Fundamental Solution for thin plates - Revision history

Mike smith at 21:14, 27 April 2010

2010-04-27T21:14:41Z

← Older revision		Revision as of 21:14, 27 April 2010
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			{{complete pages}}

	==Introduction==		==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>\mathbb{R}^2 \,</math>, which is taken to be of the form		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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	</math>		</math>
	</center>		</center>
	where <math>H_0^{(1)} \,</math> denotes Hankel functions of the first kind, which incorporates the Sommerfeld radiation condition		(where <math>H_0^{(1)} \,</math> denotes Hankel functions of the first kind) which incorporates the Sommerfeld radiation condition
	<center>		<center>
	<math>		<math>

Mike smith: /* Introduction */

2010-04-27T21:12:40Z

Introduction

← Older revision		Revision as of 21:12, 27 April 2010
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	</math>		</math>
	</center>		</center>
	where <math>u \,</math> is the displacement, and <math>\Delta = \partial_r^2 + \frac{1}{r}\partial_r, </math> in polar coordinates.		where <math>u \,</math> is the displacement, <math>k \,</math> is the wavenumber, and <math>\Delta = \partial_r^2 + \frac{1}{r}\partial_r, </math> in polar coordinates.

	==Linear Operators==		==Linear Operators==

Mike smith at 21:10, 27 April 2010

2010-04-27T21:10:14Z

← Older revision		Revision as of 21:10, 27 April 2010
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	</math>		</math>
	</center>		</center>
	where <math>\Delta = \partial_r^2 + \frac{1}{r}\partial_r, </math> in polar coordinates.		where <math>u \,</math> is the displacement, and <math>\Delta = \partial_r^2 + \frac{1}{r}\partial_r, </math> in polar coordinates.

	==Linear Operators==		==Linear Operators==
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	This is the fundamental solution for our governing equation above.		This is the fundamental solution for our governing equation above.

	=Behaviour at <math>r=0</math>=		==Behaviour at <math>r=0</math>==
	To determine the solution at <math>r \rightarrow 0</math> (where the above expression is singular), the following identities from [[Abramowitz and Stegun 1964]] are used		To determine the solution at <math>r \rightarrow 0</math> (where the above expression is singular), the following identities from [[Abramowitz and Stegun 1964]] are used
	<center><math>		<center><math>
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	<center><math>		<center><math>
	\lim_{r \rightarrow 0} G(r) = \frac{i}{8 k^2}.		\lim_{r \rightarrow 0} G(r) = \frac{i}{8 k^2}.
			</math></center>

			==Solution==
			The fundamental solution for a thin uniform plate in <math>\mathbb{R}^2 \,</math> governed by
			<center>
			<math>
			\left(\Delta^2 - k^2\right) u =0,
			</math>
			</center>
			is defined as follows
			<center><math>
			G(r) = \left\{
			\begin{align}
			&\frac{i}{8 k^2} \left[H_0^{(1)}(kr) - H_0^{(1)}(ikr) \right] &\mbox{for} \ r>0, \\
			&\frac{i}{8 k^2} &\mbox{for } \ r=0.
			\end{align}
			\right.
	</math></center>		</math></center>

Mike smith: /* Behaviour at r=0 */

2010-04-27T20:58:21Z

Behaviour at r=0

← Older revision		Revision as of 20:58, 27 April 2010
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	<center><math>		<center><math>
	\begin{align}		\begin{align}
	&H_0^{(1)}(kr) - H_0^{(1)}(ikr), \\		&H_0^{(1)}(kr) - H_0^{(1)}(ikr) \\
	=& J_0(kr) + i Y_0(kr) + \frac{2i}{\pi} K_0(kr), \\		=& 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,		=& 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}		\end{align}
	</math></center>		</math></center>
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	<center><math>		<center><math>
	\begin{align}		\begin{align}
	\lim_{r \rightarrow 0}& J_0(kr) = 1 \\		\lim_{r \rightarrow 0}& J_0(kr) = 1, \\
	\lim_{r \rightarrow 0}& \sin \left(kr \sin \theta \right) = 0 \\		\lim_{r \rightarrow 0}& \sin \left(kr \sin \theta \right) = 0, \\
	\lim_{r \rightarrow 0}& e^{-kr \sinh \theta} =1 \\		\lim_{r \rightarrow 0}& e^{-kr \sinh \theta} =1, \\
	\lim_{r \rightarrow 0}& \cos \left( kr \sinh \theta \right) = 1.		\lim_{r \rightarrow 0}& \cos \left( kr \sinh \theta \right) = 1.
	\end{align}		\end{align}
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	which then allows us to determine that		which then allows us to determine that
	<center><math>		<center><math>
	\lim_{r \rightarrow 0} G(r) = \frac{i}{8 k^2}		\lim_{r \rightarrow 0} G(r) = \frac{i}{8 k^2}.
	</math></center>		</math></center>

