First Dahlquist barrier

For a consistent zero-stable real linear $k$-step method with constant coefficients and only first-derivative evaluations, the attainable order is at most $k+1$ for odd $k$ and $k+2$ for even $k$. For an explicit method the bound is $k$. These are order barriers under the root condition for a multistep method, not bounds on multiderivative methods or on arbitrary variable-step formulations. The Second Dahlquist barrier adds the stronger bound of order two for irreducible A-stable methods.

Using Newton-Cotes closed quadrature for the integrated derivative yields the displayed linear multistep method. Its first characteristic polynomial is $\rho (z)=z^s-1$, whose unit-circle roots are all simple, proving zero-stability. Quadrature exactness gives order $s+1$ for odd $s$ and at least $s+2$ for even $s$. The first Dahlquist barrier gives equality, while the Dahlquist equivalence theorem gives convergence with consistent starts. These are finite-time convergence statements, not claims of A-stability.

For a real consistent zero-stable linear multistep method, transform its first characteristic polynomial by $z=(1+x)/(1-x)$ to $(1-x{)}^s\rho (z)=xP(x)$. The root condition for a multistep method puts all finite roots of $P$ in the closed left half-plane; real linear factors and conjugate quadratic factors consequently give nonnegative coefficients, after choosing $P(0)>0$. The order criterion requires a polynomial of degree at most $s$ to agree through degree $p-1$ with $P(x)x/(2\mathrm{arctanh}x)$. The latter quotient has strictly negative nonconstant even coefficients. Its first even coefficient beyond degree $s$ includes a strictly negative multiple of $P(0)$ and cannot vanish. This gives the stated bound.

Write $F(x)={\int }_0^1(1+x{)}^t(1-x{)}^{1-t}dt=x/\mathrm{arctanh}x$. Pairing $t$ and $1-t$ after differentiating twice givesThe power series of $(1-x^2{)}^{-3/2}$ has strictly positive even coefficients, and the other factor has nonnegative even coefficients. Thus every even coefficient of $F^{\prime \prime }$ is negative. Since $F$ is even with $F(0)=1$, the displayed sign pattern follows. This supplies a short positivity argument for the first Dahlquist barrier.