7
$\begingroup$

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 whos 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?
$\endgroup$
3
$\begingroup$

There are a few people who have liked the question, so I'm posting an answer. 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 this only applies to implicit Runge-Kutta schemes, and I'm specifically using the Pade $(2,0)$ apprixmation as stated above 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}

and 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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.