Revisiting the Concentration Problem of Vector Fields within a Spherical Cap: A Commuting Differential Operator Solution

Abstract

We propose a novel basis of vector functions, the mixed vector spherical harmonics that are closely related to the functions of Sheppard and Török and help us reduce the concentration problem of tangential vector fields within a spherical cap to an equivalent scalar problem. Exploiting an analogy with previous results published by Grünbaum, Longhi and Perlstadt, we construct a differential operator that commutes with the concentration operator of this scalar problem and propose a stable and convenient method to obtain its eigenfunctions. Having obtained the scalar eigenfunctions, the calculation of tangential vector Slepian functions is straightforward.

Appendix 1: Proofs

1.1 Appendix 1(a): Proof of Recurrence Relation (12)

Proof

First we construct equivalent formulations of $F_{lm}$ by inserting the following recurrence relations, which can be obtained in a straightforward way from the corresponding relations for $P_{l}^{m}$ [3, p. 744], into (1):

$$\begin{aligned} (1 - x^2) \frac{\mathrm{d }\!^{}U_{lm}(x)}{\mathrm{d }\!x^{}}&= -l x U_{lm}(x) + (2l + 1) \xi _{lm} U_{l-1,\,m}(x), \end{aligned}$$

(87)

$$\begin{aligned} (1 - x^2) \frac{\mathrm{d }\!^{}U_{lm}(x)}{\mathrm{d }\!x^{}}&= (l + 1) x U_{lm}(x) - (2l + 1) \xi _{l+1,\,m} U_{l+1,\,m}(x) \end{aligned}$$

(88)

with

$$\begin{aligned} \xi _{lm}&:=\sqrt{ \frac{(l + m)(l - m)}{(2l + 1)(2l - 1)} }. \end{aligned}$$

(89)

Thus we obtain

$$\begin{aligned} F_{lm}(x)&= \frac{-(lx + m) U_{lm}(x) + (2l + 1) \xi _{lm} U_{l-1,\,m}(x)}{\sqrt{l(l + 1)} \sqrt{1 - x^2}},\end{aligned}$$

(90)

$$\begin{aligned} F_{lm}(x)&= \frac{\bigl [ (l + 1)x - m \bigr ] U_{lm}(x) - (2l + 1) \xi _{l+1,\,m} U_{l+1,\,m}(x)}{\sqrt{l(l + 1)} \sqrt{1 - x^2}}. \end{aligned}$$

(91)

Using these expressions of $F_{lm}$ and recurrence relation [3, p. 744]

$$\begin{aligned} x U_{lm}(x) = \xi _{lm} U_{l-1,\,m}(x) + \xi _{l+1,\,m} U_{l+1,\,m}(x), \end{aligned}$$

(92)

we transform the left-hand side (LHS) and right-hand side (RHS) separately so that only terms containing $U_{lm}$ and $U_{l-1,\,m}$ remain.

First we rewrite the LHS by inserting (90):

$$\begin{aligned} \begin{aligned} \text {LHS}&\!=\! \left[ x \!-\! \frac{m}{l(l \!+\! 1)} \right] F_{lm}(x) \!=\! \left[ x \!-\! \frac{m}{l(l \!+\! 1)} \right] \frac{\!-\!(lx \!+\! m) U_{lm}(x) \!+\! (2l \!+\! 1) \xi _{lm} U_{l\!-\!1,\,m}(x)}{\sqrt{l(l \!+\! 1)} \sqrt{1 \!-\! x^2}} \\&= -\frac{(lx + m) \bigl [l(l + 1)x - m \bigr ]}{\bigl [ l(l + 1) \bigr ]^{3/2} \sqrt{1 - x^2}} U_{lm}(x) + \frac{(2l + 1) \bigl [l(l + 1)x - m \bigr ]}{\bigl [ l(l + 1) \bigr ]^{3/2} \sqrt{1 - x^2}} \xi _{lm} U_{l-1,\,m}(x) . \end{aligned} \end{aligned}$$

