The Wave Equation

Waves, such as light waves or sound waves, can be described by the following equation in the one dimensional case.

This equation states that the second time derivative of the displacement of the wave as proportional to the second derivative with respect to the spatial dimension. Put more simply, how the wave will change over time is related to the current shape. For the rest of this discussion, lets assume \(v\) is 1 to remove if from this calculation.

To the right we see a 1D wave with the first space derivative \(\frac{\partial u}{\partial x}\) at this particular point highlighted in blue. The slope at points nearby this point will be similar but slightly different. The second derivative \(\frac{\partial^2 u}{\partial x^2}\) tells us how quickly these slopes are changing.

What does the graph look like at a point where the second derivative is large? If its a positive number, the slopes are increasing very quickly. This forms a trough and if negative (slopes are decreasing quickly), we have a hill. Intuitively, this second derivative indicates how 'pointy' the wave form is at this point, a more pronounced trough or hill.

Lets look at the left half of the wave equation, the second time derivative \(\frac{\partial^2 u}{\partial t^2}\). The first derivative \(\frac{\partial u}{\partial t}\) tells us how a point moves up and down over time. Mechanically, taking a derivative with respect to space or time is identical, but it can help to think of this as a speed (distance over time). Using this analogy, the second derivative with respect to time is the acceleration of the point. After all, acceleration is just how fast you are changing your speed over time.

Looking back at our wave equation, the second time derivative is equal to the second space derivative. Using what we described earlier, the acceleration of the point is equal to the 'pointiness' of the wave at that point. If a point is at the top of a narrow peak, its second space derivative will be a large negative number. This means its second time derivative is also that large negative number and the point should be accelerating downward. Note that speed and acceleration can be in opposite directions. The point may still be traveling upwards at this time, but with a negative second time derivative its upward speed would be decreasing.

Code

Lets walkthrough some salient points of the code to generate the above wave simulation. Defintions of variables below.

 function diff(u) {
    let result = []
    result[0] = u[1] - u[0]
    for (let i = 1; i < u.length - 1; i++) {
        result[i] = (u[i + 1] - u[i - 1]) / 2
    }
    result[u.length - 1] = u[u.length - 1] - u[u.length - 2]
    return result
}

Now we can advance one time step. We take two derivatives of our u array to find \(\frac{\partial^2 u}{\partial x^2}\) which is also our \(\frac{\partial^2 u}{\partial t^2}\). We update our speed (dt) with our acceleration (dt2). Since our wave has two fixed points at the boundary where it can't move, we set its speed to 0 at these two points. We then update u by adding on the current speed.

 function advance() {
    let du2 = diff(diff(u))
    let dt2 = du2
    let new_dt = []
    for (let i = 0; i < total_points; i++) {
        new_dt[i] = dt[i] + dt2[i]  // Update speed by adding acceleration
    }
    new_dt[0] = 0
    new_t[total_points - 1] = 0

    dt = new_dt
    for (let i = 0; i < total_points; i++) {
        u[i] = u[i] + dt[i]   // Update position by adding speed
    }
}

Stepping forward in time by small steps will give believable results, but is prone to accumulating errors. Our adjustments at the boundaries are also not perfect. More exact methods of running this simulation are beyond the scope of this post. Ideally, it would be time reversable so that stepping forward one unit then back one unit returns you to the same state. The above code does not guarentee that.

More dimensions

The wave equation applies in more than 1 spatial dimension. We simply need to sum the second space derivatives on the right hand side. If our wave is in 2 dimensions it takes the following form.

\[\frac{\partial^2\, u}{\partial\, t^2} = v^2 \left( \frac{\partial^2\, u}{\partial\, x^2}\ + \frac{\partial^2\, u}{\partial\, y^2} \right) \]

An example of a 2D wave might be the surface of a pool of water. X and Y are the coordinates over the surface and the displacement is the height of the water at that point. Our code would be very similar to before, but u is now a 2-D array. One value for each x,y pair. We would also need to take spacial derivatives with respect to two different axes. Derivative with respect to x is given below. Note how we move along the x axis with x+1 and x-1 offsets but always use just y. diffy would be the opposite.

function diffx(u) {
    let result = []
    for(let x=0; x<pointsx; x++) {
        result[x] = []
    }

    for(let y=0; y<u[0].length; y++) {
        result[0][y] = u[1][y] - u[0][y]

        for(let x=1; x<u.length-1; x++) {
            result[x][y] = (u[x+1][y] - u[x-1][y]) / 2
        }

        result[pointsx-1][y] = u[pointsx-1][y] - u[pointsx-2][y]
    }

    return result
} 

Here we choose to render the displacement of the wave as a color rather than requiring a 3d rendered view.