# Zero-stability of multistep methods

## Contents

# 6.8. Zero-stability of multistep methods¶

For one-step methods such as Runge–Kutta, Theorem 6.2.7 guarantees that the method converges and that the global error is of the same order as the local truncation error. For multistep methods, however, a new wrinkle is introduced.

It is straightforward to check that the two-step method LIAF, defined by

is third-order accurate. Let’s apply it to the ridiculously simple IVP \(u'=u\), \(u(0)=1\), whose solution is \(e^t\). We’ll measure the error at the time \(t=1\).

```
dudt = (u,t) -> u
û = exp
a,b = 0.0,1.0;
n = [5,10,20,40,60]
err = []
t,u = [],[]
for n in n
h = (b-a)/n
t = [ a + i*h for i in 0:n ]
u = [1; û(h); zeros(n-1)]
f = [dudt(u[1],t[1]); zeros(n)]
for i in 2:n
f[i] = dudt(u[i],t[i])
u[i+1] = -4*u[i] + 5*u[i-1] + h*(4*f[i]+2*f[i-1])
end
push!( err, abs(û(b) - u[end]) )
end
pretty_table( [n (b-a)./n err],["n","h","error"] )
```

```
┌────┬───────────┬────────────┐
│ n │ h │ error │
├────┼───────────┼────────────┤
│ 5 │ 0.2 │ 0.0160452 │
│ 10 │ 0.1 │ 2.84548 │
│ 20 │ 0.05 │ 1.6225e6 │
│ 40 │ 0.025 │ 9.3442e18 │
│ 60 │ 0.0166667 │ 1.74013e32 │
└────┴───────────┴────────────┘
```

The error starts out promisingly, but things explode from there. A graph of the last numerical attempt yields a clue.

```
plot(t,abs.(u),m=3,label="",
xlabel=L"t",yaxis=(:log10,L"|u(t)|"),title="LIAF solution")
```

It’s clear that the solution is growing exponentially in time.

The source of the exponential growth in Demo 6.8.1 is not hard to identify. Recall that we can rewrite (6.8.1) as \(\rho(\mathcal{Z})u_{i-1}=h \sigma(\mathcal{Z})u_{i-1}\) using the forward shift operator \(\mathcal{Z}\):

(See (6.6.4), using \(k=2\) here.) Next, suppose that \(h\) is negligible in (6.8.2). Then the numerical solution of LIAF is roughly defined by

The graph in Demo 6.8.1 strongly suggests that for small \(h\), \(|u_i|\approx c \alpha^i\) for some \(\alpha>1\) as \(m\) gets large. So we are motivated to try defining

for all \(i\) and see if we can prove that it is an exact solution. The beauty of this choice is that for all \(i\),

Hence (6.8.3) becomes

Therefore, as \(h\to 0\), the two roots of \(z^2+4z+5\) will each correspond to an approximate solution in the form (6.8.4) of the LIAF method. These roots are \(z=1\) and \(z=-5\), and the growth curve at the end of Demo 6.8.1 is approximately \(|(-5)^i|\).

## Zero-stability¶

Here is the crucial property that LIAF lacks.

A multistep method is **zero-stable** if, as \(h\to 0\), every numerical solution produced by the method remains bounded throughout \(a\le t_i \le b\).

Without zero-stability, any truncation or roundoff error will get exponentially amplified and eventually overwhelm convergence to the exact solution.

The following theorem concisely summarizes when we can expect zero-stability.

A linear multistep method is zero-stable if and only if every root \(r\) of the generating polynomial \(\rho(z)\) satisfies \(|r|\le 1\), and any root \(r\) with \(|r|=1\) is simple.

(Partial proof, when all roots of \(\rho\) are simple.) As explained above, the values produced by the numerical method approach solutions of the difference equation \(\rho(\mathcal{Z})u_{i-k+1}=0\). We consider only the case where the roots \(r_1,\ldots,r_k\) of \(\rho(z)\). Then \(u_i=(r_j)^i\) is a solution of \(\rho(\mathcal{Z})u_i=0\) for each \(j=1,\ldots,k\). By linearity,

is a solution for any values of \(c_1,\ldots,c_k\). These constants are determined uniquely by the starting values \(u_0,\ldots,u_{k-1}\) (we omit the proof). Now, if all the roots satisfy \(|r_j|\le 1\), then

independently of \(h\) and \(i\). This proves zero-stability. Conversely, if some \(|r_j|>1\), then \(|u_i|\) cannot be bounded above by a constant independent of \(i\). Since \(b=t_i\), \(i\to\infty\) at \(t=b\) as \(h\to 0\), so zero-stability cannot hold.

A nonsimple root of \(\rho\) introduces a modification of (6.8.4) that is considered in Exercise 4.

A \(k\)-step Adams method has \(\rho(z) = z^k - z^{k-1} = z^{k-1}(z-1)\). Hence 1 is a simple root and 0 is a root of multiplicity \(k-1\). So the Adams methods are all stable.

The method \(u_{i+1} = 2u_i - u_{i-1} + h(f_i-f_{i-1})\) is first-order accurate. But \(\rho(z)=(z-1)^2\), which has a double root at \(z=1\), so it is not zero-stable.

## Dahlquist theorems¶

It turns out that lacking zero-stability is the only thing that can go wrong for a consistent multistep method.

A linear multistep method converges as \(h\to 0\) if and only if it is consistent and zero-stable.

The Dahlquist equivalence theorem is one of the most important and celebrated in the history of numerical analysis. It can be proved more precisely that a zero-stable, consistent method is convergent in the same sense as Theorem 6.2.7, with the error between numerical and exact solutions being of the same order as the local truncation error, for a wide class of problems.

You may have noticed that the Adams and BD formulas use only about half of the available data from the past \(k\) steps, i.e., they have many possible coefficients set to zero. For instance, a \(k\)-step AB method uses only the \(f_j\)-values and has order \(k\). The order could be made higher by also using \(u_j\)-values, like the LIAF method does for \(k=2\). Also like the LIAF method, however, such attempts are doomed by instability.

The order of accuracy \(p\) of a stable \(k\)-step linear multistep method satisfies

The lesson of Theorem 6.8.7 is that accuracy is not the only important feature, and trying to optimize for it leads to failure. New lessons on the same theme appear in Section 11.3.

## Exercises¶

✍ Show that the LIAF method (6.8.1) has order of accuracy equal to 3.

✍ / ⌨ Verify that the order of accuracy of the given multistep method is at least 1. Then apply Theorem 6.8.3 to determine whether it is zero-stable.

**(a)**BD2**(b)**BD3**(c)**\(u_{i+1}=u_{i-1}+2hf_i\)**(d)**\(u_{i+1} = -u_i +u_{i-1} + u_{i-2} + \frac{2h}{3}(4f_i+f_{i-1}+f_{i-2})\)**(e)**\(u_{i+1} = u_{i-3} + \frac{4h}{3} ( 2f_i - f_{i-1} + 2f_{i-2})\)**(f)**\(u_{i+1} = -2u_i + 3u_{i-1} + h (f_{i+1}+2f_i+f_{i-1})\)✍ A Fibonacci sequence is defined by \(u_{i+1}=u_i+u_{i-1}\), where \(u_0\) and \(u_1\) are seed values. Using the proof of Theorem 6.8.3, find \(r_1\) and \(r_2\) such that \(u_i=c_1(r_1)^i+c_2(r_2)^i\) for all \(i\).

✍

**(a)**Suppose that \(\rho(r) = \rho'(r) = 0\). Show that \(u_i = i r^i\) is a solution of the difference equation \(\rho(\mathcal{Z})u_i=0\).**(b)**Explain why the result of part (a) implies that a non-simple root \(r\) with \(|r|=1\) makes it impossible for a multistep method to be zero-stable.