text:
\section{Introduction}
Let $(M_1, g_1)$ and $(M_2, g_2)$ be two Riemannian manifolds and $f:M_1\to(0, \infty)$ and $\pi_1:M_1\times M_2\to M_1,~\pi_2:M_1\times M_2\to M_2$, the projections map given by $\pi_1(p, q)=p$ and $\pi_2(p, q)=q$  for any $(p, q)\in M_1\times M_2$. Then, the warped product $M=M_1\times_fM_2$ is the product manifold $M_1\times M_2$ equipped with the Riemannian structure such that
\begin{align}
\label{1}
g(X, Y)=g_1(\pi_{1}{*}X, \pi_{1}{*}Y)+(f\circ\pi_1)^2g_2(\pi_{2}{*}X, \pi_{2}{*}Y)
\end{align}
for any $X, Y$ tangent to $M$, where $*$ is the symbol for the tangent maps. The function $f$ is called the warping function of $M$. In particular a warped product manifold is said to be {\it{trivial}} or Riemannian product manifold if the warping function is constant.

 Let $M=M_1\times _fM_2$ be a warped product. For a vector field $X$ tangent to $M_1$, the lift of $X$ on $M=M_1\times _fM_2$ is the tangent vector field $\tilde X$ on $M=M_1\times _fM_2$ whose value at each $(p, q)$ is the lift $X_p$ to $(p, q)$. Thus, the lift of $X$ is the unique vector field on $M=M_1\times _fM_2$ that is $\pi_{M_1}$-related to $X$ and $\pi_{M_2}$-related to the zero vector field on $M_2$. The set of all such lifts of vector fields on $M_1$ is denoted by $\mathcal{L}(M_1)$. Similarly, we denote by $\mathcal{L}(M_2)$ the lifts of vector fields from vector fields tangent to $M_2$.

 Then for unit vector fields $X, Y\in\mathcal{L}(M_1)$ and $Z\in\mathcal{L}(M_2)$, we have
\begin{align}
\label{2}
\nabla_XZ=\nabla_ZX=X(\ln f)Z,\,\,\,\,g(\nabla_XY, Z)=0
\end{align}
which implies that (\cite{O}, page 210)
\begin{align}
\label{3}
K(X\wedge Z)=\frac{1}{f}\big\{(\nabla_XX)f-X^2f\big\}.
\end{align}
If we choose the local orthonormal frame $e_1,\cdots,e_n$ such that $e_1,\cdots, e_{n_1}$ are tangent to $M_1$ and $e_{n_1+1},\cdots,e_n$ are tangent to $M_2$, then we have
\begin{align}
\label{4}
\frac{\Delta f}{f}=\sum_{i=1}^{n_1}K(e_i\wedge e_j)
\end{align}
for each $j=n_1+1,\cdots, n$. For the most up-to-date survey on warped product manifolds and submanifolds, we refer to B.-Y. Chen's books \cite{book,book17} and his survey article \cite{C5}.

Recently, M.-I. Munteanu established an inequality in  \cite{Mun} for the squared norm of the second fundamental form of a contact CR-warped product submanifold in Sasakian space form along a similar line of B.-Y. Chen \cite{C3,C4}. Further, a similar inequality has been obtained for contact CR-warped products in Kenmotsu space forms  by Arslan et al. in \cite{Ar}. On the other hand,  warped product submanifolds of cosymplectic manifolds were studied in  \cite{KK}, \cite{MK} and \cite{U2,U3,U4}.

Motivated by these work done in this spirit, we establish in this paper  the following inequality.\\

\begin{theorem} Let $\tilde M(c)$ be a $(2m+1)$-dimensional cosymplectic space form with constant sectional curvature $c$ and $M=M_T\times _fM_\perp$ be a warped product submanifold of $\tilde M(c)$. Then we have
\begin{enumerate}
\item[{(i)}] The squared norm of the second fundamental form $\sigma$ of $M$ satisfies
\begin{align}
\label{5}
\|\sigma\|^2\geq 2q\left(\|\nabla(\ln f)\|^2-\Delta(\ln f)+\frac{pc}{2}\right),
\end{align}
where $\dim M_T=2p+1,\,\dim M_\perp=q$ and $\nabla(\ln f)$ is the gradient of $\ln f$ and $\Delta$ is the Laplacian operator of $M_T$.
\item[{(ii)}] If the equality sign holds in (i), then $M_T$ is totally geodesic in $\tilde M(c)$ and $M_\perp$ is a totally umbilical submanifold of $\tilde M(c)$.
\end{enumerate}
\end{theorem}

The paper is organized as follows: Section 2 is devoted to preliminaries. In Section 3, first we develop some basic results for later use and then we prove Theorem 1.1. In the last section, we prove two geometric inequalities  as an application of Theorem 1.1 by considering compact invariant factor $M_T$.

\section{Preliminaries}
Let $\tilde M$ be a $(2m+1)$-dimensional almost contact manifold with almost contact structure $(\varphi,\xi, \eta)$, i.e., a structure vector field $\xi$, a $(1, 1)$ tensor field $\varphi$ and a $1$-form $\eta$ on $\tilde M$ such that $\varphi^2X=-X+\eta(X)\xi,~\eta(\xi)=1$, for any vector field $X$ on $\tilde M$ \cite{BL}. There always exists a compatible Riemannian metric $g$ satisfying $g(\varphi X,\varphi Y)=g(X, Y)-\eta(X)\eta(Y)$, for any vector field $X, Y$ tangent to $\tilde M$. Thus the manifold $\tilde M$ is said to be almost contact metric manifold and $(\varphi, \xi, \eta, g)$ is its almost contact metric structure. It is clear that $\eta(X)=g(X, \xi)$. The fundamental 2-form $\Phi$ on $\tilde M$ is defined as $\Phi(X, Y)=g(X, \varphi Y)$, for any vector fields $X,~Y$ tangent to $\tilde M$. The manifold $\tilde M$ is said to be {\it{almost cosymplectic}} if the forms $\eta$ and $\Phi$ are closed, i.e., $d\eta=0$ and $d\Phi=0$, where $d$ is an exterior differential operator. An almost cosyplectic and normal manifold is {\it{cosymplectic}}. It is well known that an almost contact metric manifold $\tilde M$ is cosymplectic if and only if $\tilde\nabla_X\varphi$ vanishes identically, where $\tilde\nabla$ is the Levi-Civita connection on $\tilde M$ \cite{L}.

A cosymplectic manifold $\tilde M$ with constant $\varphi$-sectional curvature is called a {\it{cosymplectic space form}} and denoted by $\tilde M(c)$. Then the Riemannian curvature tensor $\tilde R$ is given by
\begin{align}
\label{6}
\tilde R(X, Y; Z, W)=&\frac{c}{4}\big\{g(X, W )g(Y, Z)-g(X, Z)g(Y, W )+g(X, \varphi W )g(Y, \varphi Z)\notag\\
&-g(X, \varphi Z)g(Y, \varphi W )-2g(X, \varphi Y )g(Z, \varphi W)\notag\\
&-g(X, W )\eta(Y )\eta(Z)+g(X, Z)\eta(Y )\eta(W )\notag\\
&-g(Y, Z)\eta(X)\eta(W )+g(Y, W )\eta(X)\eta(Z)\big\}.
\end{align}

Let $M$ be a $n$-dimensional Riemannian manifold isometrically immersed in a Riemannian manifold $\tilde M$. Then, the Gauss and Weingarten formulae are respectively given by $\tilde\nabla_X Y=\nabla_X Y+\sigma(X,Y)$ and $\tilde\nabla_XN=-A_NX+\nabla^\perp_XN$, for any $X,Y$ tangent to $M$, where $\nabla$ is the induced Riemannian connection on $M$, $N$ is a vector field normal to $M$, $\sigma$ is the second fundamental form of $M$, $\nabla^\perp$ is the
normal connection in the normal bundle $ TM^\perp$ and $A_N$ is the shape operator of the second fundamental form. They are related as $g(A_NX,Y)=g(\sigma(X,Y),N)$, where $g$ denotes the Riemannian metric on $\tilde M$ as well as the metric induced on $M$.

Let $M$ be an $n$-dimensional submanifold of an almost contact metric $(2m+1)$-manifold $\tilde M$ such that restricted to $M$, the vectors $e_1,\cdots, e_n$ are tangent to $M$ and hence $e_{n+1},\cdots e_{2m+1}$ are normal to $M$. Then, the mean curvature vector $\vec H$ is defined by $\vec{H}=\frac{1}{n} tr\sigma=\frac{1}{n}\sum_{i,j=1}^n\sigma(e_i, e_i)$, where $\{e_1,\cdots, e_n\}$ is a local orthonormal frame of the tangent bundle $TM$ of $M$. A submanifold $M$ is called minimal in $\tilde M$ if its mean curvature vector vanishes identically and $M$ is totally geodesic in $\tilde M$, if $\sigma(X, Y)=0$, for all $X, Y$ tangent to $M$. If $\sigma(X,Y) =g(X,Y)H$ for all $X, Y$ tangent to $M$, then $M$ is {\it totally umbilical} submanifold of $\tilde M$.

For any $X$ tangent to $M$, we decompose $\varphi X$ as $\phi X=PX+FX$, where $PX$ and $FX$ are the tangential and normal components of $\varphi X$, respectively. For a submanifold $M$ of an almost contact manifold $\bar M$, if $F$ is identically zero then $M$ is $invariant$ and if $P$ is identically zero then $M$ is {\it{anti-invariant}}.

