text:
\section{Introduction}

Let $(M,g)$ be a compact 3-dimensional Riemannian manifold. Let $\sigma$ denote a symmetric trace- and divergence-free (TT) tensor of type $(0,2)$, and let $\tau$ be a function on $M$. Lichnerowicz \cite{Li44} and Choquet-Bruhat and York \cite{CBY80} developed the so-called conformal method to transform \textit{seed data} $(M, g, \sigma, \tau)$ into \textit{initial data} satisfying Einstein constraint equations. Consider the following system of equations for a positive function $\phi$ and a one-form $W$:
\begin{subequations}\label{SD}
\begin{align}
-8 \Delta \phi + R_g \phi &= -\frac{2}{3} \tau^2 \phi^{5} + |\sigma + L_gW|_g^2 \phi^{-7}, \label{hamiltonian} \\
\Delta_L W &= \frac{2}{3} \phi^6 d\tau. \label{momentum}
\end{align}
\end{subequations}
Here $\Delta = \nabla_i \nabla^i$ and $R_g$ are the Laplacian and the scalar curvature computed with respect to metric $g$, and $\Delta_L W$ is defined as $\Delta_L W = \mathrm{div}_g (L_g W)$, where $L_g $ is the conformal Killing operator,
\begin{equation}
\label{LWdef}
(L_g W)_{ij}=\nabla_iW_j +\nabla_j W_i -\frac{2}{3}(\mbox{div}_gW)g_{ij}.
\end{equation}
Equation (\ref{hamiltonian}) is called the \textit{Lichnerowicz equation}, and equation (\ref{momentum}) is called the \textit{vector equation}. System (\ref{SD}) is referred to as the \textit{vacuum conformal constraints}. A dual to a form $W$ satisfying the equation $L_g W = 0$ is called a conformal Killing vector field.

Suppose that a pair $(\phi,W)$ solves the vacuum conformal constraints (\ref{SD}). Define $\widetilde g = \phi^4 g$, and $K = \frac{\tau}{3} \phi^{4} g + \phi^{-2} ({\sigma} + L_g W)$. Then the triple $(M, \widetilde g, K)$ becomes an initial data set satisfying the \textit{vacuum Einstein's constraints}
\begin{subequations}\label{originalconstraints}
\begin{align}
R_{\widetilde g} - |K|^2_{\widetilde g} + (\tr_{\widetilde g} K)^2 &= 0, \\
{\mathop{\rm div}}_{\widetilde g} K - d \tr_{\widetilde g} K &= 0.
\end{align}
\end{subequations}
Note that $\mathrm{tr}_{\widetilde g} K = \tau$. Choquet-Bruhat and Geroch \cite{CBG69} showed that such initial data give rise to a unique development.

Here we briefly review the current status of the study of the existence of the solutions of the vacuum conformal constraints for closed manifolds $(M,g)$. The case of constant $\tau$ is basically understood, cf.\ \cite{Isenberg}. Of course, many important results can be still obtained assuming a constant $\tau$. Examples include studies of multiplicity of solutions in the case with a positive cosmological constant \cite{premoselli, walter, Mach_Knopik}, or obtaining foliations of important spacetimes \cite{malec, beig}. In general, the case of non-constant $\tau$ remains still open. Some results are obtained when $d \tau /\tau$ or $\sigma$ are small, cf.\ \cite{IM92,IsenbergOMurchadha,ACI08,MaxwellNonCMC}. Results for rough initial data can be found in \cite{HNT2, Maxwellrough}. Interested readers may refer to a survey paper, for instance \cite{BartnikIsenberg}.

Recently, Dahl, Gicquaud, and Humbert proved the following criterion for the existence of solutions to Eqs.\ (\ref{SD}) \cite{DGH}. Assume that $(M,g)$ has no conformal Killing vector fields and that $\sigma \not \equiv 0$, if the Yamabe constant $Y(g) \geq 0$. Then, if the \textit{limit equation}
\begin{equation}
\label{eqLimit1}
\Delta_L W = \alpha \sqrt{\frac{2}{3}} |L_g W|_g \frac{d\tau}{\tau}
\end{equation}
has no nonzero solutions for all $\alpha \in (0,1]$, the vacuum conformal constraints (\ref{SD}) admit a solution $(\phi,W)$ with $\phi>0$. Moreover, they provided an example on the sphere $\mathbb S^3$ such that the limit equation \eqref{eqLimit1} does have a nontrivial solution for some $\alpha_0 \in (0,1]$ \cite[Prop.\ 1.6]{DGH}. Unfortunately, the result of Dahl, Gicquaud, and Humbert is not an alternative criterion. In fact, there also exists an example such that both the limit equation \eqref{eqLimit1} and the vacuum conformal constraints (\ref{SD}) have nontrivial solutions \cite[Prop. 3.10]{TCN}.

There is another way to construct vacuum initial data. It is sometimes referred to as `the conformally covariant split' or, historically, `Method B.
' (See \cite[Section 4.1]{BartnikIsenberg} or the original paper \cite{York}) In this case we are trying to find a positive function $\phi$ and a one-form $W$ satisfying the so-called `conformally covariant split system:'
\begin{subequations}\label{MB}
\begin{align}
\Delta \phi -\frac{1}{8}R_g \phi + \frac{1}{8}|{\sigma}|_g^2\phi^{-7} + \frac{1}{4} \langle {\sigma}, L_g W \rangle_g \phi^{-1} - \left( \frac{1}{12}\tau^2-\frac{1}{8} |L_g W|_g^2 \right) \phi^{5} &= 0, \label{hamiltonian2} \\
\nabla_i(L_g W)^i_j-\frac{2}{3} \nabla_j\tau+6(L_g W)^i_j\nabla_i\log \phi &= 0. \label{momentum2}
\end{align}
\end{subequations}
Here $\sigma\in W^{1,p}$ is a symmetric TT-tensor of type $(0,2)$, $\tau \in W^{1,p}$ is a function on $M$, and $g \in W^{2,p}$ is a metric on $M$. We assume that $p > 3$. We devote this paper to the study of system (\ref{MB}).

Let $\tilde g = \phi^4 g$, and
\[ K = \frac{\tau}{3} \phi^{4} g + \phi^{-2} {\sigma} + \phi^{4} L_g W. \]

\begin{prop}
For $(\phi, W)$ solving system \eqref{MB}, the triple $(M, \widetilde g, K)$ becomes vacuum initial data.
\end{prop}

\begin{proof}
Denote $\widetilde{g}=\phi^{4} g$ and $K=\frac{\tau}{3} \phi^{4} g + \phi^{-2} {\sigma} + \phi^{4} L_g W$.
Combining the equations
\[ R_{\widetilde g}=\phi^{-5}(R_g \phi -8\Delta \phi), \]
\[ \tr_{\widetilde{g}}K= \frac{\tau}{3} \phi^{4} \tr_{\widetilde{g}}g+\phi^{-2} \tr_{\widetilde{g}}{\sigma}+\phi^{4} \tr_{\widetilde{g}}L_g W=\tau, \]
and
\[ |K|_{\widetilde{g}}^2=\frac{1}{3}\tau^2+\phi^{-12}|{\sigma}|_g^2 + |L_g W|_g^2 +2\phi^{-6} \langle {\sigma},L_g W \rangle_g, \]
one gets
\begin{align*}
\MoveEqLeft R_{\widetilde g}-|K|_{\widetilde g}^2+(\tr_{\widetilde{g}}K)^2 \\
&= \phi^{-5}(R_g \phi -8\Delta \phi) -\left(\frac{1}{3}\tau^2+\phi^{-12}|{\sigma}|_g^2 + |L_g W|_g^2 +2\phi^{-6} \langle {\sigma}, L_g W \rangle_g \right)+\tau^2.
\end{align*}
Equation \eqref{hamiltonian2} implies $R_{\widetilde g}-|K|_{\widetilde{g}}^2+(\tr_{\widetilde{g}}K)^2=0$. On the other hand, by virtue of the fact that ${\mathop{\rm div}}_{\widetilde{g}}(\phi^{-2}\gamma)=\phi^{-6}{\mathop{\rm div}}_g(\gamma)$ for any trace-free symmetric $(0,2)$-tensor $\gamma_{ij}$ \cite[Eqn. ($\mathcal{C}$)]{De}, we have
\begin{align*}
\widetilde{\nabla}^iK_{ij} &= \widetilde{\nabla}^i \left[ \frac{\tau}{3} \widetilde{g}_{ij} + \phi^{-2} {\sigma}_{ij} + \phi^{4} (L_g W)_{ij} \right]\\
&= \frac{\widetilde{\nabla}^i\tau}{3}\widetilde{g}_{ij}+\phi^{-6}\nabla^i{\sigma}_{ij} +\phi^{-6}\nabla^i \left[ \phi^{6} (L_g W)_{ij} \right].
\end{align*}
Thus
\[ \widetilde{\nabla}^iK_{ij}-\widetilde{\nabla}_j\tr_{\widetilde{g}}K=  \nabla^i(LW)_{ij}+ 6\phi^{-1} (\nabla^i \phi)(L_g W)_{ij}-\frac{2 \nabla_j \tau}{3}. \]
Equation \eqref{momentum2} implies ${\mathop{\rm div}}_{\widetilde{g}} K - d \tr_{\widetilde{g}} K =0$.
\end{proof}

\begin{rem}\label{scaling}
An immediate observation concerning system (\ref{MB}) is that it admits the following scaling symmetry. Suppose that system (\ref{MB}) has a solution $(\phi, W)$. Set $\hat \phi = \mu^{-\frac{1}{4}} \phi$, $\hat W = \mu^{\frac{1}{2}} W$ for some positive number $\mu \in \mathbb R^+$. Then $(\hat \phi, \hat W)$ satisfy system (\ref{MB}) with the data $\hat \sigma$ and $\hat \tau$ given by $\hat \sigma_{ij} = \mu^{-1} \sigma_{ij}$, $\hat \tau = \mu^{\frac{1}{2}} \tau$.
\end{rem}

Method B and some of its variants were discussed in \cite{De}. System (\ref{MB}) possess the conformal covariance property in the following sense. Let $(\phi,W)$ solve system (\ref{MB}). Define $\omega = \omega(\sigma,\phi,W,g) := |\sigma+\phi^6 L_g W|_g$ and $P_{g,\omega} \phi := 8\Delta_g\phi - R_g \phi + \omega^2 \phi^{-7}$. Then the Lichnerowicz equation (\ref{hamiltonian2}) can be written as
\begin{equation}
\phi^{-5}P_{g,\omega}\phi=\frac{2}{3}\tau^2,
\end{equation}
and the vector equation (\ref{momentum2}) becomes
\begin{equation}
\Delta_{g,\phi}W=\frac{2}{3}d\tau,
\end{equation}
where $\Delta_{g,\phi} W = \phi^{-6} \mathrm{div}_g (\phi^6 L_g W)$. Now we make the following conformal change:
\begin{equation*}
\widetilde{g}=\psi^4g, \quad \widetilde{\phi}=\psi^{-1}\phi, \quad \widetilde{\sigma}=\psi^{-2}\sigma, \quad \widetilde{W}=\psi^4 W,
\end{equation*}
where $\psi$ is any positive function. It is easy to show that $\widetilde{\sigma}$ is still TT with respect to the metric $\widetilde{g}$, $\widetilde{\omega} = \widetilde{\omega} (\widetilde{\sigma},\widetilde{\phi},\widetilde{W},\widetilde{g}) = \psi^{-6} \omega$, and $\Delta_{\widetilde{g},\widetilde{\phi}} \widetilde{W} = \Delta_{g,\phi}W$. Then the operator given by
\begin{equation*}
\mathcal{P}_g \begin{pmatrix} \phi \\ W \\ \end{pmatrix}
:= \begin{pmatrix} \phi^{-5}P_{g,\omega} \phi \\ \Delta_{g,\phi} W \end{pmatrix}
\end{equation*}
is conformally covariant, i.e.,
\begin{equation}
\mathcal{P}_{\widetilde{g}} \begin{pmatrix} \widetilde{\phi} \\ \widetilde{W} \end{pmatrix}
= \mathcal{P}_{g} \begin{pmatrix} \phi \\ W \end{pmatrix}
= \frac{2}{3} \begin{pmatrix} \tau^2 \\ d\tau \end{pmatrix}.
\end{equation}

If $\tau$ is constant, Eqs.\ (\ref{SD}) split in a natural way. In this case, we have $W\equiv 0$, and we are only left with the well-studied Lichnerowicz equation. Much less is mathematically known about the conformally covariant split system, although it was applied in certain studies by numerical relativists \cite{Co00}. The proof or disproof of the existence of solutions of system (\ref{MB}) is difficult. When $\tau$ is constant, $L_g W \equiv 0$ is of course a trivial solution of \eqref{momentum2}, but even in this case the existence of solutions with non-zero $L_g W$ is unclear.

