Quasinormal modes of Kerr–de Sitter black holes via the Heun function

We basically follow the convention in [11]. The Teukolsky equation is separated into the angular and the radial equations. The angular Teukolsky equation is given by equation (3.1) in [11]. We rewrite it as a more compact form:

$$\begin{equation}\begin{aligned}\hfill & \left[\frac{\mathrm{d}}{\mathrm{d}x}\left(1+\alpha {x}^{2}\right)\left(1-{x}^{2}\right)\frac{\mathrm{d}}{\mathrm{d}x}+\lambda -s\left(1-\alpha \right)-2\alpha {x}^{2}+\frac{{\left(1+\alpha \right)}^{2}}{1+\alpha {x}^{2}}\right.\hfill \\ \hfill & \quad \left.{\times}\left({c}^{2}{x}^{2}-2csx-{c}^{2}+2cm+\frac{4\alpha }{1+\alpha }smx-\frac{{\left(m+sx\right)}^{2}}{1-{x}^{2}}\right)\right]S\left(x\right)=0,\hfill \end{aligned}\end{equation} \tag{ 2.4 }$$

where x = cos  θ, s is a spin-weight of a perturbing field, m is an azimuthal number and c = aω with frequency ω. λ is a separation constant, which is determined by a requirement of the regularity for S(x) both at x = ±1. We will show that this two-point boundary value problem is solved by the Heun function.

In the flat limit Λ → 0 (i.e. α → 0), the angular equation (2.4) reduces to the so-called spin-weighted spheroidal equation [10]. The relation to the original separation constant A in [10] is λ|_α=0 = A + 2s − 2cm + c². Even in this case, the eigenvalue is not known analytically ³ . If taking a further limit a → 0, then λ → ł(ł + 1) − s(s − 1) with multipole number ł.

If we take the spinless limit a → 0 first with ω and Λ fixed (i.e. c → 0 and α → 0), then the reduced equation does not contain Λ. Therefore we conclude λ|_a=0 = ł(ł + 1) − s(s − 1).

At first glance, the differential equation (2.4) has five regular singular points at $x={\pm}1,{\pm}i/\sqrt{\alpha },\infty$. However the infinity point turns out to be a removable singularity. Its characteristic exponents are −1 and 0, and moreover the corresponding two Frobenius solutions have no logarithmic singularities. As a consequence, it is removed by a proper transformation of S(x). The angular differential equation finally leads to Heun's differential equation. This surprising fact was first shown in [11]. We follow their argument. To see the relation, we first match locations of the singular points. This is done by a Möbius transformation ⁴ :

$$\begin{equation}z=\frac{\left(1-i/\sqrt{\alpha }\right)\left(x+1\right)}{2\left(x-i/\sqrt{\alpha }\right)}.\end{equation} \tag{ 2.5 }$$

In this transformation, $x=-1,1,-i/\sqrt{\alpha },i/\sqrt{\alpha },\infty$ are mapped into z = 0, 1, z_a , ∞, z_∞, respectively, where

$$\begin{equation}{z}_{a}{:=}-\frac{{\left(1-i/\sqrt{\alpha }\right)}^{2}}{4i/\sqrt{\alpha }},\quad {z}_{\infty }{:=}\frac{1-i/\sqrt{\alpha }}{2}.\end{equation} \tag{ 2.6 }$$

We next perform the following transformation of S(x):

$$\begin{equation}S\left(x\right)={z}^{{A}_{1}}{\left(z-1\right)}^{{A}_{2}}{\left(z-{z}_{a}\right)}^{{A}_{3}}\left(z-{z}_{\infty }\right){y}_{a}\left(z\right),\end{equation} \tag{ 2.7 }$$

where

$$\begin{equation}{A}_{1}=\frac{m-s}{2},\quad {A}_{2}=-\frac{m+s}{2},\quad {A}_{3}=\frac{i}{2}\left(\frac{1+\alpha }{\sqrt{\alpha }}c-m\sqrt{\alpha }-is\right).\end{equation} \tag{ 2.8 }$$

