E. The Legendre equation

When solving the equation for the geometry in spherical coordinates, the following equation for the polar angle arises

where $\Theta(z)$ is the unknown function, $m$ is a nonnegative integer, $\gamma_1^2$ is the separation constant. Let us expand the derivative in the first term

Note that for $m$​${} = 0$ the equation takes the form of the Legendre equation

This equation has solutions bounded on the interval ${} -1$​${} < z$​${} < 1$ only for the eigenvalues $\gamma_{1k}^2$​${} = k \cdot (k + 1)$, $k$​${} \in (0..\infty)$; these solutions are given by the Rodrigues formula

where $P_k(z)$ are the Legendre polynomials. We will look for solutions of equation (E.1$\displaystyle \frac{\displaystyle d}{\displaystyle dz} \left( (1 - z^2) \cdot \frac{\displaystyle d \Theta(z)}{\displaystyle dz} \right)$​$\displaystyle {} + \left[ \gamma_1^2 - \frac{\displaystyle m^2}{\displaystyle 1 - z^2} \right] \cdot \Theta(z)$​$\displaystyle {} = 0,$​$\displaystyle {} -1$​$\displaystyle {} < z$​$\displaystyle {} < 1,$) for the same eigenvalues $\gamma_{1k}^2$​${} = k \cdot (k + 1)$ — then it can be written as

Let us make the substitution $\Theta_{km}(z)$​${} = \left( 1 - z^2 \right)^{\frac{\displaystyle m}{\displaystyle 2}} \cdot \widehat{\Theta}_{km}(z)$ into equation (E.3$\displaystyle (1 - z^2) \cdot \frac{\displaystyle d^2 \Theta_{km}(z)}{\displaystyle dz^2}$​$\displaystyle {} - 2 \cdot z \cdot \frac{\displaystyle d \Theta_{km}(z)}{\displaystyle dz}$​$\displaystyle {} + \left[ k \cdot (k + 1) - \frac{\displaystyle m^2}{\displaystyle 1 - z^2} \right]$​$\displaystyle {} \cdot \Theta_{km}(z)$​$\displaystyle {} = 0,$​$\displaystyle k$​$\displaystyle {} \in (0..\infty).$). The first derivative takes the form

To compute the second derivative, we first compute

Then the second derivative can be written as

After collecting like terms it takes the form

Let us substitute everything into equation (E.3$\displaystyle (1 - z^2) \cdot \frac{\displaystyle d^2 \Theta_{km}(z)}{\displaystyle dz^2}$​$\displaystyle {} - 2 \cdot z \cdot \frac{\displaystyle d \Theta_{km}(z)}{\displaystyle dz}$​$\displaystyle {} + \left[ k \cdot (k + 1) - \frac{\displaystyle m^2}{\displaystyle 1 - z^2} \right]$​$\displaystyle {} \cdot \Theta_{km}(z)$​$\displaystyle {} = 0,$​$\displaystyle k$​$\displaystyle {} \in (0..\infty).$), cancelling along the way $(1 - z^2)^{\frac{\displaystyle m}{\displaystyle 2}}$:

Let us group the coefficients of the derivatives:

thus, after the substitution, equation (E.3$\displaystyle (1 - z^2) \cdot \frac{\displaystyle d^2 \Theta_{km}(z)}{\displaystyle dz^2}$​$\displaystyle {} - 2 \cdot z \cdot \frac{\displaystyle d \Theta_{km}(z)}{\displaystyle dz}$​$\displaystyle {} + \left[ k \cdot (k + 1) - \frac{\displaystyle m^2}{\displaystyle 1 - z^2} \right]$​$\displaystyle {} \cdot \Theta_{km}(z)$​$\displaystyle {} = 0,$​$\displaystyle k$​$\displaystyle {} \in (0..\infty).$) takes the form

