Are zeros of a cubic polynomial lipschitz wrt the constant term?

Question

Consider the cubic equation $$ ax^3 + bx^2 + cx + d = 0\,.$$

My question is: do the zeros of this equation change on the same order as the change in its coefficients? I'm interested in $d$ specifically, so in this case, suppose $|d -d'| < \epsilon$. Then, are the zeros of $$ax^3 + bx^2 + cx + d' = 0$$ within $\epsilon$ of the zeros of the first equation? Or $O(\epsilon)$?

I know that the effect of $d$ is to raise or lower the graph of the cubic equation. I also know that to solve for the zeros of a cubic equation, one typically factors it into a product of linear and quadratic terms.

Answer 1

Consider $a=0$, $b=1$, $c=0$, $d=0$. You should be able to show that the Lipschitz condition (with respect to $d$ as you asked for) fails here since $ x = \sqrt{-d} $ is (as you must check) not Lipschitz there.

However this statement can be salvaged: you can use the inverse function theorem to show that the "root-finding" function is not only Lipschitz but smooth near single-roots. (Note, my example had to single out a double-root.)

On the other hand there's no reason to expect a global Lipschitz constant. One can only expect locally Lipschitz; the constant will have to differ as one considers different parts of the domain of coefficients.

Answer 2

The basic idea is that the roots are analytic functions of the coefficients when they are distinct. The discriminant of $a x^3 + b x^2 + c x + d$ is $-27 a^{2} d^{2}+18 a b c d -4 a \,c^{3}-4 b^{3} d +b^{2} c^{2}$, and this is $0$ when there is a multiple root.

For example, take $a = 1$, $b = 0$, $c = -3$. At $d=2$ there is a simple root of $-2$ and a double root of $1$. Here is a plot of the real roots as a function of $d$.

You will notice that as $d \to 2-$ two real roots collide, with the slope of the curve approaching $\pm \infty$ (so these roots are not Lipschitz functions of $d$ there). For $d > 2$ those two roots become complex.