Exercises 5.5 Exercises

Remark 5.5.1 Solutions and feedback.

Full solutions to these exercises can be found in the solutions to this chapter A.4 in Appendix A at the back of the book.

If you would like feedback on your work, the tutor will by default concentrate on the exercises highlighted with a yellow background in the online notes (the homework exercises), unless you specifically ask for feedback on a different exercise instead.

1. Derivation of the kinematic condition.

In this question, you will derive the kinematic boundary condition on a free surface, which ensures that fluid particles on the free surface must remain on the free surface. Consider a vector that points to a location on a 2D free surface given by

\begin{equation*} \br(s, t) = (x, \eta(x, t)). \end{equation*}

Introduce the curve $F(x, y, t) = y - \eta(x, t)\text{.}$ For any particle on the surface, it must be the case that $F(x, y, t) \equiv 0\text{.}$ By applying the material derivative show that the condition that

\begin{equation*} \DD{F}{t} = 0, \end{equation*}

for particles on the free surface is equivalent to the kinematic condition

\begin{equation*} v = \pd{\eta}{t} + u\pd{\eta}{x} \qquad \text{on $y = \eta(x, t)$}. \end{equation*}

2. Stokes waves in finite depth.

Consider small 2D water waves on the free surface of a potential (incompressible and irrotational) fluid with velocity potential $\phi(x, y,t)\text{,}$ with $\phi$ satisfying $\nabla^2 \phi = 0\text{.}$ Let the free surface be at $y = \eta(x, t)$ and assume the water is confined above a flat channel of depth $y = -h\text{.}$ Show that on the free surface, $y = \eta(x, t)\text{,}$ the kinematic equation is given by

\begin{equation*} \pd{\phi}{y}= \pd{\eta}{t} + \pd{\phi}{x}\pd{\eta}{x}, \end{equation*}

while the dynamic (Bernouilli) equation is

\begin{equation*} \pd{\phi}{t} + \frac{1}{2}|\nabla \phi|^2 + g\eta = 0. \end{equation*}

Show that when the problem is linearised by ignoring quadratic terms, then instead the boundary conditions can be simplified to be

\begin{align*} \pd{\phi}{y} \amp= \pd{\eta}{t}, \\ \pd{\phi}{t} + g\eta \amp= 0, \end{align*}

and now applied on $y = 0\text{.}$

Show that travelling harmonic waves, with $\eta = A\cos(kx - \omega t)$ and $\phi = f(y)\sin(kx - \omega t)\text{,}$ are possible and derive the associated dispersion relation,

\begin{equation*} \omega^2 = gk \tanh(kh). \end{equation*}

In deriving the above, it will be convenient for you to use the hyperbolic cosine and sine functions:

\begin{equation*} \cosh z = \frac{\e^{z} + \e^{-z}}{2} \quad \text{and} \quad \sinh z = \frac{\e^{z} - \e^{-z}}{2}, \end{equation*}

and note that the cosh function is even at $z = 0$ (and therefore has a zero derivative). The use of hyperbolic functions is used in the solution of $f(y)$ and is a common trick when solving boundary-value problems with exponentials.

3. Linear waves in a current.

We now consider linear waves in a channel of depth $h\text{,}$ as in the previous question. This time, however, assume that there is a uniform flow of speed $U$ at infinity.

(a)

Begin by writing down the boundary conditions that you expect at $y = -h$ and also far upstream, with $x \to -\infty.$

Therefore, this time, we shall consider the velocity as expressed as

\begin{equation*} u = U + \pd{\phi}{x}, \qquad v = \pd{\phi}{y}, \end{equation*}

and again linearise the necessary equations.

(b)

Linearise the associate free-surface equations, and show that this time, they produce

\begin{equation*} \pd{\phi}{y} = \pd{\eta}{t} + U \pd{\eta}{x}, \qquad \text{on $y = 0$}, \end{equation*}

and

\begin{equation*} \pd{\phi}{t} + U \pd{\phi}{x} + g\eta = 0, \qquad \text{on $y = 0$}. \end{equation*}

When considering the second of the two equations above (Bernoilli’s) return to the original Bernoulli equation:

\begin{equation*} \pd{\phi}{t} + \frac{1}{2}|\nabla \phi|^2 + \frac{p}{\rho} + gy = F(t), \end{equation*}

and choose your constant $F$ more strategically.

(c)

Show finally that the dispersion relation predicts possible waves of speeds

\begin{equation*} c_{\pm} = U \pm \left( \frac{g}{k} \tanh(kh)\right)^{1/2}. \end{equation*}

The regime where the flow speed is less than $\sqrt{gh}$ is known as subcritical. The regime where flow is greater than $\sqrt{gh}$ is supercritical. What can you anticipate about the waves in these two regimes based on the wave speeds above?

4. Fluids of different densities.

Inviscid incompressible fluid of density $\rho_2$ lies in the region $y > 0$ above a fluid with greater density $\rho_1$ for $y < 0\text{.}$ Assume there there is a small-amplitude disturbance the two fluids with the surface given by $y = \eta(x, t)\text{.}$ Assuming $\eta$ and the fluid velocities are small, assume that the linearised boundary conditions on the approximated boundary at $y = 0$ are given by

\begin{align*} \pd{\eta}{t} = \pd{\phi_1}{y} = \pd{\phi_2}{y} \amp \qquad \text{at $y = 0$}, \\ \rho_1\left( \pd{\phi_1}{t}+ g\eta\right) = \rho_2\left(\pd{\phi_2}{t}+ g\eta\right) \amp \qquad \text{at $y = 0$}. \end{align*}

(Notice that these are three conditions; two kinematic conditions for the two fluids, and a dynamic (Bernoulli) condition).

If $\eta(x, t) = A\cos(kx - \omega t)$ with $k > 0\text{,}$ derive the dispersion relation

\begin{equation*} \omega^2 = \left( \frac{\rho_1 - \rho_2}{\rho_1 + \rho_2}\right) gk. \end{equation*}

5. Fluids with surface tension.

Suppose now that there is surface tension $T$ between the two fluids of Exercise 5.5.4. In lectures, it was shown that the linear dynamic boundary condition is given by

\begin{equation*} \rho_1\left( \pd{\phi_1}{t} + g\eta\right) - \rho_2 \left( \pd{\phi_2}{t} + g\eta\right) = T \pd{^2 \eta}{x^2}, \qquad \text{on $y = 0$}. \end{equation*}

consider the specific situation where the top fluid is air, with $\rho_2 = 0\text{.}$ Then the dynamic condition is

\begin{equation*} \pd{\phi}{t} + g\eta = \frac{T}{\rho} \pd{^2 \eta}{x^2}, \qquad \text{on $y = 0$}, \end{equation*}

and only need to consider the single potential $\phi_1 = \phi\text{.}$

The linear kinematic condition should be unchanged from the usual linear kinematic condition, $\pd{\eta}{t} = v = \pd{\phi}{y}$ on $y = 0\text{.}$

For the case of deep water waves, show that the frequency $\omega$ is now related to the wavenumber $k$ by the equation

\begin{equation*} \omega^2 = gk + \frac{Tk^3}{\rho}. \end{equation*}

6. The dispersion relation (simple).

The following is a classic elementary demonstration of dispersive effects when considering the addition of two sinusoidals with two nearly-identical values of the wavenumber. It can be used to motivate the definition of group and phase velocities.

Consider two sinusoidal waves with nearly the same wavenumber and frequency:

\begin{equation*} \eta(x, t) = \sin(k_1 x - \omega_1 t) + \sin(k_2 x - \omega_2 t). \end{equation*}

(a)

Use the trig angle identity

\begin{equation*} \sin A + \sin B = 2\sin\left( \frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right), \end{equation*}

to put the wave $\eta$ into the form of a product of sinusoidal waves.

(b)

Consider now

\begin{equation*} k_{1,2} = k \pm \Delta k \quad \text{and} \quad \omega_{1,2} = \omega + \Delta \omega, \end{equation*}

i.e. each wavenumber and frequency are centred about either $k$ or $\omega\text{,}$ respectively.

In the limit that $\Delta k, \Delta \omega \ll 1$ (are small), argue that the wave consists of a rapidly oscillating waveform within a slowly-varying envelope.

(c)

show that the envelope developed in the last part exhibits maxima as

\begin{equation*} x(t) = \frac{\Delta \omega}{\Delta k} t, \end{equation*}

and therefore in the limit $\Delta k, \Delta \omega \to 0\text{,}$ the velocity of such maxima (called the group velocity) is given by

\begin{equation*} v_g = \dd{\omega}{k}. \end{equation*}

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