Uniform-step BDF2 satisfies the Dahlquist conditions

Setting $r_n=1$ in [F1] gives the displayed recurrence. Its coefficients are $({\alpha }_0,{\alpha }_1,{\alpha }_2)=(1,-4,3)$ and $({\beta }_0,{\beta }_1,{\beta }_2)=(0,0,2)$, so $\rho (\zeta )=3{\zeta }^2-4\zeta +1=(3\zeta -1)(\zeta -1)$ and $\sigma (\zeta )=2{\zeta }^2$.

The moments are ${\mu }_0=1-4+3=0$, ${\mu }_1=\sum j{\alpha }_j-\sum {\beta }_j=(-4+6)-2=0$, and ${\mu }_2=\sum j^2{\alpha }_j-2\sum j{\beta }_j=(-4+12)-8=0$. Thus [F2] gives order at least two. The roots are $1$ and $1\mathrm{/}3$, both simple, so the root condition holds.

Fix $T>0$ and set $K=[-1,T^3\mathrm{/}3+1]$, which contains the exact path $y(t)=t^3\mathrm{/}3$ with distance at least $1$ from its boundary. For $f(t,Y)=t^2$, the step has the unique explicit root $Z=(4Y_n-Y_{n-1}+2h{t}_{n+1}^2)\mathrm{/}3$. Supply the algorithm's oracle on these uniform-grid calls by computing this $Z$ and returning $(\mathtt{\text{unique-root}},Z)$ if $Z\in B_{n+1}$, or no-root otherwise; the field is everywhere defined and multiple roots are impossible. Direct substitution of $y(t)=t^3\mathrm{/}3$ gives exact unscaled defect $3y(t_{n+1})-4y(t_n)+y(t_{n-1})-2h{t}_{n+1}^2=-4h^3\mathrm{/}3$. Hence, if both preceding errors are at most $1\mathrm{/}6$, then $\mathrm{\mid }Z-y(t_{n+1})\mathrm{\mid }\le 5\mathrm{/}18+4h^3\mathrm{/}9<1\mathrm{/}2$ for sufficiently small $h$. Prescribing $B_{n+1}=\bar{B}(y(t_{n+1}),1\mathrm{/}2)\subset K$ therefore makes the algorithm select the unique root in a uniform local branch around the exact trajectory. The stated starting values are exact.

Apply [F3] with $p=2$ on a fixed interval: the method is consistent by step 1.2, satisfies the root condition, has order at least two, has exact starts, and satisfies the uniform local branch hypothesis by step 1.3. The exact solution is smooth and $f(t,Y)=t^2$ is state-Lipschitz with $L=0$ on $[0,T]\times K$. Therefore the global grid error is $O(h^2)$.