Nonlinear Stability for the Maxwell–Born–Infeld System on a Schwarzschild Background

Federico Pasqualotto

doi:10.1007/s40818-019-0075-2

Annals of PDE

December 2019, 5:19 | Cite as

Nonlinear Stability for the Maxwell–Born–Infeld System on a Schwarzschild Background

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Federico Pasqualotto

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First Online: 03 December 2019

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Abstract

In this paper we prove small data global existence for solutions to the Maxwell–Born–Infeld (MBI) system on a fixed Schwarzschild background. This system has appeared in the context of string theory and can be seen as a nonlinear model problem for the stability of the background metric itself, due to its tensorial and quasilinear nature. The MBI system models nonlinear electromagnetism and does not display birefringence. The key element in our proof lies in the observation that there exists a first-order differential transformation which brings solutions of the spin $\pm 1$ Teukolsky Equations, satisfied by the extreme components of the field, into solutions of a “good” equation (the Fackerell–Ipser Equation). This strategy was established in Pasqualotto (Ann Henri Poincaré 20(4):1263–1323, 2019) for the linear Maxwell field on Schwarzschild. We show that analogous Fackerell–Ipser equations hold for the MBI system on a fixed Schwarzschild background, which are however nonlinearly coupled. To essentially decouple these right hand sides, we set up a bootstrap argument. We use the $r^p$ method of Dafermos and Rodnianski (A new physical-space approach to decay for the wave equation with applications to black hole spacetimes. In: XVIth international congress on mathematical physics. World Science Publication, Hackensack, pp 421–432, 2010) in order to deduce decay of some null components, and we infer decay for the remaining quantities by integrating the MBI system as transport equations.

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Notes

Acknowledgements

I would like to thank Prof. Jonathan Luk for his patience and guidance, for suggesting the problem to me, and for inviting me to Stanford University to finish the project. I would also like to thank my advisor, Prof. Mihalis Dafermos, for his patience, his encouragement and his comments on preliminary versions of the manuscript. Moreover, I thank Jan Sbierski for the suggestion to look at Fritz John’s approach to local existence. Moreover, I thank John Anderson and Yakov Shlapentokh-Rothman for very valuable discussions.

Appendix A. Computation with the Projected Connection

Lemma A.1

We have the following identities:

$$\begin{aligned} \,\,/\!\!{\nabla }_L \,\,/\!\!{g}= 0, \qquad \,\,/\!\!{\nabla }_{\underline{L}}\,\,/\!\!{g}= 0, \end{aligned}$$

as well as the commutation relations

$$\begin{aligned}{}[\,\,/\!\!{\nabla }_L, \,\,/\!\!{\nabla }_{\underline{L}}] = 0, \qquad [\,\,/\!\!{\nabla }_L, r \,\,/\!\!{\nabla }] = 0, \qquad [\,\,/\!\!{\nabla }_{\underline{L}}, r\,\,/\!\!{\nabla }] = 0. \end{aligned}$$

The last two equalities are meant in the following sense: if Open image in new window

is a tensor of type (N, M) tangential to the spheres of constant r, then we have

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and similarly

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Lemma A.2

We have the following identities:

$$\begin{aligned}{}[\,\,/\!\!{\nabla }_L, \,/\!\!\!{\mathcal {L}}_{\Omega _i}] = 0, \qquad [\,\,/\!\!{\nabla }_{\underline{L}}, \,/\!\!\!{\mathcal {L}}_{\Omega _i}] = 0, \qquad [\,\,/\!\!{\nabla }, \,/\!\!\!{\mathcal {L}}_{\Omega _i}] = 0. \end{aligned}$$

Here, $\,/\!\!\!{\mathcal {L}}$ is the Lie derivative induced by the connection $\,\,/\!\!{\nabla }$.

Proof

We notice that, since $\Omega _i$ are Killing for $\,\,/\!\!{g}$, the last equality holds true by standard theory, cf. [12].

We will only prove the first equality when the derivatives are acting on a spherical one-form $\omega $. Let $e_i := \Omega _i/r$. Let $\eta $ be a spherical one form, and $X:= e_j$. We compute

$$\begin{aligned} (\,/\!\!\!{\mathcal {L}}_{\Omega _i} \eta )(X) = (\,\,/\!\!{\nabla }_{\Omega _i} \eta )(X) + \eta (\,\,/\!\!{\nabla }_{X} \Omega _i). \end{aligned}$$

(see Lemma 3.2.1 in [44]). Then,

$$\begin{aligned} \begin{aligned}&([\,\,/\!\!{\nabla }_L, \,/\!\!\!{\mathcal {L}}_{\Omega _i}] \omega )(X) \\&= L(\,/\!\!\!{\mathcal {L}}_{\Omega _i}\omega (X))-(\,/\!\!\!{\mathcal {L}}_{\Omega _i})(\,\,/\!\!{\nabla }_L X)-\Omega _i(\,\,/\!\!{\nabla }_L \omega (X)) + (\,\,/\!\!{\nabla }_L \omega )([\Omega _i, X]) \\&= L(\Omega _i (\omega (X))-\omega ([\Omega _i, X]))-\Omega _i (\omega (\,\,/\!\!{\nabla }_L X))+\omega ([\Omega _i, \,\,/\!\!{\nabla }_L X])\\&\quad -\Omega _i (L(\omega (X))-\omega (\,\,/\!\!{\nabla }_L X))+L(\omega ([\Omega _i, X]))-\omega (\,\,/\!\!{\nabla }_L [\Omega _i, X]) = 0. \end{aligned} \end{aligned}$$

Since $\,\,/\!\!{\nabla }_L X = 0$, as well as $\,\,/\!\!{\nabla }_L [\Omega _i, X] = 0$. This proves the lemma, as the quantity in the beginning is a tensor, so it suffices to verify is vanishes on a basis at each point. $\square $

Appendix B. Shortand Notation for Tensors

Let $N \in {\mathbb {N}}_{\ge 0}$, let $A_1, A_2, \ldots , A_n$ be tensors of type $(k_i, l_i)$, $i \in \{1, \ldots , n\}$ on $T {\mathcal {S}}$. Using the Schwarzschild metric, we can suppose that such tensors are all covariant of order $k_i$, so that

$$\begin{aligned} A_h = (A_h)_{\alpha ^h_{1} \ldots \alpha ^h_{k_h}}. \end{aligned}$$

Definition B.1

(Shorthand notation for tensors on ${\mathcal {S}}$) Let $m, N \in {\mathbb {N}}_{\ge 0}$, let ${\mathcal {R}} \subset {\mathcal {S}}$. Then there exists a smooth covariant tensor field T on $T {\mathcal {R}}$, such that the following holds:

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Remark B.2

We will often require T to satisfy the following type of bound on ${\mathcal {R}}$: we require the existence of a constant C depending only on ${\mathcal {R}}$, M and N such that

$$\begin{aligned} |\partial ^N T| \le C_{M, N, {\mathcal {R}}}, \end{aligned}$$

in the sense of Definition 2.18 (without loss of generality, we can suppose T to be covariant).

Remark B.3

Fix $r_{\text {in}} \in (0, 2M)$. Notice that, if we consider the region ${\mathcal {R}} := {\mathcal {S}}_e {\setminus } \{r_1 \le r_{\text {in}}\}$, the Riemann tensor $\text {Rm}\,$ satisfies (w.l.o.g. it is covariant)

$$\begin{aligned} |\partial ^N \text {Rm}\,| \le C_{M, N, r_{\text {in}}}. \end{aligned}$$

Appendix C. Positivity of a Flux of $\dot{Q}$ and the Inverse MBI Metric

This section closely follows similar calculations done by J. Speck in the paper [44] (Proposition 7.4.4).

Given ${\mathcal {F}}$ solution to the MBI system (2.7), we define the inverse MBI metric as

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(C.1)

We prove the following lemma.

Lemma C.1

(Positivity of $\dot{Q}(T, X)$) Let ${\mathcal {F}}$ be a solution to the MBI system (2.7) on a region ${\mathcal {R}} \subset {\mathcal {S}}_e$. Define the vectorfield $X^\nu $ to be

$$\begin{aligned} (b^{-1})^{\nu \mu } g_{\mu \alpha } T^\alpha . \end{aligned}$$

(C.2)

Then, as long as

$$\begin{aligned} \ell _{(\text {MBI})}> 0 \end{aligned}$$

(C.3)

on ${\mathcal {R}}$, we have that the contraction

$$\begin{aligned} \dot{Q}_{\mu \nu } T^\mu X^\nu \ge 0. \end{aligned}$$

(C.4)

Proof

Let us begin by defining the shorthand notation ${\dot{{\mathcal {F}}}} := {\dot{{\mathcal {F}}}}_{,I,0}$. We decompose the field in the electric and magnetic parts.

