# Chua's circuit: Is $x=y=z=0$ a stable equilibrium?

According to this wikipedia page, the differential equations satisfied by several physical quantities in Chua's circuit is (What each letter represents doesn't matter that much in this question) For small $$x$$, $$f(x)=-mx$$ for some positive $$m$$. Then this becomes a linear system. In order to analyze the stability of the zero solution $$x=y=z=0$$, I wrote the RHS as a matrix $$\begin{bmatrix}-a & b & 0\\ c & -c & d \\ 0 & -f & 0 \end{bmatrix}$$ where all letters are positive constants.

The characteristics equation is $$\lambda^3+(a+c)\lambda^2+(ac-bc+fd)\lambda+fda=0$$. If it has three negative roots, the equilibrium $$x=y=z=0$$ is stable due to the results here. But whether it is stable or not depends on the values of parameters. When we are actually doing this circuit experiment, initially the values of $$x,y,z$$ should be really close to zero, because we have just switched on. If $$0$$ is a stable equilibrium, then that means everything remains close to zero. That means we cannot observe the double scroll pattern (because everything is zero, so you can only see a dot!)

My question is, does that mean we must choose the values of constants so that $$x=y=z=0$$ is not a stable equilibrium? If $$0$$ is stable, then the experiment will fail, since nothing can be observed?

Are there any simplier methods other than eigenvalues that can be used to analyse the stability?

• The first row of your matrix is wrong. According to the form of the system of equations, the first row of the matrix should be $$-a'\quad a\quad 0$$ with $a'\neq a$. – Gec Jul 13 at 6:25
• @Gec Oh thank you. Let me try again. – Ma Joad Jul 13 at 6:32

It is actually important what the variables represent. $$x$$ and $$y$$ are the voltages across two capacitors, and $$z$$ is the current through an inductor. When these values are non-zero, it means there is energy stored in the system. When they are all zero, it means there is no energy stored in the circuit.

The solution where there is no energy in any of the energy storage elements is indeed a valid solution to the differential equations. This is what we call a trivial solution to the equations.

You should notice that the nonlinear element (Chua diode) has an I-V characteristic that exists in quadrants II and IV: This means that except at the point where the voltage across it is zero, it will always be delivering energy to the circuit.

When we are actually doing this circuit experiment, initially the values of x,y,z should be really close to zero, because we have just switched on.

I don't see that this follows. Depending how the nonlinear element is implemented, the act of "switching on" is likely to deliver some impulse of energy to the circuit.

Does that mean we must choose the values of constants so that x=y=z=0 is not a stable equilibrium?

Yes, you'd want the slope of the Chua diode characteristic through the origin to be steep enough that the Chua diode will be delivering more energy to the circuit than the ordinary resistor in circuit is absorbing, so that any small noise fluctuation in the operating point around zero will lead to the net stored energy in the circuit increasing, thus making the trivial solution unstable.

You probably also need to be sure that once the Chua diode voltage exceeds the value $$\pm E$$, that the energy delivered by the Chua diode is (on average) less than the energy absorbed by the ordinary resistor, to avoid the solution to the equations "blowing up" (tending toward infinite total energy).

• Thank you a lot! I have two more questions: 1. If a solution to the equations "blows up", then is it an example of resonance? 2. If the trivial solution is stable, then the experiment will certainly fail?(i.e. nothing interesting can be observed, everything is zero) – Ma Joad Jul 13 at 6:15
• Are there any simplier methods other than eigenvalues that can be used to analyse the stability? – Ma Joad Jul 13 at 6:44
• @MaJoad, 1. I don't know if "resonance" applies to a nonlinear system. 2. I don't think you can use eigenvalues analyze the stability of a nonlinear system. – The Photon Jul 13 at 17:35
• No. The system is linear near the origin. That's the point！ – Ma Joad Jul 13 at 22:06
• @MaJoad, OK, but you're not going to get any interesting behavior if you only operate it in the linear region. You have to design it to be unstable in the linear region if you want to get the behavior that makes it an interesting circuit. – The Photon Jul 13 at 22:08