We have the following Runge-Kutta method :
$$k_1=f(t_j, y_j)$$
$$k_2=f(t_j+ \frac{2}{3}h, y_j+ \frac{2}{3}hk_1)$$
$$y_{j+1}=y_j+ \frac{h}{4}(k_1+3k_2)$$
We're asked to:
- Find its Butcher Tableau
- Its order of consistency and convergence
- Its stability function and interval
Best Answer
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.