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Eigenvalue analysis proves mesh-uniform stability when it is accompanied by control of eigenvectors or an energy norm; the signs of eigenvalues alone are insufficient. For an evolution equation, the method of lines first produces $u_h^{\prime }=L_h{u}_h$. A time integrator then gives an amplification matrix $S_{h,\delta }$, for example $S_{h,\delta }=R(\delta L_h)$ for a Runge-Kutta method. The useful stability requirement on a fixed physical time interval iswhere $C_T$ is independent of the refining spatial and temporal meshes. The semidiscrete analogue is $\Vert e^{t{L}_h}{\Vert }_h\le C_T$ for $0\le t\le T$. These are bounds on propagation of initial errors and residuals, not merely statements that each finite matrix has a bounded solution over its own finite time interval.

If $L_h$ is a normal matrix in the chosen discrete inner product, it has a unitary eigenbasis. ThenFor dissipative problems it is sufficient that every $\delta {\lambda }_j$ lie in the integrator's linear stability domain. A diagonalizable family also suffices if its eigenvector matrices have uniformly bounded condition numbers: the same maximum is multiplied by $\Vert V_h\Vert \Vert V_h^{-1}\Vert$. Without that uniform bound the argument does not establish mesh stability. For finite-time estimates, an amplification modulus $1+O(\delta )$ can also be acceptable; strict contraction at every step is a stronger property.

For a constant-coefficient stencil on the full lattice or a periodic grid, von Neumann stability analysis provides the eigenvectors explicitly. A Fourier mode diagonalizes each translation-invariant spatial difference, giving a scalar symbol ${\lambda }_h(\theta )$ or a small matrix symbol for a system. The discrete Fourier transform and Parseval identity turn a uniform modal bound into an actual discrete L2 norm bound. The two-dimensional stencil in question 3 has $\mathrm{Re}{\lambda }_h=-(1-\cos \eta {)}^2/h\le 0$, so its continuous-time amplification is contractive; a later time integrator must still be checked. With boundaries one must analyze the boundary rows as well: a whole-line Fourier symbol cannot detect an unstable boundary closure.

For the heat equation $u_t=\kappa u_{xx}$ on a unit interval with zero Dirichlet data, the centred second difference has a sine eigenbasis and eigenvaluesExplicit Euler method multiplies mode $j$ by $1+\delta {\lambda }_j$. Requiring all factors to have modulus at most one yields $\kappa \delta /h^2\le 1/[2{\sin }^2((N-1)\pi /(2N))]$ on a given grid, and the sharp grid-independent sufficient limit isThese parabolic time-step restrictions express the growth of the largest-magnitude diffusion eigenvalue like $h^{-2}$. Backward Euler method instead has factor $(1-\delta {\lambda }_j{)}^{-1}$ and the Crank-Nicolson method factor $(1+\delta {\lambda }_j/2)/(1-\delta {\lambda }_j/2)$, both bounded by one for every $\delta \ge 0$. Their unconditional stability follows from A-stability, with much stronger damping of the shortest scales for backward Euler.

For $u_t=-a{u}_x$, $a>0$, explicit time stepping with a backward spatial difference givesThus the upwind finite difference scheme is contractive exactly for $0\le \nu \le 1$, a Courant–Friedrichs–Lewy condition matching the direction and speed of information propagation. A centred spatial difference has purely imaginary eigenvalues, and forward Euler gives $|S{|}^2=1+{\nu }^2{\sin }^2\theta >1$. Under a fixed positive hyperbolic Courant number, high-frequency data grow by a fixed factor in each of $O(1/h)$ steps, proving instability as the mesh is refined. This is not a claim that no exceptionally small coupled step can ever give a finite-time bound: if $\delta =O(h^2)$, the excess per step is only $O(\delta )$, but the usual hyperbolic scaling is unstable. Higher-order explicit integrators can possess nontrivial imaginary-axis stability intervals and are then suitable for skew-adjoint advection or the energy-form first-order wave equation.

