Nonlinear Shallow Water Waves

Introduction

We want to consider waves occurring at the interface of the two fluids water and air. We assume that water is incompressible, viscous effects are negligible and that the typical wave lengths are much larger than the water depth. For "shallow" water we assume no variations in the y-dirn. Let the domain of interest be denoted by [math]\displaystyle{ \Omega }[/math] with z-axis vertical and x-axis horizontal, [math]\displaystyle{ \vec{x} = (x,z) }[/math]. Let [math]\displaystyle{ h_{o} }[/math] be the at rest water depth. Let [math]\displaystyle{ h(x,t) }[/math] be the local water depth or wave height. From previous lectures we have established the eqn. for Conservation of Mass.

[math]\displaystyle{ \frac{D \rho}{D t} (\vec{x} ,t) + \rho(\vec{x} ,t)\nabla \cdot \vec{u}(\vec{x} ,t) = 0, x \in \Omega }[/math]

Since water is incompressible [math]\displaystyle{ \; \frac{D \rho}{D t} = 0 \; }[/math] i.e. [math]\displaystyle{ \;\nabla \cdot \vec{u} = 0,\; }[/math] the divergance of the velocity field is zero.

We can apply the Conservation of Momentum eqn. as follows

[math]\displaystyle{ \frac{D \vec{u}}{D t} (\vec{x} ,t) = \frac{-1}{\rho} \nabla p + g(0,-1) }[/math]

components of [math]\displaystyle{ \vec{u} = (u,v) }[/math], assuming that changes in the vertical velocity are negligible i.e.[math]\displaystyle{ \;\; \frac{D v}{D t} (\vec{x} ,t) = 0 }[/math]

thus, [math]\displaystyle{ 0 = \frac{-1}{\rho}\frac{\partial p}{\partial z} - g }[/math] or [math]\displaystyle{ \frac{\partial p}{\partial z} = -\rho g }[/math] and hence,

[math]\displaystyle{ p(\vec{x} ,t) = p_{atm} + \rho g[h(x,t) - z] }[/math] (pressure is hydrostatic)

Also, [math]\displaystyle{ \frac{D u}{D t}(\vec{x} ,t) = \frac{-1}{\rho}\frac{\partial p}{\partial x} }[/math], and since [math]\displaystyle{ \frac{\partial p}{\partial x} }[/math] is independant of [math]\displaystyle{ z }[/math], so is [math]\displaystyle{ \frac{D u}{D t} }[/math]

[math]\displaystyle{ \frac{D u}{D t} = -g\frac{\partial h(x ,t)}{\partial x} }[/math] after substituting expression for [math]\displaystyle{ p(\vec{x} ,t) }[/math]

Therefore if [math]\displaystyle{ u }[/math] has no z-dependance at the, at rest condition ([math]\displaystyle{ h_{o} }[/math]), it will remain this way. Thus,

[math]\displaystyle{ \frac{\partial u(\vec{x} ,t)}{\partial t} + u(\vec{x} ,t)\frac{\partial u(\vec{x} ,x)}{\partial x} + g \frac{\partial h(x ,t)}{\partial x} = 0 \qquad (1) }[/math]

Another equation is needed. Consider conservation of mass along a vertical cross section.

[math]\displaystyle{ \frac{d}{dt}\int_{x_1}^{x_2} \rho h(x ,t) dx = -[\rho u(\vec{x} ,t) h(x ,t)]_{x_1}^{x_2} }[/math]

Assuming quantities are smooth and [math]\displaystyle{ x_1, x_2 }[/math] are arbitrary we arrive at

[math]\displaystyle{ \frac{\partial h(x ,t)}{\partial t} + \frac{\partial}{\partial x}[u(\vec{x} ,t)h(x ,t)] = 0 \qquad (2) }[/math]

Eqns. (1) & (2) are called the non-linear shallow water equations. They determine the horizontal water velocity and the local water depth.

