# Difference between revisions of "Fundamental Solution for thin plates"

## 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 $\displaystyle{ \mathbb{R}^2 \, }$, which is taken to be of the form

$\displaystyle{ \left(\Delta^2 - k^2\right) u =0, }$

where $\displaystyle{ \Delta = \partial_r^2 + \frac{1}{r}\partial_r, }$ in polar coordinates.

## Linear Operators

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

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

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

The reciprocal of $\displaystyle{ P(\lambda) \, }$ is taken to be of the form

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

where

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

It is straightforward to compute the $\displaystyle{ c_{qn} \, }$ coefficients

\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} }

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

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

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