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With $c=Ae$ and $C=\mathrm{diag}(c)$, a Runge-Kutta method has order at least four exactly when the eight Butcher order conditionshold; powers of $c$ are componentwise. These conditions match the numerical and exact coefficients indexed by rooted trees through order four. Matching the scalar Dahlquist test equation alone is insufficient to check the nonlinear Butcher order conditions.

The two-stage Gauss--Legendre Runge-Kutta method uses nodes $c_{1,2}=1/2\mp \sqrt{3}/6$, weights $b_1=b_2=1/2$ and coefficients $a_{11}=a_{22}=1/4$, $a_{12}=1/4-\sqrt{3}/6$, $a_{21}=1/4+\sqrt{3}/6$. The Butcher order conditions give order four. Its stability function is the displayed rational function, which is A-stable. The algebraic stability matrix is zero, so positive weights also give nonlinear contractivity for dissipative vector fields.

A Lobatto IIIA method collocates at the Lobatto quadrature nodes, including both endpoints of the time interval. Its three-stage version hasThe Butcher order conditions give order four. Its stability function iswhich is A-stable but not L-stable. It is not algebraically stable: the first diagonal entry of the defining matrix is $2b_1{a}_{11}-b_1^2=-1/36$. This distinguishes linear A-stability from nonlinear algebraic stability.

The original PDF has a sign defect in the lower-left stage coefficient. Put $d=c_2-c_1$. With the printed negative sign, the row sums and the relevant Butcher order conditions areThus the claimed universal second order is false. For example $c_1=0,c_2=1$ gives, on $y^{\prime }=y$, two stages equal to $y$ and the update $Y=(1+h)y$. This is the first-order Forward Euler method. For the literal tableau, order at least two holds exactly when $c_1=1/2$ or $c_2=0$, subject to distinct allowed nodes.

For completeness the literal tableau's A-stability can also be classified algebraically. DefineThe stability function obtained from $R(z)=1+z{b}^T(I-zA{)}^{-1}e$ isIf $D\ge 0$, the denominator has no zero in the closed left half-plane: its reciprocal roots are the stage-matrix eigenvalues, whose positive sum and nonnegative product give positive real parts for nonzero eigenvalues. On the imaginary axis,Consequently for the literal tableau with uniquely solvable stages throughout the left half-plane, the complete parameter criterion isThese explicit inequalities in $c_1,c_2$ are necessary by the small and large imaginary parameters, and sufficient by the maximum modulus principle, including the bound at infinity. For $D=0$ the last inequality forces $r=0$, leaving the stable linear-fractional case. For $D<0$ there is a negative real stage pole. If one defines stability solely through an analytically continued scalar stability function, the only additional possibilities areIndeed these are precisely the cases where the left pole cancels and $R(z)=[1+(1-q)z]/[1-qz]$ is an A-stable theta method. The original stage equations are nevertheless singular at $z=-1/(q-a)$, so they do not give uniquely determined stages there. This distinction prevents a canceled rational factor from concealing an ill-defined stage solve.

The natural intended repair is to replace just the lower-left coefficient by $+c_2^2/(2d)$. This makes the method the collocation Runge-Kutta method on the two nodes. To derive the repaired entries, takeIntegration gives the other three printed entries and the weights unchanged, with the positive lower-left entry. Since $L_1+L_2=1$ and $c_1{L}_1+c_2{L}_2=t$, it follows that $Ae=c$, $b^{T}e=1$ and $b^{T}c=1/2$. Hence the corrected family is always of order at least two.

For the corrected family let $a=(c_1+c_2)/2$, $D=c_1{c}_2/2$ and $r=(1-c_1)(1-c_2)/2$. Its stability function has the same displayed rational form, now with $q=1/2$. The denominator again has no left-half-plane zero, and on the imaginary axis the squared-modulus difference is $(D^2-r^2)y^4$. Because $D,r\ge 0$, the necessary and sufficient inequality is $D\ge r$, orFor sufficiency, $|R|\le 1$ on the imaginary axis and at infinity; apply the maximum modulus principle on growing left half-discs to get the same bound inside. If $D=0$, the condition leaves the endpoint trapezoidal rule and its linear denominator. If $c_1+c_2<1$, every nonzero imaginary argument gives $|R|>1$. This is the two-node collocation A-stability criterion, explicitly distinguished from the literal defective tableau.

