Nonlinear Shallow Water Waves

Introduction

We assume that water is incompressible, viscous effects are negligible and that the typical wave lengths are much larger than the water depth. This allows us to assume Shallow Depth. We assume that the problem has not variation in either the [math]\displaystyle{ y }[/math] or [math]\displaystyle{ z }[/math] direction. The fluid is governed by two parameters, [math]\displaystyle{ u }[/math] and [math]\displaystyle{ h }[/math] the water depth (not that this is not the still water depth since the problem is nonlinear).

Equations of Motion

The equation for the conservation of mass can derived by considering a a region [math]\displaystyle{ [x,x+\Delta x] }[/math] Conservation of mass then implies that

[math]\displaystyle{ \partial_t \int_x^{x + \Delta x} h(s,t) ds = h(x,t)u(x,t) - h(x+\Delta x,t)u(x+\Delta x,t) }[/math]

If we take the limit as [math]\displaystyle{ \Delta x \to 0 }[/math] we obtain

[math]\displaystyle{ \partial_t h(x ,t) + \partial_x (h(x ,t)u(x ,t)) = 0 }[/math]

A second equation comes from conservation of momentum and is

[math]\displaystyle{ \partial_t u(x ,t) + u(x ,t) \partial_x u(x ,x) + g \partial_x h(x ,t) = 0 }[/math]

These equations are called the nonlinear shallow water equations. They determine the horizontal water velocity and the local water depth.

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

[math]\displaystyle{ 2\partial_t c(x ,t) + 2u(x ,t)\partial_x c(x ,t) + c(x, t)\partial_x u(\vec{x} ,t) = 0 }[/math]

[math]\displaystyle{ \partial_t u(x ,t) + u(x ,t)\partial_x u(x ,t) + 2c(x, t) \partial_x c(x ,t) = 0 }[/math]

These equation are almost identical to those of compressible gas dynamics. Much of our understanding of the equations for water have been found by researchers studying compressible gas dynamics.

Characteristics

The equations possess characteristics. Adding and subtracting the two equations above we obtain

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

This means that 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.

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

Simple Waves

The problem as formulated can be solved by advancing the solution along the characteristics, but this will in general be quite difficult analytically. However, there is a special class of problems, called Simple Waves in which the solution only changes on one characteristic. They are best illustrated through some examples. Note that the characteristic can meet forming a shock, which is called a bore when it occurs on the surface of the water.

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, \quad \left(x_o \gt 0,\;\; t \lt \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) = 2c_o t - 3\left(\frac{x_o}{2}\right)^{2/3}(c_o t)^{1/3},\;\; }[/math] equation for a characteristic curve.

The curves indeed fill the domain [math]\displaystyle{ \big\{x \lt 2c_o t \big\} }[/math] and all satisfy [math]\displaystyle{ \big\{\chi_+ (t) \lt 2c_o t \big\} }[/math].

Example 2

A shallow water bore. Assume the water occupies etc, etc..