Mittag-Leffler Expansion for the Floating Elastic Plate Dispersion Relation

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Introduction

We derive here the important results that the Dispersion Relation for a Floating Elastic Plate can be written in the following form using the Mittag-Leffler expansion. This results is used to calculate the Free-Surface Green Function for a Floating Elastic Plate. The Mittag-Leffler expansion is a tool for expressing functions of a complex variable. We will use the Mittag-Leffler expansion to show that the function

[math] \hat{w}\left( \gamma\right) =\frac{1}{d\left( \gamma,\omega\right) } [/math]

where

[math] d\left( \gamma,\omega\right) =\gamma^{4}+1-m\omega^{2}-\frac{\omega^{2} }{\gamma\tanh\gamma H}, [/math]

is the Dispersion Relation for a Floating Elastic Plate can be expressed by a linear sum of terms like [math]1/\left( \gamma-a\right) [/math], [math]a[/math] being a zero of [math]d\left( \gamma,\omega\right) [/math]. We first remind ourselves of the Mittag-Leffler expansion that can be found in most text books on complex analysis, and then show that it can indeed be applied to [math]\hat {w}\left( \gamma\right) [/math].

We will show that

[math] \hat{w}\left( \gamma\right) =\sum_{n=-2}^{\infty}\frac{2q_nR\left( q_n\right) }{\gamma^{2}-q_n^{2}} [/math]

where [math]R\left( q_n\right) [/math] is the residue of [math]\hat{w}\left( \gamma\right) [/math] at [math]\gamma=q_n[/math] given by

[math] R\left( q_n\right) =\frac{\omega^{2}q_n}{\omega^{2}\left( 5q_n^{4}+u\right) +H\left[ \left( q_n^{5}+uq_n\right) ^{2}-\omega^{4}\right] }. [/math]

where [math]q_n[/math] are the roots of the Dispersion Relation for a Floating Elastic Plate

[math] -q^5 \sinh(kH) - k \left(1 - m\omega^2 \right) \sinh(kH) = -\omega^2 \cosh(kH) \, [/math]

with [math]n=-1,-2[/math] corresponding to the complex solutions with positive imaginary part, [math]n=0[/math] corresponding to the imaginary solution with negative real part and [math]n\gt 0[/math] corresponding to the imaginary solutions with positive imagainary part.

Mittag-Leffler expansion

Consider a function that is regular in the whole plane except at isolated points. A set of points is called isolated if there exists an open disk around each point that contains none other of the isolated points. Such a function is known as fractional function. We show that a fractional function that has an infinite number of poles can be expressed by infinite series of polynomials.

Let [math]f\left( \gamma\right) [/math] be a fractional function that has an infinite number of poles. We note that a number of poles that are situated within a bounded region is always finite since the set of poles does not have limit-points. Indeed, if there is a limit-point [math]\gamma=c[/math] then any small circle with centre at [math]\gamma=c[/math] would contain an infinite number of poles. Once we have a finite number of poles in a confined part of the plane we can number them in the order of their non-decreasing moduli, so that denoting the poles by [math]a_{i}[/math] we have

[math] \left| a_{1}\right| \leq\left| a_{2}\right| \leq\left| a_3\right| \leq..., [/math]

where [math]\left| a_{i}\right| \rightarrow\infty[/math] as [math]i\rightarrow\infty[/math]. At every pole [math]\gamma=a_{i}[/math] the function [math]f\left( \gamma\right) [/math] will have a definite infinite part, which will be a polynomial with respect to the argument [math]1/\left( \gamma-a_{i}\right) [/math] without the constant term. We denote this polynomial term by

[math] G_{i}\left( \frac{1}{\gamma-a_{i}}\right) ,\,\,i=1,2,3,...\,. [/math]

We show that the fractional function [math]f\left( \gamma\right) [/math] can be represented by a simple infinite series of [math]G_{i}[/math] by making certain additional assumptions. Suppose that a sequence of closed contours [math]C_{n}[/math] which surround the origin exists and satisfies following conditions.