The nodes are $c=(0,1/2,1{)}^T$. Their Lagrange interpolation basis on the normalized step isIntegrating these polynomials from zero to each node gives $a_{ij}={\int }_0^{c_i}{\ell }_j(s)ds$. At $c_1=0$ the row vanishes; at $c_2=1/2$ the integrals are $(5/24,1/3,-1/24)$; at $c_3=1$ they are $(1/6,2/3,1/6)$. The endpoint integrals also give $b=(1/6,2/3,1/6{)}^T$. These are exactly the coefficients in the original PDF.

Indeed, the degree-three polynomial $q$ determined by $q(0)=y_n$ and $q^{\prime }(hs)={\sum }_j{\ell }_j(s)f(q(h{c}_j))$ satisfies $q^{\prime }(h{c}_i)=f(q(h{c}_i))$. Integrating it to the nodes gives precisely the stage equations, and integrating to the endpoint gives the update. This proves it is a collocation Runge-Kutta method, specifically the three-stage Lobatto IIIA method.

For an explicit order proof, the collocation integrals imply $A1=c$, $Ac=c^2/2$ and $A{c}^2=c^3/3$, with powers componentwise. The endpoint weights are Simpson weights, so $b^T{c}^j=1/(j+1)$ for $j=0,1,2,3$. Therefore the eight Butcher order conditions through order four are satisfied:For example the last expression is $b^T{c}^3/6$, because $A^{2}c=A{c}^2/2=c^3/6$. These conditions are obtained by matching all elementary differential terms through degree four in the exact and numerical Taylor expansions, not just by testing a scalar linear equation. Nonautonomous equations are included by adjoining $t^{\prime }=1$.

To exclude higher order, the scalar stability function computed below satisfies $R(z)-e^z=-z^5/720+O(z^6)$. The local defect is therefore nonzero at degree five even on the scalar test equation. Hence $\boxed{\text{the classical order is exactly }4}$.

The tableau as printed is not a collocation tableau. Its second stage row is $(3/4,3/4)$, whose sum is $3/2$, while the corresponding printed node is $1$. This violates the collocation tableau row-sum identity. The original PDF confirms this coefficient; it is not only a TeX transcription defect.

The required integrated basis at nodes $1/3,1$ isso direct integration givesThus replacing the last stage coefficient by $1/4$ gives the intended two-stage Radau IIA collocation method. Its quadrature moments areIt integrates quadratics exactly but not cubics, so the collocation order theorem gives order three. Independently, $b^{T}Ac=1/6$, confirming the remaining third-order Butcher order condition.

For completeness the literal printed Runge-Kutta method has only order one. For the autonomous equation $y^{\prime }=y$, the Taylor expansion of its stability function beginswhereas $e^z=1+z+z^2/2+O(z^3)$. Since $b^{T}1=1$, it is consistent, but this discrepancy already rules out order two. Therefore neither the requested equivalence nor the third-order conclusion is true without the stated coefficient repair.

The coefficients are those of the three-stage Lobatto IIIA method. Set $e=(1,1,1{)}^T$, $c=Ae=(0,1/2,1{)}^T$ and $C=\mathrm{diag}(c)$. Direct calculation gives all eight fourth-order conditions for a Runge-Kutta method:Powers of $c$ are componentwise. These Butcher order conditions prove order at least four for a smooth general nonlinear vector field; matching a scalar test equation alone would not suffice. A necessary fifth-order condition fails:Hence the method has order exactly four, with one-step local truncation error $O(h^5)$ and, under stability and smoothness hypotheses, global error $O(h^4)$.

