User talk:Sean Curry - Revision history

Sean Curry: /* With the Free Surface BC */

2010-02-08T02:14:26Z

With the Free Surface BC

← Older revision		Revision as of 02:14, 8 February 2010
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	Using		Using

	<math>\ln(r) = \int_0^\infty (e^{-\mu}- e^{-\mu\|z+f\|}\cos(\mu x)\frac{d\mu}{\mu})</math>		<math>\ln(r) = \int_0^\infty (e^{-\mu}- e^{-\mu\|z+f\|}\cos(\mu x))\frac{d\mu}{\mu}</math>

Sean Curry: /* With the Free Surface BC */

2010-02-07T22:44:59Z

With the Free Surface BC

@@ Line 127: / Line 127: @@
 So our psi functions satisfy
-   <math>\nabla^2 \psi_m = 0 \quad for \quad m = 0,1,2, \ldots</math>
+   <math>\nabla^2 \psi_m = 0 \quad for \quad m = 0,1,2, \ldots \quad \ldots (1)</math>
-   <math>\psi_m \rightarrow 0 \quad as \quad z \rightarrow -\infty</math>
+   <math>\psi_m \rightarrow 0 \quad as \quad z \rightarrow -\infty \quad \ldots (2)</math>
-   <math>\partial_z \psi_m |_{z=f} - K\psi_m|_{z=f} = f_m(x) </math>
+   <math>\partial_z \psi_m |_{z=f} - K\psi_m|_{z=f} = f_m(x) \quad \ldots (3)</math>
 == Example: 2D Wave Scattering for a Submerged Horizontal Cylinder (in Infinite Depth)==

Sean Curry: /* With the Free Surface BC */

2010-02-07T21:50:04Z

With the Free Surface BC

← Older revision		Revision as of 21:50, 7 February 2010
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	<math>\partial_z \psi_m \|_{z=f} - K\psi_m\|_{z=f} = K \frac{e^{im\theta}}{r^m} \|_{z=f} - \partial_z \left(\frac{e^{im\theta}}{r^m}\right)\|_{z=f} = \frac{e^{im\tan(-x/f)}}{(x^2+f^2)^{m/2}} \left( K + \frac{imx\ln(f)}{1-(x/f)^2} + \frac {mf}{x^2 + f^2} \right) = f_m(x) \quad \quad \quad \ldots</math> needs simplifying		<math>\partial_z \psi_m \|_{z=f} - K\psi_m\|_{z=f} = K \frac{e^{im\theta}}{r^m} \|_{z=f} - \partial_z \left(\frac{e^{im\theta}}{r^m}\right)\|_{z=f} = \frac{e^{im\tan(-x/f)}}{(x^2+f^2)^{m/2}} \left( K + \frac{imx\ln(f)}{1-(x/f)^2} + \frac {mf}{x^2 + f^2} \right) = f_m(x) \quad \quad \quad \ldots</math> needs simplifying



			So our psi functions satisfy

			<math>\nabla^2 \psi_m = 0 \quad for \quad m = 0,1,2, \ldots</math>

			<math>\psi_m \rightarrow 0 \quad as \quad z \rightarrow -\infty</math>

			<math>\partial_z \psi_m \|_{z=f} - K\psi_m\|_{z=f} = f_m(x) </math>

	== Example: 2D Wave Scattering for a Submerged Horizontal Cylinder (in Infinite Depth)==		== Example: 2D Wave Scattering for a Submerged Horizontal Cylinder (in Infinite Depth)==

Sean Curry: /* With the Free Surface BC */

2010-02-07T21:41:02Z

With the Free Surface BC

← Older revision		Revision as of 21:41, 7 February 2010
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	For m>0		For m>0

	<math>\partial_z \psi_m \|_{z=f} - K\psi_m\|_{z=f} = K \frac{e^{im\theta}}{r^m} \|_{z=f} - \partial_z \left(\frac{e^{im\theta}}{r^m}\right)\|_{z=f} = \frac{e^{im\tan(-x/f)}}{(x^2+f^2)^{m/2}} \left( K + \frac{imx\ln(f)}{1-(x/f)^2} + \frac {mf}{x^2 + f^2} \right) = f_m(x)</math>		<math>\partial_z \psi_m \|_{z=f} - K\psi_m\|_{z=f} = K \frac{e^{im\theta}}{r^m} \|_{z=f} - \partial_z \left(\frac{e^{im\theta}}{r^m}\right)\|_{z=f} = \frac{e^{im\tan(-x/f)}}{(x^2+f^2)^{m/2}} \left( K + \frac{imx\ln(f)}{1-(x/f)^2} + \frac {mf}{x^2 + f^2} \right) = f_m(x) \quad \quad \quad \ldots</math> needs simplifying

