Disc of radius c For a disk with a radius of "c", let the polar coordinates be 0 < r < c {\displaystyle 0<r<c} , and − π < θ < π {\displaystyle -\pi <\theta <\pi } r 2 u r r + r u r + u θ θ = 0 {\displaystyle r^{2}u_{rr}+ru_{r}+u_{\theta \theta }=0} u ( c , θ ) = f ( θ ) {\displaystyle u(c,\theta )=f(\theta )} , boundary condition.
u ( r , π ) = u ( r , − π ) {\displaystyle u(r,\pi )=u(r,-\pi )} continuity of potential. u θ ( r , π ) = u θ ( r , − π ) {\displaystyle u_{\theta }(r,\pi )=u_{\theta }(r,-\pi )} continuity of derivative.
u ( r , θ ) = R ( r ) Θ ( θ ) {\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:
r 2 R ″ Θ + r R ′ Θ + R Θ ″ = 0 {\displaystyle \displaystyle r^{2}R''\Theta +rR'\Theta +R\Theta ''=0}
Separate,r 2 R ″ + r R ′ R + Θ ″ Θ = 0 {\displaystyle \displaystyle {\frac {r^{2}R''+rR'}{R}}+{\frac {\Theta ''}{\Theta }}=0} r 2 R ″ + r R ′ R = − Θ ″ Θ = Constant {\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.
λ 2 > 0 {\displaystyle \displaystyle \lambda ^{2}>0} − Θ ″ Θ = λ 2 {\displaystyle \displaystyle -{\frac {\Theta ''}{\Theta }}=\lambda ^{2}} Θ ″ + λ 2 Θ = 0 {\displaystyle \displaystyle \Theta ''+\lambda ^{2}\Theta =0}
Θ ( θ ) = A c o s ( λ θ ) + B s i n ( λ θ ) {\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. u ( r , π ) = u ( r , − π ) {\displaystyle \displaystyle u(r,\pi )=u(r,-\pi )} thus Θ ( π ) = Θ ( − π ) {\displaystyle \displaystyle \Theta (\pi )=\Theta (-\pi )} and A c o s ( λ π ) + B s i n ( λ π ) = A c o s ( λ − π ) + B s i n ( λ − π ) {\displaystyle \displaystyle Acos(\lambda \pi )+Bsin(\lambda \pi )=Acos(\lambda -\pi )+Bsin(\lambda -\pi )} B s i n ( λ π ) = − B s i n ( λ π ) {\displaystyle \displaystyle Bsin(\lambda \pi )=-Bsin(\lambda \pi )} 2 B s i n ( λ π ) = 0 {\displaystyle \displaystyle 2Bsin(\lambda \pi )=0}
Either B = 0 {\displaystyle \displaystyle B=0} or s i n ( λ π ) = 0 {\displaystyle \displaystyle sin(\lambda \pi )=0}
Before choosing, apply the second boundary condition:
The continuity of the derivative provides a second condition: u θ ( r , π ) = u θ ( r , − π ) {\displaystyle \displaystyle u_{\theta }(r,\pi )=u_{\theta }(r,-\pi )} thus Θ θ ( π ) = Θ θ ( − π ) {\displaystyle \displaystyle \Theta _{\theta }(\pi )=\Theta _{\theta }(-\pi )} − A λ s i n ( λ π ) + λ B c o s ( λ π ) = − A λ s i n ( λ − π ) + B λ c o s ( λ − π ) {\displaystyle \displaystyle -A\lambda sin(\lambda \pi )+\lambda Bcos(\lambda \pi )=-A\lambda sin(\lambda -\pi )+B\lambda cos(\lambda -\pi )} − A λ s i n ( λ π ) = A λ s i n ( λ π ) {\displaystyle \displaystyle -A\lambda sin(\lambda \pi )=A\lambda sin(\lambda \pi )} 2 A λ s i n ( λ π ) = 0 {\displaystyle \displaystyle 2A\lambda sin(\lambda \pi )=0}
Either A = 0 {\displaystyle \displaystyle A=0} or s i n ( λ π ) = 0 {\displaystyle \displaystyle sin(\lambda \pi )=0}
If either A or B are zero then s i n ( λ π ) = 0 {\displaystyle \displaystyle sin(\lambda \pi )=0} also must hold. So all we need is s i n ( λ π ) = 0 {\displaystyle \displaystyle sin(\lambda \pi )=0} which implies λ = n {\displaystyle \displaystyle \lambda =n} . Remember s i n ( n π ) = 0 , n = 1 , 2 , . . . {\displaystyle \displaystyle sin(n\pi )=0,{\mbox{ }}n=1,2,...}