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# 8.2: Three cases for λ

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As usual we consider the cases $$\lambda>0$$, $$\lambda<0$$ and $$\lambda=0$$ separately. Consider the $$\Phi$$ equation first, since this has the most restrictive explicit boundary conditions (8.1.4).

 $$\lambda=-a^2<0$$

We have to solve $\Phi'' = \alpha^2\Phi,$ which has as a solution $\Phi(\phi) = A \cos \alpha \phi + B \sin \alpha \phi.$ Applying the boundary conditions, we get \begin{aligned} A &= A \cos(2\alpha\pi)+ B\sin(2\alpha\pi), \\ B \alpha &= -A \alpha\sin(2\alpha\pi)+ B\alpha\cos(2\alpha\pi).\end{aligned} If we eliminate one of the coefficients from the equation, we get $A = A \cos(2\alpha\pi)-A\sin(2\alpha\pi)^2/(1-cos(2\alpha\pi))$ which leads to $\sin(2\alpha\pi)^2=-(1-cos(2\alpha\pi))^2,$ which in turn shows $2\cos(2\alpha\pi)=2,$ and thus we only have a non-zero solution for $$\alpha=n$$, an integer. We have found $\lambda_n=n^2,\;\;\Phi_n(\phi) = A_n \cos n \phi + B_n \sin n \phi.$

 $$\lambda=0$$

We have $\Phi'' = 0 .$ This implies that $\Phi = A\phi + B.$ The boundary conditions are satisfied for $$A=0$$, $\Phi_0(\phi)=B_n.$

 $$\lambda>0$$

The solution (hyperbolic sines and cosines) cannot satisfy the boundary conditions.

Now let me look at the solution of the $$R$$ equation for each of the two cases (they can be treated as one), $\rho^2 R''(\rho) + \rho R'(\rho) -n^2R(\rho)=0.$ Let us attempt a power-series solution (this method will be discussed in great detail in a future lecture) $R(\rho) =\rho^\alpha.$ We find the equation $\rho^\alpha[\alpha(\alpha-1)+\alpha^2-n^2]= \rho^\alpha[\alpha^2-n^2]=0$ If $$n\neq 0$$ we thus have two independent solutions (as should be) $R_n(\rho) = C\rho^{-n}+D\rho^n$ The term with the negative power of $$\rho$$ diverges as $$\rho$$ goes to zero. This is not acceptable for a physical quantity (like the temperature). We keep the regular solution, $R_n(\rho) = \rho^n.$ For $$n=0$$ we find only one solution, but it is not very hard to show (e.g., by substitution) that the general solution is $R_0(\rho) = C_0 + D_0\ln(\rho).$ We reject the logarithm since it diverges at $$\rho=0$$.