Solutions of systems (\ref{SD}) (standard conformal method) and (\ref{MB}) (conformally covariant split system) are, of course, related. Suppose that system (\ref{SD}) has a solution $(\phi,W)$ for the assumed data $(M,g)$, $\sigma$, and $\tau$. Suppose further that $\hat W$ is a solution to the equation
\begin{equation}
\label{bothsystemsrelated}
L_g \hat W = \phi^{-6} L_g W.
\end{equation}
It can be easily checked that the pair $(\phi, \hat W)$ solves system (\ref{MB}) with the same assumed data $(M,g)$, $\sigma$, and $\tau$. Moreover, both solutions lead to the same $K$, and hence to the same initial data $(M, \tilde g, K)$. The subtlety of the relation between $W$ and $\hat W$, and hence between systems (\ref{SD}) and (\ref{MB}), is due to the fact that $L$ is not, in general, invertible. However, computing the divergence of both sides of Eq.\ (\ref{bothsystemsrelated}) we obtain
\[ \Delta_L \hat W = \mathrm{div}_g \left( \phi^{-6} L_g W \right). \]
Note that $L_g^* = - 2 \, \mathrm{div}_g$ is a formal adjoint of $L_g$ with respect to $L^2$ product. Thus $\Delta_L = - \frac{1}{2} L_g^\ast L_g$. In explicit terms
\[ \Delta_L W_j = \nabla^i (L_g W)_{ij} = \nabla^i \nabla_i W_j + \frac{1}{3} \nabla_j (\nabla^i W_i) + W_i R^i_j, \]
where $R^i_j$ are the components of the Ricci tensor. When $(M,g)$ has no conformal Killing vector fields, the vector Laplacian $\Delta_L = - \frac{1}{2} L_g^\ast L_g$ is bijective between certain Sobolev spaces. In this case, one can solve (\ref{bothsystemsrelated}) to obtain
\begin{equation*}
\hat W = \Delta_L^{-1}\left[ \mathrm{div}_g \left( \phi^{-6}L_g W \right) \right], \quad \mbox{or equivalently} \quad W = \Delta_L^{-1} \left[ \mathrm{div}_g \left( \phi^{6} L_g \hat W \right) \right].
\end{equation*}

Suppose that we already have vacuum initial data $(M,g,K)$ such that $\bar \tau = \tr_g K$ is constant. In this case the traceless part of $K$, $\bar{\sigma}_{ij} = K_{ij} - \frac{\tr_g K}{3}g_{ij}$ is divergence free, and  system \eqref{MB} admits a particular solution $(\bar\phi \equiv1, \bar W \equiv 0)$ for the data $(\sigma, \tau) = (\bar \sigma, \bar \tau)$. This obvious solution can be understood as transforming the seed data $(M,g,K)$ into itself. In subsequent sections, we use the implicit function theorem to deduce existence of other solutions of Eqs.\ (\ref{MB}) with $\tau \neq \tr_g K$.

The order of this paper is as follows. In Section \ref{closed} we prove existence of solutions of system (\ref{MB}) on closed manifolds $M$, but admitting a non-constant $\tau$. In Section \ref{boundarycase} we obtain similar results for a compact manifold with a boundary. The assumed boundary conditions guarantee that this boundary constitutes an apparent horizon in the obtained initial data. To a large extent, the results of Sec.\ \ref{closed} and \ref{boundarycase} constitute counterparts to the analysis given in \cite{GN14} for system (\ref{SD}) (Method A). In Section \ref{withlambda} we take into account a cosmological constant and reformulate main theorems obtained in Sec.\ \ref{closed}. Finally, Sec.\ \ref{numerics} provides simple numerical examples of solutions of system (\ref{MB}) on compact manifolds, assuming non-constant $\tau$. In these examples, we assume the manifold $M = \mathbb S^1 \times \mathbb S^2$ or $M = \mathbb S^1 \times \mathbb T^2$.

We use the standard geometric notation in this paper. The product of two symmetric tensors of type (0,2) with respect to metric $g$ is expressed, in terms of their tensor components, as $\langle a, b \rangle_g = g^{ij} g^{kl} a_{jl} b_{ik}$. The square of the norm is denoted as $|a|^2_g = \langle a, a \rangle_g$. We omit the subscript referring to the metric, when the metric is obvious from the context. In most cases the metric is understood to be that of the seed data. This also applies to the standard notation of the trace, or the divergence of a tensor, as well as the scalar curvature $R_g$ and the conformal Killing operator $L_g$.

Unless otherwise stated, all the given data on the manifold $M$ are assumed to be smooth. We make use of the standard notation $W^{k,p}$ to denote the Sobolev space of functions defined on the manifold $M$ and $W^{k,p}_+$ to denote the subset consisting of all positive $W^{k,p}$ functions. In Section \ref{boundarycase} the manifold $(M,g)$ is assumed to be compact with boundary. To distinguish the functional spaces on the manifold $M$ and those on the boundary $\partial M$, we use the notation  $W^{k,p}(M)$ and $W^{k,p}(\partial M)$ respectively.

\section{Conformally covariant split system on a closed manifold}
\label{closed}

In this section, we assume that $(M,g)$ is a closed 3-dimensional Riemannian manifold. Making use of the implicit function theorem, we construct a family of solutions of the conformally covariant split system \eqref{MB} on $M$. These solutions give rise to vacuum initial data.

\begin{thm}\label{3}
Suppose that we already have vacuum initial data $(M,g,K)$. Assume that $\bar \tau = \tr_g K = \mathrm{const}$, and that $K \neq 0$ in some region of $M$. Assume further that $(M,g)$ has no conformal Killing vector fields. Then there is a small neighborhood of $\bar \tau$ in $W^{1,p}$ such that for any $\tau$ in this neighborhood there exists $(\phi_\tau, W_\tau) \in W^{2,p}_+\times W^{2,p}$ solving the system \eqref{MB} for the data $\bar{\sigma}_{ij} = K_{ij} - \frac{\bar \tau}{3} g_{ij}$ and $\tau$.
\end{thm}

\begin{proof}
The proof is based on the implicit function theorem and the ideas are borrowed from \cite{Ch04} and \cite{GN14}. First, let us define the operator
\begin{align*}
\MoveEqLeft \mathcal{F} \colon W^{1,p} \times W^{2,p}_+\times W^{2,p} \rightarrow L^p\times L^p, \\
& \begin{pmatrix} \tau\\ \phi\\ W \end{pmatrix} \mapsto \begin{pmatrix}
\Delta \phi -\frac{1}{8}R \phi + \frac{1}{8}|\bar{\sigma}|^2\phi^{-7}+\frac{1}{4} \langle \bar{\sigma}, LW \rangle \phi^{-1} - \left( \frac{1}{12}\tau^2-\frac{1}{8}|LW|^2 \right) \phi^{5} \\
\nabla_i(LW)^i_j-\frac{2}{3}
\nabla_j \tau + 6(LW)^i_j\nabla_i\log \phi
\end{pmatrix}.
\end{align*}
It is easy to see that $\mathcal{F}$ is a $C^1$-mapping and $\mathcal{F}(\bar\tau,\bar \phi\equiv1,\bar W\equiv0)=(0,0)$. We prove that the partial derivative of $\mathcal{F}$ with respect to the variables $(\phi,W)$ is an isomorphism at $(\bar\tau,\bar \phi\equiv1,\bar W\equiv0)$. The differential at the point $(\bar\tau,\bar \phi\equiv1,\bar W\equiv0)$ is given by
\begin{align*}
\MoveEqLeft D\mathcal{F}|_{(\bar\tau,1,0)} \begin{pmatrix} \delta\phi \\ \delta W \end{pmatrix} \\
&= \begin{pmatrix}
\Delta-\frac{1}{8}R-\frac{7}{8}|\bar\sigma|^2-\frac{5}{12}\bar\tau^2 & , & \frac{1}{4} \langle \bar\sigma, L(\cdot) \rangle \\
 0 & , & \Delta_L
\end{pmatrix}
\begin{pmatrix}\delta\phi \\ \delta W\end{pmatrix},
\end{align*}
and it is triangular, meaning that the second row of the above $2 \times 2$ block matrix does not depend on $\delta \phi$. Thus, the invertibility of $D\mathcal{F}|_{(\bar\tau,1,0)}$ follows from the fact that the diagonal terms are invertible. More specifically:

\textit{Claim 1.}
\begin{align*}
\MoveEqLeft \mathcal{H} \colon W^{2,p} \rightarrow L^p, \\
& \delta\phi \mapsto (\Delta-\frac{1}{8}R-\frac{7}{8}|\bar\sigma|^2-\frac{5}{12}\bar\tau^2 )\delta \phi
\end{align*}
is invertible and

\textit{Claim 2.}
\begin{align*}
\MoveEqLeft \Delta_L \colon W^{2,p} \rightarrow L^p, \\
& \delta W \mapsto \Delta_L \delta W
\end{align*}
is also invertible.

The proof of Claim 2 is a consequence of the assumption that $(M,g)$ is closed and has no conformal Killing vector fields. The proof of Claim 1 is as follows. Note that $\mathcal{H}$ is a Fredholm operator of index $0$. It suffices to show that $\mathcal{H}$ is injective. Since $(\bar \phi \equiv 1, \bar W \equiv0)$ solves the system \eqref{MB} with the data $\bar \tau$ and $\bar \sigma$, one has
\[ -\frac{1}{8}R +\frac{1}{8}|\bar \sigma|^2-\frac{1}{12}\bar\tau^2=0. \]
Hence,
\[ \Delta-\frac{1}{8}R-\frac{7}{8}|\bar\sigma|^2-\frac{5}{12}\bar\tau^2= \Delta -|\bar\sigma|^2-\frac{1}{3}\bar\tau^2=\Delta-|K|^2. \]
Clearly, it is a negatively definite operator.

Finally, the theorem follows from the implicit function theorem.
\end{proof}

\begin{rem} For $K\equiv 0$, one can set $W\equiv0$, and the system \eqref{MB} reduces the Yamabe problem \cite{trudinger,Aubin,schoen84}.
\end{rem}

Note that the scaling symmetry discussed in Remark \ref{scaling} can be used to produce new solutions from the already obtained ones. In particular, one can obtain solutions with $\tau$ deviating from the vicinity of $\bar \tau$, at a cost of rescaling $\bar \sigma$. When the seed solution $(M, g, K)$ is a maximal slice, one can also produce new non-CMC initial data with the following scaling argument.

\begin{thm}\label{exi}
Suppose that we already have vacuum initial data $(M,g,K)$ with $\tr_gK=0$. Suppose $K\neq 0$ for some region. Assume further that $(M,g)$ has no conformal Killing vector fields. Given any $\tau \in W^{1,p}$, there is a positive constant $\eta>0$ such that for any $ \mu\in(0,\eta)$, there exists at least one solution $(\phi,W) \in W^{2,p}_+\times W^{2,p}$ of system \eqref{MB} for the data $(\hat{\sigma}=\mu^{12}K, \hat\tau=\mu^{-1}\tau)$.
\end{thm}

\begin{proof}
Since $(M,g,K)$ constitute vacuum initial data, system \eqref{MB} admits a particular solution $(\bar\phi \equiv 1, \bar W \equiv 0)$ for $\bar \tau=0$ and $\bar {\sigma}=K$.

Let us consider the following $\mu$-deformed system corresponding to Eqs.\ \eqref{MB}:
\begin{align*}
\MoveEqLeft \mathcal{G} \colon \mathbb{R} \times W^{2,p}_+\times W^{2,p} \rightarrow L^p\times L^p, \\
& \begin{pmatrix} \mu \\ \phi\\ W \end{pmatrix} \mapsto \begin{pmatrix}
\Delta \phi -\frac{1}{8}R \phi + \frac{1}{8}|K|^2\phi^{-7}+\frac{1}{4}\mu^4 \langle K, LW \rangle \phi^{-1}- \left( \mu^{10}\frac{1}{12}\tau^2-\mu^{8}\frac{1}{8}|LW|^2 \right) \phi^{5} \\
\nabla_i(LW)^i_j-\frac{2}{3}
\mu\nabla_j\tau+6(LW)^i_j\nabla_i\log \phi
\end{pmatrix}.
\end{align*}
It is easy to see that $\mathcal{G}$ is a $C^1$-mapping. The condition that $(M,g,K)$ constitute vacuum initial data with $\tr_g K = 0$ implies that $\mathcal{G}(0,1,0)=(0,0)$. We now prove that the partial derivative of $\mathcal{G}$ with respect to the variables $(\phi,W)$ is an isomorphism at $(0,\bar \phi\equiv1,\bar W\equiv0)$. The differential at the point $(0, \bar \phi \equiv 1,\bar W \equiv 0)$ is given by
\begin{align*}
\MoveEqLeft D\mathcal{G}|_{(0,1,0)} \begin{pmatrix} \delta\phi \\ \delta W \end{pmatrix} \\
&= \begin{pmatrix}
\Delta-\frac{1}{8}R-\frac{7}{8}|K|^2 & , & 0\\
 0 & , & \Delta_L
\end{pmatrix}
\begin{pmatrix}\delta\phi \\ \delta W \end{pmatrix},
\end{align*}
where $\Delta_L W=\mbox{div}_g(LW)$. Since $(0, \bar \phi \equiv 1,\bar W \equiv0)$ solves system \eqref{MB}, one has
\[ \Delta-\frac{1}{8}R-\frac{7}{8}|K|^2= \Delta -|K|^2. \]
The invertibility of the derivative $D \mathcal G|_{(0,1,0)}$ follows from the arguments stated already in the proof of Theorem \ref{3}.

By the implicit function theorem, for a sufficiently small parameter $\mu$, there exists $(\phi_{\mu}, W_\mu)$ such that $\mathcal{G}(\mu,\phi_\mu,W_\mu)=0$.

