Consider the following equation:

$\ddot{x} = -a x - b \dot{x}$

which we interpret as saying that we are trying to control $x$ by setting $\ddot{x}$.

We can rewrite this with $X = \begin{bmatrix} x \\ \dot{x} \end{bmatrix}$ and $K= \begin{bmatrix} 0 & -1 \\ a & b \end{bmatrix}$ as:

$\dot{X}=-K X$.

We have a choice of $a$ and $b$. For any value $a>0$ and $b>0$, the system is stable in the sense that both eigenvalues of K have positive real parts.

Now let's assume that the control cannot be applied instantly, i.e. $-ax-b\dot{x}$ is our target $y$ for $\ddot{x}$. The system becomes:

$\dddot{x} = -k(\ddot{x}-y) = -k \ddot{x} - a k x - b k \dot{x}$

I am interested in the values of $a$ and $b$ such that this system, ie I am interested in the values of $a$ and $b$ such that the eigenvalues of

$K = \begin{bmatrix} 0 & -1 & 0 \\ 0 & 0 & -1 \\ a k & b k & k \end{bmatrix}$ all have positive real parts.

Looking at numerical examples, it seems that the solution is just that $a < b k$, but I cannot prove it simply. This would be somewhat intuitive since I can rewrite it as the sum of the inverse of the eigenvalues of the 2x2 matrix $K$ is greater than $1\over{k}$, ie that the sum of the 2 characteristic times has to be greater than the characteristic time of the lag.

  • $\begingroup$ You should be a bit careful about how you state it because for $a=bk$, the eigenvalue equation factors as $(\lambda-k)(\lambda^2+bk)=0$, which is a borderline case as one would expect only if $k> 0, b>0$. Do you always assume that $a,b,k$ are all positive and are interested in the stability condition under that assumption? $\endgroup$
    – fedja
    Commented Oct 14, 2020 at 21:11
  • $\begingroup$ $k>0$ since 1/k is the characteristic time of the lag. $\endgroup$
    – Bernard
    Commented Oct 14, 2020 at 22:46

2 Answers 2


If $k>0$, it becomes elementary algebra. As Arthur pointed out, the equation is $P(z)=z^3-kz^2+(bk)z-ak=0$.

On the one hand, assume that all roots have positive real part. Then we either have 3 positive roots, or one positive root and two non-zero complex conjugate ones. In every case, the product of roots is positive, so we must have $ak>0$, i.e., $a>0$. Now, since the sum of the roots is $k$ (Vieta), the largest positive root should be strictly less than $k$, so $P(z)$ must preserve sign on $[k,+\infty)$ and, in particular, we must have $P(k)>0$ whence $bk>a$. Thus, $0<a<bk$ is a necessary condition.

On the other hand, assume $0<a<bk$ holds. Then, clearly, the equation has no roots on $(-\infty,0]$ ($P(z)<0$ there). Thus we either have three positive roots, which is fine for us, or one positive root and two complex conjugate ones. Again, we have $P(z)=(z-k)(z^2+bk)+(bk-a)k>0$ on $[k,+\infty)$, so the positive root $z_+$ is $<k$, whence (by Vieta again), the common real part of the two complex conjugate roots is $\frac 12(k-z_+)>0$.

Thus, indeed, the condition $0<a<bk$ is both necessary and sufficient.


The eigenvalues are the solutions of the cubic equation $$-a k + b k \lambda - k \lambda^2 + \lambda^3 = 0$$.

Given that the solutions depend continuously on the parameters $a, b, k$, on the boundary of the region of valid parameters, at least one of the root must have real part 0.

If that root is exactly 0, then either $a$ or $k$ is 0. In any case the other two roots are:

$$\frac{1}{2}(k\pm\sqrt{k^2-4 b k})$$

if $k^2 < 4 b k$ the root is imaginary and since $k \ge 0$ the real part is positive. If $k^2 > 4 b k$ we still have $\sqrt(k^2 - 4 b k)$ < k since $b > 0$ so this works too.

If a = 0 or k = 0 there are no roots with negative real part.

If 0 is not a root, then there is a pure imaginary root, but the conjugate must also be a root. Expanding:

$$(\lambda^2 + C^2)(\lambda - \lambda_0) = \lambda^3 - \lambda^2 \lambda_0^2 + \lambda C^2 - C^2 \lambda_0$$

Identifying the roots: $\lambda_0 = k$ $C^2= a$ and finally $a = b k$.


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