As I was reading a book on queue theory, the author needed to solve $p(V_b, V_a)$ where: $p(V_b, V_a) = \frac{1}{4}p(V_b+1, V_a) + \frac{1}{4}p(V_b, V_a+1) + \frac{1}{4}p(V_b-1, V_a) + \frac{1}{4}p(V_b, V_a-1) $
Using the Taylor Expansion $P(V\pm1) \approx P(V,t) \pm \partial_VP(V,t) + \frac{1}{2} \partial^2_{VV}P(V,t)$, the equation above can be simplified to $(\partial^2_{V_bV_b} + \partial^2_{V_aV_a})p(V_b,V_a) = 0 \text{ with boundary conditions: }p(V_b,0)=1, p(0,V_a)=0$.
To solve this Laplace's equation, the author suggests a change of variables to polar coordinates $(R, \theta)$.
Where $R = \sqrt{V_b^2 + V_a^2}, \text{ } \theta=arctan(\frac{V_b}{V_a})$.
The author however asserts that the Laplace's Equation above becomes: $(\partial^2_{RR} + \frac{1}{R^2}\partial^2_{\theta\theta})p(R,\theta) = 0$.
I am extremely confused as to why the author ignored the $\frac{1}{R}\partial_R$ in the Laplace equation above. Shouldn't the Laplace's Equation in polar coordinates be $(\partial^2_{RR} + \frac{1}{R}\partial_R + \frac{1}{R^2}\partial^2_{\theta\theta})p(R,\theta) = 0$ by definition?
Any help is extremely appreciated!
P.S. Added the original text below:
