text:
\section{Introduction}
In the seminar paper \cite{LY}, Li and Yau first proved a gradient estimate for the positive solution of heat equation
 \begin{equation}
\frac{\partial}{\partial t}u=\Delta u
 \end{equation}
 on a Riemannian manifold and derived the classical Harnack inequality by integrating the gradient estimate along the paths in space-time. This estimate has tremendous impact in the past twenty years. On the one hand, similar technique was employed by Hamilton in study of  Harnack inequalities for Ricci flow in \cite{H3,H5} and mean curvature flow in \cite{H7}. So this class of Harnack inequalities was called Li-Yau-Hamilton estimate in \cite{NT1}. Similar Li-Yau-Hamilton estimates for Gauss curvature flow and Yamabe flow were proved by Chow in \cite{Ch1,Ch3}. Meanwhile, the estimate for K\"{a}hler-Ricci flow was proved by Cao in \cite{C2}. Later in \cite{A}, the corresponding result was established for a general class of hypersurface flows by Andrews. Moreover, a geometric interpretation of Hamilton quantities in the Li-Yau-Hamilton estimate for Ricci flow was given by Chow and Chu in \cite{CC1}. Using this geometric approach, Chow and Knopf gave a new Li-Yau-Hamilton estimate for Ricci flow in \cite{CK}.

  On the other hand, Hamilton extended Li-Yau's gradient estimate of heat equation to the full matrix version in \cite{H4}. Precisely, he proved that a positive solution of heat equation
on a closed Riemannian manifold with nonnegative sectional curvature and parallel Ricci curvature satisfies the estimate
\begin{equation}
\nabla_i\nabla_j\ln u+\frac{1}{2t}g_{ij}\geq 0.
\end{equation}
Moreover, he derived several important monotonicity formulae out of the estimate in \cite{H6}.
 The assumption that $(M,g)$ has parallel Ricci curvature in Hamilton's estimate is rather restrictive and can be removed if the manifold is K\"{a}hler. This was observed by Cao and Ni for the K\"{a}hler manifold with fixed metric in \cite{CN} and by Chow and Ni for the K\"{a}hler manifold with time dependent metrics evolving by K\"{a}hler-Ricci flow in \cite{N}. These two estimates can be unified by the interpolation consideration originally proposed by Chow in \cite{Ch4}.
\vspace{2mm}\newline
{\bf Theorem A}(Chow-Ni) {\it Assume that $(M,g_{i\bar{j}}(t))$ has nonnegative holomorphic bisectional curvature satisfying $\epsilon$-K\"{a}hler-Ricci flow
 \begin{equation}
 \frac{\partial}{\partial t}g_{i\bar{j}}=-\epsilon R_{i\bar{j}}.
 \end{equation}
 In the case that $M$ is complete noncompact, additionally assume that the bisectional curvature is bounded. If $u$ is a positive solution of
\begin{equation}
\frac{\partial}{\partial t}u=\Delta u+\epsilon Ru,
\end{equation}
where $R$ is the scalar curvature, then
\begin{equation}
\nabla_i\nabla_{\bar{j}}\ln u+\epsilon R_{i\bar{j}}+\frac{1}{t}g_{i\bar{j}}\geq 0.
\end{equation}}
It should be pointed out that there are also other important works along this direction, see \cite{CC2,N2,N4,NT1,NT2}.

To better understand how the Li-Yau-Hamilton estimate for the Ricci flow can be perturbed or extended, which is important in studying singularities of the Ricci flow, Chow and Hamilton extended the matrix Li-Yau-Hamilton estimate on Riemannian manifolds to the constrained case in \cite{CH}.
\vspace{2mm}\newline
{\bf Theorem B}(Chow-Hamilton) {\it Let $(M,g)$ be a compact Riemannian manifold with nonnegative sectional curvature and parallel Ricci curvature. If $u$ and $v$ are positive solutions of heat equation
\begin{equation}
\frac{\partial}{\partial t}u=\Delta u,\hspace{4mm}\frac{\partial}{\partial t}v=\Delta v,
\end{equation}
with $|v|<u$, then
\begin{equation}
\nabla_i\nabla_j\ln u+\frac{1}{2t}g_{ij}> \frac{\n_ih\n_{j}h}{1-h^2},
\end{equation}
where $h=v/u$.}\vspace{2mm}

In this paper, we will extend the Li-Yau-Hamilton estimates due to Cao-Ni and Chow-Ni on K\"{a}hler manifolds to the constrained case.
\begin{theorem}
Let $(M,g_{i\bar{j}}(t))$ be a compact solution of $\epsilon$-K\"{a}hler-Ricci flow
\begin{equation}
 \frac{\partial}{\partial t}g_{i\bar{j}}=-\epsilon R_{i\bar{j}}.
 \end{equation}
 If $u$ and $v$ are solutions of the equation
\begin{equation}
\frac{\partial}{\partial t}u=\Delta u+\epsilon Ru, \hspace{6mm}\frac{\partial}{\partial t}v=\Delta v+\epsilon Rv \label{heatequ}
\end{equation}
with $|v|<u$, then we have
\begin{equation}
\n_i\n_{\bar{j}}\ln u+\frac{1}{t}g_{i\bar{j}}+\epsilon R_{i\bar{j}}> \frac{\n_ih\n_{\bar{j}}h}{1-h^2},
\end{equation}
where $h=v/u$.
\end{theorem}

Set $\epsilon=0$, and we get the constrained Li-Yau-Hamilton estimate on K\"{a}hler manifolds.
\begin{corollary}
Let $M$ be a compact K\"{a}hler manifold with nonnegative holomorphic bisectional curvature.
 If $u$ and $v$ are the solutions of the heat equation
\begin{equation}
\frac{\partial}{\partial t}u=\Delta u, \hspace{6mm}\frac{\partial}{\partial t}v=\Delta v
\end{equation}
with $|v|<u$, then we have
\begin{equation}
\n_i\n_{\bar{j}}\ln u+\frac{1}{t}g_{i\bar{j}}> \frac{\n_ih\n_{\bar{j}}h}{1-h^2},
\end{equation}
where $h=v/u$.
\end{corollary}

Taking $\epsilon=1$, we get the constrained Li-Yau-Hamilton estimate on K\"{a}hler manifolds under the K\"{a}hler-Ricci flow.
\begin{corollary}
Let $(M,g_{i\bar{j}}(t))$ be a compact solution of K\"{a}hler-Ricci flow
\begin{equation}
 \frac{\partial}{\partial t}g_{i\bar{j}}=-R_{i\bar{j}}.
 \end{equation}
 If $u$ and $v$ are the solutions of the forward conjugate heat equation
\begin{equation}
\frac{\partial}{\partial t}u=\Delta u+Ru, \hspace{6mm}\frac{\partial}{\partial t}v=\Delta v+Rv
\end{equation}
with $|v|<u$, then we have
\begin{equation}
\n_i\n_{\bar{j}}\ln u+\frac{1}{t}g_{i\bar{j}}+R_{i\bar{j}}> \frac{\n_ih\n_{\bar{j}}h}{1-h^2},
\end{equation}
where $h=v/u$.
\end{corollary}

\begin{remark}If $h=C\in(-1,1)$ is a constant, then $\nabla h=0$ and we recover the Li-Yau-Hamilton estimates in \cite{CN} and \cite{N}. Since the manifold in Theorem A is complete, we expect the result also hold on complete noncompact manifolds under suitable condition.\end{remark}

\begin{remark}
Since the constrained Li-Yau-Hamilton estimate on Riemannian manifolds helps to understand the singularities of the Ricci flow, we also expect our estimates are useful in investigating the singularities of K\"{a}hler-Ricci flow.
\end{remark}

The rest of the paper is organized as follows. We devote section 2 to the notations and basic formulas in K\"{a}hler geometry. In section 3, we prove the Theorem 1.1 by using Hamilton's maximum principle for tensors.

\section{Preliminaries}\setcounter{equation}{0}
Let $(M,g_{i\bar{j}})$ be a  K\"{a}hler manifold with K\"{a}hler metric $g_{i\bar{j}}$. The K\"{a}hler form
\begin{equation}
\omega=\frac{\sqrt{-1}}{2}g_{i\bar{j}}dz^i\wedge d\bar{z}^j
\end{equation}
is a closed real $(1,1)$-form, which is equivalent to
\begin{equation}
\frac{\partial g_{i\bar{j}}}{\partial z^k}=\frac{\partial g_{k\bar{j}}}{\partial z^i},\hspace{4mm}\frac{\partial g_{i\bar{j}}}{\partial \bar{z}^k}=\frac{\partial g_{i\bar{k}}}{\partial \bar{z}^j}.
\end{equation}

The Christoffel symbols of the metric $g_{i\bar{j}}$ is defined as
\begin{equation}
\Gamma^k_{ij}=g^{k\bar{l}}\frac{\partial g_{i\bar{l}}}{\partial z^j}, \hspace{4mm} \Gamma^{\bar{l}}_{\bar{i}\bar{j}}=g^{k\bar{l}}\frac{\partial g_{k\bar{j}}}{\partial \bar{z}^i},
\end{equation}
where $g^{i\bar{j}}=(g_{i\bar{j}})^{-1}$. It is easy to see that $\Gamma^k_{ij}$ is symmetric in $i$ and $j$ and $\Gamma^{\bar{l}}_{\bar{i}\bar{j}}$ is symmetric in $\bar{i}$ and $\bar{j}$.

The curvature tensor of the metric $g_{i\bar{j}}$ is defined by
\begin{equation}
R^j_{ik\bar{l}}=\frac{\partial\Gamma^j_{ik}}{\partial \bar{z}^l},\hspace{4mm}
R_{i\bar{j}k\bar{l}}=g_{p\bar{j}}R^p_{ik\bar{l}}.
\end{equation}
It is easy to see that $R_{i\bar{j}k\bar{l}}$ is symmetric in $i$ and $k$, in $\bar{j}$ and $\bar{l}$ and in pairs $i\bar{j}$ and $k\bar{l}$. The second Bianchi identity in K\"{a}hler case reduces to
\begin{equation}
\nabla_p R_{i\bar{j}k\bar{l}}=\nabla_k R_{i\bar{j}p\bar{l}}, \hspace{4mm}\nabla_{\bar{q}}R_{i\bar{j}k\bar{l}}=\nabla_{\bar{l}}R_{i\bar{j}p\bar{q}}.
\end{equation}

We say that $M$ has nonnegative holomorphic bisectional curvature if
\begin{equation}
R_{i\bar{j}k\bar{l}}v^iv^{\bar{j}}w^kw^{\bar{l}}\geq 0
\end{equation}
for all nonzero vectors $v$ and $w$ in the holomorphic tangent bundle $T_xM$ at $x\in M$.

