It is a one-dimensional wave on a uniform oscillating string.
A(i,t+1) = (2.0 * A(i,t)) - A(i,t-1) + (c * (A(i-1,t) - (2.0 * A(i,t)) + A(i+1,t))), where c is a constant, and A(i, t) stands for the wave amplitude at position index i, and time t. And we also suppose the first index is a neighbour to the last index, and vice versa.
My attempt: (I really have no clue to any mathematical proof) my reasoning is that if I assume the wave is a standing wave, then A(i,x) at the even multiples of quarter wavelength will always be zero. Substituting this into the above equation, and given that amplitude around the nodes is equal but opposite to each other, I get the LHS equal to the RHS. Therefore, the above equation satisifies one of the standing wave properties. However, I am not sure if the above proof is sufficient to show that the equation represents a standing wave, and I am also not sure if the above equation can simultaneously depict a standing wave and a traveling wave. Please help : )
Let's look closer: $$A_{i,t+1} = 2A_{i,t} - A_{i,t-1} + c\cdot(A_{i-1,t} - 2A_{i,t} + A_{i+1,t})$$ Now let's regroup it: $$A_{i,t+1} - 2A_{i,t} + A_{i,t-1} = c\cdot(A_{i-1,t} - 2A_{i,t} + A_{i+1,t})$$ Now it is clear that our equation is nothing but a finite difference analog of the 1D wave equation: $${\partial^2u\over\partial t^2}=c^2\cdot {\partial^2u\over\partial x^2}$$ As such, the equation doesn't care whether it depicts a standing wave or a traveling wave. Depending on the initial conditions, it can be either. You may produce a wave that runs to the right, or another that runs to the left, or combine them together and get a standing wave.
So it goes.