Define $\hat \phi_\mu=\mu^3 \phi_\mu$ and $\hat W_\mu=\mu^{-2}W_\mu$. Direct calculations show that $(\hat\phi_\mu, \hat W_\mu)$ solves system \eqref{MB} for the rescaled data $(\hat{\sigma}=\mu^{12}K, \hat\tau=\mu^{-1}\tau)$.
\end{proof}

In order to convert the above theorems into existence results, one has to assert the existence of solutions with $\tau = \bar\tau = \mathrm{const}$. Fortunately, in this case $W \equiv 0$ is a solution of \eqref{momentum2}. We only need to consider equation \eqref{hamiltonian2} with $W\equiv0$:
\begin{equation}
\label{newlich}
\Delta \phi -\frac{1}{8}R \phi + \frac{1}{8}|\bar{\sigma}|^2\phi^{-7}-\frac{1}{12}\bar\tau^2\phi^{5} = 0.
\end{equation}

As a consequence of Theorem 4.10 in \cite{Ch04}, one has
\begin{thm}\label{cmc}
If $\bar\tau$ is a constant, there exists $(\phi_{\bar\tau},W_{\bar\tau}\equiv0)$ solving equation \eqref{newlich}, provided that one of the following conditions holds:
\begin{enumerate}
\item The Yamabe constant $Y(g) > 0$ and $|\bar{\sigma}|^2 > 0$
\item $\bar\tau\neq 0$, $Y(g)=0$ and $|\bar{\sigma}|^2 > 0$
\item $\bar\tau\neq 0$ and $Y(g) < 0$
\end{enumerate}
\end{thm}

Here the Yamabe constant is defined as
\[ Y(g) = \mathrm{inf}_{u \in C^\infty(M), u \not\equiv 0} \frac{\int_M \left( |\nabla u|^2 + \frac{1}{8} R u^2 \right)dv_g}{\left( \int_M u^6 dv_g \right)^\frac{1}{3}}. \]
The set $\left(M,\phi_{\bar\tau}^4g, \frac{\bar \tau}{3} \phi_{\bar \tau}^4 g + \phi_{\bar\tau}^{-2}\bar\sigma, \bar \tau \right)$ can be regarded as the assumed CMC vacuum initial data in Theorem \ref{3} and, if $\bar \tau \equiv 0$, also in Theorem \ref{exi}.

\section{Conformally covariant split system on a compact manifold with boundary}
\label{boundarycase}

We will now turn to a physically important case, where the manifold $M$ is compact and has a black hole type boundary. Since we are working at the level of initial data, basically the only available concept of a black hole is that of a region enclosed within an apparent horizon. Consequently, the boundary conditions that we now adopt guarantee that the boundary of $M$ is a marginally trapped surface (and that the manifold $M$ constitutes the `exterior' of the black hole). It is however not surprising that requiring the boundary of $M$ to correspond to an apparent horizon is not sufficient as a prescription of boundary conditions for system (\ref{MB}). Consequently, a part of the boundary conditions have to be imposed in a more or less arbitrary manner. In this work we follow a recipe proposed in \cite{GN14}.

Let $(M,\widetilde g)$ be a compact 3-dimensional manifold with the boundary $\partial M$, and let $\widetilde \nu$ be the unit vector normal to $\partial M$. We assume that $\widetilde \nu$ is pointing `outwards' of $M$, and therefore to the `inside' of the black hole. The two null expansions of $\partial M$ are given by
\[ \Theta_\pm = \mp H_{\widetilde g} - K(\widetilde \nu, \widetilde \nu) + \tr_{\widetilde g} K, \]
where $H_{\widetilde g} = \widetilde \nabla_i \widetilde \nu^i$, and $\widetilde \nabla_i$ denotes the covariant derivative with respect to metric $\tilde g$. The condition that $\partial M$ is a marginally trapped surface can be stated as $\Theta_+ = 0$, $\Theta_- \le 0$. Let us further observe that $\frac{1}{2}\left( \Theta_- + \Theta_+ \right) = - K(\widetilde \nu, \widetilde \nu)  + \tr_{\widetilde g} K$, and $\frac{1}{2}\left( \Theta_- - \Theta_+ \right) = H_{\widetilde g}$. Consequently, the condition $\Theta_+ = 0$ yields
\begin{align}
\frac{1}{2}\Theta_- &= - K(\widetilde \nu, \widetilde \nu) + \tr_{\widetilde g} K, \label{thetam} \\
\frac{1}{2}\Theta_- &= H_{\widetilde g}. \label{thetap}
\end{align}
If the metric $\widetilde g$ and the extrinsic curvature $K$ are obtained as $\widetilde g = \phi^4 g$ and $K = \frac{\tau}{3} \phi^{4} g + \phi^{-2} {\sigma} + \phi^{4} LW$, the above conditions can be also expressed in terms of quantities related directly to $(M,g)$. A vector $\nu$, normal to $\partial M$, and normalized with respect to metric $g$, is related with $\widetilde \nu$ by $\nu^i = \phi^2 \widetilde \nu^i$. Also
\[ H_{\widetilde g} = \phi^{-2} H_g + 4 \phi^{-3} \partial_\nu \phi, \]
where $H_g = \nabla_i \nu^i$ is the mean curvature of $\partial M$ with respect to the metric $g$. Similarly,
\[ K(\widetilde \nu, \widetilde \nu)  = \frac{1}{3} \tau + \phi^{-6} \sigma(\nu,\nu) + LW(\nu,\nu). \]
Conditions (\ref{thetam}) and (\ref{thetap}) can be now rewritten as
\begin{equation}
\label{prototype}
\phi^{-6} \sigma(\nu,\nu) + LW(\nu,\nu) - \frac{2}{3} \tau + \frac{1}{2}\Theta_- = 0
\end{equation}
and
\[ \partial_\nu \phi + \frac{1}{4}H_g \phi - \frac{\Theta_-}{8} \phi^3 = 0. \]
Since Eq.\ (\ref{prototype}) is not sufficient as a boundary condition for $W$, we will actually replace it with a stronger requirement. Let $\xi$ denote a 1-form tangent to the boundary $\partial M$. We will require, as a boundary condition, that
\begin{equation}
\label{boundaryw}
\phi^{-6} \sigma(\nu,\cdot)  + LW(\nu,\cdot) - \frac{2}{3} \tau \nu^\flat + \frac{1}{2}\Theta_- \nu^\flat  - \xi = 0,
\end{equation}
where $\nu^\flat$ is the 1-form dual to the normal vector field $\nu$. Clearly, Eq.\ (\ref{prototype}) follows from Eq.\ (\ref{boundaryw}), as $\xi(\nu) = 0$.

In the remaining part of this section, we always assume that
\begin{equation}
\label{boundarysigma}
{\sigma}(\nu,\cdot)=0
\end{equation}
on $\partial M$. It is, of course, a restriction on the allowed forms of $\sigma$. On the other hand, it facilitates the proofs of the theorems of this section. In addition, assuming (\ref{boundarysigma}) ensures the $L^2$-orthogonality of $LW$ and $\sigma$. This can be easily demonstrated by a direct computation
\begin{align*}
\int_M \langle LW, \sigma \rangle dv_g &= 2 \int_M \sigma^{ij} \nabla_i W_j dv_g \\
&= 2 \int_{\partial M} \sigma_{ij} \nu^i W^j d \sigma_g = 0,
\end{align*}
where the boundary term vanishes because of assumption (\ref{boundarysigma}). Note that on compact manifolds without boundary, the $L^2$-orthogonality of $LW$ and $\sigma$ is guaranteed without any additional conditions.

In summary, we are now dealing with the following set of equations
\begin{subequations}\label{BMB}
\begin{align}
\Delta \phi -\frac{1}{8}R \phi + \frac{1}{8}|{\sigma}|^2\phi^{-7}+\frac{1}{4} \langle {\sigma}, LW \rangle \phi^{-1}- \left( \frac{1}{12}\tau^2-\frac{1}{8}|LW|^2 \right)\phi^{5} &= 0, \label{Bhamiltonian} \\
\nabla_i(LW)^i_j-\frac{2}{3} \nabla_j\tau+6(LW)^i_j\nabla_i\log \phi &= 0, \label{Bmomentum} \\
\partial_{\nu}\phi +\frac{1}{4}H\phi -\frac{\Theta_{-}}{8}\phi^3 &= 0, \label{Boundary1} \\
LW(\nu,.)-\frac{2}{3}\tau\nu^\flat+\frac{\Theta_{-}}{2}\nu^\flat-\xi &= 0, \label{Boundary2}
\end{align}
\end{subequations}
where \eqref{Boundary1} and \eqref{Boundary2} are the boundary conditions on $\partial M$. Here $g\in W^{2,p}(M)$, $\sigma \in W^{1,p}(M)$, $\tau \in W^{1,p}(M)$, $\Theta_- \in W^{1-\frac{1}{p},p}(\partial M)$, $\Theta_- \le 0$, and $\xi\in W^{1-\frac{1}{p},p}( \partial M)$ are the assumed data. In Equations (\ref{BMB}), and in the remaining part of this section, we drop, for convenience, the subscript $g$ in the symbol denoting the mean curvature of $\partial M$; we write simply $H_g \equiv H$.

Suppose that we already have vacuum initial data $(M,g,K)$, and denote $\bar \tau = \tr_g K$. Assume further that $\bar \tau$ is a constant. In this case the traceless part of $K$, $\bar{\sigma}_{ij}=K_{ij}-\frac{\bar\tau}{3}g_{ij}$ is divergence free. Then the pair $(\bar \phi \equiv 1, \bar W \equiv 0)$ is a solution of Eqs.\ (\ref{Bhamiltonian}) and (\ref{Bmomentum}). We will now require $(\bar \phi \equiv 1, \bar W \equiv 0)$ to also solve Eqs.\ (\ref{Boundary1}) and (\ref{Boundary2}). This implies that $\xi \equiv 0$ and $\Theta_- = 2 H = \frac{4}{3} \bar \tau = \mathrm{const}$.

Using the implicit function theorem, we can now assert the existence of a family of solutions to Eqs.\ (\ref{BMB}).

\begin{thm} Let $(M,g,K)$ be vacuum initial data with boundary $\partial M$ such that $\bar \tau = \tr_g K = \frac{3}{2} H = \mathrm{const} \le 0$, where $H$ denotes the mean curvature of $\partial M$. Let $\Theta_- = \frac{4}{3} \bar \tau$ and $\xi \equiv 0$ so that Eqs.\ (\ref{BMB}) admit a solution $(\bar \phi \equiv 1, \bar W \equiv 0)$. Assume further that $(M,g)$ has no conformal Killing vector fields, and $K \neq 0$ in some region of M. There is a small neighborhood of $\bar \tau$ in $W^{1,p}(M)$ such that for any $\tau$ in this neighborhood there exists a solution $(\phi_\tau,W_\tau) \in W^{2,p}_+(M)\times W^{2,p}(M)$ of system \eqref{BMB}.
\end{thm}

\begin{proof}
First, let us define a mapping
\begin{align*}
\MoveEqLeft \mathcal{F} \colon W^{1,p}(M) \times W^{2,p}_+(M)\times W^{2,p}(M) \rightarrow L^p(M)\times W^{1-\frac{1}{p},p}( \partial M) \times L^p (M) \times W^{1-\frac{1}{p},p}( \partial M), \\
& \begin{pmatrix} \tau\\ \phi\\ W \end{pmatrix} \mapsto \begin{pmatrix}
\Delta \phi -\frac{1}{8}R \phi + \frac{1}{8}|\bar \sigma|^2\phi^{-7}+\frac{1}{4} \langle \bar \sigma, LW \rangle \phi^{-1}- \left( \frac{1}{12}\tau^2-\frac{1}{8}| LW|^2 \right) \phi^{5} \\
\partial_{\nu}\phi +\frac{1}{4} H\phi -\frac{\Theta_{-}}{8}\phi^3 \\
\nabla_i(LW)^i_j-\frac{2}{3}\nabla_j\tau+6(LW)^i_j\nabla_i\log \phi \\
LW(\nu,.)-(\frac{2}{3}\tau-\frac{\Theta_{-}}{2})\nu^\flat
\end{pmatrix}.
\end{align*}
It is easy to see that $\mathcal{F}$ is a $C^1$-mapping and $\mathcal{F}(\bar{\tau},1,0)=(0,0,0,0)$. We prove that the partial derivative of $\mathcal{F}$ with respect to variables $(\phi,W)$ is an isomorphism at $(\bar{\tau},1,0)$. The differential at the point $(\bar{\tau},1,0)$ is given by
\begin{align*}
\MoveEqLeft D\mathcal{F}|_{(\bar{\tau},1,0)} \begin{pmatrix} \delta\phi \\ \delta W \end{pmatrix}\\
&= \begin{pmatrix}
\Delta-\frac{1}{8}R-\frac{7}{8}|\bar\sigma|^2-\frac{5}{12}\bar\tau^2 & , & \frac{1}{4} \langle \bar\sigma, L(\cdot) \rangle \\
\partial_{\nu}+\frac{1}{4} H-\frac{3\Theta_{-}}{8}  & , & 0\\
 0 & , & \Delta_L\\
 0 & , &L\cdot(\nu,\cdot)
\end{pmatrix}
\begin{pmatrix}\delta\phi \\ \delta W\end{pmatrix},
\end{align*}
and it is block triangular. Thus, the invertibility of $D\mathcal{F}|_{(\bar\tau,1,0)}$ follows from the fact that both diagonal block terms are invertible. More precisely:

\textit{Claim 1.} The map
\begin{align*}
\MoveEqLeft \mathcal{H}\colon W^{2,p}(M) \rightarrow L^p (M)\times W^{1-\frac{1}{p},p}( \partial M), \\
& \delta\phi \mapsto \left( \left(\Delta-\frac{1}{8}R-\frac{7}{8}|\bar\sigma|^2-\frac{5}{12}\bar\tau^2 \right)\delta \phi, \left(\partial_{\nu} +\frac{1}{4}H - \frac{3\Theta_-}{8}\right)\delta \phi \right)
\end{align*}
is invertible.

