= 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 method
$$
u^{n+1}=Q_hu^n,
$$
write $H_h(\theta)$ for its <amplification factor>. The <Parseval identity> gives
$$
\|Q_h^nu^0\|_2^2
=\int_{-\pi}^{\pi}|H_h(\theta)|^{2n}|\widehat u^0(\theta)|^2\,d\theta.
$$
Consequently the exact necessary and sufficient condition for stability on every bounded time interval $0\leq n\Delta t\leq T$ is
$$
\boxed{\operatorname*{ess\,sup}_\theta|H_h(\theta)|
\leq e^{C\Delta t}}
$$
with $C$ 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 $|H_h(\theta)|\leq1$.
For a semidiscretization $u_t=A_hu$ with scalar symbol $a_h(\theta)$, the corresponding exact criterion is
$$
\boxed{\operatorname*{ess\,sup}_\theta\operatorname{Re}a_h(\theta)\leq C.}
$$
Indeed, the Fourier multiplier of the solution operator is $e^{ta_h(\theta)}$. For systems, eigenvalues alone cease to be sufficient when the matrix symbol is <non-normal matrix>[non-normal]; the necessary and sufficient statement is the uniform bound $\|e^{tA_h(\theta)}\|\leq C_T$.
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 $u_t+cu_x=0$ gives
$$
u_j^{n+1}=u_j^{n-1}-\nu(u_{j+1}^n-u_{j-1}^n),
\qquad \nu=\frac{c\Delta t}{\Delta x},
$$
and hence
$$
G^2+2i\nu\sin\theta\,G-1=0.
$$
The roots have unit modulus when $|\nu\sin\theta|\leq1$, giving the usual <Courant number>[Courant] restriction $|\nu|\leq1$. 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.
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