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?


1 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.

  • $\begingroup$ OK, I agree with everything BUT you did not address the main question. Do you need to convert to Cartesian coordinates before you can classify? In this case, the equation in Cartesian coordinates reads $\Delta u + \text{lower order terms}=0$, so it is elliptic as you rightly state. $\endgroup$ Apr 13, 2022 at 11:26

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