\textit{Claim 2.} The map
\begin{align*}
\MoveEqLeft \mathcal{L} \colon W^{2,p}(M) \rightarrow L^p (M) \times  W^{1-\frac{1}{p},p}( \partial M), \\
& \delta W \mapsto \left(\Delta_L \delta W,L\delta W(\nu,\cdot) \right)
\end{align*}
is also invertible.

Proof of \textit{Claim 1.} Note that $\mathcal{H}$ is a Fredholm operator of index $0$. It suffices to show that $\mathcal{H}$ is injective. Since the pair $(\bar \phi \equiv 1, \bar W \equiv 0)$ solves system \eqref{MB} for the mean curvature $\bar \tau$, one has
\[ -\frac{1}{8}R +\frac{1}{8}|\bar \sigma|^2-\frac{1}{12}\bar\tau^2=0, \]
and hence
\[ \Delta-\frac{1}{8}R-\frac{7}{8}|\bar\sigma|^2-\frac{5}{12}\bar\tau^2= \Delta -|\bar\sigma|^2-\frac{1}{3}\bar\tau^2 = \Delta - |K|^2. \]
Note that for any $\delta \phi \in W^{2,p}$ we have
\[ \int_M (\delta \phi) \Delta (\delta \phi) dv_g = -\int_M |\nabla (\delta \phi)|^2 dv_g + \int_{\partial M} (\delta \phi) \partial_{\nu} (\delta \phi) d\sigma_g. \]
If $\delta \phi \in \mathrm{ker} \mathcal{H}$, then one obtains
\[  \int_M |K|^2 (\delta \phi)^2 dv_g = - \int_M |\nabla (\delta \phi)|^2 dv_g + \int_{\partial M} \left(\frac{3\Theta_{-}}{8}-\frac{1}{4} H\right) (\delta \phi)^2 d \sigma_g. \]
Since $\frac{1}{4} H = \frac{\Theta_{-}}{8}$, we get
\[ \int_M |K|^2 (\delta \phi)^2 dv_g = - \int_M |\nabla (\delta \phi)|^2 dv_g + \int_{\partial M} \frac{1}{4} \Theta_- (\delta \phi)^2 d \sigma_g \leq 0, \]
where the last inequality follows from the assumption that $\Theta_{-} \leq 0$. The condition $K \neq 0$ implies that $|K|^2 > 0$, and hence $\delta \phi \equiv 0$. Therefore $\mathcal{H}$ has a trivial kernel, and it is invertible.

The proof of \textit{Claim 2} is a consequence of the assumption that $(M,g)$ admits no conformal Killing vector fields. The existence of solutions $(\phi_\tau, W_\tau)$ for $\tau$ in a neighborhood of $\bar \tau$ follows now from the implicit function theorem.
\end{proof}

\begin{thm}\label{3-mu}
Suppose that $(M,g,K)$ satisfy the vacuum Einstein's constraint equations, and $M$ has a boundary $\partial M$ such that $H \equiv 0$ on $\partial M$. Assume that $\tr_g K = 0$ and $K \neq 0$ in some region of $M$. Assume further that $(M,g)$ has no conformal Killing vector fields. Given any data $\tau \in W^{1,p}(M)$, $\Theta_- \in W^{1-\frac{1}{p},p}(\partial M)$, $\Theta_- \le 0$, and $\xi\in W^{1-\frac{1}{p},p}( \partial M)$, there is a positive constant $\eta>0$ such that for any $\mu \in (0, \eta)$, there exists at least one solution $(\phi, W) \in W^{2,p}_+(M)\times W^{2,p}(M)$ of the system \eqref{BMB} for the data $(\hat\sigma = \mu^{12} K, \hat\tau = \mu^{-1} \tau, \Theta_-, \xi)$.
\end{thm}

\begin{proof}
Let us start with defining a map
\begin{align*}
\MoveEqLeft \mathcal{F} \colon \mathbb{R} \times W^{2,p}_+(M)\times W^{2,p}(M) \rightarrow L^p(M)\times W^{1-\frac{1}{p},p}( \partial M) \times L^p (M) \times W^{1-\frac{1}{p},p}( \partial M),\\
& \begin{pmatrix} \mu\\ \phi\\ W \end{pmatrix} \mapsto \begin{pmatrix}
\Delta \phi -\frac{1}{8}R \phi + \frac{1}{8}|K|^2 \phi^{-7} + \frac{1}{4} \mu^4 \langle K, LW \rangle \phi^{-1} - \left( \frac{\mu^{10}}{12}\tau^2 - \frac{\mu^{8}}{8} |LW|^2 \right) \phi^{5}  \\
\partial_\nu \phi - \frac{\mu^6}{8} \Theta_{-} \phi^3 \\
\nabla_i(LW)^i_j - \frac{2}{3} \mu \nabla_j \tau + 6(LW)^i_j \nabla_i \log \phi \\
LW(\nu,.) - \left( \frac{2}{3} \mu \tau - \frac{\mu^2 \Theta_{-}}{2} \right) \nu^\flat - \mu^2 \xi
\end{pmatrix}.
\end{align*}
It is easy to see that $\mathcal{F}$ is a $C^1$-mapping and $\mathcal{F}(0,1,0) = (0,0,0,0)$. We prove that the partial derivative of $\mathcal{F}$ with respect to the variables $(\phi,W)$ is an isomorphism at $(0,1,0)$. The differential at the point $(0,1,0)$ is given by
\begin{align*}
\MoveEqLeft D\mathcal{F}|_{(0,1,0)} \begin{pmatrix} \delta\phi \\ \delta W \end{pmatrix}\\
&=\begin{pmatrix}
\Delta-\frac{1}{8}R-\frac{7}{8}|K|^2 & , & 0 \\
\partial_{\nu} & , & 0\\
 0 & , & \Delta_L\\
 0 & , &L\cdot(\nu,\cdot)
\end{pmatrix}
\begin{pmatrix}\delta\phi \\ \delta W\end{pmatrix}.
\end{align*}
Note that since $\mathcal{F}(0,1,0) = (0,0,0,0)$, we have $R = |K|^2$, and thus $\Delta-\frac{1}{8}R-\frac{7}{8}|K|^2 = \Delta - |K|^2$. One can easily check that $D\mathcal{F}|_{(0,1,0)}$ is invertible.

The implicit function theorem ensures now that for a sufficiently small parameter $\mu$ there exists a pair $(\phi_{\mu},W_\mu)$ such that $\mathcal{G}(\mu,\phi_\mu,W_\mu)=0$.

Define $\hat \phi_\mu=\mu^3 \phi_\mu$ and $\hat W_\mu=\mu^{-2}W_\mu$ by rescaling. Direct calculations show that $(\hat\phi_\mu, \hat W_\mu)$ solves the system \eqref{BMB} for the data $(\hat{\sigma}=\mu^{12} K, \hat\tau = \mu^{-1} \tau,  \Theta_-, \xi)$.
\end{proof}

There exist initial data satisfying the assumptions of Theorem \ref{3-mu}. From the work of Escobar \cite{E}, we know that there exists a conformal factor $\psi>0$ such that $(M,\hat g=\psi^4 g)$ has a constant scalar curvature $R(\hat g)$ and the mean curvature $H_{\hat g}$ of $\partial M$ vanishes. Thus, we can further safely consider a seed manifold $(M,g)$ with $R(g)= \mathrm{const}$ and $H \equiv 0$ on the boundary. Let $\bar \sigma$ be a symmetric TT-tensor of type $(0,2)$. We take particular data $\bar \tau\equiv0$, $\bar\Theta_-\equiv0$, $\bar \xi\equiv0$. It is clear that $W_{\bar \tau}\equiv0$ solves \eqref{Bmomentum} and \eqref{Boundary2}.

In the presence of a boundary, the Yamabe constant is defined as
\[ Y(g,\partial M)= \mathrm{inf}_{u \in C^{\infty}(M), u \not\equiv 0} \frac{\int_M \left( |\nabla u|^2+\frac{1}{8}Ru^2 \right)dv_g + \int_{\partial M} \frac{1}{4} H u^2 d\sigma_g}{\left( \int_M u^6 dv_g \right)^{\frac{1}{3}}}. \]
If $Y(g,\partial M)>0$, $|\bar \sigma|^2>0$, and $(M,g)$ has no conformal Killing vectors, then an argument similar to that of Theorem 4.10 of \cite{Ch04} shows that there is a sub-solution and a super-solution of the equation
\begin{equation}
\Delta \phi -\frac{1}{8}R \phi + \frac{1}{8}|\bar{\sigma}|^2\phi^{-7} = 0\end{equation}
with the Neumann boundary condition $\partial_\nu \phi=0$. Therefore there exists a pair $(\phi_{\bar\tau}, W_{\bar \tau}\equiv0)$ solving system \eqref{BMB} with data $\bar \tau\equiv0$, $\bar\Theta_-\equiv0$, $\bar \xi\equiv0$ and the set $(M,\phi_{\bar\tau}^4g, \phi_{\bar\tau}^{-2}\bar\sigma)$ can be regarded as the assumed CMC vacuum initial data in Theorem \ref{3-mu}.

\section{Conformally covariant split system with the cosmological constant}
\label{withlambda}

Many of the results of preceding sections can be easily generalized to the case with a non-zero cosmological constant $\Lambda$. In what follows we discuss these generalizations. In most cases we will omit the proofs, as they are analogous to those given in Sec.\ \ref{closed}. We will, however, focus on the places, where taking into account a non-zero cosmological constant introduces a change with respect to $\Lambda = 0$ cases, and where some additional assumptions are required.

The Einstein vacuum constraint equations with a cosmological constant $\Lambda$ read
\begin{subequations}\label{constraintslambda}
\begin{align}
R_{\widetilde g} - |K|^2_{\widetilde g} + (\tr_{\widetilde g} K)^2 - 2 \Lambda &= 0, \\
{\mathop{\rm div}}_{\widetilde g} K - d \tr_{\widetilde g} K &= 0.
\end{align}
\end{subequations}
Keeping standard definitions, i.e., $LW$ defined by Eq.\ (\ref{LWdef}) and $K = \frac{\tau}{3}\phi^4 g + \phi^{-2} \sigma + \phi^4 LW$, we get the system
\begin{subequations}\label{MBlambda}
\begin{align}
\Delta \phi -\frac{1}{8}R \phi + \frac{1}{8}|{\sigma}|^2\phi^{-7}+\frac{1}{4} \langle {\sigma}, LW \rangle \phi^{-1} & \nonumber \\
- \left( \frac{1}{12}\tau^2-\frac{1}{8}| LW|^2 - \frac{1}{4} \Lambda \right)\phi^{5} &= 0,\label{hamiltonian2lambda} \\
\nabla_i(LW)^i_j-\frac{2}{3}\nabla_j\tau+6(LW)^i_j\nabla_i\log \phi &= 0. \label{momentum2lambda}
\end{align}
\end{subequations}

\begin{rem}\label{scalinglambda}
Similarly to the scaling symmetry described in Remark \ref{scaling}, system (\ref{MBlambda}) admits the following scaling. Suppose that system (\ref{MBlambda}) has a solution $(\phi, W)$. Set $\hat \phi = \mu^{-\frac{1}{4}} \phi$, $\hat W = \mu^{\frac{1}{2}} W$ for some positive number $\mu \in \mathbb R^+$. Then $(\hat \phi, \hat W)$ satisfy system (\ref{MBlambda}) with the data $\hat \sigma$, $\hat \tau$, and the cosmological constant $\hat \Lambda$ given by $\hat \sigma = \mu^{-1} \sigma$, $\hat \tau = \mu^{\frac{1}{2}} \tau$, $\hat \Lambda = \mu \Lambda$.
\end{rem}

As before, we assume that $(M,g,K)$ already satisfies Eqs.\ (\ref{constraintslambda}). Denote $\bar \tau = \mathrm{tr}_g K$, and $\bar \sigma_{ij} = K_{ij} - \frac{1}{3} \bar \tau g_{ij}$. The following result is a straightforward generalization of Theorem \ref{3}.

\begin{thm}\label{lambda1}
Suppose that we already have vacuum initial data $(M,g,K)$ satisfying Eqs.\ (\ref{constraintslambda}) with $\bar \tau = \mathrm{tr}_g K = \mathrm{const}$. Assume that $(M,g)$ has no conformal Killing vector fields. Assume further that $-|K|^2 + \Lambda \le 0$ on $M$, and $-|K|^2 + \Lambda < 0$ in some region of $M$. There is a small neighborhood of $\bar \tau$ in $W^{1,p}$ such that for any $\tau$ in this neighborhood there exists $(\phi_\tau,W_\tau) \in W^{2,p}_+ \times W^{2,p}$ solving system \eqref{MBlambda} for the data $\bar{\sigma}_{ij}=K_{ij}-\frac{\bar \tau}{3}g_{ij}$ and $\tau$.
\end{thm}

Note that $-|K|^2 + \Lambda = - |\bar \sigma|^2 - \frac{1}{3} \bar \tau^2 + \Lambda$. In order to satisfy the condition $-|K|^2 + \Lambda < 0$ it is enough to require that $- \frac{1}{3} \bar \tau^2 + \Lambda < 0$.

