.. _api_TDEM_derivation:


.. math::

    \renewcommand{\div}{\nabla\cdot\,}
    \newcommand{\grad}{\vec \nabla}
    \newcommand{\curl}{{\vec \nabla}\times\,}
    \newcommand {\J}{{\vec J}}
    \renewcommand{\H}{{\vec H}}
    \newcommand {\E}{{\vec E}}
    \newcommand{\dcurl}{{\mathbf C}}
    \newcommand{\dgrad}{{\mathbf G}}
    \newcommand{\Acf}{{\mathbf A_c^f}}
    \newcommand{\Ace}{{\mathbf A_c^e}}
    \renewcommand{\S}{{\mathbf \Sigma}}
    \newcommand{\St}{{\mathbf \Sigma_\tau}}
    \newcommand{\T}{{\mathbf T}}
    \newcommand{\Tt}{{\mathbf T_\tau}}
    \newcommand{\diag}[1]{\,{\sf diag}\left( #1 \right)}
    \newcommand{\M}{{\mathbf M}}
    \newcommand{\MfMui}{{\M^f_{\mu^{-1}}}}
    \newcommand{\MeSig}{{\M^e_\sigma}}
    \newcommand{\MeSigInf}{{\M^e_{\sigma_\infty}}}
    \newcommand{\MeSigO}{{\M^e_{\sigma_0}}}
    \newcommand{\Me}{{\M^e}}
    \newcommand{\Mes}[1]{{\M^e_{#1}}}
    \newcommand{\Mee}{{\M^e_e}}
    \newcommand{\Mej}{{\M^e_j}}
    \newcommand{\BigO}[1]{\mathcal{O}\bigl(#1\bigr)}
    \newcommand{\bE}{\mathbf{E}}
    \newcommand{\bH}{\mathbf{H}}
    \newcommand{\B}{\vec{B}}
    \newcommand{\D}{\vec{D}}
    \renewcommand{\H}{\vec{H}}
    \newcommand{\s}{\vec{s}}
    \newcommand{\bfJ}{\bf{J}}
    \newcommand{\vecm}{\vec m}
    \renewcommand{\Re}{\mathsf{Re}}
    \renewcommand{\Im}{\mathsf{Im}}
    \renewcommand {\j}  { {\vec j} }
    \newcommand {\h}  { {\vec h} }
    \renewcommand {\b}  { {\vec b} }
    \newcommand {\e}  { {\vec e} }
    \newcommand {\c}  { {\vec c} }
    \renewcommand {\d}  { {\vec d} }
    \renewcommand {\u}  { {\vec u} }
    \newcommand{\I}{\vec{I}}


Time-Domain EM Derivation
*************************

The following shows the derivation for the TDEM problem. We use the b-formulation below.
(More to come soon..!)


Sensitivity Calculation
=======================

.. math::

    \begin{align}
        \dcurl \e^{(t+1)} + \frac{\b^{(t+1)} - \b^{(t)}}{\delta t} = 0 \\
        \dcurl^\top \MfMui \b^{(t+1)} - \MeSig \e^{(t+1)} = \Me \j_s^{(t+1)}
    \end{align}

Using Gauss-Newton to solve the inverse problem requires the ability to calculate the product of the
Jacobian and a vector, as well as the transpose of the Jacobian times a vector.
The above system can be rewritten as:

.. math::

    \begin{align}
        \mathbf{A} \u^{(t+1)} + \mathbf{B} \u^{(t)}= \s^{(t+1)}
    \end{align}

where

.. math::

    \begin{align}
        \mathbf{A} =
        \left[
            \begin{array}{cc}
                \frac{1}{\delta t} \MfMui & \MfMui\dcurl \\
                \dcurl^\top \MfMui & -\MeSig
            \end{array}
        \right] \\
        \mathbf{B} =
        \left[
            \begin{array}{cc}
                -\frac{1}{\delta t} \MfMui & 0 \\
                0 & 0
            \end{array}
        \right] \\
        \u^{(k)} = \left[
        \begin{array}{c}
            \b^{(k)}\\
            \e^{(k)}
        \end{array}
        \right] \\
        \s^{(k)} = \left[
        \begin{array}{c}
            0\\
            \Me \j^{(k)}_s
        \end{array}
        \right]
    \end{align}

.. note::

    Here we have multiplied through by \\(\\MfMui\\) to make A and B symmetric!

The entire time dependent system can be written in a single matrix expression

.. math::

    \begin{align}
        \hat{\mathbf{A}} \hat{u} = \hat{s}
    \end{align}

where

.. math::

    \begin{align}
        \mathbf{\hat{A}} = \left[
        \begin{array}{cccc}
            A & 0 & & \\
            B & A & & \\
              & \ddots & \ddots & \\
              & & B & A
        \end{array}
        \right] \\
        \hat{u} = \left[
            \begin{array}{c}
                \u^{(1)} \\
                \u^{(2)} \\
                \vdots \\
                \u^{(N)}
            \end{array} \right]\\
        \hat{s} = \left[
            \begin{array}{c}
                \s^{(1)} - \mathbf{B} \u^{(0)} \\
                \s^{(2)} \\
                \vdots \\
                \s^{(N)}
            \end{array}
        \right]
    \end{align}

For the fields \\(\\u\\), the measured data is given by

.. math::

    \begin{align}
        \vec{d} = \mathbf{Q} \u
    \end{align}

The sensitivity matrix **J** is then defined as

.. math::

    \begin{align}
        \mathbf{J} = \mathbf{Q} \frac{\partial \u}{\partial \sigma}
    \end{align}


