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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 $a$.

An explicit Runge-Kutta method has $a_{ij}=0$ whenever $j\ge i$. Its stage matrix is strictly lower triangular: compute each Runge-Kutta stage using already known slopes, starting with $Y_1=y_n$. In contrast, an implicit Runge-Kutta method requires solving coupled stage equations, whose existence is addressed by stage solvability of an implicit Runge-Kutta method.

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

The implicit midpoint rule is the midpoint method with $y_{n+1}=y_n+hf(t_n+h/2,(y_n+y_{n+1})/2)$. It is a one-stage implicit Runge-Kutta 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.

A collocation Runge-Kutta method approximates an ordinary differential equation by a polynomial whose derivative agrees with the vector field at selected times. Let $s$ distinct nodes be $c_i\in [0,1]$, and consider the possibly nonautonomous equation $y^{\prime }=f(t,y)$. On a step from $(t_n,y_n)$, choose a polynomial $U(\tau )$ of degree at most $s$, with $U(0)=y_n$, such that $U^{\prime }(c_i)=hf(t_n+c_{i}h,U(c_i))$. Here $\tau =(t-t_n)/h$ is scaled time; the numerical endpoint is $y_{n+1}=U(1)$.

Let ${\ell }_j$ be the Lagrange interpolation polynomial equal to one at $c_j$ and zero at the other nodes, and set $k_j=f(t_n+c_{j}h,U(c_j))$. Interpolation of the polynomial derivative givesEvaluating at the nodes and endpoint derives the Runge-Kutta method coefficients:Thus collocation is a specific family of generally implicit RK methods, not a separate update mechanism. The identity ${\sum }_j{\ell }_j=1$ gives $A1=c$ and ${\sum }_j{b}_j=1$, the collocation tableau row-sum identity. For sufficiently small $h$, a vector field with Lipschitz constant $L$ gives a contraction of the stage map when $hL{\max }_i{\sum }_j|a_{ij}|<1$, on a suitable bounded neighborhood. The Banach fixed-point theorem then gives a unique nearby stage solution. Equivalently smooth stage equations have identity derivative in the stage variables at $h=0$, so the inverse function theorem gives a smooth nearby branch. Arbitrarily large-step solvability is a separate issue for an implicit Runge-Kutta method.

Order analysis has two levels: polynomial accuracy at stages and extra accuracy at the endpoint. Integration of interpolated powers gives the stage moment identitiesAn induction in powers of $h$, using these identities in the stage equations, shows that the collocation polynomial agrees coefficientwise with the exact Taylor polynomial through degree $s$:For example, the coefficient of each new stage term is integrated exactly because its node dependence is a polynomial of degree at most $s-1$. This proves stage order $s$ and bounded actual-time derivatives of the collocation polynomial up to degree $s$. Assuming the vector field is sufficiently smooth on a fixed compact solution neighborhood, its composition with this polynomial has uniformly bounded derivatives of all finite orders needed below.

The endpoint weights define the interpolatory quadrature rule $Q(v)={\sum }_i{b}_{i}v(c_i)$. Suppose it is exact through degree $q-1$, but not through degree $q$. Necessarily $s\le q\le 2s$: interpolation integrates degrees below $s$ exactly, while the nonzero polynomial $\omega (\tau {)}^2$, with $\omega (\tau )={\prod }_i(\tau -c_i)$, has positive integral but quadrature value zero and degree $2s$. Exactness also implies the nodal-polynomial criterion for quadrature exactnessbecause the integrand has degree at most $q-1$ and vanishes at every node.

Here is the endpoint-error argument underlying the collocation order theorem. Let $v(t)=U((t-t_n)/h)$ and $\delta (t)=v^{\prime }(t)-f(t,v(t))$ be its residual. It vanishes at the collocation times, and if $F_h(\tau )=f(t_n+h\tau ,U(\tau ))$, then $\delta ={\mathcal{I}}_s{F}_h-F_h$, where ${\mathcal{I}}_s$ is interpolation at the nodes. Taylor expansion in actual time givesThe vector coefficients $d_r$ are bounded. Each bracket has the factor $\omega (\tau )$, with quotient degree $r-s$. In particular the residual is $O(h^s)$ throughout the step.

