Unstable mode solutions to the Klein-Gordon equation in Kerr-anti-de Sitter spacetimes

For any cosmological constant $\Lambda=-3/\ell^2<0$ and any $\alpha<9/4$, we find a Kerr-AdS spacetime $(\mathcal M,g_{\mathrm{KAdS}})$, in which the Klein-Gordon equation $\Box_{g_{\mathrm{KAdS}}}\psi+\alpha/\ell^2\psi=0$ has an exponentially growing mode solution satisfying a Dirichlet boundary condition at infinity. The spacetime violates the Hawking-Reall bound $r_+^2>|a|\ell$. We obtain an analogous result for Neumann boundary conditions if $5/4<\alpha<9/4$. Moreover, in the Dirichlet case, one can prove that, for any Kerr-AdS spacetime violating the Hawking-Reall bound, there exists an open family of masses $\alpha$ such that the corresponding Klein-Gordon equation permits exponentially growing mode solutions. Our result adopts methods of Shlapentokh-Rothman (see arXiv:1302.3448) and provides the first rigorous construction of a superradiant instability for negative cosmological constant.

with cosmological constant Λ can be understood as a system of second-order partial differential equations for the metric tensor g of a four-dimensional spacetime (M, g). Solutions with negative cosmological constant have drawn considerable attention in recent years, mainly due to the conjectured instability of these spacetimes. For more details, see [And06], [DH06], [BR11], [DHS11], [DHMS12], [HLSW15] and references therein. In appropriate coordinates, (1.1) forms a system of non-linear wave equations. A first step in understanding the global dynamics of solutions to (1.1) -and thus eventually answering the question of stability -is the study of linear wave equations on a fixed background. For Λ < 0, efforts have focused on understanding the dynamics of the Klein-Gordon equation for an asymptotically AdS metric g with cosmological constant Λ = −3/ 2 and a mass term α satisfying the Breitenlohner-Freedman bound α < 9/4 [BF], which is required for well-posedness of the equation -see [War12], [Hol11] and [Vas09]. The conformally coupled case α = 2 encompasses scalar-type metric perturbations around an exact AdS spacetime [IW04]. For g being the metric of an exact AdS spacetime, the massive wave equation (1.2) allows for time-periodic solutions due to the timelike nature of null and spacelike infinity I; in particular, general solutions to (1.2), while remaining bounded, do not decay. The behaviour of solutions to (1.2) on black-hole spacetimes is very different. Given a Kerr-AdS spacetime with parameters , M and a satisfying |a| < , define the Hawking-Reall Killing vector field K := T + aΞ r 2 + + a 2 Φ, where, using Boyer-Lindquist coordinates, T = ∂ t and Φ = ∂φ; see Section 1.4 for definitions of Ξ and r + . The vector field K is the (up to normalisation) unique Killing vector field that is null on the horizon H and non-spacelike in a neigbourhood of H. It is globally timelike in the black hole exterior if the Hawking-Reall bound r 2 + > |a| is satisfied. If the bound is violated, K becomes non-timelike far away from the horizon. In [HR99], Hawking and Reall use the existence of a globally causal K for r 2 + > |a| to argue towards the stability of these spacetimes. Indeed, uniform boundedness of solutions to (1.2) in the full regime α < 9/4 was proved for r 2 + > |a| in [Hol09] and [HW12]. Moreover, in [HS11], it was shown that solutions with the fastest radial decay (Dirichlet conditions at infinity) in fact decay logarithmically in time 1 and [HS13] proves that this logarithmic bound is sharp.
For spacetimes violating the Hawking-Reall bound, the global behaviour of solutions to (1.2) has not been investigated rigorously, but it was argued in the physics literature -see [CD04], [CDLY04], [CDY06] and [DHS11] -that at least for small black holes, i. e. for |a| and |a| r + , instability of solutions to (1.2) is to be expected if r 2 + < |a| . As, in this regime, there is no Killing vector field which is globally timelike in the black hole exterior, this parallels the situation of asymptotically flat Kerr spacetimes, where superradiance is present. For the present discussion, we will understand superradiance loosely as energy extraction from a rotating black hole. We will make this more precise in Lemma 1.1.

Unstable modes and superradiance in spacetimes with Λ = 0
The study of energy extraction from black holes in asymptotically flat spacetimes has a long history in the physics literature and two different, but related mechanisms have been proposed. On the one hand, Press and Teukolsky [PT72] suggested that the leakage of energy through the horizon of a rotating black hole could be used to create a black hole bomb by placing a mirror around it. Superradiance would increase the radiation pressure on the mirror over time until it finally breaks, setting free all the energy at once. 2 On the other hand, it was argued that energy could be extracted by the aid of massive waves acting as a natural mirror. This goes back to Zel'dovich [Zel71] and was explored further by Starobinsky in [Sta73]. Numerous heuristic and numerical studies on the superradiant behaviour of solutions to the Klein-Gordon equation followed, e. g. [DDR76], [ZE79], [Det80], [Dol07] and [Dol12]. These studies found exponentially growing solutions to the massive wave equation on Kerr spacetimes.
Remarkably, this instability is not present at the level of the massless wave equation g ψ = 0, see [DRSR14b], where boundedness and decay for such solutions is proved in the full subextremal range |a| < M . Even though energy can potentially leak out of the black hole, superradiance can be overcome here as the superradiant frequencies in Fourier spaces are not trapped. In particular, in the context of scattering [DRSR14a], a quantitative bound on the maximal superradiant amplification was shown.
In accordance with the above heuristic of massive waves acting as a natural mirror for a black hole bomb, this situation changes dramatically for the Klein-Gordon equation with scalar mass µ > 0. A first rigorous construction of exponentially growing finite-energy solutions in Kerr spacetimes was given by Shlapentokh-Rothman [SR13]. The constructed solutions were modes. Mode solutions are solutions of the form ψ(t, r, ϑ,φ) = e −iωt e imφ S ml (cos ϑ)R(r) (1.4) in Boyer-Lindquist coordinates (t, r, ϑ,φ) for ω ∈ C, m ∈ Z and l ∈ Z ≥|m| , where the smooth functions S ml and R satisfy ordinary differential equations arising from the separability property of the wave equation in Boyer-Lindquist coordinates [Car]. We call a mode unstable if it is exponentially growing in time, i. e. if Im ω > 0. Shlapentokh-Rothman showed that, for any given Kerr spacetime with 0 < |a| < M , there is an open family of masses µ producing unstable modes with finite energy. The construction starts from proving existence of real modes and hence produces in particular periodic solutions. We will adopt this strategy.