Let $R$ and $\tilde R$ denote the Riemannian curvature tensors of $M$ and $\tilde M$, respectively. Then the equation of Gauss is given by
\begin{align}
\label{7}
R(X,Y;Z,W)=&\tilde R(X,Y,Z,W)+g(\sigma(X, W), \sigma(Y, Z))\notag\\
&-g(\sigma(X, Z), \sigma(Y, W)),
\end{align}
for $X, Y, Z, W$ tangent to $M$.

For the second fundamental form $\sigma$, we define the covariant derivative $\tilde\nabla\sigma$ by
\begin{align}
\label{8}
(\tilde\nabla_X\sigma)(Y, Z)=\nabla^\perp_X\sigma(Y, Z)-\sigma(\nabla_XY, Z)-\sigma(Y, \nabla_XZ)
\end{align}
for any $X, Y, Z$ tangent to $M$.

\noindent The equation of Codazzi is
\begin{align}
\label{9}
(\tilde R(X, Y)Z)^\perp=(\tilde\nabla_X\sigma)(Y, Z)-(\tilde\nabla_Y\sigma)(X, Z),
\end{align}
where $(\tilde R(X, Y)Z)^\perp$ is the normal component of $(\tilde R(X, Y)Z)$.

Let $M$ be a Riemannian $p$-manifold and $e_1,\cdots, e_p$ be an orthonormal frame fields on $M$. Then for a differentiable function $\psi$ on $M$, the Laplacian $\Delta\psi$ of $\psi$ is defined by
\begin{align}
\label{10}
\Delta\psi=\sum_{i=1}^n\big\{(\tilde\nabla_{e_i}e_i)\psi-e_ie_i\psi\big\}.
\end{align}
The scalar curvature of $M$ at a point $p$ in $M$ is given by
\begin{align}
\label{11}
\tau(p)=\sum_{1\leq i<j\leq n}K(e_i, e_j),
\end{align}
where $K(e_i, e_j)$ denotes the sectional curvature of the plane section spanned by $e_i$ and $e_j$.

Due to behaviour of the tensor field $\varphi$, there are different classes of submanifolds. We mention the following:

\begin{enumerate}
\item [(1)] A submanifold $M$ tangent to the structure vector field $\xi$ is called an invariant submanifold if $\varphi$ preserves any tangent space of $M$, i.e., $\varphi(T_pM)\subseteq T_pM$, for each $p\in M$.
\item [(2)] A submanifold $M$ tangent to the structure vector field $\xi$ is said to be an anti-invariant submanifold if $\varphi$ maps any tangent space of $M$ into the normal space, i.e., $\varphi(T_pM)\subseteq T_pM^\perp$, for each $p\in M$.
\item [(3)] A submanifold $M$ tangent to the structure vector field $\xi$ is called a contact CR-submanifold if it admits an invariant distribution $\mathcal{D}$ whose orthogonal complementary distribution $\mathcal{D}^\perp$ is anti-invariant, i.e., the tangent space of $M$ is decomposed as $TM=\mathcal{D}\oplus\mathcal{D}^\perp\oplus\langle\xi\rangle$ with $\varphi\mathcal{D}_p\subseteq \mathcal{D}_p$ and  $\varphi\mathcal{D}_p^\perp\subseteq T_pM^\perp$, for each $p\in M$, where $\langle\xi\rangle$ denotes the 1-dimensional distribution spanned by the structure vector field $\xi$.
\end{enumerate}

In this paper we study contact CR-warped product submanifolds, therefore we are concerned with the case (3). For a contact CR-submanifold $M$  of an almost contact metric manifold $\tilde M$, the normal bundle $TM^\perp$ is decomposed as
\begin{align}
\label{12}
TM^\perp=\varphi\mathcal{D}^\perp\oplus\mu,~~~\varphi\mathcal{D}^\perp\perp\mu
\end{align}
where $\mu$ is orthogonal complementary distribution of $\varphi\mathcal{D}^\perp$ which invariant normal subbundle of $TM^\perp$ with respect to $\varphi$.

\section{Proof of Theorem 1.1}
Before proving the theorem we need some basic results for warped products. A warped product submanifold $M$ is said to be a contact CR-warped product submanifold if $M=M_T\times_fM_\perp$ is the product of $M_T$ and $M_\perp$, where $M_T$ is an invariant submanifold and $M_\perp$ is an anti-invariant submanifold. We need the following results to prove our main Theorem 1.1.\\

\begin{lemma} \cite{U3} Let $M=M_T\times{_{f}M_\perp}$ be a contact CR-warped product submanifold of a cosymplectic manifold $\tilde M$. Then
\begin{align}
\label{13}
g(\sigma(X, Z), \varphi W)=-\varphi X(\ln f)g(Z, W),
\end{align}
for any $X\in\mathcal{L}(M_T)$ and $Z,~W\in\mathcal{L}(M_\perp)$.
\end{lemma}

\begin{theorem} \cite{U5} Let $M=M_T\times{_{f}M_\perp}$ be a contact CR-warped product submanifold of a cosymplectic manifold $\tilde M$. Then
\begin{enumerate}
\item [(i)] The squared norm of the second fundamental form of $M$ satisfies
\begin{align}
\label{14}
\|\sigma\|^2\geq 2q\|\nabla(\ln f)\|^2
\end{align}
where $\nabla(\ln f)$ is gradient of the function $\ln f$ and $q$ is the dimension of $M_\perp$.

\item [(ii)] If equality holds in \eqref{14}, then $M_T$ is totally geodesic submanifold of $\tilde M$ and $M_\perp$ is totally umbilical in $\tilde M$ and hence $M$ is minimal in $\tilde M$.
\end{enumerate}
\end{theorem}

\noindent Let $M=M_T\times_fM_\perp$ be a contact CR-warped product submanifold of a cosymplectic manifold $\tilde M$ such that  $\xi\in\mathcal{L}(M_T)$, where  $M_T$ and  $M_\perp$ are invariant and anti-invariant submanifolds of $\tilde M$, respectively. For any $X\in\mathcal{L}(M_T)$ and any $Z\in\mathcal{L}(M_\perp)$, we have
\begin{align*}
\sigma(\varphi X, Z)=\tilde\nabla_Z\varphi X-\nabla_Z\varphi X.
\end{align*}
Using the covariant derivative property of $\tilde\nabla\varphi$ and the cosymplectic characteristic equation with \eqref{2}, we derive
\begin{align}
\label{15}
\sigma(\varphi X, Z)&=\varphi\tilde\nabla_ZX-\varphi X(\ln f)Z\notag\\
&=\varphi\sigma(X, Z)+X(\ln f)\varphi Z-\varphi X(\ln f)Z.
\end{align}

