text:
\section{Introduction}

We consider the Cauchy problem for Whitham--Boussinesq system

\begin{equation}
\label{wt2d}
\left\{
\begin{aligned}
  \partial_t \eta +  \nabla \cdot  \mathbf v & = -\epsilon K_\mu\nabla \cdot (\eta \mathbf v),
  \\
  \partial_t \mathbf v + K_\mu\nabla \eta  &=   - \epsilon  K_\mu \nabla ( |\mathbf v |^2/2),
  \\
  (\eta, \mathbf v)(0)&=( \eta_0, \mathbf v_0),
\end{aligned}
\right.
\end{equation}
where $\eta: \mathbb{R}^{2+1}\mapsto \mathbb{R}$, $ \mathbf v:  \mathbb{R}^{2+1}\mapsto  \mathbb{R}^2$ is a curl--free vector field, i.e., $
 	\nabla \times  \mathbf v   = 0$, and
\begin{equation*}
	K_\mu:=K_\mu(D)= \frac{\tanh( \sqrt \mu|D|)}{\sqrt \mu |D|}\quad \text{with}  \ D=-i\nabla.
\end{equation*}

The system \eqref{wt2d}
describes the evolution with time of surface waves of
a liquid layer in the three dimensional physical space. The
 variables $\eta$ and $\mathbf v$
denote the surface elevation
and the fluid velocity, respectively.
The shallowness parameter $\mu$ and the nonlinearity parameter $\epsilon$ are defined by
 $$\mu=( h/\lambda)^2, \qquad \epsilon=a /h,$$
 where $h$ denotes the mean depth of the fluid layer,  $a$ is a typical amplitude of the wave, $\lambda$ a typical horizontal wavelength.

  For $\mu \ll 1 $, one has formally
$$
K_\mu(D)= 1 +\frac{\mu}{3} \Delta + \mathcal O (\mu^2),
$$
and hence one can write \eqref{wt2d}
as
\begin{equation*}
\left\{
\begin{aligned}
\partial_t \eta +  \nabla \cdot  \mathbf v & = -\epsilon \nabla \cdot (\eta \mathbf v) + \mathcal O (  \mu^2+\mu \epsilon),
  \\
  \partial_t \mathbf v + \nabla \eta +  \frac{\mu}{3} \Delta\nabla \eta   &=   - \epsilon \nabla ( |\mathbf v |^2/2) +  \mathcal O (  \mu^2+\mu \epsilon),
\end{aligned}
\right.
\end{equation*}
which is a perturbation of the Boussinesq system
\begin{equation}\label{wtbq-approx}
\left\{
\begin{aligned}
\partial_t \eta +  \nabla \cdot  \mathbf v & = -\epsilon \nabla \cdot (\eta \mathbf v) ,
  \\
  \partial_t \mathbf v + \nabla \eta +  \frac{\mu}{3} \Delta\nabla \eta   &=   - \epsilon \nabla ( |\mathbf v |^2/2) .
\end{aligned}
\right.
\end{equation}
The later system is a particular member of the (abcd) family of Boussinesq systems derived in \cite{BCS2002} as asymptotic models for water waves in the Boussinesq regime. Unfortunately, this system is linearlity ill-posed and thus cannot be used as a relevant water wave model. The system \eqref{wt2d} can be viewed as a regularization of this ill-posed system.

The system \eqref{wt2d} enjoys the Hamiltonian structure
\[
	\partial_t (\eta,  \mathbf  v)^T = \mathcal J_\mu \nabla \mathcal H_\mu(\eta ,  \mathbf v)
\]
with the skew-adjoint matrix
\[
	\mathcal J_\mu
	=
	\begin{pmatrix}
		0 & - K_\mu \partial_{x_1} & - K_\mu \partial_{x_2}
		\\
		- K_\mu \partial_{x_1} & 0 & 0
		\\
		- K_\mu \partial_{x_2} & 0 & 0
	\end{pmatrix}
	.
\]
This in particular
guarantees conservation of the energy functional
\begin{equation}
\label{Hamiltonian2}
	\mathcal E_\mu(\eta, \mathbf v)  = \frac 12 \int_{\mathbb{R}^2}
	\left(
		\eta^2 + \left| K_\mu^{-\frac12} \mathbf v \right|^2
		+ \eta |\mathbf v |^2
	\right)
	dx
	.
\end{equation}

The one dimensional version of \eqref{wt2d} that
describes the evolution with time of surface waves of
a liquid layer in the two dimensional physical space
is written as
\begin{equation}
\label{wt1d}
\left\{
\begin{aligned}
  \partial_t \eta +  \partial_x v & = -\epsilon  L_\mu \partial_x(\eta v),
  \\
  \partial_t v + L_\mu \partial_x \eta  &=   - \epsilon   L_\mu  \partial_x (  v ^2/2),
  \\
  (\eta, v)(0)&=( \eta_0, v_0) ,
\end{aligned}
\right.
\end{equation}
where $\eta, v: \mathbb{R}^{2+1}\mapsto \mathbb{R}$ and
\begin{equation*}
	L_\mu :=L_\mu (D)= \frac{\tanh( \sqrt \mu D)}{\sqrt \mu D}\quad \text{with}  \ D=-i\partial_x.
\end{equation*}
 For more details on the study of \eqref{wt2d}, \eqref{wt1d} or related equations, we refer the reader to \cite{Cr18, Dy19, DDK19, DST20, DDT22, T22}.

Recently,  together with Dinvay and Selberg \cite{DST20} we studied low regularity well-posedness of the Cauchy problems \eqref{wt2d} and \eqref{wt1d} for $\mu=\epsilon=1$. In particular,  we proved that \eqref{wt1d} is globally well-posed for initial data $( \eta_0, v_0)$ that is small in the $L^2(\mathbb{R}) \times H^{1/2} ( \mathbb{R} )$--norm. Moreover,  we showed that \eqref{wt2d} is locally well-posed for initial data $( \eta_0, \mathbf v_0) \in H^s(\mathbb{R}^2) \times H^{s+1/2} ( \mathbb{R}^2 ) \times H^{s+1/2} ( \mathbb{R}^2 )$ with $s>1/4$.

In the present paper, we are interested in the problem of long-time existence of solution to \eqref{wt2d} assuming that the nonlinearity parameter $\epsilon$ is sufficiently small.  In particular, by exploiting the dispersive nature of the system we prove that  \eqref{wt2d} is well-posed with existence time of scale $\mathcal O\left(  \mu^{3/2-} \epsilon^{-2+}\right)$ if $( \eta_0, \mathbf v_0) \in H^s(\mathbb{R}^2) \times H^{s+1/2} ( \mathbb{R}^2 ) \times H^{s+1/2} ( \mathbb{R}^2 )$ with $s>1/4$.
This in particular recovers the local well-posedness result in \cite[Theorem 2]{DST20} for $\mu=\epsilon= 1$.

There has been several studies by J-C. Saut \textit{et al.} \ \cite{LPS14,  SX19, SX20, SX12,   SWX15}  (see also  \cite{ KLPS18, MP22, E21})
 regarding long-time existence of solutions for Whitham-Boussinesq type equations with initial data of size $\mathcal O\left( 1\right)$ in some Sobolev norm, where the time of existence depends on parameter $\epsilon$. In \cite{SX20, SX12, SWX15} the analysis is based only on symmetrization and energy techniques, and do not exploit the dispersive properties of the equations. The time of existence obtained for the equations involved is at most of scale $\mathcal O\left( 1/\epsilon \right)$ in the KdV regime $\{\mu \sim \epsilon $\}, but the space of resolutions are smaller.
On the other hand, in \cite{LPS14} and \cite{SX19} the dispersive nature of the systems involved is used to study the long-time existence problem.
For instance, in \cite{LPS14} using the dispersive method the authors proved that the two dimensional dispersive Boussinesq system of the form
\begin{equation}
\label{bousq}
\left\{
\begin{aligned}
  \partial_t \eta +  \nabla \cdot (1+ \epsilon \Delta) \mathbf v & = - \epsilon  \nabla\cdot  (\eta \mathbf v),
  \\
  \partial_t \mathbf v + \nabla (1 +\epsilon \Delta) \eta  &=   -  \epsilon \nabla ( |\mathbf v |^2/2),
  \\
   (\eta, \mathbf v)(0)&=( \eta_0, \mathbf v_0) \in[ H^s(\mathbb{R}^2)]^3
\end{aligned}
\right.
\end{equation}
  is locally well-posed with existence time of scale $\mathcal O\left( 1/\sqrt \epsilon \right)$ whenever $s>3/2$.

Our main result is as follows.
\begin{theorem}
\label{theorem2d}
	Let $s > 1/4$ and $ \mu, \epsilon \in (0, 1] $. Suppose that $v_0$ is curl free, i.e., $\nabla \times \mathbf v_0 = 0$ and
the initial data has size $$
	\left(	\lVert \eta_0 \rVert _{H^s(\mathbb{R}^2)}
		+ \lVert \mathbf v_0 \rVert _{ (H^{s + 1/2} (\mathbb{R}^2))^2 } \right) \sim \mathcal D_0.
	$$
	Then there is a solution
	$$
		( \eta, \mathbf v ) \in C \left([0, T];
		H^s(\mathbb{R}^2)
		\times
		\left( H^{s + 1/2} \left( \mathbb R^2 \right) \right) ^2
		\right)
$$
	of the Cauchy problem \eqref{wt2d}
	 with  existence time $T $ given by \footnote{We use the notation $a\pm:=a \pm \delta$ for sufficiently small $\delta>0$. For any positive numbers $A$ and $B$, the notation $A \lesssim B$ stands for $A \le  CB$, where $C$ is a positive constant that is independent $\mu$, $\epsilon$ or  $T$. Moreover, we denote $A \sim B$  when $A\lesssim B$ and $B \lesssim A$. }
	$$
	T \sim  \mathcal D_0^{-2+} \mu^{3/2-} \epsilon^{-2+}   .$$
	Moreover, the solution is unique in some subspace of the above solution space and the solution depends continuously on the initial data.
\end{theorem}

\begin{remark} From Theorem \ref{theorem2d} we deduce the following:
\begin{itemize}
\item In the regime $\{  \epsilon \ll \mu \sim 1\}$,  the solution exists on a larger time scale of order $\epsilon^{-2+}$.
This is due to the presence of weak nonlinearities $\{ \epsilon \ll 1\}$, which is also regularized by the operator
$$K_\mu(D) \sim \angles{ \sqrt \mu D}^{-1} \sim \angles{  D}^{-1}  \quad \text{for} \ \ \mu\sim 1.$$

\item
 In the regime $\{ \mu \ll  \epsilon \sim 1\}$, the time of existence is of order $\mu^{3/2-}$ and hence shrinks to $0$ as $\mu \rightarrow 0+$.
This is due to the presence of strong nonlinearities $\{ \epsilon \sim 1\}$ which is not regularized, since  $$K_\mu(D) \sim  1-  \frac \mu3|D|^2  \quad \text{for} \ \ \mu\ll 1.$$
In fact, in this regime the system \eqref{wt2d} is a perturbation of the ill-posed system \eqref{wtbq-approx}.

\item
Finally, in the KdV regime $\{\mu \sim \epsilon $\}, the existence time is of order $ \epsilon^{-1/2}$.
\end{itemize}

\end{remark}

In what follows we diagonalize \eqref{wt2d}, and then reduce Theorem \ref{theorem2d} to Theorem \ref{theorem2d-reduc} below which corresponds to the diagonalized system. To this end, we define
\begin{align*}
u_\pm=\frac{\eta \mp i K_\mu^{-1/2} \mathbf R\cdot \mathbf v}{2 },
\end{align*}
where $(\eta, \mathbf v)$ is a solution to \eqref{wt2d}, and
 $\mathbf R=|D|^{-1}\nabla $ is the Riesz transform. Then we have
$$
\eta=u_++u_- \quad \mathbf v=-i \sqrt K_\mu \mathbf R (u_+- u_-).
$$
Set $$m_\mu(D):= |D|  \sqrt{K_\mu(D)}.$$
The system \eqref{wt2d} therefore transforms to
\begin{equation}
\label{wt2d-transf}
\left\{
\begin{aligned}
 (i\partial_t\mp m_\mu(D) ) u_\pm
&= \epsilon  \mathcal N^\pm_\mu(u_+, u_-) ,
\\
u_\pm (0)&= f_\pm,
\end{aligned}
\right.
\end{equation}
where
\begin{equation*}
 f_\pm =\frac{\eta_0 \mp i K_\mu^{-1/2} \mathbf  R\cdot \mathbf v_0}{2 }
\end{equation*}
and
\begin{equation}
\label{wtnonlin}
\begin{split}
\mathcal N^\pm_\mu(u_+, u_-)  &=   2^{-1}|D| K_\mu\mathbf R \cdot \left\{   (u_++u_-)    \mathbf R  \sqrt{K_\mu}(u_+ - u_-)\right\}
\\
& \qquad \pm    4^{-1}|D|  \sqrt{K_\mu} \left|  \mathbf R \sqrt{K_\mu} (u_+ - u_-)\right|^2.
\end{split}
\end{equation}

Thus, Theorem \ref{theorem2d} redues to the following:
\begin{theorem}
\label{theorem2d-reduc}
	Let $s > 1/4$, $ \mu, \epsilon \in (0, 1] $, and that the initial data has size
	$$
		\sum_\pm \lVert f_\pm \rVert _{H^s}\sim\mathcal D_0.
	$$
	Then there exists a solution
	$$
		(u_+, u_-)\in C \left([0, T];
		H^s(\mathbb{R}^2)\times H^s(\mathbb{R}^2)
		\right)
$$
	of the Cauchy problem \eqref{wt2d-transf}--\eqref{wtnonlin} with existence time $T$ as in  Theorem \ref{theorem2d}.
	Moreover, the solution is unique in some subspace of
	$ C \left([0, T];
		H^s(\mathbb{R}^2)\times H^s(\mathbb{R}^2)
		\right)$ and the solution depends continuously on the initial data.
\end{theorem}

\medskip
The paper is organized as follows: In Section 2, we prove localized dispersive and Strichartz estimates for the linear propagators associated to \eqref{wt2d-transf}.
In Section 3 and 4 we prove Theorem \ref{theorem2d-reduc} and bilinear estimates that are crucial in the proof of Theorem \ref{theorem2d-reduc}.

