$\newcommand{\W}[1]{ \; #1 \; } \newcommand{\R}[1]{ {\rm #1} } \newcommand{\B}[1]{ {\bf #1} } \newcommand{\D}[2]{ \frac{\partial #1}{\partial #2} } \newcommand{\DD}[3]{ \frac{\partial^2 #1}{\partial #2 \partial #3} } \newcommand{\Dpow}[2]{ \frac{\partial^{#1}}{\partial {#2}^{#1}} } \newcommand{\dpow}[2]{ \frac{ {\rm d}^{#1}}{{\rm d}\, {#2}^{#1}} }$
Taylor's Ode Solver: A Multi-Level AD Example and Test

Purpose
This is a realistic example using two levels of AD; see mul_level . The first level uses AD<double> to tape the solution of an ordinary differential equation. This solution is then differentiated with respect to a parameter vector. The second level uses AD< AD<double> > to take derivatives during the solution of the differential equation. These derivatives are used in the application of Taylor's method to the solution of the ODE. The example mul_level_adolc_ode.cpp computes the same values using Adolc's type adouble and CppAD's type AD<adouble>. The example ode_taylor.cpp is a simpler applications of Taylor's method for solving an ODE.

ODE
For this example the ODE's are defined by the function $h : \B{R}^n \times \B{R}^n \rightarrow \B{R}^n$ where $$h[ x, y(t, x) ] = \left( \begin{array}{c} x_0 \\ x_1 y_0 (t, x) \\ \vdots \\ x_{n-1} y_{n-2} (t, x) \end{array} \right) = \left( \begin{array}{c} \partial_t y_0 (t , x) \\ \partial_t y_1 (t , x) \\ \vdots \\ \partial_t y_{n-1} (t , x) \end{array} \right)$$ and the initial condition $y(0, x) = 0$. The value of $x$ is fixed during the solution of the ODE and the function $g : \B{R}^n \rightarrow \B{R}^n$ is used to define the ODE where $$g(y) = \left( \begin{array}{c} x_0 \\ x_1 y_0 \\ \vdots \\ x_{n-1} y_{n-2} \end{array} \right)$$

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 $$y(t, x ) = \left( \begin{array}{c} x_0 t \\ x_1 x_0 t^2 / 2 \\ \vdots \\ x_{n-1} x_{n-2} \ldots x_0 t^n / n ! \end{array} \right)$$

Derivative of ODE Solution
Differentiating the solution above, with respect to the parameter vector $x$, we notice that $$\partial_x y(t, x ) = \left( \begin{array}{cccc} y_0 (t,x) / x_0 & 0 & \cdots & 0 \\ y_1 (t,x) / x_0 & y_1 (t,x) / x_1 & 0 & \vdots \\ \vdots & \vdots & \ddots & 0 \\ y_{n-1} (t,x) / x_0 & y_{n-1} (t,x) / x_1 & \cdots & y_{n-1} (t,x) / x_{n-1} \end{array} \right)$$

An m-th order Taylor method for approximating the solution of an ordinary differential equations is $$y(t + \Delta t , x) \approx \sum_{k=0}^p \partial_t^k y(t , x ) \frac{ \Delta t^k }{ k ! } = y^{(0)} (t , x ) + y^{(1)} (t , x ) \Delta t + \cdots + y^{(p)} (t , x ) \Delta t^p$$ where the Taylor coefficients $y^{(k)} (t, x)$ are defined by $$y^{(k)} (t, x) = \partial_t^k y(t , x ) / k !$$ We define the function $z(t, x)$ by the equation $$z ( t , x ) = g[ y ( t , x ) ] = h [ x , y( t , x ) ]$$ It follows that $$\begin{array}{rcl} \partial_t y(t, x) & = & z (t , x) \\ \partial_t^{k+1} y(t , x) & = & \partial_t^k z (t , x) \\ y^{(k+1)} ( t , x) & = & z^{(k)} (t, x) / (k+1) \end{array}$$ where $z^{(k)} (t, x)$ is the k-th order Taylor coefficient for $z(t, x)$. In the example below, the Taylor coefficients $$y^{(0)} (t , x) , \ldots , y^{(k)} ( t , x )$$ are used to calculate the Taylor coefficient $z^{(k)} ( t , x )$ which in turn gives the value for $y^{(k+1)} y ( t , x)$.