The new function y_a (z) now satisfies

$$\begin{equation}{y}_{a}^{{\prime\prime}}\left(z\right)+\left(\frac{2{A}_{1}+1}{z}+\frac{2{A}_{2}+1}{z-1}+\frac{2{A}_{3}+1}{z-{z}_{a}}\right){y}_{a}^{\prime }\left(z\right)+\frac{{\rho }_{+}{\rho }_{-}z+u}{z\left(z-1\right)\left(z-{z}_{a}\right)}{y}_{a}\left(z\right)=0,\end{equation} \tag{ 2.9 }$$

where

$$\begin{equation}\begin{aligned}\hfill {\rho }_{+}& =1,\quad {\rho }_{-}=1-s-im\sqrt{\alpha }+ic\left(\sqrt{\alpha }+\frac{1}{\sqrt{\alpha }}\right),\hfill \\ \hfill u& =-\left[\frac{i\lambda }{4\sqrt{\alpha }}+\frac{1}{2}+{A}_{1}+\left(m+\frac{1}{2}\right)\left({A}_{3}-{A}_{3}^{{\ast}}\right)\right].\hfill \end{aligned}\end{equation} \tag{ 2.10 }$$

Here a 'conjugate' ${A}_{3}^{{\ast}}$ is obtained by replacing i in A₃ by −i. Now the point z = z_∞ is not a singular point as expected. Note that there is a relation: (2A₁ + 1) + (2A₂ + 1) + (2A₃ + 1) = ρ₊ + ρ₋ + 1. Therefore the differential equation (2.9) is indeed Heun's differential equation whose solutions are discussed later.

The radial Teukolsky equation is more complicated. It is given by ⁵

$$\begin{equation}\begin{aligned}\hfill & \left[{{\Delta}}_{r}^{-s}\frac{\mathrm{d}}{\mathrm{d}r}{{\Delta}}_{r}^{s+1}\frac{\mathrm{d}}{\mathrm{d}r}+\frac{{\left(1+\alpha \right)}^{2}{K}^{2}-is\left(1+\alpha \right)K{{\Delta}}_{r}^{\prime }}{{{\Delta}}_{r}}+4is\left(1+\alpha \right)\omega r\right.\hfill \\ \hfill & \quad \left.-\frac{2\alpha }{{a}^{2}}\left(s+1\right)\left(2s+1\right){r}^{2}+2s\left(1-\alpha \right)-\lambda \right]R\left(r\right)=0,\hfill \end{aligned}\end{equation} \tag{ 2.11 }$$

where

$$\begin{equation}K\left(r\right)=\omega \left({r}^{2}+{a}^{2}\right)-am.\end{equation} \tag{ 2.12 }$$

For the de Sitter black holes, Δ_r = 0 has four real roots. We label these roots r_± and r_±' so that r₋' < r₋ < r₊ < r₊'. As in the angular case, the radial Tuekolsky equation has five regular singular points r = r_±, r_±', ∞, but the singularity at infinity is apparent and removable. To go to Heun's world, we need a non-trivial transformation. We do a Möbius transformation:

$$\begin{equation}z=\frac{\left({r}_{+}^{\prime }-{r}_{-}\right)\left(r-{r}_{+}\right)}{\left({r}_{+}^{\prime }-{r}_{+}\right)\left(r-{r}_{-}\right)}.\end{equation} \tag{ 2.13 }$$

After it, the five points are mapped as

$$\begin{equation}\left({r}_{+},{r}_{+}^{\prime },{r}_{-},{r}_{-}^{\prime },\infty \right)\to \left(0,1,\infty ,{z}_{r},{z}_{\infty }\right),\end{equation} \tag{ 2.14 }$$

where

$$\begin{equation}{z}_{r}{:=}\frac{\left({r}_{+}^{\prime }-{r}_{-}\right)\left({r}_{-}^{\prime }-{r}_{+}\right)}{\left({r}_{+}^{\prime }-{r}_{+}\right)\left({r}_{-}^{\prime }-{r}_{-}\right)},\quad {z}_{\infty }{:=}\frac{{r}_{+}^{\prime }-{r}_{-}}{{r}_{+}^{\prime }-{r}_{+}}.\end{equation} \tag{ 2.15 }$$

Note that z_r > 1 for the Kerr–dS black holes. Following [11], we further transform

$$\begin{equation}R\left(r\right)={z}^{{B}_{1}}{\left(z-1\right)}^{{B}_{2}}{\left(z-{z}_{r}\right)}^{{B}_{3}}{\left(z-{z}_{\infty }\right)}^{2s+1}{y}_{r}\left(z\right),\end{equation} \tag{ 2.16 }$$

where

$$\begin{equation}{B}_{1}=\frac{i\left(1+\alpha \right)K\left({r}_{+}\right)}{{{\Delta}}_{r}^{\prime }\left({r}_{+}\right)},\quad {B}_{2}=\frac{i\left(1+\alpha \right)K\left({r}_{+}^{\prime }\right)}{{{\Delta}}_{r}^{\prime }\left({r}_{+}^{\prime }\right)},\quad {B}_{3}=\frac{i\left(1+\alpha \right)K\left({r}_{-}^{\prime }\right)}{{{\Delta}}_{r}^{\prime }\left({r}_{-}^{\prime }\right)}.\end{equation} \tag{ 2.17 }$$

Then, we finally obtain the following equation:

$$\begin{align}\hfill & {y}_{r}^{{\prime\prime}}\left(z\right)+\left(\frac{2{B}_{1}+s+1}{z}+\frac{2{B}_{2}+s+1}{z-1}+\frac{2{B}_{3}+s+1}{z-{z}_{r}}\right){y}_{r}^{\prime }\left(z\right)\hfill \\ \hfill & \quad +\frac{{\sigma }_{+}{\sigma }_{-}z+v}{z\left(z-1\right)\left(z-{z}_{r}\right)}{y}_{r}\left(z\right)=0,\hfill \end{align} \tag{ 2.18 }$$

where

$$\begin{equation}{\sigma }_{+}=2s+1,\quad {\sigma }_{-}=s+1-\frac{2i\left(1+\alpha \right)K\left({r}_{-}\right)}{{{\Delta}}_{r}^{\prime }\left({r}_{-}\right)},\end{equation} \tag{ 2.19 }$$

and

$$\begin{equation}\begin{aligned}\hfill v& =\frac{\left(1+s\right)\left(1+2s\right){r}_{-}^{\prime }}{{r}_{-}-{r}_{-}^{\prime }}+\frac{3\lambda -2s\left(3-{a}^{2}{\Lambda}\right)+{\Lambda}\left(1+s\right)\left(1+2s\right){r}_{+}\left({r}_{+}+{r}_{+}^{\prime }\right)}{{\Lambda}\left({r}_{-}-{r}_{-}^{\prime }\right)\left({r}_{+}-{r}_{+}^{\prime }\right)}\hfill \\ \hfill & \quad -\frac{2i\left(1+2s\right)\left(3+{a}^{2}{\Lambda}\right)\left({r}_{+}{r}_{-}\omega +{a}^{2}\omega -am\right)}{{\Lambda}\left({r}_{-}-{r}_{-}^{\prime }\right)\left({r}_{-}-{r}_{+}\right)\left({r}_{+}-{r}_{+}^{\prime }\right)}.\hfill \end{aligned}\end{equation} \tag{ 2.20 }$$

The expression of v looks much simpler than that in [11] for Q = 0. Note that to derive (2.18), we do not need explicit forms of r_±, r_±' in terms of M, a and Λ. We have used a trivial relation r₊ + r₊' + r₋ + r₋' = 0 and the following nice identity:

$$\begin{equation}\frac{K\left({r}_{+}\right)}{{{\Delta}}_{r}^{\prime }\left({r}_{+}\right)}+\frac{K\left({r}_{+}^{\prime }\right)}{{{\Delta}}_{r}^{\prime }\left({r}_{+}^{\prime }\right)}+\frac{K\left({r}_{-}\right)}{{{\Delta}}_{r}^{\prime }\left({r}_{-}\right)}+\frac{K\left({r}_{-}^{\prime }\right)}{{{\Delta}}_{r}^{\prime }\left({r}_{-}^{\prime }\right)}=0.\end{equation} \tag{ 2.21 }$$

Again the equation (2.18) is Heun's equation.

