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An algebraically stable Runge-Kutta method is B-stable: for a dissipative vector field, a well-defined step cannot increase the distance between two starting values. This assumes that the compared stage equations have solutions, as required by stage solvability of an implicit Runge-Kutta method. The Runge-Kutta contractivity identity expresses the distance change as a nonpositive weighted term coming from the dissipative vector field minus a nonnegative quadratic form involving the matrix defining algebraic stability. Positive weights also rule out stage poles in the open left half-plane on the scalar test equation, giving A-stability there; zero-weight redundant stages need separate solvability attention.

A one-sided Lipschitz condition controls the growth of distances between solutions of an ordinary differential equation. Differentiating the squared difference and applying the Gronwall inequality gives $\Vert y(t)-z(t)\Vert \le e^{\nu t}\Vert y(0)-z(0)\Vert$. The case $\nu \le 0$ is a dissipative vector field. Ordinary Lipschitz continuity implies such a bound, but the converse fails: $f(y)=-y^3$ has one-sided constant zero on the real line without being globally Lipschitz continuous.

For a Runge-Kutta method with real stage matrix $A=(a_{ij})$ and weights $b_i$, defineThe method has algebraic stability of a Runge-Kutta method ifThe matrix condition means $z^{T}Mz\ge 0$ for every real vector $z$. Its significance is nonlinear contractivity: an algebraically stable method, on well-defined stage solutions, is B-stable for a dissipative vector field satisfying $⟨f(t,y)-f(t,z),y-z⟩\le 0$. This is distinct from scalar A-stability, which concerns the linear test equation. The contractivity proof in part (b) explains the matrix criterion directly.

The stated hypotheses are precisely the algebraic criterion in part (a). To prove its stability content, compare two Runge-Kutta method steps for the same dissipative vector field. Let $d=y_n-z_n$ be the input difference, $D_i=Y_i-Z_i$ the stage differences, and $F_i=f(t_n+c_{i}h,Y_i)-f(t_n+c_{i}h,Z_i)$. ThenExpanding the squared norm of $d_{+}$ givesIn each term of the linear sum, replace $d$ by $D_i-h{\sum }_j{a}_{ij}{F}_j$. Symmetrizing the resulting double sum proves the Runge-Kutta contractivity identity:Dissipativity makes $⟨D_i,F_i⟩\le 0$, and the weights are nonnegative. For the second term, positive semidefiniteness permits a factorization $M=C^{T}C$, givingFor $h\ge 0$ both contributions in the distance-change identity are nonpositive. Therefore $\Vert y_{n+1}-z_{n+1}\Vert \le \Vert y_n-z_n\Vert$. This proves the contractivity assertion behind algebraic stability, usually called the Butcher contractivity theorem. It assumes the compared implicit stage equations have solutions; the matrix condition by itself should not be substituted for a separate stage-solvability argument.

The Butcher contractivity theorem states that an algebraically stable Runge-Kutta method is B-stable: for a dissipative vector field and any positive step for which the stage problems are solved, its step does not increase the distance between two numerical solutions. The required condition on the dissipative vector field isfor all compared states at each common time. The inner product can be real Euclidean or complex Hermitian. The stage solvability of an implicit Runge-Kutta method is assumed in this comparison, not a conclusion from a distance estimate alone.

Let $d$ be the difference of starting values, $D_i$ the differences of corresponding stages and $F_i$ the differences of their vector-field values. ThenExpanding the squared norm of the update givesSubstitute $d=D_i-h{\sum }_j{a}_{ij}{F}_j$ in the linear term. Symmetrizing the double sum proves the Runge-Kutta contractivity identityEach inner product in the first sum is nonpositive by the defining inequality for a dissipative vector field, and its weight is nonnegative. The final quadratic form is nonnegative: resolve the vectors $F_i$ into components and apply positive semidefiniteness of $M$ to each component vector. Hence $\boxed{\Vert d_{+}\Vert \le \Vert d\Vert }$, which is exactly B-stability. No linearization of the vector field has been used.