	== Example: 2D Wave Scattering for a Submerged Horizontal Cylinder (in Infinite Depth)==		== Example: 2D Wave Scattering for a Submerged Horizontal Cylinder (in Infinite Depth)==

Sean Curry: /* With the Free Surface BC */

2010-02-07T21:39:37Z

With the Free Surface BC

@@ Line 96: / Line 96: @@
 <math>\phi_m = \frac{e^{im\theta}}{r^m} + \psi_m </math>
-<math>\phi_0 = \ln\frac{r}{a} + \psi_0 </math>
+<math>\phi_0 = \ln\left(\frac{r}{a}\right) + \psi_0 </math>
 Where
 <math>\nabla^2 \psi_m = 0 \quad for \quad m = 0,1,2, \ldots</math>
 == Example: 2D Wave Scattering for a Submerged Horizontal Cylinder (in Infinite Depth)==

Sean Curry: /* 2D Exterior Neumann Problem */

2010-02-07T21:14:35Z

2D Exterior Neumann Problem

@@ Line 85: / Line 85: @@
 <math>\phi = C(0) + D(0) \ln(r) + \sum_{m=1}^\infty \left( a(m)\frac{\cos(m\theta)}{r^m} +  b(m)\frac{\sin(m\theta)}{r^m}  \right) </math>
-Which can also be expressed simply in terms of complex exponentials
+Which can also be expressed simply in terms of complex exponentials.
 == Example: 2D Wave Scattering for a Submerged Horizontal Cylinder (in Infinite Depth)==

Sean Curry: /* 2D Exterior Neumann Problem */

2010-02-04T03:50:15Z

2D Exterior Neumann Problem

← Older revision		Revision as of 03:50, 4 February 2010
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	<math>For \quad \Theta: \quad \partial_\theta^2 \Theta = -m^2\Theta </math>		<math>For \quad \Theta: \quad \partial_\theta^2 \Theta = -m^2\Theta </math>

	<math>\Rightarrow ~~\quad~~ \Theta = A(m) \cos(m\theta) + B(m) \sin(m\theta) \quad for \quad m \neq 0 </math>		<math>\Rightarrow \Theta = A(m) \cos(m\theta) + B(m) \sin(m\theta) \quad for \quad m \neq 0 </math>

	<math>Requiring \quad \Theta(\theta + 2\pi) = \Theta(\theta) \quad gives \quad m \in Z</math>		<math>\mathrm{When} \quad m = 0 \quad \mathrm{the \quad obvious \quad solution \quad is} \quad \Theta = A(0) + B(0)\theta </math>

			<math>Requiring \quad \Theta(\theta + 2\pi) = \Theta(\theta) \quad gives \quad m \in Z \quad and \quad B(0) = 0</math>


			<math>For \quad R: \quad r^2\partial_r^2 R + r\partial_rR = m^2 R </math>

			we substitute R = r^n giving:

			<math>n(n-1)r^n + nr^n = m^2r^n \quad \Rightarrow \quad n^2 = m^2 \quad \Rightarrow \quad n= \pm m</math>

			This gives two independent solutions for all m except m = 0, where we only get the constant solution. Noting that:

			<math> \nabla^2 \left( \ln(r) \right) = \frac{-1}{r^2} + \frac{1}{r^2} = 0 \quad . </math> Hence, for m=0, we can write:

	<math>~~\mathrm{When} \quad m~~ = 0 \~~quad \mathrm{the \quad obvious \quad solution \quad is} \quad \Theta = A~~(0) + B(0)\~~theta~~ </math>		<math>R = C\left(0\right) + D(0) \ln(r)</math>

			and for <math>m\neq 0:</math>

			<math>R = C\left(m\right)r^m + D(m)r^{-m}</math>


			<math>Since\quad\|\nabla\phi\| \rightarrow 0 \quad as \quad r \rightarrow \infty, </math>