We should comment on the scalar perturbation s = 0. The scalar perturbation considered here is the conformally coupled massless scalar perturbation. If one considers other perturbations, like the minimally coupled massless scalar, the resultant differential equation does not reduce to Heun's equation. As shown in [20], for the massive scalar perturbation, Heun's equation appears if and only if the mass of scalar is chosen as a special value.

In the previous subsections we showed that the Teukolsky equations (2.4) and (2.11) are transformed into Heun's differential equations (2.9) and (2.18), respectively. Here we look at their local solutions at singular points.

Let us consider Heun's equation (1.1). There are a few conflicts of letters in the previous subsections, but they will cause no confusions. As already mentioned, Heun's equation has four regular singular points at z = 0, 1, a, ∞. For each point, we can construct Frobenius solutions. They are convergent inside a circle, whose radius is generically determined by the distance from the nearest singularity. Since the exponents at z = 0 are 0 and 1–γ, let Hℓ(a, q; α, β, γ, δ; z) be the regular solution at z = 0. We normalize it as

$$\begin{equation}\mathrm{H}\ell \left(a,q;\alpha ,\beta ,\gamma ,\delta ;0\right)=1.\end{equation} \tag{ 2.22 }$$

We refer to this local solution as the Heun function ⁶ . It turns out that the two solutions at z = 0 are given by [4],

$$\begin{equation}\begin{aligned}\hfill {y}_{01}\left(z\right)& =\mathrm{H}\ell \left(a,q;\alpha ,\beta ,\gamma ,\delta ;z\right),\hfill \\ \hfill {y}_{02}\left(z\right)& ={z}^{1-\gamma }\mathrm{H}\ell \left(a,\left(a\delta +{\epsilon}\right)\left(1-\gamma \right)+q;\alpha +1-\gamma ,\beta +1-\gamma ,2-\gamma ,\delta ;z\right).\hfill \end{aligned}\end{equation} \tag{ 2.23 }$$

Unless γ is an integer, these two are independent. Since in our application, the singular point z = a is always located outside the unit circle: |a| > 1, the radius of convergence of these local solutions is unity in general.

Similarly, two solutions at z = 1 are given by

$$\begin{align}\hfill {y}_{11}\left(z\right)& =\mathrm{H}\ell \left(1-a,\alpha \beta -q;\alpha ,\beta ,\delta ,\gamma ;1-z\right),\hfill \\ \hfill {y}_{12}\left(z\right)& ={\left(1-z\right)}^{1-\delta }\mathrm{H}\ell \left(1-a,\left(\left(1-a\right)\gamma +{\epsilon}\right)\left(1-\delta \right)+\alpha \beta -q;\right.\hfill \\ \hfill & \quad \alpha \left.+1-\delta ,\beta +1-\delta ,2-\delta ,\gamma ;1-z\right).\hfill \end{align} \tag{ 2.24 }$$

The other solutions near z = a and z = ∞ can be also expressed in terms of Hℓ. They are not needed in our analysis.

In two-point boundary value problems between z = 0 and z = 1, it is useful to consider connection relations among the local solutions in different domains. Let us write them as

$$\begin{equation}\begin{aligned}\hfill {y}_{01}\left(z\right)& ={C}_{11}{y}_{11}\left(z\right)+{C}_{12}{y}_{12}\left(z\right),\hfill \\ \hfill {y}_{02}\left(z\right)& ={C}_{21}{y}_{11}\left(z\right)+{C}_{22}{y}_{12}\left(z\right),\hfill \end{aligned}\end{equation} \tag{ 2.25 }$$

where we have assumed that the solutions y₁₁(z) and y₁₂(z) are linearly independent. In the angular eigenvalue problem, this assumption is often violated. Nevertheless the Wronskian method still works.

To our knowledge, unlike the hypergeometric function, no analytically explicit representations of the connection coefficients C_ij are known for the Heun solutions. However, formally these are represented by Wronskians of local solutions, for instance, C₁₂ = W[y₀₁, y₁₁]/W[y₁₂, y₁₁] etc. Therefore, if we know the local solutions, we can evaluate the connection coefficients. This point has just become doable in the recent update of Mathematica.

