ASTR/ATOC-5410: Fluid Instabilities, Waves, and Turbulence October 30, 2016, Axel Brandenburg

Handout 14b: Nonlinear Water Waves (cont’d)

The KdV equation can be derived under the restrictions of shallow water (wavelength longer ℓ long compared with depth h) and small (but finite) amplitude a compared with h, i.e.,

a ≪ h ≪ ℓ.  (1)

One assumes an inviscid irrotational flow u = ∇ϕ, so the governing equation is the Bernoulli equation,

∂_{t}ϕ + \frac{1}{2}(∇ϕ)^{2} + P/ρ + gz = 0,  ∇^{2}ϕ = 0.  (2)

We first need to discuss boundary conditions.

1 Boundary conditions

At rest, the water surface is assumed to be at z =, so the bottom is at z = −h, and the normal velocity vanishes there, so

ϕ_{,z} = 0  (at z = −h).  (3)

Next, the surface is assumed to be at z = ζ. Since the pressure vanishes zero, we have

∂_{t}ϕ^{s} + \frac{1}{2}(∇ϕ^{s})^{2} + gζ = 0.  (4)

where the superscript s refers to the surface. The location of the surface is described by the function ζ = ζ(x, t), and we assume that

Dζ/Dt = u_{z} ≡ ϕ_{,z}.  (5)

Since Dζ/Dt = ∂_{t}ζ + u_{x}ζ_{,x} = ∂_{t}ζ + ϕ_{,x}ζ_{,x}, we can also write

∂_{t}ζ + ϕ_{,x}ζ_{,x} = ϕ_{,z}  (at the surface).  (6)

We note in passing that the linearized form of the two equations can be combined to ∂_{t}^{2}ϕ + gϕ_{,z} = 0, which is an equation we have encountered in handout 11, see Eq. (16) of that handout.

2 Linear wave solutions

Figure 1: Dispersion relation. The dashed line shows the result without the tanh factor. Assuming wave-like solutions of the form

ϕ = f(z) sin(kx − ωt),  (7)

which satisfy ∇^{2}ϕ = 0, the f has to be of the form f = f_{1}e^{kz} + f_{2}e^{−kz}. To obey f_{,z} = 0, we have to have f_{1}ke^{kz} − f_{2}ke^{−kz} = 0 at z = −h, and thus f_{1}ke^{−kh} − f_{2}ke^{kh} = 0, or f_{2}/f_{1} = e^{−2kh} and thus

f(z) = A (e^{kz} + e^{−kz−2kh}) = Ae^{−kh} (e^{k(z+h)} + e^{−k(z+h)}) , (8)

and therefore f(z) = 2Ae^{−kh} cosh k(z + h). Inserting ϕ = 2Ae^{−kh} cosh k(z + h) sin(kx − ωt) into ∂_{t}^{2}ϕ + gϕ_{,z} = 0, and noting that f_{,z} = 2Ake^{−kh} sinh k(z + h), we have

−ω^{2} 2Ae^{−kh} cosh k(z + h) sin(kx − ωt) + 2Agk e^{−kh} cosh k(z + h) sin(kx − ωt) = 0,  (9)

and therefore ω^{2} = gk tanh k(z + h) at the surface at z = 0, so

ω^{2} = gk tanh kh.  (10)

This is shown in Figure 1 and compared with ω^{2} = gk.

3 Perturbative nonlinear wave equation

We now make the following ansatz for ϕ, which must obey ∇^{2}ϕ = 0,

ϕ = ∑_{n=0}^{∞} z^{n}ϕ_{n}(x, t).  (11)

To work with this, it is useful to write out the first few terms:

ϕ = ϕ_{0} + zϕ_{1} + z^{2}ϕ_{2} + z^{3}ϕ_{3} + z^{4}ϕ_{4} + ...  (12)

In order that this satisfies ∇^{2}ϕ = 0, let us write down the second x and z derivatives,

ϕ_{,xx} = ϕ_{0,xx} + zϕ_{1,xx} + z^{2}ϕ_{2,xx} + z^{3}ϕ_{3,xx} + z^{4}ϕ_{4,xx} + ...  (13)
ϕ_{,zz} = 2ϕ_{2} + 3 · 2 zϕ_{3} + 4 · 3 z^{2}ϕ_{4} + ...  (14)

Matching equal powers of z leads to the following recursive relations

ϕ_{0,xx} + 2 ϕ_{2} = 0,  (15)
ϕ_{1,xx} + 3 · 2 ϕ_{3} = 0,  (16)
ϕ_{2,xx} + 4 · 3 ϕ_{4} = 0,  (17)

or, more generally

ϕ_{n,xx} + (n + 2)(n + 1) ϕ_{n+2} = 0.  (18)

Next, making use of the bottom boundary condition u_{z} = 0, i.e., ϕ_{,z} = 0, we find ϕ_{1} = 0, and, because of Equation (18), all odd terms vanish, i.e., ϕ_{3} = ϕ_{5} = ... = 0. With this, we can now write ϕ as

ϕ = ϕ_{0} + z^{2} ϕ_{2} + z^{4} ϕ_{4} + ...  (19)

Inserting the recursive relations, we have ϕ_{2} = −\frac{1}{2}ϕ_{0,xx} and

ϕ_{4} = −\frac{ϕ_{2,xx}}{4 · 3} = +\frac{ϕ_{0,xxxx}}{4!} .  (20)

Let us use φ ≡ ϕ_{0} as a shorthand, and so

ϕ = φ − \frac{1}{2!} z^{2} φ′′ + \frac{1}{4!} z^{4} φ^{(iv)} − ...  (21)

With this, we find

u_{x} = ϕ_{,x} = φ′ − \frac{1}{2!} z^{2} φ′′′ + \frac{1}{4!} z^{4} φ^{(v)} − ...  (22)

and

u_{z} = ϕ_{,z} = −z φ′′ + \frac{1}{3!} z^{3} φ^{(iv)} − ...  (23)

This has to be inserted back into the full time-dependent equation (2). Furthermore, to obey the ordering (1), we define ϵ = A/h for the amplitude A and δ = (h/ℓ)^{2} for the height. Solving order by order, one arrives eventually at the equation

ζ_{,t} + cζ_{,x} + \frac{3}{2} \frac{c}{h}ζζ_{,x} + \frac{1}{6} ch^{2} ζ_{,xxx} = 0,  (24)

to lowest order. Here, c = \sqrt{gh} is the wave speed, and cζ_{,x} is just an advection term that can be removed by going into a comoving frame to obtain the KdV equation in its usual form. Higher derivatives would occur at higher order expansions.

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