I'm interested in knowing which type of wave a tsunami is, in a strictly mathematical sense.

I'm gonna cite from Wikipedia for this:

As the tsunami approaches the coast and the waters become shallow, wave shoaling compresses the wave and its speed decreases below 80 kilometres per hour (50 mph). Its wavelength diminishes to less than 20 kilometres (12 mi) and its amplitude grows enormously – in accord with Green's law.

and also

Except for the very largest tsunamis, the approaching wave does not break, but rather appears like a fast-moving tidal bore.

While the first property strongly hints at a tsunami being a shallow water wave, the latter fact, namely that they usually don't break would rather suggest for the tsunami being a soliton-solution of a Korteveeg-de-Vries-type equation.
I thought the latter property would not be inherent to shallow water waves although Wikipedia, again seems to argue differently, however only with a picture.

So maybe someone can clarify:

  • Which equations govern the surface profile of a tsunami? Has this model been tested against satellite measurments?
  • 1
    $\begingroup$ There are two main phases of the Tsunami. While traveling in the open ocean the tsunami will behave like a shallow-water wave. However, because it is so long, as it shoals it will behave more like a bore rather than a shallow-water wave. $\endgroup$ Commented Nov 25, 2016 at 23:22
  • $\begingroup$ @IsopycnalOscillation: Well I can see as much from wikipedia. That doesn't answer the question however. $\endgroup$ Commented Nov 26, 2016 at 14:05
  • $\begingroup$ The equation to model a shallow water wave is simply a sinusoidal function $\eta = \eta_0 \sin(k x - \omega t)$, where $\eta$ is the surface water displacement, $\eta_0$ is the wave amplitude, $k$ is the wavenumber and $\omega$ is the wave frequency. $x$ and $t$ are the horizontal direction and time, respectively. When the wave is a bore the Froude number is very large (typically larger than 1) so you most likely need numerical modeling. $\endgroup$ Commented Nov 26, 2016 at 21:15
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    $\begingroup$ I don't see any sense in this reasoning in the question: "that they usually don't break would rather suggest for the tsunami being a soliton-solution of a Korteveeg-de-Vries-type equation". The breaking of a wave depends on the transition from deep to shallow conditions. A tsunami is not a soliton. $\endgroup$
    – DrGC
    Commented Nov 28, 2016 at 15:58
  • $\begingroup$ @DrGC: Then give me the exact set of governing equations and we're done here. I don't see any sense in ranting here. $\endgroup$ Commented Nov 28, 2016 at 21:31

1 Answer 1


Short answer: Tsunami models use the shallow-wave mathematical approach, because their wavelength is usually much larger than the relevant water depth determining their propagation.

Long answer: Tsunamis are often called 'tidal waves' to highlight the idea that their characteristic time response is closer to tydes than to the standard wind waves we are used to see in a beach. The full spectrum of water wave periods is shown in this figure (tsunamis being classified as long-period waves, together with waves caused by meteorological events): enter image description hereClassification of the spectrum of ocean waves according to wave period. Redrawn from Figure 1 in: Walter H. Munk (1950) "Origin and generation of waves". Proceedings 1st International Conference on Coastal Engineering, Long Beach, California. ASCE, pp. 1–4.

This image shows that the wave frequency depends on the cause behind the perturbation of the water surface.

Oceanographers generally treat ocean waves as free surface waves in an ideal fluid [2].

Mathematically, tsunamis are surface gravity waves approached as shallow-water waves (see chapter 4.1.5 in here), meaning that the wavelength λ > 20H (H being the water depth). The involved shallow-water-wave mathematical approximations oppose to the deep-water waves (λ < H, see section 4.1.4 in the same [ref 3]).

Whereas for shallow waves the wave frequency and the wave speed are (ref. 3):

$ω = \sqrt {g k 2H}$ with $k = 2\pi / \lambda$

$c = \sqrt {g H}$

for deep-water waves, instead:

$ω = \sqrt {g k}$

$c = \sqrt {\frac{g \lambda}{2\pi}}$

As the wave approaches the shore and H decreases, the wave velocity decreases and energy is preserved by increasing the wave amplitude.

Detailed information here: 3.

  • $\begingroup$ I think it would be beneficial if you could address the question about the temporal evolution of the tsunami. $\endgroup$ Commented Nov 28, 2016 at 8:21
  • $\begingroup$ i tried to improve it $\endgroup$
    – DrGC
    Commented Nov 29, 2016 at 17:13
  • $\begingroup$ The link after "Detailed information here" appears to be broken. $\endgroup$ Commented Oct 14, 2023 at 15:51

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