Runge–Kutta methods applied to the test equation y' = k\cdot y take the form y_{n+1} = \phi(hk)\cdot y_n, and, by induction, y_n = \bigl(\phi(hk)\bigr)^n\cdot y_0. The function \phi is called the
stability function. Thus, the condition that y_n \to 0 as n \to \infty is equivalent to |\phi(hk)| . This motivates the definition of the
region of absolute stability (sometimes referred to simply as
stability region), which is the set \bigl\{z \in \mathbb{C} \,\big|\,|\phi(z)|. The method is A-stable if the region of absolute stability contains the set \bigl\{ z \in \Complex \,\big|\, \operatorname{Re}(z) , that is, the left half plane.
Example: The Euler methods Consider the Euler methods above. The explicit
Euler method applied to the test equation y' = k\cdot y is y_{n+1} = y_n + h\cdot f(t_n, y_n) = y_n + h\cdot(ky_n) = y_n + h\cdot k\cdot y_n = (1+h\cdot k) y_n. Hence, y_n = (1 + hk)^n\cdot y_0 with \phi(z) = 1 + z. The region of absolute stability for this method is thus \bigl\{ z \in \mathbb{C} \,\big|\, |1+z| which is the disk depicted on the right. The Euler method is not A-stable. The motivating example had k = -15. The value of
z when taking step size h = \tfrac14 is z = -15\times\tfrac14 = -3.75, which is outside the stability region. Indeed, the numerical results do not converge to zero. However, with step size h = \tfrac18, we have z = -1.875 which is just inside the stability region and the numerical results converge to zero, albeit rather slowly.
Example: Trapezoidal method Consider the
trapezoidal method y_{n+1} = y_n + \tfrac{1}{2}h\cdot \bigl(f(t_n,y_n)+f(t_{n+1},y_{n+1})\bigr), when applied to the test equation y' = k\cdot y, is y_{n+1} = y_n + \tfrac{1}{2}h\cdot \left(ky_n+ky_{n+1}\right). Solving for y_{n+1} yields y_{n+1}=\frac{1+\frac{1}{2}hk}{1-\frac{1}{2}hk}\cdot y_n. Thus, the stability function is \phi(z)=\frac{1+\frac12 z}{1-\frac12 z} and the region of absolute stability is \left\{ z \in \mathbb{C} \ \left|\ \left| \frac{1+\frac12 z}{1-\frac12 z} \right| This region contains the left half-plane, so the trapezoidal method is A-stable. In fact, the stability region is identical to the left half-plane, and thus the numerical solution of y' = k\cdot y converges to zero if
and only if the exact solution does. Nevertheless, the trapezoidal method does not have perfect behavior: it does damp all decaying components, but rapidly decaying components are damped only very mildly, because \phi(z) \to 1 as z \to -\infty . This led to the concept of
L-stability: a method is L-stable if it is A-stable and |\phi(z)| \to 0 as z \to \infty . The trapezoidal method is A-stable but not L-stable. The
implicit Euler method is an example of an L-stable method.
General theory The stability function of a
Runge–Kutta method with coefficients \mathbf A and \mathbf b is given by \phi(z) = \frac{\det\left(\mathbf{I} - z\mathbf{A} + z\mathbf{e} \mathbf{b}^\mathsf{T}\right)}{\det \left(\mathbf{I} - z\mathbf{A}\right)}, where \mathbf e denotes the vector with all ones. This is a
rational function (one
polynomial divided by another). Explicit Runge–Kutta methods have a
strictly lower triangular coefficient matrix \mathbf A and thus, their stability function is a polynomial. It follows that explicit Runge–Kutta methods cannot be A-stable. The stability function of implicit Runge–Kutta methods is often analyzed using
order stars. The order star for a method with stability function \phi is defined to be the set \bigl\{ z \in \Complex \,\big|\, |\phi(z)| > |e^z| \bigr\}. A method is A-stable if and only if its stability function has no poles in the left-hand plane and its order star contains no purely imaginary numbers. == Multistep methods ==