# Numerical solution of ordinary differential equations, multistep method

I try to solve the following question, but I have no clue why we have $x'$ in the RHS:

The formula

$x_{n+1} = (1-A)x_n + A{x_{n-1}} + \frac{h}{12}[(5-A)x'_{n+1}+8(1+A)x'_n + (5A-1)x'_{n-1}]$

is known to be exact for all polynomials of degree $m$ or less for all $A$. Determine $A$ so that it will be exact for all polynomials of degree $m+1$. Find $A$ and $m$.

Any tips or advise to solve this is very welcomed.

This is a question in chapter 8.4 of the book "Numerical Analysis, 3rd edition - 2002 , David Kincaid".

This is the context:

This question is about solving ODEs using multistep method. We assume that we are going to solve an ODE such that:

$\begin{cases} x' = f(t,x) \\ x(t_0) = x_0 \end{cases}$

We integrate $x'$ :

$\int_{t_N}^{t_{n+1}} x'(t)dt = x(t_{n+1}) - x(t_n)$

and then: $x(t_{n+1}) = x(t_n) + \int_{t_N}^{t_{n+1}} f(t,x(t))dt$

Using Adams-Bashforth Formula we approximate the integration that appears in RHS:

$\int_{t_N}^{t_{n+1}} f(t,x(t))dt \approx h[Af_n + Bf_{n-1} + Cf_{n-2} + ...]$

Replacing the integration with its approximation we get:

$x(t_{n+1}) = x(t_n) + h[Af_n + Bf_{n-1} + Cf_{n-2} + ...]$

Now it becomes clear what $x_{n+1}$ is. It is the next node and is calculated by the value of previous nodes. But when the approximation of integration itself has the term $x_{n+1}$ which makes it an 'implicit' method. which is the case in the problem mentioned above.

$x_{n+1} = Ax_n + Bx_{n-1} + Cx_{n+1}$

I guess the author has replaced $Cx_{n+1}$ with [$(5-A)x'_{n+1}+8(1+A)x'_n + (5A-1)x'_{n-1}]$ probably by using Simpson's rule (equally spaced nodes), what I don't get is that why we have $x'$ there. Also how to find $A$ and $m$ ?

Or maybe the original problem has this form:

$x_{n+1} = Ax_n + Bx_{n-1} + Cx_{n-2}$

• This question is lacking context. What is $x_n$? What is $x_n^\prime$? – parsiad Apr 24 '16 at 16:39
• Ok I added a brief context that explain what this question is about. – Ehsan Apr 24 '16 at 16:55

You did write it down in the context: $x_n'$ is short for $f(t_n,x_n)$ when using the formula as numerical method and $x'(t_n)$ while evaluating the order of the method.
• Oh my god! Am I blind or what ?! you are right! ok replacing $x'_n$ with $f(t_n, x_n)$ will answer the first part of my question. Now how to find $m$? and $A$? – Ehsan Apr 24 '16 at 17:14
• You know that for $A=1$ you get the Simpson method with error order $O(h^4)$. Thus you have to try the method with $f(t,x)=t^k$ for $k=1,2,…,6$ to find where it is exact and where the value of $A$ has influence on the exactness. – Dr. Lutz Lehmann Apr 24 '16 at 17:17
• I keep getting $A=0$. – Ehsan Apr 24 '16 at 17:45
• So $A = 0$ is correct? in that case $m=2$, and therefore $m + 1 = 3$ also is exact. – Ehsan Apr 24 '16 at 18:46