Capillary-gravity dispersion and conserved linear surface energy

Let ${\omega }_0=\sqrt{gk\tanh (kh)}>0$ and let $c_{\sigma }=\sqrt{1+\sigma k^2\mathrm{/}(\rho g)}>0$, whose existence and positivity follow from [F1] and [F2]. Set $\omega =c_{\sigma }{\omega }_0$. Then ${\omega }^2=(1+\sigma k^2\mathrm{/}(\rho g))gk\tanh (kh)=(gk+\sigma k^3\mathrm{/}\rho )\tanh (kh)$, and the radicand in the Statement is positive, so its positive square root is this $\omega$.

First, [F7] and repeated differentiation show that every derivative of $S,C$ is one of $S,C,-S,-C$, and every derivative of $\sinh ,\cosh$ alternates between those two continuous functions. All four functions are therefore $C^{\mathrm{\infty }}$. By [F3] and [F5], compositions with the affine phases and products with $G(z)$ are smooth on real coordinates, extending across both vertical endpoints; the denominator $\cosh (kh)$ is a fixed positive number. The $2\pi$-periodicity in [F7] makes the formulas unchanged under $\tilde{x}\mapsto \tilde{x}+n\lambda$ for every integer $n$, and independence of $y$ gives period $L_y$. Hence the smooth formulas and all their derivatives descend to the horizontal torus. Write ${\theta }_0(t^{\mathrm{\prime }},\tilde{x})=k\tilde{x}-{\omega }_0{t}^{\mathrm{\prime }}$, and take the gravity mode ${\eta }_0,{\varphi }_0$ from [F1]. Define $t^{\mathrm{\prime }}=c_{\sigma }t$, $\eta (t,x,y)={\eta }_0(t^{\mathrm{\prime }},x,y)$, and $\varphi (t,x,y,z)=c_{\sigma }{\varphi }_0(t^{\mathrm{\prime }},x,y,z)$. The chain rule [F3] gives ${\eta }_t=c_{\sigma }({\eta }_0{)}_{t^{\mathrm{\prime }}}=c_{\sigma }({\varphi }_0{)}_z={\varphi }_z$; harmonicity and the bottom condition persist because this transformation only rescales time and amplitude.

Let $Q=[0,\lambda ]\times [0,L_y]\times [-h,0]$. Apply [F6] successively to the rectangular cell integrals and use [F5] to differentiate the squared gradients; the fields and time derivatives are continuous on the compact cell, so the hypotheses hold. Apply [F4] with $u={\varphi }_t$ and $v=\varphi$. The interior term ${\varphi }_t\mathrm{\Delta }\varphi$ vanishes; opposite periodic faces cancel; the bottom term vanishes by ${\varphi }_z(-h)=0$. Hence $K^{\mathrm{\prime }}(t)=\rho {\int }_0^{L_y}{\int }_0^{\lambda }{\varphi }_t(t,x,y,0){\varphi }_z(t,x,y,0)dxdy$.

Differentiate the two surface terms in $E$ using [F5] and [F6]. For the gradient term apply [F4] on $Q$ to $u={\eta }_t(x,y)$ and $v=\eta (x,y)$, extended constantly in $z$; the top and bottom normal derivatives vanish and the paired periodic-face terms cancel. Dividing the identity by $h$ gives ${\int }_0^{L_y}{\int }_0^{\lambda }{\mathrm{\nabla }}_H\eta \cdot {\mathrm{\nabla }}_H{\eta }_{t}dxdy=-{\int }_0^{L_y}{\int }_0^{\lambda }{\eta }_t{\mathrm{\Delta }}_H\eta dxdy$, so $V^{\mathrm{\prime }}(t)={\int }_0^{L_y}{\int }_0^{\lambda }(\rho g\eta -\sigma {\mathrm{\Delta }}_H\eta ){\eta }_{t}dxdy$.

The gravity mode is independent of $y$, so ${\mathrm{\Delta }}_H{\eta }_0=-k^2{\eta }_0$. Its gravity dynamic condition is $({\varphi }_0{)}_{t^{\mathrm{\prime }}}+g{\eta }_0=0$. Therefore ${\varphi }_t+g\eta -(\sigma \mathrm{/}\rho ){\mathrm{\Delta }}_H\eta =c_{\sigma }^2({\varphi }_0{)}_{t^{\mathrm{\prime }}}+g{\eta }_0+(\sigma k^2\mathrm{/}\rho ){\eta }_0=(-g{c}_{\sigma }^2+g+\sigma k^2\mathrm{/}\rho ){\eta }_0=0$, since $c_{\sigma }^2=1+\sigma k^2\mathrm{/}(\rho g)$. Thus the transformed fields satisfy the capillary dynamic condition.

Since $\omega =c_{\sigma }{\omega }_0$ and ${\omega }_0^2=gk\tanh (kh)$, the coefficient of $G(z)S({\theta }_{\sigma })$ in $\varphi =c_{\sigma }{\varphi }_0(c_{\sigma }t)$ is $ag{c}_{\sigma }\mathrm{/}{\omega }_0=a\omega \mathrm{/}(k\tanh (kh))$. Also ${\theta }_0(c_{\sigma }t,\tilde{x})={\theta }_{\sigma }$. This proves the displayed mode formula, including $a=0$. At $\sigma =0$, $c_{\sigma }=1$ and the same formulas are exactly the gravity mode [F1].

The kinematic condition makes ${\varphi }_z(t,x,y,0)={\eta }_t$. The dynamic condition makes $\rho {\varphi }_t+\rho g\eta -\sigma {\mathrm{\Delta }}_H\eta =0$. Adding the derivatives in steps 1.3 and 1.4 therefore gives $E^{\mathrm{\prime }}(t)=K^{\mathrm{\prime }}(t)+V^{\mathrm{\prime }}(t)=0$. The integrands are continuous in time and space on the compact cells, so their uniform continuity makes $E$ continuous on $[0,T]$; steps 1.3 and 1.4 give differentiability on $(0,T)$. The interval $[0,T]$ is order-convex, and [F8] applied to the vanishing derivative proves that $E$ is constant on the whole interval, including its endpoints. Each term in $E$ is nonnegative because $\rho ,g>0$ and $\sigma \ge 0$, so $E$ is nonnegative.