the numerical domain of dependence
Ask which earlier information a particular answer rests on. For a wave equation the true value at a point (x, t) depends only on the initial data within a certain physical region — for a wave of speed c, the interval [x - c*t, x + c*t], because no influence travels faster than c. This triangular reach back into the past is the (true) domain of dependence. The numerical version asks the same question of the SCHEME: which initial grid values can possibly affect a given computed value?
Trace it backwards through the stencil. If your explicit scheme updates u_j^{n+1} from u_{j-1}^n, u_j^n, u_{j+1}^n, then the new value depends on three points one step back, those depend on five points two steps back, and so on — after n steps the computed value at node j draws on a fan of initial cells spanning roughly j - n to j + n, i.e. a spatial width of about n*h on each side over a time n*k. The slope of this discrete triangle is h/k: the fastest speed at which information can propagate through the grid is exactly h/k cells per unit time. This grid triangle is the numerical domain of dependence.
The two triangles must fit together correctly, and that is the content of the CFL condition. If the numerical domain of dependence does NOT contain the true one — that is, if the physical wave reaches the point from data the scheme never looked at (c > h/k, i.e. Courant number c*k/h > 1) — then refining the grid cannot help; the scheme is missing essential information and must be unstable or wrong. So the requirement 'numerical domain of dependence contains the analytical one' is the geometric heart of CFL and a clean way to see WHY explicit schemes have a time-step limit. The same picture explains why implicit schemes escape it: there, every point couples to the whole grid in one step, so the numerical domain of dependence is the entire domain and always contains the true one.
Leapfrog with h = 0.1 lets information cross at most one cell per step, grid speed h/k. If the true wave speed is c = 1 and k = 0.05, then h/k = 2 > c — the numerical triangle is wider than the physical one, so it contains it: stable. Shrink the grid speed below c (take k = 0.2, h/k = 0.5 < 1) and the physical wave outruns the grid: CFL violated, scheme unstable.
Stable only if the grid's triangle of influence covers the physical one.
Containing the true domain of dependence is NECESSARY for an explicit scheme but not sufficient: a scheme can have a wide enough numerical domain and still be unstable (FTCS for advection covers the physical domain yet blows up). Implicit schemes always satisfy the containment trivially because their numerical domain is the whole grid — which is exactly why they have no CFL limit.