# Chetaev function

 Emmanuil E. Shnol (2007), Scholarpedia, 2(9):4672. doi:10.4249/scholarpedia.4672 revision #126479 [link to/cite this article]
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Curator: Emmanuil E. Shnol

Chetaev functions are the analogue of Lyapunov functions to study instability of solutions of differential equations. Below we consider the simplest case: equilibria of autonomous system of ordinary differential equations.

## Introduction

Consider a system of differential equations $\tag{1} \frac{dx}{dt} = f(x)$ (or $$\dot{x}=f(x)$$), where $$f, x \in \R^n\ ,$$ $$x=(x_1, \ldots, x_n)\ ,$$ $$f=(f_1, \ldots, f_n)\ .$$ Let $$x(t)=c$$ be a time-independent solution, i.e., $$c$$ is the equilibrium of the system (1). We may assume that $$c=0$$ so that $$f(0)=0$$ (and do not distinguish $$0\in\R$$ and $$0\in\R^n$$).

Definitions of Lyapunov and Chetaev functions include the condition of the function behavior on the solutions of the system. If $$h(x)$$ is such a function, then the function $$h(x(t))$$ must increase (or decrease) monotonically. In definitions and applications it is convenient to use the derivative $$h(x(t))$$ with respect to time. The core idea, common for both Lyapunov and Chetaev functions, is that one does not need to know the solutions $$x(t)\ .$$ Indeed, for any scalar function $$g(x)$$ $\tag{2} \dot{g}|_{(1)} \equiv \frac{d}{dt}g(x(t)) = \sum_k \frac{\partial g}{\partial x_k}(x)f_k(x).$

If the right-hand side of Eq.(2) is positive in all points $$x$$ of some set $$M\ ,$$ then on any segment of the trajectory $$x(t)$$ in the set $$M\ ,$$ $$g(x(t))$$ increases. When the system (1) is fixed, we will use $$\dot{g}$$ to denote the derivative of $$g$$ “with respect to the system (1)” (knowing that the derivative depends on a point in $$\R^n\ ,$$ and not on $$t\ !$$).

## Stability and Instability

“Inverting” the Lyapunov’s definition (see Stability), let us give the direct definition of instability.

Definition: The equilibrium $$x=0$$ is unstable, if for some $$r>0$$ and any positive $$\varepsilon$$ there is a solution $$x_\varepsilon(t)$$ such that

(A) $$\tag{3} |x_\varepsilon(0)| < \varepsilon \,$$
(B) $$| x_\varepsilon(t_\varepsilon)| \geq r$$ for some $$t_\varepsilon > 0 \ .$$

Here, $$r$$ is the same for all $$\varepsilon\ ,$$ and $$|x|$$ is some fixed norm in $$\R^n\ .$$

The original idea of N.G. Chetaev was simple. The definition of stability of the equilibrium $$x=0$$ involves some neighborhoods of the point. To prove stability, one needs to consider all trajectories in the neighborhoods. To prove instability, one does not need to do that. In particular, it is obvious that for instability it suffices to have a solution of $$x(t)$$ whose trajectory $$l$$ “escapes” $$0\ :$$ $$x(t)\rightarrow 0$$ as $$t\rightarrow -\infty\ .$$ Formally, using conditions (3): If such a trajectory exists, then $$x_\varepsilon(0)$$ can be any point $$a$$ on the trajectory $$l$$ for which $$|a|<\varepsilon\ ,$$ and set $$r=|b|\ ,$$ where $$b$$ is some fixed point on $$l\ .$$ However, to prove the existence of at least one trajectory escaping $$0$$ in a nonlinear system could be quite challenging (Arnold and Ilyashenko 1988; Hartman 1954).

In addition to stability theorems, A.M. Lyapunov also established instability theorems that avoided considering individual solutions. Here is of one of them:

Lyapunov Theorem Let $$L(x)$$ has the following properties when $$|x|\leq r\ .$$

• (A) $$L(0)=0\ ,$$ $$L$$ is positive arbitrary close to $$0\ .$$
• (B) $$\dot{L}(x)>0$$ when $$x\neq 0\ .$$

Then the equilibrium $$x=0$$ of Eq.(1) is unstable.

In the proof of the theorem, $$L(x)$$ is used only in the set $$G$$ where $$L(x)>0\ :$$ one does not need the whole neighborhood of the equilibrium to prove instability. This remark leads to the following definition.

