# 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

Answer: This is the backward Euler Method