\medskip
\noindent \textbf{Notation}.
We fix an even smooth function $\chi \in C_0^{\infty}(\mathbb R)$ such that
\begin{equation*}
 0 \le \chi \le 1, \quad
\chi_{|_{[-1,1]}}=1 \quad \mbox{and} \quad  \mbox{supp}(\chi)
\subset [-2,2]
\end{equation*}
 and set
$$
\beta(s)
=\chi\left(s\right)-\chi \left(2s\right).
 $$
 For a dyadic number
  $\lambda \in  2^\mathbb{Z}$,  we set $\beta_{\lambda}(s):=\beta\left(s/\lambda\right)$, and thus $\supp \beta_\lambda=
\{ s\in \mathbb{R}: \lambda/ 2 \le |s| \le 2\lambda \}$.
Now define the frequency projection $P_\lambda$ via
\begin{align*}
\widehat{P_{\lambda} f}(\xi)  = \beta_\lambda(|\xi|)\widehat { f}(\xi) .
 \end{align*}
We sometimes write $f_\lambda:=P_\lambda f $, so that
\[ f=\sum_{\lambda  } f_\lambda ,\]
where summations throughout the paper are done over dyadic numbers in $ 2^\mathbb{Z}$.

The Sobolev space $H^s(\mathbb{R}^2)$ is defined via the norm
$$
\norm{f}_{H^s} \sim \left[ \sum_{\lambda }  \angles{\lambda}^{2s}\norm{ f_\lambda}^2_{L^2_x} \right]^\frac 12,
$$
 where $\angles{\xi}:= \left(1 +|\xi|^2\right)^{1/2}.$ If $B$ is a space of functions on $\mathbb{R}^2$, $T>0$ and $1\le p\le\infty$, we define the spaces $L^p\big((0,T) : B \big)$ and $L^p\big( \mathbb R : B\big)$ respectively via the norms
$$
\|f\|_{L^p_TB_x} = \left( \int_0^T \|f(\cdot,t)\|_{B}^p dt \right)^{\frac1p} \quad  \textrm{and} \quad \|f\|_{L^p_tB_x} = \left( \int_{\mathbb R} \|f(\cdot,t)\|_{B}^p dt \right)^{\frac1p} \, ,
$$
when $1 \le p < \infty$, with the usual modifications when $p=+\infty$.
\section{localised dispersive and Strichartz estimates}

First we derive lower bound estimates for the phase function
$$
m_\mu(r)=r \sqrt{K_\mu(r)}
$$
and its first two derivatives. Clearly,
$K_\mu(r) \sim \angles{ \sqrt \mu r}^{-1}$,
and hence
\begin{equation}
\label{m-est}
|m_\mu(r)| \sim |r| \angles{ \sqrt \mu r}^{-1/2}.
\end{equation}

\begin{lemma}\label{lm-mest}
 For all $r>0$, we have
\begin{align}
\label{m'-est}
0<m_\mu'(r)& \sim \angles{\sqrt \mu r}^{-1/2},
\\
\label{m''-est}
0<-m_\mu ''(r)&\sim   \mu r \angles{\sqrt \mu r}^{-5/2}.
\end{align}
\end{lemma}

\begin{proof}
Since $$
 m_{\mu}(r)= \frac{1}{\sqrt{\mu}} m_{1}(\sqrt{\mu} r)$$
 \eqref{m'-est} and \eqref{m''-est} reduce to proving
\begin{align*}
0<m_{1}'(r)& \sim  \angles{r}^{-1/2},
\\
0< - m_{1}''(r)|&\sim    |r|  \angles{ r}^{-5/2}
\end{align*}
both of which are proved in \cite[Lemma 8]{DST20}.

\end{proof}

Van der Corput's Lemma will be useful in the derivation of the dispersive estimate in Lemma \ref{lm-dispest} below.
\begin{lemma}[Van der Corput's Lemma, \cite{S93} ]\label{lm-corput}
 Assume
$g \in C^1(a, b)$, $\psi\in  C^2(a, b)$ and $|\psi''(r)|  \gtrsim A$ for all $r\in (a, b)$. Then
\begin{align}
\label{corput}
\Bigabs{\int_a^b e^{i t  \psi(r)}   g(r) \, dr}& \le C  (At)^{-1/2}  \left[ |g(b)| + \int_a^b |g'(r)| \, dr \right] ,
\end{align}
for some constant $C>0$ that is independent of $a$, $b$ and $t$.
\end{lemma}

Now we follow a similar argument\footnote{Inequality \eqref{dispest} is derived in \cite{DST20} in the case $\mu=1$ and $\lambda >1$.} as in \cite[Lemma 9]{DST20}
to derive
a  frequency localized  $L^1-L^{\infty}$ decay estimate for the linear propagator associated to \eqref{wt2d-transf}:
$$
\mathcal S_{m_{\mu}}( \pm  t) := e^{\mp it m_{\mu}(D)}.
$$

\begin{lemma}[Localised dispersive estimate] \label{lm-dispest}
Let $0 < \mu \le 1$ and $\lambda\in 2^\mathbb{Z}$.
Then
\begin{align}
\label{dispest}
\| \mathcal S_{m_{\mu}}(\pm  t)P_{\lambda} f  \|_{L^\infty_x(\mathbb{R}^2)} &\lesssim  \min\left( \lambda^2, \ \mu^{-\frac12}  \angles{\sqrt \mu \lambda}^{\frac 32} t^{-1} \right)  \| f\|_{L_x^1(\mathbb{R}^2)} \end{align}
for all $f \in \mathcal{S}(\mathbb{R}^2)$.
 \end{lemma}
 \begin{proof}
 Without loss of generality we may assume $t>0$ and $\pm=-$. Now
we can write
\[
\left[ \mathcal S_{m_\mu}(- t) f_\lambda \right](x)
= (I_{\lambda, \mu} (\cdot, t)\ast f)(x),
\]
where
\begin{equation}\label{Idef}
 I_{\lambda, \mu} (x, t)
=\lambda^2 \int_{\mathbb{R}^2} e^{i \lambda x \cdot \xi+ it {m_\mu}(\lambda \xi)}  \beta(\xi) \, d\xi .
\end{equation}
By Young's inequality
\begin{equation}\label{younginq}
\| S_{m_\mu}(-t) f_\lambda  \|_{L^\infty_x(\mathbb{R}^2)} \le \| I_{\lambda, \mu (\cdot, t)} \|_{L^\infty_x(\mathbb{R}^2)}  \|f\|_{L_x^1(\mathbb{R}^2)},
\end{equation}
and therefore, \eqref{dispest} reduces to proving
\begin{equation*}
\|I_{\lambda, \mu} (\cdot, t) \|_{L^\infty_x(\mathbb{R}^2)}  \lesssim  \min\left( \lambda^2, \ \mu^{-\frac12}  \angles{\sqrt \mu \lambda}^{\frac 32} t^{-1} \right) .
\end{equation*}
But clearly,
\begin{equation*}
\|I_{\lambda, \mu} (\cdot, t) \|_{L^\infty_x(\mathbb{R}^2)}
\lesssim  \lambda^2,
\end{equation*}
and so we reduce further to proving
\begin{equation}
\label{Iest-1}
\|I_{\lambda, \mu} (\cdot, t) \|_{L^\infty_x(\mathbb{R}^2)}  \lesssim   \mu^{-\frac12}  \angles{\sqrt \mu \lambda}^{\frac 32} t^{-1} .
\end{equation}

To prove \eqref{Iest-1} first observe that $I_{\lambda, \mu} (x, t)$ is radially symmetric w.r.t $x$, as it is
 the inverse Fourier transform of the radial function $e^{it {m_\mu}(\lambda \xi)}  \beta(\xi) $, and so we may
 set $x =(|x|, 0)$.
Therefore, using polar coordinates we can write
\begin{equation}\label{I-eq}
I_{\lambda, \mu} (x, t) =2\pi \lambda^2 \int_{1/2}^2  e^{it m_\mu(\lambda r) }J_0( \lambda r |x|) r\beta(r) \, dr,
\end{equation}
where $J_k(r)$ is the Bessel function:
$$
J_k(r)=\frac{ (r/2)^k}{(k+1/2) \sqrt{\pi}} \int_{-1}^1  e^{ir s} \left(1-s^2\right)^{k-1/2} \, ds \quad \text{for} \ k>-1/2.
$$
  The Bessel function $J_k(r)$ satisfies the following properties for $k>-1/2$ and $r>0$ (See   \cite[Appendix B]{G08} and \cite{S93}):
\begin{align}
\label{Jm1}
J_k (r) &\le Cr^{k} ,
\\
\label{Jm2}
J_k(r)& \le C r^{-1/2} ,
\\
\label{Jm3}
\partial_r \left[ r^{-k} J_k(r)\right] &= -r^{-k} J_{k+1}(r)
\end{align}
Moreover, we can write
\begin{equation}
\label{J0est}
J_0(s)= e^{is} h(s)  +e^{-is}\bar h(s)
\end{equation}
for some function $h$ satisfying the estimate
\begin{equation}
\label{h-est}
| \partial_r ^j h(r)|\le C_j \angles{r}^{-1/2-j}  \quad \text{for all} \ j\ge 0.
\end{equation}

We prove \eqref{Iest-1} by treating the cases $  |x|\lesssim \lambda^{-1}$ and $  |x|\gg  \lambda^{-1}$ separately.
\subsubsection{Case 1: $  |x|\lesssim  \lambda^{-1}$}
By \eqref{Jm1} and \eqref{Jm3} we have for all $ r\in (1/2, 2)$ the estimate
\begin{equation}
\label{J0derv-est}
\left| \partial_r^j J_0( \lambda r |x|) \right|\lesssim 1  \quad \text{for $j=0,1$}.
\end{equation}
We integrate by parts \eqref{I-eq} to obtain
\begin{align*}
 I_{\lambda, \mu} (x, t)
&=-2\pi i\lambda t^{-1}  \int_{1/2}^2  \frac{d}{dr}\left\{ e^{it m_\mu (\lambda r) }\right\}  [ m_\mu'(\lambda r) ]^{-1} J_0( \lambda r |x|) \tilde\beta(r) \, dr
\\
&=2\pi i\lambda t^{-1}  \int_{1/2}^2   e^{it m_\mu (\lambda r) }[m_\mu'(\lambda r) ]^{-1} \partial_r\left[ J_0( \lambda r |x|) \tilde\beta(r) \right] \, dr
\\
 & \qquad -2\pi i\lambda t^{-1}  \int_{1/2}^2   e^{it m_\mu (\lambda r) } [m_\mu'(\lambda r) ]^{-2} \lambda m_\mu''(\lambda r) J_0( \lambda r |x|) \tilde\beta(r)  \, dr.
\end{align*}
Then applying Lemma \ref{lm-mest}
and \eqref{J0derv-est}
we obtain
\begin{equation}
\label{I-est1-2d}
\begin{split}
|  I_{\lambda, \mu} (x, t) | &\lesssim \lambda  t^{-1} \left( \angles{ \sqrt \mu \lambda }^{\frac 12}+ \mu\lambda^2 \angles{ \sqrt \mu \lambda }^{-\frac32}  \right)
\\
&
\lesssim \lambda \angles{ \sqrt \mu \lambda }^{\frac12} t^{-1}  \lesssim \mu^{-\frac12}  \angles{\sqrt \mu \lambda}^{\frac 32} t^{-1} .
\end{split}
\end{equation}

\subsubsection{Case 2: $  |x|\gg  \lambda^{-1}$ }
Using \eqref{J0est} in \eqref{I-eq} we write
\begin{align*}
 I_{\lambda, \mu} (x, t)
 &=2\pi \lambda^2 \left\{\int_{1/2}^2  e^{it \phi^+_{\lambda,\mu} (r)  }  h(\lambda r |x|)  \tilde \beta(r) \, dr +  \int_{1/2}^2  e^{-it \phi^-_{\lambda, \mu} (r)  } \bar  h(\lambda r |x|)  \tilde\beta(r) \, dr \right\},
\end{align*}
where
$$
\phi^\pm_{\lambda, \mu} (r)=    \lambda r|x|/t  \pm  m_\mu (\lambda r) .
$$
Set $H_{\lambda}( |x|, r) =h(\lambda r |x|)  \tilde \beta(r)$. In view of \eqref{h-est} we have
\begin{equation}
\label{Hest}
| H_{\lambda}( |x|, r)  | +  |\partial_r  H_{\lambda}( |x|, r)  | \lesssim   (\lambda |x|)^{-1/2}
\end{equation}
for all $ r\in (1/2, 2)$,
where we also used the fact $\lambda |x|\gg 1$.

Now
we write
$$
 I_{\lambda, \mu} (x, t)
= I^+_{\lambda, \mu} (x, t)
+  I^-_{\lambda, \mu} (x, t) ,
$$
where
\begin{align*}
 I^+_{\lambda, \mu} (x, t)
 &=2\pi \lambda^2 \int_{1/2}^2  e^{it \phi^+_{\lambda, \mu} (r)  } H_{\lambda}( |x|, r)  \, dr ,
 \\
I^-_{\lambda, \mu} (x, t) &= 2\pi \lambda^2
   \int_{1/2}^2  e^{-it \phi^-_{\lambda, \mu} (r)  }\bar H_{\lambda}( |x|, r)  \, dr .
\end{align*}

Observe that
$$
\partial_r \phi^\pm_{\lambda, \mu} (r)=  \lambda  \left[ |x|/t \pm  m_\mu'(\lambda r) \right],\qquad \partial_r^2\phi^\pm_{\lambda, \mu} (r)=     \pm  \lambda^2 m_\mu''(\lambda r),
$$
and hence by Lemma \ref{lm-mest},
\begin{equation}
\label{phi'+:est}
|\partial_r \phi^+_{\lambda, \mu} (r)|\gtrsim  \lambda\angles{ \sqrt \mu \lambda }^{-1/2}
\qquad
|\partial^2_r \phi^\pm_{\lambda, \mu} (r)| \sim  \mu  \lambda^3\angles{  \sqrt\mu \lambda }^{-5/2}
\end{equation}
for all $ r\in (1/2, 2)$, where we also used the fact that $m'$ is positive.