The Ricci tensor of the metric $g_{i\bar{j}}$ is obtained by taking the trace of the curvature tensor:
\begin{equation}
R_{i\bar{j}}=g^{k\bar{l}}R_{{i\bar{j}k\bar{l}}}
\end{equation}
and the scalar curvature is
\begin{equation}
R=g^{i\bar{j}}R_{i\bar{j}}.
\end{equation}

Finally, we give the commutation formulas for covariant differentiations in K\"{a}hler geometry. Covariant differentiations of the same type can be commuted freely, e.g.,
\begin{equation}
\nabla_k\nabla_jv_i=\nabla_j\nabla_kv_i,\hspace{4mm}\nabla_{\bar{k}}\nabla_{\bar{j}}v_i=\nabla_{\bar{j}}\nabla_{\bar{k}}v_i.
\end{equation}
But we shall need following formulas when commuting covariant derivatives of different types:
\begin{equation}
\nabla_k\nabla_{\bar{j}}v_i=\nabla_{\bar{j}}\nabla_kv_i-R_{k{\bar{j}}i{\bar{l}}}v_l.
\end{equation}

\section{Constrained Li-Yau-Hamilton estimate}
\setcounter{equation}{0}
In this section, we will give the proof the Theorem 1.1. First we give some lemmas used in the proof. It should be pointed out that all the computations are taken in normal coordinate.
\begin{lemma} Set $L=\ln u$, then the evolution equation of $\n_{i}\n_{\bar{j}}L$ is
\begin{equation}
\begin{split}
\frac{\partial}{\partial t}\nabla_i\nabla_{\bar{j}}L=&\Delta\left(\nabla_i\nabla_{\bar{j}}L\right)+R_{l\bar{j}i\bar{k}}\nabla_k\nabla_{\bar{l}}L
+R_{i\bar{j}k\bar{l}}{\nabla_{l}L\nabla_{\bar{k}}L}\\&+\nabla_{k}L\nabla_{\bar{k}}\nabla_i\nabla_{\bar{j}}L+\nabla_{\bar{k}}L\nabla_k\nabla_i\nabla_{\bar{j}}L
\\&+\nabla_{\bar{j}}\nabla_{k}L\nabla_i\nabla_{\bar{k}}L+\nabla_i\nabla_{k}L\nabla_{\bar{j}}\nabla_{\bar{k}}L
\\&-\frac{1}{2}\left(R_{l\bar{j}}\nabla_i\nabla_{\bar{l}}L+R_{i\bar{l}}\nabla_l\nabla_{\bar{j}}L\right)\\&+\epsilon\left(\Delta R_{i{\bar{j}}}+R_{i{\bar{j}}k{\bar{l}}}{ R_{{\bar{k}}l}}-R_{i{\bar{p}}}R_{p{\bar{j}}}\right).
\end{split}
\end{equation}
\end{lemma}
Proof: By
\begin{equation}
\frac{\partial}{\partial t}L=\frac{\partial}{\partial t}\ln u=\frac{u_t}{u}=\frac{1}{u}\left(\Delta u+\epsilon Ru\right)=\frac{\Delta u}{u}+\epsilon R,
\end{equation}
and
\begin{equation}
\Delta L=\n_i\n_{\bar{i}}L=\n_i\frac{\n_{\bar{i}}u}{u}=\frac{\Delta u}{u}-\frac{|\nabla u|^2}{u^2},
\end{equation}
we get the evolution equation of $L$
\begin{equation}
\frac{\partial}{\partial t}L=\Delta L+\frac{|\nabla u|^2}{u^2}+\epsilon R=\Delta L+|\n L|^2+\epsilon R.
\end{equation}
We claim that if a function $A$ satisfies the equation
\begin{equation}
\frac{\partial}{\partial t}A=\Delta {A}+B,
\end{equation}
then $\nabla_i\nabla_{\bar{j}}A$ satisfies
\begin{equation}
\begin{split}
\frac{\partial}{\partial t}\left(\nabla_i\nabla_{\bar{j}}A\right)=&\Delta\left(\nabla_i\nabla_{\bar{j}}A\right)+R_{i\bar{j}l\bar{k}}\nabla_k\nabla_{\bar{l}}A\\&-\frac{1}{2}\left(R_{l\bar{j}}\nabla_i\nabla_{\bar{l}}A+R_{i\bar{l}}\nabla_l\nabla_{\bar{j}}A\right)
+\nabla_i\nabla_{\bar{j}}B.
\end{split}
\end{equation}
In fact a straightforward computation leads to
\begin{equation}
\begin{split}
\frac{\partial}{\partial t}\left(\nabla_i\nabla_{\bar{j}}A\right)&=\nabla_i\nabla_{\bar{j}}\left(\frac{\partial}{\partial t}A\right)\\
&=\nabla_i\nabla_{\bar{j}}\left(\Delta {A}+B\right)\\
&=\nabla_i\nabla_{\bar{j}}\left(\Delta {A}\right)+\nabla_i\nabla_{\bar{j}}B. \label{evol1}
\end{split}
\end{equation}
It is sufficient to compute
 $$\nabla_i\nabla_{\bar{j}}\left(\Delta {A}\right)=\frac{1}{2}\nabla_i\nabla_{\bar{j}}\left(\nabla_k\nabla_{\bar{k}}A\right)+\nabla_i\nabla_{\bar{j}}\left(\nabla_{\bar{k}}\nabla_{k}A\right).$$
 By using communication formula, we get
\begin{equation}
\begin{split}
\nabla_i\nabla_{\bar{j}}\left(\nabla_k\nabla_{\bar{k}}A\right)=&\nabla_i\left(\nabla_k\nabla_{\bar{j}}\nabla_{\bar{k}}A-R_{l\bar{j}}\nabla_{\bar{l}}A\right)\\
=&\nabla_k\nabla_i\nabla_{\bar{k}}\nabla_{\bar{j}}A-\nabla_i{R_{l\bar{j}}}\nabla_{\bar{l}}A-R_{l\bar{j}}\nabla_i\nabla_{\bar{l}}A\\
=&\nabla_k\left(\nabla_{\bar{k}}\nabla_i\nabla_{\bar{j}}A+R_{i\bar{k}l\bar{j}}\nabla_{\bar{l}}A\right)-\nabla_i{R_{l\bar{j}}}\nabla_{\bar{l}}A-R_{l\bar{j}}\nabla_i\nabla_{\bar{l}}A\\
=&\nabla_k\nabla_{\bar{k}}\nabla_i\nabla_{\bar{j}}A+R_{i\bar{k}l\bar{j}}\nabla_k\nabla_{\bar{l}}A+\nabla_k{R_{i\bar{k}l\bar{j}}}\nabla_{\bar{l}}A
-\nabla_i{R_{l\bar{j}}}\nabla_{\bar{l}}A-R_{l\bar{j}}\nabla_i\nabla_{\bar{l}}A\\
=&\nabla_k\nabla_{\bar{k}}\nabla_i\nabla_{\bar{j}}A+R_{i\bar{j}l\bar{k}}\nabla_k\nabla_{\bar{l}}A-R_{l\bar{j}}\nabla_i\nabla_{\bar{l}}A  \label{evol2}
\end{split}
\end{equation}
and
\begin{equation}
\begin{split}
\nabla_i\nabla_{\bar{j}}\left(\nabla_{\bar{k}}\nabla_{k}A\right)=&\nabla_i\left(\nabla_{\bar{k}}\nabla_{\bar{j}}\nabla_{k}A\right)\\=&\nabla_i\left(\nabla_{\bar{k}}\nabla_{k}\nabla_{\bar{j}}A\right)\\
=&\nabla_{\bar{k}}\nabla_i\nabla_k\nabla_{\bar{j}}A+R_{i\bar{k}l\bar{j}}\nabla_k\nabla_{\bar{l}}A-R_{i\bar{k}k\bar{l}}\nabla_l\nabla_{\bar{j}}A\\
=&\nabla_{\bar{k}}\nabla_k\left(\nabla_i\nabla_{\bar{j}}A\right)+R_{i\bar{j}l\bar{k}}\nabla_k\nabla_{\bar{l}}A-R_{i\bar{l}}\nabla_l\nabla_{\bar{j}}A. \label{evol3}
\end{split}
\end{equation}
Inserting (\ref{evol2}) and (\ref{evol3}) into (\ref{evol1}), we get the evolution equation of $\nabla_i\nabla_{\bar{j}}A$.

Applying this formula to $A=L=\ln u$, we obtain
\begin{equation}
\begin{split}
\frac{\partial}{\partial t}\left(\nabla_i\nabla_{\bar{j}}L\right)=&\Delta\left(\nabla_i\nabla_{\bar{j}}L\right)+R_{i\bar{j}l\bar{k}}\nabla_k\nabla_{\bar{l}}L\\
&-\frac{1}{2}\left(R_{l\bar{j}}\nabla_i\nabla_{\bar{l}}L+R_{i\bar{l}}\nabla_l\nabla_{\bar{j}}L\right)+\nabla_i\nabla_{\bar{j}}|\nabla L|^2
+\epsilon \nabla_i\nabla_{\bar{j}}R.
\end{split}
\end{equation}
By Lemma 1.1 in \cite{N}, we get
\begin{equation}
 \nabla_i\nabla_{\bar{j}} R=\Delta R_{i{\bar{j}}}+R_{i{\bar{j}}k{\bar{l}}}R_{{\bar{k}}l}-R_{i{\bar{p}}}R_{p{\bar{j}}}.
\end{equation}
Now the lemma follows from the following computation
\begin{equation}
\begin{split}
\nabla_i\nabla_{\bar{j}}|\nabla L|^2=&\nabla_i\nabla_{\bar{j}}\left(\nabla_{k}L\nabla_{\bar{k}}L\right)\\=&\nabla_i\left(\nabla_{\bar{j}}\nabla_{k}L\nabla_{\bar{k}}L+\nabla_{k}L\nabla_{\bar{j}}\nabla_{\bar{k}}L\right)
\\=&\nabla_i\nabla_{\bar{j}}\nabla_{k}L\nabla_{\bar{k}}L+\nabla_{\bar{j}}\nabla_{k}L\nabla_i\nabla_{\bar{k}}L+\nabla_i\nabla_{k}L\nabla_{\bar{j}}\nabla_{\bar{k}}L
+\nabla_{k}L\nabla_i\nabla_{\bar{j}}\nabla_{\bar{k}}L
\\=&\nabla_{\bar{k}}L\nabla_k\left(\nabla_i\nabla_{\bar{j}}L\right)+\nabla_{\bar{j}}\nabla_{k}L\nabla_i\nabla_{\bar{k}}L\\&+\nabla_i\nabla_{k}L\nabla_{\bar{j}}\nabla_{\bar{k}}L
+\nabla_{k}L\nabla_{\bar{k}}\left(\nabla_i\nabla_{\bar{j}}L\right)+R_{i\bar{j}k\bar{l}}\nabla_{l}L\nabla_{\bar{k}}L.
\end{split}
\end{equation}

