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Convergence of a numerical method on $[0,T]$ means that its grid error obeys ${\max }_{0\le nh\le T}|y_n-y(nh)|\to 0$ as $h\to 0$, with consistent initial values.

First, the implicit step of Backward Euler method is well-defined for $hL<1$: the map $z\mapsto y_n+hf(t_{n+1},z)$ is a contraction mapping on $\mathbb{R}$, so the Banach fixed-point theorem gives a unique solution. Let $M={\max }_{[0,T]}|y^{\prime \prime }|$, finite under the stated smoothness. The exact solution's step defect isso $|d_n|\le M{h}^2/2$. Subtract the exact and numerical step equations. The Lipschitz condition givesWith exact initial data, sum this geometric progression to obtainFor $hL\le 1/2$ and $nh\le T$, $-\log (1-hL)\le 2hL$, so the bracket is at most $e^{2LT}-1$. Thus the global error is $O(h)$, uniformly on $[0,T]$, and Backward Euler method converges with order one. A vanishing initial error adds only $(1-hL{)}^{-n}|e_0|$.

Order one is sharp: for $y^{\prime }=y$, $y(0)=1$ and $nh=T$, $y_n=(1-h{)}^{-n}=e^T(1+Th/2+O(h^2))$.

For the Dahlquist test equation $y^{\prime }=\lambda y$, the stability function is $R(z)=1/(1-z)$, $z=h\lambda$. Therefore its linear stability domain isThe strict inequality gives decay; equality gives bounded amplification. The whole closed left half-plane lies in this region, so Backward Euler method is A-stable.

A stiff differential equation can contain rapidly decaying modes as well as a slowly changing solution of interest. An explicit scheme may need a very small step solely to keep those already small fast modes from numerical growth. A-stability addresses this restriction through the Dahlquist test equation $y^{\prime }=\lambda y$, whose exact solution decays when $\mathrm{Re}\lambda <0$.

For a one-step method, write $y_{n+1}=R(z)y_n$, $z=h\lambda$. Its linear stability domain consists of the points where the update is defined and $|R(z)|\le 1$. A-stability means that this domain contains the closed left half-plane. This is a scalar linear stability property, not an order condition, an existence theorem for an implicit solve, or a general nonlinear contractivity assertion.

For a rational approximation to the exponential, consistency requires $R(z)=1+z+O(z^2)$, and order $p$ requires $R(z)-e^z=O(z^{p+1})$. Poles must be absent from the relevant half-plane. The Forward Euler method has $R=1+z$ and the disk $|1+z|\le 1$, so negative real modes require $h|\lambda |\le 2$; it is not A-stable. The Backward Euler method has $R=(1-z{)}^{-1}$, and $|1-z|\ge 1$ on the left half-plane, so it is A-stable. The trapezoidal rule and the implicit midpoint rule both have, on the scalar linear problem,They are second-order and A-stable despite being different nonlinear methods. No nonconstant polynomial approximation is A-stable, because its modulus grows without bound on the negative real axis. In particular every nontrivial explicit Runge-Kutta method has a bounded-step stability restriction.

A-stability prevents growth but need not eliminate extremely stiff modes: the trapezoidal factor tends to minus one as $z\to -\infty$. L-stability additionally requires $R(z)\to 0$ within the left half-plane. Backward Euler is L-stable; trapezoidal and implicit midpoint are not. The distinction explains persistent alternating transients in otherwise stable Crank-Nicolson calculations. Accuracy still constrains useful step sizes even for an A-stable method.

A linear multistep method is characterized by polynomials $\rho ,\sigma$, and its scalar modes satisfy $\rho (\zeta )-z\sigma (\zeta )=0$. There is generally no single scalar amplification factor: every root must lie in the closed unit disk and unit roots must be simple. At $z=0$, this is the zero-stability root condition. Consistency plus zero-stability gives convergence by the Dahlquist equivalence theorem, with suitable starting values. A-stability demands the amplification root condition throughout the left half-plane, including the zero-step condition. The Second Dahlquist barrier says that an irreducible A-stable linear multistep method has order at most two, and an explicit linear multistep method cannot be A-stable.

Examples include backward Euler, trapezoidal, and the second-order backward differentiation formulaIts zero-step roots are $1,1/3$. Its unit-circle boundary quotient has real part $(1-\cos \theta {)}^2\ge 0$, and the leading coefficient is nonzero on the left half-plane, giving A-stability by the same continuation argument as in Question 2. Conversely the third-order member of that question fails A-stability. A formal high order without zero-stability, or after silently cancelling a problematic unit factor, does not evade the barrier.

