Linear multistep methods and Dahlquist equivalence — Examples

Expanding gives $\rho (\zeta )={\zeta }^3-2{\zeta }^2+\frac{5}{4}\zeta -\frac{1}{4}$, so $\rho (1)=0$ and ${\rho }^{\mathrm{\prime }}(1)=\frac{1}{4}=\sigma (1)$. Hence [F1] gives consistency.

The factorization displays roots $1$ and $1\mathrm{/}2$ with multiplicities one and two, respectively. The unit root is simple; the repeated root has modulus strictly below one. Thus the root condition holds.

Apply [F2] to the root condition in step 1.2. The companion powers are uniformly bounded. In the scalar recurrence, the repeated-root mode $n{2}^{-n}$ is bounded as well, since $n{2}^{-n}\to 0$.

Expanding gives $\rho (\zeta )={\zeta }^3-{\zeta }^2+\zeta -1$. Thus $\rho (1)=0$, ${\rho }^{\mathrm{\prime }}(1)=2=\sigma (1)$, and [F1] proves consistency. Since $i^2=-1$ by [F5], the factorization displays three distinct roots $1,i,-i$, all of modulus one, so the root condition holds.

By [F2] and [F4], $i=\exp (i\pi \mathrm{/}2)$ and $\mathrm{\mid }i\mathrm{\mid }=\mathrm{\mid }-i\mathrm{\mid }=1$. Repeated use of [F3] gives $i^n=\exp (in\pi \mathrm{/}2)$, so $i^n$ has modulus one by [F2]; also $(-i{)}^n=(-1{)}^n{i}^n$ has modulus one. Each is a recurrence mode because substituting $Y_n={\lambda }^n$ multiplies the homogeneous recurrence by $\rho (\lambda )$.

Apply [L1] to the root condition in step 1.1. The full homogeneous companion recurrence is bounded, including these periodic modes; no decay of unit-circle modes is asserted.

Here $\rho (1)=0$ and ${\rho }^{\mathrm{\prime }}(1)=0=\sigma (1)$, so [F1] proves consistency.

The characteristic polynomial has a root $1$ of multiplicity two, so it violates the root condition in Linear multistep methods, consistency, order, and the root condition.

Apply the recurrence to $y^{\mathrm{\prime }}=0$, whose exact solution is $y=0$, with $Y_0=0$ and $Y_1=h$. The sequence $Y_n=nh$ satisfies the recurrence because its second difference is zero.

On the grid $h=T\mathrm{/}N$, the starting error is $h\to 0$, but $\mathrm{\mid }{Y}_N-y(T)\mathrm{\mid }=Nh=T$. Thus convergence fails even for starting errors of order $h$, and the proposed claim is false.

Set $z=h\lambda$ and $Y_n=1$. By [F1] and [L1], the Euler predictor is $P=1+z$. The PECE correction is $Y_{n+1}=1+\frac{z}{2}(1+P)=1+z+z^2\mathrm{/}2$; the final evaluation in PECE stores $\lambda Y_{n+1}$ and does not alter this value.

The fully implicit trapezoidal corrector instead solves $Y_{n+1}=1+\frac{z}{2}(1+Y_{n+1})$, so $(1-z\mathrm{/}2)Y_{n+1}=1+z\mathrm{/}2$ and its multiplier is $(1+z\mathrm{/}2)\mathrm{/}(1-z\mathrm{/}2)$ when $z\mathrm{\ne }2$.

At $z=-3$, the multipliers are $5\mathrm{/}2$ and $-1\mathrm{/}5$, respectively. Define scalar test stability at $z$ by the condition that the multiplier's modulus is at most one. Hence PECE is unstable there and the fully implicit Adams method is stable there, so their stability differs.

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 $O(h^2)$.

Summing the increments gives $m^{-2}+m^{-1}+(m-2)m^{-1}+m^{-1}-m^{-2}=1$, and every increment is positive for $m\ge 2$. The maximum is $H_m=m^{-1}$.

The first ratio is $r_1=h_1\mathrm{/}{h}_0=m$ and the starter increment is ${\mathrm{\Delta }}_1=1\mathrm{/}m\to 0$.

For $n\ge 2$ the grid has $r_n\le 1$, so [L1] gives $0<q(r_n)=r_n^2\mathrm{/}(1+2r_n)\le r_n\mathrm{/}2\le 1\mathrm{/}2$. The increments from ${\mathrm{\Delta }}_2$ onward are positive and their total is at most ${\mathrm{\Delta }}_2{\sum }_{j\ge 0}{2}^{-j}=2{\mathrm{\Delta }}_2<1$. Since $Y_1=1\mathrm{/}m\le 1\mathrm{/}2$, every computed value is less than $3\mathrm{/}2$. For $f=0$ the step equation has exactly one root because its coefficient $1+2r_n$ is positive, so all roots lie in the prescribed balls and the algorithm completes.

By [L1], ${\mathrm{\Delta }}_2=\frac{m^2}{1+2m}\frac{1}{m}=\frac{m}{1+2m}\to \frac{1}{2}$. Therefore $Y_2=Y_1+{\mathrm{\Delta }}_2\to \frac{1}{2}$, while $t_2=h_0+h_1\to 0$ and the exact solution remains zero.

Thus a vanishing starter perturbation produces an error bounded away from zero even though $H_m\to 0$. The claimed ratio-free uniform zero-stability fails.