We have the differential wave equation
$$\dfrac{\partial^2}{\partial{r}^2}(r \psi) = \dfrac{1}{v^2} \dfrac{\partial^2}{\partial{t}^2}(r\psi).$$
If we assume cylindrical symmetry, then we have that
$$\psi(\vec{r}) = \psi(r, \theta, z) = \psi(r).$$
It is then said that the differential wave equation becomes
$$\dfrac{1}{r} \dfrac{\partial}{\partial{r}} \left( r – \dfrac{\partial{\psi}}{\partial{r}} \right) = \dfrac{1}{v^2} \dfrac{\partial^2{\psi}}{\partial{t}^2}.$$
How was this new differential wave equation derived? The above is from chapters 2.9-2.10 (pages 38-39) of Optics, fifth edition, by Hecht.
Best Answer
Your first equation (2.69, page 38) is in spherical coordinates (coming from section 2.9). The second (2.77, page 39) is in cyllindrical (section 2.10). I do not suggest converting from spherical to cyllindrical; for one, the $r$s don't even mean the same thing. Furthermore, the first equation is derived under the assumption of spherical symmetry (c.f. page 37). A a function with cyllindrical symmetry does not have spherical symmetry, and vice versa. So you cannot start from (2.69) to derive (2.77).
Instead convert from the easy looking and fully general (i.e. no symmetry assumptions) cartesian formula
$$\frac{\partial^{2} \psi}{\partial x^{2}}+\frac{\partial^{2} \psi}{\partial y^{2}}+\frac{\partial^{2} \psi}{\partial z^{2}}=\frac{1}{v^{2}} \frac{\partial^{2} \psi}{\partial t^{2}}\tag{2.60, page 36}$$
or equivalently the "coordinate-free" version $$ \nabla^{2} \psi=\frac{1}{v^{2}} \frac{\partial^{2} \psi}{\partial t^{2}} \tag{2.62, page 36}$$ to either spherical (page 37, figure 2.26) or cyllindrical coordinates (page 39, figure 2.30), as in e.g.
In the case of section 2.10, the above leads you to $$\nabla^{2} \psi=\frac{1}{r} \frac{\partial}{\partial r}\left(r \frac{\partial \psi}{\partial r}\right)+\frac{1}{r^{2}} \frac{\partial^{2} \psi}{\partial \theta^{2}}+\frac{\partial^{2} \psi}{\partial z^{2}}\tag{2.76, page 39}$$ For $\psi$ independent of $\theta,z$, you thus get the equation $$ \frac{1}{r} \frac{\partial}{\partial r}\left(r \frac{\partial \psi}{\partial r}\right) =\frac{1}{v^{2}} \frac{\partial^{2} \psi}{\partial t^{2}}\tag{**not** 2.77, page 39}$$ where I can only guess the minus sign in (2.77) is a mistake (in a 5th edition? Is anyone reading?) Second source agrees its a typo.