classification of PDEs that are in Cylindrical coordiantes

Question

Please bear with me - it has been years since I've touched PDEs and recently gotten back into the swing due to having started on a graduate program in physics.

Hints only

The general form for a second order linear PDE is given by

(1) $A\frac{\partial^2 \phi}{\partial x^2} + B\frac{\partial^2 \phi}{\partial x \partial y} + C\frac{\partial^2 \phi}{\partial y^2} + D\frac{\partial \phi}{\partial x} + E\frac{\partial \phi}{\partial y} + F\phi + G = 0$

where $A, B, C$ are function of $x, y$.

I am asked to classify the equation (whether elliptic, parabolic or hyperbolic)

(2) $\frac{\partial^2 \phi}{\partial z^2} + \frac{\partial \phi}{\partial z} + \frac{\partial^2 \phi}{\partial r^2} + \left ( \frac{1}{r} \right )\frac{\partial \phi}{\partial \theta} = 0$

Thinking aloud:

The equation in $(2)$ appears to be in cylindrical coordinates. So it seems the general course of strategy is to convert them to cartesian before making comparison with $(1)$.

Am I thinking in the right direction?

Answer 1

The classification of the type of a second order linear PDE is based on the second order terms in it.

In the PDE (2), $$ A = 1, \ \ B = 0, \ \ C = 1 $$ where $A$, $B$ and $C$ are the coefficients of ${\partial^2 \phi \over \partial z^2}$, ${\partial^2 \phi \over \partial z \partial r}$ and ${\partial^2 \phi \over \partial r^2}$ respectively.

Hence, $$ \Delta = B^2 - 4 A C = 0 - 4 = -4 < 0 $$

Thus, the given PDE (2) is elliptic.