<p>Well, I can see one thing. Note that you can rearrange the equation you derived to look like this: <span class="math-container">$$\frac{\partial h'}{\partial t}+\bar{u}\frac{\partial h'}{\partial x}+\bar{v}\frac{\partial h'}{\partial y}+h'\frac{\partial \bar{u}}{\partial x}+h'\frac{\partial \bar{v}}{\partial y}=-\bar{h}\left(\frac{\partial u'}{\partial x}+\frac{\partial v'}{\partial y}\right)$$</span>. Observe that the left hand side can be grouped together using the product rule. <span class="math-container">$$\frac{\partial h'}{\partial t}+\frac{\partial \bar{u}h'}{\partial x}+\frac{\partial \bar{v}h'}{\partial y}=-\bar{h}\left(\frac{\partial u'}{\partial x}+\frac{\partial v'}{\partial y}\right)$$</span></p>
<p>Now there are a couple of assumptions that could be made (I don't have mine near me as I write this). I suggest you read what is in your textbook. It could be aomething along the lines of the anelastic assumption, it could be that there is no mean flow, etc. However, the assumption should yield <span class="math-container">$\frac{\partial \bar{u}h'}{\partial x}+\frac{\partial \bar{v}h'}{\partial y}\approx0$</span> or <span class="math-container">$\frac{\partial \bar{u}h'}{\partial x}+\frac{\partial \bar{v}h'}{\partial y}=0$</span>.</p>

CC BY-SA 4.0
2021-08-15T04:06:45.687Z
I see your point, but I can't find any assumption that tells me I can take $\frac{\partial \overline{u} h'}{\partial x} + \frac{\partial \overline{v} h'}{\partial y} = 0$, can you tell me what's the meaning of no mean flow?
 Hou
CC BY-SA 4.0
2021-08-15T07:05:09.637Z
oh, I know where did I make a mistake, I should take $\overline{u} = 0$ and $\overline{v} = 0$, because it assumes that the basic states of the fluid are at rest.
 Hou
CC BY-SA 4.0
2021-08-16T08:49:25.517Z
@gansub yes！I find that problem now!
 Hou
CC BY-SA 4.0
2021-08-14T23:03:32.373Z
2021-08-14T23:03:32.373Z
BarocliniCplusplus
<p>The shallow water equations are:</p>
<p><span class="math-container">\begin{aligned}
&amp;\frac{du}{dt} = -g\frac{\partial h}{\partial x} + fv\\
&amp;\frac{dv}{dt} = -g\frac{\partial h}{\partial y} - fu\\
&amp;\frac{dh}{dt} = -h(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y})
\end{aligned}</span>
where h is the height (or depth) of the shallow water, <span class="math-container">$u$</span> is the eastward velocity, <span class="math-container">$v$</span> is the northward velocity, g is the acceleration of gravity and f is the coriolis parameter.</p>
<p>my problem is how to linearize these equations. From the linearization theorem, I know that I should take
<span class="math-container">\begin{aligned}
&amp;u = \overline{u} + u'\\
&amp;v = \overline{v} + v'\\
&amp;h = \overline{h} + h'
\end{aligned}</span>
then take them into the original equations, but I just do not know how. From my understanding, take the equation of <span class="math-container">$h$</span> as an example, the result should be
<span class="math-container">\begin{equation}
\frac{\partial}{\partial t}(\overline{h} + h') + (\overline{u} + u')\frac{\partial}{\partial x}(\overline{h} + h') + (\overline{v} + v')\frac{\partial}{\partial y}(\overline{h} + h') = -(\overline{h} + h')(\frac{\partial \overline{u}}{\partial x} + \frac{\partial \overline{v}}{\partial y} + \frac{\partial u'}{\partial x} + \frac{\partial v'}{\partial y})
\end{equation}</span>
by using the equation
<span class="math-container">\begin{aligned}
\frac{d\overline{h}}{dt} = -\overline{h}(\frac{\partial \overline{u}}{\partial x}+\frac{\partial \overline{v}}{\partial y})
\end{aligned}</span>
and ignore the terms like <span class="math-container">$u'\frac{\partial h'}{\partial x}$</span> and <span class="math-container">$v'\frac{\partial h'}{\partial y}$</span>, the result I obtained is
<span class="math-container">\begin{equation}
\frac{\partial h'}{\partial t} + \overline{u}\frac{\partial h'}{\partial x} + \overline{v}\frac{\partial h'}{\partial y} = -\overline{h}(\frac{\partial u'}{\partial x} + \frac{\partial v'}{\partial y}) - h'(\frac{\partial \overline{u}}{\partial x} + \frac{\partial \overline{v}}{\partial y})
\end{equation}</span>
but in my textbook, the result is just
<span class="math-container">\begin{equation}
\frac{\partial h'}{\partial t}  = -\overline{h}(\frac{\partial u'}{\partial x} + \frac{\partial v'}{\partial y}) 
\end{equation}</span>
so I want to ask that where did I make a mistake? Actually, I got the same problem in the linearization of the equations of <span class="math-container">$u$</span> and <span class="math-container">$v$</span>.</p>

CC BY-SA 4.0
2021-08-16T21:22:31.383Z
Try and see whether you can regroup the extra terms into the extra terms in the v and u-equations, after adding the equations the extra terms should add up to zero.
AtmosphericPrisonEscape
<meteorology>
How to linearize the shallow-water equations?
 Hou
# How to linearize the shallow-water equations?