Unstable modes and superradiance in Kerr-AdS spacetimes
Let us return to the Kerr-AdS case and connect the existence of unstable modes to superradiance. Recall that the energy-momentum tensor for the Klein-Gordon equation (1.2) is given by T µν := Re ∇ µ ψ∇ ν ψ − 1 2 g µν |∇ψ| 2 − α 2 |ψ| 2 and that, for each vector field X, we obtain a current J X µ := T µν X ν .
While in Kerr spacetimes, T = ∂ t (see Section 1.4) is the (up to normalisation) unique timelike Killing field at infinity, the family of vector fields T + λΦ with Φ = ∂φ is timelike near infinity in Kerr-AdS spacetimes if and only if − −2 ( + a) < λ < −2 ( − a) . (1.5) Hence, in this range of values for λ, the conserved current J T +λΦ µ encapsulates the energy density of the scalar field measured by different (rotating) observers at infinity. The vector field T + λΦ becomes spacelike or null at the horizon.
Recall that the Hawking-Reall vector field K is tangent to the null generators of the horizon H. Therefore the energy density radiated through the horizon is measured by (1.6) A non-trivial mode solution radiates energy away from the horizon if and only if the expression (1.6) is negative for all λ in the range (1.5). The thusly characterised frequencies ω form the superradiant regime.
Proof. Since J T +λΦ α K α = 0 at the horizon for ε = 0, it suffices to differentiate (1.6) with respect to ε and evaluate at ε = 0. We see that the derivative is negative if and only if This, however, can be easily checked to hold using (1.5) and r 2 + < |a| .
Remark 1.2. If r 2 + > |a| , then K induces an energy density at infinity and J K µ K µ ≥ 0, in accordance with the intuition of not being in the superradiant regime if the Hawking-Reall bound is satisfied.
We will show that our constructed growing mode solutions -as the modes of [SR13]satisfy the assumptions of Lemma 1.1. This corroborates our interpretation that the unstable modes are a linear manifestation of the superradiant properties of Kerr-AdS spacetimes.
This expression imposes some restrictions on the range of |a| in terms of r + as shown in the following Lemma 1.3. If r + < , (1.9) If r + ≥ , |a| can take any value in [0, ).
Therefore, under the restriction of Lemma 1.3, there is a bijection between Kerr-AdS spacetimes with parameters ( , M, a) and spacetimes with parameters ( , r + , a). Henceforth we will use the shorthand notations M KAdS ( , M, a) and M KAdS ( , r + , a) to denote Kerr-AdS spacetimes with parameters ( , M, a) and ( , r + , a) respectively. The restriction of Lemma 1.3 can be seen in the above figure.
Given ( , M, a), a chart covering all of the domain of outer communication is given by Boyer-Lindquist coordinates (t, r, ϑ,φ) ∈ R × (r + , ∞) × S 2 . The metric in these coordinates is where Σ = r 2 + a 2 cos 2 ϑ, Since Boyer-Lindquist coordinates break down at r = r + , we introduce Kerr-AdS-star coordinates (t * , r, ϑ, ϕ). These are related to Boyer-Lindquist coordinates by t * := t + A(r) and ϕ :=φ + B(r), In these coordinates, the metric extends smoothly through r = r + . One sees that the boundary r = r + of the Boyer-Lindquist patch is null and we shall call it the event horizon H. Finally, we introduce the tortoise coordinate r * which is related to r by with r * (+∞) = π/2. We will denote the derivative with respect to r * by .

Statement of the results
The analysis in this paper yields two types of instability results: A. Given a cosmological constant Λ and a mass α, there is a Kerr-AdS spacetime for this Λ in which (1.2) has a growing solution.
B. Given a Kerr-AdS spacetime violating the Hawking-Reall bound, there is a range for the scalar mass such that, in this spacetime, (1.2) has a growing solution.
To make this more precise, recall that mode solutions are Fourier modes that take the form ψ(t, r, ϑ,φ) = e −iωt e imφ S ml (cos ϑ)R(r) in Boyer-Lindquist coordinates (t, r, ϑ,φ) for ω ∈ C, m ∈ Z and l ∈ Z ≥|m| . Define u(r) := (r 2 + a 2 ) 1/2 R(r). Use S mode (α, ω, m, l) to denote the set of all mode solutions with parameters ω, m, l to the Klein-Gordon equation with scalar mass α. Set We require that all mode solutions are smooth. For the S ml this is ensured automatically by the definition -see Section 2.1. Hence we only need to impose a regularity condition on the function u, given parameters , r + , a, m and ω.
Definition 1.4 (Horizon regularity condition). A smooth function f : (r + , ∞) → C satisfies the horizon regularity condition if f (r) = (r − r + ) ξ for a smooth function as well as a constant . (1.10) Henceforth we will only call a mode ψ a mode solution to (1.2) if its radial part R (and hence u) satisfies the horizon regularity condition. At infinity, we will study two different boundary conditions for u.
Mode solutions satisfying these boundary conditions are analogous to the modes considered in [SR13].
We are able to show the following result.
Theorem 1.6. Given a cosmological constant Λ = −3/ 2 , a black hole radius 0 < r + < and a scalar mass parameter α 0 ∈ (−∞, 9/4), there are a spacetime parameter a satisfying the regularity condition |a| < , mode parameters m and l and a δ > 0 such that there are a smooth curve with α(0) = α 0 and ω R (0) = Ξam r 2 + + a 2 (1.11) and corresponding mode solutions in S mode (α(ε), ω R (ε) + iε, m, l) satisfying the horizon regularity condition and Dirichlet boundary conditions. For all ε ∈ (0, δ), these modes satisfy Remark 1.7. The u in the theorem has finite energy and hence the spacetime parameters of the theorem must violate the Hawking-Reall bound as explained in the previous sections; this is explained further in Lemma 2.18 and Remark 3.9. By Lemma 1.3, we know that the a can be located anywhere in the range We remark that our result does not restrict to small |a|. In fact, we can enforce |a| to be as close to as we wish by choosing r + / < 1 large.
Lemma 1.1 implies that the constructed modes are superradiant and indicates that the instability is driven by energy leaking through the horizon.
Our next theorem builds on the first, but allows for the construction of an unstable superradiant mode with Dirichlet boundary conditions for each given α < 9/4. Theorem 1.8. Let > 0 and α < 9/4. Then there is an M KAdS ( , r + , a) and a superradiant ψ ∈ S mode (α, ω R + iε, m, l) for an ω R ∈ R and ε > 0 satisfying Dirichlet boundary conditions.
Remark 1.10. These results also apply to the massless wave equation, which is an important difference to the asymptotically flat case. Furthermore, although this will not be pursued explicitly in this paper, one can also show the analogue of Shlapentokh-Rothman's result in our setting by only adapting the proof slightly.
1. Conversely, in the asymptotically flat Kerr case of [SR13], it is also possible to prove an analogue of Theorem 1.8 instead of only the analogue of Theorem 1.6, using our strategy explained in the next section. 2. To contrast our case to the asymptotically flat setting, we add three observations. First, in [SR13], the curve ε → (µ(ε), ω R (ε) + iε) must satisfy µ(0) 2 > ω R (0) 2 . There is no equivalent condition for Kerr-AdS spacetimes as the instability is not driven by the interplay of frequency and mass, but by the violation of the Hawking-Reall bound. Second, in both cases, ∂ω R /∂ε < 0 for small ε, so ω R (0) can be seen as the upper bound of the superradiant regime. Third, the result in Kerr holds for all m = 0, l ≥ |m|. In contrast, our result is a statement about large m = l.
Let us conclude this section with a general remark on boundedness. From [HW12], we know that solutions to the Klein-Gordon equation with Dirichlet boundary conditions remain bounded for all r 2 + > |a| . A similar statement holds for Neumann boundary conditions under more restrictive assumptions on the parameters. For r 2 + = |a| , one can easily repeat the proof of the second theorem of [HS11] to see that there are no periodic solutions. One can potentially also extend the decay result of [HS11] to r 2 + = |a| . Our results do not rule out boundedness in the entire parameter range in which r 2 + < |a| since we did not show that for any given Kerr-AdS spacetime and any α, there are unstable mode solutions; they do, however, impose restrictions on the ranges of spacetime parameters and masses α in which boundedness could potentially hold. It is believed that, using more refined spectral estimates, our results can be shown to hold in the full regime r 2 + < |a| , but we will not pursue this further.