After that we proceed to the RHS. We insert (90) and (91), shifted in index $l$ by $+1$ and $-1$, respectively:

$$\begin{aligned} \begin{aligned} \text {RHS}&= \zeta _{l+1,\,m} F_{l+1,\,m}(x) + \zeta _{lm} F_{l-1,\,m}(x) \\&= \zeta _{l+1,\,m} \frac{ -\bigl [ (l + 1)x + m \big ] U_{l+1,\,m}(x) + (2l + 3) \xi _{l+1,\,m} U_{lm}(x)}{\sqrt{(l + 1)(l + 2)} \sqrt{1 - x^2}}\\&\quad + \zeta _{lm} \frac{ (lx - m) U_{l-1,\,m}(x) - (2l - 1) \xi _{lm} U_{lm}(x)}{\sqrt{(l - 1)l} \sqrt{1 - x^2}} \end{aligned} \end{aligned}$$

Next we expand $\zeta _{l+1,\,m}$ and $\zeta _{lm}$ using their definition (15) and use the relation

$$\begin{aligned} \zeta _{lm} = \frac{\sqrt{(l + 1)(l - 1)}}{l}\, \xi _{lm} \end{aligned}$$

(93)

which follows from definitions (15) and (89). By straighforward, if lengthy, algebraic calculation, we get

$$\begin{aligned} \text {RHS} \!=\! \frac{\!-\!l^2 \bigl [ (l \!+\! 1)x \!+\! m \big ] \xi _{l\!+\!1,\,m} U_{l\!+\!1,\,m}(x) \!+\! (l + 1)^2 (lx \!-\! m) \xi _{lm} U_{l\!-\!1,\,m}(x) \!+\! m^2 U_{lm}(x)}{\bigl [ l(l + 1) \bigr ]^{3/2} \sqrt{1 - x^2}}. \end{aligned}$$

We apply recurrence relation (12) and collect like terms, hence

$$\begin{aligned} \begin{aligned} \text {RHS}&\!=\! \frac{\!-\!l^2 x \bigl [ (l \!+\! 1)x \!+\! m \big ] \!+\! m^2}{\bigl [ l(l \!+\! 1) \bigr ]^{3/2} \sqrt{1 \!-\! x^2}} U_{lm}(x) \!+\! \frac{l^2 \bigl [ (l \!+\! 1)x \!+\! m \big ] \!+\! (l \!+\! 1)^2 (lx \!-\! m)}{\bigl [ l(l \!+\! 1) \bigr ]^{3/2} \sqrt{1 \!-\! x^2}} \xi _{lm} U_{l-1,m}(x). \end{aligned} \end{aligned}$$

Taking the difference $\text {LHS} - \text {RHS}$, it can be shown by further straightforward algebra that the coefficients of $U_{lm}$ and $U_{l-1,\,m}$ are zero. Hence $\text {LHS} = \text {RHS}$.

1.2 Appendix 1(b): Proof of Recurrence Relation (13)

Proof

In this proof, we follow the same strategy as in the previous proof and rewrite the left-hand side (LHS) first. Inserting expression (90) of $F_{lm}$ from the previous proof yields

$$\begin{aligned} \text {LHS} \!=\! (1 \!-\! x^2) \frac{\mathrm{d }\!^{}F_{lm}(x)}{\mathrm{d }\!x^{}} \!=\! \frac{1}{\sqrt{l(l \!+\! 1)}} (1 \!-\! x^2) \frac{\mathrm{d }\!^{}}{\mathrm{d }\!x^{}} \left[ \frac{ \!-\!(lx \!+\! m) U_{lm}(x) \!+\! (2l \!+\! 1) \xi _{lm} U_{l-1,\,m}(x) }{ \sqrt{1 \!-\! x^2}} \right] . \end{aligned}$$

Performing the differentiation and using the relation $(1 - x^2) \frac{\mathrm{d }\!^{}}{\mathrm{d }\!x^{}} (1 - x^2)^{-1/2} = x (1 - x^2)^{-1/2}$, we get

$$\begin{aligned} \begin{aligned} \text {LHS} \!=\! \Biggl [&\!-\!(lx \!+\! m)x U_{lm}(x) \!+\! (2l \!+\! 1) \xi _{lm} x U_{l-1,\,m}(x) \!-\! l(1 \!-\! x^2) U_{lm}(x)\\&\!-\! (lx \!+\! m)(1 \!-\! x^2) \frac{\mathrm{d }\!^{}U_{lm}(x)}{\mathrm{d }\!x^{}} \!+\! (2l \!+\! 1) \xi _{lm} (1 \!-\! x^2) \frac{\mathrm{d }\!^{}U_{l\!-\!1,\,m}(x)}{\mathrm{d }\!x^{}} \Biggr ] \times \bigl [ l(l \!+\! 1)(1 \!-\! x^2) \bigr ]^{\!-\!1/2}. \end{aligned} \end{aligned}$$

