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Consider the nonstiff scalar initial value problem $y^{\prime }=-y$, $y(0)=1$, on $0\le t\le 8$. Its exact solution is the exponential decay $y(t)=e^{-t}$. Use step-doubling error estimation with the explicit Euler method: from the same numerical starting value $y_n$, form one coarse step $C_n$ and two half steps giving $F_n$:Accept $y_{n+1}=F_n$ when the absolute and relative error tolerances test $E_n=|{\hat{e}}_n|/s_n\le 1$ passes, with $s_n=10^{-5}+10^{-3}\max (|y_n|,|F_n|)$. There is no Richardson extrapolation in the accepted value. Both approximations have order of a numerical method one, while this local error estimator has step exponent $r=2$. Consequently the adaptive step-size controller uses a square root, not a cube root:Start with $h_0=1$. A rejected trial retains both $t_n$ and $y_n$, caps $q_n$ at one, and retries. A zero estimate uses $q_n=5$. Before every trial, cap $h_n$ at $8-t_n$. Stop with a failure if floating-point arithmetic gives $t_n+h_n=t_n$, or after 100000 trials. These rules specify the example completely; they are not a prescription for every ordinary differential equation solver.

The first two trials, $h=1$ and $h=0.2$, have $E\approx 247.525$ and $E\approx 9.90099$ and are rejected. The next trial has $h\approx 0.0572049$ and $E\approx 0.81$ and is accepted. For positive decreasing numerical values, $s_n=10^{-5}+10^{-3}{y}_n$, so the local error estimator givesWhen the relative term dominates, this suggests nearly constant accepted steps, approximately $2(0.9)\sqrt{10^{-3}}$. When $y_n$ falls below the crossover size $10^{-5}/10^{-3}=0.01$, the absolute term increasingly permits longer steps. The sharp final decrease in $h$ only lands on $t=8$; it does not indicate a sudden loss of smoothness.

The actual signed local defect from the same numerical input isHere $O$ is Big O notation, obtained from Taylor's theorem for the exponential function. Thus ${\hat{e}}_n$ agrees with the leading term as $h_n\to 0$, but is not identical to ${\ell }_n$. At $h=1$, $y_n=1$, the estimate is $0.25$ while the actual local defect is $e^{-1}-1/4\approx 0.117879$. Their closeness improves for the smaller accepted steps. Had we added ${\hat{e}}_n$ using Richardson extrapolation, the accepted formula would instead be $y_n(1-h_n+h_n^2/2)$, of order of a numerical method two; the same raw difference would still scale as $h_n^2$ and would not estimate that improved value's leading $h_n^3$ defect.

The global error $g_n=y_n-e^{-t_n}$ obeys an exact recurrence relation involving the actual local defects:Each previous global error contracts, yet the local defects keep adding. In the plotted run, 101 steps are accepted and two trials are rejected. Every accepted $E_n$ is at most approximately $0.81$, while the largest $|g_{n+1}|/s_n$ is approximately $38.6927$. At $t=8$, the numerical value is approximately $0.00027184218$, compared with the exact $e^{-8}\approx 0.00033546263$, giving absolute error approximately $6.36204\times 10^{-5}$. The right-hand plot deliberately scales actual global error by the same $s_n$ used for local acceptance; it is not a new global error tolerance test.

Tightening both absolute and relative error tolerances by a factor ten gives 315 accepted steps and endpoint absolute error approximately $2.14922\times 10^{-5}$; tightening both by a factor one hundred gives 991 accepted steps and endpoint absolute error approximately $6.94171\times 10^{-6}$. This is a convergence comparison for this particular initial value problem. The endpoint error decreases by about a factor three per tenfold tightening, rather than by ten: a first-order Euler method accumulates defects across many steps, while its local error estimator varies quadratically with $h$. A variable-step global ODE error bound needs propagation information as well as accepted local tests.

