A.1 Solution to first Wiener–Hopf equation – $D^{(1)}$

In this section, we solve the integral equation (A 3) for $n=1$

(A 8)$$\begin{eqnarray}\displaystyle f^{(1)}(x)=4\unicode[STIX]{x03C0}\int _{-\infty }^{\infty }D^{(1)}(\unicode[STIX]{x1D6FE})j(\unicode[STIX]{x1D6FE})\text{e}^{-\text{i}\unicode[STIX]{x1D6FE}x}\,\text{d}\unicode[STIX]{x1D6FE}, & & \displaystyle\end{eqnarray}$$

subject to (A 4a) and (A 4b). Taking a Fourier transform of (A 8) in $x$ gives

(A 9)$$\begin{eqnarray}\displaystyle F_{-}^{(1)}(\unicode[STIX]{x1D6FE})+F_{+}^{(1)}(\unicode[STIX]{x1D6FE})=4\unicode[STIX]{x03C0}D_{+}^{(1)}(\unicode[STIX]{x1D6FE})j(\unicode[STIX]{x1D6FE}), & & \displaystyle\end{eqnarray}$$

where

(A 10a,b)$$\begin{eqnarray}\displaystyle F_{-}^{(1)}(\unicode[STIX]{x1D6FE})=\frac{1}{2\unicode[STIX]{x03C0}}\int _{-\infty }^{0}f^{(1)}(x)\text{e}^{\text{i}\unicode[STIX]{x1D6FE}x}\,\text{d}x,\quad F_{+}^{(1)}(\unicode[STIX]{x1D6FE})=\frac{1}{2\unicode[STIX]{x03C0}}\int _{0}^{\infty }f^{(1)}(x)\text{e}^{\text{i}\unicode[STIX]{x1D6FE}x}\,\text{d}x, & & \displaystyle\end{eqnarray}$$

(A 10c)$$\begin{eqnarray}\displaystyle D_{+}^{(1)}(\unicode[STIX]{x1D6FE})=\int _{0}^{\infty }\unicode[STIX]{x1D6E5}_{0}[\unicode[STIX]{x1D719}^{(1)}](x)\text{e}^{\text{i}\unicode[STIX]{x1D6FE}x}\,\text{d}x. & & \displaystyle\end{eqnarray}$$

The subscripts $\pm$ here indicate that the functions are analytic in the upper or lower half planes, denoted by ${\mathcal{L}}^{\pm }$. We may employ (A 4a) to obtain

$$\begin{eqnarray}\displaystyle F_{+}^{(1)}(\unicode[STIX]{x1D6FE})=\frac{w_{0}}{\unicode[STIX]{x03C0}\text{i}(\unicode[STIX]{x1D6FE}+k_{x}/\unicode[STIX]{x1D6FD}^{2})}+(\unicode[STIX]{x1D707}_{0}-\text{i}\unicode[STIX]{x1D707}_{1}\unicode[STIX]{x1D6FE}-\unicode[STIX]{x1D707}_{2}\unicode[STIX]{x1D6FE}^{2})D^{(1)}(\unicode[STIX]{x1D6FE}). & & \displaystyle \nonumber\end{eqnarray}$$

Consequently, the Wiener–Hopf equation (A 9) may be expressed as

(A 11)$$\begin{eqnarray}\displaystyle F_{-}^{(1)}(\unicode[STIX]{x1D6FE})+\frac{w_{0}}{\unicode[STIX]{x03C0}\text{i}(\unicode[STIX]{x1D6FE}+k_{x}/\unicode[STIX]{x1D6FD}^{2})}=4\unicode[STIX]{x03C0}D_{+}^{(1)}(\unicode[STIX]{x1D6FE})K(\unicode[STIX]{x1D6FE}), & & \displaystyle\end{eqnarray}$$

where

(A 12)$$\begin{eqnarray}\displaystyle K(\unicode[STIX]{x1D6FE})=j(\unicode[STIX]{x1D6FE})-\frac{1}{4\unicode[STIX]{x03C0}}(\unicode[STIX]{x1D707}_{0}-\text{i}\unicode[STIX]{x1D707}_{1}\unicode[STIX]{x1D6FE}-\unicode[STIX]{x1D707}_{2}\unicode[STIX]{x1D6FE}^{2}). & & \displaystyle\end{eqnarray}$$

The multiplicative splitting $K=K_{+}K_{-}$ is performed in appendix B. This splitting enables us to write

(A 13)$$\begin{eqnarray}\displaystyle \frac{F_{-}^{(1)}(\unicode[STIX]{x1D6FE})}{K_{-}(\unicode[STIX]{x1D6FE})}+\frac{w_{0}}{\unicode[STIX]{x03C0}\text{i}(\unicode[STIX]{x1D6FE}+k_{x}/\unicode[STIX]{x1D6FD}^{2})K_{-}(\unicode[STIX]{x1D6FE})}=4\unicode[STIX]{x03C0}D_{+}^{(1)}(\unicode[STIX]{x1D6FE})K_{+}(\unicode[STIX]{x1D6FE}). & & \displaystyle\end{eqnarray}$$

We now additively factorise the left side of (A 13). We apply pole removal (Noble Reference Noble1958) to obtain the additive splitting

$$\begin{eqnarray}\displaystyle \frac{1}{(\unicode[STIX]{x1D6FE}+k_{x}/\unicode[STIX]{x1D6FD}^{2})K_{-}(\unicode[STIX]{x1D6FE})} & = & \displaystyle \underbrace{\frac{1}{(\unicode[STIX]{x1D6FE}+k_{x}/\unicode[STIX]{x1D6FD}^{2})K_{-}(-k_{x}/\unicode[STIX]{x1D6FD}^{2})}}_{+}\nonumber\\ \displaystyle & & \displaystyle +\,\underbrace{\frac{1}{(\unicode[STIX]{x1D6FE}+k_{x}/\unicode[STIX]{x1D6FD}^{2})K_{-}(\unicode[STIX]{x1D6FE})}-\frac{1}{(\unicode[STIX]{x1D6FE}+k_{x}/\unicode[STIX]{x1D6FD}^{2})K_{-}(-k_{x}/\unicode[STIX]{x1D6FD}^{2})}}_{-},\nonumber\end{eqnarray}$$

where the underbrace $\pm$ denotes that the function is analytic in ${\mathcal{L}}^{\pm }$ respectively. Therefore, (A 11) becomes

(A 14)$$\begin{eqnarray}\displaystyle & & \displaystyle \frac{F_{-}^{(1)}(\unicode[STIX]{x1D6FE})}{K_{-}(\unicode[STIX]{x1D6FE})}+\frac{w_{0}}{\unicode[STIX]{x03C0}\text{i}(\unicode[STIX]{x1D6FE}+k_{x}/\unicode[STIX]{x1D6FD}^{2})K_{-}(\unicode[STIX]{x1D6FE})}-\frac{w_{0}}{\unicode[STIX]{x03C0}\text{i}(\unicode[STIX]{x1D6FE}+k_{x}/\unicode[STIX]{x1D6FD}^{2})K_{-}(-k_{x}/\unicode[STIX]{x1D6FD}^{2})}\nonumber\\ \displaystyle & & \displaystyle \qquad \quad =4\unicode[STIX]{x03C0}D_{+}^{(1)}(\unicode[STIX]{x1D6FE})K_{+}(\unicode[STIX]{x1D6FE})-\frac{w_{0}}{\unicode[STIX]{x03C0}\text{i}(\unicode[STIX]{x1D6FE}+k_{x}/\unicode[STIX]{x1D6FD}^{2})K_{-}(-k_{x}/\unicode[STIX]{x1D6FD}^{2})}.\end{eqnarray}$$

We may now apply the standard Wiener–Hopf argument: since the left and right sides of (A 14) are analytic in ${\mathcal{L}}^{\mp }$ respectively, and they agree on a strip, each side defines the analytic continuation of the other. Therefore, equation (A 14) defines an entire function, $E_{1}(\unicode[STIX]{x1D6FE})$. As $|\unicode[STIX]{x1D6FE}|\rightarrow \infty$ in ${\mathcal{L}}^{-}$, the left side of (A 14) decays. This can be seen by considering the growth of $K_{-}$, which is explicated in appendix B. Similarly, as $|\unicode[STIX]{x1D6FE}|\rightarrow \infty$ in ${\mathcal{L}}^{+}$, the right side of (A 14) vanishes. Therefore, $E_{1}(\unicode[STIX]{x1D6FE})$ is bounded in the entire plane and must be constant according to Liouville’s theorem. Moreover, since $E_{1}(\unicode[STIX]{x1D6FE})$ decays, this constant must be zero. Finally, we rearrange the right side of (A 14) to obtain the solution to the first Wiener–Hopf problem as

(A 15)$$\begin{eqnarray}\displaystyle D^{(1)}(\unicode[STIX]{x1D6FE})=\frac{w_{0}}{(2\unicode[STIX]{x03C0})^{2}\text{i}(\unicode[STIX]{x1D6FE}+k_{x}/\unicode[STIX]{x1D6FD}^{2})K_{-}(-k_{x}/\unicode[STIX]{x1D6FD}^{2})K_{+}(\unicode[STIX]{x1D6FE})}. & & \displaystyle\end{eqnarray}$$

A.2 Solution to second Wiener–Hopf equation – $D^{(2)}$

In this section we solve the integral equation (A 3) for $n=2$,

(A 16a)$$\begin{eqnarray}\displaystyle f^{(2)}(x)=4\unicode[STIX]{x03C0}\int _{-\infty }^{\infty }D^{(2)}(\unicode[STIX]{x1D6FE})j(\unicode[STIX]{x1D6FE})\text{e}^{-\text{i}\unicode[STIX]{x1D6FE}x}\,\text{d}\unicode[STIX]{x1D6FE}, & & \displaystyle\end{eqnarray}$$

subject to (A 5a) and (A 5b). Taking a Fourier transform of (A 16a) and applying (A 5a) yields

(A 17)$$\begin{eqnarray}\displaystyle (\unicode[STIX]{x1D707}_{0}-\text{i}\unicode[STIX]{x1D707}_{1}\unicode[STIX]{x1D6FE}-\unicode[STIX]{x1D707}_{2}\unicode[STIX]{x1D6FE}^{2})D^{(1)}(\unicode[STIX]{x1D6FE})+F_{+}^{(2)}(\unicode[STIX]{x1D6FE})=4\unicode[STIX]{x03C0}(D_{-}^{(2)}(\unicode[STIX]{x1D6FE})+D_{+}^{(2)}(\unicode[STIX]{x1D6FE}))j(\unicode[STIX]{x1D6FE}), & & \displaystyle\end{eqnarray}$$

where

(A 18a)$$\begin{eqnarray}\displaystyle & \displaystyle F_{+}^{(2)}(\unicode[STIX]{x1D6FE})=\frac{1}{2\unicode[STIX]{x03C0}}\int _{2}^{\infty }f^{(2)}(x)\text{e}^{\text{i}\unicode[STIX]{x1D6FE}x}\,\text{d}x=\frac{\text{e}^{2\text{i}\unicode[STIX]{x1D6FE}}}{2\unicode[STIX]{x03C0}}\int _{0}^{\infty }f^{(2)}(x+2)\text{e}^{\text{i}x\unicode[STIX]{x1D6FE}}\,\text{d}x=\text{e}^{2\text{i}\unicode[STIX]{x1D6FE}}F_{+}^{\ast (2)}(\unicode[STIX]{x1D6FE}),\qquad & \displaystyle\end{eqnarray}$$