Defining the function \\(\\c(m,\\u)\\) to be

.. math::

    \begin{align}
        \vec{c}(m,\u) = \hat{\mathbf{A}} \vec{u} - \vec{q} = \vec{0}
    \end{align}

then

.. math::

    \begin{align}
        \frac{\partial \vec{c}}{\partial m} \partial m
        + \frac{\partial \vec{c}}{\partial \u} \partial \vec{u} = 0
    \end{align}

or

.. math::

    \begin{align}
        \frac{\partial \vec{u}}{\partial m} = -\left(\frac{\partial \vec{c}}{\partial \u} \right)^{-1} \frac{\partial \vec{c}}{\partial m}
    \end{align}


Differentiating, we find that

.. math::

    \begin{align}
        \frac{\partial \vec{c}}{\partial \hat{u}} = \hat{\mathbf{A}}
    \end{align}

and

.. math::

    \begin{align}
        \frac{\partial \vec{c}}{\partial \sigma} = \mathbf{G}_\sigma =
        \left[
            \begin{array}{c}
                g_\sigma^{(1)}\\
                g_\sigma^{(2)}\\
                \vdots \\
                g_\sigma^{(N)}
            \end{array}
        \right]
    \end{align}

with

.. math::

    \begin{align}
        g_\sigma^{(n)} =
        \left[
            \begin{array}{c}
                \mathbf{0} \\
                - \diag{\e^{(n)}} \Ace \diag{\vec{V}}
            \end{array}
        \right]
    \end{align}


Implementing **J** times a vector
=================================

Multiplying **J** onto a vector can be broken into three steps


* Compute \\(\\vec{p} = \\mathbf{G}m\\)
* Solve \\(\\hat{\\mathbf{A}} \\vec{y} = \\vec{p}\\)
* Compute \\(\\vec{w} = -\\mathbf{Q} \\vec{y}\\)

.. math::

    \begin{align}
        \vec{p}^{(n)} = \left[
            \begin{array}{c}
                \vec{p}_b^{(n)} \\
                \vec{p}_e^{(n)}
            \end{array}
        \right] \\
        \vec{p}_b^{(n)} = 0 \\
        \vec{p}_e^{(n)} = - \diag{\e^{(n)}} \Ace \diag{V} m
    \end{align}


For all time steps:

.. math::

    \begin{align}
        \frac{1}{\delta t} \MfMui\vec{y}_{b}^{(t+1)} + \MfMui\dcurl \vec{y}_{e}^{(t+1)}
        - \frac{1}{\delta t} \MfMui \vec{y}_{b}^{(t)}
        = \vec{p}_b^{(t+1)} \\
        \dcurl^\top \MfMui \vec{y}_b^{(t+1)} - \MeSig \vec{y}_e^{(t+1)} = \vec{p}_e^{(t+1)}
    \end{align}

and

.. math::

    \begin{align}
        \left( \MfMui \dcurl \MeSig^{-1} \dcurl^\top \MfMui + \frac{1}{\delta t} \MfMui \right) \vec{y}_{b}^{(t+1)} =
        \frac{1}{\delta t} \MfMui \vec{y}_b^{(t)}
        + \MfMui \dcurl \MeSig^{-1} \vec{p}_e^{(t+1)} + \vec{p}_b^{(t+1)} \\
        \vec{y}_e^{(t+1)} = \MeSig^{-1} \dcurl^\top \MfMui \vec{y}_b^{(t+1)} - \MeSig^{-1} \vec{p}_e^{(t+1)}
    \end{align}

.. note::

    For the first time step, \\\(t=0\\\), the term: \\\(\\frac{1}{\\delta t} \\MfMui \\vec{y}_b^{(0)}\\\) is zero.




Implementing **J** transpose times a vector
===========================================

Multiplying \\(\\mathbf{J}^\\top\\) onto a vector can be broken into three steps


* Compute \\(\\vec{p} = \\mathbf{Q}^\\top \\vec{v}\\)
* Solve \\(\\hat{\\mathbf{A}}^\\top \\vec{y} = \\vec{p}\\)
* Compute \\(\\vec{w} = -\\mathbf{G}^\\top y\\)


.. math::

    \mathbf{\hat{A}}^\top = \left[
        \begin{array}{cccc}
            A & B & & \\
              & \ddots & \ddots & \\
              & & A & B \\
              & & 0 & A
        \end{array}
    \right]

For the all time-steps (going backwards in time):


.. math::

    A \vec{y}^{(t)} + B \vec{y}^{(t+1)} = \vec{p}^{(t)}


.. math::

    \begin{align}
        \frac{1}{\delta t} \MfMui\vec{y}_{b}^{(t)} + \MfMui\dcurl \vec{y}_{e}^{(t)}
        - \frac{1}{\delta t} \MfMui \vec{y}_{b}^{(t+1)}
        = \vec{p}_b^{(t)} \\
        \dcurl^\top \MfMui \vec{y}_b^{(t)} - \MeSig \vec{y}_e^{(t)} = \vec{p}_e^{(t)}
    \end{align}

and

.. math::

    \begin{align}
        \left( \MfMui \dcurl \MeSig^{-1} \dcurl^\top \MfMui + \frac{1}{\delta t} \MfMui \right) \vec{y}_{b}^{(t)} =
        \frac{1}{\delta t} \MfMui \vec{y}_b^{(t+1)}
        + \MfMui \dcurl \MeSig^{-1} \vec{p}_e^{(t)} + \vec{p}_b^{(t)} \\
        \vec{y}_e^{(t)} = \MeSig^{-1} \dcurl^\top \MfMui \vec{y}_b^{(t)} - \MeSig^{-1} \vec{p}_e^{(t)}
    \end{align}


.. note::

    For the last time step, \\\(t=N\\\), the term: \\\(\\frac{1}{\\delta t} \\MfMui \\vec{y}_b^{(N+1)}\\\) is zero.