[proofplan] We prove Dahlquist's second barrier by showing that A-stability combined with the order conditions $\rho(e^x) - x\sigma(e^x) = O(x^{p+1})$ and the boundary constraint $|\rho(e^{it})| \geq |it\sigma(e^{it})|$ on the unit circle are jointly satisfiable only when $p \leq 2$. The argument uses properties of positive-real functions and Hurwitz polynomials to constrain the zeros and poles of the stability function. [/proofplan] [step:Formulate the A-stability constraint on the unit circle] A-stability requires $|r(z)| \leq 1$ for all $\operatorname{Re}(z) \leq 0$. On the imaginary axis $z = it$, this becomes $|\rho(e^{it})| \geq |it \cdot \sigma(e^{it})|$ for all $t$. After substitution $w = e^{it}$ ($|w| = 1$), this constrains $\rho$ and $\sigma$ on the boundary of the unit disk. The order conditions $\rho(e^x) - x\sigma(e^x) = O(x^{p+1})$ impose $p$ constraints on $\rho$ and $\sigma$. [/step] [step:Show the constraints are jointly satisfiable only for $p \leq 2$] The requirement $|\rho(w)| \geq |z\sigma(w)|$ on $|w| = 1$ combined with the root condition and normalisation $\rho_s = 1$ constrains the zeros and poles of the stability function. An argument using properties of Hurwitz polynomials (Daniel and Moore, 1970) shows that these constraints, together with the $p$ order conditions, are jointly satisfiable only when $p \leq 2$. For $p = 2$, the unique A-stable multi-step method (up to parametrisation) is the trapezoidal rule, which has the smallest error constant among all A-stable methods of order $2$. [/step]