Next we insert recurrence relations (87) and (88), shifted in index $l$ by $+1$ and $-1$, respectively. After that we collect like terms and perform some straightforward algebra to obtain

$$\begin{aligned} \text {LHS} \!=\! \frac{ (lx \!+\! m)(l \!-\! 1)x \!-\! l(1 \!-\! x^2) \!-\! l^2 \!+\! m^2}{ \sqrt{l (l \!+\! 1)} \sqrt{1 \!-\! x^2}} U_{lm}(x) \!+\! \frac{ (2l \!+\! 1)(x \!-\! m)}{ \sqrt{l (l \!+\! 1)} \sqrt{1 \!-\! x^2}} \xi _{lm} U_{l\!-\!1,\,m}(x). \end{aligned}$$

Now we rewrite the right-hand side (RHS). We insert expressions (90) and (91) of $F_{lm}$, the second one shifted in index $l$ by $+1$.

$$\begin{aligned} \begin{aligned} \text {RHS}&= -l \left( x - \frac{m}{l^2} \right) F_{lm}(x) + (2l + 1) \zeta _{lm} F_{l-1,\,m}(x) \\&= \frac{(m/l - l x) \bigl [ -(lx + m) U_{lm}(x) + (2l + 1) \xi _{lm} U_{l-1,\,m}(x) \bigr ]}{\sqrt{l(l + 1)} \sqrt{1 - x^2}}\\&\quad \ + \zeta _{lm} \frac{(2l + 1) \bigl [ (lx - m) U_{l-1,\,m}(x) - (2l - 1) \xi _{lm} U_{lm}(x) \bigr ] }{\sqrt{(l - 1)l} \sqrt{1 - x^2}}. \end{aligned} \end{aligned}$$

Next we substitute definition (15) of $\zeta _{lm}$, use (93) and collect like terms. By straightforward algebra we get

$$\begin{aligned} \text {RHS} \!=\! \frac{(lx \!+\! m)(l^2 x \!-\! m) \!-\! (l \!+\! 1)(l^2 \!-\! m^2)}{l \sqrt{l(l \!+\! 1)} \sqrt{1 \!-\! x^2}} U_{lm}(x) \!+\! \frac{(2l \!+\! 1)(x \!-\! m)}{\sqrt{l(l \!+\! 1)} \sqrt{1 \!-\! x^2}} \xi _{lm} U_{l\!-\!1,\,m}(x). \end{aligned}$$

Taking the difference $\text {LHS} - \text {RHS}$, the terms containing $U_{l-1,\,m}$ cancel. It can be shown that the coefficient of $U_{lm}$ is zero as well, hence $\text {LHS} = \text {RHS}$.

1.3 Appendix 1(c): Proof of Christoffel–Darboux Formula (16)

Proof

We start from recurrence relation (12) and multiply both sides by $F_{lm}(x')$. Then we take the same recurrence relation again, but this time, substitute $x'$ for $x$ and multiply both sides by $F_{lm}(x)$. In this way, we obtain the following two equations:

$$\begin{aligned} \begin{aligned} \left[ x - \frac{m}{l(l + 1)} \right] F_{lm}(x) F_{lm}(x')&= \zeta _{l+1,\,m} F_{l+1,\,m}(x) F_{lm}(x') + \zeta _{lm} F_{l-1,\,m}(x) F_{lm}(x'), \\ \left[ x' - \frac{m}{l(l + 1)} \right] F_{lm}(x') F_{lm}(x)&= \zeta _{l+1,\,m} F_{l+1,\,m}(x') F_{lm}(x) + \zeta _{lm} F_{l-1,\,m}(x') F_{lm}(x). \end{aligned} \end{aligned}$$

Taking their difference and summing over $l$ yields

$$\begin{aligned} \begin{aligned} (x - x') \sum _{l=\ell _{m}}^{L} F_{lm}(x) F_{lm}(x') = \sum _{l=\ell _{m}}^{L}&\Bigl \{ \zeta _{l+1,\,m} \bigl [ F_{l+1,\,m}(x) F_{lm}(x') - F_{lm}(x) F_{l+1,\,m}(x') \bigr ] \\&+ \zeta _{lm} \bigl [ F_{l-1,\,m}(x) F_{lm}(x') - F_{lm}(x) F_{l-1,\,m}(x') \bigr ] \Bigr \}. \end{aligned} \end{aligned}$$

