Examples

Two-step Adams method

Two-step Adams method

We seek a parabola $g(x)=a{x}^2+bx+c$ with three properties:

$$\begin{array}{rl}g(t_n)& =y_n\\ g^{\prime }(t_n)& =f_n\\ g^{\prime }(t_{n-1})& =f_{n-1}.\end{array}$$

The graph $y=g(x)$ of this parabola will pass through the previous step $(t_n,y_n)$ and match the slope of the ODE at the two previous steps $(t_n,y_n)$ and $(t_{n-1},y_{n-1})$.

To find the coefficients of $g(x)$, first solve the system:

$$\begin{array}{rl}a{t}_n^2+b{t}_n+c& =y_n\\ 2a{t}_n+b& =f_n\\ 2a{t}_{n-1}+b& =f_{n-1}.\end{array}$$

(Observe that the second and third equations are independent of the first, so they should be solved initially and the result used to compute $c$ from the first equation.) Obtain:

$$\begin{array}{rl}a& =\frac{f_n-f_{n-1}}{2h}\\ b& =\frac{f_{n-1}{t}_n-f_n{t}_{n-1}}{h}\\ c& =y_n+\frac{f_n-f_{n-1}}{2h}{t}_n^2-f_n{t}_n.\end{array}$$

Then the next step value $y_{n+1}$ is given by evaluating the polynomial $g(x)$ at $t_{n+1}$:

$$\begin{array}{rl}{y}_{n+1}& =g(t_n+h)\\ & =a(t_n+h{)}^2+b(t_n+h)+c\\ & =(a{t}_n^2+b{t}_n+c)+2a{t}_{n}h+a{h}^2+bh\\ & =y_n+h(f_n-b)+a{h}^2+bh\\ & =y_h+h{f}_n+\frac{1}{2}h(f_n-f_{n-1})\\ & =y_n+\frac{3}{2}h{f}_n-\frac{1}{2}h{f}_{n-1}.\end{array}$$

This is the Adams formula for $y_{n+1}$ in terms of $f_n$ and $f_{n-1}$.