Calculating the Regularization Parameter

It is the purpose of the regularization parameter, $\epsilon$, to weight the regularization residual so that the iterative solver does not focus on one goal while ignoring the other. For example, an $\epsilon$ that is too large will insure that the missing data is filled but it may be too smooth. On the other hand, an $\epsilon$ that is too small will not fill in much data and will tend to leave the acquisition footprint behind.

$\epsilon$ is used in fitting goal (2) to balance the two goals. For now, I applied Jon Claerbout's idea 1991. Within the conjugate solver routine, the gradient determines in which direction to minimize the residual.

$$\begin{eqnarray} \ {\bf \Delta m} = {\bf L'}\frac{d}{dt}{\bf W'r_d} + \epsilon {\bf A' r_m}\ \end{eqnarray}$$ (3)

Finding an $\epsilon$ which balances both goals we try:

$$\begin{eqnarray} \ \epsilon = \frac{\vert{\bf L'}\frac{d}{dt}{\bf W'r_d}\vert}{\vert{\bf A'r_m}\vert}\ \end{eqnarray}$$ (4)

In initial tests, I placed these equations in the solver and calculated a new $\epsilon$ for each iteration with the first $\epsilon$ value equal to 1. After about 15 iterations it converged to an almost constant value. This calculated $\epsilon$ was slightly lower than the $\epsilon$ that I found by trial and error. Figure 6 was generated using an $\epsilon$ of 0.4 whereas the calculated $\epsilon$ for that figure, returned by the above equation, was approximately 0.3.

The calculated $\epsilon$ does not seem to work on the entire merged data set probably as a result of different data densities. The northern sparsely sampled region needs a different $\epsilon$ than the southern densely sampled region. In this case, a scalar $\epsilon$ value is not sufficient.