# 1.5 Zero Stability and the Dahlquist Equivalence Theorem

## 1.5.1 Consistency

Measurable Outcome 1.8

As given in Section 1.3.4, an \(s\)-step multi-step method can be written as,

\[v^{n+1} + \sum _{i=1}^ s \alpha _ i v^{n+1-i} - {\Delta t}\sum _{i=0}^ s \beta _ i f^{n+1-i} = 0,\] | (1.67)

where the forcing terms have been moved to the left-hand side. Substituting the exact solution, \(u(t)\), into the left-hand side will produce a remainder which is in fact the opposite of the truncation error (see Equation 1.49),

\[u^{n+1} + \sum _{i=1}^ s \alpha _ i u^{n+1-i} - {\Delta t}\sum _{i=0}^ s \beta _ i f^{n+1-i} = u^{n+1} - N(u^{n+1},u^ n, \ldots , {\Delta t}) = -\tau \label{equ:lte_ consistency}\] | (1.68)

If we only require that \(\tau \rightarrow 0\) (i.e. \(\tau = O({\Delta t})\)) as \({\Delta t}\rightarrow 0\), the method will not generally be consistent with the ODE. To see why, note that in the limit of \({\Delta t}\rightarrow 0\), the forcing terms will vanish since they are scaled by \({\Delta t}\). Thus, \(\tau \rightarrow 0\) would place a constraint only on the \(\alpha\)'s. Let's look at that constraint on the \(\alpha\)'s to build some insight. Substituting Taylor series about \(t=t^ n\) for the values of \(u\) gives,

\[u^{n+1} + \sum _{i=1}^ s \alpha _ i u^{n+1-i} = \left(1 + \sum _{i=1}^ s \alpha _ i\right)u^ n + O({\Delta t}).\] | (1.69)

Thus, for \(\tau \rightarrow 0\) as \({\Delta t}\rightarrow 0\) requires,

\[1 + \sum _{i=1}^ s \alpha _ i = 0. \label{equ:consistency_ con1}\] | (1.70)

This constraint can be interpreted as requiring a constant solution, i.e. \(u(t) =\) constant, to be a valid solution of the multi-step method. Clearly, this is not enough to guarantee consistency with the ODE since the ODE requires \(u_ t = f(u,t)\).

To achieve a consistent discretization, we force \(\tau /{\Delta t}\rightarrow 0\) (i.e. \(\tau = O({\Delta t}^2)\)). This stronger constraint can be shown to enforce that the ODE is satisfied in the limit of \({\Delta t}\rightarrow 0\):

| \(\displaystyle \frac{\tau }{{\Delta t}}\) | \(\displaystyle =\) | \(\displaystyle \frac{N(u^{n+1},u^ n, \ldots , {\Delta t}) - u^{n+1}}{{\Delta t}}\) |  | (1.71)
|  | \(\displaystyle =\) | \(\displaystyle \frac{N(u^{n+1},u^ n, \ldots , {\Delta t}) - \left(u^ n + {\Delta t}u_ t^ n + O({\Delta t}^2)\right)}{{\Delta t}}\) |  | (1.72)
|  | \(\displaystyle =\) | \(\displaystyle \frac{N(u^{n+1},u^ n, \ldots , {\Delta t}) - u^ n}{{\Delta t}} - u_ t^ n + O({\Delta t})\) |  | (1.73)
|  | \(\displaystyle =\) | \(\displaystyle \frac{N(u^{n+1},u^ n, \ldots , {\Delta t}) - u^ n}{{\Delta t}} - f(u^ n,t^ n) + O({\Delta t})\) |  | (1.74)

Thus, in the limit as \({\Delta t}\rightarrow 0\), to obtain \(\tau /{\Delta t}\rightarrow 0\) then the slope of the numerical method (i.e. the first term) must be equal to the forcing at \(t^ n\). In other words, the multi-step discretization would satisfy the governing equation in the limit. An equivalent way to write this consistency constraint is,

\[\lim _{{\Delta t}\rightarrow 0} \frac{1}{{\Delta t}}\left[ u^{n+1} + \sum _{i=1}^ s \alpha _ i u^{n+1-i}\right] - \sum _{i=0}^ s \beta _ i f^{n+1-i} = u_ t(t^ n) - f(u(t^ n),t^ n) = 0. \label{equ:consistency_ constraint}\] | (1.75)

In terms of the local accuracy, consistency requires that the multi-step method be at least first-order \((p=1)\) since \(\tau = O({\Delta t}^{p+1})\) and consistency requires that \(\tau /{\Delta t}= O({\Delta t}^ p)\) must go to zero (i.e. \(p \geq 1\)).

Exercise Which of the following numerical methods is consistent?

Exercise 1
&nbsp; \(v^{n+1}=v^ n + \Delta t f^{n+1}\) &nbsp;
&nbsp; \(v^{n+1} = v^ n + \frac{1}{2} \Delta t f^{n+1}\) &nbsp;
&nbsp; \(v^{n+1} = v^ n + \frac{1}{2}\Delta t f^ n\) &nbsp;
&nbsp; \(v^{n+1} = v^{n-1}\) &nbsp;

Answer: This is the backward Euler Method