Rewrite them in terms of the local wave speed [math]\displaystyle{ c(x, t) = \sqrt{gh(x, t)} }[/math] as follows:

[math]\displaystyle{ 2\frac{\partial c(x ,t)}{\partial t}+ 2u(\vec{x} ,t)\frac{\partial c(x ,t)}{\partial x} + c(x, t)\frac{\partial u(\vec{x} ,t)}{\partial x} = 0 \qquad (3) }[/math]

[math]\displaystyle{ \frac{\partial u(\vec{x} ,t)}{\partial t}+ u(\vec{x} ,t)\frac{\partial u(\vec{x} ,t)}{\partial x} + 2c(x, t)\frac{\partial c(x ,t)}{\partial x} = 0 \qquad (4) }[/math]

Compare these equations with those of compressible gas dynamics with [math]\displaystyle{ (\gamma = 2, k = g/2) }[/math] of previous lectures.

To make further progress with solving our equations we use the Method of Characteristics. Adding and subtracting (3) from (4) gives (5).

[math]\displaystyle{ \frac{\partial (u \pm 2c)}{\partial t}+ (u \pm c)\frac{\partial (u \pm 2c)}{\partial x} = 0 \qquad (5) }[/math]

Hence the functions [math]\displaystyle{ R_{\pm} (u ,c) = u \pm 2c }[/math], are the Riemannian invariants.

We have:

on the [math]\displaystyle{ C_+ }[/math] characteristic, given by [math]\displaystyle{ \frac{d \chi_+}{d t} = u + c = u + \sqrt{gh} }[/math],

the [math]\displaystyle{ C_+ }[/math] invariant [math]\displaystyle{ R_+ = u + 2c = u + 2\sqrt{gh} }[/math], is a constant.

and on the [math]\displaystyle{ C_- }[/math] characteristic, given by [math]\displaystyle{ \frac{d \chi_-}{d t} = u - c = u - \sqrt{gh} }[/math],

the [math]\displaystyle{ C_- }[/math] invariant [math]\displaystyle{ R_- = u - 2c = u - 2\sqrt{gh} }[/math], is a constant.

Examples

We consider two examples involving shocks.

Example 1

The dam break problem. Assume the water occupies the region[math]\displaystyle{ \big\{x \lt 0 ; 0 \lt z \lt h_0 \big\} }[/math] initially held back by a dam at [math]\displaystyle{ x = 0 }[/math]. At [math]\displaystyle{ t = 0 }[/math], the dam is removed (breaks). What is the height of the water [math]\displaystyle{ h(x,t) }[/math] for [math]\displaystyle{ t \gt 0? }[/math] Thus consider (1) and (2) subject to initial conditions

[math]\displaystyle{ h(x,0) = \begin{cases} h_o, & x \lt 0 \\ 0, & x \gt 0 \end{cases}\;,\; \; u(x ,0) = 0. }[/math]

On the characteristic that originates at [math]\displaystyle{ t = 0 }[/math] for [math]\displaystyle{ x \lt 0 }[/math]. [math]\displaystyle{ R_\pm = u \pm 2\sqrt{gh} = \pm 2\sqrt{gh_o} = \pm 2c_o \; , }[/math] where [math]\displaystyle{ c_o = \sqrt{gh_o}\; }[/math] is the initial (linear) wave speed.

Therefore, if a [math]\displaystyle{ C_+ }[/math] and a [math]\displaystyle{ C_- }[/math] characteristic from this region intersect, then [math]\displaystyle{ u + 2\sqrt{gh} = 2c_o , \;\; u - 2\sqrt{gh} = -2c_o }[/math] and hence, [math]\displaystyle{ u = 0 }[/math] and [math]\displaystyle{ h = h_o }[/math].

Moreover, [math]\displaystyle{ \frac{d \chi_\pm}{d t} = u \pm \sqrt{gh} = \pm c_o }[/math] so these characteristics are straight lines and they must lie in the region [math]\displaystyle{ \big\{x \lt -c_o t \big\} }[/math].

