# Why is the only $k$ step method with stiff decay BDF?

I'm studying for a test and I'd like to know how justify why the only $k$-step method of order $k$ with stiff decay is BDF. By definition of stiff decay(Ascher & Petzold) a method has stiff decay if

$$|y_n-g(t_n)|\rightarrow \ 0,\qquad \text{as }h_nRe(\lambda)\rightarrow -\infty,$$

where

$$y'=\lambda(y-g(t)),$$

and $g(t)$ is an arbitrary bounded function. Assuming stiff decay and considering the definition of the general LMM I don't see why this forces $\beta_j=0$ for $j>0$. Thanks for your time.

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I do not have that book. Can you tell us what BDF and LMM are? –  Amzoti Aug 8 '13 at 0:19
@Amzoti My apologies LMM refers to linear multistep methods, those of the form $\sum\limits_{j=0}^k\alpha_jy_{n-j} =\sum\limits_{j=0}^k\beta_j f(t_{n-j},y_{n-j})$. BDF stands for backward differentiation formulae, linear multistep methods where $\beta_j=0$ for $j>0$. –  Dave Aug 8 '13 at 1:27
There should also be an $h$ in front of the RHS summation. –  Dave Aug 9 '13 at 2:29

Applying a linear multistep method to the equation $y' = \lambda(y-g(t))$ yields $$\sum_{j=0}^k \alpha_jy_{n−j} = \sum_{j=0}^k h \lambda \beta_j(y_{n-j}-g(t_{n−j})),$$ which we can rewrite as $$\sum_{j=0}^k \beta_j(y_{n-j}-g(t_{n−j})) - \frac1{h\lambda} \sum_{j=0}^k \alpha_jy_{n−j} = 0.$$ In the limit that $h\lambda \to -\infty$, the second term goes to zero, so this becomes $$\sum_{j=0}^k \beta_j(y_{n-j}-g(t_{n-j})) = 0,$$ which we can re-arrange as $$y_n - g(t_n) = \frac{1}{\beta_0} \sum_{j=1}^k \beta_j(y_{n-j}-g(t_{n-j})).$$ Since this needs to be zero for all values of $y_{n-j} - g(t_{n-j})$, the coefficients $\beta_j$ ($j=1,\dots,k$) need to be zero.
Okay that's where I'm confused then because when applied to the general method we have $$\sum\limits_{j=0}^k \alpha_jy_{n−j}=\sum\limits_{j=0}^k h \lambda \beta_j(y_{n-j}-g(t_{n−j})).$$ So where do the left hand side terms go? –  Dave Aug 9 '13 at 2:27