The absolute stability boundary remains a separate constraint. On the Dahlquist test equation $y^{\prime }=-\lambda y$ with $\lambda >0$, the accepted two-half-step formula has stability function $R(-\lambda h)=(1-\lambda h/2{)}^2$. It strictly damps perturbations only for $0<\lambda h<4$. Its local error estimator is ${\lambda }^2{h}^2{y}_n/4$, which can pass an absolute-tolerance test at tiny $y_n$ even outside this damping interval. In a stiff two-mode linear system, a fast mode can therefore constrain the step long after its visible amplitude has become small. An adaptive step-size controller does not turn an explicit Euler method into an A-stable method. Official solver documentation likewise distinguishes explicit Runge-Kutta methods for nonstiff problems from implicit Runge-Kutta methods and backward differentiation formulas for stiff equations.^[ref]

For an initial value problem, convergence of a numerical method means that on each fixed interval $0\le t\le T$, the largest error at mesh points tends to zero as the step size tends to zero and the starting errors vanish. Order of a numerical method describes how rapidly an exact-start local defect decreases; stability of a numerical method controls propagation of that defect and of errors already present. These are related but different properties. A high-order expansion alone cannot ensure that accumulated error remains small.

For a one-step map $Y_{n+1}={\Phi }_h(t_n,Y_n)$, define the exact-start defect $d_{n+1}=y(t_{n+1})-{\Phi }_h(t_n,y(t_n))$. Order $p$ means $d_{n+1}=O(h^{p+1})$ for sufficiently smooth solutions. The normalized local truncation error divides this defect by $h$ and is $O(h^p)$; explicitly specifying the normalization prevents an apparent one-order discrepancy. Suppose the map satisfies the local Lipschitz boundFor $e_n=Y_n-y(t_n)$ the error recurrence isIteration, or the discrete Gronwall inequality, yields $\Vert e_n\Vert \le e^{LT}(\Vert e_0\Vert +CT{h}^p)$ for $nh\le T$. Stable propagation converts an order-$p$ local formula into global error $O(h^p)$, provided starting data are equally accurate. Consistency is the weaker requirement that the normalized local defect tend to zero.

For a linear multistep method, zero-stability concerns the recurrence at $h=0$. Its roots must have modulus at most one, and unit-modulus roots must be simple. Roots outside the circle amplify errors exponentially in the number of steps; repeated unit roots introduce polynomial growth. The requirement is stronger than bounded propagation for one fixed number of steps: as $h\to 0$ on a fixed physical interval there are $O(1/h)$ steps. Under the usual Lipschitz assumptions and consistent starting values, the Dahlquist equivalence theorem says consistency plus zero-stability is equivalent to convergence. Its sufficiency combines a bounded discrete recurrence propagator with the sum of local defects, while necessity follows by exciting an unstable homogeneous root or by observing the persistent error of an inconsistent formula.

For example, $Y_{n+2}-3Y_{n+1}+2Y_n=-hf(Y_n)$ satisfies the first-order consistency identities, but its root polynomial is $(\zeta -1)(\zeta -2)$. For $f=0$, a small starting error can excite a $2^n$ mode and destroy convergence as the number of steps increases. In contrast, simple parasitic unit roots such as the root minus one of a two-step centered method need not destroy zero-stability: they remain bounded, though they can create unwanted oscillations and affect practical behavior. Accurate starting procedures and suppression of accumulated parasitic error remain relevant.

Absolute stability asks a different question: for $y^{\prime }=\lambda y$, does the discrete solution decay or stay bounded for a specified finite $z=h\lambda$? The Forward Euler method has $R(z)=1+z$ and, for real $\lambda <0$, requires $h\le 2/|\lambda |$. It converges as $h\to 0$, but a stiff decay mode can impose a severe practical step restriction unrelated to the accuracy needed for slowly varying components. The Backward Euler method has $R(z)=1/(1-z)$, hence is A-stable and L-stable. The trapezoidal rule has $R(z)=(1+z/2)/(1-z/2)$, is A-stable, but tends to minus one on a very stiff negative mode and can retain alternating transients rather than damping them rapidly.

Zero-stability therefore does not imply unrestricted absolute stability; the three-step family analyzed earlier in this paper is a particularly strong illustration. Conversely a stable large step need not be accurate. This distinction matters for a stiff equation. The Second Dahlquist barrier limits A-stable linear multistep methods to order at most two, whereas implicit Runge-Kutta methods can have much higher order and still be A-stable. That barrier concerns the specified method class, not every numerical time integrator.

Finally, the mathematical order depends on smoothness: nonsmooth solutions or forcing can lower observed rates. Roundoff, imperfect implicit solves and inaccurate derivative evaluations introduce additional defects. Smaller steps increase the number of such perturbations; truncation error alone does not describe arbitrarily fine computation. The useful organizing principle is consistency supplies vanishing defects, stability bounds their accumulation, convergence follows, and order quantifies its rate under the stated regularity and starting-error assumptions.