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: $$-\infty\le x\le\infty,\ -L/2\le y\le L/2.$$ 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 $L\gg1$. The $L\gg1$ 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 $L\gg1$ 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 $t\geq 0$ 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.