2
$\begingroup$

Suppose we have two monic polynomials with real coefficients \begin{align*} & p_1(t) = t^n + a_{n-1} t^{n-1} + \dots + a_0 \\ & p_2(t) = t^n + b_{n-1} t^{n-1} + \dots + b_0. \end{align*} Further assume $p_1(t)$ and $p_2(t)$ are both Hurwitz stable, i.e., the roots are all lying on the open left half plane of $\mathbb C$; equivalently, real parts of all roots are $<0$.

I am wondering whether or not there is some $M < 0$ (existence of such $M$ is enough) such that if the real parts of all roots smaller than $M$, then the convex combination of the two polynomials is Hurwitz stable.

Intuitively, I am thinking by making real parts of roots "negative" enough, the convex combination would stay in the left half plane.

p.s. Routh-Hurwitz Theorem gives sufficient condition to determine whether a polynomial is Hurwitz stable. But the rule seems complicated.

$\endgroup$
6
  • $\begingroup$ It doesn't really help, but it is straightforward to see that the convex combination has no real non negative zeros. $\endgroup$
    – copper.hat
    Commented Jul 14, 2018 at 4:45
  • $\begingroup$ Note that $p_k(t) \to \infty $ as $ t \to \infty$, and there are no zeros in $t \ge 0$, hence $p_k(t) >0$ for all $t \ge 0$. Hence any convex combination is also strictly positive. $\endgroup$
    – copper.hat
    Commented Jul 14, 2018 at 5:30
  • $\begingroup$ I thought I had an answer using $\det$ and the associated companion matrix but forgot that the $\det$ may be non real which blew that approach. $\endgroup$
    – copper.hat
    Commented Jul 14, 2018 at 5:33
  • $\begingroup$ Thanks. I am curious what approach you were considering. If you put companion form, why would the $\det$ be non real since the coefficients are real in my setting? $\endgroup$
    – user1101010
    Commented Jul 14, 2018 at 5:39
  • $\begingroup$ I deleted a non answer. I was thinking that the two $\det$s were not zero in the right half plane hence the convex combination would not be. However, this only works if the $\det$ was real. $\endgroup$
    – copper.hat
    Commented Jul 14, 2018 at 5:47

1 Answer 1

Reset to default
4
+50
$\begingroup$

As shown in this answer, the sum of two stable polynomials does not have to be stable. The example given there is that, for $a>11$, the polynomial $(x+1)^3 + (x+a)$ has roots in the right half plane, although $(x+1)^3$ and $x+a$ are stable (and even have their roots real!).

This is easily adapted to give a counterexample to your question: For any positive real $M$ and any $a>11$, the polynomials $(x+M)^3$ and $(x+aM)$ have roots with real part $<M$, but their sum has roots with positive real part.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .