When a differential equation damps every disturbance, a numerical method can still manufacture growth. The scalar test equation \begin{align*} y'(t)=\lambda y(t), \qquad \operatorname{Re}(\lambda)<0 \end{align*} has solutions that decay as $t$ increases, yet an unfortunate time discretisation may turn the same problem into an iteration whose magnitude grows. A-stability is the demand that this particular failure never occur for any time step when the continuous model has exponential decay. The issue is not accuracy alone. A method may have a small local error for small time steps, but stiff equations force the product $z=k\lambda$ to wander far into the complex plane, where the method's amplification rule rather than its Taylor expansion controls the computation. A-stability asks for a geometric promise: the whole open left half-plane should be inside the method's region of numerical decay. [example: Forward Euler on a Damped Equation] Consider $y'(t)=-100y(t)$ with $y(0)=1$. The function $y(t)=e^{-100t}$ satisfies \begin{align*} y'(t)=-100e^{-100t}=-100y(t) \end{align*} and \begin{align*} y(0)=e^0=1. \end{align*} Since $|e^{-100t}|=e^{-100t}$ for $t\ge0$, the exact solution decreases from $1$ toward $0$ as $t$ increases. Forward Euler uses the update $Y_{n+1}=Y_n+kf(t_n,Y_n)$. Here $f(t,y)=-100y$, so \begin{align*} Y_{n+1}=Y_n+k(-100Y_n) \end{align*} and collecting the two terms containing $Y_n$ gives \begin{align*} Y_{n+1}=(1-100k)Y_n. \end{align*} For $k=0.03$, the multiplier is \begin{align*} 1-100k=1-100(0.03)=1-3=-2. \end{align*} Thus $Y_{n+1}=-2Y_n$. Iterating this recurrence gives $Y_1=-2Y_0$, $Y_2=(-2)Y_1=(-2)^2Y_0$, and by induction \begin{align*} Y_n=(-2)^nY_0. \end{align*} Taking absolute values gives \begin{align*} |Y_n|=|(-2)^nY_0|=|-2|^n|Y_0|=2^n|Y_0|. \end{align*} Thus the numerical values grow in magnitude even though the exact solution decays. The failure is governed by the dimensionless product $z=k\lambda=0.03(-100)=-3$, not by the time step $k$ alone. [/example] This example isolates the central question of the chapter: when does a time-stepping method respect every exponentially stable scalar mode, regardless of the step size? The answer passes through the linear stability function, its stability domain, and the geometry of the complex left half-plane. ## Definition The opening example suggests the primary demand before any supporting vocabulary is introduced: after applying the method to a decaying scalar mode, the numerical amplification factor should never have modulus greater than $1$. We therefore define A-stability first as a half-plane condition on the method's amplification function. [definition: A-Stability] A one-step numerical method with stability function $R:U\subset\mathbb{C}\to\mathbb{C}$ is A-stable if \begin{align*} \{z\in\mathbb{C}: \operatorname{Re}(z)<0\}\subset U \end{align*} and \begin{align*} |R(z)|\le 1 \qquad \text{for every } z\in\mathbb{C} \text{ with } \operatorname{Re}(z)<0. \end{align*} [/definition] This is the page's central concept. It says that every dimensionless product $z=k\lambda$ coming from a continuously decaying scalar equation lies in the method's numerical decay region. The remaining definitions in this section unpack the objects that appear in the statement: the scalar test equation, the stability function, the stability domain, and the left half-plane. A-stability begins with a deliberate simplification. Rather than test a method on every nonlinear differential equation at once, we test it on the scalar equation whose exact solution is controlled by one complex number. This equation detects whether a method turns continuous exponential damping into discrete amplification. This motivates isolating the model equation itself before discussing any particular numerical method. Once this test problem is fixed, the complex parameter $\lambda$ will measure continuous growth or decay, and the time step will enter only through the dimensionless product $k\lambda$. [definition: Dahlquist Test Equation] The Dahlquist test equation is the initial value problem \begin{align*} y:I\to\mathbb{C}, \qquad y'(t)=\lambda y(t), \qquad y(0)=y_0, \end{align*} where $I\subset\mathbb{R}$ is an interval with $0\in I$, $\lambda\in\mathbb{C}$, and $y_0\in\mathbb{C}$. [/definition] For the exact solution, decay is governed by the sign of $\operatorname{Re}(\lambda)$. If $\operatorname{Re}(\lambda)<0$, then $|e^{\lambda t}|=e^{\operatorname{Re}(\lambda)t}$ decreases with $t$. But A-stability is not a property of the differential equation alone; it is a property of what the numerical method does to this equation after continuous time is replaced by the grid $t_n=nk$. To decide whether the method preserves damping, we need the numerical analogue of $e^{k\lambda}$: the single complex multiplier that carries $Y_n$ to $Y_{n+1}$. For many standard one-step methods, the numerical values on the test equation satisfy a scalar recurrence whose multiplier depends only on the dimensionless quantity $z=k\lambda$. Naming this multiplier turns a time-stepping rule into a complex function, so the next definition is the device that converts a numerical method into a geometric stability problem in the complex plane. [definition: Stability Function] Let a one-step