1. §7-4 Lyapunov Direct Method
There are two Lyapunov methods for stability analysis.
The first method usually requires the analytical solution of the differential equation. It is an indirect method.
In the second method, it is not necessary to solve the differential equation. Instead, a so-called Lyapunov function is constructed to check the motion stability. Therefore, it is said to be direct method.
Lyapunov direct method is the most effective method for studying nonlinear and time-varying systems and is a basic method for stability analysis and control law desgin.

2. Example: Consider the stability of zero-solution of the following system
First of all, consider a positive definite function
It is clear that, and
The derivative of v is computed as
Because v(x) is positive definite and is negative definite, x converges to zero. Note that the analytical solution of the differential equation is not required.

3. Example: Consider the small-damping vibration system
Study the stability at equilibrium x1=0, x2=0.
Similar to the above example,
construct a function called v-function
It is easy to verify that the function is positive definite.

4. Generally speaking, it is difficult or sometimes impossible to solve a differential equation. Hence, it is almost impossible to get the analytical solution of v. However, the derivative of v can be computed as

5. When x1 and x2 are not zero at the same time, the inequality dv/dt<0 always holds, which means that the motion trajectory move into its inner from outside of the ellipse v=C
Hence, the system is stable at the origin.
The above examples show that with the help of a special function, we can determine the stability of the zero-solution without solving the differential equation.

6. The following figure is a sketch map of v(x) = Ci > 0 where v(x) is in a positive definite quadratic form. From the figure we can see that it is a set of close curves. C7
v(x) is usually in quadratic form and v(x) =C（>0) represents a set of hyperspheres. C6 C5 C4 C3 C2 C1

7. 1. Definition of sign functions
First of all, consider a real-valued function v(x,t) defined on , where Assume that v(x,t) is a continuous single-valued function with
. For example
Definition 7-12
Suppose v is the function of x, when
and v(x)=0 has nonzero solutions , then v(x) is said to be a constantly positive (constantly negative) function.
Example: is a constantly positive function.

8. If for and x=0 if and only if v(x)=0, then v(x) is said to be positive definite (negative definite).
The constantly positive (constantly negative) function is also said to be positive (negative) semidefinite function and they are both said to be constant sign functions.
Example: is a positive definite function.
(2) v(x,t) is said to be constantly positive (constantly negative) if holds for
Example: is a constantly positive function.
Note that in the example

9. v(x,t) is said to be positive definite if for any ,
, , where w(x) is positive definite.
v(x,t) is said to be negative definite if for any ,
, , where w(x) is positive definite.
Example:
is positive definite
and w(x) can be chosen as
Positive definite function and negative definite function are both said to be fixed-sign function.
(3) The function that is not constant-sign function or fixed-sign function is said to be variable-sign function.
Example: is a variable-sign function.

10. ε Example: v(x1, x2) = x1x2 x2 + x1 Example: is a－ +
Example: is a
positive definite function for
Example: is a constantly positive
(positive semidefinite) function for

11. (4) v(x,t) is said to be decrescent if there exists a positive definite function w(x) such that.
Example: is not decrescent.
is decrescent.

12. 2. Main theorem
The differential equations considered in this section are in the following form or
We mainly study the stability of the above system at the equilibrium state x = 0.

13. First of all, compute the derivative of v(x,t) in t along the trajectory of Equation (7-39):
If v=v(x), the derivative of v in t along the trajectory of Equation (7-39) is given by

14. Theorem 7-20*（Lyapunov,1892):
Suppose v(x,t) is positive definite and its derivative along the solution of Equation (7-39):
satisfies
then the zero-solution of Equation (7-39) is stable (in the sense of Lyapunov).

15. Remark:
This is a sufficient condition;
If t does not appear in the expression of f, then
where

16. Geometric Interpretation
Because v(x) is positive definite, v(x)=C is a set of close curves reducing to the origin as and every curve is surrounded by another one. If is positive semidefinite, the value of v(x) along the trajectory can only diminish or maintain, which means that the system is stable at the origin.

17. Example: Consider the following system
Let Then
. From Theorem
7-20, the system is stable in the sense of Lyapunov. Because , the trajectory of x must be a isoline. In fact, the solution of the above equation is

18. Theorem 7-21* Suppose v(x) is positive definite (negative definite) and its derivative along the trajectory of Equation (7-39) dx/dt=f(x), f(0)=0 satisfies
then the zero solution of Equation (7-39) is asymptotically stable.
Geometric interpretation:
Because v(x) is positive definite, v(x)=C is a set of close curves reducing to origin as and each curve is surrounded by another one. If dv/dt is negative definite for any x, the value of v(x) diminish.

19. This indicates that limt0x(t)=0, i. e
This indicates that limt0x(t)=0, i.e. the origin (zero solution) is asymptotically stable.

20. Example: Consider the stability of equilibrium x1=x2=0 of the following small-damping oscillation system
If we choose v(x) as , then
From Theorem 7-20, we can only determine that the system is stable in the sense of Lyapunov, although the system is asymptotically stable.

21. Theorem 7-22** Suppose v(x) is positive (negative) definite, the derivative of v(x) along the trajectory of Equation (7-39)
satisfies
If is not identical along the nonzero-solution of Equation (7-39), then the zero-solution of (7-39) is asymptotically stable.

