Education Journalist | Study Abroad Lead
Lyapunov functions arе a powerful tool used in the analysis of dynamical systеms and particularly in stability theory. Thе basic idеa bеhind Lyapunov functions is to еstablish a critеrion for thе stability of a systеm by еxamining thе bеhavior of a scalar function along thе trajectories of thе system.
- The Lyapunov function should bе non nеgativе for all points in thе domain of intеrеst and with strict positivity for points away from thе еquilibrium.
- It should bе zеro only at thе equilibrium point(s) of intеrеst.
- Along thе trajectories of the system, thе Lyapunov function should decrease or increase.
- If it decreases, it indicates stability and if it incrеasеs it impliеs instability.
| Table of Content |
Key Terms: Scalar, Matrix, Function, Stability, Theorem, Lyapunov Function, Origin, Equilibrium, Differential Equations
Lyapunov Function
[Click Here for Sample Questions]
A scalar function called a Lyapunov function may be used in phase space to show that an equilibrium point is stable. Assume that V(X) is a function that is continuously differentiable near the origin, U. V(X) is considered a Lyapunov function for an independent system X' = f(x) if the following criteria are satisfied.
- V(X) >0 for all X in U {0}
- (dV/dt) ≤ 0 for all X in U
- V(0) = 0
Also check:
| Related Topics | ||
|---|---|---|
| Function Notation Formula | What is disjoint set | Equal equivalent sets |
| Types of Relations | Venn Diagrams | Real Valued Functions |
Lyapunov Function Examples
[Click Here for Previous Year Questions]
Differential equations and systems are tested for stability using the Lyapunov function approach. In the parts that follow, we shall concentrate on autonomous systems.
X’ = f(x) or
(dxi)/dt = fi (x1, x2, …xn)
Here, i = 1, 2, …n
With x ≡ 0, which is the zero equilibrium
Assume that V(X) = V(x1, x2,... n) is a continuously differentiable function that is close to the origin, U. Let V(0) in the origin and V(X) be the values for all X in U\ {0}. These are examples of the type's functions.
V(x1, x2) = ax12 + bx22, V(x1, x2) = ax12 + bx24, a, b>0
The following formula is used to get the function V(X)'s total derivative with respect to time t:
V(x1, x2) = ax12 + bx22, V(x1, x2) = ax12 + bx24, a, b>0
This expression is similar to a scalar (dot) product of two vectors in that it may be expressed as:
The first vector is always going in the direction of the greatest rise in V(X), since it is the gradient of V(X).
- V(x) is a function that, given |X|→∞, usually grows with distance from the origin.
- The second vector in the scalar product is the velocity vector.
- At all times, it is tangent to the phase trajectory.
- Examine the case when the derivative of V(X) is negative near the origin, U.
- This suggests that there is a bigger angle (φ) than 90 degrees between the velocity vector and gradient.
A function with two variables is shown schematically in the following diagram:
- A phase trajectory will gravitate toward the origin if the derivative dV/dt is always negative, signifying that the system is stable.
- The system is unstable if the derivative dV/dt is positive because the trajectory deviates from the origin.
A function with two variables
Theorems of Lyapunov Stability
[Click Here for Sample Questions]
The following are the Lyapunov Stability Theorems:
Theorem of Stability in the Lyapunov Sense
When an autonomous system's zero solution, X = 0, has a Lyapunov function V(X) in its neighborhood U, then the system's equilibrium point, X = 0, is Lyapunov stable.
Theorem of Asymptotic Stability
An autonomous system's equilibrium point, X = 0, is asymptotically stable if a Lyapunov function V(X) with a negative definite derivative (dV/dt) < 0 for every X ∈ U {0}
- It exists in the neighborhood U of the system's zero solution, X = 0.
- As can be seen, the total derivative dV/dt must be precisely negative in the vicinity of the origin in order for the zero solution to have asymptotic stability.
Theorem of Lyapunov Instability
Let us assume that there is a continuously differentiable function V(x) in the vicinity U of the zero solution X = 0.
- V(0) = 0 and dV/dt > 0.
- The zero solution X = 0 will be unstable if there exist places in the neighborhood U where V(X) > 0.
Lyapunov Function Advantages
[Click Here for Previous Year Questions]
Few advantages of Lyapunov Advantages are:
- There is no need for exact equations, it works with local information.
- It analyzes various stability types, not just "stable/unstable."
- It handles weird system behaviours like circles.
- It is easy to visualize: lower energy = closer to equilibrium.
- It helps design control systems that stabilize things.
Lyapunov Function Disadvantages
[Click Here for Sample Questions]
Few disadvantages of Lyapunov Functions are mentioned below:
- It is hard to find the right function for each system.
- It might say a system is stable in a smaller area than it actually is.
- It can be computationally expensive for complex systems.
- It doesn't tell everything about the system's behavior, just stability.
Things to Remember
- Lyapunov functions are used in control theory to design stabilizing control systems for unstable systems.
- Unlike solving complex equations for the entire system's behavior, Lyapunov functions only need local information about the system's state near its stable point.
- They can reveal different types of stability, such as asymptotic stability or exponential stability .
- Lyapunov functions have limitations, they can be conservative, meaning they might underestimate the actual region of stability.
