= Solution
For a fixed-step <linear multistep method> with a nonzero leading coefficient, applied to an <ordinary differential equation> with a locally <Lipschitz continuous> vector field, the <Dahlquist equivalence theorem> states that
$$
\boxed{\text{convergence}\ \Longleftrightarrow\
\text{consistency and zero-stability}.}
$$
Here convergence is uniform on each fixed finite time interval as the <step size> tends to zero, for every set of starting values tending to the corresponding exact values. <Consistency of a numerical method> means that the normalized local defect tends to zero; for the <characteristic polynomials of a linear multistep method> it gives $\rho(1)=0$ and $\rho'(1)=\sigma(1)$. <Zero-stability> is equivalent to the <root condition for a multistep method>: every <root of a polynomial> $\xi$ of $\rho$ satisfies $|\xi|\leq1$, and roots with $|\xi|=1$ are simple. Order $p\geq1$, together with <zero-stability> and starting errors $O(h^p)$, gives <global error> $O(h^p)$ under the usual smoothness assumptions.
To prove the <necessity of the root condition for multistep convergence>, use the test problem $y'=0$, $y(0)=0$. Its discrete error satisfies $\rho(E)e_n=0$, where $Ee_n=e_{n+1}$. Set $h=T/N$. If $|\xi|>1$, the exact recurrence solution
$$
e_n=|\xi|^{-N}\xi^n
$$
has starting errors tending to zero at every fixed starting index, but $|e_N|=1$. Thus convergence fails.
If a unit-modulus root has multiplicity $m\geq2$, the recurrence instead admits
$$
e_n=N^{-(m-1)}n^{m-1}\xi^n.
$$
Again every fixed starting error tends to zero while $|e_N|=1$. These polynomial-times-exponential solutions follow from the repeated factor $(E-\xi)^m$. Complex modes can be interpreted as a real two-component system, or their real and imaginary parts can be used. Therefore \b[convergence requires precisely the <root condition for a multistep method>]. Repeated roots strictly inside the unit disk are allowed, since their polynomial factors are dominated by exponential decay.
Back to article page
