Difference between revisions of "Fundamental Solution for thin plates"

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(Created page with 'On this page, we aim to derive the Green's function for a thin uniform plate.')
 
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On this page, we aim to derive the Green's function for a thin uniform plate.
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==Introduction==
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On this page, we aim to derive the Green's function for a thin uniform plate.  This derivation relies heavily on concepts discussed in [[Boyling 1996]].  We seek the fundamental solution for the Biharmonic equation in <math>\mathbb{R}^2 \,</math>, which is taken to be of the form
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<center>
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<math>
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\left(\Delta^2 - k^2\right) u =0,
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</math>
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</center>
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where <math>\Delta = \partial_r^2 + \frac{1}{r}\partial_r, </math> in polar coordinates.
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==Linear Operators==
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We consider the operator <math>P(\Delta) = \left(\Delta^2 - k^4\right) = \left(\Delta - (ik)^2\right) \left(\Delta -k^2\right)</math>, whereas Boyling considers more general linear operators of the form
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<center>
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<math>
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P(\lambda) = c(\lambda - \lambda_1)^{m_1}(\lambda - \lambda_2)^{m_2} \ldots (\lambda - \lambda_p)^{m_p},
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</math>
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</center>
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which for our case, is identical when <math>c=1 \,</math>, <math>\lambda_1 = (ik)^2 \,</math>, <math>\lambda_2  = k^2 \,</math>, <math>m_1 = m_2 = 1 \,</math>, and <math>m_3=m_4=\ldots =0 \,</math>.
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The reciprocal of <math>P(\lambda) \,</math> is taken to be of the form
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<center>
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<math>
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\frac{1}{P(\lambda)} = \sum_{q=1}^{p} \sum_{n=1}^{m_q} \frac{c_{qn}}{(\lambda - \lambda_q)^n}.
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</math>
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</center>
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where
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<center>
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<math>
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c_{qn} = \lim_{\lambda \rightarrow \lambda_q} \frac{1}{(m_q - n)!} \frac{\partial^{m_q - n}}{\partial \lambda^{m_q-n}} \left[\frac{(\lambda - \lambda_q)^{m_q}}{P(\lambda)}\right].
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</math>
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</center>
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It is straightforward to compute the <math>c_{qn} \,</math> coefficients
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<center><math>
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\begin{align}
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c_{11} &= \lim_{\lambda \rightarrow \lambda_q} \frac{1}{(0)!} \frac{\partial^{0}}{\partial \lambda^{0}} \left[\frac{(\lambda - \lambda_1)}{P(\lambda)}\right] =  \lim_{\lambda \rightarrow (ik)^2} \frac{1}{\lambda - \lambda_2} = - \frac{1}{2 k^2}, \\
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c_{21} &= \frac{1}{2 k^2},
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\end{align}
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</math></center>
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as well as determining a corresponding quantity <math>P_{qn} \,</math>
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such that <math>\sum_{q=1}^{p} \sum_{n=1}^{m_q} c_{qn}P_{qn} = 1</math>, where
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<center>
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<math>
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P_{qn}(\lambda) \left[\lambda - \lambda_q\right]^n = P(\lambda), \quad \mbox{for all q,n}.
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</math>
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</center>
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That is, <math>P_{11} = (\lambda - k^2) \,</math>  and <math>P_{21} = (\lambda - (ik)^2) \,</math>.

Revision as of 20:37, 27 April 2010

Introduction

On this page, we aim to derive the Green's function for a thin uniform plate. This derivation relies heavily on concepts discussed in Boyling 1996. We seek the fundamental solution for the Biharmonic equation in [math]\displaystyle{ \mathbb{R}^2 \, }[/math], which is taken to be of the form

[math]\displaystyle{ \left(\Delta^2 - k^2\right) u =0, }[/math]

where [math]\displaystyle{ \Delta = \partial_r^2 + \frac{1}{r}\partial_r, }[/math] in polar coordinates.

Linear Operators

We consider the operator [math]\displaystyle{ P(\Delta) = \left(\Delta^2 - k^4\right) = \left(\Delta - (ik)^2\right) \left(\Delta -k^2\right) }[/math], whereas Boyling considers more general linear operators of the form

[math]\displaystyle{ P(\lambda) = c(\lambda - \lambda_1)^{m_1}(\lambda - \lambda_2)^{m_2} \ldots (\lambda - \lambda_p)^{m_p}, }[/math]

which for our case, is identical when [math]\displaystyle{ c=1 \, }[/math], [math]\displaystyle{ \lambda_1 = (ik)^2 \, }[/math], [math]\displaystyle{ \lambda_2 = k^2 \, }[/math], [math]\displaystyle{ m_1 = m_2 = 1 \, }[/math], and [math]\displaystyle{ m_3=m_4=\ldots =0 \, }[/math].

The reciprocal of [math]\displaystyle{ P(\lambda) \, }[/math] is taken to be of the form

[math]\displaystyle{ \frac{1}{P(\lambda)} = \sum_{q=1}^{p} \sum_{n=1}^{m_q} \frac{c_{qn}}{(\lambda - \lambda_q)^n}. }[/math]

where

[math]\displaystyle{ c_{qn} = \lim_{\lambda \rightarrow \lambda_q} \frac{1}{(m_q - n)!} \frac{\partial^{m_q - n}}{\partial \lambda^{m_q-n}} \left[\frac{(\lambda - \lambda_q)^{m_q}}{P(\lambda)}\right]. }[/math]

It is straightforward to compute the [math]\displaystyle{ c_{qn} \, }[/math] coefficients

[math]\displaystyle{ \begin{align} c_{11} &= \lim_{\lambda \rightarrow \lambda_q} \frac{1}{(0)!} \frac{\partial^{0}}{\partial \lambda^{0}} \left[\frac{(\lambda - \lambda_1)}{P(\lambda)}\right] = \lim_{\lambda \rightarrow (ik)^2} \frac{1}{\lambda - \lambda_2} = - \frac{1}{2 k^2}, \\ c_{21} &= \frac{1}{2 k^2}, \end{align} }[/math]

as well as determining a corresponding quantity [math]\displaystyle{ P_{qn} \, }[/math] such that [math]\displaystyle{ \sum_{q=1}^{p} \sum_{n=1}^{m_q} c_{qn}P_{qn} = 1 }[/math], where

[math]\displaystyle{ P_{qn}(\lambda) \left[\lambda - \lambda_q\right]^n = P(\lambda), \quad \mbox{for all q,n}. }[/math]

That is, [math]\displaystyle{ P_{11} = (\lambda - k^2) \, }[/math] and [math]\displaystyle{ P_{21} = (\lambda - (ik)^2) \, }[/math].