\subsubsection*{\underline{Estimate for  $I^+_{\lambda, \mu} (x, t)$ } }
By integration by parts we have
\begin{align*}
 I^+_{\lambda, \mu} (x, t)
&=2\pi i t^{-1} \lambda^2  \int_{1/2}^2  e^{it \phi^+_{\lambda, \mu} (r)  }  \left\{
\frac{\partial_r H_{\lambda}( |x|, r)  }{\partial_r \phi^+_{\lambda, \mu} (r) } -   \frac{\partial^2_r \phi^+_{\lambda, \mu} (r)   H_{\lambda}( |x|, r)  }{\left[\partial_r \phi^+_{\lambda, \mu} (r) \right]^{2}}\right\}\, dr.
\end{align*}
Then using \eqref{Hest} and \eqref{phi'+:est} we have
\begin{equation}
\label{I-est2-2d}
\begin{split}
| I^+_{\lambda, \mu} (x, t)|
&\lesssim  t^{-1} \lambda^2 \left( \lambda^{-1}\angles{ \sqrt \mu \lambda }^{\frac12}  + \mu \lambda \angles{ \sqrt \mu \lambda }^{-\frac 32}  \right)  (\lambda|x|)^{-\frac 12}
\\
&\lesssim   \lambda \angles{ \sqrt \mu \lambda }^{\frac12}  t^{-1} \lesssim  \mu^{-\frac12}  \angles{\sqrt \mu \lambda}^{\frac 32} t^{-1},
\end{split}
\end{equation}
where we also used the assumption $  \lambda |x|\gg  1$.

\subsubsection*{\underline{Estimate for $I^-_{\lambda, \mu} (x, t)$}}
We treat the the non-stationary and stationary cases separately. In the non-stationary
case, where
$$ |x | \ll   \angles{ \sqrt \mu \lambda }^{-1/2} t \quad \text{or} \quad |x| \gg   \angles{ \sqrt \mu \lambda } ^{-1/2} t, $$ we have
$$
|\partial_r \phi^-_{\lambda, \mu} (r)|\gtrsim  \lambda\angles{ \sqrt \mu \lambda }^{-1/2},
$$
and hence $I^-_{\lambda, \mu} (x, t)$ can be estimated in exactly the same way as $I^+_{\lambda, \mu} (x, t)$ above, and satisfies the same bound.

So it remains to treat the stationary case:
$$
|x | \sim  \angles{ \sqrt \mu \lambda }^{-1/2} t.
$$
 In this case,
we use Lemma \ref{lm-corput}, \eqref{phi'+:est} and\eqref{Hest} to obtain
\begin{equation}\label{Iest-station}
\begin{split}
| I^-_{\lambda, \mu} (x, t)
&\lesssim \lambda^2 \left( \mu  \lambda^3\angles{  \sqrt\mu \lambda }^{-5/2}  t \right)^{-\frac12}\left[ |H_\lambda^- (x, 2) |+ \int_{1/2}^2 | \partial_r  H_\lambda^- (x, r)| \, dr\right]
\\
&\lesssim  \mu^{-\frac12}  \lambda^{\frac12 }  \angles{\sqrt \mu \lambda}^{\frac54}   t^{-\frac12} \cdot (\lambda |x|)^{-\frac 12}
\\
& \lesssim      \mu^{-\frac12}  \angles{\sqrt \mu \lambda}^{\frac 32} t^{-1}  ,
\end{split}
\end{equation}
where we also used the fact that $H_\lambda^- (x, 2) =0$.

 \end{proof}

\begin{lemma}[Localised Strichartz estimates]\label{lm-LocStr}
 Let $0 < \mu \le 1$ and $\lambda\in 2^\mathbb{Z}$.
 Assume that the pair $(q, r)$ is \emph{Strichartz admissible} in the sense that
  \begin{equation} \label{admissible}
 q> 2, \ r\ge 2 \quad \text{and} \quad \frac1r+\frac1q=\frac12 \, .
 \end{equation}
 Then
 \begin{align}
\label{Strest1d}
\norm{ \mathcal S_{m_\mu}(\pm t) f_{\lambda}}_{ L^{q}_{t} L^{r}_{ x} (\mathbb{R}^{2+1}) } \lesssim \mu ^{-\frac1{2q}} \angles{\sqrt\mu \lambda }^{\frac3{2q}}
\norm{  f_{\lambda}}_{ L^2_{ x}(\mathbb{R}^2 )} ,
\\
\label{Strest1d-inh}
\norm{ \int_0^t \mathcal S_{m_\mu}(\pm(t-s)) F_\lambda (s) \, ds}_{ L^{q}_{t} L^{r}_{ x} (\mathbb{R}^{2+1}) } \lesssim \mu ^{-\frac1{2q}} \angles{\sqrt\mu \lambda }^{\frac3{2q}}
\norm{ F_{\lambda}}_{ L_t^1L^2_{ x}(\mathbb{R}^{2+1} )}
\end{align}
for all  $f \in \mathcal{S}(\mathbb{R}^2)$ and  $F \in \mathcal{S}(\mathbb{R}^{2+1})$.
\end{lemma}
 \begin{proof}
  We shall use the Hardy-Littlewood-Sobolev inequality which
asserts that
\begin{equation}
\label{HLSineq}
\norm{|\cdot |^{-\gamma}\ast f}_{L^a(\mathbb{R})} \lesssim \ \norm{ f}_{L^b(\mathbb{R})}
\end{equation}
whenever $1 < b < a < \infty$ and $0 < \gamma< 1$ obey the scaling condition
$$
\frac1b=\frac1a +1-\gamma.
$$

We prove only \eqref{Strest1d} since \eqref{Strest1d-inh}  follows from \eqref{Strest1d}  by the standard $TT^*$--argument.
First note that \eqref{Strest1d} holds true for the pair $(q, r)=(\infty, 2)$
as this is just the energy inequality.  So we may assume $q\in(2, \infty)$.

Let $q'$ and $r'$ be the conjugates of $q$ and $r$, respectively, i.e., $q'=\frac q{q-1}$ and  $r'=\frac r{r-1}$.
 By the standard $TT^*$--argument, \eqref{Strest1d} is equivalent to the estimate
\begin{equation}
\label{TTstar}
\norm{ TT^\ast F }_{L^{q}_{t} L^{r}_{ x} (\mathbb{R}^{2+1}) } \lesssim \mu ^{-\frac1{q}} \angles{\sqrt\mu \lambda }^{\frac3{q}}
\norm{ F  }_{ L^{q'}_{ t} L_x^{r'}(\mathbb{R}^{2+1} )},
\end{equation}
where
\begin{equation}\label{TTastF}
\begin{split}
 TT^\ast F (x, t)&= \int_{\mathbb{R}}  \int_\mathbb{R} e^{i  x  \xi\mp i(t-s) {m_\mu}(  \xi)}  \beta^2_\lambda (\xi)   \widehat{F}( \xi, s)\, ds  d\xi
 \\
 &= \int_\mathbb{R}  K_{\lambda,  t-s} \ast F( \cdot,  s) \, ds,
 \end{split}
\end{equation}
with
\begin{align*}
 K_{\lambda ,t}(x)&= \int_{\mathbb{R}}  e^{i  x  \xi\mp it  {m_\mu}(  \xi)}  \beta^2_\lambda (\xi)   \, d\xi.
\end{align*}
Since $$K_{\lambda, t} \ast g (x)= \mathcal S_{m_\mu}(t)  P_\lambda g_\lambda (x)$$
 it follows from \eqref{dispest} that
\begin{equation}\label{kest1}
\|K_{\lambda, t} \ast g \|_{L_x^{\infty}(\mathbb{R})} \lesssim  \mu ^{-\frac12} \angles{\sqrt\mu \lambda }^{\frac32}  |t|^{-1}  \|g\|_{L_x^{1}(\mathbb{R})}.
\end{equation}
On the other hand, we have by Plancherel
\begin{equation}\label{kest2}
\|K_{\lambda, t} \ast g \|_{L_x^{2}(\mathbb{R})} \lesssim    \|g\|_{L_x^{2}(\mathbb{R})}.
\end{equation}
So interpolation between \eqref{kest1} and \eqref{kest2} yields
\begin{equation}\label{kest3}
\|K_{\lambda, t} \ast g \|_{L_x^{r}(\mathbb{R})} \lesssim  \left[  \mu ^{-\frac12} \angles{\sqrt\mu \lambda }^{\frac32}\right]^{1- \frac2r }  |t|^{-\left(1-\frac 2r\right)} \|g\|_{L_x^{r'}(\mathbb{R}^2)}
\end{equation}
 for all $  r \in[2, \infty].$

Applying Minkowski's inequality to \eqref{TTastF}, and then  \eqref{kest3} and  \eqref{HLSineq}
with $(a, b)=(q , q' )$ and $\gamma= 1-2/r=2/q$,
 we obtain
 \begin{align*}
\norm{TT^\ast F }_{L^{q}_{t} L^{r}_{ x} (\mathbb{R}^{2+1})}
&\le \norm{   \int_\mathbb{R} \norm{ K_{\lambda, t-s,} \ast
   F(s, \cdot) }_{L_x^r (\mathbb{R}^2)}  \, ds}_{L^{q}_t(\mathbb{R})}
  \\
 &\lesssim \mu ^{-\frac1{q}} \angles{\sqrt\mu \lambda }^{\frac3{q}}   \norm{  \int_\mathbb{R}  |t-s|^{-\frac2q }
  \norm{ F(s, \cdot) }_{ L_x^{r'}(\mathbb{R}^2)}  \, ds }_{L_t^{q}(\mathbb{R})}
   \\
 &\lesssim  \mu ^{-\frac1{q}} \angles{\sqrt\mu \lambda }^{\frac3{q}}    \norm{
  \norm{ F }_{L_x^{r'}(\mathbb{R}^2) }  }_{L^{q'}_{ t} (\mathbb{R})}
    \\
 &=  \mu ^{-\frac1{q}} \angles{\sqrt\mu \lambda }^{\frac3{q}}
  \norm{ F  }_{ L^{q'}_{ t} L_x^{r'}(\mathbb{R}^{2+1})} \, ,
\end{align*}
which is the desired estimate \eqref{TTstar}.
 \end{proof}

 \section{Proof of Theorem \ref{theorem2d-reduc}}
The bilinear terms in \eqref{wtnonlin} can be written as
 \begin{equation*}
\begin{split}
\mathcal N^\pm_\mu(u_+, u_-)  &=   \frac12 \sum_{\pm_1, \pm_2} \pm_2 D K_\mu \mathbf R \cdot  \left( u_{\pm_1}  \mathbf R   \sqrt{K_\mu}u_{\pm_2} \right)
\\
& \qquad \qquad \pm  \frac14 \sum_{\pm_1, \pm_2}  (\pm_1) (\pm_2) D \sqrt{K_\mu }  \left( \mathbf R   \sqrt{K_\mu }   u_{\pm_1}   \cdot \mathbf R  \sqrt{K_\mu} u_{\pm_2} \right) ,
\end{split}
\end{equation*}
where $\pm_1$ and $\pm_2$ are independent signs.
Then the Duhamel's representation of \eqref{wt2d-transf} is given by
\begin{equation}
\label{uinteq}
\begin{split}
  u_\pm(t) &= \mathcal S_{m_\mu}(\pm t) f_\pm \mp \frac{ i  \epsilon}2 \sum_{\pm_1, \pm_2}  (\pm_2)  \mathcal A^\pm_{\mu} (u_{\pm_1},  u_{\pm_2})(t)
  \\
  & \qquad \qquad \qquad \mp \frac{ i  \epsilon}4   \sum_{\pm_1, \pm_2}   (\pm_1) (\pm_2) \mathcal B^\pm_{\mu} (u_{\pm_1},  u_{\pm_2})(t),
  \end{split}
   \end{equation}
 where
 \begin{equation}
\label{AB}
 \begin{split}
\mathcal A^\pm_{\mu} (u,v)(t)&:=\int_{0}^{t}    \mathcal S_{m_\mu}( \pm(t-t') D K_\mu  \mathbf R \cdot \left( u  \mathbf R \sqrt{K_\mu} v \right)(t') \,dt' ,
\\
\mathcal  B^\pm_{\mu}  (u,v)(t)&:= \int_{0}^{t} \mathcal S_{m_\mu}( \pm(t-t') D \sqrt{K_\mu} \left(   \mathbf R  \sqrt{K_\mu} u  \cdot   \mathbf R  \sqrt{K_\mu} v \right)(t') \,dt'  .
\end{split}
\end{equation}

Now let $(q, r)$ with $q, \, r >2$ be a Strichartz admissible pair.
We define the contraction space, $X^s_T$, via the norm
$$
\norm{u}_{X^s_T}= \left[ \sum_{\lambda}  \angles{\lambda}^{2s}\norm{ u}^2_{X_\lambda} \right]^\frac 12,
$$
where
$$
\norm{u}_{X_\lambda} =
  \left[ \norm{  P_\lambda u}^2_{ L_T^{\infty}L_x^{2} } +\mu ^{\frac1{q}} \angles{\sqrt\mu \lambda }^{-\frac3{q}}   \norm{ P_\lambda u}^2_{ L_T^{q} L_x^{r}  }  \right]^\frac12.
$$
Observe that
\begin{equation}
\label{Xlam-est}
\begin{split}
\norm{ P_\lambda  u}_{ L_T^{\infty}L_x^{2} }  &\le \norm{  u}_{X_\lambda} ,
\\
\norm{  P_\lambda u}_{L_T^{q} L_x^{r }  }  &\le  \mu ^{-\frac1{2q}} \angles{\sqrt\mu \lambda }^{\frac3{2q}}\norm{  u}_{X_\lambda} .
\end{split}
\end{equation}
Moreover,
$$
  X^s_T \subset L_T^\infty H^s.
$$

 We estimate the linear part of \eqref{uinteq} using
 \eqref{Strest1d} as follows:
 \begin{equation}
\label{homest}
\begin{split}
 \norm{\mathcal S_{m_\mu}(\pm t)  f_\pm}_{X^s_T} &= \left[ \sum_{\lambda }  \angles{ \lambda}^{2s} \norm{  \mathcal S_{m_\mu}(\pm t) f_\pm }^2_{ X_\lambda}  \right]^\frac 12
 \\
 &\lesssim  \left[ \sum_{\lambda} \angles{ \lambda}^{2s} \norm{  P_\lambda  f_\pm}^2_{ L_x^{2} }  \right]^\frac 12
 \\
 & \sim \norm{   f_\pm}_{ H^{s} } .
\end{split}
\end{equation}
So Theorem \ref{theorem2d-reduc} reduces to proving the following bilinear estimates whose proof will be given in the next section.
\begin{lemma}\label{lm-keybiest}
Let $0<\alpha\ll 1$, $0< \mu\le 1$, $s\ge 1/4+ \alpha$ and $T>0$. Then for
all $u, \ v \in X^s_T$, we have
\begin{align}
\label{Keybiest1}
 \norm{ \mathcal A^\pm_{\mu} (u,v) }_{X^s_T} &\lesssim  T^{\frac12+\alpha}  \mu^{-\frac 34+ \frac\alpha 2} \norm{u}_{X^s_T}  \norm{v}_{X^s_T}  ,
 \\
 \label{Keybiest2}
 \norm{ \mathcal B^\pm_{\mu} (u,v)   }_{X^s_T} &\lesssim  T^{\frac12+\alpha}  \mu^{-\frac 34+ \frac\alpha 2} \norm{u}_{X^s_T}  \norm{v}_{X^s_T}  ,
\end{align}
where $\mathcal A^\pm_{\mu}$ and $\mathcal B^\pm_{\mu}$ are as in \eqref{AB}.

