# Deriving shallow water equations from Euler's equations

I would like to derive the one-dimensional shallow water equations from Eulers's equations. This works perfectly for the conservation of mass. Especially the meaning of the longitudinal fluid velocity $\bar u$ in the shallow water equations becomes clear. It can be interpreted as average of the longitudinal velocity in Euler's equations over the height above ground level.

But, in Euler's balance of momentum the longitudinal velocity occurs squared and the average of the squared velocity is not necessarily equal to the square of the averaged velocity. I am stuck at this point.

Next, there follows what I have so far.

Assumptions:

• propagation in the direction of the $x$-axis (unit vector $\vec{e}_x$)
• $y$-axis points upwards (unit vector $\vec{e}_y$)
• everything is constant in $z$-direction (unit vector $\vec{e}_z$; this direction is mostly left out). Volume integrals become area integrals and surface integrals become line integrals. If the path runs with positive circulation the outer surface normal can be calculated via $d\vec{r}\times\vec{e}_z$.
• incompressible medium; The density $\rho(x,y,t)$ is constant.
• The ground at $y=0$ is flat. The water height is $h(x,t)$. Region of water: $0\leq y \leq h(x,t)$, region of air: $h(x,t) < y$.
• The static relative pressure (w.r.t. atmospheric pressure) is $p(x,y,t)=g\rho(h(x,y,t)-y)$ where $g$ is the gravitation constant.
• No friction.

We describe the fluid motion through the height above ground $h(x,t)$ and the velocity filed $\vec{v}(x,y,t)$ in the fluid region.

At first the "working" case of Euler's mass balance. A fluid motion satisfies Euler's mass balance if for all parts $A$ of the fluid cross-section area in the $(x,y)$-plane there holds the equation $$\partial_t\int_{A} \rho dA + \int_{\partial A} \rho\, \vec{v}(x,y,t)\cdot d\vec{r}\times\vec{e}_z = 0.$$ Euler's mass balance leads to the mass balance of the shallow water equations if one restricts the choice of areas to stripes $A:=\{(x,y)\in[x_1,x_2]\times\mathbb{R}\mid 0\leq y \leq h(x,t)\}$ for $x_1 < x_2$.

On the bottom $y=0$ the velocity $v(x,0,t)$ is parallel to the path element $d\vec{r}$, the spat product in the path integral over $\partial A$ is zero and thus the contribution of this section of $\partial A$ to the path integral in Euler's mass balance is zero.

Because of the height changing with time there is actually a normal component of the velocity at the top. But, this is already considered in the time-dependent area in the area integral.

It is easier to consider the fluid mass between the half planes $A_x:=\{(x,y,z)\in\mathbb{R}^3\mid y\geq 0\}$ (note the fixed $x$) at $x=x_1$ and $x=x_2$. The growth of this fluid mass together with the out-flow of the fluid mass through the planes $A_{x_1}$, $A_{x_2}$ must balance to zero. This directly leads directly to the formula $$\partial_t \int_{x_1}^{x_2}h(x,t)dx + \left[\int_{0}^{h(x,t)}u(x,y,t)dy\right]_{x=x_1}^{x_2} = 0$$ where $u(x,y,t)=\vec{e}_x\cdot \vec{v}(x,y,t)$ is the $x$-component of the fluid velocity $\vec{v}$. Thereby, we have already divided through the constant density $\rho$. With the averaged longitudinal velocity $$\bar u(x,t):=\frac1{h(x,t)}\int_0^{h(x,t)}u(x,y,t)dy$$ the last formula gives after differentiation w.r.t. $x_2$ and renaming $x_2\mapsto x$ the mass balance of the shallow water equations:

## $$\partial_t h(x,t) + \partial_x \bigl(h(x,t) \bar u(x,t)\bigr) =0$$

Now, the more difficult case of the momentum balance. The momentum balance from Euler's equations is satisfied if for all parts $A$ of the fluid cross-section area in the $(x,y)$-plane the equation $$\partial_t \int_{A} \rho \vec{v} d A + \int_{\partial A} \rho \vec{v} \vec{v}\cdot d \vec{r}\times \vec{e}_z = \int_{\partial A} p d \vec{r}\times \vec{e}_z = \int_{\partial A} g\rho \left(h(x,t)-y\right) d \vec{r}\times \vec{e}_z$$ is satisfied.

We get rid of the constant density and only consider the $x$-component of this formula by multiplying it with $\frac1{\rho}\vec{e}_x$ and we restrict the area to sections $A=\{(x,y)\in[x_1,x_2]\times\mathbb{R}\mid 0\leq y\leq h(x,t)\}$ with $x_1<x_2$. For simplification we again integrate over the half planes $A_{x_1}$ and $A_{x_2}$ $$\partial_t\int_{x_1}^{x_2} \int_0^{h(x,t)} u(x,y,t) d y\, d x + \left[ \int_{y=0}^{h(x,t)} u^2(x,y,t)d y \right]_{x=x_1}^{x_2}= \left[ \frac12 gh(x,t)^2 \right]_{x=x_1}^{x_2}$$ Substituting $\int_0^h u dy=h\cdot \bar u$ yields $$\partial_t \int_{x_1}^{x_2} h(x,t)\bar u(x,t) d x + \left[ \int_{y=0}^{h(x,t)} u^2(x,y,t)d y \right]_{x=x_1}^{x_2}= \left[ \frac12 gh(x,t)^2 \right]_{x=x_1}^{x_2}.$$ Here, I am stuck. The substitution $\int_0^h u^2 dy = h {\bar u}^2$ is possible if $u(x,y,t)$ is independent of the height $y$.

If this is really the case what is the reasoning for this assumption? How do we come to know the vertical distribution of $u$?

If this assumption is true then we arrive at the impulse balance for the shallow water equations through differentiation w.r.t. $x_2$ and renaming $x_2\mapsto x$: $$\partial_t\bigl(h\bar u\bigr)(x,t) + \partial_x \left(\left((h\bar u)(x,t)\right)^2 - \frac 12 g h(x,t)^2\right)=0$$

There is a similar derivation at http://www.whoi.edu/fileserver.do?id=136564&pt=10&p=85713. But, there it is just assumed that $u$ does not depend on $y$.

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To be more explicit, consider some type of shallow water wave, like a gravity wave, with wavenumber $k$ (units length$^{-1}$) and a fluid of height $h$. The shallow water equations are suitable when you have most of your energy in waves that satisfy $kh << 1$. Otherwise, you have to use the full fluid equations.