(A 18b)$$\begin{eqnarray}\displaystyle & \displaystyle D_{+}^{(2)}(\unicode[STIX]{x1D6FE})=\frac{1}{2\unicode[STIX]{x03C0}}\int _{2}^{\infty }\unicode[STIX]{x1D6E5}_{0}[\unicode[STIX]{x1D719}^{(2)}](x)\text{e}^{\text{i}\unicode[STIX]{x1D6FE}x}\,\text{d}x=\frac{\text{e}^{2\text{i}\unicode[STIX]{x1D6FE}}}{2\unicode[STIX]{x03C0}}\int _{0}^{\infty }\unicode[STIX]{x1D6E5}_{0}[\unicode[STIX]{x1D719}^{(2)}](x+2)\text{e}^{\text{i}x\unicode[STIX]{x1D6FE}}\,\text{d}x=\text{e}^{2\text{i}\unicode[STIX]{x1D6FE}}D_{+}^{\ast (2)}(\unicode[STIX]{x1D6FE}), & \displaystyle \nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

(A 18c)$$\begin{eqnarray}\displaystyle & \displaystyle D_{-}^{(2)}(\unicode[STIX]{x1D6FE})=\frac{1}{2\unicode[STIX]{x03C0}}\int _{-\infty }^{2}\unicode[STIX]{x1D6E5}_{0}[\unicode[STIX]{x1D719}^{(2)}](x)\text{e}^{\text{i}\unicode[STIX]{x1D6FE}x}\,\text{d}x=\frac{\text{e}^{2\text{i}\unicode[STIX]{x1D6FE}}}{2\unicode[STIX]{x03C0}}\int _{-\infty }^{0}\unicode[STIX]{x1D6E5}_{0}[\unicode[STIX]{x1D719}^{(2)}](x+2)\text{e}^{\text{i}x\unicode[STIX]{x1D6FE}}\,\text{d}x=\text{e}^{2\text{i}\unicode[STIX]{x1D6FE}}D_{-}^{\ast (2)}(\unicode[STIX]{x1D6FE}). & \displaystyle \nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

$\text{e}^{2\text{i}\unicode[STIX]{x1D6FE}}$

(A 19)$$\begin{eqnarray}\displaystyle F_{+}^{\ast (2)}(\unicode[STIX]{x1D6FE})=4\unicode[STIX]{x03C0}(D_{-}^{\ast (2)}(\unicode[STIX]{x1D6FE})+D_{+}^{\ast (2)}(\unicode[STIX]{x1D6FE}))K(\unicode[STIX]{x1D6FE}), & & \displaystyle\end{eqnarray}$$

and we may use the multiplicative splitting of $K$ to write

(A 20)$$\begin{eqnarray}\displaystyle \frac{F_{+}^{\ast (2)}(\unicode[STIX]{x1D6FE})}{K_{+}(\unicode[STIX]{x1D6FE})}=4\unicode[STIX]{x03C0}(D_{-}^{\ast (2)}(\unicode[STIX]{x1D6FE})+D_{+}^{\ast (2)}(\unicode[STIX]{x1D6FE}))K_{-}(\unicode[STIX]{x1D6FE}). & & \displaystyle\end{eqnarray}$$

We may use the downstream boundary condition for this problem (A 5b) to write

(A 21)$$\begin{eqnarray}\displaystyle D_{+}^{\ast (2)}(\unicode[STIX]{x1D6FE})=-\frac{P^{\ast (2)}}{\unicode[STIX]{x1D6FE}+\unicode[STIX]{x1D714}/\unicode[STIX]{x1D6FD}^{2}}-\frac{1}{2\unicode[STIX]{x03C0}}\int _{0}^{\infty }\unicode[STIX]{x1D6E5}_{0}[\unicode[STIX]{x1D719}^{(1)}](x+2)\text{e}^{\text{i}\unicode[STIX]{x1D6FE}x}\,\text{d}x. & & \displaystyle\end{eqnarray}$$

where $P^{\ast (2)}=P^{(2)}\text{e}^{2\text{i}k_{x}/\unicode[STIX]{x1D6FD}^{2}}$. To calculate the remaining integral we use the inversion formula for the Fourier transform

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6E5}_{0}[\unicode[STIX]{x1D719}^{(1)}](x)=\int _{-\infty -\text{i}\unicode[STIX]{x1D70F}_{1}}^{\infty -\text{i}\unicode[STIX]{x1D70F}_{1}}D^{(1)}(\unicode[STIX]{x1D6FE})\text{e}^{-\text{i}\unicode[STIX]{x1D6FE}x}\,\text{d}\unicode[STIX]{x1D6FE}, & & \displaystyle \nonumber\end{eqnarray}$$

for some $\unicode[STIX]{x1D70F}_{1}>0$. By substituting this representation into our desired integral, we obtain

$$\begin{eqnarray}\displaystyle \frac{1}{2\unicode[STIX]{x03C0}}\int _{0}^{\infty }\unicode[STIX]{x1D6E5}_{0}[\unicode[STIX]{x1D719}^{(1)}](x+2)\text{e}^{\text{i}\unicode[STIX]{x1D6FE}x}\,\text{d}x=\frac{1}{2\unicode[STIX]{x03C0}}\int _{0}^{\infty }\int _{-\infty -\text{i}\unicode[STIX]{x1D70F}_{1}}^{\infty -\text{i}\unicode[STIX]{x1D70F}_{1}}D^{(1)}(\unicode[STIX]{x1D6FE}_{1})\text{e}^{-\text{i}\unicode[STIX]{x1D6FE}_{1}(x+2)}\,\text{d}\unicode[STIX]{x1D6FE}_{1}\text{e}^{\text{i}\unicode[STIX]{x1D6FE}x}\,\text{d}x. & & \displaystyle \nonumber\end{eqnarray}$$

Rearranging the order of integration and computing the resulting $x$-integral results in

(A 22)$$\begin{eqnarray}\displaystyle \frac{1}{2\unicode[STIX]{x03C0}}\int _{0}^{\infty }\unicode[STIX]{x1D6E5}_{0}[\unicode[STIX]{x1D719}^{(1)}](x+2)\text{e}^{\text{i}\unicode[STIX]{x1D6FE}x}\,\text{d}x=\frac{1}{2\unicode[STIX]{x03C0}\text{i}}\int _{-\infty -\text{i}\unicode[STIX]{x1D70F}_{1}}^{\infty -\text{i}\unicode[STIX]{x1D70F}_{1}}\frac{D^{(1)}(\unicode[STIX]{x1D6FE}_{1})\text{e}^{-2\text{i}\unicode[STIX]{x1D6FE}_{1}}}{\unicode[STIX]{x1D6FE}_{1}-\unicode[STIX]{x1D6FE}}\,\text{d}\unicode[STIX]{x1D6FE}_{1}. & & \displaystyle\end{eqnarray}$$

Since $d<2$, we may close the remaining integral in ${\mathcal{L}}^{-}$. There are no branches in the integrand so the integral consists of the residues of simple poles at $\unicode[STIX]{x1D6FE}=-k_{x}/\unicode[STIX]{x1D6FD}^{2}$, $\unicode[STIX]{x1D703}_{n}^{-}$. Consequently, inserting (A 15) into the above integral yields

$$\begin{eqnarray}\displaystyle & & \displaystyle \frac{1}{2\unicode[STIX]{x03C0}\text{i}}\int _{-\infty -\text{i}\unicode[STIX]{x1D70F}_{1}}^{\infty -\text{i}\unicode[STIX]{x1D70F}_{1}}\frac{1}{(\unicode[STIX]{x1D6FE}+k_{x}/\unicode[STIX]{x1D6FD}^{2})K_{-}(-k_{x}/\unicode[STIX]{x1D6FD}^{2})K_{+}(\unicode[STIX]{x1D6FE})}\,\frac{\text{e}^{-2\text{i}\unicode[STIX]{x1D6FE}_{1}}}{\unicode[STIX]{x1D6FE}_{1}-\unicode[STIX]{x1D6FE}}\,\text{d}\unicode[STIX]{x1D6FE}_{1}\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{\text{e}^{2\text{i}k_{x}/\unicode[STIX]{x1D6FD}^{2}}}{(\unicode[STIX]{x1D6FE}+k_{x}/\unicode[STIX]{x1D6FD}^{2})K(-k_{x}/\unicode[STIX]{x1D6FD}^{2})}+\mathop{\sum }_{n=0}^{\infty }\frac{\text{e}^{-2\text{i}\unicode[STIX]{x1D703}_{n}^{-}}}{(\unicode[STIX]{x1D703}_{n}^{-}+k_{x}/\unicode[STIX]{x1D6FD}^{2})K_{-}(-k_{x}/\unicode[STIX]{x1D6FD}^{2})K_{+}^{\prime }(\unicode[STIX]{x1D703}_{n}^{-})(\unicode[STIX]{x1D6FE}-\unicode[STIX]{x1D703}_{n}^{-})},\nonumber\end{eqnarray}$$

where the derivatives of $K_{\pm }$ evaluated at the duct modes $\unicode[STIX]{x1D703}_{n}^{\mp }$ are given by

$$\begin{eqnarray}\displaystyle K_{\pm }^{\prime }(\unicode[STIX]{x1D703}_{n}^{\mp }) & = & \displaystyle \frac{K^{\prime }(\unicode[STIX]{x1D703}_{n}^{\mp })}{K_{\mp }(\unicode[STIX]{x1D703}_{n}^{\mp })}=\frac{1}{4\unicode[STIX]{x03C0}K_{\mp }(\unicode[STIX]{x1D703}_{n}^{\mp })}\left(-\frac{(\unicode[STIX]{x1D703}_{n}^{\mp }/\unicode[STIX]{x1D701}(\unicode[STIX]{x1D703}_{n}^{\mp }))\sin (s\unicode[STIX]{x1D701}(\unicode[STIX]{x1D703}_{n}^{\mp }))+s\unicode[STIX]{x1D703}_{n}^{\mp }\cos (s\unicode[STIX]{x1D701}(\unicode[STIX]{x1D703}_{n}^{\mp }))}{\cos (s\unicode[STIX]{x1D701}(\unicode[STIX]{x1D703}_{n}^{\mp }))-\cos (d\unicode[STIX]{x1D703}_{n}^{\mp }+\unicode[STIX]{x1D70E}^{\prime })}\right.\nonumber\\ \displaystyle & & \displaystyle \quad -\,\frac{\unicode[STIX]{x1D703}_{n}^{\mp }s\sin (s\unicode[STIX]{x1D701}(\unicode[STIX]{x1D703}_{n}^{\mp }))^{2}+d\unicode[STIX]{x1D701}(\unicode[STIX]{x1D703}_{n}^{\mp })\sin (d\unicode[STIX]{x1D703}_{n}^{\mp }+\unicode[STIX]{x1D70E}^{\prime })\sin (s\unicode[STIX]{x1D701}(\unicode[STIX]{x1D703}_{n}^{\mp }))}{(\cos (s\unicode[STIX]{x1D701}(\unicode[STIX]{x1D703}_{n}^{\mp }))-\cos (d\unicode[STIX]{x1D703}_{n}^{\mp }+\unicode[STIX]{x1D70E}^{\prime }))^{2}}\nonumber\\ \displaystyle & & \displaystyle \quad +\left.\text{i}\unicode[STIX]{x1D707}_{1}+2\unicode[STIX]{x1D707}_{2}\unicode[STIX]{x1D703}_{n}^{\mp }\right).\nonumber\end{eqnarray}$$

