[Math] absolute stability / inequality

Question

i want to find the amountof $\theta \in[0,1]$ where it is absolute stable whith $y'=\lambda y$ ,$\lambda \in \mathbb C$ or $\lambda \in \mathbb R$ for
$$u_{j+1}=u_j+h[\theta f(t_j,u_j)+(1-\theta )f(t_{j+1},u_{j+1})]$$ i got
$$u_{j+1}=u_j+h[\theta \lambda u_j+(1-\theta)\lambda u_{j+1}]$$
$$u_{j+1}=u_j(1+h\theta \lambda)+u_{j+1}h(\lambda -\theta \lambda)$$
$$u_{j+1}=u_j \dfrac{(1+h\theta \lambda)}{(1+h\lambda(\theta -1)}$$right? for the absolute stability it needs to be
$$\dfrac{|1+h\theta \lambda|}{|1+h\lambda (\theta -1)|}<1$$
$$|1+h\theta \lambda|<|1+h\lambda (\theta -1)|$$
so how can i calculate the inequality for the right $\theta$ and it must be $Re(\lambda)<0$.
I think $\theta \in [0,1/2)$ is right, but cant proof it

Answer 1

For stability we need $$\dfrac{|1+h\theta \lambda|}{|1+h\lambda (\theta -1)|}<1$$ or instead \begin{align} &\Bigl(\dfrac{1+h\theta \lambda}{1+h\lambda (\theta -1)}\Bigr)^2<1\\ \Leftrightarrow &(1+h\theta \lambda)^2< (1+h\lambda (\theta -1))^2 \\ \Leftrightarrow & 1+2h\theta \lambda+(h\theta \lambda)^2<1+2h\lambda(\theta-1) +(h\lambda(\theta-1))^2 \\ \Leftrightarrow &(h\theta \lambda)^2 < -2h\lambda + h^2\lambda^2 (\theta^2-2\theta +1)\\ \Leftrightarrow & 0 < -2h\lambda -2h^2\lambda^2\theta +h^2\lambda^2 \\ \Leftrightarrow & 2 h^2\lambda^2 \theta < h^2\lambda^2 - 2h\lambda \\ \Leftrightarrow & \theta < \frac12 -\frac{1}{h\lambda} \end{align} I was a bit confused, because it seems you knew the answer was $\theta \in [0,\frac12)$, but a random check on the following example made me optimistic that I was correct (of course this is only one single example). \begin{align} u_0 = 1, \ \lambda = -0.5, \ h=0.1 \end{align} Then for $\theta =\frac12-\frac{1}{0.1\cdot(-0.5)}=20.5$ we get $u_1=-1.00000000000001$ (using Octave). But using $\theta = 20.499$ we get $u_1=-0.999600079984009$, which is of course no good approximation, but the method will converge to $0$. Feel free to correct my answer if there is anything wrong.