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
\[ \begin{aligned} & \mathscr{H}_{m+1}\left[\left(\partial_r-\frac{m}{r}\right) f_m\right]=-k \hat{f}_m \\ & \mathscr{H}_{m-1}\left[\left(\partial_r+\frac{m}{r}\right) f_m\right]=k \hat{f}_m \end{aligned}\tag{6}\label{ref6} \]
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 \(\partial_y f\)
\[ \begin{aligned} & {\widehat{\left(\partial_x f\right)_n}}(k)=\frac{k}{2}\left(\hat{f}_{n-1}(k)-\hat{f}_{n+1}(k)\right) \\ & {\widehat{\left(\partial_y f\right)}}_n(k)=\frac{k}{2 i}\left(\hat{f}_{n-1}(k)+\hat{f}_{n+1}(k)\right) \end{aligned}\tag{10} \]
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 \(k\). \[ \begin{array}{c} -k^2 \hat{\varphi}_n+\partial_z^2 \hat{\varphi}_n=0 , \quad z \leq 0 \\ \left.\begin{array}{r} -i \omega \hat{\varphi}_n+\frac{U k}{2}\left(\hat{\varphi}_{n-1}-\hat{\varphi}_{n+1}\right)+\left(g+\frac{T}{\rho} k^2\right) \hat{\eta}_n+\hat{p}_n=0 \\ -i \omega \hat{\eta}_n+\frac{U k}{2}\left(\hat{\eta}_{n-1}-\hat{\eta}_{n+1}\right)-k \hat{\varphi}_n=0 \end{array}\right\} \text { on } z=0 \end{array}\tag{15}\label{ref15} \]
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 \[ \sum_n\left(q_{n-1}-q_{n+1}\right) e^{i n \beta}=2 i \sin \beta \widetilde{q}\tag{22} \]
Recap
- 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
- requires \(f(r, \theta)\) to be \(2 \pi\)-periodic in \(\theta\)
- 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
\[ \int_0^{\infty} r | f_m(r)| d r<\infty \quad \text { (or } f_m \in \mathscr{L}^2((0, \infty) ; r d r)\tag{28} \]
- \(f(r, \theta)\) 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 \[ P_0(r)=\bar{p} H(r-l), \quad P_m(r)=0 \quad(m \neq 0) \]
Its order-\(0\) Hankel transform
\[ \begin{aligned} \hat{p}_0(k) & =\bar{p} \int_0^{\infty} r J_0(k r) H(r-l) d r \\ & =\bar{p} \int_l^{\infty} r J_0(k r) d r \end{aligned} \]
Notes
Statement:
\[ \hat{p}_0(k)=\bar{p} \int_0^{\infty} r J_0(k r) d r \]
we have \[ H(r-l)= \begin{cases}1, & r \geqslant l \\ 0, & r<l\end{cases} \] therefore
\[ \begin{aligned} & \bar{p} \int_0^{\infty} r J_0(k r) H(r-l) d r \\ = & \bar{p}\left[\int_0^l r J_0(k r) H(r-l) d r+\int_l^{\infty} r J_0(k r) H(r-l) d r\right] \\ = & \bar{p} \int_l^{\infty} r J_0(k r) d r \end{aligned} \]
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 \(\widetilde{p}(k)\)
\[ \begin{aligned} & \hat{\eta}_m=-k \tilde{p}(k)(-i)^m c_m(k) \\ & \hat{\Phi}_m(k)=i \tilde{p}(k)(-i)^m\left[\omega c_m(k)-\frac{U k}{2}\left(c_{m-1}(k)+c_{m+1}(k)\right)\right] \end{aligned} \]
