The acoustic wave equation

The starting point for the discussion of the ray tracing is given by the acoustic wave equation, which in the case of a watercolumn with a constant density can be written as

$$\mbox{\boldmath$\nabla$}^2 p - \frac{1}{c^2}\frac{\partial{^2 p}}{\partial{t^2}} = s(\mbox{$\mathbf{r}$}_0,t) ,$$ (2.1)

where $p(\mbox{$\mathbf{r}$},t)$ stands for the pressure of the acoustic wave, $s(\mbox{$\mathbf{r}$}_0,t)$ represents the signal transmitted by the acoustic source, $\mbox{$\mathbf{r}$}_0$ represents the source position and $\mbox{\boldmath$\nabla$}$ represents the nabla differential operator. Applying a Fourier transform to both sides of Eq.(2.1) one can obtain the so called Helmholtz equation:

$$\left[ \mbox{\boldmath$\nabla$}^2 + \frac{\omega ^2}{c^2} \r... ...{$\mathbf{r}$},\omega) = S(\mbox{$\mathbf{r}$}_0,\omega ) ,$$ (2.2)

where

$$P(\mbox{$\mathbf{r}$},\omega ) = \displaystyle{ \int\limits_{... ...fty}^{\infty}} p(\mbox{$\mathbf{r}$},t) e^{-i \omega t} dt ,$$

and

$$S(\mbox{$\mathbf{r}$}_0,\omega ) = \displaystyle{ \int\limits... ...y}^{\infty}} s(\mbox{$\mathbf{r}$}_0,t) e^{-i \omega t} dt .$$

Let's consider a plane wave-like approximation to the solution of Eq.(2.2) and write that [4]

$$P(\omega ) = A e^{-i\omega \tau} ,$$ (2.3)

where $A$ represents a slowly changing wave amplitude, and $\omega \tau$ stands for a rapidly evolving phase; the surfaces with constant $\omega \tau$ represent the wavefronts; analogously, the surfaces with constant $\tau$ are called timefronts. By placing Eq.(2.3) into the homogeneous form of Eq.(2.2), considering the high frequency approximation

$$\frac{ \mbox{\boldmath$\nabla$}^2A }{ A } \ll k^2$$

(where $k = \omega /c$) and separating the real and imaginary terms of the equation, one can obtain the Eikonal equation:

$$\left( \mbox{\boldmath$\nabla$}\tau \right) ^2 = \frac{1}{c^2}$$ (2.4)

and the transport equation:

$$2\left( \mbox{\boldmath$\nabla$}A \cdot \mbox{\boldmath$\nabla$}\tau \right) + A \mbox{\boldmath$\nabla$}^2 \tau = 0 .$$ (2.5)

The following sections will describe the solution of Eqs.(2.4)-(2.5).