Difference between revisions of "Waves Incident at an Angle"

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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  

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].