For the Dahlquist test equation $y^{\prime }=\lambda y$ with $\mathrm{Re}\lambda \le 0$, this also bounds the amplification factor by one wherever the stages are well defined. With strictly positive weights, no stage pole can occur in the open left half-plane. Indeed, if $(I-zA)v=0$ for nonzero $v$, thenwhen $\mathrm{Re}z<0$, contradicting $M⪰0$. This proves the usual A-stability corollary for the scalar stability function; boundedness from the open half-plane makes any boundary pole in that rational function removable. Zero-weight redundant stages require their own solvability treatment. The nonlinear B-stability statement is stronger than A-stability, and the two notions should not be conflated.

The criterion for algebraic stability of a Runge-Kutta method is $b_i\ge 0$ and positive semidefiniteness ofThe Runge-Kutta contractivity identity implies that this criterion is sufficient for B-stability, namely contractivity for a dissipative vector field, whenever the stage equations are defined. The implication from algebraic stability to B-stability is the relevant nonlinear theorem; A-stability alone is a linear property.

All weights here are positive, butA positive semidefinite matrix cannot have this negative quadratic value. Hence the method is not algebraically stable. Failure of this sufficient criterion alone would not be a proof of failure of B-stability; no such converse is needed for the question.

A Runge-Kutta method advances $y^{\prime }=f(t,y)$ through stagesThe Butcher tableau records $A,b,c$, normally with $c=Ae$. Linear absolute stability tests the amplification of a linear decay mode; nonlinear B-stability compares distances between two solutions of a dissipative vector field. Both describe propagation of perturbations, but they impose different conditions.

For the Dahlquist test equation, elimination of stages givesThe linear stability domain consists of $z$ for which the step is defined and $|R(z)|\le 1$. Poles or singular stage systems must be excluded, even if an output formula has a formal cancellation. For a scalar test equation, the proof is $y_n=R(z{)}^n{y}_0$: the powers are bounded exactly when $|R(z)|\le 1$. For $y^{\prime }=Ly$, the update is $R(hL)$. If $L$ is a normal matrix, unitary diagonalization of a normal matrix makes the operator norm equal to the largest scalar amplification modulus. For a diagonalizable matrix, powers are bounded by the condition number of the eigenvector matrix times the maximal scalar power. These constants must remain uniform for mesh-dependent systems. A defective unit-modulus eigenvalue can instead produce a growing Jordan block; the scalar spectral test alone does not control a general matrix.

An A-stable method accepts the entire closed left half-plane. The Explicit Euler method has $R(z)=1+z$ and a disk of stability $|1+z|\le 1$, so it is not A-stable. No nonconstant polynomial stability function can be A-stable, since it is unbounded on the negative real axis; this excludes all consistent explicit Runge-Kutta methods. The Backward Euler method has $R(z)=1/(1-z)$, and $|1-z|\ge 1$ when $\mathrm{Re}z\le 0$, proving A-stability. It is also L-stable, because $R(z)\to 0$ for large left-half-plane $z$.

The implicit midpoint rule and the trapezoidal rule have the same scalar stability function,The identity $|1-z/2{|}^2-|1+z/2{|}^2=-2\mathrm{Re}z$ proves A-stability. On the imaginary axis their amplification has unit modulus, and along the negative real axis it tends to $-1$. Thus neither is L-stable; very stiff decay can persist as alternating numerical values. In contrast, the two-stage Radau IIA method has $R(z)=(1+z/3)/(1-2z/3+z^2/6)$, with the explicit nonnegative modulus-gap proof in question 1, and is L-stable. This illustrates why L-stability matters beyond A-stability for a stiff differential equation.

