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Use $N$ distinct periodic grid points $x_m=m/N$, $0\le m<N$, and $h=1/N$. Values at the two ends of the interval represent the same point, so the endpoint must not be stored twice. With indices interpreted modulo $N$, setThis replaces the Dirichlet matrix by the periodic discrete Laplacian. The normalized Fourier modes $N^{-1/2}{e}^{2\pi ijm/N}$ form an orthonormal basis and have amplification factorsEach satisfies $0<g_j\le 1$ for every $\mu >0$. The Parseval identity therefore gives $\Vert U^{n+1}{\Vert }_h\le \Vert U^n{\Vert }_h$ in the discrete L2 norm, uniformly in the grid and time step. Thus all positive Courant numbers remain stable. The constant mode has $g_0=1$, expressing preservation of the spatial mean; all nonconstant modes decay. The formula with modulo indices also handles small grids, where two neighbours can coincide.

Let ${\delta }_x^2{u}_m=u_{m-1}-2u_m+u_{m+1}$. For an exact smooth solution of the diffusion equation, the second-order central difference givesSince the Courant number $\mu =k/h^2$ is fixed, $h^2=O(k)$. Taylor expansion in time givesUsing $u_t=u_{xx}$ and $u_{tt}=u_{xxt}$ cancels the leading terms. Because $k{h}^2=O(k^2)$, the residual, and hence the local truncation error in the convention of the question, is

Keep the Courant number $\mu =\Delta t/\Delta x$ fixed and write $h=\Delta x$. An exact solution of $u_t=u_x$ is $u(x,t)=g(x+t)$. Substituting it into the update gives the defectwhere $s=x+t$ and $g^{(3)}$ is evaluated at $s$. The constant, first-derivative and second-derivative terms cancel. Dividing by $\Delta t=\mu h$ gives a second-order consistency error. Hence for fixed $0<\mu <1$, $\mu \ne 1/2$, the scheme has order two.

At the special value $\mu =1/2$, the update reduces to $u_m^{n+1}=u_{m+1}^{n-1}$. This is exactly the transport over two time steps, since $2\Delta t=\Delta x$, so the defect vanishes identically. With exact data at both starting levels it transports their values exactly; errors in those starting levels are still present, and the even and odd time subsequences evolve separately.

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.

The first scheme hasAt $\theta =\pi$,Since the Courant number obeys $\mu \ge 0$, stability therefore holds exactly for

Insert the Fourier mode $u_m^n=G^n{e}^{im\theta }$ into the scheme. Von Neumann stability analysis givesFor every physical Courant number $\mu \ge 0$, one has $0<G(\theta )\le 1$. Thus the method is unconditionally stable: the range is $\boxed{\mu \ge 0}$.