22. v(x)>0 Theorem 7-20* v(x)>0 stable Theorem 7-21*
Asymptotically stable
Theorem 7-22**
v(x)>0
It is not zero at all time. asymptotically stable.
Theorem 7-23* Suppose there is a v(x) satisfying:
(1) For ‖x‖< , there exists a domain which may involve some sub-domains uj such that v>0 and the bound of uj is composed of v=0 and ‖x ‖= .
(2) In some sub-domain, the derivative of v along the trajectory of Equation (7-39) satisfies then the zero-solution of Equation (7-39) is unstable.

23. ε x2 u2 u1 x1 u3 The geometric interpretation of Theorem 7-23*:
v>0： uj (j=1,2,3) ε u2 u1 u3

24. 3. v function in quadratic form of linear system
Theorem 7-25 The zero-solution of the time-invariant system is asymptotically stable if and only if for any positive definite matrix N=NT the Lyapunov equation
（7-44）
has a unique and positive definite solution M=MT.
Why should we study this problem?
The matrix A is often not precisely known in the control law design. Theorem 7-25 has great significance in control law design.

25. Example 7-9 Consider the following system
When the system is asymptotically stable, the matrix A satisfies the condition （7-44）.
Let N=I. From (7-44), we have

26. The determinant of A1 is
If detA10, the equations have unique solutions
From Sylvester criterion, M is positive definite if

27. From (2) it follows that
From (1) and (3), we can obtain that
Therefore, the matrix A should satisfy the conditions (3) and (4) when the system is asymptotically stable.

28. Note that Theorem 7-25 does not mean
“A is asymptotically stable and M is positive definite, then N in Equation (7-44) must be positive definite.”
Example 7-10
It is clear that the eigenvalues of A have negative real parts and M is positive definite. However, from Equation (7-44)
is not positive definite.

29. Theorem If N of Equation (7-44) in Theorem 7-25 is positive semidefinite symmetric matrix, and xTNx is not identically zero along any nonzero solution of =Ax, then the matrix equation
ATX+XA=N （7-46）
has a positive definite symmetric solution if and only if
is asymptotically stable.
Remark: The condition the xTNx is not identically zero along any nonzero solution of can not be omitted.
Example 1: A is asymptotically stable and N is positive semidefinite, then M may not be positive definite.

30. If xTNx is identically zero along any nonzero solution of , it is easy to compute
However, If
is a nonzero solution,
then xTNx is identically zero along a nonzero solution of the differential equation, which does not satisfy the condition of the theorem.

31. Example 2 N is positive semidefinite and M is positive definite, then A may not be asymptotically stable.
Analysis：1. Let xTNx=0.
Because xTNx=x12, xTNx=x12 is identically zero
2. Let C=[1 0], then N=CTC and (A, C) is unobservable.

32. Example 3
xTNx is not identically zero along any nonzero solution of the differential equation and (A, C) is observable, which satisfies the condition of the theorem.

33. Conclusion: “xTNx is not identically zero along any nonzero solution of the differential equation” can be replaced by the condition that (A, C) is observable, where N= CTC.
Theorem 7-26* The zero-solution of time-invariant differential equation is asymptotically stable if and only if under the condition that (A, N) is observable, where N is positive semidefinite, Lyapunov equation (7-44)
（7—44）
has a unique positive definite solution M.

34. Proof
Because N is positive semidefinite, it can be decomposed as
N=CTC
It is easy to prove that (A, C) is observable if (A, N) is observable.
Necessity: The proof is similar to Theorem Because the zero-solution of the system is asymptotically stable, M determined by
is a unique positive definite solution and satisfies Equation (7-44). Here, N is an arbitrarily positive semidefinite matrix such that (A，N) is observable.

35. Sufficiency: If under the condition that N is positive semidefinite such that (A, N) is observable, the solution M of Equation (7-44) is positive definite. Now, we need to prove that the system is asymptotically stable. Consider v(x)=xTMx dv/dt = xTNx. Hence, we only need to prove that xTNx0  x00.
The derivatives of the two sides are computed as

36. Which implies that is zero, i. e
Which implies that is zero, i.e. dv/dt is not identically zero along the nonzero solution of the differential equation. From Theorem 7-21**, the system is asymptotically stable Q.E.D
Remark: From “(A,C) is observable if and only if  ={0}”.

37. 4. About Lyapunov function
A scalar function is said to be Lyapunov function if by employing the function, the stability of a system can be determined without computing the analytical solution of the differential equation;
The construction of the v function is a complex problem. Even if the v function exists in theory, it is still a hard work to find an analytic expression. It is not practical to find a general method to construct the v function . However, for linear systems, there are some methods to construct their v functions.
It should be emphasized that the conditions of Theorem 7-20*—7-22** are all sufficient conditions. This means that a system may be stable even if a v function cannot be constructed.

38. Before concluding a system is unstable, we must find a v function satisfying the unstable theorem.
In this section, the construction of quadratic form Lyapunov function for linear systems is introduced, (refer to Theorem 7-25, Theorem 7-26 and Theorem 7-26*).
Lyapunov equation (7-44)：
ATM+MA=N，
appears in lots of problems of system theory;
After modified, the Lyapunov functions of linear systems can be used for a kind of nonlinear systems.

39. A better Lyapunov function has less conservative and yields a better result.
For time-varying function v(x, t), the theorems are also different from the theorems of the systems with fixed coefficients. When the stability theorems on time-varying systems are used, we should pay more attention to “decrescent”(1) or “Class-K function bound”(2).