Therefore, substitution of (A 22) into (A 21) yields

(A 23)$$\begin{eqnarray}\displaystyle D_{+}^{\ast (2)}(\unicode[STIX]{x1D6FE}) & = & \displaystyle \frac{w_{0}\text{e}^{2\text{i}k_{x}/\unicode[STIX]{x1D6FD}^{2}}}{(2\unicode[STIX]{x03C0})^{2}\text{i}(\unicode[STIX]{x1D6FE}+k_{x}/\unicode[STIX]{x1D6FD}^{2})K(-k_{x}/\unicode[STIX]{x1D6FD}^{2})}-\mathop{\sum }_{n=0}^{\infty }\frac{{\mathcal{A}}_{n}\text{e}^{-2\text{i}\unicode[STIX]{x1D703}_{n}^{-}}}{\text{i}(\unicode[STIX]{x1D703}_{n}^{-}+\unicode[STIX]{x1D714}/\unicode[STIX]{x1D6FD}^{2})(\unicode[STIX]{x1D6FE}-\unicode[STIX]{x1D703}_{n}^{-})},\nonumber\\ \displaystyle & & \displaystyle -\,\frac{P^{\ast (2)}}{\unicode[STIX]{x1D6FE}+\unicode[STIX]{x1D714}/\unicode[STIX]{x1D6FD}^{2}},\end{eqnarray}$$

where

$$\begin{eqnarray}\displaystyle {\mathcal{A}}_{n}=\frac{w_{0}(\unicode[STIX]{x1D703}_{n}^{-}+\unicode[STIX]{x1D714}/\unicode[STIX]{x1D6FD}^{2})}{(2\unicode[STIX]{x03C0})^{2}(\unicode[STIX]{x1D703}_{n}^{-}+k_{x}/\unicode[STIX]{x1D6FD}^{2})K_{-}(-k_{x}/\unicode[STIX]{x1D6FD}^{2})K_{+}^{\prime }(\unicode[STIX]{x1D703}_{n}^{-})}. & & \displaystyle \nonumber\end{eqnarray}$$

We use the notation

$$\begin{eqnarray}\displaystyle \tilde{K}_{-}(\unicode[STIX]{x1D6FE},\unicode[STIX]{x1D702}^{-})\triangleq \frac{K_{-}(\unicode[STIX]{x1D6FE})}{\unicode[STIX]{x1D6FE}-\unicode[STIX]{x1D702}^{-}}, & & \displaystyle \nonumber\end{eqnarray}$$

so that substitution of (A 23) into the Wiener–Hopf equation (A 17) yields

(A 24)$$\begin{eqnarray}\displaystyle \frac{F_{+}^{\ast (2)}(\unicode[STIX]{x1D6FE})}{4\unicode[STIX]{x03C0}K_{+}(\unicode[STIX]{x1D6FE})} & = & \displaystyle K_{-}(\unicode[STIX]{x1D6FE})D_{-}^{\ast (2)}(\unicode[STIX]{x1D6FE})-P^{\ast (2)}\tilde{K}_{-}(\unicode[STIX]{x1D6FE},-\unicode[STIX]{x1D714}/\unicode[STIX]{x1D6FD}^{2})\nonumber\\ \displaystyle & & \displaystyle +\,\frac{w_{0}\text{e}^{2\text{i}k_{x}/\unicode[STIX]{x1D6FD}^{2}}}{(2\unicode[STIX]{x03C0})^{2}\text{i}}\,\frac{\tilde{K}_{-}(\unicode[STIX]{x1D6FE},-k_{x}/\unicode[STIX]{x1D6FD}^{2})}{K(-k_{x}/\unicode[STIX]{x1D6FD}^{2})}-\mathop{\sum }_{n=0}^{\infty }\frac{{\mathcal{A}}_{n}\text{e}^{-2\text{i}\unicode[STIX]{x1D703}_{n}^{-}}}{\text{i}(\unicode[STIX]{x1D703}_{n}^{-}+\unicode[STIX]{x1D714}/\unicode[STIX]{x1D6FD}^{2})}\tilde{K}_{-}(\unicode[STIX]{x1D6FE},\unicode[STIX]{x1D703}_{n}^{-}).\qquad\end{eqnarray}$$

We note the additive splitting

(A 25)$$\begin{eqnarray}\displaystyle \tilde{K}_{-}(\unicode[STIX]{x1D6FE},\unicode[STIX]{x1D702}^{-})=[\tilde{K}_{-}(\unicode[STIX]{x1D6FE},\unicode[STIX]{x1D702})]_{+}+[\tilde{K}_{-}(\unicode[STIX]{x1D6FE},\unicode[STIX]{x1D702})]_{-}, & & \displaystyle\end{eqnarray}$$

where

$$\begin{eqnarray}\displaystyle [\tilde{K}_{-}(\unicode[STIX]{x1D6FE},\unicode[STIX]{x1D702})]_{+}=\frac{K_{-}(\unicode[STIX]{x1D702}^{-})}{\unicode[STIX]{x1D6FE}-\unicode[STIX]{x1D702}^{-}},\quad [\tilde{K}_{-}(\unicode[STIX]{x1D6FE},\unicode[STIX]{x1D702})]_{-}=\frac{K_{-}(\unicode[STIX]{x1D6FE})-K_{-}(\unicode[STIX]{x1D702}^{-})}{\unicode[STIX]{x1D6FE}-\unicode[STIX]{x1D702}^{-}}. & & \displaystyle \nonumber\end{eqnarray}$$

Substituting these splittings into (A 24) yields

(A 26)$$\begin{eqnarray}\displaystyle & & \displaystyle \frac{F_{+}^{\ast (2)}(\unicode[STIX]{x1D6FE})}{4\unicode[STIX]{x03C0}K_{+}(\unicode[STIX]{x1D6FE})}+P^{\ast (2)}[\tilde{K}_{-}(\unicode[STIX]{x1D6FE},-\unicode[STIX]{x1D714}/\unicode[STIX]{x1D6FD}^{2})]_{+}-\frac{w_{0}\text{e}^{2\text{i}k_{x}/\unicode[STIX]{x1D6FD}^{2}}}{(2\unicode[STIX]{x03C0})^{2}\text{i}}\,\frac{[\tilde{K}_{-}(\unicode[STIX]{x1D6FE},-k_{x}/\unicode[STIX]{x1D6FD}^{2})]_{+}}{K(-k_{x}/\unicode[STIX]{x1D6FD}^{2})}\nonumber\\ \displaystyle & & \displaystyle \quad +\,\mathop{\sum }_{n=0}^{\infty }\frac{{\mathcal{A}}_{n}\text{e}^{-2\text{i}\unicode[STIX]{x1D703}_{n}^{-}}}{\text{i}(\unicode[STIX]{x1D703}_{n}^{-}+\unicode[STIX]{x1D714}/\unicode[STIX]{x1D6FD}^{2})}[\tilde{K}_{-}(\unicode[STIX]{x1D6FE},\unicode[STIX]{x1D703}_{n}^{-})]_{+}=K_{-}(\unicode[STIX]{x1D6FE})D_{-}^{\ast (2)}(\unicode[STIX]{x1D6FE})-P^{\ast (2)}[\tilde{K}_{-}(\unicode[STIX]{x1D6FE},-\unicode[STIX]{x1D714}/\unicode[STIX]{x1D6FD}^{2})]_{-}\nonumber\\ \displaystyle & & \displaystyle \quad +\,\frac{w_{0}\text{e}^{2\text{i}k_{x}/\unicode[STIX]{x1D6FD}^{2}}}{(2\unicode[STIX]{x03C0})^{2}\text{i}}\,\frac{[\tilde{K}_{-}(\unicode[STIX]{x1D6FE},-k_{x}/\unicode[STIX]{x1D6FD}^{2})]_{-}}{K(-k_{x}/\unicode[STIX]{x1D6FD}^{2})}-\mathop{\sum }_{n=0}^{\infty }\frac{{\mathcal{A}}_{n}\text{e}^{-2\text{i}\unicode[STIX]{x1D703}_{n}^{-}}}{\text{i}(\unicode[STIX]{x1D703}_{n}^{-}+\unicode[STIX]{x1D714}/\unicode[STIX]{x1D6FD}^{2})}[\tilde{K}_{-}(\unicode[STIX]{x1D6FE},\unicode[STIX]{x1D703}_{n}^{-})]_{-}.\end{eqnarray}$$

In a similar way to § A.1, we now apply the typical Wiener–Hopf argument. We enforce the unsteady Kutta condition (Ayton, Gill & Peake Reference Ayton, Gill and Peake2016) which restricts the pressure at the trailing edge to be finite. Consequently, and using the results of appendix B, the left side of (A 26) decays as $|\unicode[STIX]{x1D6FE}|\rightarrow \infty$ in ${\mathcal{L}}^{+}$ and the left side of (A 26) tends to an unknown constant as $|\unicode[STIX]{x1D6FE}|\rightarrow \infty$ in ${\mathcal{L}}^{-}$. Applying analytic continuation and Liouville’s theorem determines that this constant must be zero. Accordingly, the coefficient of $\unicode[STIX]{x1D6FE}^{-1}$ on the left side of (A 26) must vanish so that

$$\begin{eqnarray}\displaystyle P^{\ast (2)}=\frac{w_{0}\text{e}^{2\text{i}k_{x}/\unicode[STIX]{x1D6FD}^{2}}}{(2\unicode[STIX]{x03C0})^{2}\text{i}K_{+}(-k_{x}/\unicode[STIX]{x1D6FD}^{2})}\,\frac{1}{K_{-}(-\unicode[STIX]{x1D714}/\unicode[STIX]{x1D6FD}^{2})}-\mathop{\sum }_{n=0}^{\infty }\frac{{\mathcal{A}}_{n}\text{e}^{-2\text{i}\unicode[STIX]{x1D703}_{n}^{-}}}{\text{i}(\unicode[STIX]{x1D703}_{n}^{-}+\unicode[STIX]{x1D714}/\unicode[STIX]{x1D6FD}^{2})}\,\frac{K_{-}(\unicode[STIX]{x1D703}_{n}^{-})}{K_{-}(-\unicode[STIX]{x1D714}/\unicode[STIX]{x1D6FD}^{2})}. & & \displaystyle \nonumber\end{eqnarray}$$

So that, after substituting in the downstream representation (A 23) and the above expression for the pressure constant, the right side of (A 26) yields

(A 27)$$\begin{eqnarray}\displaystyle D^{(2)}(\unicode[STIX]{x1D6FE}) & = & \displaystyle \frac{w_{0}(\unicode[STIX]{x1D714}-k_{x})\text{e}^{2\text{i}(\unicode[STIX]{x1D6FE}+k_{x}/\unicode[STIX]{x1D6FD}^{2})}}{(2\unicode[STIX]{x03C0}\unicode[STIX]{x1D6FD})^{2}\text{i}(\unicode[STIX]{x1D6FE}+k_{x}/\unicode[STIX]{x1D6FD}^{2})(\unicode[STIX]{x1D6FE}+\unicode[STIX]{x1D714}/\unicode[STIX]{x1D6FD}^{2})K_{+}(-k_{x}/\unicode[STIX]{x1D6FD}^{2})}\,\frac{1}{K_{-}(\unicode[STIX]{x1D6FE})}\nonumber\\ \displaystyle & & \displaystyle -\,\mathop{\sum }_{n=0}^{\infty }\frac{{\mathcal{A}}_{n}\text{e}^{2\text{i}(\unicode[STIX]{x1D6FE}-\unicode[STIX]{x1D703}_{n}^{-})}}{\text{i}(\unicode[STIX]{x1D6FE}+\unicode[STIX]{x1D714}/\unicode[STIX]{x1D6FD}^{2})(\unicode[STIX]{x1D6FE}-\unicode[STIX]{x1D703}_{n}^{-})}\,\frac{K_{-}(\unicode[STIX]{x1D703}_{n}^{-})}{K_{-}(\unicode[STIX]{x1D6FE})}.\end{eqnarray}$$

It should be noted that the only poles of $D^{(2)}$ in ${\mathcal{L}}^{+}$ are at the zeros of $K_{-}$.

