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Mathematics LibreTexts

8.2: Three cases for λ

  • Page ID
    8339
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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 \(\rho=0\).