Template:Frequency domain equations for a floating plate

If we make the assumption of Frequency Domain Problem that everything is proportional to [math]\displaystyle{ \exp (-\mathrm{i}\omega t)\, }[/math] the equations become

[math]\displaystyle{ -\mathrm{i}\omega\zeta = \partial_z\phi , \ z=0; }[/math] [math]\displaystyle{ \rho g\zeta - \mathrm{i}\omega\rho \phi = D \partial_x^4 \eta -\omega^2 \rho_i h \zeta, \ z=0; }[/math] [math]\displaystyle{ \Delta \phi = 0,\,\,-h\lt z\lt 0 }[/math] [math]\displaystyle{ \partial_z \phi = 0,\,\,z=-h, }[/math]

where [math]\displaystyle{ \zeta }[/math] is the surface displacement and [math]\displaystyle{ \phi }[/math] is the velocity potential in the frequency domain.

These equations can be simplified by defining [math]\displaystyle{ \alpha = \omega^2/g }[/math], [math]\displaystyle{ \beta = D/\rho g }[/math] and [math]\displaystyle{ \gamma = \rho_i h/\rho }[/math] to obtain

[math]\displaystyle{ \Delta \phi = 0, \;\;\; -h \lt z \leq 0, }[/math] [math]\displaystyle{ \partial_z \phi = 0, \;\;\; z = - h, }[/math] [math]\displaystyle{ \beta \partial_x^4 \zeta + \left( 1 - \gamma\alpha \right) \zeta = -\mathrm{i} \sqrt{\alpha}\phi, \;\; z = 0. }[/math] [math]\displaystyle{ -\mathrm{i}\omega\zeta = \partial_z\phi , \ z=0; }[/math]