Formulas for amplitudes (3D case)

We shall propose that points ${\bf r}^\ast$, ${\bf r}$ and ${\bf r}^\ast_0$ lie on the same ray $\gamma$ of eiconal $\tau^{(-)}$ (Figure ). Then

$$\tau_0({\bf r}_0)-T({\bf r},{\bf r}_0)=t^\ast+\epsilon+{1\ov... ...{{\bf r}^\ast_0} -{\bf A}_{{\bf r}})({\bf r}_0-{\bf r}^\ast_0),$$

$$\epsilon = T({\bf r}^\ast,{\bf r}^\ast_0)-T({\bf r},{\bf r}^\ast_0),$$

$${\bf A}_{{\bf r}} = {{\partial^2 T({\bf r},{\bf r}_0)} \over {\partial x_i \partial x_k}} \ \ \ \ i,k=1,2$$

$${\bf B}_{{\bf r}^\ast_0} = {{\partial^2 \tau_0({\bf r}_0)} \over {\partial x_i \partial x_k}} \ \ \ \ i,k=1,2$$

After substituting into the equation (101) at ${\bf r}={\bf r}^\ast$ and $t=t^\ast$ and performing rather rigorous transformations, we obtain that

$$\lim_{{\bf r}\rightarrow {\bf r}^\ast} U^{(\pm)} ({\bf r},t)... ..._0 \over \sqrt{D^{(\pm)}_\ast}} \equiv A^{(\pm)} ({\bf r}^\ast)$$

where $w^\ast = w({\bf r}^\ast,{\bf r}^\ast_0), A^\ast_0 = A_0({\bf r}^\ast_0)$,

$$D^{(\pm)}_\ast = \det ({\bf A}_{{{\bf r}}^\ast} \pm {\bf B}_{{\bf r}^\ast_0}).$$

In Figure different locations of fronts $\tau^{(-)}=t_k\ (k=1,2, t_2\gt t^\ast \gt t_1)$ are shown for reverse (a) and direct (b) continuation.

It is easy to derive from the pictures that

$$U^{(\pm)}({\bf r}^\ast,t)\stackrel{0}\sim A^{(\pm)}({\bf r}^\ast) H[\pm(t-t^\ast)]$$

and finally

$$u^{(\pm)}({\bf r},t) \stackrel{q+1} \sim \pm A^{(\pm)}({\bf r}) R_{q+1}[\pm (t-\tau^{(\pm)} ({\bf r}))].$$