\noindent{\it{Proof of Theorem 1.1.}}  Let $M=M_T\times_fM_\perp$ be a contact CR-warped product submanifold of a cosymplectic space form $\tilde M(c)$ such that  $\xi$ is tangential to $M_T$. Then, for the unit normal vectors $X\in\mathcal{L}(M_T)$ and $Z\in\mathcal{L}(M_\perp)$, from \eqref{9}, we have
\begin{align*}
\tilde R(X, \varphi X, Z, \varphi Z)=g((\nabla_X^\perp\sigma)(\varphi X, Z), \varphi Z)-g((\nabla_{\varphi X}^\perp\sigma)(X, Z), \varphi Z).
\end{align*}
Then from \eqref{8}, we derive
\begin{align}
\label{16}
\tilde R(X, \varphi X, Z, \varphi Z)&=g(\nabla_X^\perp\sigma(\varphi X, Z)-\sigma(\nabla_X\varphi X, Z)-\sigma(\varphi X, \nabla_XZ), \varphi Z)\notag\\
&-g(\nabla_{\varphi X}^\perp\sigma(X, Z)-\sigma(\nabla_{\varphi X}X, Z)-\sigma(X, \nabla_{\varphi X}Z), \varphi Z).
\end{align}
Now, we compute the following terms as follows
\begin{align*}
g(\nabla_X^\perp\sigma(\varphi X, Z), \varphi Z)&=Xg(\sigma(\varphi X, Z), \varphi Z)-g(\sigma(\varphi X, Z), \nabla_X^\perp\varphi Z)\\
&=Xg(\tilde\nabla_Z\varphi X, \varphi Z)-g(\sigma(\varphi X, Z), \tilde\nabla_X\varphi Z).
\end{align*}
Using the cosymplectic characteristic equation and the compatible metric property and the fact that $\xi$ is tangent to $M_T$, we derive
\begin{align*}
g(\nabla_X^\perp\sigma(\varphi X, Z), \varphi Z)=Xg(\tilde\nabla_ZX, Z)-g(\sigma(\varphi X, Z), \varphi\tilde\nabla_XZ).
\end{align*}
Then by Gauss formula and the relation \eqref{2}, we obtain
\begin{align*}
g(\nabla_X^\perp\sigma(\varphi X, Z), \varphi Z)&=X(X(\ln f))g(Z, Z)-X(\ln f)g(\sigma(\varphi X, Z), \varphi Z)\\
&-g(\sigma(\varphi X, Z), \varphi\sigma(X, Z)).
\end{align*}
Then from \eqref{13} and \eqref{15} , we get
\begin{align*}
g(\nabla_X^\perp\sigma(\varphi X, Z), \varphi Z)&=X^2(\ln f)g(Z, Z)+2X(\ln f)g(\nabla_XZ, Z)\\
&-(X(\ln f))^2g(Z, Z)-\|\sigma(X, Z)\|^2-\varphi X(\ln f)g(\sigma(X, Z), \varphi Z).
\end{align*}
Again using \eqref{2} and \eqref{13}, we derive
\begin{align}
\label{17}
g(\nabla_X^\perp\sigma(\varphi X, Z), \varphi Z)&=X^2(\ln f)g(Z, Z)+(X(\ln f))^2g(Z, Z)\notag\\
&-\|\sigma(X, Z)\|^2+(\varphi X(\ln f))^2g(Z, Z).
\end{align}
Since $M_T$ is invariant and totally geodesic in $M_T\times_fM_\perp$ \cite{Bi,C1}, then by the cosymlectic characteristic equation, we have
\begin{align}
\label{18}
g(\sigma(\nabla_X\varphi X, Z), \varphi Z)=g(\sigma(\varphi\nabla_XX, Z), \varphi Z).
\end{align}
Also, from \eqref{13}, we have $g(h(\varphi X, Z), \varphi W)=X(\ln f)g(Z, W)$, thus with the help of this fact \eqref{18} becomes
\begin{align}
\label{19}
g(\sigma(\nabla_X\varphi X, Z), \varphi Z)=(\nabla_XX\ln f)g(Z, Z).
\end{align}
Similarly, we obtain the following
\begin{align*}
g(\sigma(\varphi X, \nabla_XZ), \varphi Z)=(X\ln f)g(\sigma(\varphi X, Z), \varphi Z).
\end{align*}
Then from \eqref{13}, we get
\begin{align}
\label{20}
g(\sigma(\varphi X, \nabla_XZ), \varphi Z)=(X\ln f)^2g(Z, Z).
\end{align}
Now, interchanging $X$ by $\varphi X$ in \eqref{17}, \eqref{19} and \eqref{20}, then the following relations hold respectively
\begin{align}
\label{21}
-g(\nabla_{\varphi X}^\perp\sigma(X, Z), \varphi Z)&=((\varphi X)^2\ln f)g(Z, Z)+(\varphi X\ln f)^2 g(Z, Z)\notag\\
&-\|\sigma(\varphi X, Z)\|^2+(X\ln f)^2g(Z, Z),
\end{align}
\begin{align}
\label{22}
g(\sigma(\nabla_{\varphi X}X, Z), \varphi Z)=-(\nabla_{\varphi X}\varphi X\ln f)g(Z, Z)
\end{align}
and
\begin{align}
\label{23}
g(\sigma(X, \nabla_{\varphi X}Z), \varphi Z)=-(\varphi X\ln f)^2g(Z, Z).
\end{align}
Now, we compute
\begin{align*}
\|\sigma(\varphi X, Z)\|^2=g(\sigma(\varphi X, Z), \sigma(\varphi X, Z)).
\end{align*}
Then from \eqref{15}, we derive
\begin{align*}
\|\sigma(\varphi X, Z)\|^2&=g(\varphi \sigma(X, Z), \varphi \sigma(X, Z))+2(\varphi X\ln f)g(\sigma(X, Z), \varphi Z)\\
&+(X\ln f)^2g(Z, Z)+(\varphi X\ln f)^2g(Z, Z).
\end{align*}
Then by the property of compatible metric and \eqref{13}, we obtain
\begin{align}
\label{24}
\|\sigma(\varphi X, Z)\|^2=\|\sigma(X, Z)\|^2-(\varphi X\ln f)^2g(Z, Z)+(X\ln f)^2g(Z, Z).
\end{align}
With the help of \eqref{24}, the relation \eqref{21} becomes
\begin{align}
\label{25}
g(\nabla_{\varphi X}^\perp\sigma(X, Z), \varphi Z)=&-((\varphi X)^2\ln f)g(Z, Z)+\|\sigma(X, Z)\|^2\notag\\
&-2(\varphi X\ln f)^2g(Z, Z).
\end{align}
Then from \eqref{16},  \eqref{17},  \eqref{19},  \eqref{20} and  \eqref{25}, we derive
\begin{align}
\label{26}
\tilde R(X, \varphi X, Z, \varphi Z)&=(X^2\ln f)g(Z, Z)-(\nabla_XX\ln f)g(Z, Z)\notag\\
&+((\varphi X)^2\ln f)g(Z, Z)-(\nabla_{\varphi X}\varphi X\ln f)g(Z, Z)\notag\\
&-2\|\sigma(X, Z)\|^2+2(\varphi X\ln f)^2g(Z, Z).
\end{align}
Also, from \eqref{6} and \eqref{7}, we have
\begin{align}
\label{27}
\tilde R(X, \varphi X, Z, \varphi Z)=\frac{c}{2}\big(\eta^2(X)-\|X\|^2\big)g(Z, Z),
\end{align}
for any unit tangent vector $X\in\mathcal{L}(M_T)$ and $Z\in\mathcal{L}(M_\perp)$. Then from \eqref{26} and \eqref{27}, we get
\begin{align}
\label{28}
2\|\sigma(X, Z)\|^2&=(X^2\ln f)g(Z, Z)-(\nabla_XX\ln f)g(Z, Z)+((\varphi X)^2\ln f)g(Z, Z)\notag\\
&-(\nabla_{\varphi X}\varphi X\ln f)g(Z, Z)+2(\varphi X\ln f)^2g(Z, Z)\notag\\
&-\frac{c}{2}\big(\eta^2(X)-\|X\|^2\big)g(Z, Z).
\end{align}
Now, consider the orthonormal frame fields of $M_T$ and $M_\perp$ as follows: $\{X_1,\cdots, X_p, X_{p+1}=\varphi X_1,\cdots, X_{2p}=\varphi X_p, X_{2p+1}=\xi\}$ and $\{Z_1,\cdots, Z_q\}$ are the frame fields of the tangent spaces of $M_T$ and $M_\perp$, then summing over $i=1,\cdots,2p+1$ and $j=1,\cdots, q$ in \eqref{28}, thus we have
\begin{align*}
2\sum_{i=1}^{2p+1}\sum_{j=1}^{q}\|\sigma(X_i, Z_j)\|^2&=-\sum_{i=1}^{2p+1}\sum_{j=1}^{q}\left((\nabla_X{_i}X_i(\ln f))-X_i^2(\ln f)\right)g(Z_j, Z_j)\\
&-\sum_{i=1}^{2p+1}\sum_{j=1}^{q}\left((\nabla_{\varphi X_i}\varphi X_i(\ln f))-(\varphi X_i)^2(\ln f)\right)g(Z_j, Z_j)\\
&+2\sum_{i=1}^{2p+1}\sum_{j=1}^{q}(\varphi X_i(\ln f))^2g(Z_j, Z_j)\\
&-\sum_{i=1}^{2p+1}\sum_{j=1}^{q}\frac{c}{2}\big(\eta^2(X_i)-\|X_i\|^2\big)g(Z_j, Z_j).
\end{align*}
Then, from the definition of the gradient and \eqref{10}, we derive
\begin{align*}
2\sum_{i=1}^{2p+1}\sum_{j=1}^{q}\|\sigma(X_i, Z_j)\|^2=-2q\Delta(\ln f)+2q\|\varphi\nabla(\ln f)\|^2+pqc
\end{align*}
or
\begin{align}
\label{29}
\|\sigma\|^2_{T\perp}=-q\Delta(\ln f)+q\|\nabla(\ln f)\|^2+\frac{pqc}{2}\notag\\
\end{align}
Thus, from Theorem 1.1 of  \cite{U5}, we have $\|\sigma\|^2\geq2\|\sigma\|^2_{T\perp}$ only and left all other terms in the right hand side in the inequality of that theorem. Then using \eqref{29} in this relation we get inequality (i) with equality sign holds if and only if
\begin{align}
\label{30}
\sigma(\mathcal{L}(M_T), \mathcal{L}(M_T))=\sigma(\mathcal{L}(M_\perp), \mathcal{L}(M_\perp))=0.
\end{align}
i.e., $M$ is both $M_T$ and $M_\perp$-totally geodesic. The equality case holds just like Theorem 3.1. Hence, the proof is complete.

\section{Some Applications of Theorem 1.1}
For a warped product CR-submanifold $M_T\times_fM_\perp$ of a cosymplectic space form, if the holomorphic submanifold $M_T$ is compact, then we have the following useful results.

\begin{theorem} Let $M_T\times_fM_\perp$ be a warped product CR-submanifold of a cosymplectic space form $\tilde M(c)$ such that $M_T$ is a compact invariant submanifold of $\tilde M(c)$. Then
\begin{enumerate}
\item[{(i)}] For any $s\in M_\perp$, we have
\begin{align}
\label{31}
\int_{M_T\times\{s\}}\|\sigma\|^2\,dV_T\geq pqc\,\rm{vol}(M_T),
\end{align}
where $dV_T$ and $\rm{vol}(M_T)$ are the volume element and the volume of $M_T$, respectively and $2p+1=\dim M_T,\,q=\dim M_\perp$.
\item[{(ii)}] The equality sign holds in (i) identically if and only if $M$ is a Riemannian product of $M_T$ and $M_\perp$, i.e., the warping function $f$ is constant on $M$ .
\end{enumerate}
\end{theorem}
\begin{proof} For a warped product submanifold $M_T\times_fM_\perp$  with compact $M_T$, from Theorem 1.1, we have
\begin{align}
\label{32}
\int_{M_T\times\{s\}}\|\sigma\|^2\,dV_T\geq 2q\int_{M_T\times\{s\}}\left(\|\nabla(\ln f)\|^2-\Delta(\ln f)+\frac{pc}{2}\right)\,dV_T.
\end{align}
Since $M_T$ is compact, it follows from Hopf's Lemma that
\begin{align}
\label{33}
\int_{M_T\times\{s\}}\|\sigma\|^2\,dV_T\geq 2q\int_{M_T\times\{s\}}\left(\|\nabla(\ln f)\|^2\right)\,dV_T+pqc\,\rm{vol}(M_T).
\end{align}
Thus, inequality \eqref{33} implies inequality \eqref{31}, with the equality sign holding if and only if  (1) $f$ is constant i.e., $M$ is Riemannian product and (2) the equality $\|\sigma\|^2=pqc$ holds identically. Hence, the theorem is proved completely.
\end{proof}

Next, let us assume $f$ is non-constant. Then the minimal principle on $\lambda_1$ yields (see \cite{Be} page 186, \cite{C4})
\begin{align}
\label{34}
\int_{M_T}\|\nabla(\ln f)\|^2\,dV_T\geq \lambda_1\int_{M_T}\left(\ln f\right)^2\,dV_T
\end{align}
with equality holding if and only if $\Delta\ln f=\lambda_1\ln f$ holds.

