Runge Kutta, Gaus Legendre Stability

Question

I am looking for the stability function of the Gaus Legendre algorithm of order $4$.

I already have a solution but im not sure if its right:

$$R(z)=\frac{\frac{1}{12}z²+z+1} {\frac{1}{12}z²+1}$$

Answer 1

Apply the method to $y'=f(y)=zy$, think of $z$ as a sufficiently small constant, use step size $1$. Then the Gauß-Legendre method reads as \begin{align} k_1&=z\left(y_0+\frac14k_1+\left[\frac14-\frac{\sqrt3}6\right]k_2\right) \\ k_1&=z\left(y_0+\left[\frac14+\frac{\sqrt3}6\right]k_1+\frac14k_2\right) \\ y_1&=y_0+\frac12(k_1+k_2) \end{align} This is now to be solved, made easier by exploiting the symmetry of the coefficient table \begin{align} k_1-k_2&=-\frac{\sqrt3}6z(k_1+k_2) \\ k_1+k_2&=z\left(2y_0+\frac12(k_1+k_2)+\frac{\sqrt3}6(k_1-k_2)\right) \\ &=z\left(2y_0+\frac12(k_1+k_2)-\frac1{12}z(k_1+k_2)\right) \\\implies k+1+k_2&=\frac{2zy_0}{1-\frac12z+\frac1{12}z^2}\\ y_1&=y_0+\frac{zy_0}{1-\frac12z+\frac1{12}z^2}\\ &=\frac{1+\frac12z+\frac1{12}z^2}{1-\frac12z+\frac1{12}z^2}y_0=R(z)y_0 \end{align} so that indeed $R(z)$ is the symmetric 2-2 Pade approximant of $\exp(z)$. Somewhere you must have mixed up $y_0$ and $y_1$ close to the end of your computation.