PZJuly working place: The shallow water wave equation and tsunami propagation

The shallow water wave equation and tsunami propagation: "

As we are all now very much aware, tsunamis are water waves that start in the deep ocean, usually because of an underwater earthquake, and then propagate towards shore. Initially, tsunamis have relatively small amplitude (a metre or so is typical), which would seem to render them as harmless as wind waves. And indeed, tsunamis often pass by ships in deep ocean without anyone on board even noticing.

However, being generated by an event as large as an earthquake, the wavelength of the tsunami is huge – 200 kilometres is typical (in contrast with wind waves, whose wavelengths are typically closer to 100 metres). In particular, the wavelength of the tsunami is far greater than the depth of the ocean (which is typically 2-3 kilometres). As such, despite the depth of the ocean, the dynamics of tsunami are essentially governed by the shallow water equations. One consequence of these equations is that the speed of propagation {v}

of a tsunami can be approximated by the formula

$\displaystyle v \approx \sqrt{g b} \ \ \ \ \ (1)$

where

is the depth of the ocean, and ${g \approx 9.8 ms^{-2}}$ is the force of gravity. As such, tsunamis in deep water move very fast – speeds such as 500 kilometres per hour (300 miles per hour) are quite typical; enough to travel from Japan to the US, for instance, in less than a day. Ultimately, this is due to the incompressibility of water (and conservation of mass); the massive net pressure of a very broad and deep wave of water forces the profile of the wave to move horizontally at vast speeds. (Note though that this is the phase velocity of the tsunami wave, and not the velocity of the water particles themselves, which are far slower.)

As the tsunami approaches shore, the depth {b}

of course decreases, causing the tsunami to slow down, at a rate proportional to the square root of the depth. Unfortunately, wave shoaling then forces the amplitude to increase at the opposite rate,

$\displaystyle A \propto \frac{1}{\sqrt{b}} \ \ \ \ \ (2)$

at least until the amplitude becomes comparable to the water depth (at which point the assumptions that underlie the above approximate results break down). If one starts with a tsunami whose initial amplitude was ${A_0}$ at depth ${b_0}$ and computes the point at which the amplitude {A}

and depth

become comparable using the proportionality relationship (2), one soon one finds that at this point, amplitude of a tsunami (and the depth of the water) is about ${A_0^{2/3} b_0^{1/3}}$ . Thus, for instance, a tsunami with initial amplitude of one metre at a depth of 2 kilometres can end up with a final amplitude of about 12 metres near shore, while still traveling at about ten metres per second (35 kilometers per hour, or 22 miles per hour), and we have all now seen the impact that can have when it hits shore.

While tsunamis are far too massive of an event to be able to control, we can at least model them mathematically, allowing one to predict their impact at various places along the coast with high accuracy. (For instance, here is a video of the NOAA’s model of the March 11 tsunami, which has matched up very well with subsequent measurements.) The full equations and numerical methods used to perform such models are somewhat sophisticated, but by making a large numebr of simplifying assumptions, it is relatively easy to come up with a rough model that already predicts the basic features of tsunami propagation, such as the velocity formula (1) and the amplitude proportionality law (2). I give this (standard) derivation below the fold. The argument will largely be heuristic in nature; there are very interesting analytic issues in actually justifying many of the steps below rigorously, but I will not discuss these matters here.

— 1. The shallow water wave equation —

The ocean is, of course, a three-dimensional fluid, but to simplify the analysis we will consider a two-dimensional model in which the only spatial variables are the horizontal variable {x}

and the vertical variable {z}

, with

being equilibrium sea level. We model the ocean floor by a curve

$\displaystyle z = - b(x),$

thus

measures the depth of the ocean at position {x}

. At any time {t}

and position {x}

, the height of the water (compared to sea level {z=0}

) will be given by an unknown height function {h(t,x)}

; thus, at any time {t}

, the ocean occupies the region

$\displaystyle \Omega_t := \{ (x,z): -b(x) < z < h(t,x) \}.$

Now we model the motion of water inside the ocean by assigning at each time {t}

and each point ${(x,z) \in \Omega_t}$ in the ocean, a velocity vector

$\displaystyle \vec u(t,x,z) = (u_x(t,x,z), u_z(t,x,z)).$

We make the basic assumption of incompressibility, so that the density ${\rho}$ of water is constant throughout ${\Omega_t}$ .