Now, we give the following result.

\begin{theorem} Let $M=M_T\times_fM_\perp$ be a warped product CR-submanifold of a cosymplectic space form $\tilde M(c)$ with compact $M_T$. If the warping function $f$ is non-constant, then, for each $s\in M_\perp$, we have
\begin{align}
\label{35}
\int_{M_T\times\{s\}}\|\sigma\|^2\,dV_T\geq 2q\lambda_1\int_{M_T}\left(\ln f\right)^2\,dV_T+pqc\,\rm{vol}(M_T)
\end{align}
where $dV_T$, $\lambda_1$ and $\rm{vol}(M_T)$ are the volume element, the first positive eigenvalue of the Laplacian $\Delta$ and the volume of $M_T$, respectively.\\

\noindent Moreover, the equality sign of \eqref{35} holds identically if and only if we have:
\begin{enumerate}
\item[(i)] $\Delta\ln f=\lambda_1\ln f$
\item[(ii)] $M$ is both $M_T$-totally geodesic and $M_\perp$-totally geodesic.
\end{enumerate}
\end{theorem}
\begin{proof} By combining \eqref{33} and \eqref{34}, we get inequality \eqref{35}. From the above discussion, we know that the equality sign of \eqref{35} holds identically if and only if we have (i) $\Delta\ln f=\lambda_1\ln f$ and (ii) the warped product is both $M_T$ and $M_\perp$ totally geodesic.
\end{proof}

Another motivation of Theorem 1.1 is to give the expression of Dirichlet energy of the warping function in physics, which is defined of a function $\psi$ on a compact manifold $M$ as follows
\begin{align}
\label{36}
E(\psi)=\frac{1}{2}\int_{M}\|\nabla\psi\|^2\,dV
\end{align}
where $\nabla\psi$ is the gradient of $\psi$ and $dV$ is the volume element.

Now, we give the expression of Dirichlet energy of the warping function for a contact CR-warped product $M_T\times_fM_\perp$ in a cosymplectic space form $\tilde M(c)$ with compact invariant submanifold $M_T$. For any $s\in M_\perp$, from Theorem 1.1, we have
\begin{align}
\label{37}
\int_{M_T\times\{s\}}\|\sigma\|^2\,dV_T\geq2q\int_{M_T\times\{s\}}\left(\|\nabla\ln f\|^2+\frac{pc}{2}\right)\,dV_T.
\end{align}
From \eqref{36} and \eqref{37}, we find
\begin{align*}
E(\ln f)\leq\frac{1}{4q}\int_{M_T\times\{s\}}\|\sigma\|^2\,dV_T-\frac{pc}{4}\,\rm{vol}(M_T)
\end{align*}
which is the Dirichlet energy $E(\ln f)\,(0\leq E(\ln f)<\infty)$ of the warping function.\\

\noindent{\bf{Acknowledgements.}} The authors would like to express their hearty thanks to anonymous referees for their valuable suggestions and comments towards the improvement of this manuscript, especially, Section 4.

redpajama_set_name: RedPajamaArXiv


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PGA Cheat Sheet: Valspar Championship
Geoff Ulrich
The elite field didn't disappoint last week, as we saw four or five elite players go at it on Sunday, with Adam Scott eventually being the last man standing. All in all, we saw the players who had already perform well this season continue to perform well, and we saw continued slumps from Jason Day, Jordan Spieth, and Patrick Reed. It will be interesting to see if any of these three rebound with a big week before Augusta. Finally, before moving on to the Valspar, we have to mention Steven Bowditch, three rounds in the 80's, a 10 on one hole and he still made nearly $50,000 in prize money. Don't you wish you were a pro golfer?
More PGA Content for the Valspar Championship
PGA Cheat Sheet
PGA Love/Hate
PGA Targets
After two weeks with the same winner we'll be guaranteed to get some new blood this week as Adam Scott is not in the field. This event is likely the weakest we've seen all year, but there are still big names in attendance. Jordan Spieth is here to defend his title, while top Euro players Henrik Stenson and Danny Willet are also back in play this week. The cut is also back this week and will take place after Friday's round like usual with the top 70 players and ties all making the weekend. With 140 players in the field and a fairly thin group of elite talent, there's definitely going to be some interesting decisions to make at the bottom of the field for DFS purposes. Many players will be priced higher than usual, and the top players' ownership won't be super spread out. Deciding who is value and who is overpriced this week will be key.
Copperhead Course—Palm Harbour, Florida
Par 71, 7,300 yards
Copperhead has hosted this event since its inception in 2000, but it is important to note that the event underwent a date change in 2007 and has tended to play tougher since. The other thing to note this week is that the course underwent huge renovations of greens and fairways last summer, and while the design of the course didn't change much (if at all), the effect this new grass will have is unknown.
…if the wind gets up, -7 might be enough to win."
The course itself has tended to play quite tough and a winning score similar to the last few weeks could be expected; although, if the wind gets up, -7 might be enough to win. Copperhead is unique in that it's a par 71 but features five (yes 5) par three's and four par fives. The course isn't brutally long but plays long due to the number of doglegs on the course, which often forces players into playing less than driver off the tee. On top of the tighter fairways and trees making life tough, water is also a big factor here and comes into play on 11 of the 18 holes. Unlike last week, though, this isn't a course where players will be looking to grip it and rip it. Rather, they will be looking to place their ball off the tee as strategically as possible. With some long par 4's and at least three par 3's measuring over 200 yards, pressure will be put on the players' approach games on almost every hole.
Copperhead also features it's own brutally hard signature finishing stretch "the snake pit," which consists of two long, winding par 4's and a hard to hit par 3. This three-hole stretch should once again ensure an exciting finish and keep any DFS player from celebrating their players' success until they've put the ball in the hole on 18.
Last 5 winners
Jordan Spieth—2015
John Senden—2014
Kevin Streelman—2013
Luke Donald—2012
Gary Woodland—2011
Winning Trends
Each of the past five winners of the Valspar Championship had recorded at least one top ten on the season before their victory here
Four of the past five winners ranked 36th or better in strokes gained tee to the green in the year of their victory
Par 3 Scoring
All-Around Ranking
Strokes Gained: Tee To Green is generally just a good indicator of which golfers are outplaying their competition off the tee and on approaches into the green. On this course, past winners and players who have played well here have tended to be inside the top 50 in this category for the year. This course puts a lot of pressure on players off the tee and on approaches, so there's no doubt this category is a good way to start research this week, as players will need to be solid tee to green to have a shot.
Copperhead is a unique course because it's a par 71 but still has four par 5's. The reason for this is that it also contains five par 3's of varying lengths. I really like looking for players who are solid in both of these categories (Par 3 and Par 5 scoring) this week, as the challenge will generally be to hold water on the tough par 4's and then try and score on the rest of the holes. Past winners of this event have also tended to rank well in par 3 scoring particularly, and a solid week on the five 'one-shotters' will be needed.
It's not a shock to see that players will need to be sharp in a variety of different areas to succeed this week."
The last stat I looked at this week is the All-Around Ranking. The all-around ranking measures how well players rank in several stats (Birdie Leaders, Sand Saves, Greens in Regulation, Driving Distance and Driving Accuracy to name a few). Past winners at Valspar have ranked well in this category, and last year's top three finishers all finished 43rd or better in the all-around rankings last year. This course challenges players with difficult approaches, tee shots, and greens that can be tricky to read at times. It's not a shock to see that players will need to be sharp in a variety of different areas to succeed this week.
Vegas Values
Odds to win are one factor to think about when picking players (but not the only thing, so be careful putting too much weight on them). This section is going to detail a few of the players who have the best fantasy value comparative to their odds of winning this week.
DK Price
Jason Kokrak 60-1 $7,800 C. Howell III $7,900 and 70-1
Danny Lee $8,000 and 70-1
Sean O'Hair $7,800 and 70-1
Chez Reavie 90-1 $7,300 Ian Poulter $7,400 and 100-1
Kevin Chappell $7,400 and 110-1
Russell Henley $7,300 and 125-1
Luke Donald has played the Valspar five times and finished 6th or better in four attempts. His form hasn't been great this year, but his past history on this course is strong.
Jason Dufner might have a more impressive record on this tough course than anyone. He's played this event 7 times and never finished outside the top 30 here.
Harris English has success playing junior golf on Copperhead and also has two top tens in his last two visits here, he likes the course too.
John Senden has played here nine years in a row. While he's missed the cut a couple times, he's also recorded two runner-up finishes and a win on Copperhead. He's definitely someone who can get it done on this course.
There's a ton of big price increases this week due to the watered down nature of this field. Some players — like Patrick Reed — might see their ownership go way down as a result. I think you can use the price difference to your advantage this week, especially if you're looking for some really under-owned players. The other thing to keep note of is the weather. Wind is in the forecast for Thursday and Friday, and one wave of tee-times (early-late or late-early) could get a big advantage this week. Stacking players from one wave over the other could be advantageous, and I'd definitely research the weather on Wednesday night to see if this is a viable play.
Top Performing Studs From Recent Weeks
Top Performing Value Plays From Recent Weeks
Luke List
Chez Reavie
Colt Knost
Webb Simpson – $8,900
It can sometimes be a bit of an adventure with Webb, especially on the greens, but there's no denying he's started the season well. Simpson has recorded three top 20's in three starts and ranks 13th in strokes gained tee to green and 4th in greens in regulation. In short, he is hitting it beautifully, but he needs his putter to get semi-hot. I also like the fact Simpson has had success on this course before and will be coming in rested. He's top twenty in par 3 scoring, and tee to green, he might be playing better than anyone in the field. I think Webb can ensnare the field this week and land his first victory with a short putter.
Cadillac Championships: Henrik Stenson (28th)
Honda Classic: Brooks Koepka (26th)
Northern Trust Open: Bubba Watson (Win)
Pebble Beach: J.B. Holmes (11th)
WMO Phoenix: Justin Thomas (MC)
Farmers Pick: Jimmy Walker (T4)
CareerBuilder Pick: Graham DeLaet (T42)
Sony Open Pick: Danny Lee (T33)
MY SLEEPER
George McNeil – $6,400
Sleeper Qualification: must be under $7000 in salary and projected by me to be less than 10% owned
McNeil has not played well to start the season, but he finally showed some life in his last start at the Honda Classic. McNeil is also a Florida guy and has played this course a ton, which is likely the reason he's got such a strong record here. Last year McNeil had some success on similar, "tighter" style courses and was in contention last year at Colonial. I think he makes for an interesting play at only $6400 and is someone who could certainly grind you out four decent rounds given his experience here, with the possibility of ending up inside the top 25 for the week. This would be a huge result given his cheap price.
Cadillac Championships: Daniel Berger (28th)
Honda Classic: Jon Curran (MC)
Northern Trust Open: Scott Pinckney (MC)
Pebble Beach: Alex Prugh (MC)
WMO Phoenix: Jason Kokrak (MC)
Farmers Sleeper: Robert Garrigus (T50)
CareerBuilder Sleeper: Chez Reavie (T17)
Sony Open Sleeper: Daniel Summerhays (T13)
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On a Canonical form for Maxwell Equations and Convergence of Finite Element Scheme for Vlasov--Maxwell system.
This work is a swift introduction to the nature of governing laws involved in the Maxwell equations. We then approximate a "one and one-half" dimensional relativistic Vlasov-Maxwell (VM) system using streamline diffusion finite element method. In this geometry d'Alembert representation for the fields functions guarantees the existence of a unique solution of the Maxwell equations. The VM system is then approximated using the streamline diffusion finite element method. In this part we derive some stability inequalities and optimal a priori error estimates due to the maximal available regularity of the exact solution.