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ans similarly we define $\dot{E}$, $\dot{B}$ using the tensor ${\dot{{\mathcal {F}}}}$. We note that $E, B, \dot{E}, \dot{B}$ are parallel to the foliation of ${\mathcal {S}}_e$ by surfaces of constant t. We therefore denote by $\langle \cdot , \cdot \rangle $ the (positive definite) inner product induced by the Schwarzschild metric on that foliation. With this notation, we calculate the contractions:

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(C.5)

Let us now denote by latin letters $a, b, c, \ldots $ indices relative to tensors tangent to the surfaces $\{t = \text {const}\}$. Recall:

$$\begin{aligned} 2\ell _{(\text {MBI})}^2(1+\mathbf{F }) X^\mu = 2 \ell _{(\text {MBI})}^2 (1+\mathbf{F })(b^{-1})^{\mu \nu } T_\nu = 2 \ell _{(\text {MBI})}^2(1+\mathbf{F })T^\mu - 2 \ell _{(\text {MBI})}^2 {\mathcal {F}}^{\mu \kappa }{\mathcal {F}}_{T\kappa }. \end{aligned}$$

(C.6)

We then have

$$\begin{aligned} 2\ell _{(\text {MBI})}^2(1+\mathbf{F }) \dot{Q}_{\mu \nu }T^\mu X^\nu = 2 \ell _{(\text {MBI})}^2(1+(1-\mu )^{-1}|B|^2) \dot{Q}_{TT} - 2 \ell _{(\text {MBI})}^2 \dot{Q}_{Ta} {\mathcal {F}}^{a \kappa } {\mathcal {F}}_{T \kappa }. \end{aligned}$$

(C.7)

Recalling the form of $\dot{Q}$, Equation (7.2), we have

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(C.8)

We then compute the first term in (C.7):

$$\begin{aligned} \begin{aligned} 2 \ell _{(\text {MBI})}^2 \dot{Q}_{TT}&= \ell _{(\text {MBI})}^2 (|\dot{E}|^2 + |\dot{B}|^2)\\&\quad + (1-\mu )^{-1}(\langle E, \dot{E}\rangle ^2 - \langle B, \dot{B}\rangle ^2)\\&\quad +(1+(1-\mu )^{-1}(|B|^2-|E|^2))(1-\mu )^{-1}(\langle B, \dot{E} \rangle ^2 - \langle E, \dot{B} \rangle ^2)\\&\quad +2(1-\mu )^{-2}\langle E, B\rangle (\langle B, \dot{B}\rangle \langle E, \dot{B}\rangle + \langle E ,\dot{E} \rangle \langle B, \dot{E} \rangle ). \end{aligned} \end{aligned}$$

We compute the second term in (C.7):

$$\begin{aligned} \begin{aligned} \ell _{(\text {MBI})}^2 \dot{Q}_{T a} {\mathcal {F}}^{a\kappa } {\mathcal {F}}_{T \kappa }&= - (1-\mu )^{-2}\langle B, \dot{B} \rangle ^2 \left( |E|^2 + (1-\mu )^{-1}\langle E, B\rangle \right) \\&\quad + (1-\mu )^{-2}\langle B , \dot{B}\rangle \langle B, \dot{E}\rangle (1+(1-\mu )^{-1}|B|^2)\langle E,B \rangle \\&\quad + 2(1-\mu )^{-2} \langle B, \dot{B}\rangle \langle E,\dot{B} \rangle (1+ (1-\mu )^{-1}|B|^2)\langle E,B \rangle \\&\quad +(1-\mu )^{-1} \langle B,\dot{B} \rangle \langle E, \dot{E} \rangle (1+(1-\mu )^{-1}|B|^2)\\&\quad -(1-\mu )^{-2}\langle E,\dot{B} \rangle \left\{ (1+(1-\mu )^{-1}(|B|^2-|E|^2))|B|^2+(1-\mu )^{-1}\langle E, B\rangle ^2 \right\} \\&\quad -(1-\mu )^{-1}\langle E, \dot{B} \rangle \langle B, \dot{E} \rangle (1+(1-\mu )^{-1}(|B|^2-|E|^2))(1+(1-\mu )^{-1}|B|^2)\\&\quad -(1-\mu )^{-2}\langle E,\dot{B} \rangle \langle E, \dot{E} \rangle \langle E, B\rangle (1+(1-\mu )^{-1}|B|^2) \end{aligned} \end{aligned}$$

(C.9)

These expressions are exactly the same as the ones appearing in the paper [44], with the formal substitutions

$$\begin{aligned} E \rightarrow (1-\mu )^{-\frac{1}{2}} E, \qquad B \rightarrow (1-\mu )^{-\frac{1}{2}} B, \qquad \dot{E} \rightarrow \dot{E}, \qquad \dot{B} \rightarrow \dot{B}. \end{aligned}$$

We proceed to show nonnegativity (in the components of $\dot{E}$, $\dot{B}$) for the resulting quadratic form:

$$\begin{aligned} 2\ell _{(\text {MBI})}^2(1+\mathbf{F }) \dot{Q}_{\mu \nu }T^\mu X^\nu = 2 \ell _{(\text {MBI})}^2(1+(1-\mu )^{-1}|B|^2) \dot{Q}_{TT} - 2 \ell _{(\text {MBI})}^2 \dot{Q}_{Ta} {\mathcal {F}}^{a \kappa } {\mathcal {F}}_{T \kappa }. \end{aligned}$$

We choose vectorfields $e_1$, $e_2$, $e_3$ such that $E \in \text {span} \{e_1\}$, $B \in \text {span}\{e_1, e_2\}$, $\langle e_i, e_3 \rangle = \delta _{i3}$, $|e_i| =1$. We decompose:

$$\begin{aligned} E&= E_1 e_1,\\ B&= B_1 e_1 + B_2 e_2,\\ \dot{E}&= \dot{E}_1 e_1 + \dot{E}_2 e_2 + \dot{E}_3 e_3,\\ \dot{B}&= \dot{B}_1 e_1 + \dot{B}_2 e_2 + \dot{B}_3 e_3. \end{aligned}$$

Due to the orthogonality of $e_1, e_2, e_3$, the only terms containing $E_3$ and $B_3$ are the terms arising from the first term in $2\ell _{(\text {MBI})}^2(1+\mathbf{F }) \dot{Q}_{\mu \nu }T^\mu X^\nu $, i.e. the “linear” term

$$\begin{aligned} \ell _{(\text {MBI})}^2 (1 + (1-\mu )^{-1}|B|^2)(E_3^2+B_3^2), \end{aligned}$$

(C.10)

which is manifestly positive in $E_3, B_3$ if $\ell _{(\text {MBI})}> 0$ on ${\mathcal {R}}$.

We then proceed to calculate the components of the matrix A such that

$$\begin{aligned}&(B_1, B_2, E_1, E_2) A (B_1, B_2, E_1, E_2)^t = 2\ell _{(\text {MBI})}^2(1+\mathbf{F }) \dot{Q}_{\mu \nu }T^\mu X^\nu \nonumber \\&\quad - \ell _{(\text {MBI})}^2 (1 + (1-\mu )^{-1}|B|^2)(E_3^2+B_3^2). \end{aligned}$$

(C.11)

Here, the superscript $^t$ denotes transposition. We have that A is obviously a symmetric matrix, with entries

$$\begin{aligned} A_{11}&= \left( \frac{B_2^2-E_1^2}{1-\mu }+1\right) \left( \frac{B_2^2 E_1^2}{(1-\mu )^2}+\ell _{(\text {MBI})}^2\right) \\ A_{12}&= -\frac{B_1 B_2 \left( \frac{B_2^2 E_1^2}{(1-\mu )^2}+\ell _{(\text {MBI})}^2\right) }{1-\mu }\\ A_{13}&= \frac{B_1 B_2^2 E_1 \left( \frac{B_1^2+B_2^2}{1-\mu }+1\right) }{(\mu -1)^2}\\ A_{14}&= (1 - \mu )^{-1} B_2 E_1 \left( 1 + \frac{B_1^2 + B_2^2}{1-\mu }\right) \left( 1 + \frac{B_2^2 - E_1^2}{1-\mu }\right) \\ A_{22}&= \left( \frac{B_1^2}{1-\mu }+1\right) \left( \frac{B_2^2 E_1^2}{(1-\mu )^2}+\ell _{(\text {MBI})}^2\right) \\ A_{23}&=-\frac{B_2 E_1 \left( \frac{B_1^2}{1-\mu }+1\right) \left( \frac{B_1^2+B_2^2}{1-\mu }+1\right) }{1-\mu } \\ A_{24}&=-\frac{B_1 B_2^2 E_1 \left( \frac{B_1^2+B_2^2}{1-\mu }+1\right) }{(1-\mu )^2} \\ A_{33}&= \left( \frac{B_1^2}{1-\mu }+1\right) \left( \frac{B_1^2+B_2^2}{1-\mu }+1\right) ^2\\ A_{34}&= \frac{B_1 B_2 \left( \frac{B_1^2+B_2^2}{1-\mu }+1\right) ^2}{1-\mu }\\ A_{44}&= \left( \frac{B_1^2+B_2^2}{1-\mu }+1\right) ^2 \left( \frac{B_2^2-E_1^2}{1-\mu }+1\right) . \end{aligned}$$

