 Solution

Let $A,b$ be the coefficients of the $\nu$-stage Runge-Kutta method. On the Dahlquist test equation $y^{\prime }=\lambda y$, its stability function isBoth polynomials have degree at most $\nu$, with denominator normalized to one at zero. Order $2\nu$ implies $R(z)-e^z=O(z^{2\nu +1})$. A rational function with these degree bounds and this accuracy is uniquely the diagonal Padé approximant $[\nu /\nu {]}_{e^z}$. To see uniqueness directly, if $P/Q$ and $\tilde{P}/\tilde{Q}$ have the required accuracy, then $P\tilde{Q}-\tilde{P}Q$ is a polynomial of degree at most $2\nu$ vanishing to order at least $2\nu +1$ at zero, hence identically zero.

The diagonal exponential Padé approximant can be writtenUse the permitted Padé property that the zeros of $P_{\nu }(-z)$ lie strictly in the right half-plane and numerator and denominator are coprime. Thus $R$ has no poles in the closed left half-plane. Its real coefficients give $P_{\nu }(-iy)=\overline{P_{\nu }(iy)}$, so $|R(iy)|=1$. Also $R(z)\to (-1{)}^{\nu }$ at infinity. Applying the maximum modulus principle on left half-disks and letting their radii increase proves $|R(z)|\le 1$ for $\mathrm{Re}z\le 0$. There are no hidden stage poles there either: the irreducible Padé denominator has degree $\nu$, exhausting the possible degree of $\det (I-zA)$, so the latter cannot contain an additional canceled factor. Every such maximal-order Runge-Kutta method is A-stable. This is maximal-order Runge-Kutta methods are A-stable; it does not assert L-stability, since the stability function does not tend to zero at infinity.