Ministry of Propaganda - 04/March/2003: "Tax return over the phone?"
Tax return over the phone?
Apparently it's already available in the US and Canada (Can anyone comment on that?), now the Inland Revenue is planning to introduce tax return filing over the phone in the UK. I'm not sure how this will work and how long you would have to spend on the phone to get this done. I can't really see me doing this if it goes "Press 1 to..., press 2 to..., press 3 to...", I think I'll go for the Online tax return this year.


text:
Juniors Brierley, McMahon and Bacon Ride to Victory
Round 9 of the 2017 Yamaha Australian Off-Road Championship (AORC) kicked-off today as the talented Junior competitors competed in their first Enduro for the AORC in 2017. There was plenty of anticipation to see which riders would come out on top in front of the cheering crowd in a sea of team colours. The Juniors began their day out from the bLU cRU loop.
Despite the technical nature of the Kyogle track, the Juniors adapted phenomenally well. Juniors that excelled and won for their respective J4, J3 and J2 classes were Joshua Brierley, Korey McMahon and Kyron Bacon.
In J4 it would once again be the phenomenal Joshua Brierley on his Husqvarna bike to win an incredible Enduro race for his class. Brierley finished with a compiled time of 25:33.656. Brierley's pace was undeniable as he finished 26 seconds ahead of his nearest challenger. The first test from the bLU cRU loop gave Joshua the edge as he laid down an incredibly quick time, leaving the other competitors to play catch up. Up next was Joshua's brother Mitchell Brierley, also Husqvarna mounted. The two brothers have enjoyed several one-two finishes this year and this race continued that trend. Mitchell's total Enduro time amounted to a 24:08.069. The final podium placer for J4 was Corey Hammond. The KTM rider managed to finish with a total accumulated time of 24:15.837.
One of the most fascinating wins was achieved by Korey McMahon in the J3 class. Mcmahon who also competes in Motocross, has had a excellent Saturday Enduro to finish with a total time of 24:27.827. Finishing a further 23 seconds behind was Ben Bertinazzi on his KTM, the result marking his best place finish for his class so far this year. Bertinazzi fought hard to register a time of 24:50.868. Rounding up the podium positions was the very consistent Ned Chapman, on his Yamaha. Chapman adapted to the Kyogle track well to attain a podium position with his time of 25:21.476 to complete the Enduro. Championship leader and KTM rider Jayden Rudd finished off the podium for the first time this season in seventh place.
Honda mounted Kyron Bacon was the standout rider for J2 and finished first for his class. Young Bacon is emerging as a young rising star in the Off-Road scene and the win will be very helpful in his tight championship battle with Wil Riordan. Bacon had an accumulated Enduro time of 25:33.656. Second position belonged to Riordan who took 25:48.431 to finish the Enduro. KTM rider Riordan finished a further 15 seconds behind Bacon. To complete the podium would be brother Gus Riordan, who is also KTM mounted. His total time was a 26:38.095.
Provisional Results J4
1. Josuhua Brierley – 23:42.980
2. Mitchell Brierley – 24:08.069
3. Corey Hammond – 24:15.837
4. Nathan Howe – 24:44.766
5. Ryan Thompson – 24:47.351
6. Riley Nancarrow – 25:54.498
7. Connor Gee – 26:01.115
8. Brock Brady – 28:02.540
9. Emi Ruprecht – 28:49.620
10. Mitch Chapman – 30:29.698
1. Korey McMahon – 24:27.827
2. Ben Bertinazzi – 24:50.868
3. Ned Chapman – 25:21.476
4. Matthew Pye – 25:37.671
5. Blake Hollis – 25:48.798
6. Raife Dooley – 25:54.841
7. Jayden Rudd – 26:03.480
8. Tom Teed – 27:06.638
9. Ben Teed – 27:25.632
10.Jared Watson – 28:06.998
1. Kyron Bacon – 25:33.656
2. Wil Riordan – 25:48.431
3. Gus Riordan – 26:38.095
4. Mackenzie Johnson – 27:10.367
5. Harrison Foster – 27:28.657
6. Jasper Franklin – 27:44.231
7. Zac Perry – 28:11.339
8. Cooper Holroyd – 29:27.278
9. Oscar Collins – 29:40.508
10. Levii Read – 30:20.154
Keep up with the Yamaha AORC action by visiting aorc.org.au including news and reports on the AORC website at www.aorc.org.au, following the Facebook, @aorc on Instagram and don't forget to add 'motorcyclingaus' on Snapchat. Stay tuned to see which riders come out on top.
Stay up to date with all the happenings of the 2017 Yamaha Australian Off-Road Championship by visiting the AORC website at www.aorc.org.au or by following the AORC on Twitter and Facebook as well as @aorc on Instagram.
This is the link for the Entry List for Rounds 9 & 10
Don't forget to check out the AORC's website at aorc.org.au
Remaining Rounds of 2017 Yamaha Australian Off-Road Championship Calendar
Round 9 & 10: Kyogle, NSW 22 – 23 July 2017
Round 11 & 12: Omeo, VIC 5 – 6 August 2017
Photo courtesy of John Pearson Media: Josuhua Brierley at Kyogle


