${\displaystyle \displaystyle u(r,\theta )=R(r)\Theta (\theta )}$

The solution as a product of two independent functions. By substitution into the above PDE we have:

${\displaystyle \displaystyle r^{2}R''\Theta +rR'\Theta +R\Theta ''=0}$

Separate,

${\displaystyle \displaystyle {\frac {r^{2}R''+rR'}{R}}+{\frac {\Theta ''}{\Theta }}=0}$
${\displaystyle \displaystyle {\frac {r^{2}R''+rR'}{R}}=-{\frac {\Theta ''}{\Theta }}={\mbox{Constant}}}$

The constant may be greater than , equal to or less than zero.

• ${\displaystyle \displaystyle \lambda ^{2}>0}$
  

${\displaystyle \displaystyle -{\frac {\Theta ''}{\Theta }}=\lambda ^{2}}$

${\displaystyle \displaystyle \Theta ''+\lambda ^{2}\Theta =0}$

${\displaystyle \displaystyle \Theta (\theta )=Acos(\lambda \theta )+Bsin(\lambda \theta )}$
Use the continuity conditions and try to determine something more about A, B and λ.
${\displaystyle \displaystyle u(r,\pi )=u(r,-\pi )}$ thus ${\displaystyle \displaystyle \Theta (\pi )=\Theta (-\pi )}$ and ${\displaystyle \displaystyle Acos(\lambda \pi )+Bsin(\lambda \pi )=Acos(\lambda -\pi )+Bsin(\lambda -\pi )}$
${\displaystyle \displaystyle Bsin(\lambda \pi )=-Bsin(\lambda \pi )}$
${\displaystyle \displaystyle 2Bsin(\lambda \pi )=0}$
Either ${\displaystyle \displaystyle B=0}$ or ${\displaystyle \displaystyle sin(\lambda \pi )=0}$
Before choosing, apply the second boundary condition:

The continuity of the derivative provides a second condition:
${\displaystyle \displaystyle u_{\theta }(r,\pi )=u_{\theta }(r,-\pi )}$ thus ${\displaystyle \displaystyle \Theta _{\theta }(\pi )=\Theta _{\theta }(-\pi )}$
${\displaystyle \displaystyle -A\lambda sin(\lambda \pi )+\lambda Bcos(\lambda \pi )=-A\lambda sin(\lambda -\pi )+B\lambda cos(\lambda -\pi )}$
${\displaystyle \displaystyle -A\lambda sin(\lambda \pi )=A\lambda sin(\lambda \pi )}$
${\displaystyle \displaystyle 2A\lambda sin(\lambda \pi )=0}$
Either ${\displaystyle \displaystyle A=0}$ or ${\displaystyle \displaystyle sin(\lambda \pi )=0}$
If either A or B are zero then ${\displaystyle \displaystyle sin(\lambda \pi )=0}$ also must hold. So all we need is ${\displaystyle \displaystyle sin(\lambda \pi )=0}$ which implies ${\displaystyle \displaystyle \lambda =n}$. Remember ${\displaystyle \displaystyle sin(n\pi )=0,{\mbox{ }}n=1,2,...}$