Dafermos and Rodnianski’s r^p-weighted approach to decay for wave equations

Dafermos and Rodnianski introduced an ${r^p}$ -weighted vector field method to obtain boundedness and decay of solutions to wave equations on a Lorentzian background, avoiding to use global vector field multipliers and commutators with weights in ${t}$ and thus proving to be more robust in dealing with black holes such as in Schwarzschild and Kerr background metrics. The approach has found numerous applications; see, for instance, Dafermos-Rodnianski, Moschidis, or Keir for many insightful discussions. In this blog post, I will give some basic details of this approach, based on lecture notes of S. Klainerman, focusing mostly on the flat Minkowski spacetime.

1. Waves on a Lorentzian manifold

I start with some basic notations on general four-dimensional Lorentzian manifolds, but will focus solely on the flat Minkowski case in the next section. Let ${(\mathcal{M},g)}$ be a Lorentzian manifold associated with a Lorentzian metric ${g = g_{\alpha\beta} dx^\alpha dx^\beta}$ with signature ${(-,+,+,+)}$ ; namely, ${g_{\alpha\beta} = g(\partial_\alpha,\partial_\beta)}$ is defined as the inner product on the tangent bundles ${\mathcal{T}\mathcal{M}\times \mathcal{T}\mathcal{M}}$ . Consider the scalar wave equation on ${(\mathcal{M},g)}$ :

$\displaystyle \Box \phi = 0\ \ \ \ \ (1)$

where ${\Box = \nabla^\alpha \nabla_\alpha = g^{\alpha\beta}\nabla_\alpha \nabla_\beta}$ , with ${g^{\alpha\beta}}$ being the matrix inverse of ${g_{\alpha\beta}}$ . Occasionally, we write ${\Box_g}$ to emphasize the underlying metric ${g}$ . Here, ${\nabla}$ denotes the Levi-Civita connection associated to the metric. Recall, for instance, that for any functions ${f}$ and vector fields ${X = X^\alpha \partial_\alpha}$ , the covariant derivatives are computed by ${\nabla_\alpha f = \partial_\alpha f}$ and ${\nabla_\alpha X^\beta = \partial_\alpha X^\beta + \Gamma^\beta_{\alpha \gamma} X^\gamma}$ , where ${\Gamma^\gamma_{\alpha\beta}}$ denotes the Christoffel symbols

$\displaystyle \Gamma^\gamma_{\alpha\beta} = \frac12 g^{\gamma \delta} (\partial_\alpha g_{\delta \beta} + \partial_\beta g_{\alpha \delta} - \partial_\delta g_{\alpha \beta}) .\ \ \ \ \ (2)$

Recall also the connection between contravariant and covariant derivatives given by lowering and raising indexes ${\nabla^\alpha = g^{\alpha\beta}\nabla_\beta}$ and ${\nabla_\alpha = g_{\alpha\beta}\nabla^\beta}$ . In particular, as a direct consequence of the Levi-Civita connection, the covariant derivative of the metric vanishes

$\displaystyle \nabla_\gamma g_{\alpha \beta} = 0.\ \ \ \ \ (3)$

On the other hand, for any tensor ${T^\alpha_\beta}$ , we compute ${ \nabla_\mu T^\alpha_\beta = \partial_\mu T^\alpha_\beta + \Gamma^\alpha_{\mu \sigma}T^\sigma_\beta - \Gamma^\sigma_{\mu\beta} T^\alpha_\sigma.}$ For any vector field ${X}$ , we also denote by ${\mathcal{L}_X}$ the Lie derivatives on ${(\mathcal{M},g)}$ . Recall that ${\mathcal{L}_X(f) = X(f)}$ and ${\mathcal{L}_XY = [X,Y]}$ for vectors ${Y}$ . On the other hand, for any vector ${X}$ , the deformation tensor ${\pi^{(X)} := \mathcal{L}_X g}$ is computed by

$\displaystyle (\mathcal{L}_X g)(Y,Z) = X(g (Y,Z)) - g(\mathcal{L}_X Y,Z) - g(Y, \mathcal{L}_X Z)\ \ \ \ \ (4)$

or in local coordinates ${\pi^{(X)}_{\alpha\beta} =(\mathcal{L}_X g)_{\alpha\beta} = \nabla_\alpha X_\beta + \nabla_\beta X_\alpha}$ .

1.1. Divergence theorem

The divergence theorem is a basic tool in deriving energy estimates for the wave equations (1). Let ${\Sigma_0, \Sigma_1}$ be spacelike hypersurfaces on ${(\mathcal{M},g)}$ with common boundary, enclosing a spacetime region ${\mathcal{D}}$ , and let ${n_{\Sigma_0}, n_{\Sigma_1}}$ be their future-directed unit normal vectors. Then, for any one-form ${P_\alpha}$ , the divergence theorem reads

$\displaystyle \int_{\Sigma_1} P_\alpha n_{\Sigma_1}^\alpha + \int_{\mathcal{D}} \nabla^\alpha P_\alpha = \int_{\Sigma_0} P_\alpha n^\alpha_{\Sigma_0} \ \ \ \ \ (5)$

where the integrals are taken with respect to the induced volume form. The volume form of spacetime is

$\displaystyle \sqrt{|\det g|} dx^0\cdots dx^3.$

On the other hand, when part of the hypersurface ${\Sigma_1}$ is null, the future-directed unit “normal” ${n_{\Sigma_1}}$ is actually tangent and chosen as a generator for (the null part) of ${\Sigma_1}$ .

1.2. Klainerman’s vector field method

Klainerman introduced the vector field method (e.g., see the original papers here and here) to derive boundedness and decay for solutions to wave equations. This involves using vector fields as multipliers (e.g., Morawetz) and vector fields with good commutators. Let us first recall

Lemma 1 Introduce the energy-momentum tensor

$\displaystyle \mathcal{T}_{\alpha \beta} [\phi]= \nabla_\alpha \phi \nabla_\beta \phi - \frac12 g_{\alpha\beta} \nabla^\gamma \phi \nabla_\gamma \phi .\ \ \ \ \ (6)$

Then, there holds

$\displaystyle \nabla^\beta \mathcal{T}_{\alpha\beta} = \partial_\alpha \phi \Box\phi .$

Proof: Recalling ${\Box\phi = \nabla^\alpha \nabla_\alpha \phi}$ and using (3), we compute

$\displaystyle \begin{aligned} \nabla^\beta \mathcal{T}_{\alpha\beta} &= \nabla_\alpha \phi \Box \phi + \nabla_\beta \phi \nabla^\beta \nabla_\alpha \phi - \frac12 g_{\alpha \beta} ( \nabla^\beta \nabla^\gamma \phi \nabla_\gamma \phi + \nabla^\gamma \phi \nabla^\beta \nabla_\gamma \phi ) \\ &= \nabla_\alpha \phi \Box \phi + \nabla_\beta \phi \nabla^\beta \nabla_\alpha \phi - g_{\alpha \beta} \nabla^\beta \nabla^\gamma \phi \nabla_\gamma \phi \\ &= \partial_\alpha \phi \Box \phi \end{aligned}$

in which we have repeatedly used ${\nabla^\alpha \nabla_\beta \phi = \nabla_\beta \nabla^\alpha \phi}$ , since ${\phi}$ is a scalar. $\Box$