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text:
African victims of modern slavery rescued in Interpol-backed operation
An operation coordinated by international police organisation Interpol aimed at strengthening border controls in West Africa has resulted in the rescue of nearly 40 victims of modern slavery.
The majority of those identified as having been trafficked for the purposes of forced labour were children, who were being taken to work in illicit gold mines in Guinea, Interpol said.
Adults who had paid people smugglers to get them to Europe via Libya were also among those rescued.
The eight-day Operation Adwenpa III, which took place in October, was supported by more than 100 officers at 23 sites across key land border and airports in eight countries – Benin, Burkina Faso, Côte d'Ivoire, Guinea, Mali, Niger, Senegal and Togo.
As well as resulting in the discovery and rescue of human trafficking victims, the operation also saw law enforcement officers seize 300kgs of counterfeit pharmaceutical drugs at Abidjan airport.
The seizure led to the search of a house that was used as a warehouse and packaging site for other counterfeit medicines, and the arrest of six people.
Harold O'Connell, Director of Interpol's Capacity Building and Training Unit, commented: "The operation shows the range of activities in which organised crime is involved.
"In addition to helping protect communities, these interventions are part of a longer term solution where information sharing and cross-border cooperation between various law enforcement agencies is essential."
Last year, the operation resulted in the arrest of human traffickers and migrant smugglers, along with the seizure of counterfeit goods, a large amount of cash, drugs and a number of stolen vehicles.
As well as apprehending organised criminals, Operation Adwenpa, which is funded by the German government, is also intended to provide training on how to strengthen border control to local law enforcement officials from various agencies.
According to the UN Office on Drugs and Crime, West Africa is vulnerable to organised crime, people trafficking, drug smuggling and terrorism due to its geographic location, poorly-controlled borders and long coastline.
In recent years, the problem has been exacerbated by the high number of economic migrants seeking to leave the region for Europe, many of whom pay people smuggling gangs to facilitate their journeys.
A recent investigation conducted by CNN found that trafficking gangs are auctioning off migrants in Libya, many of whom have travelled from West Africa, for as little as $400.
The broadcaster obtained mobile phone footage showing two African men being sold off to work on farms.
People smugglers have resorted to selling their human cargo in the North African country after European authorities launched a crackdown on migrant boats attempting to cross the Mediterranean, sending many back to the beaches from which they came.
Speaking after visiting Libya, Mohammed Abdiker, International Organisation for Migration (IOM) Director of Operation and Emergencies, said: "The situation is dire. The more IOM engages inside Libya, the more we learn that it is a vale of tears for many migrants.
"Some reports are truly horrifying and the latest reports of 'slave markets' for migrants can be added to a long list of outrages."
40 people traffickers held in Interpol-backed crackdown on African smugglers
Public and private sectors must work together to prevent human trafficking, Interpol conference told
People smugglers charging migrants to travel in vehicle engine compartments, Europol warns
Egyptian human trafficking probe results in 75 arrests
Illegal US massage parlours run by human traffickers make $2.5 billion a year, study finds
Related Topics:Africahuman traffickingInterpolmodern slaverypeople smuggling
Opium production rise to record level in Afghanistan
New York named US capital of illicit tobacco


If you're typical, your mom probably never told you much of anything about budgeting. If she didn't say something it was probably just that you needed to have a budget. Your dad might have been a bit better but again if you're typical, he probably didn't sit down and discuss the facts of budgeting with you as he or your mom might have discussed the facts of life.
It's true that many families do very well without a budget. These people generally fall into one of two categories. They either have so much money that they don't need a budget or they've sort of learned how to budget on the fly based on their life experiences. For example, many people in their 40s and 50s instinctually know how much they can spend and save without having to sit down and formalize things.
However, most of us do need a budget. This will be especially true if you are struggling with money and debts. We're surrounded today with messages about how if we just buy this latest electronic gadget or a new wardrobe, we will be as happy as a fiddler at a bluegrass concert. If we fall victim to these siren songs, it's easy to wake up one day and find that our finances have become hopelessly hashed up. This is where a budget can become a white knight to the rescue.
Actually the first step in establishing a budget isn't to create a budget. It's to your goals – both short-term and long-term. The reason for this is because most financial experts agree that it's practically impossible to stay on a budget unless you have goals.
When you have goals, you can see that you're making progress towards achieving them, which can be a great motivator for staying on that budget.
The second step in making a budget is to a learn where your money's been going. This means tracking your spending for probably a month. We know you are thinking, "oh, what a drag." However, thanks to today's smart phone apps, tracking your spending is about as tough as rolling down a hill. Many of these apps, such as Mint and Expensify will automatically categorize your spending for you so that you can see exactly where your money's going.
If you would like some additional help in making your budget, this video has some good information.
Of course, learning where your money's going is only half the battle. The other half is going over each category with a fine tooth comb to see where you could make cuts. This is especially important if you're struggling with debt. The reason why you're having a problem with debt is simple. You're spending more than you earn. So your first goal must be to reduce your spending to the point where it's less than your income. Once you achieve this goal, the next step is to go over that budget again, find areas where you can make even more cuts so that you're now spending a fair amount less than you earn and can put the difference into savings or into paying down your debts.
One way to attack this is to divide all your categories into two groups – fixed and variable. While we can't tell you precisely where you could make cuts, we have found that the easiest categories to find "leakage," or places where you could reduce your spending are variable ones such as food, entertainment, dining out and clothing. However, don't ignore those fixed costs such as housing and transportation. If you put your mind to it, you should be able to reduce your spending in these as well. It may not be pleasant but it can be done. For example, you could reduce your housing costs by moving to a smaller house or apartment. If you own your own home, you could sell it and buy a smaller, more energy-efficient house. You might also be able to reduce your transportation costs by selling your car and buying a more efficient one, by getting into a car pool or by taking public transportation several days a week. If you're having a problem deciding how to categorize your spending, here is a list that could help.
Family obligations – are you paying for child care, elder care or child support?
Debt payments – your monthly payments on credit cards, etc.
Miscellaneous – stuff you spend money on that doesn't fit into any of the categories listed here.
Most people who start budgeting soon learn that they've made errors and maybe even big ones. As an example of this, suppose you thought you could limit your grocery spending to $400 a month and you actually spent $550. Don't be discouraged. Learn from your mistake, make adjustments to your budget and move on. You might be able to find other categories where you could make up that $150 difference. Or you might find ways to spend less money on food. For that matter, you shouldn't think of your budget as something carved in stone. Instead, think of it more as a game plan. No football coach would institute a game plan and then never deviate from it no matter the circumstances. He would watch the game and then make changes to fit what was happening on the field. You should view your budget the same way.


text:
Grammys find balance between awards and performances
Posted 8:57 am, February 6, 2019, by CNN Wires
The Grammys return this weekend, facing some of the same challenges as those confronting its movie-industry counterpart, the Oscars, but with at least one significant advantage: the music world's top prize has already found a balance between handing out awards and delivering entertainment.
Like the Oscars, ratings for last year's Grammys fell sharply, dropping by roughly 24% in total viewers (to 19.8 million) and in key young-adult demographics that advertisers sell against compared to the previous year.
Also like the Oscars, the presentation will shake up aspects of the broadcast. Alicia Keys will serve as host, after CBS late-night star James Corden filled that role back-to-back years; and there will be an expanded field of competitors, from five to eight, in the four major categories: album of the year, record of the year, song of the year and best new artist.
Related: See who's nominated for a Grammy
In one key respect, though, the Grammys have seemingly become the enviable model of a modern awards ceremony, largely because the awards themselves co-exist more organically with entertainment and live performances. Although the Recording Academy presents more than 80 awards each year in a wide variety of categories, the telecast that airs on CBS features only about 10 of those presentations, squarely putting the emphasis on a concert-like atmosphere that highlights unique combinations of talent and tributes.
Other major awards, by contrast, present two dozen or more trophies, as well as honorary awards. With the pressure to keep viewers engaged, that has produced tension about streamlining the process by racing through it or even presenting awards off air during commercial breaks in order to squeeze what is theoretically more engaging content into the presentations.
Unlike other major award events, the Grammys have considerable entertainment value baked into the format, which is significantly different than stars (and in the case of the Oscars, technical talent) parading to the stage giving speeches. The Grammy format has "evolved over the last decade or so, as audience desire has changed and shifted," said Jack Sussman, CBS' executive vice president of specials, music and live events.
Sussman acknowledged that the overall surplus of award shows is a likely factor in declining ratings, diluting the experience. But he stressed that a few major awards presented by industry peers — the Oscars, Emmys and Tonys being the others — still possess a level of prestige that sets them apart.
Although the number of on-air awards has declined, the Grammys remain committed to providing a broader celebration of the year in music, as well as tributes that offer performance opportunities. This year, the latter includes segments devoted to Motown's 60th anniversary (the subject of a stand-alone special that will be taped next week and broadcast in April) and Dolly Parton.
According to Sussman, the balance that must be struck involves highlighting the year's award-worthy artistry — with Kendrick Lamar and Drake leading the way in terms of total nominations — and still putting on a show that's mindful of the viewer at home.
"You need to maintain that integrity, and give the audience something that entertains them and keeps them watching," he said. "And they're not mutually exclusive."
The 61st Annual Grammy Awards air Feb. 10 on CBS.
Sting performs at Variety the Children's Charity Dinner with the Stars
Stanley Cup Playoffs: Fans gear up for Bruins vs. Blues Game 4 at Enterprise Center
Zac Brown Band has a message for its haters
Taylor Swift headlining Amazon Prime Day concert
Stevie Wonder announces he'll be having kidney surgery in September
Stevie Wonder announces he'll be having kidney surgery during London concert
New Orleans musician Dr. John dies at 77
BTS' new music video breaks YouTube record for most views in 24 hours
PICTURES: Heart with Sheryl Crow and Lucie Silvas
Legendary actress and singer Doris Day dead at 97
Cartoonist loses job after illustration of Trump goes viral
'I'm lost without you': Lauren London remembers late boyfriend Nipsey Hussle in heartbreaking post


mix together the eggs, milk and butter all at once. using a wooden or silicon spoon stir until just mixed. do not over beat. the batter must be lumpy.
spoon the mixture into the cupcake paper moulds in half and let in the oven to bake for 15 to 20 minutes.
You can decorate it with buttercream.


