Past exam of the mathematics course of the University of Cambridge / 2010 / ii / Paper 4 / 39A (Numerical Analysis)

For an explicit Runge-Kutta method, write its strictly lower-triangular stage matrix as $A$, weight vector as $b$, and $1=(1,\dots ,1{)}^T$. Applied to the scalar test equation, its stage vector satisfiesSince $A^s=0$, inversion is a finite Neumann series. Putting $z=\lambda h$ givesThis proves degree at most $s$. The printed statement of degree exactly $s$ needs a nondegeneracy assumption: a two-stage explicit scheme with $b=(1,0)$ is simply Euler's method with an unused second stage, and has polynomial $1+z$ of degree one.

Order $p$ applied to this particular differential equation forces the local stability polynomial to agree with the exact exponential through degree $p$:As its degree is at most $s$, the nonzero coefficient $1/p!$ cannot be present for $p>s$. ThereforeThe implication is a necessary stability-polynomial condition; it does not assert that every such polynomial automatically gives the full nonlinear order conditions.

An A-stable method must satisfy $|P_s(z)|\le 1$ for every $z$ with nonpositive real part. A consistent explicit Runge-Kutta method has $P_s^{\prime }(0)=1$, so its stability polynomial is nonconstant. Along the negative real axis the leading term of any nonconstant polynomial has magnitude tending to infinity. HenceThe positive order assumption in the question supplies the required consistency.