Variable coefficients usually prevent Fourier diagonalization, but may preserve a self-adjoint spatial operator. The positive-face diffusion discretization in question 5 is an example. Its discrete energy identity makes $A+B$ symmetric negative definite, so a unitary eigenbasis still exists even without a closed eigenvalue formula. More generally, a dissipative operator $L_h$ in a discrete energy inner product obeyswhich directly proves a uniform semigroup estimate. The dissipative Cayley-transform contraction also follows without normality: for $\alpha \ge 0$,Taking $v=(I-\alpha L_h{)}^{-1}u$ proves contraction of the Crank–Nicolson factor. The contractivity of split Crank-Nicolson diffusion then follows from multiplying two contraction bounds; simultaneous eigenvectors of the two directional matrices are unnecessary.

The negative spectra do not imply uniform semidiscrete stability example shows what can go wrong. LetBoth eigenvalues are $-1$ for every mesh, but applying the propagator to the second unit vector at $t=1$ gives norm at least $1/(eh)$. The evolution is therefore not mesh-uniformly stable. For a discrete matrix, a unit-modulus eigenvalue with a nontrivial Jordan block gives powers growing polynomially in the step number. Even strictly interior eigenvalues can allow large transient growth when eigenvectors are badly conditioned. A pseudospectrum or numerical range of an operator, and especially a direct energy estimate, can reveal behavior missed by eigenvalues.

A resolvent estimate provides another way to see the needed uniformity. If $\Vert S_h^n\Vert \le C$ for every $n$, then for $|z|>1$,This shows directly that a uniform power estimate imposes more than the location of the spectrum. Conversely, resolvent-to-power estimates in varying dimension must themselves be uniform; a bound with constants depending on the number of grid points is not enough.

For a linear multistep method, each spatial eigenvalue gives an amplification polynomial, and all roots must satisfy the root condition for a multistep method, including simplicity of any unit-modulus root. Merely following the root approximating $e^{\delta \lambda }$ misses parasitic modes. The temporal and spatial analyses must therefore be combined with the appropriate zero-stability and uniform modal bounds.

Finally stability connects to convergence. If the exact grid samples have normalized residual ${\tau }^n$ and the error obeys $e^{n+1}=S_h{e}^n+\delta {\tau }^n$, iteration givesA residual tending uniformly to zero then gives convergence on $[0,T]$. For a well-posed linear initial-value problem and a consistent linear finite-difference approximation, this is the forward implication of the Lax equivalence theorem. The practical conclusion is to analyze the full operator including boundaries, select a time method whose stability region contains the relevant scaled spectrum, and verify that eigenvectors or an energy estimate make the resulting bounds uniform as the grid is refined.

Stability asks whether perturbations already present in data, arithmetic, or an earlier numerical step remain controlled under subsequent evolution. It complements consistency of a numerical method: consistency says that an exact smooth solution nearly satisfies one numerical step, while stability prevents the accumulated local defects from being amplified without bound. For a well-posed differential equation these two properties are what make convergence possible.

For a linear multistep method, zero-stability concerns the limit $h\to 0$. The roots of its first characteristic polynomial must lie in the closed unit disk, and every root on the unit circle must be simple. The Dahlquist equivalence theorem states that a consistent linear multistep method converges precisely when it is zero-stable. The parasitic root $5/11$ in Question 1 decays, whereas a repeated root at $1$ would turn small defects into secular growth.

Absolute stability instead fixes $z=h\lambda$ for the Dahlquist test equation. A one-step method has amplification factor $R(z)$ and absolute-stability regionA-stability means that $\mathcal{S}$ contains the entire closed left half-plane, matching every decaying scalar linear problem. L-stability additionally requires $R(z)\to 0$ as $z\to -\infty$, which suppresses unresolved fast transients in a stiff differential equation. Backward Euler is L-stable; the trapezoidal rule is A-stable but approaches $-1$ and can retain stiff oscillations; forward Euler is stable only in the disk $|1+z|\le 1$.