Principal at SL Global Partners, an international consortium of corporate finance attorneys. Areas of specialization include: Global Capital Markets, Cross-Border Transactions, Islamic Finance & Alternative Investment Structures, FinTech, International Regimes, SEC Regulations and Corporate Governance.
Prior experience practicing over a decade at AMLaw 50 Firms in DC, London, Dubai and Shanghai, completing cross-border transactions, international securities offerings, private placements and public offerings on foreign exchanges in Europe, the Middle East and Asia and SEC-registered offering on the NYSE and NASDAQ, . Advisory Board Member to the Dubai International Financial Centre (DIFC).
I concentrate my practice on secured and unsecured credit transactions, mergers and acquisitions, equity and debt financings, ongoing securities law requirements of public companies and other general corporate matters.
(202) 672-5300	3000 K Street, N.W.
I am a partner and business lawyer with Foley & Lardner LLP where I primarily represent clients in mergers and acquisitions, finance transactions, joint ventures and related complex commercial transactions. I am a member of the firm's Transactional & Securities Practice as well as the Energy Industry Team. I have particular expertise representing clients in various aspects of the energy industry and communications industry including utilities, private equity funds, financial institutions, investors and project developers.
(202) 618-3722	200 Pennslyvania Avenue, N.W.
(202) 282-5988	1700 K Street N.W.
(202) 756-8000	500 North Capitol Street, N.W.
Use Justia to research and compare Washington attorneys so that you can make an informed decision when you hire your counsel.


text:
Powering Properties With Renewable Energy
WWD Global Stock Tracker Up 0.1%
Vince shares shed 19.3% following earnings report.
By Arnold J. Karr Plus Icon
Arnold J. Karr
Recent Articles by Arnold J. Karr
PVH's Calvin Klein Jeans Getting Their 'Cool,' and Sales, Back
Guess Q2 Beats Consensus, Trails Year-Ago Results
M-commerce Outpacing E-commerce Growth
December 2, 2014, 6:09pm
The WWD Global Stock Tracker regained some of the ground lost Monday with a 0.1 percent uptick to 108.07.
The tracker — an international group of 100 publicly traded fashion, retail and beauty companies — had started the week and the month with a 0.9 percent decline.
The increase came despite a 19.3 percent drop, to $29.48, from Vince Holding Corp., which began the day by reporting a third-quarter profit against a year-ago loss and a 20 percent sales increase. Still, analysts expressed concern about lower-than-expected gross margins.
Stocks showing strength were led by Youngor Group Co., up 5.3 percent to 10.03 yuan, or $1.63 at current exchange, and I.T Ltd., up 3.9 percent to 2.39 Hong Kong dollars, or 31 cents. Express Inc. shares were up 2.6 percent to $14.61 and Elizabeth Arden Inc. gained 2.2 percent to move to $17.67, while Avon Products Inc. ended the day 1.8 percent higher at $9.78.
Express is scheduled to report third-quarter results after the markets close Thursday.
The second largest decline among tracker stocks after Vince's came from Quiksilver Inc., down 5.7 percent to $1.81. Salvatore Ferragamo Italia SpA shares gave back 4.6 percent to 20.77 euros, or $25.90, while Sears Holdings Corp.'s were down 4.4 percent to $33.91. Dickson Concepts lost 4.3 percent to close at 3.81 Hong Kong dollars, or 49 cents, and American Apparel Inc. slipped 3.7 percent to 53 cents.
Among major indices, the Dow Jones Industrial Average and S&P 500 both expanded 0.6 percent to 17,879.55 and 2,066.55, respectively, while the S&P 500 Retailing Industry Group was up 0.3 percent to 999.09, narrowly missing its third close above 1,000 since last Monday.
People & Companies


text:
Q: Cannot accept popup playwright i tried the below code, i want to accept the AlertPopup to confirm the delete of a selected user :
async dialogListener() {
  this.page.on('dialog', async (dialog) => { dialog.accept(); })
}

but the Alert popup is automatically dismissed and the user is not deleted from the database.
Could you please help me with that

redpajama_set_name: RedPajamaStackExchange


text:
SAQP Judges Biographies
Follow SAQP
The following 4 Judges will make their recommendations to the Scottish Ministers:
Iain Connelly – Architect and former President of the Royal Incorporation of Architects Scotland.
Alistair MacDonald – Planner and former Convenor of the Royal Town Planning Institute.
Dr Lori McElroy MBE – Director of Building Energy Performance and Research, BRE
Craig McLaren – Director of Scotland of Ireland in the Royal Town Planning Institute
As with previous years, we continue to welcome PAS as a guest Judge to review the community engagement aspect, where relevant, within your entries. We are delighted, therefore, that David McAllister (Manager -Training) will represent PAS again this year.
Iain Connelly
After qualifying, Iain joined Sir Frank Mears & Partners in 1978. He moved to Torrance Oliver Architects, Dunfermline in 1979 where he worked on residential, industrial and commercial projects. In 1983 he joined Dunfermline District Council, progressing to Principal Architect with responsibility for the award winning restoration of Abbot House, Dunfermline and the new Dalgety Bay Library. In 1996 Iain became Area Manager for Fife Council, leading a large, multi-disciplinary, practice. In March 2002 he became Head of Property Services with over 200 staff. Iain represented Fife Council on the Society of Chief Architects in Local Authorities and was its first Scotland President (2004-2006). On Royal Incorporation of Architects Scotland (RIAS) Council since 2004, Iain sat on the Consultancy Committee and was Vice-Convenor of the Government and Consultancy Committee. Throughout his career he has campaigned to promote to promote the benefits of well-designed buildings. Iain was President of the Royal Incorporation from 2013 to 2015.
Alistair MacDonald
A former Convenor of the RTPI in Scotland, Alistair has been a member of the Royal Town Planning Institute since 1983. He is the former Head of Planning and Building Control for Glasgow City Council and was Chair of Heads of Planning Scotland. He is the founding Director of ALS Planning Ltd and remains a Trustee of Glasgow City Heritage Trust serving on the Grants Committee. Since 2013 he has chaired (on behalf of the RTPI) the Joint Historic Environment Group, a collaboration between COSLA and the Scottish Government. He became a Director of Glasgow Humane Society in 2015.
Dr Lori McElroy (MBE)
Dr Lori McElroy is an Environmental Engineer with a Masters in Design Philosophy from Glasgow School of Art and a PhD in Mechanical Engineering from the University of Strathclyde. Following on from leading the Sust. Programme for A&DS, Lori recently joined BRE as Director of Building Energy Performance and Research for Scotland, Wales and the South West of England. Lori is a past Board member of A&DS, a past panel member of the Building Standards Advisory Committee and has contributed to numerous panels and steering groups including chairing the working groups on the development of Section 7 (Sustainability) and participation on the recently reconvened Sullivan Panel for Scottish Building Standards and participation on the Historic Scotland advisory group for the National Centre for Conservation. She was a member of the Sustainable Housing Strategy Group and is actively involved as a founding partner of Retrofit Scotland.
Craig McLaren
Craig is Director of Scotland of Ireland in the Royal Town Planning Institute. He took up post with RTPI in January 2011 and works with around 2,500 RTPI members in Scotland and Ireland to shape and inform policy and practice to ensure that planning helps to create great places for people. Prior to this Craig spent almost 8 years setting up, and then managing, the Scottish Centre for Regeneration in Scottish Government and its agency Communities Scotland, as its Director. Between 1997 and 2003 Craig was Chief Executive of the Scottish Urban Regeneration Forum (SURF), the independent network promoting best practice in regeneration. He was Scottish Planning Policy Officer with the RTPI between 1994 and 1997, and before this he worked as a planner and economic development officer for the London Borough of Kingston upon Thames. Craig is a member of the Royal Town Planning Institute and the Institute of Economic Development and is an Academician of the Institute of Urbanism. He is a Trustee of Resilient Scotland Ltd and Scotland's Towns Partnership. Craig was recently asked by Scottish Government to Chair the National Waking Strategy Implementation Group.
SAQP 2016
People's Choice Award
Personal Achievement in Planning
Key stages for the Awards
Judges Biographies
SAQP Winner 2016


text:
Q: multiple routes with same url I currently have the following in my RouteConfig.cs:
            routes.MapRoute(
            name: "Default",
            url: "{controller}/{action}/{id}",
            defaults: new { controller = "Home", action = "Index", id = UrlParameter.Optional }
        );

        routes.MapRoute(
            name: "Identity",
            url: "{controller}/{action}/{id}",
            defaults: new { controller = "Profile", action = "Identity" }
        );

I'm trying to use the following in a Action method:
return RedirectToRoute("Identity", new { @id = id });

but it doesn't seem to go there, when I check with Fiddler, i see the request is going back to the same page i was currently on, it appears to be hitting the Default route.  Is there anyway to force it to get the Identity one, even if their urls are the same, I want to use that to force users to the other controller when needed.

A: MVC will stop looking for a route as soon as a match is found. In your case, it will look in the following order: 1.) Default 2.) Identity.
If you want to create a specific route with the same pattern, you can achieve it with the following code:
routes.MapRoute(
    name: "Identity",
    url: "Profile/{action}/{id}",
    defaults: new { controller = "Profile", action = "Identity" }
);

routes.MapRoute(
    name: "Default",
    url: "{controller}/{action}/{id}",
    defaults: new { controller = "Home", action = "Index", id = UrlParameter.Optional }
);

A: You do not need to 'Force' the user to another controller. It is a good practice to use the RedirectToAction method so you can supply the controller and action you want to bring the user to.
The solution supplied by @thiag0 will not work.
Use the following
return RedirectToAction("Identity", "Profile", new { id = 5 });

and in your Profile controller, make sure you can accept the parameter
    public ActionResult Identity(int id)
    {
        return View();
    }

and in RouteConfig.cs
            routes.MapRoute(
            name: "Default",
            url: "{controller}/{action}/{id}",
            defaults: new { controller = "Home", action = "Index", id = UrlParameter.Optional }
            );

            routes.MapRoute(
            name: "Identity",
            url: "{controller}/{action}/{id}",
            defaults: new { controller = "Profile", action = "Identity"}
            );


At RateWaste, we strive to bring you the best service at the best price available. We have a wide selection of dumpster sizes in stock in Ceylon, MN, with flexible & timley pickup and delivery. Give us a call today for all your dumpster rental and roll off needs.
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Tourism, the main way to obtain economy and is experiencing a down turn in the recent periods. Recession could be the main reason intended for decreased number of visitors to Guam and alter in the climatic conditions adds to the idea. Why is environmental surroundings of the condition suffering severe effects of smog? The answer can be none other than the careless acts of citizens of the condition i.electronic. improper disposal of the trash.
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The ht-2117 multimedia spk 2117 for computer (black) Speakers & Headset is manufactured by 5 CORE and was added around July 2017.
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Winter Kids Girls Boys Ski Suit Set Waterproof Children SnowSuit 2T 4T 6T Children Romper Overall Windproof Jumpsuit Girls Boys Ski Suit Set padded with cotton ,with windproof and waterproof material. Girls Boys Ski Suit Set is Warm and soft for cold winter. Girls Boys Ski Suit Set for kids from 90cm to 150cm, 12M to 2 4 6 8 10 years old child. 1.


