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ODE Inverse Problem Definitions: Source Code

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

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

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

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

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

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

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

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

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

Source
The following source code implements the ODE inversion method proposed above:
# include <cppad/ipopt/solve.hpp>

namespace {

// 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

// Evaluation of the objective f(x), and constraints g(x)
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;

// 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

// solve the problem
options, xi, xl, xu, gl, gu, fg_eval, solution
);
//
// Check some of the solution values
//
}