For an implicit Runge-Kutta method, the stages on the test equation satisfy $Y=e{y}_n+zAY$. Whenever these stages are solvable, its stability function isThis rational function is used to test scalar A-stability; the Butcher order conditions separately establish nonlinear order. The Lobatto IIIA method in Question 3 is an A-stable fourth-order example with $R$ the diagonal degree-two Padé approximant. More generally the $s$-stage Gauss--Legendre Runge-Kutta method has order $2s$ and is A-stable, with the diagonal Padé approximation to $e^z$; its stiff-limit factor is nonzero. The $s$-stage Radau IIA method has order $2s-1$ and is L-stable. Thus Runge-Kutta stages permit arbitrarily high A-stable order, unlike irreducible linear multistep formulas.

For a normal matrix, scalar amplification bounds transfer through unitary diagonalization. For a non-normal matrix, eigenvalue information alone does not assert Euclidean contraction: transient growth and eigenvector conditioning can matter. Direct energy method arguments, such as the dissipative Cayley-transform proof in Question 1, use the symmetric part and are more informative in that setting.

For nonlinear dissipative vector fields, the stronger B-stability property asks that distances between two numerical solutions not increase. A standard sufficient condition for Runge-Kutta methods is algebraic stability: $b_i\ge 0$ andis positive semidefinite. It follows from the Runge-Kutta contractivity identity when the implicit stages exist. The three-stage Lobatto IIIA method is A-stable but has the first diagonal entry of that matrix equal to $-1/36$, so scalar A-stability should not be confused with that sufficient nonlinear condition. Stage solvability of an implicit Runge-Kutta method remains a separate issue and computational cost of the implicit equations must be considered. A-stability removes a scalar decay-mode stability restriction; order, stiff damping, nonlinear stability and solvability remain distinct properties.

Consider an initial value problem $y^{\prime }=f(t,y)$ on $[0,T]$, with a Lipschitz continuous vector field in the solution variable and enough smoothness for the stated error expansions. Convergence of a numerical method meanswith consistent starting values. An order of a numerical method $p$ quantifies this approximation: with sufficiently accurate starting data and the required stability, the global error is $O(h^p)$. Stability of a numerical method controls the amplification and accumulation of perturbations; it is the link between a small local defect and small global error.

For a one-step formula $Y_{n+1}={\Phi }_h(t_n,Y_n)$, define the unscaled local truncation errorThe method has local order $p$ if $\Vert d_{n+1}\Vert \le C{h}^{p+1}$ uniformly along sufficiently smooth solutions. If the defect is divided by $h$, its order is instead $O(h^p)$; the convention must be specified. Assume also the perturbation estimateThe global error $e_n=Y_n-y(t_n)$ then satisfiesIterating, or applying the discrete Gronwall inequality, givesThere are $O(1/h)$ steps on a fixed interval, so stable propagation turns $O(h^{p+1})$ local defects into $O(h^p)$ accumulated error. Starting error $O(h^p)$ is needed to retain order $p$. This explains why matching an exact Taylor expansion is insufficient unless errors are controlled during repeated steps.

For a fixed-coefficient linear multistep method, writeThe consistency of a numerical method conditions are $\rho (1)=0$ and ${\rho }^{\prime }(1)=\sigma (1)$. The higher order conditions for a linear multistep method arewith right side zero for $q=0$; exact order $p$ means first failure at $q=p+1$. They come from inserting the exact solution and comparing its Taylor expansion.

Errors at $h=0$ follow the homogeneous recurrence determined by $\rho$. Zero-stability is equivalent to the root condition for a multistep method: all roots of a polynomial have modulus at most one, and every unit-modulus root is simple. Roots outside the disk give exponential growth in the number of steps; repeated unit roots give polynomial growth in that number. Repeated interior roots are harmless because their polynomial factors are dominated by decay. Under the usual Lipschitz continuity and initialization assumptions, the Dahlquist equivalence theorem givesFor a zero-stable order-$p$ linear multistep method, local defects and starting errors of the stated orders yield global error $O(h^p)$ by a recurrence bound and the discrete Gronwall inequality. Every required starting value matters.

A concrete failure of “consistency implies convergence” isHere $\rho =(\zeta -1)(\zeta -2)$ and $\sigma =-1$, so $\rho (1)=0$ and ${\rho }^{\prime }(1)=\sigma (1)=-1$: the method is consistent, of order one. But its root $2$ violates zero-stability. For $y^{\prime }=0$ with exact solution zero and starting values $Y_0=0$, $Y_1=h^2$, the computed solution is $Y_n=h^2(2^n-1)$. These starting errors vanish, yet at $n\approx T/h$ the error diverges. Local accuracy cannot compensate for an unstable recurrence.

