Mittag-Leffler Expansion for the Floating Elastic Plate Dispersion Relation

From WikiWaves
Jump to navigationJump to search


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


[math]\displaystyle{ 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]\displaystyle{ 1/\left( \gamma-a\right) }[/math], [math]\displaystyle{ a }[/math] being a zero of [math]\displaystyle{ 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]\displaystyle{ \hat {w}\left( \gamma\right) }[/math].

We will show that

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

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

[math]\displaystyle{ 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]\displaystyle{ q_n }[/math] are the roots of the Dispersion Relation for a Floating Elastic Plate

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

with [math]\displaystyle{ n=-1,-2 }[/math] corresponding to the complex solutions with positive imaginary part, [math]\displaystyle{ n=0 }[/math] corresponding to the imaginary solution with negative real part and [math]\displaystyle{ 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]\displaystyle{ 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]\displaystyle{ \gamma=c }[/math] then any small circle with centre at [math]\displaystyle{ \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]\displaystyle{ a_{i} }[/math] we have

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

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

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

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

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

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

[math]\displaystyle{ \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]\displaystyle{ \gamma^{\prime}=\gamma }[/math] and poles [math]\displaystyle{ \gamma^{\prime}=a_{i} }[/math]. We can calculate the residue at the pole [math]\displaystyle{ \gamma^{\prime}=\gamma }[/math] by

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

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

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

[math]\displaystyle{ -\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]\displaystyle{ \omega_{n} }[/math] and [math]\displaystyle{ G_{i} }[/math] is a polynomial of [math]\displaystyle{ 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]\displaystyle{ R }[/math], we have

[math]\displaystyle{ 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]\displaystyle{ R }[/math] increases, and the RHS[math]\displaystyle{ \rightarrow0 }[/math] as [math]\displaystyle{ R\rightarrow\infty }[/math]. Indeed,

[math]\displaystyle{ \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]\displaystyle{ \left| \cdot\right| }[/math] tends to zero as [math]\displaystyle{ 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]\displaystyle{ \gamma^{\prime}=\gamma }[/math] is [math]\displaystyle{ \omega_{n}\left( \gamma\right) }[/math], the sum of the rest is formula (4). Thus, we have an expression for the integral (1),

[math]\displaystyle{ \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]\displaystyle{ \gamma=0 }[/math] we have

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

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

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

[math]\displaystyle{ 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]\displaystyle{ \hat{w}\left( \gamma\right) }[/math], the polynomial term is

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


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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

[math]\displaystyle{ \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]


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

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

[math]\displaystyle{ \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]\displaystyle{ \hat{w}\left( \gamma\right) }[/math]. Note that [math]\displaystyle{ R\left( q\right) =-R\left( -q\right) }[/math], since [math]\displaystyle{ \hat{w}\left( \gamma\right) }[/math] is an even function and

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

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

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

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

[math]\displaystyle{ 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]\displaystyle{ q_{n}H }[/math] tends to [math]\displaystyle{ n\pi }[/math] as [math]\displaystyle{ n }[/math] becomes large, which makes [math]\displaystyle{ \tan q_{n}H }[/math] become small.