Source

// =========================================================================
// define types for each level
namespace { // BEGIN empty namespace

// -------------------------------------------------------------------------
// class definition for C++ function object that defines ODE
class Ode {
private:
// copy of a that is set by constructor and used by g(y)
public:
// constructor
{ }
// the function g(y) is evaluated with two levels of taping
{     size_t n = a2y.size();
size_t i;
a2g[0] = a1x_[0];
for(i = 1; i < n; i++)
a2g[i] = a1x_[i] * a2y[i-1];

return a2g;
}
};

// -------------------------------------------------------------------------
// Routine that uses Taylor's method to solve ordinary differential equaitons
// and allows for algorithmic differentiation of the solution.
Ode                            G       ,  // function that defines the ODE
size_t                         order   ,  // order of Taylor's method used
size_t                         nstep   ,  // number of steps to take
const a1type&                  a1dt    ,  // Delta t for each step
const CPPAD_TESTVECTOR(a1type)& a1y_ini)  // y(t) at the initial time
{
// some temporary indices
size_t i, k, ell;

// number of variables in the ODE
size_t n = a1y_ini.size();

// copies of x and g(y) with two levels of taping

// y, y^{(k)} , z^{(k)}, and y^{(k+1)}

// initialize x
for(i = 0; i < n; i++)
a1y[i] = a1y_ini[i];

// loop with respect to each step of Taylors method
for(ell = 0; ell < nstep; ell++)
{     // prepare to compute derivatives using a1type
for(i = 0; i < n; i++)
a2y[i] = a1y[i];

// evaluate ODE in a2type
a2z = G(a2y);

// define differentiable version of a1g: y -> z
// that computes its derivatives using a1type objects

// Use Taylor's method to take a step
a1y_k            = a1y;     // initialize y^{(k)}
a1type   a1dt_kp = a1dt;  // initialize dt^(k+1)
for(k = 0; k <= order; k++)
{     // evaluate k-th order Taylor coefficient of y
a1z_k = a1g.Forward(k, a1y_k);

for(i = 0; i < n; i++)
{     // convert to (k+1)-Taylor coefficient for x
a1y_kp[i] = a1z_k[i] / a1type(k + 1);

// add term for to this Taylor coefficient
// to solution for y(t, x)
a1y[i]    += a1y_kp[i] * a1dt_kp;
}
// next power of t
a1dt_kp *= a1dt;
// next Taylor coefficient
a1y_k   = a1y_kp;
}
}
return a1y;
}
} // END empty namespace
// ==========================================================================
// Routine that tests alogirhtmic differentiation of solutions computed
// by the routine taylor_ode.
bool mul_level_ode(void)
{     bool ok = true;
double eps = 100. * std::numeric_limits<double>::epsilon();

// number of components in differential equation
size_t n = 4;

// some temporary indices
size_t i, j;

// parameter vector in both double and a1type
for(i = 0; i < n; i++)
a1x[i] = x[i] = double(i + 1);

// declare the parameters as the independent variable

// arguments to taylor_ode
Ode G(a1x);                // function that defines the ODE
size_t   order = n;      // order of Taylor's method used
size_t   nstep = 2;      // number of steps to take
a1type   a1dt  = double(1.);     // Delta t for each step
// value of y(t, x) at the initial time
for(i = 0; i < n; i++)
a1y_ini[i] = 0.;

// integrate the differential equation
a1y_final = taylor_ode(G, order, nstep, a1dt, a1y_ini);

// define differentiable fucntion object f : x -> y_final
// that computes its derivatives in double

// check function values
double check = 1.;
double t     = double(nstep) * Value(a1dt);
for(i = 0; i < n; i++)
{     check *= x[i] * t / double(i + 1);
ok &= CppAD::NearEqual(Value(a1y_final[i]), check, eps, eps);
}

// evaluate the Jacobian of h at a
// There appears to be a bug in g++ version 4.4.2 because it generates
// a warning for the equivalent form

// check Jacobian
for(i = 0; i < n; i++)
{     for(j = 0; j < n; j++)
{     double jac_ij = jac[i * n + j];
if( i < j )
check = 0.;
else     check = Value( a1y_final[i] ) / x[j];
ok &= CppAD::NearEqual(jac_ij, check, eps, eps);
}
}
return ok;
}

Input File: example/general/mul_level_ode.cpp