mirror of
https://github.com/wassname/simpeg.git
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Rowan Cockett
e622dfb5cc
Conflicts: .coveragerc .gitignore .travis.yml docs/api_Utils.rst docs/conf.py docs/index.rst requirements.txt setup.py
342 lines
8.3 KiB
ReStructuredText
342 lines
8.3 KiB
ReStructuredText
.. _api_TDEM_derivation:
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.. math::
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\renewcommand{\div}{\nabla\cdot\,}
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\newcommand{\grad}{\vec \nabla}
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\newcommand{\curl}{{\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}{{\mathbf C}}
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\newcommand{\dgrad}{{\mathbf G}}
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\newcommand{\Acf}{{\mathbf A_c^f}}
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\newcommand{\Ace}{{\mathbf A_c^e}}
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\renewcommand{\S}{{\mathbf \Sigma}}
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\newcommand{\St}{{\mathbf \Sigma_\tau}}
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\newcommand{\T}{{\mathbf T}}
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\newcommand{\Tt}{{\mathbf T_\tau}}
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\newcommand{\diag}[1]{\,{\sf diag}\left( #1 \right)}
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\newcommand{\M}{{\mathbf M}}
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\newcommand{\MfMui}{{\M^f_{\mu^{-1}}}}
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\newcommand{\MeSig}{{\M^e_\sigma}}
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\newcommand{\MeSigInf}{{\M^e_{\sigma_\infty}}}
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\newcommand{\MeSigO}{{\M^e_{\sigma_0}}}
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\newcommand{\Me}{{\M^e}}
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\newcommand{\Mes}[1]{{\M^e_{#1}}}
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\newcommand{\Mee}{{\M^e_e}}
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\newcommand{\Mej}{{\M^e_j}}
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\newcommand{\BigO}[1]{\mathcal{O}\bigl(#1\bigr)}
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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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\newcommand {\c} { {\vec c} }
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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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Time-Domain EM Derivation
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*************************
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The following shows the derivation for the TDEM problem. We use the b-formulation below.
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(More to come soon..!)
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Sensitivity Calculation
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=======================
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.. math::
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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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Using Gauss-Newton to solve the inverse problem requires the ability to calculate the product of the
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Jacobian and a vector, as well as the transpose of the Jacobian times a vector.
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The above system can be rewritten as:
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.. math::
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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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.. math::
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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} \MfMui & \MfMui\dcurl \\
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\dcurl^\top \MfMui & -\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} \MfMui & 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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.. note::
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Here we have multiplied through by \\(\\MfMui\\) to make A and B symmetric!
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The entire time dependent system can be written in a single matrix expression
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.. math::
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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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.. math::
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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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For the fields \\(\\u\\), the measured data is given by
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.. math::
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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 **J** is then defined as
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.. math::
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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 \\(\\c(m,\\u)\\) to be
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.. math::
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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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.. math::
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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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.. math::
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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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.. math::
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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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.. math::
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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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.. math::
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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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Implementing **J** times a vector
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=================================
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Multiplying **J** onto a vector can be broken into three steps
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* Compute \\(\\vec{p} = \\mathbf{G}m\\)
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* Solve \\(\\hat{\\mathbf{A}} \\vec{y} = \\vec{p}\\)
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* Compute \\(\\vec{w} = -\\mathbf{Q} \\vec{y}\\)
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.. math::
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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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\vec{p}_b^{(n)} \\
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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}_b^{(n)} = 0 \\
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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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For all time steps:
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.. math::
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\begin{align}
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\frac{1}{\delta t} \MfMui\vec{y}_{b}^{(t+1)} + \MfMui\dcurl \vec{y}_{e}^{(t+1)}
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- \frac{1}{\delta t} \MfMui \vec{y}_{b}^{(t)}
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= \vec{p}_b^{(t+1)} \\
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\dcurl^\top \MfMui \vec{y}_b^{(t+1)} - \MeSig \vec{y}_e^{(t+1)} = \vec{p}_e^{(t+1)}
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\end{align}
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and
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.. math::
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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}^{(t+1)} =
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\frac{1}{\delta t} \MfMui \vec{y}_b^{(t)}
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+ \MfMui \dcurl \MeSig^{-1} \vec{p}_e^{(t+1)} + \vec{p}_b^{(t+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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.. note::
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For the first time step, \\\(t=0\\\), the term: \\\(\\frac{1}{\\delta t} \\MfMui \\vec{y}_b^{(0)}\\\) is zero.
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Implementing **J** transpose times a vector
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===========================================
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Multiplying \\(\\mathbf{J}^\\top\\) onto a vector can be broken into three steps
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* Compute \\(\\vec{p} = \\mathbf{Q}^\\top \\vec{v}\\)
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* Solve \\(\\hat{\\mathbf{A}}^\\top \\vec{y} = \\vec{p}\\)
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* Compute \\(\\vec{w} = -\\mathbf{G}^\\top y\\)
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.. math::
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\mathbf{\hat{A}}^\top = \left[
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\begin{array}{cccc}
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A & B & & \\
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& \ddots & \ddots & \\
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& & A & B \\
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& & 0 & A
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\end{array}
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\right]
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For the all time-steps (going backwards in time):
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.. math::
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A \vec{y}^{(t)} + B \vec{y}^{(t+1)} = \vec{p}^{(t)}
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.. math::
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\begin{align}
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\frac{1}{\delta t} \MfMui\vec{y}_{b}^{(t)} + \MfMui\dcurl \vec{y}_{e}^{(t)}
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- \frac{1}{\delta t} \MfMui \vec{y}_{b}^{(t+1)}
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= \vec{p}_b^{(t)} \\
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\dcurl^\top \MfMui \vec{y}_b^{(t)} - \MeSig \vec{y}_e^{(t)} = \vec{p}_e^{(t)}
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\end{align}
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and
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.. math::
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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}^{(t)} =
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\frac{1}{\delta t} \MfMui \vec{y}_b^{(t+1)}
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+ \MfMui \dcurl \MeSig^{-1} \vec{p}_e^{(t)} + \vec{p}_b^{(t)} \\
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\vec{y}_e^{(t)} = \MeSig^{-1} \dcurl^\top \MfMui \vec{y}_b^{(t)} - \MeSig^{-1} \vec{p}_e^{(t)}
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\end{align}
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.. note::
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For the last time step, \\\(t=N\\\), the term: \\\(\\frac{1}{\\delta t} \\MfMui \\vec{y}_b^{(N+1)}\\\) is zero.
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