Lemma 2 For any vector field ${X}$ and function ${w}$ , introduce the one-form

$\displaystyle P_\alpha [X,w] = \mathcal{T}_{\alpha\beta}[\phi]X^\beta + \frac12 w \phi \nabla_\alpha \phi - \frac14\nabla_\alpha w \phi^2 \ \ \ \ \ (7)$

for the momentum tensor ${\mathcal{T}_{\alpha\beta}}$ as in (6). Then, for any vector field ${Y}$ , there hold

$\displaystyle \begin{aligned}\nabla^\alpha P_\alpha [X,w] &=\frac12\Big( \mathcal{T}_{\alpha\beta} {\pi^{(X)}}^{\alpha\beta} + w g(d\phi,d\phi) \Big) + X_w(\phi) \Box \phi - \frac14\Box w \phi^2, \\ P_\alpha Y^\alpha&= X(\phi)Y(\phi) - \frac12 g(X,Y) g(d\phi,d\phi)+ \frac12 w \phi Y(\phi) - \frac14Y(w)\phi^2 ,\end{aligned}\ \ \ \ \ (8)$

in which ${X_w: = X + \frac12 w}$ and ${\pi^{(X)} = \mathcal{L}_X g}$ denotes the deformation tensor of ${X}$ .

Proof: We first compute

$\displaystyle \begin{aligned} \nabla^\alpha ( \mathcal{T}_{\alpha\beta}[\phi]X^\beta) &= \nabla^\alpha (\mathcal{T}_{\alpha\beta}[\phi] )X^\beta + \mathcal{T}_{\alpha\beta} \nabla^\alpha X^\beta \\& = \Box \phi X(\phi) + \frac12 \mathcal{T}_{\alpha\beta} (\nabla^\alpha X^\beta + \nabla^\beta X^\alpha ) \\& = \Box \phi X(\phi) + \frac12 \mathcal{T}_{\alpha\beta} {\pi^{(X)}}^{\alpha\beta} . \end{aligned}$

As for the remaining terms, we simply use the product rule of ${\nabla_\alpha}$ and recall that ${g(d\phi,d\phi) =\nabla^\beta \phi \nabla_\beta \phi }$ . The lemma follows. $\Box$

Remark 1 For the obvious reason of multiplying the wave equation by ${X_w}$ in (8), the vector field ${X_w}$ is often referred to as a multiplier.

Remark 2 By definition, we note that

$\displaystyle \mathcal{T}_{\alpha\beta} {\pi^{(X)}}^{\alpha\beta} + w \nabla^\beta \phi \nabla_\beta \phi = {\pi^{(X)}}(d\phi,d\phi) + \Big( w - \frac12{\pi^{(X)}}^\alpha_\alpha \Big)g(d\phi,d\phi) .\ \ \ \ \ (9)$

Therefore, the role of ${w}$ in (7) is to treat the trace of the deformation tensor ${{\pi^{(X)}}}$ , up to lower order terms as in (8).

Remark 3 For Killing vector fields ${X}$ , ${{\pi^{(X)}} =0}$ . As a consequence of Lemmas 2, the energy current ${J_\alpha^X [\phi]= \mathcal{T}_{\alpha\beta}[\phi]X^\beta}$ is divergence free for solutions ${\phi}$ to the wave equation ${\Box \phi =0}$ . Thus, for any two spacelike hypersurfaces ${\Sigma_0, \Sigma_1}$ with common boundary, the divergence theorem (17) yields

$\displaystyle \int_{\Sigma_1} J^X_\alpha[\phi] n_{\Sigma_1}^\alpha = \int_{\Sigma_0} J^X_\alpha[\phi] n^\alpha_{\Sigma_0} \ \ \ \ \ (10)$

which gives the control on the energy flux on ${\Sigma_1}$ . This serves as a basic energy estimate to the wave equation.

For instance, on the flat Minkowski spacetime, ${T = \partial_t}$ is Killing, since ${\mathcal{L}_T g = 0}$ . The hypersurface ${\Sigma(t) = \{t\} \times \mathbb{R}^3}$ has ${T}$ as the future directed unit normal, since ${g(T,T) = -1}$ . In addition, from the definition, we compute

$\displaystyle \begin{aligned} J^T [\phi] \cdot T &= |T\phi|^2 - \frac12 g(T,T) g(d\phi,d\phi) = |\partial_t \phi|^2 + \frac12 (-|\partial_t\phi|^2 + |\nabla\phi|^2) = \frac12 |\partial\phi|^2 . \end{aligned}\ \ \ \ \ (11)$

That is, the divergence theorem (10) yields a control in ${L^\infty_tL^2_x}$ norm on ${\partial\phi}$ ; see Section 2 below. To control higher order derivatives, one uses commutators.

Lemma 3 For any vector field ${X = X^\alpha \partial_\alpha}$ , there holds

$\displaystyle [\Box , X] \phi = - {\pi^{(Y)}}^{\alpha \beta} \nabla_\alpha \nabla_\beta \phi - (2 \nabla^\beta {\pi^{(X)}}_{\alpha\beta} - \nabla_\alpha ({\pi^{(X)}}^\beta_\beta) )\nabla^\alpha \phi .$

Proof: Recalling the property of Lie derivatives ${\mathcal{L}_X}$ , we compute

$\displaystyle \begin{aligned} X(\Box \phi) &= X(g^{\alpha\beta} \nabla_\alpha\nabla_\beta \phi) = \mathcal{L}_X(g^{\alpha\beta} \nabla_\alpha\nabla_\beta \phi) \\ & = {\pi^{(Y)}}^{\alpha\beta} \nabla_\alpha \nabla_\beta \phi + g^{\alpha\beta}\mathcal{L}_X (\nabla_\alpha \nabla_\beta \phi) \end{aligned}$

in which, recalling ${\nabla_\alpha\nabla_\beta = \partial_\alpha \partial_\beta + \Gamma^\gamma_{\alpha\beta}\partial_\gamma}$ , we compute

$\displaystyle \mathcal{L}_X (\nabla_\alpha \nabla_\beta \phi) = \nabla_\alpha \mathcal{L}_X \nabla_\beta \phi + \Big( \nabla_\alpha (\mathcal{L}_X g)_{\alpha\gamma} + \nabla_\beta (\mathcal{L}_X g)_{\alpha \gamma} - \nabla_\gamma (\mathcal{L}_X g)_{\alpha\beta} \Big) \nabla^\gamma \phi.$

with ${\mathcal{L}_X \nabla_\beta \phi = X(\nabla_\beta \phi) + \nabla_\beta X^\gamma \partial_\gamma \phi = \nabla_\beta (X\phi)}$ . $\Box$

