# water wave and fluids dispersion relation

Small-amplitude water waves travel on the free surface $y = \eta(x, t)$ of an incompressible inviscid fluid of uniform depth $h$. Derive the linearised boundary conditions

$$\text{ at }y=0\quad \frac{\partial\varphi}{\partial t}=\frac{\partial\eta}{\partial t},\quad \frac{\partial\varphi}{\partial t}+g\eta=0$$

and write down the boundary condition satisfied by the velocity potential $\varphi$ at the rigid boundary $y = −h$. ￼￼￼Show that waves of the form are possible, $$\eta(x, t) = A\cos(kx − \omega t)$$ and find the wave speed $c = \omega/k$ in terms of $k$.

If $kh\ll 1$, show that $c$ is approximately equal to $$\sqrt{gh}\left(1 − \frac{1}{6}k^2h^2\right).$$

For the finite depth, I got $\text { at }y=-h,\quad\frac{\partial\varphi}{\partial y}=0$

My idea:

I get the boundary conditions and show that waves are possible when $$\varphi=f(y)\sin(kx-\omega t)$$ Check the boundary condition to find out that when the dispersion relation $$\omega^2=gk\tanh(kh)$$(maybe this is wrong?) then the $c$ can be get in terms of $k$.

However I am confused about the approximation of $c.$ Since $kh<<1$ (it is long and shallow water), $\tanh(kh) \approx kh$ but I didn't get from here to the result and what is the $\dfrac{1}{6}$(does it come from $k=\dfrac{2\pi}{\lambda}$).

Thank you so much!

• Can anyone please give me some hint? – stedmoaoa May 14 '17 at 20:32
• please check the boundary conditions – Yuri Negometyanov May 20 '17 at 13:22

Starting from your dispersion relation, $$c = \sqrt{g} \sqrt{\frac{\tanh kh}{k}}$$ So we'd like to find a series expansion for $\tanh kh$ for small $kh$. $$\frac{\tanh t}{t} = \frac{t -\frac{1}{3} t^3 + \dots}{t} \\ = 1 - \frac{1}{3}t^2 + \dots \\ \sqrt{\frac{\tanh t}{t}} = 1 - \frac{1}{6}t^2 + ...$$ from the binomial theorem. Hence, $$\sqrt{\frac{\tanh kh} {kh}} \approx 1 - \frac{1}{6}k^2h^2 \\ \sqrt{g} \sqrt{\frac{\tanh kh}{k}} \approx \sqrt{gh} \left(1-\frac{1}{6}k^2h^2\right)$$
The step you may have been missing was to take the next term in the series expansion of $\tanh$, not just the linear one.
• Thank you so much! It really makes sense. And can you maybe check my $w^2$ – stedmoaoa May 20 '17 at 13:26