Stability study of a model for the Klein–Gordon equation in Kerr space-time

Abstract

The current early stage in the investigation of the stability of the Kerr metric is characterized by the study of appropriate model problems. Particularly interesting is the problem of the stability of the solutions of the Klein–Gordon equation, describing the propagation of a scalar field of mass $\mu$ in the background of a rotating black hole. Rigorous results prove the stability of the reduced, by separation in the azimuth angle in Boyer–Lindquist coordinates, field for sufficiently large masses. Some, but not all, numerical investigations find instability of the reduced field for rotational parameters $a$ extremely close to $1$. Among others, the paper derives a model problem for the equation which supports the instability of the field down to $a/M \approx 0.97$.

Appendix

Lemma 6.1

Let $\mu \in [0,\infty )$. By

$$\begin{aligned} g(\lambda ) := i \lambda + \left(\mu ^2 - \lambda ^2\right)^{1/2} \end{aligned}$$

for every $\lambda \in \mathbb{R } \times (-\infty ,0)$, where $(\mu ^2 - \lambda ^2)^{1/2}$ denotes the square root with strictly positive real part, there is defined a biholomorphic map

$$\begin{aligned} g : \mathbb{R } \times (-\infty ,0) \rightarrow \left(\mathbb{C } \setminus B_{\mu }(0)\right) \cap \left((0,\infty ) \times \mathbb{R }\right) \end{aligned}$$

with inverse

$$\begin{aligned} g^{-1} : (\mathbb{C } \setminus B_{\mu }(0)) \cap \left((0,\infty ) \times \mathbb{R }\right) \rightarrow \mathbb{R } \times (-\infty ,0) \end{aligned}$$

given by

$$\begin{aligned} g^{-1}(z) = - \, \frac{i}{2z} \, \big (z^2 - \mu ^2\big ) \end{aligned}$$

for every $z \in (\mathbb{C } \setminus B_{\mu }(0)) \cap ((0,\infty ) \times \mathbb{R })$. In addition,

$$\begin{aligned} \left[\mu ^2 - \left(g^{-1}(z)\right)^2 \right]^{1/2} = \frac{1}{2 z} \, \left(z^2 + \mu ^2\right) \end{aligned}$$

for every $z \in (\mathbb{C } \setminus B_{\mu }(0)) \cap ((0,\infty ) \times \mathbb{R })$ (Fig. 5).

Fig. 5

Sketch of domain and range of the map $g$ from Lemma 6.1

Proof

First, we note that $g$ is well-defined. For this, let $\lambda \in \mathbb{R } \times (-\infty ,0)$ and $\lambda _1 := \mathrm Re (\lambda ), \lambda _2 := \mathrm Im (\lambda ) < 0$. Then

$$\begin{aligned} \mu ^2 - \lambda ^2 = \mu ^2 - \lambda _1^2 + \lambda _2^2 - 2 i \lambda _1 \lambda _2. \end{aligned}$$

Hence $\mu ^2 - \lambda ^2$ is real iff $\lambda _1 = 0$. In the latter case,

$$\begin{aligned} \mathrm Re \,\left(\mu ^2 - \lambda ^2\right) = \mu ^2 + \lambda _2^2 > 0. \end{aligned}$$

Therefore, $\mu ^2 - \lambda ^2 \in \mathbb{C } \setminus ((-\infty ,0] \times \{0\})$, and there is precisely one square root of $\mu ^2 - \lambda ^2$ with strictly positive real part. Further,

$$\begin{aligned} \mathrm Re \,\left[\,i \lambda +\left(\mu ^2 - \lambda ^2\right)^{1/2}\,\right] = - \lambda _2 + \mathrm Re \, \left(\mu ^2 - \lambda ^2\right)^{1/2} > 0. \end{aligned}$$

In particular, if

$$\begin{aligned} z := i \lambda + \left(\mu ^2 - \lambda ^2\right)^{1/2} \end{aligned}$$

is such that $|z| \le \mu$, then

$$\begin{aligned} z^2 - 2 i z \lambda - \lambda ^2 = (z - i \lambda )^2 = \mu ^2 - \lambda ^2. \end{aligned}$$

Hence

$$\begin{aligned} z^2 - \mu ^2 = 2 i z \lambda \end{aligned}$$

and

$$\begin{aligned} \lambda = - \frac{i}{2} \left(z - \frac{\mu ^2}{z}\right). \end{aligned}$$

The latter implies that

$$\begin{aligned} \lambda _2 = \mathrm{Im } \lambda = - \frac{\mathrm{Re }(z)}{2} \left( 1 - \frac{\mu ^2}{|z|^2} \right) \end{aligned}$$

and hence that $\mathrm Re (z) \le 0$. As a consequence, $|z| > \mu$.

For the second step, let $z \in (\mathbb{C } \setminus B_{\mu }(0)) \cap ((0,\infty ) \times \mathbb{R })$, $x := \mathrm Re (z),$ and $y := \mathrm Im (z)$. Then

$$\begin{aligned}&- \frac{i}{2} \left(z - \frac{\mu ^2}{z}\right) = - \frac{i}{2} \left(x + iy - \frac{\mu ^2}{x+iy} \right) = - \frac{i}{2} \left[x + iy - \frac{\mu ^2}{|z|^2} \, (x-iy) \right] \\&\quad = \frac{1}{2} \left[y \left(1 + \frac{\mu ^2}{|z|^2}\right) - i x \left(1 - \frac{\mu ^2}{|z|^2}\right) \right] \in \mathbb{R } \times (-\infty ,0) \end{aligned}$$

and

$$\begin{aligned} g\left(- \frac{i}{2} \left(z - \frac{\mu ^2}{z}\right) \right)&= \frac{1}{2} \left(z - \frac{\mu ^2}{z}\right) + \left[\mu ^2 + \frac{1}{4} \left(z - \frac{\mu ^2}{z}\right)^2\right]^{1/2} \\&= \frac{1}{2} \left(z - \frac{\mu ^2}{z}\right) + \left[ \frac{1}{4} \left(z + \frac{\mu ^2}{z}\right)^2 \right]^{1/2}. \end{aligned}$$

Since

$$\begin{aligned} \frac{1}{2} \left(z + \frac{\mu ^2}{z}\right)&= \frac{1}{2} \left[x + iy + \frac{\mu ^2}{|z|^2} \, (x-iy) \right] \\&= \frac{1}{2} \left[x \left(1 + \frac{\mu ^2}{|z|^2}\right) + iy \left(1 - \frac{\mu ^2}{|z|^2}\right)\right] \in (0,\infty ) \times \mathbb{R }, \end{aligned}$$

the latter implies that

$$\begin{aligned} g\left(- \frac{i}{2} \left(z - \frac{\mu ^2}{z}\right) \right)&= \frac{1}{2} \left(z - \frac{\mu ^2}{z}\right) + \left[ \frac{1}{4} \left(z + \frac{\mu ^2}{z}\right)^2 \right]^{1/2} \\&= \frac{1}{2} \left(z - \frac{\mu ^2}{z}\right) + \frac{1}{2} \left(z + \frac{\mu ^2}{z}\right) = z. \end{aligned}$$

In particular, we conclude that by

$$\begin{aligned} h(z) := - \, \frac{i}{2} \left(z - \frac{\mu ^2}{z} \right) = - \frac{i}{2z}\, \left(z^2 - \mu ^2\right) \end{aligned}$$

for every $z \in (\mathbb{C } \setminus B_{\mu }(0)) \cap ((0,\infty ) \times \mathbb{R })$, there is defined a map

$$\begin{aligned} h : (\mathbb{C } \setminus B_{\mu }(0)) \cap \left((0,\infty ) \times \mathbb{R }\right) \rightarrow \mathbb{R } \times (-\infty ,0) \end{aligned}$$

such that

$$\begin{aligned} (g \circ h)(z) = z \end{aligned}$$

for every $z \in (\mathbb{C } \setminus B_{\mu }(0)) \cap ((0,\infty ) \times \mathbb{R })$. Therefore, $g$ is surjective.

Further, if $\mu = 0$,

$$\begin{aligned} (h \circ g)(\lambda ) = - \, \frac{i}{2} \, \left[\, i \lambda + \left(- \lambda ^2\right)^{1/2}\,\right] = - \, \frac{i}{2} \, \left(\, i \lambda + i \lambda \, \right) = \lambda \end{aligned}$$

and if $\mu \ne 0$,

$$\begin{aligned} (h \circ g)(\lambda )&= - \, \frac{i}{2} \left[ i \lambda + \left(\mu ^2 - \lambda ^2\right)^{1/2} - \frac{\mu ^2}{i \lambda + \left(\mu ^2 - \lambda ^2\right)^{1/2}} \right] \\&= - \, \frac{i}{2} \left[i \lambda + \left(\mu ^2 - \lambda ^2\right)^{1/2} - \mu ^2 \, \frac{i \lambda - \left(\mu ^2 - \lambda ^2\right)^{1/2}}{-\lambda ^2 - \left(\mu ^2 - \lambda ^2\right)} \right] \\&= - \, \frac{i}{2} \left[i \lambda + \left(\mu ^2 - \lambda ^2\right)^{1/2} + i \lambda - \left(\mu ^2 - \lambda ^2\right)^{1/2}\right] = \lambda \end{aligned}$$

for every $\lambda \in \mathbb{R } \times (-\infty ,0)$. Therefore, $g$ is injective and altogether bijective with inverse $h$. Finally, it follows for $\lambda \in \mathbb{R } \times (-\infty ,0)$ that

$$\begin{aligned}&-\! \left(\mu ^2 - \lambda ^2\right)^{1/2} = i \lambda - g(\lambda ) = i h(g(\lambda )) - g(\lambda ) = - i \, \frac{i}{2} \left(g(\lambda ) - \frac{\mu ^2}{g(\lambda )} \right) - g(\lambda ) \\&= - \frac{1}{2} \left(g(\lambda ) + \frac{\mu ^2}{g(\lambda )} \right) = - \frac{1}{2 g(\lambda )} \, \left[(g(\lambda ))^2 + \mu ^2\right] \end{aligned}$$

and hence for every $z \in (\mathbb{C } \setminus B_{\mu }(0)) \cap ((0,\infty ) \times \mathbb{R })$ that

$$\begin{aligned} \left[\mu ^2 - \left(g^{-1}(z)\right)^2\right]^{1/2} = \frac{1}{2 z} \, \left(z^2 + \mu ^2\right). \end{aligned}$$

$\square$