For nonlinear problems, assume a real or complex Euclidean inner product and the dissipativity hypothesisfor all relevant states at each common time. Exact solutions satisfyA B-stable Runge-Kutta method reproduces this nonexpansion of distances whenever its stages exist. Put $d=y_n-{\tilde{y}}_n$, $D_i=Y_i-{\tilde{Y}}_i$, and $G_i=f(t_n+c_{i}h,Y_i)-f(t_n+c_{i}h,{\tilde{Y}}_i)$. Then $D_i=d+h{\sum }_j{a}_{ij}{G}_j$. Expansion of the output squared norm, followed by substituting $d=D_i-h{\sum }_j{a}_{ij}{G}_j$ in its linear terms, proves the Runge-Kutta contractivity identityIf $b_i\ge 0$ and $M=(m_{ij})$ is a positive semidefinite matrix, the first sum is nonpositive by dissipativity, and the second is nonnegative: expand each coordinate of the stage vectors to express it as a sum of nonnegative quadratic forms. These are exactly the conditions for algebraic stability of a Runge-Kutta method. Hence algebraic stability implies B-stability. Applying B-stability to the linear dissipative problem $y^{\prime }=\lambda y$, or its two-dimensional real form, also proves A-stability whenever stages are well defined.

The Backward Euler method has $b=A=1$, hence $M=1$; the implicit midpoint rule has $b=1$, $A=1/2$, hence $M=0$. Both are algebraically stable and B-stable. For the two-stage Radau IIA method,a positive semidefinite matrix, while both weights are positive. It therefore combines third-order accuracy, B-stability, and L-stability. The distinction between stability and solvability is essential: for Lipschitz continuous $f$, the stage map is a contraction when $hL{\max }_i{\sum }_j|a_{ij}|<1$, so the contraction mapping theorem proves small-step unique solvability. The preceding algebraic stability proof is an estimate for existing stages; it is not a theorem asserting stage existence for arbitrary step sizes and domains.

Linear A-stability is insufficient for nonlinear B-stability. The trapezoidal rule fails B-stability, even though its scalar stability function is identical to that of the B-stable implicit midpoint rule. Its Runge-Kutta method coefficients areFailure of this sufficient algebraic stability test alone would not prove failure of B-stability. An actual counterexample does: take $f(y)=-y^3$ and $h=2$. Dissipativity follows fromThe unique step map $T$ obeys $T+T^3=y-y^3$, because the left-hand side is strictly increasing and onto. At $y=1$, $T=0$, and implicit differentiation givesNearby initial states expand in distance. This proves a true nonlinear failure rather than merely failure of a coefficient criterion.

Finally, stability of a numerical method on finite intervals connects these contractivity properties to accuracy. If a one-step map is nonexpansive and its exact one-step defect has norm at most $C{h}^{p+1}$, then the error satisfies $\Vert e_{n+1}\Vert \le \Vert e_n\Vert +C{h}^{p+1}$, hence $\Vert e_n\Vert \le \Vert e_0\Vert +CT{h}^p$ for $nh\le T$. A Lipschitz step bound $1+Kh$ gives the same order using the discrete Gronwall inequality. For a stiff differential equation, the value of B-stability is that the propagation estimate need not grow with a large negative dissipative rate. The accuracy constant still needs appropriate smoothness and uniform derivative bounds; stability alone does not establish a mesh-uniform error order.

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.

A Runge-Kutta method with stage matrix $A$, weights $b$ and step size $h$ isIts order measures accuracy as $h\to 0$, while stability concerns how errors or stiff components propagate. Linear A-stability, strong stiff damping through L-stability, and nonlinear B-stability address different questions.

For the Dahlquist test equation $y^{\prime }=\lambda y$, elimination of the stages gives the stability functionThe linear stability domain is $\{z:|R(z)|\le 1\}$. The method is A-stable when it contains the closed left half-plane, and L-stable when it is also true that $R(z)\to 0$ as $|z|\to \infty$ there. These conditions describe, respectively, stability for every non-growing scalar linear mode and damping of very rapidly decaying modes.