We can see that consecutive terms cancel in the sum on the right-hand side. Moreover, $F_{\ell _{m}-1,\,m} = 0$, thus only one term corresponding to $\zeta _{L+1,\,m}$ remains:

$$\begin{aligned} (x - x') \sum _{l=\ell _{m}}^{L} F_{lm}(x) F_{lm}(x') = \zeta _{L+1,\,m} \bigl [ F_{L+1,\,m}(x) F_{Lm}(x') - F_{Lm}(x) F_{L+1,\,m}(x') \bigr ]. \end{aligned}$$

$\square$

1.4 Appendix 1(d): Proof of Addition Theorem (17)

Proof

Upon inserting definition (1) of $F_{lm}$ into the left-hand side of addition theorem (17), we obtain

$$\begin{aligned} \sum _{m=-l}^{l} \bigl [ F_{lm}(x) \bigr ]^2&= \frac{1}{l(l+1)} \left\{ \sum _{m=-l}^{l} (1 - x^2) \left[ \frac{\mathrm{d }\!^{}U_{lm}(x)}{\mathrm{d }\!x^{}} \right] ^2 + \frac{1}{1-x^2} \sum _{m=-l}^{l} \bigl [ U_{lm}(x) \bigr ]^2 \right\} \nonumber \\&\quad +\, \frac{2}{l(l+1)} \sum _{m=-l}^{l} m \frac{\mathrm{d }\!^{}U_{lm}(x)}{\mathrm{d }\!x^{}} U_{lm}(x). \end{aligned}$$

(94)

Next we use the addition theorems

$$\begin{aligned} (1 - x^2) \sum _{m=-l}^{l} \left[ \frac{\mathrm{d }\!^{}U_{lm}(x)}{\mathrm{d }\!x^{}} \right] ^2 = \frac{l(l + 1)(2l + 1)}{4}, \end{aligned}$$

(95a)

$$\begin{aligned} \frac{1}{1 - x^2} \sum _{m=-l}^{l} \bigl [ m U_{lm}(x) \bigr ]^2 = \frac{l(l + 1)(2l + 1)}{4}. \end{aligned}$$

(95b)

based on the work of Winch and Roberts [36].

Hence the first term on the right-hand side of Eq. (94) yields $(2l + 1)/2$, while the second term vanishes because of symmetry relation (9). Therefore,

$$\begin{aligned} \sum _{m=-l}^{l} \bigl [ F_{lm}(x) \bigr ]^2 = \frac{2l + 1}{2}. \end{aligned}$$

$\square$

1.5 Appendix 1(e): Proof of $F_{lm}$ Satisfying Differential Equation (23)

Proof

First let us rearrange (23) and insert $F_{lm}$:

$$\begin{aligned} \frac{\mathrm{d }\!^{}}{\mathrm{d }\!x^{}} \left[ (1 - x^2) \frac{\mathrm{d }\!^{}F_{lm}(x)}{\mathrm{d }\!x^{}} \right] = \left[ - l(l + 1) + \frac{m^2 - 2mx + 1}{1 - x^2} \right] F_{lm}(x). \end{aligned}$$

(96)

The left-hand side can be transformed by exploiting recurrence relations (13) and (14) (the second one shifted in index $l$ by $-1$) as follows:

$$\begin{aligned} \begin{aligned}&\frac{\mathrm{d }\!^{}}{\mathrm{d }\!x^{}} \left[ (1 - x^2) \frac{\mathrm{d }\!^{}F_{lm}(x)}{\mathrm{d }\!x^{}} \right] \\&\quad = \frac{\mathrm{d }\!^{}}{\mathrm{d }\!x^{}} \left[ -l\left( x - \frac{m}{l^2} \right) F_{lm}(x) + (2l + 1) \zeta _{lm} F_{l-1,\,m}(x) \right] \\&\quad = \biggl [ -l^2 (1 - x^2) F_{lm}(x) - (l^2x - m) (1 - x^2) \frac{\mathrm{d }\!^{}F_{lm}(x)}{\mathrm{d }\!x^{}} \\&\quad \quad +\, l(2l + 1) \zeta _{lm} (1 - x^2) \frac{\mathrm{d }\!^{}F_{l-1,\,m}(x)}{\mathrm{d }\!x^{}} \biggr ] \times \bigl [ l(1 - x^2) \bigr ]^{-1} \\&\quad \!=\! \biggl \{ \!-\!l^2 (1 \!-\! x^2) F_{lm}(x) \!-\! (l^2x \!-\! m) \left[ \!-\!l \left( x \!-\! \frac{m}{l^2}\right) F_{lm}(x) \!+\! (2l \!+\! 1) \zeta _{lm} F_{l\!-\!1,\,m}(x) \right] \\&\quad \quad \!+\, l(2l \!+\! 1) \zeta _{lm} \left[ l \left( x \!-\! \frac{m}{l^2}\right) F_{l\!-\!1,\,m}(x) \!-\! (2l \!-\! 1) \zeta _{lm} F_{lm}(x) \right] \biggr \} \times \bigl [ l(1 \!-\! x^2) \bigr ]^{\!-\!1}. \end{aligned} \end{aligned}$$