The complete Butcher tableau in the PDF suppliesThese are the two Gauss nodes and the corresponding Gauss--Legendre Runge-Kutta method. We can verify its order directly rather than infer it from its name. Write $e=(1,1{)}^T$, $C=\mathrm{diag}(c)$ and let powers of $c$ be componentwise. The Butcher order conditions through order four areFor these coefficients $Ae=c$, $Ac=c^2/2$ and $c^2=c-e/6$. Also $b^T{c}^j=1/(j+1)$ for $j=0,1,2,3$, as follows by averaging the two values $1/2\pm s$. These identities give all eight displayed Butcher order conditions; for example $b^{T}CAc=b^T{c}^3/2=1/8$, $b^{T}A{c}^2=b^T(c^2/2-c/6)=1/12$, and $b^T{A}^{2}c=b^{T}A{c}^2/2=1/24$. They establish order at least four for smooth nonlinear ordinary differential equations.

To exclude order five, apply the method to the Dahlquist test equation. Solving its stage equations gives the stability functionThe exact solution has coefficient $1/120$ at degree five. Hence the method has order exactly four; its one-step defect is $O(h^5)$.

The Lagrange interpolation polynomials for these nodes areTheir integrals give the Lobatto IIIA method tableauTo verify its nonlinear order, not just its scalar linear order, set $e=(1,1,1{)}^T$, $c=Ae$ and $C=\mathrm{diag}(c)$. Direct multiplication verifies the fourth-order conditions for a Runge-Kutta method:Powers of $c$ are componentwise. These eight Butcher order conditions establish order at least four for general smooth ODEs. On the Dahlquist test equation, elimination of the stages givesA nonzero fifth-order step defect rules out order five. Thus the method has exactly order four.

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 tableau is the three-stage Lobatto IIIA method. To check its order of a Runge-Kutta method directly, write $e=(1,1,1{)}^T$, $c=Ae=(0,1/2,1{)}^T$, and $C=\mathrm{diag}(c)$. The eight fourth-order conditions for a Runge-Kutta method evaluate toPowers of $c$ here mean componentwise powers. For example, $Ac=(0,1/8,1/2{)}^T$, $A{c}^2=(0,1/24,1/3{)}^T$, and $A^{2}c=(0,1/48,1/6{)}^T$, which make the last three checks immediate. These Butcher order conditions establish order at least four for general nonlinear ordinary differential equations.

It is not order five: part (b)'s stability function has expansionThe fifth coefficient already fails on the Dahlquist test equation. The order is exactly four.

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.

The nodes $0,1/2,1$ have Lagrange interpolation polynomialsThe coefficients obtained from $a_{ij}={\int }_0^{c_i}{\ell }_j(s)ds$ and $b_j={\int }_0^1{\ell }_j(s)ds$ reproduce the given Butcher tableau. In particular, the second row's last entry is $-1/24$, which is missing from the TeX transcription. Thus this is the three-stage Lobatto IIIA method.

One applicable collocation theorem is that $s$-stage Lobatto IIIA collocation has order $2s-2$. Alternatively, the fourth-order conditions for a Runge-Kutta method give a direct verification. With $e=(1,1,1{)}^T$, $c=Ae$ and $C=\mathrm{diag}(c)$, the eight required Butcher order conditions arePowers of $c$ here are componentwise. Substitution satisfies all eight. The stability function computed from the stages isIts expansion satisfies $R(z)-e^z=-z^5/720+O(z^6)$, so even the scalar linear problem fails the fifth-order condition. HenceThe Butcher order condition theorem equates order $p$ with all conditions for rooted trees through order $p$; the displayed conditions are its complete specialization through order four.

The tableau is the three-stage Lobatto IIIA method. For its stage matrix $A$, weights $b$, nodes $c$ and $C=\mathrm{diag}(c)$, the Butcher order conditions through order four arewhere $c^{\circ j}$ means coordinatewise powers. Direct substitution verifies all eight identities. Equivalently, this is a collocation Runge-Kutta method at the three Lobatto quadrature nodes, whose weights integrate cubic polynomials exactly.

Order five would require $b^T{c}^{\circ 4}=1/5$, but here $b^T{c}^{\circ 4}=5/24$. Therefore the method has exactly order four.

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.