The Regularity of Spacetime Perturbations for a Flat Spacetime Under Sobolev Spaces and Spectral Theory

Abstract

This study introduces an analysis concerning spacetime perturbations within the context of quantum foam models. Under the framework of Sobolev spaces, $H^s\left({\mathbb{R}}^n\right)$, we establish the existence and uniqueness of solutions to a linearized wave equation considering the harmonic gauge condition. Energy estimates are derived, demonstrating the conservation of both standard and higher-order energy functionals, which ensures the stability and regularity of metric perturbations over time. In addition, a spectral analysis of the d’Alembertian operator is conducted through Fourier transform techniques. Explicit calculations of Sobolev norms further confirm that these norms remain uniformly bounded, reinforcing the stability of solutions in the Sobolev space framework.

1. Introduction

The notion of quantum foam, first introduced by Wheeler [1], posits that at the Planck scale (∼1.6 $\times {10}^{-35}$ m), spacetime undergoes significant quantum fluctuations, leading to a "foamy" structure where the classical smooth manifold description ceases to hold. To investigate this behavior, we depart from some relevant and classic texts: Wald’s exposition on general relativity [2] along with Friedlander’s analysis of the wave equation in curved spacetime [3]. Additionally, we acknowledge the idea introduced by Hawking concerning the path-integral approach to quantum gravity [4]. Together, these seminal works constitute the basics for our study as they introduce foundational principles and analytical techniques required to assess the stability and regularity of spacetime perturbations. Particularly, we are concerned with the exploration of the conditions under which small perturbations, $h_{\mu \nu }$, around a background metric, ${\overline{g}}_{\mu \nu }$, remain well-behaved, and interestingly, under the scope of Sobolev spces and spectral methods.

Hence, we state the following objectives:

• Demonstrate the existence and uniqueness of solutions to the linear wave equation within the Sobolev spaces, $H^s\left({\mathbb{R}}^n\right)$, ensuring that the solutions continuously depend on the initial data.
• Derive and verify the conservation of energy for solutions to the linearized wave equation, including the formulation and preservation of higher-order energy functionals in Sobolev spaces.
• Perform a spectral analysis of the wave operator, □, utilizing Fourier transform techniques to examine the stability and boundedness of spacetime perturbations in quantum foam models.
• Provide explicit calculations of Sobolev norms for the solutions. The intention is to show that these norms remain uniformly bounded over time, which in turn leads to showing the regularity and stability of the metric perturbations.

Although the analysis presented in this work revisits classical topics within the theory of Sobolev spaces and linear wave equations, its integration with quantum foam models introduces results concerning the stability and regularity of spacetime perturbations. In addition, the set of analysis introduced in this work provides a complete series to ensure relevant aspects of the solutions with respect to the existence, uniqueness, and energy conservation theorems. Furthermore, the spectral analysis and explicit calculations of Sobolev norms reinforce the stability of solutions and state potential avenues for exploring more complex, nonlinear perturbations in quantum foam theories. The set of analyses presented here are intended, hence, to provide a unique reference result paper in which the properties of solutions are considered within the framework of well-known mathematical theories, confirming, in this way, their exact applications in the theory of quantum foam.

2. The Formulation of the Perturbation Problem

To start our analysis, we consider small perturbations, $h_{\mu \nu }$, around a background metric, ${\overline{g}}_{\mu \nu }$:

$$g_{\mu \nu }={\overline{g}}_{\mu \nu }+h_{\mu \nu },\parallel h_{\mu \nu }\parallel \ll 1.$$

(1)

For our analysis, we take the background metric ${\overline{g}}_{\mu \nu }$ to be the Minkowski metric ${\overline{g}}_{\mu \nu }=diag(-1,1,1,1)$, representing flat spacetime.

The Einstein field equations in a vacuum are as follows:

$$R_{\mu \nu }-{\displaystyle \frac{1}{2}}{g}_{\mu \nu }R=0.$$

(2)

By expanding the Ricci tensor $R_{\mu \nu }$ and the Ricci scalar R to the first order in $h_{\mu \nu }$, we obtain the linearized Einstein equations. The following lines within this section are well known from the standard literature, and we simply reproduce the different steps for this self-contained article. The detailed steps are presented as follows.

Firstly, the inverse metric $g^{\mu \nu }$ can be expanded to the first order:

$$g^{\mu \nu }={\overline{g}}^{\mu \nu }-h^{\mu \nu }+\mathcal{O}\left(h^2\right),$$

(3)

where indices are raised and lowered using ${\overline{g}}^{\mu \nu }$ and ${\overline{g}}_{\mu \nu }$.

The determinant g expands as follows:

$$\sqrt{-g}=\sqrt{-\overline{g}}\left(1+{\displaystyle \frac{1}{2}}h+\mathcal{O}\left(h^2\right)\right),$$

(4)

where $h={\overline{g}}^{\mu \nu }{h}_{\mu \nu }$.

Since $\overline{g}=det\left({\eta }_{\mu \nu }\right)=-1$, we have $\sqrt{-\overline{g}}=1$.

Secondly, the Christoffel symbols are given as follows:

$${\mathrm{\Gamma }}_{\mu \nu }^{\lambda }={\displaystyle \frac{1}{2}}{g}^{\lambda \sigma }\left({\partial }_{\mu }{g}_{\sigma \nu }+{\partial }_{\nu }{g}_{\sigma \mu }-{\partial }_{\sigma }{g}_{\mu \nu }\right).$$

(5)

Expanding this to the first order gives the following:

$${\mathrm{\Gamma }}_{\mu \nu }^{\lambda }={\displaystyle \frac{1}{2}}{\overline{g}}^{\lambda \sigma }\left({\partial }_{\mu }{h}_{\sigma \nu }+{\partial }_{\nu }{h}_{\sigma \mu }-{\partial }_{\sigma }{h}_{\mu \nu }\right)+\mathcal{O}\left(h^2\right).$$

(6)

Thirdly, the Ricci tensor is as follows:

$$R_{\mu \nu }={\partial }_{\lambda }{\mathrm{\Gamma }}_{\mu \nu }^{\lambda }-{\partial }_{\nu }{\mathrm{\Gamma }}_{\mu \lambda }^{\lambda }+{\mathrm{\Gamma }}_{\lambda \sigma }^{\lambda }{\mathrm{\Gamma }}_{\mu \nu }^{\sigma }-{\mathrm{\Gamma }}_{\nu \sigma }^{\lambda }{\mathrm{\Gamma }}_{\mu \lambda }^{\sigma }.$$

(7)

To the first order, the quadratic terms in $\mathrm{\Gamma }$ can be neglected, yielding the following:

$$R_{\mu \nu }={\partial }_{\lambda }{\mathrm{\Gamma }}_{\mu \nu }^{\lambda }-{\partial }_{\nu }{\mathrm{\Gamma }}_{\mu \lambda }^{\lambda }+\mathcal{O}\left(h^2\right).$$

(8)

By substituting the linearized Christoffel symbols, we obtain the following:

$$\begin{array}{cc}\hfill R_{\mu \nu }& ={\displaystyle \frac{1}{2}}\left({\partial }_{\lambda }{\partial }_{\mu }{h}_{\nu }^{\lambda }+{\partial }_{\lambda }{\partial }_{\nu }{h}_{\mu }^{\lambda }-{\partial }_{\lambda }{\partial }^{\lambda }{h}_{\mu \nu }-{\partial }_{\mu }{\partial }_{\nu }h\right)+\mathcal{O}\left(h^2\right),\hfill \end{array}$$

(9)

where $h={\overline{g}}^{\alpha \beta }{h}_{\alpha \beta }$.

The Ricci scalar is as follows:

$$R=g^{\mu \nu }{R}_{\mu \nu }={\overline{g}}^{\mu \nu }{R}_{\mu \nu }-h^{\mu \nu }{\overline{R}}_{\mu \nu }+\mathcal{O}\left(h^2\right).$$

(10)

Since the background is flat (${\overline{R}}_{\mu \nu }=0$), we have the following:

$$R={\overline{g}}^{\mu \nu }{R}_{\mu \nu }={\partial }^{\mu }{\partial }^{\nu }{h}_{\mu \nu }-\square h,$$

(11)

where $\square ={\overline{g}}^{\mu \nu }{\partial }_{\mu }{\partial }_{\nu }=-{\partial }_t^2+{\nabla }^2$.

Then, by substituting $R_{\mu \nu }$ and R into the Einstein equations, we obtain the following:

$$\square h_{\mu \nu }-{\partial }_{\mu }{\partial }^{\lambda }{h}_{\lambda \nu }-{\partial }_{\nu }{\partial }^{\lambda }{h}_{\lambda \mu }+{\partial }_{\mu }{\partial }_{\nu }h+{\overline{g}}_{\mu \nu }\left({\partial }^{\lambda }{\partial }^{\sigma }{h}_{\lambda \sigma }-\square h\right)=0.$$

(12)

The linearized Einstein equations possess gauge freedom due to the invariance under infinitesimal coordinate transformations:

$$x^{\mu }\to x^{\prime \mu }=x^{\mu }+{ϵ}^{\mu }\left(x\right),$$

(13)

where ${ϵ}^{\mu }\left(x\right)$ is an arbitrary infinitesimal vector field.

Under such a transformation, the metric perturbation transforms as follows:

$$h_{\mu \nu }\to h_{\mu \nu }^{\prime }=h_{\mu \nu }-{\partial }_{\mu }{ϵ}_{\nu }-{\partial }_{\nu }{ϵ}_{\mu }.$$

(14)

We can exploit this freedom to impose the harmonic gauge (also known as the de Donder gauge).

$${\partial }^{\nu }{\overline{h}}_{\mu \nu }=0,$$

(15)

where the trace-reversed perturbation ${\overline{h}}_{\mu \nu }$ is defined as follows:

$${\overline{h}}_{\mu \nu }=h_{\mu \nu }-{\displaystyle \frac{1}{2}}{\overline{g}}_{\mu \nu }h.$$

(16)

In the harmonic gauge, the linearized Einstein equations simplify to the following:

$$\square {\overline{h}}_{\mu \nu }=0.$$

(17)

In addition to the harmonic gauge, commonly-used alternative gauges include the Lorenz gauge, which enforces ${\partial }^{\nu }{\overline{h}}_{\mu \nu }=0$, simplifying equations into a wave-like form similar to electrodynamics, and the Transverse–Traceless (TT) gauge, primarily used in the study of gravitational waves. The TT gauge imposes the conditions $h_{\mu }^{\mu }=0$ (traceless) and ${\partial }^{\nu }{h}_{\mu \nu }=0$ (transverse), effectively eliminating non-physical degrees of freedom and isolating the dynamic components of the gravitational field. Each gauge has its strengths depending on its physical context and computational requirements, but for our scenario we have chosen the harmonic gauge.

3. Mathematical Framework

Sobolev Spaces

To analyze the regularity of solutions to the wave equation, we employ Sobolev spaces $H^s\left({\mathbb{R}}^n\right)$ (for additional details, the reader is referred to the classical texts [5,6])

Definition 1 (Sobolev Space $H^s\left({\mathbb{R}}^n\right)$).

Let $s\in \mathbb{R}$. The Sobolev space $H^s\left({\mathbb{R}}^n\right)$ consists of all tempered distributions, u, such that

$${\parallel u\parallel }_{H^s}^2={\int }_{{\mathbb{R}}^n}{(1+|\mathit{\xi }|}^2{)}^s{\left|\widehat{u}\left(\mathit{\xi }\right)\right|}^2{d}^n\mathit{\xi }<\infty ,$$

(18)

where $\widehat{u}\left(\mathit{\xi }\right)$ is the Fourier transform of u.

We consider the wave operator □ acting on suitable functional spaces as it will be further specified.

4. Analysis of Spacetime Perturbations

Firstly, we introduce a theorem concerning the existence and uniqueness of solutions to a suitable Cauchy problem. The existence of solutions will actually refer to the existence of regular solutions, but we will simply mention the "existence of solutions" in line with the literature, and the reader shall interpret the regularity of those which are introduced by the use of Sobolev spaces.

4.1. Existence and Uniqueness Theorem

Theorem 1 (Existence and Uniqueness).

Let $s\ge 1$, and suppose that the initial data satisfy $f_{\mu \nu }\in H^{s+1}\left({\mathbb{R}}^3\right)$ and $g_{\mu \nu }\in H^s\left({\mathbb{R}}^3\right)$. Then, there exists a unique solution, ${\overline{h}}_{\mu \nu }\in C([0,T];H^{s+1}\left({\mathbb{R}}^3\right))$ with ${\partial }_t{\overline{h}}_{\mu \nu }\in C([0,T];H^s\left({\mathbb{R}}^3\right))$, to the Cauchy problem for the linear wave equation.

$$\begin{array}{cc}\hfill \hspace{1em}\hspace{1em}\hspace{1em}\square {\overline{h}}_{\mu \nu }& =0,in[0,T]\times {\mathbb{R}}^3,\hfill \end{array}$$

(19)

$$\begin{array}{cc}\hfill {\overline{h}}_{\mu \nu }(0,\mathbf{x})& =f_{\mu \nu }\left(\mathbf{x}\right),\mathbf{x}\in {\mathbb{R}}^3,\hfill \end{array}$$

(20)

$$\begin{array}{cc}\hfill {\partial }_t{\overline{h}}_{\mu \nu }(0,\mathbf{x})& =g_{\mu \nu }\left(\mathbf{x}\right),\mathbf{x}\in {\mathbb{R}}^3.\hfill \end{array}$$

(21)

Moreover, the solution depends continuously on the initial data.

Proof. 

We aim to prove the existence, uniqueness, and continuous dependence of solutions to the linear wave Equation (19) with the initial conditions (20) and (21) in the Sobolev space framework.

We will use the following properties:

• The Fourier transform maps $H^s\left({\mathbb{R}}^n\right)$ isomorphically onto $L_s^2\left({\mathbb{R}}^n\right)$, where $L_s^2\left({\mathbb{R}}^n\right)$ is the weighted $L^2$ space with the weight ${(1+|\mathit{\xi }|}^2{)}^s$.
• The Sobolev space $H^s\left({\mathbb{R}}^n\right)$ is a Hilbert space, with the following inner product:

  $${\langle u,v\rangle }_{H^s}={\int }_{{\mathbb{R}}^n}{(1+|\mathit{\xi }|}^2{)}^s\widehat{u}\left(\mathit{\xi }\right)\overline{\widehat{v}\left(\mathit{\xi }\right)}d\mathit{\xi }.$$
• For $s\ge 1$, $H^s\left({\mathbb{R}}^n\right)\subset L^2\left({\mathbb{R}}^n\right)$.

We take the Fourier transform in the spatial variables $\mathbf{x}\in {\mathbb{R}}^3$ of the wave Equation (19). Denote, by ${\widehat{\overline{h}}}_{\mu \nu }(t,\mathit{\xi })$, the Fourier transform of ${\overline{h}}_{\mu \nu }(t,\mathbf{x})$ with respect to $\mathbf{x}$:

$${\widehat{\overline{h}}}_{\mu \nu }(t,\mathit{\xi })={\int }_{{\mathbb{R}}^3}{\overline{h}}_{\mu \nu }(t,\mathbf{x})e^{-i\mathit{\xi }·\mathbf{x}}{d}^{3}x.$$

Similarly, the Fourier transforms of the initial data are as follows:

$$\begin{array}{c}\hfill {\widehat{f}}_{\mu \nu }\left(\mathit{\xi }\right)={\int }_{{\mathbb{R}}^3}{f}_{\mu \nu }\left(\mathbf{x}\right)e^{-i\mathit{\xi }·\mathbf{x}}{d}^{3}x,\\ \hfill {\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)={\int }_{{\mathbb{R}}^3}{g}_{\mu \nu }\left(\mathbf{x}\right)e^{-i\mathit{\xi }·\mathbf{x}}{d}^{3}x.\end{array}$$

The Laplacian operator transforms as follows:

$$\widehat{\Delta {\overline{h}}_{\mu \nu }}(t,\mathit{\xi })=-{\left|\mathit{\xi }\right|}^2{\widehat{\overline{h}}}_{\mu \nu }(t,\mathit{\xi }).$$

Therefore, the Fourier transform of the wave equation becomes an ordinary differential equation in t for each fixed $\mathit{\xi }$:

$${\partial }_t^2{\widehat{\overline{h}}}_{\mu \nu }(t,\mathit{\xi })+{\left|\mathit{\xi }\right|}^2{\widehat{\overline{h}}}_{\mu \nu }(t,\mathit{\xi })=0.$$

(22)

Equation (22) is a second-order linear homogeneous ODE with constant coefficients. The general solution is as follows:

$${\widehat{\overline{h}}}_{\mu \nu }(t,\mathit{\xi })=A_{\mu \nu }\left(\mathit{\xi }\right)e^{i\left|\mathit{\xi }\right|t}+B_{\mu \nu }\left(\mathit{\xi }\right)e^{-i\left|\mathit{\xi }\right|t}.$$

Alternatively, we can write the solution using real functions by employing sine and cosine:

$${\widehat{\overline{h}}}_{\mu \nu }(t,\mathit{\xi })={\widehat{C}}_{\mu \nu }\left(\mathit{\xi }\right)cos\left(\right|\mathit{\xi }\left|t\right)+{\widehat{D}}_{\mu \nu }\left(\mathit{\xi }\right)sin\left(\right|\mathit{\xi }\left|t\right).$$

We use the initial conditions in the Fourier space:

$$\begin{array}{cc}\hfill {\widehat{\overline{h}}}_{\mu \nu }(0,\mathit{\xi })& ={\widehat{f}}_{\mu \nu }\left(\mathit{\xi }\right),\hfill \\ \hfill {\partial }_t{\widehat{\overline{h}}}_{\mu \nu }(0,\mathit{\xi })& ={\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right).\hfill \end{array}$$

By substituting $t=0$ into the general solution, we obtain the following:

$$\begin{array}{cc}\hfill {\widehat{\overline{h}}}_{\mu \nu }(0,\mathit{\xi })& =A_{\mu \nu }\left(\mathit{\xi }\right)+B_{\mu \nu }\left(\mathit{\xi }\right)={\widehat{f}}_{\mu \nu }\left(\mathit{\xi }\right),\hfill \\ \hfill {\partial }_t{\widehat{\overline{h}}}_{\mu \nu }(0,\mathit{\xi })& =i\left|\mathit{\xi }\right|\left(A_{\mu \nu }\left(\mathit{\xi }\right)-B_{\mu \nu }\left(\mathit{\xi }\right)\right)={\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right).\hfill \end{array}$$

We solve this system for $A_{\mu \nu }\left(\mathit{\xi }\right)$ and $B_{\mu \nu }\left(\mathit{\xi }\right)$:

$$\begin{array}{cc}\hfill A_{\mu \nu }\left(\mathit{\xi }\right)& ={\displaystyle \frac{1}{2}}\left({\widehat{f}}_{\mu \nu }\left(\mathit{\xi }\right)-{\displaystyle \frac{i}{\left|\mathit{\xi }\right|}}{\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)\right),\hfill \\ \hfill B_{\mu \nu }\left(\mathit{\xi }\right)& ={\displaystyle \frac{1}{2}}\left({\widehat{f}}_{\mu \nu }\left(\mathit{\xi }\right)+{\displaystyle \frac{i}{\left|\mathit{\xi }\right|}}{\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)\right).\hfill \end{array}$$

Thus, the solution in the Fourier space is as follows:

$${\widehat{\overline{h}}}_{\mu \nu }(t,\mathit{\xi })={\widehat{f}}_{\mu \nu }\left(\mathit{\xi }\right)cos\left(\right|\mathit{\xi }\left|t\right)+{\displaystyle \frac{{\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)}{\left|\mathit{\xi }\right|}}sin\left(\right|\mathit{\xi }\left|t\right).$$

We now take the inverse Fourier transform to obtain ${\overline{h}}_{\mu \nu }(t,\mathbf{x})$:

$${\overline{h}}_{\mu \nu }(t,\mathbf{x})={\int }_{{\mathbb{R}}^3}\left({\widehat{f}}_{\mu \nu }\left(\mathit{\xi }\right)cos\left(\right|\mathit{\xi }\left|t\right)+{\displaystyle \frac{{\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)}{\left|\mathit{\xi }\right|}}sin\left(\right|\mathit{\xi }\left|t\right)\right)e^{i\mathit{\xi }·\mathbf{x}}{\displaystyle \frac{d^3\mathit{\xi }}{{\left(2\pi \right)}^3}}.$$

Now, we need to show that ${\overline{h}}_{\mu \nu }(t,\mathbf{x})\in H^{s+1}\left({\mathbb{R}}^3\right)$ and ${\partial }_t{\overline{h}}_{\mu \nu }(t,\mathbf{x})\in H^s\left({\mathbb{R}}^3\right)$ for each $t\in [0,T]$.

First, note that the Fourier transform of ${\overline{h}}_{\mu \nu }(t,\mathbf{x})$ satisfies the following:

$${(1+|\mathit{\xi }|}^2{)}^{(s+1)/2}{\widehat{\overline{h}}}_{\mu \nu }(t,\mathit{\xi })={(1+|\mathit{\xi }|}^2{)}^{(s+1)/2}\left({\widehat{f}}_{\mu \nu }\left(\mathit{\xi }\right)cos\left(\right|\mathit{\xi }\left|t\right)+{\displaystyle \frac{{\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)}{\left|\mathit{\xi }\right|}}sin\left(\right|\mathit{\xi }\left|t\right)\right).$$

Similarly, the Fourier transform of ${\partial }_t{\overline{h}}_{\mu \nu }(t,\mathbf{x})$ is as follows:

$$\widehat{{\partial }_t{\overline{h}}_{\mu \nu }}(t,\mathit{\xi })=-\left|\mathit{\xi }\right|{\widehat{f}}_{\mu \nu }\left(\mathit{\xi }\right)sin\left(\right|\mathit{\xi }\left|t\right)+{\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)cos\left(\right|\mathit{\xi }\left|t\right).$$

Then,

$${(1+|\mathit{\xi }|}^2{)}^{s/2}\widehat{{\partial }_t{\overline{h}}_{\mu \nu }}(t,\mathit{\xi })={(1+|\mathit{\xi }|}^2{)}^{s/2}\left(-\left|\mathit{\xi }\right|{\widehat{f}}_{\mu \nu }\left(\mathit{\xi }\right)sin\left(\right|\mathit{\xi }\left|t\right)+{\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)cos\left(\right|\mathit{\xi }\left|t\right)\right).$$

Since $f_{\mu \nu }\in H^{s+1}\left({\mathbb{R}}^3\right)$ and $g_{\mu \nu }\in H^s\left({\mathbb{R}}^3\right)$, we have the following:

$$\parallel f_{\mu \nu }{\parallel }_{H^{s+1}}^2={\int }_{{\mathbb{R}}^3}{(1+|\mathit{\xi }|}^2{)}^{s+1}{\left|{\widehat{f}}_{\mu \nu }\left(\mathit{\xi }\right)\right|}^2{d}^3\mathit{\xi }<\infty ,$$

$$\parallel g_{\mu \nu }{\parallel }_{H^s}^2={\int }_{{\mathbb{R}}^3}{(1+|\mathit{\xi }|}^2{)}^s{\left|{\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)\right|}^2{d}^3\mathit{\xi }<\infty .$$

We now estimate $\parallel {\overline{h}}_{\mu \nu }{\left(t\right)\parallel }_{H^{s+1}}$:

$$\begin{array}{cc}\hfill \parallel {\overline{h}}_{\mu \nu }{\left(t\right)\parallel }_{H^{s+1}}^2& ={\int }_{{\mathbb{R}}^3}{(1+|\mathit{\xi }|}^2{)}^{s+1}{\left|{\widehat{f}}_{\mu \nu }\left(\mathit{\xi }\right)cos\left(\right|\mathit{\xi }\left|t\right)+{\displaystyle \frac{{\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)}{\left|\mathit{\xi }\right|}}sin\left(\right|\mathit{\xi }\left|t\right)\right|}^2{d}^3\mathit{\xi }\hfill \\ \hfill & \le 2{\int }_{{\mathbb{R}}^3}{(1+|\mathit{\xi }|}^2{)}^{s+1}\left(|{\widehat{f}}_{\mu \nu }{\left(\mathit{\xi }\right)|}^2+{\displaystyle \frac{|{\widehat{g}}_{\mu \nu }{\left(\mathit{\xi }\right)|}^2}{{\left|\mathit{\xi }\right|}^2}}\right)d^3\mathit{\xi }.\hfill \end{array}$$

Since

$${(1+|\mathit{\xi }|}^2{)}^{s+1}{\displaystyle \frac{|{\widehat{g}}_{\mu \nu }{\left(\mathit{\xi }\right)|}^2}{{\left|\mathit{\xi }\right|}^2}}\le {(1+|\mathit{\xi }|}^2{)}^s{\left|{\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)\right|}^2,$$

for $\left|\mathit{\xi }\right|\ge 1$, and there is a similar consideration for $\left|\mathit{\xi }\right|\le 1$, we have the following:

$$\parallel {\overline{h}}_{\mu \nu }{\left(t\right)\parallel }_{H^{s+1}}^2\le C\left(\parallel f_{\mu \nu }{\parallel }_{H^{s+1}}^2+{\parallel g_{\mu \nu }\parallel }_{H^s}^2\right),$$

for some constant C independent of t.

Similarly, for ${\partial }_t{\overline{h}}_{\mu \nu }\left(t\right)$,

$$\begin{array}{cc}\hfill \parallel {\partial }_t{\overline{h}}_{\mu \nu }{\left(t\right)\parallel }_{H^s}^2& ={\int }_{{\mathbb{R}}^3}{(1+|\mathit{\xi }|}^2{)}^s{\left|-\left|\mathit{\xi }\right|{\widehat{f}}_{\mu \nu }\left(\mathit{\xi }\right)sin\left(\right|\mathit{\xi }\left|t\right)+{\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)cos\left(\right|\mathit{\xi }\left|t\right)\right|}^2{d}^3\mathit{\xi }\hfill \\ \hfill & \le 2{\int }_{{\mathbb{R}}^3}{(1+|\mathit{\xi }|}^2{)}^s\left({\left|\mathit{\xi }\right|}^2|{\widehat{f}}_{\mu \nu }{\left(\mathit{\xi }\right)|}^2+{\left|{\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)\right|}^2\right)d^3\mathit{\xi }\hfill \\ \hfill & \le 2{\int }_{{\mathbb{R}}^3}{(1+|\mathit{\xi }|}^2{)}^{s+1}|{\widehat{f}}_{\mu \nu }{\left(\mathit{\xi }\right)|}^2{d}^3\mathit{\xi }+2{\int }_{{\mathbb{R}}^3}{(1+|\mathit{\xi }|}^2{)}^s{\left|{\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)\right|}^2{d}^3\mathit{\xi }\hfill \\ \hfill & =2\parallel f_{\mu \nu }{\parallel }_{H^{s+1}}^2+2{\parallel g_{\mu \nu }\parallel }_{H^s}^2.\hfill \end{array}$$

Therefore, $\parallel {\overline{h}}_{\mu \nu }{\left(t\right)\parallel }_{H^{s+1}}$ and $\parallel {\partial }_t{\overline{h}}_{\mu \nu }{\left(t\right)\parallel }_{H^s}$ are bounded uniformly in t, implying that ${\overline{h}}_{\mu \nu }\left(t\right)\in H^{s+1}\left({\mathbb{R}}^3\right)$ and ${\partial }_t{\overline{h}}_{\mu \nu }\left(t\right)\in H^s\left({\mathbb{R}}^3\right)$.

The mappings $t\mapsto {\overline{h}}_{\mu \nu }\left(t\right)$ and $t\mapsto {\partial }_t{\overline{h}}_{\mu \nu }\left(t\right)$ are continuous from $[0,T]$ to $H^{s+1}\left({\mathbb{R}}^3\right)$ and $H^s\left({\mathbb{R}}^3\right)$, respectively, because the integrals defining the norms involve continuous functions of t (sine and cosine functions are continuous in t) and the integrals converge uniformly in t.

Now, let us focus on the uniqueness of solutions. Suppose there are two solutions, ${\overline{h}}_{\mu \nu }^{\left(1\right)}$ and ${\overline{h}}_{\mu \nu }^{\left(2\right)}$, satisfying the wave equation and the same initial conditions. Then, their difference $\delta {\overline{h}}_{\mu \nu }={\overline{h}}_{\mu \nu }^{\left(1\right)}-{\overline{h}}_{\mu \nu }^{\left(2\right)}$ satisfies the following:

$$\square \delta {\overline{h}}_{\mu \nu }=0,\delta {\overline{h}}_{\mu \nu }(0,\mathbf{x})=0,{\partial }_t\delta {\overline{h}}_{\mu \nu }(0,\mathbf{x})=0.$$

By taking the Fourier transform, we find that $\delta {\widehat{\overline{h}}}_{\mu \nu }(t,\mathit{\xi })=0$ for all t, $\mathit{\xi }$. Therefore, $\delta {\overline{h}}_{\mu \nu }(t,\mathbf{x})=0$, and the solution is unique.

Let us consider two sets of initial data, $(f_{\mu \nu },g_{\mu \nu })$ and $(f_{\mu \nu }^{\prime },g_{\mu \nu }^{\prime })$, with corresponding solutions, ${\overline{h}}_{\mu \nu }$ and ${\overline{h}}_{\mu \nu }^{\prime }$. Then, the difference $\delta {\overline{h}}_{\mu \nu }={\overline{h}}_{\mu \nu }-{\overline{h}}_{\mu \nu }^{\prime }$ satisfies the following:

$$\square \delta {\overline{h}}_{\mu \nu }=0,\delta {\overline{h}}_{\mu \nu }(0,\mathbf{x})=\delta f_{\mu \nu }\left(\mathbf{x}\right),{\partial }_t\delta {\overline{h}}_{\mu \nu }(0,\mathbf{x})=\delta g_{\mu \nu }\left(\mathbf{x}\right),$$

where $\delta f_{\mu \nu }=f_{\mu \nu }-f_{\mu \nu }^{\prime }$ and $\delta g_{\mu \nu }=g_{\mu \nu }-g_{\mu \nu }^{\prime }$.

Proceeding as before, we have the following:

$$\delta {\widehat{\overline{h}}}_{\mu \nu }(t,\mathit{\xi })=\delta {\widehat{f}}_{\mu \nu }\left(\mathit{\xi }\right)cos\left(\right|\mathit{\xi }\left|t\right)+{\displaystyle \frac{\delta {\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)}{\left|\mathit{\xi }\right|}}sin\left(\right|\mathit{\xi }\left|t\right).$$

By estimating the norms, we find the following:

$$\begin{array}{cc}\hfill \parallel \delta {\overline{h}}_{\mu \nu }{\left(t\right)\parallel }_{H^{s+1}}& \le C\left(\parallel \delta f_{\mu \nu }{\parallel }_{H^{s+1}}+{\parallel \delta g_{\mu \nu }\parallel }_{H^s}\right),\hfill \\ \hfill \parallel {\partial }_t\delta {\overline{h}}_{\mu \nu }{\left(t\right)\parallel }_{H^s}& \le C\left(\parallel \delta f_{\mu \nu }{\parallel }_{H^{s+1}}+{\parallel \delta g_{\mu \nu }\parallel }_{H^s}\right).\hfill \end{array}$$

This shows that the solution depends continuously on the initial data in the Sobolev norms.

We have established the following:

• A solution, ${\overline{h}}_{\mu \nu }$, exists in $C([0,T];H^{s+1}\left({\mathbb{R}}^3\right))$ with ${\partial }_t{\overline{h}}_{\mu \nu }\in C([0,T];H^s\left({\mathbb{R}}^3\right))$.
• The solution is unique.
• The solution depends continuously on the initial data.

Thus, the theorem is proved. □

4.2. Energy Estimates

We now derive explicit energy estimates to demonstrate the stability of the solution. For this, let us define the energy functional:

Theorem 2 (Conservation of Energy).

Let us consider the Cauchy problem (19)–(21), but now the initial data belong to the Schwartz space of rapidly decreasing smooth functions, where $f_{\mu \nu },g_{\mu \nu }\in \mathcal{S}\left({\mathbb{R}}^3\right)$.

The energy functional is defined:

$$E\left(t\right)={\displaystyle \frac{1}{2}}{\int }_{{\mathbb{R}}^3}\left(|{\partial }_t{\overline{h}}_{\mu \nu }{(t,\mathbf{x})|}^2+{|\nabla {\overline{h}}_{\mu \nu }(t,\mathbf{x})|}^2\right)d^{3}x.$$

(23)

Then, the energy $E\left(t\right)$ is conserved over time, i.e.,

$$E\left(t\right)=E\left(0\right),\forall t\in \mathbb{R}.$$

(24)

Proof. 

To prove the conservation of energy, we compute the time derivative of the energy functional $E\left(t\right)$ and show that ${\displaystyle \frac{dE}{dt}}=0$.

Firstly, note that since $f_{\mu \nu },g_{\mu \nu }\in \mathcal{S}\left({\mathbb{R}}^3\right)$ and the wave equation is linear and has constant coefficients, the solution ${\overline{h}}_{\mu \nu }(t,\mathbf{x})$ remains in $\mathcal{S}\left({\mathbb{R}}^3\right)$ for all t. This ensures that all derivatives are well defined and that we can perform integration by parts without boundary terms (due to the rapid decay at infinity).

We compute the time derivative of $E\left(t\right)$:

$$\begin{array}{cc}\hfill {\displaystyle \frac{dE}{dt}}& ={\displaystyle \frac{1}{2}}{\int }_{{\mathbb{R}}^3}{\displaystyle \frac{d}{dt}}\left(|{\partial }_t{\overline{h}}_{\mu \nu }{|}^2+{|\nabla {\overline{h}}_{\mu \nu }|}^2\right)d^{3}x\hfill \\ \hfill & ={\displaystyle \frac{1}{2}}{\int }_{{\mathbb{R}}^3}\left(2{\partial }_t{\overline{h}}_{\mu \nu }{\partial }_t^2{\overline{h}}^{\mu \nu }+2\nabla {\overline{h}}_{\mu \nu }·\nabla {\partial }_t{\overline{h}}^{\mu \nu }\right)d^{3}x\hfill \\ \hfill & ={\int }_{{\mathbb{R}}^3}\left({\partial }_t{\overline{h}}_{\mu \nu }{\partial }_t^2{\overline{h}}^{\mu \nu }+\nabla {\overline{h}}_{\mu \nu }·\nabla {\partial }_t{\overline{h}}^{\mu \nu }\right)d^{3}x.\hfill \end{array}$$

(25)

From the wave Equation (19), we have the following:

$${\partial }_t^2{\overline{h}}_{\mu \nu }-\Delta {\overline{h}}_{\mu \nu }=0.$$

(26)

Substitute ${\partial }_t^2{\overline{h}}_{\mu \nu }$ from (26) into (25):

$$\begin{array}{cc}\hfill {\displaystyle \frac{dE}{dt}}& ={\int }_{{\mathbb{R}}^3}\left({\partial }_t{\overline{h}}_{\mu \nu }(\Delta {\overline{h}}_{\mu \nu })+\nabla {\overline{h}}_{\mu \nu }·\nabla {\partial }_t{\overline{h}}_{\mu \nu }\right)d^{3}x.\hfill \end{array}$$

(27)

Now, consider the first term in (27):

$$\begin{array}{cc}\hfill I_1& ={\int }_{{\mathbb{R}}^3}{\partial }_t{\overline{h}}_{\mu \nu }(\Delta {\overline{h}}^{\mu \nu })d^{3}x\hfill \\ \hfill & ={\int }_{{\mathbb{R}}^3}{\partial }_t{\overline{h}}_{\mu \nu }\nabla ·\nabla {\overline{h}}^{\mu \nu }{d}^{3}x.\hfill \end{array}$$

(28)

By integrating by parts (noting that boundary terms vanish due to rapid decay), we obtain the following:

$$\begin{array}{cc}\hfill I_1& =-{\int }_{{\mathbb{R}}^3}\nabla {\partial }_t{\overline{h}}_{\mu \nu }·\nabla {\overline{h}}^{\mu \nu }{d}^{3}x.\hfill \end{array}$$

(29)

The second term in (27) is as follows:

$$\begin{array}{cc}\hfill I_2& ={\int }_{{\mathbb{R}}^3}\nabla {\overline{h}}_{\mu \nu }·\nabla {\partial }_t{\overline{h}}^{\mu \nu }{d}^{3}x.\hfill \end{array}$$

(30)

Notice that $I_2$ and $I_1$ are negatives of each other:

$$I_2=-I_1.$$

(31)

By combining $I_1$ and $I_2$, the total derivative of energy is as follows:

$$\begin{array}{cc}\hfill {\displaystyle \frac{dE}{dt}}& =I_1+I_2=-{\int }_{{\mathbb{R}}^3}\nabla {\partial }_t{\overline{h}}_{\mu \nu }·\nabla {\overline{h}}^{\mu \nu }{d}^{3}x+{\int }_{{\mathbb{R}}^3}\nabla {\overline{h}}_{\mu \nu }·\nabla {\partial }_t{\overline{h}}^{\mu \nu }{d}^{3}x\hfill \\ \hfill & =0.\hfill \end{array}$$

(32)

Since ${\displaystyle \frac{dE}{dt}}=0$, the energy $E\left(t\right)$ is constant in time:

$$E\left(t\right)=E\left(0\right),\forall t\in \mathbb{R}.$$

(33)

In addition, based on the process introduced, we can postulate an even higher regularity as the integration by parts is justified in $H^1\left({\mathbb{R}}^3\right)$ due to the density of $\mathcal{S}\left({\mathbb{R}}^3\right)$ in $H^1\left({\mathbb{R}}^3\right)$ and its weak derivative properties. □

4.3. Regularity in Sobolev Spaces

Theorem 3 (Higher-Order Energy Conservation).

Consider the Cauchy problem (19)–(21) and recall the initial data $f_{\mu \nu }\in H^{s+1}\left({\mathbb{R}}^3\right)$ and $g_{\mu \nu }\in H^s\left({\mathbb{R}}^3\right)$ for some integer $s\ge 1$.

Define the higher-order energy functional:

$$E_s\left(t\right)={\displaystyle \frac{1}{2}}\sum _{\left|\alpha \right|\le s}{\int }_{{\mathbb{R}}^3}\left(|{\partial }_t{\partial }^{\alpha }{\overline{h}}_{\mu \nu }{(t,\mathbf{x})|}^2+{|\nabla {\partial }^{\alpha }{\overline{h}}_{\mu \nu }(t,\mathbf{x})|}^2\right)d^{3}x,$$

(34)

where ${\partial }^{\alpha }$ denotes the partial derivative operator of multi-index α, with $\left|\alpha \right|$ denoting the order of differentiation.

Then, the higher-order energy $E_s\left(t\right)$ is conserved over time, i.e.,

$$E_s\left(t\right)=E_s\left(0\right),\forall t\in \mathbb{R}.$$

(35)

Proof. 

We will prove the conservation of $E_s\left(t\right)$ by the induction of s.

Before proceeding, recall that for $u\in H^k\left({\mathbb{R}}^3\right)$, the norm is defined as follows:

$${\parallel u\parallel }_{H^k}^2=\sum _{\left|\beta \right|\le k}{\parallel {\partial }^{\beta }u\parallel }_{L^2}^2.$$

The functions ${\overline{h}}_{\mu \nu }(t,\mathbf{x})$ and their derivatives are assumed to be sufficiently smooth to justify the calculations below, and any necessary decay at infinity ensures that the integrals converge.

When $s=0$, the energy functional reduces to the following:

$$E_0\left(t\right)={\displaystyle \frac{1}{2}}{\int }_{{\mathbb{R}}^3}\left(|{\partial }_t{\overline{h}}_{\mu \nu }{|}^2+{|\nabla {\overline{h}}_{\mu \nu }|}^2\right)d^{3}x.$$

(36)

We have previously established that $E_0\left(t\right)$ is conserved, i.e.,

$${\displaystyle \frac{d}{dt}}{E}_0\left(t\right)=0,$$

(37)

which implies that $E_0\left(t\right)=E_0\left(0\right)$ for all t.

Assume that for some integer $k\ge 0$, the higher-order energy $E_k\left(t\right)$ is conserved, i.e.,

$$E_k\left(t\right)=E_k\left(0\right),\forall t\in \mathbb{R}.$$

(38)

Our goal is to prove that $E_{k+1}\left(t\right)=E_{k+1}\left(0\right)$.

Consider the multi-indices $\alpha$ with $\left|\alpha \right|=k+1$. For each such $\alpha$, the function ${\partial }^{\alpha }{\overline{h}}_{\mu \nu }$ satisfies the wave equation because the wave operator □ commutes with partial derivatives (since the coefficients are constant):

$$\square {\partial }^{\alpha }{\overline{h}}_{\mu \nu }={\partial }^{\alpha }\square {\overline{h}}_{\mu \nu }=0.$$

(39)

Define the energy functional for ${\partial }^{\alpha }{\overline{h}}_{\mu \nu }$:

$$E_{\alpha }\left(t\right)={\displaystyle \frac{1}{2}}{\int }_{{\mathbb{R}}^3}\left(|{\partial }_t{\partial }^{\alpha }{\overline{h}}_{\mu \nu }{|}^2+{|\nabla {\partial }^{\alpha }{\overline{h}}_{\mu \nu }|}^2\right)d^{3}x.$$

(40)

Our aim is to show that ${\displaystyle \frac{d}{dt}}{E}_{\alpha }\left(t\right)=0$ for each $\left|\alpha \right|=k+1$.

Compute ${\displaystyle \frac{d}{dt}}{E}_{\alpha }\left(t\right)$:

$$\begin{array}{cc}\hfill {\displaystyle \frac{d}{dt}}{E}_{\alpha }\left(t\right)& ={\int }_{{\mathbb{R}}^3}\left({\partial }_t{\partial }^{\alpha }{\overline{h}}_{\mu \nu }·{\partial }_t^2{\partial }^{\alpha }{\overline{h}}^{\mu \nu }+\nabla {\partial }^{\alpha }{\overline{h}}_{\mu \nu }·\nabla {\partial }_t{\partial }^{\alpha }{\overline{h}}^{\mu \nu }\right)d^{3}x.\hfill \end{array}$$

(41)

In addition, we can obtain the following:

$${\partial }_t^2{\partial }^{\alpha }{\overline{h}}_{\mu \nu }=\Delta {\partial }^{\alpha }{\overline{h}}_{\mu \nu },$$

(42)

Now, substitute (42) into (41):

$$\begin{array}{cc}\hfill {\displaystyle \frac{d}{dt}}{E}_{\alpha }\left(t\right)& ={\int }_{{\mathbb{R}}^3}\left({\partial }_t{\partial }^{\alpha }{\overline{h}}_{\mu \nu }·\Delta {\partial }^{\alpha }{\overline{h}}^{\mu \nu }+\nabla {\partial }^{\alpha }{\overline{h}}_{\mu \nu }·\nabla {\partial }_t{\partial }^{\alpha }{\overline{h}}^{\mu \nu }\right)d^{3}x.\hfill \end{array}$$

(43)

Let us examine each term in (43).

The first term is discussed as follows:

$$\begin{array}{cc}\hfill I_1& ={\int }_{{\mathbb{R}}^3}{\partial }_t{\partial }^{\alpha }{\overline{h}}_{\mu \nu }·\Delta {\partial }^{\alpha }{\overline{h}}^{\mu \nu }{d}^{3}x\hfill \end{array}$$

(44)

$$\begin{array}{cc}\hfill & ={\int }_{{\mathbb{R}}^3}{\partial }_t{\partial }^{\alpha }{\overline{h}}_{\mu \nu }·{\partial }_i{\partial }_i{\partial }^{\alpha }{\overline{h}}^{\mu \nu }{d}^{3}x,\hfill \end{array}$$

(45)

where summation over $i=1,2,3$ is implied.

Integrate by parts with respect to $x_i$ (note that boundary terms vanish due to sufficient decay at infinity):

$$\begin{array}{cc}\hfill I_1& =-{\int }_{{\mathbb{R}}^3}{\partial }_i{\partial }_t{\partial }^{\alpha }{\overline{h}}_{\mu \nu }·{\partial }_i{\partial }^{\alpha }{\overline{h}}^{\mu \nu }{d}^{3}x.\hfill \end{array}$$

(46)

For the second term,

$$\begin{array}{cc}\hfill I_2& ={\int }_{{\mathbb{R}}^3}\nabla {\partial }^{\alpha }{\overline{h}}_{\mu \nu }·\nabla {\partial }_t{\partial }^{\alpha }{\overline{h}}^{\mu \nu }{d}^{3}x\hfill \end{array}$$

(47)

$$\begin{array}{cc}\hfill & ={\int }_{{\mathbb{R}}^3}{\partial }_i{\partial }^{\alpha }{\overline{h}}_{\mu \nu }·{\partial }_i{\partial }_t{\partial }^{\alpha }{\overline{h}}^{\mu \nu }{d}^{3}x.\hfill \end{array}$$

(48)

By adding $I_1$ and $I_2$, we can obtain the following:

$$\begin{array}{cc}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}{I}_1+I_2\hfill & =-{\int }_{{\mathbb{R}}^3}{\partial }_i{\partial }_t{\partial }^{\alpha }{\overline{h}}_{\mu \nu }·{\partial }_i{\partial }^{\alpha }{\overline{h}}^{\mu \nu }{d}^{3}x+{\int }_{{\mathbb{R}}^3}{\partial }_i{\partial }^{\alpha }{\overline{h}}_{\mu \nu }·{\partial }_i{\partial }_t{\partial }^{\alpha }{\overline{h}}^{\mu \nu }{d}^{3}x\hfill \end{array}$$

(49)

$$\begin{array}{l}=0.\end{array}$$

(50)

Therefore,

$${\displaystyle \frac{d}{dt}}{E}_{\alpha }\left(t\right)=I_1+I_2=0.$$

(51)

This shows that $E_{\alpha }\left(t\right)$ is conserved for each multi-index $\alpha$ with $\left|\alpha \right|=k+1$.

Define the energy $E_{k+1}\left(t\right)$ as follows:

$$E_{k+1}\left(t\right)=E_k\left(t\right)+\sum _{\left|\alpha \right|=k+1}{E}_{\alpha }\left(t\right),$$

(52)

where $E_k\left(t\right)$ is conserved by the induction hypothesis, and each $E_{\alpha }\left(t\right)$ with $\left|\alpha \right|=k+1$ is conserved as shown above.

Therefore,

$$\begin{array}{cc}\hfill \hspace{1em}\hspace{1em}{E}_{k+1}\left(t\right)& =E_k\left(t\right)+\sum _{\left|\alpha \right|=k+1}{E}_{\alpha }\left(t\right)\hfill \end{array}$$

(53)

$$\begin{array}{cc}\hfill & \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}=E_k\left(0\right)+\sum _{\left|\alpha \right|=k+1}{E}_{\alpha }\left(0\right)\hfill \end{array}$$

(54)

$$\begin{array}{cc}\hfill & =E_{k+1}\left(0\right).\hfill \end{array}$$

(55)

By mathematical induction, the higher-order energy $E_s\left(t\right)$ is conserved for all integers $s\ge 1$.

Since the higher-order energies $E_s\left(t\right)$ are conserved and the Sobolev norms can be expressed in terms of these energies (up to constants), we have the following:

$$\parallel {\overline{h}}_{\mu \nu }{\left(t\right)\parallel }_{H^{s+1}\left({\mathbb{R}}^3\right)}^2+{\parallel {\partial }_t{\overline{h}}_{\mu \nu }\left(t\right)\parallel }_{H^s\left({\mathbb{R}}^3\right)}^2\le C_s{E}_s\left(t\right)=C_s{E}_s\left(0\right),$$

(56)

where $C_s$ is a constant depending on s.

This implies that the Sobolev norms $\parallel {\overline{h}}_{\mu \nu }{\left(t\right)\parallel }_{H^{s+1}}$ and $\parallel {\partial }_t{\overline{h}}_{\mu \nu }{\left(t\right)\parallel }_{H^s}$ remain bounded for all t, and, thus, the solution ${\overline{h}}_{\mu \nu }$ maintains its regularity in $H^{s+1}\left({\mathbb{R}}^3\right)$ for all times. □

We introduce one remark:

The higher-order energies $E_s\left(t\right)$ are equivalent to the Sobolev norms of ${\overline{h}}_{\mu \nu }$ and ${\partial }_t{\overline{h}}_{\mu \nu }$. Specifically, the constants $C_1,C_2>0$ exist such that

$$C_1\left(\parallel {\partial }_t{\overline{h}}_{\mu \nu }{\left(t\right)\parallel }_{H^s}^2+{\parallel {\overline{h}}_{\mu \nu }\left(t\right)\parallel }_{H^{s+1}}^2\right)\le E_s\left(t\right)\le C_2\left(\parallel {\partial }_t{\overline{h}}_{\mu \nu }{\left(t\right)\parallel }_{H^s}^2+{\parallel {\overline{h}}_{\mu \nu }\left(t\right)\parallel }_{H^{s+1}}^2\right).$$

The conservation of higher-order energies implies that the solution ${\overline{h}}_{\mu \nu }$ retains its regularity over time and the norms do not grow unbounded, ensuring the stability of the solution in the Sobolev space $H^{s+1}\left({\mathbb{R}}^3\right)$.

5. Spectral Analysis

The d’Alembertian operator □ is a second-order linear differential operator with constant coefficients in Minkowski spacetime. In the context of functions defined on ${\mathbb{R}}^{1+3}$, the operator □ is self-adjoint when acting on suitable function spaces, such as $L^2\left({\mathbb{R}}^3\right)$ for a fixed time t. Then, consider the Fourier transform of the eigenvalue equation:

$$\square \varphi \left(\mathbf{x}\right)=\lambda \varphi \left(\mathbf{x}\right).$$

(57)

Applying the Fourier transform in $\mathbf{x}$, and assuming time-independent eigenfunctions (since □ includes time derivatives), we obtain the following:

$$-{\left|\mathit{\xi }\right|}^2\widehat{\varphi }\left(\mathit{\xi }\right)=\lambda \widehat{\varphi }\left(\mathit{\xi }\right).$$

(58)

This implies that the eigenvalues are $\lambda =-{\left|\mathit{\xi }\right|}^2$ and the eigenfunctions are plane waves, $e^{i\mathit{\xi }·\mathbf{x}}$.

Therefore, the spectrum of □ is the continuous spectrum $\sigma (\square )=(-\infty ,0]$. The Fourier transform diagonalizes linear constant coefficient differential operators. The operator □ acts as multiplication by $-{\omega }^2$, where $\omega =\left|\mathit{\xi }\right|$ is the frequency.

It is standard to show (following the same process as provided in Theorem 1) that the solution ${\overline{h}}_{\mu \nu }(t,\mathbf{x})$ to the Cauchy problem (19)–(21) has the Fourier transform given as follows:

$${\widehat{\overline{h}}}_{\mu \nu }(t,\mathit{\xi })={\widehat{f}}_{\mu \nu }\left(\mathit{\xi }\right)cos\left(\right|\mathit{\xi }\left|t\right)+{\displaystyle \frac{{\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)}{\left|\mathit{\xi }\right|}}sin\left(\right|\mathit{\xi }\left|t\right),$$

(59)

where ${\widehat{f}}_{\mu \nu }\left(\mathit{\xi }\right)$ and ${\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)$ are the Fourier transforms of the initial data $f_{\mu \nu }\left(\mathbf{x}\right)$ and $g_{\mu \nu }\left(\mathbf{x}\right)$, respectively.

From Equation (59), we observe that for each fixed $\mathit{\xi }$, the solution ${\widehat{\overline{h}}}_{\mu \nu }(t,\mathit{\xi })$ is a linear combination of $cos\left(\right|\mathit{\xi }\left|t\right)$ and $sin\left(\right|\mathit{\xi }\left|t\right)$ functions, which are bounded in time and oscillatory.

This indicates that the solution in the Fourier space exhibits harmonic oscillations with the angular frequency $\left|\mathit{\xi }\right|$.

Since the cosine and sine functions are bounded for all t, and assuming that ${\widehat{f}}_{\mu \nu }\left(\mathit{\xi }\right)$ and ${\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)$ are in $L^2\left({\mathbb{R}}^3\right)$ (which follows from $f_{\mu \nu },g_{\mu \nu }\in L^2\left({\mathbb{R}}^3\right)$), we can conclude that ${\widehat{\overline{h}}}_{\mu \nu }(t,\mathit{\xi })$ remains in $L^2\left({\mathbb{R}}^3\right)$ for all t. We recall here that the $L^2$ norm of ${\widehat{\overline{h}}}_{\mu \nu }(t,\mathit{\xi })$ is conserved due to the Plancherel theorem and the conservation of energy for the wave equation.

Taking the inverse Fourier transform, we obtain ${\overline{h}}_{\mu \nu }(t,\mathbf{x})$:

$${\overline{h}}_{\mu \nu }(t,\mathbf{x})={\int }_{{\mathbb{R}}^3}\left({\widehat{f}}_{\mu \nu }\left(\mathit{\xi }\right)cos\left(\right|\mathit{\xi }\left|t\right)+{\displaystyle \frac{{\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)}{\left|\mathit{\xi }\right|}}sin\left(\right|\mathit{\xi }\left|t\right)\right)e^{i\mathit{\xi }·\mathbf{x}}{\displaystyle \frac{d^3\mathit{\xi }}{{\left(2\pi \right)}^3}}.$$

(60)

Note that the term ${\displaystyle \frac{{\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)}{\left|\mathit{\xi }\right|}}$ may be singular at $\mathit{\xi }=0$. However, since ${\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)$ is in $H^s$, it behaves like ${\left|\mathit{\xi }\right|}^{-s}$ near $\mathit{\xi }=0$. The factor of $\left|\mathit{\xi }\right|$ in the denominator is offset by the decay of ${\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)$, ensuring that the product remains integrable. Specifically, for $s\ge 1$, the integrand ${\displaystyle \frac{|{\widehat{g}}_{\mu \nu }{\left(\mathit{\xi }\right)|}^2}{{\left|\mathit{\xi }\right|}^2}}$ multiplied by ${(1+|\mathit{\xi }|}^2{)}^{s+1}$ remains integrable near $\mathit{\xi }=0$.

Since the integrand is bounded in time, and the initial data are in suitable Sobolev spaces, ensuring the sufficient decay of ${\widehat{f}}_{\mu \nu }\left(\mathit{\xi }\right)$ and ${\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)$ as $\left|\mathit{\xi }\right|\to \infty$, the integral converges for all t; hence, ${\overline{h}}_{\mu \nu }(t,\mathbf{x})$ remains bounded.

5.1. Absence of Exponential Growth

Suppose, for the sake of argument, that the solution contains exponentially growing terms, i.e., terms of the form $e^{\lambda t}$ with $\lambda >0$. Such behavior would imply instability, as perturbations would grow unboundedly over time.

To investigate this possibility, consider the characteristic equation associated with the ODE (22):

$$r^2+{\left|\mathit{\xi }\right|}^2=0,$$

(61)

with solutions $r=\pm i\left|\mathit{\xi }\right|$. This indicates that the eigenvalues are purely imaginary, and thus the general solution involves oscillatory functions.

If we attempt to find solutions with exponential growth, we would need to consider eigenvalues with a positive real part. Suppose $r=\lambda$, with $\lambda >0$. We can substitute this into a characteristic equation:

$${\lambda }^2+{\left|\mathit{\xi }\right|}^2=0⟹{\lambda }^2=-{\left|\mathit{\xi }\right|}^2.$$

Since ${\left|\mathit{\xi }\right|}^2\ge 0$, ${\lambda }^2\le 0$, which contradicts $\lambda >0$. Therefore, there are no eigenvalues with positive real parts, and the solution cannot contain exponentially growing terms.

The boundedness and oscillatory nature of the solution implies that small initial perturbations in the metric tensor remain small for all times. This suggests that, at the linearized level, spacetime is stable under small fluctuations, and the metric does not deviate significantly from the background Minkowski spacetime due to these perturbations.

In the context of quantum foam models, this stability is significant. It indicates that the quantum fluctuations of spacetime at the Planck scale do not lead to instabilities that could disrupt the spacetime. The absence of exponentially growing modes supports the physical plausibility of the quantum foam concept, as it ensures that spacetime can sustain small fluctuations.

5.2. Operator Analysis and the Spectral Theorem

To provide an additional mathematical foundation for the stability of solutions to the wave equation, we employ the spectral theorem for unbounded self-adjoint operators (the reader is referred to the classical text [7] for additional details). We show that we obtain similar results by other resolution paths.

Let $\mathcal{H}$ be a Hilbert space, and let H be a densely defined, self-adjoint, non-negative operator on $\mathcal{H}$. Consider the abstract Cauchy problem:

$$\left\{\begin{array}{c}{\displaystyle \frac{d^2}{d{t}^2}}\psi \left(t\right)+H\psi \left(t\right)=0,\hfill \\ \psi \left(0\right)={\psi }_0\in \mathcal{D}\left(H^{1/2}\right),\hfill \\ {\psi }^{\prime }\left(0\right)={\psi }_1\in \mathcal{H}.\hfill \end{array}\right.$$

(62)

Here, $\mathcal{D}\left(H^{1/2}\right)$ denotes the domain of $H^{1/2}$, which consists of all $\varphi \in \mathcal{H}$ such that $\parallel H^{1/2}{\varphi \parallel }_{\mathcal{H}}<\infty$.

The spectral theorem states that H admits a spectral decomposition:

$$H={\int }_0^{\infty }\lambda dE\left(\lambda \right),$$

(63)

where $E\left(\lambda \right)$ is the projection-valued spectral measure associated with H.

Now, we define the following:

$$\begin{array}{cc}\hfill cos\left(t{H}^{1/2}\right)& ={\int }_0^{\infty }cos\left(t\sqrt{\lambda }\right)dE\left(\lambda \right),\hfill \end{array}$$

(64)

$$\begin{array}{cc}\hfill sin\left(t{H}^{1/2}\right)& ={\int }_0^{\infty }sin\left(t\sqrt{\lambda }\right)dE\left(\lambda \right).\hfill \end{array}$$

(65)

The solution to (62) is then given as follows:

$$\psi \left(t\right)=cos\left(t{H}^{1/2}\right){\psi }_0+H^{-1/2}sin\left(t{H}^{1/2}\right){\psi }_1.$$

(66)

In our case, the Hilbert space is $\mathcal{H}=L^2\left({\mathbb{R}}^3\right)$, and the operator $H=-\Delta$, the (negative) Laplacian on ${\mathbb{R}}^3$, which is self-adjoint and non-negative with the domain $\mathcal{D}\left(H\right)=H^2\left({\mathbb{R}}^3\right)$.

The operator $H^{1/2}$ corresponds to $|\nabla |$, defined via the Fourier transform:

$$\widehat{H^{1/2}\varphi }\left(\mathit{\xi }\right)=\left|\mathit{\xi }\right|\widehat{\varphi }\left(\mathit{\xi }\right).$$

(67)

Therefore, the solution to the wave equation in terms of the initial data is as follows:

$${\overline{h}}_{\mu \nu }\left(t\right)=cos\left(t|\nabla |\right)f_{\mu \nu }+{|\nabla |}^{-1}sin\left(t|\nabla |\right)g_{\mu \nu }.$$

(68)

In the Fourier space, this becomes the following:

$${\widehat{\overline{h}}}_{\mu \nu }(t,\mathit{\xi })=cos\left(t\left|\mathit{\xi }\right|\right){\widehat{f}}_{\mu \nu }\left(\mathit{\xi }\right)+{\displaystyle \frac{sin\left(t\left|\mathit{\xi }\right|\right)}{\left|\mathit{\xi }\right|}}{\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right),$$

(69)

which matches our earlier results.

The energy functional associated with the abstract wave equation is as follows:

$$E\left(t\right)={\displaystyle \frac{1}{2}}\left(\parallel {\psi }^{\prime }{\left(t\right)\parallel }_{\mathcal{H}}^2+{\langle H\psi \left(t\right),\psi \left(t\right)\rangle }_{\mathcal{H}}\right).$$

(70)

We show that $E\left(t\right)$ is conserved over time. By differentiating $E\left(t\right)$ with respect to t, we have

$$\begin{array}{cc}\hfill {\displaystyle \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\frac{dE}{dt}}& ={〈{\psi }^{\prime }\left(t\right),{\psi }^{\prime \prime }\left(t\right)〉}_{\mathcal{H}}+{〈H\psi \left(t\right),{\psi }^{\prime }\left(t\right)〉}_{\mathcal{H}}\hfill \end{array}$$

(71)

$$\begin{array}{cc}\hfill & \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}={〈{\psi }^{\prime }\left(t\right),-H\psi \left(t\right)〉}_{\mathcal{H}}+{〈H\psi \left(t\right),{\psi }^{\prime }\left(t\right)〉}_{\mathcal{H}}\hfill \end{array}$$

(72)

$$\begin{array}{cc}\hfill & \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}=-{〈{\psi }^{\prime }\left(t\right),H\psi \left(t\right)〉}_{\mathcal{H}}+{〈H\psi \left(t\right),{\psi }^{\prime }\left(t\right)〉}_{\mathcal{H}}\hfill \end{array}$$

(73)

$$\begin{array}{cc}\hfill & =0,\hfill \end{array}$$

(74)

since the terms cancel due to the self-adjointness of H (i.e., ${〈H\psi \left(t\right),{\psi }^{\prime }\left(t\right)〉}_{\mathcal{H}}={〈{\psi }^{\prime }\left(t\right),H\psi \left(t\right)〉}_{\mathcal{H}}$).

In our specific setting, the energy functional becomes the following:

$$E\left(t\right)={\displaystyle \frac{1}{2}}\left(\parallel {\partial }_t{\overline{h}}_{\mu \nu }{\left(t\right)\parallel }_{L^2\left({\mathbb{R}}^3\right)}^2+{\parallel \nabla {\overline{h}}_{\mu \nu }\left(t\right)\parallel }_{L^2\left({\mathbb{R}}^3\right)}^2\right).$$

(75)

The conservation of energy implies that

$$E\left(t\right)=E\left(0\right),\forall t\in \mathbb{R}.$$

(76)

This result aligns with Theorem 2 and confirms that the $L^2$ norms of ${\partial }_t{\overline{h}}_{\mu \nu }\left(t\right)$ and $\nabla {\overline{h}}_{\mu \nu }\left(t\right)$ remain constant over time.

Spectral representation also provides information about the absence of exponentially growing modes. Since H is non-negative and self-adjoint, its spectrum $\sigma \left(H\right)\subset [0,\infty )$. The functions $cos\left(t\sqrt{\lambda }\right)$ and $sin\left(t\sqrt{\lambda }\right)$ are bounded for all $\lambda \ge 0$ and $t\in \mathbb{R}$.

If H were to have eigenvalues with negative real parts (i.e., if $\lambda$ were negative), the functions $\sqrt{\lambda }$ would be imaginary, and the solutions would involve exponential growth or decay. Specifically, the solution would contain terms like $e^{\gamma t}$ with $\gamma =\sqrt{-\lambda }$. However, this is not the case, as Laplacian’s spectrum is non-negative.

Therefore, the spectral theorem confirms that all modes of the solution are oscillatory and bounded in time, and this reinforce our previous assessments.

6. Explicit Calculation of Sobolev Norms

In this section, we demonstrate that the Sobolev norm $\parallel {\overline{h}}_{\mu \nu }{\left(t\right)\parallel }_{H^{s+1}\left({\mathbb{R}}^3\right)}$ remains bounded for all time t, given the initial data $f_{\mu \nu }\in H^{s+1}\left({\mathbb{R}}^3\right)$ and $g_{\mu \nu }\in H^s\left({\mathbb{R}}^3\right)$.

Theorem 4. 

Let $s\ge 1$ be an integer, and suppose that $f_{\mu \nu }\in H^{s+1}\left({\mathbb{R}}^3\right)$ and $g_{\mu \nu }\in H^s\left({\mathbb{R}}^3\right)$. Let ${\overline{h}}_{\mu \nu }(t,\mathbf{x})$ be the solution to the linear wave equation with the initial data $f_{\mu \nu },g_{\mu \nu }$. Then, for all $t\in \mathbb{R}$,

$$\parallel {\overline{h}}_{\mu \nu }{\left(t\right)\parallel }_{H^{s+1}\left({\mathbb{R}}^3\right)}\le C\left(\parallel f_{\mu \nu }{\parallel }_{H^{s+1}\left({\mathbb{R}}^3\right)}+{\parallel g_{\mu \nu }\parallel }_{H^s\left({\mathbb{R}}^3\right)}\right),$$

(77)

where C is a constant independent of t.

Proof. 

We begin by recalling the definition of the Sobolev norm $H^{s+1}\left({\mathbb{R}}^3\right)$:

$${\parallel u\parallel }_{H^{s+1}\left({\mathbb{R}}^3\right)}^2={\int }_{{\mathbb{R}}^3}{(1+|\mathit{\xi }|}^2{)}^{s+1}{\left|\widehat{u}\left(\mathit{\xi }\right)\right|}^2{d}^3\mathit{\xi },$$

(78)

where $\widehat{u}\left(\mathit{\xi }\right)$ denotes the Fourier transform of $u\left(\mathbf{x}\right)$.

The solution ${\overline{h}}_{\mu \nu }(t,\mathbf{x})$ has the Fourier transform as provided in (59). Our goal is to estimate $\parallel {\overline{h}}_{\mu \nu }{\left(t\right)\parallel }_{H^{s+1}}$ in terms of the norms of the initial data.

We compute the following:

$$\begin{array}{cc}\hfill \parallel {\overline{h}}_{\mu \nu }{\left(t\right)\parallel }_{H^{s+1}}^2& ={\int }_{{\mathbb{R}}^3}{(1+|\mathit{\xi }|}^2{)}^{s+1}{\left|{\widehat{\overline{h}}}_{\mu \nu }(t,\mathit{\xi })\right|}^2{d}^3\mathit{\xi }\hfill \end{array}$$

(79)

By substituting the expression for ${\widehat{\overline{h}}}_{\mu \nu }(t,\mathit{\xi })$ from (59), we have the following:

$$\begin{array}{cc}\hfill {\left|{\widehat{\overline{h}}}_{\mu \nu }(t,\mathit{\xi })\right|}^2& ={\left|{\widehat{f}}_{\mu \nu }\left(\mathit{\xi }\right)cos\left(\right|\mathit{\xi }\left|t\right)+{\displaystyle \frac{{\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)}{\left|\mathit{\xi }\right|}}sin\left(\right|\mathit{\xi }\left|t\right)\right|}^2\hfill \\ \hfill & ={\left|{\widehat{f}}_{\mu \nu }\left(\mathit{\xi }\right)\right|}^2{cos}^2\left(\right|\mathit{\xi }\left|t\right)+{\left|{\displaystyle \frac{{\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)}{\left|\mathit{\xi }\right|}}\right|}^2{sin}^2\left(\right|\mathit{\xi }\left|t\right)\hfill \\ \hfill & +2\Re \left({\widehat{f}}_{\mu \nu }\left(\mathit{\xi }\right)\overline{{\displaystyle \frac{{\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)}{\left|\mathit{\xi }\right|}}}cos\left(\right|\mathit{\xi }\left|t\right)sin\left(\right|\mathit{\xi }\left|t\right)\right)\hfill \end{array}$$

(80)

The cross term involves the real part of a complex number, which can be positive or negative. To obtain an upper bound, we can take the absolute value:

$$\begin{array}{cc}\hfill \left|2\Re \left({\widehat{f}}_{\mu \nu }\left(\mathit{\xi }\right)\overline{{\displaystyle \frac{{\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)}{\left|\mathit{\xi }\right|}}}cos\left(\right|\mathit{\xi }\left|t\right)sin\left(\right|\mathit{\xi }\left|t\right)\right)\right|& \le 2\left|{\widehat{f}}_{\mu \nu }\left(\mathit{\xi }\right)\right|\left|{\displaystyle \frac{{\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)}{\left|\mathit{\xi }\right|}}\right|\left|cos\left(\right|\mathit{\xi }\left|t\right)sin\left(\right|\mathit{\xi }\left|t\right)\right|\hfill \\ \hfill & \le 2\left|{\widehat{f}}_{\mu \nu }\left(\mathit{\xi }\right)\right|\left|{\displaystyle \frac{{\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)}{\left|\mathit{\xi }\right|}}\right|·{\displaystyle \frac{1}{2}}\hfill \\ \hfill & =\left|{\widehat{f}}_{\mu \nu }\left(\mathit{\xi }\right)\right|\left|{\displaystyle \frac{{\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)}{\left|\mathit{\xi }\right|}}\right|\hfill \end{array}$$

(81)

Here, we used the following inequality:

$$\left|cos\left(\theta \right)sin\left(\theta \right)\right|\le {\displaystyle \frac{1}{2}},\forall \theta \in \mathbb{R}$$

(82)

Using (80) and (81), we can write the following:

$$\begin{array}{cc}\hfill {\left|{\widehat{\overline{h}}}_{\mu \nu }(t,\mathit{\xi })\right|}^2& \le {\left|{\widehat{f}}_{\mu \nu }\left(\mathit{\xi }\right)\right|}^2+{\left|{\displaystyle \frac{{\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)}{\left|\mathit{\xi }\right|}}\right|}^2+\left|{\widehat{f}}_{\mu \nu }\left(\mathit{\xi }\right)\right|\left|{\displaystyle \frac{{\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)}{\left|\mathit{\xi }\right|}}\right|\hfill \\ \hfill & \le {\left(\left|{\widehat{f}}_{\mu \nu }\left(\mathit{\xi }\right)\right|+\left|{\displaystyle \frac{{\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)}{\left|\mathit{\xi }\right|}}\right|\right)}^2\hfill \end{array}$$

(83)

Therefore, we have

$$\begin{array}{cc}\hfill \parallel {\overline{h}}_{\mu \nu }{\left(t\right)\parallel }_{H^{s+1}}^2& \le {\int }_{{\mathbb{R}}^3}{(1+|\mathit{\xi }|}^2{)}^{s+1}{\left(\left|{\widehat{f}}_{\mu \nu }\left(\mathit{\xi }\right)\right|+\left|{\displaystyle \frac{{\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)}{\left|\mathit{\xi }\right|}}\right|\right)}^2{d}^3\mathit{\xi }\hfill \end{array}$$

(84)

By expanding the square in (84), we obtain the following:

$$\begin{array}{cc}\hfill {\left(\left|{\widehat{f}}_{\mu \nu }\left(\mathit{\xi }\right)\right|+\left|{\displaystyle \frac{{\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)}{\left|\mathit{\xi }\right|}}\right|\right)}^2& ={\left|{\widehat{f}}_{\mu \nu }\left(\mathit{\xi }\right)\right|}^2+2\left|{\widehat{f}}_{\mu \nu }\left(\mathit{\xi }\right)\right|\left|{\displaystyle \frac{{\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)}{\left|\mathit{\xi }\right|}}\right|+{\left|{\displaystyle \frac{{\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)}{\left|\mathit{\xi }\right|}}\right|}^2\hfill \end{array}$$

(85)

Therefore, the Sobolev norm becomes

$$\begin{array}{cc}\hfill \parallel {\overline{h}}_{\mu \nu }{\left(t\right)\parallel }_{H^{s+1}}^2& \le {\int }_{{\mathbb{R}}^3}{(1+|\mathit{\xi }|}^2{)}^{s+1}\left({\left|{\widehat{f}}_{\mu \nu }\left(\mathit{\xi }\right)\right|}^2+2\left|{\widehat{f}}_{\mu \nu }\left(\mathit{\xi }\right)\right|\left|{\displaystyle \frac{{\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)}{\left|\mathit{\xi }\right|}}\right|+{\left|{\displaystyle \frac{{\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)}{\left|\mathit{\xi }\right|}}\right|}^2\right)d^3\mathit{\xi }\hfill \end{array}$$

(86)

We will estimate each term in (86) separately.

First Term:

$$\begin{array}{cc}\hfill I_1& ={\int }_{{\mathbb{R}}^3}{(1+|\mathit{\xi }|}^2{)}^{s+1}{\left|{\widehat{f}}_{\mu \nu }\left(\mathit{\xi }\right)\right|}^2{d}^3\mathit{\xi }\hfill \\ \hfill & = \parallel f_{\mu \nu }{\parallel }_{H^{s+1}}^2\hfill \end{array}$$

(87)

Second Term:

$$\begin{array}{cc}\hfill I_2& =2{\int }_{{\mathbb{R}}^3}{(1+|\mathit{\xi }|}^2{)}^{s+1}\left|{\widehat{f}}_{\mu \nu }\left(\mathit{\xi }\right)\right|\left|{\displaystyle \frac{{\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)}{\left|\mathit{\xi }\right|}}\right|d^3\mathit{\xi }\hfill \end{array}$$

(88)

We need to bound $I_2$. We will consider two cases:

Case 1: $\left|\mathit{\xi }\right|\ge 1$

In this case, we have $\left|\mathit{\xi }\right|\ge 1$, so ${\left|\mathit{\xi }\right|}^{-1}\le 1$. Also, ${(1+|\mathit{\xi }|}^2{)}^{s+1}\le 2^{s+1}{\left|\mathit{\xi }\right|}^{2(s+1)}$.

Therefore,

$$\begin{array}{cc}\hfill I_2^{\left(1\right)}& =2{\int }_{\left|\mathit{\xi }\right|\ge 1}{(1+|\mathit{\xi }|}^2{)}^{s+1}\left|{\widehat{f}}_{\mu \nu }\left(\mathit{\xi }\right)\right|\left|{\displaystyle \frac{{\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)}{\left|\mathit{\xi }\right|}}\right|d^3\mathit{\xi }\hfill \\ \hfill & \le 2{\int }_{\left|\mathit{\xi }\right|\ge 1}{2}^{s+1}{\left|\mathit{\xi }\right|}^{2(s+1)}\left|{\widehat{f}}_{\mu \nu }\left(\mathit{\xi }\right)\right|\left|{\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)\right|d^3\mathit{\xi }\hfill \end{array}$$

(89)

Case 2: $\left|\mathit{\xi }\right|\le 1$

In this case, $\left|\mathit{\xi }\right|\le 1$, so ${\left|\mathit{\xi }\right|}^{-1}\ge 1$. Also, ${(1+|\mathit{\xi }|}^2{)}^{s+1}\le 2^{s+1}$.

Therefore,

$$\begin{array}{cc}\hfill I_2^{\left(2\right)}& =2{\int }_{\left|\mathit{\xi }\right|\le 1}{(1+|\mathit{\xi }|}^2{)}^{s+1}\left|{\widehat{f}}_{\mu \nu }\left(\mathit{\xi }\right)\right|\left|{\displaystyle \frac{{\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)}{\left|\mathit{\xi }\right|}}\right|d^3\mathit{\xi }\hfill \\ \hfill & \le 2^{s+2}{\int }_{\left|\mathit{\xi }\right|\le 1}\left|{\widehat{f}}_{\mu \nu }\left(\mathit{\xi }\right)\right|\left|{\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)\right|{\displaystyle \frac{1}{\left|\mathit{\xi }\right|}}{d}^3\mathit{\xi }\hfill \end{array}$$

(90)

The integral $I_2^{\left(2\right)}$ involves ${\displaystyle \frac{1}{\left|\mathit{\xi }\right|}}$, which is singular at $\mathit{\xi }=0$. To ensure that the integral converges, we need to ensure that the functions ${\widehat{f}}_{\mu \nu }$ and ${\widehat{g}}_{\mu \nu }$ vanish sufficiently quickly at $\mathit{\xi }=0$.

However, since $f_{\mu \nu }\in H^{s+1}\left({\mathbb{R}}^3\right)$ and $g_{\mu \nu }\in H^s\left({\mathbb{R}}^3\right)$, their Fourier transforms are in $L^2$ weighted spaces, and ${\widehat{f}}_{\mu \nu }$ and ${\widehat{g}}_{\mu \nu }$ are squarely integrable.

We can use the Cauchy-Schwarz inequality to bound $I_2$.

$$\begin{array}{cc}\hfill I_2& =2{\int }_{{\mathbb{R}}^3}\sqrt{{(1+|\mathit{\xi }|}^2{)}^{s+1}}\left|{\widehat{f}}_{\mu \nu }\left(\mathit{\xi }\right)\right|·\sqrt{{(1+|\mathit{\xi }|}^2{)}^{s+1}}\left|{\displaystyle \frac{{\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)}{\left|\mathit{\xi }\right|}}\right|d^3\mathit{\xi }\hfill \\ \hfill & \le 2{\left({\int }_{{\mathbb{R}}^3}{(1+|\mathit{\xi }|}^2{)}^{s+1}{\left|{\widehat{f}}_{\mu \nu }\left(\mathit{\xi }\right)\right|}^2{d}^3\mathit{\xi }\right)}^{1/2}{\left({\int }_{{\mathbb{R}}^3}{(1+|\mathit{\xi }|}^2{)}^{s+1}{\left|{\displaystyle \frac{{\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)}{\left|\mathit{\xi }\right|}}\right|}^2{d}^3\mathit{\xi }\right)}^{1/2}\hfill \end{array}$$

(91)

Note that

$${\int }_{{\mathbb{R}}^3}{(1+|\mathit{\xi }|}^2{)}^{s+1}{\left|{\widehat{f}}_{\mu \nu }\left(\mathit{\xi }\right)\right|}^2{d}^3\mathit{\xi }={\parallel f_{\mu \nu }\parallel }_{H^{s+1}}^2$$

(92)

For the second integral, observe that

$$\begin{array}{cc}\hfill {(1+|\mathit{\xi }|}^2{)}^{s+1}{\left|{\displaystyle \frac{{\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)}{\left|\mathit{\xi }\right|}}\right|}^2& ={\displaystyle \frac{{(1+|\mathit{\xi }|}^2{)}^{s+1}}{{\left|\mathit{\xi }\right|}^2}}{\left|{\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)\right|}^2\hfill \\ \hfill & ={(1+|\mathit{\xi }|}^2{)}^{s-1}{\left|{\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)\right|}^2+\left({\displaystyle \frac{{(1+|\mathit{\xi }|}^2{)}^{s+1}}{{\left|\mathit{\xi }\right|}^2}}{-(1+|\mathit{\xi }|}^2{)}^{s-1}\right){\left|{\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)\right|}^2\hfill \end{array}$$

(93)

Since ${(1+|\mathit{\xi }|}^2{)}^{s-1}\le {(1+|\mathit{\xi }|}^2{)}^s$ for $s\ge 1$, we can write the following:

$$\begin{array}{c}\hfill {(1+|\mathit{\xi }|}^2{)}^{s-1}\le {(1+|\mathit{\xi }|}^2{)}^s\end{array}$$

(94)

Therefore,

$$\begin{array}{cc}\hfill {(1+|\mathit{\xi }|}^2{)}^{s-1}{\left|{\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)\right|}^2& \le \parallel g_{\mu \nu }{\parallel }_{H^s}^2\hfill \end{array}$$

(95)

The additional term in (93) can be bounded by noting that, for $\left|\mathit{\xi }\right|\ge 0$,

$${\displaystyle \frac{{(1+|\mathit{\xi }|}^2{)}^{s+1}}{{\left|\mathit{\xi }\right|}^2}}{-(1+|\mathit{\xi }|}^2{)}^{s-1}\le C_s{(1+|\mathit{\xi }|}^2{)}^s$$

(96)

Thus, we can bound the second integral by a multiple of $\parallel g_{\mu \nu }{\parallel }_{H^s}^2$.

Combining the above estimates, we find that

$$I_2\le C\parallel f_{\mu \nu }{\parallel }_{H^{s+1}}{\parallel g_{\mu \nu }\parallel }_{H^s}$$

(97)

Third Term:

$$\begin{array}{cc}\hfill I_3& ={\int }_{{\mathbb{R}}^3}{(1+|\mathit{\xi }|}^2{)}^{s+1}{\left|{\displaystyle \frac{{\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)}{\left|\mathit{\xi }\right|}}\right|}^2{d}^3\mathit{\xi }\hfill \\ \hfill & ={\int }_{{\mathbb{R}}^3}{\displaystyle \frac{{(1+|\mathit{\xi }|}^2{)}^{s+1}}{{\left|\mathit{\xi }\right|}^2}}{\left|{\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)\right|}^2{d}^3\mathit{\xi }\hfill \end{array}$$

(98)

As before, we can split the integral into $\left|\mathit{\xi }\right|\ge 1$ and $\left|\mathit{\xi }\right|\le 1$.

For $\left|\mathit{\xi }\right|\ge 1$,

$$\begin{array}{cc}\hfill I_3^{\left(1\right)}& ={\int }_{\left|\mathit{\xi }\right|\ge 1}{\displaystyle \frac{{(1+|\mathit{\xi }|}^2{)}^{s+1}}{{\left|\mathit{\xi }\right|}^2}}{\left|{\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)\right|}^2{d}^3\mathit{\xi }\hfill \\ \hfill & \le {\int }_{\left|\mathit{\xi }\right|\ge 1}{\left|\mathit{\xi }\right|}^{2(s+1)-2}{\left|{\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)\right|}^2{d}^3\mathit{\xi }\hfill \\ \hfill & ={\int }_{\left|\mathit{\xi }\right|\ge 1}{\left|\mathit{\xi }\right|}^{2s}{\left|{\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)\right|}^2{d}^3\mathit{\xi }\hfill \end{array}$$

(99)

For $\left|\mathit{\xi }\right|\le 1$,

$$\begin{array}{cc}\hfill I_3^{\left(2\right)}& ={\int }_{\left|\mathit{\xi }\right|\le 1}{\displaystyle \frac{{(1+|\mathit{\xi }|}^2{)}^{s+1}}{{\left|\mathit{\xi }\right|}^2}}{\left|{\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)\right|}^2{d}^3\mathit{\xi }\hfill \\ \hfill & \le 2^{s+1}{\int }_{\left|\mathit{\xi }\right|\le 1}{\displaystyle \frac{1}{{\left|\mathit{\xi }\right|}^2}}{\left|{\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)\right|}^2{d}^3\mathit{\xi }\hfill \end{array}$$

(100)

Again, the integral over $\left|\mathit{\xi }\right|\le 1$ may diverge due to the singularity at $\mathit{\xi }=0$. However, since $g_{\mu \nu }\in H^s\left({\mathbb{R}}^3\right)$, its Fourier transform ${\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)$ must satisfy certain decay properties near $\mathit{\xi }=0$.

Specifically, for $s\ge 1$, we have the following:

$${\int }_{\left|\mathit{\xi }\right|\le 1}{\left|\mathit{\xi }\right|}^{2s}{\left|{\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)\right|}^2{d}^3\mathit{\xi }<\infty$$

(101)

This implies that ${\left|{\widehat{g}}_{\mu \nu }\left(\mathit{\xi }\right)\right|}^2$ behaves like ${\left|\mathit{\xi }\right|}^{-2s}$ near $\mathit{\xi }=0$.

Therefore, the integral in (100) converges for $s\ge 1$.

Combining both cases, we have the following:

$$I_3\le C {\parallel g_{\mu \nu }\parallel }_{H^s}^2$$

(102)

By adding up the estimates for $I_1$, $I_2$, and $I_3$, we obtain the following:

$$\begin{array}{cc}\hfill \parallel {\overline{h}}_{\mu \nu }{\left(t\right)\parallel }_{H^{s+1}}^2& \le \parallel f_{\mu \nu }{\parallel }_{H^{s+1}}^2+C \parallel f_{\mu \nu }{\parallel }_{H^{s+1}}\parallel g_{\mu \nu }{\parallel }_{H^s}+C {\parallel g_{\mu \nu }\parallel }_{H^s}^2\hfill \\ \hfill & \le C\left(\parallel f_{\mu \nu }{\parallel }_{H^{s+1}}^2+{\parallel g_{\mu \nu }\parallel }_{H^s}^2\right)\hfill \end{array}$$

(103)

Here, we use the inequality $ab\le {\displaystyle \frac{1}{2}}(a^2+b^2)$ to combine the mixed term.

Thus, we have shown that

$$\parallel {\overline{h}}_{\mu \nu }{\left(t\right)\parallel }_{H^{s+1}}\le C\left(\parallel f_{\mu \nu }{\parallel }_{H^{s+1}}+{\parallel g_{\mu \nu }\parallel }_{H^s}\right)$$

(104)

where C is a constant independent of t.

This completes the proof of the theorem. □

7. Conclusions

In this paper, we have provided an analysis of the stability and regularity of spacetime perturbations within quantum foam models.

Our results confirm that the energy associated with these perturbations is conserved, and the solutions to the linearized Einstein equations are well behaved in appropriate function spaces. These results have implications for quantum foam theories, suggesting that, despite intense fluctuations at the Planck scale, spacetime can exhibit stable behavior.

Future work may extend this analysis to more general background metrics and consider the impact of non-linear effects and higher-order perturbations.