\begin{lemma}(see \cite{N}) If K\"{a}hler metrics $g_{i\bar{j}}(t)$ satisfy the $\epsilon$-K\"{a}hler-Ricci flow, then the Ricci tensor satisfies
\begin{equation}
\frac{\partial}{\partial t}\left(\epsilon R_{i\bar{j}}\right)=\epsilon^{2}\left(\Delta R_{i{\bar{j}}}+R_{i{\bar{j}}k{\bar{l}}}R_{{\bar{k}}l}-R_{i{\bar{p}}}R_{p{\bar{j}}}\right).
\end{equation}
\end{lemma}

\begin{lemma}Let $h=v/u$, then
\begin{equation}
\begin{split}
\frac{\partial}{\partial t}\left(\frac{\n_{i}h\n_{\bar{j}}h}{1-h^2}\right)
=&\Delta\left(\frac{\n_{i}h\n_{\bar{j}}h}{1-h^2}\right)+\n_{k}L\n_{\bar{k}}\left(\frac{\n_{i}h\n_{\bar{j}}h}{1-h^2}\right)+\n_{\bar{k}}L\n_{k}\left(\frac{\n_{i}h\n_{\bar{j}}h}{1-h^2}\right)\\
&-\frac{1}{1-h^2}\left(\n_{i}\n_{k}h+\frac{2h\n_{i}h\n_{k}h}{1-h^2}\right)\left(\n_{\bar{j}}\n_{\bar{k}}h+\frac{2h\n_{\bar{j}}h\n_{\bar{k}}h}{1-h^2}\right)\\
&-\frac{1}{1-h^2}\left(\n_{i}\n_{\bar{k}}h+\frac{2h\n_{i}h\n_{\bar{k}}h}{1-h^2}\right)\left(\n_{\bar{j}}\n_{k}h+\frac{2h\n_{\bar{j}}h\n_{k}h}{1-h^2}\right)\\
&+\n_{i}\n_{k}L\left(\frac{\n_{\bar{j}}h\n_{\bar{k}}h}{1-h^2}\right)+\n_{i}\n_{\bar{k}}L\left(\frac{\n_{\bar{j}}h\n_{k}h}{1-h^2}\right)\\&+\n_{\bar{j}}\n_{k}L\left(\frac{\n_{i}h\n_{\bar{k}}h}{1-h^2}\right)
+\n_{\bar{j}}\n_{\bar{k}}L\left(\frac{\n_{i}h\n_{k}h}{1-h^2}\right)
\\&-\frac{1}{2}R_{i\bar{l}}\left(\frac{\n_{\bar{j}}h\n_{l}h}{1-h^2}\right)-\frac{1}{2}R_{k\bar{j}}\left(\frac{\n_{\bar{k}}h\n_{i}h}{1-h^2}\right)\\&-\frac{2\n_{i}h\n_{\bar{j}}h}{\left(1-h^2\right)^2}|\n h|^2.
\end{split}
\end{equation}
\end{lemma}
Proof: We divide our proof in fours steps. First, we need to give the evolution equation of $h$. By the definition of $h$, we have
\begin{equation}
\frac{\partial}{\partial t}h=\frac{\partial}{\partial t}\left(\frac{v}{u}\right)=\frac{v_t}{u}-\frac{vu_t}{u^2}=\frac{1}{u^2}\left(u\Delta v-v\Delta u\right)\label{i1}
\end{equation}
and
\begin{equation}
\begin{split}
\Delta h=\nabla_k\nabla_{\bar{k}}h=&\nabla_k\frac{u\nabla_{\bar{k}}v-v\nabla_{\bar{k}}u}{u^2}
\\=&\frac{\nabla_ku\nabla_{\bar{k}}v+u\nabla_k\nabla_{\bar{k}}v-\nabla_kv\nabla_{\bar{k}}u-v\nabla_k\nabla_{\bar{k}}u}{u^2}
\\&-\frac{2\nabla_ku\left(u\nabla_{\bar{k}}v-v\nabla_{\bar{k}}u\right)}{u^3}
\\=&\frac{u\Delta v-v\Delta u}{u^2}-\frac{\nabla_{k}u\nabla_{\bar{k}}v+\nabla_kv\nabla_{\bar{k}}u}{u^2}+\frac{2v\nabla_{\bar{k}}u\nabla_{k}u}{u^3}
\\=&\frac{u\Delta v-v\Delta u}{u^2}-\nabla_{k}L\nabla_{\bar{k}}h-\nabla_{\bar{k}}L\nabla_{k}h.\label{i2}
\end{split}
\end{equation}
Combining (\ref{i1}) and (\ref{i2}), we obtain the evolution of $h$
\begin{equation}
\begin{split}
\frac{\partial}{\partial t}h=&\Delta h+\nabla_{k}L\nabla_{\bar{k}}h+\nabla_{\bar{k}}L\nabla_{k}h.
\end{split}
\end{equation}

The next thing to do in the proof is computing the evolution equation of $1-h^2$. By
\begin{equation}
\begin{split}
\frac{\partial}{\partial t}\left(1-h^2\right)=&-2h\frac{\partial}{\partial t}h\\=&-2h\left(\Delta h+\nabla_{k}L\nabla_{\bar{k}}h+\nabla_{\bar{k}}L\nabla_{k}h\right)
\end{split}
\end{equation}
and
\begin{equation}
\Delta\left(1-h^2\right)=-2|\nabla h|^2-2h\Delta h,
\end{equation}
we have that
\begin{equation}
\begin{split}
\frac{\partial}{\partial t}\left(1-h^2\right)=&\Delta\left(1-h^2\right)+2|\nabla h|^2{}-2h\left(\nabla_{k}L\nabla_{\bar{k}}h+\nabla_{\bar{k}}L\nabla_{k}h\right)\\=&\Delta\left(1-h^2\right)+2|\nabla h|^2+\nabla_{k}L\nabla_{\bar{k}}\left(1-h^2\right)+\nabla_{\bar{k}}L\nabla_{k}\left(1-h^2\right). \label{gradient evol1}
\end{split}
\end{equation}

Another step in the proof is computing the evolution equation of $\nabla_{i}h\nabla_{\bar{j}}h$. Take partial derivative with respective to $t$
\begin{equation}
\frac{\partial}{\partial t}\left(\nabla_{i}h\nabla_{\bar{j}}h\right)=\nabla_{i}\left(\frac{\partial}{\partial t}h\right)\nabla_{\bar{j}}h+\nabla_{i}h\nabla_{\bar{j}}\left(\frac{\partial}{\partial t}h\right).
\end{equation}
Direct computation yields
\begin{equation}
\begin{split}
\nabla_{i}\left(\frac{\partial}{\partial t}h\right)\nabla_{\bar{j}}h=&\nabla_{i}\left(\Delta h+\nabla_{k}L\nabla_{\bar{k}}h+\nabla_{\bar{k}}L\nabla_{k}h\right)\nabla_{\bar{j}}h
\\=&\nabla_{\bar{j}}h\Delta\left(\nabla_{i}h\right)-\frac{1}{2}R_{i\bar{l}}\nabla_{l}h\nabla_{\bar{j}}h+\nabla_i\nabla_{k}L\nabla_{\bar{k}}h\nabla_{\bar{j}}h
\\&+\nabla_{k}L\nabla_i\nabla_{\bar{k}}h\nabla_{\bar{j}}h+\nabla_i\nabla_{\bar{k}}L\nabla_{k}h\nabla_{\bar{j}}h+\nabla_{\bar{k}}L\nabla_i\nabla_{k}h\nabla_{\bar{j}}h
\end{split}\label{i3}
\end{equation}
and
\begin{equation}
\begin{split}
\nabla_{\bar{j}}\left(\frac{\partial}{\partial t}h\right)\nabla_{i}h=&\nabla_{\bar{j}}\left(\Delta h+\nabla_{k}L\nabla_{\bar{k}}h+\nabla_{\bar{k}}L\nabla_{k}h\right)\nabla_{i}h
\\=&\n_{\bar{j}}\Delta h\n_{i}h+\n_{\bar{j}}\n_{k}L\n_{\bar{k}}h\n_{i}h+\n_{k}L\n_{\bar{j}}\n_{\bar{k}}h\n_{i}h\\
&+\n_{\bar{j}}\n_{\bar{k}}L\n_{k}h\n_{i}h+\n_{\bar{k}}L\n_{\bar{j}}\n_{k}h\n_{i}h\\
=&\Delta\left(\n_{\bar{j}}h\right)\n_{i}h-\frac{1}{2}R_{k\bar{j}}\n_{\bar{k}}h\n_{i}h\\
&+\n_{\bar{j}}\n_{k}L\n_{\bar{k}}h\n_{i}h+\n_{k}L\n_{\bar{j}}\n_{\bar{k}}h\n_{i}h\\
&+\n_{\bar{j}}\n_{\bar{k}}L\n_{k}h\n_{i}h+\n_{\bar{k}}L\n_{\bar{j}}\n_{k}h\n_{i}h.
\end{split}\label{i4}
\end{equation}
Combining (\ref{i3}) and (\ref{i4}), we have that
\begin{equation}
\begin{split}
\frac{\partial}{\partial t}\left(\n_{i}h\n_{\bar{j}}h\right)=&\Delta\left(\n_{i}h\right)\n_{\bar{j}}h+\Delta\left(\n_{\bar{j}}h\right)\n_{i}h-\frac{1}{2}R_{i\bar{l}}\n_{l}h\n_{\bar{j}}h\\
&-\frac{1}{2}R_{k\bar{j}}\n_{\bar{k}}h\n_{i}h+\n_{i}\n_{k}L\n_{\bar{j}}h\n_{\bar{k}}h+\n_{i}\n_{\bar{k}}L\n_{k}h\n_{\bar{j}}h\\
&+\n_{k}L\n_{\bar{k}}\left(\n_{i}h\n_{\bar{j}}h\right)+\n_{\bar{k}}L\n_{k}\left(\n_{i}h\n_{\bar{j}}h\right)
\\&+\n_{\bar{j}}\n_{k}L\n_{\bar{k}}h\n_{i}h+\n_{\bar{j}}\n_{\bar{k}}L\n_{k}h\n_{i}h.
\end{split}
\end{equation}
The Laplacian of $\nabla_{i}h\nabla_{\bar{j}}h$ is
\begin{equation}
\begin{split}
\Delta\left(\n_{i}h\n_{\bar{j}}h\right)
=&\frac{1}{2}\left(\n_{k}\n_{\bar{k}}+\n_{\bar{k}}\n_{k}\right)\left(\n_{i}h\n_{\bar{j}}h\right)\\
=&\frac{1}{2}\n_{k}\left(\n_{\bar{k}}\n_{i}h\n_{\bar{j}}h+\n_{i}h\n_{\bar{k}}\n_{\bar{j}}h\right)
\\&+\frac{1}{2}\n_{\bar{k}}\left(\n_{k}\n_{i}h\n_{\bar{j}}h+\n_{i}h\n_{k}\n_{\bar{j}}h\right)\\
=&\Delta\left(\n_{i}h\right)\n_{\bar{j}}h+\n_{i}h\Delta\left(\n_{\bar{j}}h\right)\\&+\n_{k}\n_{i}h\n_{\bar{k}}\n_{\bar{j}}h+\n_{\bar{k}}\n_{i}h\n_{k}\n_{\bar{j}}h.
\end{split}
\end{equation}
Collecting the two equalities above gives that
\begin{equation}
\begin{split}
\frac{\partial}{\partial t}\left(\n_{i}h\n_{\bar{j}}h\right)=&\Delta\left(\n_{i}h\n_{\bar{j}}h\right)-\n_{k}\n_{i}h\n_{\bar{k}}\n_{\bar{j}}h-\n_{\bar{k}}\n_{i}h\n_{k}\n_{\bar{j}}h
\\&+\n_{i}\n_{k}L\n_{\bar{j}}h\n_{\bar{k}}h
+\n_{i}\n_{\bar{k}}L\n_{k}h\n_{\bar{j}}h\\&+\n_{\bar{j}}\n_{k}L\n_{\bar{k}}h\n_{i}h+\n_{\bar{j}}\n_{\bar{k}}L\n_{k}h\n_{i}h\\
&+\n_{k}L\n_{\bar{k}}\left(\n_{i}h\n_{\bar{j}}h\right)+\n_{\bar{k}}L\n_{k}\left(\n_{i}h\n_{\bar{j}}h\right)\\&
-\frac{1}{2}R_{i\bar{l}}\n_{\bar{j}}h\n_{l}h-\frac{1}{2}R_{k\bar{j}}\n_{\bar{k}}h\n_{i}h. \label{gradient evol2}
\end{split}
\end{equation}