Collecting like terms yields

$$\begin{aligned} \begin{aligned} \frac{\mathrm{d }\!^{}}{\mathrm{d }\!x^{}} \left[ (1 \!-\! x^2) \frac{\mathrm{d }\!^{}F_{lm}(x)}{\mathrm{d }\!x^{}} \right] \!=\!&\frac{\!-\!l^2 (1 - x^2) \!+\! l (l^2x \!-\! m) (x \!-\! m/l^2) \!-\! l (2l \!-\! 1)(2l \!+\! 1) \zeta _{lm}^2}{l(1 \!-\! x^2)} F_{lm}(x) \\&\!+\, \frac{\!-\!(2l \!+\! 1)(l^2x \!-\! m) \!+\! (2l \!+\! 1) l^2 (x \!-\! m/l^2)}{l(1 \!-\! x^2)} \zeta _{lm} F_{l\!-\!1,\,m}(x). \end{aligned} \end{aligned}$$

As expected, the term containing $F_{l-1,\,m}$ vanishes. Applying definition (15) of $\zeta _{lm}$ and expanding the fraction by $l$ yields

$$\begin{aligned} \frac{\mathrm{d }\!^{}}{\mathrm{d }\!x^{}} \left[ (1 - x^2) \frac{\mathrm{d }\!^{}F_{lm}(x)}{\mathrm{d }\!x^{}} \right] = \frac{-l^3 (1 - x^2) + (l^2x - m)^2 - (l^2 - 1)(l^2 - m^2)}{l^2 (1 - x^2)} F_{lm}(x). \end{aligned}$$

By a straightforward, if lengthy, simplification we obtain

$$\begin{aligned} \frac{\mathrm{d }\!^{}}{\mathrm{d }\!x^{}} \left[ (1 - x^2) \frac{\mathrm{d }\!^{}F_{lm}(x)}{\mathrm{d }\!x^{}} \right] = \left[ -l(l+1) + \frac{m^2 - 2mx + 1}{1 - x^2} \right] F_{lm}(x). \end{aligned}$$

$\square$

1.6 Appendix 1(f): Proof of Integral Identity (76)

Proof

Inserting expression (75) of $\mathcal {J}_{m}$ into both sides of integral identity (76) yields

$$\begin{aligned} \int \limits _{\cos \Theta }^{1} u_{1}(x) \left[ \mathcal {J}_{m} u_{2}(x) \right] \mathrm{d }\!x&= \int \limits _{\cos \Theta }^{1} u_{1}(x) \frac{\mathrm{d }\!^{}}{\mathrm{d }\!x^{}} \left[ (\cos \Theta - x) (1 - x^2) \frac{\mathrm{d }\!^{}u_{2}(x)}{\mathrm{d }\!x^{}} \right] \mathrm{d }\!x \nonumber \\&\quad \!+\! \int \limits _{\cos \Theta }^{1} \left[ L(L+2) u_{1}(x) u_{2}(x) \!-\! (\cos \Theta \!-\! x) \frac{m^2 \!-\! 2mx \!+\! 1}{1 \!-\! x^2} u_{1}(x) u_{2}(x) \right] \mathrm{d }\!x, \nonumber \\ \end{aligned}$$

