Luke's variational principle

The variation ${\displaystyle \delta {\mathcal {L}}=0}$ in the Lagrangian with respect to variations in the velocity potential Φ(x,z,t), as well as with respect to the surface elevation η(x,t), have to be zero. We consider both variations subsequently.

Consider a small variation δΦ in the velocity potential Φ.[8] Then the resulting variation in the Lagrangian is:

${\displaystyle {\begin{aligned}\delta _{\Phi }{\mathcal {L}}\,&=\,{\mathcal {L}}(\Phi +\delta \Phi ,\eta )\,-\,{\mathcal {L}}(\Phi ,\eta )\\&=\,-\,\int _{t_{0}}^{t_{1}}\iint \left\{\int _{-h({\boldsymbol {x}})}^{\eta ({\boldsymbol {x}},t)}\rho \,\left({\frac {\partial (\delta \Phi )}{\partial t}}+\,{\boldsymbol {\nabla }}\Phi \cdot {\boldsymbol {\nabla }}(\delta \Phi )+\,{\frac {\partial \Phi }{\partial z}}\,{\frac {\partial (\delta \Phi )}{\partial z}}\,\right)\;{\text{d}}z\,\right\}\;{\text{d}}{\boldsymbol {x}}\;{\text{d}}t.\end{aligned}}}$

Using Leibniz integral rule, this becomes, in case of constant density ρ:[8]

${\displaystyle {\begin{aligned}\delta _{\Phi }{\mathcal {L}}\,=\,&-\,\rho \,\int _{t_{0}}^{t_{1}}\iint \left\{{\frac {\partial }{\partial t}}\int _{-h({\boldsymbol {x}})}^{\eta ({\boldsymbol {x}},t)}\delta \Phi \;{\text{d}}z\;+\,{\boldsymbol {\nabla }}\cdot \int _{-h({\boldsymbol {x}})}^{\eta ({\boldsymbol {x}},t)}\delta \Phi \,{\boldsymbol {\nabla }}\Phi \;{\text{d}}z\,\right\}\;{\text{d}}{\boldsymbol {x}}\;{\text{d}}t\\&+\,\rho \,\int _{t_{0}}^{t_{1}}\iint \left\{\int _{-h({\boldsymbol {x}})}^{\eta ({\boldsymbol {x}},t)}\delta \Phi \;\left({\boldsymbol {\nabla }}\cdot {\boldsymbol {\nabla }}\Phi \,+\,{\frac {\partial ^{2}\Phi }{\partial z^{2}}}\right)\;{\text{d}}z\,\right\}\;{\text{d}}{\boldsymbol {x}}\;{\text{d}}t\\&+\,\rho \,\int _{t_{0}}^{t_{1}}\iint \left[\left({\frac {\partial \eta }{\partial t}}\,+\,{\boldsymbol {\nabla }}\Phi \cdot {\boldsymbol {\nabla }}\eta \,-\,{\frac {\partial \Phi }{\partial z}}\right)\,\delta \Phi \right]_{z=\eta ({\boldsymbol {x}},t)}\;{\text{d}}{\boldsymbol {x}}\;{\text{d}}t\\&-\,\rho \,\int _{t_{0}}^{t_{1}}\iint \left[\left({\boldsymbol {\nabla }}\Phi \cdot {\boldsymbol {\nabla }}h\,+\,{\frac {\partial \Phi }{\partial z}}\right)\,\delta \Phi \right]_{z=-h({\boldsymbol {x}})}\;{\text{d}}{\boldsymbol {x}}\;{\text{d}}t\\=\,&0.\end{aligned}}}$

The first integral on the right-hand side integrates out to the boundaries, in x and t, of the integration domain and is zero since the variations δΦ are taken to be zero at these boundaries. For variations δΦ which are zero at the free surface and the bed, the second integral remains, which is only zero for arbitrary δΦ in the fluid interior if there the Laplace equation holds:

${\displaystyle \Delta \Phi \,=\,0\qquad {\text{ for }}-h({\boldsymbol {x}})\,<\,z\,<\,\eta ({\boldsymbol {x}},t),}$

with Δ=∇·∇ + ∂²/∂z² the Laplace operator.

If variations δΦ are considered which are only non-zero at the free surface, only the third integral remains, giving rise to the kinematic free-surface boundary condition:

${\displaystyle {\frac {\partial \eta }{\partial t}}\,+\,{\boldsymbol {\nabla }}\Phi \cdot {\boldsymbol {\nabla }}\eta \,-\,{\frac {\partial \Phi }{\partial z}}\,=\,0.\qquad {\text{ at }}z\,=\,\eta ({\boldsymbol {x}},t).}$

Similarly, variations δΦ only non-zero at the bottom z = -h result in the kinematic bed condition:

${\displaystyle {\boldsymbol {\nabla }}\Phi \cdot {\boldsymbol {\nabla }}h\,+\,{\frac {\partial \Phi }{\partial z}}\,=\,0\qquad {\text{ at }}z\,=\,-h({\boldsymbol {x}}).}$

Considering the variation of the Lagrangian with respect to small changes δη gives:

${\displaystyle \delta _{\eta }{\mathcal {L}}\,=\,{\mathcal {L}}(\Phi ,\eta +\delta \eta )\,-\,{\mathcal {L}}(\Phi ,\eta )=\,-\,\int _{t_{0}}^{t_{1}}\iint \left[\rho \,\delta \eta \,\left({\frac {\partial \Phi }{\partial t}}+\,{\frac {1}{2}}\,\left|{\boldsymbol {\nabla }}\Phi \right|^{2}\,+\,{\frac {1}{2}}\,\left({\frac {\partial \Phi }{\partial z}}\right)^{2}+\,g\,\eta \right)\,\right]_{z=\eta ({\boldsymbol {x}},t)}\;{\text{d}}{\boldsymbol {x}}\;{\text{d}}t\,=\,0.}$

