Taylor's Ode Solver: An Example and Test

Purpose
This example solves an ordinary differential equation using Taylor's method; i.e.,   $Z(t + \Delta t) \approx Z^{(0)} (t) + \frac{ Z^{(1)} (t) }{ 1 !} \Delta t + \cdots + \frac{ Z^{(p)} (t) }{ p !} ( \Delta t )^p )$ 

ODE
The ODE is defined by the function   h : \B{R}^n \rightarrow \B{R}^n , which for this example is given by   $Z^{(1)} (t) = H[ Z(t) ] = \left( \begin{array}{c} 1 \\ Z_1 (t) \\ \vdots \\ Z_{n-1} (t) \end{array} \right)$  and the initial condition is   z(0) = 0 .

ODE Solution
The solution for this example can be calculated by starting with the first row and then using the solution for the first row to solve the second and so on. Doing this we obtain   $Z(t) = \left( \begin{array}{c} t \\ t^2 / 2 \\ \vdots \\ t^n / n ! \end{array} \right)$ 

Forward Mode
Given the Taylor coefficients for   k = 0 , \ldots , K   $z^{(k)} = \frac{ Z^{(k)} }{ k !} (t)$ we note that   $\begin{array}{rcl} Z^{(1)} (t) & = & H( z^{(0)} + z^{(1)} t + \cdots + z^{(K)} t^K ) + O( t^{K+1} ) \\ & = & h^{(0)} + h^{(1)} t + \cdots + h^{(K)} t^K + O( t^{K+1} ) \end{array}$  where   h^{(k)} is the k-th order Taylor coefficient for   H( Z(t) ) . Taking K-th order derivatives of both sides we obtain   $\begin{array}{rcl} Z^{(K+1)} (t) & = & K ! h^{(K)} \\ z^{(K+1)} & = & h^{(K)} / K \end{array}$  The code below uses this relationship to implement Taylor's method for approximating the solution of an ODE.  # include <cppad/cppad.hpp> // ========================================================================= // define types for each level namespace { // BEGIN empty namespace using CppAD::AD; CPPAD_TESTVECTOR( AD<double> ) ode( const CPPAD_TESTVECTOR( AD<double> )& Z ) { size_t n = Z.size(); CPPAD_TESTVECTOR( AD<double> ) y(n); y[0] = 1; for(size_t k = 1; k < n; k++) y[k] = Z[k-1]; return y; } } // ------------------------------------------------------------------------- // Example that uses Taylor's method to solve ordinary differential equaitons bool ode_taylor(void) { // initialize the return value as true bool ok = true; // some temporary indices size_t i, j, k; // The ODE does not depend on the arugment values // so only tape once, also note that ode does not depend on t size_t n = 5; // number of independent and dependent variables CPPAD_TESTVECTOR( AD<double> ) a_x(n), a_y(n); CppAD::Independent( a_x ); a_y = ode(a_x); CppAD::ADFun<double> H(a_x, a_y); // initialize the solution vector at time zero CPPAD_TESTVECTOR( double ) z(n); for(j = 0; j < n; j++) z[j] = 0.0; size_t order = n; // order of the Taylor method size_t n_step = 4; // number of time steps double dt = 0.5; // step size in time // Taylor coefficients of order k CPPAD_TESTVECTOR( double ) hk(n), zk(n); // loop with respect to each step of Taylor's method for(size_t i_step = 0; i_step < n_step; i_step++) { // Use Taylor's method to take a step zk = z; // initialize z^{(k)} for k = 0 double dt_kp = dt; // initialize dt^(k+1) for k = 0 for(k = 0; k < order; k++) { // evaluate k-th order Taylor coefficient of H hk = H.Forward(k, zk); for(j = 0; j < n; j++) { // convert to (k+1)-Taylor coefficient for z zk[j] = hk[j] / double(k + 1); // add term for to this Taylor coefficient // to solution for y(t, x) z[j] += zk[j] * dt_kp; } // next power of t dt_kp *= dt; } } // check solution of the ODE, // Taylor's method should have no truncation error for this case double eps = 100. * std::numeric_limits<double>::epsilon(); double check = 1.; double t = n_step * dt; for(i = 0; i < n; i++) { check *= t / double(i + 1); ok &= CppAD::NearEqual(z[i], check, eps, eps); } return ok; } 
Input File: example/ode_taylor.cpp