D02PRF resets the end point in an integration performed by
D02PFF.
D02PRF and its associated routines (
D02PFF,
D02PQF,
D02PSF,
D02PTF and
D02PUF) solve the initial value problem for a first-order system of ordinary differential equations. The routines, based on Runge–Kutta methods and derived from RKSUITE (see
Brankin et al. (1991)), integrate
where
$y$ is the vector of
$n$ solution components and
$t$ is the independent variable.
D02PRF is used to reset the final value of the independent variable,
${t}_{f}$, when the integration is already underway. It can be used to extend or reduce the range of integration. The new value must be beyond the current value of the independent variable (as returned in
TNOW by
D02PFF) in the current direction of integration. It is much more efficient to use D02PRF for this purpose than to use
D02PQF which involves the overhead of a complete restart of the integration.
If you want to change the direction of integration then you must restart by a call to
D02PQF.
Brankin R W, Gladwell I and Shampine L F (1991) RKSUITE: A suite of Runge–Kutta codes for the initial value problems for ODEs SoftReport 91-S1 Southern Methodist University
If on entry
${\mathbf{IFAIL}}={\mathbf{0}}$ or
${-{\mathbf{1}}}$, explanatory error messages are output on the current error message unit (as defined by
X04AAF).
Not applicable.
None.
This example integrates a two body problem. The equations for the coordinates
$\left(x\left(t\right),y\left(t\right)\right)$ of one body as functions of time
$t$ in a suitable frame of reference are
The initial conditions
lead to elliptic motion with
$0<\epsilon <1$.
$\epsilon =0.7$ is selected and the system of ODEs is reposed as
over the range
$\left[0,6\pi \right]$. Relative error control is used with threshold values of
$\text{1.0E\u221210}$ for each solution component and compute the solution at intervals of length
$\pi $ across the range using D02PRF to reset the end of the integration range. A high-order Runge–Kutta method (
${\mathbf{METHOD}}=-3$) is also used with tolerances
${\mathbf{TOL}}=\text{1.0E\u22124}$ and
${\mathbf{TOL}}=\text{1.0E\u22125}$ in turn so that the solutions may be compared.