Difference between revisions of "Eigenfunction Matching for a Semi-Infinite Dock"

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Equation (10) gives the required equations to solve for the
coefficients of the water velocity potential in the dock covered region.
\left(  k_{0}I_{n}^{\prime}(k_{0}a)-k_{0}\frac{K_{n}^{\prime}(k_{0}
a)}{K_{n}(k_{0}a)}I_{n}(k_{0}a)\right)  e_{n}A_{0}\delta_{0l}
=\sum_{m=-2}^{\infty}\left(  \kappa_{m}I_{n}^{\prime}(\kappa_{m}
_{m}a)\right)  B_{ml}b_{mn}\,\,\,(10)
for each <math>n</math>.
Together with equations (6) and (7)
equation (10) gives the required equations to solve for the
coefficients of the water velocity potential in the plate covered region.
=Numerical Solution=
=Numerical Solution=

Revision as of 10:45, 14 September 2006


This is the simplest problem in eigenfunction matching. It also is an easy problem to understand the Wiener-Hopf and Residue Calculus

Governing Equations

We begin with the Frequency Domain Problem for a dock which occupies the region [math]x\gt 0[/math]. The water is assumed to have constant finite depth [math]H[/math] and the [math]z[/math]-direction points vertically upward with the water surface at [math]z=0[/math] and the sea floor at [math]z=-H[/math]. The boundary value problem can therefore be expressed as

[math] \Delta\phi=0, \,\, -H\lt z\lt 0, [/math]

[math] \phi_{z}=0, \,\, z=-H, [/math]

[math] \phi_{z}=\alpha\phi, \,\, z=0,\,x\lt 0, [/math]

[math] \phi_{z}=0, \,\, z=0,\,x\gt 0, [/math]

We must also apply the Sommerfeld Radiation Condition as [math]|x|\rightarrow\infty[/math]. The subscript [math]z[/math] denotes the derivative in [math]z[/math]-direction.

Solution Method

Separation of variables

We now separate variables and write the potential as

[math] \phi(x,z)=\zeta(z)\rho(x) [/math]

Applying Laplace's equation we obtain

[math] \zeta_{zz}+\mu^{2}\zeta=0 [/math]

so that:

[math] \zeta=\cos\mu(z+H) [/math]

where the separation constant [math]\mu^{2}[/math] must satisfy the Dispersion Relation for a Free Surface

[math] k\tan\left( kH\right) =-\alpha,\quad x\lt 0\,\,\,(1) [/math]


[math] \kappa\tan(\kappa H)=0,\quad x\gt 0 \,\,\,(2) [/math]

Note that we have set [math]\mu=k[/math] under the free surface and [math]\mu=\kappa[/math] under the plate. We denote the positive imaginary solution of (1) by [math]k_{0}[/math] and the positive real solutions by [math]k_{m}[/math], [math]m\geq1[/math]. The solutions of (2) [math]\kappa_{m}=m\pi/H[/math], [math]m\geq 0[/math]. We define

[math] \phi_{m}\left( z\right) =\frac{\cos k_{m}(z+H)}{\cos k_{m}H},\quad m\geq0 [/math]

as the vertical eigenfunction of the potential in the open water region and

[math] \psi_{m}\left( z\right) = \cos\kappa_{m}(z+H),\quad m\geq 0 [/math]

as the vertical eigenfunction of the potential in the dock covered region. For later reference, we note that:

[math] \int\nolimits_{-H}^{0}\phi_{m}(z)\phi_{n}(z) d z=A_{m}\delta_{mn} [/math]


[math] A_{m}=\frac{1}{2}\left( \frac{\cos k_{m}H\sin k_{m}H+k_{m}H}{k_{m}\cos ^{2}k_{m}H}\right) [/math]


[math] \int\nolimits_{-H}^{0}\phi_{n}(z)\psi_{m}(z) d z=B_{mn} [/math]


[math] B_{mn}=\frac{k_{n}\sin k_{n}H\cos\kappa_{m}H-\kappa_{m}\cos k_{n}H\sin \kappa_{m}H}{\left( \cos k_{n}H\cos\kappa_{m}H\right) \left( k_{n} ^{2}-\kappa_{m}^{2}\right) } [/math]


[math] \int\nolimits_{-H}^{0}\psi_{m}(z)\psi_{n}(z) d z=C_{m}\delta_{mn} [/math]


[math] C_{m}=\frac{1}{2}H,\quad,m\neq 0 \quad \mathrm{and} \quad C_0 = H [/math]

Therefore the potential can be expanded as

[math] \phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\phi_{m}(z), \;\;x\lt 0 [/math]


[math] \phi(x,z)=\sum_{m=0}^{\infty}b_{m} e^{-\kappa_{m}x}\psi_{m}(z), \;\;x\gt 0 [/math]

where [math]a_{m}[/math] and [math]b_{m}[/math] are the coefficients of the potential in the open water and the plate covered region respectively.