A different meaning of stability concerns a fixed step on decaying equations. The Dahlquist test equation $y^{\prime }=\lambda y$ introduces $z=h\lambda$. A one-step method gives $Y_{n+1}=R(z)Y_n$ and is absolutely stable when $|R(z)|\le 1$. For a linear multistep method, the corresponding amplification polynomial of a multistep method is $\rho (\zeta )-z\sigma (\zeta )$, and every root must satisfy the bounded root condition, not just the root approximating $e^z$. Strict asymptotic decay requires the roots to have modulus less than one for $\mathrm{Re}z<0$. This distinction matters at reducible methods with an undamped parasitic mode, as in question 1(c).

The Forward Euler method has order one and $R(z)=1+z$, so its linear stability domain is $|1+z|\le 1$. For $y^{\prime }=-\kappa y$, with $\kappa >0$, it requires $0\le h\kappa \le 2$. It converges as $h\to 0$, yet an excessively large step can produce growing oscillations on an exactly decaying problem. The Backward Euler method has the same order, but $R(z)=(1-z{)}^{-1}$ makes it A-stable and L-stable. For a stiff differential equation, fast decay can impose a severe explicit time-step restriction even when accuracy of the slow dynamics would permit a much larger step.

The trapezoidal rule has order two and is A-stable, but is not L-stable, since its stability function tends to $-1$ along the negative real axis. The Second Dahlquist barrier limits an irreducible A-stable linear multistep method to order at most two. This does not limit implicit Runge-Kutta methods: question 4's Lobatto IIIA method is fourth-order and A-stable. Conversely, question 1's fourth-order member is convergent but not A-stable. These examples show that order of a numerical method, zero-stability and A-stability answer different questions.

For nonlinear dissipative problems, B-stability controls distances between numerical solutions; algebraic stability of a Runge-Kutta method is a sufficient coefficient criterion. Question 4 shows that A-stability alone does not imply that coefficient criterion. Variable time steps require additional analysis of step ratios and error propagation, and stiff problems may require uniform-in-stiffness error estimates beyond the fixed-problem convergence result. The useful chain is local accuracy plus the appropriate perturbation bound, giving global accuracy; absolute and nonlinear stability then determine which practical time steps preserve the intended dynamics.

Apply the numerical method to the scalar test equation $y^{\prime }=\lambda y$, with $z=h\lambda$. Its linear stability domain consists of those $z\in \mathbb{C}$ for which the discrete solution stays bounded as the step count tends to infinity. For a one-step method with stability function $R$, this isIf asymptotic decay is required, use $|R(z)|<1$ instead; this only changes the boundary conventions in part (b). A method is A-stable if its stability domain contains the left half-plane, so every decaying linear test equation is stable for every positive step size. For a multistep method, the corresponding definition uses the root condition for a multistep method for every characteristic root, including simplicity of unit-modulus roots.

Expand the second stage by the Taylor theorem, writing $f=f(t_n,y_n)$ and $D_{y}f$ for the derivative in the state variable:The update therefore agrees with the exact solution's Taylor series:Thus the one-step local truncation error is $O(h^3)$, and the method has order at least two under the usual smoothness and Lipschitz continuity assumptions.

For $f=\lambda y$, the two-stage Runge-Kutta method givesOn the real axis, $R(z)=\frac{1}{2}(z+1{)}^2+\frac{1}{2}\ge \frac{1}{2}$, so $|R(z)|\le 1$ is equivalent to $(z+1{)}^2\le 1$. HenceThe decay convention gives $(-2,0)$. Since negative real values such as $z=-3$ lie outside the linear stability domain, the method is not A-stable. Its order is exactly two, since $R(z)$ lacks the cubic term of $e^z$.

Assume $f$ is sufficiently smooth, for example $C^3$ near the exact trajectory. Write $f,f^{\prime },f^{\prime \prime }$ at the current value, with $f^{\prime }$ a linear map and $f^{\prime \prime }$ a bilinear map when the ordinary differential equation is vector valued. Taylor expansion of the stages givesThe exact flow has Taylor expansionMatching the terms through $h^3$ in the Runge-Kutta method gives the sufficient third-order conditions for an explicit Runge-Kutta methodFor scalar $f$, replace $f^{\prime \prime }[f,f]$ by $f^{\prime \prime }{f}^2$ and $(f^{\prime }{)}^{2}f$ by $(f^{\prime }{)}^{2}f$ in the scalar sense. The resulting local error is $O(h^4)$; under a local Lipschitz continuity assumption and bounded derivatives on a finite time interval, the standard one-step error recurrence gives global error $O(h^3)$, hence third order.

