# Template:Separation of variables for a free surface

### Separation of variables for a free surface

We use separation of variables

We express the potential as

$\phi(x,z) = X(x)Z(z)\,$

and then Laplace's equation becomes

$\frac{X^{\prime\prime}}{X} = - \frac{Z^{\prime\prime}}{Z} = k^2$

The separation of variables equation for deriving free surface eigenfunctions is as follows:

$Z^{\prime\prime} + k^2 Z =0.$

subject to the boundary conditions

$Z^{\prime}(-h) = 0$

and

$Z^{\prime}(0) = \alpha Z(0)$

We can then use the boundary condition at $z=-h \,$ to write

$Z = \frac{\cos k(z+h)}{\cos kh}$

where we have chosen the value of the coefficent so we have unit value at $z=0$. The boundary condition at the free surface ($z=0 \,$) gives rise to:

$k\tan\left( kh\right) =-\alpha \,$

which is the Dispersion Relation for a Free Surface

The above equation is a transcendental equation. If we solve for all roots in the complex plane we find that the first root is a pair of imaginary roots. We denote the imaginary solutions of this equation by $k_{0}=\pm ik \,$ and the positive real solutions by $k_{m} \,$, $m\geq1$. The $k \,$ of the imaginary solution is the wavenumber. We put the imaginary roots back into the equation above and use the hyperbolic relations

$\cos ix = \cosh x, \quad \sin ix = i\sinh x,$

to arrive at the dispersion relation

$\alpha = k\tanh kh.$

We note that for a specified frequency $\omega \,$ the equation determines the wavenumber $k \,$.

Finally we define the function $Z(z) \,$ as

$\chi_{m}\left( z\right) =\frac{\cos k_{m}(z+h)}{\cos k_{m}h},\quad m\geq0$

as the vertical eigenfunction of the potential in the open water region. From Sturm-Liouville theory the vertical eigenfunctions are orthogonal. They can be normalised to be orthonormal, but this has no advantages for a numerical implementation. It can be shown that

$\int\nolimits_{-h}^{0}\chi_{m}(z)\chi_{n}(z) \mathrm{d} z=A_{n}\delta_{mn}$

where

$A_{n}=\frac{1}{2}\left( \frac{\cos k_{n}h\sin k_{n}h+k_{n}h}{k_{n}\cos ^{2}k_{n}h}\right).$