Stellar Fountain Records celebrating the 5th anniversary in 2016. With 200 release behind their back the Stellar-guys built a decent sound, which is continously looking for fresh inspirations and new, exciting faces, while stay true to the underground progressive roots. The result can be checked on numerous beatport chartings, 90 of the label's releases were featured on beatport top 100, regular appereances on must have progressive house top 10 selections. The label's priority to support their local artists and give them the chance to spin live on parties as dj's.


Could Everton secure a bargain"
Marco Silva and Everton are reportedly interested in signing £14million rated Chelsea striker Michy Batshuayi after his Valencia loan was cut short.
The Belgian has had a torrid time during the current season.
The Chelsea man had an impressive loan at Borussia Dortmund midway through last season, scoring 7 goals in 10 games.
However, Dortmund decided against paying what Chelsea wanted for him and that left the player searching for another loan move.
He then chose Valencia as his club current for this season, a side who would test his ability and allow him to continue competing at the highest level.
Though, it was a stark contrast to his Dortmund days.Get notified!
The player has only scored two goals in the current campaign and it was announced on Thursday that his loan had been cut short and he would head back to London.
Now, several clubs are fighting to sign him and Everton are one of them.
Is Batshuayi worth paying £14million for"
That is according to several outlets on Thursday.
All of the Mail, Mirror and the Telegraph have reported that Everton are interested in a deal.
The Mirror does claim the player is rated at £14million which could actually end up being a steal considering his performances in the Bundesliga.
The Toffees are also in need of a striker with all of Dominic Calvert-Lewin, Cenk Tosun and Oumar Niasse not quite cutting it.
Richarlison has been used up front at several times this season due to that lack of quality options.
Everton are not the only ones interested in Batshuayi though.
The Mirror's report does also cite Fulham, Crystal Palace and Batshuayi's former club Marseille as potential suitors.
Despite his poor loan spell, the Belgian will have plenty of clubs to choose from this month.
Safe to say, he will be hoping to finally find a permanent home.


November 9, 2015 was the 40th anniversary of the big tournament in Greensboro to crown a new United States heavyweight champion after the title had been vacated by Johnny Valentine following the tragic plane crash in Wilmington, NC a month earlier.
"The Tournament" was a pivotal night for Jim Crockett promotions and as part of the events of that fall set the direction of the company for years to come. It is one of the most famous "big events" in Jim Crockett Promotions history and featured some of the best talent from across the NWA.
Below are links to the five anniversary posts we made the week of the 40th anniversary, including some amazing vintage audio clips from TV at the time, as well as first hand accounts from Paul Jones and Terry Funk.
Detailed results of the entire 16-man, 15 match tournament, including final brackets. Also, a look at the five judges appointed by the NWA.
Back in 2005 when we celebrated the 30th anniversary of the tournament, I spoke with both Paul Jones and Terry Funk about that grueling night in Greensboro.
Vintage audio recordings made with a portable cassette tape recorder in 1975. Highlights from matches including Harley Race vs. Paul Jones, Blackjack Mulligan vs. Dusty Rhodes, and Terry Funk vs. Paul Jones in the final. Commentary by the legendary Ed Capral who had just joined Jim Crockett Promotions to host the new program "Wide World Wrestling."
Images of the original event poster and newspaper articles on the tournament.
More vintage audio recordings from the TV shows of special interviews in the weeks that followed the tournament. Blackjack Mullgian expresses outrage over how Tim Woods helped Dusty Rhodes defeat him in their quarter final match. Plus Paul Jones tells fans he let them down by letting Terry Funk beat him and take the title out of the area. But he promises to bring it home in the rematch!
If you would like to view all five updates on the same page, click here.


WASHINGTON, DC -- NAB released a statement today in response to a press release issued by the musicFIRST Coalition, an organization backed by the Recording Industry Association of America. In their release, musicFIRST, which is seeking legislation in Congress that would require local radio stations to begin paying a new fee for music aired free to listeners, attempted to trivialize the cost-burden of the new fee on small market radio stations.
"NAB is delighted by the growing bipartisan opposition in Congress to RIAA's attempted fleecing of America's hometown radio stations. Five thousand dollars may not sound like a lot of money to a fat-cat foreign record label mogul from Paris, France. But in Paris, Texas, a $5,000 loss in revenue could threaten a station's ability to carry AMBER Alerts, emergency weather warnings, and high school football games."
NAB announced earlier today that nine additional lawmakers, including House Ways and Means Committee Chairman Charles Rangel (NY-15), have signed on to a bipartisan U.S. House of Representatives resolution opposing the introduction of "any new performance fee, tax, royalty, or other charge" on local radio stations. The growing chorus of support for The Local Radio Freedom Act now stands at 158.


When our trusty Consumer Reports landed in our mailbox last week, I was interested to see a great story on supermarkets. I pored over the story and the always great visuals, learning that my beloved Trader Joe's scored really well, and that the other two (non-Whole Foods) chains in the area were at the bottom of the pile, scoring only marginally better than the epicurean gem, Wal-Mart.
I was not surprised. I grew up shopping at Stop and Shop and Shaw's, and with some notable exceptions, I was unimpressed. I've always hated Shaw's–terrible produce, sad meats, uninteresting food. Great for a bargain shopper and definitely serves a purpose, but it hasn't served mine. Stop and Shop is a bit better but it varies by store. For example, the S+S in Bristol, RI is excellent compared to the one near my house (which has a meat section about the size of a Ford Taurus, yet boasts a potato chip and soda aisle that could double as a main street thru town).
I used my best third grade handwriting for you.
My go to shopping spot is Trader Joe's. It is 3.5 minutes from my house, it is cheap, it is interesting. I do not like their produce for the most part, and I only occasionally buy meat there, but it works. I shop there probably once every ten days.
Oh, Dave's. I love Dave's. It's a local chain that focuses on high quality foods, local (when possible) and gorgeous (always) produce and meats, and super fresh seafood. But I don't shop there as frequently as I used to. The locations are sort of weirdly spread out for me to be able to shop on my way home from work, which is the only time I really shop.
The rest are the premise for this blog: the farmer's market (we love the Coastal Growers Market, both winter and summer), our two CSAs, and the milk man. I actually rely on those sources a lot more than bricks-and-mortar stores because they are local, they are cost-effective, the quality is through the roof, and they provide a really rich and satisfying experience. The experience is incredibly important to me, and we both enjoy the connection that we make when we engage in these food experiences.
A few final words on the CR supermarket report: 1) I am apparently categorized as an impulse shopper because I go in knowing what I want and which brand I will buy. Apparently it is opposite day at CR; 2) I love Publix, which scored well. Jealous of my Florida family!; 3) I do occasionally shop at Whole Foods, but it's rare, and it has little to do with the price. I've just been able to replicate the WF experience and quality elsewhere; and finally 4) I don't eat seafood, and Steve does. Steve, therefore, is responsible for the seafood for the house and does so by either catching it himself or going to a local fish market manned by a fisherman's son that we know from childhood. That kid sells a great fish, I'm told!
Where do you shop? What do you buy where?
It's better that way, lol.
Now I'm reading it that way too! Totally apropo.
I miss the little markets and bakeries and pastry shops of BCN. Nothing like samosas from a shoebox on the counter at the market withthe grandma sitting in the back of the store! Talk about a food experience!
Or ordering a sandwich platter and getting toast topped with baby snakes….
I start first and foremost with our CSA. Then I move to our community market that represents other local food vendors and finally to our locally owned grocery store. Then to TJ's for the exotic stuff that I can't get elsewhere. I love that you brought consumer reports into this post.


text:
Kenyam is a town and district in Nduga Regency, Highland Papua, Indonesia. It is the capital of the regency. Its population is 8148.

History
On 31 December 2015, President Joko Widodo, accompanied by First Lady Iriana, visited Kenyam, Nduga Regency, Papua Province, to review the construction of the road that will connect Nduga and Wamena. The trip of  Widodo and First Lady Iriana and his entourage from Wamena to Kenyam, Nduga Regency was taken for 2 hours by changing transportation modes from Hercules VVIP A 1314 and Heli Super Puma TNI AU. On that occasion President Jokowi also conveyed, in addition to building a road leading to Wamena, in the Nduga region Mumugu's large port will also be opened. With the opening of this large port, the logistics and materials from and to Mumugu can be distributed using land routes that have penetrated all districts in Papua.

Climate
Kenyam has a tropical rainforest climate (Af) with heavy rainfall year-round.

References

Populated places in Highland Papua
Regency seats of Highland Papua

redpajama_set_name: RedPajamaWikipedia