On the linear test equation $y^{\prime }=\lambda y$, putting $z=h\lambda$ gives the stability functionThus $\boxed{|R(-5/2)|=47/48<1}$, so $-5/2$ lies inside the linear stability domain. The four order conditions are compatible, for example $a_1=1/2,a_2=-1,a_3=2$ and $(b_1,b_2,b_3)=(1/6,2/3,1/6)$.

A Runge-Kutta method advances $y^{\prime }=f(t,y)$ through stagesThe Butcher tableau records $A,b,c$, normally with $c=Ae$. Linear absolute stability tests the amplification of a linear decay mode; nonlinear B-stability compares distances between two solutions of a dissipative vector field. Both describe propagation of perturbations, but they impose different conditions.

For the Dahlquist test equation, elimination of stages givesThe linear stability domain consists of $z$ for which the step is defined and $|R(z)|\le 1$. Poles or singular stage systems must be excluded, even if an output formula has a formal cancellation. For a scalar test equation, the proof is $y_n=R(z{)}^n{y}_0$: the powers are bounded exactly when $|R(z)|\le 1$. For $y^{\prime }=Ly$, the update is $R(hL)$. If $L$ is a normal matrix, unitary diagonalization of a normal matrix makes the operator norm equal to the largest scalar amplification modulus. For a diagonalizable matrix, powers are bounded by the condition number of the eigenvector matrix times the maximal scalar power. These constants must remain uniform for mesh-dependent systems. A defective unit-modulus eigenvalue can instead produce a growing Jordan block; the scalar spectral test alone does not control a general matrix.

An A-stable method accepts the entire closed left half-plane. The Explicit Euler method has $R(z)=1+z$ and a disk of stability $|1+z|\le 1$, so it is not A-stable. No nonconstant polynomial stability function can be A-stable, since it is unbounded on the negative real axis; this excludes all consistent explicit Runge-Kutta methods. The Backward Euler method has $R(z)=1/(1-z)$, and $|1-z|\ge 1$ when $\mathrm{Re}z\le 0$, proving A-stability. It is also L-stable, because $R(z)\to 0$ for large left-half-plane $z$.

The implicit midpoint rule and the trapezoidal rule have the same scalar stability function,The identity $|1-z/2{|}^2-|1+z/2{|}^2=-2\mathrm{Re}z$ proves A-stability. On the imaginary axis their amplification has unit modulus, and along the negative real axis it tends to $-1$. Thus neither is L-stable; very stiff decay can persist as alternating numerical values. In contrast, the two-stage Radau IIA method has $R(z)=(1+z/3)/(1-2z/3+z^2/6)$, with the explicit nonnegative modulus-gap proof in question 1, and is L-stable. This illustrates why L-stability matters beyond A-stability for a stiff differential equation.

For nonlinear problems, assume a real or complex Euclidean inner product and the dissipativity hypothesisfor all relevant states at each common time. Exact solutions satisfyA B-stable Runge-Kutta method reproduces this nonexpansion of distances whenever its stages exist. Put $d=y_n-{\tilde{y}}_n$, $D_i=Y_i-{\tilde{Y}}_i$, and $G_i=f(t_n+c_{i}h,Y_i)-f(t_n+c_{i}h,{\tilde{Y}}_i)$. Then $D_i=d+h{\sum }_j{a}_{ij}{G}_j$. Expansion of the output squared norm, followed by substituting $d=D_i-h{\sum }_j{a}_{ij}{G}_j$ in its linear terms, proves the Runge-Kutta contractivity identityIf $b_i\ge 0$ and $M=(m_{ij})$ is a positive semidefinite matrix, the first sum is nonpositive by dissipativity, and the second is nonnegative: expand each coordinate of the stage vectors to express it as a sum of nonnegative quadratic forms. These are exactly the conditions for algebraic stability of a Runge-Kutta method. Hence algebraic stability implies B-stability. Applying B-stability to the linear dissipative problem $y^{\prime }=\lambda y$, or its two-dimensional real form, also proves A-stability whenever stages are well defined.