method be applied with time step $k>0$ to the Dahlquist test equation. If the resulting numerical solution satisfies \begin{align*} Y_{n+1}=R(z)Y_n, \qquad z=k\lambda, \end{align*} for a function $R:U\subset\mathbb{C}\to\mathbb{C}$, then $R$ is called the stability function of the method. [/definition] Once the multiplier $R(z)$ has been extracted, the next question is no longer about the original differential equation but about where this multiplier is safe to use. A stiff problem can send $z=k\lambda$ across a large part of the complex plane, so we need a named region that separates numerical damping from numerical growth. Values with $|R(z)|<1$ decay, values with $|R(z)|=1$ preserve magnitude, and values with $|R(z)|>1$ grow; the useful information is the geometry of the first two cases. [definition: Linear Stability Domain] Let $R:U\subset\mathbb{C}\to\mathbb{C}$ be the stability function of a one-step method. The linear stability domain of the method is \begin{align*} D_R=\{z\in U: |R(z)|<1\}. \end{align*} The closed linear stability domain is \begin{align*} \overline{D}_R=\{z\in U: |R(z)|\le 1\}. \end{align*} [/definition] The stability domain is the picture behind the definition of A-stability: it records where the method damps. To turn that picture into the A-stability test, we also need the corresponding picture for the continuous equation. The Dahlquist test equation decays exactly when $z=k\lambda$ lies in one specific half of the complex plane, so the target region deserves its own name before we compare it with $\overline{D}_R$. [definition: Left Half-Planes] The open left half-plane is the set \begin{align*} \mathbb{C}_{-}=\{z\in\mathbb{C}: \operatorname{Re}(z)<0\}. \end{align*} Its closure is the closed left half-plane \begin{align*} \overline{\mathbb{C}_{-}}=\{z\in\mathbb{C}: \operatorname{Re}(z)\le 0\}. \end{align*} [/definition] With this notation, the decay condition in the definition of A-stability concerns $\mathbb{C}_{-}$: equivalently, $\mathbb{C}_{-}\subset U$ and $\mathbb{C}_{-}\subset\overline{D}_R$. Some sources build the boundary into the convention and state A-stability using $\operatorname{Re}(z)\le 0$ instead. On this page, $\mathbb{C}_{-}$ always denotes the open half-plane, while $\overline{\mathbb{C}_{-}}$ denotes the closed half-plane including the imaginary axis. When a quoted result uses the closed convention, it should be read as the corresponding boundary-strengthened form, not as a change in notation. This is a parent-level stability notion for numerical time discretisations: it is related to, but not the same as, general stability of fixed points in dynamical systems or stability of a time-stepping method on a particular problem class. A-stability is intentionally a worst-case condition. It does not say that every unstable differential equation becomes stable, and it does not measure accuracy. It says that the method will not destroy scalar linear decay merely because the time step is large relative to a fast negative mode. ## Stability Functions and Stability Domains ### Explicit and Implicit First Steps The definition becomes useful only once we can compute $R(z)$. The first examples show that different methods may have identical order of accuracy for small $z$ while having dramatically different stability geometry for large negative $z$. Forward Euler is the baseline explicit method. It predicts the next value by following the current slope, so on the test equation it multiplies by a first-order Taylor approximation to $e^z$. [example: Forward Euler Stability Domain] Forward Euler uses $Y_{n+1}=Y_n+kf(t_n,Y_n)$. For the Dahlquist test equation $y'=\lambda y$, we have $f(t,y)=\lambda y$, so \begin{align*} Y_{n+1}=Y_n+k\lambda Y_n. \end{align*} Both terms contain $Y_n$, hence \begin{align*} Y_{n+1}=(1+k\lambda)Y_n. \end{align*} With $z=k\lambda$, this becomes \begin{align*} Y_{n+1}=(1+z)Y_n. \end{align*} Therefore the stability function is \begin{align*} R(z)=1+z. \end{align*} The closed stability condition is $|R(z)|\le1$, so for Forward Euler it is \begin{align*} |1+z|\le1. \end{align*} This is exactly the set of complex numbers $z$ whose distance from $-1$ is at most $1$, because \begin{align*} |1+z|=|z-(-1)|. \end{align*} Thus the closed stability domain is the closed disc centred at $-1$ with radius $1$. On the negative real axis, write $z=-r$ with $r\ge0$. The stability condition becomes \begin{align*} |1-r|\le1. \end{align*} This inequality is equivalent to \begin{align*} -1\le1-r\le1. \end{align*} Subtracting $1$ from all three parts gives \begin{align*} -2\le-r\le0. \end{align*} Multiplying by $-1$ reverses the inequalities, giving \begin{align*} 0\le r\le2. \end{align*} Hence the intersection with the negative real axis is $[-2,0]$. For $y'=-ay$ with $a>0$, the scaled variable is \begin{align*} z=k\lambda=k(-a)=-ka. \end{align*} Since $z=-r$ with $r=ka$, the condition $0\le r\le2$ becomes \begin{align*} 0\le ka\le2. \end{align*} For positive step size $k>0$, stability therefore requires \begin{align*} 0<ka\le2. \end{align*} Forward Euler is not A-stable, because its closed stability domain is only the bounded disc $|1+z|\le1$, while the open left half-plane contains negative [real numbers](/page/Real%20Numbers) such as $z=-3$, and \begin{align*} |1+(-3)|=|-2|=2>1. \end{align*} [/example] The failure is geometric: the left half-plane is unbounded, while Forward Euler only controls a bounded disc. Fast decay modes force $z$ far to the left, outside that disc, where the method alternates and grows. Backward Euler makes the opposite choice. It evaluates the slope at the unknown future value, turning the update into an implicit equation. That implicitness changes the stability function from a polynomial to a rational function with much better behaviour on the left half-plane. [example: Backward Euler Stability Domain] Backward Euler uses the future slope, so applying it to $y'=\lambda y$ gives \begin{align*} Y_{n+1}=Y_n+k\lambda Y_{n+1}. \end{align*} Move the term containing $Y_{n+1}$ on the right to the left: \begin{align*} Y_{n+1}-k\lambda Y_{n+1}=Y_n. \end{align*} Factoring out $Y_{n+1}$ gives \begin{align*} (1-k\lambda)Y_{n+1}=Y_n. \end{align*} With $z=k\lambda$, this is \begin{align*} (1-z)Y_{n+1}=Y_n. \end{align*} If $z\ne1$, division by $1-z$ gives \begin{align*} Y_{n+1}=\frac{1}{1-z}Y_n. \end{align*} Thus the stability function is \begin{align*} R(z)=\frac{1}{1-z}. \end{align*} Now let $z=x+iy$ with $x=\operatorname{Re}(z)$ and $y=\operatorname{Im}(z)$. Then \begin{align*} 1-z=1-(x+iy)=(1-x)-iy. \end{align*} For a complex number $a+ib$, its squared modulus is $a^2+b^2$, so \begin{align*} |1-z|^2=(1-x)^2+(-y)^2. \end{align*} Since $(-y)^2=y^2$, this becomes \begin{align*} |1-z|^2=(1-x)^2+y^2. \end{align*} If $\operatorname{Re}(z)<0$, then $x<0$, so $1-x>1$. Hence \begin{align*} (1-x)^2>1. \end{align*} Because $y^2\ge0$, we get \begin{align*} |1-z|^2=(1-x)^2+y^2>1. \end{align*} Therefore $|1-z|>1$, and so \begin{align*} |R(z)|=\left|\frac{1}{1-z}\right|=\frac{1}{|1-z|}<1. \end{align*} Every $z$ in the open left half-plane is therefore damped by Backward Euler, so Backward Euler is A-stable. [/example] This calculation is the first real payoff of the definition. Implicitness is not merely a computational inconvenience; it can expand the stability region from a bounded disc to a half-plane-sized region. ### Boundary Behaviour The trapezoidal rule takes an average of the old and new slopes. It is second order and implicit, and its stability function sits exactly on the boundary for purely imaginary $z$. [example: Trapezoidal Rule Stability Domain] The trapezoidal rule applied to the Dahlquist test equation $y'=\lambda y$ gives \begin{align*} Y_{n+1}=Y_n+\frac{k}{2}\lambda(Y_n+Y_{n+1}). \end{align*} Distributing the factor $\frac{k\lambda}{2}$ gives \begin{align*} Y_{n+1}=Y_n+\frac{k\lambda}{2}Y_n+\frac{k\lambda}{2}Y_{n+1}. \end{align*} Move the term containing $Y_{n+1}$ on the right to the left: \begin{align*} Y_{n+1}-\frac{k\lambda}{2}Y_{n+1}=Y_n+\frac{k\lambda}{2}Y_n. \end{align*} Factoring both sides gives \begin{align*} \left(1-\frac{k\lambda}{2}\right)Y_{n+1}=\left(1+\frac{k\lambda}{2}\right)Y_n. \end{align*} With $z=k\lambda$, this becomes \begin{align*} \left(1-\frac{z}{2}\right)Y_{n+1}=\left(1+\frac{z}{2}\right)Y_n. \end{align*} If $z\ne2$, division by $1-z/2$ gives \begin{align*} Y_{n+1}=\frac{1+z/2}{1-z/2}Y_n. \end{align*} Thus the stability function is \begin{align*} R(z)=\frac{1+z/2}{1-z/2}. \end{align*} Now let $z=x+iy$ with $x=\operatorname{Re}(z)<0$ and $y=\operatorname{Im}(z)$. Then \begin{align*} 1+\frac{z}{2}=1+\frac{x+iy}{2}=\left(1+\frac{x}{2}\right)+i\frac{y}{2}. \end{align*} Using $|a+ib|^2=a^2+b^2$, we get \begin{align*} \left|1+\frac{z}{2}\right|^2=\left(1+\frac{x}{2}\right)^2+\left(\frac{y}{2}\right)^2. \end{align*} Since $\left(\frac{y}{2}\right)^2=\frac{y^2}{4}$, this is \begin{align*} \left|1+\frac{z}{2}\right|^2=\left(1+\frac{x}{2}\right)^2+\frac{y^2}{4}. \end{align*} Similarly, \begin{align*} 1-\frac{z}{2}=1-\frac{x+iy}{2}=\left(1-\frac{x}{2}\right)-i\frac{y}{2}. \end{align*} Again using $|a+ib|^2=a^2+b^2$, \begin{align*} \left|1-\frac{z}{2}\right|^2=\left(1-\frac{x}{2}\right)^2+\left(-\frac{y}{2}\right)^2. \end{align*} Since $\left(-\frac{y}{2}\right)^2=\frac{y^2}{4}$, this becomes \begin{align*} \left|1-\frac{z}{2}\right|^2=\left(1-\frac{x}{2}\right)^2+\frac{y^2}{4}. \end{align*} To compare the two squared moduli, subtract the first from the second: \begin{align*} \left|1-\frac{z}{2}\right|^2-\left|1+\frac{z}{2}\right|^2=\left(1-\frac{x}{2}\right)^2-\left(1+\frac{x}{2}\right)^2. \end{align*} Expanding the two squares gives \begin{align*} \left(1-\frac{x}{2}\right)^2=1-x+\frac{x^2}{4}. \end{align*} Also, \begin{align*} \left(1+\frac{x}{2}\right)^2=1+x+\frac{x^2}{4}. \end{align*} Therefore \begin{align*} \left(1-\frac{x}{2}\right)^2-\left(1+\frac{x}{2}\right)^2=\left(1-x+\frac{x^2}{4}\right)-\left(1+x+\frac{x^2}{4}\right). \end{align*} Canceling the equal terms $1$ and $x^2/4$ gives \begin{align*} \left(1-x+\frac{x^2}{4}\right)-\left(1+x+\frac{x^2}{4}\right)=-2x. \end{align*} Since $x<0$, we have $-2x>0$. Hence \begin{align*} \left|1-\frac{z}{2}\right|^2>\left|1+\frac{z}{2}\right|^2. \end{align*} Both moduli are nonnegative, so taking square roots preserves the strict inequality: \begin{align*} \left|1-\frac{z}{2}\right|>\left|1+\frac{z}{2}\right|. \end{align*} In particular, $1-z/2\ne0$ for every $z$ with $\operatorname{Re}(z)<0$, and \begin{align*} |R(z)|=\left|\frac{1+z/2}{1-z/2}\right|=\frac{|1+z/2|}{|1-z/2|}<1. \end{align*} Thus