The velocity changes over time according to Newton’s second law {F=ma}

. To apply this law to fluids, we consider an infinitesimal amount of water as it flows along the velocity field ${\vec u}$ . Thus, at time {t}

, we assume that this amount of water occupies some infinitesimal area {dA}

and some position ${\vec x(t) = (x(t), z(t))}$ , where we have

$\displaystyle \frac{d}{dt} \vec x(t) = \vec u( t, \vec x(t) ).$

Because of incompressibility, the area {dA}

stays constant, and the mass of this infinitesimal portion of water is ${m = \rho dA}$ . There will be two forces on this body of water; the force of gravity, which is ${(0, -mg) = (0, -\rho) dA}$ , and the force of the pressure field {p(t,x,z)}

, which is given by ${-\nabla p dA}$ . Newton’s law ${m \frac{du}{dt} = F}$ then gives

$\displaystyle m \frac{d}{dt} \vec u(t, \vec x(t) ) = - \nabla p dA + (0,-mg)$

which simplifies to the incompressible Euler equation

$\displaystyle \frac{\partial}{\partial t} \vec u + (\vec u \cdot \nabla) \vec u = - \frac{1}{\rho} \nabla p + (0,-g).$

At present, the pressure is not given. However, we can simplify things by making the assumption of (vertical) hydrostatic equilibrium, i.e. the vertical effect ${ - \frac{1}{\rho} \frac{\partial}{\partial z} p}$ of pressure cancels out the effect {-g}

of gravity. We also assume that the pressure is zero on the surface {z=h(t,x)}

of the water. Together, these two assumptions force the pressure to be the hydrostatic pressure

$\displaystyle p = \rho g ( h(t,x) - z ). \ \ \ \ \ (3)$

The incompressible Euler equation now simplifies to

$\displaystyle \frac{\partial}{\partial t} \vec u + (\vec u \cdot \nabla) \vec u = - g (\frac{\partial}{\partial x} h, 0). \ \ \ \ \ (4)$

We next make the shallow water approximation that the wavelength of the water is far greater than the depth of the water. In particular, we do not expect significant changes in the velocity field in the {z}

variable, and thus make the ansatz

$\displaystyle \vec u(t,x,z) = \vec u(t,x).$

Taking the {x}

component of (4), and abbreviating ${u_x}$ as {u}

, we obtain the first shallow water wave equation

$\displaystyle \frac{\partial}{\partial t} u + u \frac{\partial}{\partial x} u = - g \frac{\partial}{\partial x} h. \ \ \ \ \ (5)$

The next step is to play off the incompressibility of water against the finite depth of the ocean. Consider an infinitesimal slice

$\displaystyle \{ (x,z) \in \Omega_t: x_0 \leq x \leq x_0+dx \}$

of the ocean at some time {t}

and position ${x_0}$ . The total mass of this slice is roughly

$\displaystyle \rho (h(t,x_0) + b(x_0)) dx$

and so the rate of change of mass of this slice over time is

$\displaystyle \rho \frac{\partial h}{\partial t}(t,x_0) dx.$

On the other hand, the rate of mass entering this slice on the left ${x=x_0}$ is

$\displaystyle \rho u(t,x_0) (h(t,x_0) + b(x_0))$

and the rate of mass exiting on the right ${x=x_0+dx}$ is

$\displaystyle \rho u(t,x_0+dx) (h(t,x_0+dx) + b(x_0+dx)).$

Putting these three facts together, we obtain the equation

$\displaystyle \rho \frac{\partial h}{\partial t}(t,x_0) dx = \rho u(t,x_0) (h(t,x_0) + b(x_0))$

$\displaystyle - \rho u(t,x_0+dx) (h(t,x_0+dx) + b(x_0+dx))$

which simplifies after Taylor expansion to the second shallow water wave equation

$\displaystyle \frac{\partial}{\partial t} h + \frac{\partial}{\partial x}( u (h+b) ) = 0. \ \ \ \ \ (6)$

The equations (5), (6) are nonlinear in the unknowns {u, h}

. However, one can approximately linearise them by making the hypothesis that the amplitude of the wave is small compared to the depth of the water:

$\displaystyle |h| \ll b. \ \ \ \ \ (7)$

This already simplifies (6) to (approximately)

$\displaystyle \frac{\partial}{\partial t} h + \frac{\partial}{\partial x}( u b ) = 0. \ \ \ \ \ (8)$

As for (5), we argue that the second term on the left-hand side is negligible, leading to

$\displaystyle \frac{\partial}{\partial t} u = - g \frac{\partial}{\partial x} h. \ \ \ \ \ (9)$

To explain heuristically why we expect this to be the case, let us make the ansatz that {h}

and

have amplitude {A}

respectively, and propagate at some phase velocity {v}

and wavelength ${\lambda}$ ; let us also make the (reasonable) assumption that {b}

varies much slower in space than {u}

does (i.e. that {b}

is roughly constant at the scale of the wavelength ${\lambda}$ , so we may (for a first approximation) replace ${\frac{\partial}{\partial x}(ub)}$ by ${b \frac{\partial}{\partial x} u}$ . Heuristically, we then have

$\displaystyle \frac{\partial}{\partial x} u = O(V/\lambda)$

$\displaystyle \frac{\partial}{\partial x} h = O(A/\lambda)$

$\displaystyle \frac{\partial}{\partial t} u = O(vV/\lambda)$

$\displaystyle \frac{\partial}{\partial t} h = O(vA/\lambda)$

and equation (8) then suggests

$\displaystyle vA/\lambda \approx V b/\lambda. \ \ \ \ \ (10)$

From (7) we expect ${A \ll b}$ , and thus ${v \gg V}$ ; the wave propagates much faster than the velocity of the fluid. In particular, we expect ${u \frac{\partial}{\partial x} u = O(V^2 / \lambda)}$ to be much smaller than ${\frac{\partial}{\partial t} u = O( v V/\lambda)}$ , which explains why we expect to drop the second term in (5) to obtain (9).

If we now insert the above ansatz into (9), we obtain

$\displaystyle v V/\lambda \approx g A / \lambda;$

combining this with (10), we already get the velocity relationship (1).

To get the relation (2), we have to analyse the ansatz a bit more carefully. Specifically, we take

$\displaystyle h(t,x) = A(t,x) \sin( \phi( t, x ) )$

and

$\displaystyle u(t,x) = V(t,x) \sin( \phi( t, x ) )$

where

are slowly varying amplitude functions, and ${\phi}$ is a more rapidly varying function whose relationship with the wavelength ${\lambda}$ and velocity {v}

is (heuristically, at least) given by the formulae

$\displaystyle \frac{\partial}{\partial x} \phi = \frac{2\pi}{\lambda}$

$\displaystyle \frac{\partial}{\partial t} \phi = - v \times \frac{2\pi}{\lambda}.$

From the equation (8) we thus (approximately) get a Hamilton-Jacobi type equation

$\displaystyle \frac{\partial}{\partial t} A + \frac{\partial}{\partial x}( V b ) = 0.$

Appling (10) we conclude a transport equation for the amplitude {A}

$\displaystyle \frac{\partial}{\partial t} A + \frac{\partial}{\partial x}( v A ) = 0.$

From (1), the velocity depends only on space and not on time, and so we have

$\displaystyle \frac{\partial}{\partial t} (vA) + v \frac{\partial}{\partial x}( v A ) = 0.$

In other words, the quantity {vA}

is transported along the tsunami at the phase velocity {v}

, and so is constant over time (as one follows the tsunami):

$\displaystyle \frac{d}{dt}(vA) = 0.$

Thus

varies in inverse proportion to {v}

, giving (2).

Remark 1 It becomes difficult to retain the sinusoidal ansatz once the amplitude exceeds the depth, as it leads to the absurd conclusion that the troughs of the wave lie below the ocean floor. However, a remnant of this effect can actually be seen in real-life tsunamis, namely that if the tsunami starts with a trough rather than a crest, then the water at the shore draws back at first (sometimes for hundreds of metres), before the crest of the tsunami hits. As such, the sudden withdrawal of water of a shore is an important warning sign of an immediate tsunami.

Filed under: expository, math.AP, math.MP Tagged: incompressible Euler equations, shallow water equation, tsunami