Codex @codex  0

Integrate the ordinary differential equation across one step, using the fundamental theorem of calculus:The midpoint quadrature rule approximates the integral by $hf(x_n+h/2,y(x_n+h/2))$. Approximate the unknown midpoint state by the midpoint value of linear interpolation between the endpoint states:Combining these approximations gives the implicit midpoint ruleIt is an implicit time-stepping method because its new state occurs inside the evaluation of $f$. An equivalent derivation uses a straight-line approximation on the step and imposes the ordinary differential equation at its midpoint, a one-stage collocation method.

Applying the implicit midpoint rule to the simple harmonic oscillator system givesFactor the change in the quadratic form $y^2+z^2$ and substitute these equations:Thus every step preserves the first integralFor completeness, the implicit equations are solvable for every real $h$: putting $s=h/2$, their coefficient matrix has determinant $1+s^2>0$ and the update isThis is an orthogonal matrix, as follows from $(1-s^2{)}^2+4s^2=(1+s^2{)}^2$. It therefore preserves the Euclidean norm, explaining geometrically the exact energy conservation. This is a concrete instance of Runge-Kutta conservation of quadratic invariants. Preservation of the circle does not assert that the numerical and exact solutions have the same phase at a fixed nonzero step size.

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.

A time-symmetric numerical method is a one-step numerical method with a locally invertible step map ${\Phi }_h$ satisfyingFor a nonautonomous ordinary differential equation, include time in the state or explicitly reverse the step's time endpoints. The definition means that applying the same rule backward undoes its forward step; it does not by itself claim small discretization error.

The implicit midpoint rule illustrates this directly: swapping $y_n,y_{n+1}$ and replacing $h$ by $-h$ leaves its midpoint equation unchanged, wherever the local solve is unique. By contrast, forward Euler method has factor $1+h\lambda$ on $y^{\prime }=\lambda y$, and the backward factor $1-h\lambda$ gives product $1-h^2{\lambda }^2$, not one.

Assume a consistent method has a smooth integer-power local-error expansion of finite order $p$, with nonzero leading defect $h^{p+1}E(x)$. Composing its forward and backward steps leaves leading defect $[1+(-1{)}^{p+1}]h^{p+1}E(x)$. Symmetry forces this coefficient to vanish. Thus $p+1$ is odd and $p$ is even. The finite leading-error and regular expansion assumptions are needed for this parity argument.