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%\chapter{Parcel method: static instability}
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%%\textbf{Conception:}

{\it Parcel method} is a method of testing for instability.
In this method, we assume that a displacement is made from a steady state, and only the parcel or parcels displaced are affected, while the environment remaining unchanged.
Here, using the parcel method, we study a problem related to static stability in the gravity field.

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\section{Theoretical Background}

In this chapter, the incompressible ideal fluid is considered (Yoden, 2000).
We assume that the Z-axis is positive upward for the coordinate system and the density distribution for the fluid is $\bar{\rho}(z)$.
While a force of gravity $-\bar{\rho}g$ is applied to each unit volume of the fluid (parcel), a pressure gradient force that has the same magnitude as the gravity is applied to the parcel.
This leads to the {\it hydrostatic stability} given by
\begin{equation}
\frac{\partial p}{\partial z} = -\bar{\rho}g.
\label{9.1}
\end{equation}

Taking this parcel out from the fluid layer and putting a parcel with density $\rho$ instead of it, we find that this parcel is applied by a force of buoyancy of $(\bar{\rho}-\rho)g$.
The direction of the buoyancy is upward when $\bar{\rho}>\rho$, and downward when $\bar{\rho}<\rho$.

We assume that a parcel of the unit volume makes a displacement of $\delta z$ along the z-direction, while the field surrounding the parcel is not disturbed by the displacement of the parcel.
If the density difference between the parcel and the surrounding field is $\delta \rho$ and $\delta \rho = -(d\rho/dz)\delta z$, since the buoyancy force is $-g\delta \rho$, the dynamic equation for the parcel can be given by
\begin{equation}
\rho\frac{d^{2}}{dt^{2}}(\delta z) = g\frac{d\bar{\rho}}{dz}\delta z.
\label{9.2}
\end{equation}
When $N^{2}\equiv -g\frac{d\bar{\rho}}{dz}$, (\ref{9.2}) can be written as follows.
\begin{equation}
\frac{d^{2}}{dt^{2}}(\delta z) + N^{2}\delta z = 0
\label{9.3}
\end{equation}
Firstly, when $N^{2}>0$, the general solution for (\ref{9.3}) is given by
\begin{equation}
\delta z = Ae^{iNt} + Be^{-iNt},
\label{9.4}
\end{equation}
where coefficient A and B are defined by the initial conditions (the initial location and velocity of the parcel), respectively.
Under these conditions, the parcel will vibrate at equilibrium position at angular frequency N.
Therefore, when the density decreases vertically as $d\rho/dz<0$, the field is stable; this kind of status is called {\it static stability}, and this kind of stratification is called {\it stable stratification}.
The angular frequency of N is called {\it buoyancy frequency} or {\it brunt vaisala frequency}, which is used as an index for the magnitude of the density stratification.

Secondly, when $N^{2}<0$, the general solution for (\ref{9.3}) becomes
\begin{equation}
\delta z =Ae^{Nt} + Be^{-Nt}.
\label{9.5}
\end{equation}
Under this condition, buoyancy force increases the displacement of the parcel.
%Since the parcel will exponentially move away from the equilibrium position as the time lapsed, when the density increases vertically as $d\rho/dz<0$, the field is unstable
When the density increases vertically as $d\rho/dz<0$, the parcel will exponentially move away from the equilibrium position as the time lapsed.
The field is thus unstable; this kind of status is called {\it static instability}, while this kind of stratification is called {\it unstable stratification}.
Under such condition, generally, it is not that only one parcel makes displacement and the left part keeps the unchanged status, but all of the elements of the flow affect by each other and make mass motion.
This kind of mass motion is generally called {\it convection}.

On the other hand, when $N^{2}=0$, the general solution for $\displaystyle \frac{\delta^{2}}{\delta t^{2}}(\delta z) = 0$ becomes
\begin{equation}
\delta z = c_{1}t+c_{2},
%\mbox{(c1 and c2 are coefficient of determination)}
\label{9.6}
\end{equation}
where c1 and c2 are decided by initial conditions of the parcel.
Since the density of the flow does not vary with the height, we find that the parcel will move at a constant velocity.
Obviously, when the initial velocity is equal to 0, the parcel stays at the initial position; this kind of status is called {\it static neutral}.

