Jtdylktuy

Consider the wave equation:
$$frac{partial^2 u}{partial t^2}=frac{partial^2u}{partial y^2},$$
with wave speed equal to 1.
Consider a domain with strings arranged like so:

where each vertical line denotes a string. The domain is given by:
$$-inftyle xleinfty, -L/2le yle L/2.$$
Assume the strings are fixed at both ends. If the initial condition is given by:
$$u(x,y,0)=e^{-(x^2+y^2)},$$
and it is assumed the wave is travelling in the positive $y$ direction, then the full solution is given by approximately:
$$u(x,y,t)=sum_{n=0}^{infty}(-1)^ne^{-{x^2+[y-(-1)^n(t-nL)]^2}},$$
using d'Alembert's formula, provided $Lgg1$. The $Lgg1$ requirement is needed to ensure the exponentials with their centres initially outside the domain contribute a negligible amount to the domain initially. I could have easily replaced the Gaussians with rectangular functions and this would mean the $Lgg1$ requirement would not be needed. My question is what would the solution be if I rotated the strings such that the domain looked like this:

I naively hoped that if I rotated the coordinate system
to form:
$$s = frac{1}{sqrt{2}}(x+y), A=frac{1}{sqrt{2}}(x-y)$$
then the solution would be:
$$u(s,A,t)=sum_{n=0}^{infty}(-1)^ne^{-{[s-(-1)^n(t-nL_s)]^2+A^2}},$$
however, this suggests the solution would retain its Gaussian shape, but from looking at a simulation I know this is not the case:

Note that now the strings have been rotated, the system is described by:
$$frac{partial^2 u}{partial t^2}=frac{partial^2u}{partial s^2}.$$

To formulate an analytic solution we first write down the equations and boundary conditions. We are working in the space $(s, A, t) $ where $tgeq 0$ and
$$
A-L_s <s<A+L_s
$$
In this domain
$$
frac{partial^2 u}{partial s^2 } = frac{partial^2 u}{partial t^2 }
$$
and on its boundary
$$
u=0.
$$
At time $t=0$ we know the solution is
$$
u(s, A, 0) = f(s,A)
$$
and this wave moves in the positive $s$ direction. To write out this solution fully we extend the function $f$ outside of the domain to have value $0$, so that we can write the solution as
$$
u(s, A, 0) = f(s-t, A) - f(2(A+L_s)-[s-t],A) + f(2(A-L_s) - [2(A+L_s)-[s-t]], A) - f(2(A+L_s)-[2(A-L_s) - [2(A+L_s)-[s-t]]], A) +ldots
$$
where each term is a subsequent reflection. Simplifying
$$
u(s, A, 0) = sum_{n=0}^infty f(s-t-4nL_s, A) - f(-s+t-(4n+2)L_s+2A, A)
$$
The distortion you are seeing comes from the $2A$ term in the first coordinate of the second $f$, and is the result of angling the domain edge.