The math is sound. The method is not.
We factor the quadratic denominator \(a+b \cos \gamma+c \cos ^2 \gamma\) and reduce the computation of its Fourier coefficients to those of \(1 /(\cos \gamma-\lambda)\). Solving the associated second-order recurrence for the Fourier coefficients yields a closed-form expression in terms of \(\lambda_{ \pm}\)and \(q_{ \pm}=\lambda_{ \pm}+i \sqrt{1-\lambda_{ \pm}^2}\), leading to the final explicit formula for \(c_m(k)\) in the steady ship-wave setting.
Statement
Let \[ D(k,\gamma) = a + b\cos\gamma + c\cos^2\gamma, \] with \(c\neq 0\) and discriminant \(b^2 - 4ac \neq 0\). Let \(\lambda_{\pm}\) be the roots of \[ c\lambda^2 + b\lambda + a = 0, \qquad \lambda_{\pm} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2c}. \]
Let \(c_m\) be the Fourier coefficients defined by \[ \frac{1}{D(k,\gamma)} = \sum_{m\in\mathbb Z} c_m(k)\,e^{im\gamma}. \]
Then, for the steady problem (so that \(\lambda_{\pm}\in(-1,1)\), see the discussion below), the coefficients can be expressed as \[ c_m(k)=-\frac{1}{c(\lambda_+ - \lambda_-)}\left[\frac{q_+^{|m|}}{\sqrt{\lambda_+^2 - 1}}-\frac{q_-^{|m|}}{\sqrt{\lambda_-^2 - 1}}\right],\tag{1}\label{ref1} \] where \[ q_{\pm} = \lambda_{\pm} + i\sqrt{1-\lambda_{\pm}^2} = e^{i\theta_{\pm}}, \qquad \lambda_{\pm} = \cos\theta_{\pm},\; \theta_{\pm}\in (0,\pi). \]
(Remark: depending on the branch choice for \(\sqrt{\lambda^2-1}\), an overall minus sign may be absorbed into the square roots; here we use the convention consistent with the classical Fourier/Chebyshev formula for \(1/(\cos\gamma-\lambda)\).)
Range of \(\lambda\) in the steady case
In the steady case \(\omega = 0\), the dispersion relation for linear capillary–gravity waves gives \[ \omega_0^2 = gk + \frac{T}{\rho}k^3, \] and the phase speed \[ c_p = \frac{\omega_0}{k}. \]
In the present notation, one finds (after non-dimensionalisation) that \[ \lambda = \frac{\sqrt{4\left(gk + \frac{T}{\rho}k^3\right) U^2 k^2}}{2U^2 k^2} = \frac{2\omega_0 U k}{2 U^2 k^2} = \frac{\omega_0}{Uk} = \frac{c_p}{U}. \]
For a steady ship-wave pattern, the stationarity condition implies \[ U\cos\theta = c_p, \] where \(\theta\) is the angle between the wavevector and the direction of motion. Hence \[ \lambda = \frac{c_p}{U} = \cos\theta, \qquad |\lambda| = |\cos\theta| < 1. \]
This justifies the assumption \[ \lambda_\pm \in (-1,1) \] for the steady problem.
Reduction to \(\displaystyle F(\gamma) = \frac{1}{\cos\gamma - \lambda}\)
We begin with \[ D(k,\gamma) = a + b\cos\gamma + c\cos^2\gamma. \tag{2} \] Factorising the quadratic in \(\cos\gamma\) gives \[ D(k,\gamma) = c\bigl(\cos\gamma - \lambda_+\bigr)\bigl(\cos\gamma - \lambda_-\bigr), \tag{3} \] and hence the partial fraction decomposition \[ \frac{1}{a + b\cos\gamma + c\cos^2\gamma} = \frac{1}{c(\lambda_+ - \lambda_-)} \left( \frac{1}{\cos\gamma - \lambda_+} - \frac{1}{\cos\gamma - \lambda_-} \right). \tag{4}\label{ref4} \]
Thus it suffices to compute the Fourier coefficients of \[ F(\gamma) = \frac{1}{\cos\gamma - \lambda}, \qquad \lambda\in(-1,1), \tag{5} \] in the form \[ F(\gamma) = \sum_{m\in\mathbb Z} f_m(\lambda)\,e^{im\gamma}. \tag{6} \]
Once the coefficients \(f_m(\lambda)\) are known, the coefficients \(c_m(k)\) follow immediately by substituting \(\lambda = \lambda_\pm\) in \(\eqref{ref4}\).