every point in the open left half-plane is damped by the trapezoidal rule, so the trapezoidal rule is A-stable. [/example] Backward Euler and the trapezoidal rule show two different flavours of A-stability. Backward Euler damps very stiff modes strongly because $R(z)\to0$ as $z\to\infty$ along the negative real axis. The trapezoidal rule remains bounded but tends to $-1$, so it may preserve highly oscillatory remnants of very stiff decay. ## Stiff Decay and Step-Size Restrictions ### Diagonal Modes A-stability is most valuable when the equation contains several time scales. If one component decays at rate $1$ and another at rate $10^6$, accuracy may be dictated by the slow component, while stability of an explicit method is dictated by the fast component. The result is an artificial restriction on $k$. To state this pressure in a form that also covers systems, we move from the scalar test equation to diagonal linear systems. The scalar stability function then acts on each eigenmode. [definition: Stiff Linear Diagonal System] A stiff linear diagonal system is an initial value problem \begin{align*} y:I\to\mathbb{C}^m, \qquad y'(t)=Ay(t), \qquad y(0)=y_0, \end{align*} where $I\subset\mathbb{R}$ is an interval with $0\in I$, $y_0\in\mathbb{C}^m$, $A\in\mathbb{C}^{m\times m}$ is diagonal, every eigenvalue $\lambda_j$ of $A$ satisfies $\operatorname{Re}(\lambda_j)<0$, and the magnitudes $|\lambda_j|$ occur on widely separated scales. [/definition] The phrase "widely separated" is descriptive rather than a sharp invariant; in applications it means that stability and accuracy impose different step-size demands. The next statement records how scalar A-stability transfers to diagonal systems. [quotetheorem:7970] The theorem explains the role of the left half-plane: every decaying eigenmode is accepted, no matter how far to the left it lies after scaling by $k$. For standard Runge-Kutta methods and the usual linear one-step methods, the componentwise decoupling assumption is automatic on diagonal systems because each component satisfies the same scalar test equation with its own $\lambda_j$. This is why A-stability is a natural condition for stiff systems whose linearised dynamics have eigenvalues with large negative real parts. ### Artificial Stability Limits The contrast with explicit Euler is stark in a two-scale system. One component might be resolved well by a moderate step size while another component forces a much smaller step merely to prevent numerical growth. [example: Two Decay Rates and an Artificial Restriction] Consider the two-component diagonal system \begin{align*} y_1'(t)=-y_1(t), \qquad y_2'(t)=-1000y_2(t). \end{align*} For Forward Euler, the first component satisfies \begin{align*} Y_{n+1,1}=Y_{n,1}+k(-Y_{n,1}). \end{align*} Factoring out $Y_{n,1}$ gives \begin{align*} Y_{n+1,1}=(1-k)Y_{n,1}. \end{align*} The second component satisfies \begin{align*} Y_{n+1,2}=Y_{n,2}+k(-1000Y_{n,2}). \end{align*} Factoring out $Y_{n,2}$ gives \begin{align*} Y_{n+1,2}=(1-1000k)Y_{n,2}. \end{align*} Thus the amplification factors are $1-k$ and $1-1000k$. For the first factor, stability requires \begin{align*} |1-k|\le 1. \end{align*} This is equivalent to \begin{align*} -1\le 1-k\le 1. \end{align*} Subtracting $1$ from all three parts gives \begin{align*} -2\le -k\le 0. \end{align*} Multiplying by $-1$ reverses the inequalities, so \begin{align*} 0\le k\le 2. \end{align*} For the second factor, stability requires \begin{align*} |1-1000k|\le 1. \end{align*} This is equivalent to \begin{align*} -1\le 1-1000k\le 1. \end{align*} Subtracting $1$ from all three parts gives \begin{align*} -2\le -1000k\le 0. \end{align*} Multiplying by $-1$ reverses the inequalities: \begin{align*} 0\le 1000k\le 2. \end{align*} Dividing by $1000>0$ gives \begin{align*} 0\le k\le \frac{2}{1000}. \end{align*} Since a time step has $k>0$, both Forward Euler stability conditions together require \begin{align*} 0<k\le \frac{2}{1000}=0.002. \end{align*} The first component alone would allow $0<k\le2$, but the rapidly decaying second component forces $k\le0.002$. For Backward Euler, the first component satisfies \begin{align*} Y_{n+1,1}=Y_{n,1}+k(-Y_{n+1,1}). \end{align*} Moving the future term to the left gives \begin{align*} (1+k)Y_{n+1,1}=Y_{n,1}. \end{align*} Hence \begin{align*} Y_{n+1,1}=\frac{1}{1+k}Y_{n,1}. \end{align*} Similarly, the second component satisfies \begin{align*} Y_{n+1,2}=Y_{n,2}+k(-1000Y_{n+1,2}). \end{align*} Moving the future term to the left gives \begin{align*} (1+1000k)Y_{n+1,2}=Y_{n,2}. \end{align*} Hence \begin{align*} Y_{n+1,2}=\frac{1}{1+1000k}Y_{n,2}. \end{align*} For every $k>0$, we have $1+k>1$ and $1+1000k>1$, so \begin{align*} \left|\frac{1}{1+k}\right|<1. \end{align*} Also, \begin{align*} \left|\frac{1}{1+1000k}\right|<1. \end{align*} Backward Euler therefore damps both components for every positive step size, while Forward Euler is restricted by the fast decay rate rather than by the slow time scale. [/example] This does not make large time steps accurate for every purpose. If the slow component needs detailed resolution, accuracy still limits $k$. A-stability removes a stability restriction tied to fast decaying modes; it does not replace