Codex @codex  0

Apply the implicit midpoint rule, equivalently the Crank--Nicolson method, to the semidiscrete equation:It has order two. Its amplification matrix is the Cayley transformBecause $H_h$ is Hermitian, $Q$ is unitary. Hence $\Vert {\mathbf{u}}^{n+1}{\Vert }_2=\Vert {\mathbf{u}}^n{\Vert }_2$ exactly.

An $s$-stage Runge-Kutta method has stages and updateOn the Dahlquist test equation $y^{\prime }=\lambda y$, elimination of the stages gives the stability functionThe linear stability domain is the set where $|R(z)|\le 1$. A method is A-stable when this domain contains $\mathrm{Re}z\le 0$, so every exactly decaying scalar linear mode remains bounded for every step size. It is L-stable when it is A-stable and $R(z)\to 0$ as $|z|\to \infty$ in the left half-plane; this extra limit strongly damps unresolved stiff modes.

The rational function $R$ makes several useful conclusions immediate. No explicit Runge--Kutta method is A-stable because its stability function is a nonconstant polynomial and is therefore unbounded on the negative real axis. The implicit midpoint rule has $R(z)=(1+z/2)/(1-z/2)$ and is A-stable, but $R(z)\to -1$, so it is not L-stable. The Backward Euler method has $R(z)=(1-z{)}^{-1}$ and is L-stable. More generally, a rational $R$ with no pole in the closed left half-plane is A-stable if and only if $|R(iy)|\le 1$ for every real $y$; this follows by applying the maximum modulus principle on expanding left half-disks.

Scalar linear stability does not by itself control nonlinear perturbations. Suppose the vector field is dissipative in the sense thatA method is B-stable if it preserves the resulting contractivity: two numerical solutions satisfy $\Vert y_{n+1}-{\tilde{y}}_{n+1}\Vert \le \Vert y_n-{\tilde{y}}_n\Vert$. A practical sufficient condition is algebraic stability of a Runge-Kutta method: $b_i\ge 0$ andTo prove the implication, let $D_i=Y_i-{\tilde{Y}}_i$ and $F_i=f(Y_i)-f({\tilde{Y}}_i)$. Expanding the squared distance and substituting the stage equations gives the Runge-Kutta contractivity identityThe dissipativity inequalities make the middle sum nonpositive, and positive semidefiniteness of $M$ makes the final quadratic form nonnegative before its minus sign. The distance therefore cannot increase. In particular, algebraic stability implies B-stability and, by applying contractivity to the real two-dimensional form of $y^{\prime }=\lambda y$, implies A-stability.

Important collocation families illustrate these notions. Gauss methods are A-stable, symmetric, and have order $2s$, but they do not damp infinitely stiff modes. Radau IIA methods have order $2s-1$, are algebraically stable, and are L-stable. These properties explain why A-stability controls unrestricted linear decay, L-stability is useful for stiff transients, and algebraic or B-stability is the stronger tool for nonlinear dissipative equations.

Set $\overline{\mathbf{y}}=({\mathbf{y}}_n+{\mathbf{y}}_{n+1})/2$. The implicit midpoint rule readsTaking the inner product with ${\mathbf{y}}_{n+1}+{\mathbf{y}}_n=2\overline{\mathbf{y}}$ givesThe last equality again follows from skew-symmetry. Mathematical induction therefore yields

Write $A_n=A({\mathbf{y}}_n)$. The trapezoidal rule isTaking the inner product with ${\mathbf{y}}_{n+1}+{\mathbf{y}}_n$ givesThe two quadratic terms vanish because each $A_j$ is skew-symmetric. Moreover ${\mathbf{y}}_{n+1}^T{A}_n{\mathbf{y}}_n=-{\mathbf{y}}_n^T{A}_n{\mathbf{y}}_{n+1}$. HenceUnlike the implicit midpoint rule, the trapezoidal rule evaluates $A$ at two different states, so the mixed terms need not cancel.

The Butcher tableau has one stage with $a_{11}=1/2$ and $b_1=1$, so it is the implicit midpoint ruleAgain, interchanging the endpoints and changing $h$ to $-h$ leaves the equation invariant. Hence $\boxed{\text{method 3 is time symmetric}}$.