The Backward Euler method has $b=A=1$, hence $M=1$; the implicit midpoint rule has $b=1$, $A=1/2$, hence $M=0$. Both are algebraically stable and B-stable. For the two-stage Radau IIA method,a positive semidefinite matrix, while both weights are positive. It therefore combines third-order accuracy, B-stability, and L-stability. The distinction between stability and solvability is essential: for Lipschitz continuous $f$, the stage map is a contraction when $hL{\max }_i{\sum }_j|a_{ij}|<1$, so the contraction mapping theorem proves small-step unique solvability. The preceding algebraic stability proof is an estimate for existing stages; it is not a theorem asserting stage existence for arbitrary step sizes and domains.

Linear A-stability is insufficient for nonlinear B-stability. The trapezoidal rule fails B-stability, even though its scalar stability function is identical to that of the B-stable implicit midpoint rule. Its Runge-Kutta method coefficients areFailure of this sufficient algebraic stability test alone would not prove failure of B-stability. An actual counterexample does: take $f(y)=-y^3$ and $h=2$. Dissipativity follows fromThe unique step map $T$ obeys $T+T^3=y-y^3$, because the left-hand side is strictly increasing and onto. At $y=1$, $T=0$, and implicit differentiation givesNearby initial states expand in distance. This proves a true nonlinear failure rather than merely failure of a coefficient criterion.

Finally, stability of a numerical method on finite intervals connects these contractivity properties to accuracy. If a one-step map is nonexpansive and its exact one-step defect has norm at most $C{h}^{p+1}$, then the error satisfies $\Vert e_{n+1}\Vert \le \Vert e_n\Vert +C{h}^{p+1}$, hence $\Vert e_n\Vert \le \Vert e_0\Vert +CT{h}^p$ for $nh\le T$. A Lipschitz step bound $1+Kh$ gives the same order using the discrete Gronwall inequality. For a stiff differential equation, the value of B-stability is that the propagation estimate need not grow with a large negative dissipative rate. The accuracy constant still needs appropriate smoothness and uniform derivative bounds; stability alone does not establish a mesh-uniform error order.

For a fully discrete linear partial differential equation evolution, write $U^{n+1}=C_{h,k}{U}^n$ in a specified discrete norm. Finite-time stability of a numerical method meanswith $C_T$ independent of the allowed meshes. For a multistep method, augment the state with its time-history values and include a stable starter. Bounds may grow as $e^{\omega T}$ when the continuous problem has growth; demanding decay for all time would be a stronger assertion. The treatment of initial data, forcing, Dirichlet boundary conditions, and the norm is part of the hypothesis, not a detail supplied by an interior calculation.

For constant coefficients on an infinite or periodic uniform grid, Von Neumann stability analysis substitutes the Fourier mode $U_j^n=G(\theta {)}^n{e}^{ij\theta }$. The discrete Fourier transform and the Parseval identity turn a uniform multiplier estimate into a discrete L2 norm estimate. In a one-step scalar scheme, $|G(\theta )|\le 1$ proves contractivity, while $|G(\theta )|\le 1+\omega k$ gives finite-time stability. In a multilevel scheme, every root of a polynomial in its amplification polynomial of a multilevel finite difference scheme matters, as does uniform control of the associated companion matrix.

For the Forward Euler diffusion scheme with $\mu =k/h^2$,Thus $|G|\le 1$ for all frequencies exactly whenSufficiency follows since $G\in [-1,1]$; necessity follows from the highest-frequency mode $\theta =\pi$ on even periodic grids. This gives the usual parabolic mesh restriction $k\le h^2/2$. For the Backward Euler diffusion scheme,so every $\mu \ge 0$ is stable. The Crank--Nicolson method hasagain of modulus at most one for all $\mu \ge 0$, but stiff modes approach $-1$ rather than zero. These are PDE manifestations of the A-stability and L-stability distinctions for time integration.

For advection $u_t+a{u}_x=0$, assume $a>0$ and set $\nu =ak/h$. The upwind finite difference scheme is $U_j^{n+1}=(1-\nu )U_j^n+\nu U_{j-1}^n$, withIt is contractive when $0\le \nu \le 1$. This also has a direct maximum-norm proof: each newest value is a convex combination of two old values. The resulting Courant–Friedrichs–Lewy condition expresses that the numerical domain of dependence covers the physical one. For negative $a$, the upwind direction must be reversed. By contrast, Forward Euler method time stepping with the centered first finite difference has $G=1-i\nu \sin \theta$ and $|G|>1$ for nonzero $\nu \sin \theta$. Under fixed nonzero $\nu$ refinement it is unstable, since a fixed nontrivial mode grows geometrically over $T/k$ steps. This is not a claim of instability under every imaginable mesh coupling: $k=O(h^2)$ instead bounds its spurious finite-time growth, since $\log |G|\le a^2{k}^2/(2h^2)$.

