Codex @codex  0

A first integral of $y^{\prime }=f(t,y)$ is a differentiable function $I(t,y)$ whose value is constant along solutions through every time and state in its domain. The chain rule makes this equivalent to the displayed pointwise identity. For $I(y)=y^{T}Sy$ with symmetric $S$, it becomes $y^{T}Sf(t,y)=0$. This identity at the implicit stage is what proves exact Runge-Kutta conservation of quadratic invariants for the implicit midpoint rule. Invariance asserted only on the image of one initial-time flow needs that image to contain the stage under consideration.

Let $Q(y)=y^{T}Sy$. In the usual pointwise meaning of an ODE invariant, differentiation along an exact solution through an arbitrary state givesFor a nonautonomous equation this identity must hold at the stage time as well as at the initial time. It follows directly from invariance for solutions started at any time and state, or from the stated initial-time invariance when the exact flow onto the stage-time domain is invertible. This is the first-integral condition used below.

Put $Y=(y_n+y_{n+1})/2$ and $t_{\ast }=t_n+h/2$. Since $S$ is symmetric,Induction over all well-defined implicit steps givesNo positive-definiteness assumption on $S$ was used. Nor is uniqueness of the stage essential to this algebraic identity: any stage solution satisfying the method and the pointwise invariant condition preserves $Q$. This is the one-stage case of Runge-Kutta conservation of quadratic invariants.

Define the stage states $Y_i=y_n+h{\sum }_j{a}_{ij}{k}_j$, so $k_i=P{Y}_i$. Skew symmetry gives $Y_i^T{k}_i=Y_i^{T}P{Y}_i=0$, and henceInsert this into the preceding norm identity and symmetrize the double sum, using $k_i^T{k}_j=k_j^T{k}_i$. The result is the Runge-Kutta conservation of quadratic invariants identityThuswhich proves the sufficient condition for every well-defined step and then for all $n$ by induction. For example the implicit midpoint rule, with $a_{11}=1/2$ and $b_1=1$, has $M_{11}=0$ and preserves this norm exactly. The condition is sufficient, and need not be necessary for a particular matrix or a particular solution.

A collocation Runge-Kutta method approximates an ordinary differential equation by a polynomial whose derivative agrees with the vector field at selected times. Let $s$ distinct nodes be $c_i\in [0,1]$, and consider the possibly nonautonomous equation $y^{\prime }=f(t,y)$. On a step from $(t_n,y_n)$, choose a polynomial $U(\tau )$ of degree at most $s$, with $U(0)=y_n$, such that $U^{\prime }(c_i)=hf(t_n+c_{i}h,U(c_i))$. Here $\tau =(t-t_n)/h$ is scaled time; the numerical endpoint is $y_{n+1}=U(1)$.

Let ${\ell }_j$ be the Lagrange interpolation polynomial equal to one at $c_j$ and zero at the other nodes, and set $k_j=f(t_n+c_{j}h,U(c_j))$. Interpolation of the polynomial derivative givesEvaluating at the nodes and endpoint derives the Runge-Kutta method coefficients:Thus collocation is a specific family of generally implicit RK methods, not a separate update mechanism. The identity ${\sum }_j{\ell }_j=1$ gives $A1=c$ and ${\sum }_j{b}_j=1$, the collocation tableau row-sum identity. For sufficiently small $h$, a vector field with Lipschitz constant $L$ gives a contraction of the stage map when $hL{\max }_i{\sum }_j|a_{ij}|<1$, on a suitable bounded neighborhood. The Banach fixed-point theorem then gives a unique nearby stage solution. Equivalently smooth stage equations have identity derivative in the stage variables at $h=0$, so the inverse function theorem gives a smooth nearby branch. Arbitrarily large-step solvability is a separate issue for an implicit Runge-Kutta method.

Order analysis has two levels: polynomial accuracy at stages and extra accuracy at the endpoint. Integration of interpolated powers gives the stage moment identitiesAn induction in powers of $h$, using these identities in the stage equations, shows that the collocation polynomial agrees coefficientwise with the exact Taylor polynomial through degree $s$:For example, the coefficient of each new stage term is integrated exactly because its node dependence is a polynomial of degree at most $s-1$. This proves stage order $s$ and bounded actual-time derivatives of the collocation polynomial up to degree $s$. Assuming the vector field is sufficiently smooth on a fixed compact solution neighborhood, its composition with this polynomial has uniformly bounded derivatives of all finite orders needed below.