In Mathematica, the Heun function [1] is implemented as

$$\begin{equation}\mathtt{H}\mathtt{e}\mathtt{u}\mathtt{n}\mathtt{G}\left[\mathtt{a},\mathtt{q},\boldsymbol{\alpha },\boldsymbol{\beta },\boldsymbol{\gamma },\boldsymbol{\delta },\mathtt{z}\right].\end{equation} \tag{ 2.26 }$$

Recall that = 1 + α + β − γ − δ is not an independent parameter. The function 'HeunG' satisfies Heun's differential equation (1.1) and has the same normalization (2.22). Therefore locally 'HeunG' is exactly the same as Hℓ. Globally 'HeunG' gives analytic continuation of Hℓ. Though sometimes multi-valuedness causes a problem in numerical computations, all our computations are done inside the convergence circle. There is no multi-value problem.

As a warm-up, we begin with non-rotating asymptotically dS₄ black holes. In the non-rotating cases, the angular problem is trivialized, and we have

$$\begin{equation}\lambda {\vert }_{a=0}=l{}\left(l{}+1\right)-s\left(s-1\right).\end{equation} \tag{ 3.1 }$$

We can take the limit a → 0 smoothly in (2.17), (2.19) and (2.20). Also, for a = 0, the inner horizon is always at r₋ = 0.

Now two solutions near the event horizon z = 0 (r = r₊) are given by (2.23) with the identification

$$\begin{equation}\begin{aligned}\hfill a& ={z}_{r},\quad q=-v,\quad \alpha ={\sigma }_{+},\quad \beta ={\sigma }_{-},\hfill \\ \hfill \gamma & =2{B}_{1}+s+1,\quad \delta =2{B}_{2}+s+1,\quad {\epsilon}=2{B}_{3}+s+1.\hfill \end{aligned}\end{equation} \tag{ 3.2 }$$

Similarly, the solutions near the de Sitter horizon z = 1 (r = r₊') are given by (2.24) with the same correspondence.

To compute the QNM frequencies, we have to see asymptotic behaviors. Near z = 0, the two solutions behave as y₀₁ ∼ z⁰, y₀₂ ∼ z^1−γ . Therefore the original radial eigenfunctions show

$$\begin{equation}\begin{aligned}\hfill {R}_{01}\left(r\right){:=}{z}^{{B}_{1}}{\left(z-1\right)}^{{B}_{2}}{\left(z-{z}_{r}\right)}^{{B}_{3}}{\left(z-{z}_{\infty }\right)}^{2s+1}{y}_{01}\left(z\right)& \sim {\left(r-{r}_{+}\right)}^{i\omega {r}_{+}^{2}/{{\Delta}}_{r}^{\prime }\left({r}_{+}\right)},\hfill \\ \hfill {R}_{02}\left(r\right){:=}{z}^{{B}_{1}}{\left(z-1\right)}^{{B}_{2}}{\left(z-{z}_{r}\right)}^{{B}_{3}}{\left(z-{z}_{\infty }\right)}^{2s+1}{y}_{02}\left(z\right)& \sim {\left(r-{r}_{+}\right)}^{-s-i\omega {r}_{+}^{2}/{{\Delta}}_{r}^{\prime }\left({r}_{+}\right)}.\hfill \end{aligned}\end{equation} \tag{ 3.3 }$$

We conclude that in our convention R₀₂(r) satisfies the QNM boundary condition at the event horizon. Similarly, near z = 1, we have

$$\begin{equation}\begin{aligned}\hfill {R}_{11}\left(r\right){:=}{z}^{{B}_{1}}{\left(z-1\right)}^{{B}_{2}}{\left(z-{z}_{r}\right)}^{{B}_{3}}{\left(z-{z}_{\infty }\right)}^{2s+1}{y}_{11}\left(z\right)& \sim {\left({r}_{+}^{\prime }-r\right)}^{i\omega {{r}_{+}^{\prime }}^{2}/{{\Delta}}_{r}^{\prime }\left({r}_{+}^{\prime }\right)},\hfill \\ \hfill {R}_{12}\left(r\right){:=}{z}^{{B}_{1}}{\left(z-1\right)}^{{B}_{2}}{\left(z-{z}_{r}\right)}^{{B}_{3}}{\left(z-{z}_{\infty }\right)}^{2s+1}{y}_{12}\left(z\right)& \sim {\left({r}_{+}^{\prime }-r\right)}^{-s-i\omega {{r}_{+}^{\prime }}^{2}/{{\Delta}}_{r}^{\prime }\left({r}_{+}^{\prime }\right)}.\hfill \end{aligned}\end{equation} \tag{ 3.4 }$$