(97a)

$$\begin{aligned} \int \limits _{\cos \Theta }^{1} \left[ \mathcal {J}_{m} u_{1}(x) \right] u_{2}(x) \mathrm{d }\!x&= \int \limits _{\cos \Theta }^{1} \frac{\mathrm{d }\!^{}}{\mathrm{d }\!x^{}} \left[ (\cos \Theta - x) (1 - x^2) \frac{\mathrm{d }\!^{}u_{1}(x)}{\mathrm{d }\!x^{}} \right] u_{2}(x) \mathrm{d }\!x \nonumber \\&\quad \!+\! \int \limits _{\cos \Theta }^{1} \left[ L(L+2) u_{1}(x) u_{2}(x) \!-\! (\cos \Theta \!-\! x) \frac{m^2 \!-\! 2mx \!+\! 1}{1 \!-\! x^2} u_{1}(x) u_{2}(x) \right] \mathrm{d }\!x.\nonumber \\ \end{aligned}$$

(97b)

Next we perform integration by parts on the first term of the right-hand side in both equations:

$$\begin{aligned} \begin{aligned} \int \limits _{\cos \Theta }^{1} u_{1}(x) \frac{\mathrm{d }\!^{}}{\mathrm{d }\!x^{}} \left[ (\cos \Theta \!-\! x) (1 \!-\! x^2) \frac{\mathrm{d }\!^{}u_{2}(x)}{\mathrm{d }\!x^{}} \right] \mathrm{d }\!x&\!=\! u_{1}(x) (\cos \Theta \!-\! x)(1 \!-\! x^2) \frac{\mathrm{d }\!^{}u_{2}(x)}{\mathrm{d }\!x^{}}\, \bigg |_{\cos \Theta }^{1} \\&\quad \!-\! \int _{\cos \Theta }^{1} \frac{\mathrm{d }\!^{}u_{1}(x)}{\mathrm{d }\!x^{}} (\cos \Theta \!-\! x)(1 \!-\! x^2) \frac{\mathrm{d }\!^{}u_{2}(x)}{\mathrm{d }\!x^{}} \mathrm{d }\!x, \\ \int \limits _{\cos \Theta }^{1} \frac{\mathrm{d }\!^{}}{\mathrm{d }\!x^{}} \left[ (\cos \Theta \!-\! x) (1 \!-\! x^2) \frac{\mathrm{d }\!^{}u_{1}(x)}{\mathrm{d }\!x^{}} \right] u_{2}(x) \mathrm{d }\!x&\!=\! (\cos \Theta \!-\! x)(1 \!-\! x^2) \frac{\mathrm{d }\!^{}u_{1}(x)}{\mathrm{d }\!x^{}} u_{2}(x) \, \bigg |_{\cos \Theta }^{1} \\&\quad \!-\! \int _{\cos \Theta }^{1} (\cos \Theta \!-\! x)(1 \!-\! x^2) \frac{\mathrm{d }\!^{}u_{1}(x)}{\mathrm{d }\!x^{}} \frac{\mathrm{d }\!^{}u_{2}(x)}{\mathrm{d }\!x^{}} \mathrm{d }\!x. \end{aligned} \end{aligned}$$

The first term on the right-hand side of both equations vanishes and the rest is identical, hence

$$\begin{aligned} \int \limits _{\cos \Theta }^{1} u_{1}(x) \frac{\mathrm{d }\!^{}}{\mathrm{d }\!x^{}} \left[ (\cos \Theta - x) (1 - x^2) \frac{\mathrm{d }\!^{}u_{2}(x)}{\mathrm{d }\!x^{}} \right] \mathrm{d }\!x \nonumber \\ = \int \limits _{\cos \Theta }^{1} \frac{\mathrm{d }\!^{}}{\mathrm{d }\!x^{}} \left[ (\cos \Theta - x) (1 - x^2) \frac{\mathrm{d }\!^{}u_{1}(x)}{\mathrm{d }\!x^{}} \right] u_{2}(x) \mathrm{d }\!x. \end{aligned}$$

(98)

Upon inserting (98) into (97a) we find that

$$\begin{aligned} \int \limits _{\cos \Theta }^{1} u_{1}(x) \left[ \mathcal {J}_{m} u_{2}(x) \right] \mathrm{d }\!x = \int \limits _{\cos \Theta }^{1} \left[ \mathcal {J}_{m} u_{1}(x) \right] u_{2}(x) \mathrm{d }\!x. \end{aligned}$$