The leapfrog advection scheme illustrates a multilevel subtlety. Its amplification polynomial of a multilevel finite difference scheme isFor $|\nu |<1$, both roots of a polynomial have unit modulus and separation at least $2\sqrt{1-{\nu }^2}$. A uniformly conditioned eigenbasis of the companion matrix then proves power boundedness of a two-level Fourier scheme for arbitrary history data. At $|\nu |=1$ and a grid admitting $\theta =\pi /2$, the roots coincide on the unit circle, and a Jordan block produces $n{\xi }^n$ growth. Thus the endpoint fails the ordinary arbitrary-history root condition for a multistep method; checking only that both moduli equal one misses the instability. Restricting the starter or filtering the parasitic mode is an additional hypothesis.

The energy method handles variable coefficients and finite boundaries for which Fourier stability analysis may be unavailable. For the Backward Euler diffusion scheme with homogeneous Dirichlet boundary conditions, define $\Vert U{\Vert }_h^2=h{\sum }_j|U_j{|}^2$. The discrete summation by parts identity isTake the inner product of $U^{n+1}-U^n=k{L}_h{U}^{n+1}$ with $2U^{n+1}$. The elementary identity $2\mathrm{Re}⟨x-y,x⟩=\Vert x{\Vert }^2-\Vert y{\Vert }^2+\Vert x-y{\Vert }^2$ givesThis proves unconditional contractivity including the actual boundary treatment. More generally, a dissipative operator $L_h$ gives $(I-k{L}_h{)}^{-1}$ of operator norm at most one: with $V=(I-k{L}_h)U$, dissipativity implies $\mathrm{Re}⟨V,U⟩\ge \Vert U{\Vert }^2$, and the Cauchy-Schwarz inequality gives $\Vert U\Vert \le \Vert V\Vert$. In finite dimensions this also proves invertibility. For the second-order backward differentiation formula, the BDF2 discrete energy identity used in question 4 controls both time levels and proves unconditional diffusion stability. A repeated root strictly inside the disk is harmless here; the energy argument supplies the uniform bound without a singular eigenvector formula.

A second technique is eigenvalue stability analysis of a finite difference method. Under the method of lines, a time integrator advances $U^{\prime }=L_{h}U$ by $R(k{L}_h)$. If $L_h$ is a normal matrix, the scalar linear stability domain criterion on all $k{\lambda }_j(L_h)$ controls its powers exactly in the corresponding L2 norm. For the centered Dirichlet discrete Laplacian, its eigenvalues lie between $-4/h^2$ and zero; intersecting this interval with a time integrator's stability interval determines its mesh restriction. A uniform bound on the diagonalization of a matrix is needed for nonnormal diagonalizable systems. Eigenvalues alone can be misleading. For example,has both eigenvalues inside the disk for $0<k<1$, but the upper-right entry of $C_{h,k}^n$ is $nk(1-k{)}^{n-1}/h$. Taking $k=h$, $nk\to T>0$, this grows like $T{e}^{-T}/h$. Hence stable scalar eigenvalues do not give mesh-uniform stability. Boundary closures can introduce precisely the extra growth that an interior Fourier symbol overlooks, so boundary stability of a finite-difference method must be checked separately.

For nonlinear spatial discretizations, a useful replacement for Fourier stability analysis is a convexity argument. If the Forward Euler method map $E_k(U)=U+kF(U)$ is nonexpansive in a chosen norm for $k\le k_{\mathrm{FE}}$, the second-order strong stability preserving Runge-Kutta methodis also nonexpansive under the same restriction. Indeed, for two inputs, the triangle inequality gives $\Vert U_{\mathrm{new}}-{\tilde{U}}_{\mathrm{new}}\Vert \le \frac{1}{2}\Vert U-\tilde{U}\Vert +\frac{1}{2}\Vert E_k(Y)-E_k(\tilde{Y})\Vert \le \Vert U-\tilde{U}\Vert$. Its Taylor expansion is $U+kF(U)+k^2{F}^{\prime }(U)F(U)/2+O(k^3)$, proving order two. Such strong stability preserving Runge-Kutta methods transfer suitable forward-step bounds without relying on a linear spectral calculation.

