Difference between revisions of "Waves Incident at an Angle"
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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 | + | [[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 |
Revision as of 00:43, 21 August 2008
We can consider the problem when the waves are incident at an angle [math]\displaystyle{ \theta }[/math]. Thus the incident potential can be expressed as follow:
[math]\displaystyle{ \phi^I=e^{-k_0(\cos \theta x + \sin \theta y)} }[/math]
In this case the wavenumber in the [math]\displaystyle{ y }[/math]-direction is [math]\displaystyle{ k_y = \sin\theta k_0 }[/math] where [math]\displaystyle{ k_0 }[/math] is the travelling mode, the pure imaginary root of the Dispersion Relation for a Free Surface (note that [math]\displaystyle{ k_y }[/math] is imaginary).
We assume here that the object has infinite length in the [math]\displaystyle{ 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]\displaystyle{ \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]\displaystyle{ z }[/math] remains unchanged. However, we obtain
[math]\displaystyle{ \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]
where [math]\displaystyle{ k_n }[/math] is a root of the dispersion equation. This simplifies as
[math]\displaystyle{ k_y^2+\frac{1}{X}\frac{d^2X}{dx^2}=k_n^2 }[/math]
which permits to obtain the followinf differential equation for [math]\displaystyle{ X }[/math]
[math]\displaystyle{ \frac{d^2X}{dx^2}-(k_n^2-k_y^2)X=0 }[/math]
If we introduce the new variable [math]\displaystyle{ k_x^2=k_n^2-k_y^2 }[/math], we obtain a similar differential equation to solve in the [math]\displaystyle{ x }[/math]-direction as for the null incident angle problem, so that we can develop a solution for [math]\displaystyle{ X }[/math] of the same form that the one with no angle, taking care of replacing the wave number [math]\displaystyle{ k_n }[/math] by [math]\displaystyle{ \hat{k}_{n} = \sqrt{k_n^2 - k_y^2} }[/math].