Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 63 / 4

For $y^{\prime }=f(t,y)$ and step $h$, a collocation Runge-Kutta method constructs a polynomial $P$ of degree at most $s$ on $[t_n,t_n+h]$ satisfyingThe step value is $y_{n+1}=P(t_n+h)$. Thus the differential equation holds exactly at the prescribed collocation nodes, while the initial value fixes the integration constant. For a smooth vector field and small enough $h$, the implicit stage equations have the local solution branch continuing the zero-step initial state; arbitrary large steps need not have a unique nonlinear solution.

Let ${\ell }_j(\tau )={\prod }_{k\ne j}(\tau -c_k)/(c_j-c_k)$ be the Lagrange interpolation basis. Set $Y_i=P(t_n+c_{i}h)$ and $F_j=f(t_n+c_{j}h,Y_j)$. Since $P^{\prime }$ is a polynomial of degree at most $s-1$, the collocation conditions giveIntegrate from zero to $c_i$ and to one to obtainThese are the stage and update equations of a Runge-Kutta method with nodes $c_i$. Conversely, stages satisfying these equations reconstruct $P$ by integrating the displayed interpolation polynomial, so the equivalence works in both directions. Because ${\sum }_j{\ell }_j=1$, they also satisfy the collocation tableau row-sum identity ${\sum }_j{a}_{ij}=c_i$.

The collocation order theorem identifies the classical order with the order of its interpolatory quadrature. If ${\sum }_i{b}_{i}g(c_i)$ integrates every polynomial of degree at most $p-1$ exactly, but fails for some polynomial of degree $p$, then the collocation method has order $p$ for sufficiently smooth differential equations and the local implicit stage branch. With $s$ nodes, $s\le p\le 2s$.

Equivalently, put ${\pi }_s(\tau )={\prod }_i(\tau -c_i)$. If its first $k$ moments vanish,and the next does not, then $p=s+k$. Polynomial division explains the quadrature part: any polynomial through degree $s+k-1$ is a remainder of degree below $s$ plus ${\pi }_s$ times a polynomial of degree below $k$. The remainder is integrated exactly by interpolation, and the remaining integral vanishes. Conversely a nonzero next moment is an explicit failed quadrature test. The differential-equation order assertion is the collocation theorem being stated here. In particular Gauss collocation methods achieve $2s$, Radau IIA methods achieve $2s-1$, and endpoint-including Lobatto IIIA methods achieve $2s-2$ for $s\ge 2$.

The tableau as printed is not a collocation tableau. Its second stage row is $(3/4,3/4)$, whose sum is $3/2$, while the corresponding printed node is $1$. This violates the collocation tableau row-sum identity. The original PDF confirms this coefficient; it is not only a TeX transcription defect.

The required integrated basis at nodes $1/3,1$ isso direct integration givesThus replacing the last stage coefficient by $1/4$ gives the intended two-stage Radau IIA collocation method. Its quadrature moments areIt integrates quadratics exactly but not cubics, so the collocation order theorem gives order three. Independently, $b^{T}Ac=1/6$, confirming the remaining third-order Butcher order condition.

For completeness the literal printed Runge-Kutta method has only order one. For the autonomous equation $y^{\prime }=y$, the Taylor expansion of its stability function beginswhereas $e^z=1+z+z^2/2+O(z^3)$. Since $b^{T}1=1$, it is consistent, but this discrepancy already rules out order two. Therefore neither the requested equivalence nor the third-order conclusion is true without the stated coefficient repair.