This appendix describes the solution to the problem defined in Section 18.3. The solution for the jump in potential across a blade that is part of a rectilinear cascade, as shown in Fig. 18.5, in response to a harmonic gust defined as w_oexp(−iωt+ik₁x+ik₂y+ik₃z) is given by

$\mathrm{\Delta }\hat{\varphi }\left(x\right)={\displaystyle {\int }_{-\infty }^{\infty }D\left(\alpha ,k_1,\sigma \right)e^{-i\alpha x}d\alpha }$

where it is shown in Chapter 18, Ref. [12] that

$D\left(\alpha ,k_1,\sigma \right)=\frac{-i{w}_o}{{\left(2\pi \right)}^2\left(\alpha +k_1\right)J_{+}\left(\alpha \right)J_{-}\left(-k_1\right)}-\left\{{\displaystyle \sum _n\frac{\left(A_n+C_n\right)e^{i\left(\alpha -{\delta }_n\right)c}}{i\left(\omega +\alpha U\right)\left(\alpha -{\delta }_n\right)}\left[\frac{J_{-}\left({\delta }_n\right)}{J_{-}\left(\alpha \right)}\right]}\right\}-\left\{{\displaystyle \sum _m\frac{B_m}{\left(\alpha -{\varepsilon }_m\right)}\left[\frac{J_{+}\left({\varepsilon }_m\right)}{J_{+}\left(\alpha \right)}\right]}\right\}$

where

$A_o=\frac{w_o\left(\omega -k_{1}U\right)}{{\left(2\pi \right)}^{2}j\left(-k_1\right)}{\delta }_o=-k_1$

$A_n=\frac{w_o\left(\omega +{\delta }_{n}U\right)}{{\left(2\pi \right)}^2\left({\delta }_n+k_1\right)J_{+}^{\prime }\left({\delta }_n\right)J_{-}\left(-k_1\right)}{\delta }_n=\kappa M+{\theta }_{n-1}n>0$

with

$\begin{array}{l}{J}_{+}^{\prime }\left({\delta }_n\right)=\frac{j^{\prime }\left({\delta }_n\right)}{J_{-}\left({\delta }_n\right)}\\ j^{\prime }\left({\delta }_n\right)=\frac{\left(\kappa M-{\delta }_n\right)h{\beta }^2}{4\pi \left(1-cos\left({\delta }_{n}d+\sigma \right)cos\left(\left(n-1\right)\pi \right)\right)}\left\{\begin{array}{cc}\hfill 2\hfill & \hfill n=1\hfill \\ \hfill 1\hfill & \hfill n>1\hfill \end{array}\right.\end{array}$

and the coefficients B_m are obtained as the solutions to the equations

$\left\{B_m\right\}={\left[\left[\mathbf{1}\right]-\left[{\mathbf{F}}_{mn}\right]\left[{\mathbf{L}}_{mn}\right]\right]}^{-1}\left\{\left[{\mathbf{F}}_{mn}\right]\left\{A_n\right\}+A_o{G}_m\right\}$

where

$F_{mn}=-\frac{e^{i\left({\varepsilon }_m-{\delta }_n\right)c}}{i\left(\omega +{\varepsilon }_{m}U\right)\left({\varepsilon }_m-{\delta }_n\right)}\left[\frac{J_{-}\left({\delta }_n\right)}{J_{-}^{\prime }\left({\varepsilon }_m\right)}\right]$

$G_m=-\frac{e^{i\left({\varepsilon }_m-{\delta }_o\right)c}}{i\left(\omega +{\varepsilon }_{m}U\right)\left({\varepsilon }_m-{\delta }_o\right)}\left[\frac{J_{-}\left({\delta }_o\right)}{J_{-}^{\prime }\left({\varepsilon }_m\right)}\right]$

$L_{mn}=\frac{i\left(\omega +{\delta }_{m}U\right)}{\left({\varepsilon }_n-{\delta }_m\right)}\left[\frac{J_{+}\left({\varepsilon }_n\right)}{J_{+}^{\prime }\left({\delta }_m\right)}\right]$

The coefficients C_n are obtained from

$\begin{array}{c}{C}_o=0\\ C_n={\displaystyle \sum _m\frac{i\left(\omega +{\delta }_{n}U\right)}{\left({\varepsilon }_m-{\delta }_n\right)}\left[\frac{J_{+}\left({\varepsilon }_m\right)}{J_{+}^{\prime }\left({\delta }_n\right)}\right]B_m}n>0\end{array}$

The split functions are defined as

$J_{+}\left(\alpha \right)=\frac{{\kappa }_e\beta sin\left({\kappa }_{e}h\beta \right)}{4\pi \left(cos\left({\kappa }_{e}h\beta \right)-cos\left(\rho \right)\right)}\frac{{\displaystyle \prod _{m=0}^{\infty }\left(1-\left(\alpha -k_{o}M\right)/{\theta }_m\right)}}{{\displaystyle \prod _{m=-\infty }^{\infty }\left(1-\left(\alpha -k_{o}M\right)/{\eta }_m^{-}\right)}}{e}^{\Phi }$

$J_{-}\left(\alpha \right)=\frac{{\displaystyle \prod _{m=0}^{\infty }\left(1-\left(\alpha -k_{o}M\right)/{\vartheta }_m\right)}}{{\displaystyle \prod _{m=-\infty }^{\infty }\left(1-\left(\alpha -k_{o}M\right)/{\eta }_m^{+}\right)}}{e}^{-\Phi }$

The function Φ must be chosen so that both J₊ and J₋ have algebraic growth as α tends to infinity and is given by

$\Phi =\frac{-i\left(\alpha -k_{o}M\right)}{\pi }\left\{h\beta log\left(2cos{\chi }_e\right)+{\chi }_{e}d\right\}$

The singularities of these functions are given by

${\theta }_m=-\sqrt{{\kappa }_e^2-{\left(\frac{m\pi }{\beta h}\right)}^2}{\vartheta }_m=\sqrt{{\kappa }_e^2-{\left(\frac{m\pi }{\beta h}\right)}^2}$

${\eta }_m^{\pm }=-f_{m}sin{\chi }_e\pm cos{\chi }_e\sqrt{{\kappa }_e^2-f_m^2}{f}_m=\frac{\sigma +k_{o}Md-2\pi m}{\sqrt{d^2+{\left(h\beta \right)}^2}}$

where tan χ_ɛ=d/hβ. Finally, we have used the variables

$M=U_e/c_o{\beta }^2=1-M^2{k}_o=\omega /c_o{\beta }^2{\kappa }_e^2=k_o^2-{\left(k_3/\beta \right)}^2\rho =\sigma +k_{o}Md$

so

$j\left(\alpha \right)=\frac{\zeta }{4\pi }\left\{\frac{sin\left(\zeta h\right)}{cos\left(\zeta h\right)-cos\left(\left(\alpha -k_{o}M\right)d+\rho \right)}\right\}\zeta =\beta \sqrt{{\kappa }_e^2-{\left(\alpha -k_{o}M\right)}^2}$