(Asked by: username_0 on 2021-08-14)

The shallow water equations are:

\begin{aligned}
&\frac{du}{dt} = -g\frac{\partial h}{\partial x} + fv\\
&\frac{dv}{dt} = -g\frac{\partial h}{\partial y} - fu\\
&\frac{dh}{dt} = -h(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y})
\end{aligned}
where h is the height (or depth) of the shallow water, $u$ is the eastward velocity, $v$ is the northward velocity, g is the acceleration of gravity and f is the coriolis parameter.

my problem is how to linearize these equations. From the linearization theorem, I know that I should take
\begin{aligned}
&u = \overline{u} + u'\\
&v = \overline{v} + v'\\
&h = \overline{h} + h'
\end{aligned}
then take them into the original equations, but I just do not know how. From my understanding, take the equation of $h$ as an example, the result should be
\begin{equation}
\frac{\partial}{\partial t}(\overline{h} + h') + (\overline{u} + u')\frac{\partial}{\partial x}(\overline{h} + h') + (\overline{v} + v')\frac{\partial}{\partial y}(\overline{h} + h') = -(\overline{h} + h')(\frac{\partial \overline{u}}{\partial x} + \frac{\partial \overline{v}}{\partial y} + \frac{\partial u'}{\partial x} + \frac{\partial v'}{\partial y})
\end{equation}
by using the equation
\begin{aligned}
\frac{d\overline{h}}{dt} = -\overline{h}(\frac{\partial \overline{u}}{\partial x}+\frac{\partial \overline{v}}{\partial y})
\end{aligned}
and ignore the terms like $u'\frac{\partial h'}{\partial x}$ and $v'\frac{\partial h'}{\partial y}$, the result I obtained is
\begin{equation}
\frac{\partial h'}{\partial t} + \overline{u}\frac{\partial h'}{\partial x} + \overline{v}\frac{\partial h'}{\partial y} = -\overline{h}(\frac{\partial u'}{\partial x} + \frac{\partial v'}{\partial y}) - h'(\frac{\partial \overline{u}}{\partial x} + \frac{\partial \overline{v}}{\partial y})
\end{equation}
but in my textbook, the result is just
\begin{equation}
\frac{\partial h'}{\partial t} = -\overline{h}(\frac{\partial u'}{\partial x} + \frac{\partial v'}{\partial y})
\end{equation}
so I want to ask that where did I make a mistake? Actually, I got the same problem in the linearization of the equations of $u$ and $v$.

**Tags:** meteorology

> **Comments:**
> **username_1**: Try and see whether you can regroup the extra terms into the extra terms in the v and u-equations, after adding the equations the extra terms should add up to zero.  

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## 1 answers

### Answer by username_2 on 2021-08-14. Score: 1

Well, I can see one thing. Note that you can rearrange the equation you derived to look like this: $$\frac{\partial h'}{\partial t}+\bar{u}\frac{\partial h'}{\partial x}+\bar{v}\frac{\partial h'}{\partial y}+h'\frac{\partial \bar{u}}{\partial x}+h'\frac{\partial \bar{v}}{\partial y}=-\bar{h}\left(\frac{\partial u'}{\partial x}+\frac{\partial v'}{\partial y}\right)$$. Observe that the left hand side can be grouped together using the product rule. $$\frac{\partial h'}{\partial t}+\frac{\partial \bar{u}h'}{\partial x}+\frac{\partial \bar{v}h'}{\partial y}=-\bar{h}\left(\frac{\partial u'}{\partial x}+\frac{\partial v'}{\partial y}\right)$$

Now there are a couple of assumptions that could be made (I don't have mine near me as I write this). I suggest you read what is in your textbook. It could be aomething along the lines of the anelastic assumption, it could be that there is no mean flow, etc. However, the assumption should yield $\frac{\partial \bar{u}h'}{\partial x}+\frac{\partial \bar{v}h'}{\partial y}\approx0$ or $\frac{\partial \bar{u}h'}{\partial x}+\frac{\partial \bar{v}h'}{\partial y}=0$.

> **Comments:**
> **username_0**: I see your point, but I can't find any assumption that tells me I can take $\frac{\partial \overline{u} h'}{\partial x} + \frac{\partial \overline{v} h'}{\partial y} = 0$, can you tell me what's the meaning of no mean flow?  
> **username_0**: oh, I know where did I make a mistake, I should take $\overline{u} = 0$ and $\overline{v} = 0$, because it assumes that the basic states of the fluid are at rest.  
> **username_0**: @gansub yes！I find that problem now!  

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