For $0\leq \theta \leq 1$ we have given the method
$$
x_n-x_{n-1}=h(\theta f(x_,t_n) + (1-\theta)f(x_{n-1},t_{n-1}))
$$
Now I should formulate this as a Rune-Kutta-Method and show that it is A-stable if $\theta\leq \frac{1}{2}$.
(1.) For the first part I would start by looking at the definition of Runge-Kutta-Methods. Runge-Kutta-methods with $s$ stepts and with constant lenght $h$ for
$$
x'(t)=f(x(t),t), \quad x(t_0)=x_0
$$
have the form
$$
X_i=f(x_{n-1}+h \sum_{j=1}^s a_{ij}X_j,t_{n-1}+hc_i), \quad i=1,…,s \\ x_n=x_{n-1}+h\sum_{i=1}^s b_iX_i
$$
We have also given the assumption that $c_i=\sum_{j=1}^s a_{ij}$. Now from this I would follow that our given method $x_n = x_{n-1} + h((1-\theta)f(x_{n-1},t_{n-1})+ \theta f(x_,t_n))$ has
$$
s=2, \quad b_1 = (1-\theta), \quad b_2=\theta, \quad X_1=f(x_{n-1},t_{n-1}), \quad X_2=f(x_n,t_n)
$$
Now for finding $c_1,c_2$ I would write
$$
(i) \quad f(x_{n-1},t_{n-1})=X_1 = f(x_{n-1}+h(a_{11}X_1 + a_{12}X_2),t_{n-1}+hc_1)\\
(ii) \quad f(x_n,t_n)= X_2 = f(x_{n-1}+h(a_{21}X_1 + a_{22}X_2),t_{n-1}+hc_2)
$$
Now from (i) it follows that $c_1=0$ and from (ii) that $c_2=1$ since $t_n=t_{n-1}+h$. Furthermore from (i) and the assumption $c_i=\sum_{j=1}^s a_{ij}$ which here ist $c_1=a_{11}+a_{12}$ I would have say
$$
x_{n-1} = x_{n-1}+h(a_{11}X_1 + a_{12}X_2) \Leftrightarrow a_{11}X_1 + a_{12}X_2 = 0 \Leftrightarrow a_{11} = a_{12} = 0
$$
Is that correct?
Then also from (ii)
$$
x_n=x_{n-1}+h(a_{21}X_1 + a_{22}X_2)
$$
and using the definition of the given method
$$
x_n = x_{n-1} + h((1-\theta)f(x_{n-1},t_{n-1})+ \theta f(x_,t_n))
$$
I would get $a_{21}=(1-\theta),a_{22}=\theta$
So this would an implicit Runge-Kutta-Method with
$$
b_1=a_{21}=1-\theta, \quad b_2=a_{22}=\theta, \quad a_{11} = a_{12} = 0, \quad c_1=0, \quad c_2=1
$$
Is that correct?
(2.) For this part I would then use $A=(a_{ij})_{i,j=1}^s$ and look at the stability function
$$
R(z)=1+zb^T(I-zA)^{-1} \mathbb{1}
$$
and check when $|R(z)|\leq 1$
Is that ok?
\Edit: I get
$$
R(z)=\frac{1+z(1-\theta)}{1-z\theta}=1+ \frac{z}{1-z\theta}
$$
Is that correct?
But I am struggeling to show that $|R(z)| \leq 1$ for $\theta \geq \frac{1}{2}$
Best Answer
$$ R(z)=\frac{1+z(1-\theta)}{1-z\theta}=1+ \frac{z}{1-z\theta} $$ From the second term one observes that for $z=x\to-\infty$ or generally for $z\to\infty$ on the Riemann sphere, $$R(\infty)=1-\frac1θ.$$ So to remain inside the unit circle, $θ>\frac12$. Use the form $\frac{1+θz}{1-θz}$ to transform a point with $Re(z)\le 0$ to a point inside the unit circle. $$ R(z)=1-\frac1{2θ}+\frac1{2θ}\frac{1+θz}{1-θz}. $$ So on the negative half plane we get $$ |R(z)|\le\left|1-\frac1{2θ}\right|+\frac1{2θ}=1. $$