Therefore R₁₁(r) is a preferred solution. The QNM boundary condition then requires C₂₂ = 0 in (2.25). This leads to the vanishing condition of the Wronskian:

$$\begin{equation}{C}_{22}\left(\omega \right)=\frac{W\left[{y}_{02},{y}_{11}\right]}{W\left[{y}_{12},{y}_{11}\right]}=0.\end{equation} \tag{ 3.5 }$$

In the evaluation of the Wronskian, we choose a value of z so that it is located in both convergence circles at z = 0 and z = 1. Typically z = 1/2 is a good point for this purpose. The ratio of the Wronskian is independent of z. Solving this equation, we get the QNM frequencies ω.

For given (M, Λ) and (s, ł), we compute all the quantities in (3.2), and evaluate the Wronskian as a function of ω by using 'HeunG'. The result for (s, ł) = (2, 2) is shown in table 1. In this approach, the eigenvalues are obtained in a few seconds. To find a solution to C₂₂(ω) = 0, we have to choose an initial value of 'FindRoot' carefully. Such an initial value is estimated by using semi-analytic results or by referring to a previous value if parameters are varied slightly. Other numerical codes such as [22, 23] are also useful for this purpose. To make table 1, we slightly varied the cosmological constant, and used a previous eigenvalue as a next initial value.

Strictly speaking, the results here are obtained by the non-rotating limit of the Teukolsky equation. The result in [24] is based on the Regge–Wheeler equation. Therefore our test also confirms the isospectrality of these distinct master equations.

We should recall the scalar perturbations. The result in [24] is for the minimally coupled massless scalar perturbation, while the result in this note is for the conformally coupled massless scalar perturbation. As mentioned in the previous section, the minimally coupled scalar perturbation cannot be mapped into Heun's equation, as was shown in [18]. It is possible if and only if the massless scalar is conformally coupled to the Ricci scalar.

As a further benchmark, we compare with a high-precision method recently proposed in [23]. In this approach, numerical precisions get better for smaller overtone numbers and for larger multipole numbers. We set (s, ℓ) = (2, 5), and compute the lowest overtone frequency for M²Λ = 1/100, 1/10. The two distinct approaches showed at least 50-digit agreements. A numerical limitation occurs in the approach of [23], but the method in this note does not seem to show such a limitation.

The extension to the Kerr black holes is very simple. The radial problem is almost the same as the Schwarzschild case. We consider the angular problem. Two solutions near z = 0 (x = −1) and z = 1 (x = 1) are given by (2.23) and (2.24), respectively. We need the following identification:

$$\begin{equation}\begin{aligned}\hfill a& ={z}_{a},\quad q=-u,\quad \alpha ={\rho }_{+},\quad \beta ={\rho }_{-},\hfill \\ \hfill \gamma & =2{A}_{1}+1,\quad \delta =2{A}_{2}+1,\quad {\epsilon}=2{A}_{3}+1.\hfill \end{aligned}\end{equation} \tag{ 3.6 }$$

Let us see the boundary conditions. Near x = −1 (z = 0), the original eigenfunctions behave as

$$\begin{equation}\begin{aligned}\hfill {S}_{01}\left(x\right)& {:=}{z}^{{A}_{1}}{\left(z-1\right)}^{{A}_{2}}{\left(z-{z}_{a}\right)}^{{A}_{3}}\left(z-{z}_{\infty }\right){y}_{a\mathrm{01}}\left(z\right)\sim {\left(1+x\right)}^{\left(m-s\right)/2},\hfill \\ \hfill {S}_{02}\left(x\right)& {:=}{z}^{{A}_{1}}{\left(z-1\right)}^{{A}_{2}}{\left(z-{z}_{a}\right)}^{{A}_{3}}\left(z-{z}_{\infty }\right){y}_{a\mathrm{02}}\left(z\right)\sim {\left(1+x\right)}^{-\left(m-s\right)/2}.\hfill \end{aligned}\end{equation} \tag{ 3.7 }$$