We now denote by $M_k$ the k-th principal minor of the matrix A: $M_k := (A_{ij})_{(i,j)\in \{1, \ldots , k\}\times \{1, \ldots , k\}}$. Let us calculate the determinants of such minors:

$$\begin{aligned} \det (M_1)&= \left( \frac{B_2^2-E_1^2}{1-\mu }+1\right) \left( \frac{B_2^2 E_1^2}{(1-\mu )^2}+\ell _{(\text {MBI})}^2\right) , \end{aligned}$$

(C.12)

$$\begin{aligned} \det (M_2)&=\frac{\ell _{(\text {MBI})}^2 \left( B_2^2 E_1^2+\ell _{(\text {MBI})}^2 (1-\mu )^2\right) ^2}{(1-\mu )^4} ,\end{aligned}$$

(C.13)

$$\begin{aligned} \det (M_3)&= \frac{\ell _{(\text {MBI})}^4 \left( B_1^2+1-\mu \right) \left( B_1^2+B_2^2+1-\mu \right) ^2 \left( B_2^2 E_1^2+\ell _{(\text {MBI})}^2 (1-\mu )^2\right) }{(1-\mu )^5},\end{aligned}$$

(C.14)

$$\begin{aligned} \det (M_4)&= \frac{\ell _{(\text {MBI})}^8 \left( B_1^2+B_2^2+1-\mu \right) ^4}{(1-\mu )^4} . \end{aligned}$$

(C.15)

Since ${\mathcal {R}} \subset {\mathcal {S}}_e$, and since we are assuming $\ell _{(\text {MBI})}> 0$, the expressions in (C.13), (C.14), (C.15) are manifestly positive.

Concerning $\det (M_1)$, we distinguish two cases.

If $1-(1-\mu )^{-1}E_1 > 0$, clearly $\det (M_1) > 0$,
If $1-(1-\mu )^{-1}E_1 \le 0$, we notice
$$\begin{aligned} \frac{B_2^2-E_1^2}{1-\mu }+1 = \ell _{(\text {MBI})}^2 - (1-\mu )^{-1}B_1^2(1-(1-\mu )^{-1}E_1) \ge \ell _{(\text {MBI})}^2 > 0. \end{aligned}$$

This concludes the proof of the Lemma. $\square $

Appendix D. Calculations to Deduce the Fackerell–Ipser Equations

In this appendix, we collect useful calculations which are used to derive the form of the Fackerell–Ipser Equations in Propositions 10.1 and 10.2 . We restrict here to the case of the linear Maxwell system. In the proofs of Propositions 10.1 and 10.2, we extend the reasoning to the MBI case.

Let’s first prove a simple lemma about commutation of derivatives.

Lemma D.1

We have the equation

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Proof

Commuting derivatives,

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This implies the claim. $\square $

D.1. The Wave Equation, Maxwell Case

Let’s derive the wave equation for $\rho $ and $\sigma $ suppressing the nonlinear term.

Lemma E.1

Let ${\mathcal {F}}$ satisfy the Maxwell system

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We then have

$$\begin{aligned}&- r^{-2} L {\underline{L}}(r^2 \rho ) +(1-\mu ) \,\,/\!\!\!{\Delta }\rho = 0, \end{aligned}$$

(D.1)

$$\begin{aligned}&- r^{-2} L {\underline{L}}(r^2 \sigma ) +(1-\mu ) \,\,/\!\!\!{\Delta }\sigma = 0. \end{aligned}$$

(D.2)

Here, $\sigma $ and $\rho $ are the middle components defined in Equation (2.29).

Proof

We calculate:

$$\begin{aligned} \begin{aligned}&\nabla _\mu \nabla ^\mu (L^\nu {\underline{L}}^\alpha {\mathcal {F}}_{\nu \alpha }) \\&= (\nabla _\mu \nabla ^\mu L^\nu ) {\underline{L}}^\alpha {\mathcal {F}}_{\nu \alpha } + (\nabla ^\mu L^\nu )(\nabla _\mu {\underline{L}}^\alpha ) {\mathcal {F}}_{\nu \alpha } + (\nabla ^\mu L^\nu ) {\underline{L}}^\alpha \nabla _\mu {\mathcal {F}}_{\nu \alpha }\\&\quad + (\nabla _\mu L^\nu )(\nabla ^\mu {\underline{L}}^\alpha ) {\mathcal {F}}_{\nu \alpha } + L^\nu (\nabla _\mu \nabla ^\mu {\underline{L}}^\alpha ) {\mathcal {F}}_{\nu \alpha } + L^\nu (\nabla ^\mu {\underline{L}}^\alpha ) \nabla _\mu {\mathcal {F}}_{\nu \alpha }\\&\quad + (\nabla _\mu L^\nu ) {\underline{L}}^\alpha \nabla ^\mu {\mathcal {F}}_{\nu \alpha } + L^\nu (\nabla _\mu {\underline{L}}^\alpha ) \nabla ^\mu {\mathcal {F}}_{\nu \alpha } + L^\nu {\underline{L}}^\alpha \nabla ^\mu \nabla _\mu {\mathcal {F}}_{\nu \alpha } \\&= (a) +(b)+(c)\\&\quad + (d) +(e) +(f) \\&\quad + (g)+ (h) + (i). \end{aligned} \end{aligned}$$

We first consider $(c) + (f) + (g) + (h)$. We obtain

$$\begin{aligned} \begin{aligned}&(c) + (f) + (g) + (h) \\&= (\nabla ^\mu L^\nu ) {\underline{L}}^\alpha \nabla _\mu {\mathcal {F}}_{\nu \alpha } + L^\nu (\nabla ^\mu {\underline{L}}^\alpha ) \nabla _\mu {\mathcal {F}}_{\nu \alpha }+ (\nabla _\mu L^\nu ) {\underline{L}}^\alpha \nabla ^\mu {\mathcal {F}}_{\nu \alpha } + L^\nu (\nabla _\mu {\underline{L}}^\alpha ) \nabla ^\mu {\mathcal {F}}_{\nu \alpha } \\&= 2 (\nabla ^\mu L^\nu ) {\underline{L}}^\alpha \nabla _\mu {\mathcal {F}}_{\nu \alpha } + 2L^\nu (\nabla ^\mu {\underline{L}}^\alpha ) \nabla _\mu {\mathcal {F}}_{\nu \alpha } \\&= 2 (\nabla ^\mu L)^L \nabla _\mu {\mathcal {F}}_{L {\underline{L}}} + 2 (\nabla ^\mu L)^\theta \nabla _\mu {\mathcal {F}}_{\theta {\underline{L}}} + 2 (\nabla ^\mu L)^\varphi \nabla _\mu {\mathcal {F}}_{\varphi {\underline{L}}} \\&\quad + 2(\nabla ^\mu {\underline{L}})^{\underline{L}}\nabla _\mu {\mathcal {F}}_{L {\underline{L}}}+ 2(\nabla ^\mu {\underline{L}})^\theta \nabla _\mu {\mathcal {F}}_{L \theta }+ 2(\nabla ^\mu {\underline{L}})^\varphi \nabla _\mu {\mathcal {F}}_{L \varphi } \\&= 2 (\nabla _L L)^L \nabla ^L {\mathcal {F}}_{L {\underline{L}}} + 2(\nabla _{\underline{L}}{\underline{L}})^{\underline{L}}\nabla ^{\underline{L}}{\mathcal {F}}_{L {\underline{L}}} + (\text {angular terms}). \end{aligned} \end{aligned}$$

The first two terms in the last line of the previous equation then read:

$$\begin{aligned} \begin{aligned} 2 (\nabla _L L)^L \nabla ^L {\mathcal {F}}_{L {\underline{L}}} + 2(\nabla _{\underline{L}}{\underline{L}})^{\underline{L}}\nabla ^{\underline{L}}{\mathcal {F}}_{L {\underline{L}}} = \frac{2M}{r^2} (1-\mu )^{-1}(\nabla _L {\mathcal {F}}_{L {\underline{L}}}- \nabla _{\underline{L}}{\mathcal {F}}_{L {\underline{L}}}). \end{aligned} \end{aligned}$$

