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\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 }[/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], 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], 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, \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..