Finally, we give the evolution of $\frac{\n_{i}h\n_{\bar{j}}h}{1-h^2}$. With the help of (\ref{gradient evol1}) and (\ref{gradient evol2}), we have that
\begin{equation}
\begin{split}
\frac{\partial}{\partial t}\left(\frac{\n_{i}h\n_{\bar{j}}h}{1-h^2}\right)=&\frac{1}{1-h^2}\frac{\partial}{\partial t}{\left(\n_{i}h\n_{\bar{j}}h\right)}-\frac{\n_{i}h\n_{\bar{j}}h}{\left(1-h^2\right)^2}\frac{\partial}{\partial t}\left(1-h^2\right){}\\=&\frac{1}{1-h^2}[\Delta\left(\n_{i}h\n_{\bar{j}}h\right)-\n_{k}\n_{i}h\n_{\bar{k}}\n_{\bar{j}}h-\n_{\bar{k}}\n_{i}h\n_{k}\n_{\bar{j}}h{}\\&+\n_{i}\n_{k}L\n_{\bar{j}}h\n_{\bar{k}}h
+\n_{i}\n_{\bar{k}}L\n_{k}h\n_{\bar{j}}h+\n_{\bar{j}}\n_{k}L\n_{\bar{k}}h\n_{i}h\\&+\n_{\bar{j}}\n_{\bar{k}}L\n_{i}h\n_{k}h+\n_{k}L\n_{\bar{k}}(\n_{i}h\n_{\bar{j}}h){}\\
&+\n_{\bar{k}}L\n_{k}\left(\n_{i}h\n_{\bar{j}}h\right)-\frac{1}{2}R_{i\bar{l}}\n_{\bar{j}}h\n_{l}h-\frac{1}{2}R_{k\bar{j}}\n_{\bar{k}}h\n_{i}h]{}\\
&-\frac{\n_{i}h\n_{\bar{j}}h}{\left(1-h^2\right)^2}[\Delta\left(1-h^2\right)+2|\n h|^2+\n_{k}L\n_{\bar{k}}\left(1-h^2\right) \\&+\n_{\bar{k}}L\n_{k}\left(1-h^2\right)].
\end{split}
\end{equation}
Straightforward computation yields
\begin{equation}
\begin{split}
\Delta\left(\frac{\n_{i}h\n_{\bar{j}}h}{1-h^2}\right)=&\frac{\Delta\left(\n_{i}h\n_{\bar{j}}h\right)}{1-h^2}-\frac{\n_{i}h\n_{\bar{j}}h}{\left(1-h^2\right)^2}\Delta\left(1-h^2\right)-\frac{\n_{k}\left(\n_{i}h\n_{\bar{j}}h\right)\n_{\bar{k}}\left(1-h^2\right)}{\left(1-h^2\right)^2}\\&
-\frac{\n_{\bar{k}}\left(\n_{i}h\n_{\bar{j}}h\right)\n_{k}\left(1-h^2\right)}{\left(1-h^2\right)^2}+\frac{2\n_{i}h\n_{\bar{j}}h[\n_{k}\left(1-h^2\right)\n_{\bar{k}}\left(1-h^2\right)]}{\left(1-h^2\right)^3}\\
=&\frac{\Delta\left(\n_{i}h\n_{\bar{j}}h\right)}{1-h^2}-\frac{\n_{i}h\n_{\bar{j}}h}{\left(1-h^2\right)^2}\Delta\left(1-h^2\right)\\
&+\frac{2h\n_{k}\left(\n_{i}h\n_{\bar{j}}h\right)\n_{\bar{k}}h+2h\n_{\bar{k}}\left(\n_{i}h\n_{\bar{j}}h\right)\n_{k}h}{\left(1-h^2\right)^2}\\&+\frac{8h^2\n_{i}h\n_{\bar{j}}h|\n h|^2}{\left(1-h^2\right)^3},
\end{split}
\end{equation}
so we have that
\begin{equation}
\begin{split}
\frac{\partial}{\partial t}\left(\frac{\n_{i}h\n_{\bar{j}}h}{1-h^2}\right)
=&\Delta\left(\frac{\n_{i}h\n_{\bar{j}}h}{1-h^2}\right)+\frac{1}{1-h^2}(-\n_{k}\n_{i}h\n_{\bar{k}}\n_{\bar{j}}h\\&-\n_{\bar{k}}\n_{i}h\n_{k}\n_{\bar{j}}h+\n_{i}\n_{k}L\n_{\bar{j}}h\n_{\bar{k}}h\\
&+\n_{i}\n_{\bar{k}}L\n_{\bar{j}}h\n_{k}h+\n_{\bar{j}}\n_{k}L\n_{\bar{k}}h\n_{i}h\\
&+\n_{\bar{j}}\n_{\bar{k}}L\n_{i}h\n_{k}h-\frac{1}{2}R_{i\bar{l}}\n_{\bar{j}}h\n_{l}h-\frac{1}{2}R_{k\bar{j}}\n_{\bar{k}}h\n_{i}h)
\\&-\frac{2\n_{i}h\n_{\bar{j}}h}{\left(1-h^2\right)^2}|\n h|^2+\n_{k}L\n_{\bar{k}}\left(\frac{\n_{i}h\n_{\bar{j}}h}{1-h^2}\right)\\
&+\n_{\bar{k}}L\n_{k}\left(\frac{\n_{i}h\n_{\bar{j}}h}{1-h^2}\right)-\frac{8h^2\n_{i}h\n_{\bar{j}}h|\n h|^2}{\left(1-h^2\right)^3}\\
&-\frac{2h\n_{k}\left(\n_{i}h\n_{\bar{j}}h\right)\n_{\bar{k}}h+2h\n_{\bar{k}}\left(\n_{i}h\n_{\bar{j}}h\right)\n_{k}h}{\left(1-h^2\right)^2}.
\end{split}
\end{equation}
Rearranging terms leads to
\begin{equation}
\begin{split}
\frac{\partial}{\partial t}\left(\frac{\n_{i}h\n_{\bar{j}}h}{1-h^2}\right)
=&\Delta\left(\frac{\n_{i}h\n_{\bar{j}}h}{1-h^2}\right)+\n_{k}L\n_{\bar{k}}\left(\frac{\n_{i}h\n_{\bar{j}}h}{1-h^2}\right)+\n_{\bar{k}}L\n_{k}\left(\frac{\n_{i}h\n_{\bar{j}}h}{1-h^2}\right)\\
&-\frac{1}{1-h^2}\left(\n_{i}\n_{k}h+\frac{2h\n_{i}h\n_{k}h}{1-h^2}\right)\left(\n_{\bar{j}}\n_{\bar{k}}h+\frac{2h\n_{\bar{j}}h\n_{\bar{k}}h}{1-h^2}\right)\\
&-\frac{1}{1-h^2}\left(\n_{i}\n_{\bar{k}}h+\frac{2h\n_{i}h\n_{\bar{k}}h}{1-h^2}\right)\left(\n_{\bar{j}}\n_{k}h+\frac{2h\n_{\bar{j}}h\n_{k}h}{1-h^2}\right)\\
&+\frac{1}{1-h^2}(\n_{i}\n_{k}L\n_{\bar{j}}h\n_{\bar{k}}h+\n_{i}\n_{\bar{k}}L\n_{\bar{j}}h\n_{k}h+\n_{\bar{j}}\n_{k}L\n_{i}h\n_{\bar{k}}h\\&+\n_{\bar{j}}\n_{\bar{k}}L\n_{i}h\n_{k}h
-\frac{1}{2}R_{i\bar{l}}\n_{\bar{j}}h\n_{l}h-\frac{1}{2}R_{k\bar{j}}\n_{\bar{k}}h\n_{i}h)\\
&-\frac{2\n_{i}h\n_{\bar{j}}h}{\left(1-h^2\right)^2}|\n h|^2.
\end{split}
\end{equation}
This complete the proof of lemma 3.