where we use the subscript a to distinguish the solutions from those for the radial problem. Hence we should choose S₀₁(x) for m > s or S₀₂(x) for m < s. For m = s, both are equivalent. Similarly, the behavior near x = 1 is given by

$$\begin{equation}\begin{aligned}\hfill {S}_{11}\left(x\right)& {:=}{z}^{{A}_{1}}{\left(z-1\right)}^{{A}_{2}}{\left(z-{z}_{a}\right)}^{{A}_{3}}\left(z-{z}_{\infty }\right){y}_{a\mathrm{11}}\left(z\right)\sim {\left(1-x\right)}^{-\left(m+s\right)/2},\hfill \\ \hfill {S}_{12}\left(x\right)& {:=}{z}^{{A}_{1}}{\left(z-1\right)}^{{A}_{2}}{\left(z-{z}_{a}\right)}^{{A}_{3}}\left(z-{z}_{\infty }\right){y}_{a\mathrm{12}}\left(z\right)\sim {\left(1-x\right)}^{\left(m+s\right)/2}.\hfill \end{aligned}\end{equation} \tag{ 3.8 }$$

The regularity condition depends on the sign of m + s. In summary, we have four branches to classify the regularity conditions of S(x) depending on values of s and m. The same classification is found in the asymptotically flat case in [14].

Let us consider the case (s, ł) = (−2, 2) for concreteness. Since |m| ⩽ 2, the regularity condition requires S₀₁(x) and S₁₁(x) at both sides. Their linear dependence leads to C₁₂ = 0. However, in the angular problem, the two exponents 0 and 1–δ at z = 1 are integers. Therefore the linear independence of y_a11(z) and y_a12(z) is not guaranteed. Recall that our interested boundary condition is just the regularity of S(x) at x = ±1. This condition is simply satisfied by the linear dependence of S₀₁(x) and S₁₁(x), which leads to

$$\begin{equation}W\left[{y}_{a\mathrm{01}},{y}_{a\mathrm{11}}\right]=0.\end{equation} \tag{ 3.9 }$$

Keep in mind that this condition may be modified, depending on the values of s and m. In the radial problem, we impose the same condition as in the previous subsection:

$$\begin{equation}\frac{W\left[{y}_{r\mathrm{02}},{y}_{\mathrm{11}}\right]}{W\left[{y}_{r\mathrm{12}},{y}_{r\mathrm{11}}\right]}=0.\end{equation} \tag{ 3.10 }$$

These two conditions determine ω and λ simultaneously.

Let us compare our results with known numerical QNM frequencies. For this purpose, we follow a setup in [25]. We fix the mass M by the following condition

$$\begin{equation}\frac{{{\Delta}}_{r}^{\prime }\left({r}_{+}\right)}{2{r}_{+}}{\left\vert \right.}_{a=0,{\Lambda}=3}=\frac{4}{10},\end{equation} \tag{ 3.11 }$$

where Δ_r is defined by (2.3), and r₊ corresponds to the event horizon, which is the second largest (real) root of Δ_r = 0. The equation (3.11) simply leads to

$$\begin{equation}10{r}_{+}^{3}-3{r}_{+}+5M=0.\end{equation} \tag{ 3.12 }$$

Also, since r₊ is a solution to Δ_r = 0, we have, for a = 0 and Λ = 3,

$$\begin{equation}{r}_{+}\left({r}_{+}^{3}-{r}_{+}+2M\right)=0.\end{equation} \tag{ 3.13 }$$

These two conditions fix M. We obtain the following physically allowed unique value:

$$\begin{equation}M=\frac{7}{15\sqrt{15}}=0.120\enspace 492\enspace 815\dots .\end{equation} \tag{ 3.14 }$$

For this mass value and Λ = 3, we compute the QNM frequencies ⁷ . The result is shown in table 2. Our result is consistent with table I in [25] within numerical errors.