# Two-Step backward differentiation

## Homework Statement

By using Taylor expansion, derive the following two-step backward differentiation which has second
order accuracy:
$$\frac{3y_{j+1}-4y_j+y_{j-1}}{2h}=f(t_{j+1},y_{j+1})$$

## Homework Equations

Taylor expansion
ODE
$$y^{\prime}=f(t,y) , y(0)=\alpha$$

## The Attempt at a Solution

I find the expansion for $$y_{j+1}=y_j+hy^{\prime}_j+\frac{h^2}{2!}y^{\prime \prime}+\cdots$$
and
$$y_{j-1}=y_j-hy^{\prime}_j+\frac{h^2}{2!}y^{\prime \prime}+\cdots$$

This is where I get stuck. If I multiply $y_{j+1}$ by 3 and add $y_{j-1}$ I get the needed left hand side but the right hand side is $f(t_{j},y_{j})=y^{\prime}_j$. How can I have an expansion that includes$f(t_{j+1},y_{j+1})$ that will yield the LHS of the derivation? Am I going about this all wrong? This problem seems relatively simple yet I think I am missing an important step.

Chestermiller
Mentor
Your equation doesn't have second order accuracy. It only has first order accuracy. Do your expansion around tj+1 and yj+1, going 1 step back and 2 steps back. You need to end up with an equation that has a zero coefficient for the second derivative.

D H
Staff Emeritus
It does have second order accuracy, Chester. This is an implicit integration technique as both the left and right hand sides involve yj+1.

stvoffutt, what this means is that you should be doing your expansions about yj+1 rather than yj.

Chestermiller
Mentor
It does have second order accuracy, Chester. This is an implicit integration technique as both the left and right hand sides involve yj+1.

stvoffutt, what this means is that you should be doing your expansions about yj+1 rather than yj.
I think you might have misunderstood what I said. The original equation does have second order accuracy, but stvoffutt's derived formula does not. My suggested method is the same as yours for deriving the original formula (featuring 2nd order accuracy).

D H
Staff Emeritus
Actually, stvoffutt wasn't able to derive a formula. That was his problem.

Chestermiller
Mentor
Actually, stvoffutt wasn't able to derive a formula. That was his problem.
Actually, it was clear to me that that the formula he came up with was:

$$\frac{3y_{j+1}-4y_j+y_{j-1}}{2h}=f(t_{j},y_{j})$$

This was the source of his concern. It didn't match the given equation. Also, he didn't realize that, as a consequence of the method that he used, this formula is not 2nd order accurate. The method that both you and I suggested will yield the desired formula, and the derivation will automatically lead to 2nd order accuracy (by requiring that the coefficient of the second derivative in the final equation is zero).

So should I expand like this?
$$y_{j+2}=y_{j+1}+2hy^{\prime}_{j+1}+\frac{4h^2}{2!}y^{\prime \prime}_{j+1}+\cdots$$
and do the same thing for $y_{j-2}$?

D H
Staff Emeritus
No. There is no yj+2 anywhere in the problem statement. You don't need it. You should expand yj and yj-1 in terms of yj+1.

I'm not sure how to expand y_j in terms of y_j+1. Can you point me in the right direction?

D H
Staff Emeritus
Sure you do. You did it right here:
$$y_{j-1}=y_j-hy^{\prime}_j+\frac{h^2}{2!}y^{\prime \prime}+\cdots$$

$y_j=y_{j+1}+hy^{\prime}_{j+1}+\frac{h^2}{2!}y^{\prime \prime}_{j+1}+O(h^3)$?

D H
Staff Emeritus
No. That's yj+2. You want yj.

$y_j=y_{j-1}-hy^{\prime}_{j-1}+\frac{h^2}{2!}y^{\prime \prime}_{j-1}+O(h^3)$?
I'm sorry. I am having a hard time trying to keep all of this straight. This is my first course in numerical analysis.

D H
Staff Emeritus