \\ &\quad {}+O \bigl(h^{12} \bigr). \end{aligned} $$
(10)
Then the local truncation error of the classical form of (7) (the same structure with constant coefficients), namely \(\mathrm{LTE} _{\mathrm{Class}}\), will be
$$ \mathrm{LTE}_{\mathrm{Class}}= \frac{326\text{,}687}{5\text{,}670\text{,}000}h^{10}y^{(10)}_{n}+O \bigl(h^{12} \bigr). $$
(11)
To study the efficiency of the new scheme for solving (1), we have to have its Schrödinger error. To this end, we can transfer (1) to
Now, as per Ixaru and Rizea’s paper [9], \(f(x)\) can be changed to \(f(x)=g(x)+G\), in which \(g(x)=V(x)V_{c}\), and \(V_{c}\) is the constant approximation of the potential and \(G=v^{2}=V_{c}E\) (see [9]).
Theorem 2.1
The Schrödinger error of the new method increases as the second power of G.
Proof
We know that, for any eighth algebraicorder linear multistep method, the general form of the LTEs is given by
$$ \mathrm{LTE}=h^{10}\sum_{i=0}^{j}N_{i}G^{i}, $$
where, in the classical case, \(N_{i}\) are constant numbers and in the frequency dependent cases, \(N_{i}\) are functions of v and G, and j is the maximum power of G. Since \(G=V_{c}E\), we can assume two cases according to the values of E:

1.
\(G = V_{c}  E \approx 0\).

2.
\(G\gg 0\) or \(G\ll 0\), where \(G\) has a large value.
If \(G = V_{c}  E \approx 0\), then we have \(G^{r}=0\), \(r=1,2,\ldots \) . Obviously, the method with asymptotic form of LTE, which includes the minimum power of G, is the most accurate one. Now, \(G = V_{c}  E \approx 0\) implies \(\mathrm{LTE}=h^{10}N_{0}\), where \(N_{0}\) in (10) and (11) is \(\frac{326\text{,}687}{5\text{,}670\text{,}000} [\omega ^{10}y_{n}+5\omega ^{8}y^{(2)}_{n}+10 \omega ^{6}y^{(4)}_{n}+10\omega ^{4}y^{(6)}_{n} +5\omega ^{2}y^{(8)}_{n}+y^{(10)}_{n} ]\) and \(\frac{326\text{,}687}{5\text{,}670\text{,}000}y^{(10)}_{n}\), respectively. For the case \(G\gg 0\) or \(G\ll 0\), firstly, we should calculate higher derivatives of y. By simple calculation, we have
$$\begin{aligned}& y^{(2)}(x)= \bigl(V(x)V_{c}+G\bigr)y(x), \\& y^{(4)}(x)= \biggl(\frac{d^{2}}{dx^{2}}V(x) \biggr)y(x)+2 \biggl( \frac{d}{dx}V(x) \biggr) \biggl(\frac{d}{dx}y(x) \biggr)+ \bigl(V (x) V_{c} + G \bigr) \biggl(\frac{d^{2}}{dx^{2}}y(x) \biggr), \end{aligned}$$
etc. Therefore, by substituting the above derivatives in LTE, we can get the Schrödinger error for the new scheme. Hence, the principal term of the Schrödinger error of the new method is
$$ \mathrm{Err}_{\mathrm{Sch}}=\frac{326\text{,}687}{354\text{,}375} \biggl( \frac{d^{4}}{dx^{4}}g(x) \biggr)y(x)h^{10}G^{2}, $$
and thus, for this method, the error increases as the second power of G. □
Periodicity analysis
In order to confidently implement a numerical method on problems, we must have accurate information about how the method behaves. More specifically, we need to know the stability or instability, convergence or divergence of the method and even the maximum step length. For such studies, we must calculate the stability function of the method and then generate the stability region. This helps us to easily compare the proposed method with other methods. To generate the stability area of the scheme, we apply the main method to the problem
$$ y''(x)=\phi ^{2}y(x) $$
(12)
and obtain its CE. Note that the frequency used in (12) is different from the frequency used in (3). If we assume \(s=\phi h\) and \(v=\omega h\), then we will produce yaxes with v and xaxis with s in the stability region. Therefore, the proposed method can be used for problems that have two frequencies. Applying (7) to (12), we will have
$$ \sum_{i=1}^{4}A_{i}(s,v) (y_{n+i}+y_{ni} )+A_{0}(s,v)y_{n}=0, $$
(13)
in which \(v=\omega h\), \(s=\phi h\), and \(A_{i}(s,v)\) are functions of s and v. To save space, \(A_{i}(s,v)\) are given in Appendix B.