Incident potential

The incident potential is a wave of amplitude [math]A[/math] in displacement travelling in the positive [math]x[/math]-direction. The incident potential can therefore be written as

[math] \phi^{\mathrm{I}} =e^{-k_{0}x}\phi_{0}\left( z\right) [/math]

An infinite dimensional system of equations

The potential and its derivative must be continuous across the transition from open water to the plate covered region. Therefore, the potentials and their derivatives at [math]x=0[/math] have to be equal. We obtain

[math] \phi_{0}\left( z\right) + \sum_{m=0}^{\infty} a_{m} \phi_{m}\left( z\right) =\sum_{m=0}^{\infty}b_{m}\psi_{m}(z) [/math]


[math] -k_{0}\phi_{0}\left( z\right) +\sum _{m=0}^{\infty} a_{m}k_{m}\phi_{m}\left( z\right) =-\sum_{m=0}^{\infty}b_{m}\kappa_{m}\psi _{m}(z) [/math]

for each [math]n[/math]. We solve these equations by multiplying both equations by [math]\phi_{l}(z)[/math] and integrating from [math]-H[/math] to [math]0[/math] to obtain:

[math] A_{0}\delta_{0l}+a_{l}A_{l} =\sum_{n=0}^{\infty}b_{m}B_{ml}\,\,\,(8) [/math]


[math] -k_{0}A_{0}\delta_{0l}+a_{l}k_{l}A_l =-\sum_{m=0}^{\infty}b_{m}\kappa_{m}B_{ml} \,\,\,(9) [/math]

If we mutiple equation (8) by [math]-k_l[/math] and add this to equation (9) we obtain

[math] -2k_{0}A_{0}\delta_{0l} =-\sum_{m=0}^{\infty}b_{m}(k_l + \kappa_{m})B_{ml} \,\,\,(9) [/math]

Equation (10) gives the required equations to solve for the coefficients of the water velocity potential in the dock covered region.

Numerical Solution

To solve the system of equations (10) together with the boundary conditions (6 and 7) we set the upper limit of [math]l[/math] to be [math]M[/math]. We also set the angular expansion to be from [math]n=-N[/math] to [math]N[/math]. This gives us

[math] \phi(r,\theta,z)=\sum_{n=-N}^{N}\sum_{m=0}^{M}a_{mn}K_{n}(k_{m}r)e^{i n\theta }\phi_{m}(z), \;\;r\gt a [/math]


[math] \phi(r,\theta,z)=\sum_{n=-N}^{N}\sum_{m=-2}^{M}b_{mn}I_{n}(\kappa _{m}r)e^{i n\theta}\psi_{m}(z), \;\;r\lt a [/math]

Since [math]l[/math] is an integer with [math]0\leq l\leq M[/math] this leads to a system of [math]M+1[/math] equations. The number of unknowns is [math]M+3[/math] and the two extra equations are obtained from the boundary conditions for the free plate (6) and (7). The equations to be solved for each [math]n[/math] are

[math] \left( k_{0}I_{n}^{\prime}(k_{0}a)-k_{0}\frac{K_{n}^{\prime}(k_{0} a)}{K_{n}(k_{0}a)}I_{n}(k_{0}a)\right) e_{n}A_{0}\delta_{0l} =\sum_{m=-2}^{M}\left( \kappa_{m}I_{n}^{\prime}(\kappa_{m}a)-k_{l} \frac{K_{n}^{\prime}(k_{l}a)}{K_{n}(k_{l}a)}I_{n}(\kappa_{m}a)\right) B_{ml}b_{mn} [/math]

[math] \sum_{m=-2}^{M}(\beta\kappa_{m}^{4}+1-\alpha\gamma)^{-1}b_{mn} \left( \kappa_{m}^{2}I_{n}(\kappa_{m}a)-\frac{1-\nu}{a}\left( \kappa _{m}I_{n}^{\prime}(\kappa_{m}a)-\frac{n^{2}}{a}I_{n}(\kappa_{m}a)\right) \right) =0 [/math]


[math] \sum_{m=-2}^{M}(\beta\kappa_{m}^{4}+1-\alpha\gamma)^{-1}b_{mn} \left( \kappa_{m}^{3}I_{n}^{\prime}(\kappa_{m}a)+n^{2}\frac{1-\nu}{a^{2} }\left( \kappa_{m}I_{n}^{\prime}(\kappa_{m}a)+\frac{1}{a}I_{n}(\kappa _{m}a)\right) \right) =0 [/math]

It should be noted that the solutions for positive and negative [math]n[/math] are identical so that they do not both need to be calculated. There are some minor simplifications which are a consequence of this which are discussed in more detail in Zilman and Miloh 2000.