			<math>\partial_r R \rightarrow 0 \quad as \quad r \rightarrow \infty </math>

			<math>\Rightarrow C(m)= 0 \quad\forall m \neq 0</math>


			Hence the general solution can be expressed as:

			<math>\phi = C(0) + D(0) \ln(r) + \sum_{m=1}^\infty \left( a(m)\frac{\cos(m\theta)}{r^m} + b(m)\frac{\sin(m\theta)}{r^m} \right) </math>

	~~<math> \nabla^2 \left( \ln(r) \right) = \frac{-1}{r^2} + \frac{1}{r^2} = 0 \quad . </math> Hence, for m=0, we~~ can ~~write:~~		Which can also be expressed simply in terms of complex exponentials

	== Example: 2D Wave Scattering for a Submerged Horizontal Cylinder (in Infinite Depth)==		== Example: 2D Wave Scattering for a Submerged Horizontal Cylinder (in Infinite Depth)==

Sean Curry: /* 2D Exterior Neumann Problem */

2010-02-04T02:49:43Z

2D Exterior Neumann Problem

← Older revision		Revision as of 02:49, 4 February 2010
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	Outside a disk of radius a centered at the origin in an infinite domain:		Outside a disk of radius a centered at the origin in an infinite domain:

	<math>\nabla^2 phi = \left[ \partial_r^2 + \frac{1}{r}\partial_r + \frac{1}{r^2} \right] \phi= 0 </math>		<math>\nabla^2 \phi = \left[ \partial_r^2 + \frac{1}{r}\partial_r + \frac{1}{r^2} \partial_\theta^2 \right] \phi= 0 \quad for \quad r>a </math>

Sean Curry: /* 2D Exterior Neumann Problem */

2010-02-04T02:10:23Z

2D Exterior Neumann Problem

← Older revision		Revision as of 02:10, 4 February 2010
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	== 2D Exterior Neumann Problem ==		== 2D Exterior Neumann Problem ==

			Outside a disk of radius a centered at the origin in an infinite domain:

			<math>\nabla^2 phi = \left[ \partial_r^2 + \frac{1}{r}\partial_r + \frac{1}{r^2} \right] \phi= 0 </math>


			<math>\partial_n \phi \|_{r=a} = 0</math>

			<math>\|\nabla\phi\| \rightarrow 0 \quad as \quad r \rightarrow \infty </math>




			<math>Let \quad \phi = R(r)\Theta(\theta)</math>

			Seperating variables we get:


			<math>\frac{r^2}{R}\partial_r^2 R + \frac{r}{R}\partial_r R = - \frac{1}{\Theta}\partial_\theta^2 \Theta = m^2 </math>


			<math>For \quad \Theta: \quad \partial_\theta^2 \Theta = -m^2\Theta </math>

			<math>\Rightarrow \quad \Theta = A(m) \cos(m\theta) + B(m) \sin(m\theta) \quad for \quad m \neq 0 </math>

			<math>Requiring \quad \Theta(\theta + 2\pi) = \Theta(\theta) \quad gives \quad m \in Z</math>

			<math>\mathrm{When} \quad m = 0 \quad \mathrm{the \quad obvious \quad solution \quad is} \quad \Theta = A(0) + B(0)\theta </math>







			<math> \nabla^2 \left( \ln(r) \right) = \frac{-1}{r^2} + \frac{1}{r^2} = 0 \quad . </math> Hence, for m=0, we can write:

	== Example: 2D Wave Scattering for a Submerged Horizontal Cylinder (in Infinite Depth)==		== Example: 2D Wave Scattering for a Submerged Horizontal Cylinder (in Infinite Depth)==

Sean Curry: /* Multipole Expansions */

2010-02-03T23:49:54Z

Multipole Expansions

← Older revision		Revision as of 23:49, 3 February 2010
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	- The feild equation		- The feild equation

	<math>\nabla \phi = 0 </math>throughout the fluid domain		<math>\nabla^2 \phi = 0 </math>throughout the fluid domain

	- The free surface boundary condition		- The free surface boundary condition
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	A linear combination of these multipoles can then be made to satisfy the body boundary conditions.		A linear combination of these multipoles can then be made to satisfy the body boundary conditions.


			== 2D Exterior Neumann Problem ==

	== Example: 2D Wave Scattering for a Submerged Horizontal Cylinder (in Infinite Depth)==		== Example: 2D Wave Scattering for a Submerged Horizontal Cylinder (in Infinite Depth)==