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Write the Runge-Kutta method coefficients as $A\in {\mathbb{R}}^{\nu \times \nu }$, $b\in {\mathbb{R}}^{\nu }$, and let $e$ be the all-ones column. On the test equation $y^{\prime }=\lambda y$, with $z=h\lambda$, the stage vector satisfies $Y=e{y}_n+zAY$ and the next value is $y_{n+1}=y_n+z{b}^{T}Y$. Hence its stability function isBoth polynomials have degree at most $\nu$, and the denominator has value one at zero. Order of a Runge-Kutta method $2\nu$ implies that this test-equation approximation agrees with $e^z$ through degree $2\nu$:By uniqueness of the diagonal Padé approximant, $R=[\nu /\nu {]}_{e^z}$. Explicitly,The permitted diagonal Padé approximant property says this rational function has no poles in the closed left half-plane and satisfies $|R(z)|\le 1$ there. ThereforeThere is also no hidden stage-solve singularity from a canceled denominator: the Padé approximant numerator and denominator are coprime and each has degree $\nu$. The original determinant denominator already has degree at most $\nu$, so, with its normalization at zero, it equals $P_{\nu }(-z)$ and cannot contain an additional canceled factor. This proves maximal-order Runge-Kutta methods are A-stable.

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.

Writing $f,f^{\prime },f^{\prime \prime }$ at $y_n$, the implicit stage has the expansionThus one step isThe exact Taylor expansion has third-order termThe coefficients agree through order two for every real $a$, while the $f^{\prime \prime }{f}^2$ coefficient prevents order three for any $a$. Hence the order of a Runge-Kutta method is(The linear special case has one extra matched term when $a=1/3$, but the order for general nonlinear equations remains two.)