Dispersion relation
Contents
1. Phase velocity
\[ \eta = a\sin\left(\frac{2\pi}{T}t - \frac{2\pi}{\lambda}x\right) \]
With \(\omega = \dfrac{2\pi}{T}\) (angular frequency) and \(k = \dfrac{2\pi}{\lambda}\) (wavenumber):
\[ \eta = a\sin(\omega t - kx) \]
The argument of the sine wave is the phase:
\[ \varphi = \omega t - kx \]
(A constant added to the phase, \(\omega t - kx + \alpha\), only shifts the wave along \(x\); it does not change its speed.)
The sine repeats every \(2\pi\):
\[ \sin(\varphi + 2\pi n) = \sin(\varphi) \]
\(\rightarrow\) the elevation \(\eta\) depends only on the phase. A crest is a point of fixed phase; it stays a crest as long as its phase does not change.
Follow one crest:
- at time \(t_1\) it is at \(x_1\)
- at time \(t_2\) it is at \(x_2\)
Same crest \(\rightarrow\) same phase at \((x_1, t_1)\) and \((x_2, t_2)\):
\[ \omega t_1 - kx_1 = \omega t_2 - kx_2 \]
\[ k(x_2 - x_1) = \omega(t_2 - t_1) \]
\[ \frac{x_2 - x_1}{t_2 - t_1} = \frac{\omega}{k} \]
\[ x_2 = x_1 + \frac{\omega}{k}(t_2 - t_1) \]
The crest moves a distance \(\dfrac{\omega}{k}(t_2 - t_1)\) in a time \(t_2 - t_1\) \(\rightarrow\) its speed is:
\[ c = \frac{x_2 - x_1}{t_2 - t_1} = \frac{\omega}{k} \quad \rightarrow \quad \text{phase velocity} \]
With \(\omega = \dfrac{2\pi}{T}\) and \(k = \dfrac{2\pi}{\lambda}\):
\[ c = \frac{2\pi/T}{2\pi/\lambda} = \frac{\lambda}{T} \]
\(\rightarrow\) the crest travels one wavelength in one period.
Example:
\[ \eta_1 = \sin(\omega t - kx), \qquad \eta_2 = \sin(kx - \omega t) \]
Which moves in which direction?
\(\eta_1\): constant phase \(\omega t - kx\) \(\rightarrow\) \(x_2 = x_1 + \dfrac{\omega}{k}(t_2 - t_1)\) \(\rightarrow\) \(x\) increases with time \(\rightarrow\) moves in \(+x\).
\(\eta_2\): constant phase \(kx - \omega t\):
\[ kx_1 - \omega t_1 = kx_2 - \omega t_2 \quad \rightarrow \quad x_2 = x_1 + \frac{\omega}{k}(t_2 - t_1) \]
\(\rightarrow\) also moves in \(+x\). \(\;\sin(kx - \omega t) = -\sin(\omega t - kx)\): the same wave, shifted by half a wavelength.
A wave moving in \(-x\) has the phase \(\omega t + kx\):
\[ \omega t_1 + kx_1 = \omega t_2 + kx_2 \quad \rightarrow \quad x_2 = x_1 - \frac{\omega}{k}(t_2 - t_1) \]
2. Continuity and bed condition
\[ \vec{v}(x, z, t) = u(x, z, t)\,\vec{i} + w(x, z, t)\,\vec{k} \]
\[ \frac{\partial u}{\partial x} + \frac{\partial w}{\partial z} = 0 \qquad\qquad \nabla^2\phi = 0 \]
\[ w(x, z = -h, t) = \frac{\partial\phi}{\partial z}(x, z = -h, t) = 0 \]
3. Kinematic boundary condition
The free surface moves with the fluid, so the vertical component of a fluid particle on the water’s surface equals the rate of change of the free-surface elevation at that location:
\[ \frac{D}{Dt}\big(z - \eta(x, t)\big) = 0 \qquad \text{on } z = \eta(x, t) \]
\[ \frac{\partial(z - \eta)}{\partial t} + u\,\frac{\partial(z - \eta)}{\partial x} + w\,\frac{\partial(z - \eta)}{\partial z} = 0 \]
\[ \frac{\partial z}{\partial t} = 0 \qquad \frac{\partial z}{\partial x} = 0 \qquad \frac{\partial z}{\partial z} = 1 \]
\[ -\frac{\partial\eta}{\partial t} - u\,\frac{\partial\eta}{\partial x} + w = 0 \]
\[ w = \frac{\partial\eta}{\partial t} + u\,\frac{\partial\eta}{\partial x} \quad \rightarrow \quad \text{kinematic boundary condition} \]
This expresses the physical condition that a fluid particle at the surface should remain there at all times → kinematic condition.
