Determine the constants $a$ and $b$ in terms of the parameter $α$ in such a way that the multistep method
$y_{n+1}=-ay_n - αy_{n-1} + hby'_n$
have the largest order possible. For what ranges of values of $α$ is the stable method?
I am a little confused solving this problem, I do not know if what I have done is fine, I hope someone will tell me if I am going in the right direction, thank you very much.
To find the parameters in such a way that the multistep method has the largest possible order, do I have to pay attention to the global truncation error? A multistep method has the general form
$$y_{n+1}=\sum_{j=0}^{p}a_jy_{n-j}+h\sum_{j=-1}^{p}b_jf(x_{n-j},y_{n-j})$$ for all $n\geq p$.
And we know that for $\tau(h)=O(h^m)$ it is necessary and sufficient that
$$\sum_{j=0}^{p}(-j)^ia_j+i\sum_{j=-1}^{p}(-j)^{i-1}b_j=1$$
For all $i=1,2,...,m$
What else could I do? I think that from here I can get conditions for $a$ and $b$ but I do not know what the $a_j$ and $b_j$ are. Here $p$ is the step size that I think is $2$.
For the second question I know that if the multistep method meets the root condition then it is stable. To see if it fulfills the root condition I have to take out the associated polynomial that is $\rho(r)=r^{p+1}-\sum_{j=0}^{p}a_jr^{p-j}$ and its roots. We know that in this case $p=2, a_0=-a$ and $a_1=-\alpha$, with which we have left that $\rho(r)=r^3+ar^2+\alpha r$ and so $r_0=0, r_1=\frac{-a+\sqrt{a^2-4\alpha}}{2}, r_2=\frac{-a-\sqrt{a^2-4\alpha}}{2}$ are the roots.
What I have to prove is that $|r_j|\leq 1$ for all $j=0,1,2$ and that if $|r_j|=1$ is for that $\rho'(r_j)\neq 0$ but I do not know how to do it, could someone help me please?
1) For the first part of the question, you are correct that you need to look at the global truncation error $\tau(h)$. Since you already know the necessary and sufficient condition for $\tau(h) = \mathcal{O}(h^m)$ $$ \sum_{j=0}^p (-j)^ia_j + i\sum_{j=-1}^p (-j)^{i-1}b_j = 1 \ \ \textrm{ for any $i=1,2,\dots,m,$} \tag{GTC}$$ you just need to find the coefficients $a_j, b_j$, substitute these into (GTC) and solve linear equations in terms of these coefficients.
It should be clear that $p=1$. To see the required coefficients, let us expand the general form of multistep method with $p=1$ and compare it to the given multistep method: \begin{align*} y_{n+1} & = \sum_{j=0}^p a_jy_{n-j} + h\sum_{j=-1}^p b_jf(x_{n-j},y_{n-j}) \\ & = \sum_{j=0}^p a_jy_{n-j} + h\sum_{j=-1}^p b_jy_{n-j}' \\ & = a_0y_n + a_1y_{n-1} + h\left(b_{-1}y_{n+1}' + b_0y_n' + b_1y_{n-1}'\right) \\ & = -ay_n -\alpha y_{n-1} + hby_n'. \end{align*} We see that \begin{align*} a_0 & = -a \\ a_1 & = -\alpha \\ b_{-1} & = 0 \\ b_0 & = b \\ b_1 & = 0. \end{align*} In particular, we need three equations since we only have three unknowns $a,\alpha,b$. Two equations comes from requiring the given method to be consistent.
The last equation comes from (GTC) with $i=2$.
2) You are correct again, that you need to look at the root condition. You are almost correct except that $p$ should be 1 instead of 2. So the associated polynomial is \begin{align*} \rho(r) & = r^{p+1} - \sum_{j=0}^p a_jr^{p-j} \\ & = r^2 - \sum_{j=0}^1 a_jr^{1-j} \\ & = r^2 + ar + \alpha \end{align*} which has roots $$ r = \frac{-a\pm\sqrt{a^2 - 4\alpha}}{2}. $$ Clearly this depends on both $a$ and $\alpha$ but since the question only asks for the range of $\alpha$, this means that $a$ is related to $\alpha$ in some way. Indeed, from the first part, one of the consistency conditions gives $$ \sum_{j=0}^1 a_j = a_0 + a_1 = -a - \alpha = 1 \implies a = -1-\alpha. $$ Substituting this into the roots $r$ then yields \begin{align*} r & = \frac{(1 + \alpha)\pm \sqrt{(1+\alpha)^2 - 4\alpha}}{2} \\ & = \frac{(1+\alpha)\pm \sqrt{1 + 2\alpha + \alpha^2 - 4\alpha}}{2} \\ & = \frac{(1+\alpha)\pm \sqrt{1 - 2\alpha + \alpha^2}}{2} \\ & = \frac{(1+\alpha)\pm \sqrt{(1-\alpha)^2}}{2} \\ & = \frac{(1+\alpha)\pm (1-\alpha)}{2}. \end{align*} The roots are $$ r_+ = \frac{1+\alpha+1-\alpha}{2} = 1 \ \ \textrm{ and } \ \ r_- = \frac{1+\alpha-(1-\alpha)}{2} = \alpha. $$ The final step is to check this against the root conditions and these should spit out ranges for $\alpha$, and I will let you finish this.