lines 6-205 of file: example/ipopt_solve/ode_inverse.cpp

{xrst_begin ipopt_solve_ode_inverse.cpp}
{xrst_spell
   np
   nz
   trapezoidal
}

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:

.. math::
   :nowrap:

   \begin{eqnarray}
      \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{eqnarray}

with the initial conditions

.. math::

   y_0 (0 , a) = ( a_0 , 0 )^\R{T}

Our forward problem is stated as follows:
Given :math:`a \in \B{R}^3`
determine the value of :math:`y ( t , a )`,
for :math:`t \in R`, that solves the initial value problem above.

Measurements
************
Suppose we are also given measurement times :math:`s \in \B{R}^5`
and  a measurement vector :math:`z \in \B{R}^4`
and for :math:`i = 0, \ldots, 3`, we model :math:`z_i` by

.. math::

   z_i = y_1 ( s_{i+1} , a) + e_i

where :math:`e_{i-1} \sim {\bf N} (0 , \sigma^2 )`
is the measurement noise,
and :math:`\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:

.. math::
   :nowrap:

   \begin{eqnarray}
      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{eqnarray}

Simulation Parameter Values
===========================

.. list-table::
   :widths: auto

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

Simulated Measurement Values
============================
The simulated measurement values are given by the equation

.. math::
   :nowrap:

   \begin{eqnarray}
   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{eqnarray}

for :math:`i = 0, \ldots , 3`.

Inverse Problem
***************
The maximum likelihood estimate for :math:`a` given :math:`z`
solves the following optimization problem

.. math::
   :nowrap:

   \begin{eqnarray}
   {\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{eqnarray}

Trapezoidal Approximation
*************************
We are given a number of approximation points per measurement interval
:math:`np` and define the time grid :math:`t \in \B{R}^{4 \cdot np + 1}`
as follows:
:math:`t_0 = s_0` and
for :math:`i = 0 , 1 , 2, 3`, :math:`j = 1 , \ldots , np`

.. math::

   t_{i \cdot np + j} = s_i + (s_{i+1} - s{i}) \frac{i}{np}

We note that for :math:`i = 1 , \ldots , 4`,
:math:`t_{i \cdot np} = s_i`.
This example uses a trapezoidal approximation to solve the ODE.
Given :math:`a \in \B{R}^3` and :math:`y^{k-1} \approx y( t_{k-1} , a )`,
the a trapezoidal method approximates :math:`y ( t_j , a )`
by the value :math:`y^k \in \B{R}^2` ) that solves the equation

.. math::

   y^k  =  y^{k-1} + \frac{G( y^k , a ) + G( y^{k-1} , a ) }{2} * (t_k - t_{k-1})

where :math:`G : \B{R}^2 \times \B{R}^3 \rightarrow \B{R}^2` is defined by

.. math::
   :nowrap:

   \begin{eqnarray}
      G_0 ( y , a ) & = & - a_1 * y_0
      \\
      G_1 ( y , a ) & = & + a_1 * y_0  - a_2 * y_1
   \end{eqnarray}

Solution Method
***************
We use constraints to embed the
forward problem in the inverse problem.
To be specific, we solve the optimization problem

.. math::
   :nowrap:

   \begin{eqnarray}
   {\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{eqnarray}

The code below we using the notation
:math:`x \in \B{3 + (4 \cdot np + 1) \cdot 2}` defined by

.. math::

   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:
{xrst_literal
   // BEGIN C++
   // END C++
}

{xrst_end ipopt_solve_ode_inverse.cpp}