The Forward Euler method has $R(z)=1+z$, so it fails A-stability, for example at $z=-3$. In fact no nontrivial explicit Runge-Kutta method is A-stable: its stability function is a nonconstant polynomial, which is unbounded along the negative real axis. The Backward Euler method has $R(z)=1/(1-z)$. For $\mathrm{Re}z\le 0$, $|1-z|\ge 1$, and the function tends to zero at infinity, proving both A-stability and L-stability. The implicit midpoint rule has $R(z)=(1+z/2)/(1-z/2)$. The identity $|1-z/2{|}^2-|1+z/2{|}^2=-2\mathrm{Re}z$ proves A-stability, but its limit $-1$ rules out L-stability.

For nonlinear equations, a dissipative vector field satisfies $⟨f(x)-f(y),x-y⟩\le 0$. The corresponding exact solutions contract because the time derivative of their squared distance is twice that inner product. A B-stable method preserves this contraction for every $h>0$, whenever its stages are well defined.

Let $d=y_n-{\tilde{y}}_n$, $D_i=Y_i-{\tilde{Y}}_i$, and $F_i=f(Y_i)-f({\tilde{Y}}_i)$. Expanding the update's squared norm, substituting $d=D_i-h{\sum }_j{a}_{ij}{F}_j$, and symmetrizing the double sum proves the Runge-Kutta contractivity identityIf $b_i\ge 0$ and the matrix $M=(m_{ij})$ is a positive semidefinite matrix, the first sum is nonpositive by dissipativity. The second double sum is nonnegative: sum the quadratic form of $M$ over each coordinate of the $F_i$. Thus algebraic stability implies B-stability. Applied to complex scalar linear equations with the real part of the Hermitian inner product, the same argument also implies A-stability, where the stages are defined.

The Backward Euler method has $b_1=a_{11}=1$, hence $M=(1)$, and is algebraically stable. One can prove its nonlinear contraction directly: $d_{n+1}=d+h(f(y_{n+1})-f({\tilde{y}}_{n+1}))$ gives $\Vert d_{n+1}{\Vert }^2\le ⟨d,d_{n+1}⟩\le \Vert d\Vert \Vert d_{n+1}\Vert$. The implicit midpoint rule has $a_{11}=1/2$, $b_1=1$ and $M=0$, so it is B-stable even though it is not L-stable.

Linear A-stability does not by itself imply nonlinear B-stability. The ODE trapezoidal rule has the same stability function as the implicit midpoint rule, but take the dissipative scalar field $f(y)=-\max (y,0)$ and $h=6$. For any positive starting value, its trapezoidal update is $y_{n+1}=y_n+3(f(y_n)+f(y_{n+1}))=-2y_n<0$. Thus two positive starting values have their distance doubled, violating B-stability. Its algebraic stability matrix also has a negative first diagonal entry. The field is globally Lipschitz continuous, so this is a well-defined nonlinear counterexample.

These examples distinguish accuracy, scalar stiff-mode stability, nonlinear contraction, and rapid stiff-mode damping. For implicit methods, solvability of the stage equations remains an additional requirement; a formal stability function alone does not establish it for arbitrary nonlinear problems.

For differences $D_i$ between corresponding stages and differences $F_i$ between their vector fields, a Runge--Kutta step satisfiesTo derive the identity, expand the squared norm of $d_{n+1}=d_n+h{\sum }_i{b}_i{F}_i$ and substitute $d_n=D_i-h{\sum }_j{a}_{ij}{F}_j$ in the linear term. Symmetrizing the double sum gives the displayed coefficients $m_{ij}$. For a dissipative vector field, the first sum is nonpositive when $b_i\ge 0$, and the second sum is nonnegative when the matrix $(m_{ij})$ is a positive semidefinite matrix. Thus algebraic stability implies B-stability.

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.

The trapezoidal rule is A-stable but is not B-stable. For the dissipative vector field $f(y)=-y^3$ at step size one, its uniquely defined step map $T$ obeys $T+T^3/2=y-y^3/2$. At $y=\sqrt{2}$, $T=0$ and implicit differentiation gives $T^{\prime }=-2$, so nearby states expand rather than contract. This is a concrete nonlinear obstruction, rather than merely failure of the sufficient algebraic stability criterion.