Fourier coefficients of \(\displaystyle \frac{1}{\cos\gamma - \lambda}\)
Derivation of the recurrence
The defining identity is \[ (\cos\gamma - \lambda) F(\gamma) = 1. \tag{7} \] Insert the Fourier expansions \[ \cos\gamma = \tfrac12\left(e^{i\gamma} + e^{-i\gamma}\right), \qquad F(\gamma) = \sum_{m\in\mathbb Z} f_m e^{im\gamma}. \tag{8} \]
Then \[ \begin{aligned} \cos\gamma\,F(\gamma) &= \frac{1}{2}\sum_m f_m e^{i(m+1)\gamma} + \frac{1}{2}\sum_m f_m e^{i(m-1)\gamma} \\ &= \frac{1}{2}\sum_m\bigl(f_{m-1} + f_{m+1}\bigr)e^{im\gamma}, \end{aligned} \tag{9} \] so that \[ (\cos\gamma - \lambda)F(\gamma) = \sum_m\left[ \tfrac12\bigl(f_{m-1} + f_{m+1}\bigr) - \lambda f_m \right]e^{im\gamma}. \tag{10} \]
On the other hand, the constant function 1 has the Fourier expansion \[ 1 = \sum_m \delta_{m0}e^{im\gamma}, \tag{11} \] where \(\delta_{m0}\) is the Kronecker delta. Equating coefficients of \(e^{im\gamma}\) for each \(m\in\mathbb Z\) gives \[ \frac{1}{2}\bigl(f_{m-1} + f_{m+1}\bigr) - \lambda f_m = \delta_{m0}. \tag{12} \]
This is a second-order linear recurrence with a single inhomogeneous point at \(m=0\):
- For \(m\neq 0\) (homogeneous recurrence), \[ \frac{1}{2}\bigl(f_{m+1} + f_{m-1}\bigr) - \lambda f_m = 0. \tag{13}\label{ref13} \]
- For \(m=0\) (inhomogeneous condition), \[ \frac{1}{2}\bigl(f_{-1} + f_1\bigr) - \lambda f_0 = 1. \tag{14}\label{ref14} \]
Notes
§ Evenness of the sequence \(\{f_m\}\)
The function \(F\) is even in \(\gamma\): \[ F(-\gamma) = \frac{1}{\cos(-\gamma) - \lambda} = \frac{1}{\cos\gamma - \lambda} = F(\gamma). \]
Using the Fourier expansion, \[ F(\gamma) = \sum_{m\in\mathbb Z} f_m e^{im\gamma}, \] we also have \[ F(-\gamma) = \sum_{m\in\mathbb Z} f_m e^{-im\gamma} = \sum_{m\in\mathbb Z} f_{-m} e^{im\gamma}, \] after reindexing \(m\mapsto -m\).
Comparing coefficients of \(e^{im\gamma}\) gives \[ f_{-m} = f_m \quad \text{for all } m\in\mathbb Z. \] Thus the sequence \(\{f_m\}\) is even, and it suffices to determine \(f_m\) for \(m\ge 0\).
Solution of the homogeneous recurrence
For \(m\neq 0\), consider the homogeneous recurrence \(\eqref{ref13}\). Seek a solution of the form \[ f_m = \xi^{|m|}. \]
Substituting into \(\eqref{ref13}\) for \(m\ge 1\) gives \[ \frac{1}{2}\left(\xi^{m+1} + \xi^{m-1}\right) - \lambda \xi^m = 0 \quad\Longrightarrow\quad \xi^{m-1}\left(\xi^2 - 2\lambda\xi + 1\right)=0. \tag{15} \] Thus \(\xi\) must satisfy the quadratic \[ \xi^2 - 2\lambda\xi + 1 = 0, \] whose roots are \[ \xi_{\pm} = \lambda \pm \sqrt{\lambda^2 - 1}. \tag{16} \]
For \(\lambda\in(-1,1)\), set \[ \lambda = \cos\theta_0, \quad \theta_0\in(0,\pi), \qquad \sqrt{\lambda^2 - 1} = i\sin\theta_0, \tag{17} \] so that \[ \xi_{\pm} = \cos\theta_0 \pm i\sin\theta_0 = e^{\pm i\theta_0}. \tag{18} \]
Hence, for \(m\neq 0\), the general homogeneous solution is \[ f_m = A e^{i|m|\theta_0} + B e^{-i|m|\theta_0}. \tag{19} \] Writing \(q := e^{i\theta_0}\), we can express this more compactly as \[ f_m = A q^{|m|} + B q^{-|m|}, \qquad q = \lambda + i\sqrt{1-\lambda^2}. \tag{20} \]
Imposing the inhomogeneous condition at \(m=0\)
We now impose the inhomogeneous condition \(\eqref{ref14}\). Using the evenness \(f_{-m} = f_m\), we have \[ f_0 = A + B, \qquad f_1 = A q + B q^{-1}, \qquad f_{-1} = f_1. \tag{21} \]
Substituting into \(\eqref{ref14}\) gives \[ f_1 - \lambda f_0 = 1 \quad\Longrightarrow\quad A q + B q^{-1} - \lambda(A+B) = 1. \tag{22}\label{ref22} \]
It is convenient to use the identity \[ q + q^{-1} = 2\lambda, \tag{23} \] which follows from \(\lambda = \cos\theta_0\) and \(q = e^{i\theta_0}\).
