# 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:

[math] \phi^I=e^{-k_0(\cos \theta x + \sin \theta y)} [/math]

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

[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]

where [math]k_n[/math] is a root of the dispersion equation. This simplifies as

[math] k_y^2+\frac{1}{X}\frac{d^2X}{dx^2}=k_n^2 [/math]

which permits to obtain the following differential equation for [math]X[/math]

[math] \frac{d^2X}{dx^2}-(k_n^2-k_y^2)X=0 [/math]

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].