Cubic equation with random coefficients: Probability of one real root

Question

Let's say we have an equation: $$\frac{x^3}{3}-a^2x+b=0$$ where $a$ and $b$ are randomly picked from an interval (0,1). Let $N$ be a number of real roots of the equation (so $N$ can be $1$ or $3$). How do I find the probability of $N$ being $1$?

I'm assuming that the probability of picking a number from (0,1) is uniformly distributed, so: $$a,b\in\mathcal{U}(0,1)$$ How do I go from there?

Answer 1

Comment: According to Wikipedia on 'Cubic function', the equation $\alpha x^3 + \beta x^2 + \gamma x + \delta = 0$ has only one real root, if $$\Delta = 18\alpha\beta\delta\gamma - 4\beta^3\delta + \beta^2\gamma^2 - 4 \alpha\gamma^3 - 27\alpha^2\delta^2 < 0.$$

For our cubic equation $\frac 1 3x^2 - A^2x + B = 0,$ we have $\alpha = 1/3, \beta=0, \gamma=-A^2,$ and $\delta = B,$ so the discriminant becomes $\Delta = \frac 4 3 A^6 - 3B^2.$ If $A$ and $B$ are independently distributed as $\mathsf{Unif}(0,1),$ then it should not be difficult to find $P(\Delta < 0).$

Because OP has shown no engagement, I expect this Question might be closed soon. In this comment, I will show only the approximate value $P(\Delta < 0) = 0.875,$ obtained from an easy simulation in R. (A million iterations should give 3-place accuracy; three such simulations agree to three places.)

a = runif(10^6);  b = runif(10^6); Dlt = .75*a^6 - 3*b^2
mean(Dlt < 0)
## 0.875421

Here are four of the one million cubic curves in the simulation above. All of them happen to have one real root (each at $x < -1);$ one of them 'almost' had three real roots.

Answer 2

The cubic equation $f(x)=\frac{x^3}{3}-a^2x+b=0$ has two real roots (a single root and a double root) when it has two turning points, and one of these corresponds to a zero value of the cubic, i.e. if there is an $x$ such that $f(x)=f'(x)=0$. Since $f'(x)=x^2-a^2$ and $f'(x)=0$ when $x=\pm |a|$, that leads to the possibilities:

• $b=\tfrac23 |a|^3$ with the double root at $x=|a|$ and single root at $x=-2|a|$
• $b=-\tfrac23 |a|^3$ with the double root at $x=-|a|$ and single root at $x=2|a|$

In this question we are looking for one real root, which quick investigation suggests happens when $b$ is outside these two possibilities i.e. $\mathbb P( b<-\tfrac23 |a|^3) + \mathbb P( b > \tfrac23 |a|^3)$.

Here $a$ and $b$ are independently distributed uniformly on $[0,1]$ so $|a|=a$ and $\mathbb P( b< -\frac23 |a|^3)=0$. So overall the probability is $$\mathbb P\left( \frac23 a^3 < b \le 1 \right)=\int\limits_0^{1} \left(1-\frac23 a^3\right) \,da= \frac56 \approx 0.833$$