Convert Linear Multistep Method into one-step method

Question

When given an nth-order ordinary differential equation, it is possible to make a simple algebraic manipulation to convert the differential equation into a first-order "autonomous" differential equation. Is there a similar transformation from a linear multistep method into a one-step method or is this in general not possible? For example, consider

$y_{n+1} = y_{n} + \frac{h}{2} \left( f(y_n) + f(y_{n+1}) \right)$.

We can "convert" this into an explicit method by

$y_{n+1} = y_{n} + \frac{h}{2} \left( f(y_n) + f(y_n + hf(y_n)) \right)$.

This makes the method go from two-step into one-step (and in fact, converts it from implicit to explicit). Can this be generalized?

Answer 1

I believe the following construction works.Consider $s$-step linear multistep method

\begin{align*} y_{n+s} = a_{s-1}y_{n+s-1} + ...+a_0y_{n}+h(b_{s}f(y_{n+s})+...+b_0f(y_n)). \end{align*}

Define

\begin{align*} A = \begin{bmatrix} a_{s-1} & a_{s-2} & ... & a_{0} \\ 1 & 0 & ... & 0 \\ 0 & 1 & ... & 0 \\ . \\ 0 & 0 & ... & 1 \end{bmatrix} \end{align*}

\begin{align*} B = \begin{bmatrix} b_{s} & b_{s-1} & ... & b_{0} \\ 0 & 0 & ... & 0 \\ 0 & 0 & ... & 0 \\ . \\ 0 & 0 & ... & 0 \end{bmatrix} \end{align*}

\begin{align*} z_n = \begin{bmatrix} y_{n+s-1} \\ . \\ . \\ . \\ y_n \end{bmatrix} \end{align*}

\begin{align*} F(z_n) = \begin{bmatrix} f(y_{n+s-1}) \\ . \\ . \\ . \\ f(y_n) \end{bmatrix}. \end{align*}

The multistep method now reads \begin{align*} z_{n+1} = A z_n + h B F(z_n). \end{align*} This is a one-step method.