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).

The theory we present here is discussed in Stoker 1957, Billingham and King 2000 and Johnson 1997.

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 or a hydraulic jump when it occurs on the surface of the water.

The dam break problem

Assume the water occupies the region [math]\displaystyle{ {x \lt 0 ; 0 \lt z \lt h_0 } }[/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] The initial condition is therefore

[math]\displaystyle{ h(x,0) = \begin{cases} h_0, & x \lt 0 \\ 0, & x \gt 0 \end{cases} }[/math]

[math]\displaystyle{ 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_0} = \pm 2c_0 }[/math]

where [math]\displaystyle{ c_0 = \sqrt{gh_0} }[/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_0 , \;\mathrm{and}\; u - 2\sqrt{gh} = -2c_0 }[/math]

and hence, [math]\displaystyle{ u = 0 }[/math] and [math]\displaystyle{ h = h_0 }[/math]. Moreover,

[math]\displaystyle{ \frac{d \chi_\pm}{d t} = u \pm \sqrt{gh} = \pm c_0 }[/math]

so these characteristics are straight lines in the region [math]\displaystyle{ \big\{x \lt -c_0 t \big\} }[/math] (the undisturbed region).

The [math]\displaystyle{ \;C_+ }[/math] characteristic leave the region a[math]\displaystyle{ \big\{x \lt -c_0 t \big\} }[/math] and enter [math]\displaystyle{ \big\{x \gt -c_0 t \big\} }[/math]. For now we will assume that these characteristics fill the domain (and show that this is true shortly). For [math]\displaystyle{ \big\{x \gt -c_0 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 of the [math]\displaystyle{ C_- }[/math] characteristics [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_0 }[/math], [math]\displaystyle{ u }[/math] and [math]\displaystyle{ h }[/math] must be constant on each [math]\displaystyle{ C_- }[/math] characteristic. This also means that the [math]\displaystyle{ C_- }[/math] characteristics must be straight lines.

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]

which in turn implies that

[math]\displaystyle{ u - \sqrt{gh} = \frac{x}{t} }[/math]

We also have [math]\displaystyle{ R_+ = u + 2\sqrt{gh} = 2c_0 }[/math] from the [math]\displaystyle{ C_+ }[/math] characteristics. We can solve these equations at each point in [math]\displaystyle{ \big\{x \gt -c_0 t \big\} }[/math]. Solving for [math]\displaystyle{ u }[/math] and [math]\displaystyle{ h }[/math] gives

[math]\displaystyle{ h(x, t) = \frac{h_0}{9}\left(2 - \frac{x}{c_0 t}\right)^2 , \qquad u(x, t) = \frac{2}{3} \left (c_0 + \frac{x}{t} \right ) }[/math]

From this we can see that [math]\displaystyle{ h = 0 }[/math] for [math]\displaystyle{ x = 2c_0 t, }[/math] suggesting that the [math]\displaystyle{ C_+ }[/math] characteristic only exist in the region [math]\displaystyle{ \big\{x \gt 2c_0 t \big\} }[/math].

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_0 t \big\} }[/math]. For [math]\displaystyle{ \big\{x \lt -c_0 t \big\} }[/math], the [math]\displaystyle{ \;C_+ }[/math] characteristics are straight lines with slope [math]\displaystyle{ c_0 }[/math] and are given by

[math]\displaystyle{ \chi_+ (t) = -x_0 + c_0 t, \quad \left(x_0 \gt 0,\;\; t \lt \frac{x_0}{2c_0}\right) }[/math]

When [math]\displaystyle{ t = \frac{x_0}{2c_0},\;\;\chi_+ (t) = -c_0 t }[/math] so that for

[math]\displaystyle{ t \gt \frac{x_0}{2c_0},\;\;\frac{d\chi_+ (t)}{d t} = u + \sqrt{gh} }[/math]

and substituting the solution we found for [math]\displaystyle{ h }[/math] and [math]\displaystyle{ u }[/math]

[math]\displaystyle{ \frac{d\chi_+ (t)}{d t} = \frac{4}{3}c_0 + \frac{\chi_+ (t)}{3t} }[/math]

Solving this ODE subject to [math]\displaystyle{ \chi_+ \left(\frac{x_0}{2c_0}\right) = -\frac{x_0}{2} }[/math] gives

[math]\displaystyle{ \chi_+ (t) = 2c_0 t - 3\left(\frac{x_0}{2}\right)^{2/3}(c_0 t)^{1/3},\;\; }[/math]

equation for a characteristic curve.

The curves indeed fill the domain [math]\displaystyle{ \big\{x \lt 2c_0 t \big\} }[/math] and all satisfy [math]\displaystyle{ \big\{\chi_+ (t) \lt 2c_0 t \big\} }[/math].

Shocks

For a unique solution two exist there must be a single [math]\displaystyle{ C_+ }[/math] and [math]\displaystyle{ C_- }[/math] characteristic through each point.

Example 2

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