Boundedness of the conformal hyperboloidal energy for a wave-Klein–Gordon model

Abstract

We consider the global evolution problem for a model which couples together a nonlinear wave equation and a nonlinear Klein–Gordon equation, and was introduced by P.G. LeFloch and Y. Ma and, independently, by Q. Wang. By revisiting the Hyperboloidal Foliation Method, we establish that a weighted energy of the solutions remains (almost) bounded for all times. The new ingredient in the proof is a hierarchy of fractional Morawetz energy estimates (for the wave component of the system) which is defined from two conformal transformations. The optimal case for these energy estimates corresponds to using the scaling vector field as a multiplier for the wave component.

A Proof of Technical Results

Proof of Lemma 4.1

To establish the Eq. (4.4) we expand $X^{i} = {\overline{X}}^i+ \omega ^i \omega _j X^j$

$$\begin{aligned} \partial ^B_\gamma X^\gamma= & {} \partial ^B_u X^u + \partial ^B_i(\omega ^{i})\omega _{j} X^j + \omega ^{i}\partial ^B_i(\omega ^{i}X^j) + \partial ^B_i\overline{X}^i \\= & {} \partial ^B_u X^u + \frac{2X^{r}}{r}+ \partial ^B_r X^r+ \partial ^B_i\overline{X}^i. \end{aligned}$$

Using this together with the basic identity ${}^{(X)}{\widehat{\pi }}^{\alpha \beta } = - {\mathcal {L}}_{X} \eta ^{-1}- \eta ^{\alpha \beta }\partial _\gamma X^\gamma$, in (3.10) gives the result. To calculate all the components of ${\mathcal {A}}$ we use Eq. (4.4) together with the Eq. (3.10) to get

$$\begin{aligned} {\mathcal {A}}^{\alpha \beta }= & {} \frac{1}{2}\Big ( \partial _\gamma ^B(X^\alpha ) \eta ^{\gamma \beta } + \partial _\gamma ^B(X^\beta ) \eta ^{\alpha \gamma } -X( \eta ^{\alpha \beta }) \\{} & {} -({\partial _u^B} X^u + \partial _r^B X^r )\eta ^{\alpha \beta } \Big ) - \Big ( \frac{X^r}{r}-X \ln \Omega ) \Big )\eta ^{\alpha \beta }. \end{aligned}$$

We now compute all the components of this tensor. For ${\mathcal {A}}^{uu}$ we use the fact that $\eta ^{u\alpha } =-1$ if $\alpha =r$, and $\eta ^{u\alpha } =0$ otherwise. This gives

$$\begin{aligned} {\mathcal {A}}^{uu}&= \frac{1}{2}\Big ( \partial _\gamma ^B(X^u) \eta ^{\gamma u} + \partial _\gamma ^B(X^u) \eta ^{\gamma u} -X( \eta ^{uu}) -({\partial _u^B} X^u + \partial _r^B X^r )\eta ^{uu} \Big ) - \Big ( \frac{X^r}{r}\\ {}&\quad -X \ln \Omega \Big ) )\eta ^{uu} = - \partial _r^B X^u. \end{aligned}$$

For ${\mathcal {A}}^{rr}$ we use the fact that $\eta ^{ur} =-1$, $\eta ^{rr} =1$, and $\eta ^{r\alpha } =0$ otherwise. This yields

$$\begin{aligned} {\mathcal {A}}^{rr}&= \frac{1}{2}\big ( \partial _\gamma ^B(X^r) \eta ^{\gamma r} + \partial _\gamma ^B(X^r) \eta ^{\gamma r} -X( \eta ^{rr}) -({\partial _u^B} X^u + \partial _r^B X^r )\eta ^{rr} \big ) - \Big ( \frac{X^r}{r}-X \ln \Omega \Big ) )\eta ^{rr} \\&= \frac{1}{2}( 2\partial _r^BX^r -2\partial _u^BX^r )-({\partial _u^B} X^u + \partial _r^B X^r ) \big ) - \Big ( \frac{X^r}{r}-X \ln \Omega \Big ) ) \\&= \frac{1}{2}\partial _r^B X^r - \frac{1}{2}{\partial _u^B} X^u- {\partial _u^B} X^r - \Big ( \frac{X^r}{r}-X \ln \Omega \Big ) ). \end{aligned}$$

Similarly, for ${\mathcal {A}}^{ur}$ we get

$$\begin{aligned} {\mathcal {A}}^{ur}&= \frac{1}{2}\big ( \partial _\gamma ^B(X^u) \eta ^{\gamma r} + \partial _\gamma ^B(X^r) \eta ^{\gamma u} -X( \eta ^{ur}) -({\partial _u^B} X^u + \partial _r^B X^r )\eta ^{ur} \big ) - \Big ( \frac{X^r}{r}-X \ln \Omega \Big ) )\eta ^{ur} \\&= \frac{1}{2} \Big (- \partial _u^B X^u+ \partial _r^B X^u - \partial _r^B X^r +{\partial _u^B} X^u + \partial _r^B X^r \Big ) +\Big ( \frac{X^r}{r}-X \ln \Omega \Big ) \\&\quad = \frac{1}{2}\partial _r^B X^u+\Big ( \frac{X^r}{r}-X \ln \Omega \Big ). \end{aligned}$$

