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Apply the method to the Dahlquist test equation $y^{\prime }=\lambda y$ and set $z=h\lambda$. Its amplification roots satisfyAt $z=0$ the roots are $1$ and $5/11$. A root can leave the unit disk only through $\xi =e^{i\phi }$. Substitution gives the boundary equationDirect separation into real and imaginary parts shows that both branches satisfy $\mathrm{Re}z\ge 0$; equality occurs on the branch through $z=0$. Hence no root crosses the unit circle in $\mathrm{Re}z<0$. Moreover both roots tend to zero as $z\to \infty$ in the left half-plane. Consequently the method is A-stable, and it also strongly damps the infinitely stiff limit.

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.

An $s$-stage Runge-Kutta method has stages and updateOn the Dahlquist test equation $y^{\prime }=\lambda y$, elimination of the stages gives the stability functionThe linear stability domain is the set where $|R(z)|\le 1$. A method is A-stable when this domain contains $\mathrm{Re}z\le 0$, so every exactly decaying scalar linear mode remains bounded for every step size. It is L-stable when it is A-stable and $R(z)\to 0$ as $|z|\to \infty$ in the left half-plane; this extra limit strongly damps unresolved stiff modes.

The rational function $R$ makes several useful conclusions immediate. No explicit Runge--Kutta method is A-stable because its stability function is a nonconstant polynomial and is therefore unbounded on the negative real axis. The implicit midpoint rule has $R(z)=(1+z/2)/(1-z/2)$ and is A-stable, but $R(z)\to -1$, so it is not L-stable. The Backward Euler method has $R(z)=(1-z{)}^{-1}$ and is L-stable. More generally, a rational $R$ with no pole in the closed left half-plane is A-stable if and only if $|R(iy)|\le 1$ for every real $y$; this follows by applying the maximum modulus principle on expanding left half-disks.

Scalar linear stability does not by itself control nonlinear perturbations. Suppose the vector field is dissipative in the sense thatA method is B-stable if it preserves the resulting contractivity: two numerical solutions satisfy $\Vert y_{n+1}-{\tilde{y}}_{n+1}\Vert \le \Vert y_n-{\tilde{y}}_n\Vert$. A practical sufficient condition is algebraic stability of a Runge-Kutta method: $b_i\ge 0$ andTo prove the implication, let $D_i=Y_i-{\tilde{Y}}_i$ and $F_i=f(Y_i)-f({\tilde{Y}}_i)$. Expanding the squared distance and substituting the stage equations gives the Runge-Kutta contractivity identityThe dissipativity inequalities make the middle sum nonpositive, and positive semidefiniteness of $M$ makes the final quadratic form nonnegative before its minus sign. The distance therefore cannot increase. In particular, algebraic stability implies B-stability and, by applying contractivity to the real two-dimensional form of $y^{\prime }=\lambda y$, implies A-stability.

Important collocation families illustrate these notions. Gauss methods are A-stable, symmetric, and have order $2s$, but they do not damp infinitely stiff modes. Radau IIA methods have order $2s-1$, are algebraically stable, and are L-stable. These properties explain why A-stability controls unrestricted linear decay, L-stability is useful for stiff transients, and algebraic or B-stability is the stronger tool for nonlinear dissipative equations.

For a numerical evolution, numerical stability means that perturbations remain controlled over a fixed finite interval by a bound independent of sufficiently small discretization parameters. With perturbations ${\delta }_n$ to starting data and added step defects $d_n$, a typical estimate iswhere $C_T$ is independent of $h$. For a linear homogeneous recurrence, a uniform bound on powers of its update matrix supplies this control. von Neumann stability analysis and the energy method provide complementary tools.

Finite-interval numerical stability differs from absolute stability at a fixed step size. Applying the Euler method to $y^{\prime }=-2y$ multiplies a perturbation by $1-2h$ each step. At $h=0.2$ this is $0.6$, so perturbations shrink; at $h=1.2$ it is $-1.4$, so they alternate and grow. Nevertheless the Euler method converges as $h\to 0$ for this problem: failure at a coarse step does not contradict the convergence theorem. Absolute stability asks whether a selected step preserves boundedness of the Dahlquist test equation over arbitrarily many steps. MIT's analysis uses strict decay, $|1+h\lambda |<1$, rather than the boundary-inclusive boundedness convention.^[ref]

The strict decay convention for a linear stability domain demands that every numerical mode tends to zero on the Dahlquist test equation. For a one-step stability function it requires $|R(z)|<1$; for a multistep method it requires every amplification root to have modulus below one. This differs from a boundedness convention that allows simple unit-circle roots, and from mesh-uniform finite-time numerical stability. Cambridge numerical analysis notes use the decay convention. Explicitly specifying the convention is essential for methods with undamped modes.