[Kelvin Wedge] DFT Method Main

The math is sound. The method is not.

We consider the governing equations for water waves in the context of an inviscid fluid of infinite depth. The fundamental system of equations are

\[ \begin{array}{c} \partial_x^2 \varphi+\partial_y^2 \varphi+\partial_z^2 \varphi=0 , \quad z \leq 0 \\ \left.\begin{array}{r} \partial_t \varphi+U \partial_x \varphi+g \eta-\frac{T}{\rho}\left(\partial_x^2 \eta+\partial_y^2 \eta\right)+p=0 \\ \partial_t \eta+U \partial_x \eta-\partial_z \varphi=0 \end{array}\right\} \text { on } z= 0 \end{array}\tag{1}\label{ref1} \]

Decompose in azimuthal modes

Define the order\(-m\) Hankel transform (Fourier-Bessel)

\[ \begin{aligned} & \hat{f}_m(k, t)=\int_0^{\infty} r J_m(k r) f_m(r, t) d r \\ & f_m(r, t)=\int_0^{\infty} k J_m(k r) \hat{f}_m(k, t) d k \end{aligned}\tag{2}\label{ref2} \] Laplace's equation in \(z\) gives the usual vertical structure: if the fluid is \(z<0\) and fields decay as \(z \rightarrow-\infty\).

\[ \begin{aligned} & \hat{\varphi}_m(k, z, t)=\hat{\Phi}_m(k, t) e^{k z} \\ \Rightarrow & \widehat{\left(\partial_z \varphi\right)_m}(k, 0, t)=k \hat{\Phi}_m(k, t) \end{aligned}\tag{3} \]

Ladder identities for single derivatives

Introduce the complex combinations \(D_{ \pm}:=\partial_x+i \partial_y\). in polar form. \[ D_{ \pm}=e^{ \pm i \theta}\left(\partial_r \pm \frac{i}{r} \partial_\theta\right)\tag{4} \]

Acting on a single azimuthal mode \(f_m(r) e^{i m \theta}\)

\[ \begin{aligned} & D_{+}\left[f_m e^{i m \theta}\right]=e^{i(m+1) \theta}\left(\partial_r f_m-\frac{m}{r} f_m\right) \\ & D_{-}\left[f_m e^{i m \theta}\right]=e^{i(m-1) \theta}\left(\partial_r f_m+\frac{m}{r} f_m\right) \end{aligned}\tag{5}\label{ref5} \] Under the Hankel transform these become pure algebra shifts

\[ \begin{aligned} & \mathscr{H}_{m+1}\left[\left(\partial_r-\frac{m}{r}\right) f_m\right]=-k \hat{f}_m \\ & \mathscr{H}_{m-1}\left[\left(\partial_r+\frac{m}{r}\right) f_m\right]=k \hat{f}_m \end{aligned}\tag{6}\label{ref6} \]

[[kelvin-wedge-operator-ladder-relations-for-hankel-transforms-of-bessel-type]]

Continue eqn \(\eqref{ref5}\) and \(\eqref{ref6}\)

\[ \begin{aligned} \mathscr{H}_{m+1}\left[\left(\partial_r-\frac{m}{r}\right) f_m\right] & =-k \int_0^{\infty} f_m(r) J_m(k r) r d r \\ & =-k \hat{f}_m(k) \end{aligned}\tag{7} \]

Similarly:

\[ \mathscr{H}_{m+1}\left[\left(\partial_r+\frac{m}{r}\right) f_m\right]=k \hat{f}_m(k)\tag{8} \]

Since

\[ \partial_x=\frac{1}{2}\left(D_{+}+D_{-}\right) \quad\text {and} \quad\partial_y=\frac{1}{2 i}\left(D_{+}-D_{-}\right)\tag{9} \]

The mode-n Hankel coefficients of \(\partial_x f\) and \(\partial_y f\)

\[ \begin{aligned} & {\widehat{\left(\partial_x f\right)_n}}(k)=\frac{k}{2}\left(\hat{f}_{n-1}(k)-\hat{f}_{n+1}(k)\right) \\ & {\widehat{\left(\partial_y f\right)}}_n(k)=\frac{k}{2 i}\left(\hat{f}_{n-1}(k)+\hat{f}_{n+1}(k)\right) \end{aligned}\tag{10} \]

[[kelvin-wedge-cartesian-derivatives-in-azimuthal-fourier-hankel-space]]

Rewrite the free surface system \(\eqref{ref1}\) with time-harmonic form \[ \begin{aligned} & \eta(r, \theta, t) \backsim \eta(r, \theta) e^{-i \omega t} \\ & \varphi(r, \theta, t) \backsim \varphi(r, \theta) e^{-i \omega t} \\ & p(r, \theta, t) \backsim p(r, \theta) e^{-i \omega t} \end{aligned}\tag{11}\label{ref11} \] It's straightforward to show in azimuthal harmonics that \[ \begin{aligned} & \eta(r, \theta) \backsim \sum_{m \in \mathbb{Z}} \eta_m(r) e^{i m \theta} \\ & \varphi(r, \theta) \backsim \sum_{m \in \mathbb{Z}} \varphi_m(r) e^{i m \theta} \\ & p(r, \theta) \backsim \sum_{m \in \mathbb{Z}} P_m(r) e^{i m \theta} \end{aligned}\tag{12}\label{ref12} \] The expression for laplace's equation in a polar coordinate system is given by \[ \nabla^2 \varphi=\left(\partial_r^2+\frac{1}{r} \partial_r+\frac{1}{r^2} \partial_\theta^2\right) \varphi+\partial_z^2 \varphi=0\tag{13}\label{ref13} \] Substituting \(\eqref{ref13}\) into \(\eqref{ref12}\) leads to

\[ \left(\partial_r^2+\frac{1}{r} \partial_r-\frac{n^2}{r^2}\right) \varphi_n+\partial_z^2 \varphi_n=0\tag{14}\label{ref14} \] Based on equations \(\eqref{ref1}\) and \(\eqref{ref14}\), applying the hankel transform to the order-\(n\) coefficients yields a tridiagonal coupling in \(m\) for each \(k\). \[ \begin{array}{c} -k^2 \hat{\varphi}_n+\partial_z^2 \hat{\varphi}_n=0 , \quad z \leq 0 \\ \left.\begin{array}{r} -i \omega \hat{\varphi}_n+\frac{U k}{2}\left(\hat{\varphi}_{n-1}-\hat{\varphi}_{n+1}\right)+\left(g+\frac{T}{\rho} k^2\right) \hat{\eta}_n+\hat{p}_n=0 \\ -i \omega \hat{\eta}_n+\frac{U k}{2}\left(\hat{\eta}_{n-1}-\hat{\eta}_{n+1}\right)-k \hat{\varphi}_n=0 \end{array}\right\} \text { on } z=0 \end{array}\tag{15}\label{ref15} \]

[[kelvin-wedge-hankel-diagonalization-of-the-radial-bessel-operator]]

For an infinitely deep fluid occupying \(z<0\), with decay as \(z \rightarrow-\infty\), the solution of the first equation of \(\eqref{ref15}\) gives

\[ \hat{\varphi}_n(k, z)=\hat{\Phi}_n(k) e^{k z}, \quad z \leq 0\tag{16} \]

so at the surface \(z=0\)

\[ \hat{\varphi}_n(k, 0)=\hat{\Phi}_n(k)\tag{17} \]

Therefore the control equations may be written.

\[ \begin{array}{c} \hat{\varphi}_n(k, z)=\hat{\Phi}_n(k) e^{k z}, \quad z \leq 0 \\ \left.\begin{array}{c} -i \omega \hat{\Phi}_n+\frac{U k}{2}\left(\hat{\Phi}_{n-1}-\hat{\Phi}_{n+1}\right)+\left(g+\frac{T}{\rho} k^2\right) \hat{\eta}_n+\hat{p}_n=0 \\ -i \omega \hat{\eta}_n+\frac{U k}{2}\left(\hat{\eta}_{n-1}-\hat{\eta}_{n+1}\right)-k \hat{\Phi}_n=0 \end{array}\right\} z=0 \end{array}\tag{18} \] we diagonalize the \(m\)-coupling (DFT over \(m\) )

