Math 575-Lecture 25 Deep Water Waves, Dispersion Relation

Home , Dispersion (water waves), Free surface

Math 575-Lecture 25

The full 2D equations describing the water are given by the Euler equations 1 ∂ u + u∂ u + v∂ u = − ∂ p, t x y ρ x 1 ∂ v + u∂ v + v∂ v = − ∂ p − g, t x y ρ y

∂xu + ∂yv = 0. We aim to ﬁnd the equations for the surface waves. Assume that the water is deep and the ﬂows are irrotational: u = ∇φ. Then, the Euler equations are reduced to the Bernoulli equations 1 ∂ φ + |u|2 + (p/ρ + gy) = 0, t 2 and

∆φ = 0.

At the surface y = h(x, t), we have

v = ht + uhx, at y = h(x, t). and if we impose p = 0, 1 ∂ φ + (u2 + v2) + gh = 0, at y = h(x, t). t 2 Since the water is deep, h = h0 + η, with = η/h0 1. This is called the small amplitude assumption. By throwing away a totally time dependent constant in φ, we are able to keep the leading order eﬀects:

2 v(x, h0 + η) = ht + O( ), 2 ∂tφ(x, h0 + η) + gη + O( ) = 0 Further, by Taylor expansion, we ﬁnd to leading order,

∂yφ(x, h0, t) = v(x, h0) = ηt, ∂tφ(x, h0, t) + gη = 0

1 The deep water is then described by

∂tφ(x, t) + gη = 0, y = h0

ηt = ∂yφ(x, t), y = h0,

φxx + φyy = 0. Inserting the ansatz η = A sin(kx − ωt), we ﬁnd the dispersion relation ω2 = g|k|. The energy is tranported with group velocity dω v = g dk If k ∼ 0, the wavelength is big, the approximation we did p = 0 at y = h is reasonable and the equations we have are good. Such water waves are called the gravity waves. If k ∼ ∞, p = 0 at y = h is no longer good. The surface tension matters . The ﬁrst equation should be revised to T ∂ φ(x, t) + gη − η = 0. t ρ xx These waves are called ‘capillary waves’. This is the case when a drop falls onto the surface of lake. The dispersion relation is then ω2 = g|k| + T k2/ρ.

Shallow water equations

Consider the water above the ground y = 0. The free surface is given by y = h(x, t). h0 is the typical value of h. We are not assuming that the amplitude of the wave is small. Instead, we assume h0/L 1 where L is the typical length of the domain (or wavelength). This is the shallowness assumption. Consider the Euler equations again: 1 ∂ u + u∂ u + v∂ u = − ∂ p, t x y ρ x 1 ∂ v + u∂ v + v∂ v = − ∂ p − g, t x y ρ y

∂xu + ∂yv = 0.

2 0 0 0 0 As before, we scale x, y by x = x/L and y = y/h0. u = u/U and v = v/(Uh0/L), 0 √ t = t/(L/U), U = gh0. We assume we are considering the regime where all prime numbers are O(1). By ﬁnding the leading order terms, we have 1 ∂ u + u∂ u + v∂ u = − ∂ p, t x y ρ x 1 0 = − ∂ p − g, ρ y

∂xu + ∂yv = 0.

By the second equation: p = p0 − ρg(y − h(x, t)) Hence, the ﬁrst equation is reduced to

∂tu + u∂xu + v∂yu = −ghx

Consider that u(x, y, 0) = u0(x), i.e. it does not depend on y. Then, according to the above equation, u(x, y, t) = u(x, t) for all time. Hence,

∂tu + u∂xu = −ghx

By the incompressibility condition,

v = −∂xuy, because v = 0 for y = 0. Lastly, consider a ﬂuid particle y = h(x, t). It then implies v = ht + uhx for y = h(x, t), or ht + uhx + ∂xuh(x, t) = 0. We then have the Shallow water equations

ut + uux + ghx = 0,

ht + uhx + uxh(x, t) = 0.

In conservation law form

2 ut + (u /2 + gh)x = 0,

ht + (uh)x = 0. or

(ρht) + (ρhu)x = 0, g (ρh2) + (ρhu2) + ( (ρh)2) = 0. t x 2ρ x

3 √ Letting c = gh, we can rewrite the equations into

(∂t + (u + c)∂x)(u + 2c) = 0,

(∂t + (u − c)∂x)(u − 2c) = 0.

Such kind of nonlinear hyperbolic systems can develop discontinuity, called ‘shock’ even though the initial value is smooth.