Inversion Module

Equation Index

Authors

Affiliations

Douglas Oldenburg

University of British Columbia

Lindsey Heagy

University of British Columbia

April 20, 2021

s=\frac{1}{v}

\rho = 1/\sigma

\mu = \mu_0 (1 + \kappa)

\mathcal{F}[m] = d

d_j=\mathcal{F}_j[m]

\mathcal{F}[\alpha f+\beta g]=\alpha\mathcal{F}[f]+\beta\mathcal{F}[g]

d_j=\int^b_ag_j(x)m(x)dx

y = m \otimes w

y(t_j)=\int^{\infty}_{-\infty}m(\tau)w(t_j-\tau)d\tau

\vec{B}(\vec{r}_j)=\frac{\mu_0}{4\pi}\int\frac{\vec{J}(\vec{r}\prime)\times\hat{r}}{\left|\vec{r}_j-\vec{r}\prime\right|^2}

d_j^{(1)}=\int^b_a g_j(x)m_1(x)dx

d_j^{(2)}=\int^b_a g_j(x)m_2(x)dx

\begin{aligned}d_j=\int^b_a g_j(x)[\alpha & m_1(x) +\beta m_2(x)]\\ d_j=\alpha &d_j^{(1)}+\beta d_j^{(2)}\end{aligned}

d_j=\int^b_ag_j(x)m^2(x)dx

\nabla\cdot\sigma\nabla V=-I\delta(\vec{r}-\vec{r}\prime)

V^2_{rms}(t_j)=\frac{1}{t_j}\int^{t_{max}}_0 v^2_{int}(u)H(t_j-u)du

V^2_{rms}(t)=\frac{1}{t}\int^{t_{max}}_0 v_{int}^2(u)H(t-u)du

v_{int}=V_{rms}(t)\left(1+\frac{2tV'_{rms}(t)}{V_{rms}(t)}\right)^{\frac{1}{2}}

V_{rms}(t)=v_0+a\sin(2\pi ft)

V'_{rms}(t)=2\pi af\cos(2\pi ft)

d^{obs}=d^{true}+\delta d

\mathcal{F}^{-1}[d^{true}]=m_c

\mathcal{F}^{-1}[d^{obs}]=\mathcal{F}^{-1}[d^{true}+\delta d]=m_c+\delta m

\delta v_{int}\propto \sqrt{\frac{fta}{v_0}}

P(m|d^{obs})\propto P(d^{obs}|m)P(m)

\begin{aligned}&\text{Minimize}~~\phi=\phi_d+\beta\phi_m\\ &\text{subject to}~~m_l<m<m_u,\end{aligned}

g_j(x)= e^{p_jx}\cos(2\pi q_jx)

\begin{aligned} d_j&=\int_0^{x_1}g_j(x)m_1dx +\int_{x_1}^{x_2}g_j(x)m_2dx+\dots \\ &=\sum^M_{i=1}\left(\int_{x_{k-1}}^{x_k}g_j\left(x\right)dx\right)m_i\\ &\\ d_j &= \mathbf g_j \mathbf m\end{aligned}

\begin{aligned} \mathbf{d} = \mathbf{G}\mathbf{m} = \begin{bmatrix} d_1\\ \vdots\\ d_{N} \end{bmatrix}\end{aligned}

G_{jk} = \int_{x_{k-1}}^{x_k} g_j(x) dx

\mathbf{d}^{obs}=\mathbf{d}+\mathbf{n}

\epsilon_j = \%|d_j| + \nu_j

\phi_d=\sum^N_{j=1}\left(\frac{d_j-d_j^{obs}}{\epsilon_j}\right)^2

\eta_j = \frac {d_j - d_j^{obs}} {\epsilon_j}

E[\chi^2] = N

Var[\chi^2] = 2N

\phi_d^*=N

\phi_m=\int (m-m^{ref})^2 dx

\phi_m=\int \left(\frac{d(m-m^{ref})}{dx}\right)^2 dx

\phi_m=\alpha_s\int (m-m^{ref})^2 dx+\alpha_x\int (\frac{d(m-m^{ref})}{dx})^2 dx

\phi_m=\|\mathbf{m - m^{ref}}\|^2=\sum_{i=1}^M(m_i-m^{ref}_i)^2

\phi_m=\|\frac{d\mathbf{m}}{dx}\|^2=\sum_{i=1}^{M-1}(m_{i+1}-m_i)^2

\phi(m) = \phi_d(m)+\beta\phi_m(m)

\phi=T+\beta F

\mathbf{g}_j(\mathbf{x}) = e^{p_j\mathbf{x}} cos (2 \pi q_j \mathbf{x}) \Delta x

\begin{aligned} \mathbf{G} = \begin{bmatrix} \mathbf{g}_1\\ \vdots\\ \mathbf{g}_{N} \end{bmatrix} \end{aligned}

Time domain cyl forward