Mike smith at 20:55, 27 April 2010

2010-04-27T20:55:09Z

← Older revision		Revision as of 20:55, 27 April 2010
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	=Behaviour at <math>r=0</math>=		=Behaviour at <math>r=0</math>=
			To determine the solution at <math>r \rightarrow 0</math> (where the above expression is singular), the following identities from [[Abramowitz and Stegun 1964]] are used
			<center><math>
			\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 < 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) \| < \frac{\pi}{2}
			\end{align}
			</math></center>
			from pages [http://www.math.auckland.ac.nz/~hcoh001/as/page_358.htm 358], [http://www.math.auckland.ac.nz/~hcoh001/as/page_375.htm 375], [http://www.math.auckland.ac.nz/~hcoh001/as/page_376.htm 376], and [http://www.math.auckland.ac.nz/~hcoh001/as/page_360.htm 360] respectively.


			Let us consider
			<center><math>
			\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></center>
			where introducing the limit yields
			<center><math>
			\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></center>
			Consequently
			<center><math>
			\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></center>
			which then allows us to determine that
			<center><math>
			\lim_{r \rightarrow 0} G(r) = \frac{i}{8 k^2}
			</math></center>

Mike smith at 20:45, 27 April 2010

2010-04-27T20:45:28Z

← Older revision		Revision as of 20:45, 27 April 2010
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	</center>		</center>
	That is, <math>P_{11} = (\lambda - k^2) \,</math> and <math>P_{21} = (\lambda - (ik)^2) \,</math>.		That is, <math>P_{11} = (\lambda - k^2) \,</math> and <math>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:
			<center>
			<math>
			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>
			</center>

			Using the Green's function for the [http://en.wikipedia.org/wiki/Helmholtz_equation Helmholtz Equation]
			<center>
			<math>
			G_{1,\beta} = \frac{i}{4} H_0^{(1)} (\beta r),
			</math>
			</center>
			where <math>H_0^{(1)} \,</math> denotes Hankel functions of the first kind, which incorporates the Sommerfeld radiation condition
			<center>
			<math>
			\lim_{r \rightarrow \infty} r^{1/2} \left(\frac{\partial}{\partial r} - ik \right)G = 0,
			</math>
			</center>
			we obtain the final expression
			<center><math>
			\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></center>
			This is the fundamental solution for our governing equation above.

			=Behaviour at <math>r=0</math>=

Mike smith at 20:37, 27 April 2010

2010-04-27T20:37:48Z

← Older revision		Revision as of 20:37, 27 April 2010
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	On this page, we aim to derive the Green's function for a thin uniform plate.		==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>\mathbb{R}^2 \,</math>, which is taken to be of the form
			<center>
			<math>
			\left(\Delta^2 - k^2\right) u =0,
			</math>
			</center>
			where <math>\Delta = \partial_r^2 + \frac{1}{r}\partial_r, </math> in polar coordinates.

			==Linear Operators==
			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

			<center>
			<math>
			P(\lambda) = c(\lambda - \lambda_1)^{m_1}(\lambda - \lambda_2)^{m_2} \ldots (\lambda - \lambda_p)^{m_p},
			</math>
			</center>

			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>.

			The reciprocal of <math>P(\lambda) \,</math> is taken to be of the form
			<center>
			<math>
			\frac{1}{P(\lambda)} = \sum_{q=1}^{p} \sum_{n=1}^{m_q} \frac{c_{qn}}{(\lambda - \lambda_q)^n}.
			</math>
			</center>
			where
			<center>
			<math>
			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>
			</center>
			It is straightforward to compute the <math>c_{qn} \,</math> coefficients
			<center><math>
			\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></center>
			as well as determining a corresponding quantity <math>P_{qn} \,</math>
			such that <math>\sum_{q=1}^{p} \sum_{n=1}^{m_q} c_{qn}P_{qn} = 1</math>, where
			<center>
			<math>
			P_{qn}(\lambda) \left[\lambda - \lambda_q\right]^n = P(\lambda), \quad \mbox{for all q,n}.
			</math>
			</center>
			That is, <math>P_{11} = (\lambda - k^2) \,</math> and <math>P_{21} = (\lambda - (ik)^2) \,</math>.

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