Lyapunov’s stability analysis technique is very common and dominant. The main deficiency, which severely limits its utilization, in reality, is the complication linked with the development of the Lyapunov function which is needed by the technique.

The system dynamics must be described by a state-space model. It is a description in terms of a set of first-order differential equations. For example, a nonlinear system might be described by a set of n first-order nonlinear differential equations.

$\overset{.}{\mathop{x}}\,{{}_{i}}={{f}_{i}}({{x}_{1}},{{x}_{2}},\cdots ,{{x}_{n}},t)\text{ i=1,}\cdots \text{,n (1)}$

In the form of a state-space model as

$\overset{.}{\mathop{x}}\,=f(x,t)$

Where

\[x=\left( \begin{align} & {{x}_{1}} \\ & \vdots \\& {{x}_{n}} \\\end{align} \right)\]

$\overset{.}{\mathop{x}}\,=\left( \begin{align}& {{\overset{.}{\mathop{x}}\,}_{1}} \\& \vdots \\& {{\overset{.}{\mathop{x}}\,}_{n}} \\\end{align} \right)$

$f(x,t)=\left[ \begin{align}& {{f}_{1}}({{x}_{1}},\cdots ,{{x}_{n}},t) \\& \vdots \\& {{f}_{n}}({{x}_{1}},\cdots ,{{x}_{n}},t) \\\end{align} \right]$

The vector x is the state vector, and its elements are state variables. The origin x=0(x_{1}=⋯=x_{n}=0) of the state space will be assumed to be an equilibrium solution, where f_{i}=0, i=1, ⋯, 0. The state-model description of a given system is not unique but depends on which variables are chosen as state variables.

The Lyapunov function, V(x_{1}, ⋯, x_{n}), is a scalar function of the state variables. To motivate the following and to make the stability theorems plausible, let V be selected to be

$V(x)={{\left\| x \right\|}^{2}}=\sum\limits_{i=1}^{n}{x_{i}^{2}}$

Here, ∥x∥ is the Euclidean norm of x, the length of the vector x, and the distance to the origin of the state space. V is evidently positive and V (0) =0. Now let

\[\overset{.}{\mathop{V}}\,=\frac{dV}{dt}=\frac{\partial V}{\partial {{x}_{1}}}{{\overset{.}{\mathop{x}}\,}_{1}}+\cdots +\frac{\partial V}{\partial {{x}_{n}}}{{\overset{.}{\mathop{x}}\,}_{n}}\]

Be calculated by substituting (1). If $\overset{.}{\mathop{V}}\,$ were found to be always negative, with $\overset{.}{\mathop{V}}\,$ (0) =0, then apparently V decreases continuously, and the state must end up in the origin of the state space, implying asymptotic stability.

It might be the case that $\overset{.}{\mathop{V}}\,$ is negative ONLY in a very small region around the origin. Therefore the following perceptions are suitable:

- If a system returns to x=0 after any size of disturbance then it is globally asymptotically stable.
- A system is locally asymptotically stable if it does so after an adequately small disturbance.
- A system is stable if, for any size of disturbance, the solution remains inside a definite region.

In order to build up these conceptions, the following statements are employed for the sign of V (and $\overset{.}{\mathop{V}}\,$):

- V is a positive (negative) definite in a region, which includes the origin if it is positive (negative) all over except that V (0) =0.
- V is a positive (negative) semi-definite, which includes the origin if it has a consistent positive (negative) sign but is zero as well at points other than the origin.
- V is indefinite if both the signs lie in the region which includes the origin.

**Lyapunov stability theorem**

If there exists a:

- Positive-definite V,
- and V→∞ as ∥x∥ →∞,

then it means that a system is asymptotically stable inside the region where $\overset{.}{\mathop{V}}\,$ is negative definite and is stable in the case of $\overset{.}{\mathop{V}}\,$ being negative semidefinite. The attributes are global if the region expands all over the state space.

** Lyapunov Instability theorem**

If there exists a V, such that:

- $\overset{.}{\mathop{V}}\,$ is negative definite,
- and $\overset{.}{\mathop{V}}\,$ →∞ as ∥x∥ →∞

then it means that a system is unstable inside the region where V is not positive (semi-) definite.

**Remarks**

- The reason for two theorems is that if the origin is unstable it will be impossible to find a V-function that satisfies the stability theorem. But one satisfying the instability theorem will exist and, if it can be found, will prove instability.
- The V-function is not unique, and different choices, in general, will indicate different stability regions. As this implies, a system, in general, does not become unstable where $\overset{.}{\mathop{V}}\,$ changes sign. The theorems give only sufficient conditions, and the predicted stability boundaries are usually quite conservative.
- Asymptotic stability can often be proved even if $\overset{.}{\mathop{V}}\,$ is only semi-definite: if the curve on which $\overset{.}{\mathop{V}}\,$ =0 is found not to satisfy the system equations, it is not a trajectory, so the state cannot remain on this curve, and hence the system must be asymptotically stable.

**Example: Nonlinear Spring-Mass-Damper System**

A spring-mass-damper system for which the damping force is proportional to the third power of the velocity is described by the differential equation

\[\overset{..}{\mathop{y}}\,+0.5\overset{.}{\mathop{y}}\,{{\text{ }}^{\text{3}}}+y=0\]

Let mass position y and velocity $\overset{.}{\mathop{y}}\,$ be chosen as state variables:

x_{1}=y, x_{2}=$\overset{.}{\mathop{y}}\,$

Then,

${{\overset{.}{\mathop{x}}\,}_{1}}={{x}_{2}}$

And, from the differential equation,

${{\overset{.}{\mathop{x}}\,}_{2}}=\overset{..}{\mathop{y}}\,=-0.5\overset{.}{\mathop{y}}\,{{\text{ }}^{\text{3}}}-y=-0.5x_{2}^{3}-{{x}_{1}}$

Are the state equations. Attempt the positive-definite V-function

$V=x_{1}^{2}+x_{2}^{2}\text{ (V}\to \infty \text{ as }\left\| x \right\|\to \infty )$

By differentiating, we obtain the following

$\overset{.}{\mathop{V}}\,=2{{x}_{1}}{{\overset{.}{\mathop{x}}\,}_{1}}+2{{x}_{2}}{{\overset{.}{\mathop{x}}\,}_{2}}=2{{x}_{1}}{{x}_{2}}-x_{2}^{4}-2{{x}_{1}}{{x}_{2}}=-x_{2}^{4}$

$\overset{.}{\mathop{V}}\,$ is negative semi-definite, so the system is globally stable. But since the x_{1}-axis (x_{2}=0) is not a trajectory, the system is also globally asymptotically stable, by remark 3. However, usually $\overset{.}{\mathop{V}}\,$ for this simple V-function would be indefinite, and would not prove either stability or instability.

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