4. Dynamic condition
The pressure at the free surface is atmospheric. This needs the unsteady Bernoulli equation, which follows from Newton’s second law applied to a fluid element (Euler’s equations).
4.1 Euler’s equations
Fluid element: sides \(dx\) and \(dz\), unit width.
\[ m = \rho\,dx\,dz \]
Forces in \(x\) (inviscid: pressure only):
- left face, at \(x\): \(\;p\,dz\) in \(+x\)
- right face, at \(x + dx\): \(\;\left(p + \dfrac{\partial p}{\partial x}dx\right)dz\) in \(-x\)
\[ F_x = p\,dz - \left(p + \frac{\partial p}{\partial x}dx\right)dz = -\frac{\partial p}{\partial x}\,dx\,dz \]
Forces in \(z\): pressure on the bottom and top faces, and gravity:
\[ F_z = p\,dx - \left(p + \frac{\partial p}{\partial z}dz\right)dx - \rho g\,dx\,dz = -\frac{\partial p}{\partial z}\,dx\,dz - \rho g\,dx\,dz \]
Acceleration: in a time \(dt\) the element moves from \((x, z)\) to \((x + u\,dt,\ z + w\,dt)\):
\[ du = u(x + u\,dt,\ z + w\,dt,\ t + dt) - u(x, z, t) \]
\[ du = \left[u(x, z, t) + \frac{\partial u}{\partial x}\,u\,dt + \frac{\partial u}{\partial z}\,w\,dt + \frac{\partial u}{\partial t}\,dt\right] - u(x, z, t) \]
\[ a_x = \frac{du}{dt} = \frac{\partial u}{\partial t} + u\frac{\partial u}{\partial x} + w\frac{\partial u}{\partial z} \]
Same steps for \(w\):
\[ a_z = \frac{dw}{dt} = \frac{\partial w}{\partial t} + u\frac{\partial w}{\partial x} + w\frac{\partial w}{\partial z} \]
\(m\,a = F\), divided by \(\rho\,dx\,dz\):
\[ \frac{\partial u}{\partial t} + u\frac{\partial u}{\partial x} + w\frac{\partial u}{\partial z} = -\frac{1}{\rho}\frac{\partial p}{\partial x} \]
\[ \frac{\partial w}{\partial t} + u\frac{\partial w}{\partial x} + w\frac{\partial w}{\partial z} = -\frac{1}{\rho}\frac{\partial p}{\partial z} - g \]
4.2 Unsteady Bernoulli equation
Irrotational flow: \(\;\dfrac{\partial u}{\partial z} = \dfrac{\partial w}{\partial x}\), \(\;u = \dfrac{\partial\phi}{\partial x}\), \(\;w = \dfrac{\partial\phi}{\partial z}\)
\(x\)-equation. Replace \(\dfrac{\partial u}{\partial z}\) by \(\dfrac{\partial w}{\partial x}\):
\[ u\frac{\partial u}{\partial x} + w\frac{\partial u}{\partial z} = u\frac{\partial u}{\partial x} + w\frac{\partial w}{\partial x} \]
Chain rule, \(\dfrac{\partial}{\partial x}\left(\dfrac{1}{2}u^2\right) = u\dfrac{\partial u}{\partial x}\) and \(\dfrac{\partial}{\partial x}\left(\dfrac{1}{2}w^2\right) = w\dfrac{\partial w}{\partial x}\):
\[ u\frac{\partial u}{\partial x} + w\frac{\partial w}{\partial x} = \frac{\partial}{\partial x}\left[\frac{1}{2}\left(u^2 + w^2\right)\right] \]
Time derivative: \(\;\dfrac{\partial u}{\partial t} = \dfrac{\partial}{\partial t}\dfrac{\partial\phi}{\partial x} = \dfrac{\partial}{\partial x}\dfrac{\partial\phi}{\partial t}\)
Every term is now an \(x\)-derivative; \(gz\) does not depend on \(x\), so it can be added inside:
\[ \frac{\partial}{\partial x}\left[\frac{p}{\rho} + \frac{\partial\phi}{\partial t} + \frac{1}{2}\left(u^2 + w^2\right) + gz\right] = 0 \]
\(z\)-equation. Replace \(\dfrac{\partial w}{\partial x}\) by \(\dfrac{\partial u}{\partial z}\), then the chain rule as above:
\[ u\frac{\partial w}{\partial x} + w\frac{\partial w}{\partial z} = u\frac{\partial u}{\partial z} + w\frac{\partial w}{\partial z} = \frac{\partial}{\partial z}\left[\frac{1}{2}\left(u^2 + w^2\right)\right] \]
\(\dfrac{\partial w}{\partial t} = \dfrac{\partial}{\partial z}\dfrac{\partial\phi}{\partial t}\) \(\;\) and \(\;\) \(g = \dfrac{\partial(gz)}{\partial z}\):
\[ \frac{\partial}{\partial z}\left[\frac{p}{\rho} + \frac{\partial\phi}{\partial t} + \frac{1}{2}\left(u^2 + w^2\right) + gz\right] = 0 \]
The bracket varies with neither \(x\) nor \(z\) \(\rightarrow\) it depends on \(t\) only:
\[ \frac{p}{\rho} + \frac{\partial\phi}{\partial t} + \frac{1}{2}\left(u^2 + w^2\right) + gz = C(t) \]
Remove \(C(t)\): \(\;\Phi = \phi - \displaystyle\int C(t)\,dt\) gives the same \(u\) and \(w\), and \(\dfrac{\partial\Phi}{\partial t} = \dfrac{\partial\phi}{\partial t} - C(t)\). Write \(\phi\) for \(\Phi\):
\[ \frac{p}{\rho} + \frac{\partial\phi}{\partial t} + \frac{1}{2}\left(u^2 + w^2\right) + gz = 0 \]
4.3 Dynamic boundary condition
At the free surface, \(z = \eta\):
\[ \frac{p}{\rho} + \frac{\partial\phi}{\partial t} + \frac{1}{2}\left(u^2 + w^2\right) + g\eta = 0 \]
The pressure there is atmospheric, taken as zero (gauge): \(\;p = 0\).
