14.2: Lypanunov's Indirect Method- Analyzing the Linearization

Suppose the system

\[\dot{x}=f(x) \label{14.8}\]

has an equilibrium point at \(\bar{x} = 0\) (an equilibrium at any other location can be dealt with by a preliminary change of variables to move that equilibrium to the origin). Assume we can write

\[f(x)=A x+h\tag{x}\]

where

\[\lim _{\|x\| \rightarrow 0} \frac{\|h(x)\|}{\|x\|}=0\nonumber\]

i.e. \(h(x)\) denotes terms that are higher order than linear, and \(A\) is the Jacobian matrix associated with the linearization of (14.8) about the equilibrium point. The linearized system is thus given by

\[\dot{x}=A x \label{14.9}\]

We might expect that if Equation \ref{14.9} is asymptotically stable, then in a small neighborhood around the equilibrium point, the system in Equation \ref{14.8} behaves like Equation \ref{14.9} and will be stable. This is made precise in the following theorem.

Theorem 14.3

If the system describe by Equation \ref{14.9} is asymptotically stable, then the equilibrium point of system (Equation \ref{14.8}) at the origin is (locally) asymptotically stable.

Proof

If system (14.9) is asymptotically stable, then for any \(Q > 0\), there exists \(P > 0\) such that

\[A^{T} P+P A=-Q\nonumber\]

and \(V (x) = x^{T} P x\) is a Lyapunov function for system (14.9). Consider \(V (x)\) as a Lyapunov function candidate for system (14.8). Then

\[\begin{aligned}
\dot{V}(x) &=x^{T}\left(A^{T} P+P A\right) x+2 x^{T} P h(x) \\
& \leq-\lambda_{\min }(Q)\|x\|^{2}+2\|x\| \cdot\|h(x)\| \cdot \lambda_{\max }(P) \\
& \leq-\left[\lambda_{\min }(Q)-2 \lambda_{\max }(P) \frac{\|h(x)\|}{\|x\|}\right] \cdot\|x\|^{2}
\end{aligned}\nonumber\]

From the assumption on \(h\), for every \(\epsilon > 0\), there exists \(r > 0\) such that

\[\|h(x)\|<\epsilon\|x\|, \quad \forall\|x\|<r\nonumber\]

This implies that \(\dot{V}\) is strictly negative for all \(\|x\|<r\), where \(r\) is chosen for

\[\epsilon<\frac{\lambda_{\min }(Q)}{2 \lambda_{\max }(P)}\nonumber\]

This concludes the proof.

Notice that asymptotic stability of the equilibrium point of the system (14.8) can be concluded from the asymptotic stability of the linearized system (14.9) only when the eigenvalues of \(A\) have negative real parts. It can also be shown that if there is any eigenvalue of A in the right half plane, i.e. if the linearization is exponentially unstable, then the equilibrium point of the nonlinear system is unstable. The above theorem is inconclusive if there are eigenvalues on the imaginary axis, but none in the right half plane. The higher-order terms of the nonlinear model can in this case play a decisive role in determining stability; for instance, if the linearization is polynomially (rather than exponentially) unstable, due to the presence of one or more Jordan blocks of size greater than 1 for eigenvalues on the imaginary axis (and the absence of eigenvalues in the right half plane), then the higher-order terms can still cause the equilibrium point to be stable.

It turns out that stronger versions of the preceding theorem hold if \(A\) has no eigenvalues on the imaginary axis: not only the stability properties of the equilibrium point, but also the local behavior of (14.8) can be related to the behavior of (14.9). We will not discuss these results further here. Similar results hold for discrete-time systems.

Example 14.4

The equations of motion for a pendulum with friction are

\[\begin{aligned}
\dot{x_{1}} &=x_{2} \\
\dot{x_{2}} &=-x_{2}-\sin x_{1}
\end{aligned}\nonumber\]

The two equilibrium points of the system are at \((0, 0)\) and \((\pi, 0)\). The linearized system at the origin is given by

\[\begin{array}{l}
\dot{x_{1}}=x_{2} \\
\dot{x_{2}}=-x_{1}-x_{2}
\end{array}\nonumber\]

or

\[\dot{x}=\left[\begin{array}{cc}
0 & 1 \\
-1 & 0
\end{array}\right] x=Ax\nonumber\]

This \(A\) has all its eigenvalues in the OLHP. Hence the equilibrium point at the origin is asymptotically stable. Note, however, that if there were no damping, then the linearized system would be

\[\dot{x}=\left[\begin{array}{cc}
0 & 1 \\
-1 & 0
\end{array}\right] x\nonumber\]

and the resulting matrix \(A\) has eigenvalues on the imaginary axis. No conclusions can be drawn from this situation using Lyapunov linearization methods. Lyapunov's direct method, by contrast, allowed us to conclude stability even in the case of zero damping, and also permitted some detailed global conclusions in the case with damping.

The linearization around the equilibrium point at (\(\pi, 0\)) is

\[\begin{array}{l}
\dot{z_{1}}=z_{2} \\
\dot{z_{2}}=+z_{1}-z_{2}
\end{array}\nonumber\]

where \(z_{1}=x_{1}-\pi\) and \(z_{2}=x_{2}\), so these variables denote the (small) deviations of \(x_{1}\) and \(x_{2}\) from their respective equilibrium values. Hence

\[A=\left[\begin{array}{cc}
0 & 1 \\
1 & -1
\end{array}\right] x=A x,\nonumber\]

which has one eigenvalues in the RHP, indicating that this equilibrium point is unstable.