error analysis. ## Barriers for Explicit Methods The examples suggest a structural divide. Explicit one-step methods usually have polynomial stability functions, and polynomials cannot remain bounded by $1$ on the whole left half-plane unless they are constant. Since a nonconstant method must approximate $e^z$ near $0$, this blocks A-stability. To state the obstruction cleanly, we isolate the polynomial form that appears for explicit Runge-Kutta methods and other explicit one-step schemes on the test equation. [definition: Polynomial Stability Function] A stability function $R:\mathbb{C}\to\mathbb{C}$ is polynomial if there exist $s\in\mathbb{N}\cup\{0\}$ and coefficients $a_0,a_1,\dots,a_s\in\mathbb{C}$ such that \begin{align*} R(z)=\sum_{j=0}^{s}a_jz^j \end{align*} for every $z\in\mathbb{C}$. [/definition] The next issue is whether a polynomial amplification rule can ever cover the entire left half-plane. This theorem answers that design question and explains why explicit methods can enlarge stability intervals but cannot achieve A-stability through a nonconstant polynomial $R$. [quotetheorem:7971] This theorem gives a precise reason explicit methods struggle with stiff decay. They may have bounded stability intervals on the negative real axis, and those intervals can be enlarged by design, but they cannot contain the entire left half-plane. [example: A Long Stability Interval Is Still Not A-Stability] Suppose a method has a polynomial stability function $R$ and its negative real stability interval contains $[-L,0]$ for some $L>0$. This means that every real $z$ with $-L\le z\le0$ satisfies \begin{align*} |R(z)|\le1. \end{align*} For the scalar decay equation $y'=-ay$ with $a>0$, the test-equation parameter is $\lambda=-a$, so the dimensionless stability variable is \begin{align*} z=k\lambda=k(-a)=-ka. \end{align*} The condition $z\in[-L,0]$ is therefore \begin{align*} -L\le -ka\le0. \end{align*} Multiplying all three parts by $-1$ reverses the inequalities: \begin{align*} L\ge ka\ge0. \end{align*} Equivalently, \begin{align*} 0\le ka\le L. \end{align*} Since $a>0$, division by $a$ gives \begin{align*} 0\le k\le \frac{L}{a}. \end{align*} For positive time steps, the permitted stability range is therefore \begin{align*} 0<k\le \frac{L}{a}. \end{align*} Now fix any $k>0$. To keep $z=-ka$ inside $[-L,0]$, the inequality above requires \begin{align*} a\le \frac{L}{k}. \end{align*} If $a>L/k$, then multiplying by $k>0$ gives \begin{align*} ka>L. \end{align*} Multiplying by $-1$ reverses the inequality: \begin{align*} -ka<-L. \end{align*} Thus \begin{align*} z=-ka<-L, \end{align*} so $z$ lies to the left of the interval $[-L,0]$. A long negative real stability interval can allow larger steps for moderately stiff decay rates, but because $a$ can be made larger than $L/k$ for any fixed $k>0$, it cannot give the all-step-size left-half-plane guarantee required by A-stability. [/example] The obstruction is not a complaint about explicit methods in nonstiff regimes. It identifies the specific feature A-stability demands: boundedness over an unbounded half-plane. Rational stability functions from implicit methods can have that geometry. ## Rational Stability and Implicit Methods Implicit methods produce equations for $Y_{n+1}$ that often lead to rational stability functions. Rational functions can decay at infinity, stay bounded on half-planes, or have poles that must be excluded. A-stability is therefore a question about both the numerator and the denominator. A method can fail A-stability in two different ways: the amplification may exceed $1$ somewhere in $\mathbb{C}_{-}$, or the stability function may not even be defined throughout $\mathbb{C}_{-}$. The latter failure is easy to miss if only points on the negative real axis are tested. [definition: Pole-Free Left Half-Plane] Let $R=P/Q$ be a rational stability function with polynomials $P,Q:\mathbb{C}\to\mathbb{C}$ and $Q$ not identically zero. The function $R$ is pole-free on the open left half-plane if \begin{align*} Q(z)\ne0 \qquad \text{for every } z\in\mathbb{C}_{-}. \end{align*} [/definition] The definition separates the domain obstruction from the size obstruction. The next criterion combines them, giving a practical test for rational stability functions: no poles in the wrong half-plane, and numerator no larger than denominator there. In that statement, $\mathbb{C}[z]$ denotes the ring of polynomials in the variable $z$ with complex coefficients. [quotetheorem:7972] The criterion turns A-stability into a complex-analytic inequality. For low-order methods the inequality can be checked by hand; for families of methods it becomes a design constraint. [example: The Implicit Midpoint Rule] The implicit midpoint rule applied to the Dahlquist test equation $y'=\lambda y$ uses the slope at the midpoint approximation $(Y_n+Y_{n+1})/2$, so \begin{align*} Y_{n+1}=Y_n+k\lambda\left(\frac{Y_n+Y_{n+1}}{2}\right). \end{align*} Distributing the factor $k\lambda/2$ gives \begin{align*} Y_{n+1}=Y_n+\frac{k\lambda}{2}Y_n+\frac{k\lambda}{2}Y_{n+1}. \end{align*} Move the term containing $Y_{n+1}$ on the right to the left: \begin{align*} Y_{n+1}-\frac{k\lambda}{2}Y_{n+1}=Y_n+\frac{k\lambda}{2}Y_n. \end{align*} Factoring both sides gives \begin{align*} \left(1-\frac{k\lambda}{2}\right)Y_{n+1}=\left(1+\frac{k\lambda}{2}\right)Y_n. \end{align*} With $z=k\lambda$, this becomes \begin{align*} \left(1-\frac{z}{2}\right)Y_{n+1}=\left(1+\frac{z}{2}\right)Y_n. \end{align*} If $z\ne2$, then $1-z/2\ne0$, so division by $1-z/2$ gives \begin{align*} Y_{n+1}=\frac{1+z/2}{1-z/2}Y_n. \end{align*} Thus the stability function is \begin{align*} R(z)=\frac{1+z/2}{1-z/2}. \end{align*} The denominator vanishes exactly when \begin{align*} 1-\frac{z}{2}=0. \end{align*} Adding $z/2$ to both sides gives \begin{align*} 1=\frac{z}{2}. \end{align*} Multiplying by $2$ gives \begin{align*} z=2. \end{align*} Since $\operatorname{Re}(2)=2>0$, this pole is not in $\mathbb{C}_{-}$. Now let $z=x+iy$ with $x=\operatorname{Re}(z)<0$. Then \begin{align*} 1+\frac{z}{2}=\left(1+\frac{x}{2}\right)+i\frac{y}{2}. \end{align*} Using $|a+ib|^2=a^2+b^2$, \begin{align*} \left|1+\frac{z}{2}\right|^2=\left(1+\frac{x}{2}\right)^2+\frac{y^2}{4}. \end{align*} Similarly, \begin{align*} 1-\frac{z}{2}=\left(1-\frac{x}{2}\right)-i\frac{y}{2}. \end{align*} Again using $|a+ib|^2=a^2+b^2$, \begin{align*} \left|1-\frac{z}{2}\right|^2=\left(1-\frac{x}{2}\right)^2+\frac{y^2}{4}. \end{align*} Subtracting the two squared moduli gives \begin{align*} \left|1-\frac{z}{2}\right|^2-\left|1+\frac{z}{2}\right|^2=\left(1-\frac{x}{2}\right)^2-\left(1+\frac{x}{2}\right)^2. \end{align*} Expanding the first square gives \begin{align*} \left(1-\frac{x}{2}\right)^2=1-x+\frac{x^2}{4}. \end{align*} Expanding the second square gives \begin{align*} \left(1+\frac{x}{2}\right)^2=1+x+\frac{x^2}{4}. \end{align*} Therefore \begin{align*} \left(1-\frac{x}{2}\right)^2-\left(1+\frac{x}{2}\right)^2=\left(1-x+\frac{x^2}{4}\right)-\left(1+x+\frac{x^2}{4}\right). \end{align*} Canceling the equal terms $1$ and $x^2/4$ leaves \begin{align*} \left(1-x+\frac{x^2}{4}\right)-\left(1+x+\frac{x^2}{4}\right)=-2x. \end{align*} Since $x<0$, we have $-2x>0$, and hence \begin{align*} \left|1-\frac{z}{2}\right|^2>\left|1+\frac{z}{2}\right|^2. \end{align*} Both moduli are nonnegative, so taking square roots gives \begin{align*} \left|1-\frac{z}{2}\right|>\left|1+\frac{z}{2}\right|. \end{align*} Therefore, for every $z\in\mathbb{C}_{-}$, \begin{align*} |R(z)|=\left|\frac{1+z/2}{1-z/2}\right|=\frac{|1+z/2|}{|1-z/2|}<1. \end{align*} The implicit midpoint rule is therefore A-stable: its stability function is defined throughout the open left half-plane and strictly damps every point there. [/example] The implicit midpoint rule also illustrates a limitation. Since $R(z)\to -1$ as $z\to\infty$ along the negative real axis, it does not crush extremely stiff modes. This motivates a stronger notion. ## L-Stability and Stiff Damping A-stability prevents growth of decaying modes, but some applications need more: very fast modes should disappear numerically after a few steps, because the exact solution has already made them negligible. A-stable methods that approach a unit-modulus value at infinity may leave persistent oscillations in components that should be gone. This sharper damping property is most naturally expressed along the far negative real axis, or more generally as $z$ tends to infinity within the left half-plane. [definition: L-Stability] An A-stable one-step method with stability function $R:U\subset\mathbb{C}\to\mathbb{C}$ is L-stable if, for every sequence $(z_n)_{n\in\mathbb{N}}$ in $\mathbb{C}_{-}$ with $|z_n|\to\infty$, \begin{align*} R(z_n)\to0. \end{align*} [/definition] For rational stability functions, this is the familiar condition that the value at infinity is $0$. Checking the negative real ray often detects the issue in examples, but the definition asks for damping throughout the far left half-plane. The remaining issue is whether this stronger damping requirement is actually achieved by the basic implicit method used for stiff scalar decay. For Backward Euler, the possible obstruction is not growth on the left half-plane; it is whether the rational factor $R(z)=(1-z)^{-1}$ also vanishes as the stiffness parameter sends $z$ far into that half-plane. [quotetheorem:7973] Backward Euler is the model example of robust damping: for $R(z)=(1-z)^{-1}$, the factor tends to zero as $z=-x$ with $x\to\infty$. This can be desirable in parabolic equations, chemical kinetics, and relaxation systems where fast transients should not pollute the long-time computation. The trapezoidal rule shows why A-stability alone may not be the end of the story. [example: A-Stable but Not L-Stable] For the trapezoidal rule, the stability function is \begin{align*} R(z)=\frac{1+z/2}{1-z/2}. \end{align*} On the negative real ray, put $z=-x$ with $x>0$. Then \begin{align*} R(-x)=\frac{1+(-x)/2}{1-(-x)/2}. \end{align*} Since $1+(-x)/2=1-x/2$ and $1-(-x)/2=1+x/2$, this becomes \begin{align*} R(-x)=\frac{1-x/2}{1+x/2}. \end{align*} Multiplying numerator and denominator by $2$ gives \begin{align*} R(-x)=\frac{2-x}{2+x}. \end{align*} Dividing numerator and denominator by $x>0$ gives \begin{align*} R(-x)=\frac{2/x-1}{2/x+1}. \end{align*} As $x\to\infty$, we have $2/x\to0$, so \begin{align*} \lim_{x\to\infty}R(-x)=\frac{0-1}{0+1}=-1. \end{align*} Thus the trapezoidal rule is A-stable, but it is not L-stable: along the sequence $z_x=-x$ in the left half-plane with $x\to\infty$, the amplification factor tends to $-1$ instead of $0$. Very stiff negative modes are therefore not amplified, but their limiting numerical multiplier has modulus $|-1|=1$ and alternates sign. [/example] L-stability is therefore a refinement rather than a replacement. A-stability answers the half-plane containment question; L-stability asks for additional damping at the far edge of that half-plane. ## Dahlquist Barriers and Multistep Methods A-stability was introduced in the study of stiff ordinary differential equations, and its most famous limitation is a barrier theorem for linear multistep methods. The theorem explains why high-order A-stable multistep schemes are rare. To state the result, we recall the class of methods it concerns. These methods use several previous values, so their stability analysis involves characteristic polynomials rather than a single amplification factor. [definition: Linear Multistep Method] Let $I\subset\mathbb{R}$ be an interval, let $V$ be a finite-dimensional complex [vector space](/page/Vector%20Space), and let $f:I\times V\to V$ be a function. A linear multistep method with step number $s\in\mathbb{N}$ is a scheme of the form \begin{align*} \sum_{j=0}^{s}\alpha_jY_{n+j}=k\sum_{j=0}^{s}\beta_j f(t_{n+j},Y_{n+j}), \end{align*} where $\alpha_j,\beta_j\in\mathbb{C}$, $\alpha_s\ne0$, $Y_{n+j}\in V$, and the method is used to approximate solutions of $y'=f(t,y)$ on a grid $t_n=t_0+nk$. [/definition] For multistep methods, A-stability is still a half-plane demand, but the phrase is interpreted through the root condition for the test equation. The same idea remains: every $z$ with negative real part must lead to a stable recurrence. [definition: A-Stability for Linear Multistep Methods] Let a linear multistep method have coefficient polynomials $\rho,\sigma:\mathbb{C}\to\mathbb{C}$ defined by \begin{align*} \rho(\xi)=\sum_{j=0}^{s}\alpha_j\xi^j, \qquad \sigma(\xi)=\sum_{j=0}^{s}\beta_j\xi^j. \end{align*} The method is A-stable if, for every $z\in\mathbb{C}_{-}$, all roots $\xi\in\mathbb{C}$ of the characteristic equation \begin{align*} \rho(\xi)-z\sigma(\xi)=0 \end{align*} satisfy $|\xi|\le1$, and every root with $|\xi|=1$ is simple. [/definition] Before the half-plane test can matter, a multistep method must pass a more basic recurrence test at $z=0$. Repeated or exterior roots of the homogeneous recurrence can amplify parasitic starting-value modes even when the differential equation contributes no damping or growth. The additional simplicity condition in the A-stability definition controls such roots for $z\in\mathbb{C}_{-}$; the next definition imposes the corresponding baseline condition at the origin. To separate genuine stiff-decay behaviour from instability already present at zero right-hand side, we need a baseline condition at $z=0$. This condition is needed before A-stability can be used as evidence that the method itself is fit for convergence analysis. [definition: Zero-Stability for Linear Multistep Methods] Let a linear multistep method have coefficient polynomial $\rho:\mathbb{C}\to\mathbb{C}$. The method is zero-stable if every root $\xi\in\mathbb{C}$ of \begin{align*} \rho(\xi)=0 \end{align*} satisfies $|\xi|\le1$, and every root with $|\xi|=1$ is simple. [/definition] Once consistency, zero-stability, and A-stability are all on the table, a natural design question becomes unavoidable: can a multistep method have this robust half-plane stability while also achieving arbitrarily high order? This is the exact obstruction that shapes the classical catalogue of stiff multistep solvers. The theorem below gives the negative answer. It explains why the multistep methods used for stiff problems stop at low order when full A-stability is required, and why higher-order stiff solvers usually leave the linear multistep class. [quotetheorem:492] The barrier explains the prominence of the trapezoidal rule and backward differentiation formula of order $2$ in stiff computation. If high order together with large stability regions is needed, one usually turns to implicit Runge-Kutta methods, specialised rational approximations, or problem-adapted exponential and splitting methods. [example: Why the Barrier Matters in Practice] Consider two method choices for the same differential equation. A nonstiff problem has no very large negative decay rate, so the scaled stability variables $z=k\lambda$ may remain in a moderate part of the complex plane when $k$ is chosen for accuracy. In that case, a high-order explicit or multistep method can be useful because its smaller local error near $z=0$ is the dominant issue. For a stiff decay mode $y'=-ay$ with $a>0$, the scaled variable is \begin{align*} z=k\lambda=k(-a). \end{align*} Thus \begin{align*} z=-ka. \end{align*} If a method is stable only on a bounded negative real interval $[-L,0]$, then stability of this mode requires \begin{align*} -L\le -ka\le 0. \end{align*} Multiplying all three parts by $-1$ reverses the inequalities: \begin{align*} L\ge ka\ge 0. \end{align*} Equivalently, \begin{align*} 0\le ka\le L. \end{align*} Since $a>0$, division by $a$ gives \begin{align*} 0\le k\le \frac{L}{a}. \end{align*} For positive time steps, this becomes \begin{align*} 0<k\le \frac{L}{a}. \end{align*} As the