# Proving strong lyapunov function

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?

• Is it possible that there is a typo in $y'=cd-df(y)$? I suspect that $cd$ should be $cx$. Aug 6 at 7:24
• That is correct it should be cx Aug 6 at 16:30
• 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 $f$. Aug 6 at 18:17
• 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. Aug 6 at 23:24

From

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

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

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

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 $$q$$.

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

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.