  1. None of poles of [math]f\left( \gamma\right)[/math] are on the contours [math]C_{n},\,n=1,2,3,...[/math]
  2. Every contour [math]C_{n}[/math] lies inside the contour [math]C_{n+1}[/math].
  3. Let [math]l_{n}[/math] be length of the contour [math]C_{n}[/math] and [math]\delta_{n}[/math] be its shortest distance from the origin then [math]\delta_{n}\rightarrow\infty[/math] as [math]n\rightarrow\infty[/math] , i.e., the contours [math]C_{n}[/math] widen indefinitely in all directions as [math]n[/math] increases.
  4. A positive number [math]m[/math] exists such that
    [math]\frac{l_{n}}{\delta_{n}}\leq m\quad \mathrm{for}\quad n=1,2,3,....[/math]

We now suppose that given such a sequence of contours, there exists a positive number [math]M,[/math] such that on any contour [math]C_{n}[/math] our fractional function [math]f\left(\gamma\right) [/math] satisfies [math]\left| f\left( \gamma\right) \right| \leq M[/math]. Consider the integral

[math] \frac{1}{2\pi\mathrm{i}}\int_{C_{n}}\frac{f\left( \gamma^{\prime }\right) }{\gamma^{\prime}-\gamma}d\gamma^{\prime}\,\,\, (1) [/math]

where the point [math]\gamma[/math] lies inside [math]C_{n}[/math] and is other than [math]a_{i}[/math] (the poles inside [math]C_{n}.[/math]) We also consider the sum of the polynomials for the poles [math]a_{i}[/math], inside [math]C_{n}[/math],

[math] \omega_{n}\left( \gamma\right) =\sum_{\left( C_{n}\right) }G_{i}\left( \frac{1}{\gamma-a_{i}}\right) . \,\,\,(2) [/math]

The integrand of ((1)) has a pole [math]\gamma^{\prime}=\gamma[/math] and poles [math]\gamma^{\prime}=a_{i}[/math]. We can calculate the residue at the pole [math]\gamma^{\prime}=\gamma[/math] by

[math] \left. \frac{f\left( \gamma^{\prime}\right) }{\left( \gamma^{\prime }-\gamma\right) ^{\prime}}\right| _{\gamma^{\prime}=\gamma}=\left. f\left( \gamma^{\prime}\right) \right| _{\gamma^{\prime}=\gamma}=f\left( \gamma\right) . [/math]

The residues at the poles [math]\gamma^{\prime}=a_{i}[/math] are, by the definition (2), the same as the residues of the function

[math] \frac{\omega_{n}\left( \gamma^{\prime}\right) }{\gamma^{\prime}-\gamma }.\,\,\, (3) [/math]

We note that all poles of this function are situated inside [math]C_{n}[/math]. We now show that the sum of residues of function (3) at the poles [math]a_{i}[/math] is

[math] -\omega_{n}\left( \gamma\right) =-\sum_{\left( C_{n}\right) }G_{i}\left( \frac{1}{\gamma-a_{i}}\right) .\,\,\, (4) [/math]

Since the definition of [math]\omega_{n}[/math] and [math]G_{i}[/math] is a polynomial of [math]1/\left( \gamma-a_{i}\right) ,[/math] the order of the denominator of function (3) is at least two units higher than that of the numerator of function (3). Hence, for a circle with a sufficiently large radius [math]R[/math], we have

[math] 2\pi\mathrm{i}\sum_{\left( C_{n}\right) }Res _{\gamma^{\prime}=a_{i}}\frac{\omega_{n}\left( \gamma^{\prime}\right) }{\gamma^{\prime}-\gamma}=\oint_{C_{R}}\frac{\omega_{n}\left( \gamma^{\prime }\right) }{\gamma^{\prime}-\gamma}d\gamma^{\prime}. [/math]

The LHS of this does not change as the radius [math]R[/math] increases, and the RHS[math]\rightarrow0[/math] as [math]R\rightarrow\infty[/math]. Indeed,