5. Linearised conditions
For small amplitudes: \(a \ll \lambda\)
\[ \frac{\partial\eta}{\partial x} = 0 \quad \rightarrow \quad \text{linearised kinematic conditions} \]
\[ \boxed{\;w = \frac{\partial\eta}{\partial t}\;} \;\xrightarrow{\;\text{assume } a\,\ll\,1\;}\; w(x, z, t) \rightarrow \boxed{\;w(x, 0, t) = \frac{\partial\eta(x, t)}{\partial t}\;} \]
Linearised dynamic condition:
\[ \frac{\partial\phi}{\partial t} + g\eta = 0 \quad \rightarrow \quad \frac{\partial\phi}{\partial t} = -g\eta \]
6. Summary of conditions
\[ \frac{\partial^2\phi}{\partial x^2} + \frac{\partial^2\phi}{\partial z^2} = 0, \qquad -h \le z \le \eta, \quad \text{subject to:} \tag{1} \]
\[ \frac{\partial\phi(x, 0, t)}{\partial t} = -g\,\eta(x, t) \tag{I} \]
\[ \frac{\partial\eta}{\partial t} = w(x, 0, t) \tag{II} \]
\[ \frac{\partial\phi}{\partial z}(x, z = -h, t) = 0 \tag{III} \]
7. Solution
Assume we can write:
\[ \phi = A(z)\sin(\omega t - kx) \]
Insert into (1). First the \(x\)-derivatives. \(A(z)\) does not depend on \(x\); chain rule on \(\sin(\omega t - kx)\), with \(\dfrac{\partial}{\partial x}(\omega t - kx) = -k\):
\[ \frac{\partial\phi}{\partial x} = A(z)\cos(\omega t - kx)\cdot(-k) = -kA(z)\cos(\omega t - kx) \]
\[ \frac{\partial^2\phi}{\partial x^2} = -kA(z)\cdot\big(-\sin(\omega t - kx)\big)\cdot(-k) = -k^2A(z)\sin(\omega t - kx) \]
Then the \(z\)-derivatives. \(\sin(\omega t - kx)\) does not depend on \(z\):
\[ \frac{\partial\phi}{\partial z} = A'(z)\sin(\omega t - kx), \qquad \frac{\partial^2\phi}{\partial z^2} = A''(z)\sin(\omega t - kx) \]
Add them, as in (1):
\[ \left[-k^2A(z) + A''(z)\right]\sin(\omega t - kx) = 0 \tag{2} \]
\(\sin(\omega t - kx)\) is not zero for all \(x\) and \(t\) \(\;\rightarrow\;\) the bracket must be zero:
\[ A''(z) - k^2A(z) = 0 \]
(2) is an ODE for \(A(z)\): second order, linear, constant coefficients.
Try an exponential. \(A(z) = e^{mz}\), with \(m\) unknown:
\[ A'(z) = m\,e^{mz}, \qquad A''(z) = m^2e^{mz} \]
Substitute:
\[ m^2e^{mz} - k^2e^{mz} = 0 \;\rightarrow\; e^{mz}\left(m^2 - k^2\right) = 0 \]
\(e^{mz}\) is never zero \(\;\rightarrow\;\) \(m^2 - k^2 = 0\) \(\;\rightarrow\;\) \(m = +k\) or \(m = -k\).
Two solutions, \(e^{kz}\) and \(e^{-kz}\). Any sum of the two, each times a constant, is also a solution:
\[ A(z) = D_1e^{kz} + D_2e^{-kz} \]
Check: \(A''(z) = k^2D_1e^{kz} + (-k)^2D_2e^{-kz} = k^2\left(D_1e^{kz} + D_2e^{-kz}\right) = k^2A(z)\) ✓
Write it about the bed. \(e^{kz} = e^{-kh}\,e^{k(z + h)}\) and \(e^{-kz} = e^{kh}\,e^{-k(z + h)}\):
\[ A(z) = D_1e^{-kh}\,e^{k(z + h)} + D_2e^{kh}\,e^{-k(z + h)} \]
\(D_1e^{-kh}\) and \(D_2e^{kh}\) are just constants; call them \(E_1\) and \(E_2\):
\[ A(z) = E_1e^{k(z + h)} + E_2e^{-k(z + h)} \]
Apply (III). \(\dfrac{\partial\phi}{\partial z} = A'(z)\sin(\omega t - kx) = 0\) at \(z = -h\) for all \(x\) and \(t\) \(\;\rightarrow\;\) \(A'(-h) = 0\).