Finally, consistency of a numerical method explains what stability buys. For a linear Hadamard well-posed problem, the Lax equivalence theorem equates convergence of a consistent discretization with its stability, under the stated approximation-space and norm hypotheses. Directly, if the error satisfies $e^{n+1}=C_{h,k}{e}^n+k{\tau }^n$, iteration gives the discrete Duhamel principleThus vanishing normalized local truncation error and initial error yield convergence. With a nonlinear Lipschitz step estimate, the discrete Gronwall inequality plays the same role. A stable but inconsistent stencil, such as the literal defective formulas in questions 3 and 4 under their usual refinement, is not rescued by any amplification bound. The decisive checks are a mesh-uniform evolution bound, a valid boundary closure, and consistency with the actual PDE.

An $s$-stage Runge-Kutta method for $y^{\prime }=f(t,y)$ is defined byThe Butcher tableau records $A=(a_{ij})$, $b$ and $c$; internal consistency usually sets $c=Ae$, and first-order consistency requires $b^{T}e=1$. A strictly lower triangular $A$ gives an explicit method, while other stage dependencies generally require an implicit Runge-Kutta method. The stage solvability of an implicit Runge-Kutta method must be checked separately: if $f$ is globally Lipschitz continuous in $y$ with constant $L$, the stage fixed-point map is a contraction in the maximum stage norm whenever $hL{\max }_i{\sum }_j|a_{ij}|<1$. This is a sufficient small-step condition, not a restriction intrinsic to the definitions of A-stability or B-stability.

Linear stability begins with the Dahlquist test equation $y^{\prime }=\lambda y$. Solving the stages givesThe linear stability domain consists of test values for which the stage system is well defined and $|R(z)|\le 1$. A-stability means this includes the closed left half-plane. For $\mathrm{Re}\lambda <0$, this gives stability with no scalar decay-mode step restriction. For a normal matrix $L$ in $y^{\prime }=Ly$, unitary diagonalization of a normal matrix reduces the norm estimate to these scalar factors. For a non-normal matrix, eigenvalues alone do not establish a uniform norm bound: eigenvector conditioning and transient amplification matter. Mesh-uniform estimates for discretized PDEs must control operator powers, not only their spectra.

Every consistent explicit Runge-Kutta method has a nonconstant polynomial stability function and therefore cannot be A-stable, since that polynomial is unbounded on the negative real axis. For example, Forward Euler method has $R(z)=1+z$, stability disk $|1+z|\le 1$, and restriction $0\le hq\le 2$ for $\lambda =-q<0$. For a pure imaginary mode $\lambda =i\omega \ne 0$, $|1+ih\omega |>1$, so it is unstable for every positive step. The classical fourth-order Runge-Kutta method instead hasIt is stable on the imaginary axis exactly for $|q|\le 2\sqrt{2}$, an illustrative conditional stability range for oscillatory evolution.

Backward Euler method has $R(z)=(1-z{)}^{-1}$ and is A-stable. L-stability adds $R(z)\to 0$ as $|z|\to \infty$ in the left half-plane; backward Euler satisfies this and strongly damps unresolved rapidly decaying modes in a stiff differential equation. The implicit midpoint rule and the trapezoidal rule both have $R(z)=(1+z/2)/(1-z/2)$, so they are A-stable but not L-stable. They preserve the modulus of a pure imaginary test mode, but their stiff-decay limit is $-1$, leaving oscillatory numerical remnants. The three-stage Lobatto IIIA method has the fourth-order rational function found in Question 2 and likewise lacks L-stability. Thus high order and A-stability do not themselves imply efficient stiff damping.

Nonlinear stability measures differences between solutions of the same equation. A one-sided Lipschitz condition isDifferentiating the squared difference and applying the Gronwall inequality gives $\Vert y(t)-\tilde{y}(t)\Vert \le e^{\nu (t-t_0)}\Vert y(t_0)-\tilde{y}(t_0)\Vert$. A dissipative vector field has $\nu \le 0$ and is contractive. A B-stable method preserves this property for every positive step size for which the stages are well defined:This includes time-dependent vector fields when dissipativity holds at each common time argument. Applying it to scalar linear dissipative fields shows that B-stability implies A-stability when the linear stages are well defined; the converse fails.

The standard sufficient theorem is algebraic stability implies B-stability, subject to stage solvability. Define $B=\mathrm{diag}(b)$ and $M=BA+A^{T}B-b{b}^T$. Algebraic stability of a Runge-Kutta method means $b_i\ge 0$ and $M$ is a positive semidefinite matrix. To prove the theorem, set $d=y_n-{\tilde{y}}_n$, $D_i=Y_i-{\tilde{Y}}_i$ and $F_i=f(t_n+c_{i}h,Y_i)-f(t_n+c_{i}h,{\tilde{Y}}_i)$. Expanding the output difference and using $D_i=d+h{\sum }_j{a}_{ij}{F}_j$ yields the Runge-Kutta contractivity identityDissipativity makes the first sum nonpositive, while positive semidefiniteness makes the last quadratic form nonnegative. The latter follows by factoring $M=Q^{T}Q$ and writing it as a sum of squared norms of linear combinations of the $F_i$. This proves the claimed contraction. It is a sufficient theorem; absence of algebraic stability is not, by itself, a proof of failure of B-stability for every representation.

