Mild-slope equation

When ${\displaystyle \eta =a\,\exp(i\theta )}$ is used in the mild-slope equation, the result is, apart from a factor ${\displaystyle \exp(i\theta )}$:

$${\displaystyle c_{p}\,c_{g}\,\left(\Delta a\,+\,2i\,\nabla a\cdot \nabla \theta \,-\,a\,\nabla \theta \cdot \nabla \theta \,+\,i\,a\,\Delta \theta \right)\,+\,\nabla \left(c_{p}\,c_{g}\right)\cdot \left(\nabla a\,+\,i\,a\,\nabla \theta \right)\,+\,k^{2}\,c_{p}\,c_{g}\,a\,=\,0.}$$

Now both the real part and the imaginary part of this equation have to be equal to zero:

$${\displaystyle {\begin{aligned}&c_{p}\,c_{g}\,\Delta a\,-\,c_{p}\,c_{g}\,a\,\nabla \theta \cdot \nabla \theta \,+\,\nabla \left(c_{p}\,c_{g}\right)\cdot \nabla a\,+\,k^{2}\,c_{p}\,c_{g}\,a\,=\,0\quad {\text{and}}\\&2\,c_{p}\,c_{g}\,\nabla a\cdot \nabla \theta \,+\,c_{p}\,c_{g}\,a\,\Delta \theta \,+\,\nabla \left(c_{p}\,c_{g}\right)\cdot \left(a\,\nabla \theta \right)\,=\,0.\end{aligned}}}$$

The effective wavenumber vector ${\displaystyle {\boldsymbol {\kappa }}}$ is defined as the gradient of the wave phase:

$${\displaystyle {\boldsymbol {\kappa }}=\nabla \theta }$$

and its

vector length

is

$${\displaystyle \kappa =|{\boldsymbol {\kappa }}|.}$$

Note that ${\displaystyle {\boldsymbol {\kappa }}}$ is an

is zero:

$${\displaystyle \nabla \times {\boldsymbol {\kappa }}=0.}$$

Now the real and imaginary parts of the transformed mild-slope equation become, first multiplying the imaginary part by ${\displaystyle a}$:

$${\displaystyle {\begin{aligned}&\kappa ^{2}\,=\,k^{2}\,+\,{\frac {\nabla (c_{p}\,c_{g})}{c_{p}\,c_{g}}}\cdot {\frac {\nabla a}{a}}\,+\,{\frac {\Delta a}{a}}\quad {\text{and}}\\&c_{p}\,c_{g}\,\nabla \left(a^{2}\right)\cdot {\boldsymbol {\kappa }}\,+\,c_{p}\,c_{g}\,\nabla \cdot {\boldsymbol {\kappa }}\,+\,a^{2}\,{\boldsymbol {\kappa }}\cdot \nabla \left(c_{p}\,c_{g}\right)\,=\,0.\end{aligned}}}$$

The first equation directly leads to the eikonal equation above for ${\displaystyle \kappa \,}$, while the second gives:

$${\displaystyle \nabla \cdot \left({\boldsymbol {\kappa }}\,c_{p}\,c_{g}\,a^{2}\right)\,=\,0,}$$

which—by noting that ${\displaystyle c_{p}=\omega /k}$ in which the angular frequency ${\displaystyle \omega }$ is a constant for time-harmonic motion—leads to the wave-energy conservation equation.

Luke's

${\displaystyle \rho }$

${\displaystyle z=\zeta (x,y,t)}$

${\displaystyle z=-h(x,y),}$

${\displaystyle \delta {\mathcal {L}}=0}$

$${\displaystyle {\mathcal {L}}=\int _{t_{0}}^{t_{1}}\iint L\,{\text{d}}x\,{\text{d}}y\,{\text{d}}t,}$$