For ${\mathcal {A}}^{bc}$ we use the hypothesis that our vector field is radial and get

$$\begin{aligned} {\mathcal {A}}^{bc}&= \frac{1}{2}\Big ( \partial _\gamma ^B(X^b) \eta ^{\gamma c} + \partial _\gamma ^B(X^c) \eta ^{\gamma b} -X( \eta ^{bc}) -({\partial _u^B} X^u + \partial _r^B X^r \Big )\eta ^{bc} \big ) - \Big ( \frac{X^r}{r}-X \ln \Omega \Big ) )\eta ^{bc} \\&= \Big (\frac{X^{r}}{r} - \frac{1}{2}{\partial _u^B} X^u- \frac{1}{2}\partial _r^B X^r - \Big ( \frac{X^r}{r}-X \ln \Omega \Big ) \Big )\frac{\delta ^{bc}}{r^{2}}. \end{aligned}$$

$\square$

Proof of Lemma 4.2

By Lemma 4.3, the conformal potential V identically vanishes for the choices of weight ${}^{I}\Omega$ and ${}^{I\!I}\Omega$. Therefore we have $Q_{\alpha \beta }[\Phi ]= \widetilde{Q}_{\alpha \beta }[\Phi ],$ where Q is energy-momentum tensor in Eq. (3.1). Expanding this tensor in polar Bondi coordinates and using the Eq. (4.1) for the metric coefficients gives:

$$\begin{aligned}&Q_{uu}[\Phi ] = (\partial ^B_{u}\Phi )^{2}+ \frac{1}{2} (\partial ^B_{r}\Phi )^{2} + \frac{1}{2}| {\, {/\!\!\! \nabla }} \Phi |^{2}- \partial ^B_{u}\Phi \partial ^B_{r}\Phi , \\&Q_{ur}[\Phi ] = \frac{1}{2} (\partial ^B_{r}\Phi )^{2} + \frac{1}{2}| {\, {/\!\!\! \nabla }} \Phi |^{2}, \qquad Q_{rr}[\Phi ] = (\partial ^B_{r}\Phi )^{2}. \end{aligned}$$

Since the rescaled normal is $N{'} =(1-{r \over t} )\partial ^B_{u}+ {r \over t} \partial ^B_{r}$ we use these identities to get:

$$\begin{aligned}&{}^{(X)}\!\widetilde{P}_{\alpha } N{'}^\alpha =X^{u}(1-{r \over t} )Q_{uu}[\Phi ]+X^{u}{r \over t} Q_{ur}[\Phi ] +X^{r}(1-{r \over t} )Q_{ru}[\Phi ]\\ {}&\quad +X^{r}{r \over t} Q_{rr}[\Phi ] + \frac{\Lambda }{\Omega ^2} {\mathcal {B}}\Phi ^2\\&=X^{u}\big [(1-{r \over t} )\big ((\partial ^B_{u}\Phi )^2+ \frac{1}{2}(\partial ^B_{r}\Phi )^2+ \frac{1}{2}| {\, {/\!\!\! \nabla }} \Phi |^2 - \partial ^B_u\Phi \partial _r^B\Phi \big ) +{r \over t} \big (\frac{1}{2}(\partial ^B_{r}\Phi )^{2} + \frac{1}{2}| {\, {/\!\!\! \nabla }} \Phi |^{2} \big )\big ] \\&\quad + X^{r}\big [(1-{r \over t} )(\frac{1}{2} (\partial ^B_{r}\Phi )^{2} + \frac{1}{2}| {\, {/\!\!\! \nabla }} \Phi |^{2})+ {r \over t} (\partial ^B_{r}\Phi )^{2} \big ]+ \frac{\Lambda }{\Omega ^2} {\mathcal {B}}\Phi ^2. \end{aligned}$$

Combining terms yields the Eq. (4.6) immediately. To prove a coercive lower bound we apply Young inequality with $\epsilon$ to the unsigned term $- \partial ^B_u\Phi \partial _r^B\Phi$ in the second equation. This gives us:

$$\begin{aligned} {}^{(X)}\! \widetilde{P}_{\alpha } N{'}^\alpha&\ge X^{u}(1-{r \over t} ) \big [(1- \frac{1}{2\epsilon ^{2}})(\partial ^B_{u}\Phi )^2\nonumber \\&\quad + (\frac{1}{2}- \frac{1}{2}\epsilon ^{2})(\partial ^B_{r}\Phi )^2\big ] + \frac{1}{2}X^{u}({r \over t} (\partial ^B_{r}\Phi )^2+| {\, {/\!\!\! \nabla }} \Phi |^2) \nonumber \\&\qquad \qquad + \frac{1}{2}X^{r}\big [(1+ {r \over t} ) (\partial ^B_{r}\Phi )^{2} +(1-{r \over t} )| {\, {/\!\!\! \nabla }} \Phi |^{2}) \big ]+ \frac{\Lambda }{\Omega ^2} {\mathcal {B}}\Phi ^2 . \end{aligned}$$

(A.1)

Choosing $\epsilon$ satisfying the property $\frac{1}{2}<\epsilon ^{2}<1$ and using the positivity of $X^{u}$ yields the coercive bound

$$\begin{aligned}&{}^{(X)}\! \widetilde{P}_{\alpha } N{'}^\alpha \gtrsim _{\epsilon } X^{u}(1-{r \over t} ) (\partial ^B_{u}\Phi )^2 + \big (X^{u}+X^{r}(1+ {r \over t} )\big )(\partial ^B_{r}\Phi )^2\\&\quad + \big (X^{u}+X^{r}(1-{r \over t} )\big )| {\, {/\!\!\! \nabla }} \Phi |^2 + \frac{\Lambda }{\Omega ^2} {\mathcal {B}}\Phi ^2, \end{aligned}$$

where the implicit constant depends only on $\epsilon$. The upper bound follows trivially from applying Young inequality to the right-hand side of the Eq. (4.6). This completes the derivation of (4.8). $\square$