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Added some notes.
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Dave Marchant
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\documentclass[]{article}
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\renewcommand{\div}{\nabla\cdot\,}
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\newcommand{\grad}{\ensuremath {\vec \nabla}}
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\newcommand{\curl}{\ensuremath{{\vec \nabla}\times\,}}
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\newcommand {\J} { {\vec J} }
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\renewcommand {\H} { {\vec H} }
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\newcommand {\E} { {\vec E} }
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\newcommand{\dcurl}{\ensuremath{{\mathbf C}}}
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\newcommand{\dgrad}{\ensuremath{{\mathbf G}}}
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\newcommand{\Acf}{\ensuremath{{\mathbf A_c^f}}}
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\newcommand{\Ace}{\ensuremath{{\mathbf A_c^e}}}
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\renewcommand{\S}{\ensuremath{{\mathbf \Sigma}}}
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\newcommand{\St}{\ensuremath{{\mathbf \Sigma_\tau}}}
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\newcommand{\T}{\ensuremath{{\mathbf T}}}
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\newcommand{\Tt}{\ensuremath{{\mathbf T_\tau}}}
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\newcommand{\diag}[1]{\, {\sf diag}\left( #1 \right)}
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%Common mass matricies
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\newcommand{\M}{\ensuremath{{\mathbf M}}}
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\newcommand{\MfMui}{\ensuremath{{\M^f_{\mu^{-1}}}}}
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\newcommand{\MeSig}{\ensuremath{{\M^e_\sigma}}}
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\newcommand{\MeSigInf}{\ensuremath{{\M^e_{\sigma_\infty}}}}
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\newcommand{\MeSigO}{\ensuremath{{\M^e_{\sigma_0}}}}
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\newcommand{\Me}{\ensuremath{{\M^e}}}
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\newcommand{\Mes}[1]{\ensuremath{{\M^e_{#1}}}}
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\newcommand{\Mee}{\ensuremath{{\M^e_e}}}
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\newcommand{\Mej}{\ensuremath{{\M^e_j}}}
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\newcommand{\BigO}[1]{\ensuremath{\mathcal{O}\bigl(#1\bigr)}}
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% ********** TDIP paper
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\newcommand{\bE}{\mathbf{E}}
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\newcommand{\bH}{\mathbf{H}}
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\newcommand{\B}{\vec{B}}
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\newcommand{\D}{\vec{D}}
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\renewcommand{\H}{\vec{H}}
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\newcommand{\s}{\vec{s}}
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\newcommand{\bfJ}{\bf{J}}
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\newcommand{\vecm}{\vec m}
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\renewcommand{\Re}{\mathsf{Re}}
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\renewcommand{\Im}{\mathsf{Im}}
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\renewcommand {\j} { {\vec j} }
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\newcommand {\h} { {\vec h} }
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\renewcommand {\b} { {\vec b} }
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\newcommand {\e} { {\vec e} }
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\renewcommand {\d} { {\vec d} }
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\renewcommand {\u} { {\vec u} }
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\newcommand{\I}{\vec{I}}
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\usepackage{pslatex,palatino,avant,graphicx,color,amsmath}
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% \usepackage[margin=2cm]{geometry}
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\begin{document}
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\title{TEM}
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\section{Sensitivity Calculation}
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\begin{subequations}
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\begin{align}
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\dcurl \e^{(t+1)} + \frac{\b^{(t+1)} - \b^{(t)}}{\delta t} = 0 \\
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\dcurl^\top \MfMui \b^{(t+1)} - \MeSig \e^{(t+1)} = \Me \j_s^{(t+1)}
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\end{align}
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\end{subequations}
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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
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\begin{align}
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\mathbf{A} \u^{(t+1)} + \mathbf{B} \u^{(t)}= \s^{(t+1)}
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\end{align}
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where
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\begin{subequations}
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\begin{align}
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\mathbf{A} =
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\left[
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\begin{array}{cc}
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\frac{1}{\delta t} \mathbf{I} & \dcurl \\
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\dcurl^\top & -\MeSig
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\end{array}
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\right] \\
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\mathbf{B} =
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\left[
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\begin{array}{cc}
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-\frac{1}{\delta t} \mathbf{I} & 0 \\
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0 & 0
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\end{array}
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\right] \\
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\u^{(k)} = \left[
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\begin{array}{c}
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\b^{(k)}\\
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\e^{(k)}
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\end{array}
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\right] \\
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\s^{(k)} = \left[
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\begin{array}{c}
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0\\
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\Me \j^{(k)}_s
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\end{array}
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\right]
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\end{align}
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\end{subequations}
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The entire time dependent system can be written in a single matrix expression
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\begin{align}
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\hat{\mathbf{A}} \hat{u} = \hat{s}
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\end{align}
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where
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\begin{subequations}
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\begin{align}
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\mathbf{\hat{A}} = \left[
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\begin{array}{cccc}
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A & 0 & & \\
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B & A & & \\
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& \ddots & \ddots & \\
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& & B & A
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\end{array}
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\right] \\
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\hat{u} = \left[
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\begin{array}{c}
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\u^{(1)} \\
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\u^{(2)} \\
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\vdots \\
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\u^{(N)}
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\end{array} \right]\\
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\hat{s} = \left[
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\begin{array}{c}
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\s^{(1)} - \mathbf{B} \u^{(0)} \\
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\s^{(2)} \\
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\vdots \\
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\s^{(N)}
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\end{array}
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\right]
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\end{align}
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\end{subequations}
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For the fields $\u$, the measured data is given by
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\begin{align}
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\vec{d} = \mathbf{Q} \u
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\end{align}
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The sensitivity matrix $\mathbf{J}$ is then defined as
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\begin{align}
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\mathbf{J} = \mathbf{Q} \frac{\partial \u}{\partial \sigma}
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\end{align}
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Defining the function $\vec{c}(m,\vec{u})$ to be
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\begin{align}
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\vec{c}(m,\u) = \hat{\mathbf{A}} \vec{u} - \vec{q} = \vec{0}
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\end{align}
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then
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\begin{align}
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\frac{\partial \vec{c}}{\partial m} \partial m
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+ \frac{\partial \vec{c}}{\partial \u} \partial \vec{u} = 0
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\end{align}
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or
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\begin{align}
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\frac{\partial \vec{u}}{\partial m} = -\left(\frac{\partial \vec{c}}{\partial \u} \right)^{-1} \frac{\partial \vec{c}}{\partial m}
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\end{align}
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Differentiating, we find that
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\begin{align}
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\frac{\partial \vec{c}}{\partial \hat{u}} = \hat{\mathbf{A}}
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\end{align}
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and
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\begin{align}
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\frac{\partial \vec{c}}{\partial \sigma} = \mathbf{G}_\sigma =
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\left[
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\begin{array}{c}
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g_\sigma^{(1)}\\
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g_\sigma^{(2)}\\
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\vdots \\
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g_\sigma^{(N)}
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\end{array}
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\right]
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\end{align}
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with
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\begin{subequations}
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\begin{align}
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g_\sigma^{(n)} =
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\left[
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\begin{array}{c}
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\mathbf{0} \\
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- \diag{\e^{(n)}} \Ace \diag{\vec{V}}
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\end{array}
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\right]
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\end{align}
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\end{subequations}
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\subsection{Implementing $\mathbf{J}$ times a vector}
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Multiplying $\mathbf{J}$ onto a vector can be broken into three steps
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\begin{enumerate}
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\item Compute $\vec{p} = \mathbf{G}m$
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\item Solve $\hat{\mathbf{A}} \vec{y} = \vec{p}$
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\item Compute $\vec{w} = -\mathbf{Q} \vec{y}$
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\end{enumerate}
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\begin{subequations}
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\begin{align}
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\vec{p}^{(n)} = \left[
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\begin{array}{c}
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0 \\
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\vec{p}_e^{(n)}
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\end{array}
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\right] \\
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\vec{p}_e^{(n)} = - \diag{\e^{(n)}} \Ace \diag{V} m
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\end{align}
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\end{subequations}
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\paragraph{First time step}
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\begin{subequations}
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\begin{align}
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\frac{1}{\delta t} \vec{y}_{b}^{(1)} + \dcurl \vec{y}_{e}^{(1)} = 0 \\
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\dcurl^\top \MfMui \vec{y}_b^{(1)} - \MeSig \vec{y}_e^{(1)} = \vec{p}_e^{(1)}
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\end{align}
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\end{subequations}
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\begin{subequations}
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\begin{align}
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\left( \MfMui \dcurl \MeSig^{-1} \dcurl^\top \MfMui + \frac{1}{\delta t} \MfMui \right) \vec{y}_{b}^{(1)} = \MfMui \dcurl \MeSig^{-1} \vec{p}_e^{(1)} \\
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\vec{y}_e^{(1)} = \MeSig^{-1} \dcurl^\top \MfMui \vec{y}_b^{(1)} - \MeSig^{-1} \vec{p}_e^{(1)}
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\end{align}
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\end{subequations}
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\paragraph{Remaining time steps}
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\begin{subequations}
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\begin{align}
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\dcurl \vec{y}_{e}^{(t+1)} + \frac{1}{\delta t} \vec{y}_{b}^{(t+1)} - \frac{1}{\delta t} \vec{y}_{b}^{(t)} = 0 \\
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\vec{y}_e^{(t+1)} = \MeSig^{-1} \dcurl^\top \MfMui \vec{y}_b^{(t+1)} - \MeSig^{-1} \vec{p}_e^{(t+1)}
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\end{align}
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\end{subequations}
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\begin{subequations}
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\begin{align}
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\left( \MfMui \dcurl \MeSig^{-1} \dcurl^\top \MfMui + \frac{1}{\delta t} \MfMui \right) \vec{y}_{b}^{(1)} = \frac{1}{\delta t} \MfMui \vec{y}_b^{(t)} + \MfMui \dcurl \MeSig^{-1} \vec{p}_e^{(1)} \\
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\vec{y}_e^{(t+1)} = \MeSig^{-1} \dcurl^\top \MfMui \vec{y}_b^{(t+1)} - \MeSig^{-1} \vec{p}_e^{(t+1)}
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\end{align}
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\end{subequations}
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\subsection{Implementing $\mathbf{J}^\top$ onto a vector}
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\end{document}
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