# Mittag-Leffler Expansion for the Floating Elastic Plate Dispersion Relation

## 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

$\hat{w}\left( \gamma\right) =\frac{1}{d\left( \gamma,\omega\right) }$

where

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

is the Dispersion Relation for a Floating Elastic Plate can be expressed by a linear sum of terms like $1/\left( \gamma-a\right)$, $a$ being a zero of $d\left( \gamma,\omega\right)$. 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 $\hat {w}\left( \gamma\right)$.

We will show that

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

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

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

where $q_n$ are the roots of the Dispersion Relation for a Floating Elastic Plate

$-q^5 \sinh(kH) - k \left(1 - m\omega^2 \right) \sinh(kH) = -\omega^2 \cosh(kH) \,$

with $n=-1,-2$ corresponding to the complex solutions with positive imaginary part, $n=0$ corresponding to the imaginary solution with negative real part and $n\gt 0$ 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 $f\left( \gamma\right)$ 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 $\gamma=c$ then any small circle with centre at $\gamma=c$ 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 $a_{i}$ we have

$\left| a_{1}\right| \leq\left| a_{2}\right| \leq\left| a_3\right| \leq...,$

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

$G_{i}\left( \frac{1}{\gamma-a_{i}}\right) ,\,\,i=1,2,3,...\,.$

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

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

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

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

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

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

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

$\left. \frac{f\left( \gamma^{\prime}\right) }{\left( \gamma^{\prime }-\gamma\right) ^{\prime}}\right| _{\gamma^{\prime}=\gamma}=\left. f\left( \gamma^{\prime}\right) \right| _{\gamma^{\prime}=\gamma}=f\left( \gamma\right) .$

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

$\frac{\omega_{n}\left( \gamma^{\prime}\right) }{\gamma^{\prime}-\gamma }.\,\,\, (3)$

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

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

Since the definition of $\omega_{n}$ and $G_{i}$ is a polynomial of $1/\left( \gamma-a_{i}\right) ,$ 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 $R$, we have

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

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

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

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

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

Also, when $\gamma=0$ we have

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

Subtracting Equation (5) from Equation (6) gives

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

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

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

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

Finally, we have formula for $f\left( \gamma\right)$,

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

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

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

For the expansion formula of $\hat{w}\left( \gamma\right)$, the polynomial term is

$G_{i}\left( \frac{1}{\gamma-q_{i}}\right) =\frac{R\left( q_{i}\right) }{\gamma-q_{i}}.$

## Expansion of the Dispersion Relation for a Floating Elastic Plate

Now we show that the function

$\hat{w}\left( \gamma\right) =\frac{1}{d\left( \gamma,\omega\right) }$

where

$d\left( \gamma,\omega\right) =\gamma^{4}+1-m\omega^{2}-\frac{\omega^{2} }{\gamma\tanh\gamma H},$
satisfies the

conditions for the Mittag-Leffler expansion.

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

Before beginning we recall that $q_n$ are the roots of the Dispersion Relation for a Floating Elastic Plate

$-q^5 \sinh(kH) - k \left(1 - m\omega^2 \right) \sinh(kH) = -\omega^2 \cosh(kH) \,$

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

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

$\frac{1}{\left( q_{n}+u\right) q_{n}}=\tan\left( q_{n}H\right) ,$

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

For any $n\gt N$ we have

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

where $C$ is a constant determined by $u$. When $y=\epsilon_{n}$ we have

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

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

$\left| \frac{e^{2xH}-1}{e^{2xH}+1}\right| \leq1$

to show this.) For large $\left| \gamma\right|$ we have

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

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

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

for any $n\geq N$. Note that the same relationship holds for $y=-\epsilon_{n}$.

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

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

for any $y$, $n\geq N$, where $E_{N}$ is defined as

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

Therefore

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

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

Hence, the expansion of $\hat{w}\left( \gamma\right)$ becomes, from $\hat{w}\left( 0\right) =0$,

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

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

$-\left( \gamma-q\right) \hat{w}\left( \gamma\right) =\left( -\gamma+q\right) \hat{w}\left( -\gamma\right) =\left( \gamma+q\right) \hat{w}\left( \gamma\right) .$

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

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

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

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

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

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

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

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