We also need following result, which is an easy consequence of Cao's differential Harnack inequality for the K\"{a}hler-Ricci flow \cite{C2}.
\begin{lemma} Let $(M,g_{i\bar{j}}(t))$ be a solution of $\epsilon$-K\"{a}hler-Ricci flow with nonnegative bisectional curvature and $u$ be a solution of (9), then there holds
\begin{eqnarray*}
Y_{i{\bar{j}}}&=&\Delta R_{i{\bar{j}}}+R_{i{\bar{j}}{k}{\bar{l}}}R_{{\bar{k}}l}-\left(\frac{\n_{\bar{k}}L}{\epsilon}\n_{k}R_{i{\bar{j}}}+\frac{\n_{k}L}{\epsilon}\n_{\bar{k}}R_{i{\bar{j}}}\right)
+R_{i{\bar{j}}l{\bar{k}}}\frac{\n_{k}L}{\epsilon}\frac{\n_{\bar{l}}L}{\epsilon}+\frac{R_{i{\bar{j}}}}{\epsilon t}\geq 0.
\end{eqnarray*}
\end{lemma}

With the help of the preceding four lemmas we are now in the position to give the proof of the theorem.\newline

{\bf Proof of the Theorem 1.1}
Let \begin{equation}
P_{i{\bar{j}}}=\n_{i}\n_{\bar{j}}L+\epsilon R_{i\bar{j}}-\frac{\n_{i}h\n_{\bar{j}}h}{1-h^2}.
\end{equation}
Combine the lemmas together to get
\begin{equation}
\begin{split}
\frac{\partial}{\partial t}P_{i{\bar{j}}}=&\Delta P_{i{\bar{j}}} +\nabla_{k}L\nabla_{\bar{k}}P_{i{\bar{j}}}+\nabla_{\bar{k}}L\nabla_{k}P_{i{\bar{j}}}+R_{i{\bar{j}}k{\bar{l}}}P_{l{\bar{k}}}+\epsilon^2Y_{i{\bar{j}}}-\epsilon^2R_{i\bar{p}}R_{p\bar{j}}
\\&-\frac{1}{2}\left(R_{l{\bar{j}}}P_{i{\bar{l}}}+R_{i{\bar{l}}}P_{l{\bar{j}}}\right)+\frac{1}{2}R_{l{\bar{j}}}\left(\epsilon R_{i{\bar{l}}}\right)+\frac{1}{2}R_{i{\bar{l}}}\left(\epsilon R_{l{\bar{j}}}\right)-\epsilon R_{i\bar{p}}R_{p\bar{j}}\\&+\left(\nabla_{\bar{j}}\nabla_{k}L-\frac{\n_{\bar{j}}h\n_{k}h}{1-h^2}\right)\left(\nabla_{i}\nabla_{\bar{k}}L-\frac{\n_{i}h\n_{\bar{k}}h}{1-h^2}\right)
\\&+\left(\n_{i}\n_{k}L-\frac{\n_{i}h\n_{k}h}{1-h^2}\right)\left(\n_{\bar{j}}\n_{\bar{k}}L-\frac{\n_{\bar{j}}h\n_{\bar{k}}h}{1-h^2}\right)\\&+\frac{1}{1-h^2}\left(\n_{i}\n_{k}h+\frac{2h\n_{i}h\n_{k}h}{1-h^2}\right)\left(\n_{\bar{j}}\n_{\bar{k}}h+\frac{2h\n_{\bar{j}}h\n_{\bar{k}}h}{1-h^2}\right)\\
&+\frac{1}{1-h^2}\left(\n_{i}\n_{\bar{k}}h+\frac{2h\n_{i}h\n_{\bar{k}}h}{1-h^2}\right)\left(\n_{\bar{j}}\n_{k}h+\frac{2h\n_{\bar{j}}h\n_{k}h}{1-h^2}\right).
\end{split}
\end{equation}
Since the three matrices in the last three terms are nonnegative definite, we have that
\begin{equation}
\begin{split}
\frac{\partial}{\partial t}P_{i{\bar{j}}}\geq&\Delta P_{i{\bar{j}}}+\nabla_{\bar{k}}L\nabla_{k}P_{i{\bar{j}}}+\nabla_{k}L\nabla_{\bar{k}}P_{i{\bar{j}}}+R_{i{\bar{j}}k{\bar{l}}}P_{l{\bar{k}}}
\\&+\left(\nabla_{\bar{j}}\nabla_{k}L-\frac{\n_{\bar{j}}h\n_{k}h}{1-h^2}\right)\left(\nabla_{i}\nabla_{\bar{k}}L-\frac{\n_{i}h\n_{\bar{k}}h}{1-h^2}\right)
\\&+\epsilon^2Y_{i{\bar{j}}}-\epsilon^2R_{i\bar{p}}R_{p\bar{j}}-\frac{\epsilon R_{i{\bar{j}}}}{ t}-\frac{1}{2}\left(R_{l{\bar{j}}}P_{i{\bar{l}}}+R_{i{\bar{l}}}P_{l{\bar{j}}}\right).
\end{split}
\end{equation}
Therefore, by Lemma 4, we get that
\begin{equation}
\begin{split}
\frac{\partial}{\partial t}\left(P_{i{\bar{j}}}+\frac{g_{i{\bar{j}}}}{t}\right)
=&\frac{\partial}{\partial t}P_{i{\bar{j}}}+\frac{1}{t}\frac{\partial}{\partial t}g_{i{\bar{j}}}-\frac{1}{t^2}g_{i{\bar{j}}}\\\geq&\Delta\left(P_{i{\bar{j}}}+\frac{g_{i{\bar{j}}}}{t}\right)+\n_{\bar{k}}L\n_{k}\left(P_{i{\bar{j}}}+\frac{g_{i{\bar{j}}}}{t}\right)
\\
&+\n_{k}L\n_{\bar{k}}\left(P_{i{\bar{j}}}+\frac{g_{i{\bar{j}}}}{t}\right)+R_{i{\bar{j}}k{\bar{l}}}\left(P_{l{\bar{k}}}+\frac{g_{l{\bar{k}}}}{t}\right)\\&-R_{i{\bar{j}}k{\bar{l}}}\frac{g_{l{\bar{k}}}}{t}
-\frac{1}{2}\left[R_{l{\bar{j}}}\left(P_{i{\bar{l}}}+\frac{g_{i{\bar{l}}}}{t}\right)+R_{i{\bar{l}}}\left(P_{l{\bar{j}}}+\frac{g_{l{\bar{j}}}}{t}\right)\right]
\\&+\left(\nabla_{\bar{j}}\nabla_{k}L-\frac{\n_{\bar{j}}h\n_{k}h}{1-h^2}\right)\left(\nabla_{i}\nabla_{\bar{k}}L-\frac{\n_{i}h\n_{\bar{k}}h}{1-h^2}\right)\\
&-\epsilon^2R_{i\bar{p}}R_{p\bar{j}}-\frac{2\epsilon R_{i{\bar{j}}}}{ t}-\frac{1}{t^2}g_{i{\bar{j}}}.
\end{split}
\end{equation}
Let
\begin{equation}
\begin{split}
Q_{i{\bar{j}}}=P_{i{\bar{j}}}+\frac{g_{i{\bar{j}}}}{t},
\end{split}
\end{equation}
then $Q_{i{\bar{j}}}$ satisfies that
\begin{equation}
\begin{split}
\frac{\partial}{\partial t}Q_{i{\bar{j}}}\geq&\Delta Q_{i{\bar{j}}}+\left<\n L,\n Q_{i{\bar{j}}}\right>+\left<\n Q_{i{\bar{j}}},\n L\right>+R_{i{\bar{j}}k{\bar{l}}}Q_{l{\bar{k}}}\\&-\frac{1}{2}\left(R_{l{\bar{j}}}Q_{i{\bar{l}}}+R_{i{\bar{l}}}Q_{l{\bar{j}}}\right)
+P_{i{\bar{k}}}P_{k{\bar{j}}}-\epsilon R_{k{\bar{j}}}P_{i{\bar{k}}}\\&-\epsilon R_{i{\bar{k}}}\left(P_{k{\bar{j}}}-\epsilon R_{k{\bar{j}}}\right)-\epsilon^2R_{i\bar{p}}R_{p\bar{j}}-\frac{2\epsilon R_{i{\bar{j}}}}{ t}-\frac{1}{t^2}g_{i{\bar{j}}}\\
=&\Delta Q_{i{\bar{j}}}+\left<\n L,\n Q_{i{\bar{j}}}\right>+\left<\n Q_{i{\bar{j}}},\n L\right>+R_{i{\bar{j}}k{\bar{l}}}Q_{l{\bar{k}}}\\&-\left(\frac{1}{2}+\epsilon\right)\left(R_{l{\bar{j}}}Q_{i{\bar{l}}}+R_{i{\bar{l}}}Q_{l{\bar{j}}}\right)
+\left(P_{k{\bar{j}}}+\frac{g_{k{\bar{j}}}}{t}\right)\left(P_{i{\bar{k}}}-\frac{g_{i{\bar{k}}}}{t}\right)
\\=&\Delta Q_{i{\bar{j}}}+\left<\n L,\n Q_{i{\bar{j}}}\right>+\left<\n Q_{i{\bar{j}}},\n L\right>+R_{i{\bar{j}}k{\bar{l}}}Q_{l{\bar{k}}}\\&-\left(\frac{1}{2}+\epsilon\right)\left(R_{p{\bar{j}}}Q_{i{\bar{p}}}+R_{i{\bar{p}}}Q_{p{\bar{j}}}\right)
+\left(Q_{i{\bar{k}}}-\frac{2}{t}g_{i{\bar{k}}}\right)Q_{k{\bar{j}}}. \label{Q}
\end{split}
\end{equation}
According to the tensor maximum principle of Hamilton \cite{H1,H2}, to prove $Q_{i{\bar{j}}}\geq 0$, it suffices to show that
\begin{equation}
R_{i{\bar{j}}k{\bar{l}}}Q_{l{\bar{k}}}-\left(\frac{1}{2}+\epsilon\right)\left(R_{p{\bar{j}}}Q_{i{\bar{p}}}+R_{i{\bar{p}}}Q_{p{\bar{j}}}\right) \label{B}
+\left(Q_{i{\bar{k}}}-\frac{2}{t}g_{i{\bar{k}}}\right)Q_{k{\bar{j}}}
\end{equation}
satisfies null-eigenvector condition.
By the well-known result of Bando \cite{B} and Mok \cite{M2}, the nonnegativity of holomorphic bisectional curvature is preserved along the $\epsilon$-K\"{a}hler-Ricci flow. So each term of (\ref{B}) is nonnegative when evaluated at any null-eigenvector of $Q_{i\bar{j}}$. This completes the proof of the theorem.