\begin{proof}
The only difference with respect to the proof of Theorem \ref{3} is that now the function $\mathcal H \colon W^{2,p} \to L^p$ is defined by
\[ \delta \phi \mapsto \left[ \Delta - \frac{1}{8} R - \frac{7}{8} |\bar \sigma|^2 - \frac{5}{4} \left( \frac{\bar \tau^2}{3} - \Lambda \right)  \right] \delta \phi. \]
For $\phi \equiv 1$, $W \equiv 0$, we have from Eq.\ (\ref{hamiltonian2lambda})
\[ -\frac{1}{8}R + \frac{1}{8} |\bar \sigma|^2 - \frac{1}{4}\left( \frac{\bar \tau^2}{3} - \Lambda \right) = 0.  \]
Thus
\[ \Delta - \frac{1}{8} R - \frac{7}{8} |\bar \sigma|^2 - \frac{5}{4} \left( \frac{\bar \tau^2}{3} - \Lambda \right) = \Delta - |\bar \sigma|^2 - \frac{1}{3} \bar \tau^2 + \Lambda, \]
which is a negatively definite map, provided that
\[ - |\bar \sigma|^2 - \frac{1}{3} \bar \tau^2 + \Lambda \le 0 \]
on $M$, and
\[ - |\bar \sigma|^2 - \frac{1}{3} \bar \tau^2 + \Lambda < 0 \]
in some region of $M$. Noting that $- |\bar \sigma|^2 - \frac{1}{3} \bar \tau^2 = - |K|^2$, we conclude that $\mathcal H$ has a trivial kernel.
\end{proof}

Theorem \ref{exi} can be generalized in two ways. One can generate data corresponding to $\Lambda \neq 0$ from a seed-initial data with $\mathrm{tr}_g K = 0$ and $\Lambda = 0$, i.e., initial data satisfying Eqs.\ (\ref{originalconstraints}). The other possibility is to start with seed-initial data that already satisfy constraints (\ref{constraintslambda}) with $\mathrm{tr}_g K = 0$ and some nonzero value of $\Lambda$. These data can be then used to generate another set of initial data corresponding to some mean curvature $\hat \tau \neq 0$ and a different value of the cosmological constant $\hat \Lambda$. We will discuss these two cases in what follows.

\begin{thm}\label{exi_lambda_a}
Suppose that $(M, g, K)$ satisfy the constraint equations (\ref{originalconstraints}), and $\mathrm{tr}_g K = 0$. Let $K \neq 0$ in some region, and let $(M,g)$ admit no conformal Killing vector fields. Given any $\tau \in W^{1,p}$ and $\Lambda$, there is a positive constant $\eta > 0$ such that for any $\mu \in (0,\eta)$ there exists a solution $(\phi, W) \in W^{2,p}_+ \times W^{2,p}$ of system (\ref{MBlambda}) for the data $\hat \sigma = \mu^{12} K$, $\hat \tau = \mu^{-1} \tau$, and the cosmological constant $\hat \Lambda = \mu^{-2} \Lambda$.
\end{thm}

\begin{proof}
The function $\mathcal{G}$ appearing in the proof of Theorem \ref{exi} is now defined as
\begin{align*}
\MoveEqLeft \mathcal{G} \colon \mathbb{R} \times W^{2,p}_+ \times W^{2,p} \rightarrow L^p\times L^p, \\
& \begin{pmatrix} \mu \\ \phi\\ W \end{pmatrix} \mapsto \begin{pmatrix}
\Delta \phi -\frac{1}{8}R \phi + \frac{1}{8}|K|^2\phi^{-7}+\frac{1}{4}\mu^4 \langle K, LW \rangle \phi^{-1} - \left( \frac{\mu^{10} \tau^2}{12} - \frac{\mu^8}{8}| LW|^2 - \frac{\mu^{10}}{4} \Lambda \right) \phi^{5} \\
\nabla_i(LW)^i_j-\frac{2}{3}
\mu\nabla_j\tau+6(LW)^i_j\nabla_i\log \phi
\end{pmatrix}.
\end{align*}
The assumptions of the theorem imply that $\mathcal G(0,1,0) = (0,0)$. The differential of $\mathcal G$ with respect to $(\phi, W)$ at the point $(0, \bar \phi \equiv 1, \bar W \equiv 0)$ reads
\[ D\mathcal{G}|_{(0, 1, 0)} \begin{pmatrix} \delta\phi \\ \delta W \end{pmatrix}
 = \begin{pmatrix}
\Delta-\frac{1}{8}R-\frac{7}{8}|K|^2 & , & 0\\
 0 & , & \Delta_L
\end{pmatrix}
\begin{pmatrix}\delta\phi \\ \delta W\end{pmatrix}, \]
where $\Delta_L W=\mbox{div}_g(LW)$. There is no change with respect to the proof of Theorem \ref{exi} at this point. We conclude that for sufficiently small $\mu$, there exist $\phi_\mu$ and $W_\mu$ solving $\mathcal G(\mu, \phi_\mu, W_\mu) = 0$.

The scaling argument works analogously to the one used in Theorem \ref{exi}. We define $\hat \phi_\mu = \mu^3 \phi_\mu$, $\hat W_\mu = \mu^{-2} W_\mu$. Then the pair $(\hat \phi_\mu, \hat W_\mu)$ solves system (\ref{MBlambda}) for the data $(\hat \sigma = \mu^{12} K, \hat \tau = \mu^{-1} \tau)$, and the cosmological constant $\hat \Lambda = \mu^{-2} \Lambda$.
\end{proof}

\begin{thm}\label{exi_lambda_b}
Suppose that $(M, g, K)$ satisfy the constraint equations (\ref{constraintslambda}) with a non-zero cosmological constant $\Lambda$, and $\mathrm{tr}_g K = 0$. Assume that $(M,g)$ admit no conformal Killing vector fields. Assume further that $- |K|^2 + \Lambda \le 0$ on $M$ and $-|K|^2 + \Lambda < 0$ in some region of $M$. Given any $\tau \in W^{1,p}$, there is a positive constant $\eta > 0$ such that for any $\mu \in (0,\eta)$ there exists a solution $(\phi, W) \in W^{2,p}_+ \times W^{2,p}$ of system (\ref{MBlambda}) for the data $\hat \sigma = \mu^{12} K$, $\hat \tau = \mu^{-1} \tau$, and the cosmological constant $\hat \Lambda = \mu^{-12} \Lambda$.
\end{thm}

\begin{proof}
The proof is essentially a variant of the proof of Theorem \ref{exi_lambda_a}. The map $\mathcal G$ is defined as
\begin{align*}
\MoveEqLeft \mathcal{G} \colon \mathbb{R} \times W^{2,p}_+ \times W^{2,p} \rightarrow L^p\times L^p, \\
& \begin{pmatrix} \mu \\ \phi\\ W \end{pmatrix} \mapsto \begin{pmatrix}
\Delta \phi -\frac{1}{8}R \phi + \frac{1}{8}|K|^2\phi^{-7}+\frac{1}{4}\mu^4 \langle K, LW \rangle \phi^{-1} - \left( \frac{\mu^{10} \tau^2}{12} - \frac{\mu^8}{8}| LW|^2 - \frac{1}{4} \Lambda \right) \phi^{5} \\
\nabla_i(LW)^i_j-\frac{2}{3}
\mu\nabla_j \tau+6(LW)^i_j\nabla_i\log \phi
\end{pmatrix},
\end{align*}
i.e., with no $\mu$-dependent coefficient in front of the $\Lambda$ term. Injectivity of $D\mathcal G$ (with respect to $(\phi, W)$) at the point $(\mu = 0, \bar \phi \equiv 1, \bar W \equiv 0)$ is ensured by requiring that $-|K|^2 + \Lambda \le 0$ on $M$, and $-|K|^2 + \Lambda < 0$ in some region of $M$, together with the condition that $(M,g)$ admits no conformal Killing vector fields (cf. the proof of Theorem \ref{lambda1}). The implicit function theorem then yields that for a sufficiently small $\mu$, there exist $\phi_\mu$ and $W_\mu$ solving $\mathcal G(\mu, \phi_\mu, W_\mu) = 0$.

The scaling argument can be now used to infer the existence of solutions of system (\ref{MBlambda}), but at a cost of rescaling both $K$ and $\Lambda$. We define $\hat \phi_\mu = \mu^3 \phi_\mu$, $\hat W_\mu = \mu^{-2} W_\mu$. Then $\hat \phi_\mu$ and $\hat W_\mu$ solve system (\ref{MBlambda}) for the data $\hat \sigma = \mu^{12} K$, $\hat \tau = \mu^{-1} \tau$ and the cosmological constant $\hat \Lambda = \mu^{-12} \Lambda$.
\end{proof}

In the same spirit it is possible to generalize the results of Sec.\ \ref{boundarycase} referring to manifolds with black hole boundaries. We omit this discussion here; the corresponding theorems can be easily formulated and proved by combining the techniques used in this section and in Sec.\ \ref{closed}.

\section{Numerical examples}
\label{numerics}

In this section we give simple numerical examples of solutions of system (\ref{MB}) with non-constant $\tau$. Our examples are inspired by a recent numerical study of non-CMC solutions obtained in the framework of the standard conformal approach \cite{Dilts_numerical}. Numerical techniques used in this paper are essentially adapted from \cite{Mach_Knopik}. We should emphasize that the spectrum of solutions of systems introduced in this section could be complex and should probably be a subject of a separate investigation, similar to \cite{Dilts_numerical} or \cite{Mach_Knopik}. The main aim of providing our numerical examples in this work is to demonstrate the applicability of the conformally covariant split method in numerical studies.

We start by considering the manifold $M = \mathbb S^1 \times \mathbb S^2$ endowed with the metric
\[ g = d \alpha^2 + d\theta^2 + \sin^2 \theta d \varphi^2. \]
Here $\alpha$ is the coordinate on $\mathbb S^1$, and $(\theta,\varphi)$ are coordinates on $\mathbb S^2$. The scalar curvature associated with metric $g$ reads $R = 2$. We assume the tensor $\sigma$ in the form
\[ \sigma_{ij} = b \left( \begin{array}{ccc}
-2 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & \sin^2 \theta
\end{array} \right) \]
[in coordinates $(\alpha, \theta, \varphi)$], where $b$ is a constant. An elementary calculation shows that $\sigma$ is both trace-less and divergence-free. We have $|\sigma|^2 = 6b^2$.

For the vector field $W^i$ we assume an ansatz $W^i = (f(\alpha),0,0)$. It follows that
\[ (LW)_{ij} = \frac{2}{3} f^\prime \left( \begin{array}{ccc}
2 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & - \sin^2 \theta
\end{array} \right), \]
where the prime denotes the derivative with respect to $\alpha$. It is easy to show that $|LW|^2 = \frac{8}{3} (f^\prime)^2$ and $\langle \sigma, LW \rangle  = -4 b f^\prime$. We also have $\nabla_i (LW)^i_j = \left( \frac{4}{3} f^{\prime \prime}, 0, 0 \right)$.

We will assume further that the conformal factor $\phi$ depends only on $\alpha$. This gives
\[ (LW)^i_j \nabla_i \log \phi = \left( \frac{4}{3} f^\prime (\log \phi)^\prime, 0, 0 \right). \]
It follows from Eq.\ (\ref{momentum2}) that the above choices are only consistent with $\partial_\theta \tau = \partial_\varphi \tau = 0$. From now on we assume that the mean curvature $\tau$ depends only on $\alpha$. Equation (\ref{momentum2}) then yields
\begin{equation}
\label{momentumf}
f^{\prime \prime} - \frac{1}{2} \tau^\prime + 6 \phi^{-1} f^\prime \phi^\prime = 0.
\end{equation}

Equation (\ref{hamiltonian2}) can be written as
\begin{equation}
\label{hamiltonianf}
\phi^{\prime \prime} - \frac{1}{4} \phi + \frac{3}{4} b^2 \phi^{-7} - b f^\prime \phi^{-1} -  \frac{1}{3}\left( \frac{\tau^2}{4} - (f^\prime)^2 \right) \phi^5 = 0.
\end{equation}

Note that the function $f$ appears in Eqs.\ (\ref{momentumf}) and (\ref{hamiltonianf}) only through its derivatives. Consequently, we will now define $w = f^\prime$, and write Eqs.\ (\ref{momentumf}) and (\ref{hamiltonianf}) as
\begin{subequations}\label{systemS1S2}
\begin{align}
\phi^{\prime \prime} - \frac{1}{4} \phi + \frac{3}{4} b^2 \phi^{-7} - b w \phi^{-1} -  \frac{1}{3}\left( \frac{\tau^2}{4} - w^2 \right) \phi^5 &= 0, \label{systemS1S2hamiltonian} \\
w^\prime - \frac{1}{2} \tau^\prime + 6 \phi^{-1} w \phi^\prime &= 0. \label{systemS1S2momentum}
\end{align}
\end{subequations}
The above set of equations has to be solved assuming that $\phi$ and $w$ are both functions on $\mathbb S^1$. The condition that $w = f^\prime$, where $f$ is also a regular function on $\mathbb S^1$ yields, in particular, that
\begin{equation}
\label{whasnoconstantterm}
\int_0^{2\pi} w d\alpha = 0.
\end{equation}

System (\ref{systemS1S2}) has an elementary solution for $\tau = \mathrm{const}$. The condition $\tau^\prime = 0$ and Eq.\ (\ref{systemS1S2momentum}) give $w = C \phi^{-6}$, where $C$ is an integration constant. Since $\phi^{-6}$ is non-negative, it follows from Eq.\ (\ref{whasnoconstantterm}) that in fact $C = 0$ and $w \equiv 0$. Equation (\ref{systemS1S2hamiltonian}) is now reduced to
\begin{equation}
\label{systemS1S2zerow}
\phi^{\prime \prime} - \frac{1}{4} \phi + \frac{3}{4} b^2 \phi^{-7} -  \frac{1}{12} \tau^2 \phi^5 = 0.
\end{equation}
Let $b \neq 0$. It is easy to see that the above equation admits a positive solution $\phi = \phi_0 = \mathrm{const}$. Assuming $\phi = \phi_0$, we get from Eq.\ (\ref{systemS1S2zerow})
\[ \mathrm{LHS} \equiv -x^2 + 3 b^2 - \frac{1}{3} \tau^2 x^3 = 0, \]
where $x = \phi_0^4$. Clearly, for $x = 0$ we have $\mathrm{LHS} = 3b^2 \ge 0$. For $x \to \infty$ one gets $\mathrm{LHS} \to - \infty$. Thus there exist a positive solution $\phi = \phi_0$ for all constants $b \neq 0$ and $\tau$. It can be easily expressed by a Cardano formula. It can be also shown that a positive solution to Eq.\ (\ref{systemS1S2zerow}) (not necessarily constant) is unique \cite{Isenberg}. Thus the solution $\phi = \phi_0$ is the only positive one.