A.3 Solution to third and fourth Wiener–Hopf equations – $D^{(3)},D^{(4)}$

Since the integral equations for $\unicode[STIX]{x1D6E5}_{0}[\unicode[STIX]{x1D719}^{(3)}]$ and $\unicode[STIX]{x1D6E5}_{0}[\unicode[STIX]{x1D719}^{(4)}]$ are coupled, we must solve for them simultaneously. Taking a Fourier Transform of (A 3) and applying the boundary conditions for (A 6a) and (A 7a) for $n=3,4$ gives

(A 28)$$\begin{eqnarray}\displaystyle & \displaystyle F_{-}^{(3)}(\unicode[STIX]{x1D6FE})=4\unicode[STIX]{x03C0}[D_{-}^{(3)}(\unicode[STIX]{x1D6FE})+D_{+}^{(3)}(\unicode[STIX]{x1D6FE})]K(\unicode[STIX]{x1D6FE}), & \displaystyle\end{eqnarray}$$

(A 29)$$\begin{eqnarray}\displaystyle & \displaystyle F_{+}^{\ast (4)}(\unicode[STIX]{x1D6FE})=4\unicode[STIX]{x03C0}[D_{-}^{\ast (4)}(\unicode[STIX]{x1D6FE})+D_{+}^{\ast (4)}(\unicode[STIX]{x1D6FE})]K(\unicode[STIX]{x1D6FE}), & \displaystyle\end{eqnarray}$$

where $F_{-}^{(3)}$, $D_{-}^{(3)}$ and $D_{+}^{(3)}$ are defined in an analogous way to (A 10a,b) and (A 10c), and $F_{+}^{\ast (4)}$, $D_{+}^{\ast (4)}$ and $D_{-}^{\ast (4)}$ are defined in an analogous way to (A 18a), (A 18b) and (A 18c). Using a similar approach to § A.2, the upstream boundary condition (A 6b) may be expressed as

$$\begin{eqnarray}\displaystyle D_{-}^{(3)}(\unicode[STIX]{x1D6FE})=\frac{1}{2\unicode[STIX]{x03C0}\text{i}}\int _{-\infty +\text{i}\unicode[STIX]{x1D70F}_{0}}^{\infty +\text{i}\unicode[STIX]{x1D70F}_{0}}\frac{D^{(2)}(\unicode[STIX]{x1D6FE}_{1})+D^{(4)}(\unicode[STIX]{x1D6FE}_{1})}{\unicode[STIX]{x1D6FE}_{1}-\unicode[STIX]{x1D6FE}}\,\text{d}\unicode[STIX]{x1D6FE}_{1}. & & \displaystyle \nonumber\end{eqnarray}$$

Consequently, we may express $D_{-}^{(3)}$ in terms of its poles $\unicode[STIX]{x1D703}_{n}^{+}$ as

(A 30)$$\begin{eqnarray}\displaystyle D_{-}^{(3)}(\unicode[STIX]{x1D6FE})=-\mathop{\sum }_{n=0}^{\infty }\frac{{\mathcal{B}}_{n}}{\unicode[STIX]{x1D6FE}-\unicode[STIX]{x1D703}_{n}^{+}}, & & \displaystyle\end{eqnarray}$$

where ${\mathcal{B}}_{n}$ are the residues of $D_{-}^{(2)}(\unicode[STIX]{x1D6FE}_{1})+D_{-}^{(4)}(\unicode[STIX]{x1D6FE}_{1})$ at $\unicode[STIX]{x1D6FE}=\unicode[STIX]{x1D703}_{n}^{+}$. The residues of $D_{-}^{(4)}$ are currently unknown, but the residues of $D_{-}^{(2)}$ are given by

$$\begin{eqnarray}\displaystyle D_{r,k}^{(2)} & = & \displaystyle \frac{\text{e}^{2\text{i}\unicode[STIX]{x1D703}_{k}^{+}}}{K_{-}^{\prime }(\unicode[STIX]{x1D703}_{k}^{+})}\left\{\frac{-w_{0}(\unicode[STIX]{x1D714}-k_{x})\text{e}^{2\text{i}k_{x}/\unicode[STIX]{x1D6FD}^{2}}}{(2\unicode[STIX]{x03C0}\unicode[STIX]{x1D6FD})^{2}\text{i}(\unicode[STIX]{x1D703}_{k}^{+}+k_{x}/\unicode[STIX]{x1D6FD}^{2})(\unicode[STIX]{x1D703}_{k}^{+}+\unicode[STIX]{x1D714}/\unicode[STIX]{x1D6FD}^{2})K_{+}(-k_{x}/\unicode[STIX]{x1D6FD}^{2})}\right.\nonumber\\ \displaystyle & & \displaystyle +\left.\mathop{\sum }_{n=0}^{\infty }\frac{\text{i}{\mathcal{A}}_{n}\text{e}^{\text{i}(\unicode[STIX]{x1D703}_{k}^{+}-\unicode[STIX]{x1D703}_{n}^{-})}K_{-}(\unicode[STIX]{x1D703}_{n}^{-})}{(\unicode[STIX]{x1D703}_{k}^{+}+\unicode[STIX]{x1D714}/\unicode[STIX]{x1D6FD}^{2})(\unicode[STIX]{x1D703}_{k}^{+}-\unicode[STIX]{x1D703}_{n}^{-})}\right\}.\nonumber\end{eqnarray}$$

We may now substitute (A 30) into (A 28) to obtain the Wiener–Hopf equation

$$\begin{eqnarray}\displaystyle \frac{F_{-}^{(3)}(\unicode[STIX]{x1D6FE})}{4\unicode[STIX]{x03C0}K_{-}(\unicode[STIX]{x1D6FE})}+\mathop{\sum }_{n=0}^{\infty }\frac{{\mathcal{B}}_{n}}{\unicode[STIX]{x1D6FE}-\unicode[STIX]{x1D703}_{n}^{+}}K_{+}(\unicode[STIX]{x1D703}_{n}^{+})=D_{+}^{(3)}(\unicode[STIX]{x1D6FE})K_{+}(\unicode[STIX]{x1D6FE})-\mathop{\sum }_{n=0}^{\infty }\frac{{\mathcal{B}}_{n}}{\unicode[STIX]{x1D6FE}-\unicode[STIX]{x1D703}_{n}^{+}}(K_{+}(\unicode[STIX]{x1D6FE})-K_{+}(\unicode[STIX]{x1D703}_{n}^{+})). & & \displaystyle \nonumber\end{eqnarray}$$

The edge conditions are identical to those applied in § A.1, and we employ the typical Wiener–Hopf argument to obtain

$$\begin{eqnarray}\displaystyle D_{+}^{(3)}(\unicode[STIX]{x1D6FE})=\mathop{\sum }_{n=0}^{\infty }\frac{{\mathcal{B}}_{n}}{\unicode[STIX]{x1D6FE}-\unicode[STIX]{x1D703}_{n}^{+}}\left\{1-\frac{K_{+}(\unicode[STIX]{x1D703}_{n}^{+})}{K_{+}(\unicode[STIX]{x1D6FE})}\right\}. & & \displaystyle \nonumber\end{eqnarray}$$

Combining this solution with (A 30) yields

(A 31)$$\begin{eqnarray}\displaystyle D^{(3)}(\unicode[STIX]{x1D6FE})=-\mathop{\sum }_{n=0}^{\infty }\frac{{\mathcal{B}}_{n}}{\unicode[STIX]{x1D6FE}-\unicode[STIX]{x1D703}_{n}^{+}}\,\frac{K_{+}(\unicode[STIX]{x1D703}_{n}^{+})}{K_{+}(\unicode[STIX]{x1D6FE})}. & & \displaystyle\end{eqnarray}$$

We proceed to the solution for $D^{(4)}$. In a similar way to § A.2, we may invert the Fourier transform for the downstream boundary condition (A 7b) to write

$$\begin{eqnarray}\displaystyle D_{+}^{\ast (4)}(\unicode[STIX]{x1D6FE}) & = & \displaystyle -\frac{P^{\ast (4)}}{\text{i}(\unicode[STIX]{x1D6FE}+\unicode[STIX]{x1D714}/\unicode[STIX]{x1D6FD}^{2})}\nonumber\\ \displaystyle & & \displaystyle -\,\frac{1}{2\unicode[STIX]{x03C0}\text{i}}\int _{-\infty -\text{i}\unicode[STIX]{x1D70F}_{1}}^{\infty -\text{i}\unicode[STIX]{x1D70F}_{1}}\mathop{\sum }_{n=0}^{\infty }\frac{{\mathcal{B}}_{n}}{(\unicode[STIX]{x1D6FE}_{1}-\unicode[STIX]{x1D6FE})(\unicode[STIX]{x1D6FE}_{1}-\unicode[STIX]{x1D703}_{n}^{+})}\left\{1-\frac{K_{+}(\unicode[STIX]{x1D703}_{n}^{+})}{K_{+}(\unicode[STIX]{x1D6FE}_{1})}\right\}\text{e}^{-2\text{i}\unicode[STIX]{x1D6FE}_{1}}\,\text{d}\unicode[STIX]{x1D6FE}_{1}.\nonumber\end{eqnarray}$$

This integral can be closed in ${\mathcal{L}}^{-}$ to obtain

(A 32)$$\begin{eqnarray}\displaystyle D_{+}^{\ast (4)}(\unicode[STIX]{x1D6FE})=-\frac{P^{\ast (4)}}{\unicode[STIX]{x1D6FE}+\unicode[STIX]{x1D714}/\unicode[STIX]{x1D6FD}^{2}}-\mathop{\sum }_{n=0}^{\infty }\frac{{\mathcal{C}}_{n}\text{e}^{-2\text{i}\unicode[STIX]{x1D703}_{n}^{-}}}{\text{i}(\unicode[STIX]{x1D703}_{n}^{-}+\unicode[STIX]{x1D714}/\unicode[STIX]{x1D6FD}^{2})(\unicode[STIX]{x1D6FE}-\unicode[STIX]{x1D703}_{n}^{-})}, & & \displaystyle\end{eqnarray}$$

where

$$\begin{eqnarray}\displaystyle {\mathcal{C}}_{n}=\mathop{\sum }_{m=0}^{\infty }\frac{\text{i}(\unicode[STIX]{x1D703}_{n}^{-}+\unicode[STIX]{x1D714}/\unicode[STIX]{x1D6FD}^{2})}{(\unicode[STIX]{x1D703}_{m}^{+}-\unicode[STIX]{x1D703}_{n}^{-})}\,\frac{K_{+}(\unicode[STIX]{x1D703}_{m}^{+})}{K_{+}^{\prime }(\unicode[STIX]{x1D703}_{n}^{-})}\boldsymbol{\cdot }{\mathcal{B}}_{m}. & & \displaystyle \nonumber\end{eqnarray}$$