A particularly convenient choice of initial values \((f_0,f_1)\) satisfying \(\eqref{ref22}\) and consistent with the homogeneous recurrence is \[ f_0 = \frac{2}{q - q^{-1}}, \qquad f_1 = \frac{2q}{q - q^{-1}}. \tag{24}\label{ref24} \]
Indeed, with these choices, \[ \begin{aligned} f_1 - \lambda f_0 &= \frac{2q}{q - q^{-1}} - \lambda\frac{2}{q - q^{-1}} \\ &= \frac{2(q-\lambda)}{q - q^{-1}} \\ &= \frac{2\left[q - \tfrac12(q+q^{-1})\right]}{q - q^{-1}} \\ &= 1. \end{aligned} \tag{25} \] Thus \(\eqref{ref24}\) satisfies the inhomogeneous condition, and the recurrence \(\eqref{ref13}\) then determines a unique sequence \(\{f_m\}_{m\in\mathbb Z}\).
By construction one checks that this sequence is given explicitly by \[ f_m(\lambda) = \frac{2q^{|m|}}{q - q^{-1}}, \qquad m\in\mathbb Z. \tag{26} \]
Rewriting in the \(\sqrt{\lambda^2-1}\) form
Using \(\lambda = \cos\theta_0\) and \(q=e^{i\theta_0}\), \[ q - q^{-1} = 2i\sin\theta_0, \quad \sin\theta_0 = \sqrt{1-\lambda^2}, \quad \sqrt{\lambda^2 - 1} = i\sqrt{1-\lambda^2} = i\sin\theta_0. \tag{27} \]
Hence \[ \frac{2}{q - q^{-1}} = \frac{2}{2i\sin\theta_0} = \frac{1}{i\sin\theta_0} = -\frac{1}{\sqrt{\lambda^2 - 1}}, \tag{28} \] and therefore \[ f_m(\lambda) = \frac{2q^{|m|}}{q - q^{-1}} = -\frac{q^{|m|}}{\sqrt{\lambda^2 - 1}}. \tag{29} \]
In particular, the Fourier coefficients are \[ \frac{1}{2\pi}\int_0^{2\pi} \frac{e^{-im\gamma}}{\cos\gamma - \lambda}\,d\gamma = f_m(\lambda) = -\frac{q^{|m|}}{\sqrt{\lambda^2 - 1}}, \qquad q = \lambda + i\sqrt{1-\lambda^2} = e^{i\arccos\lambda}, \tag{30} \] for \(\lambda\in(-1,1)\), interpreted via analytic continuation (or as a principal value in \(\gamma\)).
Application to \(a + b\cos\gamma + c\cos^2\gamma\)
Returning to \[ a + b\cos\gamma + c\cos^2\gamma = c(\cos\gamma - \lambda_+)(\cos\gamma - \lambda_-), \] with \[ \lambda_{\pm} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2c}, \] the partial fraction decomposition (4) gives \[ \frac{1}{a + b\cos\gamma + c\cos^2\gamma} = \frac{1}{c(\lambda_+ - \lambda_-)} \left( \frac{1}{\cos\gamma - \lambda_+} - \frac{1}{\cos\gamma - \lambda_-} \right). \]
Let \(F_m(\lambda)\) denote the \(m\)-th Fourier coefficient of \(1/(\cos\gamma - \lambda)\): \[ F_m(\lambda) = \frac{1}{2\pi}\int_0^{2\pi} \frac{e^{-im\gamma}}{\cos\gamma - \lambda}\,d\gamma = -\frac{q(\lambda)^{|m|}}{\sqrt{\lambda^2 - 1}}, \tag{31} \] where \[ q(\lambda) = \lambda + i\sqrt{1-\lambda^2}. \]
Then the coefficients \(c_m(k)\) of \(1/D(k,\gamma)\) are \[ c_m(k) = \frac{1}{c(\lambda_+ - \lambda_-)} \left( F_m(\lambda_+) - F_m(\lambda_-) \right), \] that is, \[ c_m(k) = -\frac{1}{c(\lambda_+ - \lambda_-)} \left[ \frac{q_+^{|m|}}{\sqrt{\lambda_+^2 - 1}} - \frac{q_-^{|m|}}{\sqrt{\lambda_-^2 - 1}} \right]\tag{32} \] with \[ q_{\pm} = \lambda_{\pm} + i\sqrt{1-\lambda_{\pm}^2}. \]
This is precisely the formula stated in \(\eqref{ref1}\), up to the overall sign convention in the choice of the square-root branches.