Outline of the proof
The difficulty lies in the construction of the radial part u, for which we use the strategy of [SR13], which, as our present work shows, can be applied to more general settings than Kerr spacetimes. The technique contains two main steps.
I. Construct u corresponding to a real frequency ω 0 ∈ R.
II. Obtain a mode solution corresponding to a complex ω with Im ω > 0 by varying spacetime and mode parameters.
We note that both steps are completely independent of each other, in particular step II does not rely on the method by which the periodic mode solution was constructed, but only requires existence of such a mode. Let us first only deal with Dirichlet boundary conditions. To complete step I, u needs to satisfy the radial ODE for the given boundary condition -see Section 2.1. Lemma 2.20 then already restricts ω 0 to ω + := maΞ/(r 2 + + a 2 ). It is important to note that the boundary value problem does not admit nontrivial solutions in general.
Proof. Define Q(r) := Re (u u), note that Q(r + ) = Q(∞) = 0 and integrate dQ/dr. Hence, in a first step in Section 2.3, we will find spacetime and mode parameters such that V − ω 2 0 < 0 on some subinterval of (r + , ∞) for given and α 0 by a careful analysis of the shape of the potential V in Lemma 2.13. This requires proving an asymptotic estimate for the eigenvalues of the modified oblate spheroidal harmonics (Lemma 2.12). The spacetime parameters will necessarily violate the Hawking-Reall bound.
The radial ODE is the Euler-Lagrange equation of the functional The functional is not bounded below, so we need to impose a norm constraint, which we choose to be f /r L 2 (r + ,∞) = 1. Then Lemma 3.3 gives a coercivity-type estimate. To carry out the direct method of the calculus of variations, we use the weighted Sobolev spaces that arise naturally from the functional -see Section 3.1. This setting of the minimisation problem then guarantees that the minimiser satisfies the correct boundary conditions. We remark that we will directly work with the functional (1.13) instead of regularising first at the horizon and then taking the limit, as in [SR13]. Then, in Lemma 3.7, we obtain an ODE with a Lagrange multiplier ν a ≤ 0 that depends continuously on the spacetime parameter a.
By varying a, we find anâ such that νâ = 0 (Proposition 3.8) and hence a solution to the radial ODE.
For the parameters from step I, A(α 0 , ω 0 ) = 0. By varying ω and α simultaneously in Section 3.2, the implicit function theorem yields a curve ε → (ω R (ε) + iε, α(ε)) with ω R (0) = ω 0 and α(0) = α 0 such that A(α(ε), ω(ε)) = 0. along the curve. As Im ω(ε) > 0 for ε > 0, these modes grow exponentially whilst satisfying Dirichlet boundary conditions. In Section 3.3, we show that ∂ω R ∂ε (0) < 0 and ∂α ∂ε (0) > 0, (1.14) which proves Theorem 1.6. A careful analysis of the domain of the implicit function theorem in Section 3.4 yields Theorem 1.8. Here, the analysis heavily exploits several continuity properties in the parameters. A difficulty is caused byâ being defined as the infimum of an open set. For Corollary 1.9, one observes that, by Lemma 2.13, once the Hawking-Reall bound is violated, one can always make the potential V negative on some interval by choosing |m| sufficiently large. This yields periodic modes for very small violation of the Hawking-Reall bound and hence growing modes by repeating the above argument.
The situation is more complicated if u satisfies the Neumann boundary condition. Since, in this case, u ∼ r −1/2+κ as r → ∞, L a is not well-defined and hence cannot be used to produce periodic modes. To carry out the construction of step I, we use twisted derivatives as introduced in [War12] and used extensively in [HW12]. To find the minimiser via the variational argument, we also need to modify our function spaces and use twisted weighted Sobolev spaces. All details are given in Section 4.1.
The main technical problems, however, arise in the second part of the argument. The underlying reason is that the proofs for step II rely severely on establishing monotonicity properties for the functional when varying α. Since the twisting necessarily depends on α, proving monotonicity in α is more involved and indeed the monotonicity properties shown in the Neumann case are weaker; nevertheless, the ideas introduced in Section 4.2 are sufficiently robust not only to construct the growing modes, but also to be applicable to showing (1.14) and to transition from Theorem 1.14 to Theorem 1.15. It is also in the Neumann case, where the independence of steps I and II -alluded to above -is exploited.
If α ≤ 0, S ml satisfies the angular ODE for λ m (ω, α, a) ∈ C. If α > 0, the angular ODE takes the form (2.2) Using these modified oblate spheroidal harmonics, one obtains that, for fixed m and l, u := √ r 2 + a 2 R satisfies the radial ODE Here Θ(x) = 1 if x > 0 and zero otherwise. We will use the shorthandṼ := V − ω 2 . Recall that denotes an r * -derivative.
To indicate the dependence upon a, we will often write V a andṼ a for V andṼ respectively.

