Objective function

Let $\tau_0$ be an estimate of the relative time-shift obtained through the non-linear optimization. We assume the true solution is  

$$\tau=\tau_0 + \Delta \tau,$$ (5)

where $\Delta \tau$ is a residual time-shift. We will estimate $\Delta \tau$by solving a linear optimization problem. First, we linearize the model. and Then we substitute these expressions into equation (3). The new objective function becomes a quadratic function of the unknown $\Delta \tau$: 

$$\begin{array} {lll} \hat{E}(\Delta \tau) & = & {1 \over N} \... ...\tau_0)+u^\prime_1(t-\tau_0)\Delta \tau]^2 \right\}.\end{array}$$ (6)

The optimal estimation of $\Delta \tau$ is obtained by finding the minimizer of $\hat{E}(\Delta \tau)$. The standard least squares technique gives

 

$$\Delta \tau = -\displaystyle{{\sum^N_t\left\{ {1 \over \Vert... ...\over \Vert u^2_2(t) \Vert}[u^\prime_1(t-\tau_0)]^2 \right\}}}.$$ (7)

This solution looks similar in expression to the solution of the plane-wave destructor. It actually has two distinguished features. First it includes weighting functions. Second, it does not require to compute the partial derivatives of data with respect to the spatial axis, which is difficult to do in high accuracy when spatial-sample interval is large. The second feature comes from the fact that we estimate the relative time-shifts instead of the dips. The computation of the time-derivatives of data is not a problem as we have assumed that data is adequately sampled in the time axis.

Thus, the overall optimal relative time-shift is

 

$$\tau_{opt}=\tau_0+\Delta \tau$$ (8)

and the optimal dip is

$$p_{opt}={\tau_{opt} \over \Delta x}.$$ (9)

Clearly the linear optimization can be run iteratively. Examples with synthetic and field data show that first iteration provides a solution of sufficient accuracy.