d02prc resets the end point in an integration performed by
d02pfc and
d02pgc.
d02prc and its associated functions (
d02pfc,
d02pgc,
d02phc,
d02pjc,
d02pqc,
d02psc,
d02ptc and
d02puc) solve the initial value problem for a first-order system of ordinary differential equations. The functions, 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.
d02prc 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
d02pfc or
d02pgc) in the current direction of integration. It is much more efficient to use
d02prc for this purpose than to use
d02pqc 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
d02pqc.
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
Not applicable.
Background information to multithreading can be found in the
Multithreading documentation.
None.
This example integrates a two body problem. The equations for the coordinates
$(x\left(t\right),y\left(t\right))$ 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
$[0,6\pi ]$. 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
d02prc to reset the end of the integration range. A high-order Runge–Kutta method (
${\mathbf{method}}=\mathrm{Nag\_RK\_7\_8}$) 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.