Fundamental Solution for thin plates

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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{ u \, }[/math] is the displacement, [math]\displaystyle{ k \, }[/math] is the wavenumber, and [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].

Determining the Green's function

In order to compute the Green's function for polynomials in the Laplacian, we express our unknown Green's function in terms of the fundamental solution of Helmholtz's equation:

[math]\displaystyle{ G(r) = \sum_{q=1}^{p} \sum_{n=1}^{m_q} c_{qn} G_{n,\lambda_q}(r) = c_{11} G_{1,\lambda_1}+ c_{21} G_{1,\lambda_2}. }[/math]

Using the Green's function for the Helmholtz Equation

[math]\displaystyle{ G_{1,\beta} = \frac{i}{4} H_0^{(1)} (\beta r), }[/math]

(where [math]\displaystyle{ H_0^{(1)} \, }[/math] denotes Hankel functions of the first kind) which incorporates the Sommerfeld radiation condition

[math]\displaystyle{ \lim_{r \rightarrow \infty} r^{1/2} \left(\frac{\partial}{\partial r} - ik \right)G = 0, }[/math]

we obtain the final expression

[math]\displaystyle{ \begin{align} G(r) &= \left(-\frac{1}{2 k^2}\right)\left(\frac{i}{4}H_0^{(1)}(ikr)\right) + \left(\frac{1}{2 k^2}\right)\left(\frac{i}{4}H_0^{(1)}(kr)\right) \\ &= \frac{i}{8 k^2} \left[H_0^{(1)}(kr) - H_0^{(1)}(ikr) \right]. \end{align} }[/math]

This is the fundamental solution for our governing equation above.

Behaviour at [math]\displaystyle{ r=0 }[/math]

To determine the solution at [math]\displaystyle{ r \rightarrow 0 }[/math] (where the above expression is singular), the following identities from Abramowitz and Stegun 1964 are used

[math]\displaystyle{ \begin{align} H_0^{(1)}(z) &= J_0(z) + i Y_0(z), \\ K_0(z) &= \frac{\pi i}{2} H_0^{(1)}(iz), \quad -\pi \lt arg(z) \leq \pi/2, \\ K_0(z) &= \int_{0}^{\infty} \cos \left[z\sinh \theta \right]d \theta, \\ Y_0(z) &= \frac{1}{\pi}\int_0^\pi \sin \left[z \sin \theta \right] d \theta - \frac{1}{\pi} \int_0^\infty 2 e^{-z \sinh \theta} d \theta, \ | arg(z) | \lt \frac{\pi}{2} \end{align} }[/math]

from pages 358, 375, 376, and 360 respectively.

Let us consider

[math]\displaystyle{ \begin{align} &H_0^{(1)}(kr) - H_0^{(1)}(ikr) \\ =& J_0(kr) + i Y_0(kr) + \frac{2i}{\pi} K_0(kr) \\ =& J_0(kr) + \frac{i}{\pi}\int_0^\pi \sin \left(kr \sin \theta \right) d \theta - \frac{i}{\pi} \int_0^\infty 2 e^{-kr \sinh \theta} d \theta + \frac{2i}{\pi} \int_{0}^{\infty} \cos \left( kr \sinh \theta \right)d \theta \end{align} }[/math]

where introducing the limit yields

[math]\displaystyle{ \begin{align} \lim_{r \rightarrow 0}& J_0(kr) = 1, \\ \lim_{r \rightarrow 0}& \sin \left(kr \sin \theta \right) = 0, \\ \lim_{r \rightarrow 0}& e^{-kr \sinh \theta} =1, \\ \lim_{r \rightarrow 0}& \cos \left( kr \sinh \theta \right) = 1. \end{align} }[/math]

Consequently

[math]\displaystyle{ \lim_{r \rightarrow 0} \left( H_0^{(1)}(kr) - H_0^{(1)}(ikr) \right) = \lim_{r \rightarrow 0} J_0(kr) -\frac{2i}{\pi}\int_{0}^{\infty} 1 d\theta + \frac{2i}{\pi}\int_{0}^{\infty} 1 d\theta = 1, }[/math]

which then allows us to determine that

[math]\displaystyle{ \lim_{r \rightarrow 0} G(r) = \frac{i}{8 k^2}. }[/math]

Solution

The fundamental solution for a thin uniform plate in [math]\displaystyle{ \mathbb{R}^2 \, }[/math] governed by

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

is defined as follows

[math]\displaystyle{ G(r) = \left\{ \begin{align} &\frac{i}{8 k^2} \left[H_0^{(1)}(kr) - H_0^{(1)}(ikr) \right] &\mbox{for} \ r\gt 0, \\ &\frac{i}{8 k^2} &\mbox{for } \ r=0. \end{align} \right. }[/math]