A generator of families of two-step numerical methods with free parameters and minimal phase-lag

Abstract

This research work is oriented in the behaviour of oscillating systems. In order to study such problems, we deal with the solution of ordinary second order differential equations. A generator of families of numerical methods is developed in our effort to solve equations of that type. The families created have constant coefficients and free parameters. We calculate the free parameters taking into consideration the condition of minimal phase-lag. The new methods are applied to the problem of the time independent Schrödinger equation and their results are presented. We also examine the properties of stability and minimum local truncation error.

Appendix

1.1 Proof for $A_0$

For $b=3$ we obtain

$$\begin{aligned} A_{0_3}=-2+v^2-\frac{1}{12} v^4 +\frac{5}{6} a_3 v^6 +\frac{5}{3} a_3 a_2 v^8 + \frac{10}{3} a_3 a_2 a_1 v^{10} \end{aligned}$$

(73)

We assume that the formula is true for the value $b=k$

$$\begin{aligned} A_{0_k}=-2+v^2-\frac{1}{12} v^4 +\frac{5}{3} \displaystyle \sum _{q=3}^{k+2} \left( 2^{q-4} v^{2q} \displaystyle \prod _{i=k-q+3}^{k} a_i \right) \end{aligned}$$

(74)

For $b=k+1$

$$\begin{aligned} A_{0_{k+1}}&=-2+v^2-\frac{1}{12} v^4 +\frac{5}{3} \displaystyle \sum _{q=3}^{(k+1)+2} \left( 2^{q-4} v^{2q} \prod _{i=(k+1)-q+3}^{k+1} a_i \right) \nonumber \\&=-2+v^2-\frac{1}{12} v^4 +\frac{5}{3} \left( \displaystyle \sum _{q=3}^{k+2} \left( 2^{q-4} v^{2q} \displaystyle \prod _{i=k-q+4}^{k+1} a_i \right) +2^{k-1} v^{2(k+3)} \displaystyle \prod _{i=1}^{k+1} a_i \right) \end{aligned}$$

(75)

If $A_{0_k}$ is written as follows

$$\begin{aligned} A_{0_k}=\Omega _0 +\frac{5}{3} \displaystyle \sum _{q=3}^{k+2} \left( \Omega _1 \displaystyle \prod _{i=k-q+3}^{k} a_i \right) \end{aligned}$$

(76)

then $A_{0_{k+1}}$ is written in the following way

$$\begin{aligned} A_{0_{k+1}}= & {} \Omega _0 +\frac{5}{3} \left( \sum _{q=3}^{k+2} \left( \Omega _1 a_{k+1} \prod _{i=k-q+4}^{k} a_i \right) + \Omega _2 a_1 a_2 \ldots a_{k+1} \right) , \nonumber \\= & {} \Omega _0 +\frac{5}{3} \left( a_{k+1} \sum _{q=3}^{k+2} \left( \Omega _1 \prod _{i=k-q+4}^{k} a_i \right) + \Omega _2 \Theta \right) \end{aligned}$$

(77)

where $\Omega _0(v)=-2+v^2-\frac{1}{12} v^4$, $\Omega _1(q)=2^{q-4} v^{2q}$, $\Omega _2(k+3)= 2^{k-1} v^{2(k+3)}$ and $\Theta = a_1 a_2\ldots a_{k+1}$. When the meter in $\prod _{i=k-q+4}^{k} a_i$ decreases then the value of product is equal to 1.

We easily notice that

$$\begin{aligned} A_{0_{k+1}}=A_{0_k}+ (a_{k+1}-1) \Psi _1 +\Psi _2, \end{aligned}$$

(78)

where $\Psi _1=\frac{5}{3} \sum _{q=3}^{k+2} \left( \Omega _1 \prod _{i=k-q+3}^{k} a_i \right)$ and $\Psi _2=\Omega _2 \times \Theta$

1.2 Proof for the formula of the free parameters $a_i$

For $b=5$ and $i=5 \Rightarrow a_5=-\frac{1}{480}$, which is true.

We assume that the formula is true for $i=k$

$$\begin{aligned} a_{b-k}=- \frac{1}{364+4 \displaystyle \sum \nolimits _{j=3}^{k-2} (17+4j)} \end{aligned}$$

(79)

We will prove the formula for $i=k+1$

$$\begin{aligned} a_{b-k-1}= & {} - \frac{1}{364+4 \displaystyle \sum \nolimits _{j=3}^{k+1-2} (17+4j)} \nonumber \\= & {} - \frac{1}{364+4 \displaystyle \sum \nolimits _{j=3}^{k-1} (17+4j)} \nonumber \\= & {} - \frac{1}{364+4 \displaystyle \sum \nolimits _{j=3}^{k-2} \left[ 17+4j \right] +4(17+4(k-1))} \nonumber \\\end{aligned}$$

(80)

$$\begin{aligned} aden_{b-(k+1)}-aden_{b-k}= & {} 4(17+4(k-1)), \end{aligned}$$

(81)

$$\begin{aligned} \frac{1}{aden_{b-(k+1)}}= & {} \frac{1}{aden_{b-k}+4(17+4(k-1))} \end{aligned}$$

(82)

with $aden_i$ being the denominator of the free parameter $a_i$. Everytime we add the quantity $17+4(k-1)$ to the denominator of a parameter $a_i$ we obtain the denominator of the next parameter.

1.3 Proof for the local truncation error

For $b=2$ we obtain

$$\begin{aligned} LTE_3=\frac{1}{239{,}500{,}800} q^{(12)}(t), \end{aligned}$$

(83)

which is true.

For $b=k-1$ we assume the formula is

$$\begin{aligned} LTE_k=\frac{q^{(2k+6)}(t)}{239{,}500{,}800 \displaystyle \prod \nolimits _{j=3}^{k-1} \left[ 182 + 58 (j-3) + 4 (j-3) (j-4) \right] } \end{aligned}$$

(84)

In order to prove the formula for $b=k$

$$\begin{aligned} LTE_{k+1}= & {} \frac{q^{(2k+8)}(t)}{239{,}500{,}800 \displaystyle \prod \nolimits _{j=3}^{k} \left[ 182 + 58 (j-3) + 4 (j-3) (j-4) \right] }\nonumber \\\Leftrightarrow & {} LTE_{k+1}=\frac{q^{(2k+8)}(t)}{den_{k+1}} \Leftrightarrow LTE_{k+1}=\frac{q^{(2k+8)}(t)}{den_{k} \times Z(k)}, \end{aligned}$$

(85)

where $den_i$ is the denominator in $LTE_{i}$ and $Z(k)=2(91+29(k-3)+2(k-3)(k-4))$, we will differentiate twice and divide by Z(k)

$$\begin{aligned} (LTE_k)''= & {} \frac{1}{Z(k)} \left( \frac{q^{(2k+6)}(t)}{den_k} \right) ''\nonumber \\\Rightarrow & {} \frac{q^{(2k+8)}(t)}{A ( 182 + 58 (k-3) + 4 (k-3) (k-4) ) \displaystyle \prod \nolimits _{j=3}^{k-1} \left[ 182 + 58 (j-3) + 4 (j-3) (j-4) \right] } \nonumber \\= & {} \frac{q^{(2k+8)}(t)}{A \displaystyle \prod _{j=3}^{k} \left[ 182 + 58 (j-3) + 4 (j-3) (j-4) \right] }, A=23{,}9500{,}800 \end{aligned}$$

(86)