Wave Scattering By A Vertical Circular Cylinder

• 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 = \mathbf{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) \sim 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.