Wavenumber 2D Eikonal equations

The system given by Eq.(A.4) can be written in cylindrical coordinates as[14]:
$\displaystyle \frac{d}{dr}\cos\theta$ $\textstyle =$ $\displaystyle \frac {1}{k} \frac{\partial{k}}{\partial{r}} \; \frac {\sin^2\theta}{\cos\theta} - \frac {1}{k} \frac{\partial{k}}{\partial{z}}\sin\theta  ,$ (A.5)
$\displaystyle \frac{d}{dr}\sin\theta$ $\textstyle =$ $\displaystyle \frac {1}{k} \frac{\partial{k}}{\partial{z}}\cos\theta - \frac {1}{k} \frac{\partial{k}}{\partial{r}}\sin\theta  .$ (A.6)

If integration over $z$ is required the system becomes:
$\displaystyle \frac{d}{dz}\cos\theta$ $\textstyle =$ $\displaystyle \frac {1}{k} \frac{\partial{k}}{\partial{r}}\sin\theta - \frac {1}{k} \frac{\partial{k}}{\partial{z}}\cos\theta  ,$ (A.7)
$\displaystyle \frac{d}{dz}\sin\theta$ $\textstyle =$ $\displaystyle \frac {1}{k} \frac{\partial{k}}{\partial{z}} \; \frac {\cos^2\theta}{\sin\theta} - \frac {1}{k} \frac{\partial{k}}{\partial{r}}\cos\theta  .$ (A.8)

Independently of integrating over $r$ or $z$ it holds valid in both cases that

\begin{displaymath}\frac{dz}{dr} = \frac {\sin\theta}{\cos\theta}  . \end{displaymath}



Orlando Camargo Rodríguez 2012-06-21