Helmholtz Differential Equation Polar Coordinates

In 2-D Polar Coordinates, attempt Separation of Variables by writing

$$F(r, \theta) = R(r)\Theta(\theta),$$ (1)

then the Helmholtz Differential Equation becomes

$${d^2R\over dr^2}\Theta + {1\over r}{dR\over dr}\Theta + {1\over r^2} {d^2\Theta \over d\theta^2}R +k^2R\Theta = 0.$$ (2)

Divide both sides by $R\Theta$

$$\left({{r^2\over R}{d^2R\over dr^2}+ {r\over R}{dR\over dr}}... ...({{1\over \Theta } {d^2\Theta \over d\theta^2}+k^2}\right)= 0.$$ (3)

The solution to the second part of (3) must be periodic, so the differential equation is

$${d^2\Theta\over d\theta^2}{1\over\Theta}= -(k^2+m^2),$$ (4)

which has solutions

$$\Theta(\theta) = c_1e^{i\sqrt{k^2+m^2}\,\theta}+c_2e^{-i\sqrt{k^2+m^2}\,\theta}$$

$$= c_3\sin(\sqrt{k^2+m^2}\,\theta)+c_4\cos(\sqrt{k^2+m^2}\,\theta).$$ (5)

Plug (4) back into (3)

$$r^2R''+rR'-m^2R = 0.$$ (6)

This is an Euler Differential Equation with $\alpha \equiv 1$ and $\beta \equiv -m^2$. The roots are $r = \pm m$. So for $m = 0$, $r = 0$ and the solution is

$$R(r) = c_1+c_2\ln r.$$ (7)

But since $\ln r$ blows up at $r = 0$, the only possible physical solution is $R(r) = c_1$. When $m > 0$, $r = \pm m$, so

$$R(r) = c_1r^m+c_2r^{-m}.$$ (8)

But since $r^{-m}$ blows up at $r = 0$, the only possible physical solution is $R_m(r) = c_1r^m$. The solution for $R$ is then

$$R_m(r) = c_mr^m$$ (9)

for $m = 0$, 1, ...and the general solution is

$$F(r, \theta) = \sum_{m=0}^\infty [a_mr^m\sin (\sqrt{k^2+m^2}\,\theta)+b_mr^m\cos(\sqrt{k^2+m^2}\,\theta)].$$ (10)

References

Morse, P. M. and Feshbach, H. Methods of Theoretical Physics, Part I. New York: McGraw-Hill, pp. 502-504, 1953.