# In the numerical solution of the Wave Equation, using finite differences, where do I obtain the spatial values from?

In trying to implement a simplistic numerical solver for wave equations, I have run into a conceptual problem that I haven't been able to solve.

Consider a one-dimensional wave equation of a quantity $y$

$$\frac{\partial^2y(x,t)}{\partial t^2} = c^2 \frac{\partial^2y(x,t)}{\partial x^2}$$

With a periodic boundary condition, i.e.: $y(x+L)=y(x)$, with $L$ being the length of the system.

The point usually made in books, lecture notes, internet tutorials, etc., is that the wave equation needs to be discretized and rearranged:

$$y(x,t+\Delta t) = 2(1 - \lambda) y(x,t) - y(x,t-\Delta t) + \lambda \{y(x+\Delta x, t) + y(x-\Delta x,t)\}$$

With $\lambda = (\frac{c\Delta t}{\Delta x})^2$ a constant prefactor, that must not become greater than 1.

Now, the scheme is an explicit one and simply solves the equation forward in time. I am quite familiar with the idea behind it and how it works (one might compare this to the Euler method). One needs two initial conditions (said to be at time $y(x,0)$ and $y(x,-\Delta t$) to start solving it.

The point I don't understand, though, is: Assuming I have the two points, $y(x,t)$ and $y(x,t-\Delta t)$, where do I obtain the extra two required spatial points from? In other words, where do I obtain $y(x+\Delta x, t)$ and $y(x-\Delta x,t)$ from?

These, of course, cannot be given in advance, since they depend on $\Delta x$ but I also do not see a way to calculate these points simply, since they are part of the equation.

All guidelines I have looked at seem to treat this problem as if these points were already known or trivial to calculate, but I don't understand how this is to be done. Any help is greatly appreciated.

You start with $y(x,0)$ and $y(x,-\Delta t)$ for all values of $x$ (or all values of $x$ on your numerical grid, at least).
Then, when you need to compute the value of $y(x,\Delta t) = y(x, 0 + \Delta t)$ for a particular value of $x$, you need to know, e.g. $y(x + \Delta x,0)$ and $y(x -\Delta x,0)$. But you already know these values, as we discussed in the first paragraph. Thus, you can compute $y(x,\Delta t)$ for every $x$ in your grid, and you are now setup to compute $y(x,2 \Delta t)$ with these values.