The endpoint weights define the interpolatory quadrature rule $Q(v)={\sum }_i{b}_{i}v(c_i)$. Suppose it is exact through degree $q-1$, but not through degree $q$. Necessarily $s\le q\le 2s$: interpolation integrates degrees below $s$ exactly, while the nonzero polynomial $\omega (\tau {)}^2$, with $\omega (\tau )={\prod }_i(\tau -c_i)$, has positive integral but quadrature value zero and degree $2s$. Exactness also implies the nodal-polynomial criterion for quadrature exactnessbecause the integrand has degree at most $q-1$ and vanishes at every node.

Here is the endpoint-error argument underlying the collocation order theorem. Let $v(t)=U((t-t_n)/h)$ and $\delta (t)=v^{\prime }(t)-f(t,v(t))$ be its residual. It vanishes at the collocation times, and if $F_h(\tau )=f(t_n+h\tau ,U(\tau ))$, then $\delta ={\mathcal{I}}_s{F}_h-F_h$, where ${\mathcal{I}}_s$ is interpolation at the nodes. Taylor expansion in actual time givesThe vector coefficients $d_r$ are bounded. Each bracket has the factor $\omega (\tau )$, with quotient degree $r-s$. In particular the residual is $O(h^s)$ throughout the step.

Write $\varphi (t_{\ast },t,z)$ for the exact flow from time $t$ and state $z$ to $t_{\ast }=t_n+h$. Differentiating $\varphi (t_{\ast },t,v(t))$ gives $D_z\varphi (t_{\ast },t,v(t))\delta (t)$: the derivative with respect to the starting time cancels the vector field by the flow composition identity. Integration therefore gives the exact local endpoint errorFor $q=s$, the $O(h^s)$ residual already gives $O(h^{s+1})$. For $q>s$, Taylor-expand the smooth matrix weight as $\Psi (t_n+h\tau )={\sum }_{\ell =0}^{q-s-1}{h}^{\ell }{\Psi }_{\ell }{\tau }^{\ell }+O(h^{q-s})$. Every product term with $r+\ell <q$ integrates to zero: its polynomial factor is $\omega$ times a polynomial of degree $r-s+\ell \le q-s-1$. All remaining terms are $O(h^q)$ before the time integration, including both remainders. HenceThis is collocation endpoint superconvergence by quadrature orthogonality. For the global assertion, the stage contraction estimate gives a one-step Lipschitz bound $1+Ch$ on a fixed solution neighborhood. Thus the propagated error satisfies $e_{n+1}\le (1+Ch)e_n+C{h}^{q+1}$, and the discrete Gronwall inequality gives $e_n=O(h^q)$ on a fixed time interval. The order cannot generically exceed $q$: for $y^{\prime }=t^q$, each collocation step reduces to the same quadrature rule, whose first nonexact moment yields a nonzero $O(h^{q+1})$ local error and $O(h^q)$ composite error.

The usual node families illustrate the result. Gauss collocation methods use interior Gaussian nodes and have $q=2s$, the maximum possible order. Right-endpoint Radau nodes give the Radau IIA method of order $2s-1$. Including both endpoints with Lobatto quadrature gives the Lobatto IIIA method of order $2s-2$ for $s\ge 2$. One-stage Gaussian collocation is the second-order implicit midpoint rule, while collocation at the right endpoint is the first-order Backward Euler method. The two-stage Gaussian nodes $1/2\pm \sqrt{3}/6$ give $b=(1/2,1/2)$ anda fourth-order method. Node placement, rather than just stage count, controls the superconvergent endpoint order.

There is also a structural link with Question 2. For Gaussian nodes, let $L_i(\tau )={\int }_0^{\tau }{\ell }_i(v)dv$. The derivative of $L_i{L}_j$ has degree at most $2s-1$, so Gaussian quadrature is exact on it. The integral is $b_i{b}_j$, while evaluating at the nodes gives $b_i{a}_{ij}+b_j{a}_{ji}$. Thus Gaussian collocation satisfies $M=0$ and has Runge-Kutta conservation of quadratic invariants. For the scalar test equation, any RK method has stability function $R(z)=1+z{b}^T(I-zA{)}^{-1}1$. For implicit midpoint this is $(1+z/2)/(1-z/2)$; the identity $|1-z/2{|}^2-|1+z/2{|}^2=-2\mathrm{Re}z$ proves A-stability. High order, stage solvability, invariant preservation and damping are distinct design considerations, so no stability conclusion should be inferred from collocation order alone.