In particular, for any vector fields ${Z}$ with ${[\Box,Z]=0}$ , or more generally, for any conformal Killing vector fields: ${[\Box,Z] = \lambda \Box}$ , we can apply the divergence theorem (10) again for ${Z\phi}$ , yielding ${L^2}$ estimates on higher derivatives of ${\phi}$ . For instance, again in Minkowski spacetime, examples for such a ${Z}$ include translation vector fields ${Z = \partial_\alpha}$ and rotations ${Z = \Omega_{\alpha\beta} = x_\alpha\partial_\beta - x_\beta \partial_\alpha}$ , noting ${x_0 = -x^0}$ and ${x_i = x^i}$ . One can also use the scaling vector field ${S = x^\alpha\partial_\alpha}$ in the energy estimates, as it is conformally Killing: ${[\Box, S] = 2\Box}$ . Note that these commutators ${\Omega_{\alpha\beta}}$ and ${S}$ are weighted in space and time. This yields weighted ${L^2}$ estimates on ${\phi}$ and higher derivatives of ${\phi}$ , which in turn give pointwise decay estimates on ${\phi}$ or ${\psi = \partial\phi}$ through Klainerman-Sobolev’s inequalities:

$\displaystyle | \psi(t,x)| \lesssim (1+t+|x|)^{-1}(1+|t-|x||)^{-1/2} \sum_{|\alpha|\le 3} \| \Gamma^\alpha \psi\|_{L^2(\mathbb{R}^3)}, \ \ \ \ \ (12)$

for ${x\in \mathbb{R}^3}$ , where ${\Gamma^\alpha}$ denote ${\partial_\alpha,\Omega_{\alpha\beta},}$ and ${S}$ . This is the celebrated vector field method introduced and developed by Klainerman in the 80s (references cited above). We note that these vector fields are global in spacetime. In the next section, we perform the Dafermos-Rodnianski’s ${r^p}$ -weighted approach to derive decay for solutions to the wave equations.

2. Waves on Minkowski spacetime ${\mathbb{R}^{1+3}}$

2.1. Coordinates

On the flat Minkowski spacetime ${\mathbb{R}^{1+3}}$ , the Minkowski metric reads

$\displaystyle g = -dt^2 + (dx^i)^2\ \ \ \ \ (13)$

and so the wave operator is simply ${\Box = -\partial_t^2 + \partial_i^2}$ . In polar coordinates, the metric reads

$\displaystyle g = - dt^2 + dr^2 + g\hspace{-.07in} / \hspace{0in}_{ab} d\theta^ad\theta^b\ \ \ \ \ (14)$

where ${g\hspace{-.07in} / \hspace{0in}_{ab}}$ denotes the induced metric on ${\mathbb{S}^2}$ , whose nonzero components are ${g\hspace{-.07in} / \hspace{0in}_{11} = r^2}$ and ${g\hspace{-.07in} / \hspace{0in}_{22} = r^2\sin^2 \theta}$ . That is, ${g\hspace{-.07in} / \hspace{0in}_{ab} d\theta^ad\theta^b =r^2 d\mathbb{S}^2= r^2 (d\theta^2 + \sin^2 \theta d\varphi^2)}$ , with ${\theta^a = \theta,\varphi}$ . Hence,

$\displaystyle \Box = -\partial_t^2 + \partial_r^2 + \Delta\hspace{-.1in} / \hspace{.04in} \phi + 2r^{-1} \partial_r$

with the induced Laplacian ${\Delta\hspace{-.1in} / \hspace{.04in} = r^{-2}\Big( \frac{1}{\sin\theta} \partial_\theta(\sin\theta \partial_\theta ) + \frac{1}{\sin^2\theta} \partial_\varphi^2\Big)}$ . Finally, we introduce null coordinates

$\displaystyle u = \frac12(t-r), \qquad v = \frac12 (t+r)$

and so ${t = v+u}$ and ${r = v-u}$ . In these coordinates,

$\displaystyle g = -4dudv + g\hspace{-.07in} / \hspace{0in}_{ab} d\theta^ad\theta^b.\ \ \ \ \ (15)$

Note that ${g_{uv} = -2}$ and ${g^{uv} = -\frac12}$ , while ${g_{uu} = g_{vv} =0}$ . We also denote

$\displaystyle e_a = \partial_{\theta^a},\qquad e_3 = \underbar{L}= \partial_u = \partial_t - \partial_r, \qquad e_4 = L = \partial_v = \partial_t + \partial_r,$

for ${a=1,2}$ .

2.2. Divergence theorem

Consider the traditional spacetime slab region

$\displaystyle \mathcal{D}(t_0,t_1) = [t_0,t_1] \times \mathbb{R}^3\ \ \ \ \ (16)$

whose boundary consists of ${ \Sigma(t_0) = \{ t = t_0\} \times \mathbb{R}^3}$ and ${\Sigma(t_1) = \{ t = t_1\} \times \mathbb{R}^3, }$ having ${n_\Sigma = \partial_t}$ as their future directed unit normal. The divergence theorem then reads

$\displaystyle \int_{\Sigma(t_1)} P\cdot \partial_t + \int_{\mathcal{D}(t_0,t_1)} \nabla^\alpha P_\alpha = \int_{\Sigma(t_0)} P\cdot \partial_t,\ \ \ \ \ (17)$

recalling ${P}$ is a one-form and so ${P\cdot \partial_t}$ is a scalar function. The volume form on ${\mathcal{D}(t_0,t_1)}$ is ${dt dx}$ , while the induced volume form on ${\Sigma(t)}$ is ${dx}$ .

On the other hand, consider a less traditional spacetime region

$\displaystyle \mathcal{D}(t_0,t_1) = D_L (t_0,t_1) \cup \mathcal{D}_R (t_0,t_1)\ \ \ \ \ (18)$

which is bounded by hypersurfaces ${\Sigma(t) = \Sigma_L(t) \cup \Sigma_R(t)}$ , separating by the timelike hypersurface ${|x| = R}$ . The left piece is defined by ${\Sigma_L(\tau) = \{ t = \tau\} \times \{ |x|\le R\}}$ and the right piece is defined by

$\displaystyle \Sigma_R(t) =\Big\{ u = u_\tau, ~v \ge v_\tau, ~ \omega \in \mathbb{S}^2\Big\}$

where ${(u,v)}$ denotes the null coordinates: ${u = \frac12 (t-r), v = \frac12 (t+r)}$ , and so ${u_\tau = \frac12 (\tau - R), v_\tau = \frac12 (\tau+R)}$ . The hypersurface ${\Sigma_R(t)}$ is null, since the geodesic generator ${L = \partial_t + \partial_r}$ is null: ${g(L,L) =0}$ . We also set ${\mathcal{I}^+(t_0,t_1)}$ the future null infinity defined by ${\{ u\in [u_{t_0}, u_{t_1}], v = +\infty, \omega \in \mathbb{S}^2\} }$ . The divergence theorem on ${\mathcal{D}(t_0,t_1)}$ then reads

$\displaystyle \int_{\Sigma(t_1)} P\cdot n_{\Sigma(t_1)} + \int_{\mathcal{I}^+(t_0,t_1)} P\cdot n_{\mathcal{I}^+} + \int_{\mathcal{D}(t_0,t_1)} \nabla^\alpha P_\alpha = \int_{\Sigma(t_0)} P\cdot n_{\Sigma(t_2)},\ \ \ \ \ (19)$

where ${n_\Sigma = \partial_t}$ on ${\Sigma = \Sigma_L}$ and ${n_\Sigma = L = \partial_t + \partial_r}$ on ${\Sigma = \Sigma_R}$ , recalling that ${L}$ is the generator along the null hypersurface ${\Sigma_R}$ . Similarly, ${\underline{L} = \partial_t - \partial_r}$ is the generator along the future null infinity ${\mathcal{I}^+}$ , and so we take ${n_{\mathcal{I}^+} = \underline{L}}$ . The volume form on ${\mathcal{D}_R}$ is ${r^2 du dv d\mathbb{S}^2}$ , while the induced volume form on ${\Sigma_R}$ is ${r^2dvd\mathbb{S}^2}$ . The integral over the future null infinity ${\mathcal{I}^+}$ is understood as the limit of integrals over the null hypersurfaces ${v = V}$ , taking ${V \rightarrow \infty}$ , with the induced volume form ${r^2 du d\mathbb{S}^2}$ .

Lemma 4 For vector fields ${X = aL + b \underline L}$ and ${n = cL + d\underline L}$ , the energy current ${J^X_\alpha = \mathcal{T}_{\alpha\beta} X^\beta}$ satisfies

$\displaystyle J^X \cdot n = a c |L\phi|^2 + bd |\underline L\phi|^2 + (ad+bc)|\nabla\hspace{-.118in} / \hspace{.04in} \phi|^2.$

In particular, ${J^X \cdot n_{\Sigma} \gtrsim |\partial\phi|^2}$ for future-directed timelike vector field ${X}$ and spacetime hypersurfaces ${\Sigma}$ .

Proof: By definition, we compute

$\displaystyle \begin{aligned} J^X_\alpha n^\alpha &= X(\phi)n(\phi) - \frac12 g(X,n) g(d\phi,d\phi) \\ &= a c |L\phi|^2 + bd |\underline L\phi|^2 + (ad + bc) L\phi \underline L \phi + (ad + bc) g(d\phi,d\phi). \end{aligned}$

Using ${g(d\phi,d\phi) = - L\phi \underline L\phi + |\nabla\hspace{-.118in} / \hspace{.04in} \phi|^2}$ , we obtain the lemma. $\Box$

2.3. Basic energy identity with ${X = T}$

Proposition 5 For all ${t_0\le t_1}$ , there holds

$\displaystyle \mathcal{E}[\phi](t_1) + \underline{\mathcal{E}}^\infty[\phi](t_0,t_1) = \mathcal{E}[\phi](t_0) - 2\int_{\mathcal{D}(t_0,t_1)} T\phi \Box \phi \ \ \ \ \ (20)$

where the energy fluxes are defined by

$\displaystyle \begin{aligned} \mathcal{E}[\phi](t): &= \int_{\Sigma_L (t)}|\partial\phi|^2 + \int_{\Sigma_R(t)} (|L\phi|^2 + |\nabla\hspace{-.118in} / \hspace{.04in} \phi|^2) , \\ \underline{\mathcal{E}}^\infty[\phi](t_0,t_1): &= \int_{\mathcal{I}^+(t_0,t_1)} (|\underline{L}\phi|^2 + |\nabla\hspace{-.118in} / \hspace{.04in} \phi|^2) . \end{aligned} \ \ \ \ \ (21)$

Remark 4 Note that no control on the derivatives of ${\phi}$ in direction ${\underline{L}}$ on the null hypersurface ${\Sigma_R(t)}$ and in direction ${L}$ on the future null infinity ${\mathcal{I}^+}$ .

Proof: Since ${T}$ is Killing, Lemma 2 applying to the one-form ${P_\alpha [T] = \mathcal{T}_{\alpha\beta}[\phi]T^\beta}$ gives ${\nabla^\alpha P_\alpha = T\phi \Box \phi }$ . Thus, in view of the divergence theorem (19), it suffices to calculate ${P\cdot n}$ on each the boundary. Using Lemma 4 and recalling ${T = \frac12 (L + \underline L)}$ , we obtain

$\displaystyle P\cdot T = \frac12 |\partial\phi|^2, \qquad P\cdot L = \frac12 (|L\phi|^2 + |\nabla\hspace{-.118in} / \hspace{.04in} \phi|^2), \qquad P\cdot \underline{L} = \frac12 (| \underline{L}\phi|^2 + |\nabla\hspace{-.118in} / \hspace{.04in} \phi|^2).$

The proposition follows. $\Box$

2.4. Energy estimates with ${X = f(r)\partial_r}$

In this section, we shall derive energy estimates using multiplier ${X = f(r)\partial_r}$ and ${w = w(r)}$ for some chosen functions ${f(r)}$ and ${w(r)}$ . The estimates rely on the divergence identity, obtained from Lemma 2. We first compute the deformation tensor of ${X = f(r)\partial_r}$ .

Lemma 6 The deformation tensor of ${X = f(r)\partial_r}$ satisfies

$\displaystyle \begin{aligned} {\pi^{(X)}}^\alpha_\alpha &=4r^{-1} f(r) + 2f'(r) \\ {\pi^{(X)}}(d\phi,d\phi) &= 2f' (r)|\partial_r\phi|^2 + 2 r^{-1} f(r)|\nabla\hspace{-.118in} / \hspace{.04in} \phi|^2 . \end{aligned}\ \ \ \ \ (22)$

Proof: Writing ${X = fY}$ with ${Y = \partial_r}$ , we note that

$\displaystyle {\pi^{(X)}}_{\alpha\beta} = f{\pi^{(Y)}}_{\alpha\beta} + \nabla_\alpha f Y_\beta + \nabla_\beta f Y_\alpha , \qquad {\pi^{(X)}}^\alpha_\alpha = f{\pi^{(Y)}}^\alpha_\alpha + 2Y(f).\ \ \ \ \ (23)$

Recall that ${f = f(r)}$ . It thus suffices to calculate the deformation tensor of ${Y = \partial_r}$ whose only nonzero component is ${Y^r = Y_r = 1}$ . We work with the polar coordinates and write the Minkowski metric as

$\displaystyle g = - dt^2 + dr^2 + g\hspace{-.07in} / \hspace{0in}_{ab} d\theta^ad\theta^b$

where ${g\hspace{-.07in} / \hspace{0in}_{ab}}$ denotes the induced metric on ${\mathbb{S}^2}$ , with nonzero components ${g\hspace{-.07in} / \hspace{0in}_{11} = r^2}$ and ${g\hspace{-.07in} / \hspace{0in}_{22} = r^2\sin^2 \theta}$ . By definition, we compute

$\displaystyle {\pi^{(Y)}}_{\alpha\beta} = Y^\mu \partial_\mu g_{\alpha\beta} - g([Y, \partial_\alpha],\partial_\beta) - g(\partial_\alpha, [Y,\partial_\beta]) = \partial_r g_{\alpha\beta}.\ \ \ \ \ (24)$

This yields ${{\pi^{(Y)}}_{ij} = 0}$ and ${{\pi^{(Y)}}_{ab} = \partial_r g\hspace{-.07in} / \hspace{0in}_{ab} = \frac2r g\hspace{-.07in} / \hspace{0in}_{ab}}$ . Finally, recalling ${{\pi^{(X)}}^{\mu\nu} = g^{\mu\alpha} {\pi^{(X)}}_{\alpha\beta} g^{\beta \nu}}$ , we have ${{\pi^{(X)}}^{rr} = {\pi^{(X)}}_{rr} = 2 f'(r)}$ and ${{\pi^{(X)}}^{ab} = {\pi^{(X)}}_{ab}=\frac2r f(r)g\hspace{-.07in} / \hspace{0in}^{ab}}$ . The lemma follows. $\Box$

Proposition 7 For ${X = f(r) \partial_r}$ and ${w = 2 r^{-1}f(r)}$ , let ${P_\alpha[X,w]}$ be defined as in (7). Then, there holds

$\displaystyle Div {\bf P}= \frac12 f' (|\partial_t \phi|^2 + |\partial_r\phi|^2) +\frac12 (2r^{-1}f - f') |\nabla\hspace{-.118in} / \hspace{.04in}\phi|^2 - \frac1{2} r^{-1}f'' \phi^2 + X_w(\phi) \Box\phi. \ \ \ \ \ (25)$

Proof: First note that ${\Box w = 2r^{-1} \partial_r^2(r r^{-1} f) = 2 r^{-1}f''}$ . In addition, using Lemma 6, we have

$\displaystyle \begin{aligned} {\pi^{(X)}}(d\phi,d\phi) &= 2f' (r)|\partial_r\phi|^2 + 2 r^{-1} f(r)|\nabla\hspace{-.118in} / \hspace{.04in} \phi|^2 \\ \Big( w - \frac12{\pi^{(X)}}^\alpha_\alpha \Big)g(d\phi,d\phi) & = \Big(w - f' - 2 r^{-1}f) \Big)g(d\phi,d\phi) = -f'(r) g(d\phi ,d\phi) \end{aligned}$

in which we used ${w = 2r^{-1}f}$ . Recalling ${g(d\phi,d\phi) = - |\partial_t\phi|^2 + |\partial_r\phi|^2 + |\nabla\hspace{-.118in} / \hspace{.04in} \phi|^2}$ and using (8) and (9), we obtain the proposition. $\Box$

Proposition 8 For ${X = f(r)\partial_r}$ and ${w = 2r^{-1}f(r)}$ , let ${P_\alpha[X,w]}$ be defined as in (7). Then, there hold

$\displaystyle \begin{aligned} {\bf P} \cdot T &= f T\phi (\partial_r + r^{-1})\phi \\ {\bf P} \cdot L & = \frac12 f (| \widehat L\phi|^2 - |\nabla\hspace{-.118in} / \hspace{.04in} \phi|^2)- \frac12 r^{-1} f' \phi^2 \\ {\bf P} \cdot \underline{L} & = \frac12 f ( - | \widehat{\underline{L}}\phi|^2 + |\nabla\hspace{-.118in} / \hspace{.04in} \phi|^2) + \frac12 r^{-1} f' \phi^2 \end{aligned} \ \ \ \ \ (26)$

where ${\widehat L = L + r^{-1}}$ and ${\widehat{\underline L} =\underline{L} - r^{-1}}$ .

Proof: Recalling Lemma 4, we have

$\displaystyle J^X \cdot Y = a c |L\phi|^2 + bd |\underline L\phi|^2 + (ad+bc)|\nabla\hspace{-.118in} / \hspace{.04in} \phi|^2,$

for ${X = aL + b \underline L}$ and ${Y = cL + d \underline L}$ . Thus, writing ${\partial_r = \frac12 (L - \underline L)}$ , we compute

$\displaystyle \begin{aligned}{\bf P} \cdot L &= \frac12 f(|L\phi|^2 - |\nabla\hspace{-.118in} / \hspace{.04in} \phi|^2) + \frac12 w \phi L(\phi) - \frac14L(w)\phi^2 \\ &= \frac12 f (| \widehat L\phi|^2 - |\nabla\hspace{-.118in} / \hspace{.04in} \phi|^2)- \frac12 r^{-1} f' \phi^2 \end{aligned}$

in which we used ${\widehat L = L + r^{-1}}$ and ${Lw = 2\partial_r(r^{-1}f) = 2(r^{-1}f' - r^{-2}f)}$ . Other computations are similar, upon noting that ${T = \frac12 (L + \underline L)}$ . $\Box$

Theorem 9 (Morawetz estimates I) For ${\mathcal{D}(t_0,t_1) = [t_0,t_1]\times \mathbb{R}^3}$ , there holds

$\displaystyle \int_{\mathcal{D}(t_0,t_1)}\Big[ \langle r\rangle^{-1-\delta} (|\partial\phi|^2 + r^{-2}|\phi|^2) + r^{-1}|\nabla\hspace{-.118in} / \hspace{.04in}\phi|^2 \Big] \lesssim\int_{\Sigma(t_0)}|\partial\phi|^2 + \int_{\mathcal{D}(t_0,t_1) } \langle r\rangle^{1+\delta} |\Box\phi|^2.$

Proof: Choosing ${f =1 - \langle r\rangle^{-\delta}}$ with ${\delta >0}$ and integrating the divergence equation over ${\mathcal{D}(t_0,t_1)}$ defined as in (18), we obtain at once from the previous two propositions

$\displaystyle \begin{aligned} &\int_{\mathcal{D}(t_0,t_1)}\Big[ \langle r\rangle^{-1-\delta} (|\partial\phi|^2 + r^{-2}|\phi|^2) + r^{-1}|\nabla\hspace{-.118in} / \hspace{.04in}\phi|^2 \Big] \\&\lesssim \int_{\mathcal{D}(t_0,t_1)} (\partial_r \phi + r^{-1}\phi) \Box\phi - \int_{\Sigma(t_1)} P\cdot T + \int_{\Sigma(t_0)} P\cdot T. \end{aligned}$

Since ${f}$ is bounded, we have

$\displaystyle \int_{\Sigma(t)} P\cdot T \lesssim \int_{\Sigma(t)} \partial_t \phi (\partial_r \phi+ r^{-1}\phi) \lesssim \mathcal{E}[\phi](t) + \int_{\Sigma(t)} r^{-2}|\phi|^2$

which is bounded by ${\mathcal{E}[\phi](t)}$ , upon using the Hardy type inequality ${\int_{\Sigma(t)} \langle r\rangle^{-2} |\phi|^2 \lesssim \mathcal{E}[\phi](t)}$ . The theorem follows. $\Box$

Theorem 10 (Morawetz estimates II) For ${\mathcal{D}(t_0,t_1) = \mathcal{D}_L(t_0,t_1)\cup \mathcal{D}_R(t_0,t_1)}$ defined as in (18), there holds

$\displaystyle \int_{\mathcal{D}(t_0,t_1)}\Big[ \langle r\rangle^{-1-\delta} (|\partial\phi|^2 + r^{-2}|\phi|^2) + r^{-1}|\nabla\hspace{-.118in} / \hspace{.04in}\phi|^2 \Big] \lesssim \mathcal{E}[\phi](t_0) + \int_{\mathcal{D}(t_0,t_1) } \langle r\rangle^{1+\delta} |\Box\phi|^2.$

Proof: Take ${f =1 - \langle r\rangle^{-\delta}}$ as in the previous theorem. It thus suffices to bound the boundary terms on ${\Sigma_R(t)}$ and on the null infinity ${\mathcal{I}^+(t_0,t_1)}$ . Since ${f\le 1}$ , we have

$\displaystyle \begin{aligned} \int_{\Sigma_R(t)}{\bf P} \cdot L & \lesssim \int_{\Sigma_R(t)} (| \widehat L\phi|^2 + |\nabla\hspace{-.118in} / \hspace{.04in} \phi|^2 + \langle r\rangle^{-2-\delta} \phi^2) \\ & \lesssim \int_{\Sigma_R(t)} (| L\phi|^2 + |\nabla\hspace{-.118in} / \hspace{.04in} \phi|^2 + \langle r\rangle^{-2} \phi^2) \end{aligned}$

which is bounded by the energy flux ${\mathcal{E}[\phi](t)}$ . Similarly, we compute

$\displaystyle - \int_{\mathcal{I}^+(t_0,t_1)} P\cdot \underline{L} \lesssim \int_{\mathcal{I}^+(t_0,t_1)} (|{\underline L} \phi |^2 + r^{-2} |\phi|^2) \lesssim \underline{\mathcal{E}}^\infty[\phi](t_0,t_1) + \int_{\mathcal{I}^+(t_0,t_1)} r^{-2}|\phi|^2.$

We show that the last term in the above in fact vanishes. Indeed, by definition, for any ${V}$ , we write

$\displaystyle \begin{aligned}\int_{\underline{N}_V(t_0,t_1)} r^{-2}|\phi|^2 &= \int_{u_{t_0}}^{u_{t_1}} \phi^2(u,V,\omega) \; du d\mathbb{S}^2 \le \int_{u_{t_0}}^{u_{t_1}} \Big(\int_V^\infty L\phi (u,v,\omega) \; dv\Big)^2 \; du d\mathbb{S}^2 \\ &\le \int_{u_{t_0}}^{u_{t_1}} \int_V^\infty |L\phi|^2 \;r^2 dvdu d\mathbb{S}^2 \Big( \int_V^\infty r^{-2} \; dv \Big) \\ &\lesssim V^{-1} \int_{t_0}^{t_1} \mathcal{E}[\phi](t)\; dt \end{aligned}$

which vanishes in the limit of ${V \rightarrow \infty}$ . The theorem follows. $\Box$

2.5. Energy estimates with ${X = f(r)\partial_v}$

We now derive energy estimates with multiplier ${X = f(r)\partial_v}$ . First, we have the following.

Lemma 11 The deformation tensor of ${X = f(r)\partial_v}$ satisfies

$\displaystyle \begin{aligned}{\pi^{(X)}}^\alpha_\alpha &=4r^{-1} f(r) + 2f'(r)\\ {\pi^{(X)}}(d\phi,d\phi) &= f'(r) | \partial_v \phi|^2 -f' (r) \partial_u\phi \partial_v\phi + 2 r^{-1} f|\nabla\hspace{-.118in} / \hspace{.04in} \phi|^2. \end{aligned}\ \ \ \ \ (27)$

Proof: Again we first compute it for ${Y = \partial_v}$ , whose nonzero components are ${Y^v=1}$ and ${Y_u = -2}$ . We work with the null coordinates ${u,v}$ , recalling the metric ${ g = - 4 du dv + g\hspace{-.07in} / \hspace{0in}_{ab} d\theta^ad\theta^b}$ . By definition, we compute

Recalling ${\partial_v = \partial_t + \partial_r, }$ the above yields ${{\pi^{(Y)}}_{ij} = 0}$ and ${{\pi^{(Y)}}_{ab} = \partial_r g\hspace{-.07in} / \hspace{0in}_{ab} = \frac2r g\hspace{-.07in} / \hspace{0in}_{ab}}$ . This yields

$\displaystyle {\pi^{(X)}}_{33} = 4f'(r), \qquad {\pi^{(X)}}_{34} = -2f'(r), \qquad {\pi^{(X)}}_{ab} = 2r^{-1} f(r)g\hspace{-.07in} / \hspace{0in}_{ab},$

upon using (23). Finally, using again ${{\pi^{(X)}}^{\mu\nu} = g^{\mu\alpha} {\pi^{(X)}}_{\alpha\beta} g^{\beta \nu}}$ , we have ${{\pi^{(X)}}^{34} = -f'/2}$ and ${{\pi^{(X)}}^{44} = f'}$ . Hence, we compute

$\displaystyle \begin{aligned} \nabla_\mu \phi \nabla_\nu \phi {\pi^{(X)}}^{\mu\nu} & = {\pi^{(X)}}^{44} | \nabla_4 \phi|^2 + 2 {\pi^{(X)}}^{34} \nabla_3 \phi \nabla_4\phi + \nabla_a \phi \nabla_b \phi {\pi^{(X)}}^{ab} \\ & = f'(r) | \partial_v \phi|^2 -f' (r) \partial_u\phi \partial_v\phi + 2 r^{-1} f|\nabla\hspace{-.118in} / \hspace{.04in} \phi|^2. \end{aligned}$

which ends the proof of the lemma. $\Box$

Proposition 12 For ${X = f(r) \partial_v}$ and ${w = 2 r^{-1}f(r)}$ , the one-form ${P_\mu[X,w]}$ defined as in (7) satisfies

$\displaystyle Div{\bf P}= \frac12 f' |\partial_v \phi|^2 + \frac12 (2r^{-1}f - f') |\nabla\hspace{-.118in} / \hspace{.04in}\phi|^2 - \frac1{2} r^{-1}f'' \phi^2 + X_w(\phi) \Box\phi. \ \ \ \ \ (29)$

Proof: Using Lemma 11, we have

$\displaystyle \begin{aligned} {\pi^{(X)}}(d\phi,d\phi) &= f'(r) | \partial_v \phi|^2 -f' (r) \partial_u\phi \partial_v\phi + 2 r^{-1} f|\nabla\hspace{-.118in} / \hspace{.04in} \phi|^2 \\ \Big( w - \frac12{\pi^{(X)}}^\alpha_\alpha \Big)g(d\phi,d\phi) & = -f'(r) g(d\phi ,d\phi) = f' \Big( |\partial_t\phi|^2 - |\partial_r\phi|^2 - |\nabla\hspace{-.118in} / \hspace{.04in} \phi|^2 \Big), \end{aligned}$

in which we note that ${ |\partial_t\phi|^2 - |\partial_r\phi|^2 = \partial_u \phi \partial_v \phi}$ . Adding up these identities, we obtain the proposition. $\Box$

Observe that the third term in the divergence identity from Proposition 12 has a bad sign when ${f'' >0}$ (for instance for ${f = r^p}$ with ${p> 1}$ ). To treat this term, we introduce

$\displaystyle {\bf P}^M = {\bf P} + \frac14 {\bf M} \phi^2$

for some one-form ${{\bf M}}$ to be determined. We then have

$\displaystyle \begin{aligned}Div {\bf P}^M &= Div {\bf P} + \frac14 Div {\bf M} \phi^2 + \frac12 \phi M_\alpha \nabla^\alpha \phi \\ &= \frac12 f' |L\phi|^2 + \frac12 (2r^{-1}f - f') |\nabla\hspace{-.118in} / \hspace{.04in}\phi|^2 + f(\partial_v \phi + r^{-1}\phi) \Box\phi \\&\quad + \frac12 \phi M_\alpha \nabla^\alpha \phi+ \frac1{2} ( \frac12Div {\bf M} - r^{-1}f'') \phi^2 .\end{aligned}$

Thus, choosing ${{\bf M} = 2r^{-1} f'(r)L}$ and noting that ${M^r = 2r^{-1} f'(r)}$ , we have

$\displaystyle \begin{aligned} Div {\bf M} &= \nabla_\alpha M^\alpha = \partial_\alpha M^\alpha + \Gamma^\alpha_{\alpha \gamma} M^\gamma = \partial_r M^r + \frac12 g\hspace{-.07in} / \hspace{0in}^{ab} \partial_r g\hspace{-.07in} / \hspace{0in}_{ab} M^r \\ & = 2 \partial_r(r^{-1}f'(r)) + 4 r^{-2} f'(r) = 2 r^{-1}f''(r) + 2 r^{-2} f'(r) \end{aligned}$

and so ${( \frac12Div {\bf M} - r^{-1}f'') \phi^2 = r^{-2}f'(r) \phi^2}$ , taking care of the bad term ${f''(r)\phi^2}$ . Finally, we compute

$\displaystyle \phi M_\alpha \nabla^\alpha \phi = \phi M^\alpha \nabla_\alpha \phi = 2r^{-1} f'(r) \phi L\phi .$

Collecting terms, we obtain the following

Proposition 13 For ${X = f(r) \partial_v}$ , ${w = 2 r^{-1}f(r)}$ , and ${{\bf M} = 2r^{-1}f'(r)L}$ , define the one-form

$\displaystyle P^M_\alpha [X,w] = \mathcal{T}_{\alpha\beta}[\phi]X^\beta + \frac12 w \phi \nabla_\alpha \phi - \frac14\nabla_\alpha w \phi^2 + \frac14 M_\alpha \phi^2. \ \ \ \ \ (30)$

Then, there holds

$\displaystyle \begin{aligned}Div {\bf P}^M &= \frac12 f' |\widehat L\phi|^2 + \frac12 (2r^{-1}f - f') |\nabla\hspace{-.118in} / \hspace{.04in}\phi|^2 +f \widehat L \phi \Box\phi \end{aligned} \ \ \ \ \ (31)$

for ${\widehat L = L + r^{-1}}$ .

Proposition 14 For ${X = f(r)L}$ , ${w = 2r^{-1}f(r)}$ , and ${{\bf M} = 2r^{-1}f'(r)L}$ , let ${{\bf P}^M[X,w]}$ be defined as in (30). Then there hold

$\displaystyle \begin{aligned} {\bf P}^M \cdot T &=\frac12 f (|\widehat L\phi|^2 + |\nabla\hspace{-.118in} / \hspace{.04in}\phi|^2) - \frac12 r^{-2} \partial_r (rf\phi^2) \\ {\bf P}^M \cdot L & = f |\widehat L\phi|^2 - \frac12 r^{-2}L(r f \phi^2) \\ {\bf P}^M \cdot \underline{L} & = f |\nabla\hspace{-.118in} / \hspace{.04in}\phi|^2 +\frac12 r^{-2} \underline L( rf \phi^2) . \end{aligned} \ \ \ \ \ (32)$

Proof: Recalling Lemma 4, we have

$\displaystyle J^X \cdot Y = a c |L\phi|^2 + bd |\underline L\phi|^2 + (ad+bc)|\nabla\hspace{-.118in} / \hspace{.04in} \phi|^2,$

for ${X = aL + b \underline L}$ and ${Y = cL + d \underline L}$ . Thus, for ${X = f L}$ and ${w = 2 r^{-1} f}$ , we compute

$\displaystyle \begin{aligned}{\bf P}^M \cdot L &= f |L\phi|^2 + \frac12 w \phi L\phi - \frac14 Lw\phi^2 \\&= f |L\phi|^2 + r^{-1} f \phi L\phi + \frac12 r^{-2} f(r)\phi^2 - \frac12 r^{-1} f' \phi^2 \\&= f |\widehat L\phi|^2 - r^{-1} f \phi L\phi - \frac12 r^{-2} f(r)\phi^2 - \frac12 r^{-1} f' \phi^2 \\ &= f |\widehat L\phi|^2 - \frac12 r^{-2}L(r f \phi^2) .\end{aligned}$

Similarly, using ${g(L,\underline L) = -2}$ , we have

$\displaystyle \begin{aligned} {\bf P}^M \cdot \underline L &= f |\nabla\hspace{-.118in} / \hspace{.04in}\phi|^2 + \frac12 w \phi \underline L \phi+ \frac14\Big(2r^{-1}f'(r) g(L,\underline L) - \underline L w\Big)\phi^2 \\ &= f |\nabla\hspace{-.118in} / \hspace{.04in}\phi|^2 + r^{-1} f \phi \underline L \phi - \frac12\Big(r^{-1}f'(r) + r^{-2} f\Big)\phi^2 \\ &= f |\nabla\hspace{-.118in} / \hspace{.04in}\phi|^2 +\frac12 r^{-2} \underline L( rf \phi^2) . \end{aligned}$

Finally, writing ${T = \frac12 (L + \underline L)}$ , we obtain the proposition. $\Box$

Theorem 15 ( ${r^p}$ -weighted estimates) For ${p\in [0,2]}$ , there holds

$\displaystyle \begin{aligned} \int_{\Sigma(t_1)} &r^p |\widehat L\phi|^2 + \int_{\Sigma_L(t_1)} r^p |\nabla\hspace{-.118in} / \hspace{.04in}\phi|^2 + \int_{\mathcal{I}^+(t_0,t_1)} r^p |\nabla\hspace{-.118in} / \hspace{.04in}\phi|^2 + \int_{\mathcal{D}(t_0,t_1)} r^{p-1} ( p|\widehat L\phi|^2 + (2-p) |\nabla\hspace{-.118in} / \hspace{.04in}\phi|^2 ) \\& \lesssim \int_{\Sigma(t_0)} r^p |\widehat L\phi|^2 + \int_{\Sigma_L(t_0)} r^p |\nabla\hspace{-.118in} / \hspace{.04in}\phi|^2 + \int_{\mathcal{D}(t_0,t_1)} \frac1p r^{p+1} |\Box\phi|^2. \end{aligned}$

Proof: Again we use the divergence theorem (19) on ${\mathcal{D}(t_0,t_1) = \mathcal{D}_L(t_0,t_1) \cup \mathcal{D}_R(t_0,t_1)}$ as in (18), giving