Function $$V(x)$$ is called Chetaev function for the equilibrium point $$x=0$$ of system (1), if the following conditions are satisfied:

• (A) $$V(0)=0\ ,$$ $$V(x)$$ can take positive values arbitrary close to $$0\ .$$ That is, the equilibrium $$0$$ is on the boundary of the open set $$G$$ defined by the condition $$V(x)>0\ .$$
• (B) When $$|x|\leq r\ ,$$ the inequality $$\dot{V}>0$$ ($$r>0$$ is fixed) is satisfied in the set $$G\ .$$

Theorem (N.G. Chetaev) If Chetaev function exists, then the equilibrium $$x=0$$ is unstable.

Indeed, let the initial point $$x(0) \in G$$ and $$|x(0)|<\varepsilon\ .$$ From the condition A, $$V(x(0))>0\ .$$ From the condition B, $$V(x(t))$$ increases and the trajectory $$x(t)$$ cannot cross the boundary of $$G$$ where $$V=0\ .$$ Then, there is some $$t_\star>0$$ where $$|x(t_\star)|=r\ .$$ Since $$\varepsilon$$ is arbitrary (and $$r$$ is fixed), the equilibrium $$x=0$$ is unstable. Minor technical remarks convert this informal consideration into a proof.

Remark 1. If the conditions of Lyapunov theorem are satisfied, then the restriction of the function $$L(x)$$ in the region $$L(x)>0$$ could be taken as the Chetaev function. In this case, a stronger version of conditions of Chetaev theorem will be satisfied: On the boundary of region $$L(x)>0$$ in the neighborhood $$|x|\leq r$$ (except $$0$$) we have the strong inequality $$\dot{V}(x)>0\ .$$ However, it is easier to find Chetaev function than to find Lyapunov function.

Remark 2. In many cases one could use quadratic forms as Lyapunov and Chetaev functions.

Remark 3. If the region $$G$$ consists of a few connected parts having $$x=0$$ as the connecting point, then it suffices to consider only one of the parts.

### Example

Consider the two-dimensional system $\tag{4} \dot{x}_1 = x_2 + Q_2(x), \ \dot{x}_2 = ax^2_1+bx_1x_2+x_2^2+Q_3(x); \ |Q_2|\leq C|x|^2, |Q_3|\leq C|x|^3$

Proposition. The equilibrium $$x=0$$ of the system (4) is unstable when $$a \neq 0$$ regardless of the values of the parameters $$b$$ and $$c$$ and the form of the remaining terms $$Q_k\ .$$

Suppose $$a>0\ .$$ Let us consider the quadrant $$\{x_1>0, x_2>0\}\ .$$ It is easy to check that for some small $$r>0$$ the function $$V(x)=x_1x_2$$ satisfies all the conditions of Chetaev theorem: All terms in the expression for $$\dot{V}$$ are dominated by the main two$x_2^2+ax_1^3\ .$ (The only non-obvious estimation is $$x_1^2x_2 = o(x_1^3+x_2^2)\ .$$) When $$a<0\ ,$$ we need to use the quadrant $$\{x_1<0, x_2<0\}$$ (see Remark 3 above).

The function $$V$$ in the Chetaev theorem plays dual role. Firstly, the condition $$V(x)>0$$ sets the “instability sector” $$G$$ adjacent to the point $$0\ ,$$ from which trajectories cannot escape. Secondly, it defines the divergence from the equilibrium in the sector with respect to growth of $$V$$ on each trajectory. It makes sense to separate these two roles.

## Sector of Instability – “attracting cone”

Let $$K$$ be a closed cone with side surface $$S\ ,$$ vertex at $$x=0\ ,$$ and an arbitrary shape of cross-section. Let us make the following assumptions:

A+. Trajectories through $$S$$ can only go inside $$K\ :$$ if $$x(0)\in S\ ,$$ then $$x(t)\subset K$$ for all sufficiently small positive $$t\ .$$
B1. Any trajectory inside $$K$$ (different from $$x=0$$) diverge from $$0\ :$$ $$|x(t)|$$ increased monotonically and, in finite time, leave $$K\ .$$

Then, the equilibrium $$x=0$$ is unstable. Indeed, from the condition A+, the trajectory $$x(t)$$ can leave the cone $$K$$ only through its base $$\Gamma\ .$$ In the definition of instability (3), we can take $$x_\varepsilon(t)$$ to be any solution of (1) with $$x(0)\in K$$ and $$0<|x(0)|<\varepsilon\ ,$$ and take $$r$$ to be $$\mbox{min } |x|$$ on $$\Gamma\ .$$

Notice that the choice of the metric in the phase space is irrelevant here, so far as we define only “the distance to the origin 0”. We can use any non-negative function that equals zero only when $$x=0\ .$$

B. Let function $$V\ ,$$ defined in the cone $$K\ ,$$ have the following properties:
1. $$V(0)=0$$
2. $$V(x)$$ and $$\dot{V}(x)$$ are positive for all $$x \neq 0\ .$$

From the conditions A+ and B, the equilibrium $$x=0$$ is unstable.