For nonlinear dissipative systems, scalar absolute stability can be insufficient. B-stability controls distances between numerical solutions of contractive differential equations. For Runge–Kutta methods, algebraic stability—nonnegative weights and positive semidefiniteness of $BA+A^{T}B-b{b}^T$—is a useful sufficient condition for B-stability. Question 2 shows how a single negative diagonal entry disproves it.

After spatial discretization of a partial differential equation, stability can be studied through the semidiscrete matrix. If its Hermitian part is nonpositive, the energy method proves contractivity without diagonalizing it. A skew-Hermitian generator instead conserves norm, as in the discrete Schrodinger problem of Question 3. Eigenvalues alone can be misleading for a non-normal matrix: transient growth may be large even when every eigenvalue lies in the left half-plane, so matrix norms, logarithmic norms, resolvent bounds, or a pseudospectrum may be needed.

For constant-coefficient Cauchy problems, von Neumann stability analysis inserts Fourier modes and bounds their amplification factors. Question 4 gives a typical result: centered diffusion contributes a negative real symbol and centered advection an imaginary symbol. Semidiscrete evolution is stable, yet forward Euler imposes both the parabolic restriction $\Delta t=O(\Delta x^2)$ and an advection-diffusion restriction. This illustrates a Courant–Friedrichs–Lewy condition: a stable spatial approximation need not remain stable under an arbitrary time stepper.

For a well-posed linear initial-value problem and a consistent finite-difference approximation, the Lax equivalence theorem identifies stability with convergence. Its hypotheses matter: it does not by itself cover nonlinear equations, inconsistent boundary closures, changing norms, or non-smooth solutions. Boundaries also defeat a naive whole-line Fourier argument; an energy estimate, normal-mode boundary analysis, or a discrete semigroup bound must include the boundary treatment.

Practical stability analysis therefore starts from the structure of the differential equation. Conservation laws favor unitary or symplectic methods, diffusion favors A- or L-stable implicit methods, monotone transport may require strong-stability-preserving time stepping and upwind fluxes, and stiff splitting requires attention to both the factors and their commutators. Stability does not guarantee accuracy: a heavily damped method may be stable while erasing the solution, and a stable computation with $h\lambda$ near the edge of its stability region may have an unacceptable phase error.

Let $M={\max }_{S}u$ and suppose $u(z_0)=M$. The sub-mean inequality on every closed disk contained in $S$ givesEquality holds throughout. If $u$ were strictly below $M$ at one point of the circle, upper semicontinuity would make it uniformly below $M$ on a small arc, contradicting equality of the average. Thus $u=M$ on every sufficiently small circle centered at $z_0$, and hence throughout a neighborhood of $z_0$.

The set $\{u=M\}$ is therefore open. It is also closed because upper semicontinuity makes $\{u<M\}$ open. Since the domain is connected and $\{u=M\}$ is nonempty, it is all of $S$. This proves the maximum principle for subharmonic functions.

LetThe resolvent set is open, and the resolvent of an element is operator-valued holomorphic on each of its components. Fix $z_0$ in the resolvent set. For any $\delta >0$, choose unit vectors $x,y$ such thatThe scalar function $f(z)=⟨R(z)x,y⟩$ is holomorphic. Its modulus is subharmonic, soLetting $\delta \downarrow 0$ proves that the resolvent norm is subharmonic on the resolvent component containing $z_0$.

Use the convention that the reciprocal resolvent norm is zero on the spectrum. Suppose a bounded component $U$ ofcontained no spectral point. A spectral point in its boundary would belong to the same pseudospectral component, so $\overline{U}$ is contained in the resolvent set. On $\partial U$ one has $\Vert R(z)\Vert \le {ϵ}^{-1}$, whereas inside $U$ one has $\Vert R(z)\Vert >{ϵ}^{-1}$. Continuity on the compact set $\overline{U}$ makes the resolvent norm attain a maximum at an interior point. The subharmonic maximum principle would make it constant, contradicting its boundary values. Hence every bounded component of the pseudospectrum contains spectrum: