# I Derivation of second order Adams-Bashforth

1. May 9, 2017

### goggles31

My notes state that the method is constructed based on the idea:
yk+1=yk+∫f(x,y)dx where the integral is taken from xk to xk+1
We can estimate the integral by considering
∫f(x)dx (from xk to xk+1) =c0fk+c1fk-1
To simplify the equation, we move xk to the origin such that
∫f(x)dx (from 0 to h) =c0f(0)+c1f(-h)

Starting from below, I start to get confused.
It says "Replace f(x) with the polynomials, we have"
f(x)=1 : h=c0(1)+c1(1)
f(x)=x : h2/2=c0(0)+c1(-h)
Solving for c0 and c1,
c0=3h/2 and c1=-h/2 giving
yk+1=yk+h/2*(3fk-fk-1) + O(h3)

I've gone through the working and understand where the numbers come from, but I have no idea why they replace f(x) with 1 and x. Why not x2 or x3? I've gone through some books but they derive this with the Taylor's series and I also understand that. I just don't understand the part I mentioned.

2. May 9, 2017

### eys_physics

As I understand the point is to choose the constants $c_0$ and $c_1$ such that the integral $\int_0^h f(x) dx$ is exact if f(x) is a polynomial of degree $n\leq 1$, i.e. for all constants and linear polynomials. All such polynomials can be written in the form $p_1(x)=a\cdot 1+b\cdot x$, for some constants $a$ and $b$. We can thus consider $1$ and $x$ as basis functions. If we determine $c_0$ and $c_1$ such that $\int_0^h 1 dx$ and $\int_0^h x dx$ are reproduced exactly, $\int_0^h p_1(x) dx$ can be expressed in terms of the aforementioned integrals over $1$ and $x$ .

3. May 9, 2017

### FactChecker

Those are the two simplest linearly independent examples of f(x) that the method needs to work for. So they are the easiest ones to use to see what c0 and c1 should be. It would be more complicated to use x2 or x3. I haven't thought it through, but those may be so complicated that they won't give you the solution for c0 and c1, but I think they will work also.

PS. Even if x2 and x3 work, they may give a different answer and not be the Adams-Bashforth method.

4. May 9, 2017

### FactChecker

Yes. And it is wise to match the lower order polynomials before worrying about matching higher orders. So that makes good sense.