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For the two-stage implicit Runge-Kutta method withthe stability function isFor $z=x+iy$,The denominator has no zero in the closed left half-plane, so the method is A-stable for every real $a$.

The implicit midpoint rule advances an autonomous ordinary differential equation byIt is a second-order, A-stable, time-symmetric numerical method.

An implicit time-stepping method determines a new solution value through an equation involving that same value. A small-step contraction mapping argument often establishes local solvability when the vector field is Lipschitz continuous. Examples include the Backward Euler method, implicit Runge-Kutta methods and implicit linear multistep methods. Solvability of the update is a separate issue from stability of a numerical method.

For an implicit Runge-Kutta method, algebraic stability means that $b_i\ge 0$ and the symmetric matrixis positive semidefinite. Here $b_1=b_2=1/2$ and $A+A^T=e{e}^T/2$, soThe method is algebraically stable. This also explains why the property matters for nonlinear problems. If a vector field satisfies $⟨f(v)-f(w),v-w⟩\le 0$, the Runge-Kutta contractivity identity for two solutions, with stage differences $D_i$ and vector-field differences $F_i$, isThe first sum is nonpositive and the second vanishes. Thus, whenever the implicit stage equations have the relevant solutions, their updates are contractive in the Hilbert space norm.

Let $p$ be the collocation polynomial of degree at most $s$ over one step, with $p(t_n)=y_n$, and let $Y_i=p(t_n+c_{i}h)$. Its derivative, of degree at most $s-1$, is fixed by its values at the distinct nodes. Lagrange interpolation therefore givesIntegrating from zero to $\tau$ yieldsAt $\tau =c_i$ this gives the stage equations of an implicit Runge-Kutta method; at $\tau =1$ it gives the update. ConsequentlyConversely, stages satisfying these equations define the integrated polynomial displayed above. It takes the stage values at the nodes and has the required derivative there, so it solves the collocation equations. This proves equivalence for every common solution branch, not merely equality on the scalar test equation. Also ${\sum }_j{a}_{ij}=c_i$, since the Lagrange polynomials sum to one. Stage existence or uniqueness requires the usual implicit-solvability assumptions; for a Lipschitz vector field a sufficiently small step gives a contraction. This is the collocation Runge-Kutta method construction.

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.

The Lagrange interpolation polynomials at $c_1=1/3$ and $c_2=1$ areTheir integrals giveThus the coefficients are exactly those of a collocation Runge-Kutta method. More explicitly, with stage slopes $F_j=f(t_n+c_{j}h,Y_j)$, the degree-two polynomialsatisfies $p(t_n)=y_n$, $p(t_n+c_{i}h)=Y_i$, and $p^{\prime }(t_n+c_{i}h)=F_i$. These are the collocation Runge-Kutta method equations, and $y_{n+1}=p(t_n+h)$. For an implicit Runge-Kutta method, the nearby stages exist uniquely for sufficiently small $h$ under Lipschitz continuity; see stage solvability of an implicit Runge-Kutta method.

Write $e=(1,1{)}^T$, $c=Ae$, and interpret powers of $c$ componentwise. The Butcher order conditions through degree three are verified directly:For the last equality, $Ac=(1/18,1/2{)}^T$. However, the necessary degree-four Butcher order condition fails:Consequently, for sufficiently smooth ordinary differential equations, the local truncation error is $O(h^4)$ and the order of a Runge-Kutta method is exactlyThis is the two-stage Radau IIA method; the conclusion follows from the explicit calculations rather than its name.

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.

For the test equation $f(y)=\lambda y$, put $z=h\lambda$. The stage equations of this implicit Runge-Kutta method areSubstitution into the update gives the stability functionFor $z=x+iy$, direct expansion givesThis is nonnegative whenever $x\le 0$, and hence $|R(z)|\le 1$ throughout the closed left half-plane, provided the denominator has no zero there.

If $a=0$, the denominator vanishes only at $z=2$. If $a\ne 0$, its zeros solveReal roots are positive, while a complex-conjugate pair has positive real part because the sum of the roots is $1/(2a^2)>0$. Thus there are no poles in the closed left half-plane for any real $a$. Thereforeis the complete set of A-stable parameters, as summarized by the A-stability of a symmetric two-stage implicit Runge-Kutta family.

For a vector field uniformly Lipschitz continuous in the state with constant $L$, the stage fixed-point map of an implicit Runge-Kutta method is a contraction in the maximum stage norm if $hL{\max }_i{\sum }_j|a_{ij}|<1$. The contraction mapping theorem then gives unique stages. This is a sufficient small-step condition; a dissipative vector field may permit solvability for larger steps. An algebraic stability or B-stability estimate compares existing stage solutions and should not be mistaken for an unqualified stage-existence theorem.