I was reading through Hairer's book for stiff ODEs, and while mainly looking at the theory, I ended up reading chapter IV.8 on implementation details, just to write a toy program to see things in action. In said section of the book, the author proposes an embedded scheme for the RadauIIA method, such that: $$ \hat{y_1}-y_1 =\gamma_0 h f(x_0, y_0) + \sum_i (\hat{b}_i-b_i)hf(x_0+c_i h,g_i) = \gamma_0 h f(x_0, y_0) + \sum_i e_i z_i $$ concluding that for $\hat{y}_1 - y_1 = \mathcal{O}(h^4)$, it is needed that $$ \left(e_1, e_2, e_3\right)=\frac{\gamma_0}{3}(-13-7\sqrt{6},-13+7\sqrt{6},-1). $$ This implies that $e_i = \hat{b}_i-b_i$, therefore: $$ \hat{b}_i = b_i + e_i = \left( \dfrac{16-\sqrt{6}}{36}, \dfrac{16+\sqrt{6}}{36} , \dfrac{1}{9}\right) + e_i $$ The vector $b_i$ is well known for the 3-stage RadauIIA method (see e.g. here). In fact, for the (augmented) embedded method, these are 4-element vectors, with $e_0 = \gamma_0$ and $b_0 = 0$.
My questions are:
- Any insights on how to get such values for $e_i$?
- Looking at the first part of the error term, $\gamma_0 h f(x_0, y_0)$, implies $c_0 = 0$ and $a_{1j} = 0$. No more function evaluations and/or stages are computer. Is it then safe to assume that the Butcher tableau of the embedded method has the representation: $$ \begin{array}{c|c c} 0 & 0 & \{0\}^T \\ c & \{0\} & A \\ \hline & 0 & b^T \\ \hline & \gamma_0 & \hat{b}^T \\ \end{array} $$
- Does this new method need not satisfy the simplifying assumptions, namely: $$ B(p) = \sum_i^s b_i c_i^{q-1}=\dfrac{1}{q}, \qquad q=1, ..., p \\ C(\eta) = \sum_i^s a_{ij} c_j^{q-1}=\dfrac{c_i^q}{q}, \qquad i=1,...,s, \quad q=1, ..., \eta $$ $C(\eta)$ is satisfied, since the new elements (assuming the previous tableaux is correct) are all zeros, but $B(p)$ is not (with $\gamma_0 \ne 0$). However a different choice of $\hat{b}_i$ can be used in order to satisfy it.
Thank you.
EDIT: Point 2) has been edited since it was partially answered in the paragraphs preceding it.