\[ A'(z) = kE_1e^{k(z + h)} - kE_2e^{-k(z + h)} \]
At \(z = -h\): \(\;z + h = 0\), \(\;e^0 = 1\):
\[ A'(-h) = kE_1 - kE_2 = 0 \;\rightarrow\; E_1 = E_2 \]
\[ A(z) = E_1\left[e^{k(z + h)} + e^{-k(z + h)}\right] \]
\(\cosh X = \dfrac{e^{X} + e^{-X}}{2}\) \(\;\rightarrow\;\) \(e^{X} + e^{-X} = 2\cosh X\), with \(X = k(z + h)\):
\[ A(z) = 2E_1\cosh\big(k(z + h)\big) \]
Write \(C_1 = 2E_1\):
\[ A(z) = C_1\cosh\big(k(z + h)\big) \]
This is the general solution \(A(z) = C_1\cosh(kz + C_2)\) with \(C_2 = kh\), the value that satisfies (III).
\[ \phi(x, z, t) = C_1\cosh\big(k(z + h)\big)\sin(\omega t - kx) \tag{3} \]
8. Dispersion relation
\[ \eta(x, t) = -\frac{1}{g}\frac{\partial}{\partial t}\phi(x, z = 0, t) = -\frac{\omega}{g}\,C_1\cosh(kh)\cos(\omega t - kx) \tag{4} \]
\[ \frac{\partial\eta}{\partial t} = +\frac{\omega^2}{g}\,C_1\cosh(kh)\sin(\omega t - kx) \tag{5} \]
Also,
\[ w(x, z = 0, t) = \frac{\partial\phi}{\partial z}(x, z = 0, t) \]
\[ = \frac{\partial}{\partial z}\Big(C_1\cosh\big(k(z + h)\big)\sin(\omega t - kx)\Big)\bigg|_{z = 0} \]
\[ = C_1\,k\sinh(kh)\sin(\omega t - kx) \tag{6} \]
But (5) and (6) must be equal, then
\[ \frac{\omega^2}{g}\,C_1\cosh(kh)\sin(\omega t - kx) = C_1\,k\sinh(kh)\sin(\omega t - kx) \]
\[ \boxed{\;\omega^2 = gk\tanh(kh)\;} \tag{7} \]
9. Velocity components
From (4), set the amplitude of \(\eta\) equal to \(a\):
\[ a = -\frac{\omega}{g}\,C_1\cosh(kh) \quad \rightarrow \quad C_1 = -\frac{ga}{\omega\cosh(kh)} \]
\[ \eta(x, t) = a\cos(\omega t - kx) \tag{8} \]
Plug \(C_1\) into (3):
\[ \phi(x, z, t) = -\frac{ga}{\omega}\,\frac{\cosh\big(k(z + h)\big)}{\cosh(kh)}\,\sin(\omega t - kx) \tag{9} \]
9.1 Horizontal velocity
\[ u = \frac{\partial\phi}{\partial x} = -\frac{ga}{\omega}\,\frac{\cosh\big(k(z + h)\big)}{\cosh(kh)}\,\frac{\partial}{\partial x}\sin(\omega t - kx) \]
\[ \frac{\partial}{\partial x}\sin(\omega t - kx) = -k\cos(\omega t - kx) \]
\[ u(x, z, t) = \frac{gak}{\omega}\,\frac{\cosh\big(k(z + h)\big)}{\cosh(kh)}\,\cos(\omega t - kx) \tag{10} \]
9.2 Vertical velocity
\[ w = \frac{\partial\phi}{\partial z} = -\frac{ga}{\omega}\,\frac{\sin(\omega t - kx)}{\cosh(kh)}\,\frac{\partial}{\partial z}\cosh\big(k(z + h)\big) \]
\[ \frac{\partial}{\partial z}\cosh\big(k(z + h)\big) = k\sinh\big(k(z + h)\big) \]
