Use Taylor's method to determine the constants a and b in the 4th order Adams-Bashforth method.

Question

Use Taylor's method to determine the constants a and b in the 4th order Adams-Bashforth method;

$$w_{i+1}=w_i + \frac{h}{24}(55f_i +af_{i-1} + 37f_{i-2} + bf_{i-3}).$$

I am not sure how to begin this question, can someone explain please?

Answer 1

Define the error function

$$L = x(t_{n+1}) − x(t_n) − \frac{h}{24}\Big(55x'(t_n) +ax'(t_{n−1}) +37x'(t_{n−2})+b x'(t_{n−3})\Big)$$

Expand $L$ in the following way

$$L = d_0x(t_n) + d_1hx'(t_n) + d_2h^2x''(t_n) + \cdots$$ by using the following expansions, for example,

$$x'(t_{n−3}) = x'(t_n) + (−3h)x'(t_n) + \cdots$$

Since by assumption, it's a $4$th-order method, we will require $d_0 = d_1 = \cdots = d_4 = 0$.

For coefficient of $h$, we have $1-\frac{1}{24}(55+a+37+b)=0$

For coefficient of $h^2$, we have $\frac{1}{2}-\frac{1}{24}(a(-1)+37(-2)+b(-3))=0$

For coefficient of $h^3$, we have $\frac{1}{6}-\frac{1}{24}(a\frac{1}{2}(-1)^2+37\frac{1}{2}(-2)^2+b\frac{1}{2}(-3)^2)=0$

For coefficient of $h^4$, we have $\frac{1}{24}-\frac{1}{24}(a\frac{1}{6}(-1)^3+37\frac{1}{6}(-2)^3+b\frac{1}{6}(-3)^3)=0$

Solve above system, you get $a=-59,b=-9$.