# 1.5 Zero Stability and the Dahlquist Equivalence Theorem

## 1.5.2 Stability

The remaining issue to determine is whether the solutions to the multi-step method can grow unbounded as $${\Delta t}\rightarrow 0$$ for finite time $$T$$. Consider again the $$s$$-step multi-step method:

 $v^{n+1} + \sum _{i=1}^ s \alpha _ i v^{n+1-i} = {\Delta t}\sum _{i=0}^ s \beta _ i f^{n+1-i}.$ (1.76)

In the limit of $${\Delta t}\rightarrow 0$$, the multi-step approximation will satisfy the following recurrence relationship,

 $v^{n+1} + \sum _{i=1}^ s \alpha _ i v^{n+1-i} = 0. \label{equ:stability_ recurrence}$ (1.77)

This recurrence relationship can be viewed a providing the characteristic or unforced behavior of the multi-step method. In terms of stability, the question is whether or not the solutions to Equation 1.77 can grow unbounded.

## Definition of (Zero) Stability

A multi-step method is stable (also known as zero stable) if all solutions to

 $v^{n+1} + \sum _{i=1}^ s \alpha _ i v^{n+1-i} = 0,$ (1.78)

are bounded as $$n\rightarrow \infty$$.

To determine if a method if stable, we assume that the solution to the recurrence has the following form,

 $v^ n = v^0 z^ n,$ (1.79)

where the superscript in the $$z^ n$$ term is in fact a power. Note: $$z$$ can be a complex number. If the recurrence relationship has solutions with $$|z|>1$$, then the multi-step method would be unstable. For our purposes, a multi-step method with a root of $$|z|=1$$ is zero-stable provided the root is not repeated. (Note that in general, we need to be careful with the case of $$|z|=1$$.)

## Stability of Most Accurate Two-Step Method

In Section 1.3.4 , the most accurate two-step, explicit method was found to be,

 $v^{n+1} + 4v^ n - 5v^{n-1} = {\Delta t}\left(4f^ n + 2f^{n-1}\right).$ (1.80)

We will determine if this algorithm is stable. The recurrence relationship is,

 $v^{n+1} + 4v^ n - 5v^{n-1} = 0.$ (1.81)

Then, substitution of $$v^ n = v^0 z^ n$$ gives,

 $z^{n+1} + 4z^ n - 5z^{n-1} = 0.$ (1.82)

Factoring this relationship gives,

 $z^{n-1}\left(z^2 + 4z - 5\right) = z^{n-1}(z-1)(z+5) = 0.$ (1.83)

Thus, the recurrence relationship has roots at $$z=1$$, $$z = -5$$, and $$z=0$$ ($$n-1$$ of these roots). The root at $$z=-5$$ will grow unbounded as $$n$$ increases so this method is not stable. By the Dahlquist Equivalence Theorem, this means the method is not convergent (even though it has local accuracy $$p=3$$ and is therefore consistent).

To demonstrate the lack of convergence for this method (due to its lack of stability), we again consider the solution of $$u_ t = -u^2$$ with $$u(0) = 1$$. These results are shown in Figure 1.7.

Figure 1.7: Most-accurate explicit, two-step multi-step method applied to $$\dot{u} = -u^2$$ with $$u(0) = 1$$ with $${\Delta t}= 0.1$$ (upper plot) and $$0.05$$ (lower plot).

These results clearly show the instability. Note that the solution oscillates as is expected since the large parasitic root is negative ($$z = -5$$). Furthermore, decreasing $${\Delta t}$$ from $$0.1$$ to $$0.05$$ only causes the instability to manifest itself in shorter time (though the same number of steps). Clearly, though the method is consistent, it will not converge because of this instability.