\newcommand{\W}[1]{ \; #1 \; }
\newcommand{\R}[1]{ {\rm #1} }
\newcommand{\B}[1]{ {\bf #1} }
\newcommand{\D}[2]{ \frac{\partial #1}{\partial #2} }
\newcommand{\DD}[3]{ \frac{\partial^2 #1}{\partial #2 \partial #3} }
\newcommand{\Dpow}[2]{ \frac{\partial^{#1}}{\partial {#2}^{#1}} }
\newcommand{\dpow}[2]{ \frac{ {\rm d}^{#1}}{{\rm d}\, {#2}^{#1}} }This is cppad-20221105 documentation. Here is a link to its
current documentation
.
Evaluate a Function Defined in Terms of an ODE
Syntax # include <cppad/speed/ode_evaluate.hpp> ode_evaluate(x, p, fp)
Purpose
This routine evaluates a function
f : \B{R}^n \rightarrow \B{R}^n
defined by
f(x) = y(x, 1)
where
y(x, t)
solves the ordinary differential equation
\begin{array}{rcl}
y(x, 0) & = & x
\\
\partial_t y (x, t ) & = & g[ y(x,t) , t ]
\end{array}
where
g : \B{R}^n \times \B{R} \rightarrow \B{R}^n
is an unspecified function.
Inclusion
The template function ode_evaluate
is defined in the CppAD namespace by including
the file cppad/speed/ode_evaluate.hpp
(relative to the CppAD distribution directory).
Operation Sequence
The type
Float
must be a NumericType
.
The
Floatoperation sequence
for this routine does not depend on the value of the argument
x
,
hence it does not need to be retaped for each value of
x
.
fabs
If
y
and
z
are
Float
objects, the syntax
y = fabs(z)
must be supported. Note that it does not matter if the operation
sequence for fabs depends on
z
because the
corresponding results are not actually used by ode_evaluate;
see fabs in Runge45
.
x
The argument
x
has prototype
const CppAD::vector<Float>& x
It contains he argument value for which the function,
or its derivative, is being evaluated.
The value
n
is determined by the size of the vector
x
.
p == 0
In this case a numerical method is used to solve the ode
and obtain an accurate approximation for
y(x, 1)
.
This numerical method has a fixed
that does not depend on
x
.
p = 1
In this case an analytic solution for the partial derivative
\partial_x y(x, 1)
is returned.
fp
The argument
fp
has prototype
CppAD::vector<Float>& fp
The input value of the elements of
fp
does not matter.
Function
If
p
is zero,
fp
has size equal to
n
and contains the value of
y(x, 1)
.
Gradient
If
p
is one,
fp
has size equal to
n^2
and for
i = 0 , \ldots 1
,
j = 0 , \ldots , n-1
\D{y[i]}{x[j]} (x, 1) = fp [ i \cdot n + j ]