Notes
Statement: \[ \hat{\Phi}_m(k)=i \tilde{p}(k)(-i)^m\left[\omega c_m(k)-\frac{U k}{2}\left(c_{m-1}(k)+c_{m+1}(k)\right)\right] \]
we have \[ \widetilde{\Phi}(k, \beta)=\frac{i \Omega(\beta) \widetilde{p}(k, \beta)}{D(k, \beta)} \]
we start from the definition:
\[ \begin{aligned} \hat{\Phi}_m(k) & =\frac{1}{2 \pi} \int_0^{2 \pi} \tilde{\Phi}(k, \beta) e^{-i m \beta} d \beta \\ & =\frac{1}{2 \pi} \int_0^{2 \pi} \frac{i \Omega(\beta) \tilde{p}(k)}{D(k, \beta)} e^{-i m \beta} d \beta \end{aligned} \]
factor out the \(i \tilde{p}(k)\)
\[ \hat{\Phi}_m(k)=i \widetilde{p}_{(k)} \frac{1}{2 \pi} \int_0^{2 \pi} \frac{\Omega(\beta) e^{-i m \beta}}{D(k, \beta)} d \beta \]
change variable \(\beta \rightarrow \gamma\)
let,
\[ \gamma=\beta-\frac{\pi}{2} \]
Also
\[ e^{-i m \beta}=e^{-i m\left(\gamma+\frac{\pi}{2}\right)}=(-i)^m e^{-i m \gamma} \]
And
\[ \Omega(\beta)=\omega-U k \sin \beta=\omega-U k \cos \gamma \]
split the integral into \(\omega\)-part and \(\cos \gamma\) part
\[ \frac{1}{2 \pi} \int_0^{2 \pi} \frac{(\omega-U k \cos \gamma) e^{-i m \gamma}}{a+b \cos \gamma+c \cos ^2 \gamma} d \gamma=\omega I_1-U k I_2 \]
where
\[ \begin{aligned} & I_1:=\frac{1}{2 \pi} \int_0^{2 \pi} \frac{e^{-i m \gamma}}{a+b \cos \gamma+c \cos ^2 \gamma} d \gamma \\ & I_2:=\frac{1}{2 \pi} \int_0^{2 \pi} \frac{\cos \gamma e^{-i m \gamma}}{a+b \cos \gamma+c \cos ^2 \gamma} d \gamma \end{aligned} \] By definition of \(c_m(k), I_1=c_m(k)\), for \(I_2\), use the standard identity \[ \cos \gamma \ e^{-i m \gamma}=\frac{1}{2}\left(e^{-i(m+1) \gamma}+e^{-i(m-1) \gamma}\right) \]
Therefore
\[ \begin{aligned} I_2 & =\frac{1}{2 \pi} \int_0^{2 \pi} \frac{\cos \gamma e^{-i m \gamma}}{a+b \cos \gamma+c \cos ^2 \gamma} d \gamma \\ & =\frac{1}{2}\left[\frac{1}{2 \pi} \int_0^{2 \pi} \frac{e^{-i(m+1) \gamma}}{a+b \cos \gamma+c \cos ^2 \gamma} d \gamma+\frac{1}{2 \pi} \int_0^{2 \pi} \frac{e^{-i(m-1) \gamma}}{a+b \cos \gamma+c \cos ^2 \gamma} d \gamma\right] \\ & =\frac{1}{2}\left(c_{m-1}(k)+c_{m+1}(k)\right) \end{aligned} \]
put it back into \(\widehat{\Phi}_m\)
\[ \begin{aligned} \hat{\Phi}_m(k) & =i \tilde{p}(k)(-i)^m\left(\omega I_1-U k I_2\right) \\ & =i \tilde{p}(k)(-i)^m\left[\omega c_m-\frac{U k}{2}\left(c_{m-1}+c_{m+1}\right)\right] \end{aligned} \]
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
\[ \begin{aligned} \Phi(r, \theta) & =i \sum_m(-i)^m e^{i m \theta} \int_0^{\infty} k J_m(k r) \tilde{p}(k)\left[\omega c_m-\frac{U k}{2}\left(c_{m-1}+c_{m+1}\right)\right] d k \\ &=i \bar{p} l \sum_m(-i)^m e^{i m \theta} \int_0^{\infty} J_m(k r) J_1(k l)\left[\omega c_m-\frac{U k}{2} \left(c_{m-1}+c_{m+1}\right)\right] d k \end{aligned} \]