${\displaystyle L}$

$${\displaystyle L=-\rho \,\left\{\int _{-h(x,y)}^{\zeta (x,y,t)}\left[{\frac {\partial \Phi }{\partial t}}+\,{\frac {1}{2}}\left(\left({\frac {\partial \Phi }{\partial x}}\right)^{2}+\left({\frac {\partial \Phi }{\partial y}}\right)^{2}+\left({\frac {\partial \Phi }{\partial z}}\right)^{2}\right)\right]\;{\text{d}}z\;+\,{\frac {1}{2}}\,g\,(\zeta ^{2}\,-\,h^{2})\right\},}$$

where ${\displaystyle \Phi (x,y,z,t)}$ is the velocity potential, with the flow velocity components being ${\displaystyle \partial \Phi /\partial {x},}$ ${\displaystyle \partial \Phi /\partial {y}}$ and ${\displaystyle \partial \Phi /\partial {z}}$ in the ${\displaystyle x}$, ${\displaystyle y}$ and ${\displaystyle z}$ directions, respectively. Luke's Lagrangian formulation can also be recast into a Hamiltonian formulation in terms of the surface elevation and velocity potential at the free surface.^[10] Taking the variations of ${\displaystyle {\mathcal {L}}(\Phi ,\zeta )}$ with respect to the potential ${\displaystyle \Phi (x,y,z,t)}$ and surface elevation ${\displaystyle \zeta (x,y,t)}$ leads to the

${\displaystyle \Phi }$

${\displaystyle z=\zeta (x,y,t)}$

${\displaystyle z=-h(x,y).}$

Linear wave theory

In case of linear wave theory, the vertical integral in the Lagrangian density ${\displaystyle L}$ is split into a part from the bed ${\displaystyle z=-h}$ to the mean surface at ${\displaystyle z=0,}$ and a second part from ${\displaystyle z=0}$ to the free surface ${\displaystyle z=\zeta }$. Using a Taylor series expansion for the second integral around the mean free-surface elevation ${\displaystyle z=0,}$ and only retaining quadratic terms in ${\displaystyle \Phi }$ and ${\displaystyle \zeta ,}$ the Lagrangian density ${\displaystyle L_{0}}$ for linear wave motion becomes

$${\displaystyle L_{0}=-\rho \,\left\{\zeta \,\left[{\frac {\partial \Phi }{\partial t}}\right]_{z=0}\,+\,\int _{-h}^{0}{\frac {1}{2}}\left[\left({\frac {\partial \Phi }{\partial x}}\right)^{2}+\left({\frac {\partial \Phi }{\partial y}}\right)^{2}+\left({\frac {\partial \Phi }{\partial z}}\right)^{2}\right]\;{\text{d}}z\;+\,{\frac {1}{2}}\,g\,\zeta ^{2}\,\right\}.}$$

The term ${\displaystyle \partial \Phi /\partial {t}}$ in the vertical integral is dropped since it has become dynamically uninteresting: it gives a zero contribution to the Euler–Lagrange equations, with the upper integration limit now fixed. The same is true for the neglected bottom term proportional to ${\displaystyle h^{2}}$ in the potential energy.

The waves propagate in the horizontal ${\displaystyle (x,y)}$ plane, while the structure of the potential ${\displaystyle \Phi }$ is not wave-like in the vertical ${\displaystyle z}$-direction. This suggests the use of the following assumption on the form of the potential ${\displaystyle \Phi :}$

$${\displaystyle \Phi (x,y,z,t)=f(z;x,y)\,\varphi (x,y,t)}$$

$${\displaystyle f(0;x,y)=1}$$

${\displaystyle z=0.}$

Here ${\displaystyle \varphi (x,y,t)}$ is the velocity potential at the mean free-surface level ${\displaystyle z=0.}$ Next, the mild-slope assumption is made, in that the vertical shape function ${\displaystyle f}$ changes slowly in the ${\displaystyle (x,y)}$-plane, and horizontal derivatives of ${\displaystyle f}$ can be neglected in the flow velocity. So:

$${\displaystyle {\begin{pmatrix}{\dfrac {\partial \Phi }{\partial {x}}}\\[2ex]{\dfrac {\partial \Phi }{\partial {y}}}\\[2ex]{\dfrac {\partial \Phi }{\partial {z}}}\end{pmatrix}}\,\approx \,{\begin{pmatrix}f{\dfrac {\partial \varphi }{\partial {x}}}\\[2ex]f{\dfrac {\partial \varphi }{\partial {y}}}\\[2ex]{\dfrac {\partial {f}}{\partial {z}}}\,\varphi \end{pmatrix}}.}$$