We now turn to the non-CMC case, and construct simple numerical solutions of system (\ref{systemS1S2}). We employ the following spectral numerical scheme. We express both functions $\phi$ and $w$ truncating a Fourier series:
\[ \phi = \frac{a_0}{2} + \sum_{k = 1}^N a_k \cos (k \alpha) + \sum_{k=1}^N b_k \sin(k \alpha), \]
\[ w = \sum_{k = 1}^N c_k \cos (k \alpha) + \sum_{k=1}^N d_k \sin(k \alpha). \]
Note that there is no constant term in the expression for $w$, due to condition (\ref{whasnoconstantterm}). These expansions are substituted into Eqs.\ (\ref{systemS1S2}). Equation (\ref{systemS1S2hamiltonian}) is then projected on functions $\cos(k \alpha)$, $k = 0, \dots, N$ and $\sin(k \alpha)$, $k = 1, \dots, N$. Equation (\ref{systemS1S2momentum}) is projected on  $\cos(k \alpha)$ and $\sin(k \alpha)$, $k = 1, \dots, N$. This yields $4N + 1$ nonlinear conditions on the coefficients $a_k$, $k = 0, \dots, N$ and $b_k$, $c_k$, $d_k$ with $k = 1, \dots N$. These conditions are then solved using a standard Newton-Raphson procedure.

The integrals
\[ \frac{1}{\pi} \int_0^{2 \pi} d \alpha f(\alpha) \cos (k \alpha), \quad \frac{1}{\pi} \int_0^{2 \pi} d \alpha f(\alpha) \sin (k \alpha) \]
are performed using standard formulas
\[ \frac{1}{\pi} \int_0^{2 \pi} d \alpha f(\alpha) \cos (k \alpha) = \frac{2}{\tilde N} \sum_{j = 1}^{\tilde N} f(\alpha_j) \cos (k \alpha_j), \]
\[ \frac{1}{\pi} \int_0^{2 \pi} d \alpha f(\alpha) \sin (k \alpha) = \frac{2}{\tilde N} \sum_{j = 1}^{\tilde N} f(\alpha_j) \sin (k \alpha_j), \]
where $\tilde N = 2N +1$, and $\alpha_j = 2 \pi (j -1)/\tilde N$, $j = 1, \dots, \tilde N$. Our numerical calculations were performed using Wolfram Mathematica \cite{mathematica}.

In the following examples we assume $\tau = 1 + a \cos(\alpha)$. This choice allows one to search for solutions for $\phi$ and $w$, for which $b_k = d_k = 0$, $k = 1, \dots, N$. Note that in such a case Eq.\ (\ref{systemS1S2momentum}) has to be projected on the set of functions $\sin(k \alpha)$, $k = 1, \dots, N$.

\begin{figure}[t]
\includegraphics[width=0.6\textwidth]{S1S2_a12_b1.pdf}
\caption{\label{S1S2b1a12}A sample numerical solution of system (\ref{systemS1S2}) with $\tau = 1 + a \cos(\alpha)$, $a = 1/2$, $b = 1$. The solid line depicts the conformal factor $\phi$. The dashed line shows the graph of $w$.}
\end{figure}

\begin{figure}[t]
\includegraphics[width=0.6\textwidth]{S1S2_a2_b3.pdf}
\caption{\label{S1S2b3a2}Same as in Fig.\ \ref{S1S2b1a12}, but for $a = 2$, $b = 3$.}
\end{figure}

Sample solutions obtained for system (\ref{systemS1S2}) with $\tau = 1 + a \cos(\alpha)$ are shown in Figs.\ \ref{S1S2b1a12} and \ref{S1S2b3a2}. Figure \ref{S1S2b1a12} shows a solution corresponding to $a = 1/2$ and $b = 1$. Note that in this case $\tau$ is strictly positive. Figure \ref{S1S2b3a2} depicts a solution obtained for $a = 2$, $b = 3$. This means that the mean curvature $\tau$ changes its sign.

As usual for spectral methods, the minimum acceptable value of $N$ depends on the equation to be solved, and the desired accuracy. In the specific examples shown in Figs.\ \ref{S1S2b1a12} and \ref{S1S2b3a2} we chose $N = 25$ and $N = 55$, respectively. The quality of solutions can be assessed by computing the left-hand sides of Eqs.\ (\ref{systemS1S2}) for the obtained numerical solutions. For the solution shown in Fig.\ \ref{S1S2b1a12} these values are of the order of $10^{-15}$. For the solution depicted in Fig.\ \ref{S1S2b3a2} we get the left-hand sides of Eqs.\ (\ref{systemS1S2}) not exceeding a value of the order of $10^{-14}$.

\begin{figure}[th]
\includegraphics[width=0.6\textwidth]{S1T2_a12_b1.pdf}
\caption{\label{S1T2b1a12}A sample numerical solution of system (\ref{systemS1T2}) with $\tau = 1 + a \cos(\alpha)$, $a = 1/2$, $b = 1$. The solid line depicts the conformal factor $\phi$. The dashed line shows the graph of $w$.}
\end{figure}

\begin{figure}[th]
\includegraphics[width=0.6\textwidth]{S1T2_a1_b1.pdf}
\caption{\label{S1T2b1a1}Same as in Fig.\ \ref{S1T2b1a12}, but for $a = b = 1$.}
\end{figure}

There is another example that can be obtained in the similar fashion, and it is again motivated by a system analyzed in \cite{Dilts_numerical}. We take $M = \mathbb S^1 \times \mathbb T^2$, where the torus $\mathbb T^2 = \mathbb S^1 \times \mathbb S^1$. Let $(\alpha, \beta, \gamma)$ be the coordinates on $M$, each spanning the range $[0,2 \pi)$. We assume the metric $g$ to be flat, $g_{ij} = \mathrm{diag}(1,1,1)$, and $\sigma_{ij} = b \, \mathrm{diag}(-2, 1, 1)$, where $b$ is a constant. Clearly, $\sigma$ is a TT tensor. As before, $|\sigma|^2 = 6 b^2$. Let us also assume that $W^i = (f(\alpha),0,0)$. It follows that $(LW)_{ij} = \frac{2}{3}f^\prime \mathrm{diag}(2,-1,-1)$, $|LW|^2  = \frac{8}{3}(f^\prime)^2$, $\langle \sigma, LW \rangle  = - 4 b f^\prime$. Assuming that $\phi$ and $\tau$ depend only on $\alpha$, we also have $\nabla_i (LW)^i_j = \left(\frac{4}{3}f^{\prime \prime},0,0 \right)$, and
\[ (LW)^i_j \nabla_i \log \phi = \left( \frac{4}{3} f^\prime (\log \phi)^\prime, 0, 0\right). \]
It turns out that systems (\ref{MB}) written for the two cases $\mathbb S^1 \times \mathbb S^2$ and $\mathbb S^1 \times \mathbb T^2$ differ only in the term proportional to the scalar curvature $R$. We have $R = 0$ for the $\mathbb S^1 \times \mathbb T^2$ case. The analogue of Eqs.\ (\ref{systemS1S2}) can be now written as
\begin{subequations}\label{systemS1T2}
\begin{align}
\phi^{\prime \prime} + \frac{3}{4} b^2 \phi^{-7} - b w \phi^{-1} -  \frac{1}{3}\left( \frac{\tau^2}{4} - w^2 \right) \phi^5 &= 0, \label{systemS1T2hamiltonian} \\
w^\prime - \frac{1}{2} \tau^\prime + 6 \phi^{-1} w \phi^\prime &= 0, \label{systemS1T2momentum}
\end{align}
\end{subequations}
where again $w \equiv f^\prime$.

If $\tau^\prime = 0$, we again obtain a solution with $w = 0$. In this case Eq.\ (\ref{systemS1T2hamiltonian}) reads
\begin{equation}
\label{S1T2constt}
\phi^{\prime \prime} + \frac{3}{4} b^2 \phi^{-7} - \frac{\tau^2}{12} \phi^5 = 0.
\end{equation}
If in addition $\tau \neq 0$, there is a unique positive solution $\phi = \phi_0 = \mathrm{const}$, where
\[ \phi_0 = \left| \frac{3b}{\tau} \right|^\frac{1}{6}. \]
If, in turn, $\tau = 0$, we get from Eq.\ (\ref{S1T2constt})
\[ \phi^{\prime \prime} = - \frac{3}{4}b^2 \phi^{-7}. \]
It follows from the maximum principle that the above equation has no positive solution for $b \neq 0$.

Figures \ref{S1T2b1a12} and \ref{S1T2b1a1} show examples of solutions of system (\ref{systemS1T2}) obtained for $\tau = 1 + a \cos(\alpha)$. Figure \ref{S1T2b1a12} corresponds to $a = 1/2$, $b = 1$ (the same set of parameters, as the one chosen for the solution of system (\ref{systemS1S2}) shown in Fig.\ \ref{S1S2b1a12}). Another solution, obtained assuming $a = b = 1$, is shown in Fig.\ \ref{S1T2b1a1}. In both cases we set the series cutoff parameter $N = 25$. The precision with which Eqs.\ (\ref{systemS1T2}) are satisfied is of the order of $10^{-15}$.

Both systems (\ref{systemS1S2}) and (\ref{systemS1T2}) can be easily generalized to include the cosmological constant. System (\ref{systemS1S2}) ($M = \mathbb S^1 \times \mathbb S^2$) with the cosmological constant reads
\begin{subequations}\label{systemS1S2lambda}
\begin{align}
\phi^{\prime \prime} - \frac{1}{4} \phi + \frac{3}{4} b^2 \phi^{-7} - b w \phi^{-1} - \left( \frac{\tau^2}{12} - \frac{1}{3} w^2 - \frac{1}{4} \Lambda \right) \phi^5 &= 0, \label{systemS1S2hamlambda} \\
w^\prime - \frac{1}{2} \tau^\prime + 6 \phi^{-1} w \phi^\prime &= 0. \label{systemS1S2momlambda}
\end{align}
\end{subequations}
System (\ref{systemS1T2}) ($M = \mathbb S^1 \times \mathbb T^2$) can be generalized as
\begin{subequations}\label{systemS1T2lambda}
\begin{align}
\phi^{\prime \prime} + \frac{3}{4} b^2 \phi^{-7} - b w \phi^{-1} - \left( \frac{\tau^2}{12} - \frac{1}{3} w^2 - \frac{1}{4} \Lambda \right) \phi^5 &= 0, \label{systemS1T2hamlambda} \\
w^\prime - \frac{1}{2} \tau^\prime + 6 \phi^{-1} w \phi^\prime &= 0. \label{systemS1T2momlambda}
\end{align}
\end{subequations}

While numerical solutions of the above systems can be easily obtained using our method, the properties of these systems can be remotely different, especially for large $\Lambda > 0$. This can be seen immediately by inspecting the cases with $\tau = \mathrm{const}$ and $w = 0$. In this case system (\ref{systemS1S2lambda}) yields
\[ \phi^{\prime \prime} - \frac{1}{4} \phi + \frac{3}{4} b^2 \phi^{-7} - \left( \frac{\tau^2}{12} - \frac{1}{4} \Lambda \right) \phi^5 = 0. \]
Assume that $b \neq 0$. If $\frac{1}{3}\tau^2 - \Lambda \ge 0$, there exists a unique positive solution. This solution is constant $\phi = \phi_0 = \mathrm{const}$, and it can be obtained as a solution to the cubic equation
\[ - x^2 + 3 b^2 - \left( \frac{\tau^2}{3} - \Lambda \right) x^3 = 0, \]
where $x = \phi_0^4$. If in turn $\frac{1}{3}\tau^2 - \Lambda < 0$, solutions are no longer unique. This case was analyzed in detail in \cite{chrusciel_gicquaud}, where a corresponding bifurcation structure of solutions was also described.

System (\ref{systemS1T2lambda}) with $\tau = \mathrm{const}$ and $w = 0$ yields
\begin{equation}
\label{S1T2lambdaw0}
\phi^{\prime \prime} + \frac{3}{4} b^2 \phi^{-7} - \left( \frac{\tau^2}{12} - \frac{1}{4} \Lambda \right) \phi^5 = 0.
\end{equation}
Assume again that $b \neq 0$. If $\frac{1}{3}\tau^2 - \Lambda > 0$, there exist a unique positive solution $\phi_0$. It is constant, and it is given by
\[ \phi_0 = \left( \frac{3 b^2}{\frac{\tau^2}{3} - \Lambda} \right)^\frac{1}{12}. \]
If $\frac{1}{3}\tau^2 - \Lambda \le 0$, Eq.\ (\ref{S1T2lambdaw0}) has no positive solutions (this can again be easily seen from the maximum principle).

\section*{Acknowledgments}
The authors would like to thank the referees for very useful comments and suggestions. This research was partially supported by the National Natural Science Foundation of China grant No.\ 11671089 and the Polish National Science Centre grant No.\ 2017/26/A/ST2/00530. Part of this work was done while PM was visiting the School of Mathematical Sciences, Fudan University. He would like to thank this institution for the hospitality and financial support.

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text:
Home / Tugs make owner proud
Tugs make owner proud
By	Ken Hocke on	 October 25, 2011
Last week I journeyed to Main Iron Works in Houma, La., where the venerable shipyard is building two new 100'x34′ tugs — Chesapeake Coast and Discovery Coast — for Dann Marine Towing in Chesapeake City, Md.
Robert Dann Jr. met me in the galley of the Chesapeake Coast, a look of pride on his face. He took me for a tour of the boat, relishing the opportunity to show me what had caused that look.
This was another of those moments I often write about that raises the question: Would I rather be roaming around in a new, state-of-the-art tugboat that few outside of the shipyard have seen or would I rather be in the office?
No contest. (That's always the answer.)
When we got to the wheelhouse at the end of the tour, I realized that Dann had enjoyed it as much as, or even more so, than I. Talk about a guy who loves his work. He also has great respect for Main Iron Works. "We've had a relationship with them for about 40 years," he said. "Coming here, I know we're going to get a great boat."
Dann said he expected to take the Chesapeake Coast on sea trials sometime this week. Sure, it's work, but for Dann, it's also fun. "These two will go into our fleet in Maryland, but they can operate anywhere," he said.
Look for Kathy Bergren Smith's article about Dann Towing and the Chesapeake Coast and Discovery Coast in the upcoming December issue of WorkBoat.
Ken Hocke
Ken Hocke has been the senior editor of WorkBoat since 1999. He was the associate editor of WorkBoat from 1997 to 1999. Prior to that, he was the editor of the Daily Shipping Guide, a transportation daily in New Orleans. He has written for other publications including The Times-Picayune. He graduated from Louisiana State University with an arts and sciences degree, with a concentration in English, in 1978.

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text:
LibertyShares
K2 Advisors : Why We Like Activist Hedge Strategies
Given the unpredictability of today's financial markets, many investors are looking to reduce the impact of market volatility on their portfolios. Hedge-fund strategies—a type of alternative investment strategy—may help by potentially offering additional diversification, new sources of return and reduced risk. K2 Advisors seeks to add value through active portfolio management, tactical allocation and diversification across four main hedge strategies: long-short equity, relative value, global macro and event driven. Here, K2's Rob Christian discusses why the team favors activist hedge strategies as a subset of the event-driven strategy, and how attitudes toward activist investing have changed.
Robert Christian
Co-Head of Investment Research and Management
K2 Advisors®
Rob Christian
K2 Advisors®, Franklin Templeton Solutions®
When describing hedge-fund performance—and more specifically hedge-fund alpha1 —words like "generating," "producing," and "creating" are often used. These expressions are somewhat misleading, in my view. They imply that hedge funds, by their own sheer will and determination, can somehow conjure alpha on demand. For the majority of hedge funds this is not the case. Technically speaking alpha is not generated, nor is it produced, manufactured or built. Indeed, for the majority of successful hedge-fund managers, the most that can be expected of alpha is that a fair amount of it may be captured over the course of an investment horizon. That is the key—alpha is largely captured. A good market day for a hedge fund means that the fleeting alpha ghost has had the misfortune of stepping into a cleverly constructed trap—if only momentarily. The nature of alpha is elusive, and that is why investors will pay for it.
The Exception to the Rule
Like all good rules however, there are exceptions. In my view, activist hedge funds represent one such exception. It could be said that activist managers in some ways represent the only strategy that generates alpha to some degree. While not always successful, activist funds seek to unlock "hidden" value in the companies they invest in.
They look for stocks that are trading at a discount to intrinsic value, and then instead of waiting passively for the market to recognize this perceived hidden value, the activist hedge fund will proactively work to expose—or elevate—that value to the market. In this way the strategy quite literally attempts to generate or harvest alpha proactively by promoting the catalysts that may increase stock value.
Activism: A Primer
Briefly, the strategy of activist investing involves taking a minority position in a public company, with the goal of helping unlock said company's value by influencing management to modify its capital allocation activities. That is, run their businesses in a more efficient and profitable manner. Tactics include financial restructurings, operational turnarounds and/or strategic initiatives.
Implicit in the strategy is the notion that the targeted company is undervalued and generating earnings below its potential (based on the hedge fund manager's assessment) and the activist has a plan to help unlock this potential.
More than three decades ago, some of the same players in the activist space today may have grabbed headlines under different—perhaps less flattering—characterizations as part of the leveraged buy-out (LBO) frenzy of the late 1980s and early 1990s. Indeed, some of the same iconic names that today are held up as defenders of the average investor, often praised for holding management teams accountable, were in days of yore condemned as "robber barons," "corporate raiders" and "barbarians at the gate."
Not anymore. The "barbarians" of the 1980's no longer crash the gates as much as they wish to repair them.
The game today is more about diplomacy than aggression, as activist investors have found it easier (and clearly more profitable) to work with corporate management to help implement positive change, versus trying to take full control of a company and possibly liquidate its prized assets. In this way, the spoils are shared among many.
Today it is about building venerable brands versus destroying them, and activists no longer need to take over a company to influence action. Often it is possible for a hedge fund with a minority stake (perhaps just over 5%) to initiate change with cooperation from management, possibly by nominating a member of the hedge fund's team to the company's board of directors.
As a high-profile activist stated to Bloomberg, "the least effective way to do activism these days is to come in with a big club. The best result is to have a board hear your ideas, realize they are good and then acquiesce…we prefer to work with management in making changes that increase shareholder value, but sometimes you have to go to the mat."2
Commonly Deployed Activist Tactics
Create value by restructuring a company's balance sheet, altering the leverage structure in an attempt to impact/unlock value. Tactics include reducing excess cash by increasing dividends, share buyback programs, retiring or refinancing outstanding debt.
Operational Changes
Create value by increasing operating margins. Tactics may include modifying product prices, focusing on core brands/products, changing raw materials suppliers/distribution channels, streamlining logistics, eliminating unprofitable franchises or business units and restructuring management compensation.
Strategic Moves
Create value by creating strategic moves, which may include merger-and-acquisition & transactions, joint ventures, and/or the spin-off of incompatible divisions or product lines. Activist hedge funds often maintain a wide network of corporate relationships that may be helpful in implementing such maneuvers.
Shareholder Support
Perhaps the most significant of all the factors contributing to the rise of activist investing today is the role and support of the shareholder.
All of the tools once used to avert activists in the boardroom, from "poison pills" that derail takeovers to golden parachutes that enrich management, are today generally viewed as negatives for shareholders.
Investor sentiment over excessive compensation levels and lax oversight has shifted moods in favor of activists and away from corporate management. Many investors view the lure of stock options and other cozy perks often available to senior management in public companies as a disincentive to long-term strategic thinking.
Attractive Opportunity
The market and economic environment is perhaps more favorable to activist investment strategies today than it has been in some time. With the outlook for growth muted, corporate balance sheets flush with cash, financing rates still very low, and an increasingly hospitable public attitude toward activist investors—the opportunity set appears robust.
Huge cash balances may create a significant drag on corporate earnings, and many companies will likely choose to allocate this cash through strategic M&A or via stock buybacks and dividends—both significant areas of focus for activist funds.
Management teams may lack the skill set or incentives required to successfully undertake such strategic maneuvers, and this is where we believe hedge funds may find opportunity.
In addition, there remains a substantial amount of uninvested private-equity (PE) capital on the sidelines, much of which is likely to be invested over the next several years (as investment periods expire). We believe this activity should increase opportunities for activist investors to participate in buyouts and other strategic acquisition deals.
Some Final Thoughts on Alpha
When thinking about the unique value I see in activist hedge-fund strategies, I find it helpful to remind myself of the nature and composition of alpha. Alpha is a byproduct of an inefficient market. Mathematically speaking, alpha represents the abnormal rate of return on a security or portfolio in excess of what would be predicted by an equilibrium model like the capital asset pricing model (CAPM). So it is the coefficient—or residual of the expected return of a security—based on the security's market sensitivity, or beta.
In an efficient market, the expected value of the alpha coefficient would be zero; however, we believe markets are naturally inefficient. As such, if a security returns more than what would be expected given its beta (market sensitivity or risk), it has positive alpha. If it returns less than its beta predicts, it has negative alpha. In other words, alpha is the portion of return that is the result of factors other than exposure to the market. Activist hedge funds seek to not only capture, but cultivate them.
Learn more about K2's strategy and watch videos featuring K2's Brooks Ritchey on our website.
Comments, opinions and analyses expressed herein are for informational purposes only and should not be considered individual investment advice or recommendations to invest in any security or to adopt any investment strategy. Because market and economic conditions are subject to rapid change, comments, opinions and analyses are rendered as of the date of the posting and may change without notice. The material is not intended as a complete analysis of every material fact regarding any country, region, market, industry, investment or strategy.
This information is intended for US residents only.
To get insights from Franklin Templeton Investments delivered to your inbox, subscribe to the Beyond Bulls & Bears blog.
For timely investing tidbits, follow us on Twitter @FTI_US and on LinkedIn.
All investments involve risks, including possible loss of principal. The identification of attractive investment opportunities is difficult and involves a significant degree of uncertainty and there is no assurance any such alternative investment strategies will be successful. An investment in these strategies is subject to various risks, such as those market risks common to entities investing in all types of securities, including market volatility. It is always possible that any trade could generate a loss if the manager's expectations do not come to pass.
The market values of securities held in the K2 liquid alternatives portfolios will go up or down, sometimes rapidly or unpredictably Hedge strategy outlooks are determined relative to other hedge strategies and do not represent an opinion regarding absolute expected future performance or risk. Conviction sentiment determined by the K2 Research Group is based on a variety of factors and may change from time to time.
For more information on any of K2's alternative strategies funds, contact your financial advisor or download a free prospectus. Investors should carefully consider a fund's investment goals, risks, sales charges and expenses before investing. The prospectus contains this and other information. Please read the prospectus carefully before investing or sending money.
1. Alpha refers to a risk-adjusted measure of the value that an active portfolio manager adds to or subtracts from a portfolio's return.
2. Source: Bloomberg, "The Good Barbarian: How Icahn, Ackman, and Loeb Became Shareholder Heroes," May 14, 2012.
Posted in AlternativesTagged activist investing, alternative investing, alternative investment strategies, hedge funds, K2 Advisors, liquid alts funds, Rob Christian
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This was originally intended for Oculus Rift DK2 for an installation at Art Contemporary Los Angeles in January 2016.
UPDATE: 11/11/17 - I updated it for Oculus CV1 using SDK 1.17 and Ue4. Should work on Vive, too. Let me know!
Put the red dot into the vortex. Totems come out. AI players interact with them and tell you stories. History builds up around you.
This was made for a swivel chair but you can also sit back and reorient yourself with the mouse if you want.


text:
Egy síknegyed vagy kvadráns a koordinátageometriában egy Descartes-féle koordináta-rendszerrel koordinátázott sík két koordinátatengely által határolt része. Maguk a koordinátatengelyek nem részei egyik síknegyednek sem.

A szokásos konvenciók szerint az első kvadráns a jobb felső kvadráns (ahol mindkét koordináta pozitív). Innen elindulva az óramutató járásával szembeni forgásirányban növekvő számokat kapják a síknegyedek. A jelöléshez használhatnak római, vagy arab számokat is, így a számozás rendre I, II, III, IV; illetve 1, 2, 3, 4.

Előfordulhatnak más számozások is.

Kapcsolat a trigonometriával
A trigonometriában a szögfüggvények (szinusz, koszinusz, tangens, kotangens, szekáns, koszekáns) előjele attól függ, hogy a szög melyik síknegyedbe esik.

Egy teljes fordulat alatt egy-egy szögfüggvény két síknegyedben is ugyanazt az előjelet veszi fel. Például, ha , akkor α lehet a harmadik vagy a negyedik negyedben. Tehát π < α < 2 · π illetve 180° < α < 360°

A hajózásban és a geodéziában gyakran van szükség kvadránstáblára, illetve kvadránsalkalmazásra, hogy meghatározzanak egy irányt két pont koordinátái alapján.

A fenti táblázat kibővítve a féltengelyekkel:

Fogászat

A fogászatban egy kvadráns  egy állkapocsfél. Így a teljes fogazat négy kvadránsból áll. A számozás a síknegyedek számozását követi a páciens nézőpontjából: az első negyed a jobb felső állkapocs; a második a bal felső állkapocs; a harmadik a bal alsó állkapocs; a negyedik a jobb alsó állkapocs. Az FDI-fogsémában egy fogat a kvadráns száma és az utána írt azonosító jelöl.

