How to derive an Implicit Runge-Kutta method from Pade approximation

Question

I was reading some work by Butcher and I came across Pade approximations and the correlation between them and stability functions for some Implicit Runge-Kutta methods. For example, in this Pade table for the exponential function, we see that the $(2,1)$ Pade approximation

$$\frac{1+\frac{1}{3}z}{1-\frac{2}{3}z+\frac{1}{6}z^{2}}$$

corresponds to the stability function for the RADAU IIA method. Similarly, the $(2,2)$ Pade approximation

$$\frac{1+\frac{1}{2}z+\frac{1}{12}z^{2}}{1-\frac{1}{2}z+\frac{1}{12}z^{2}} $$

corresponds to the stability function of the 2-stage Gauss Runge-Kutta method. However, I'm not sure if other Pade approximations correspond to stability functions; For example, I can't find the numerical method whose stability function corresponds to the $(2,0)$ Pade approximation

$$\frac{1}{1-z+\frac{1}{2}z^{2}}$$

So my question is two fold (I apologise for asking two questions in one post, it seems like a waste to create two new posts though):

1. Is it always possible to generate an Implicit Runge-Kutta scheme whos stability function corresponds to a particular Pade approximation? And if so
2. How do we construct such an implicit scheme?

Answer 1

The only way I could figure out how to do this is to just work backwards, starting with the stability function for implicit Runge-Kutta schemes. I'm not sure if there is a slicker way to do this, this is just the most obvious.

Note that the following only applies to implicit Runge-Kutta schemes. Also, I'll be using the Pade $(2,0)$ approximation (as stated in the question) as an example.

Using the fact that the Pade approximation is given by

\begin{align} R(z) &= \frac{P(z)}{Q(z)} \\ &= \frac{1}{1-z+\frac{z^{2}}{2}} \end{align}

and the stability function for an implicit Runge-Kutta is given by ($\vec e$ is a vector of ones)

\begin{align} R(z) &= \frac{P(z)}{Q(z)} \\ &= \frac{\det(I-zA+z \vec e b^{T})}{\det(I-zA)} \end{align}

we just need to construct matrices such that

\begin{align} \det(I - zA + z \vec e b^{T}) &= 1 \\ \det(I - zA) &= 1 - z+\frac{z^{2}}{2} \end{align}

Setting

$$A = \begin{pmatrix} a_{1} & a_{2} \\ a_{3} & a_{4} \end{pmatrix}$$

gives us

\begin{align} \det(I - zA) &= 1 - z(a_{1} + a_{4}) + z^{2}(a_{1} a_{4} - a_{2} a_{3}) \\ &= 1 - z + \frac{z^{2}}{2} \end{align}

which gives us conditions on our coefficients

\begin{align} a_{1} + a_{4} &= 1 \\ a_{1} a_{4} - a_{2} a_{3} &= \frac{1}{2} \end{align}

Now, as far as I'm aware, we can arbitrarily choose values for the $a_{i}$ such that they satisfy the above underdetermined system (for example, $a_{1} = a_{4} = \frac{1}{2}$, $a_{2} = \frac{1}{3}$, $a_{3} = -\frac{3}{4}$).

We then solve $\det(I - zA + z \vec e b^{T})$ in a similar manner, by taking

$$\vec e b^{T} = \begin{pmatrix} b_{1} & b_{2} \\ b_{1} & b_{2} \end{pmatrix}$$

constructing $I - zA + z \vec e b^{T}$ (using the previous matrix we constructed, $I - zA$, with the components of $A$ now known from the $a_{i}$), taking the determinant and solving the corresponding system of equations in $b_{1}$ and $b_{2}$. The matrix $A$ and vector $b^{T}$ can now be written in a Butcher tableau.