\end{lemma}

Indeed, given that Lemma \ref{lm-keybiest} holds,
we solve the integral equations \eqref{uinteq} by contraction mapping techniques
as follows. Define the mapping
\[
  (u_+, u_-) \mapsto \left (\Phi^+_\mu (u_+, u_-), \Phi_\mu^-(u_+, u_-)\right),
\]
where $\Phi^\pm_\mu(u_+, u_-)$ is given by the right hand side of \eqref{uinteq}.

Now given initial data with norm
$$\sum_\pm \| f_\pm\|_{ H^s}\le \mathcal D_0 , $$
we look for a solution in the set
\[
  E_T = \left\{  (u_\pm  \in X^s_T
  \colon  \sum_\pm \| u_\pm\|_{ X^s_T} \le 2 C \mathcal D_0 \right\}.
\]
Then by \eqref{homest} and Lemma \ref{lm-keybiest} we have
\begin{align*}
\sum_\pm  \norm{\Phi^\pm_\mu(u_+, u_-)}_{ X^s_T} &\le C\sum_\pm  \norm{   f_\pm}_{ H^{s} } + C T^{\frac12+\alpha}  \mu^{-\frac 34+ \frac\alpha 2} \epsilon \left(\sum_\pm \| u_\pm\|_{ X^s_T}  \right)^2
\\
&\le C \mathcal D_0 + C T^{\frac12+\alpha}  \mu^{-\frac 34 + \frac\alpha 2}  \epsilon ( 2C \mathcal D_0)^2
\\
&\le 2 C \mathcal D_0,
\end{align*}
provided that
\begin{equation}\label{T}
T\le \left( 8 C^2 \mathcal D_0\right)^{-2+ \delta} \mu^{\frac32-\delta } \epsilon^{-2+\delta },
\end{equation}
where $\delta = \frac {4\alpha } {1+2\alpha} \ll 1$.

Similarly, for two pair of solutions $(u_+, u_-)$ and $(v_+, v_-)$ in $E_T$ with the same data, one can derive the difference estimate
\begin{align*}
 &\sum_\pm  \norm{ \Phi^\pm_\mu(u_+, u_-)- \Phi^\pm_\mu(v_+, v_-)}_{ X^s_T}
 \\
 & \quad \le C T^{\frac12+\alpha}  \mu^{-\frac 34+ \frac\alpha 2}  \epsilon \left(\sum_\pm \| u_\pm\|_{ X^s_T}+ \| v_\pm\|_{ X^s_T}  \right) \left(\sum_\pm \| u_\pm- v_\pm\|_{ X^s_T}  \right)
  \\
 & \quad \le 4 C^2 T^{\frac12+\alpha}  \mu^{-\frac 34+ \frac\alpha 2}  \epsilon \mathcal D_0 \left(\sum_\pm \| u_\pm- v_\pm\|_{ X^s_T}  \right)
 \\
 & \quad \le \frac12   \left(\sum_\pm \| u_\pm- v_\pm\|_{ X^s_T}  \right),
\end{align*}
where in the last inequality we used \eqref{T}.

Therefore, $( \Phi^+_\mu, \Phi^-_\mu)$ is a contraction on $E_T$ and therefore it has a unique fixed point $(u_+, u_-) \in E_T$ solving the integral equation \eqref{uinteq}--\eqref{AB} on $\mathbb{R}^2\times [0, T]$, where $$T \sim  \mathcal D_0^{-2+}\mu^{\frac32-}  \epsilon^{-2+}.$$ Uniqueness in the space $X^s_T \times X^s_T$ and continuous dependence on the initial data can be shown in a similar way, by the difference estimates. This concludes the proof of Theorems \ref{theorem2d-reduc}.

\section{Proof of Lemma \ref{lm-keybiest}}

First we prove key bilinear estimates in Lemma \ref{lm-biest} below that will be crucial in the proof of Lemma  \ref{lm-keybiest}.
By Bernstein inequality, we have for all $r_1, r_2 \in \mathbb{R}$ and $p\ge 1$,
\begin{equation}\label{Bern-est}
 \| |D|^{r_1} K^{r_2}_\mu \mathbf R f_{\lambda} \|_{L^p_x}   \lesssim  \lambda^{r_1} \angles{ \sqrt \mu \lambda}^{-r_2}  \|f_{\lambda}\|_{L^p_x},
\end{equation}
where $\mathbf R=|D|^{-1}\nabla $ is the Riesz transform.

For $\lambda_j \in2^\mathbb{Z}$ ($j=0,1, 2)$,
we denote
$$ \bm \lambda =(\lambda_0, \lambda_1, \lambda_2), \quad \lambda_1 \wedge \lambda_2=\min (\lambda_1, \lambda_2).
$$
\begin{lemma}
\label{lm-biest}
Let $0<\alpha\ll 1$, $0< \mu\le 1$ and $T>0$. Then for all $u\in X_{\lambda_1 }$ and $v \in X_{\lambda_2 }$ we have
\begin{align}
\label{keybiles11}
\norm{ |D| K_\mu   P_{\lambda_0}
 \mathbf R \cdot \left(   u_{\lambda_1} \mathbf R \sqrt  K_\mu v_{\lambda_2} \right) }_{L_{T,x}^2}
& \lesssim  C_{\mu, T}( \bm \lambda)
\norm{u}_{  X_{\lambda_1 } }
\norm{v}_{  X_{\lambda_2 } },
\\
\label{keybiles12}
\norm{ |D| \sqrt {K_\mu   }
 P_{\lambda_0}\left(   \mathbf R  \sqrt  K_\mu  u_{\lambda_1}  \cdot \mathbf R \sqrt  K_\mu v_{\lambda_2} \right) }_{L_{T,x}^2}
& \lesssim  \widetilde C_{\mu, T}(\bm \lambda)
\norm{u}_{  X_{\lambda_1 } }
\norm{v}_{  X_{\lambda_2 } },
\end{align}
 where
\begin{align}
\label{C}
C_{\mu, T}(\bm \lambda)  &=   T^{\alpha}  \mu^{-\frac14+\frac\alpha 2} \frac{\lambda_0
(\lambda_1 \wedge \lambda_2)^{2\alpha} \angles{\sqrt\mu (\lambda_1 \wedge \lambda_2) }^{\frac34-\frac{3\alpha}2} }{ \angles{\sqrt{\mu} \lambda_0}  \angles{\sqrt{\mu} \lambda_2} ^{\frac12}}   ,
\\
\label{Ctilde}
\widetilde C_{\mu, T}(\bm \lambda)  &=  T^{\alpha}  \mu^{-\frac14+\frac\alpha 2} \frac{\lambda_0 (\lambda_1 \wedge \lambda_2)^{2\alpha}  \angles{\sqrt\mu (\lambda_1 \wedge \lambda_2) }^{\frac34-\frac{3\alpha}2}   }{\prod_{j=0}^2 \angles{\sqrt{\mu} \lambda_j} ^{\frac12}}.
\end{align}
\end{lemma}
\begin{proof}

We only prove \eqref{keybiles11} since the proof for \eqref{keybiles12} is similar.
By symmetry we may assume $\lambda_1 \le \lambda_2$. Let
$$
\frac1q=\frac12-\alpha, \quad \frac1r = \alpha
$$
so that $(q,r)$ is Strichartz admissible.
Then by H\"{o}lder, Bernstein inequality, \eqref{Bern-est} and \eqref{Xlam-est} we obtain
\begin{align*}
\text{LHS} \ \eqref{keybiles11} & \lesssim \lambda_0\angles{\sqrt{\mu} \lambda_0}^{-1} \norm{   \mathbf R \cdot( u_{\lambda_1} \mathbf R \sqrt  K_\mu v_{\lambda_2} ) }_{L_{T,x}^2}
\\
&\lesssim   \lambda_0\angles{\sqrt{\mu} \lambda_0}^{-1}
 \angles{\sqrt{\mu} \lambda_2}^{-\frac12}  \cdot  T^{\frac12-\frac1q} \lambda_{1}^{\frac 2r
}  \norm{u_{\lambda_1}}_{ L_T^q L_x^r  } \norm{  v_{\lambda_2}}_{ L_T^\infty L_x^2 }
\\
&\lesssim T^\alpha\mu^{-\frac14 + \frac\alpha 2}  \frac{\lambda_0\lambda_1^{2\alpha} \angles{\sqrt\mu \lambda_1}^{\frac34 -\frac{3\alpha} 2} }{ \angles{\sqrt{\mu} \lambda_0}  \angles{\sqrt{\mu} \lambda_2}^{\frac12} }
   \norm{u}_{X_{\lambda_1} }
  \norm{ v }_{X_{\lambda_2}}
\end{align*}
which proves \eqref{keybiles11}.
\end{proof}

Now we are ready to prove Lemma  \ref{lm-keybiest}:  \eqref{Keybiest1} \&  \eqref{Keybiest2}. To this end, we decompose
 $ u=\sum_{\lambda} u_{\lambda}$ and $v=\sum_{\lambda} v_{\lambda}.$
Note that by denoting
$$
a_{\lambda}:=  \norm{u }_{  X_{\lambda} }, \quad  b_{\lambda}:=  \norm{v}_{  X_{\lambda }} $$
we can write
\begin{equation}\label{Xs-redf}
\norm{u}_{  X^s_T }= \norm{ \left(  \angles{\lambda}^{s}  a_{\lambda } \right)}_{l^2_{\lambda}}, \quad
\norm{v}_{  X^s_T }= \norm{ \left(  \angles{\lambda}^{s}  b_{\lambda } \right)}_{l^2_{\lambda}}.
\end{equation}

We shall make a frequent use of of the following dyadic summation estimate, for $\mu, \lambda \in 2^\mathbb{Z}$ and $ c_1, c_2,  p>0$:
\begin{equation*}
\sum_{\mu \sim \lambda } a_\mu \sim a_\lambda, \qquad
\sum_{c_1 \lesssim \lambda \lesssim c_2} \lambda^p \lesssim
\begin{cases}
& c_2^p  \quad \text{if} \ p>0,
\\
& c_1^p  \quad \text{if} \ p<0.
\end{cases}
\end{equation*}

\subsection{Proof of \eqref{Keybiest1} }
Using \eqref{Strest1d-inh} and H\"{o}lder, we have
  \begin{align*}
\norm{   \mathcal A^\pm_{\mu} (u,v)  }_{X^s_T} ^2
& \lesssim \sum_{\lambda_0}  \angles{\lambda_0}^{2s} \norm{   |D| K_\mu    \mathbf R \cdot P_{\lambda_0}  \left( u  \mathbf R \sqrt{K_\mu} v \right) }^2_{ L_T^1L_x^{2} }
\\
  & \lesssim T \sum_{\lambda_0}  \angles{\lambda_0}^{2s} \norm{   |D| K_\mu    \mathbf R \cdot P_{\lambda_0}  \left( u  \mathbf R \sqrt{K_\mu} v \right) }^2_{ L_{T, x}^{2} } .
 \end{align*}
 But the dyadic decomposition
 \begin{equation}\label{Iloc}
 \norm{   |D| K_\mu    \mathbf R \cdot P_{\lambda_0}  \left( u  \mathbf R \sqrt{K_\mu} v \right) }_{ L_{T, x}^{2} }
 \lesssim \sum_{\lambda_1, \lambda_2}   \norm{   |D| K_\mu    \mathbf R \cdot P_{\lambda_0}  \left( u_{\lambda_1}  \mathbf R \sqrt{K_\mu} v_{\lambda_2}  \right) }_{ L_{T, x}^{2} } .
 \end{equation}
Now let $\lambda_\text{min}  $, $\lambda_\text{med}$  and $ \lambda_{\max}$ denote the minimum, median and the maximum of $\{\lambda_0, \lambda_1, \lambda_2\}$, respectively.
 By checking the support properties in Fourier space of the bilinear term on the right hand side of \eqref{Iloc} one can see that this term vanishes unless $\bm \lambda=(\lambda_0, \lambda_1, \lambda_2) \in   \Lambda,$
where
\begin{equation*}
  \Lambda= \{\bm \lambda : \  \lambda_\text{med} \sim \lambda_{\max}  \}.
  \end{equation*}
Thus, we have a non-trivial contribution in \eqref{Iloc} if $\bm \lambda \in \cup_{j=0}^2  \Lambda_j,$ where
\begin{align*}
\Lambda_0 &= \{\bm \lambda : \ \lambda_0 \lesssim \lambda_1\sim  \lambda_2 \},
\\
\Lambda_1&= \{\bm \lambda : \ \lambda_2 \ll  \lambda_1\sim  \lambda_0 \},
\\
\Lambda_2 &= \{\bm \lambda : \ \lambda_1 \ll \lambda_2 \sim  \lambda_0 \}.
\end{align*}
 By using these facts, and applying \eqref{keybiles11} to the right hand side of \eqref{Iloc}, we get
 \begin{align*}
\norm{   \mathcal A^\pm_{\mu} (u,v)  }_{X^s_T} ^2
& \lesssim T  \sum_{j=0}^2   \mathcal I^{(j)}_{\mu, T},
 \end{align*}
 where
\begin{equation}
\label{Gammadefj}
  \mathcal I^{(j)}_{\mu, T}= \sum_{\lambda_0}  \angles{\lambda_0}^{2s} \left[  \sum_{\lambda_1, \lambda_2: \ \bm \lambda \in   \Lambda_j }    C_{\mu, T}( \bm \lambda)   a_{\lambda_1 } b_{\lambda_2 }    \right]^2
\end{equation}
with $C_{\mu, T}(\bm \lambda)$ as in \eqref{C}.

