Adams-Bashforth and Adams-Moulton methods

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DefinitionsEdit

Linear multistep methods are used for the numerical solution of ordinary differential equations, in particular the initial value problem

 

The Adams-Bashforth methods and Adams-Moulton methods are described on the Linear multistep method page.

DerivationEdit

There are (at least) two ways that can be used to derive the Adams-Bashforth methods and Adams-Moulton methods. We will demonstrate the derivations using polynomial interpolation and using Taylor's theorem for the two-step Adams-Bashforth method.

Derive the two-step Adams-Bashforth method by using polynomial interpolationEdit

From the w:Fundamental theorem of Calculus we can get

 

 

 

 

 

(1)

Set

 

 

 

 

 

(2)

To get the value of  , we can use an interpolating polynomial   as an approximation of  . The interpolation polynomial in the Lagrange form is a linear combination

 

where

 

Then the interpolation can be

 

Thus, (2 ) becomes

 

 

 

 

 

(3)

Integrating and simplifying, the right hand side of equation (3 ) becomes

 

Since  ,   and   are equally spaced, then  . Therefore, the value of   is

 

Putting this value back to (1 ) yields

 

Thus, the equation   is the two-step Adams-Bashforth method.

Derive the two-step Adams-Bashforth method by using Taylor's theoremEdit

To simplify, let's set  . Then the general form of Adams-Bashforth method is

 

 

 

 

 

(4)

where  . For the two-step Adams-Bashforth method, let's set  . Then (4 ) becomes

 

By using Taylor's theorem, expand   at   to get

 

Thus, the simplified form is

 

 

 

 

 

(5)

Expanding   at   yields

 

 

 

 

 

(6)

Subtracting (5 ) from (6 ) and then requiring the h^2 term to cancel makes  . The two-step Adams-Bashforth method is then

 

Since

 

the local truncation error is of order   and thus the method is second order. (See w: Linear multistep method#Consistency and order and w:Truncation error)

ExercisesEdit

1. Derive three-step Adams-Bashforth method by using polynomial interpolation

2. Derive the second-order Adams-Moulton method by using Taylor's theorem

Predictor–corrector methodEdit

To solve an ordinary differential equation (ODE), a w:Predictor–corrector method is an algorithm that can be used in two steps. First, the prediction step calculates a rough approximation of the desired quantity, typically using an explicit method. Second, the corrector step refines the initial approximation using another means, typically an implicit method.

Here mainly discuss about using Adams-bashforth and Adams-moulton methods as a pair to construct a predictor–corrector method.

Example: Adams predictor–corrector methodEdit

Let's start from the two-step Adams method. The prediction step is to use two-step Adams-bashforth:

 

Then, by using two-step Adams-moulton the corrector step can be:

 

Also, by using four-step Adams-bashforth and Adams-moulton methods together, the predictor-corrector formula is:

 

Note, the four-step Adams-bashforth method needs four initial values to start the calculation. It needs to use other methods, for example Runge-Kutta, to get these initial values.

Matlab programsEdit

Four-step Adams predictor-corrector method:

function [x y]=maadams4(dyfun,xspan,y0,h)
% use four-step Adams predictor-corrector method to solve an ODE y'=f(x,y), y(x0)=y0
% inputs: dyfun -- the function f(x,y), as an inline
% xspan -- the interval [x0,xn]
% y0 -- the initial value
% h -- the step size
% output: x, y -- the node and the value of y
x=xspan(1):h:xspan(2);
% use Runge-Kutta method to get four initial values
[xx,yy]=marunge(dyfun,[x(1),x(4)],y0,h);
y(1)=yy(1);y(2)=yy(2);
y(3)=yy(3);y(4)=yy(4);
for n=4:(length(x)-1)
p=y(n)+h/24*(55*feval(dyfun,x(n),y(n))-59*feval(dyfun,x(n-1),y(n-1))+37*feval(dyfun,x(n-2),y(n-2))-9*feval(dyfun,x(n-3),y(n-3)));
y(n+1)=y(n)+h/24*(9*feval(dyfun,x(n+1),p)+19*feval(dyfun,x(n),y(n))-5*feval(dyfun,x(n-1),y(n-1))+feval(dyfun,x(n-2),y(n-2)));
end
x=x';y=y';

Runge-Kutta method:

function [x y]=marunge(dyfun,xspan,y0,h)
x=xspan(1):h:xspan(2);
y(1)=y0;
for n=1:(length(x)-1)
 k1=feval(dyfun,x(n),y(n));
 k2=feval(dyfun,x(n)+h/2,y(n)+h/2*k1);
 k3=feval(dyfun,x(n)+h/2,y(n)+h/2*k2);
 k4=feval(dyfun,x(n+1),y(n)+h*k3);
 y(n+1)=y(n)+h*(k1+2*k2+2*k3+k4)/6;
end
x=x';y=y'

ReferencesEdit