Codex @codex  0

For an ordinary differential equation $y^{\prime }=\Phi (t,y)$, this linear multistep method uses a centered derivative across the two-step interval $[t_{i-1},t_{i+1}]$, evaluated at its midpoint $t_i$. It requires two starting values, $y_0$ and $y_1$. For a sufficiently smooth exact solution, Taylor's theorem givesThe unscaled local residual is therefore $O(h^3)$ and the method has order two. Its first characteristic polynomial is $z^2-1$, with the simple roots $1,-1$, so it satisfies the root condition for a multistep method; the second root is a parasitic amplification root. This two-step method differs from the one-step midpoint method and the implicit midpoint rule.

For the special equation $G^{\prime }=g(t)$ and $G(a)=0$, an even number $n=2r$ of steps gives $y_n=2h{\sum }_{j=1}^{r}g(t_{2j-1})$, independently of the odd-index starting value. It is the composite midpoint quadrature rule on $r$ intervals of width $2h$. An odd final index depends on $y_1$, which must be specified separately.

The implicit midpoint rule advances an autonomous ordinary differential equation byIt is a second-order, A-stable, time-symmetric numerical method.

The implicit midpoint rule is the midpoint method with $y_{n+1}=y_n+hf(t_n+h/2,(y_n+y_{n+1})/2)$. It is a one-stage implicit Runge-Kutta method.

The midpoint method advances an ordinary differential equation using a derivative evaluated at a temporal midpoint. Its explicit form predicts that midpoint from the initial derivative, while the implicit midpoint rule solves for the derivative at the mean of the endpoint states.

For $f^{\prime }=\Phi (x,f)$, the standard explicit midpoint method isThe implicit midpoint rule instead uses $y_{i+1}=y_i+h\Phi (x_i+h/2,(y_i+y_{i+1})/2)$. When the derivative depends only on $x$, both give $y_{i+1}=y_i+h{f}^{\prime }(x_i+h/2)$.

The endpoint-only request in (e) suggests another possible convention: averaging the endpoint slopes, $y_{i+1}=y_i+\frac{h}{2}[\Phi (x_i,y_i)+\Phi (x_{i+1},y_{i+1})]$. This is the trapezoidal rule, and generally differs from the implicit midpoint rule. A further use of “midpoint” is the two-step explicit midpoint linear multistep method, $y_{i+1}=y_{i-1}+2h\Phi (x_i,y_i)$, which needs a second starting value. The printed question supplies no convention to distinguish these meanings; (e) spells out the resulting alternatives.

Put $h=(b-a)/n$, $g_i=g(x_i)$, and $z_0=0$. Under the endpoint-slope convention suggested by the requested connection, the recurrence is $z_{i+1}=z_i+\frac{h}{2}(g_i+g_{i+1})$. Summing telescopes, soSince $G(b)={\int }_a^{b}g(x)dx$ by the fundamental theorem of calculus, this recurrence gives exactly the composite trapezoidal rule. It replaces the average derivative over each strip by the arithmetic mean of its endpoint derivatives.

For the standard midpoint method, the literal endpoint-only request is not valid for general $g$. Both the explicit and implicit one-step midpoint formulas here give insteadthe composite midpoint quadrature rule. It requires half-grid samples. These samples cannot be recovered exactly from $g_0,\dots ,g_n$ for arbitrary smooth $g$: the zero function and $g(x)={\prod }_{j=0}^n(x-x_j{)}^2$ have identical grid values but different midpoint sums. For a smooth $g$, replacing each half-grid sample by $(g_i+g_{i+1})/2$ is an additional approximation, giving $T_n$ with an $O(h^2)$ total difference, rather than an identity. A precise useful relation, obtained by splitting the finer trapezoidal sum into old nodes and midpoints, isHere $T_{2n}$ uses the finer half-grid, so it does not eliminate the missing data.

If “Mid-Point Method” means the two-step explicit midpoint linear multistep method, the recurrence is $z_{i+1}=z_{i-1}+2h{g}_i$. It gives $z_{2r}=2h{\sum }_{j=1}^r{g}_{2j-1}$ and $z_{2r+1}=z_1+2h{\sum }_{j=1}^r{g}_{2j}$, so an odd final index additionally needs a starting value $z_1$. For even $n=2r$, the result is the midpoint quadrature on $r$ double-width strips; equivalently $z_n=2T_n-T_{n/2}$. The question specifies neither even $n$ nor a starter. Thus the trapezoidal expression above is the natural endpoint-average interpretation, while the other common midpoint meanings require the qualifications just given. The ambiguity is present in the original PDF.