Write $\varphi (t_{\ast },t,z)$ for the exact flow from time $t$ and state $z$ to $t_{\ast }=t_n+h$. Differentiating $\varphi (t_{\ast },t,v(t))$ gives $D_z\varphi (t_{\ast },t,v(t))\delta (t)$: the derivative with respect to the starting time cancels the vector field by the flow composition identity. Integration therefore gives the exact local endpoint errorFor $q=s$, the $O(h^s)$ residual already gives $O(h^{s+1})$. For $q>s$, Taylor-expand the smooth matrix weight as $\Psi (t_n+h\tau )={\sum }_{\ell =0}^{q-s-1}{h}^{\ell }{\Psi }_{\ell }{\tau }^{\ell }+O(h^{q-s})$. Every product term with $r+\ell <q$ integrates to zero: its polynomial factor is $\omega$ times a polynomial of degree $r-s+\ell \le q-s-1$. All remaining terms are $O(h^q)$ before the time integration, including both remainders. HenceThis is collocation endpoint superconvergence by quadrature orthogonality. For the global assertion, the stage contraction estimate gives a one-step Lipschitz bound $1+Ch$ on a fixed solution neighborhood. Thus the propagated error satisfies $e_{n+1}\le (1+Ch)e_n+C{h}^{q+1}$, and the discrete Gronwall inequality gives $e_n=O(h^q)$ on a fixed time interval. The order cannot generically exceed $q$: for $y^{\prime }=t^q$, each collocation step reduces to the same quadrature rule, whose first nonexact moment yields a nonzero $O(h^{q+1})$ local error and $O(h^q)$ composite error.

The usual node families illustrate the result. Gauss collocation methods use interior Gaussian nodes and have $q=2s$, the maximum possible order. Right-endpoint Radau nodes give the Radau IIA method of order $2s-1$. Including both endpoints with Lobatto quadrature gives the Lobatto IIIA method of order $2s-2$ for $s\ge 2$. One-stage Gaussian collocation is the second-order implicit midpoint rule, while collocation at the right endpoint is the first-order Backward Euler method. The two-stage Gaussian nodes $1/2\pm \sqrt{3}/6$ give $b=(1/2,1/2)$ anda fourth-order method. Node placement, rather than just stage count, controls the superconvergent endpoint order.

There is also a structural link with Question 2. For Gaussian nodes, let $L_i(\tau )={\int }_0^{\tau }{\ell }_i(v)dv$. The derivative of $L_i{L}_j$ has degree at most $2s-1$, so Gaussian quadrature is exact on it. The integral is $b_i{b}_j$, while evaluating at the nodes gives $b_i{a}_{ij}+b_j{a}_{ji}$. Thus Gaussian collocation satisfies $M=0$ and has Runge-Kutta conservation of quadratic invariants. For the scalar test equation, any RK method has stability function $R(z)=1+z{b}^T(I-zA{)}^{-1}1$. For implicit midpoint this is $(1+z/2)/(1-z/2)$; the identity $|1-z/2{|}^2-|1+z/2{|}^2=-2\mathrm{Re}z$ proves A-stability. High order, stage solvability, invariant preservation and damping are distinct design considerations, so no stability conclusion should be inferred from collocation order alone.

A-stability removes the scalar linear decay restriction on the step size; it does not replace accuracy, nonlinear solvability or a suitable system norm. Its importance is clearest for a stiff differential equation, where rapidly decaying modes coexist with slowly varying quantities of interest. On the Dahlquist test equation $y^{\prime }=\lambda y$, the exact solution decays for $\mathrm{Re}\lambda <0$. A stable time integrator should not turn that decay into numerical growth.

For a one-step method write $y_{n+1}=R(z)y_n$, where $z=h\lambda$ and $R$ is its stability function. Its linear stability domain consists of parameters with $|R(z)|\le 1$ and a well-defined update. The method is A-stable when this domain contains the closed left half-plane. A consistent order-$p$ one-step method has $R(z)=e^z+O(z^{p+1})$ near zero, but that local agreement alone says nothing about stability at large negative $z$.

For a Runge-Kutta method with stage matrix $A$, weights $b$ and $1=(1,\dots ,1{)}^T$, eliminating the stages on the test equation givesAn explicit finite-stage method has a polynomial $R$. A nonconstant polynomial is unbounded along the negative real axis, so no consistent explicit Runge-Kutta method is A-stable. The explicit Euler method, with $R(z)=1+z$, is stable in the disk $|1+z|\le 1$ and on the negative real axis only for $-2\le z\le 0$. A stiff mode with a large negative eigenvalue therefore forces an unnecessarily small step even after its exact contribution has almost disappeared.

