Modified two‐step hybrid methods for the numerical integration of oscillatory problems

Tsitouras³

2018

Self Cite

“…The well known orbital problem of Stiefel and Bettis follows:

z^{''} + z = 0.001 e^{i t}, z (0) = 1, z^{'} (0) = 0.9995 i, z \in C .

For this problem the theoretical solution represents motion on the perturbation of a circular orbit in the complex plane and is given by as follows:

\begin{matrix} z false(t false) & = u false(t false) + i v false(t false), u, v \in double-struckR, \\ u false(t false) & = \cos false(t false) + 0.0005 x \sin false(t false), \\ v false(t false) & = \sin false(t false) - 0.0005 x \cos false(t false) . \end{matrix}

The point z ( t ) spirals outwards so, at time t , its distance from the orbit is,

g (t) = {[u^{2} (t) + v^{2} (t)]}^{\frac{1}{2}} = {[1 + {(0.0005 t)}^{2}]}^{\frac{1}{2}} .

We solve the equivalent couple real problems,

\begin{matrix} u^{''} + u & = 0.0001 \cos false(t false), u false(0 false) = 1, u^{'} false(0 false) \end{matrix}

…”

Section: Numerical Resultsmentioning

confidence: 99%

“…The well known orbital problem of Stiefel and Bettis 40 follows: For this problem the theoretical solution represents motion on the perturbation of a circular orbit in the complex plane and is given by as follows [41][42][43] :…”

Section: The Orbital Problem Of Stiefel and Bettismentioning

confidence: 99%

Trigonometric–fitted hybrid four–step methods of sixth order for solving

Медведев

Tsitouras³

2018

Self Cite

“…The problem of Stiefel and Bettis defines an orbit according to equations:

z^{''} + z = 0.001 e^{i t}, z (0) = 1, z^{'} (0) = 0.9995 i, z \in C .

The theoretical solution of this problem can be found in the previous studies,()

\begin{matrix} z false(t false) & = i v false(t false) + u false(t false) u, v \in double-struckR \\ u false(t false) & = 0.0005 t \sin false(t false) + \cos false(t false), \\ v false(t false) & = - 0.0005 t \cos false(t false) + \sin false(t false) . \end{matrix}

At time t , the distance of z ( t ) from the orbit is,

g (t) = {[u^{2} (t) + v^{2} (t)]}^{\frac{1}{2}} = {[1 + {(0.0005 t)}^{2}]}^{\frac{1}{2}} .

Solving the equivalent real problem,()

\begin{matrix} u^{''} + u & = 0.0001 \cos false(t false), u false(0 false) = 1, u^{'} false(0 false) = 0, \end{matrix}

…”

Section: Numerical Performancementioning

confidence: 99%

Hybrid, phase–fitted, four–step methods of seventh order for solving x″(t) = f(t,x)

Медведев

Tsitouras³

2019

Self Cite

A four‐step method of seventh algebraic order is presented. It is tuned for addressing the special second order initial value problem. The new method is hybrid, explicit, and uses three stages per step. In addition is phase fitted. In consequence it uses variable coefficients that depend on the magnitude of the step‐size. We also present numerical tests on a set of standard problems that illustrate the efficiency of the derived method over older ones given in the relevant literature.

“…Our group is involved in the research of numerical solution of IVPs . Thus, a series of works have been published about various methods, such as Runge‐Kutta (RK), Runge‐Kutta‐Nyström (RKN), Numerov‐type, finite differences, or other multistep methods …”

Section: Introductionmentioning

confidence: 99%

Phase‐fitted, six‐step methods for solving x′′ = f(t,x)

Liu

Hsu

et al. 2019

Self Cite

An explicit, six‐step method of sixth order is presented and tuned for the numerical solution of x′′ = f(t,x). This method is explicit, hybrid, and uses two function evaluations (stages) per step. Its coefficients are varied and depend on the step size. This variance comes from the demand of the method to nullify the phase errors produced when solving the standard simple oscillator. The first and second derivative of this error vanish also. Numerical tests in a set of relevant problems illustrate the efficiency of the newly derived method.