DFT: Discrete Fourier Transform

for each fixed \(k\), define the angular spectra \[ \begin{aligned} & \tilde{\Phi}(k, \beta)=\sum_{n \in \mathbb{Z}} \hat{\Phi}_n(k) e^{i n \beta} \\ & \tilde{\eta}(k, \beta)=\sum_{n \in \mathbb{Z}} \hat{\eta}_n(k) e^{i n \beta} \\ & \tilde{p}(k, \beta)=\sum_{n \in \mathbb{Z}} \hat{p}_n(k) e^{i n \beta} \end{aligned}\tag{19} \]

shifts becomes multipliers

\[ \begin{aligned} \sum_n \hat{q}_{n+1} e^{i n \beta} & =\sum_m \hat{q}_m e^{i(m-1) \beta} \\ & =e^{-i \beta} \sum_m \hat{q}_m e^{i m \beta} \\ & =e^{-i \beta} \tilde{q}(k, \beta) \\ \sum_n \hat{q}_{n-1} e^{i n \beta} & =e^{i \beta} \sum_m \hat{q}_m e^{i m \beta} \\ & =e^{i \beta} \tilde{q}(k, \beta) \end{aligned}\tag{20} \]

where \[ \tilde{q}=\sum_{m \in \mathbb Z} \hat{q}_m e^{i m \beta}\tag{21} \] Hence \[ \sum_n\left(q_{n-1}-q_{n+1}\right) e^{i n \beta}=2 i \sin \beta \widetilde{q}\tag{22} \]

[[kelvin-wedge-diagonalizing-azimuthal-mode-coupling-via-a-discrete-fourier-transform]]

Recap

1. physical-space angular Fourier series (over physical angle \(\theta\) )

\[ f(r, \theta)=\sum_{m \in \mathbb{Z}} f_m(r) e^{i m \theta}, f_m(r)=\frac{1}{2 \pi} \int_0^{2 \pi} f(r, \theta) e^{-i m \theta} d \theta\tag{23} \]

This decomposes \(f\) by azimuthal harmonics act each radius \(r\) 2. After that, for each \(m\) we take the order- \(m\) Hankel transform in \(r\)

\[ \hat{f}_m(k)=\int_0^{\infty} r J_m(k r) f_m(r) d r\tag{24} \]

Now we have the sequence \(\left\{\hat{f}_m(k)\right\}_{m \in \mathbb{Z}}\) for each \(k\) 3. To diagonalize the \(m \leftrightarrow m \pm 1\) coupling, we perform a discrete Fourier transform over index \(m\). That's what

\[ \tilde{f}(k, \beta)=\sum_{m \in \mathbb{Z}} \hat{f}_m(k) e^{i m \beta}\tag{25} \]

Hear \(\beta\) is an auxiliary angle in \(k\)-space (a spectral angle) not the physical angle.

Overall, we just move through spaces

\[ f(r, \theta) \xrightarrow{\text { Fourier in } \theta}\left\{f_m(r)\right\}_m \xrightarrow{\text {Hankel in } r}\{\hat{f}(m)\}_m \xrightarrow{\text { DFT in } m} \tilde{f}(k, \beta)\tag{26}\label{ref26} \]

where \(\left\{f_m(r)\right\}_m\) and \(\{\hat{f}(m)\}_m\) indicate the whole sequence indexed by \(m \in \mathbb{Z}\).

Reconstruction of \(f(r, \theta)\) from \(\tilde{f}(k, \beta)\).

Through a sequence of inverse operations, reversing the spatial transformations in \(\eqref{ref26}\), we ultimately obtain \(f(r, \theta)\) from \(\tilde{f}(k, \beta)\). \[ \begin{aligned} f(r, \theta) & =\sum_{m \in \mathbb{Z}} e^{i m \theta}\left[\int_0^{\infty} k J_m(k r)\left(\frac{1}{2 \pi} \int_0^{2 \pi} \tilde{f}(k, \beta) e^{-i m \beta} d \beta\right) d k\right] \\ & =\frac{1}{2 \pi} \int_0^{\infty} \int_0^{2 \pi}\left(\sum_{m \in \mathbb{Z}} e^{i m(\theta-\beta)} J_m(k r)\right) k \tilde{f}(k, \beta) d \beta d k \end{aligned}\tag{27} \]

