- Mccamy-Fuchs analytical solution of the scattering of regular waves by a vertical circular cylinder.
This important flow accepts a closed-form analytical solution for arbitrary values of the wavelength [math]\displaystyle{ \lambda\, }[/math]. This was shown to be the case by Mccamy-Fuchs using separation of variables
[math]\displaystyle{ \Phi_I = \mathfrak{Re} \left\{\phi_I e^{i\omega t} \right \} \, }[/math]
[math]\displaystyle{ \phi_I = \frac{i g A}{\omega} \frac{\cosh K(Z+H)}{\cosh K H} e^{-iKX} }[/math]
Let the diffraction potential be:
[math]\displaystyle{ \phi_7 = \frac{i g A}{\omega} \frac{\cosh K(Z+H)}{\cos K H} \psi(X,Y) }[/math]
- For [math]\displaystyle{ \phi_7\, }[/math] to satisfy the 3D Laplace equation, it is easy to show that [math]\displaystyle{ \psi\, }[/math] must satisfy the Helmholtz equation:
[math]\displaystyle{ \left( \frac{\partial^2}{\partial X^2} + \frac{\partial^2}{\partial y^2} + K^2 \right) \psi = 0\, }[/math]
In polar coordinates:
[math]\displaystyle{
\begin{Bmatrix}
x=R\cos\theta \\
y=R\sin\theta
\end{Bmatrix} j \quad \psi(R,\theta)
}[/math]
The Helmholtz equation takes the form:
[math]\displaystyle{ \left( \frac{\partial^2}{\partial R^2} + \frac{1}{R} \frac{\partial}{\partial R} + \frac{1}{R^2} \frac{\partial^2}{\partial\theta^2} + K^2 \right) \psi = 0 \, }[/math]
On the cylinder:
[math]\displaystyle{ \frac{\partial\phi_7}{\partial n} = - \frac{\partial\phi_I}{\partial n} \, }[/math]
or
[math]\displaystyle{ \frac{\partial\psi}{\partial R} = - \frac{\partial}{\partial R} \left( e^{-iKX} \right) = -\frac{\partial}{\partial R} \left( e^{-iKE\cos R} \right) }[/math]
Here we make use of the familiar identity:
[math]\displaystyle{ e^{-iKR\cos\theta} = \sum_{m=0}^{infty} \epsilon_m J_m ( K R } \cos m \theta }[/math]
[math]\displaystyle{ \epsilon_m = \begin{Bmatrix}
1, & m = 0 \\
2(-i)^m, & m \gt 0
\end{Bmatrix} }[/math]
Try:
[math]\displaystyle{ \psi(R,\theta) = \sum_{m=0}^{infty} A_m F_m ( K R ) \cos m \theta \, }[/math]
Upon substitution in Helmholtz's equation we obtain:
[math]\displaystyle{ \left( \frac{\partial^2}{\partial R^2} + \frac{1}{R} \frac{\partial}{\partial R} - \frac{m^2}{R^2} + K^2 \right) F_m ( K R ) = 0 }[/math]
This is the Bessel equation of order m accepting as solutions linear combinations of the Bessel functions
[math]\displaystyle{ \begin{Bmatrix}
J_m ( K R } \\
Y_m ( K R )
\end{Bmatrix} }[/math]
The proper linear combination in the present problem is suggested by the radiation condition that [math]\displaystyle{ \psi\, }[/math] must satisfy:
As [math]\displaystyle{ R \to \infty\, }[/math]:
[math]\displaystyle{ \psi(R,\theta) \nsim e^{-iKR + i\omega t} \, }[/math]
Also as [math]\displaystyle{ R \to \infty\, }[/math]:
[math]\displaystyle{ J_m ( K R ) \sim \left( \frac{2}{\pi K R} \right)^{1/2} \cos \left( K R - \frac{1}{2} m \pi - \frac{\pi}{4} \right) }[/math]
[math]\displaystyle{ Y_m ( K R ) \sim \left( \frac{2}{\pi K R} \right)^{1/2} \sin \left( K R - \frac{1}{2} m \pi - \frac{\pi}{4} \right) }[/math]
Hence the Hankel function:
[math]\displaystyle{ H_m^{(2)} ( K R } = J_m ( K R ) - i Y_m ( K R ) \, }[/math]
[math]\displaystyle{ \sim \left( \frac{2}{\pi K R} \right)^{1/2} e^{-i \left( K R - \frac{1}{2} m \pi - \frac{\pi}{4} \right)} }[/math]
Satisfies the far field condition required by [math]\displaystyle{ \psi(R,\theta) \, }[/math]. So we set:
[math]\displaystyle{ \psi(r,\theta) = \sum_{m=0}^{\infty} \epsilon_m A_m H_m^{(2)} ( K R ) \cos m \theta }[/math]
With the constants [math]\displaystyle{ A_m \, }[/math] to be determined. The cylinder condition requires:
[math]\displaystyle{ \left. \frac{\partial\psi}{\partial R} \right|_{R=a} = - \frac{\\partial}{\partial R} \sum_{m=0}^{\infty} \epsilon_m J_m ( K R ) \left.\cos m \theta \right|_{r=a} }[/math]
It follows that:
[math]\displaystyle{ A_m {H_m^{(2)}}^' (K a) = - J_m^' (K a) \, }[/math]
or:
[math]\displaystyle{ A_m = - \frac{J_m^' ( K a ) }{{H_m^{(2)}}^' (K a)} \, }[/math]
where [math]\displaystyle{ (')\, }[/math] denotes derivatives with respect to the argument. The solution for the total velocity potential follows in the form
[math]\displaystyle{ (\psi+x)(r,\theta) = \sum_{m=0}^{\infty} \epsilon_m \left[ J_m (K R) - \frac{J_m^'(K a)}{{H_m^{(2)}}^'(K a)} H_m^{(2)} (K a) \right] \cos m \theta }[/math]
And the total original potential follows:
[math]\displaystyle{ \phi = \phi_I + \phi_7 = \frac{i g A}{\omega} \frac{\cosh K (Z+H)}{\cosh K H } (\psi+x) (r,\theta) }[/math]
In the limit as [math]\displaystyle{ H \to \infty \quad \frac{\cosh K (Z+H)}{K H} \longrightarrow e^{K Z} \, }[/math] and the series expansion solution survives.
Surge exciting force
The total complex potential, incident and scattered was derived above. The hydrodynamic pressure follows from Bernoulli:
[math]\displaystyle{ P = \mathfrak{Re} \left\{ \mathbf{P} e^{i\omega t} \right\} \, }[/math]
[math]\displaystyle{ \mathbf{P} = - i\omega \rho \left( \phi_I + \phi_7 \right) \, }[/math]
The Surge exciting force is given by
[math]\displaystyle{ X_1 = \iint_{S_B} P n_1 dS = \mathbf{Re \left\{ \mathbf{X}_1 e^{i\omega t} \right\} }[/math]
[math]\displaystyle{ \mathbf{X}_1 = \rho \int_{-\infty}^0 dZ \int_0^{2\pi} a d\theta \left( - i \omega \frac{i g A}{\omega} \right) e^{K Z} n_1 (\psi + x)_{R=a} }[/math]
Simple algebra in this case of water of infinite depth leads to the expression.
