/* --------------------------------------------------------------------------
CppAD: C++ Algorithmic Differentiation: Copyright (C) 2003-16 Bradley M. Bell

CppAD is distributed under the terms of the
             Eclipse Public License Version 2.0.

This Source Code may also be made available under the following
Secondary License when the conditions for such availability set forth
in the Eclipse Public License, Version 2.0 are satisfied:
      GNU General Public License, Version 2.0 or later.
---------------------------------------------------------------------------- */
/*
$begin ipopt_solve_ode_inverse.cpp$$
$spell
$$

$section ODE Inverse Problem Definitions: Source Code$$


$head Purpose$$
This example demonstrates how to invert for parameters
in a ODE where the solution of the ODE is numerically approximated.

$head Forward Problem$$
We consider the following ordinary differential equation:
$latex \[
\begin{array}{rcl}
    \partial_t y_0 ( t , a ) & = & - a_1 * y_0 (t, a )
    \\
    \partial_t y_1 (t , a )  & = & + a_1 * y_0 (t, a ) - a_2 * y_1 (t, a )
\end{array}
\] $$
with the initial conditions
$latex \[
    y_0 (0 , a) = ( a_0 , 0 )^\R{T}
\] $$
Our forward problem is stated as follows:
Given $latex a \in \B{R}^3$$
determine the value of $latex y ( t , a ) $$,
for $latex t \in R$$, that solves the initial value problem above.

$head Measurements$$
Suppose we are also given measurement times $latex s \in \B{R}^5$$
and  a measurement vector $latex z \in \B{R}^4$$
and for $latex i = 0, \ldots, 3$$, we model $latex z_i$$ by
$latex \[
    z_i = y_1 ( s_{i+1} , a) + e_i
\] $$
where $latex e_{i-1} \sim {\bf N} (0 , \sigma^2 )$$
is the measurement noise,
and $latex \sigma > 0$$ is the standard deviation of the noise.

$subhead Simulation Analytic Solution$$
The following analytic solution to the forward problem is used
to simulate a data set:
$latex \[
\begin{array}{rcl}
    y_0 (t , a) & = & a_0 * \exp( - a_1 * t )
    \\
    y_1 (t , a) & = &
    a_0 * a_1 * \frac{\exp( - a_2 * t ) - \exp( -a_1 * t )}{ a_1 - a_2 }
\end{array}
\] $$

$subhead Simulation Parameter Values$$
$table
$latex \bar{a}_0 = 1$$ $pre $$ $cnext
    initial value of $latex y_0 (t, a)$$
$rnext
$latex \bar{a}_1 = 2$$ $pre $$ $cnext
    transfer rate from compartment zero to compartment one
$rnext
$latex \bar{a}_2 = 1$$ $pre $$ $cnext
    transfer rate from compartment one to outside world
$rnext
$latex \sigma = 0$$ $pre $$ $cnext
    standard deviation of measurement noise
$rnext
$latex e_i = 0$$ $pre $$ $cnext
    simulated measurement noise, $latex i = 1 , \ldots , Nz$$
$rnext
$latex s_i = i * .5$$ $pre $$ $cnext
    time corresponding to the $th i$$ measurement,
    $latex i = 0 , \ldots , 3$$
$tend

$subhead Simulated Measurement Values$$
The simulated measurement values are given by the equation
$latex \[
\begin{array}{rcl}
z_i
& = &  e_i + y_1 ( s_{i+1} , \bar{a} )
\\
& = &
\bar{a}_0 * \bar{a}_1 *
    \frac{\exp( - \bar{a}_2 * s_i ) - \exp( -\bar{a}_1 * s_i )}
        { \bar{a}_1 - \bar{a}_2 }
\end{array}
\] $$
for $latex i = 0, \ldots , 3$$.


