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TAKING THE STEP

We adopt the convention that components of a vector $ \mathbf{ u}$ range over the values of $ (u_t)$ , likewise for other vectors. Given the gradient direction $ \Delta\mathbf{ u}$ we need to know the residual change $ \Delta\mathbf{r}$ and a distance $ \alpha$ to go: $ \alpha\Delta\mathbf{r}$ and $ \alpha\Delta\mathbf{u}$ .

A two-term example demonstrates a required linearization.

$\displaystyle e^{\alpha\Delta U}$ $\displaystyle =$ $\displaystyle e^{\alpha(\Delta u_1 Z + \Delta u_2 Z^2)}$ (18)
$\displaystyle e^{\alpha\Delta U}$ $\displaystyle =$ $\displaystyle 1 + \alpha (\Delta u_1 Z +\Delta u_2 Z^2) + \alpha^2(\cdots)$ (19)
$\displaystyle {\rm FT}^{-1}\ e^{\alpha\Delta U}$ $\displaystyle =$ $\displaystyle (1 , \alpha \Delta u_1 , \alpha\Delta u_2 ) + \alpha^2(\cdots)$ (20)
$\displaystyle {\rm FT}^{-1}\ e^{\alpha\Delta U}$ $\displaystyle =$ $\displaystyle (1 , \alpha \Delta \mathbf{ u} ) + \alpha^2(\cdots)$ (21)

With that background, neglecting $ \alpha^2$ , and knowing the gradient $ \Delta\mathbf{ u}$ , let us work out the forward operator to find $ \Delta \mathbf{q}$ . Let ``$ \ast$ '' denote convolution.
$\displaystyle \mathbf{r} + \alpha \Delta \mathbf{r}$ $\displaystyle =$ $\displaystyle {\rm FT}^{-1}( D e^{ U +\alpha \Delta U})$ (22)
  $\displaystyle =$ $\displaystyle {\rm FT}^{-1} (De^ U e^{\alpha \Delta U})$ (23)
  $\displaystyle =$ $\displaystyle {\rm FT}^{-1} (De^ U )
\ast
{\rm FT}^{-1} ( e^{\alpha \Delta U})$ (24)
  $\displaystyle =$ $\displaystyle \mathbf{r} \ast ( 1 , \alpha \Delta \mathbf{u})$ (25)
  $\displaystyle =$ $\displaystyle \mathbf{r} + \alpha \, \mathbf{r} \ast \Delta \mathbf{u}$ (26)
$\displaystyle \Delta \mathbf{ r}$ $\displaystyle =$ $\displaystyle \mathbf{ r} \ \ast\ \Delta \mathbf{u}$ (27)
$\displaystyle \Delta q_t$ $\displaystyle =$ $\displaystyle g_t\ \Delta r_t$ (28)

It is pleasing that $ \Delta\mathbf{r}$ is proportional to $ \mathbf{r}$ . This might mean we can deal with a wide dynamic range within $ r_t$ . The convolution, a physical process, occurs in the physical domain which is only later gained to the statistical domain $ q_t$ . Naturally, the convolution may be done as a product in the frequency domain.

To minimize $ H(\mathbf{q}+\alpha\Delta\mathbf{q})$ express it as a Taylor series approximation to quadratic order. Minimizing yields

$\displaystyle \alpha$ $\displaystyle =$ $\displaystyle -\ \sum_t \Delta q_t H_t' \ \ /\ \ \sum_t (\Delta q_t)^2 H_t''$ (29)

Update $ q_t = q_t + \alpha \Delta q_t$ and $ U=U+\alpha \Delta U$ , optionally (Newton method) iterate (because the locations of the many Taylor series have changed slightly with the change in $ \mathbf{q}$ ).


next up previous [pdf]

Next: ALGORITHM Up: Claerbout et al.: Log Previous: THE GRADIENT

2012-05-10