[math] \left| \oint_{C_{R}}\frac{\omega_{n}\left( \gamma^{\prime}\right) } {\gamma^{\prime}-\gamma}d\gamma^{\prime}\right| \leq\oint_{C_{R}}\left| \gamma^{\prime}\frac{\omega_{n}\left( \gamma^{\prime}\right) } {\gamma^{\prime}-\gamma}\frac{1}{\gamma^{\prime}}d\gamma^{\prime}\right| \leq\max_{\left| \gamma^{\prime}\right| =R}\left| \gamma^{\prime} \frac{\omega_{n}\left( \gamma^{\prime}\right) }{\gamma^{\prime}-\gamma }\right| \frac{2\pi R}{R} [/math]

and the term [math]\left| \cdot\right| [/math] tends to zero as [math]R\rightarrow\infty[/math]. Thus, the sum of residues at poles within a finite distance is zero. Since we know that the residue of (3) at [math]\gamma^{\prime}=\gamma[/math] is [math]\omega_{n}\left( \gamma\right) [/math], the sum of the rest is formula (4). Thus, we have an expression for the integral (1),

[math] \frac{1}{2\pi\mathrm{i}}\int_{C_{n}}\frac{f\left( \gamma^{\prime }\right) }{\gamma^{\prime}-\gamma}d\gamma^{\prime}=f\left( \gamma\right) -\sum_{\left( C_{n}\right) }G_{i}\left( \frac{1}{\gamma-a_{i}}\right) . \,\,\,(5) [/math]

Also, when [math]\gamma=0[/math] we have

[math] \frac{1}{2\pi\mathrm{i}}\int_{C_{n}}\frac{f\left( \gamma^{\prime }\right) }{\gamma^{\prime}}d\gamma^{\prime}=f\left( 0\right) -\sum_{\left( C_{n}\right) }G_{i}\left( -\frac{1}{a_{i}}\right) . (6) [/math]

Subtracting Equation (5) from Equation (6) gives

[math] \frac{\gamma}{2\pi\mathrm{i}}\int_{C_{n}}\frac{f\left( \gamma^{\prime }\right) }{\gamma^{\prime}\left( \gamma^{\prime}-\gamma\right) } d\gamma^{\prime}=f\left( \gamma\right) -f\left( 0\right) -\sum_{\left( C_{n}\right) }\left[ G_{i}\left( \frac{1}{\gamma-a_{i}}\right) -G_{i}\left( -\frac{1}{a_{i}}\right) \right] . [/math]

We now prove that the integrand on the LHS of this expression tends to zero as [math]n\rightarrow\infty[/math]. Since, [math]\left| \gamma^{\prime}\right| \geq\delta _{n},\,\,\left| \gamma^{\prime}-\gamma\right| \geq\left| \gamma^{\prime }\right| -\left| \gamma\right| \geq\delta_{n}-\left| \gamma\right| ,[/math] we have

[math]\begin{matrix} \left| \int_{C_{n}}\frac{f\left( \gamma^{\prime}\right) }{\gamma^{\prime }\left( \gamma^{\prime}-\gamma\right) }d\gamma^{\prime}\right| & \leq \frac{Ml_{n}}{\delta_{n}\left( \delta_{n}-\left| \gamma\right| \right) }\\ & \lt \frac{Mm}{\delta_{n}-\left| \gamma\right| }.\,\,\, (7) \end{matrix}[/math]

Since [math]\delta_{n}\rightarrow\infty[/math] as [math]n\rightarrow\infty[/math] and {\bf condition 4}, the integral in inequality (7) tends to zero as [math]n[/math] increases.

Finally, we have formula for [math]f\left( \gamma\right) [/math],

[math] f\left( \gamma\right) =f\left( 0\right) +\lim_{n\rightarrow\infty} \sum_{\left( C_{n}\right) }\left[ G_{i}\left( \frac{1}{\gamma-a_{i} }\right) -G_{i}\left( -\frac{1}{a_{i}}\right) \right] . [/math]

Since, the contour [math]C_{n}[/math] will widen indefinitely as [math]n[/math] increases, the second term is a sum over all poles, so we have [math]f\left( \gamma\right) [/math] in the form of an infinite series

[math] f\left( \gamma\right) =f\left( 0\right) +\sum_{i=1}^{\infty}\left[ G_{i}\left( \frac{1}{\gamma-a_{i}}\right) -G_{i}\left( -\frac{1}{a_{i} }\right) \right] . [/math]

