- The Butcher Tableau is
$$
\renewcommand\arraystretch{1.2}
\begin{array}
{c|cccc}
0\\
\frac{2}{3} & \frac{2}{3}\\
\hline
& \frac{1}{4} &\frac{3}{4}
\end{array}
$$
- Similar to Find order of consistency of IVP with Butcher Tableau and Find $\alpha$ such that $y_{j+1}=y_j+\frac{h}{2 \alpha}k_1 + h(1- \frac{1}{2 \alpha})k_2$ has order of consistency 2
Let $$\Phi(h) = \frac{1}{4}f(y(t_j)) + \frac{3}{4}f(y(t_j) + \frac{2}{3}hf(y(t_j)))$$
Let $$M := f(y(t_j) + \frac{2}{3}hf(y(t_j)))$$
Then $$\dot{\Phi}(h)= \frac{2}{3}\frac{3}{4}f \dot{M} $$ and $$ \ddot{\Phi}(h)= \frac{1}{3}f^2 \ddot{M}$$
Let $$\dot{y}(t) = f(t, y(t))$$ then
$$\ddot{y}(t) = \frac{df}{dy}(y(t))f(y(t))$$ and
$$\dddot{y}(t) = \frac{d^2f}{dy^2}(y(t))f(y(t))^2 + (\frac{df}{dy}(y(t)))^2f(y(t))$$.
We see that $\dot{y}(t_j) = \Phi(0)$ and $\ddot{y}(t_j) = 2\dot{\Phi}(0)$ but $\dddot{y}(t_j) \neq 3\ddot{\Phi}(0)$ so the order of consistency $p=2$ and thus the order of convergence is also $2$.
3.
$$k_1=f(t_j, y_j)= \lambda y_j$$
$$k_2=f(t_j+ \frac{2}{3}h, y_j+ \frac{2}{3}hk_1)= \lambda (y_j + \frac{2}{3}hk_1 ) = \lambda (y_j + \frac{2}{3}h \lambda y_j )$$
$$y_{j+1}=y_j+ \frac{h}{4}(k_1+3k_2) =y_j+ \frac{h}{4}((\lambda y_j) + 3 (\lambda (y_j + \frac{2}{3}h \lambda y_j ))) = y_j(1 + h\lambda + \frac{1}{2}h^2 \lambda^2) = y_j(1 + z + \frac{1}{2}z^2)$$
(We let $z := h \lambda$)
The stability function is thus $g(z)= 1 + z + \frac{1}{2}z^2$
For the stablity interval: $|g(z)| < 1$, that is $z + \frac{1}{2}z^2 < 0$ or $z(2 + z) < 0$ so $-2 < z < 0$ is the interval of stability.