Note. The base of the cone does not have to be flat. For example, one can use a part of the sphere $$|x|=R$$ cut by the side surface of the cone, or one can use the surface $$V(x)=\mbox{const}$$ (See Figure 1).

## Sector of Instability – “repelling cone”

Let us keep condition B, but replace A+ by the opposite condition:

A-. Trajectories through $$S$$ can only go outside of $$K\ :$$ If $$x(0)\in S\ ,$$ then a segment of the trajectory $$x(t)$$ is outside $$K$$ when $$t$$ is small but positive (See Figure 12a).

Now, trajectories near the equilibrium $$0$$ can leave the cone $$K$$ and it is not clear whether we can guarantee instability. However, we can reverse the direction of time (Figure 12b) and choose the value $$V_0>0$$ less than $$\mbox{min} V(x)$$ at the base of the cone, so that the set $$V(x)=V_0$$ does not intersect the base of the cone $$K\ .$$ We see that all trajectories with $$V(x(0))=V_0$$ approach the equilibrium $$x=0$$ as $$t\rightarrow -\infty\ ;$$ the existence of at least one such trajectory implies instability of the equilibrium.

## Regions of Instability of a general type

Let $$G$$ be a bounded region (open connected set) and $$K$$ be its closure. Suppose that the equilibrium $$x=0$$ belongs to the boundary of $$K$$ and that a part of this boundary, $$S\ ,$$ on which $$0<|x|\leq r\ ,$$ consists entirely of “points of entry” or “points of exit”. That is, one of the conditions above, A+ or A-, is satisfied. Let the condition B be satisfied: A function $$V(x)$$ is defined in $$K$$ such that $$V$$ and $$\dot{V}$$ are non-negative and equal to zero only at the point $$x=0\ .$$ Then the equilibrium $$x=0$$ is unstable. The proof is essentially the same as in the case of $$K$$ being a cone.

N.G. Chetaev understood that these two extreme cases guaranteed instability soon after he had proven his theorem (provided above) and published the corresponding paper in 1938. However, it is easier to deal with functions and not sets. For this, Chetaev introduced a second function, $$\Phi$$ that defines the domain of $$V\ .$$

## Pairs of Chetaev Functions

Consider two functions $$\Phi$$ and $$V\ .$$ Suppose that the following conditions are satisfied when $$|x|\leq r\ .$$

1. The equilibrium $$x=0$$ belongs to the boundary of a set $$G$$ defined by $$\Phi(x)>0$$ (so that $$\Phi(0)=0$$).
2. $$\dot{\Phi}(x)>0$$ for all points of the boundary of $$G$$ except the point $$x=0\ .$$
3. $$\dot{V}(x)>0$$ for all $$x \in \bar{G}\ ,$$ $$x \neq 0$$ (here, $$\bar{G}$$ is the closure of the set $$G$$).

Let us call such $$\Phi$$ and $$V$$ the pair of Chetaev functions of the first type.

Notice that the conditions 1-3 do not exclude the case when $$\Phi$$ and $$V$$ are the same. If $$\Phi(x) \equiv V(x)\ ,$$ then the condition 2 follows from 3, and we arrive to the definition of the Chetaev function $$V(x)$$ (with stronger conditions on the boundary of $$G$$).

Let us change the sign in the condition 2.

2-. $$\dot{\Phi}(x)<0$$ for all points of the boundary of $$G$$ except the point $$x=0\ .$$

Let us call such $$\Phi$$ and $$V$$ the pair of Chetaev functions of the second type.

If an equilibrium $$x=0$$ of the system (1) has a pair of Chetaev functions (of the first or second type), then the equilibrium is unstable.

If the boundary of $$G$$ has “points of entry” and “points of exit”, then we cannot guarantee instability (even if the condition 3 is satisfied). However, if these points are arranged in a certain way, one can still prove instability of the equilibrium. Such special results on instability use topological theorems and they could be useful in complex cases. (Rouche, Habets, and Laloy 1979; Khazin and Shnol 1991).

Notes.

• The review article of Arnold and Ilyashenko (1988) [Ch. 1, sect. 4.3] refers to Chetaev function as the function $$V$$ from the pair of Chetaev functions of the first type, and the “attracting” region $$G$$ is defined geometrically without using this function.
• We have considered time-independent solutions of the system (1), but definition of Chetaev Function could be given for any solution $$x(t)$$ (see Chetaev 1961).