View source for Waves Incident at an Angle

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 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 
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}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, 
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 following 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>.