The Backward Euler method has $R(z)=1/(1-z)$. For $\mathrm{Re}z\le 0$, $|1-z{|}^2=1-2\mathrm{Re}z+|z{|}^2\ge 1$, so it is A-stable. The trapezoidal rule hasand is also A-stable. The implicit midpoint method has the same scalar stability function, although its nonlinear update differs from the trapezoidal rule. The first method has order one and the latter two order two. Their implicit solves trade greater work per step for freedom from the scalar stiff stability restriction.

Damping of very stiff modes separates A-stability from L-stability. For a one-step method, the latter additionally requires $R(z)\to 0$ as $z$ tends to infinity in the left half-plane. Backward Euler method satisfies this. The trapezoidal and midpoint rules instead have $R(z)\to -1$ along the negative real axis: a very stiff mode can alternate in sign with little damping even though its amplitude remains bounded. L-stability is often preferable for diffusion and rapid relaxation, while retaining oscillatory modes can be desirable in conservative dynamics.

For a linear multistep method, writeIts test-equation amplification roots satisfy $\rho (\zeta )-z\sigma (\zeta )=0$. Absolute stability requires every root to lie in the unit disk or on its boundary with all unit-modulus roots simple; all parasitic modes matter. At $z=0$ this is the zero-stability root condition. Consistency requires $\rho (1)=0$ and ${\rho }^{\prime }(1)=\sigma (1)$, and the Dahlquist equivalence theorem combines consistency and zero-stability to characterize convergence. Thus a claim of stability obtained by canceling a common factor is not a proof about the original recurrence and its arbitrary consistent starting data.

Explicit linear multistep methods cannot be A-stable either. If ${\beta }_k=0$ and ${\alpha }_k\ne 0$, some lower coefficient of $\rho -z\sigma$ becomes unbounded as $z\to -\infty$, while the leading coefficient stays fixed. If all roots remained in the unit disk, every coefficient-to-leading-coefficient ratio, an elementary symmetric function of those roots, would remain bounded. This contradiction forces an unstable root.

An important implicit example is BDF2 method, whose characteristic equation is $(3-2z){\zeta }^2-4\zeta +1=0$. A unit root $\zeta =e^{i\theta }$ givesThe leading coefficient vanishes only at $z=3/2$, outside the left half-plane. Near a small negative $z$, both roots are inside the disk; no unit-circle crossing can occur in the connected open left half-plane. On the imaginary axis the boundary equality can occur only at $z=0$, where the roots are $1,1/3$. Hence BDF2 is A-stable and its two roots tend to zero for very large negative $z$, giving strong stiff-mode damping. This is a complete boundary-locus argument, not just a boundary plot.

The Second Dahlquist barrier says that a convergent A-stable ordinary linear multistep method has order at most two. This theorem does not limit implicit Runge-Kutta methods or multiderivative multistep methods. In particular the A-stable third-order two-step multiderivative method in question 1 uses $y^{\prime \prime }$ and satisfies different order conditions. Among collocation Runge-Kutta methods, the standard family results give Gauss collocation methods order $2s$ with diagonal $[s/s]$ Padé approximants to $e^z$ as stability functions, and Radau IIA methods order $2s-1$ with subdiagonal $[s-1/s]$ approximants. Both families are A-stable; the Gauss functions have a nonzero limit at infinity, whereas Radau IIA is L-stable. The corrected two-stage right-Radau method, for example, haswhich matches the exponential through cubic order and tends to zero at infinity. Its poles $2\pm i\sqrt{2}$ lie in the right half-plane. For $z=-r+iy$, $r\ge 0$, direct subtraction givesproving its A-stability directly.