 So \eqref{Keybiest1} reduces to proving
\begin{equation}
\label{GammaEstj}
 \mathcal I^{(j)}_{\mu, T} \lesssim  T^{2\alpha}  \mu^{-\frac 32+ \alpha} \norm{u}^2_{  X^s_T }\norm{v}^2_{ X^s_T } \quad (j=0,1,2).
\end{equation}
Further, observe that $\Lambda_j\subset \cup_{k=0}^2  \Lambda_{jk},$ where
\begin{align*}
\Lambda_{j0} &= \{\bm \lambda \in \Lambda_j : \  \lambda_{\max}  \lesssim \mu^{-1/2} \},
\\
\Lambda_{j1} &= \{\bm \lambda \in \Lambda_j : \  \lambda_{\min} \lesssim \mu^{-1/2} \ \& \
 \lambda_\text{med}  \gg \mu^{-1/2}\},
 \\
 \Lambda_{j2}& = \{\bm \lambda \in \Lambda_j : \  \lambda_{\min}  \gg \mu^{-1/2}\}.
\end{align*}
So
$$ \mathcal I^{(j)}_{\mu, T} \lesssim \sum_{k=0}^2 \mathcal I^{(jk)}_{\mu, T}, $$
 where
\begin{equation}
\label{Gammadefjk}
  \mathcal I^{(jk)}_{\mu, T}= \sum_{\lambda_0}  \angles{\lambda_0}^{2s} \left[  \sum_{\lambda_1, \lambda_2: \ \bm \lambda \in   \Lambda_{jk} }    C_{\mu, T}( \bm \lambda)   a_{\lambda_1 } b_{\lambda_2 }    \right]^2.
\end{equation}
Thus, \eqref{GammaEstj} reduces further to proving
\begin{equation}
\label{GammaEst-jk}
\mathcal I^{(jk)}_{\mu, T}  \lesssim  T^{2\alpha}  \mu^{-\frac 32+ \alpha} \norm{u}^2_{  X^s_T }\norm{v}^2_{ X^s_T } \quad (j,k=0, 1,2).
\end{equation}

We prove \eqref{GammaEst-jk} as follows.
\subsubsection{Estimates for $\mathcal I^{(0k)}_{\mu, T} $ ($k=0,1,2$)  }
In the case $\bm \lambda \in \Lambda_{00} $, we have $$C_{\mu, T}(\bm \lambda)  \lesssim T^{\alpha}  \mu^{-\frac1{4} + \frac \alpha 2} \lambda_0 \lambda_2^{2\alpha} , $$ and hence
\begin{align*}
 \mathcal I^{(00)}_{\mu, T}
&
\lesssim  T^{2\alpha}
\mu^{-\frac12+ \alpha}  \sum_{ \lambda_0 \lesssim \mu^{-1/2}}  \lambda_0^2 \left( \sum_{ \lambda_1\sim \lambda_2\lesssim  \mu^{-1/2}}
\angles{\lambda_1}^{s}  a_{\lambda_1 }  \cdot \angles{\lambda_2}^{2\alpha} b_{\lambda_2 } \right)^2
\\
&
\lesssim   T^{2\alpha}  \mu^{-\frac 32+ \alpha} \norm{u}^2_{  X^s_T }\norm{v}^2_{ X^s_T }
\end{align*}
for all $s\ge 2\alpha$,
where to obtain the second line we used Cauchy Schwarz inequality in $\lambda_1\sim \lambda_2$ and \eqref{Xs-redf}.

If $\bm \lambda \in \Lambda_{01} $, we have $$C_{\mu, T}(\bm \lambda)  \lesssim T^{\alpha}   \mu^{-\frac{1}{8}-\frac\alpha 4}  \lambda_0 \lambda_2^{\frac14+\frac\alpha 2}, $$ and hence
\begin{align*}
\mathcal I^{(01)}_{\mu, T}
&
\lesssim
T^{2\alpha}
\mu^{-\frac14- \frac \alpha 2}    \sum_{ \lambda_0 \lesssim \mu^{-1/2}}  \lambda_0^2\left( \sum_{ \lambda_1\sim \lambda_2\gg \mu^{-1/2}}    \angles{\lambda_1}^{s}
a_{\lambda_1 }  \cdot \ \lambda_2^{\frac14+\frac\alpha 2} b_{\lambda_2 } \right)^2
\\
&
\lesssim
T^{2\alpha}
\mu^{-\frac54- \frac \alpha 2}   \norm{u}^2_{  X^s_T }\norm{v}^2_{ X^s_T }
\end{align*}
for all $s\ge 1/4+ \alpha /4$.

Finally, if $\bm \lambda \in \Lambda_{02} $, we have $$C_{\mu, T}(\bm \lambda)  \lesssim T^{\alpha}  \mu^{-\frac{5}{8}-\frac\alpha 4}   \lambda_2^{\frac14+\frac\alpha 2} ,$$ and hence
\begin{align*}
\mathcal I^{(02)}_{\mu, T}
&
\lesssim
T^{2\alpha}
\mu^{-\frac54- \frac \alpha 2}
  \sum_{ \lambda_0 \gg \mu^{-1/2}}  \angles{\lambda_0}^{\frac12+ \alpha -2s} \left( \sum_{ \lambda_1\sim \lambda_2\gg \mu^{-1/2}}    \angles{\lambda_1}^{s}
a_{\lambda_1 }  \cdot \angles{\lambda_2}^{s}  b_{\lambda_2 } \right)^2
\\
&
\lesssim  T^{2\alpha}
\mu^{-\frac54- \frac \alpha 2}   \norm{u}^2_{  X^s_T }\norm{v}^2_{ X^s_T }
\end{align*}
for all $s> 1/4+\alpha/ 2$.

\subsubsection{Estimates for $\mathcal I^{(1k)}_{\mu, T} $ ($k=0,1,2$) } \label{secGamma1k}
If $\bm \lambda \in \Lambda_{10} $, we have $$C_{\mu, T}(\bm \lambda)  \lesssim T^{\alpha}  \mu^{-\frac1{4} + \frac \alpha 2} \lambda_0 \lambda_2^{2\alpha} .$$
Hence
\begin{align*}
\mathcal I^{(10)}_{\mu, T}
&
\lesssim T^{2\alpha}
\mu^{-\frac12+ \alpha} \sum_{ \lambda_0 \lesssim \mu^{-1/2}}   \angles{\lambda_0}^{2s}\lambda_0^2 \left( \sum_{ \lambda_2 \ll \lambda_1 \sim \lambda_0\lesssim \mu^{-1/2} }
a_{\lambda_1 } \cdot  \lambda_2^{2\alpha}  b_{\lambda_2 } \right)^2
\\
& \lesssim T^{2\alpha}
\mu^{-\frac 32+ \alpha }  \sum_{ \lambda_0 \lesssim \mu^{-1/2}}   \angles{\lambda_0}^{2s} a_{\lambda_0}^2 \left( \sum_{ \lambda_2  \lesssim \mu^{-1/2} }
  \lambda_2^{2\alpha}  b_{\lambda_2 } \right)^2
\\
&
\lesssim T^{2\alpha}
\mu^{-\frac 32+ \alpha } \norm{u}^2_{  X^s_T }\norm{v}^2_{ X^s_T },
\end{align*}
where to get the last two inequalities we used
$\sum_{\lambda_1 \sim \lambda_0} a_{\lambda_1} \sim a_{\lambda_0} $ and by Cauchy Schwarz
\begin{align*}
 \sum_{ \lambda_2  \lesssim \mu^{-1/2} }
  \lambda_2^{2\alpha}  b_{\lambda_2 } & \le \left( \sum_{ \lambda_2  \le 1}
  +  \sum_{ \lambda_2  > 1}  \right)
  \lambda_2^{2\alpha}  b_{\lambda_2 } \\
  &\lesssim  \norm{ \left(  b_{\lambda_2 } \right)}_{l^2_{\lambda_2}}+ \norm{ \left(  \angles{\lambda_2}^{s}  b_{\lambda_2 } \right)}_{l^2_{\lambda_2}} \lesssim \norm{v}_{ X^s_T }
\end{align*}
for all $s>2\alpha $.

If $\bm \lambda \in \Lambda_{11} $, we have $$C_{\mu, T}(\bm \lambda)  \lesssim T^{\alpha}  \mu^{-\frac3{4} + \frac \alpha 2}  \lambda_2^{2\alpha} ,  $$
and hence using the previous argument
\begin{align*}
\mathcal I^{(11)}_{\mu, T}
&
\lesssim
T^{2\alpha}
\mu^{-\frac 32+ \alpha }  \sum_{ \lambda_0 \gg \mu^{-1/2}}   \angles{\lambda_0}^{2s}  \left( \sum_{  \lambda_1 \sim \lambda_0\gg \lambda_2, \ \lambda_2 \lesssim \mu^{-1/2}  }
a_{\lambda_1 }  \lambda_2^{2\alpha}    b_{\lambda_2 } \right)^2
\\
&
\lesssim  T^{2\alpha}
\mu^{-\frac 32+ \alpha }  \norm{u}^2_{  X^s_T }\norm{v}^2_{ X^s_T }
\end{align*}
for all $s>2\alpha $.

Finally, if $\bm \lambda \in \Lambda_{12} $, we have $$C_{\mu, T}(\bm \lambda)  \lesssim T^{\alpha}  \mu^{-\frac 58 -  \frac {\alpha} 4} \lambda_2^{\frac14+ \frac\alpha2}  ,  $$
and hence using the previous argument
\begin{align*}
\mathcal I^{(12)}_{\mu, T}
&
\lesssim
T^{2\alpha}
\mu^{-\frac 54-\frac \alpha 2 }  \sum_{ \lambda_0 \gg \mu^{-1/2}}   \angles{\lambda_0}^{2s}  \left( \sum_{  \lambda_1 \sim \lambda_0\gg \lambda_2\gg  \mu^{-1/2}  }
a_{\lambda_1 }  \lambda_2^{ \frac14 + \frac\alpha 2}    b_{\lambda_2 } \right)^2
\\
&
\lesssim  T^{2\alpha}
\mu^{-\frac 32+ \alpha }\norm{u}^2_{  X^s_T }\norm{v}^2_{ X^s_T }
\end{align*}
for all $s> 1/4 + \alpha /2$.

\subsubsection{Estimates for $\mathcal I^{(2k)}_{\mu, T} $ ($k=0,1,2$)  }
If $\bm \lambda \in \Lambda_{20} $, we have $$C_{\mu, T}(\bm \lambda)  \lesssim T^{\alpha}  \mu^{-\frac1{4} + \frac \alpha 2} \lambda_0 \lambda_1^{2\alpha}  $$
and hence
 arguing as in the preceding subsection we obtain
\begin{align*}
 \mathcal I^{(20)}_{\mu, T}
&
\lesssim T^{2\alpha}
\mu^{-\frac12+ \alpha} \sum_{ \lambda_0 \lesssim \mu^{-1/2}}   \angles{\lambda_0}^{2s}\lambda_0^2 \left( \sum_{ \lambda_1 \ll \lambda_2 \sim \lambda_0\lesssim \mu^{-1/2} }
 \lambda_1^{2\alpha}  a_{\lambda_1 }   b_{\lambda_2 } \right)^2
 \\
&
\lesssim T^{2\alpha}
\mu^{-\frac 32+ \alpha }\norm{u}^2_{  X^s_T }\norm{v}^2_{ X^s_T }
\end{align*}
for all $s>2\alpha $.

Next if $\bm \lambda \in \Lambda_{21} $, then $$C_{\mu, T}(\bm \lambda)  \lesssim T^{\alpha}  \mu^{-1 + \frac \alpha 2} \lambda_0^{-\frac12}  \lambda_1^{2\alpha}  , $$
and hence
\begin{align*}
 \mathcal I^{(21)}_{\mu, T}
&
\lesssim T^{2\alpha}
\mu^{-2+ \alpha }  \sum_{ \lambda_0 \gg \mu^{-1/2}}   \angles{\lambda_0}^{2s}    \lambda_0^{-1} \left( \sum_{ \lambda_2 \sim \lambda_0\gg \lambda_1, \ \lambda_1 \lesssim \mu^{-1/2}  }
 \lambda_1^{2\alpha}  a_{\lambda_1 }    b_{\lambda_2 } \right)^2
\\
&
\lesssim  T^{2\alpha}
\mu^{-\frac 32+ \alpha }  \norm{u}^2_{  X^s_T }\norm{v}^2_{ X^s_T }
\end{align*}
for all $s>2\alpha $.

Finally, if $\bm \lambda \in \Lambda_{21} $, then $$C_{\mu, T}(\bm \lambda)  \lesssim T^{\alpha}  \mu^{-\frac 58 -  \frac {\alpha} 4}  \lambda_1^{\frac14+ \frac\alpha2}  , $$
and hence
\begin{align*}
  \mathcal I^{(22)}_{\mu, T}
&
\lesssim
T^{2\alpha}
\mu^{-\frac 54-\frac \alpha 2 }  \sum_{ \lambda_0 \gg \mu^{-1/2}}   \angles{\lambda_0}^{2s}  \left( \sum_{ \lambda_2 \sim \lambda_0\gg \lambda_1\gg  \mu^{-1/2}  }   \lambda_1^{ \frac14 + \frac\alpha 2}
a_{\lambda_1 }   b_{\lambda_2 } \right)^2
\\
&
\lesssim T^{2\alpha}
\mu^{-\frac 54-\frac \alpha 2 } \norm{u}^2_{  X^s_T }\norm{v}^2_{ X^s_T }
\end{align*}
for all $s> 1/4 + \alpha / 2$.

