# Waves Incident at an Angle

We can consider the problem when the waves are incident at an angle $\theta$. Thus the incident potential can be expressed as follow:

$\phi^I=e^{-k_0(\cos \theta x + \sin \theta y)}$

In this case the wavenumber in the $y$-direction is $k_y = \sin\theta k_0$ where $k_0$ is the travelling mode, the pure imaginary root of the Dispersion Relation for a Free Surface (note that $k_y$ is imaginary).

We assume here that the object has infinite length in the $y$-direction, so the solution does not vary in that direction except over a period. This means that the potential is now of the form $\phi(x,y,z)=e^{k_y y}X(x)Z(z)$. The separation of variables and the application of Laplace's equation clearly shows that the dependance of $z$ remains unchanged. However, we obtain

$\frac{1}{e^{k_yy}X(x)}(k_y^2e^{k_yy}X(x)+e^{k_yy}\frac{d^2X}{dx^2})=k_n^2=-\frac{1}{Z(z)}\frac{d^2Z}{dz^2}$

where $k_n$ is a root of the dispersion equation. This simplifies as

$k_y^2+\frac{1}{X}\frac{d^2X}{dx^2}=k_n^2$

which permits to obtain the following differential equation for $X$

$\frac{d^2X}{dx^2}-(k_n^2-k_y^2)X=0$

If we introduce the new variable $k_x^2=k_n^2-k_y^2$, we obtain a similar differential equation to solve in the $x$-direction as for the null incident angle problem, so that we can develop a solution for $X$ of the same form that the one with no angle, taking care of replacing the wave number $k_n$ by $\hat{k}_{n} = \sqrt{k_n^2 - k_y^2}$.