$head Inverse Problem$$
The maximum likelihood estimate for $latex a$$ given $latex z$$
solves the following optimization problem
$latex \[
\begin{array}{rcl}
{\rm minimize} \;
    & \sum_{i=0}^3 ( z_i - y_1 ( s_{i+1} , a ) )^2
    & \;{\rm w.r.t} \; a \in \B{R}^3
\end{array}
\] $$

$head Trapezoidal Approximation$$
We are given a number of approximation points per measurement interval
$latex np$$ and define the time grid $latex t \in \B{R}^{4 \cdot np + 1}$$
as follows:
$latex t_0 = s_0$$ and
for $latex i = 0 , 1 , 2, 3$$, $latex j = 1 , \ldots , np$$
$latex \[
    t_{i \cdot np + j} = s_i + (s_{i+1} - s{i}) \frac{i}{np}
\] $$
We note that for $latex i = 1 , \ldots , 4$$,
$latex t_{i \cdot np} = s_i$$.
This example uses a trapezoidal approximation to solve the ODE.
Given $latex a \in \B{R}^3$$ and $latex y^{k-1} \approx y( t_{k-1} , a )$$,
the a trapezoidal method approximates $latex y ( t_j , a )$$
by the value $latex y^k \in \B{R}^2$$ ) that solves the equation
$latex \[
y^k  =  y^{k-1} + \frac{G( y^k , a ) + G( y^{k-1} , a ) }{2} * (t_k - t_{k-1})
\] $$
where $latex G : \B{R}^2 \times \B{R}^3 \rightarrow \B{R}^2$$ is defined by
$latex \[
\begin{array}{rcl}
    G_0 ( y , a ) & = & - a_1 * y_0
    \\
    G_1 ( y , a ) & = & + a_1 * y_0  - a_2 * y_1
\end{array}
\] $$

$head Solution Method$$
We use constraints to embed the
forward problem in the inverse problem.
To be specific, we solve the optimization problem
$latex \[
\begin{array}{rcl}
{\rm minimize}
& \sum_{i=0}^3 ( z_i - y_1^{(i+1) \cdot np} )^2
& \; {\rm w.r.t} \; a \in \B{R}^3
    \; y^0 \in \B{R}^2 , \ldots , y^{3 \cdot np -1} \in \B{R}^2
\\
{\rm subject \; to}
    0 = y^0 - ( a_0 , 0 )^\R{T}
    \\
    & 0 = y^k  -  y^{k-1} -
    \frac{G( y^k , a ) + G( y^{k-1} , a ) }{2}  (t_k - t_{k-1})
    & \; {\rm for} \; k = 1 , \ldots , 4 \cdot np
\end{array}
\] $$
The code below we using the notation
$latex x \in \B{3 + (4 \cdot np + 1) \cdot 2}$$ defined by
$latex \[
    x = \left(
        a_0, a_1, a_2 ,
        y_0^0, y_1^0,
        \ldots ,
        y_0^{4 \cdot np}, y_1^{4 \cdots np}
    \right)
\] $$

$head Source$$
The following source code
implements the ODE inversion method proposed above:
$srcfile%example/ipopt_solve/ode_inverse.cpp%0%// BEGIN C++%// END C++%1%$$

$end
------------------------------------------------------------------------------
*/
// BEGIN C++
# include <cppad/ipopt/solve.hpp>

namespace {
    using CppAD::AD;

    // value of a during simulation a[0], a[1], a[2]
    double a_[] =                   {2.0,  1.0, 0.5};
    // number of components in a
    size_t na_ = sizeof(a_) / sizeof(a_[0]);

    // function used to simulate data
    double yone(double t)
    {   return
            a_[0]*a_[1] * (exp(-a_[2]*t) - exp(-a_[1]*t)) / (a_[1] - a_[2]);
    }

    // time points were we have data (no data at first point)
    double s_[] = {0.0,   0.5,        1.0,          1.5,         2.0 };

    // Simulated data for case with no noise (first point is not used)
    double z_[] = {yone(s_[1]), yone(s_[2]), yone(s_[3]), yone(s_[4])};
    size_t nz_  = sizeof(z_) / sizeof(z_[0]);

    // number of trapozoidal approximation points per measurement interval
    size_t np_  = 40;


    class FG_eval
    {
    private:
    public:
        // derived class part of constructor
        typedef CPPAD_TESTVECTOR( AD<double> ) ADvector;

        // Evaluation of the objective f(x), and constraints g(x)
        void operator()(ADvector& fg, const ADvector& x)
        {   CPPAD_TESTVECTOR( AD<double> ) a(na_);
            size_t i, j, k;

            // extract the vector a
            for(i = 0; i < na_; i++)
                a[i] = x[i];