\subsection{Proof of \eqref{Keybiest2} }
Arguing as in the preseding subsection we use \eqref{Strest1d-inh},  H\"{o}lder,  \eqref{keybiles12} and \eqref{Ctilde} to obtain
 \begin{align*}
\norm{   \mathcal B^\pm_{\mu} (u,v)  }_{X^s_T} ^2
& \lesssim T \sum_{j=0}^2   \mathcal J^{(j)}_{\mu, T},
 \end{align*}
 where
\begin{equation}\label{Gammatilde}
 \mathcal J^{(j)}_{\mu, T}= \sum_{\lambda_0}  \angles{\lambda_0}^{2s} \left[  \sum_{\lambda_1, \lambda_2 \in \Lambda_j}    \tilde C_{\mu, T}(\bm\lambda)   a_{\lambda_1 } b_{\lambda_2 }    \right]^2,
 \end{equation}
 with $\widetilde C_{\mu, T}(\bm \lambda)$ as in \eqref{Ctilde}.

 By symmetry it suffices to estimate $\mathcal J^{(0)}_{\mu, T}$ and $\mathcal J^{(1)}_{\mu, T}$.
Now if $\bm \lambda \in \Lambda_{0} $ or  $\bm \lambda \in \Lambda_{1} $, then from \eqref{C}--\eqref{Ctilde} we have
$$
 \widetilde C_{\mu, T}(\bm\lambda) \lesssim   C_{\mu, T}(\bm \lambda),
$$
and hence
$$
 \mathcal J^{(j)}_{\mu, T} \lesssim  \mathcal I^{(j)}_{\mu, T}\lesssim T^{2\alpha}  \mu^{-\frac 32+ \alpha}  \norm{u}^2_{  X^s_T }\norm{v}^2_{ X^s_T } \quad (j=0,1),
$$
where to get the second inequality we used \eqref{GammaEstj}.

\vspace{8mm}

\noindent \textbf{Acknowledgments}
The author would like to thank the anonymous referee for useful comments on an earlier version of this article.

redpajama_set_name: RedPajamaArXiv


From design duo Anderssen & Voll, the inviting Muuto Rest Three-Seater Sofa.
Its relaxed, voluptuous form combined with its solid wood frame makes the Rest the perfect place to do just that. Settle in on the comfortable Kvadrat fabric, available in three variations and multiple colourways.


text:
/**
 * ProjectExperimentDAO.java
 * @author Vagisha Sharma
 * Apr 2, 2009
 * @version 1.0
 */
package org.yeastrc.experiment;

import java.sql.Connection;
import java.sql.ResultSet;
import java.sql.SQLException;
import java.sql.Statement;
import java.util.ArrayList;
import java.util.List;

import org.yeastrc.db.DBConnectionManager;
import org.yeastrc.project.Project;

/**
 *
 */
public class ProjectExperimentDAO {

    private static ProjectExperimentDAO instance;

    private ProjectExperimentDAO() {}

    public static ProjectExperimentDAO instance() {
        if(instance == null)
            instance = new ProjectExperimentDAO();
        return instance;
    }

    public List<Integer> getExperimentIdsForProjects(List<Project> projects) throws SQLException {
        if(projects == null || projects.size() == 0)
            return new ArrayList<Integer>(0);

        String projIdStr = "";
        for(Project proj: projects) {
            projIdStr += ","+proj.getID();
        }
        projIdStr = projIdStr.substring(1); // remove first comma

        Connection conn = null;
        Statement stmt = null;
        ResultSet rs = null;

        try {

            String sql = "SELECT experimentID FROM tblProjectExperiment WHERE projectID in ("+projIdStr+")";

            conn = DBConnectionManager.getConnection(DBConnectionManager.MAIN_DB);
            stmt = conn.createStatement();
            rs = stmt.executeQuery(sql);

            List<Integer> experimentIds = new ArrayList<Integer>();
            while (rs.next()) {
                experimentIds.add( rs.getInt("experimentID"));
            }
            return experimentIds;

        } finally {

            if (rs != null) {
                try { rs.close(); rs = null; } catch (Exception e) { ; }
            }

            if (stmt != null) {
                try { stmt.close(); stmt = null; } catch (Exception e) { ; }
            }

            if (conn != null) {
                try { conn.close(); conn = null; } catch (Exception e) { ; }
            }
        }
    }

    public List<Integer> getExperimentIdsForProject(int projectId) throws SQLException {

        Connection conn = null;
        Statement stmt = null;
        ResultSet rs = null;

        try {

            String sql = "SELECT experimentID FROM tblProjectExperiment WHERE projectID="+projectId;

            conn = DBConnectionManager.getConnection(DBConnectionManager.MAIN_DB);
            stmt = conn.createStatement();
            rs = stmt.executeQuery(sql);

            List<Integer> experimentIds = new ArrayList<Integer>();
            while (rs.next()) {
                experimentIds.add( rs.getInt("experimentID"));
            }
            return experimentIds;

        } finally {

            if (rs != null) {
                try { rs.close(); rs = null; } catch (Exception e) { ; }
            }

            if (stmt != null) {
                try { stmt.close(); stmt = null; } catch (Exception e) { ; }
            }

            if (conn != null) {
                try { conn.close(); conn = null; } catch (Exception e) { ; }
            }
        }
    }

    public void deleteProjectExperiment(int experimentId) throws SQLException {
        Connection conn = null;
        Statement stmt = null;

        try {

            String sql = "DELETE FROM tblProjectExperiment WHERE experimentID="+experimentId;

            conn = DBConnectionManager.getConnection(DBConnectionManager.MAIN_DB);
            stmt = conn.createStatement();
            stmt.executeUpdate(sql);

        } finally {

            if (stmt != null) {
                try { stmt.close(); stmt = null; } catch (Exception e) { ; }
            }

            if (conn != null) {
                try { conn.close(); conn = null; } catch (Exception e) { ; }
            }
        }
    }

    public List<Integer> getProjectIdsForExperiments(List<Integer> experimentIds) throws SQLException {
        if(experimentIds == null || experimentIds.size() == 0)
            return new ArrayList<Integer>(0);

        String exptIdStr = "";
        for(Integer exptId: experimentIds) {
            exptIdStr += ","+exptId;
        }
        exptIdStr = exptIdStr.substring(1); // remove first comma

        Connection conn = null;
        Statement stmt = null;
        ResultSet rs = null;

        try {

            String sql = "SELECT distinct(projectID) FROM tblProjectExperiment WHERE experimentID in ("+exptIdStr+")";

            conn = DBConnectionManager.getConnection(DBConnectionManager.MAIN_DB);
            stmt = conn.createStatement();
            rs = stmt.executeQuery(sql);

            List<Integer> projectIds = new ArrayList<Integer>();
            while (rs.next()) {
                projectIds.add( rs.getInt("projectID"));
            }
            return projectIds;

        } finally {

            if (rs != null) {
                try { rs.close(); rs = null; } catch (Exception e) { ; }
            }

            if (stmt != null) {
                try { stmt.close(); stmt = null; } catch (Exception e) { ; }
            }

            if (conn != null) {
                try { conn.close(); conn = null; } catch (Exception e) { ; }
            }
        }
    }

    public int getProjectIdForExperiment(int experimentId) throws SQLException {

        Connection conn = null;
        Statement stmt = null;
        ResultSet rs = null;

        try {

            String sql = "SELECT projectID FROM tblProjectExperiment WHERE experimentID = "+experimentId+" ";

            conn = DBConnectionManager.getConnection(DBConnectionManager.MAIN_DB);
            stmt = conn.createStatement();
            rs = stmt.executeQuery(sql);

            if (rs.next()) {
                return rs.getInt("projectID");
            }

        } finally {

            if (rs != null) {
                try { rs.close(); rs = null; } catch (Exception e) { ; }
            }

            if (stmt != null) {
                try { stmt.close(); stmt = null; } catch (Exception e) { ; }
            }

            if (conn != null) {
                try { conn.close(); conn = null; } catch (Exception e) { ; }
            }
        }
        return 0;
    }

}

redpajama_set_name: RedPajamaGithub


text:
Brent Cobb chats with Will Houk ahead of his Lake Tahoe gig
by Will Houk
Brent Cobb is a Grammy-Nominated singer/songwriter who hails from Georgia. Last year he released the album Providence Canyon, which he recorded with his cousin Dave Cobb at the famed RCA Studio A in Nashville Tennessee. Brent Cobb's music is a mix of styles ranging from Classic Country and Rock to the funk sounds of the 1970s.
Will Houk spoke with Cobb by phone about his his latest record and the touring he's been doing the last couple of years with artists like Chris Stapleton and Whiskey Myers. In this interview Brent discusses his musical styles and a song he wrote for Bradley Cooper's movie "A Star Is Born". Cobb also gives details on his new single with Jade Bird, Feet Off the Ground. Brent Cobb and Them will play the Crystal Bay Club on October 27th at 8pm. Opening act will be Haley Whitters.
Listen to audio of the interview.
https://nevadacapitalnews.org/wp-content/uploads/2019/10/Brent-Cobb-Interview-v1-.mp3
Will Houk: I wanted to start off talking a little bit about the most recent album Providence Canyon. The title track is also the album opener. It's a nice reflective tune. Can you tell us about the origins of that song?
Brent Cobb: "The origin of Providence Canyon is a little canyon about 45 minutes west where I grew up. We always called it the little Grand Canyon. I grew up going to this place and camping and then it's just a beautiful little spot to go to a peaceful place. I originally started writing the song Providence Canyon reminiscing about growing up, going into the canyon. It's sort of turned into more of a spiritual, more literal providence canyon in my own life, you know, not just the actual canyon … just feels like a spiritual haven, you know? That's kind of where the song came from."
Cobb recorded Providence Canyon in 4 days in the famous RCA Studio A in Nashville. A few notable artists have recorded songs in the famous studio to include a who's who of popular music: the Beach Boys, Tony Bennett, Joe Cocker, Alan Jackson, Waylon Jennings, B.B. King, Chet Atkins, Loretta Lynn, The Monkees, Dolly Parton, Leon Russell, William Shatner, George Strait, and Brent Cobb.
WH: "I was listening recently to the podcast Walking The Floor, and you're talking a little bit about the guys who recorded on the album. The drummer was that Chris Powell? Is that correct?"
BC: "That's right."
WH: "You guys go way back and playing music together …"
BC: "Yeah. When I first left Georgia, first place I moved to was LA (Los Angeles). Chris is Leroy's Powell's brother. Do you know Leroy?"
WH: "I don't know him."
BC: "Leroy was one of the original members of Shooter Jennings and the 357s. Chris is Leroy's younger brother. He was playing with Leroy and Leroy's side band at that time. And yeah, he was just one of the best drummers that I ever met. Still is … he's just fantastic drummer, but he's just kinda been like an older brother to me."
WH: "A couple of the songs on the album, a number of them have a real kind of Country Funk kind of sound to them and "Morning's Gonna Come" and "30-06″ are a couple of them. I'm interested to hear a little more on the influences for that sound stuff you heard growing up and that kind of stuff."
BC: "I grew up on stuff like "Bad Bad Leroy Brown" you know … all that stuff from the '60s and '70s, even the ballads back then, it seemed like they were playing sort of a backbeat. Nowadays a lot of the songwriter stuff is just super, you know, not very exciting. That sounds bad I shouldn't say it that way, but there's no, there's no backbeat. So I've always been drawn to that.
"Even in '06 we were trying to show, to make a country funk album. We wouldn't call it that, but then this compilation called "Country Funk" of different artists from the '70s that came out a few years ago and it was… we were just really listening to a lot of old Larry John Wilson and all that, Bobby Charles … just a bunch of these great classic Country Funky stuff. So yeah, when we got back into studio for Providence Canyon we just really wanted to try to capture that … kind of feels like rural America?"
WH: "The song King of Alabama's is a pretty cool tribute to the life and music of Wayne Mills, and I just I'd like to hear a little bit about what was it about his music that was moving to you that that kind of drew you to him."
BC: "His music was moving but also just as a person he's just a great guy … again, even like an older brother sort of when I first moved to Nashville in '08 … and you know learning and getting to know people and that sort of thing, but the deep part of that story in his music, are you familiar with any of this music?"
WH: "I'm not super familiar with it. Just kind of cursory level."
BC: "The crazy part of that story is the last album that he ever released. It's called The Last Honky Tonk. It's got the songs on there … it was called One of These Days I'm Flying to the Moon … one of these days, I have nothing to prove … one these days I get out of this place … one of these days… and all the songs on that album kinda lend themselves to that. And then of course he died. It's just crazy. They put that out to the world before he went."
WH: "You have a new song out with Jade Bird called Feet Off The Ground, and I was interested to hear about how that collaboration came about?"
BC: "Man, Jade is such a beautiful spirit. It was my first trip to the UK two years ago, and Jade opened a show for me at the Slaughtered Lamb. We had the same at booking agency. I'd never met her. We met that night, and then we're going to get together to write the next day. I got this crazy text message like four pages long as if someone was describing a character to me. I didn't know that that's what was happening. It was just this wild random text message from a number I didn't have on my phone.
"So I responded, thanks for this, but I'm not sure who this is. It was Bradley Cooper. Apparently, Dave had given him my phone number because he's working on "A Star is Born." And he thought maybe I'd be able to write some stuff for the movie. I already had Feet Off The Ground sort of lying around. I had the melody. I had the first verse. And I thought, man that might be a really cool song for that movie.
"The next day when Jade and I got together, first time we really hung out, I told her what was happening and played a little bit of Feet Off the Ground that I already had. And then we finished it there in the hotel room. It never made the movie, but it's just … it's been a really cool song. We recorded it. We actually recorded it while we were recording Providence Canyon, and it didn't make my record either. We thought why not put it out and try it again? Let's do it."
WH: "So you've been on that road touring quite a bit this year, and you're going to be playing up a Lake Tahoe in October. Where have you been out on the road so far?"
BC: "Oh, man, we've been all over the country. We started out on the West Coast. We started "Sucker For A Good Time Tour" out west and then we picked up with Whiskey Myers. We opened for them for about a month. It was in like Wyoming and Montana and Colorado. Then we did this Stapleton tour again this year, and we were out with them for about a month and more of the Southeast. And now we are currently in the Northeast. Making our way back out west."
WH: "Do you have any plans for the next record? Are you just kind of touring right now or are you still songwriting?"
BC: "We're touring. Then we're going to go in the studio in December and start working on the new album. I got most of it wrote. I like to have most of my albums wrote and then go into studio and like, write the last two or three in the moment. We're going to try to have something out sometime, like the spring of next year."
Previous articleNatchez Elementary Turnaround Part 2, Overcoming Odds and Expectations
Next articleNorthern California city braces for another multi-day blackout, Nevada City mayor calls on Governor Newsom to declare a state of emergency, microgrids
Poem: Though the Gods
Buried! The 1982 Alpine Meadows Avalanche
Lake Tahoe Basin, Summer 2021 – Breathing Smoke and Waiting for Fire
A Trustworthy Friend

redpajama_set_name: RedPajamaCommonCrawl


An airline flight attendant, Ashgan Attia, told friends she felt reborn when she survived a hijacking that left 60 people dead in Malta last November.
But her ''second life'' was short.
Miss Attia, 24 years old, died Tuesday night when an Air Sinai Fokker-27 crashed in a sandstorm just short of Cairo Airport, killing all 5 crew members and 17 of the 21 passengers.
She had been a flight attendant just three months when her plane was hijacked to Malta by Arab gunmen.
When the gunmen started killing passengers, Egypt sent commandos to storm the plane. The hijackers threw grenades, which caused the airliner to catch fire, and 60 people died.