The angular terms, instead, become

$$\begin{aligned} \begin{aligned}&= 2 \frac{1-\mu }{r} (\nabla ^\theta {\mathcal {F}}_{\theta {\underline{L}}}+\nabla ^\varphi {\mathcal {F}}_{\varphi {\underline{L}}} )-2 \frac{1-\mu }{r} (\nabla ^\theta {\mathcal {F}}_{L\theta }+\nabla ^\varphi {\mathcal {F}}_{L\varphi } ) \\&{\mathop {=}\limits ^{(*)}} 2 \frac{1-\mu }{r} (-\nabla ^L {\mathcal {F}}_{L {\underline{L}}}+ \nabla ^{\underline{L}}{\mathcal {F}}_{L {\underline{L}}}) = \frac{1}{r} (\nabla _{\underline{L}}{\mathcal {F}}_{L {\underline{L}}}- \nabla _L {\mathcal {F}}_{L {\underline{L}}}) \\&= - \frac{1-\mu }{r} (1-\mu )^{-1}( \nabla _L {\mathcal {F}}_{L {\underline{L}}}-\nabla _{\underline{L}}{\mathcal {F}}_{L {\underline{L}}}). \end{aligned} \end{aligned}$$

Therefore,

$$\begin{aligned} \begin{aligned}&(c) + (f) + (g) + (h) = \frac{2M}{r^2} (1-\mu )^{-1}(\nabla _L {\mathcal {F}}_{L {\underline{L}}}- \nabla _{\underline{L}}{\mathcal {F}}_{L {\underline{L}}}) \\&\quad - \frac{1-\mu }{r} (1-\mu )^{-1}( \nabla _L {\mathcal {F}}_{L {\underline{L}}}-\nabla _{\underline{L}}{\mathcal {F}}_{L {\underline{L}}})\\&= (1-\mu )^{-1}\left( \frac{2M}{r^2} - \frac{1}{r} + \frac{2M}{r^2} \right) ( \nabla _L {\mathcal {F}}_{L {\underline{L}}}-\nabla _{\underline{L}}{\mathcal {F}}_{L {\underline{L}}}). \end{aligned} \end{aligned}$$

Let us notice that

$$\begin{aligned} \begin{aligned}&\nabla _\theta \partial _\theta = \frac{r}{2} {\underline{L}}- \frac{r}{2} L,\\&\nabla _\varphi \partial _\varphi = \sin ^2 \theta \frac{r}{2} {\underline{L}}- \sin ^2 \theta \frac{r}{2} L- \sin \theta \cos \theta \partial _\varphi . \end{aligned} \end{aligned}$$

(D.3)

Now, consider

$$\begin{aligned}&\nabla ^\mu \nabla _\mu {\underline{L}}= g^{L {\underline{L}}} (\nabla _L \nabla _{\underline{L}}{\underline{L}}- \nabla _{\nabla _{L}{\underline{L}}}{\underline{L}}) + g^{{\underline{L}}L} (\nabla _{\underline{L}}\nabla _L {\underline{L}}- \nabla _{\nabla _{{\underline{L}}} L}{\underline{L}}) \\&\quad + g^{\theta \theta } (\nabla _\theta \nabla _\theta {\underline{L}}- \nabla _{\nabla _\theta \partial _\theta } {\underline{L}}) + g^{\varphi \varphi } \left( \nabla _\varphi \nabla _\varphi {\underline{L}}- \nabla _{\nabla _\varphi \partial _\varphi }{\underline{L}}\right) \\&= - \frac{1}{2} (1-\mu )^{-1} \left( -\nabla _L\left( \frac{2M}{r^2} {\underline{L}}\right) \right) \\&\quad + r^{-2}\left( -\frac{1-\mu }{r}\nabla _\theta \partial _\theta - \nabla _{ \frac{r}{2} {\underline{L}}- \frac{r}{2} L}{\underline{L}}\right) \\&\qquad + r^{-2} \sin ^{-2} \theta \left( -\frac{1-\mu }{r}\nabla _\varphi \partial _\varphi - \nabla _{\sin ^2 \theta \frac{r}{2} {\underline{L}}- \sin ^2 \theta \frac{r}{2} L- \sin \theta \cos \theta \partial _\varphi }{\underline{L}}\right) \\&= \frac{1}{2} (1-\mu )^{-1} \left( -2 \frac{2M}{r^3} \right) (1-\mu ){\underline{L}}+ \\&\qquad r^{-2}\left( - \frac{1-\mu }{r} \left( \frac{r}{2} {\underline{L}}- \frac{r}{2} L \right) + \frac{r}{2} \frac{2M}{r^2} {\underline{L}}\right) \\&\quad + r^{-2} \sin ^{-2} \theta \left( - \frac{1-\mu }{r} \left( \sin ^2 \theta \frac{r}{2} {\underline{L}}- \sin ^2 \theta \frac{r}{2} L- \sin \theta \cos \theta \partial _\varphi \right) \right. \\&\left. \qquad - \frac{r}{2} \sin ^2 \theta \nabla _{\underline{L}}{\underline{L}}+ \sin \theta \cos \theta \nabla _\varphi {\underline{L}}\right) =\\&\quad - \frac{2M}{r^3} {\underline{L}}+ r^{-2}\left( - \frac{1-\mu }{2}{\underline{L}}+ \frac{1-\mu }{2} L + \frac{M}{r} {\underline{L}}\right) \\&\qquad + r^{-2} \left( - \frac{1-\mu }{2}{\underline{L}}+ \frac{1-\mu }{2} L + \frac{r}{2} \frac{2M}{r^2} {\underline{L}}\right) \\&=- \frac{2M}{r^3} {\underline{L}}+ 2r^{-2}\left( - \frac{1-\mu }{2}{\underline{L}}+ \frac{1-\mu }{2} L + \frac{M}{r} {\underline{L}}\right) \\&= - \frac{1-\mu }{r^2} {\underline{L}}+ \frac{1-\mu }{r^2} L. \end{aligned}$$

Similarly,

$$\begin{aligned}&\nabla ^\mu \nabla _\mu L = g^{L {\underline{L}}} (\nabla _L \nabla _{\underline{L}}L - \nabla _{\nabla _{L}{\underline{L}}}L) + g^{{\underline{L}}L} (\nabla _{\underline{L}}\nabla _L L - \nabla _{\nabla _{{\underline{L}}} L}L) \\&\quad + g^{\theta \theta } (\nabla _\theta \nabla _\theta L - \nabla _{\nabla _\theta \partial _\theta } L) + g^{\varphi \varphi } \left( \nabla _\varphi \nabla _\varphi L - \nabla _{\nabla _\varphi \partial _\varphi }L\right) \\&= -\frac{1}{2} (1-\mu )^{-1} \nabla _{\underline{L}}\left( \frac{2M}{r^2}L \right) \\&\quad + r^{-2}\left( \frac{1-\mu }{r}\nabla _\theta \partial _\theta - \nabla _{ \frac{r}{2} {\underline{L}}- \frac{r}{2} L}L\right) \\&\qquad + r^{-2} \sin ^{-2} \theta \left( \frac{1-\mu }{r}\nabla _\varphi \partial _\varphi - \nabla _{\sin ^2 \theta \frac{r}{2} {\underline{L}}- \sin ^2 \theta \frac{r}{2} L- \sin \theta \cos \theta \partial _\varphi }L\right) \\&= -\frac{1}{2} (1-\mu )^{-1} \nabla _{\underline{L}}\left( \frac{2M}{r^2}L \right) +\\&\qquad r^{-2}\left( \frac{1-\mu }{r} \left( \frac{r}{2} {\underline{L}}- \frac{r}{2} L \right) + \frac{r}{2} \frac{2M}{r^2} L\right) \\&\quad + r^{-2} \sin ^{-2} \theta \left( \frac{1-\mu }{r} \left( \sin ^2 \theta \frac{r}{2} {\underline{L}}- \sin ^2 \theta \frac{r}{2} L- \sin \theta \cos \theta \partial _\varphi \right) \right. \\&\left. \qquad + \frac{r}{2} \sin ^2 \theta \nabla _L L + \sin \theta \cos \theta \nabla _\varphi L\right) =\\&\quad - \frac{2M}{r^3} L + r^{-2}\left( \frac{1-\mu }{2}{\underline{L}}- \frac{1-\mu }{2} L + \frac{M}{r} L \right) + r^{-2} \left( \frac{1-\mu }{2}{\underline{L}}- \frac{1-\mu }{2} L + \frac{r}{2} \frac{2M}{r^2} L\right) \\&=- \frac{2M}{r^3} {\underline{L}}+ 2r^{-2}\left( \frac{1-\mu }{2}{\underline{L}}- \frac{1-\mu }{2} L + \frac{M}{r} L \right) \\&= \frac{1-\mu }{r^2} {\underline{L}}- \frac{1-\mu }{r^2} L. \end{aligned}$$

Now,

$$\begin{aligned} (a) + (e) = (\nabla _\mu \nabla ^\mu L^\nu ) {\underline{L}}^\alpha {\mathcal {F}}_{\nu \alpha } + L^\nu (\nabla _\mu \nabla ^\mu {\underline{L}}^\alpha ) {\mathcal {F}}_{\nu \alpha } = -2 \frac{1-\mu }{r^2} {\mathcal {F}}(L, {\underline{L}}) \end{aligned}$$

Finally,

$$\begin{aligned} \begin{aligned}&(b) + (d) =2 (\nabla _\mu L^\nu )(\nabla ^\mu {\underline{L}}^\alpha ) {\mathcal {F}}_{\nu \alpha } \\&= 2 (\nabla _L L)^L (\nabla ^L {\underline{L}})^{\underline{L}}{\mathcal {F}}(L,{\underline{L}}) = (1-\mu )^{-1} \frac{4M^2}{r^4} {\mathcal {F}}(L, {\underline{L}}). \end{aligned} \end{aligned}$$

Putting everything together, we obtain

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Let us now define

$$\begin{aligned} f := {\mathcal {F}}(L, {\underline{L}}). \end{aligned}$$

Now, we calculate

$$\begin{aligned} \begin{aligned}&\square _g (f) = 2 g^{L {\underline{L}}} L {\underline{L}}f + g^{\theta \theta } \nabla _\theta \nabla _\theta f - \nabla _{\nabla _\theta \partial _\theta }f + g^{\varphi \varphi } \nabla _\varphi \nabla _\varphi f - \nabla _{\nabla _\varphi \partial _\varphi }f \\&= - (1-\mu )^{-1} L {\underline{L}}f +\,\,/\!\!\!{\Delta }f + r^{-1} L f - r^{-1} {\underline{L}}f. \end{aligned} \end{aligned}$$

Also,

$$\begin{aligned} \begin{aligned} \nabla _L {\mathcal {F}}_{L {\underline{L}}}-\nabla _{\underline{L}}{\mathcal {F}}_{L {\underline{L}}} = L f - {\underline{L}}f - {\mathcal {F}}(\nabla _L L, {\underline{L}}) + {\mathcal {F}}(L, \nabla _{\underline{L}}{\underline{L}}) = L f - {\underline{L}}f - \frac{4M}{r^2} f. \end{aligned} \end{aligned}$$

We also calculate:

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We finally have:

$$\begin{aligned} \begin{aligned}&- (1-\mu )^{-1} L {\underline{L}}f +\,\,/\!\!\!{\Delta }f + r^{-1} L f - r^{-1} {\underline{L}}f \\&\quad = (1-\mu )^{-1} \left( \frac{2M}{r^2} - \frac{1}{r} + \frac{2M}{r^2} \right) \left( L f - {\underline{L}}f - \frac{4M}{r^2} f\right) +\\&\qquad -2 \frac{1-\mu }{r^2} f +(1-\mu )^{-1} \frac{4M^2}{r^4}f -4 M r^{-3} f. \end{aligned} \end{aligned}$$

This implies

$$\begin{aligned}&- (1-\mu )^{-1} L {\underline{L}}f +\,\,/\!\!\!{\Delta }f = (1-\mu )^{-1}\left( \frac{2M}{r^2} - \frac{2}{r} + \frac{4M}{r^2} \right) \left( L f - {\underline{L}}f\right) \\&\quad - (1-\mu )^{-1} \left( \frac{2M}{r^2} - \frac{1}{r} + \frac{2M}{r^2} \right) \frac{4M}{r^2} f +\\&\quad -2 \frac{1-\mu }{r^2} f +(1-\mu )^{-1} \frac{4M^2}{r^4}f -4 M r^{-3} f, \\&\quad - (1-\mu )^{-1} L {\underline{L}}f +\,\,/\!\!\!{\Delta }f = -(1-\mu )^{-1}\frac{2}{r} \left( 1- \frac{3M}{r}\right) \left( L f - {\underline{L}}f\right) \\&\quad - (1-\mu )^{-1} \left( \frac{2M}{r^2} - \frac{1}{r} + \frac{2M}{r^2} \right) \frac{4M}{r^2} f +\\&\quad -2 \frac{1-\mu }{r^2} f +(1-\mu )^{-1} \frac{4M^2}{r^4}f -4 M r^{-3} f, \\&- (1-\mu )^{-1} L {\underline{L}}f +\,\,/\!\!\!{\Delta }f = -(1-\mu )^{-1}\frac{2}{r} \left( 1- \frac{3M}{r}\right) \left( L f - {\underline{L}}f\right) \\&\quad - (1-\mu )^{-1} \frac{2}{r} \left( 1- \frac{4M}{r} +6 \frac{M^2}{r^2}\right) f. \end{aligned}$$

The last display is equivalent to the equation

$$\begin{aligned} r^{-2}(- L {\underline{L}}( r^2(1-\mu )^{-1} f)+(1-\mu ) \,\,/\!\!\!{\Delta }((1-\mu )^{-1}f)) = 0, \end{aligned}$$

which is the equation

$$\begin{aligned} - r^{-2} L {\underline{L}}(r^2 \rho ) +(1-\mu ) \,\,/\!\!\!{\Delta }\rho = 0, \end{aligned}$$

(D.4)

having defined $\rho := \frac{1}{2} (1-\mu )^{-1} {\mathcal {F}}({\underline{L}}, L)$.

By Hodge duality (see Appendix 1), the reasoning for $\sigma $ is the same. $\square $

Appendix E. Derivation of the Transport Equations Satisfied by $\alpha $ and ${\underline{\alpha }}$

Recall that $(\theta ^A, \theta ^B)$ is a system of local coordinates on ${\mathbb {S}}^2$. Let $\partial _{\theta ^A}, \partial _{\theta ^B}$ the associated local vector fields. To shorten notation, contraction with $\partial _{\theta ^A}$ is denoted by a capital subscript A.

Lemma E.1

Assume ${\mathcal {F}}$ satisfies the MBI system (2.7). Then:

$$\begin{aligned} \boxed { \,\,/\!\!{\nabla }_{\underline{L}}(r \alpha _A) = r(1-\mu )(\,\,/\!\!{\nabla }_A \rho + \,\,/\!\!{\varepsilon }_{AB}\,\,/\!\!{\nabla }^B \sigma ) + r(1-\mu )({H_{\Delta }})_{\mu A \kappa \lambda } \nabla ^{\mu } {\mathcal {F}}^{\kappa \lambda }, } \end{aligned}$$

(E.1)

and

$$\begin{aligned} \boxed { \,\,/\!\!{\nabla }_L(r {\underline{\alpha }}_A) = r(1-\mu )(- \,\,/\!\!{\nabla }_A \rho + \,\,/\!\!{\varepsilon }_{AB}\,\,/\!\!{\nabla }^B \sigma ) +r(1-\mu )({H_{\Delta }})_{\mu A \kappa \lambda } \nabla ^{\mu } {\mathcal {F}}^{\kappa \lambda }. } \end{aligned}$$

(E.2)

Proof

By the MBI system (2.7), we have

$$\begin{aligned} \nabla _{A} {\mathcal {F}}_{{\underline{L}}L} + \nabla _{L} {\mathcal {F}}_{A {\underline{L}}} + \nabla _{{\underline{L}}} {\mathcal {F}}_{L A} = 0. \end{aligned}$$

Now, $[\partial _{\theta ^A}, L] = [\partial _{\theta ^A}, {\underline{L}}] = 0$. Since $\nabla _L \partial _{\theta ^A} = \frac{1-\mu }{r} \partial _{\theta ^A}$, and $\nabla _{\underline{L}}\partial _{\theta ^A} = - \frac{ 1-\mu }{r} \partial _{\theta ^A}$, the previous display implies

$$\begin{aligned} 2(1-\mu )\,\,/\!\!{\nabla }_{A} \rho + \frac{1-\mu }{r} {\mathcal {F}}_{A {\underline{L}}} - \frac{1-\mu }{r} {\mathcal {F}}_{L A} + \,\,/\!\!{\nabla }_L {\underline{\alpha }}_A - \,\,/\!\!{\nabla }_{\underline{L}}\alpha _A= 0 \end{aligned}$$

which in turn implies

$$\begin{aligned} 2(1-\mu )\,\,/\!\!{\nabla }_{A} \rho + \frac{1}{r} \,\,/\!\!{\nabla }_L(r {\underline{\alpha }}_A) - \frac{1}{r} \,\,/\!\!{\nabla }_{\underline{L}}(r \alpha _A)= 0. \end{aligned}$$

(E.3)

Similarly, from the second equation of the MBI system (2.7), we deduce

$$\begin{aligned} \nabla ^L {\mathcal {F}}_{AL} + \nabla ^{\underline{L}}{\mathcal {F}}_{A{\underline{L}}} + \nabla ^B {\mathcal {F}}_{AB} = ({H_{\Delta }})_{\mu A \kappa \lambda } \nabla ^{\mu } {\mathcal {F}}^{\kappa \lambda }. \end{aligned}$$

This implies

$$\begin{aligned} - \frac{1}{2} (1-\mu )^{-1} \left( \frac{1}{r} \,\,/\!\!{\nabla }_{\underline{L}}(r \alpha _A) + \frac{1}{r} \,\,/\!\!{\nabla }_L(r {\underline{\alpha }}_A)\right) + \,\,/\!\!{\varepsilon }_{AB} \,\,/\!\!{\nabla }^B \sigma = ({H_{\Delta }})_{\mu A \kappa \lambda } \nabla ^{\mu } {\mathcal {F}}^{\kappa \lambda }, \end{aligned}$$

which implies

$$\begin{aligned} -\left( \frac{1}{r} \,\,/\!\!{\nabla }_{\underline{L}}(r \alpha _A) + \frac{1}{r} \,\,/\!\!{\nabla }_L(r {\underline{\alpha }}_A)\right) + 2 (1-\mu )\,\,/\!\!{\varepsilon }_{AB} \,\,/\!\!{\nabla }^B \sigma = 2 (1-\mu )({H_{\Delta }})_{\mu A \kappa \lambda } \nabla ^{\mu } {\mathcal {F}}^{\kappa \lambda }. \end{aligned}$$

(E.4)

Summing now (E.3) and (E.4), we obtain

$$\begin{aligned} \frac{1}{r} \,\,/\!\!{\nabla }_{\underline{L}}(r \alpha _A) = (1-\mu )(\,\,/\!\!{\nabla }_A \rho + \,\,/\!\!{\varepsilon }_{AB}\,\,/\!\!{\nabla }^B \sigma ) + (1-\mu )({H_{\Delta }})_{\mu A \kappa )\lambda } \nabla ^{\mu } {\mathcal {F}}^{\kappa \lambda }. \end{aligned}$$

Which is,

$$\begin{aligned} \boxed { \,\,/\!\!{\nabla }_{\underline{L}}(r \alpha _A) = r(1-\mu )(\,\,/\!\!{\nabla }_A \rho + \,\,/\!\!{\varepsilon }_{AB}\,\,/\!\!{\nabla }^B \sigma ) + r(1-\mu )({H_{\Delta }})_{\mu A \kappa \lambda } \nabla ^{\mu } {\mathcal {F}}^{\kappa \lambda }. } \end{aligned}$$

(E.5)

On the other hand, subtracting (E.4) from (E.3), we obtain:

$$\begin{aligned} \boxed { \,\,/\!\!{\nabla }_L(r {\underline{\alpha }}_A) = r(1-\mu )(- \,\,/\!\!{\nabla }_A \rho + \,\,/\!\!{\varepsilon }_{AB}\,\,/\!\!{\nabla }^B \sigma ) +r(1-\mu )({H_{\Delta }})_{\mu A \kappa \lambda } \nabla ^{\mu } {\mathcal {F}}^{\kappa \lambda }. } \end{aligned}$$

(E.6)

$\square $

Appendix G. The Nonzero Components of the Riemann Tensor

We define the Riemann tensor $\text {Rm}\,$ as as 4-covariant tensor field such that, for any four vectorfields A, B, C, D we have the following:

$$\begin{aligned} \text {Rm}\,(A,B,C,D) = g(A, \nabla _C \nabla _D B - \nabla _D \nabla _C B - \nabla _{[C,D]}B). \end{aligned}$$

(G.1)

Then, we have the commutation relations (valid for two-forms F):

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We notice that, in the $(u,v,\theta , \varphi )$ coordinates on Schwarzschild, the only nonzero components of the Riemann tensor are:

$$\begin{aligned}&\text {Rm}\,_{uvuv} = - 8M (1-\mu )^2r^{-3}, \qquad&\text {Rm}\,_{\theta u \theta v} = \frac{2M}{r}(1-\mu ), \\&\text {Rm}\,_{\varphi u \varphi v} = \frac{2M}{r} (1-\mu ) \sin ^2 \theta , \qquad&\text {Rm}\,_{\theta \varphi \theta \varphi } = 2M r \sin ^2 \theta . \end{aligned}$$

Appendix H. Poincaré and Sobolev Lemmas

We collect here a few useful results used in the paper. We begin by a standard Poincaré estimate on the sphere ${\mathbb {S}}^2$:

Lemma H.1

(Poincaré inequality for functions on ${\mathbb {S}}^2$) Let F be a smooth function on ${\mathbb {S}}^2$ with vanishing integral average on ${\mathbb {S}}^2$. Then, we have the inequality

$$\begin{aligned} \int _{{\mathbb {S}}^2} |d F|^2 \text{ d }{{\mathbb {S}}^2}\ge 2 \int _{{\mathbb {S}}^2} |F|^2 \text{ d }{{\mathbb {S}}^2}. \end{aligned}$$

We also recall the following standard result.

Lemma H.2

(Sobolev estimate for scalar functions on the sphere) Let $(\mathbb {S^2}, g_{{\mathbb {S}}^2})$ be the two-sphere with the standard metric, let $\nabla $ be the associated Levi–Civita connection, and let f be a smooth function $f: {\mathbb {S}}^2 \rightarrow {\mathbb {R}}$. Let ${\bar{f}} := \frac{1}{4 \pi } \int _{{\mathbb {S}}^2} f \text{ d }{{\mathbb {S}}^2}$ be the spherical average of f. There exists a universal constant C such that

$$\begin{aligned} \sup _{{\mathbb {S}}^2}|f - {\bar{f}}|^2 \le C \int _{{\mathbb {S}}^2} |\nabla \nabla f|^2 \text{ d }{{\mathbb {S}}^2}. \end{aligned}$$

(H.1)

The following Lemma is also standard.

Lemma H.3

(1-d trivial inequality) Let $[a,b] \subset {\mathbb {R}}$, with $a < b$. Let $f:[a,b] \rightarrow {\mathbb {R}}$ a smooth function with zero integral on [a, b]. Then there holds:

$$\begin{aligned} \sup _{[a,b]}|f| \le C \int _{[a,b]} |f'(x)| \text{ d }x \end{aligned}$$

(H.2)

We furthermore recall the following Lemma:

Lemma H.4

(Sobolev inequality involving only certain derivatives) Let $f: (\Sigma _{t_1^*} = \{t^* = t_1^*\}) \rightarrow {\mathbb {R}}$. Let $R> r_c > 2M$. Let ${\bar{f}}$ be again the mean of f over the spheres:

$$\begin{aligned} {\bar{f}} := \frac{1}{4 \pi } \int _{{\mathbb {S}}^2} f(\omega ) \text{ d }{{\mathbb {S}}^2}(\omega ). \end{aligned}$$

(H.3)

There exists a constant $C_{r_c, R}$ such that

$$\begin{aligned} \sup _{r_1 \in [r_c, R], \omega \in {\mathbb {S}}^2}|f(t_1^*, r_1, \omega )-{\bar{f}} (t_1^*, r_1)|^2 \le C \int _{r_c}^R \int _{{\mathbb {S}}^2}(|\,\,/\!\!{\nabla }\,\,/\!\!{\nabla }f|^2 + |\,\,/\!\!{\nabla }_{\partial _{r_1}} \,\,/\!\!{\nabla }\,\,/\!\!{\nabla }f|^2) \text{ d }{{\mathbb {S}}^2} \text{ d }r_1 \end{aligned}$$

(H.4)

Proof

Let

$$\begin{aligned} F(r_1) := \int _{{\mathbb {S}}^2}|(\,\,/\!\!{\nabla }\,\,/\!\!{\nabla }f)(t^*_1, r_1, \omega )|^2 \text{ d }{{\mathbb {S}}^2}(\omega )- \frac{1}{R-r_c} \int _{r_c}^R \int _{{\mathbb {S}}^2}|(\,\,/\!\!{\nabla }\,\,/\!\!{\nabla }f)(t^*_1, r_1, \omega )|^2 \text{ d }{{\mathbb {S}}^2}(\omega ) \text{ d }r_1. \end{aligned}$$

The preceding lemma then shows

$$\begin{aligned} \sup _{r_1 \in [r_c, R]} |F(r_1)| \le \int _{r_c}^R |\partial _{r_1}F(s)| \text{ d }s. \end{aligned}$$

Since $\,\,/\!\!{\nabla }_L \,\,/\!\!{g}= \,\,/\!\!{\nabla }_{\underline{L}}\,\,/\!\!{g}= 0$, f is smooth, and the Cauchy–Schwarz inequality, we have

$$\begin{aligned} \left| \partial _{r_1} \int _{{\mathbb {S}}^2} \,\,/\!\!{g}^{AB} \,\,/\!\!{g}^{CD} (\,\,/\!\!{\nabla }_A \,\,/\!\!{\nabla }_C f) (\,\,/\!\!{\nabla }_B \,\,/\!\!{\nabla }_D f) \text{ d }{{\mathbb {S}}^2}\right| \lesssim \int _{{\mathbb {S}}^2} (|\,\,/\!\!{\nabla }\,\,/\!\!{\nabla }f|^2+|\,\,/\!\!{\nabla }_{\partial _{r_1}} \,\,/\!\!{\nabla }\,\,/\!\!{\nabla }f|^2) \text{ d }{{\mathbb {S}}^2}. \end{aligned}$$

Combining this with the previous Sobolev lemma on spheres H.2, we have the claim. $\square $

Appendix I. The Theorem of Cauchy–Kowalevskaya

Let B be an open connected set in ${\mathbb {R}}^n$, and $f: B \rightarrow {\mathbb {R}}$ be analytic. We say that f belongs to the class ${\mathscr {A}}_{M, c_0}(B)$ for some $M, c_0 >0$ if, on B, we have, for all multi-indices $\alpha $,

$$\begin{aligned} |D^\alpha f| \le M {c_0}^{-k}, \end{aligned}$$

(I.1)

where $|\alpha | = k$ and D denotes partial differentiation.

We state here the form of the Cauchy–Kowalevskaya Theorem useful for our purposes.

Theorem I.1

(Cauchy–Kowalevskaya Theorem) Let $B \subset {\mathbb {R}}^n$ be an Euclidean ball. Suppose that W is the graph of an analytic function $\phi $ over B:

$$\begin{aligned} W = \{x \in {\mathbb {R}}^{n+1}, x = (\phi (x_1, \ldots , x_n), x_1, \ldots , x_n), \text { for } (x_1, \ldots , x_n) \in B\}. \end{aligned}$$

Suppose furthermore that we have a quasilinear system of PDEs in Cauchy–Kowalevskaya form:

$$\begin{aligned} \partial _{x^0} U = \sum _{i=1}^n {\mathcal {M}}_i(x, U) \partial _{x^i} U + F(x, U), \end{aligned}$$

(I.2)

where the unknown U is a vector in ${\mathbb {R}}^k$, $x = (x^1, \ldots , x^n)$, and furthermore ${\mathcal {M}}_i : {\mathbb {R}}^{n+k} \rightarrow {\mathbb {R}}^{k^2}$ are matrices with analytic coefficients in the variables x, U. Also, $F: {\mathbb {R}}^{n+k} \rightarrow {\mathbb {R}}^k$ is an analytic function of all its arguments. We impose analytic initial data $U_0$ for U on W.

Then, for every point $q \in B$, let $p := (\phi (q), q) \in {\mathbb {R}}^{n+1}$. In these conditions, there is a radius $r_p > 0$ such that the system of equations

$$\begin{aligned} \left\{ \begin{array}{ll} \partial _t U = \sum _{i=1}^n {\mathcal {M}}_i(x, U) \partial _{x^i} U + F(x, U) &{} \text { on } B(p,r_p) \\ U = U_0 &{} \text { on } W \cap B(p,r_p). \end{array} \right. \end{aligned}$$

(I.3)

admits an analytic solution $U: B(p,r_p) \rightarrow {\mathbb {R}}^k$. Here, $B(p, r_p)$ denotes the $(n+1)$-dimensional Euclidean ball of center p and radius $r_p$.

Furthermore, let $c_0, M$ be positive numbers. The radius $r_p$ depends only on $c_0$, M if all the following requirements are satisfied:

$$\begin{aligned} \begin{aligned}&U_0 \in {\mathscr {A}}_{M,c_0}(B), \\&{\mathcal {M}}_i, F \in {\mathscr {A}}_{M,c_0}(B \times {\mathbb {R}}^{2n}) \text { for } i = 1, \ldots , n,\\&\phi \in {\mathscr {A}}_{M,c_0}(B), \\&r_p \le d(q, \partial B). \end{aligned} \end{aligned}$$

(I.4)

Here, d denotes the Euclidean distance in ${\mathbb {R}}^n$, and $\text { diam}$ denotes the Euclidean diameter.

Remark I.1

The last requirement ensures that the ball $B(p, r_p)$ does not “overshoot” the ball B.

Appendix J. Addendum: Stationary Solutions and a Heuristic Calculation of the Asymptotics of the Spherical Averages of $\sigma $ and $\rho $

The aim of this Section is to elaborate on how the spherical averages of $\sigma $ and $\rho $ evolve dynamically, both in the linear Maxwell theory and in the MBI theory. We first describe charged (stationary) solutions to the Maxwell system on Schwarzschild, then we discuss charged solutions to the MBI system on Schwarzschild. Finally, we sketch a derivation of the asymptotics of the spherical averages of $\sigma $ and $\rho $, as a function of the initial data, under some reasonable (but not justified) assumptions on the decay of the null components.

J.1. Charged Solutions to the Linear Maxwell Equations on Schwarzschild

In the case of the linear Maxwell theory, there exist solutions to the Maxwell Equations on the Schwarzschild spacetime that are regular, stationary and nonzero (in particular, they do not decay in time). It can be proved that all such solutions are given by the following expression for the field tensor ${\mathcal {F}}$:

$$\begin{aligned} {\mathcal {F}}= \frac{\rho _0}{r^2} dt \wedge dr^* + \frac{\sigma _0}{r^2}\sin \theta (d \theta \wedge d \phi ), \end{aligned}$$

with $\rho _0$ and $\sigma _0$ real numbers.

In order to prove time-decay, then, one has to exclude the presence of these charged solutions. In the case of linear Maxwell, it is easy to eliminate such issue as the Equations are linear, hence it suffices to subtract the corresponding charged components in order to obtain decay of the field tensor. See [41].

J.2. Charged Solutions to the MBI System on Schwarzschild

A similar phenomenon appears in the MBI case, and indeed there exist stationary solutions to the MBI system on Schwarzschild. Here, we limit our calculation to stationary solutions in spherical symmetry, such that $\alpha = {\underline{\alpha }}=0$ (cf. the “hairy ball theorem”).

Proposition J.1

Let ${\mathcal {F}}$ be a spherically symmetric smooth stationary solution to the MBI system (2.7) on Schwarzschild, such that the null components $\alpha = {\underline{\alpha }}= 0$ identically. Then, there exist numbers $\sigma _0, \rho _0 \in {\mathbb {R}}$ such that

$$\begin{aligned} \sigma (t^*, r_1, \theta , \varphi ) = \sigma (r) = \frac{\sigma _0}{r^2}, \qquad \rho (t^*, r_1, \theta , \varphi ) = \rho (r) = \frac{\rho _0}{\sqrt{\rho _0^2+\sigma _0^2+ r^4}}.\qquad \end{aligned}$$

(J.1)

Proof

Let us start from the form of Equations (9.6) to obtain the form of the stationary solutions of the MBI system.

Open image in new window

(J.2)

Open image in new window

(J.3)

In these equations, set $\alpha = {\underline{\alpha }}= 0$, and assume that $\rho , \sigma $ are functions of r only. Then, the equations in the first column readily give the existence of $\sigma _0 \in {\mathbb {R}}$ such that $\sigma = \sigma _0 r^{-2}$ identically. We then proceed to sum the equations in the right column of displays (J.2) and (J.3), to obtain

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By our choice of $\alpha = {\underline{\alpha }}= 0$, we have

$$\begin{aligned} \begin{aligned} \mathbf{F }&= \sigma ^2 - \rho ^2, \qquad \mathbf{G }&= -\rho \sigma . \end{aligned} \end{aligned}$$

(J.4)

By Equation (13.1), then,

$$\begin{aligned} H^{\mu \nu \kappa \lambda }_{\Delta } \nabla _{\mu }{\mathcal {F}}_{\kappa \lambda } = \frac{1}{2\ell _{(\text {MBI})}^2} \nabla _\mu (\mathbf{F }-\mathbf{G }^2) \left( - {\mathcal {F}}^{\mu \nu } + \mathbf{G }\,{^\star \!}{\mathcal {F}}^{\mu \nu } \right) - \nabla _\mu \mathbf{G }\,{^\star \!}{\mathcal {F}}^{\mu \nu }. \end{aligned}$$

(J.5)

Hence

$$\begin{aligned} 2(1-\mu )\partial _r (r^2 \rho ) = \frac{r^2 (1-\mu )}{(1+\sigma ^2)(1-\rho ^2)} \partial _r\left( (1+\sigma ^2)(1-\rho ^2) \right) (1+\sigma ^2)\rho - 2 r^2 (1-\mu )\partial _r(\rho \sigma ) \sigma . \end{aligned}$$

By the fact that $\sigma r^2 = \sigma _0$, we then have

$$\begin{aligned} 2\partial _r( \rho r^2 (1+\sigma ^2)) = \frac{r^2 }{(1+\sigma ^2)(1-\rho ^2)} \partial _r\left( (1+\sigma ^2)(1-\rho ^2) \right) (1+\sigma ^2)\rho . \end{aligned}$$

Integrating the previous display yields the existence of $\rho _0 \in {\mathbb {R}}$ such that

$$\begin{aligned} \rho ^2 r^4 (1+\sigma ^2) = \rho _0^2 (1-\rho ^2). \end{aligned}$$

(J.6)

Given the form of $\sigma $, this implies the claim. $\square $

J.3. Asymptotic Behaviour of the Spherical Averages of $\rho $ and $\sigma $ When the Initial Charge is Nonzero

Recall that, in the above proof of global stability for the MBI system on Schwarzschild, we needed to deduce asymptotic bounds for spherical averages of $\rho $ and $\sigma $ along the evolution. To do that, we imposed the charge to vanish at spacelike infinity, on $\Sigma _{t_0^*}$. This was then propagated along the evolution in Section 9. We remark that such is an essential element of our proof, since we make large use of Poincaré estimates, which in turn require information on the spherical means.