\begin{center}
\bf Acknowledgements
\end{center}

The first named author would like to thank Professors Q.-K. Lu, S.-K. Wang and K.
Wu for valuable discussions and comments. He would also like to thank Professor H.-W. Xu for helpful suggestions. This work partially supported by the Fundamental Research Funds for Central Universities (Grant No. 2012QNA40).

redpajama_set_name: RedPajamaArXiv


text:
Thymin (T, Thy, 5-Methyluracil) ist eine der vier Nukleinbasen in der DNA, zusammen mit Adenin, Cytosin und Guanin. In der RNA steht an seiner Stelle Uracil. Es ist eine heterocyclische organische Verbindung mit einem Pyrimidingrundgerüst und drei Substituenten (Sauerstoffatome an den Positionen 2 und 4, Methylgruppe an Position 5). Die Nukleoside von Thymin sind das Desoxythymidin in der DNA und das seltene Ribothymidin in der RNA (z. B. in der tRNA). In der Watson-Crick-Basenpaarung bildet es zwei Wasserstoffbrücken mit Adenin.

Darstellung und Gewinnung
Eine Isolierung kann aus Rinderhirnen oder Kabeljau-Rogen erfolgen.

Eine synthetische Darstellung gelingt durch Cyclisierung von N-Ethoxycarbonyl-3-methoxy-2-methylacrylamid in wässriger Ammoniaklösung.

Eine weitere Synthese geht von 3-Methyläpfelsäure aus, welche in rauchender Schwefelsäure decarboxyliert und mit Harnstoff kondensiert wird.

Eigenschaften

Physikalische Eigenschaften
Thymin bildet glänzende, bitter schmeckende, nadelförmige oder prismenförmige Kristalle, die bei 335–337 °C unter Zersetzung schmelzen. Die Verbindung löst sich gut in heißem Wasser, in Alkohol und Ether ist die Löslichkeit gering. In alkalischen Medien löst es sich unter Salzbildung infolge einer Enolatbildung abgeleitet von der Enolform 5-Methyl-2,4-pyrimidindiol.

Chemische Eigenschaften
Prinzipiell kann Thymin in sechs tautomeren Strukturen vorliegen. Die Lactamform (1) wird aber gegenüber den Enolformen bevorzugt.

Geschichte und biologische Bedeutung
1893 berichtete der spätere Nobelpreisträger Albrecht Kossel von einer Entdeckung. Aus den Thymusdrüsen des Kalbes hatte er mit dem Assistenten Albert Neumann Nukleinsäure gewonnen und mit Schwefelsäure behandelt. Es bildete sich ein gut kristallisiertes Spaltprodukt, für das – abgeleitet von der Thymusdrüse – der Name Thymin vorgeschlagen wurde.

Thymin kann Bestandteil der DNA oder verschiedener Nukleoside und Nukleotide sein.

Nukleoside
Über das N1-Atom des Ringes kann Thymin an das C1-Atom der Desoxyribose N-glycosidisch gebunden werden; man spricht dann von einem Nukleosid, dem Desoxythymidin. Bei der Bindung an Ribose entsteht das Nukleosid Ribothymidin.

{| class="wikitable left" style="text-align:center"
|-
|
|
|-
| Desoxythymidin, dT
| Ribothymidin, T
|}

Nukleotide
Über die Phosphorylierung des Thymidins am C5-Atom der Ribose gelangt man zu den wichtigen Nukleotiden Desoxythymidinmonophosphat (dTMP), Desoxythymidindiphosphat (dTDP) und Desoxythymidintriphosphat (dTTP).

Bestandteil der DNA
In der DNA-Doppelhelix bildet Thymin über die 4-Oxogruppe und die N3–H-Gruppe zwei Wasserstoffbrücken mit der zugehörigen Adenin-Base des komplementären Stranges aus.

Vergleich von Thymin und Uracil
In der DNA tritt Thymin an die Stelle von Uracil. Uracil kann relativ einfach durch Desaminierung und Hydrolyse aus Cytosin entstehen, wodurch dann die Basensequenz geändert (mutiert) und die in der Nukleotidsequenz genetisch codierte Information womöglich verändert wird.

Thymin hingegen unterscheidet sich vom Uracil durch eine zusätzliche Methylgruppe und kann so auch nicht ohne weiteres aus Cytosin entstehen. In der DNA vorhandenes Uracil kann somit als Mutation erkannt und durch Basenexzisionsreparatur gegen Cytosin ausgetauscht werden.

Thymindimer
Bei Thymindimeren handelt es sich um eine DNA-Mutation, welche durch UV-Strahlung induziert wird. Dabei verbinden sich zwei auf einem DNA-Strang nebeneinanderliegende Thymin-Basen über eine [2+2]-Cycloaddition kovalent zu einem Dimer, das ein relativ stabiles Cyclobutan-Derivat ist.

Besonders anfällig für eine solche Mutation sind Hautzellen, die dem Sonnenlicht ausgesetzt sind. Aus diesem Grund werden Thymindimere als eine wesentliche Ursache für die Entstehung von Hautkrebs diskutiert.

Verwendung
Thymin dient als Ausgangsstoff für einige Arzneistoffe wie z. B. Zidovudin, Telbivudin und Clevudin.

