I am working on the following problem:

Let $x'=-ax+bf(y)$ and $y'=cx-df(y)$ with $f(0)=0$, $yf(y)>0$ for $y \neq 0$ and $a,b,c,d>0$. I need to show that the following function (for suitable values of p and q) is a strong Lyapunov function for the zero solutions of my system.

$$ V = px^2/2+q\int_{0}^{y} f(u) \,du $$

I had no problem showing that $V>0$ for all nonzero input as it's easy to show that the integral is positive using the fact that $yf(y)>0$ and $f(0) = 0$

I then go on to try to show that the derivative of $V$ with respect to time is negative definite, which is where I run into problems.

$$\dot V = px\dot x + q\dot y f(y)$$

via the chain rule, but when I plug in the defined values of $\dot x$ and $\dot y$ I cannot figure out how to show that this expression is always less than zero. Any advice as to how to start this process?

  • $\begingroup$ Is it possible that there is a typo in $y'=cd-df(y)$? I suspect that $cd$ should be $cx$. $\endgroup$ Aug 6 at 7:24
  • $\begingroup$ That is correct it should be cx $\endgroup$
    – Half_Baked
    Aug 6 at 16:30
  • $\begingroup$ If $f(y)=y$, then the system is linear and the stability is determined by the eigenvalues of the system matrix. If the off-diagonal elements are large enough, the eigenvalues become real with opposite sign. Then no Lyapunov function will exist. So you will need at least $ad-bc>0$. This seems also to be necessary and sufficient for general $f$. $\endgroup$ Aug 6 at 18:17
  • $\begingroup$ Assuming $ad-bc>0$, should I just be plugging in the x and y time derivatives into the derivative of the Lyapunov? And then presumably reorganizing terms. $\endgroup$
    – Half_Baked
    Aug 6 at 23:24

2 Answers 2



$$\dot{V} = p x \dot{x} + q f(y) \dot{y},$$

the substitutions are made

$$\dot{V} = p x \left[- a x + b f(y)\right] + q f(y) \left[c x - d f(y)\right]$$

$$\dot{V} = - a p x^{2} + b p x f(y) + c q x f(y) - d q f^{2}(y)$$

and be rewritten in a compact form

$$\dot{V} = - \left[\matrix{x & f(y)}\right] \left[\matrix{a p & - b p \cr - c q & d q}\right] \left[\matrix{x \cr f(y)}\right]$$

$$\dot{V} = - \mathbf{x}^{T} \mathbf{P} \mathbf{x}.$$

To ensure that $\dot{V} < 0$ for $\mathbf{x} \neq \mathbf{0}$, then the matrix $\mathbf{P} = \left[\matrix{a p & - b p \cr - c q & d q}\right]$ has to be positive-definite.

To ensure that the matrix $\mathbf{P}$ is positive-definite, and this boils down to having the condition

$$a p d q - b p c q > 0$$

Edit: Since $\mathbf{P}$ has to be symmetric for the positive-definite property, the values for $p$ and $q$ can be selected to be $p = \frac{1}{b}$ and $q = \frac{1}{c}$ so that

$$\mathbf{P} = \left[\matrix{\frac{a}{b} & -1 \cr -1 & \frac{d}{c}}\right]$$

Then, the inequality becomes

$$\frac{ad}{bc} - 1 > 0$$

and if the condition is satisfied, all of the eigenvalues of $\mathbf{P}$ are positive. It is also possible to select other values for $p$ and $q$.

  • $\begingroup$ Is that argument also valid for non-symmetric matrices? "Positive definite" usually implies that the matrix was symmetric from the start. $\endgroup$ Aug 12 at 5:37
  • $\begingroup$ @LutzLehmann, thanks for the matrix property. Edited to show the selections for $p$ and $q$. $\endgroup$
    – Xega Xam
    2 days ago
  • $\begingroup$ Great answer, I hadn’t considered a matrix representation of the derivative $\endgroup$
    – Half_Baked

After insertion you get \begin{align} \dot V &= -pax^2 + (pb+qc)xf(y)-qdf(y)^2 \\ 4pa\dot V&=-\Bigl(2pax-(pb+qc)f(y)\Bigr)^2+\Bigl[(pb+qc)^2-4paqd\Bigr]f(y)^2 \\ &=-\Bigl(2pax-(pb+qc)f(y)\Bigr)^2+\Bigl[(pb-qc)^2-4(ad-bc)pq\Bigr]f(y)^2 \end{align} Now the aim is to have the coefficient of the second term be negative. The square can be reduced to zero using $p=c$, $q=b$. Then the sign condition only depends on if $ad-bc>0$.

In the case that the second coefficient is negative, slight variations of $p$ and $q$ will keep that quality, so the values used above are not unique.


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