# Multistep methods¶

In Runge–Kutta methods we start at $$u_i$$ to find $$\mathbf{u}_{i+1}$$, taking multiple $$f$$-evaluations (stages) to achieve high accuracy. In contrast, multistep methods boost accuracy by employing more of the history of the solution, taking information from time steps $$i-1$$, $$i-2$$, etc. For the discussion in this and following sections, we introduce the shorthand notation

$f_i = f(t_i,u_i).$

A $$k$$-step multistep (or linear multistep) method is given by the difference equation

(188)$\mathbf{u}_{i+1} = a_{k-1}u_i + \cdots + a_0 u_{i-k+1} + h ( b_kf_{i+1} + \cdots + b_0 f_{i-k+1}), \quad i=k-1,\ldots,n-1,$

where the $$a_j$$ and the $$b_j$$ are constants. If $$b_k=0$$, the method is explicit; otherwise, it is implicit. In order to use (188), we also need some way of generating the starting values

$u_1=\alpha_1, \quad \ldots \quad u_{k-1}=\alpha_{k-1},$

which are otherwise undefined. In practice the starting values are often found using an RK formula.1

The difference formula (188) defines $$\mathbf{u}_{i+1}$$ in terms of known values of the solution and its derivative from the past. In the explicit case with $$b_k=0$$, equation (188) immediately gives a formula for the unknown quantity $$\mathbf{u}_{i+1}$$ in terms of values at time level $$t_i$$ and earlier. Thus only one new evaluation of $$f$$ is needed to make a time step, provided that we store the recent history. For an implicit method, however, $$b_k\neq 0$$ and (188) has the form

$\mathbf{u}_{i+1} - hb_kf(t_{i+1},\mathbf{u}_{i+1}) = F(u_i,u_{i-1},\ldots,u_{i-k+1}).$

Now the unknown $$\mathbf{u}_{i+1}$$ that we seek appears inside the function $$f$$. In general this equation is a nonlinear rootfinding problem for $$\mathbf{u}_{i+1}$$ and is not solvable in a finite number of steps by a formula. The implementation of both explicit and implicit multistep formulas is discussed in detail in Implementation of multistep methods.

As with RK formulas, a multistep method is entirely specified by the values of a few constants. Table 4 and Table 5 present some of the most well-known and important formulas. The Adams–Bashforth (AB) methods are explicit, while Adams–Moulton (AM) and backward differentiation formulas (BD) are implicit. The tables also list the methods’ order of accuracy, to be defined shortly. We adopt the convention of referring to a multistep method by appending its order of accuracy to a two-letter name abbreviation, e.g., the “AB3 method.”

Table 4 Coefficients of Adams multistep formulas. All have $$a_{k-1}=1$$ and $$a_{k-2} = \cdots = a_0 = 0$$.

name/order

steps $$k$$

$$b_k$$

$$b_{k-1}$$

$$b_{k-2}$$

$$b_{k-3}$$

$$b_{k-4}$$

AB1

1

0

1

(Euler)

AB2

2

0

$$\frac{3}{2}$$

$$-\frac{1}{2}$$

AB3

3

0

$$\frac{23}{12}$$

$$-\frac{16}{12}$$

$$\frac{5}{12}$$

AB4

4

0

$$\frac{55}{24}$$

$$-\frac{59}{24}$$

$$\frac{37}{24}$$

$$-\frac{9}{24}$$

AM1

1

1

(Backward Euler)

AM2

1

$$\frac{1}{2}$$

$$\frac{1}{2}$$

(Trapezoid)

AM3

2

$$\frac{5}{12}$$

$$\frac{8}{12}$$

$$-\frac{1}{12}$$

AM4

3

$$\frac{9}{24}$$

$$\frac{19}{24}$$

$$-\frac{5}{24}$$

$$\frac{1}{24}$$

AM5

4

$$\frac{251}{720}$$

$$\frac{646}{720}$$

$$-\frac{264}{720}$$

$$\frac{106}{720}$$

$$-\frac{19}{720}$$

Table 5 Coefficients of backward differentiation formulas. All have $$b_k\neq 0$$ and $$b_{k-1} = \cdots = b_0 = 0$$.

name/order

steps $$k$$

$$a_{k-1}$$

$$a_{k-2}$$

$$a_{k-3}$$

$$a_{k-4}$$

$$b_k$$

BD1

1

1

(Backward Euler)

1

BD2

2

$$\frac{4}{3}$$

$$-\frac{1}{3}$$

$$\frac{2}{3}$$

BD3

3

$$\frac{18}{11}$$

$$-\frac{9}{11}$$

$$\frac{2}{11}$$

$$\frac{6}{11}$$

BD4

4

$$\frac{48}{25}$$

$$-\frac{36}{25}$$

$$\frac{16}{25}$$

$$-\frac{3}{25}$$

$$\frac{12}{25}$$

There is a simple shorthand notation for a multistep method, the generating polynomials