Local analysis of the radial ODE
To see which boundary conditions are appropriate for u, we perform a local analysis of the radial ODE near the horizon r = r + and at infinity, using the following theorem about regular singularities, which we cite from [Tes12], but it can also be found in [SR13] or [Olv74].
Suppose f and g are meromorphic and have poles of order (at most) one and two, respectively, at z 0 ∈ C. Let f 0 (ν) and g 0 (ν) be the coefficients of pole of order one and two, respectively, in the Laurent expansions. Let s 1 (ν) and s 2 (ν) be the two solutions of the indicial equation If s 2 (ν) − s 1 (ν) / ∈ N 0 , a fundamental system of solutions is given by where the functions j are holomorphic and satisfy j (z 0 , ν) = 1. If s 2 (ν) − s 1 (ν) = m ∈ N 0 , a fundamental system is given by The constant c may be zero unless m = 0.
In both cases, the radius of convergence of the power series of j is at least equal to the minimum of the radii of convergence of the Laurent series of f and g.

The horizon
Adopting the notation of the previous section, and, after expressing the radial ODE (2.3) with r-derivatives, we have Thus we obtain Therefore if ξ = 0, a local basis of solutions u (or R) is given by for holomorphic functions ϕ i satisfying ϕ i (r + ) = 1. For ξ = 0, a local basis is given by for ϕ 1 (r + ) = 1 and some constant c.
Lemma 2.2. If u extends smoothly to the horizon, then there is a smooth function : Proof. Boyer-Lindquist coordinates break down at the horizon, so we need to change to Kerr-star coordinates. Then the solution ψ takes the form Hence R extends smoothly to the horizon if there is a smooth function f such that Therefore the claim reduces to showing that which proves the claim.
Corollary 2.3. Assume u satisfies the horizon regularity condition. Then a local basis of solutions to the ODE at the horizon is given by for a holomorphic function defined around r = r + .
This asymptotic analysis at the horizon motivates the horizon regularity condition of Definition 1.4.

Infinity
The radial ODE has a regular singularity at r = ∞. To analyse it using the Theorem 2.1, we rewrite equation (2.4) by introducing x := 1/z. This yields For the radial ODE, we have x = 1/r. We obtain The indicial equation becomes which is solved by s ± = 1 2 ± 9 4 − α. Set Then, for α / ∈ E, a local basis of solutions near infinity is given by with functions 1 , 2 , smooth at ∞ and satisfying 1 (∞) = 2 (∞) = 1. For α ∈ E, a local basis is given by If u extends smoothly to r = r + and we specify a boundary value u(r + ), then the arguments of Section 2.2.3 show that C 3 has to be zero.
Corollary 2.5. If u satisfies the horizon regularity condition and the Neumann boundary condition at infinity, then, for 5/4 < α < 9/4, If u satisfies the horizon regularity condition and the Dirichlet boundary condition at infinity, then, for all α < 9/4, Remark 2.6. The asymptotics near infinity do not change if we add ν(r 2 + a 2 )/(r 2 ∆ − ) to g as in Section 3.1.

Uniqueness of solutions and dependence on parameters
As one would expect, specifying one of the boundary conditions at infinity and choosing a value of u at r = r + determines the solution to the radial ODE uniquely, which is being made more precise in the following standard lemma.
Lemma 2.7. Let C 0 ∈ C. Then there is a unique classical solution to (2.3) on (r + , ∞) satisfying u(r + ) = C 0 and extending smoothly to r = r + .
The continuous dependence of the solution u on parameters is also well-known: Lemma 2.8. Let u 0 be a unique solution to (2.3) for a certain set of parameters (α 0 , ω 0 , a 0 ) with fixed u(r + ) satisfying either the Dirichlet or Neumann boundary condition. Let there be a neighbourhood of these parameters such that for all (α, ω, a) in said neighbourhood, there is a unique solution u α,ω,a with the same boundary conditions. Fix anr ∈ (r + , ∞). Then is smooth.
Lemma 2.9. A and B are smooth in α, ω and a.
Proof. Note that A and B are independent of r and apply Lemma 2.8.

Detailed analysis of the potential
From the analysis in [HS11] we know that the angular ODE has countably many simple eigenvalues λ ml , labelled by l = |m|, |m| + 1, . . . for any given m ∈ Z, and corresponding real-valued eigenfunction S ml . For later use, we need a bound from below which can be found in [HS11], where it is proved under the assumption of the Hawking-Reall bound. We give the slight extension to our regime.
Lemma 2.10. Let ω ∈ R. For |a| < , the eigenvalues satisfy where C ,a > 0 depends on and a only and Proof. We focus on the second inequality since the first one can be obtained similarly. Let Then where we have already used that the mass term is always nonnegative. We want to show that P c ≥ 0. We have the decomposition where P + c is the ω + -part (i. e. the part for ω = ω + ) with Interpreting the bracket as a function in ϑ, we see that it has critical points only at ϑ = 0, π/2, π. Hence P + c ≥ 0. Therefore, we know To obtain the estimate, one only needs to integrate by parts on the sphere. TheP term yields where we compared with spherical harmonics via the min-max principle.
We will also need an asymptotic upper bound on the ground state eigenvalue λ mm . By the min-max principle, we know that Lemma 2.11. Let n ∈ N.
Proof. By equation (2.6), we already have lim m→∞ λ mm /m 2 ≥ Ξ 2 . To prove the result, we compute Lemma 2.13. Let N, L > 0. Then, given > 0 and α < 9/4. Moreover assume the spacetime parameters r + and a satisfy r 4 + − a 2 2 (r 2 + + a 2 ) 2 < −N. (2.7) Then there is an m 0 > 0 such that for all mode parameters |m| ≥ m 0 and l = m, we have Proof. Let us first rewrite the potential: Note that Moreover λ−Ξ 2 m 2 > 0 by (2.6). Therefore, to obtain negativity, we will violate the Hawking-Reall bound in the boxed term. First we can choose |m| large such that λ/Ξ 2 m 2 − 1 is sufficiently small by Lemma 2.12. Since the term in the last line is decaying, we can find an R 1 such that the last term is bounded on [R 1 , R 1 + L]. By making |m| possibly larger, the terms of the first line are also bounded on the interval. Let There is an R 1 such that Putting everything together, the lemma follows.
Remark 2.14. The same proof yields the analogous negativity results for V − ω 2 + F , where F is any continuous function on (r + , ∞) that is independent of m. This will be used in Section 4.1.
Define the functional We often suppress some of the indices and write L a and V a in view of Section 3.1.
Lemma 2.15. Choose (r + , a, ) and (m, l) as in Lemma 2.13. Then there is a function f ∈ C ∞ 0 (r + , ∞) such that Proof. We have the following estimate for the functional if f is supported in (R 1 , R 2 ): Choose an f such that f is 1 on [R 1 + L/4, R 2 − L/4] and 0 outside of (R 1 , R 2 ). Furthermore we require that df dr ≤ 2 4 L . Hence .
If necessary, we can increase m further to make the expression negative. We want to conclude this section by showing that L a is always non-negative if the Hawking-Reall bound is satisfied. We borrow the following Hardy inequality from [HS11].
Lemma 2.17. For any r cut ≥ r + , we have for a smooth function f with f r 1/2 = o(1) at infinity that Proof. We include a proof for the sake of completeness. Integrating by parts and applying the Cauchy-Schwarz inequality yields The lemma follows by estimating r cut ≥ r + .
An analogue of Lemma 2.15 can be proved for the twisted functional used in Section 4.1. For 0 < κ < 1, definẽ withṼ h a as in Section 4.1. Lemma 2.19. Choose (r + , a, ) and (m, l) as in Lemma 2.13. Then there is a function
Proof. We wish to show (i). First let us only deal with the Dirichlet branch. Then u is decaying at infinity. Define the microlocal energy current Q T := Im u u .
For the Neumann branch of the solution we observe that From the boundary condition, we immediately get Q T (∞) = 0 as well and the rest follows as above. Part (ii) follows immediately from (r 2 + + a 2 )ω = Ξam.

