Problem in Adams Bashforth 3rd order method

Question

We know the Adams Bashforth 3rd order method is: $$ y^{n+3}-y^{n+2}=h\left(\frac{23}{12}f^{n+2}-\frac{4}{3}f^{n+1}+\frac{5}{12}f^{n}\right) $$ I need to proof that the interval of absolute stability is $-1<hl<0$.
My attempt:
Let $f^{n+i}=hly^{n+i}$, Then i take the characteristic polynomial and i arrive on a 3rd order Polynomial $$ y^{n+3}+\left(-1-\frac{23hl}{12}\right)y^{n+2}+\frac{4hl}{3}y^{n+1}-\frac{5hl}{12}y^n=0$$ And let $z=y^n$ $$ z^3+\left(-1-\frac{23hl}{12}\right)z^{2}+\frac{4hl}{3}z^{1}-\frac{5hl}{12}=0$$ which it can't be solved with simple computation. Any ideas?

Answer 1

The question is about when all roots of $$ z^3-z^2=\frac{hl}{12}(23z^2-16z+5) $$ are inside the unit circle. For $h\approx 0$ one can use perturbation series to get an impression, especially whether the root at $z=1$ for $h=0$ moves inside the circle. Unsurprisingly it follows the first terms in the expansion of $\exp(hl)$, so that stability is given for sufficiently small $h$.

As a next step one can refine the factorization on the left side similar to a partial fraction decomposition of the right side, $$ z^2(z-1)=\frac{hl}{12}(12z^2+(11z-5)(z-1))\\ \left(z^2-\frac{hl}{12}(11z-5)\right)\left(z-1-hl\right)=\frac{(hl)^2}{12}(11z-5) $$ The second factor tells that for $l<0$ the corresponding root will cross into the unit disk. The first factor tells that for $l<0$ the roots of this quadratic will be real (until the root sets cross). So following the root paths for increasing $h>0$, stability gets violated after a root path crosses $\pm 1$. $z=+1$ is only possible for $hl=0$, so it remains to insert $z=-1$, $$ -2=\frac{hl}{12}(23+16+5)=\frac{11}{3}hl\iff hl=-\frac{6}{11} $$ This predicts instability for $hl=-0.55<-\frac6{11}=-\frac{54}{99}$, with one root close to $z=-1.0076$ contradicting the claim.

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The point of double roots and the ensuing split into a complex-conjugate pair is close to this, but outside the stability region at about $hl=-0.5489738$.