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]