Existence of real mode solutions
We now fix > 0, α < 9/4 and 0 < r + < . Recall the variational functional for ω = ω + = amΞ/(r 2 + + a 2 ). Define By Lemma 2.15, there is an m 0 such that A is non-empty for all |m| ≥ m 0 and l = m. Fix m and l henceforth. If the bound r 2 + ≥ |a| is satisfied, then L a (f ) ≥ 0 for all compactly supported f by Lemma 2.18. Hence A is bounded below by a strictly positive infimum. Moreover A is open as a → L a (f ) is continuous for any fixed f .
Remark 3.1. We restrict ourselves to a > 0, but we could have defined the set A to also include negative values of a.
Our aim is to show that L a has a minimiser for a ∈ A. We will apply the natural steps of the direct method of the calculus of variations. First, we will specify an appropriate function space, then we will show that the functional obeys a coercivity condition and that the functional is weakly lower semicontinuous. The existence of a minimiser follows by an application of compactness results.
For U ⊆ (r + , ∞) define the weighted norm This is clearly a Hilbert space with the natural inner product (·, ·) L 2 (U ) . For U ⊆ (r + , ∞), we define the weighted Sobolev space H 1 via the norm Note that for U ⊆ (r + , ∞) compact, the H 1 norm is equivalent to the standard Sobolev norm. As usual, let H 1 0 (U ) be the completion of C ∞ 0 (U ) under · H 1 (U ) .
Lemma 3.2. Let u ∈ H 1 0 (r + , ∞). Then is u is also in C(r + + 1, ∞) (after possibly changing it on a set of measure zero) and Proof. Establishing the embedding H 1 0 (r + + 1, ∞) ⊆ C(r + + 1, ∞) is standard. Now take a sequence (u m ) in C ∞ 0 such that u m → u in H 1 0 and pointwise almost everywhere. Choose an R such that (u m ) converges pointwise there. For any β < 1/2, we have Therefore, the claim follows.
To establish a coercivity-type inequality, we use the Hardy inequality of Lemma 2.17: Lemma 3.3. Let a ∈ A be fixed. There exist constants r + < B 0 < B 1 < ∞ and constants C 0 , C 1 , C 2 > 0, such that, for sufficiently large m, we have for all smooth functions f with f r 1/2 = o(1) at infinity that Here we can choose C 2 = 1 if α < 2.
Remark 3.4. Note that the dependence of the expression on α is viaṼ in L a . Recall from Section 2.1 thatṼ Proof. First, we have to study the potential again: Thus for sufficiently large |m|, the expression is greater than zero. Furthermore, note the asymptotics as r → ∞. We will deal with the cases α < 2 and α ≥ 2 separately. First, let α < 2. The function r 2 +a 2 ∆ −Ṽ is only nonpositive on an interval [R 1 , R 2 ]. Choose constants such that r + < B 1 < R 1 < R 2 < B 2 < ∞. Set C 0 to be the minimum of r 2 +a 2 ∆ −Ṽ on (r + , ∞)\[B 1 , B 2 ] and set −C 1 to be its minimum on [B 1 , B 2 ]. This immediately yields the result. Now let α ≥ 2. There exist R 1 , R 2 such that r 2 +a 2 ∆ −Ṽ is positive on (r + , R 1 ) and on (R 2 , ∞) for an ε > 0 because of (3.1). Hence by Lemma 2.17. Choose B 1 , B 2 as before. Let C be the minimum of r 2 +a 2 ∆ −Ṽ on (r + , B 1 ). Let εC 0 /2 be the minimum of C and ε/(8 2 ). Moreover, set −εC 1 /2 to be the minimum of and hence the inequality.
Lemma 3.5. The functional L a is weakly lower semicontinuous in H 1 (r + , ∞) when restricted to functions of untit L 2 norm.
Proof. As the functional is convex in the derivative, the statement is standard and a proof can be extracted from [Eva10,§8]. We note that the boundedness from below comes from the norm constraint. The r weight deals with (r + , ∞) having non-finite measure.
Lemma 3.6. Let a ∈ A. Then there exists an f a ∈ H 1 0 (r + , ∞) with unit L 2 (r + , ∞) norm such that L a achieves its infimum over Proof. By Lemma 3.3, holds for all f ∈ H 1 0 . From this, it is evident that We can choose a minimising sequence of functions of compact support by density. Thus let {f a,n } be a sequence of smooth functions, compactly supported in (r + , ∞) with f a,n L 2 = 1, such that L a (f a,n ) → ν a .
The bound (3.2) implies that f a,n H 1 is uniformly bounded. Thus by the Banach-Alaoglu theorem, it has a weakly convergent subsequence in H 1 0 (r + , ∞). Recall a simple version of Rellich-Kondrachov: H 1 [a, b] embeds compactly into L 2 [a, b]. Hence by the equivalence of norms, the subsequence has a strongly in L 2 convergent subsequence on compact subsets of (r + , ∞). Relabelling, we have a sequence {f a,n } that converges to f a weakly in H 1 0 and strongly in L 2 on compact subsets of (r + , ∞). The space H 1 0 is a linear (hence convex) subspace of H 1 that is norm-closed. Every convex subset that is norm closed is weakly closed. Therefore, f a ∈ H 1 0 . We claim that f a L 2 = 1. We have i. e. the norm must concentrate either near the horizon or near infinity. Suppose first that f a,n L 2 (r + ,r + +δ) ≥ 1 > 0 for infinitely many f a,n and any δ > 0. By (3.2), we have for r ∈ (r + , r + + 1): r 2 df a,n dr 2 dr 1/2 ≤ C 1 + log 1 r − r + for a constant C > 0. Since r → | log(r − r + )| is integrable on compact subsets of [r + , ∞), we obtain f a,n L 2 (r + ,r + +δ) → 0 as δ → 0, a contradiction. Hence we only need to exclude the case that the norm is bounded away from zero for large r. Thus, suppose that f a,n L 2 (R 0 ,∞) ≥ 2 > 0 for infinitely many f a,n and any R 0 > 0. However, for a constant C > 0 by (3.2) and any R 0 , a contradiction. This shows that By the infimum property, we have ν a ≤ L a (f a ).