As explained at the beginning of this section, the stability function of the new method has two parameters, s and v, each of which is derived from a frequency. We are looking for Pstability of the method to solve more problems with the proposed method. The colored parts in Fig. 1, which is the figure for the stability area of the method, show the stability parts, and the white parts show the instability region of the method. The method will be Pstable when the entire \(sv\) plane is colored. But this is when we talk about twofrequency issues. Since the equation discussed in this paper has one frequency, the concept of Pstability changes to the one of singular Pstability. It is enough that the bisector of the first quarter belongs to the colored part. Hence, the method is singular Pstable. Now we prove the property of singular Pstability in the next theorem algebraically.
Theorem 2.2
The explicit eightstep scheme (7) is singularly Pstable.
Proof
If we take \(s=v\), then CE of (7) can be written as
$$ \mathrm{CE}=\lambda ^{8}2\lambda ^{4} \bigl(8\cos ^{4} v8\cos ^{2} v+1 \bigr)+1. $$
Since \(\cos (4v)=8\cos ^{4} v8\cos ^{2} v+1\), CE of the new scheme may be rewritten as
$$ \mathrm{CE}=\lambda ^{8}2\lambda ^{4}\cos (4v)+1. $$
(14)
Note that (14) is a polynomial of degree eight with real coefficients. So, it has eight roots namely \(\lambda _{i}\), \(i=0,1,2,\ldots ,7\). According to these roots, \(\mathrm{CE}=0\) is equivalent to
$$ \prod_{k=0}^{3} (\lambda \lambda _{2k} ) (\lambda  \lambda _{2k+1} )=0. $$
If we assume \(\lambda _{2k}=\exp (I\frac{2k\pi }{m} )\exp (Iv)\) (see [22]), where \(k=0,1,2,3\) and \(I=\sqrt{1}\), then
$$ \begin{aligned} &(\lambda _{2k} )^{8}2\cos (mv) (\lambda _{2k} )^{4}+1 \\ &\quad = \biggl[\exp \biggl(I\frac{2k\pi }{m} \biggr)\exp (Iv) \biggr]^{8}2 \cos (mv) \biggl[\exp \biggl(I\frac{2k\pi }{m} \biggr) \exp (Iv) \biggr]^{4}+1 \\ &\quad =\exp (I2mv)2\cos (mv)\exp (Imv)+1 \\ &\quad =0. \end{aligned} $$
(15)
Also, if for \(k=0,1,2,3\) we set \(\lambda _{2k+1}=\exp (I\frac{2k\pi }{m} )\exp (Iv)\), then we have
$$ \begin{aligned} &(\lambda _{2k+1} )^{8}2\cos (mv) (\lambda _{2k+1} )^{4}+1 \\ &\quad = \biggl[\exp \biggl(I\frac{2k\pi }{m} \biggr)\exp (Iv) \biggr]^{8}2 \cos (mv) \biggl[\exp \biggl(I\frac{2k\pi }{m} \biggr) \exp (Iv) \biggr]^{4}+1 \\ &\quad =\exp (I2mv)2\cos (mv)\exp (Imv)+1 \\ &\quad =0. \end{aligned} $$
(16)
So, from (15) and (16), the characteristic roots of the new method are obtained as
$$\begin{aligned}& \lambda _{0}=\exp (Iv), \qquad \lambda _{1}=\exp (Iv),\qquad \lambda _{2}=I \exp (Iv),\qquad \lambda _{3}=I\exp (iv), \\& \lambda _{4}=\exp (Iv), \qquad \lambda _{5}=\exp (Iv),\qquad \lambda _{6}=I \exp (Iv),\qquad \lambda _{7}=I\exp (iv). \end{aligned}$$
Clearly, for these roots we have \(\lambda _{0}=\lambda _{1}=1\) and \(\lambda _{i}\leq 1\), \(i=2,3,\ldots ,7\). Accordingly, when \(s=v\), the periodicity interval of the new scheme is equal to \((0,\infty )\), and the method is Pstable. Therefore, it is singularly Pstable. □
Remark 2.3
The characteristic equation (14) is the same as the one obtained in [18, Theorem 5]. To explain this, since to prove the Pstability property of a linear multistep method, we have to show that the characteristic roots have modulus less (or equal) than one. In other words, they must lie in or on the unit circle. Although the structure of the proposed method in [18] is different from the method presented in this paper, we know the characteristic equation of an eightstep linear multistep method is equal to (8), and also its two variable characteristic equation is (13). Singularly Pstability means Pstability when \(s=v\). To generate a system of equations for calculating the coefficients of the method, we have assumed that \(A_{1}(v)\), \(A_{2}(v)\), and \(A_{3}(v)\) are equal to zero and the remaining equations are obtained from vanishing phaselag and some of its derivatives. Now, by substituting \(s=v\) in \(A_{i}(s,v)\), \(i=0,1,2,3,4\), we have \(y_{n+4}+B(v)y_{n}+y_{n4}=0\), where \(B(v)=2(8\cos ^{4} v8\cos ^{2} v+1)\). Hence, the characteristic polynomial will be \(\lambda ^{8}+B(v)\lambda ^{4}+1=0\), and then all characteristic roots are equal or less than one.