lines 5-146 of file: include/cppad/example/atomic_four/lin_ode/reverse_2.xrst

{xrst_begin atomic_four_lin_ode_reverse_2}

Atomic Linear ODE Second Order Reverse
######################################

x^1 Partial
***********
We need to compute

.. math::

   \R{partial\_x} [ j  * q + 1 ]
   =
   \sum_{i=0}^{m-1} \R{partial\_y} [ i * q + 1]
      ( \partial y_i^1 ( x^0 , x^1 ) / \partial x_j^1 )

where :math:`q = 2` and :math:`j = 0 , \ldots , n-1`.
Using the :ref:`reverse_identity-name` we have

.. math::

   \partial y_i^1 ( x^0 , x^1 ) / \partial x_j^1
   =
   \partial y_i^0 ( x^0 ) / \partial x_j^0

.. math::

   \R{partial\_x} [ j  * q + 1 ]
   =
   \sum_{i=0}^{m-1} \R{partial\_y} [ i * q + 1]
      ( \partial y_i^0 ( x^0 ) / \partial x_j^0 )

which is the same as the first order
:ref:`theory<atomic_four_lin_ode_reverse.hpp@First Order Theory>` with

.. math::

   w_i = \R{partial\_y} [ i * q + 1]

x^0 Partial
***********
We also need to compute

.. math::

   \R{partial\_x} [ j  * q + 0 ]
   =
   \sum_{i=0}^{m-1} \R{partial\_y} [ i * q + 0]
      ( \partial y_i^0 ( x^0 ) / \partial x_j^0 )
   +
   \R{partial\_y} [ i * q + 1]
      ( \partial y_i^1 ( x^0 , x^1 ) / \partial x_j^0 )

Note that we can solve for

.. math::

   y^1 ( x^0 , x^1 ) = z^1 ( r , x^0 , x^1 )

using the following extended ODE; see
:ref:`forward theory<atomic_four_lin_ode_forward.hpp@Theory>` .

.. math::

   \left[ \begin{array}{c}
   z^0_t (t, x^0 ) \\
   z^1_t (t, x^0 , x^1 )
   \end{array} \right]
   =
   \left[ \begin{array}{cc}
   A^0 & 0   \\
   A^1 & A^0
   \end{array} \right]
   \left[ \begin{array}{c}
   z^0 (t, x^0 ) \\
   z^1 (t, x^0 , x^1 )
   \end{array} \right]
   \; , \;
   \left[ \begin{array}{c}
   z^0 (0, x^0 ) \\
   z^1 (0, x^0 , x^1 )
   \end{array} \right]
   =
   \left[ \begin{array}{c}
   b^0 \\
   b^1
   \end{array} \right]

Note that
:math:`A^0`, :math:`b^0` are components of :math:`x^0`
and
:math:`A^1`, :math:`b^1` are components of :math:`x^1`.
We use the following notation

.. math::

   \bar{x} = \left[ \begin{array}{c}
      x^0 \\ x^1
   \end{array} \right]
   \W{,}
   \bar{z}(t , \bar{x} ) = \left[ \begin{array}{c}
      z^0 (t, x^0) \\ z^1 ( t, x^0 , x^1 )
   \end{array} \right]
   \W{,}
   \bar{A} =
   \left[ \begin{array}{cc}
   A^0 & 0   \\
   A^1 & A^0
   \end{array} \right]
   \W{,}
   \bar{b} = \left[ \begin{array}{c}
      b^0 \\ b^1
   \end{array} \right]

Using this notation we have

.. math::

   \bar{z}_t ( t , \bar{x} ) = \bar{A} \bar{z} (t, \bar{x} )
   \W{,}
   \bar{z} (0, \bar{x} ) = \bar{b}

Define :math:`\bar{w} \in \B{R}^{m + m}` by

.. math::

   \bar{w}_i = \R{partial\_y}[ i * q + 0 ]
   \W{,}
   \bar{w}_{m + i} = \R{partial\_y}[ i * q + 1 ]

For this case, we can compute

.. math::

   \partial_\bar{x} \bar{w}^\R{T} \bar{z}(r, \bar{x} )

which is same as the first order case but with the extended variables
and extended ODE.
We will only use the components of
:math:`\partial_\bar{x}` that correspond to partials w.r.t. :math:`x^0`.

{xrst_end atomic_four_lin_ode_reverse_2}