Unfaulting method

Similar to fault zone replacement, we can also undo faulting in a seismic image to align seismic reflectors across faults. Following the theory part of the unfaulting method in Wu et al. (2016), we solve a similar regularized linear equation to get the shift vector $\mathbf s (\mathbf x)$ where $\mathbf s \equiv (s_1, s_2, s_3)$ and coordinates $\mathbf x \equiv (x_1, x_2, x_3)$ (1, 2 and 3 are indices representing different spatial dimensions). Shift vector $\mathbf s (\mathbf x)$ means that we can unfault the image if the sample at $\mathbf x$ moves to $\mathbf x+\mathbf s(\mathbf x)$.

If $\mathbf x_a$ is a left point of a fault, we can use local similarity to estimate its fault slip vector $\mathbf t(\mathbf x_a)$, which can provide us with a simple linear equation:

$\displaystyle \mathbf s(\mathbf x_a + \mathbf t(\mathbf x_a)) - \mathbf s(\mathbf x_a) = \mathbf t(\mathbf x_a) \; .$ (4)

This is the linear equation only applying to those samples alongside faults. For other samples away from faults, we expect unfaulting shifts to vary slowly and continuously along structural directions. This constraint can be added to the inverse problem by using a gradient operator as a regularization term (Wu et al., 2016) or by using a structure-oriented smoothing operator as a shaping operator in the framework of shaping regularization (Fomel, 2007b). We choose the latter, and solve for the shift vector $\mathbf s (\mathbf x)$ using the method previously utilized by Xue et al. (2016).

After we get the shift vector $\mathbf s (\mathbf x)$, we unfault the original image $f(\mathbf x)$ to a fault-free image $\tilde{f} (\mathbf x)$ by doing an inverse interpolation:

$\displaystyle \tilde{f} (\mathbf x + \mathbf s (\mathbf x)) = f(\mathbf x) \; .$ (5)

Then we can estimate dip from the new image and perform predictive painting to get a painting result $\tilde{p} (\mathbf x)$, which can be converted to the painting result $p(\mathbf x)$ of the original image $f(\mathbf x)$ by doing a forward interpolation:

$\displaystyle p(\mathbf x) = \tilde{p} (\mathbf x + \mathbf s (\mathbf x)) \; .$ (6)


2019-05-06