Boundary conditions

1. requires \(f(r, \theta)\) to be \(2 \pi\)-periodic in \(\theta\)
2. near \(r=0: \quad f_m(r)=\mathscr{O}\left(|r|^m\right)\), so that \(r f_m(r)\) is bounded and boundary vanish; near \(r \rightarrow \infty:\quad f_m(r)\) decays fast enough that

\[ \int_0^{\infty} r | f_m(r)| d r<\infty \quad \text { (or } f_m \in \mathscr{L}^2((0, \infty) ; r d r)\tag{28} \]

3. \(f(r, \theta)\) is sufficiently smooth in \(\theta\)

Water waves

At this point, we turn to the ship wave problem, from the previous analysis, multiplying both sides of equation \(\eqref{ref26}\) by \(e^{i n \beta}\) and then summing over \(n\) on both sides yields:

\[ \begin{aligned} &\begin{aligned} -i \omega \sum_n \hat{\Phi}_n e^{i n \beta}+\frac{U k}{2}\left(\sum_n \right .&\left. \hat{\Phi}_{n-1} e^{i n \beta}-\sum_n \hat{\Phi}_{n+1} e^{i n \beta}\right) \\ &+\left(g+\frac{T}{\rho} k^2\right) \sum_n \hat{\eta}_n e^{i n \beta}+ \sum_n \hat{p}_n e^{i n \beta}=0 \end{aligned}\\ &-i \omega \sum_n \hat{\eta}_n e^{i n \beta}+ \frac{U k}{2}\left(\sum_n \hat\eta_{n-1} e^{i n \beta}-\sum_n \hat\eta_{n+1} e^{i n \beta}\right)-k \sum_n \hat{\Phi}_n e^{i n \beta}=0 \end{aligned}\tag{29} \] Therefore, it gives

\[ \begin{array}{l} -i \omega \widetilde{\Phi}+U k i \sin \beta \widetilde{\Phi}+\left(g+\frac{T}{\rho} k^2\right) \widetilde{\eta}+\widetilde{p}=0 \\ -i \omega \tilde{\eta}+U k i \sin \beta \tilde{\eta}-k \widetilde{\Phi}=0 \end{array}\tag{30} \]

where \(\widetilde{\eta}=\widetilde{\eta}(k, \beta), \widetilde{\Phi}=\widetilde{\Phi}(k, \beta)\).

therefore

\[ \begin{aligned} & \widetilde{\eta}=-\frac{k \tilde{p}}{g k+\frac{T}{\rho} k^3-\Omega^2} \\ & \widetilde{\Phi}=\frac{i \Omega \tilde{p}}{g k+\frac{T}{\rho} k^3-\Omega^2} \end{aligned}\tag{31} \]

where \(\Omega=\omega-U k \sin \beta\), it follows that

\[ \begin{aligned} & \hat{\eta}_m(k)=\frac{1}{2 \pi} \int_0^{2 \pi}-\frac{k \tilde{p}}{g k+\frac{T}{\rho} k^3-\Omega^2} e^{-i m \beta} d \beta \\ & \hat{\Phi}_m(k)=\frac{1}{2 \pi} \int_0^{2 \pi} \frac{i \Omega \tilde{p}}{g k+\frac{T}{\rho} k^3-\Omega^2} e^{-i m \beta} d \beta \end{aligned}\tag{32} \]

Shift \(\gamma=\beta-\frac{\pi}{2}\), so \(\sin \beta=\cos \gamma\) write

\[ \begin{aligned} D(k, \beta) & =g k+\frac{T}{\rho} k^3-(\omega-U k \sin \beta)^2 \\ & =a(k)+b(k) \cos \gamma+c(k) \cos ^2 \gamma \\ & =D(k, \gamma) \end{aligned}\tag{33} \]

where \[ a=g k+\frac{T}{\rho} k^3-\omega^2, \quad b=2 \omega U k,\quad c=-U^2 k^2\tag{34} \]

