# How to linearize the shallow-water equations?

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 $$$$\frac{\partial}{\partial t}(\overline{h} + h') + (\overline{u} + u')\frac{\partial}{\partial x}(\overline{h} + h') + (\overline{v} + v')\frac{\partial}{\partial y}(\overline{h} + h') = -(\overline{h} + h')(\frac{\partial \overline{u}}{\partial x} + \frac{\partial \overline{v}}{\partial y} + \frac{\partial u'}{\partial x} + \frac{\partial v'}{\partial y})$$$$ by using the equation \begin{aligned} \frac{d\overline{h}}{dt} = -\overline{h}(\frac{\partial \overline{u}}{\partial x}+\frac{\partial \overline{v}}{\partial y}) \end{aligned} and ignore the terms like $$u'\frac{\partial h'}{\partial x}$$ and $$v'\frac{\partial h'}{\partial y}$$, the result I obtained is $$$$\frac{\partial h'}{\partial t} + \overline{u}\frac{\partial h'}{\partial x} + \overline{v}\frac{\partial h'}{\partial y} = -\overline{h}(\frac{\partial u'}{\partial x} + \frac{\partial v'}{\partial y}) - h'(\frac{\partial \overline{u}}{\partial x} + \frac{\partial \overline{v}}{\partial y})$$$$ but in my textbook, the result is just $$$$\frac{\partial h'}{\partial t} = -\overline{h}(\frac{\partial u'}{\partial x} + \frac{\partial v'}{\partial y})$$$$ so I want to ask that where did I make a mistake? Actually, I got the same problem in the linearization of the equations of $$u$$ and $$v$$.

• 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. Aug 16, 2021 at 21:22

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 $$\frac{\partial \bar{u}h'}{\partial x}+\frac{\partial \bar{v}h'}{\partial y}=0$$.

• 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
Aug 15, 2021 at 4:06
• 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
Aug 15, 2021 at 7:05
• @Hou That is incorrect. You do not assume that the base state of the fluid are zero but a constant.
– user1066
Aug 16, 2021 at 8:47
• @gansub yes！I find that problem now!
–  Hou
Aug 16, 2021 at 8:49