About MedImmune: A leading global biologics R&D powerhouse, MedImmune comprises nearly half of AstraZeneca's overall R&D portfolio and currently has more than 120 research projects and product candidates. The company's focus includes three core therapeutic areas: Oncology; Respiratory, Inflammation and Autoimmune; and Cardiovascular and Metabolic Disease.
Why Gaithersburg: The company was born here, and acquired by Astra Zeneca in 2007. Today, Gaithersburg is home to MedImmune's headquarters with 2500 employees.


The trend of large pharmaceutical companies divesting parts of their R&D infrastructure for better cost containment, revenue growth, and shorter clinical trial cycles has increased business for Contract Research Organizations (CROs). However, these collaborative partnerships are creating a need for flexible arrangements and integrated end-to-end solutions. This is particularly true as companies are embracing adaptive study designs that require data visualization, analytics, and transparency across the clinical trial continuum. To be successful, CROs need to align their services to support their customers to efficiently navigate regulatory pathways and shorten the clinical trial lifecycle, all without compromising quality. By adopting unique technology solutions, CROs have the ability to develop reputations as leaders in moving new drugs and devices from conception to FDA/EMA marketing approval with limited sponsor dependence.


text:
Ryan de Laureal
The truth about Trump's golden boy
When Donald Trump made Jared Kushner a senior adviser, he brought the family business--corruption--to the White House, writes Ryan de Laureal.
Click for more options
close | x
PERHAPS NO figure embodies the bizarre state of U.S. politics better than Jared Kushner.
With Trump's victory in November, the bland and unremarkable 36-year-old real estate tycoon went from being a virtual nobody--the husband of Trump's daughter Ivanka and, like Trump himself, the pampered heir to his father's business empire--to a key powerbroker in the new administration.
In his new role as senior adviser, Kushner has influence over some of the most significant areas of world politics--tasks normally reserved for people who spent their careers on them--from U.S. policy in the Middle East to solving the U.S. opioid epidemic to restructuring the American executive state.
But more important than the "experience" question, Kushner is a shining example of the web of hypocrisy, contradiction and intrigue surrounding Trump himself and his promises to rejuvenate American capitalism.
Despite Kushner's frequent portrayal in the media as a soft-spoken, moderating influence among his father-in-law's advisers, the more you investigate Kushner, the more a shady and Trumpian character emerges.
Jared Kushner (Navy Petty Officer 2nd Class Dominique A. Pineiro | Wikimedia Commons)
In May, Kushner became the latest member of Trump's inner circle to be implicated in backroom dealings with Russia. Federal and congressional investigators announced that they were looking into Kushner's meeting in mid-December with Sergey Gorkov, a Russian banker whose financial institution has ties to Russian intelligence under Vladimir Putin and is under sanction from the U.S.
As the investigation into the Trump campaign's suspected dealings with Russia continues, the Democrats have kept a narrow focus on these allegations, to the exclusion of Trump's other crimes. But as inadequate as the Democrats' "opposition" has been, it's also easy to see why Trump and Co. could be suspected of corruption.
And Jared Kushner is a great example of how the Trump family business is corruption.
AS A landlord, Kushner has a history of harassing low-income tenants and evicting them from his properties. As the owner of the New York Observer, he used the newspaper as a platform to pursue personal vendettas against his rivals.
Kushner also has a history of partnerships with corrupt foreign businessmen and officials that is particularly hypocritical for somebody serving a president who flaunts "buy American and hire American" policies and who rode to power on a wave of nationalistic fervor by employing rhetorical attacks against Muslims, Mexicans and China.
There are Kushner's business deals with an Israeli bank under criminal investigation by the Justice Department, and with Chinese business officials linked to the country's ruling elite.
And now there's his meeting with a Russian banker in the months before Trump's inauguration--in an apparent attempt to create "back channel" communications with Moscow, away from the prying eyes of U.S. intelligence officials.
As head of Kushner Companies, Kushner obtained millions of dollars through a U.S. government visa program that allows foreign investors to buy their way to permanent residency in the U.S. The program's extension was part of Trump's first major piece of legislation.
The program, called the EB-5 Immigrant Investor Program, was introduced in the early 1990s and is a favorite of real estate tycoons like Trump and Kushner, allowing them to attract wealthy foreigners to bankroll luxury building projects at extremely low interest rates.
Because EB-5 recipients must pledge a minimum of $500,000 to $1 million toward a U.S. business venture, the program is officially billed as a way to attract foreign capital to the U.S. and create jobs. But the actual requirements for job creation on an EB-5 project are extremely low--only 10 permanent jobs need to be created.
As a result, the primarily beneficiaries of the program aren't U.S. workers, but developers like Kushner and wealthy investors in China and elsewhere.
Could there be a starker refutation of Trump's con?
WHILE HUNDREDS of poor and desperate migrants die every year attempting to cross the U.S. border, while the rest are vilified by xenophobes like Trump and scapegoated for the failures of U.S. capitalism, the president and his closest associates are busy extending a program from which they have personally gained--one under which thousands of foreign millionaires are granted residency visas every year with little controversy, including nearly 3,000 from January to March of this year alone.
For comparison, this is almost the same number of visas that were issued over a three-year period, from January 2008 through June 2011, for unskilled workers who fall under the EB-3 visa for foreign workers. This is the only type of visa open to working-class foreigners who are not related to a U.S. citizen and who can't apply for some other type of special visa status.
Unlike the EB-5 program, in which foreign investors apply for residency directly, workers aren't allowed to file their own paperwork to obtain an EB-3. Instead, they must be sponsored by an employer, which must obtain a "foreign labor certification" from the Department of Labor, a complex process that can take months.
In order to be approved, the employer must demonstrate that that no domestic worker can be found to perform the job in question, and even if certification is granted, the employer must still file separate paperwork with U.S. Citizenship and Immigration Services and fork over a $700 filing fee. Even then, there is a good chance the visa won't be approved.
The difference between these visa programs is a perfect illustration of the reality of the U.S. immigration system, where the vast majority of the world's population is essentially blocked from ever obtaining any form of legal residency in the U.S., while for the rich it is virtually guaranteed.
And it exposes the hypocrisy of con men like Trump and Kushner who use scapegoating rhetoric about "immigrants taking away American jobs" and immigration laws being too lenient, yet manipulate the system to strike overseas business deals and exploit current immigration laws for their own gain.
Holding out a helping hand to the banksters
Trump's war on birth control
Ashley Smith
The bitter fruit of Trump's China-bashing
Donald Trump's new cold war rivalry with China has put Chinese foreign students at U.S. universities squarely in the crosshairs.
By a federal worker
Will labor learn the lessons of the shutdown?
With the threat of another shutdown coming later this month, federal workers and their unions have to get ready for another fight.
Who shut down the shutdown?
It wasn't the Democrats, but federal workers, who finally put real pressure on the Trump administration to reopen the government.
No more food banks, we need paychecks
Hundreds came out to protest the government shutdown, holding up empty plates to symbolize the workers who are going hungry.
No paychecks, no peace
With Trump's lockout entering its fifth week, federal workers need to look for solutions to bring the government shutdown to an end.
Why we remember 1917
One hundred years ago, the final act of the Russian Revolution shook the world--by setting an example of mass democracy that still inspires today.
Welcome to the new normal
My mom has found herself among the long-term unemployed who are struggling in the suburbs where the American Dream is supposed to thrive.
Anti-imperialism and the Syrian Revolution
The Arab Spring uprising in Syria has tested the left by posing a blunt question: Are you on the side of a dictator--or that of a popular uprising?
Alpana Mehta
They want to rule the streets. We can't let them.
One year after Charlottesville, far-right organizers have a permit to "Unite the Right" in Washington, D.C., this August. They can't go unopposed.
Elizabeth Lalasz and Sean Larson
Learning lessons from the Oklahoma walkouts
Teachers in Oklahoma won a $6,000 raise by standing strong, but their walkouts ended with dissension and confusion after union leaders retreated.


Purchase the recurring Endorphin & Neurotransmitter Therapy for an Individual person/pet to a Group of 7. The recurring subscription must be used for a minimum of 4 months. Recommended subscription is 6 to 12 months. Click here for more information about the Endorphin and Neurotransmitter Therapy.


Distinguished by its drying capabilities, the men's Q Speed Jacquard Short Sleeve performance top offers the highest level of moisture management with premium NB Dry X technology. The jacquard mesh fabric is extremely breathable while passing NB Ice standards for exceptional cooling and drying, meaning this top works overtime to keep you comfortable and focused during your toughest workouts. Plus, the athletic fit features modern elbow-length sleeves that make it perfect for both running on the road or hitting the gym. Iridescent reflective detail on all sides adds unique style whlie helping to up visibility during early morning and nighttime runs.


text:
Autohaus Hamilton - End of Year Drive 2014
December 18, 2014 in Events, Autohaus Hamilton News
We're not going to lie, we were equally nervous as excited for our final event of the year. When we put the word out through our Track & Drive Mailing List (to join – click here) we had a huge rush of reservations, within the first week we had 50 Porsches and by the second we were sitting on 80!
That's when we had to make the call and capacity was reached!
When the 14th of December came, sixty-eight Porsche's made the effort to come along, joined with a handful of other makes and models. Fortunately and thanks to the car gods we were greeted with a spectacular morning, which was a shock after a week of lightning storms! What better addition to the weekend than an easy Sunday morning to gather and drive.
Porsche fanatics came from as far north as Newcastle, south as Canberra and west as Bathurst; all to meet in the Royal National Park on the southern border of Sydney. 1 Porsche, 2 Porsche, 3 Porsche, 4! As the minutes ticked over, the sound of flat 6's filled the National Park bouncing off the cliff-faces and sending chills down the spines of all who could hear. We found ourselves trying to 'Guess the Porsche' by distant exhaust notes, then finally seeing them pop out of the bushes and come across a water level bridge.
There was a strong presence of Police through-out the weekend on Sydney's roads and that didn't change even this early on a Sunday. But, as the police rolled by there was nothing but thumbs up and smiles ear to ear! It was time to round up the group, give a quick briefing… then ladies and gentleman – start your engines!
Our first leg of the drive was simple, head south through the Royal National Park towards The Scarborough Hotel which had kindly opened early for us to enjoy breakfast by the sea! This was by far the easiest leg of the drive; with no real places to get lost there were 68 Porsche's stretched kilometres down the road!
We had a few extra minutes, so we squeezed in one quick stop at Stanwell Tops, a very scenic look-out just a few minutes before our breakfast destination. We managed to pull all 68 cars over and basically grid-locked all parking available, with spirits so high and adrenaline pumping everyone was soaking up the morning sun and already telling stories about the last 45kms of twists and turns.
Matt (Matthew Everingham Photography) joined us on the day to cover the drive and live the Porsche. This stop-over was too good to be true for eyes like his and with such a great opportunity to take a few cars aside and frame a shot, he didn't waste any time! Matt has done photography for many well-known publishments and international blogs; he's also just recently won the 2014 Canon AIPP Advertising Photographer of the Year! (Congratulations and Thank-you!)
As a few cars stayed back to work on the shot with Matt, the rest of us continued down the road and started to fill the 2 car-parks, along with both sides of street parking at The Scarborough Hotel. The hotel living up to their promised word and reserving the best view in town coupled with all our breakfast and coffee ready to consume within minutes!
Breakfast was great, it was really nice to see so many couples, friends, fathers joined with daughters and sons enjoying a shared passion. With breakfast over we swiftly head back up to the carpark where a quick map briefing was held. The gist was basic… "We are going to hit traffic, we are going to hit lights and there will be people getting lost" but, on the up side "Guys, take it as it comes! Remember all these roads are great and even if you're lost for a second, enjoy the road and when you can stop safely take a quick look at the map. If you're too far gone, punch in the final destination into your GPS and we'll see you there!"
With so much excitement, cars were already started and moving before you could even count to ten! Before you knew it there were small groups of Porsche's all headed in the same direction but on different roads, some lost… some found. At the next check point, about 40 cars arrived. The other 20 were… on a different route. As the southern decent continued, more cars were found and more cars were lost.
But here's the kicker, we all found our way to the final destination just as briefed! There was so much laughter at Briars Country Lodge, jokes about all the wrong turns made and random places small groups of cars found each other. Suddenly all the 'stress' was laughed away and nothing but good reports given. It just goes to show with the right frame of mind, the right steering wheel in your hands and no matter if you're lost or found, Living the Porsche is an enjoyment like no other.
Before all this sinks in, there's also a video! Thanks to Lachlan from Petrol Fuelled Passion for joining in on the fun and some-how compressing 250km's into a punchy 4 minute video.
What if perfect really did exist? Speedster.
Build Update - 964 Skunkworks & 1973 911 Targa
In The Haus - February


The motorhome is parked at American Legion Post #19 in Yuma, Arizona. I expect to be here for a while.
Mary remains at home in California tending to medical issues.
The motorhome is now parked at the Fraternal Order of Eagles Airee #3695 in Espanola about 25 miles north of Santa Fe. I expect to be here for several days.
Yesterday I drove the motorhome about 50 miles from Los Alamos to Espanola dropping down from about 7,200 to about 5,600 feet of elevation.
I went into the bar for happy hour and got acquainted with about a dozen very friendly people. My host had suggested I park near the picnic shed where there is electricity available. My tester says there is something wrong with their electrical circuit, so I'm running strictly on the solar panels. So long as there is lots of sunshine, I'll have plenty of electricity.
I always like to show the view from my dinette window. While this one is not particularly exciting, the price is right! The Bronco is parked in the shade of the picnic shed.
I'll begin my explorations of the local area today.
Enjoying historical locations is another joy of the full-time RVing lifestyle!
Forecast for today is partly sunny and 76 degrees.