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\section{Exercises}

Since the motion of the parcel is totally decided by its initial displacement, its initial velocity, and density gradient of the surrounding field, we can repeat the experiment by changing these values.
The condition-setting parameters for experiments have already been prepared in the program "C10" in GFD Menu.
After opening the program, we can adjust the density gradient, initial position and initial velocity for the parcel to watch the results.

\begin{figure}[t]
  \begin{center}
    \includegraphics[width=10cm,keepaspectratio,clip]{fig/C09_01.epsf}
    \caption{Result of stasta}
    \label{fig:C09_01}
  \end{center}
\end{figure}

\begin{description}
\item[Example 1] stable stratification (positive displacement)
\end{description}
Set the coefficients as follows:
\begin{center}
\begin{tabular}{cc}
0.6  &  0.0 \\
Initial position &  Initial velocity
\end{tabular}
\begin{tabular}{ccccc}
  -0.1   & -0.075   &  -0.05   & -0.025   &   0.0 \\
dr/dz(1) & dr/dz(2) & dr/dz(3) & dr/dz(4) & dr/dz(5)
\end{tabular}
\end{center}
When the density gradient $d\bar{\rho}/dz<0$, the parcel will vibrate.
The vibration period is shorter as the stratification is stronger (left side).
Because the initial velocity is 0, in neutral density stratification, the parcel will stop at the point where the displacement is finished.

\begin{description}
\item[Example 2] unstable stratification (positive displacement)
\end{description}
Set the coefficients as follows:
\begin{center}
\begin{tabular}{cc}
0.3  &  0.0 \\
Initial position &  Initial velocity
\end{tabular}
\begin{tabular}{ccccc}
   0.1   &  0.075   &   0.05   &  0.025   &   0.0 \\
dr/dz(1) & dr/dz(2) & dr/dz(3) & dr/dz(4) & dr/dz(5)
\end{tabular}
\end{center}

When a density gradient is assumed, the parcel will exponentially moves away from its equilibrium position toward the positive direction.
The speed of the motion of the parcel is in proportional to the degree of the instability.
 
\begin{description}
\item[Example 3] stable stratification (toss up)
\end{description}
Set the coefficients as follows:
\begin{center}
\begin{tabular}{cc}
0.0  &  0.4 \\
Initial position &  Initial velocity
\end{tabular}
\begin{tabular}{ccccc}
  -0.1   & -0.075   &  -0.05   & -0.025   &   0.0 \\
dr/dz(1) & dr/dz(2) & dr/dz(3) & dr/dz(4) & dr/dz(5)
\end{tabular}
\end{center}

Status of stratification is the same as shown in Example 1.
When the parcel is initially tossed up toward the positive direction, the result is the same as seen in Example 1, because the vibration period is only decided by the stratification.
% the vibration period is only decided by the stratification.
%This result is the same as seen in Example 1.
When initial velocities are the same, the amplitude depends on the status of stratification.
The weaker the stratification is, the larger the amplitude is.
In neutral density stratification (right side), the parcel moves away from equilibrium position with the initial velocity.

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\section{Appendix}

Compare the results by assuming a positive displacement, a negative displacement, a toss up, and a toss down which are as shown in above examples.
Although the motion of parcel in field of unstable stratification is unrealistic, it should be a meaningful subject for an experiment when we consider the fundamental of dynamics.
There are series that only give a displacement to the parcel without initial velocity and series that make a positive displacement and then toss down.
When we consider the motion of mass point in gravity potential field, it becomes meaningful for practice.

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\section*{References}

\begin{description}
\itemsep   -1.55mm
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\item Shigeo Yoden, 2000: {\it Note of meteorology lecture I} (In Japanese)
\end{description}

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