Notes
Statement: \[ c_{2 n+1}(k)=0 \]
This follows from a symmetry (periodicity) argument. Define the integrand \[ f_m(\gamma)=\frac{e^{-i m \gamma}}{A^{\prime}-B^{\prime} \cos 2 \gamma} \]
where
\[ \begin{aligned} & A^{\prime}=g k+\frac{T}{\rho} k^3-\frac{U k^2}{2} \\ & B^{\prime}=\frac{U^2 k^2}{2} \end{aligned} \]
it readily to see
\[ \begin{aligned} g k+\frac{T}{\rho k^3} & -U^2 k^2 \cos ^2 \gamma \\ & =A^{\prime}(k)-B^{\prime}(k) \cos 2 \gamma \end{aligned} \]
Now evaluate \(c_m\) by splitting the integral into \([0, \pi]\) and \([\pi, 2 \pi]\) \[ c_m=\frac{1}{2 \pi}\left(\int_0^\pi f_m(\gamma) d \gamma+\int_\pi^{2 \pi} f_m(\gamma) d \gamma\right) \]
In the second integral, substitute \(\gamma=\gamma^{\prime}+\pi\) so
\[ \gamma^{\prime} \in[0, \pi] \]
Hence,
\[ \int_\pi^{2 \pi} f_m(\gamma) d \gamma=\int_0^\pi \frac{e^{-i m\left(\gamma^{\prime}+\pi\right)}}{A^{\prime}-B^{\prime} \cos 2\left(\gamma^{\prime}+\pi\right)} d \gamma^{\prime}=\int_0^\pi \frac{(-1)^m e^{-i m \gamma^{\prime}}}{A^{\prime}-B^{\prime} \cos 2 \gamma} d \gamma^{\prime} \]
Therefore
\[ c_m=\frac{1}{2 \pi}\left[1+(-1)^m\right] \int_0^\pi \frac{e^{-i m \gamma}}{A^{\prime}-B^{\prime} \cos 2 \gamma} d \gamma \]
If \(m\) is odd then \((-1)^m=-1\) so \(c_m=0\). That is
\[ c_{2 n+1}(k)=0 \] Let \(m=2 n\) Then
\[ c_m=c_{2 n}=\frac{1}{2 \pi} \int_0^{2 \pi} \frac{e^{-i 2 n \gamma}}{A^{\prime}-B^{\prime} \cos 2 \gamma} d \gamma \]
now make the substitution \(a=2 \gamma\), then we have
\[ d \alpha=2 d \gamma, \quad \alpha \in[0,4 \pi] \]
Then,
\[ c_{2 n}(k)=\frac{1}{2 \pi} \int_0^{4 \pi} \frac{e^{-i n \alpha}}{A^{\prime}-B^{\prime} \cos \alpha} \frac{1}{2} d \alpha \]
Because \(\cos \alpha\) and \(e^{-i n \alpha}\) are \(2 \pi\)-periodic, the integral over \([0,4 \pi]\) is twice the integral over \([0,2 \pi]\), so
\[ \int_0^{4 \pi} \frac{e^{-i n \alpha}}{A^{\prime}-B^{\prime} \cos \alpha} d \alpha=2 \int_0^{2 \pi} \frac{e^{-i n \alpha}}{A^{\prime}-B^{\prime} \cos \alpha} d \alpha \]
Therefore
\[ c_{2 n}(k)=\frac{1}{2 \pi} \cdot \frac{1}{2} \cdot 2 \int_0^{2 \pi} \frac{e^{-i n \alpha}}{A^{\prime}-B^{\prime} \cos \alpha} d \alpha=\frac{1}{2 \pi} \int_0^{2 \pi} \frac{e^{-i n \alpha}}{A^{\prime}-B^{\prime} \cos \alpha} d \alpha \]
so we have reduced to the classical Fourier coefficients of \(1 /\left(A^{\prime}-B^{\prime} \cos \theta\right)\)