For the expansion formula of [math]\hat{w}\left( \gamma\right) [/math], the polynomial term is

[math] G_{i}\left( \frac{1}{\gamma-q_{i}}\right) =\frac{R\left( q_{i}\right) }{\gamma-q_{i}}. [/math]

Expansion of the Dispersion Relation for a Floating Elastic Plate

Now we show that the function

[math] \hat{w}\left( \gamma\right) =\frac{1}{d\left( \gamma,\omega\right) } [/math]

where

[math] d\left( \gamma,\omega\right) =\gamma^{4}+1-m\omega^{2}-\frac{\omega^{2} }{\gamma\tanh\gamma H}, [/math]

satisfies the

conditions for the Mittag-Leffler expansion.

Define a sequence of square contours [math]C_{n}[/math], square with its four corners at [math]\epsilon_{n}-\mathrm{i}\epsilon_{n}[/math], [math]\epsilon_{n}+\mathrm{i} \epsilon_{n}[/math], [math]-\epsilon_{n}+\mathrm{i}\epsilon_{n}[/math] and [math]-\epsilon_{n}-\mathrm{i}\epsilon_{n}[/math], where [math]\epsilon_{n}=\left( n+\frac{1}{2}\right) \pi/H,\,n=N,N+1,...[/math]. We start by showing that [math]\left| \hat{w}\left( \gamma\right) \right| [/math] is bounded on any [math]C_{n}[/math] in order to follow the proof of Mittag-Leffler expansion given in the previous subsection.

Before beginning we recall that [math]q_n[/math] are the roots of the Dispersion Relation for a Floating Elastic Plate

[math] -q^5 \sinh(kH) - k \left(1 - m\omega^2 \right) \sinh(kH) = -\omega^2 \cosh(kH) \, [/math]

with [math]n=-1,-2[/math] corresponding to the complex solutions with positive imaginary part, [math]n=0[/math] corresponding to the imaginary solution with negative real part and [math]n\gt 0[/math] corresponding to the imaginary solutions with positive imagainary part.

For the sake of simplicity, write [math]u=1-m\omega^{2}[/math]. When [math]Im \gamma[/math] is large the poles of [math]\hat{w}[/math] are almost [math]\pm\mathrm{i} n\pi/H.[/math] In fact, the poles [math]\left\{ \mathrm{i}q_{n}\right\} _{n=1,2,...}[/math], [math]q_{n}\in\mathbb{R}[/math] of [math]\hat{w}[/math] satisfy

[math] \frac{1}{\left( q_{n}+u\right) q_{n}}=\tan\left( q_{n}H\right) , [/math]

so [math]\gamma_{n}\rightarrow\pm n\pi/H[/math] as [math]n[/math] increases. Thus, by choosing a large [math]N[/math], the contour [math]C_{n}[/math] is always a certain distance away from the poles for any [math]n\geq N[/math]. We prove the boundedness of [math]\left| \hat{w}\right| [/math] by showing that [math]\left| \hat{w}\left( x+\mathrm{i}y\right) \right| [/math] is bounded for [math]y=\pm\epsilon_{n}[/math], [math]n=N,N+1,...[/math], and [math]x,y\in\mathbb{R}[/math], and then for [math]x=\pm\epsilon_{n},\,n=N,N+1,...,[/math] [math]y\in\left[ -\epsilon_{n},\epsilon_{n}\right] [/math].

For any [math]n\gt N[/math] we have

[math] \left| \gamma^{4}+u\right| \gt \left| \gamma\right| ^{4}+C=\left| x+\mathrm{i}y\right| ^{4}+C [/math]
[math] \geq\epsilon_{n}^{4}+C\,\,\mathrm{for}\, \mathrm{any}\,\, x\in\mathbb{R},\,y=\epsilon _{n}, [/math]

where [math]C[/math] is a constant determined by [math]u[/math]. When [math]y=\epsilon_{n}[/math] we have

[math] \left| \frac{1}{\gamma\tanh\left( \gamma H\right) }\right| =\left| \frac{e^{2xH}e^{\mathrm{i}2yH}+1}{\left( x+\mathrm{i}y\right) \left( e^{2xH}e^{\mathrm{i}2yH}-1\right) }\right| [/math]
[math] =\frac{\left| e^{2xH}e^{\mathrm{i}2yH}+1\right| }{\left| x+\mathrm{i}y\right| \left| e^{2xH}e^{\mathrm{i}2yH}-1\right| } [/math]
[math] =\frac{\left| e^{2xH}-1\right| }{\left| x+\mathrm{i}y\right| \left| e^{2xH}+1\right| }\leq\frac{1}{\left| x+\mathrm{i}y\right| }\leq\frac{1}{\epsilon_{n}} [/math]

for any [math]x\in\mathbb{R}[/math]. (We used [math]\exp\left( \mathrm{i}\left( 2n+1\right) \pi\right) =-1[/math] and

[math] \left| \frac{e^{2xH}-1}{e^{2xH}+1}\right| \leq1 [/math]

to show this.) For large [math]\left| \gamma\right| [/math] we have

[math] \left| \gamma^{4}+u-\frac{\omega^{2}}{\gamma\tanh\left( \gamma H\right) }\right| \geq\left| \gamma^{4}+u\right| -\left| \frac{\omega^{2}} {\gamma\tanh\left( \gamma H\right) }\right| . [/math]

Since the RHS of this inequality is positive from the previous equations,

[math] \left| \hat{w}\left( \gamma\right) \right| \leq\frac{1}{\left| \gamma ^{4}+u\right| -\left| \frac{\omega^{2}}{\gamma\tanh\left( \gamma H\right) }\right| }\leq\frac{1}{\epsilon_{n}^{4}+C-\frac{\omega^{2}}{\epsilon_{n}} } [/math]

for any [math]n\geq N[/math]. Note that the same relationship holds for [math]y=-\epsilon_{n}[/math].

For [math]\gamma[/math] on the line segment [math]\epsilon_{n}-\mathrm{i}\epsilon_{n}[/math] to [math]\epsilon_{n}+\mathrm{i}\epsilon_{n}[/math] we use the fact that

[math] \frac{\left| e^{2xH}e^{\mathrm{i}2yH}+1\right| }{\left| x+\mathrm{i}y\right| \left| e^{2xH}e^{\mathrm{i}2yH}-1\right| }\leq\frac{1}{\left| x+\mathrm{i}y\right| }\frac{1+\left| e^{-2xH}\right| }{1-\left| e^{-2xH}\right| }\leq\frac{E_{N}}{\epsilon_{N}} [/math]

for any [math]y[/math], [math]n\geq N[/math], where [math]E_{N}[/math] is defined as

[math] \frac{1+\left| e^{-2xH}\right| }{1-\left| e^{-2xH}\right| }\leq \frac{1+\left| e^{-2\epsilon_{N}H}\right| }{1-\left| e^{-2\epsilon_{N} H}\right| }=E_{N}\,\,=, = \frac{1}{\left| x+\mathrm{i}y\right| }\leq\frac{1}{\epsilon_{N}}. [/math]

Therefore

[math] \frac{1}{\left| \gamma^{4}+u\right| -\left| \frac{\omega^{2}}{\gamma \tanh\left( \gamma H\right) }\right| }\leq\frac{1}{\epsilon_{N}^{4} +C-\frac{\omega^{2}E_{N}}{\epsilon_{N}}} [/math]

for any [math]n\geq N[/math]. The same proof can be applied for the line segment [math]-\epsilon_{n}-\mathrm{i}\epsilon_{n}[/math] to [math]-\epsilon_{n} +\mathrm{i}\epsilon_{n}[/math]. We have proved that [math]\left| \hat{w}\left( \gamma\right) \right| [/math] is bounded on all sides of the contours [math]C_{n},\,n\geq N[/math] where [math]N[/math] is chosen to be large so that the contours are a certain distance away from all the poles of [math]\hat{w}[/math].

Hence, the expansion of [math]\hat{w}\left( \gamma\right) [/math] becomes, from [math]\hat{w}\left( 0\right) =0[/math],

[math] \hat{w}\left( \gamma\right) =\sum_{n=-2}^{\infty}\left[ \frac{R\left( q_n\right) }{\gamma-q_n}+\frac{R\left( q_n\right) }{q_n}\right] =\sum_{n=-2}^{\infty} \left[ \frac{2q_nR\left( q_n\right) }{\gamma^{2}-q_n^{2}}+\frac{2R\left( q_n\right) }{q_n}\right] . [/math]

Note that the summation on the first line is over all poles of [math]\hat{w}\left( \gamma\right) [/math]. Note that [math]R\left( q\right) =-R\left( -q\right) [/math], since [math]\hat{w}\left( \gamma\right) [/math] is an even function and

[math] -\left( \gamma-q\right) \hat{w}\left( \gamma\right) =\left( -\gamma+q\right) \hat{w}\left( -\gamma\right) =\left( \gamma+q\right) \hat{w}\left( \gamma\right) . [/math]

Note that the term [math]\sum2R\left( q\right) /q[/math] is zero. Indeed, expansion of the function [math]\hat{w}\left( \gamma\right) \gamma[/math] which has the same analytic properties and poles as the function [math]\hat{w}[/math] and residues [math]R\left( q\right) q[/math] at [math]\gamma=q[/math]. Hence, [math]\hat{w}\left( \gamma\right) \gamma[/math] is expanded as,

[math] \hat{w}\left( \gamma\right) \gamma=\sum_{n=-2}^{\infty}\left[ \frac{q_nR\left( q_n\right) }{\gamma-q_n}+\frac{q_nR\left( q_n\right) }{q_n}\right] = \sum_{n=-2}^{\infty} \frac{2\gamma q_nR\left( q_n\right) }{\gamma^{2}-q_n^{2}}. [/math]

The fact that [math]\sum2R\left( q\right) /q[/math] is zero can also be confirmed by using the contour integration of the function [math]\hat{w}\left( \gamma\right) /\gamma[/math]. The function [math]\hat{w}\left( \gamma\right) /\gamma[/math] is an odd function and has the same poles as the function [math]\hat{w}\left( \gamma\right) [/math] with the residues [math]R\left( q\right) /q[/math]. Notice that [math]\gamma=0[/math] is not a singular point of [math]\hat{w}\left( \gamma\right) /\gamma[/math]. Hence, the integration over the real axis is zero and [math]\hat{w}\left( \gamma\right) /\gamma\rightarrow0[/math] on the semi-arc with order of [math]A^{-3}[/math] as [math]A\rightarrow\infty[/math].

The residues [math]R\left( q\right) [/math] can be calculated using the usual formula. Since each of the poles of [math]\hat{w}\left( \gamma\right) [/math] is simple, the residue [math]R\left( q\right) [/math] at a pole [math]q[/math] can be found using the expression

[math]\begin{matrix} R\left( q\right) & =\left[ \left. \frac{d}{d\gamma}d\left( \gamma ,\omega\right) \right| _{\gamma=q}\right] ^{-1}\\ & =\left[ 4q^{3}+\omega^{2}\left( \frac{qH+\tanh qH-qH\tanh^{2}qH} {q^{2}\tanh^{2}qH}\right) \right] ^{-1}. \end{matrix}[/math]

As each pole [math]q[/math] is a root of the dispersion equation, we may substitute [math]\tanh qH=\omega^{2}/\left( q^{5}+uq\right) [/math], where for brevity we have defined [math]u=\left( 1-m\omega^{2}\right) [/math]. The residue may then be given as the rational function of the pole

[math] R\left( q\right) =\frac{\omega^{2}q}{\omega^{2}\left( 5q^{4}+u\right) +H\left[ \left( q^{5}+uq\right) ^{2}-\omega^{4}\right] }. [/math]

This form avoids calculation of the hyperbolic tangent which becomes small at the imaginary roots. This causes numerical round-off problems since [math]q_{n}H[/math] tends to [math]n\pi[/math] as [math]n[/math] becomes large, which makes [math]\tan q_{n}H[/math] become small.