After truncation, we may write this system of equations in matrix form

(A 33)$$\begin{eqnarray}\displaystyle \boldsymbol{{\mathcal{C}}}=\unicode[STIX]{x1D647}\boldsymbol{{\mathcal{B}}}, & & \displaystyle\end{eqnarray}$$

where

$$\begin{eqnarray}\displaystyle \{\unicode[STIX]{x1D647}\}_{n,m}=\frac{\text{i}(\unicode[STIX]{x1D703}_{n}^{-}+\unicode[STIX]{x1D714}/\unicode[STIX]{x1D6FD}^{2})}{(\unicode[STIX]{x1D703}_{m}^{+}-\unicode[STIX]{x1D703}_{n}^{-})}\,\frac{K_{+}(\unicode[STIX]{x1D703}_{m}^{+})}{K_{+}^{\prime }(\unicode[STIX]{x1D703}_{n}^{-})}. & & \displaystyle \nonumber\end{eqnarray}$$

By applying the notation introduced in (A 25), we may express the Wiener–Hopf equation (A 29) in the form

(A 34)$$\begin{eqnarray}\displaystyle & & \displaystyle \frac{F_{+}^{\ast (4)}(\unicode[STIX]{x1D6FE})}{4\unicode[STIX]{x03C0}K_{+}(\unicode[STIX]{x1D6FE})}+P^{\ast (4)}[\tilde{K}_{-}(\unicode[STIX]{x1D6FE},-\unicode[STIX]{x1D714}/\unicode[STIX]{x1D6FD}^{2})]_{+}+\mathop{\sum }_{n=0}^{\infty }\frac{[\tilde{K}_{-}(\unicode[STIX]{x1D6FE},\unicode[STIX]{x1D703}_{n}^{-})]_{+}}{\text{i}(\unicode[STIX]{x1D703}_{n}^{-}+\unicode[STIX]{x1D714}/\unicode[STIX]{x1D6FD}^{2})}{\mathcal{C}}_{n}\text{e}^{-2\text{i}\unicode[STIX]{x1D703}_{n}^{-}}\nonumber\\ \displaystyle & & \displaystyle \qquad =D_{-}^{\ast (4)}(\unicode[STIX]{x1D6FE})K_{-}(\unicode[STIX]{x1D6FE})-P^{\ast (4)}[\tilde{K}_{-}(\unicode[STIX]{x1D6FE},-\unicode[STIX]{x1D714}/\unicode[STIX]{x1D6FD}^{2})]_{-}-\mathop{\sum }_{n=0}^{\infty }\frac{[\tilde{K}_{-}(\unicode[STIX]{x1D6FE},\unicode[STIX]{x1D703}_{n}^{-})]_{-}}{\text{i}(\unicode[STIX]{x1D703}_{n}^{-}+\unicode[STIX]{x1D714}/\unicode[STIX]{x1D6FD}^{2})}{\mathcal{C}}_{n}\text{e}^{-2\text{i}\unicode[STIX]{x1D703}_{n}^{-}}.\quad \qquad\end{eqnarray}$$

Employing the unsteady Kutta condition in (A 34) yields

$$\begin{eqnarray}\displaystyle P^{\ast (4)}=-\mathop{\sum }_{n=0}^{\infty }\frac{{\mathcal{C}}_{n}\text{e}^{-2\text{i}\unicode[STIX]{x1D703}_{n}^{-}}}{\text{i}(\unicode[STIX]{x1D703}_{n}^{-}+\unicode[STIX]{x1D714}/\unicode[STIX]{x1D6FD}^{2})}\,\frac{K_{-}(\unicode[STIX]{x1D703}_{n}^{-})}{K_{-}(-\unicode[STIX]{x1D714}/\unicode[STIX]{x1D6FD}^{2})}. & & \displaystyle \nonumber\end{eqnarray}$$

Finally, applying the downstream boundary condition (A 32) and rearranging (A 34) yields

(A 35)$$\begin{eqnarray}\displaystyle D^{(4)}(\unicode[STIX]{x1D6FE})=-\mathop{\sum }_{n=0}^{\infty }\frac{{\mathcal{C}}_{n}\text{e}^{2\text{i}(\unicode[STIX]{x1D6FE}-\unicode[STIX]{x1D703}_{n}^{-})}}{\text{i}(\unicode[STIX]{x1D703}_{n}^{-}+\unicode[STIX]{x1D714}/\unicode[STIX]{x1D6FD}^{2})(\unicode[STIX]{x1D6FE}-\unicode[STIX]{x1D703}_{n}^{-})}\,\frac{K_{-}(\unicode[STIX]{x1D703}_{n}^{-})}{K_{-}(\unicode[STIX]{x1D6FE})}. & & \displaystyle\end{eqnarray}$$

We are now able to calculate the residues of $D^{(3)}$ as

$$\begin{eqnarray}\displaystyle {\mathcal{B}}_{n}=D_{r,n}^{(2)}-\mathop{\sum }_{m=0}^{\infty }\frac{{\mathcal{C}}_{n}\text{e}^{2\text{i}(\unicode[STIX]{x1D703}_{n}^{+}-\unicode[STIX]{x1D703}_{m}^{-})}}{\text{i}(\unicode[STIX]{x1D703}_{n}^{+}+\unicode[STIX]{x1D714}/\unicode[STIX]{x1D6FD}^{2})(\unicode[STIX]{x1D703}_{n}^{+}-\unicode[STIX]{x1D703}_{m}^{-})}\,\frac{K_{-}(\unicode[STIX]{x1D703}_{m}^{-})}{K_{-}^{\prime }(\unicode[STIX]{x1D703}_{n}^{+})}, & & \displaystyle \nonumber\end{eqnarray}$$

or, in matrix form,

(A 36)$$\begin{eqnarray}\displaystyle \boldsymbol{{\mathcal{B}}}=\boldsymbol{D}_{r}+\unicode[STIX]{x1D641}\boldsymbol{{\mathcal{C}}}, & & \displaystyle\end{eqnarray}$$

where

$$\begin{eqnarray}\displaystyle \{\unicode[STIX]{x1D641}\}_{n,m}=\frac{\text{i}\text{e}^{2\text{i}(\unicode[STIX]{x1D703}_{n}^{+}-\unicode[STIX]{x1D703}_{m}^{-})}}{(\unicode[STIX]{x1D703}_{n}^{+}+\unicode[STIX]{x1D714}/\unicode[STIX]{x1D6FD}^{2})(\unicode[STIX]{x1D703}_{n}^{+}-\unicode[STIX]{x1D703}_{m}^{-})}\,\frac{K_{-}(\unicode[STIX]{x1D703}_{m}^{-})}{K_{-}^{\prime }(\unicode[STIX]{x1D703}_{n}^{+})}. & & \displaystyle \nonumber\end{eqnarray}$$

We can combine (A 33) and (A 36) to obtain

(A 37)$$\begin{eqnarray}\displaystyle (\unicode[STIX]{x1D644}-\unicode[STIX]{x1D641}\unicode[STIX]{x1D647})\boldsymbol{{\mathcal{B}}}=\boldsymbol{D}_{r}. & & \displaystyle\end{eqnarray}$$

This matrix equation has important physical implications. If the matrix $\unicode[STIX]{x1D644}-\unicode[STIX]{x1D641}\unicode[STIX]{x1D647}$ is not singular then it can be inverted to find $\boldsymbol{{\mathcal{B}}}$, and then $\boldsymbol{{\mathcal{C}}}$ by (A 33). On the other hand, if $\unicode[STIX]{x1D644}-\unicode[STIX]{x1D641}\unicode[STIX]{x1D647}$ is singular then there is no solution to the inhomogeneous linear system. However, in such a scenario there is a zero eigenvalue, so the homogeneous linear system ($\boldsymbol{D}_{r}=\mathbf{0}$) has a non-trivial solution. Physically speaking, this means that at certain frequencies there exist non-trivial solutions for the scattering by an incident wave of zero amplitude. Such solutions correspond to the so-called ‘Parker modes’ discovered by Parker (Reference Parker1967), who observed resonance caused by the shedding frequency of the wake coinciding with the natural frequency of the cascade. Koch (Reference Koch1983) extended Parker’s results to calculate the resonant frequencies for staggered cascades in a mean flow by using a Wiener–Hopf approach much like the present work. Woodley & Peake (Reference Woodley and Peake1999) also deployed the Wiener–Hopf technique to calculate the resonant frequency of tandem cascades. While we do not investigate this further now, the present method could be used to investigate how surface treatments can be used to avoid resonance; the aforementioned references only considered case 0 cascades, whereas our approach can consider more general boundary conditions.

B.1 Factorisation of poles of $K$

Previous work (Peake Reference Peake1992; Glegg Reference Glegg1999) has factorised the poles of $K$ into the form

$$\begin{eqnarray}\displaystyle K_{d}(\unicode[STIX]{x1D6FE})=K_{d}(0)\mathop{\prod }_{m=-\infty }^{\infty }\left(1-\frac{\unicode[STIX]{x1D6FE}}{\unicode[STIX]{x1D706}_{m}^{+}}\right)\left(1-\frac{\unicode[STIX]{x1D6FE}}{\unicode[STIX]{x1D706}_{m}^{-}}\right), & & \displaystyle \nonumber\end{eqnarray}$$

where

(B 4)$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D706}_{m}^{\pm }=-f_{m}\sin (\unicode[STIX]{x1D712})\pm \cos (\unicode[STIX]{x1D712})\unicode[STIX]{x1D701}(f_{m}),\quad f_{m}=\frac{\unicode[STIX]{x1D70E}^{\prime }-2\unicode[STIX]{x03C0}m}{\unicode[STIX]{x1D6E5}}. & & \displaystyle\end{eqnarray}$$

The asymptotic behaviour of these poles is

(B 5)$$\begin{eqnarray}\displaystyle & \displaystyle \unicode[STIX]{x1D706}_{m}^{+}\sim \unicode[STIX]{x1D706}_{R}^{(0)+}m+\unicode[STIX]{x1D706}_{R}^{(2)+}+o(1), & \displaystyle\end{eqnarray}$$

(B 6)$$\begin{eqnarray}\displaystyle & \displaystyle \unicode[STIX]{x1D706}_{-m}^{+}\sim \unicode[STIX]{x1D706}_{L}^{(0)+}m+\unicode[STIX]{x1D706}_{L}^{(2)+}+o(1), & \displaystyle\end{eqnarray}$$

as $m\rightarrow \infty$ where

(B 7)$$\begin{eqnarray}\left.\begin{array}{@{}c@{}}\displaystyle \unicode[STIX]{x1D706}_{R}^{(0)+}=\frac{2\unicode[STIX]{x03C0}\text{i}\text{e}^{-\text{i}\unicode[STIX]{x1D712}}}{\unicode[STIX]{x1D6E5}},\quad \unicode[STIX]{x1D706}_{R}^{(2)+}=\frac{-\text{i}\text{e}^{-\text{i}\unicode[STIX]{x1D712}}}{\unicode[STIX]{x1D6E5}}\unicode[STIX]{x1D70E}^{\prime },\\ \displaystyle \unicode[STIX]{x1D706}_{L}^{(0)+}=\frac{2\unicode[STIX]{x03C0}\text{i}\text{e}^{\text{i}\unicode[STIX]{x1D712}}}{\unicode[STIX]{x1D6E5}},\quad \unicode[STIX]{x1D706}_{L}^{(2)+}=\frac{\text{i}\text{e}^{\text{i}\unicode[STIX]{x1D712}}}{\unicode[STIX]{x1D6E5}}\unicode[STIX]{x1D70E}^{\prime }.\end{array}\right\}\end{eqnarray}$$

The subscripts $R$ and $L$ indicate that the pole is in the right or left side of ${\mathcal{L}}^{+}$ respectively.

