Local truncation error of a given multi-step method

Question

This is admittedly a homework problem, so I've also attached my best attempt at a solution.

Question (abbreviated, from Numerical Analysis, Burden and Faires):

Given the multistep method $w_{i+1} = -\frac{3}{2}w_{i} + 3w_{i-1} - \frac{1}{2}w_{i-2} + 3hf(t_{i}, w_{i})$, find the local truncation error.

My attempt so far:

I expanded the true solution $y(t)$ in a 3rd order Taylor series centered at $t_{i}$ (here I'm using the shorthand $y_{i} = y(t_{i})$) $$y(t) = y_{i} + y'_{i}(t-t_{i}) + \frac{1}{2}y''_{i}(t-t_{i})^2 + \frac{1}{6}y'''_{i}(t-t_{i})^3 + \frac{1}{24}y^{(4)}(\xi(t))(t-t_{i})^4$$ for some $\xi(t) \in (t_{i}, t)$. In addition, we have $$f(t_{i}, y_{i}) = y'_{i}.$$ By the definition of local truncation error, $\tau_{i+1}(h)$, we have $$h\tau_{i+1}(h) = y_{i+1} + \frac{3}{2}y_{i} - 3w_{i-1} + \frac{1}{2}y_{i-2} - 3hf(t_{i}, w_{i}).$$ By plugging in the Taylor expansion of $y$ for $y_{i+1}, y_{i-1}, y_{i-2}$ in the equation above, I arrived at the new equality $$h\tau_{i+1}(h) = \frac{1}{24}y^{(4)}(\xi(t_{i+1}))h^4 - \frac{1}{8}y^{(4)}(\xi(t_{i-1}))h^4 + \frac{1}{3}y^{(4)}(\xi(t_{i-2}))h^4,$$ where the $\xi(t)$ values are not necessarily equal.

The solution at the back of the book has $h\tau_{i+1}(h) = \frac{1}{4}y^{(4)}(\xi(t))h^4$. Coincidentally, $\frac{1}{24} - \frac{1}{8} + \frac{1}{3} = \frac{1}{4}$, so I seem to be close, but I don't think I'm aware of a result beyond simple IVT to combine these error terms. Could anyone help me finish this problem?

As an aside, I stumbled onto an alternate solution here that uses interpolating polynomials, but I don't understand how this solution works either.

Any help would be much appreciated!

Answer 1

As you have seen, you lose too much information using first the mean value theorem for the individual remainder terms and then trying to compose the total remainder. Thus you have first to combine a more exact form of the integral form of the remainder terms and then combine them.

Thus use $$ y(t+h)=y(t)+y'(t)h+\frac12y''(t)h^2+\frac16y'''(t)h^3+\frac16\int_0^1y^{(4)}(t+sh)(1-s)^3\,ds \,h^4 $$ to get, for instance $$ h\tau_{i+1}(h)=\frac16\int_0^1y^{(4)}(t+sh)(1-s)^3\,ds \,h^4-\frac12\int_{-1}^0y^{(4)}(t+sh)(1+s)^3\,ds \,h^4+\frac1{12}\int_0^2y^{(4)}(t+sh)(2-s)^3 $$ This can formally be written as $$ h^4\,\int_{-1}^2 \underbrace{(w_1(s)+w_2(s)+w_3(s))}_{=w(s)}y^{(4)}(t+sh)\,ds. $$ Now you need to check that the piecewise polynomial, but not necessarily continuous function $w$ is not negative over the interval $[-1,2]$. After applying the mean value theorem of integration, the remaining weight integral can be split into the single integrals $\int_{-1}^2 w_k(s)\,ds$ which have the already known values $\frac1{24},-\frac18,\frac13$. Only that now they apply all to the 4th derivative at the same midpoint that now can lie anywhere in $(t-h,t+2h)$, so they can be summed up without problem.