As a result:

$${\displaystyle L_{0}=-\rho \left\{\zeta {\frac {\partial \varphi }{\partial t}}\,+{\frac {1}{2}}\,F\left[\left({\frac {\partial \varphi }{\partial {x}}}\right)^{2}+\left({\frac {\partial \varphi }{\partial {y}}}\right)^{2}\right]\,+{\frac {1}{2}}\,G\varphi ^{2}+{\frac {1}{2}}\,g\zeta ^{2}\right\},}$$

$${\displaystyle {\begin{aligned}F&=\int _{-h}^{0}f^{2}\,{\text{d}}z\\G&=\int _{-h}^{0}\left({\frac {{\text{d}}f}{{\text{d}}z}}\right)^{2}{\text{d}}z.\end{aligned}}}$$

The Euler–Lagrange equations for this Lagrangian density ${\displaystyle L_{0}}$ are, with ${\displaystyle \xi (x,y,t)}$ representing either ${\displaystyle \varphi }$ or ${\displaystyle \zeta :}$

$${\displaystyle {\frac {\partial {L_{0}}}{\partial \xi }}-{\frac {\partial }{\partial {t}}}\left({\frac {\partial {L_{0}}}{\partial (\partial \xi /\partial {t})}}\right)-{\frac {\partial }{\partial {x}}}\left({\frac {\partial {L_{0}}}{\partial (\partial \xi /\partial {x})}}\right)-{\frac {\partial }{\partial {y}}}\left({\frac {\partial {L_{0}}}{\partial (\partial \xi /\partial {y})}}\right)=0.}$$

Now ${\displaystyle \xi }$ is first taken equal to ${\displaystyle \varphi }$ and then to ${\displaystyle \zeta .}$ As a result, the evolution equations for the wave motion become:^[4]

$${\displaystyle {\begin{aligned}{\frac {\partial \zeta }{\partial t}}\,&+\nabla \cdot \left(F\,\nabla \varphi \right)-G\varphi =0\quad {\text{and}}\\{\frac {\partial \varphi }{\partial t}}\,&+\,g\zeta =0,\end{aligned}}}$$

The next step is to choose the shape function ${\displaystyle f}$ and to determine ${\displaystyle F}$ and ${\displaystyle G.}$

Vertical shape function from Airy wave theory

Since the objective is the description of waves over mildly sloping beds, the shape function ${\displaystyle f(z)}$ is chosen according to Airy wave theory. This is the linear theory of waves propagating in constant depth ${\displaystyle h.}$ The form of the shape function is:^[4]

$${\displaystyle f={\frac {\cosh \left(k\left(z+h\right)\right)}{\cosh(kh)}},}$$

${\displaystyle k(x,y)}$

${\displaystyle x}$

${\displaystyle y}$

${\displaystyle h(x,y)}$

$${\displaystyle \omega _{0}^{2}=gk\tanh(kh).}$$

Here ${\displaystyle \omega _{0}}$ a constant angular frequency, chosen in accordance with the characteristics of the wave field under study. Consequently, the integrals ${\displaystyle F}$ and ${\displaystyle G}$ become:^[4]

$${\displaystyle {\begin{aligned}F&=\int _{h}^{0}f^{2}\,{\text{d}}z={\frac {1}{g}}c_{p}c_{g}\quad {\text{and}}\\G&=\int _{h}^{0}\left({\frac {\partial f}{\partial z}}\right)^{2}{\text{d}}z={\frac {1}{g}}\left(\omega _{0}^{2}-k^{2}c_{p}c_{g}\right).\end{aligned}}}$$