Verwandte Verbindungen

Weblinks

Einzelnachweise

Nucleinbase
Pyrimidindion

redpajama_set_name: RedPajamaWikipedia


text:
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B&B's Katherine Kelly Lang Talks On 35 Years As Brooke: The Men, The Booze, The Rivalries
Photo: CBS
For 35 years, Katherine Kelly Lang has portrayed The Bold and the Beautiful's longest-running female lead as Brooke Logan. Lang, who along with co-star John McCook (Eric Forrester), are the only two remaining original cast members of the CBS daytime drama.
Last week, B&B celebrated their 35th anniversary with a special Brooke-standalone episode which featured the character and the key men in her life through the years. The episode featured the returns of Jack Wagner as Nick, along with Winsor Harmon as Thorne.
While the show took a look-back at some Brooke history, Kelly is also involved in a current major storyline which finds Brooke seemingly losing the love of her life, Ridge (Thorsten Kaye) back to Taylor Hayes (Krista Allen) who has returned to town.
All of this followed Brooke's New Year's Eve debacle where she fell off the wagon, thanks to Sheila's (Kimberlin Brown) plot, and drank way too much alcohol and wound up making out with Deacon (Sean Kanan).
Photo: CBX
Now in a special conversation on You Tube's Michael Fairman Channel, Kelly chats with Michael about her thoughts on Brooke's current predicament (thanks to Sheila!), what it was like to have the 35th anniversary episode centered around her character and working with Wagner, Harmon, McCook, as well as Don Diamont (Bill), and of course, Thorsten Kaye for the milestone show.
In addition, Lang reveals what she thought of some of Brooke's other hook-up foibles including the time Brooke confused Oliver Jones for Ridge, while they were donning masks at a masquerade party.
For years, Brooke's nemesis, Stephanie Forrester (Susan Flannery) dubbed her, "The Slut of the Valley". The character often still has that dubious title thrown at her by various characters on the canvas and most recently, Sheila Carter. Kelly shares what she thinks of the use of that phrase through the years, and much more in a candid and insightful interview.
Check out the full conversation below, and then let us know, who has been your favorite romantic pairing for Brooke through the years? What do you think will happen now that Brooke has a feeling that something else transpired to make her drink again? Share your thoughts in the comment section.
Make sure to 'subscribe' to the Michael Fairman Channel for more celebrity features.
Related Topics:bill spencerbrad bellBrooke Logan ForrestercbsDeacon SharpeDon DiamontFeaturedInterviewjack wagnerkatherine kelly langKimberlin BrownMichael FairmanMichael Fairman ChannelNick MaroneRidge Forrestersean kananSheila CarterStephanie Forrestersusan flanneryThe Bold and the BeautifulThorne Forresterthorsten kayevideo interviewWinsor Harmon
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THE BOLD AND THE BEAUTIFUL: Sheila Shoots Finn after Aiming for Steffy in Cliffhanger Shocker
I loved the SLs with Brooke and Nick Marone and Ridge when played by Ronn Moss. I like Thorsten Kaye but he always seemed too old and rough around the edges compared to Brooke's softness.
B&B's Don Diamont On Bill's Shocking Relationship with Sheila: "He Can Just Be Who He Is. He Can Be His Worst Self"
THE BOLD AND THE BEAUTIFUL: Sheila and Bill are in Cahoots as History Comes Back to Bite Steffy
Just a few weeks ago on the January 4th episode of General Hospital, viewers were left devastated when Dr. Britt Westbourne succumbed to a fatal poisoning after a confrontation with 'The Hook'.
While it was revealed that Daytime Emmy winner, Kelly Thiebaud, was exiting the ABC soap opera at the end of 2022, many thought the story would see Britt leaving Port Charles to deal with her declining health due to Huntington's Disease in private, in order to keep the door open for a possible return down the line. However, that was not meant to be. When a report from Deadline surfaced back in August of 2022, it indicated that Kelly was returning to her primetime role on ABC's Station 19, which caused some confusion and rumors of just what would Thiebaud's ultimate status be with the daytime drama series.
Now, in a very special exit interview for You Tube's Michael Fairman Channel, Kelly chats with Michael Fairman and clarifies what led to her decision to leave her role as Dr. Britt Westbourne.
Talking with Fairman direct from London, Kelly, who won a Daytime Emmy in 2022 for the Outstanding Supporting Actress in a Drama Series, reveals how it came to pass that GH would kill-off the character of Britt by becoming another victim of 'The Hook', and more.
Photo: KThiebaudIG
During the conversation, Thiebaud shares her thoughts on working with her former GH co-stars including: Steve Burton (ex-Jason), Parry Shen (Brad), Kirsten Storms (Maxie), Kathleen Gati (Liesl), Cassandra James (Terry), Roger Howarth (Austin) and Josh Kelly (Cody), as well as some of her favorite memories throughout her run on the series which began in 2012.
Kelly took the character from being the manipulative 'Britch' as an interloper in the romance of Patrick (Jason Thompson) and Sabrina (Teresa Castillo), all the way to becoming a well-loved character to root for.
Courtesy/ABC
While, GH viewers witnessed Britt's shocking death scenes, Kelly weighs-in on taping those moments, as well as Britt's emotional goodbye birthday party, and having the opportunity to portray a character who was trying to navigate her life, while suffering from the effects of Huntington's Disease.
In addition, Thiebaud gives a very special 'thank you' and shout out to the fans for their enduring support as she looks forward to new roles and new adventures ahead.
Photo: NATAS
Check out our heartfelt farewell interview with Kelly below, and for upcoming celebrity features, interviews, event coverage, music videos, and more, make sure to 'subscribe' to the Michael Fairman Channel.
So, what did you think about Kelly's decision to depart GH? Will you miss her as Britt? Share your thoughts on our conversation in the comment section below.
Y&R's Michael Graziadei Talks Daniel's Past and Present, Reuniting with His Co-Stars, and Life as Dad to Twins
Longtime viewers of CBS' The Young and the Restless are excited that Michael Graziadei has returned to the top-rated daytime drama during its 50th anniversary season as Daniel Romalotti.
Graziadei made his on-screen re-emergence right around Thanksgiving, as in story, Daniel came home to surprise his mother, Phyllis (Michelle Stafford), but also for a business idea he wanted to present to good friend, Devon (Bryton James) and Daniel's former girlfriend, Lily Winters (Christel Khalil). Seems Daniel has a mega-universe gaming platform he's created that he wants Chancellor-Winters to potentially be a part of. But there must be more to the story than meets the eye, right? Will Daniel find himself drawn back to Lily, especially since her relationship with Billy (Jason Thompson) has hit a major rough spot/road block named Chelsea (Melissa Claire Egan)?
Courtesy/CBS
In a new virtual interview exclusively for You Tube's Michael Fairman Channel, Graziadei sits-down to talk about coming home to Genoa City with a look at Daniel's past, present and future. Michael first appeared as Y&R's Daniel from 2004-2013, made a guest stop back to the show in 2016, and now at the end of 2022, he is back full-time with the soap opera.
During the conversation, Graziadei shares what it's been like working with beloved cast mates including: Michelle Stafford, Christel Khalil, Camryn Grimes (Mariah) and more.
Michael also shared his enthusiasm at getting a chance to reconnect not only with Michelle Stafford, but Daniel's adoptive father, Michael Damian (Danny) as well, when they all appear in scenes together for the upcoming holiday episodes.
Just what have been Michael's favorite Daniel storylines of all-time? Was it Daniel's part in Cassie's death (Camryn Grimes) and its aftermath? When they gave Daniel a porn addiction storyline? Graziadei weighs-in.
In addition, Michael reveals the wonders of being a new dad to twins boys in real-life, Oliver Bear and Arlo. He opens up about how he and his lady love, Lauren found out they were having twins, the naming process they went through, and what life is like now at-home.
Photo: GoodGriefCharlieBrownIG
During Graziadei's time away from Y&R, he appeared in numerous TV and film projects. Most recently, Michael was seen in Netflix's The Lincoln Lawyer, where he shares what it was like to take on a legal role for the first time in his career, plus what it was like to act and kiss Oscar-winner, Jessica Lange in American Horror Story (a role he played while still at Y&R during his first go-round).
Courtesy/Netflix
You can watch the full interview with Graziadei below.
For more celebrity interviews and features make sure to 'subscribe' to the Michael Fairman Channel.
Now let us know, are you glad Daniel is back in GC? Do you hope Lily and Daniel get back together all of these years later? What has been your favorite Daniel storyline of all time … and your favorite part of our conversation with him? Share your thoughts in the comment section.
Y&R's Camryn Grimes and Fiancé Brock Powell Talk On Their Roles as Mrs. Claus and Santa in 'Mickey Saves Christmas'
Two-time Daytime Emmy winner, Camryn Grimes (Mariah, The Young and the Restless) has been involved in a very special passion project in more ways than one! Now, everyone will be able can see it, and hear it, when Grimes voices the role of Mrs. Claus opposite her real-life fiancé and voiceover star, Brock Powell as Santa in Mickey Saves Christmas.
The new stop-motion holiday special premieres Sunday, November 27th (7:00-7:30 p.m. EST) on ABC, Disney Channel, Disney Junior and Disney XD, and Monday, November 28 on Disney+ and Hulu. It features Disney icons: Mickey, Minnie and the gang, who are sent to the North Pole on a quest to save Christmas and find the true meaning of the holiday.
Camryn and Brock Powell became engaged earlier this year on January 7th, which also happened to be Camryn's 32nd birthday. How the two met all comes back to the world of voiceover and Disney, as you will learn.
Michael Fairman TV chatted with the real-life lovebirds to preview Mickey Saves Christmas, how they wound up voicing the important voices of Christmas, Santa and Mrs. Claus, and just what Brock thinks of Camryn's Genoa City alter-ego Mariah being married to Tessa (Cait Fairbanks), and more. Is Brock #TeamTeriah? Check out what they shared with us below.
Photo: Disney+
The two of you were actually introduced by Bill Farmer, the voice of Goofy! How did that happen?
BROCK: So, Bill Farmer is my mentor, and through him, that was the first time I actually met Camryn.
CAMRYN: I knew Bill growing up, and in 2020 when I wanted to get into voiceovers, I reached out to him and that's how this all started and I guess, the rest is history.
Now you are voicing Santa and Mrs. Claus in Mickey Save Christmas. How did it come about that you both landed parts on the project?
BROCK: I got to play the role of Santa in Mickey and Minnie Wish Upon a Christmas, which was the first time I got to voice Santa for Disney, and now I'm returning to the role this time. Camryn received the audition randomly for Ms. Claus on the day we were getting engaged, and it was perfect timing because I had to get her out of the house so I could set up the proposal here. So, it really was a Christmas miracle because I was sweating!
CAMRYN: We were cast separately. The production team had no idea that Brock and I were a couple. When I went in to record, they were like, "Here's the voice of Santa for you to record to, and I was like, "Yeah, I know… that's my fiancé!" They were like, "What??" (Laughs)
Photo: Sarah Irene Decristoforo
Did the engagement go off without a hitch?
BROCK: Yes, so after the audition process, Camryn returned home. I proposed, and there was a whole Disney/Pixar theme going on in the backyard. It was a really special day. To have her say, "Yes", and also to have this crazy opportunity for her to audition for the role of Mrs. Claus happening at the same time … and then weeks later finding out she booked the role, it was all amazing. We were now going to be working on this Christmas special together. I think anytime you get to voice an iconic character, it's special. However, if you'd told me that I would get to return to the role I love, with the person I love playing opposite me … I think I'm done. I just don't know how you can ever top it!
Camryn, what did it mean to you to be in this film as Mrs. Claus, because fans saw this post of yours with a photo of you outside the Anaheim Convention Center where Disney's D23 convention was happening. It made everyone speculate just what you were working on, which you shared was a lifelong dream.
CAMRYN: Yes! Y&R and CBS is my home, but I've always wanted to do something with Disney. My family raised me on all things Disney and all things Christmas. They were huge into the holiday, and so this kind of just checks every box. It's not only Disney, it's Christmas, it's stop-motion animation, and it's Mrs. Claus. I think my family was more excited for me to book this then anything I've ever done. I think there were tears! So sentimentally, this project really meant a lot to me.
Photo: BrockPowellIG
Camryn, did Brock coach you on doing voiceover work, or give you any tips?
CAMRYN: I mean, I live with my voiceover coach. (Laughs) It's amazing!
BROCK: You know, I do coach for voiceovers and Camryn is my star pupil now. She's a talented actress, obviously. The biggest misconception is that, while it starts with the voice, Camryn, as well as most of the amazing voice actors I know, they have access to this childhood imagination which I think makes their performances possible. So, while you're watching Mickey Saves Christmas, even though we recorded everything separately, what you're looking for is that energy and every character connecting. You're inventing the circumstances, you're inventing your scene partner, where we have to close our eyes and embody a character we may, or may not, have a reference for. I think watching Camryn and hearing Camryn's work on this, and a couple of other projects she's worked on, I'm just so impressed at how coming from daytime television where there is a very stringent set routine and at the level that she excels at there, to then come into my world and then just kill it behind the microphone. She was able to let all of that go and have the bravery to try something new. I think a lot of people are going to love what Camryn has done here. I'm most proud of her to be willing to look silly which is what is required in voiceover.
CAMRYN: That's true, and that's probably the most fun thing about it.
Now, what can you tease us that happens in the story of Mickey Saves Christmas?
CAMRYN: It's kind of built into the title a little bit, but I think what you're going to get out of this is a deep sense of community and coming together to fix a problem and being able to lend a helping hand. I feel, it's just a really lovely film about chosen family.
BROCK: It's about family, it's about friends, it's about putting a lot of pressure on the holidays be perfect. I think everyone involved in this special from the animators to the music team, everything in it helped make this a love letter to the holidays. The film really just talks about that Christmas is who you spend it with, and not necessarily how it turns out. I think it's a really positive reminder that the most important part about Christmas is being with the ones we love. These are characters that Camryn and I love, not just the ones we voiced. To be with these actors, to be with these characters. it is just is going to be so special with everyone watching on November 27. It's going to be the first tine for all us watching it together,
Now Brock, do you watch Camryn on The Young and the Restless?
BROCK: Yes, and I will say sometimes when Camryn is at work, our puppy does sit and watch mommy on TV, too!
Is Brock, #TeamTeriah on Y&R, as well?
CAMRYN: Yes, Brock is all about #TeamTeriah. It was actually really beautiful kind of getting engaged around the same time, Mariah and Tessa did on the show.
In another post of yours on social media, I saw that you had picked a spot for your wedding to Brock.
CAMRYN: Yes. We don't actually have a set date (we have three options), but we did pick a location. So, we will be figuring that all out very soon.
Check out a clip from 'Mickey Saves Christmas' featuring Camryn as Mrs. Clause and Brock as Santa below. Then let us know, will you be watching the special? What did you think of Camryn getting to work on this project and opposite the love-of-her-life, Brock Powell? Share your thoughts in the comment section.