B.2 Factorisation of zeros of $K$

We now outline the procedure for factorising the zeros of $K_{n}$. In contrast to previous analyses for case 0 plates (Peake Reference Peake1992; Glegg Reference Glegg1999), no analytic expression for the zeros is available. In contrast, asymptotic expansions for the large zeros are still tractable, which is useful in two regards. First, the asymptotic expansions provide a good initial guess for a Newton solver to determine some of the actual roots. Second, the asymptotic expansions can be used to bound the growth of the factorised Wiener–Hopf kernel in ${\mathcal{L}}^{\pm }$. Understanding this growth is an important step in the Wiener–Hopf procedure, specifically the determination of the relevant entire function.

We focus on the zeros located in the first quadrant of the complex $\unicode[STIX]{x1D6FE}$-plane. These roots are labelled as $\unicode[STIX]{x1D703}_{n,R}^{+}$. The asymptotic behaviour of the roots in the other quadrants can be determined by a similar procedure.

Recall the definition of the branch cut of $\unicode[STIX]{x1D701}$

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D701}=\sqrt{\unicode[STIX]{x1D714}^{2}w^{2}-\unicode[STIX]{x1D6FE}^{2}}=\text{e}^{\text{i}\unicode[STIX]{x1D713}_{1}/2}\text{e}^{\text{i}\unicode[STIX]{x1D713}_{2}/2}|\unicode[STIX]{x1D714}^{2}w^{2}-\unicode[STIX]{x1D6FE}^{2}|^{1/2}, & & \displaystyle \nonumber\end{eqnarray}$$

where

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D713}_{1}=\arg (\unicode[STIX]{x1D714}w-\unicode[STIX]{x1D6FE}),\quad \unicode[STIX]{x1D713}_{2}=\arg (\unicode[STIX]{x1D714}w+\unicode[STIX]{x1D6FE}), & & \displaystyle \nonumber\end{eqnarray}$$

and

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x03C0}/2<\unicode[STIX]{x1D713}_{1}<5\unicode[STIX]{x03C0}/2,\quad -\unicode[STIX]{x03C0}/2<\unicode[STIX]{x1D713}_{2}<3\unicode[STIX]{x03C0}/2. & & \displaystyle \nonumber\end{eqnarray}$$

Since the $\unicode[STIX]{x1D703}_{n,R}^{+}$ are in the first quadrant, we have

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D701}(\unicode[STIX]{x1D703}_{n,R}^{+})\sim \text{i}\unicode[STIX]{x1D703}_{n,R}^{+},\quad \text{as }n\rightarrow \infty . & & \displaystyle \nonumber\end{eqnarray}$$

This leads us to determine the following asymptotic behaviours:

$$\begin{eqnarray}\displaystyle & \displaystyle \sin (s\unicode[STIX]{x1D701}(\unicode[STIX]{x1D703}_{n,R}^{+}))\sim -\frac{1}{2\text{i}}\text{e}^{s\unicode[STIX]{x1D703}_{n,R}^{+}}, & \displaystyle \nonumber\\ \displaystyle & \displaystyle \cos (s\unicode[STIX]{x1D701}(\unicode[STIX]{x1D703}_{n,R}^{+}))\sim {\textstyle \frac{1}{2}}\text{e}^{s\unicode[STIX]{x1D703}_{n,R}^{+}},\quad \cos (d\unicode[STIX]{x1D703}_{n,R}^{+}+\unicode[STIX]{x1D70E}^{\prime })\sim {\textstyle \frac{1}{2}}\text{e}^{-\text{i}(d\unicode[STIX]{x1D703}_{n,R}^{+}+\unicode[STIX]{x1D70E}^{\prime })}. & \displaystyle \nonumber\end{eqnarray}$$

We now substitute these representations into (B 1) to obtain asymptotic expansions for the roots $\unicode[STIX]{x1D703}_{n,R}^{+}$. Each case must be considered separately, although the asymptotic behaviours are similar.

Case I boundary condition

For the no-mean-flow boundary condition, the asymptotic behaviour of the roots obeys

(B 8)$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D703}_{n,R}^{+}\sim -\unicode[STIX]{x1D707}_{0}(1-\exp [-\text{i}((d-\text{i}s)\unicode[STIX]{x1D703}_{n,R}^{+}+\unicode[STIX]{x1D70E}^{\prime })]). & & \displaystyle\end{eqnarray}$$

We seek an asymptotic expansion of the first quadrant roots of the classical form

(B 9)$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D703}_{n,R}^{+}\sim \mathop{\sum }_{m=0}^{\infty }a_{R}^{(m)}\unicode[STIX]{x1D703}_{R}^{(m)}(n), & & \displaystyle\end{eqnarray}$$

where $\unicode[STIX]{x1D703}_{R}^{(m+1)}(n)=o(\unicode[STIX]{x1D703}_{R}^{(m)}(n))$ as $n\rightarrow \infty$. In (B 8) we require the linear (left) and exponential (right) terms to match. However, since $|\unicode[STIX]{x1D703}_{n,R}^{+}|\rightarrow \infty$ as $n\rightarrow \infty$, the exponential term will generally grow at a faster rate than the linear term. Consequently, the real part of the argument of the exponential must be asymptotically small compared to the imaginary part. We therefore expand the roots into real and imaginary parts as $\unicode[STIX]{x1D703}_{n,R}^{+}=\unicode[STIX]{x1D703}_{n,R}^{r+}+\text{i}\unicode[STIX]{x1D703}_{n,R}^{i+}$ and write

(B 10)$$\begin{eqnarray}\displaystyle d\unicode[STIX]{x1D703}_{n,R}^{i+}-s\unicode[STIX]{x1D703}_{n,R}^{r+}=G(n), & & \displaystyle\end{eqnarray}$$

where $G(n)=o(\unicode[STIX]{x1D703}_{R}^{(0)}(n))$. Rearranging yields

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D703}_{n,R}^{+}=\unicode[STIX]{x1D703}_{n,R}^{+}+\text{i}\unicode[STIX]{x1D703}_{n,R}^{i+}=\left(1+\text{i}\frac{s}{d}\right)\unicode[STIX]{x1D703}_{n,R}^{r+}+\text{i}\frac{G(n)}{d}. & & \displaystyle \nonumber\end{eqnarray}$$

Since the arguments of the left- and right-hand sides of (B 8) must match, we obtain an expression for the imaginary part of the argument of the exponential:

(B 11)$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D70E}^{\prime }+d\unicode[STIX]{x1D703}_{n,R}^{r+}+s\unicode[STIX]{x1D703}_{n,R}^{i+}=\unicode[STIX]{x1D712}-2n\unicode[STIX]{x03C0}+o(1). & & \displaystyle\end{eqnarray}$$

Applying the asymptotic expansion (B 9) and taking the leading-order terms of (B 10) and (B 11) yields

(B 12)$$\begin{eqnarray}\displaystyle \left.\begin{array}{@{}c@{}}\displaystyle \unicode[STIX]{x1D703}_{R}^{(0)}(n)=n,\\ \displaystyle a_{R}^{(0)}=\frac{2\unicode[STIX]{x03C0}\text{i}\text{e}^{-\text{i}\unicode[STIX]{x1D712}}}{\unicode[STIX]{x1D6E5}}.\end{array}\right\} & & \displaystyle\end{eqnarray}$$

We may now substitute the expansion for $\unicode[STIX]{x1D703}_{n,R}^{+}$ so far into (B 8) to obtain

$$\begin{eqnarray}\displaystyle \frac{2\unicode[STIX]{x03C0}\text{i}\text{e}^{-\text{i}\unicode[STIX]{x1D712}}}{\unicode[STIX]{x1D6E5}}n+o(n)\sim -\unicode[STIX]{x1D707}_{0}(1-\exp [G(n)-\text{i}\unicode[STIX]{x1D712}]). & & \displaystyle \nonumber\end{eqnarray}$$

We now match leading-order terms to obtain

$$\begin{eqnarray}\displaystyle \frac{2\unicode[STIX]{x03C0}}{\unicode[STIX]{x1D6E5}}n=-\unicode[STIX]{x1D707}_{0}\exp [G(n)], & & \displaystyle \nonumber\end{eqnarray}$$

so that

$$\begin{eqnarray}\displaystyle G(n)=\log (n)+\log \left(\frac{-2\unicode[STIX]{x03C0}}{\unicode[STIX]{x1D707}_{0}\unicode[STIX]{x1D6E5}}\right). & & \displaystyle \nonumber\end{eqnarray}$$

Similarly, taking the leading-order terms in (B 10) and (B 11) yields

(B 13)$$\begin{eqnarray}\displaystyle \left.\begin{array}{@{}c@{}}\displaystyle \unicode[STIX]{x1D703}_{R}^{(1)}(n)=\log (n),\\ \displaystyle a_{R}^{(1)}=-\frac{\text{e}^{-\text{i}\unicode[STIX]{x1D712}}}{\unicode[STIX]{x1D6E5}},\end{array}\right\} & & \displaystyle\end{eqnarray}$$

and

(B 14)$$\begin{eqnarray}\displaystyle \left.\begin{array}{@{}c@{}}\displaystyle \unicode[STIX]{x1D703}_{R}^{(2)}(n)=1,\\ \displaystyle a_{R}^{(2)}=\frac{\text{i}\text{e}^{-\text{i}\unicode[STIX]{x1D712}}}{\unicode[STIX]{x1D6E5}}\left(\frac{\unicode[STIX]{x03C0}}{2}+\unicode[STIX]{x1D712}+\text{i}\log \left(\frac{-2\unicode[STIX]{x03C0}}{\unicode[STIX]{x1D707}_{0}\unicode[STIX]{x1D6E5}}\right)-\unicode[STIX]{x1D70E}^{\prime }\right).\end{array}\right\} & & \displaystyle\end{eqnarray}$$

Substitution of ((B 12), (B 13), (B 14)) into the asymptotic expansion (B 9) yields

(B 15)$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D703}_{n,R}^{+}\sim \frac{\text{e}^{-\text{i}\unicode[STIX]{x1D712}}}{\unicode[STIX]{x1D6E5}}\left(2\unicode[STIX]{x03C0}\text{i}n-\log (n)+\text{i}\left(\frac{\unicode[STIX]{x03C0}}{2}+\unicode[STIX]{x1D712}-\unicode[STIX]{x1D70E}^{\prime }\right)-\log \left(\frac{-2\unicode[STIX]{x03C0}}{\unicode[STIX]{x1D707}_{0}\unicode[STIX]{x1D6E5}}\right)+o(1)\right). & & \displaystyle\end{eqnarray}$$

