Determine whether the roots of a cubic equation have positive real part

Question

Consider the cubic equation

$a x^3 + b x^2 + cx + d =0 $,

where all coefficients depend on three parameters $$a=a(i, j, k), b=b(i, j, k),\cdots$$ and $a, b, c \in \mathbb{R}$ for all $(i, j, k)$. The explicit form of the dependence is known (a polynomial with given coefficients), it is also known that $a \neq 0$. How to determine at what values of the variables $(i, j, k)$ the real parts of the roots of this equation are positive?

Is there an alternative to Cardano's formula? I tried to use it, and I got quite complicated (with respect to $i, j, k$) expressions. I believe that final expressions should be simple, since we don't need to know the specific values of the roots.

Answer 1

The Routh–Hurwitz criterion for real cubics states that:

The third-order polynomial $P(s)=s^{3}+a_{2}s^{2}+a_{1}s+a_{0}$ has all roots in the open left half-plane if and only if $a_{2}$, $a_{1}$, and $a_{0}$ are positive and $a_{2}a_{1} \gt a_{0}\,$.

The roots of $f(x) = a x^3 + b x^2 + cx + d = 0$ have positive real parts if and only if the roots of $f(-x) = - a x^3 + b x^2 - cx + d = 0$ have negative real parts. Dividing by $-a \ne 0$ and applying the criterion with $a_2 = - \frac{b}{a}$, $a_1 = \frac{c}{a}$, $a_0 = - \frac{d}{a}$ results in the following necessary and sufficient conditions.

• $- \frac{b}{a} \gt 0\,$, $\;\frac{c}{a} \gt 0\,$, $\;- \frac{d}{a} \gt 0\,$ $\iff ab \lt 0\,$, $\;ac \gt 0\,$, $\;ad \lt 0\,$;

• $- \frac{bc}{a^2} \gt - \frac{d}{a} \iff ad \gt bc\,$.

Answer 2

Right, so we don`t really have a closed form for polynomials with order greater than two, which is given by the Bhaskara formula. For order $n \ge 3$ you might want to look into the general solution, i.e.:

Take the condition that $a \ne 0$, and factor the three roots $r_1$, $r_2$ and $r_3$ into the following form:

$a(x-r_1)(x-r_2)(x-r_3)=0$

So, notice that this reduces back to our initial equation where $b$, $c$ and $d$ are given as functions of $a$, $r_1$, $r_2$ and $r_3$.

$a(x-r_1)(x-r_2)(x-r_3)=a x^3 + a (r_1 + r_2 + r_3) x^2 + a (r_1 r_2 + r_2 r_3 + r_3 r_1) x + a r_1 r_2 r_3$

Finally, one of the roots is real, let's choose one - noting we don't lose any generality in our solution - and say $r_1 \in \mathbb{R}$, then we know that:

$r_2 = e+fi$ and $r_3 = e-fi$

I'm unclear on what your parameters are, but you do realize the final expressions are not "simple". Now, you must study the sign of $r_1$ and $e$.