Section A.4 Water waves

Solutions Solutions

5.5 Exercises

5.5.1. Derivation of the kinematic condition.

Solution.

By the definition of the material derivative, we have that

\begin{equation*} \DD{(y - \eta)}{t} = \pd{(y - \eta)}{t} + (u, v) \cdot \nabla(y - \eta) = -\pd{\eta}{t} + (u, v) \cdot (-\eta_x, 1). \end{equation*}

Equating the above to zero for the case of $y = \eta(x, t)$ yields the desired result.

5.5.2. Stokes waves in finite depth.

Solution.

Note that the finite-depth linear wave theory was derived as part of the problem class, so you can examine the video there for more details. Firstly, we look to linearise the two surface boundary conditions,

\begin{gather*} \pd{\phi}{y}= \pd{\eta}{t} + \pd{\phi}{x}\pd{\eta}{x},\\ \pd{\phi}{t} + \frac{1}{2}|\nabla \phi|^2 + g\eta = 0, \end{gather*}

which are imposed on $y = \eta(x)\text{.}$ The quadratic terms can be ignored by assuming that the fluid is linearised about a small disturbance. Hence $\eta$ is assumed to be small. Similarly, the velocity terms are expanded about $y = 0\text{.}$ For example,

\begin{equation*} \pd{\phi}{x}\biggr\rvert_{y = \eta} = \pd{\phi}{x}\biggr\rvert_{y = 0} + \pd{^2\phi}{x^2}\biggr\rvert_{y = 0}\eta + \ldots. \end{equation*}

Thus we can approximate this quantity to leading order by

\begin{equation*} \pd{\phi}{x}\biggr\rvert_{y = \eta} \sim \pd{\phi}{x}\biggr\rvert_{y = 0}. \end{equation*}

Therefore, both kinematic and dynamic conditions are simplified in this manner, yielding

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

now imposed on $y = 0\text{.}$

We substitute now the ansatzes

\begin{equation*} \eta = A\cos(kx - \omega t), \qquad \phi = f(y)\sin(kx - \omega t), \end{equation*}

into the governing equations.

Written in terms of $f\text{,}$ we must now solve the following system:

\begin{gather*} f'' + k^2 f = 0, \\ f'(-h) = 0, \\ f'(0) = (\omega)A, \\ -(\omega) f(0) + gA = 0. \end{gather*}

Now instead of writing the solution for $f$ in terms of the two exponentials $\e^{\pm ky}\text{,}$ it is better to write the general solution of the second-order ODE as

\begin{equation*} f(y) = B \cosh[k(y - y_0)], \end{equation*}

where $y_0$ is constant. We can then see that if $y_0 = -h\text{,}$ then the boundary condition $f'(-h) = 0$ is satisfied.

The last two boundary conditions, evaluated on $y = 0$ yield

\begin{equation} \begin{pmatrix} \omega & -k \sinh (kh) \\ g & -(\omega)\cosh(kh) \end{pmatrix} \begin{pmatrix} A \\ B \end{pmatrix} = 0.\tag{5.5.1} \end{equation}

For there to be non-trivial solutions, we thus need

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

and therefore there are two possible wave speeds, namely

\begin{equation*} c = \mp \left(\frac{g}{k}\tanh(kh)\right)^{1/2}, \end{equation*}

and both are simply negatives of one another; so represent left and rightwards propagating waves of the same speed.

5.5.3. Linear waves in a current.

5.5.3.a

Solution.

We begin with the governing equations as shown in Theorem 5.1.1.

On the channel bottom, we expect that the vertical velocity is zero, hence

\begin{equation*} \pd{\phi}{y} = 0, \qquad y = -h, \end{equation*}

while at upstream infinity, we expect the the velocity to be uniform and of speed $U\text{.}$ Hence in terms of the potential,

\begin{equation*} \phi \sim Ux, \qquad x \to -\infty. \end{equation*}

Indeed, this suggests that we should linearise about the uniform flow, with

\begin{equation*} \phi = Ux + \hat{\phi}, \end{equation*}

where $\hat{\phi}$ is the approximation, which we expect to be small (as is standard in linear problems, we often ignore the hat notation).

5.5.3.b

Solution.

The linearisation process for the kinematic condition remains largely the same, except we would produce:

\begin{equation*} \pd{\eta}{t} + \left[ U + \pd{\hat\phi}{x}\right] \pd{\eta}{x} = \pd{\hat{\phi}}{y}. \end{equation*}

Again quadratic terms are ignored, and we see that the requisite condition becomes (dropping the hat):

\begin{equation*} \pd{\eta}{t} + U\pd{\eta}{x} = \pd{\phi}{y}, \qquad y = \eta(x, t). \end{equation*}

The linearisation of Bernoulli’s equation is done similarly.

When considering the Bernoulli equation,

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

it is prudent to take the choice of $F(t)$ so that the inertial term is removed. Specifically, if we take

\begin{equation*} F(t) = \frac{U^2}{2} + \frac{p_\text{atm}}{\rho}, \end{equation*}

this would remove the leading term from inertia.

The only difference with the usual linearisation process is the inertial term:

\begin{equation*} |\nabla \phi|^2 = \left\lvert \left(U + \pd{\hat{\phi}}{x}\right)^2 + \left(\pd{\hat{\phi}}{y}\right)^2 \right\rvert = 2U \pd{\hat\phi}{x} + \ldots \end{equation*}

so now Bernoulli’s equation yields (dropping hats):

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

where we have placed the boundary condition on $y = 0$ as standard.

5.5.3.c

Solution.

At this point, the linear solution follows similar steps to the case of finite depth with zero current. We have $\eta = A\cos(kx - \omega t)$ and $\phi = f(y) \sin(kx - \omega t)\text{.}$

Written in terms of $f\text{,}$ we must now solve the following system:

\begin{gather*} f'' + k^2 f = 0, \\ f'(-h) = 0, \\ f'(0) = (\omega - Uk)A, \\ -(\omega - Uk) f(0) + gA = 0, \end{gather*}

The first two equations yield the typical solution for the case of finite depth:

\begin{equation*} f(y) = B \cosh[k(y + h)]. \end{equation*}

The last two conditions yield

\begin{equation} \begin{pmatrix} \omega - Uk & -k \sinh (kh) \\ g & -(\omega - Uk)\cosh(kh) \end{pmatrix} \begin{pmatrix} A \\ B \end{pmatrix} = 0.\tag{5.5.2} \end{equation}

For there to be non-trivial solutions, we thus need

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

and therefore there are two possible wave speeds, namely

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

In the deep water limit, if $U < \sqrt{g/k}$ this yields subcritical solutions, and it is possible to have a wave pattern that is non-trivial upstream of the flow.

5.5.4. Fluids of different densities.

Solution.

The kinematic condition(s) must be exactly the same form as the usual kinematic condition, since the density of the fluids play no role. They arise from setting $\DD{(y - \eta)}{t} = 0,$ for both fluids. This gives,

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

after linearising, and as imposed on $i = 1, 2\text{.}$

The dynamic condition expresses the fact that the pressures are equal at the interface between the two fluids. Therefore at the interface, we have (ignoring quadratic terms), from Theorem 3.5.9

\begin{align*} \pd{\phi_1}{t} + \frac{p_1}{\rho_1} + g\eta \amp= B_1, \\ \pd{\phi_2}{t} + \frac{p_2}{\rho_2} + g\eta \amp= B_2. \end{align*}

However at the interface, we need the pressures to be equal. Therefore we can set $p_1 = p_2\text{.}$ Subtracting the two equations then yields

\begin{equation*} \rho_1\left( \pd{\phi_1}{t}+ g\eta\right) + \left( \frac{p}{\rho_1} - \frac{p}{\rho_2} - B_1 + B_2\right) = \rho_2\left(\pd{\phi_2}{t}+ g\eta\right) \end{equation*}

as applied on $y = 0\text{.}$ We are free to pick the constants, so we pick them so as to completely zero the expression in the second group of brackets. Thus

\begin{equation*} \rho_1\left( \pd{\phi_1}{t}+ g\eta\right) = \rho_2\left(\pd{\phi_2}{t}+ g\eta\right) \end{equation*}

on $y = 0\text{.}$

So now the natural question is to ask how the linearisation procedure changes with the two fluids of different densities. Gain, we set $\eta = A\cos(kx - \omega t)$ and $\phi_i = f_i(y) \sin(kx - \omega t)\text{.}$ The equation for $f$ is similarly given by $f'' = k^2 f,$ and this time, we must select different exponential powers so as to ensure the fluid is bounded either as $y \to \infty$ or as $y \to -\infty\text{.}$ This essentially yields

\begin{align*} f_1 = B \e^{ky}, \amp \qquad y < 0, \\ f_2 = C \e^{-ky}, \amp \qquad y > 0, \end{align*}

for constants $B$ and $C\text{.}$

The remaining boundary conditions are:

\begin{gather*} f_1'(0) = (\omega)A, \\ f_2'(0) = (\omega)A, \\ \rho_1(-\omega f_1(0) + gA) = \rho_2(-\omega f_2(0) + gA), \end{gather*}

thus

\begin{gather*} Bk = \omega A, \\ -Ck = \omega A, \\ \rho_1(-\omega B + gA) = \rho_2(-\omega C + gA), \end{gather*}

which in matrix form is

\begin{equation*} \begin{pmatrix} \omega \amp -k \amp 0 \\ \omega \amp 0 \amp k \\ g(\rho_1 - \rho_2) \amp -\omega\rho_1 \amp \omega \rho_2 \end{pmatrix} \begin{pmatrix} A \\ B \\ C \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}. \end{equation*}

Expanding the determinant about the first row yields

\begin{align*} \omega \begin{vmatrix} 0 \amp k \\ -\omega \rho_1 \amp \omega \rho_2 \end{vmatrix} + k \begin{vmatrix} \omega \amp k \\ g(\rho_1 - \rho_2) \amp \rho_2 \omega \end{vmatrix}\\ = \omega^2 k (\rho_1 + \rho_2) - k^2 g (\rho_1 - \rho_2) = 0, \end{align*}

which gives the desired result.

5.5.5. Fluids with surface tension.

Solution.

This question was done in the problem class recording on 11 November 2025, accessed here.

Very little changes in the linearisation procedure with surface tension, as the only difference is the Bernoulli equation, which is linearised to give

\begin{equation*} \pd{\phi}{t} + g\eta = \frac{T}{\rho} \pd{^2 \eta}{x^2}, \end{equation*}

evaluated on $y = 0\text{.}$

Thus in this equation, we use the same ansatzes as for the previous few questions. The system of equations is

\begin{gather*} f'' + k^2 f = 0, \\ f(-\infty) \text{ is bounded}, \\ f'(0) = (\omega)A, \\ -(\omega) f(0) + [g + (T/\rho)k^2] A = 0, \end{gather*}

The derivation must be exactly the same as for infinite-depth gravity waves, and the only difference is that instead of $g\text{,}$ we can substitute $g + (T/\rho)k^2$ when we impose the boundary conditions and apply the determinant condition.

Previously for deep-water gravity waves, we had $\omega^2 = gk\text{.}$ So now we must have

\begin{equation*} \omega^2 = \left(g + \frac{T}{\rho} k^2\right) k, \end{equation*}

as desired.

5.5.6. The dispersion relation (simple).

5.5.6.a

Solution.

\begin{equation*} \eta(x, t) = 2\sin \left[ \frac{(k_1 + k_2)x - (\omega_1 + \omega_2)t}{2}\right] \cos \left[ \frac{(k_1 - k_2)x - (\omega_1 - \omega_2)t}{2}\right] \end{equation*}

5.5.6.b

Solution.

Substituting these expressions, we have

\begin{equation*} \eta = 2\sin(kx - \omega t)\cos(\Delta k x - \Delta \omega t). \end{equation*}

The point here is that $\Delta k, \Delta \omega \ll 1\text{,}$ so the cosine term is very nearly 1. This yields the standard travelling sinusoidal wave. However, over long timescales, notably when $t$ is large, the cosine term is no longer constant. This yields a slowly varying envelope or amplitude multiplying the sine,

5.5.6.c

Solution.

This part is quite straightforward. The maxima of the cosines are found at multiples of $2\pi$ in the argument. For example, the first maxima lies at $\Delta k x - \Delta \omega t = 0,$ hence $x(t) = (\Delta k/\Delta \omega)t.$

This is indeed equivalent to a particle that moves with velocity $\Delta k/\Delta \omega\text{.}$ In the limit the perturbations to the quantities tends to zero, we can informally note that this leads to the notion of the group velocity,

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

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