\[ w(x, z, t) = -\frac{gak}{\omega}\,\frac{\sinh\big(k(z + h)\big)}{\cosh(kh)}\,\sin(\omega t - kx) \tag{11} \]
9.3 Using \(\omega^2 = gk\tanh(kh)\)
\[ gk = \frac{\omega^2}{\tanh(kh)} = \frac{\omega^2\cosh(kh)}{\sinh(kh)} \]
\[ \frac{gk}{\omega\cosh(kh)} = \frac{\omega^2\cosh(kh)}{\omega\cosh(kh)\sinh(kh)} = \frac{\omega}{\sinh(kh)} \]
\[ u(x, z, t) = a\omega\,\frac{\cosh\big(k(z + h)\big)}{\sinh(kh)}\,\cos(\omega t - kx) \tag{12} \]
\[ w(x, z, t) = -a\omega\,\frac{\sinh\big(k(z + h)\big)}{\sinh(kh)}\,\sin(\omega t - kx) \tag{13} \]
Check (II): from (13) at \(z = 0\), \(\;w(x, 0, t) = -a\omega\sin(\omega t - kx)\). From (8), \(\;\dfrac{\partial\eta}{\partial t} = -a\omega\sin(\omega t - kx)\) ✓
Check (III): from (13) at \(z = -h\), \(\;\sinh(0) = 0 \;\rightarrow\; w(x, -h, t) = 0\) ✓
9.4 Deep water
\(kh\) large \(\;\rightarrow\; \tanh(kh) \rightarrow 1\):
\[ \omega^2 = gk \tag{14} \]
\[ \cosh X = \frac{e^{X} + e^{-X}}{2}, \qquad \sinh X = \frac{e^{X} - e^{-X}}{2} \]
For \(kh\) large, \(e^{-kh} \rightarrow 0\) and \(e^{-k(z + h)} \rightarrow 0\):
\[ \frac{\cosh\big(k(z + h)\big)}{\sinh(kh)} \rightarrow \frac{e^{k(z + h)}}{e^{kh}} = e^{kz}, \qquad \frac{\sinh\big(k(z + h)\big)}{\sinh(kh)} \rightarrow \frac{e^{k(z + h)}}{e^{kh}} = e^{kz} \]
\[ u(x, z, t) = a\omega\,e^{kz}\cos(\omega t - kx) \tag{15} \]
\[ w(x, z, t) = -a\omega\,e^{kz}\sin(\omega t - kx) \tag{16} \]
10. Pressure
Dynamic condition at any point in the fluid (\(z\) in place of \(\eta\)):
\[ \frac{p}{\rho} + \frac{\partial\phi}{\partial t} + \frac{1}{2}\left(u^2 + w^2\right) + gz = 0 \]
For small amplitudes: \(\frac{1}{2}\left(u^2 + w^2\right) \approx 0\)
\[ p = -\rho gz - \rho\,\frac{\partial\phi}{\partial t} \]
From (9):
\[ \frac{\partial\phi}{\partial t} = -\frac{ga}{\omega}\,\frac{\cosh\big(k(z + h)\big)}{\cosh(kh)}\,\omega\cos(\omega t - kx) = -ga\,\frac{\cosh\big(k(z + h)\big)}{\cosh(kh)}\,\cos(\omega t - kx) \]
\[ p(x, z, t) = \underbrace{-\rho gz}_{\text{hydrostatic}} + \underbrace{\rho ga\,\frac{\cosh\big(k(z + h)\big)}{\cosh(kh)}\,\cos(\omega t - kx)}_{\text{dynamic}} \tag{17} \]
Check at \(z = 0\): \(\;p = \rho ga\cos(\omega t - kx) = \rho g\eta\) from (8) ✓
10.1 Deep water
\(\dfrac{\cosh\big(k(z + h)\big)}{\cosh(kh)} \rightarrow \dfrac{e^{k(z + h)}}{e^{kh}} = e^{kz}\)
\[ p(x, z, t) = -\rho gz + \rho ga\,e^{kz}\cos(\omega t - kx) \tag{18} \]