            // compute the object f(x)
            fg[0] = 0.0;
            for(i = 0; i < nz_; i++)
            {   k = (i + 1) * np_;
                AD<double> y_1 = x[na_ + 2 * k + 1];
                AD<double> dif = z_[i] - y_1;
                fg[0]         += dif * dif;
            }

            // constraint corresponding to initial value y(0, a)
            // Note that this constraint is invariant with size of dt
            fg[1] = x[na_+0] - a[0];
            fg[2] = x[na_+1] - 0.0;

            // constraints corresponding to trapozoidal approximation
            for(i = 0; i < nz_; i++)
            {   // spacing between grid point
                double dt = (s_[i+1] - s_[i]) / static_cast<double>(np_);
                for(j = 1; j <= np_; j++)
                {   k = i * np_ + j;
                    // compute derivative at y^k
                    AD<double> y_0  = x[na_ + 2 * k + 0];
                    AD<double> y_1  = x[na_ + 2 * k + 1];
                    AD<double> G_0  = - a[1] * y_0;
                    AD<double> G_1  = + a[1] * y_0 - a[2] * y_1;

                    // compute derivative at y^{k-1}
                    AD<double> ym_0  = x[na_ + 2 * (k-1) + 0];
                    AD<double> ym_1  = x[na_ + 2 * (k-1) + 1];
                    AD<double> Gm_0  = - a[1] * ym_0;
                    AD<double> Gm_1  = + a[1] * ym_0 - a[2] * ym_1;

                    // constraint should be zero
                    fg[1 + 2*k ] = y_0  - ym_0 - dt*(G_0 + Gm_0)/2.;
                    fg[2 + 2*k ] = y_1  - ym_1 - dt*(G_1 + Gm_1)/2.;

                    // scale g(x) so it has similar size as f(x)
                    fg[1 + 2*k ] /= dt;
                    fg[2 + 2*k ] /= dt;
                }
            }
        }
    };
}
bool ode_inverse(void)
{   bool ok = true;
    size_t i;
    typedef CPPAD_TESTVECTOR( double ) Dvector;

    // number of components in the function g
    size_t ng = (np_ * nz_ + 1) * 2;
    // number of independent variables
    size_t nx = na_ + ng;
    // initial vlaue for the variables we are optimizing w.r.t
    Dvector xi(nx), xl(nx), xu(nx);
    for(i = 0; i < nx; i++)
    {   xi[i] =   0.0; // initial value
        xl[i] = -1e19; // no lower limit
        xu[i] = +1e19; // no upper limit
    }
    for(i = 0; i < na_; i++)
        xi[0] = 1.5;   // initial value for a

    // all the difference equations are constrainted to be zero
    Dvector gl(ng), gu(ng);
    for(i = 0; i < ng; i++)
    {   gl[i] = 0.0;
        gu[i] = 0.0;
    }
    // object defining both f(x) and g(x)
    FG_eval fg_eval;

    // options
    std::string options;
    // Use sparse matrices for calculation of Jacobians and Hessians
    // with forward mode for Jacobian (seems to be faster for this case).
    options += "Sparse  true        forward\n";
    // turn off any printing
    options += "Integer print_level 0\n";
    options += "String  sb        yes\n";
    // maximum number of iterations
    options += "Integer max_iter    30\n";
    // approximate accuracy in first order necessary conditions;
    // see Mathematical Programming, Volume 106, Number 1,
    // Pages 25-57, Equation (6)
    options += "Numeric tol         1e-6\n";

    // place to return solution
    CppAD::ipopt::solve_result<Dvector> solution;

    // solve the problem
    CppAD::ipopt::solve<Dvector, FG_eval>(
        options, xi, xl, xu, gl, gu, fg_eval, solution
    );
    //
    // Check some of the solution values
    //
    ok &= solution.status == CppAD::ipopt::solve_result<Dvector>::success;
    //
    double rel_tol    = 1e-4;  // relative tolerance
    double abs_tol    = 1e-4;  // absolute tolerance
    for(i = 0; i < na_; i++)
        ok &= CppAD::NearEqual( a_[i],  solution.x[i],   rel_tol, abs_tol);

    return ok;
}
// END C++