Források

Fordítás

Jegyzetek

Analitikus geometria

redpajama_set_name: RedPajamaWikipedia


In this article you'll discover five ways LinkedIn Publisher can help you grow your business.
Discover five ways to use LinkedIn Publisher to grow your business.
Make sure you include a call to action in each LinkedIn Publisher post, whether it's to get more information from another link, contact you, or sign up for a free ebook or white paper. A call to action that speaks directly to your potential, ideal, or current clients will engage your audience and direct them to your funnel.
Fabrizio Poli, who brokers private jets, offers downloadable white papers to his audience.
Include additional information of value with calls to action in your LinkedIn Publisher posts.
People who are successful at generating leads with LinkedIn Publisher do a combination of their own content and sharing links to external resources that both inform and build their credibility. They also include a phone number, email address, or calendar link at the end of the post, so prospects can easily contact them.
Fabrizio makes himself available in his posts by sharing his contact information and including a shortcut to booking a 15-minute call.
Share a calendar link at the end of your LinkedIn Publisher posts.
Sharing a calendar link can be more effective for conversions than just your contact info alone. People may not be ready to talk to you now, but might be open to scheduling a time to speak with you in the future.
Visuals can make or break whether people click on your LinkedIn Publisher articles. Cohesive, colorful, well-branded images will draw people to your article and drive them to open it.
Human-resources professional Liz Ryan creates original images for every LinkedIn Publisher post for her company Human Workplace. The images are reflective of the brand, and users know by the visual that it's a Liz Ryan post. She uses humor and color to tell the story of her posts, which generates positive sentiment. Plus, readers get a sense of what the article is about just by looking at the image.
Liz Ryan's LinkedIn Publisher images reflect her brand, while giving readers a sense of the article.
Liz's approach to HR is unique in and of itself, and she uses consistently branded visuals (in her background image, within the post, on her website, in her media, and more) to get noticed.
For your LinkedIn Publisher articles, you can use your own photos or stock photos, or create custom graphics with Canva or PicMonkey. If you're able to, consider hiring a graphic artist to create something special just for your brand.
To add an image to the top of your LinkedIn Publisher post, click the camera icon in the header and upload your image. LinkedIn recommends a 700 x 400 image for the header. Be sure to add images throughout your post, as well.
Add images that align with your brand to each LinkedIn post.
If you're already creating valuable content in the form of blog posts, newsletters, and other articles, you can repurpose content for LinkedIn Publisher. But you'll also want to create original content for your LinkedIn readers.
One reason to publish original, unique content on LinkedIn audience is to use it as a testing platform. After you publish original content to LinkedIn Publisher, use the feedback to refine the article before you share it on your blog, as an ebook, or on a different platform.
John White, Chief Marketing Officer of Social Marketing Solutions, offers a nice blend of existing content that he has repurposed for LinkedIn Publisher, as well as original content specifically for his LinkedIn audience. John's most-viewed post, with more than 600,000 views, 600 likes, and 1,900 comments, was repurposed from his master's thesis.
John White repurposed his master's thesis into a LinkedIn post, which got massive visibility and engagement.
Unlike what you hear about blogging, on LinkedIn Publisher the more words the better. The magic number seems to be 900-1,400 words. If you're repurposing a blog post that only has 300 or 400 words, consider expanding your content with more examples and case studies. Or you could combine a few blog posts into one big article.
When creating or repurposing blog content, make use of all of Publisher's formatting features. It's best to start with unformatted copy and then apply headers, block quotes, bold, italics, underline, alignment, and bullets (and numbered lists).
Use formatting features to make your blog posts easy to read.
You can also add hyperlinks, images, and video and embed content with the embed code feature. Don't forget you can add up to three tags to make it easier for people to find your posts.
Add interest to your post with images and video.
The key to creating good content is to have a clear idea of what you have to offer, know your audience and what they need, and then create solutions for them. Remember, good content gets noticed. John has gained 29,000 followers since he started posting. He was at around 1,800 followers before writing his first LinkedIn Publisher post.
You can't just post an article on LinkedIn Publisher and expect people to find it. There's no guarantee you'll attract a large audience. To help improve the visibility of your content and position yourself as a thought leader who garners more business, promote your Publisher content on your other social networks.
When photography expert Matt Sweetwood first started posting on LinkedIn, he was getting 200 to 400 views per article. After he started actively sharing his LinkedIn Publisher posts on Twitter, he drove his average to 2,000 to 6,000 views per post. The increased visibility resulted in more consulting and speaking opportunities for him. Matt also @tags the Twitter handles of people he thinks would be interested in each article.
Actively share your LinkedIn posts on your social networks to increase visibility and reach.
Think of your LinkedIn posts as you would a product you're promoting. Share them on all of your social media sites and share them more than once.
To make it easier for readers to share your Publisher posts, add click to tweet links and LinkedIn sharing links within your posts to make your Publisher articles easy to share.
In order to add a LinkedIn sharing link, you must first publish the article. Then click the icon to share your article on LinkedIn and copy the URL from the pop-up window.
Copy the URL of your LinkedIn post.
Go back and select Edit post. Insert the URL as a hyperlink within the text and save.
Add a hyperlink to make it easier for others to share your articles.
Now it will be easier for readers to share your article with their audiences, increasing your article's reach.
When writing a post for LinkedIn Publisher, think about which thought leaders in your industry might be a good fit for the topic, and consider incorporating them in your article. This tactic is a great way to improve engagement, because the people quoted or referenced in your articles are likely to share them with their own networks.
Author Dave Kerpen, founder and CEO of Likeable Media, frequently promotes thought leaders in his LinkedIn Publisher posts, quoting and sharing their thoughts and recommendations. This makes his articles more interesting to read. For this article, Dave asked influencers to share their insights into how someone could land his or her dream job at their company.
Publish posts that underline your connections.
For your articles, you can get quotes from industry experts, interview influencers, or write compilation posts with your favorite articles or techniques from influencers.
Compile posts with tips from other thought leaders.
For instance, compile articles by others that have already been published into a list post, such as "Top 10 Posts on X" or "5 Tools for Y." Write a paragraph describing what's in each article, add a photo, and then link to the original post.
Another method is to reach out and ask relevant influencers for a quote or even a paragraph on a particular topic. Be sure to include a headshot and link back to their website.
Whether you reference previously published materials or connect with experts directly, let people know when you include them in a LinkedIn Publisher post, so they can share it, too.
Doing compilation posts with industry influencers is an excellent way to expand your network. You spread the word about who you are and what you do to other influencers, as well as their audience.
To stand out on LinkedIn Publisher, you'll want to consistently create valuable content with great visuals. Add appropriate calls to action and then share your posts regularly on all of your social networks.
When people get to know you through your articles, it increases their like, know, and trust factor. And people do business with those people.
What do you think? Do you post on LinkedIn Publisher? What types of articles do you write? Which posts get the best response? How do you stand out? Please share your thoughts in the comments.
Tips on ways you can promote your business using LinkedIn Publisher.


BANGUI, Central African Republic (AFP) - At least 12 people, including children and a pregnant woman, have been killed close to the capital Bangui by Christian vigilantes, a military source said on Wednesday, amid rising fears of sectarian massacres in the Central African Republic.
The vicious attack against the Peuhl Muslim minority, in which a pregnant woman was disembowelled, came as the United Nations is about to approve the deployment of an African Union force to restore order in the lawless country.
"The assailants were seen not far from the camp and they returned in the night to attack the herdsmen who were there with just knives," a military source said of the attack about 95km from the north of Bangui.
"Among the victims were children and a disembowelled pregnant woman. Some managed to escape and there are more than 10 wounded who have been transferred to Bangui," the source added.
The Peuhl people are traditionally pastoralist, and are a minority in the impoverished, landlocked nation of 4.5 million people which was plunged into chaos after the March coup by the mostly Muslim Seleka rebels.
The Seleka have since been disbanded, but rights groups say the motley crew of ex-rebels have taken to looting and burning villages, killing inhabitants who fail to flee.
Locals in the majority Christian country have responded by forming vigilante groups known as "anti-balaka" (anti-machete). The ensuing clashes have taken on a religious undertone, raising fears of sectarian massacres.


Downtempo sample - downward bending dub effect loop, 130 bpm.
Electric guitar bossa nova 130 BPM Bb.
A light elegant and slow bossa nova guitar at 130 BPM in Bb.
Bass guitar bossa nova 130 BPM D.
Smooth lead bass guitar riff in bossa nova style played at 130 BPM in the key of C.
Light Electric guitar bossa nova 130 BPM Bb.
A soft and short three note bossa nova guitar riff at 130 BPM in Bb.
Quick and light bass riff bossa nova style played at 130 BPM in the key of Bb.
Bass bossa nova 130 BPM B.


text:
Fair Game Extra Statehouse
Walters: Scott, Legislature Still Talking Past Each Other
by John Walters
Administration Secretary Susanne Young addresses three Senate committees
Three Vermont Senate committees met Wednesday morning to begin work on the legislature's third version of a budget bill, the first two having been vetoed by Gov. Phil Scott. By day's end, the three panels had approved a proposal that moves slightly in Scott's direction — but only slightly.
"I would not predict a veto," Administration Secretary Susanne Young said after the hearing. "But this new proposal is not substantially different from the last one."
The most recent budget, H.13, would have held residential property tax rates level but would have not stopped a 5.5-cent statutory increase in nonresidential rates from going forward. The committees' new plan reduces the nonresidential rate by one penny.
The Appropriations, Finance and Education committees convened jointly due to severe time constraints. If there's no budget on July 1, state government would be forced to shut down. And so far, the Scott administration has refused to disclose contingency plans for that eventuality.
The administration's top fiscal officials came to the hearing with a new "proposal" consisting of a single-page flowchart. Its essence is this: The governor and legislature would split $33.2 million in unanticipated revenue. Lawmakers could use their half however they wish, and Scott would use his half to lower property tax rates.
Scott would still fully intend to keep property tax rates level, which would require another $14 million, according to his analysis. "The governor is willing to risk that the $14 million will be there" in additional surplus revenue, Young testified. If not, she said, the administration would find enough savings in general state spending to cover the difference by the end of the year.
Related Harsh Words: Dueling Rhetoric on the Budget Battle
Harsh Words: Dueling Rhetoric on the Budget Battle
Committee members were skeptical of the new administration plan, to the extent they addressed it at all. The plan fails to address their fundamental objections: The legislative majority opposes the use of onetime money to fund ongoing programs, it isn't willing to book speculative future savings, it doesn't want to use general fund revenue to lower property tax rates and it's concerned about creating a bigger education fund gap in the next fiscal year.
"What is different about this plan?" Senate Appropriations Committee chair Jane Kitchel (D-Caledonia) asked Stephen Klein, chief of the legislature's Joint Fiscal Office.
"I don't see any change," Klein responded.
"This is the same proposal with a different set of contingencies?" queried Sen. Ginny Lyons (D-Chittenden).
"Correct," Klein replied.
Legislative fiscal analyst Mark Perrault testified that Scott's proposal would lead to a repeat of this year's standoff over property taxes. "Even if you're able to agree on onetime money, you're creating an issue for [fiscal year] 2020," he said.
Scott's plan, he said, would create a shortfall of $50 million to $55 million in the education fund. That would require a property tax rate increase of 7 to 8 cents — even without any increases in school spending at all.
"When I see a $55 million [gap], I ask why the governor is OK with creating a crisis next year," said Senate Education Committee chair Phil Baruth (D/P-Chittenden). "I can't see how anybody with good fiscal management could choose the governor's option."
The responsible course, agreed Senate Finance Committee chair Ann Cummings (D-Washington), would be to collect enough property tax revenue to pay for school budgets approved by voters on Town Meeting Day. But in the end, Cummings and her fellow Dem/Prog caucus members agreed to keep residential tax rates level and give Scott a one-cent reduction in nonresidential rates.
"I'll hold my nose and vote 'yes,'" Cummings said, "in the interest of compromise."
It's an ironic turn of events. Democratic lawmakers are preaching fiscal discipline, while the Republican administration is willing to "risk" future revenue increases and a fiscal cliff in the education fund next year.
The three committees voted to let their three chairs craft a detailed bill that will go before the full Senate on Thursday. Approval is all but certain; it would then go to the House. And face a likely gubernatorial veto.
Seriously: You Serious?
by Bryan Parmelee June 24, 2018
by John Walters June 20, 2018
Walters: Vermont House Fails to Override Scott's Budget Veto
Scott Vetoes Vermont Budget as July 1 Shutdown Looms
by Sasha Goldstein June 15, 2018
Walters: Governor and Legislature Still at Odds Over Taxes
Grand Old Parting? Riot at U.S. Capitol Exposes Rifts in the VTGOP
Vermont's Leaders Take Oaths, Unite in Condemnation of Trump
Vermont Gov. Phil Scott Calls for Trump to 'Resign or Be Removed From Office'
Backstory: Most Justifiable Parking Violation
Fair Game Extra Statehouse Susanne Young Phil Scott Jane Kitchel Ann Cummings Phil Baruth Ginny Lyons Mark Perrault Stephen Klein budget Web Only


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We have been in the forefront of delivering healthcare services in Bangladesh, through the largest Healthcare network of establishments for last 22 years. We are serving millions of people every year. Our Vision is to establish an environment around the healthcare sector through the highest level of innovative and technical support which will add value to our society and act as a driven force of our humanity. With the committed staff, we promise to elevate our own benchmarks in healthcare sector.
Looking ahead, we are determined to achieve our goal of building beautiful Bangladesh.
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Load your family into the 2016 GMC Terrain. With fewer than 25,000 miles on the odometer, this 4 door sport utility vehicle prioritizes comfort, safety and convenience. Smooth gearshifts are achieved thanks to the efficient 4 cylinder engine, and for added security, dynamic Stability Control supplements the drivetrain. Top features include cruise control, variably intermittent wipers, an automatic dimming rear-view mirror, air conditioning, telescoping steering wheel, power door mirrors, remote keyless entry, and power windows. Audio features include an AM/FM radio, and 8 speakers, providing excellent sound throughout the cabin. Passenger security is always assured thanks to various safety features, such as: head curtain airbags, front side impact airbags, traction control, brake assist, ignition disabling, onStar, and 4 wheel disc brakes with ABS. Sophisticated all wheel drive assures superb handling in any weather condition. It also arrives with a Carfax history report, indicating just one previous owner. You will have a pleasant shopping experience that is fun, informative, and never high pressured. Please don't hesitate to give us a call.