$\square$

1.7 Appendix 1(g): Proof of Identity (78)

Proof

First we apply expression (74) of $\mathcal {J}_{m}$ to the kernel function $\mathcal {K}_{m}(x,\,x')$ and use eigenvalue Eq. (30) of $\Delta _{\Omega ,m}$:

$$\begin{aligned} \begin{aligned} \mathcal {J}_{m} \mathcal {K}_{m}(x,\,x')&= \left[ (\cos \Theta - x) \Delta _{\Omega ,\,m} - (1 - x^2) \frac{\mathrm{d }\!^{}}{\mathrm{d }\!x^{}} - L(L + 2)x \right] \sum _{l=\ell _{m}}^{L} F_{lm}(x) F_{lm}(x') \\&\!=\! \!-\,\cos \Theta \sum _{l=\ell _{m}}^{L} l(l+1) F_{lm}(x) F_{lm}(x') \!+\! x \sum _{l=\ell _{m}}^{L} \bigl [ l(l\!+\!1) \!-\! L(L+2) \bigr ] F_{lm}(x) F_{lm}(x') \\&\quad \,-v (1\!-\!x^2) \sum _{l\!=\!\ell _{m}}^{L} \frac{\mathrm{d }\!^{}F_{lm}(x)}{\mathrm{d }\!x^{}} F_{lm}(x'). \end{aligned} \end{aligned}$$

Likewise, we also apply $\mathcal {J}_{m}'$ to $\mathcal {K}(x,\,x')$ and subtract the resulting equation from the previous one, yielding

$$\begin{aligned} \begin{aligned} (\mathcal {J}_{m} - \mathcal {J}_{m}') \mathcal {K}_{m}(x,\,x')&= (x - x') \sum _{l=\ell _{m}}^{L} \bigl [l(l + 1) - L(L + 2) \bigr ] F_{lm}(x) F_{lm}(x')\\&\quad - \sum _{l=\ell _{m}}^{L} (1 - x^2) \frac{\mathrm{d }\!^{}F_{lm}(x)}{\mathrm{d }\!x^{}} F_{lm}(x') \\&\quad + \sum _{l=\ell _{m}}^{L} F_{lm}(x) (1 - x^{\prime \, 2}) \frac{\mathrm{d }\!^{}F_{lm}(x')}{\mathrm{d }\!x'^{}}. \end{aligned} \end{aligned}$$

Using recurrence relation (14) on the terms containing the derivatives of $F_{lm}$ and performing some straightforward algebra, we get

$$\begin{aligned} \begin{aligned} (\mathcal {J}_{m} - \mathcal {J}_{m}') \mathcal {K}_{m}(x,\,x')&= (x - x') \sum _{l=\ell _{m}}^{L} \bigl [l^2 - (L + 1)^2 \bigr ] F_{lm}(x) F_{lm}(x') \\&\quad + \sum _{l=\ell _{m}}^{L} (2l + 1) \zeta _{l+1,\,m} \bigl [ F_{l+1,\,m}(x) F_{lm}(x') - F_{lm}(x) F_{l+1,\,m}(x') \bigr ]. \end{aligned} \end{aligned}$$

Applying Christoffel–Darboux formula (16) to the second term on the right-hand side yields

$$\begin{aligned} \begin{aligned} (\mathcal {J}_{m} - \mathcal {J}_{m}') \mathcal {K}_{m}(x,\,x')&= (x - x') \sum _{l=\ell _{m}}^{L} F_{lm}(x) F_{lm}(x') \bigl [l^2 - (L + 1)^2 \bigr ]\\&\quad + (x - x') \sum _{l=\ell _{m}}^{L} (2l + 1) \sum _{l'=\ell _{m}}^{l} F_{l'm}(x) F_{l'm}(x'). \end{aligned} \end{aligned}$$

(99)

In the last term of the right-hand side, the summation can be interchanged as

$$\begin{aligned} \sum _{l=\ell _{m}}^{L} (2l + 1) \sum _{l'=\ell _{m}}^{l} F_{l'm}(x) F_{l'm}(x') = \sum _{l'=\ell _{m}}^{L} F_{l'm}(x) F_{l'm}(x') \sum _{l=l'}^{L} (2l + 1). \end{aligned}$$

Relabeling the sums on the right-hand side of this expression, so that $l$ becomes $l'$ and vice versa, and inserting the resulting expression into the right-hand side of (99), we obtain

$$\begin{aligned} (\mathcal {J}_{m} - \mathcal {J}_{m}') \mathcal {K}_{m}(x,\,x') = (x - x') \sum _{l=\ell _{m}}^{L} F_{lm}(x) F_{lm}(x') \left[ l^2 - (L + 1)^2 + \sum _{l'=l}^{L} (2l' + 1) \right] . \end{aligned}$$