We define the \(\gamma\) - Fourier coefficients

\[ c_m(k)=\frac{1}{2 \pi} \int_0^{2 \pi} \frac{e^{-i m \gamma}}{a+b \cos \gamma+c \cos ^2 \gamma} d \gamma \tag{35}\label{ref35} \]

Now we consider the form of \(p\), we assume it as \[ p=\bar{p} H(r-l) \] Where \(H\) is Heaviside function

Hankel transform of \(p=\bar{p}H(r-l)\)

\[ \begin{aligned} & p(r, \theta)=\sum_{m \in \mathbb{Z}} P_m(r) e^{i m \theta} \\ & \hat{p}_m(k)=\int_0^{\infty} r J_m(k r) P_m(r) d r \end{aligned} \] Then DFT over \(m\) to get \(\tilde{p}(k, \beta)=\sum_m \hat{p}(k) e^{i m \beta}\)

\(p\) is axisymmetric (no \(\theta-\text{dependence}\)) only \(m=0\) mode is nonzero \[ P_0(r)=\bar{p} H(r-l), \quad P_m(r)=0 \quad(m \neq 0) \]

[[kelvin-wedge-axisymmetric-angular-fourier-modes-and-hankel-transform-of-a-radial-heaviside-pressure-field]]

Its order-\(0\) Hankel transform

\[ \begin{aligned} \hat{p}_0(k) & =\bar{p} \int_0^{\infty} r J_0(k r) H(r-l) d r \\ & =\bar{p} \int_l^{\infty} r J_0(k r) d r \end{aligned} \]

Now decompose the Heaviside as

\[ H(r-l)=1-H(l-r) \]

The "1" part may be a uniform pressure to infinity (like hydrostatic offset, no radiating waves) the localized part is \(\bar{p} H\left(l-r\right)\). Wave generation depends on spatial variations, so we throw away the uniform part and keep only localized patch:

\[ p_{l o c}(r)=-\bar{p} H(l-r) \]

This is exactly a circular pressure patch of radius \(l\) strength \(-\bar{p}\). For this

\[ \begin{aligned} \hat{p}_0^{(l o c)}(k) & =-\bar{p} \int_0^l r J_0(k r) d r \\ & =-\bar{p} \frac{l J_1(k l)}{k} \end{aligned} \]

Here \(\bar{p}(k)\) means \(\tilde{p}(k, \beta)\) is independent of \(\beta\)

Convert back from \(\gamma\) to \(\beta: e^{-i m \beta}=(-i)^m e^{-i m \gamma}\), with axisymmetric forcing \(\widetilde{p}(k)\)

\[ \begin{aligned} & \hat{\eta}_m=-k \tilde{p}(k)(-i)^m c_m(k) \\ & \hat{\Phi}_m(k)=i \tilde{p}(k)(-i)^m\left[\omega c_m(k)-\frac{U k}{2}\left(c_{m-1}(k)+c_{m+1}(k)\right)\right] \end{aligned} \]

substitute into \(\eta, \Phi\) equation

\[ \begin{aligned} \eta(r, \theta) & =-\sum_m(-i)^m e^{i m \theta} \int_0^{\infty} k^2 J_m(k r) \tilde{p}(k) c_m(k) d k \\ & =-\sum_m(-i)^m e^{i m \theta} \int_0^{\infty} k^2 J_m(k r)\left[-\bar{p} \frac{l J_1(k l)}{k}\right] c_m(k) d k \\ & =\bar{p} l \sum_{m \in \mathbb{Z}}(-i)^m e^{i m \theta} \int_0^{\infty} k J_m(k r) J_1(k l) c_m(k) d k \end{aligned} \]

similarly

\[ \begin{aligned} \Phi(r, \theta) & =i \sum_m(-i)^m e^{i m \theta} \int_0^{\infty} k J_m(k r) \tilde{p}(k)\left[\omega c_m-\frac{U k}{2}\left(c_{m-1}+c_{m+1}\right)\right] d k \\ &=i \bar{p} l \sum_m(-i)^m e^{i m \theta} \int_0^{\infty} J_m(k r) J_1(k l)\left[\omega c_m-\frac{U k}{2} \left(c_{m-1}+c_{m+1}\right)\right] d k \end{aligned} \]