Two-Step backward differentiation

[stvoffutt]

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 includesf(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]

Your equation doesn't have second order accuracy. It only has first order accuracy. Do your expansion around t_j+1 and y_j+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]

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

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

 

[Chestermiller]

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

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

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]

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

 

[Chestermiller]

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).

 

[stvoffutt]

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]

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

 

[stvoffutt]

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]

Sure you do. You did it right here:

y_{j-1}=y_j-hy^{\prime}_j+\frac{h^2}{2!}y^{\prime \prime}+\cdots

Just relabel your indices.

 

[stvoffutt]

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

 

[D H]

No. That's y_j+2. You want y_j.

 

[stvoffutt]

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]

No! Don't just guess. You want y_j on the left, terms involving y_j+1 and it's time derivatives on the right.