This has to be zero for arbitrary δη, giving rise to the dynamic boundary condition at the free surface:

${\displaystyle {\frac {\partial \Phi }{\partial t}}+\,{\frac {1}{2}}\,\left|{\boldsymbol {\nabla }}\Phi \right|^{2}\,+\,{\frac {1}{2}}\,\left({\frac {\partial \Phi }{\partial z}}\right)^{2}+\,g\,\eta \,=\,0\qquad {\text{ at }}z\,=\,\eta ({\boldsymbol {x}},t).}$

This is the Bernoulli equation for unsteady potential flow, applied at the free surface, and with the pressure above the free surface being a constant — which constant pressure is taken equal to zero for simplicity.

The Hamiltonian formulation can be derived from Luke's Lagrangian description by using Leibniz integral rule on the integral of ∂Φ/∂t:[6]

${\displaystyle {\mathcal {L}}_{H}=\int _{t_{0}}^{t_{1}}\iint \left\{\varphi ({\boldsymbol {x}},t)\,{\frac {\partial \eta ({\boldsymbol {x}},t)}{\partial t}}\,-\,H(\varphi ,\eta$ ;{\boldsymbol {x}},t)\right\}\;{\text{d}}{\boldsymbol {x}}\;{\text{d}}t,}

with ${\displaystyle \varphi ({\boldsymbol {x}},t)=\Phi ({\boldsymbol {x}},\eta ({\boldsymbol {x}},t),t)}$ the value of the velocity potential at the free surface, and ${\displaystyle H(\varphi ,\eta$ ;{\boldsymbol {x}},t)} the Hamiltonian density — sum of the kinetic and potential energy density — and related to the Hamiltonian as:

${\displaystyle {\mathcal {H}}(\varphi ,\eta )\,=\,\iint H(\varphi ,\eta$ ;{\boldsymbol {x}},t)\;{\text{d}}{\boldsymbol {x}}.}

The Hamiltonian density is written in terms of the surface potential using Green's third identity on the kinetic energy:[9]

${\displaystyle H\,=\,{\frac {1}{2}}\,\rho \,{\sqrt {1\,+\,\left|{\boldsymbol {\nabla }}\eta \right|^{2}}}\;\;\varphi \,{\bigl (}D(\eta )\;\varphi {\bigr )}\,+\,{\frac {1}{2}}\,\rho \,g\,\eta ^{2},}$

where D(η) φ is equal to the normal derivative of ∂Φ/∂n at the free surface. Because of the linearity of the Laplace equation — valid in the fluid interior and depending on the boundary condition at the bed z=-h and free surface z=η — the normal derivative ∂Φ/∂n is a linear function of the surface potential φ, but depends non-linear on the surface elevation η. This is expressed by the Dirichlet-to-Neumann operator D(η), acting linearly on φ.

The Hamiltonian density can also be written as:[6]

${\displaystyle H\,=\,{\frac {1}{2}}\,\rho \,\varphi \,{\Bigl [}w\,\left(1\,+\,\left|{\boldsymbol {\nabla }}\eta \right|^{2}\right)-\,{\boldsymbol {\nabla }}\eta \cdot {\boldsymbol {\nabla }}\,\varphi {\Bigr ]}\,+\,{\frac {1}{2}}\,\rho \,g\,\eta ^{2},}$

with w(x,t) = ∂Φ/∂z the vertical velocity at the free surface z = η. Also w is a linear function of the surface potential φ through the Laplace equation, but w depends non-linear on the surface elevation η:[9]

${\displaystyle w\,=\,W(\eta )\,\varphi ,}$

with W operating linear on φ, but being non-linear in η. As a result, the Hamiltonian is a quadratic functional of the surface potential φ. Also the potential energy part of the Hamiltonian is quadratic. The source of non-linearity in surface gravity waves is through the kinetic energy depending non-linear on the free surface shape η.[9]

Further ∇φ is not to be mistaken for the horizontal velocity ∇Φ at the free surface:

${\displaystyle {\boldsymbol {\nabla }}\varphi \,=\,{\boldsymbol {\nabla }}\Phi {\bigl (}{\boldsymbol {x}},\eta ({\boldsymbol {x}},t),t{\bigr )}\,=\,\left[{\boldsymbol {\nabla }}\Phi \,+\,{\frac {\partial \Phi }{\partial z}}\,{\boldsymbol {\nabla }}\eta \right]_{z=\eta ({\boldsymbol {x}},t)}\,=\,{\Bigl [}{\boldsymbol {\nabla }}\Phi {\Bigr ]}_{z=\eta ({\boldsymbol {x}},t)}\,+\,w\,{\boldsymbol {\nabla }}\eta .}$

Taking the variations of the Lagrangian ${\displaystyle {\mathcal {L}}_{H}}$ with respect to the canonical variables ${\displaystyle \varphi ({\boldsymbol {x}},t)}$ and ${\displaystyle \eta ({\boldsymbol {x}},t)}$ gives:

${\displaystyle {\begin{aligned}\rho \,{\frac {\partial \eta }{\partial t}}\,&=\,+\,{\frac {\delta {\mathcal {H}}}{\delta \varphi }},\\\rho \,{\frac {\partial \varphi }{\partial t}}\,&=\,-\,{\frac {\delta {\mathcal {H}}}{\delta \eta }},\end{aligned}}}$

provided in the fluid interior Φ satisfies the Laplace equation, ΔΦ=0, as well as the bottom boundary condition at z=-h and Φ=φ at the free surface.