decay rate $a$ becomes large, the upper bound $L/a$ becomes small, so the step size is being limited by stability of the fast decaying mode rather than by the accuracy needed to resolve the slow part of the solution. The *Dahlquist Second Barrier* says that within consistent zero-stable linear multistep methods of positive order, A-stability and order greater than $2$ cannot be combined. Thus, in stiff problems, choosing a method by order alone can be misleading: the stiffness structure determines whether high order is usable at the step sizes accuracy would otherwise allow. [/example] The theorem is not a ban on high-order stable computation. It is a map of where to look: beyond A-stable linear multistep methods, not beyond numerical analysis. ## Connections with Dynamical Stability The word "stability" appears across analysis, dynamics, optimisation, and numerical computation. A-stability belongs to this family, but it is a specialised compatibility condition between continuous linear decay and discrete amplification. The fixed point viewpoint helps clarify the relation. The exact time-$k$ map of $y'=\lambda y$ is multiplication by $e^{k\lambda}$. A numerical method replaces that multiplier by $R(k\lambda)$. [definition: Discrete Linear Fixed Point Stability] Let $g:\mathbb{C}\to\mathbb{C}$ be the [linear map](/page/Linear%20Map) $g(w)=\xi w$ for some $\xi\in\mathbb{C}$. The fixed point $0$ is linearly stable for the discrete iteration $w_{n+1}=g(w_n)$ if \begin{align*} |\xi|\le1. \end{align*} [/definition] There is a small but important compatibility problem here. Continuous decay says that $k\lambda$ lies in the left half-plane, while stability of the induced discrete fixed point says that the multiplier must lie in the closed unit disc. A-stability is precisely the condition that the numerical stability function sends every continuously decaying scalar mode to a discretely stable multiplier, keeping the differential equation, the method, and the induced iteration distinct. [quotetheorem:7974] This statement keeps the conceptual hierarchy tidy. Continuous stability supplies the left half-plane, discrete fixed point stability supplies the unit disc, and A-stability demands that the numerical method map the former into the latter. [remark: Accuracy and Stability Are Different Questions] A-stability does not assert that $R(z)$ is close to $e^z$ for large $z$. It only asserts that $|R(z)|\le1$ on the left half-plane. Order conditions control the behaviour near $z=0$; A-stability controls the global shape of the stability region. [/remark] The distinction is especially important in time series and applied modelling, where stability of an estimator, stationarity of a process, and stability of a numerical integration scheme may all appear in the same discussion. A-stability concerns the numerical propagation of damped differential modes. ## Beyond and Connected Topics A-stability is usually the entry point to the [Numerical Analysis](/page/Numerical%20Analysis) of stiff ordinary differential equations. The next layer is absolute stability for specific method families, where the full stability domain is plotted or estimated and then compared with spectra of Jacobian matrices arising from linearisation. L-stability and stiff accuracy refine the A-stable condition by asking how strongly the method damps modes far inside the left half-plane. These ideas are central for parabolic PDE semi-discretisations, reaction-diffusion equations, and differential-algebraic equations. Implicit Runge-Kutta methods provide high-order examples beyond the multistep barrier. Gauss methods are A-stable and symplectic, while Radau methods often provide strong stiff damping. This is the natural direction when the order barrier for multistep methods becomes restrictive. The fixed point perspective connects A-stability to dynamical systems. Linearisation of an equilibrium gives eigenvalues of a Jacobian matrix, while a time discretisation transforms those eigenvalues through a stability function. This explains why numerical stability and stability of fixed points are adjacent but distinct notions. For stochastic and data-driven models, the language of stability also appears in [Time Series Analysis](/page/Time%20Series%20Analysis), where stationarity and decay of correlations play roles analogous to damping. The analogy is useful, but A-stability remains a deterministic condition on numerical methods for differential equations. ## References The references below separate the main numerical-analysis context from adjacent Androma background notes, so the list is not read as a chain of prerequisite citations. - Androma, [Numerical Analysis](/page/Numerical%20Analysis). - Androma, [Cambridge IB Analysis and Topology](/page/Cambridge%20IB%20Analysis%20and%20Topology). - Androma, [Time Series Analysis](/page/Time%20Series%20Analysis). - Dahlquist, *A Special Stability Problem for Linear Multistep Methods* (1963). - Hairer and Wanner, *Solving Ordinary Differential Equations II: Stiff and Differential-Algebraic Problems* (1996). - Iserles, *A First Course in the Numerical Analysis of Differential Equations* (2009). - Trefethen, *Finite Difference and Spectral Methods for Ordinary and Partial Differential Equations* (1996).