# Find order of consistency, stability function and stability interval of Runge-Kutta method

We have the following Runge-Kutta method : $$k_1=f(t_j, y_j)$$ $$k_2=f(t_j+ \frac{2}{3}h, y_j+ \frac{2}{3}hk_1)$$ $$y_{j+1}=y_j+ \frac{h}{4}(k_1+3k_2)$$

1. Find its Butcher Tableau
2. Its order of consistency and convergence
3. Its stability function and interval

1. The Butcher Tableau is $$\renewcommand\arraystretch{1.2} \begin{array} {c|cccc} 0\\ \frac{2}{3} & \frac{2}{3}\\ \hline & \frac{1}{4} &\frac{3}{4} \end{array}$$
2. 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.