In the remaining part of this Section, we would like to address the problem of determining the asymptotic behaviour of the spherical averages of $\sigma $ and $\rho $, if we assume nonzero initial charge and certain decay in time for all the components of the field. This may prove useful in a proof of global stability of the MBI system on Schwarzschild with non-vanishing initial charge.

We point out that a similar problem arises in the context of the Kerr stability conjecture, the so-called final state problem. In that case, one seeks to calculate, as a function of initial data, the parameters a and M such that the solution will be asymptotic to a Kerr black hole of parameters (a, M).

However, there is a caveat. The calculation below points to the fact that the charge for the MBI system is conserved in the nonlinear evolution, along future null infinity. We do not expect a similar statement to hold for angular momentum and mass in the context of the Kerr stability conjecture.

We prove the following Proposition.

Proposition J.2

Let ${\mathcal {F}}$ be a smooth solution to the MBI system (2.7) on Schwarzschild, satisfying the following decay assumptions: there exist $a, b, C >0$, and a smooth radial function $\rho _f(r)$ such that

$$\begin{aligned} \begin{aligned} |\rho - \rho _f (r)|&\le C \tau ^{-a},\\ |\sigma - \sigma _0 r^{-2}|&\le C \tau ^{-a}r^{-b},\\ |\alpha |&\le C \tau ^{-1}r^{-1} ,\\ |{\underline{\alpha }}|&\le C \tau ^{-\frac{1}{2}} r^{-3}. \end{aligned} \end{aligned}$$

(J.7)

Then, we let

$$\begin{aligned} \rho _0 := \frac{1}{4\pi }\lim _{r \rightarrow \infty } r^2 \int _{{\mathbb {S}}^2} \rho (t_0^*, r) \text{ d }{{\mathbb {S}}^2}. \end{aligned}$$

(In particular, the limit appearing in the right hand side of the previous display exists). In these conditions, we have

$$\begin{aligned} \rho _f(r) = \frac{\rho _0}{\sqrt{r^4 + \sigma _0^2+ \rho _0^2}}. \end{aligned}$$

(J.8)

Remark J.3

We remark that the assumption (J.7) is a reasonable one but, in a proof of stability of MBI on Schwarzschild, such assumption will need to be proved in the context of a bootstrap argument. Hence the above Proposition is very far from addressing the stability problem for MBI in the charged case.

Remark J.4

We remark that both $\rho _0$ and $\sigma _0$ can be calculated starting from initial data on ${\widetilde{\Sigma }}_{t_0^*}$. Hence the Proposition gives a way of calculating the asymptotic behaviour of spherical means as a function of initial data.

Remark J.5

Furthermore, we remark that the expression for the final charge (J.8) coincides with the form of stationary solutions found in (J.1).

Remark J.6

This Proposition is not concerned with the behaviour of the charge along null infinity, rather with the behaviour of spherical averages of $\rho $ and $\sigma $ on a region of finite r-coordinate. Nevertheless, very similar calculations indicate that the charge for $\rho $ is conserved along future null infinity.

Proof of Proposition J.2

The set of equations

$$\begin{aligned} \nabla ^\mu {^\star \!}{\mathcal {M}}_{\mu \nu } = 0, \qquad \nabla ^\mu \,{^\star \!}{\mathcal {F}}_{\mu \nu } = 0 \end{aligned}$$

implies, through the null decomposition, that the quantities

$$\begin{aligned} \int _{{\mathbb {S}}^2} \frac{1}{2(1-\mu )}{^\star \!}{\mathcal {M}}({\underline{L}}, L) r^2 \text{ d }{{\mathbb {S}}^2}= C_\rho , \qquad \int _{{\mathbb {S}}^2} \frac{1}{2(1-\mu )}\,{^\star \!}{\mathcal {F}}({\underline{L}}, L) r^2 \text{ d }{{\mathbb {S}}^2}= C_\sigma , \end{aligned}$$

(J.9)

where $C_\rho $, $C_\sigma $ are constant along the evolution. From the second equation, we obtain that

$$\begin{aligned} \int _{{\mathbb {S}}^2} \sigma r^2 \text{ d }{{\mathbb {S}}^2}= \text {const.} \end{aligned}$$

The goal now is to determine $\rho _f(r)$. Recall that

$$\begin{aligned} {^\star \!}{\mathcal {M}}_{\mu \nu } = -\ell _{(\text {MBI})}^{-1}({\mathcal {F}}_{\mu \nu }- \mathbf{G }\,{^\star \!}{\mathcal {F}}_{\mu \nu }), \end{aligned}$$

and that

$$\begin{aligned} \mathbf{F }= \sigma ^2 - \rho ^2 + 2 \alpha ^A {\underline{\alpha }}_A, \qquad \mathbf{G }= -\rho \sigma - 2 \,\,/\!\!{\varepsilon }_{AB} \alpha ^A {\underline{\alpha }}^B. \end{aligned}$$

From (J.9) and the assumptions, we obtain, taking the limit as $r \rightarrow \infty $ along points of the form $(t^*_0, r)$,

$$\begin{aligned} -C_\rho = \lim _{r \rightarrow \infty } r^2 \int _{{\mathbb {S}}^2} \rho (t_0^*, r) \text{ d }{{\mathbb {S}}^2}. \end{aligned}$$

This follows from the expression for ${\mathcal {M}}$, in which the only term that survives is the linear term (the one corresponding to ${\mathcal {F}}$) and furthermore $\ell _{(\text {MBI})}\rightarrow 1$ as $r \rightarrow \infty $. We let $\rho _0 := - \frac{1}{4\pi } C_\rho $. Now,

$$\begin{aligned} \begin{aligned} \frac{1}{2(1-\mu )} \ell _{(\text {MBI})}^{-1} ({\mathcal {F}}({\underline{L}}, L)-\mathbf{G }\,{^\star \!}{\mathcal {F}}({\underline{L}}, L)) = \frac{\rho + \rho \sigma ^2 +2 \rho \,\,/\!\!{\varepsilon }_{AB} \alpha ^A {\underline{\alpha }}^B}{\sqrt{1+\mathbf{F }-\mathbf{G }^2}} \end{aligned} \end{aligned}$$

Let us now consider the limit as r is fixed, and $t^* \rightarrow \infty $. Then,

$$\begin{aligned} \begin{aligned} \lim _{t^* \rightarrow \infty } \mathbf{F }&= \sigma _0^2 r^{-4} - \rho ^2_f(r), \qquad \lim _{t^* \rightarrow \infty } \mathbf{G }^2&= \sigma _0^2 r^{-4} \rho ^2_f(r). \end{aligned} \end{aligned}$$

We obtain eventually,

$$\begin{aligned} \int _{{\mathbb {S}}^2} ((1 + \sigma ^2_0 r^{-4})(1-\rho ^2_f))^{-\frac{1}{2}}\rho _f r^2 \text{ d }{{\mathbb {S}}^2}= 4 \pi \rho _0. \end{aligned}$$

This implies

$$\begin{aligned} (1 + \sigma ^2_0 r^{-4})^{-\frac{1}{2}}(1-\rho ^2_f)^{-\frac{1}{2}}\rho _f r^2 = \rho _0. \end{aligned}$$

By inverting the last display, we get

$$\begin{aligned} \rho _f(r) = \frac{\rho _0}{\sqrt{r^4 + \sigma _0^2+ \rho _0^2}} \end{aligned}$$

$\square $

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Federico Pasqualotto
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1.Department of MathematicsPrinceton UniversityPrincetonUSA
2.Department of Pure Mathematics and Mathematical StatisticsUniversity of CambridgeCambridgeUK

Nonlinear Stability for the Maxwell–Born–Infeld System on a Schwarzschild Background

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