Several examples make the distinctions concrete. For backward Euler, $b=1$, $A=(1)$, so $M=(1)$: it is both L-stable and B-stable. For implicit midpoint, $b=1$, $A=(1/2)$, so $M=(0)$: it is B-stable and preserves squared norms for a skew-Hermitian linear equation, but lacks stiff damping. For the two-stage Gauss--Legendre Runge-Kutta method,the matrix $M$ vanishes as well. The collocation theorem gives order four, and its stability function is exactly the same as the three-stage Lobatto IIIA function. Yet the Gauss method is algebraically stable, whereas that Lobatto method is not. Identical scalar linear stability functions need not imply identical nonlinear stability properties.

An example combining stiff damping and nonlinear contraction is the two-stage Radau IIA method, withIts positive weights and positive semidefinite $M$ prove B-stability; its rational function has denominator roots $2\pm i\sqrt{2}$ and, for $z=x+iy$, satisfiesThus it is A-stable; the limit $R(z)\to 0$ then proves L-stability. The standard Radau IIA collocation theorem gives order $2s-1$, here three. These properties explain its usefulness for stiff nonlinear equations.

Finally, A-stability alone does not imply B-stability. The trapezoidal rule fails B-stability even for the scalar dissipative equation $y^{\prime }=-y^3$. With $h=1$, its step map $T$ is uniquely defined byAt $y=\sqrt{2}$, $T(y)=0$. Implicit differentiation givesThus nearby starting values are separated by approximately twice their original distance after one step, although the exact flow is contractive. This example and the contractivity theorem show why scalar linear analysis, stiff damping and genuinely nonlinear norm estimates answer different stability questions.

A Runge-Kutta method with stage matrix $A$, weights $b$ and step size $h$ isIts order measures accuracy as $h\to 0$, while stability concerns how errors or stiff components propagate. Linear A-stability, strong stiff damping through L-stability, and nonlinear B-stability address different questions.

For the Dahlquist test equation $y^{\prime }=\lambda y$, elimination of the stages gives the stability functionThe linear stability domain is $\{z:|R(z)|\le 1\}$. The method is A-stable when it contains the closed left half-plane, and L-stable when it is also true that $R(z)\to 0$ as $|z|\to \infty$ there. These conditions describe, respectively, stability for every non-growing scalar linear mode and damping of very rapidly decaying modes.

The Forward Euler method has $R(z)=1+z$, so it fails A-stability, for example at $z=-3$. In fact no nontrivial explicit Runge-Kutta method is A-stable: its stability function is a nonconstant polynomial, which is unbounded along the negative real axis. The Backward Euler method has $R(z)=1/(1-z)$. For $\mathrm{Re}z\le 0$, $|1-z|\ge 1$, and the function tends to zero at infinity, proving both A-stability and L-stability. The implicit midpoint rule has $R(z)=(1+z/2)/(1-z/2)$. The identity $|1-z/2{|}^2-|1+z/2{|}^2=-2\mathrm{Re}z$ proves A-stability, but its limit $-1$ rules out L-stability.

For nonlinear equations, a dissipative vector field satisfies $⟨f(x)-f(y),x-y⟩\le 0$. The corresponding exact solutions contract because the time derivative of their squared distance is twice that inner product. A B-stable method preserves this contraction for every $h>0$, whenever its stages are well defined.

Let $d=y_n-{\tilde{y}}_n$, $D_i=Y_i-{\tilde{Y}}_i$, and $F_i=f(Y_i)-f({\tilde{Y}}_i)$. Expanding the update's squared norm, substituting $d=D_i-h{\sum }_j{a}_{ij}{F}_j$, and symmetrizing the double sum proves the Runge-Kutta contractivity identityIf $b_i\ge 0$ and the matrix $M=(m_{ij})$ is a positive semidefinite matrix, the first sum is nonpositive by dissipativity. The second double sum is nonnegative: sum the quadratic form of $M$ over each coordinate of the $F_i$. Thus algebraic stability implies B-stability. Applied to complex scalar linear equations with the real part of the Hermitian inner product, the same argument also implies A-stability, where the stages are defined.