(189)$\begin{split}\begin{split} \rho(z) &= z^k - a_{k-1} z^{k-1} - \cdots - a_0 :label: rho\\ \sigma(z) &= b_k z^k + b_{k-1}z^{k-1} + \cdots + b_0. \end{split}\end{split}$

For example, the AB3 method is completely specified by

$\rho(z) = z^3-z^2, \qquad \sigma(z) = \tfrac{1}{12}(23z^2-16z+5).$

In general, the polynomial $$\rho(z)$$ is monic (i.e., its leading term has a unit coefficient), and the degree of $$\rho$$ is the number of steps $$k$$. Furthermore, $$\operatorname{deg} \sigma(z)=k$$ for an implicit method and $$\operatorname{deg} \sigma(z) < k$$ for an explicit method. The connection with polynomials is straightforward, if a bit abstract. Let $$\mathcal{Z}$$ be a forward-shift operator, so that, for example, $$\mathcal{Z} t_i = t_{i+1}$$, $$\mathcal{Z}^3 u_{i-1} = u_{i+2}$$, etc. With this, the difference formula (188) can be written concisely as

(190)$\rho(\mathcal{Z}) u_{i-k+1} = h \sigma(\mathcal{Z}) f_{i-k+1}.$

## Truncation and global error¶

The definition of local truncation error (LTE) is easily extended to multistep methods. As with Runge–Kutta, we plug the exact solution $$\hat{u}$$ into the difference formula and see what is left over, dividing by $$h$$ to account for the order difference between local and global errors. Thus the local truncation error of a $$k$$-step multistep formula is defined as

(191)$\begin{split} \tau_{i+1}(h) = \frac{\hat{u}(t_{i+1}) - a_{k-1}\hat{u}(t_i) - \cdots - a_0 \hat{u}(t_{i-k+1})}{h} \\ \qquad - \bigl[ b_kf(t_{i+1},\hat{u}(t_{i+1})) + \cdots + b_0f(t_{i-k+1},\hat{u}(t_{i-k+1})) \bigr].\end{split}$

The order of accuracy of the method is the leading (lowest) exponent of $$h$$ in the series expansion of $$\tau_{i+1}(h)$$ around $$h=0$$. Although we shall not present the analysis, the conclusion for the multistep methods in this section is the same as for one-step methods: the order of accuracy in the global error is the same as for the local truncation error.

Example 69

The first-order Adams–Moulton method is also known as backward Euler, because its difference equation is

$\mathbf{u}_{i+1} = u_i + hf_{i+1},$

which is equivalent to a backward-difference approximation to $$u'(t_{i+1})$$. AM1 is characterized by $$\rho(z) = z-1$$ and $$\sigma(z) = z$$.

To derive the LTE, we use the definition:

$\begin{split}\begin{split} h\tau_{i+1}(h) &= \hat{u}(t_{i+1}) - \hat{u}(t_i) - hf\bigl(t_{i+1},\hat{u}(t_{i+1})\bigr) \\ &= \hat{u}(t_i) + h\hat{u}'(t_i) + \frac{h^2}{2}\hat{u}''(t_i) + O(h^3) - \hat{u}(t_i) -h \hat{u}'(t_{i+1}) \\ &= h\hat{u}'(t_i) + \frac{h^2}{2}\hat{u}''(t_i) + O(h^3) - h[\hat{u}'(t_i) + h\hat{u}''(t_i) + O(h^2)]\\ &= - \frac{h^2}{2}\hat{u}''(t_i) + O(h^3). \end{split}\end{split}$

Thus $$\tau_{i+1}(h)=O(h)$$ and AM1 (backward Euler) is a first-order method.

Example 70

The AB2 method has the formula

$\mathbf{u}_{i+1} = u_i + h\left(\frac{3}{2} f_i - \frac{1}{2} f_{i-1} \right).$

The generating polynomials are $$\rho(z)=z^2-z$$ and $$\sigma(z) = (3z-1)/2$$. We find that the method is second order from the LTE:

$\begin{split}\begin{split} h\tau_{i+1}(h) & = \hat{u}(t_{i+1}) - \hat{u}(t_i) - h\left[ \frac{3}{2}f(t_i,\hat{u}(t_i)) - \frac{1}{2}f(t_{i-1},\hat{u}(t_{i-1})) \right] \\ & = \hat{u}(t_i) + h\hat{u}'(t_i) + \frac{h^2}{2}\hat{u}''(t_i) + \frac{h^3}{6}\hat{u}'''(t_i) + O(h^4) \\ & \qquad - \hat{u}(t_i) - \frac{3h}{2}\hat{u}'(t_i) \\ &\qquad + \frac{h}{2} \bigl[\hat{u}'(t_i) - h\hat{u}''(t_i) + \frac{h^2}{2}\hat{u}'''(t_i) + O(h^3)\bigr] \\ & = \frac{5h^3}{12}\hat{u}'''(t_i) + O(h^4), \end{split}\end{split}$

so that $$\tau_{i+1}(h)=O(h^2)$$.

## Derivation of the formulas¶

Where do coefficients like those in Table 4 come from? There are different ways to answer that question, but Adams and BD methods have distinctive stories to tell. The derivation of Adams methods begins with the observation that

(192)$\hat{u}(t_{i+1}) = \hat{u}(t_i) + \int_{t_i}^{t_{i+1}} \hat{u}'(t) \, dt = \hat{u}(t_i) + \int_{t_i}^{t_{i+1}} f\bigl(t,\hat{u}(t)\bigr) \, dt.$

The integrand is unknown over the interval of integration. But we can approximate it by a polynomial interpolant by using the solution history. The polynomial can be integrated analytically, leading to a derivation of the coefficients $$b_0,\ldots,b_k$$.

Example 71

Let’s derive a one-step AM method using the two values $$(t_i,f_i)$$ and $$(t_{i+1},f_{i+1})$$. The interpolating polynomial is the linear function

$p(t) = f_i\frac{t_{i+1}-t}{t_{i+1}-t_i} + f_{i+1}\frac{t-t_i}{t_{i+1}-t_i}.$

Things become a little clearer with the change of variable $$s=t-t_i$$, and using $$h=t_{i+1}-t_i$$:

$\int_{t_i}^{t_{i+1}} p(t) \, dt = \int_0^h p(t_i+s) \, ds = h^{-1} \int_0^h [ (h-s)f_i + s f_{i+1} ]\, ds = \frac{h}{2}(f_i + f_{i+1}),$

which explains the entries for AM2 in Table~\ref{tab:adams}. The derivation also points out why this method is commonly called “trapezoid,” because like the trapezoid formula for a definite integral, we compute the exact integral of a piecewise linear interpolant.

In AB methods, the interpolating polynomial has degree $$k-1$$, which means that its interpolation error is $$O(h^k)$$. Upon integrating we get a local error of $$O(h^{k+1})$$, which reduces to a global error of $$O(h^k)$$. The AM interpolating polynomial is one degree larger, so its order of accuracy is one higher for the same number of steps.

The idea behind backward difference formulas is complementary to that for Adams: Interpolate solution values $$\mathbf{u}_{i+1},\ldots,u_{i-k+1}$$ by a polynomial $$q$$, and then, motivated by $$f(t,\hat{u})=\hat{u}'(t)$$, set

(193)$f_{i+1} =q'(t_{i+1}).$

With some algebra, the quantity $$q'(t_{i+1})$$ can be expressed as a linear combination of the past solution values to get the coefficients in Table 5. In summary, Adams methods are based on local integration, and BD methods are based on local differentiation (i.e., finite differences).

## Exercises¶

1. ✍ For each method, write out the generating polynomials $$\rho(z)$$ and $$\sigma(z)$$.

(a) AM2, (b) AB2, (c) BD2, (d) AM3, (e) AB3.

2. ✍ Write out by hand an equation that defines the first solution value $$u_1$$ produced by AM1 (backward Euler) for each IVP. (Reminder: This is an implicit formula.)

(a) $$u' = -2t u, \quad 0 \le t \le 2, \quad u_0 = 2, \quad h = 0.2$$

(b) $$u' = u + t, \quad 0 \le t \le 1, \quad u_0 = 2, \quad h = 0.1$$

(c) $$(1+x^3)uu' = x^2,\quad 0 \le x \le 3, \quad u_0=1, , \quad h = 0.5$$

3. ✍ Do the preceding problem for AM2 (trapezoid) instead of backward Euler.

4. ✍ For each method, find the leading term in the local truncation error using (191).

(a) AM2, (b) AB2, (c) BD2.

5. ✍/ ⌨ For each method, find the leading term in the local truncation error using (191). (Computer algebra is recommended.)

(a) AM3, (b) AB3, (c) BD4.

6. ✍ A formula for the quadratic polynomial interpolant through the points $$(s_1,y_1)$$, $$(s_2,y_2)$$, and $$(s_3,y_3)$$ is

$p(x) = \frac{(x-s_2)(x-s_3)}{(s_1-s_2)(s_1-s_3)}\,y_1 + \frac{(x-s_1)(x-s_3)}{(s_2-s_1)(s_2-s_3)}\,y_2 + \frac{(x-s_1)(x-s_2)}{(s_3-s_1)(s_3-s_2)}\,y_3.$

(a) Use (192) and a polynomial interpolant through three points to derive the coefficients of the AM3 method.

(b) Use (193) and a polynomial interpolant through three points to derive the coefficients of the BD2 method.

7. ✍ By doing series expansion about the point $$z=1$$, show for BD2 that

$\frac{\rho(z)}{\sigma(z)} = \log(z-1) + O\bigl( (z-1)^3 \bigr).$
8. ✍/ ⌨ By doing series expansion about the point $$z=1$$, show for AB3 and AM3 that

$\frac{\rho(z)}{\sigma(z)} = \log(z-1) + O\bigl( (z-1)^4 \bigr).$

(Computer algebra is recommended.)

1

If we must use an RK method to start anyway, why bother with multistep formulas at all? The answer is that multistep methods can be more efficient, even at the same order of accuracy.