 Solution

Consider first a scalar constant-coefficient Cauchy problem for a partial differential equation on the whole line. A translation-invariant spatial discretization is a convolution operator, so the discrete Fourier transform turns it into multiplication by its Fourier symbol. This reduction is exact because Fourier modes are the simultaneous generalized eigenfunctions of every translation-invariant stencil.

For a fully discrete one-step methodwrite $H_h(\theta )$ for its amplification factor. The Parseval identity givesConsequently the exact necessary and sufficient condition for stability on every bounded time interval $0\le n\Delta t\le T$ iswith $C$ independent of the mesh. Sufficiency follows directly from Parseval; necessity follows by choosing transformed initial data concentrated where the multiplier is largest. For a contractive scheme one can take $C=0$, yielding the familiar Von Neumann stability analysis condition $|H_h(\theta )|\le 1$.

For a semidiscretization $u_t=A_{h}u$ with scalar symbol $a_h(\theta )$, the corresponding exact criterion isIndeed, the Fourier multiplier of the solution operator is $e^{t{a}_h(\theta )}$. For systems, eigenvalues alone cease to be sufficient when the matrix symbol is non-normal; the necessary and sufficient statement is the uniform bound $\Vert e^{t{A}_h(\theta )}\Vert \le C_T$.

On the whole lattice, a finite stencil defines a Laurent operator, whose operator norm is the essential supremum of its symbol. Restricting the same stencil to a half-line gives a Toeplitz operator. The Cauchy symbol condition remains necessary for a stable initial-boundary scheme because data localized far from the boundary behave like the whole-line problem for a finite time. It is not sufficient: the boundary closure can support growing modes or amplify incoming modes even when every interior Fourier mode is stable.

Boundary stability therefore requires a uniform estimate for the forced half-line recurrence. After a Laplace transform in time and a Fourier transform in tangential variables, one solves a normal-direction recurrence. The Uniform Kreiss--Lopatinskii condition requires the boundary equations to determine its decaying roots with a uniformly bounded inverse. In practical terms, one imposes one independent boundary condition for each incoming characteristic or numerical mode and none for outgoing modes. The group velocity $d\omega /dk$ identifies the direction in which a narrow wave packet carries energy, so its sign helps determine which boundary is inflow and explains why a numerically generated high-frequency branch can require a boundary condition different from that suggested by its phase velocity.

For a two-step method, the Fourier substitution produces an amplification polynomial of a multilevel finite difference scheme. Every root must lie in the closed unit disk, and unit-modulus roots must be simple, uniformly in the wavenumber. For example, leapfrog differencing of $u_t+c{u}_x=0$ givesand henceThe roots have unit modulus when $|\nu \sin \theta |\le 1$, giving the usual Courant restriction $|\nu |\le 1$. At the endpoint, a wavenumber with $|\nu \sin \theta |=1$ produces a repeated unit root and violates the uniform root condition; strict $|\nu |<1$ avoids this marginal linear growth. The second root is the familiar oscillatory computational mode, illustrating why a multilevel scheme requires its full amplification polynomial rather than a single multiplier.