Similar analysis yields the asymptotic behaviour for the roots in the second quadrant as

(B 16)$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D703}_{n,L}^{+}\sim \frac{\text{e}^{\text{i}\unicode[STIX]{x1D712}}}{\unicode[STIX]{x1D6E5}}\left(2\unicode[STIX]{x03C0}\text{i}n+\log (n)+\text{i}\left(\frac{\unicode[STIX]{x03C0}}{2}+\unicode[STIX]{x1D712}+\unicode[STIX]{x1D70E}^{\prime }\right)+\log \left(\frac{-2\unicode[STIX]{x03C0}}{\unicode[STIX]{x1D707}_{0}\unicode[STIX]{x1D6E5}}\right)+o(1)\right). & & \displaystyle\end{eqnarray}$$

Case II boundary condition

In the case of a porous blade in a background flow, the asymptotic behaviour of the roots obeys the equation

(B 17)$$\begin{eqnarray}\displaystyle 1\sim \text{i}\unicode[STIX]{x1D707}_{1}(1-\exp [-\text{i}((d-\text{i}s)\unicode[STIX]{x1D703}_{n,R}^{+}+\unicode[STIX]{x1D70E}^{\prime })]). & & \displaystyle\end{eqnarray}$$

We assume an asymptotic expansion of the roots of the form (B 9). Similar reasoning to the previous section yields that the leading-order terms are also given by

(B 18a,b)$$\begin{eqnarray}\displaystyle a_{R}^{(0)}=\frac{2\unicode[STIX]{x03C0}\text{i}\text{e}^{-\text{i}\unicode[STIX]{x1D712}}}{\unicode[STIX]{x1D6E5}},\quad a_{L}^{(0)}=\frac{2\unicode[STIX]{x03C0}\text{i}\text{e}^{\text{i}\unicode[STIX]{x1D712}}}{\unicode[STIX]{x1D6E5}}. & & \displaystyle\end{eqnarray}$$

We may now solve (B 17) directly to find the coefficients of the next two orders of the asymptotic expansion as

$$\begin{eqnarray}\displaystyle & \displaystyle a_{R}^{(1)}=0,\quad a_{L}^{(1)}=0, & \displaystyle \nonumber\\ \displaystyle & \displaystyle a_{R}^{(2)}=-\frac{\text{e}^{-\text{i}\unicode[STIX]{x1D712}}}{\unicode[STIX]{x1D6E5}}\left(\log \left(1+\frac{\text{i}}{\unicode[STIX]{x1D707}_{0}}\right)+\text{i}\unicode[STIX]{x1D70E}^{\prime }\right),\quad a_{L}^{(2)}=\frac{\text{e}^{\text{i}\unicode[STIX]{x1D712}}}{\unicode[STIX]{x1D6E5}}\left(\log \left(1-\frac{\text{i}}{\unicode[STIX]{x1D707}_{0}}\right)+\text{i}\unicode[STIX]{x1D70E}^{\prime }\right). & \displaystyle \nonumber\end{eqnarray}$$

Case III boundary condition

For the case III boundary condition, the asymptotic behaviour of the roots obeys

(B 19)$$\begin{eqnarray}\displaystyle 1\sim \unicode[STIX]{x1D707}_{2}\unicode[STIX]{x1D703}_{n,R}^{+}(1-\exp [-\text{i}((d-\text{i}s)\unicode[STIX]{x1D703}_{n,R}^{+}+\unicode[STIX]{x1D70E}^{\prime })]). & & \displaystyle\end{eqnarray}$$

Similar analysis to the previous sections possesses an identical asymptotic expansion (up to the terms considered) and we have, at leading order,

(B 20a,b)$$\begin{eqnarray}\displaystyle a_{R}^{(0)}=\frac{2\unicode[STIX]{x03C0}\text{i}\text{e}^{-\text{i}\unicode[STIX]{x1D712}}}{\unicode[STIX]{x1D6E5}},\quad a_{L}^{(0)}=\frac{2\unicode[STIX]{x03C0}\text{i}\text{e}^{\text{i}\unicode[STIX]{x1D712}}}{\unicode[STIX]{x1D6E5}}. & & \displaystyle\end{eqnarray}$$

Substitution of (B 20) into (B 19) yields the coefficients of the next two orders of the asymptotic expansion as

$$\begin{eqnarray}\displaystyle & \displaystyle a_{R}^{(1)}=0,\quad a_{L}^{(1)}=0, & \displaystyle \nonumber\\ \displaystyle & \displaystyle a_{R}^{(2)}=-\frac{\text{i}\unicode[STIX]{x1D70E}^{\prime }\text{e}^{-\text{i}\unicode[STIX]{x1D712}}}{\unicode[STIX]{x1D6E5}},\quad a_{L}^{(2)}=\frac{\text{i}\unicode[STIX]{x1D70E}^{\prime }\text{e}^{\text{i}\unicode[STIX]{x1D712}}}{\unicode[STIX]{x1D6E5}}. & \displaystyle \nonumber\end{eqnarray}$$

B.3 Full factorisation of kernel, $K$

We propose a multiplicative splitting of $K$ of the form:

$$\begin{eqnarray}\displaystyle K(\unicode[STIX]{x1D6FE})=K_{+}(\unicode[STIX]{x1D6FE})K_{-}(\unicode[STIX]{x1D6FE}), & & \displaystyle \nonumber\end{eqnarray}$$

where

(B 21a)$$\begin{eqnarray}\displaystyle & \displaystyle K_{-}(\unicode[STIX]{x1D6FE})=\text{e}^{E(\unicode[STIX]{x1D6FE})}\frac{\displaystyle \mathop{\prod }_{m=1}^{\infty }(1-\unicode[STIX]{x1D6FE}/\unicode[STIX]{x1D703}_{m,R}^{+})(1-\unicode[STIX]{x1D6FE}/\unicode[STIX]{x1D703}_{m,L}^{+})}{\displaystyle \mathop{\prod }_{m=-\infty }^{\infty }(1-\unicode[STIX]{x1D6FE}/\unicode[STIX]{x1D706}_{m}^{+})}, & \displaystyle\end{eqnarray}$$

(B 21b)$$\begin{eqnarray}\displaystyle & \displaystyle K_{+}(\unicode[STIX]{x1D6FE})=\text{e}^{-E(\unicode[STIX]{x1D6FE})}\frac{\displaystyle \mathop{\prod }_{m=1}^{\infty }(1-\unicode[STIX]{x1D6FE}/\unicode[STIX]{x1D703}_{m,R}^{-})(1-\unicode[STIX]{x1D6FE}/\unicode[STIX]{x1D703}_{m,L}^{-})}{\displaystyle \mathop{\prod }_{m=-\infty }^{\infty }(1-\unicode[STIX]{x1D6FE}/\unicode[STIX]{x1D706}_{m}^{-})}K(0). & \displaystyle\end{eqnarray}$$

$E$

$K_{\pm }$

${\mathcal{L}}^{\pm }$

$E$

$E$

Asymptotic behaviour of proposed factorisation

We focus on $K_{-}$ since the asymptotic behaviour of $K_{+}$ follows in an analogous manner. We first note that the infinite product

(B 22)$$\begin{eqnarray}\displaystyle P_{1}=\mathop{\prod }_{m=1}^{\infty }\left(\frac{a_{R}^{(0)}m+a_{R}^{(2)}}{\unicode[STIX]{x1D703}_{m,R}^{+}}\right)\left(\frac{a_{L}^{(0)}m+a_{L}^{(2)}}{\unicode[STIX]{x1D703}_{m,L}^{+}}\right) & & \displaystyle\end{eqnarray}$$

exists. This is seen by employing the asymptotic expansions (B 15) and (B 16) to obtain

$$\begin{eqnarray}\displaystyle & \displaystyle \frac{a_{R}^{(0)}m+a_{R}^{(1)}}{\unicode[STIX]{x1D703}_{m,R}^{+}}=1-\frac{a_{R}^{(1)}\log (m)}{a_{R}^{(0)}m}+o(m^{-1}), & \displaystyle \nonumber\\ \displaystyle & \displaystyle \frac{a_{L}^{(0)}m+a_{L}^{(1)}}{\unicode[STIX]{x1D703}_{m,L}^{+}}=1-\frac{a_{L}^{(1)}\log (m)}{a_{L}^{(0)}m}+o(m^{-1}), & \displaystyle \nonumber\end{eqnarray}$$

respectively. Substitution into the product (B 22) yields

(B 23)$$\begin{eqnarray}\displaystyle P_{1}=\mathop{\prod }_{m=1}^{\infty }\left(1-R\frac{\log (m)}{m}+o(m^{-1})\right), & & \displaystyle\end{eqnarray}$$

where

(B 24)$$\begin{eqnarray}\displaystyle R=\frac{a_{R}^{(1)}}{a_{R}^{(0)}}+\frac{a_{L}^{(1)}}{a_{L}^{(0)}}=0, & & \displaystyle\end{eqnarray}$$

and the last identity is due to (B 12) and (B 13). Consequently, the products (B 23) and (B 22) exist via the comparison method. Moreover, we note that since

$$\begin{eqnarray}\displaystyle & \displaystyle \text{Im}[a_{R}^{(0)}m_{1}+a_{R}^{(2)}-\unicode[STIX]{x1D6FE}]>m_{1}\text{Im}[a_{R}^{(0)}], & \displaystyle \nonumber\\ \displaystyle & \displaystyle \text{Im}[a_{L}^{(0)}m_{1}+a_{L}^{(2)}-\unicode[STIX]{x1D6FE}]>m_{1}\text{Im}[a_{L}^{(0)}], & \displaystyle \nonumber\end{eqnarray}$$

we can show that the infinite product

(B 25)$$\begin{eqnarray}\displaystyle P_{2}(\unicode[STIX]{x1D6FE})=\mathop{\prod }_{m=1}^{\infty }\left(\frac{\unicode[STIX]{x1D703}_{m,R}^{+}-\unicode[STIX]{x1D6FE}}{a_{R}^{(0)}m+a_{R}^{(2)}-\unicode[STIX]{x1D6FE}}\right)\left(\frac{\unicode[STIX]{x1D703}_{m,L}^{+}-\unicode[STIX]{x1D6FE}}{a_{L}^{(0)}m+a_{L}^{(2)}-\unicode[STIX]{x1D6FE}}\right) & & \displaystyle\end{eqnarray}$$

has, at worst, algebraic growth as $\unicode[STIX]{x1D6FE}\rightarrow \infty$ in ${\mathcal{L}}^{-}$, i.e. $\exists \unicode[STIX]{x1D6FC}$ s.t. $P(\unicode[STIX]{x1D6FE})\sim \unicode[STIX]{x1D6FE}^{\unicode[STIX]{x1D6FC}}$, up to a multiplicative constant. We now use $P_{1}$ and $P_{2}$ to calculate the asymptotic behaviour of $K_{-}$ by comparison with the product