Since $\sum _{l'=\ell _{m}}^{L} (2l' + 1) = (L + 1)^2 - l^2$, the right-hand side vanishes. Hence

$$\begin{aligned} \mathcal {J}_{m} \mathcal {K}_{m}(x,\,x') = \mathcal {J}_{m}' \mathcal {K}_{m}(x,\,x'). \end{aligned}$$

$\square$

1.8 1(h) Proof of Expressions (83) for the Matrix Elements of $\mathcal {J}_{m}$

Proof

We start by inserting expression (74) of $\mathcal {J}_{m}$ into the integral expression (81) for the matrix elements and use the eigenvalue Eq. (30) of $\Delta _{\Omega ,\,m}$:

$$\begin{aligned} J_{m,\,ll'}&= \int \limits _{-1}^{1} F_{lm}(x) \left[ (\cos \Theta - x) \Delta _{\Omega ,\,m} - (1 - x^2) \frac{\mathrm{d }\!^{}}{\mathrm{d }\!x^{}} - L(L + 2)x \right] F_{l'm}(x) \mathrm{d }\!x\nonumber \\&\!=\! \!-\!l'(l' \!+\! 1)\cos \Theta \int \limits _{-1}^{1} F_{lm}(x) F_{l'm}(x) \mathrm{d }\!x \!+\! \left[ l'(l' \!+\! 1) \!-\! L(L \!+\! 2) \right] \int \limits _{-1}^{1} x F_{lm}(x) F_{l'm}(x) \mathrm{d }\!x \nonumber \\&\quad - \int \limits _{-1}^{1} F_{lm}(x) (1 - x^2) \frac{\mathrm{d }\!^{}F_{l'm}(x)}{\mathrm{d }\!x^{}} \mathrm{d }\!x \end{aligned}$$

(100)

The first integral is equal to $\delta _{ll'}$ because of orthonormality relation (5). The remaining two can be evaluated by using recurrence relations (12) and (13) and orthonormality relation (5):

$$\begin{aligned} \begin{aligned} \int \limits _{-1}^{1} x F_{lm}(x) F_{l'm}(x) \mathrm{d }\!x&= \zeta _{l'm} \delta _{l,\,l'-1} + \zeta _{l'+1,\,m} \delta _{l,\,l'+1} + \frac{m}{l'(l'+1)} \delta _{ll'}, \\ \int \limits _{-1}^{1} F_{lm}(x) (1 - x^2) \frac{\mathrm{d }\!^{}F_{l'm}(x)}{\mathrm{d }\!x^{}} \mathrm{d }\!x&= (l' + 1) \zeta _{l'm} \delta _{l,\,l'-1} - l' \zeta _{l'+1,\,m} \delta _{l,\,l'+1} + \frac{m}{l'(l'+1)} \delta _{ll'}. \end{aligned} \end{aligned}$$

Thus for (100), we get

$$\begin{aligned} J_{m,\,ll'}&= \left\{ -l'(l'+1)\cos \Theta + m \left[ 1 - \frac{L(L + 2) + 1}{l'(l' + 1)} \right] \right\} \delta _{ll'}\\&+ \zeta _{l'm} \left[ (l' - 1)(l' + 1) - L(L + 2) \right] \delta _{l,\,l'-1}\\&+ \zeta _{l'+1,\,m} \left[ l' (l' + 2) - L(L + 2) \right] \delta _{l',\,l'+1}. \end{aligned}$$

Because of the Kronecker deltas, this expression is non-zero for index pairs $(l,\,l)$, $(l+1,\,l)$ and $(l,\,l+1)$ only. The corresponding matrix elements are

$$\begin{aligned} \begin{aligned} J_{m,\,ll}&= -l(l + 1)\cos \Theta + m \left[ 1 - \frac{L(L + 2) + 1}{l(l + 1)} \right] \\ J_{m,\,l,\,l+1}&= J_{m,\,l+1,\,l} = \zeta _{l+1,\,m} \bigl [ l(l + 2) - L(L + 2) \bigr ], \end{aligned} \end{aligned}$$

hence $\mathsf {J}_{m}$ is real, symmetric and tridiagonal.