Let us take equation (E.3$\displaystyle (1 - z^2) \cdot \frac{\displaystyle d^2 \Theta_{km}(z)}{\displaystyle dz^2}$​$\displaystyle {} - 2 \cdot z \cdot \frac{\displaystyle d \Theta_{km}(z)}{\displaystyle dz}$​$\displaystyle {} + \left[ k \cdot (k + 1) - \frac{\displaystyle m^2}{\displaystyle 1 - z^2} \right]$​$\displaystyle {} \cdot \Theta_{km}(z)$​$\displaystyle {} = 0,$​$\displaystyle k$​$\displaystyle {} \in (0..\infty).$), set $m$​${} = 0$ and take into account that in this case its solutions are the Legendre polynomials $P_k(z)$ — we obtain the following equation

which we differentiate $m$ times using the Leibniz formula, defined as follows

where $\binom{m}{i}$​${} = \frac{\displaystyle m!}{\displaystyle i! \cdot (m - i)!}$ is the binomial coefficient.

Let us start with the first term of equation (E.5$\displaystyle (1 - z^2) \cdot \frac{\displaystyle d^2 P_k(z)}{\displaystyle dz^2}$​$\displaystyle {} - 2 \cdot z \cdot \frac{\displaystyle d P_k(z)}{\displaystyle dz}$​$\displaystyle {} + k \cdot (k + 1) \cdot P_k(z)$​$\displaystyle {} = 0,$), taking $v(z)$​${} = 1$​${} - z^2$

now let us take $v(z)$​${} = -2 \cdot z$, we obtain

for the case $v(z)$​${} = k \cdot (k + 1)$ everything is trivial

The binomial coefficients are respectively equal to

The result of differentiating the first term of equation (E.5$\displaystyle (1 - z^2) \cdot \frac{\displaystyle d^2 P_k(z)}{\displaystyle dz^2}$​$\displaystyle {} - 2 \cdot z \cdot \frac{\displaystyle d P_k(z)}{\displaystyle dz}$​$\displaystyle {} + k \cdot (k + 1) \cdot P_k(z)$​$\displaystyle {} = 0,$) equals

for the second term

for the third term

let us add all three terms

The left-hand side of equation (E.5$\displaystyle (1 - z^2) \cdot \frac{\displaystyle d^2 P_k(z)}{\displaystyle dz^2}$​$\displaystyle {} - 2 \cdot z \cdot \frac{\displaystyle d P_k(z)}{\displaystyle dz}$​$\displaystyle {} + k \cdot (k + 1) \cdot P_k(z)$​$\displaystyle {} = 0,$) is identically zero, hence so is its $m$-fold derivative:

The resulting equation coincides with equation (E.4$\displaystyle (1 - z^2) \cdot \frac{\displaystyle d^2 \widehat{\Theta}_{km}(z)}{\displaystyle dz^2}$​$\displaystyle {} - 2 \cdot (m + 1) \cdot z \cdot \frac{\displaystyle d \widehat{\Theta}_{km}(z)}{\displaystyle dz}$​$\displaystyle {} + (k - m) \cdot (k + m + 1) \cdot \widehat{\Theta}_{km}(z)$​$\displaystyle {} = 0.$), whence $\widehat{\Theta}_{km}(z)$​${} = \frac{\displaystyle d^m}{\displaystyle dz^m} P_k(z)$, which means the solution of equation (E.1$\displaystyle \frac{\displaystyle d}{\displaystyle dz} \left( (1 - z^2) \cdot \frac{\displaystyle d \Theta(z)}{\displaystyle dz} \right)$​$\displaystyle {} + \left[ \gamma_1^2 - \frac{\displaystyle m^2}{\displaystyle 1 - z^2} \right] \cdot \Theta(z)$​$\displaystyle {} = 0,$​$\displaystyle {} -1$​$\displaystyle {} < z$​$\displaystyle {} < 1,$) has the form

where $\gamma_{1k}^2$​${} = k \cdot (k + 1)$ are the eigenvalues, $P_k^{(m)}(z)$ are the associated Legendre polynomials. For $m$​${} > k$ the derivative $\frac{\displaystyle d^m}{\displaystyle dz^m} P_k(z)$ vanishes, so nontrivial solutions exist only for $k$​${} \ge m$.