# 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

[math] \phi(x,z) = X(x)Z(z)\, [/math]

and then Laplace's equation becomes

[math] \frac{X^{\prime\prime}}{X} = - \frac{Z^{\prime\prime}}{Z} = k^2 [/math]

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

[math] Z^{\prime\prime} + k^2 Z =0. [/math]

subject to the boundary conditions

[math] Z^{\prime}(-h) = 0 [/math]

and

[math] Z^{\prime}(0) = \alpha Z(0) [/math]

We can then use the boundary condition at [math]z=-h \, [/math] to write

[math] Z = \frac{\cos k(z+h)}{\cos kh} [/math]

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

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 [math]k_{0}=\pm ik \,[/math] and the positive real solutions by [math]k_{m} \,[/math], [math]m\geq1[/math]. The [math]k \,[/math] of the imaginary solution is the wavenumber. We put the imaginary roots back into the equation above and use the hyperbolic relations

[math] \cos ix = \cosh x, \quad \sin ix = i\sinh x, [/math]

to arrive at the dispersion relation

[math] \alpha = k\tanh kh. [/math]

We note that for a specified frequency [math]\omega \,[/math] the equation determines the wavenumber [math]k \,[/math].

Finally we define the function [math]Z(z) \,[/math] as

[math] \chi_{m}\left( z\right) =\frac{\cos k_{m}(z+h)}{\cos k_{m}h},\quad m\geq0 [/math]

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

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

where

[math] 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). [/math]