Robust & Linear Control Systems The characteristic polynomial of a control system is the following uncertain polynomial:
$$s^3 + a_2 s^2 + a_1 s + 3.5 $$
Where $a_1 \in  [1.5,4.2],a_2 ∈ [1.2,4.25]$ and $ 4.2 ≤ a_1 + a_2 ≤ 6.3$
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Is this uncertain polynomial stable?
How do I solve this problem, when I have got two values from $a_1$ and $a_2$.
 A: The Hurwitz matrix of the polynomial is:
$$ \begin{bmatrix} a_2 & 3.5 & 0 \\ 1 & a_1 & 0 \\ 0 & a_2 & 3.5 \end{bmatrix}$$
To stability this matrix must be definite-positive, this is, all subdeterminants must be positive. The conditions are:
$$a_2 >0$$
$$a_2 a_1 >3.5 $$
If the second subdeterminant is positive, it follows that the third one also is (because $3.5 \gt 0$ ). You can check the polynomial will always be stable for the values you have.
A: As @Villa proposed you can start with the Hurwitz criterion for a cubic polynomial.
$$a_1>0$$
$$a_2>0$$
$$a_1a_2>3.5.$$
The first two conditions are trivially fulfilled for the given intervals. Hence, we only need to care for $a_1a_2>3.5\implies a_2 > 3.5/a_1$. The given intervals for your coefficients will give you the feasible region for the inequalities (see below) without $a_1a_2>3.5$.
 
The only thing we need to do is to check if the corners of the lower diagonal edge given by $a_2=4.2-a_1$ are fulfilling the inequality $a_1a_2>3.5$. 
Determining the corners:
$$a_1=1.5\implies a_2=4.2-1.5=2.7 \implies P_1=(1.50,2.70)$$
$$a_2=1.20 \implies a_1=3 \implies P_2=(3.00,1.20)$$ 
Now, check if $a_1a_2>3.5$ is true for both points:
$$P_1: a_1a_2=4.05>3.5$$
$$P_2: a_1a_2=3.6>3.5$$
We can conclude that the polynomials are asymptotically stable for the given conditions. 
A: I would suggest checking the Kharitonov's theorem that gives a simple and practical answer for a similar question for polynomials of any degree.