By Lemma 3.5, we get
L a (f a ) ≤ lim inf n→∞ L a (f a,n ).

As
L a (f a,n ) → ν a , the latter equals ν a . Thus the minimum is attained by f a .
We would like to derive the Euler-Lagrange equation corresponding to this minimiser.
Proof. The proof can be extracted from [Eva10]. The analogous proof for twisted derivatives is given for Lemma 4.6.
Proposition 3.8. There is anâ and a corresponding non-zero function fâ ∈ C ∞ (r + , ∞) such that and fâ satisfies the horizon regularity condition and the Dirichlet boundary condition at infinity.
Proof. First we would like to show that ν a is continuous in a. We will use the notation ∆ a − to denote the ∆ − corresponding to a. Given a 1 and a 2 , we have Due to the continuity of L a (f ) in a, the first line is greater or equal than ν a 2 if a 1 is sufficiently close to a 2 . Since the coefficients in the second line are continuously differentiable in a, we can use the mean value theorem to obtain for some constant C > 0. We obtain an analogous inequality reversing the rôles of a 1 and a 2 . Using (3.2) and f a L 2 = 1 yields Since A = ∅, we setâ := inf A.
As stated in the introduction to this section, A is open, soâ / ∈ A. By continuity of ν a , this implies that νâ = 0. Now choose a sequence a n →â and corresponding minimisers f an ∈ H 1 0 satisfying f an L 2 = 1. Then, as in the proof of Lemma 3.6, by Lemma 3.3, f an is bounded in H 1 and there is a subsequence (also denoted (a n )) such that f an → fâ weakly in H 1 and strongly in L 2 on compact subsets for a fâ ∈ H 1 0 . Again by Lemma 3.3 and the strong L 2 convergence on compact subsets, we see that fâ is non-zero. Moreover, we have sufficient decay towards infinity by Lemma 3.2. Hence we get the desired asymptotics.
From the weak convergence of (f an ), Lemma 3.7 yields that fâ satisfies from which we obtain that fâ ∈ C ∞ . It remains to check the boundary condition at the horizon. The lower semi-continuity of convex functionals with respect to weak convergence implies that (3.4) Near r + , the local theory (Theorem 2.1) implies that there exist constants A, B and non-zero analytic functions ϕ i such that whence B = 0. Hence fâ satisfies the horizon regularity condition.
Remark 3.9. From Lemma 2.18, we already know that |â| ≥ r 2 + / . In [HS11], it is shown directly that, if the Hawking-Reall bound is satisfied, there are no periodic solutions. One can easily see that the proof generalises to the case when the Hawking-Reall bound is saturated. Thus we even obtain |â| > r 2 + / . Corollary 3.10. Assume our choice of parameters, a =â and ω = ω + . Let C 0 ∈ C. Then the radial ODE (2.3) has a unique solution satisfying u(r + ) = C 0 and the Dirichlet boundary condition at infinity.
where we have used |u(r + )| = 1. We have due to the asymptotics of the h i . We obtain Now we differentiate at ω R = ω R (0) and α = α 0 with respect to ω R and α: To extend the coefficient A(α, ω R ) = 0 to complex ω, we want to appeal to the implicit function theorem establishing This is true indeed: Lemma 3.11.

Behaviour for small ε > 0 for Dirichlet boundary conditions
From the analysis of the previous section, we have a family of mode solutions u(r, ε) to the radial ODE parameters (ω(ε), m, l, α(ε)), where The mode u satisfies the horizon regularity condition and the Dirichlet boundary condition at infinity. This proves the first part of Theorem 1.6. To prove the second part, we would like to study the behaviour of ω(ε) and α(ε) for small ε > 0.
To obtain the following statements, we potentially need to make |m| even larger than in the previous sections.
Proposition 3.12. If |m| is sufficiently large, we have
We have As already used in Section 3.1, there is an R > r + such that, for r ≥ R, By an application of Lemma 2.17, we conclude (3.11) For the sake of contradition, suppose K(r + ) ≥ 0. Then, by (3.10), K > 0 on (r + , ∞), whence we obtain strict positivity for (3.11). As |ω(0)| 2 = m 2 a 2 Ξ 2 (r 2 + + a 2 ) 2 and as, for fixed r + ,â is bounded away from zero for all m, Since ε → ω(ε) is continuous, |ω| 2 scales as m 2 for small ε, so dK/dr can be chosen as large as possible at r = r + , in particular, it can be used to overcome the potentially non-positive derivative of the remaining terms of the right hand side of (3.9) on (r + , R). Then, (3.8) implies u = 0, a contradiction. Hence K(r + ) < 0 which is equivalent to This in turn is equivalent to the claim.
In the following, we will fix an |m| ≥ m 0 such that Proposition 3.12 holds.
Remark 3.13. The choice of m could have been made right at the beginning as the choice of m 0 in Lemma 2.13 is independent of the largeness required for Proposition 3.12.
The next proposition shows that the mass α is at first increasing along the curve obtained by the implicit function theorem. The proof requires a technical lemma which is given at the end of this section.
Proposition 3.14. Let α(0) < 9/4. Then Proof. Define Then ∂ ∂r We would like to multiply this equation by u and then integrate by parts, but, at r = r + , u ε does not satisfy the boundary conditions of a mode solution. However, using u ∼ (r − r + ) −1/2−κ(ε) for all ε by Section 3.2, u ε ∼ log r(r − r + ) −1/2−κ and hence satisfies the Dirichlet boundary condition at infinity.
We know that is smooth, whence We have d dr We conclude that ∆ − r 2 + a 2 log(r − r + )Im du dr u is zero at r = r + . Thus evaluating the radial ODE at ε = 0, multiplying it by u, taking real parts and integrating by parts yields For α = 0, the derivative is given by (3.13) Noting that π 0 |S| 2 sin ϑ dϑ = 1 and using Lemma 3.15 to eliminate the dependence on λ and then Proposition 3.12, we conclude that ∂α/∂ε(0) needs to be positive to make the integrand of (3.12) negative. The restriction to α = 0 can by removed by continuity of the reflection and transmission coefficients A and B.

A continuity argument
We now deduce Theorem 1.8 from Theorem 1.6. In this section, we fix > 0 and α 0 < 9/4. In the previous sections, we have produced a curve ε → α 0 (ε) of masses with ∂α 0 (0)/∂ε > 0. This means that the constructed mode solutions will solve a radial ODE with a different scalar mass. This section formalises the intuitive idea of "following up" the curves ε → α(ε) starting at an α close to α 0 until one "hits" the desired mass, which is made possible by ∂α(0)/∂ε > 0. The proof consists simply in establishing necessary continuity and carefully choosing neighbourhoods. This can be divided into two independent steps.
1. We show that the function mapping α to the correspondingâ is left-continuous.
2. We show that, for α and correspondingâ sufficiently close to α 0 and the correspondingâ 0 , the implicit function theorem guarantees a curve, starting at α and the correspondingâ and real frequency ω + , which exists "long enough" to "hit" α 0 .
Note that, for any f ∈ C ∞ 0 ,(α, r + , a) → L α,r + ,a (f ) defines a continuous function. For a given f ∈ C ∞ 0 , define the family of sets Remark 3.16. A α,r + corresponds to the set A from Section 3.1.
We shall fix r + now. Moreover, we shall fix an m ≥ m 0 > 0.
Proof. Suppose Φ(·, r + ) was not non-increasing. Then inf A α,r + < inf A α ,r + for some α < α . Hence there is an a > 0 with inf A α,r + < a < inf A α ,r + and an f ∈ C ∞ 0 such that This contradicts that L α,r + ,a (f ) > L α ,r + ,a (f ) for all f and a if α < α .
Hence in this neighbourhood, the vector field is well-defined. It is this vector field whose integral curves describe the solutions given by the implicit function theorem as applied in Section 3.2. In particular, solving the ODE with initial conditions (α, Ω R (α), 0, Φ(α, r + )) (α ∈ I) gives the previously introduced α(ε) and ω R (ε).
Lemma 3.20. T is continuous.
The function Φ(·, r + ) induces the curve Γ : α → (α, Ω R (α), 0, Φ(α, r + )) t for α ∈ I; it is continuous on the left at α 0 . The result of the previous section says that along this curve, the implicit function theorem produces parameter curves that correspond to superradiant modes; these parameter curves are exactly the integral curves of W starting on a point of Γ. Since Γ is left-continuous, This shows Theorem 1.8

Growing mode solutions satisfying Neumann boundary conditions 4.1. Existence of real mode solutions
In this section, we will construct growing mode solutions satisfying Neumann boundary conditions. Every result has a counterpart in Section 3. In the following, whenever proofs will be short in detail, the reader can extract those from Section 3. The two novel techniques in this section are the use of twisted derivatives with appropriately modified Sobolev spaces and a new Hardy inequality (Lemma 4.2). Fix > 0 and r + . In this section, we look at the range 5/4 < α < 9/4, i. e. 0 < κ < 1, for Neumann boundary conditions.
To treat the Neumann case variationally, we need to modify the functional, so it becomes finite for Neumann modes. We achieve this by conjugating the derivatives by a power of r; more precisely, we consider the twisted derivative h d dr h −1 · , where h = r −1/2+κ . This "kills off" the highest order term of the Neumann branch. Moreover, squaring the twisted derivative term does not introduce any "mixed terms" in f and its derivative; it only produces a zeroth order term that also makes the potential finite.
Thus introduce the twisted variational functional whereṼ h a as in Appendix B, i. e.
By Lemma B.2,Ṽ h a = O(1) andṼ h a is positive near infinity for sufficiently large |m|, which shall be assumed henceforth. Moreover,Ṽ h a is chosen such that the twisted variational problem leads to the same Euler-Lagrange equation as the untwisted one.
Proof. The existence of a continuous version follows from Sobolev embedding as in Lemma 3.2. Then, there exists a sequence (f n ) ∈ C ∞ as in the definition such that f n → f in H 1 κ . Let R > R > r + and let f n (R) converge to f (R): Hence r 1/2−κ (f n (r) − f (r)) converges uniformly for all r ≥ R. Hence we even have convergence at r = ∞. Since lim r→∞ r 1/2−κ f n (r) = ∞ for all n, we obtain the result.
As in Section 3.1, choose mode parameters such that the conditions for Lemma 2.19 are satisfied. Let Note that A is non-empty, open and bounded below.
Lemma 4.3. Let a ∈ A be fixed. There exist constants r + < B 0 < B 1 < ∞ and C 0 , C 1 > 0, such that, for large enough m, we have for all smooth functions f for which the following integrals are defined that Proof. The proof follows the strategy of Lemma 3.3. The analysis of the potential goes through as in Section 3.1 as the twisting part ofṼ h a does not depend on m and has the right asymptotics. Thus we know that there is an R 1 such that for some large constant C > 0 by the Hardy inequality of Lemma 4.2. Choosing B 0 ,B 1 , C 0 and C 1 appropriately (as in the proof of Lemma 3.3), we obtain the inequality.
Proof. See the comments to Lemma 3.5.
Lemma 4.5. Let a ∈ A. There exists an f a ∈ H 1 κ (r + , ∞) with norm f a L 2 (r + ,∞) = 1 such thatL a achieves its infimum in Proof. The proof is similar to the one in Section 3.1. We obtain a minimising sequence (f a,n ) that converges weakly in H 1 κ and strongly in L 2 on compact subsets of (r + , ∞). In analogy to the Dirichlet, the f a,n are can be taken from a dense subset and can be chosen to be of the form f a,n = r −1/2+κ g a,n for g a,n smooth and compactly supported around infinity.
We will show that the norm is conserved. Suppose not. Then, for any N , there are infinitely many of the f a,n such that f a,n L 2 ((r + ,∞)\[r + +1/N,N ]) ≥ > 0.
Suppose f a,n L 2 (r + ,r + +δ) ≥ 1 > 0 for infinitely many f a,n and any δ > 0. Because of the L 2 convergence on compact subsets, there is an R such that f a,n (R) → f a (R) as n → ∞, in particular f a,n (R) is bounded for all n. By Lemma 4.3, we have for r ∈ (r + , R): , we obtain f a,n L 2 (r + ,r + +δ) → 0 as δ → 0, a contradiction. Hence we only need to exclude the case that the norm is bounded away from zero for large r. Thus, suppose that f a,n L 2 (R 0 ,∞) ≥ 2 > 0 for infinitely many f a,n and any R 0 > 0. Since f a,n (r + ) = 0, we have , which is uniformly bounded for all n. Hence As in the proof of Lemma 3.6, we have L a (f a,n ) = ν a and the rest follows.
We would like to derive the Euler-Lagrange equation corresponding to this minimiser.
The proof of Lemma 4.6 can be found in Appendix C.
Proposition 4.7. There is anâ and a corresponding non-zero function fâ ∈ C ∞ (r + , ∞) such that and f a satisfies the horizon regularity condition and the Neumann boundary condition at infinity.
Proof. As in Proposition 3.8, we find an fâ ∈ H 1 κ such that Choosing ψ(r) = r − 1 2 +κ g(r) with g having compact support around infinity and integrating by parts, we obtain ∆ − r 2 +â 2 r −1+2κ d dr r 1 2 −κ fâ g → 0 as r = ∞ for all g as in (4.1). This yields the asymptotics. Moreover, as in the proof of Proposition 3.8, we retrieve the ODE. The boundary condition at the horizon follows analogously to Section 3.1.