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[video of the week] Chantal Acda – The Sparkle In Our Flaws
"The Sparkle In Our Flaws (Radio Edit)" – taken from the new same-titled album by Chantal Acda, coming 09/25/2015 at Glitterhouse Records.
Guitar & vocals: Chantal Acda
Violin & vocals: Peter Broderick
Bass & vocals: Shahzad Ismailly
Short Bio: Working under the Sleepingdog moniker since 2006 Chantal has made three acclaimed records that closed on the "With Our Heads in the Clouds and Our Hearts in the Fields" (2011) album for which she collaborated with Adam Wiltzie (Stars of the Lid, A Winged Victory For The Sullen), and touring extensively both in Europe and North America. Besides Sleepingdog, Chantal has played in various other formations like Isbells, True Bypass, Marble Sounds and Distance, Light & Sky. She recorded her new album "The Sparkle In Our Flaws" with Peter Broderick (known from his solo work on labels such as Bella Union and Erased Tapes and from his work with a.o. Efterklang) and Shahzad Ismaily (regular collaborator of a.o. Lou Reed, Tom Waits and Bonnie Prince Billy). Besides Peter and Shahzad, you will also hear contributions from Valgeir Sigurðsson, Heather Woods Broderick, Niels Van Heertum and Eric Thielemans.
http://www.chantalacda.com/
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A Journey from Physics to Faith
"It's nice to have you here."
It's a phrase that may not seem out of the ordinary. Yet, it was those words of welcome spoken by a priest, along the presence of campus ministry, which helped a young man reconnect with his faith.
Now, Matthew Valles, who is working on his graduate degree in physics at the University of Maine in Orono, is also discerning the priesthood. He has been accepted as seminarian for the Diocese of Portland.
"I think God is calling me to seminary and to try to be the best seminarian that I can possibly be: pastorally, academically, in human formation, and in spiritual formation," says Matthew.
Originally from Barrington, Rhode Island, Matthew came to Maine in 2009 to study physics at Bates College.
He grew up Catholic and especially remembers the joy he felt from participating in community service projects when he was a teenager preparing for the sacrament of confirmation and just afterwards. They included doing yardwork at a Carmelite monastery and a mission trip to Jamaica.
"I was able to see Jesus more clearly in the people around me and in the poor. Despite having very few things, they exuded a kind of joy that was certainly contagious," he says of the Jamaica trip.
Ironically, however, while embracing Christ's call to serve, he says he started to question the teachings of the Church.
"I thought they were too rigid in some ways. And I wasn't really taking time to explore the reasons behind the teachings," he says.
It led him to fall away from the faith during his high school and early college years. Then, while at Bates, he participated in an engineering internship in Boston. When four members of the group discovered that being Catholic was something they had in common. They decided to attend Mass together.
"That was how, I think, I came back to the Church," Matthew says. "I felt something really powerful and life-giving."
It led him to begin going to weekend Mass at the churches of Prince of Peace Parish in Lewiston and, each time, Monsignor Marc Caron was the celebrant. It was after the third Mass that Monsignor Caron greeted Matthew.
"He asked me about who I was and things like this, and then, 'It's nice to have you here.' I can hear him saying it to me now," says Matthew. "God speaks to us in a way that we'll understand, and those were the right words he spoke to me at that time."
Matthew says he started to walk away but then turned back and inquired about opportunities to serve at the church.
"The overwhelming feeling I had was that of peace, and that I had really found something that I didn't want to have to find again from a distance," he says.
Matthew became an altar server at the Basilica of Saints Peter & Paul and, later, would also serve when Mass was celebrated at Bates. He also began to go to confession again.
"I started to go every few months, and my grades got better. I was still struggling with things, but I do remember things were getting better at the time," he says. "I had a renewed sense of purpose."
He says he started to feel some sort of calling, although he wasn't sure what it was.
"At the time, I was really afraid. I didn't know what it meant," he says. "I'm starting to go to Mass more, wondering about the seminary but, at the same time, thinking maybe God is calling me to be an engineer but very involved in the parish. I really wasn't sure.
After receiving his bachelor's degree from Bates, he decided to pursue a graduate degree in physics. Although not his first choice, he opted to go to the University of Maine because of an opportunity to be a teaching assistant. It turned out to not only be the right decision academically but also spiritually.
"I didn't realize it until I got there, but I was seeking a Catholic community," he says.
Matthew became active in all aspects of campus ministry. He attended the Wednesday night dinners, participated in eucharistic adoration, began altar serving, and joined, and later led, a Bible study group.
Matthew credits the guidance of Father Bill Labbe, campus chaplain and the diocesan coordinator of the Office of Campus Ministry, and the arrival of FOCUS missionaries. FOCUS, an acronym for the Fellowship of Catholic University Students, is a nationwide program that has been placing young adult missionaries on campuses since 1998.
"From talking with them and spending time with them, I started to become more intentional about seeking Jesus in the Catholic Church, not just the Catholic Church but Jesus in the Catholic Church, such that my faith has grown immensely since the time they came to the university," he says. "I'm very grateful that they are present."
Since the FOCUS missionaries arrived in the fall of 2015, Matthew says he has become increasingly devoted to prayer and to communicating with God. He also developed an increased devotion to the Blessed Mother.
"The FOCUS missionaries going to the university, under the guidance of Father Bill and this diocese, I think, led to an increase in devotions, and students turning inward into looking for guidance from the Holy Spirit and acting on campuses as disciples of Christ. This came in the form of more students, including myself, leading Bible studies, service projects at the University of Maine, a willingness, I think, to share Jesus with other people, and a greater love for Jesus and the Eucharist."
Father Bill says that Matthew became a mentor to many of the younger students.
"Matthew has been a leader in this community, just quietly a leader, very generous with his time and his talents. He is very pastoral, gentle," says Father Bill. "The way I describe Mathew is that he is just very kind, and he's kind in a way that's not ordinary."
One of Matthew's most memorable experiences was taking on the role of Jesus in Saint Paul the Apostle Parish's annual Palm Sunday Way of the Cross Community Procession, which begins at Saint Joseph Church in Brewer and concludes at Saint John Church in Bangor, 1.6 miles away.
"It was really powerful but, also, very simple," he says. "What made it really amazing was the crowd that came out and the prayer with which everyone did the things they did. I was really blessed to be a part of it. It was a really great honor."
He says one of the most moving parts for him was when the weeping women came to be by the cross.
"One of them was really crying. She was really crying. It was hard for me to see. I felt so bad," he says. "Just in that moment, it was very raw."
Another difficult moment was when he fell for the third time, and the cross bounced and landed on his hand.
"I was slowly pulling my hand back and just crying. There was nothing else I could do."
He says he was thankful that one of the men playing a legionnaire helped him up.
"I picked up this cross, and I said, 'I can do this.' Then I walked inside (Saint John Church) and saw the stained-glass window of Jesus with Mary and John," he recalls. "I saw Jesus, and as I looked up, He was smiling at me, and I was smiling at Him. And I walked the rest of the way down and then finished all of it."
Reflecting back on the experience, he says, "I will never look at a crucifix the same way again."
When he participated in the Way of the Cross, Matthew had already started the application process for seminary. He says, although he liked research and teaching, he did not feel he was making use of his gifts in the way God had intended. He says it came down to what he could and couldn't live without.
"If I don't have spirituality, I have nothing. I have absolutely nothing. I guess this was affirmed by the Way of the Cross procession. I wouldn't have been able to get through it if I didn't have the means by which I've been able to communicate with God on this journey," he says. "My prayer life feeds everything I do."
While Matthew believes his path may lead to seminary, he says he knows it's really up to God.
"I think I should let God decide more about this," he says. "And, I should let the Holy Spirit guide me."
Gaudete et Exsultate


text:
Q: Streaming mp4 from grails controller not working on iPhone I'm trying to stream a mp4 file to iOS devices (iPhone and iPad) from a Grails controller:
def streamContent() {
    def contentPath = "/path/to/file"
    File f = new File(contentPath)
    if(f.exists()) {
        response.setContentType("video/mp4")
        response.outputStream << f.newInputStream()
        response.outputStream.flush()
        response.outputStream.close()
    } else {
        render status: 404
    }
}

this code works well on desktop browsers like safari (I see the video), but when I access to the same page with an iPhone or an iPad the video won't play. Note that if I put the same video on Apache httpd and I request it from iOS devices, there is no problem. So it must be a streaming problem.
On the html page the video is embedded using HTML5 video tag:
<video width="360" height="200" controls>
    <source src="http://localhost:8080/myapp/controller/streamContent" type='video/mp4'>
</video>

A: I solved this problem by handling Partial Content and Range Requests (HTTP 206 status). It seems that mobile browsers/media players are using partial requests the avoid to much data transfer all at once.
So instead of doing a simple
response.outputStream << f.newInputStream()

I read only the requested bytes when the request is for a range of bytes:
if (isRange) {
    //start and end are requested bytes offsets
    def bytes = new byte[end-start]
    f.newInputStream().read(bytes, start, bytes.length)
    response.outputStream << bytes
    response.status = 206
} else {
    response.outputStream << f.newInputStream()
    response.status = 200
}

A: I don't have enough reputation to comment yet, but I just want to point out that the answer above is not complete, specifically that you need to include additional headers, and that the usage of f.newInputStream().read() is not being used accurately, as it won't just read a chunk at any starting point from the input stream, but it will read a chunk from the current position, so you must use save the inputStream and then use inputStream.skip() to jump to the right position.
I have a more complete answer here (where I answered my own similar question)
https://stackoverflow.com/a/23137725/2601060

redpajama_set_name: RedPajamaStackExchange


text:
Home / Short-line consolidator United Rail acquires New England Southern Railway in New Hampshire
Short-line consolidator United Rail acquires New England Southern Railway in New Hampshire
United Rail, Inc. , a short line railroad consolidator and owner of the SmartRail brand for electronic Precision Railroading and brand licensor, announced that it has acquired the New England Southern Railroad, a freight short line railroad, located in Concord, New Hampshire.
New England Southern (NES) is a short-line railroad operator and provides a variety of value-added rail transportation services including transloading, private industrial rail car switching and rail car storage. NES leases and operates approximately 22 miles of rail lines in the Concord area.
"Our plan is to consolidate the smaller and sometimes overlooked short lines and consolidate them under the United Rail banner," said Michael Barron, CEO of United Rail. He added, "We plan to aggressively grow our presence in other markets. With the inclusion of the NES under the United Rail umbrella, we now have a presence in the Northeast, an area where we feel there are great growth opportunities."
Danielle M. Mitchell, EVP Business Development for United Rail, stated, "We have been looking for some time to find a partner in this region with the same vision as we have for the consolidation of these small short lines. Now with this deal, we will have a leg up on contacting and closing these rail properties and afford the sellers a path to liquidity."
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