Notice that the [math]\displaystyle{ C_+ }[/math] characteristic leaves this region and enters [math]\displaystyle{ \big\{x \gt -c_o t \big\} }[/math]. For now we will assume that these characteristics fill the domain.

For [math]\displaystyle{ \big\{x \gt -c_o t \big\} }[/math], the [math]\displaystyle{ C_- }[/math] characteristics are given by [math]\displaystyle{ \frac{d \chi_-}{d t} = u - \sqrt{gh} \;\; (*) }[/math] and on each curve [math]\displaystyle{ R_- = u - 2\sqrt{gh} }[/math] is constant.

However, since this region is filled with [math]\displaystyle{ C_+ }[/math] characteristics where [math]\displaystyle{ R_+ = u + 2\sqrt{gh} = 2c_o }[/math], [math]\displaystyle{ u }[/math] and [math]\displaystyle{ h }[/math] must be constant on each [math]\displaystyle{ C_- }[/math] characteristic, which from (*) must be a straight line.

Since the fluid occupies [math]\displaystyle{ \big\{x \lt 0 \big\} }[/math] at [math]\displaystyle{ t = 0 }[/math], these [math]\displaystyle{ C_- }[/math] characteristics must start at the origin, with [math]\displaystyle{ \chi_-(t) = \left(u - \sqrt{gh}\right)t }[/math], therefore, [math]\displaystyle{ R_+ = u + 2\sqrt{gh} = 2c_o }[/math]

and [math]\displaystyle{ u - 2\sqrt{gh} = \frac{x}{t} }[/math] at each point in[math]\displaystyle{ \big\{x \gt -c_o t \big\} }[/math]. Solving for [math]\displaystyle{ u }[/math] and [math]\displaystyle{ h }[/math] gives

[math]\displaystyle{ h(x, t) = \frac{h_o}{9}\left(2 - \frac{x}{c_o t}\right)^2 , \qquad u(x, t) = \frac{2}{3} \left (c_o + \frac{x}{t} \right )\quad (**). }[/math]

From this, [math]\displaystyle{ h = 0 }[/math] for [math]\displaystyle{ x = 2c_o t, }[/math] so the [math]\displaystyle{ C_+ }[/math] characteristic will reach the region [math]\displaystyle{ \big\{x \gt 2c_o t \big\} }[/math] whereby [math]\displaystyle{ u = h = 0 }[/math] there.

It remains to verify that the [math]\displaystyle{ C_+ }[/math] characteristic, which originated in [math]\displaystyle{ \big\{x \lt 0 \big\} }[/math] will fill the domain [math]\displaystyle{ \big\{x \lt 2c_o t \big\} }[/math]. For [math]\displaystyle{ \big\{x \lt -c_o t \big\} }[/math],

the [math]\displaystyle{ C_+ }[/math] characteristics are straight lines with slope [math]\displaystyle{ c_o }[/math] and are given by [math]\displaystyle{ \chi_+ (t) = -x_o + c_o t }[/math], \quad \left(x_o > 0,\;\; t < \frac{x_o}{2c_o}\right)</math>.

When [math]\displaystyle{ t = \frac{x_o}{2c_o},\;\;\chi_+ (t) = -c_o t }[/math] so that for [math]\displaystyle{ t \gt \frac{x_o}{2c_o},\;\;\frac{d\chi_+ (t)}{d t} = u + \sqrt{gh} }[/math] and hence from (**), [math]\displaystyle{ \frac{d\chi_+ (t)}{d t} = \frac{4}{3}c_o + \frac{\chi_+ (t)}{3t}. }[/math]

Solving this ODE using the Integrating factor [math]\displaystyle{ e^{\int p(t) dt} }[/math] subject to [math]\displaystyle{ \chi_+ \left(\frac{x_o}{2c_o}\right) = -\frac{x_o}{2} }[/math] gives

[math]\displaystyle{ \chi_+ (t) = -x_o + c_o t }[/math]