Multistep method Numerical Analysis : IVP

Question

Given an explicit multistep method , we need to find the constants of the terms such as f(xi) , f(xi-1)..... and method order ..etc. A general approach is to write the corresponding Taylor series and compare the coefficients of terms .But I don't get how it works ?

Answer 1

Analysis

Consider $y'=f(x,y)=y$ discretize with step size $h$, $x_k=x_0+kh$. Then the multistep method $$ \sum_{k=0}^d a_k y_{n-k}=h\sum_{k=0}^db_kf(x_{n-k},y_{n-k}) $$ has discretization error for this example problem $$ \sum_{k=0}^d a_k e^{(n-k)h}-h\sum_{k=0}^d b_ke^{(n-k)h} $$ and using $e^{-h}=1-u$, the consistency order is $p$ if $$ a(1-u)+\ln(1-u)b(1-u)=O(u^{p+1}) $$

Algorithm

So compute any rational function satisfying $\frac{A(u)}{B(u)}=-\ln(1-u)+O(u^{p+1})$ and get the coefficients of the method via $a(z)=A(1-z)$, $b(z)=B(1-z)$. If necessary, adjust $b(z)$ by multiples of $(1-z)^p$.

Example

$$ -\ln(1-u)=u(1+\frac u2)+O(u^3)=\frac{u}{1-\frac u2}+O(u^3) $$ $a(z)=1-z$ and $b(z)=1-\tfrac12 (1-z)=\tfrac12(1+z)$ resulting in the trapezoid rule $$ y_k-y_{k-1}=\frac h2\Bigl(f(x_k,y_k)+f(x_{k-1},y_{k-1})\Bigr) $$ To get an explicit method, subtract $\tfrac12(1-z)^2$ to get the alternative $b(z)=\tfrac12(3z-z^2)$.

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Using Pade approximants, one may start with $$ -\ln(1-u) =u(1+\frac u2+\frac{u^2}3+\frac{u^3}4)+O(u^5) =\frac{75⋅u^2 + 18⋅u}{-7⋅u^3 - 39⋅u^2 + 66⋅u + 18}+O(u^5) $$ giving $a(z)=93 - 168⋅z + 75⋅z^2$ and $b(z)=38 + 33⋅z - 60⋅z^2 + 7⋅z^3$.