Time-Domain EM Derivation The following shows the derivation for the TDEM problem. We use the b-formulation below. (More to come soon..!)
$$\begin{aligned}
\begin{align}
\dcurl \e^{(t+1)} + \frac{\b^{(t+1)} - \b^{(t)}}{\delta t} = 0 \\\
\dcurl^\top \MfMui \b^{(t+1)} - \MeSig \e^{(t+1)} = \Me \j_s^{(t+1)}
\end{align}
\end{aligned}$$
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:
$$\begin{align}
\mathbf{A} \u^{(t+1)} + \mathbf{B} \u^{(t)}= \s^{(t+1)}
\end{align}$$
where
$$\begin{aligned}
\begin{align}
\mathbf{A} =
\left[
\begin{array}{cc}
\frac{1}{\delta t} \MfMui & \MfMui\dcurl \\\
\dcurl^\top \MfMui & -\MeSig
\end{array}
\right] \\\
\mathbf{B} =
\left[
\begin{array}{cc}
-\frac{1}{\delta t} \MfMui & 0 \\\
0 & 0
\end{array}
\right] \\\
\u^{(k)} = \left[
\begin{array}{c}
\b^{(k)}\\\
\e^{(k)}
\end{array}
\right] \\\
\s^{(k)} = \left[
\begin{array}{c}
0\\\
\Me \j^{(k)}_s
\end{array}
\right]
\end{align}
\end{aligned}$$
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
$$\begin{align}
\hat{\mathbf{A}} \hat{u} = \hat{s}
\end{align}$$
where
$$\begin{aligned}
\begin{align}
\mathbf{\hat{A}} = \left[
\begin{array}{cccc}
A & 0 & & \\\
B & A & & \\\
& \ddots & \ddots & \\\
& & B & A
\end{array}
\right] \\\
\hat{u} = \left[
\begin{array}{c}
\u^{(1)} \\\
\u^{(2)} \\\
\vdots \\\
\u^{(N)}
\end{array} \right]\\\
\hat{s} = \left[
\begin{array}{c}
\s^{(1)} - \mathbf{B} \u^{(0)} \\\
\s^{(2)} \\\
\vdots \\\
\s^{(N)}
\end{array}
\right]
\end{align}
\end{aligned}$$
For the fields \(\u\), the measured data is given by
$$\begin{align}
\vec{d} = \mathbf{Q} \u
\end{align}$$
The sensitivity matrix J is then defined as
$$\begin{align}
\mathbf{J} = \mathbf{Q} \frac{\partial \u}{\partial \sigma}
\end{align}$$
Defining the function \(\c(m,\u)\) to be
$$\begin{align}
\vec{c}(m,\u) = \hat{\mathbf{A}} \vec{u} - \vec{q} = \vec{0}
\end{align}$$
then
$$\begin{align}
\frac{\partial \vec{c}}{\partial m} \partial m
+ \frac{\partial \vec{c}}{\partial \u} \partial \vec{u} = 0
\end{align}$$
or
$$\begin{align}
\frac{\partial \vec{u}}{\partial m} = -\left(\frac{\partial \vec{c}}{\partial \u} \right)^{-1} \frac{\partial \vec{c}}{\partial m}
\end{align}$$
Differentiating, we find that
$$\begin{align}
\frac{\partial \vec{c}}{\partial \hat{u}} = \hat{\mathbf{A}}
\end{align}$$
and
$$\begin{aligned}
\begin{align}
\frac{\partial \vec{c}}{\partial \sigma} = \mathbf{G}_\sigma =
\left[
\begin{array}{c}
g_\sigma^{(1)}\\\
g_\sigma^{(2)}\\\
\vdots \\\
g_\sigma^{(N)}
\end{array}
\right]
\end{align}
\end{aligned}$$
with
$$\begin{aligned}
\begin{align}
g_\sigma^{(n)} =
\left[
\begin{array}{c}
\mathbf{0} \\\
- \diag{\e^{(n)}} \Ace \diag{\vec{V}}
\end{array}
\right]
\end{align}
\end{aligned}$$
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 \(\vec{w} = -\mathbf{Q} \vec{y}\)
$$\begin{aligned}
\begin{align}
\vec{p}^{(n)} = \left[
\begin{array}{c}
\vec{p}_b^{(n)} \\\
\vec{p}_e^{(n)}
\end{array}
\right] \\\
\vec{p}_b^{(n)} = 0 \\\
\vec{p}_e^{(n)} = - \diag{\e^{(n)}} \Ace \diag{V} m
\end{align}
\end{aligned}$$
For all time steps:
$$\begin{aligned}
\begin{align}
\frac{1}{\delta t} \MfMui\vec{y}_{b}^{(t+1)} + \MfMui\dcurl \vec{y}_{e}^{(t+1)}
- \frac{1}{\delta t} \MfMui \vec{y}_{b}^{(t)}
= \vec{p}_b^{(t+1)} \\\
\dcurl^\top \MfMui \vec{y}_b^{(t+1)} - \MeSig \vec{y}_e^{(t+1)} = \vec{p}_e^{(t+1)}
\end{align}
\end{aligned}$$
and
$$\begin{aligned}
\begin{align}
\left( \MfMui \dcurl \MeSig^{-1} \dcurl^\top \MfMui + \frac{1}{\delta t} \MfMui \right) \vec{y}_{b}^{(t+1)} =
\frac{1}{\delta t} \MfMui \vec{y}_b^{(t)}
+ \MfMui \dcurl \MeSig^{-1} \vec{p}_e^{(t+1)} + \vec{p}_b^{(t+1)} \\\
\vec{y}_e^{(t+1)} = \MeSig^{-1} \dcurl^\top \MfMui \vec{y}_b^{(t+1)} - \MeSig^{-1} \vec{p}_e^{(t+1)}
\end{align}
\end{aligned}$$
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 \(\vec{w} = -\mathbf{G}^\top y\)
$$\begin{aligned}
\mathbf{\hat{A}}^\top = \left[
\begin{array}{cccc}
A & B & & \\\
& \ddots & \ddots & \\\
& & A & B \\\
& & 0 & A
\end{array}
\right]
\end{aligned}$$
For the all time-steps (going backwards in time):
A y⃗ (t ) + B y⃗ (t + 1) = p⃗ (t )
$$\begin{aligned}
\begin{align}
\frac{1}{\delta t} \MfMui\vec{y}_{b}^{(t)} + \MfMui\dcurl \vec{y}_{e}^{(t)}
- \frac{1}{\delta t} \MfMui \vec{y}_{b}^{(t+1)}
= \vec{p}_b^{(t)} \\\
\dcurl^\top \MfMui \vec{y}_b^{(t)} - \MeSig \vec{y}_e^{(t)} = \vec{p}_e^{(t)}
\end{align}
\end{aligned}$$
and
$$\begin{aligned}
\begin{align}
\left( \MfMui \dcurl \MeSig^{-1} \dcurl^\top \MfMui + \frac{1}{\delta t} \MfMui \right) \vec{y}_{b}^{(t)} =
\frac{1}{\delta t} \MfMui \vec{y}_b^{(t+1)}
+ \MfMui \dcurl \MeSig^{-1} \vec{p}_e^{(t)} + \vec{p}_b^{(t)} \\\
\vec{y}_e^{(t)} = \MeSig^{-1} \dcurl^\top \MfMui \vec{y}_b^{(t)} - \MeSig^{-1} \vec{p}_e^{(t)}
\end{align}
\end{aligned}$$
For the last time step, \(t=N\), the term: \(\frac{1}{\delta t} \MfMui \vec{y}_b^{(N+1)}\) is zero.