$\displaystyle \int_{\Sigma(t_1)} {\bf P}^M\cdot n_{\Sigma(t_1)} + \int_{\mathcal{I}^+(t_0,t_1)} {\bf P}^M\cdot n_{\mathcal{I}^+} + \int_{\mathcal{D}(t_0,t_1)} Div {\bf P}^M= \int_{\Sigma(t_0)} {\bf P}\cdot n_{\Sigma(t_2)},$

for the one-form ${{\bf P}^M}$ defined as in (30), where ${n_{\Sigma_L} = T }$ , ${n_{\Sigma_R} = L }$ , and ${n_{\mathcal{I}^+} = \underline{L}}$ . We now use the previous propositions with

$\displaystyle f(r) = r^p\ \ \ \ \ (33)$

for ${p\in [0,2]}$ . The identity (31) yields

$\displaystyle \begin{aligned}Div {\bf P}^M & \gtrsim r^{p-1} ( p|\widehat L\phi|^2 + (2-p) |\nabla\hspace{-.118in} / \hspace{.04in}\phi|^2 ) + r^p\widehat L \phi \Box\phi \\& \gtrsim r^{p-1} ( p|\widehat L\phi|^2 + (2-p) |\nabla\hspace{-.118in} / \hspace{.04in}\phi|^2 ) - \frac1p r^{p+1}|\Box\phi|^2 . \end{aligned}$

We now check the boundary terms. Recalling the induced volume form ${r^2 dr d\mathbb{S}^2}$ on ${\Sigma_L(t)}$ , we compute

$\displaystyle \begin{aligned} \int_{\Sigma_L(t)} {\bf P}^M \cdot T &= \frac12 \int_{\Sigma_L(t)} r^p (|\widehat L\phi|^2 + |\nabla\hspace{-.118in} / \hspace{.04in}\phi|^2) - \frac12\int_{\Sigma_L(t)} r^{-2} \partial_r (r^{p+1}\phi^2) \\ &= \frac12 \int_{\Sigma_L(t)} r^p (|\widehat L\phi|^2 + |\nabla\hspace{-.118in} / \hspace{.04in}\phi|^2) - \frac12 R^{p+1}\int_{\mathbb{S}^2} \phi^2 d\mathbb{S}^2 . \end{aligned}$

On the other hand, on ${\Sigma_R(t)}$ , we compute

$\displaystyle \begin{aligned} \int_{\Sigma_R(t)} {\bf P}^M \cdot L &= \int_{\Sigma_R(t)} r^p |\widehat L\phi|^2 -\frac12 \int_{\Sigma_R(t)} r^{-2}L(r^{p+1} \phi^2) \\ &= \int_{\Sigma_R(t)} r^p |\widehat L\phi|^2 -\frac12 \int_{\frac12(t+R)}^\infty \int_{\mathbb{S}^2}\partial_v(r^{p+1} \phi^2) \; d\mathbb{S}^2 dv \\ &= \int_{\Sigma_R(t)} r^p |\widehat L\phi|^2 + \frac12 R^{p+1}\int_{\mathbb{S}^2} \phi^2 \; d\mathbb{S}^2 - \frac12 R^{p+1} \lim_{v\rightarrow \infty}\int_{\mathbb{S}^2} \phi^2(t) d\mathbb{S}^2. \end{aligned}$

Finally, we compute

$\displaystyle \begin{aligned}\int_{\mathcal{I}^+(t_0,t_1)} {\bf P}^M\cdot \underline L &= \int_{\mathcal{I}^+(t_0,t_1)} r^p |\nabla\hspace{-.118in} / \hspace{.04in}\phi|^2 +\frac12 \int_{u_{t_0}}^{u_{t_1}} \int_{\mathbb{S}^2} \partial_u ( r^{p+1} \phi^2) \; d\mathbb{S}^2 du \\ &= \int_{\mathcal{I}^+(t_0,t_1)} r^p |\nabla\hspace{-.118in} / \hspace{.04in}\phi|^2 + \frac12R^{p+1} \lim_{v\rightarrow \infty}\int_{\mathbb{S}^2} ( \phi^2 (t_1) - \phi^2(t_0)) \; d\mathbb{S}^2 . \end{aligned}$

The theorem follows from adding up these estimates. $\Box$

Theorem 16 (First decay estimates) Introduce the energy flux

$\displaystyle \begin{aligned}\mathcal{E}_0[\phi](t) &= \int_{\Sigma_L(t)} |\partial\phi|^2 + \int_{\Sigma_R(t)} (|\widehat L \phi|^2 + |\nabla\hspace{-.118in} / \hspace{.04in} \phi|^2) \\ \mathcal{E}_p[\phi](t) &= \int_{\Sigma_L(t)} r^p |\partial\phi|^2 + \int_{\Sigma_R(t)} r^p |\widehat L \phi|^2, \end{aligned}\ \ \ \ \ (34)$

for ${p>0}$ . Then, solutions ${\phi}$ to ${\Box \phi =0}$ , with compactly supported initial data, satisfy

$\displaystyle \mathcal{E}_0[\phi](t) \lesssim t^{-2} \mathcal{E}_2[\phi](t_0), \qquad \forall ~t\ge t_0.\ \ \ \ \ (35)$

Proof: Without loss of generality, we prove the theorem for ${t_0=0}$ . Applying Theorem 15 for ${p=1,2}$ , we get

$\displaystyle \begin{aligned} \mathcal{E}_p[\phi](t) + \int_{0}^{t} \mathcal{E}_{p-1}[\phi](s)\; ds \lesssim \mathcal{E}_p[\phi](0) \end{aligned}$

for all ${t\ge 0}$ . Recall also the basic energy inequality ${\mathcal{E}_0[\phi](t) \le \mathcal{E}_0[\phi](0)}$ . Thus, for any ${t > 0}$ , using the mean value theorem and the above energy inequality repeatedly, we estimate

$\displaystyle \mathcal{E}_0(t) \le \mathcal{E}_0(\tau) = 3 t^{-1} \int_{2t/3}^t \mathcal{E}_0(s)\; ds \lesssim t^{-1}\mathcal{E}_1(2t/3),$

for some ${\tau \in [2t/3,t]}$ . Similarly, for some ${\tau \in [t/3,2t/3]}$ , we have

$\displaystyle \mathcal{E}_1(2t/3) \lesssim \mathcal{E}_1(\tau) = 3 t^{-1}\int_{t/2}^{2t/3} \mathcal{E}_1(s)\; ds \lesssim t^{-1} \mathcal{E}_2(t/3) \lesssim t^{-1} \mathcal{E}_2(0).$

The theorem follows. $\Box$

Remark 5 The decay for the energy flux can be improved, up to ${t^{-5}}$ , by first commuting the wave equation with ${T}$ and ${X = f(r)\partial_v}$ , and deriving again the ${r^p}$ weighted estimates for the derivatives. Let’s skip these details; see, for instance, this paper by Y. Angelopoulos, S. Aretakis, and D. Gajic.

Remark 6 For more applications and developments on the ${r^p}$ -weighted approach, see the references mentioned in the beginning of the post.

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Dafermos and Rodnianski’s r^p-weighted approach to decay for wave equations

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