Perturbing the Neumann modes into the complex plane
In Section 4.1, we constructed real mode solutions for 5/4 < α < 9/4 satisfying Neumann conditions. For the growing radial parts, we proceed as in Section 3.1 with the difference that here, finding a mode solution is equivalent to finding a zero of B. The present case is considerably more difficult than the Dirichlet case. A first manifest difference is the asymmetry in the definitions of Dirichlet and Neumann boundary conditions since a Dirichlet mode has more decay than required by Definition 1.5. This means that if a function satisfies the Dirichlet boundary condition for a mass α 1 , it also does so for every α 2 sufficiently close to α 1 . As Definition 1.13 is tighter, this is not true in the Neumann case. Another difficulty stems from twisting as the dependence of the equations on α becomes more complicated.
By the horizon regularity condition, u ∼ (r − r + ) ξ near the horizon, u α is smooth at r = r + . However, u α does not satisfy the Neumann condition at infinity as the second term behaves as r −1/2+κ u log r. Let f : (r + , ∞) → C be C 1 and piecewise C 2 . Then the function v(r) does satisfy the Neumann boundary condition if, for large r, f (r) = log r + O(r −γ ), where γ > 0. From the radial ODE, we obtain d dr Lemma B.1 yields a twisted version for an R > r + + 1 to be determined. Note that f is continuously differentiable. For r > R, whence f is monotonic. First, we choose R sufficiently large such thatṼ h a > 0 for r > R according to Lemma B.2, whence which is non-negative since the first integral with the constant coefficient is zero and the second integral is positive. For r > R, one easily computes We can prove the following Hardy inequality: Lemma 4.8. Let u satisfy the Neumann boundary condition at infinity and let β > 0. Then Proof. We compute yielding the result. Since as R → ∞, by choosing R possibly larger, we obtain For convenience, set C 2 := 4β 2 C 1 . SinceṼ h a ∼ r −2 , we need to show that (4.7) We will deal with the two cases 0 < κ ≤ 1/2 and 1/2 < κ < 1 separately. Let us first consider 0 < κ ≤ 1/2. We choose β = 2κ. Note that in this case Lemma 4.9. Let C > 0. There is an R such that, for all r > R, Proof. It suffices to show that As this holds at r = R, it suffices to show the statement for the derivatives. Substituting x := r 2κ , we need to show Therefore, the result holds if R 4κ > 4κC.
Lemma 4.10. Let C > 0 and 1/2 < κ < 1. There is an R such that, for all r > R, Proof. As equality holds for r = R, it suffices to consider the derivatives, i. e. we would like to establish This again holds for r = R, so, after dividing the inequality by r, it suffices to prove the corresponding inequality for the derivatives, i. e.
The last term on the left hand side is always positive. Thus the left hand side is greater than which is positive for sufficienlty large R.
Therefore, for both ranges of κ, the right hand side of (4.5) is bounded below by for non-trivial u, a contradiction. Thus we have shown the following Lemma 4.11.

Behaviour for small ε > 0 for Neumann boundary conditions
The main new idea of this section can be found in the proof of Proposition 4.13, where the insights of Section 4.2 are essential to overcome the difficulties outlined at the beginning of the previous section.
From now on we fix m -see Remark 3.13. Observe that as in the proof of Proposition 3.14 dr. (4.12) The expression for Re ∂Ṽ ∂ε ε=0 can be taken from (3.13). First one can eliminate the explicit λ dependence via Lemma 3.15 and one obtains a lower bound on the right hand side using Proposition 4.12. Then suppose for the sake of contradiction that ∂α/∂ε ≤ 0. It follows immediately from Section 4.2 that the right hand side of (4.12) is positive, a contradiction.

The continuity argument for Neumann boundary conditions
To apply the continuity argument to the Neumann case, we need to take the two steps outlined in the introduction to Section 3.4. The second step merely relied on continuity properties of A and the monotonicity properties of ω(ε) and α(ε) established in Sections 3.3 and 4.3, respectively; in particular, it did not rely directly on properties of the functional. Hence this part of the argument can be carried out almost verbatim. Therefore, we only need to deal with the first step here.
Moreover, we define Φ : (5/4, 9/4) × (0, ∞) → (0, ∞), Φ(α, r + ) := inf A α,r + if A α,r + = ∅. Instead of showing monotonicity for Φ, we will define a left-continuous function Ψ that can play the rôle of Φ in the continuity argument. For each α ∈ (5/4, 9/4), there will be a value Φ(α, r + ) for a such that there is a real mode solution satisfying the Neumann boundary condition for α and this a. The function Ψ(·, r + ) will essentially look like Φ(·, r + ), but will be modified on potential jump points to achieve left-continuity. The arguments of Sections 4.2 and 4.3 (which depend only on the existence of a Neumann mode solution) can be repeated for Ψ(α, r + ) instead of Φ(α, r + ), thus we can substitue Φ by Ψ in the remainder of the proof of Section 3.4.
As we can repeat this construction for all ε > 0 and all jump points α 0 , we obtain a function Ψ(·, r + ) defined in (5/4, 9/4), whose values correspond to parameters a with periodic mode solutions.

Acknowledgements
I would like to thank my supervisors Mihalis Dafermos and Gustav Holzegel for proposing this project, helpful discussions, guidance and advice on the exposition. I would also like to thank Yakov Shlapentokh-Rothman for useful conversations and advice. I gratefully acknowledge the financial support of EPSRC, the Cambridge Trust and the Studienstiftung des deutschen Volkes.