Sectorial A-alpha stability relaxes the requirement to a sector around the negative real axis, useful when the spectrum of a stiff problem remains in that sector; higher-order backward differentiation formulas illustrate the usefulness of this weaker requirement. For a normal linear system $y^{\prime }=Ly$, unitary diagonalization reduces the numerical update to its scalar factors $R(h{\lambda }_j)$. For a nonnormal system, a scalar eigenvalue test alone does not give a uniform bound in the chosen norm. Nonlinear stability is different again: B-stability asks for contractivity when $\mathrm{Re}⟨f(u)-f(v),u-v⟩\le 0$, and algebraic stability of a Runge-Kutta method is a standard sufficient criterion. Thus A-stability addresses a precise and valuable linear test, while step selection still has to meet accuracy requirements and the actual implicit equations must be solved on the relevant branch.

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.

A Runge-Kutta method advances an initial value problem $y^{\prime }=f(t,y)$, $y(t_0)=y_0$, by sampling its vector field at several trial states within one time step. These samples are Runge-Kutta stages; a weighted combination of their slopes produces the next state. The construction improves on the single tangent line used by the Euler method while requiring evaluations of $f$, rather than explicit higher derivatives of the solution. It applies to scalar equations and systems of ordinary differential equations alike. The coefficient choices distinguish methods such as Heun's method, the classical fourth-order Runge-Kutta method and implicit Runge-Kutta methods.

For $s$ stages, write $A=(a_{ij})$ for the stage matrix, $c_i$ for relative time nodes and $b_i$ for output weights. A time step of length $h$ computesHere $k_i$ is a slope, with units of state per time; $h{k}_i$ is a state increment. Another common convention writes $K_i=h{k}_i$, so its stage states are $Y_i=y_n+{\sum }_j{a}_{ij}{K}_j$ and its output is $y_n+{\sum }_i{b}_i{K}_i$. These are equivalent conventions: including $h$ both inside $K_i$ and again in the output would count the time step twice. The textbook formulation uses the increment convention.^[ref]

Usually the Butcher tableau obeys $c_i={\sum }_j{a}_{ij}$, called internal consistency. Stage $i$ then combines slopes over an effective time $c_{i}h$. The second-order conditions with independent Runge-Kutta abscissae explain what changes when time nodes are specified independently. A Runge-Kutta stage is generally an auxiliary approximation, not a separate accepted solution value at its node. Several stages can even have the same time node but different state arguments, as in the classical fourth-order Runge-Kutta method.

The useful idea is to approximate the integral $y(t_n+h)-y(t_n)={\int }_{t_n}^{t_n+h}f(t,y(t))dt$ without knowing the intermediate exact trajectory. Trial states provide places to sample $f$, and Butcher order conditions make the resulting combination agree with the exact Taylor expansion. More stages alone do not guarantee a higher order of a numerical method: their coefficients must satisfy the appropriate conditions. Embedded Runge-Kutta pairs reuse these same samples for a second output and a practical local error estimator.

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.

As a quadrature rule, the trapezoidal rule replaces an integrand by the linear interpolation joining its two endpoint values:For nonnegative endpoint values, this is the area of a trapezoid; in general it is a signed area. The quadrature nodes are $a,b$, each with quadrature weight $(b-a)/2$. Unlike the midpoint quadrature rule, it uses endpoint samples, which can be reused between adjacent panels.

Every affine function $f(x)=\alpha x+\beta$ agrees with its linear interpolation, so the trapezoidal rule integrates it exactly. On $[0,2]$, $f(x)=2x+1$ gives $T(f)=(2/2)(1+5)=6=I(f)$. On $[0,1]$, the polynomial $f(x)=x^2$ instead gives $T(f)=1/2$, exceeding $I(f)=1/3$ by $1/6$. Its degree of precision of a quadrature rule is exactly one.

If $f\in C^2[a,b]$, its signed approximation error satisfiesA convex function lies below its endpoint chord, so the trapezoidal rule overestimates its integral; a concave function gives the reverse inequality. The error formula assumes that the second derivative exists and is a continuous function throughout the closed interval.^[ref] A short derivation is the identityobtained by two integrations by parts. The nonnegative Peano kernel has integral $(b-a{)}^3/12$, and the weighted mean value theorem for integrals gives the formula.

The trapezoidal rule for an ordinary differential equation averages vector fields at the old and new states and is A-stable.

This time-stepping use applies the same trapezoidal rule to the integral of the derivative in an ordinary differential equation. That connection explains its place under implicit Runge-Kutta methods; its quadrature behavior and the nonlinear issue in trapezoidal rule fails B-stability concern different uses of the same endpoint averaging.