Fourier Transform Solution for an Incident Wave Packet

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We present here a very general and simple method to solve for an incident wave packet in the time domian using the freqency domain solutions for an incident wave. The theory is present here in two dimensions but can be extended to three dimensions (for plane incident waves) straightforwardly.

Frequency Domain Solution

We assume small amplitude so that we can linearise all the equations (see Linear and Second-Order Wave Theory). We also assume that Frequency Domain Problem with frequency [math]\omega[/math] and we assume that all variables are proportional to [math]\exp(-\mathrm{i}\omega t)\,[/math] The water motion is represented by a velocity potential which is denoted by [math]\phi\,[/math] so that

[math] \Phi(\mathbf{x},t) = \mathrm{Re} \left\{\phi(\mathbf{x},\omega)e^{-\mathrm{i} \omega t}\right\}. [/math]

The coordinate system is the standard Cartesian coordinate system with the [math]z-[/math]axis pointing vertically up. The water surface is at [math]z=0[/math] and the region of interest is [math]-h\ltz\lt0[/math]. There is a body which occupies the region [math]\Omega[/math] and we denote the wetted surface of the body by [math]\partial\Omega[/math] We denote [math]\mathbf{r}=(x,y)[/math] as the horizontal coordinate in two or three dimensions respectively and the Cartesian system we denote by [math]\mathbf{x}[/math]. We assume that the bottom surface is of constant depth at [math]z=-h[/math].

The equations are the following

[math] \begin{align} \Delta\phi &=0, &-h\ltz\lt0,\,\,\mathbf{x} \in \Omega \\ \partial_z\phi &= 0, &z=-h, \\ \partial_z \phi &= \alpha \phi, &z=0,\,\,\mathbf{x} \in \partial \Omega_{\mathrm{F}}, \end{align} [/math]

(note that the last expression can be obtained from combining the expressions:

[math] \begin{align} \partial_z \phi &= -\mathrm{i} \omega \zeta, &z=0,\,\,\mathbf{x} \in \partial \Omega_{\mathrm{F}}, \\ \mathrm{i} \omega \phi &= g\zeta, &z=0,\,\,\mathbf{x} \in \partial \Omega_{\mathrm{F}}, \end{align} [/math]

where [math]\alpha = \omega^2/g \,[/math])

[math] \partial_n\phi = \mathcal{L}\phi, \quad \mathbf{x}\in\partial\Omega_B, [/math]

where [math]\mathcal{L}[/math] is a linear operator which relates the normal and potential on the body surface through the physics of the body.

The simplest case is for a fixed body where the operator is [math]L=0[/math] but more complicated conditions are possible.

The equation is subject to some radiation conditions at infinity. We assume the following. [math]\phi^{\mathrm{I}}\,[/math] is a plane wave travelling in the [math]x[/math] direction,

[math] \phi^{\mathrm{I}}(x,z)=A \phi_0(z) e^{\mathrm{i} k x} \, [/math]

where [math]A [/math] is the wave amplitude (in potential) [math]\mathrm{i} k [/math] is the positive imaginary solution of the Dispersion Relation for a Free Surface (note we are assuming that the time dependence is of the form [math]\exp(-\mathrm{i}\omega t) [/math]) and

[math] \phi_0(z) =\frac{\cosh k(z+h)}{\cosh k h} [/math]

In two-dimensions the Sommerfeld Radiation Condition is

[math] \left( \frac{\partial}{\partial|x|} - \mathrm{i} k \right) (\phi-\phi^{\mathrm{{I}}})=0,\;\mathrm{{as\;}}|x|\rightarrow\infty\mathrm{.} [/math]

where [math]\phi^{\mathrm{{I}}}[/math] is the incident potential.

Solution in the Time Domain using the Frequency Domain Solution

Due to linearity we can write the solution in the time domain as a superposition of the frequency domain solutions. We restrict the potential to the free surface (remembering that the surface displacement is proportional to the potential).

[math] \Phi(x,t) = \mathrm{Re} \left\{\int_0^{\infty} \bar{f}(\omega) \left\{\phi(x,\omega)e^{-\mathrm{i} \omega t}\right\} \mathrm{d} \omega \right\} [/math]

For reasons which will become apparent we change the variable of integration to [math]k[/math] the wavenumber [math]ik =k_0[/math] in our standard notation for the solutions of the Dispersion Relation for a Free Surface.


[math] \Phi(x,t) = \mathrm{Re} \left\{\int_0^{\infty} \hat{f}(k) \left\{\phi(\mathbf{x},\omega(k))e^{-\mathrm{i} \omega(k) t}\right\} \mathrm{d} k \right\} [/math]

We now consider that we have an incident wave packet which has the shape [math]f(x)[/math] at [math]t=0[/math] in the absense of any scattering. Therefore

[math] f(x) = \mathrm{Re} \left\{\int_0^{\infty} \hat{f}(k) e^{\mathrm{i} k x} \right\} \mathrm{d} k [/math]

We can therefore calculate [math]\hat{f}(k)[/math] as

[math] \hat{f}(k) = \frac{1}{\pi} \int_{-\infty}^{\infty} f(x) e^{-\mathrm{i} k x} \mathrm{d} x [/math]