Michael Kors handbags can be on many discounted sites for fractions of retail cost. That should bring a great of smiles to many faces. Cope with have to wait patiently for sales to michael kors black friday handbags. Do not driving to outlet malls in thought of finding an important deal on an out-dated handbag. Michael Kors michael kors outlet online handbags are cheaper than can have assumed.Fresh vegetables and salads combined with healthy nuts and yogurt are primarily the staples of the mediterranean diet. Melons are especially popular as are stuffed tomatoes, for some reason. An exceptional model in this particular eating routine is a Greek salad normally seen in upscale restaurants. Instead of pouring associated with money the usual American salad dressing, utilize oil and vinegar. It is fine you skip the excess fat kinds of fatty salad dressings, besides. In place of overly processed American snacks, an average Greek diet uses nuts as well as garden-fresh greens.Your individual method, to life's Michael Kors daily obstacles, always be the gateway to achievement or to fail. For that reason, success is a make an improvement of choice.Among those well-known brands, I recommend michael kors cyber monday purses since it offers a good taste however you like. It has its own unique method of creating designs that will establish a fashion trend accessible. The only nervous about its products is that hot weather sells for the much higher cost compared to its competitors. Published reviews struggling and thinking twice if it's worth money.It might sound like a stereotype, but no woman can resist the allure of handbags and footwear types. Shoes will probably be a challenge to select on your own, arrive at the men and women. Handbags are a little bit easier opt for. Just notice eliminating handbags she already enjoys using. Restrict go wrong with all the top designers like Marc Jacobs, Burberry, and Fendi. Large or small, most woman are satisfied almost all size handbags.Discounted Michael Kors handbag websites for business spots go to to get excellent bargains on handbags and also other Michael Kors furnishings. There are many on-line stores that meet the needs of the wants of individuals appreciate searching out the best charges and who adore to bargain quest. Some shops will in addition drop their prices to install up with their rivals. Along with that is truly tremendous!


Discover your favorite coupon through 23 live and hot Invitation In A Bottle coupon codes and deals. Shop at invitationinabottle.com and get extra savings on your purchase with current top Invitation In A Bottle promo codes and promotions. Here is the best promotion:Save Up To 70% Sitewide @invitation In A Bottle.


This compensator is specifically designed for the Volquartsen LLV upper. This will also fit the Volquartsen Scorpion pistol and 10/22 barrels. It is CNC-machined from an aluminum alloy to minimize weight and anodized for durability. Color matches the color of the LLV / Scorpion upper.
This compensator matches the same profile of the LLV and uses a knurled nut to secure to the muzzle threads. Spanner wrench for installation is included.
As a result of various state laws restricting the sale of high-capacity magazines, we are prohibited from selling magazines of a certain capacity in some states. These are state laws, sometimes there are county or city laws with stricter restrictions. It is your responsibility to be or become familiar with local gun laws.
California: We do not ship magazines over 10 rounds California.
Connecticut: We do not ship magazines over 10 rounds to Connecticut.
Hawaii: We do not ship magazines over 10 rounds that may fit a handgun to Hawaii.
Illinois: We do not ship magazines over 12 rounds to Chicago, IL. We do not ship magazines over 15 rounds to Aurora, IL.
Maryland: We do not ship magazines over 20 rounds to Maryland.
Massachusetts: We do not ship magazines over 10 rounds to Massachusetts.
New Jersey: We do not ship magazines over 15 rounds to New Jersey.
New York: We do not ship magazines over 10 rounds to New York.
Ohio: We do not ship magazines over 30 rounds to Ohio.
Customers who fall under one of these restrictions will have their order for such magazines cancelled. Don't blame us, blame your state legislators.


This image was made in August during our UK small ship cruise around the various islands. While the evening was overcast and some typical London drizzle, it did make for a great mood. 5D4 with 16-35 F4. 1 sec, f9 ISO 200. Tripod, live view and 2 sec timer used. Processed in Lightroom and Photoshop. I used several radial tools in LR to highlight various areas of "interest". Thanks for looking. Comments most welcome.
You did a beautiful job getting the lighting just right on this magnificent bridge. It looks fairy tale magical . Love the soft glow in the sky. An interesting blend of the old and the new.
Well done, Ed. Love the lighting on the bridge.
Those smooth colors are wonderful. The rendition may benefit from some perspective control; the left tower is listing like a drunken sailor.
Iconic London landmark, Ed. Nicely done.
This is so good. I absolutely love this!!
Thank you all for the kind comments. They are very much appreciated.
Thank you so much for the EP for this image. I can't go without saying that this recognition has a lot to do with all the learning. constructive criticism and advice I have gained from my NSN membership over the years. Thank you again for the recognition.


This statistic depicts the share of elderly population aged 65 and over suffering from depression in Spain in 2014, by age group. That year, approximately 15 percent of the elderly population aged 85 and over in Spain had depression. Depressive disorders are mental disorders characterized by long episodes of low mood. Common symptoms include low self-esteem, loss of interest in normally enjoyable activities, low energy, pain without a clear cause and, in severe cases, psychosis, delusions and hallucinations. A family history of the condition, major life changes, certain medications, chronic health problems, and substance abuse are among its most common risk factors. In 2014, over 300 thousand women aged between 75 and 84 were diagnosed with depression in Spain.


text:
Sign in to start taking action.
Not a Global Citizen yet? Sign up
Remember me Forget your password?
Thanks for signing up as a global citizen. In order to create your account we need you to provide your email address. You can check out our Privacy Policy to see how we safeguard and use the information you provide us with. If your Facebook account does not have an attached e-mail address, you'll need to add that before you can sign up.
Rerequest with Facebook
This account has been deactivated.
Please contact us at contact@globalcitizen.org if you would like to re-activate your account.
Finance & Innovation
Food & Hunger
Global Citizen Events
Global Citizen Programs
Waislitz Global Citizen Awards
Curtis Scholarship
Curtis Scholarship Terms and Conditions
Take ActionRewardsFestivalImpact
Girls & WomenHealthEducationFinance & InnovationFood & HungerWater & SanitationEnvironmentCitizenship
Global Citizen is a community of people like you
People who want to learn about and take action on the world's biggest challenges. Extreme poverty ends with you.
1. Get Informed
Check out original content and videos published every day to help you learn about the issues that mean the most to you.
Send petitions, emails, or tweets to world leaders. Call governments or join rallies. We offer a variety of ways to make your voice heard.
3. Connect with Others
Meet other Global Citizens who care about the same issues you do. Keep updated on what they're doing to change the world.
Find out more about Global Citizen
By Joe McCarthy
5 ways ocean acidification is connected to poverty
Climate change is revealing just how interconnected everything is on Earth.
Share ShareTweetEmail
Flickr: Mark Seton
I can't think of anything more calming than watching and listening to the ocean's waves, those great bands of water rising and falling.
Just plop me down on a shore, get rid of all my personal responsibilities, and I'll stay there for eternity, achieving enlightenment.
In the course of my meditation, I'd probably think about the world underneath the surface of the waves, all the organisms living in vast harmony.
All the dolphins and shellfish and algae and tuna and snails.
What are they up to, I'd wonder. How are those guys doing?
And then I'd do a quick search on my smart phone and get stressed out, ruining everything.
Apparently, "over the past 200 years, the world's seas have absorbed 150 billion metric tons of carbon from human activities," according to a recent piece in The New York Times.
That's 15 pounds of carbon per person, per week, which is enough fuel to circle the globe in a coal train 13 times.
And this carbon dissolves as carbonic acid, which at such large scales changes the water, making it acidic.
Here are 5 reasons why the world should be worried about this growing acidification, especially in relation to poverty.
The first things that acidification affects are the smallest creatures of the ocean: the shellfish, coral and other small organisms.
Acidified water makes it hard for these organisms to spawn and then grow. Oftentimes, they'll simply dissolve before they become sturdy enough to thrive.
This is bad not just because some of these organisms (shellfish) are a source of food for humans, but they also feed bigger fish such as salmon that are a primary source of food.
Image: Flickr: Alexandru Stoian
15 percent of the world receives a third of their daily protein intake from fish and more than a billion people depend on fish as their primary source of protein.
Overfishing is already imperilling the world's fish populations. Back in 2011, the UN's Food and Agriculture Organization estimated that 32 percent of fish populations were being depleted and needed to be urgently rebuilt.
So fish are being depleted from above by fishing and from below by acidification, while demand for fish grows.
The livelihoods of an estimated 540 million people are directly supported by fisheries and aquaculture.
And it's not just fishermen and women. Fish distributors, markets, restaurants and all the jobs that branch off these fields will be affected.
This is an issue in developed nations as well. In the US state of Alaska, more than 100,000 jobs are supported by fishing.
Fishing is especially important for people in developing countries who often use the sea as a way to stay ahead of poverty.
Image: Flickr: David Stanley
All in all, fish are the most-traded food commodity in the world.
Ultimately, acidification threatens to collapse the fishing industry.
Harmful and toxic algal species flourish in acidified waters.
Just like how increased mercury pollution first contaminates small creatures and then makes its way to predators, ending with humans, these toxic species will climb up the food chain.
Image: Flickr: UCAR
Marine species can be killed, or they will harbor the toxins to be later absorbed by humans.
I buy tuna supported by the World Wildlife Fund because it's ethical and has lower mercury concentration, but how will the fish industry respond to pervasive toxins from algae?
The oceans are the biggest carbon sink in the world, storing way more carbon than the Amazon rainforest. Coastal areas absorb carbon dioxide 50 times faster than some rainforests.
The oceans can't store carbon indefinitely and remain intact. Eventually they will reach a tipping point. As the ocean acidifies, it may stop releasing a form of sulfur that acts to cool the atmosphere, intensifying climate change.
Image: Flickr: U.S. Geological Survey
Also, as coral reefs dissolve, the range of places where carbon can be stowed will diminish.
For example, the Great Barrier Reef has been cut in half since the 1990s.
Climate change primarily affects the world's most vulnerable populations. The rise of droughts, floods and large storms, triggered by climate change, will be concentrated in some of the poorest regions in the world.
Land ecosystems
Humans aren't the only land creatures that depend on the oceans.
Nearly every land animal you can think of--from bears to birds--depends in part on the oceans, either directly or indirectly.
If fish populations continue to decline from acidification, overfishing, pollution and other factors, these creatures will see their diets changing.
And, of course, this in turn affects human diets by disrupting ecosystems.
The more climate change progresses, the more the world sees the delicate interconnectivity of all life on Earth.
So what can be done?
As The New York Times piece describes, much more funding has to be given to marine research teams.
When measurements become more widespread and yield more data points, better policies can be recommended and legislators will be more open to them.
You can't really argue with this kind of approach and use of hard data.
Countries have to do a better job at collaborating and sharing data, because this is a global problem.
Fisheries have to be rehabilitated, whether by fostering better conditions for marine life or by imposing fishing restrictions to allow populations to revive.
Finally, economies have to be restructured to limit CO2 emissions. World leaders are meeting this December in Paris. They have to make bold proposals to invest in green energy and environmentally friendly practices.
You can go to TAKE ACTION NOW to spur leaders to embrace Global Goal 15: Life in the Oceans.
TopicsEnvironmentOceanOcean acidification
© 2012-2019 Global Poverty Project, Inc All Rights Reserved
English (en) Deutsch (de) français (fr) español (es)


The first step for the formation of glucose from photosynthesis is the absorption of light. According to Estrella Mountain Community College's website, when sunlight hits an organism that can perform photosynthesis (such as a plant), one of three things can happen. The light energy can be given off as heat; it can be re-emitted at a different wavelength (color); or it can set off a chemical reaction. Photosynthetic organisms contain various pigments, such as chlorophyll, that can absorb and harness light to make high-energy molecules. When light gets absorbed by these pigments, it causes the release of a high-energy particle (called an electron), which can then be used to convert the light energy into chemical energy. This portion of photosynthesis is called the light reaction. because it has to occur in parts of the organism that receive sunlight.
Once high-energy electrons have been generated, the photosynthetic organism can turn these electrons into sugar. The first step is storing this energy in a more stable form. The electron gets absorbed by molecules in the organism that are able to perform a series of reactions. These reactions use the electron's energy to form a molecule called adenosine triphosphate (ATP). ATP is a molecule that is similar to DNA but is used by organisms as a temporary energy reservoir. In order to turn the energy from ATP into glucose, plants also need carbon dioxide, which they obtain from the atmosphere (or, in the case of plants that live underwater, from the surrounding water).
According to the University of Cincinnati's biology department, carbon dioxide and ATP are turned into glucose in what is called the dark reaction (because it does not directly require light). Three ATP molecules react with three molecules of carbon dioxide via a series of chemical reactions (called the Calvin cycle) to form an intermediate called glyceraldehyde 3-phosphate. Two of the glyceraldehyde 3-phosphate molecules then are combined to form a glucose molecule. Thus, six molecules of carbon dioxide and light energy are needed to make one molecule of glucose for photosynthesis.
What is the Complete Oxidation of a Glucose Molecule?