Waves Incident at an Angle - Revision history

Fmontiel at 01:51, 22 August 2008

2008-08-22T01:51:16Z

← Older revision		Revision as of 01:51, 22 August 2008
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	We assume here that the object has infinite length in the <math>y</math>-direction, so the solution		We assume here that the object has infinite length in the <math>y</math>-direction, so the solution
	does not vary in that direction except over a period. This means that the potential is now of the		does not vary in that direction except over a period. This means that the potential is now of the
	form <math>\phi(x,y,z)=e^{k_y y}W(x)Z(z)</math>. The separation of variables and the application of		form <math>\phi(x,y,z)=e^{k_y y}X(x)Z(z)</math>. The separation of variables and the application of
	Laplace's equation clearly shows that the dependance of <math>z</math> remains unchanged. However,		Laplace's equation clearly shows that the dependance of <math>z</math> remains unchanged. However,
	we obtain		we obtain

Fmontiel at 01:10, 21 August 2008

2008-08-21T01:10:05Z

← Older revision		Revision as of 01:10, 21 August 2008
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	</math>		</math>
	</center>		</center>
	which permits to obtain the ~~followinf~~ differential equation for <math>X</math>		which permits to obtain the following differential equation for <math>X</math>
	<center>		<center>
	<math>		<math>

Fmontiel at 00:43, 21 August 2008

2008-08-21T00:43:47Z

← Older revision		Revision as of 00:43, 21 August 2008
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	case the wavenumber in the <math>y</math>-direction is <math>k_y = \sin\theta k_0</math>		case the wavenumber in the <math>y</math>-direction is <math>k_y = \sin\theta k_0</math>
	where <math>k_0</math> is the travelling mode, the pure imaginary root of the		where <math>k_0</math> is the travelling mode, the pure imaginary root of the
	[[Dispersion Relation ~~For A~~ Free Surface]] (note that <math>k_y</math> is imaginary).		[[Dispersion Relation for a Free Surface]] (note that <math>k_y</math> is imaginary).

	We assume here that the object has infinite length in the <math>y</math>-direction, so the solution		We assume here that the object has infinite length in the <math>y</math>-direction, so the solution

Fmontiel at 00:43, 21 August 2008

2008-08-21T00:43:10Z

← Older revision		Revision as of 00:43, 21 August 2008
Line 8:		Line 8:
	In this		In this
	case the wavenumber in the <math>y</math>-direction is <math>k_y = \sin\theta k_0</math>		case the wavenumber in the <math>y</math>-direction is <math>k_y = \sin\theta k_0</math>
	where <math>k_0</math> is ~~as defined previously~~ (note that <math>k_y</math> is imaginary).		where <math>k_0</math> is the travelling mode, the pure imaginary root of the
			[[Dispersion Relation For A Free Surface]] (note that <math>k_y</math> is imaginary).

	We assume here that the object has infinite length in the <math>y</math>-direction, so the solution		We assume here that the object has infinite length in the <math>y</math>-direction, so the solution

Fmontiel at 00:35, 21 August 2008

2008-08-21T00:35:24Z

New page

We can consider the problem when the waves are incident at an angle <math>\theta</math>. Thus the incident
potential can be expressed as follow:
<center>
<math>
\phi^I=e^{-k_0(\cos \theta x + \sin \theta y)}
</math>
</center>
In this
case the wavenumber in the <math>y</math>-direction is <math>k_y = \sin\theta k_0</math>
where <math>k_0</math> is as defined previously (note that <math>k_y</math> is imaginary).

We assume here that the object has infinite length in the <math>y</math>-direction, so the solution
does not vary in that direction except over a period. This means that the potential is now of the
form <math>\phi(x,y,z)=e^{k_y y}W(x)Z(z)</math>. The separation of variables and the application of
Laplace's equation clearly shows that the dependance of <math>z</math> remains unchanged. However,
we obtain
<center>
<math>
\frac{1}{e^{k_yy}X(x)}(k_y^2e^{k_yy}X(x)+e^{k_yy}\frac{d^2X}{dx^2})=k_n^2=-\frac{1}{Z(z)}\frac{d^2Z}{dz^2}
</math>
</center>
where <math>k_n</math> is a root of the dispersion equation. This simplifies as
<center>
<math>
k_y^2+\frac{1}{X}\frac{d^2X}{dx^2}=k_n^2
</math>
</center>
which permits to obtain the followinf differential equation for <math>X</math>
<center>
<math>
\frac{d^2X}{dx^2}-(k_n^2-k_y^2)X=0
</math>
</center>
If we introduce the new variable <math>k_x^2=k_n^2-k_y^2</math>, we obtain a similar differential equation to solve
in the <math>x</math>-direction as for the null incident angle problem, so that we can develop a solution for
<math>X</math> of the same form that the one with no angle, taking care of replacing the wave number <math>k_n</math>
by <math>\hat{k}_{n} = \sqrt{k_n^2 - k_y^2}</math>.