(B 26)$$\begin{eqnarray}\displaystyle P(\unicode[STIX]{x1D6FE})=\mathop{\prod }_{m=1}^{\infty }{\displaystyle \frac{(1-\unicode[STIX]{x1D6FE}/\unicode[STIX]{x1D703}_{m,R}^{+})\text{e}^{\unicode[STIX]{x1D6FE}/\unicode[STIX]{x1D703}_{m,R}^{+}}(1-\unicode[STIX]{x1D6FE}/\unicode[STIX]{x1D703}_{m,L}^{+})\text{e}^{\unicode[STIX]{x1D6FE}/\unicode[STIX]{x1D703}_{m,L}^{+}}}{(1-\unicode[STIX]{x1D6FE}/(a_{R}^{(0)}m+a_{R}^{(2)}))\text{e}^{\unicode[STIX]{x1D6FE}/a_{R}^{(0)}m}(1-\unicode[STIX]{x1D6FE}/(a_{L}^{(0)}m+a_{L}^{(2)}))\text{e}^{\unicode[STIX]{x1D6FE}/a_{L}^{(0)}m}}}. & & \displaystyle\end{eqnarray}$$

This function may alternatively be written as

$$\begin{eqnarray}\displaystyle P(\unicode[STIX]{x1D6FE})=\exp \left(\unicode[STIX]{x1D6FE}\mathop{\sum }_{m=1}^{\infty }\frac{1}{\unicode[STIX]{x1D703}_{m,R}^{+}}-\frac{1}{a_{R}^{(0)}m}+\frac{1}{\unicode[STIX]{x1D703}_{m,L}^{+}}-\frac{1}{a_{L}^{(0)}m}\right)P_{1}P_{2}(\unicode[STIX]{x1D6FE}). & & \displaystyle \nonumber\end{eqnarray}$$

Since $P_{1}$ exists and $P_{2}$ has algebraic growth as $\unicode[STIX]{x1D6FE}\rightarrow$ in ${\mathcal{L}}^{-}$, we have, up to a multiplication constant,

$$\begin{eqnarray}\displaystyle P(\unicode[STIX]{x1D6FE})\sim P_{1}\unicode[STIX]{x1D6FE}^{\unicode[STIX]{x1D6FC}}\exp \left(\unicode[STIX]{x1D6FE}\mathop{\sum }_{m=1}^{\infty }\frac{1}{\unicode[STIX]{x1D703}_{m,R}^{+}}-\frac{1}{a_{R}^{(0)}m}+\frac{1}{\unicode[STIX]{x1D703}_{m,L}^{+}}-\frac{1}{a_{L}^{(0)}m}\right). & & \displaystyle \nonumber\end{eqnarray}$$

We now calculate the growth of the denominators of $P$. By applying the asymptotic behaviour of the gamma function (Peake Reference Peake1992, B7 & B8), we derive the relation

$$\begin{eqnarray}\displaystyle \mathop{\prod }_{m=1}^{\infty }\left(1-\frac{\unicode[STIX]{x1D6FE}}{am+b}\right)\exp \left[\frac{\unicode[STIX]{x1D6FE}}{am}\right]\sim C\exp \left[\frac{\unicode[STIX]{x1D6FE}}{a}({\mathcal{E}}-1-\log (-a))+\left(\frac{\unicode[STIX]{x1D6FE}}{a}-\frac{b}{a}-\frac{1}{2}\right)\log (\unicode[STIX]{x1D6FE})\right]\!, & & \displaystyle \nonumber\end{eqnarray}$$

where ${\mathcal{E}}$ is the Euler–Mascheroni constant and

$$\begin{eqnarray}\displaystyle C=\frac{-b}{\sqrt{2\unicode[STIX]{x03C0}}}\unicode[STIX]{x1D6E4}\left({\displaystyle \frac{b}{a}}\right)\left(-a\right)^{((b/a)-1/2)}. & & \displaystyle \nonumber\end{eqnarray}$$

This representation may be substituted into the denominator of (B 26). Rearranging yields the asymptotic behaviour

$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\prod }_{m=1}^{\infty }\left(1-\frac{\unicode[STIX]{x1D6FE}}{\unicode[STIX]{x1D703}_{m,R}^{+}}\right)\text{e}^{\unicode[STIX]{x1D6FE}/\unicode[STIX]{x1D703}_{m,R}^{+}}\left(1-\frac{\unicode[STIX]{x1D6FE}}{\unicode[STIX]{x1D703}_{m,L}^{+}}\right)\text{e}^{\unicode[STIX]{x1D6FE}/\unicode[STIX]{x1D703}_{m,L}^{+}}\nonumber\\ \displaystyle & & \displaystyle \quad \sim B_{2}\unicode[STIX]{x1D6FE}^{\unicode[STIX]{x1D6FC}}\exp \left[\unicode[STIX]{x1D6FE}\left(\mathop{\sum }_{m=1}^{\infty }\frac{1}{\unicode[STIX]{x1D703}_{m,R}^{+}}-\frac{1}{a_{R}^{(0)}m}+\frac{1}{\unicode[STIX]{x1D703}_{m,L}^{+}}-\frac{1}{a_{L}^{(0)}m}\right)\right]\nonumber\\ \displaystyle & & \displaystyle \quad \times \exp \left[\frac{\unicode[STIX]{x1D6FE}}{a_{R}^{(0)}a_{L}^{(0)}}(({\mathcal{E}}-1)(a_{R}^{(0)}+a_{L}^{(0)})-a_{L}^{(0)}\log (-a_{R}^{(0)})-a_{R}^{(0)}\log (-a_{L}^{(0)}))\right]\nonumber\\ \displaystyle & & \displaystyle \quad \times \exp \left[\left(\unicode[STIX]{x1D6FE}\left(\frac{1}{a_{R}^{(0)}}+\frac{1}{a_{L}^{(0)}}\right)-\left(\frac{a_{R}^{(2)}}{a_{R}^{(0)}}+\frac{a_{L}^{(2)}}{a_{L}^{(0)}}\right)-1\right)\log (\unicode[STIX]{x1D6FE})\right],\nonumber\end{eqnarray}$$

for some constant $B_{2}$. We also note that

$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\prod }_{m=-\infty }^{\infty }\left(1-\frac{\unicode[STIX]{x1D6FE}}{\unicode[STIX]{x1D706}_{m}^{+}}\right)\text{e}^{\unicode[STIX]{x1D6FE}/\unicode[STIX]{x1D706}_{m}^{+}}\sim B_{3}\left(1-\frac{\unicode[STIX]{x1D6FE}}{\unicode[STIX]{x1D706}_{m}^{+}}\right)\exp \left[\unicode[STIX]{x1D6FE}\left(\frac{1}{\unicode[STIX]{x1D706}_{0}^{+}}+\mathop{\sum }_{\substack{ m=-\infty \\ m\neq 0}}^{\infty }\frac{1}{\unicode[STIX]{x1D706}_{m}^{+}}-\frac{1}{\unicode[STIX]{x1D706}_{R}^{(0)+}|m|}\right)\right]\nonumber\\ \displaystyle & & \displaystyle \quad \times \exp \left[\frac{\unicode[STIX]{x1D6FE}}{\unicode[STIX]{x1D706}_{R}^{(0)+}\unicode[STIX]{x1D706}_{L}^{(0)+}}(({\mathcal{E}}-1)(\unicode[STIX]{x1D706}_{R}^{(0)+}+\unicode[STIX]{x1D706}_{L}^{(0)+})-\unicode[STIX]{x1D706}_{L}^{(0)+}\log (-\unicode[STIX]{x1D706}_{R}^{(0)+})-\unicode[STIX]{x1D706}_{R}^{(0)+}\log (-\unicode[STIX]{x1D706}_{L}^{(0)+}))\right]\nonumber\\ \displaystyle & & \displaystyle \quad \times \exp \left[\left(\unicode[STIX]{x1D6FE}\left(\frac{1}{\unicode[STIX]{x1D706}_{R}^{(0)+}}+\frac{1}{\unicode[STIX]{x1D706}_{L}^{(0)+}}\right)-\left(\frac{\unicode[STIX]{x1D706}_{R}^{(2)+}}{\unicode[STIX]{x1D706}_{R}^{(0)+}}+\frac{\unicode[STIX]{x1D706}_{L}^{(2)+}}{\unicode[STIX]{x1D706}_{L}^{(0)+}}\right)-1\right)\log (\unicode[STIX]{x1D6FE})\right],\nonumber\end{eqnarray}$$

for some constant $B_{3}$. Noting that $\unicode[STIX]{x1D706}_{R}^{(0)+}=a_{R}^{(0)}$ and $\unicode[STIX]{x1D706}_{L}^{(0)+}=a_{L}^{(0)}$, and using (B 7), we obtain

$$\begin{eqnarray}\displaystyle \frac{\displaystyle \mathop{\prod }_{m=1}^{\infty }(1-\unicode[STIX]{x1D6FE}/\unicode[STIX]{x1D703}_{m,R}^{+})(1-\unicode[STIX]{x1D6FE}/\unicode[STIX]{x1D703}_{m,L}^{+})}{\displaystyle \mathop{\prod }_{m=-\infty }^{\infty }(1-\unicode[STIX]{x1D6FE}/\unicode[STIX]{x1D706}_{m}^{+})}\sim B_{4}\unicode[STIX]{x1D6FE}^{c}, & & \displaystyle \nonumber\end{eqnarray}$$

for some constant $B_{4}$ and

$$\begin{eqnarray}\displaystyle c=\unicode[STIX]{x1D6FC}-\left(\frac{a_{R}^{(2)}}{a_{R}^{(0)}}+\frac{a_{L}^{(2)}}{a_{L}^{(0)}}\right). & & \displaystyle \nonumber\end{eqnarray}$$

In particular, on a correct selection of the logarithmic branches, we find the following behaviours of the asymptotic factorisations of the kernels (subject to a multiplicative constant):

1. (i) Case 0,

   $$\begin{eqnarray}\displaystyle K_{\pm }(\unicode[STIX]{x1D6FE})\sim \unicode[STIX]{x1D6FE}^{1/2}; & & \displaystyle \nonumber\end{eqnarray}$$

2. (ii) Case I,

   $$\begin{eqnarray}\displaystyle K_{\pm }(\unicode[STIX]{x1D6FE})\sim \unicode[STIX]{x1D6FE}^{1/2}; & & \displaystyle \nonumber\end{eqnarray}$$

3. (iii) Case II,

   $$\begin{eqnarray}\displaystyle K_{\pm }(\unicode[STIX]{x1D6FE})\sim \unicode[STIX]{x1D6FE}^{\mp (1/\unicode[STIX]{x03C0})\cot ^{-1}(\unicode[STIX]{x1D707}_{1})+n} & & \displaystyle \nonumber\end{eqnarray}$$
   for some non-negative integer $n$. It is worth noting that the above exponent has appeared before in the porous aerofoil literature. Baddoo, Hajian & Jaworski (Reference Baddoo, Hajian and Jaworski2019) showed that an incompressible, irrotational velocity field near a porous edge must scale like $\pm (1/\unicode[STIX]{x03C0})\cot ^{-1}(\unicode[STIX]{x1D713})+n$, where $\unicode[STIX]{x1D713}$ is the porosity parameter. It is not a coincidence that this factor also appears here: through the Wiener–Hopf analysis, the behaviour of the unsteady velocity field at the endpoints of the blade is governed by the behaviour of the factorised kernel at large $\unicode[STIX]{x1D6FE}$. Since the edge is self-similar, the singular behaviour is identical for the Laplace and Helmholtz equations.

4. (iv) Case III:

   $$\begin{eqnarray}\displaystyle K_{\pm }(\unicode[STIX]{x1D6FE})\sim \unicode[STIX]{x1D6FE}. & & \displaystyle \nonumber\end{eqnarray}$$