- Additionally, they don't provide complete information about the system's behavior, only its stability characteristics.
Read More:
| Relevant Concepts | ||
|---|---|---|
| Negative of a Vector | Coplanar Vector | Addition Of Vector |
| Unit Vector | Resolution of vectors | Vector product of two vectors |
Previous Year Questions
- Extraction of metal from the ore cassiterite involves...[JEE Advanced 2011]
- Commonly used vectors for human genome sequencing are...[NEET UG 2014]
- Interfascicular cambium and cork cambium are formed due to..
- Pneumotaxic centre is present in...[UP CPMT 2007]
- Reaction of HBr with propene in the presence of peroxide gives….[NEET UG 2004]
- Assuming the expression for the pressure exerted by the gas on the walls of the container, it can be shown that pressure is...[MHT CET 2016]
- Which among the following is the strongest acid?...[TS EAMCET 2017]
- Isopropyl alcohol on oxidation forms..
- A vector is not changed if..
- Which of the following arrangements does not represent the correct order of the property stated against it?...[JEE Main 2013]
- The major product of the following reaction is...[JEE Main 2019]
- Major product of the following reaction is..[JEE Main 2023]
- The percentage of nitrogen in urea is about..
- The electric field at a point is
- Which of the following statements is true?..[JKCET 2006]
Sample Questions
Ques. What are Lyapunov functions? (1 mark)
Ans. Scalar functions called Lyapunov functions can be used to show that the equilibrium of an ordinary differential equation is stable.
Ques. Provide the Lyapunov function's application. (1 mark)
Ans. With Lyapunov functions, the stability characteristics of equilibrium points of both linear and nonlinear systems may be investigated.
Ques. What benefits do Lyapunov functions offer? (2 marks)
Ans. Lyapunov functions can be used to identify whether a system is stable or unstable. One benefit of this approach is that it doesn't want us to know the precise answer. It is also possible to examine the stability of non-rough equilibrium points using this approach.
Ques. If \((\frac{x}{3} + 1, y - \frac{2}{3}) = (\frac{5}{3}, \frac{1}{3})\), find the values of x and y. (2 marks)
Ans. Given,
\((\frac{x}{3} + 1, y - \frac{2}{3}) = (\frac{5}{3}, \frac{1}{3})\)
As the ordered pairs are equal, the corresponding elements should also be equal.
Thus, x/3 + 1 = 5/3 and y – 2/3 = 1/3
Solving, we get
x + 3 = 5 and 3y – 2 = 1 [Taking L.C.M. and adding]
x = 2 and 3y = 3
Therefore, x = 2 and y = 1
Ques. If G = {7, 8} and H = {5, 4, 2}, find G × H and H × G. (3 marks)
Ans. Given, G = {7, 8} and H = {5, 4, 2}
We know that, The Cartesian product of two non-empty sets P and Q is given as
P × Q = {(p, q): p ∈ P, q ∈ Q}
So,
G × H = {(7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)}
H × G = {(5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8)}
Ques. State whether each of the following statements is true or false. If the statement is false, rewrite the given statement correctly. (2 marks)
(i) If P = {m, n} and Q = {n, m}, then P × Q = {(m, n), (n, m)}
(ii) If A and B are non-empty sets, then A × B is a non-empty set of ordered pairs (x, y) such that x ∈ A and y ∈ B.
(iii) If A = {1, 2}, B = {3, 4}, then A × (B ∩ Φ) = Φ
Ans. (i) The statement is false. The correct statement is
If P = {m, n} and Q = {n, m}, then
P × Q = {(m, m), (m, n), (n, m), (n, n)}
(ii) True
(iii) True
Ques. 7. Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that (2 marks)
(i) A × (B ∩ C) = (A × B) ∩ (A × C)
(ii) A × C is a subset of B × D
Ans. Given,
A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}
(i) To verify: A × (B ∩ C) = (A × B) ∩ (A × C)
Now, B ∩ C = {1, 2, 3, 4} ∩ {5, 6} = Φ
Thus,
L.H.S. = A × (B ∩ C) = A × Φ = Φ
Next,
A × B = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)}
A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}
Thus, R.H.S. = (A × B) ∩ (A × C) = Φ
Therefore, L.H.S. = R.H.S.
Hence verified
(ii) To verify: A × C is a subset of B × D
First,
A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}
And,
B × D = {(1, 5), (1, 6), (1, 7), (1, 8), (2, 5), (2, 6), (2, 7), (2, 8), (3, 5), (3, 6), (3, 7), (3, 8), (4, 5), (4, 6), (4, 7), (4, 8)}
Now, it’s clearly seen that all the elements of set A × C are the elements of set B × D. Thus, A × C is a subset of B × D.
Hence verified
Ques. Let A = {1, 2, 3, … , 14}. Define a relation R from A to A by R = {(x, y): 3x – y = 0, where x, y ∈ A}. Write down its domain, codomain and range. (3 marks)
Ans. The relation R from A to A is given as:
R = {(x, y): 3x – y = 0, where x, y ∈ A}
= {(x, y): 3x = y, where x, y ∈ A}
So,
R = {(1, 3), (2, 6), (3, 9), (4, 12)}
Now,
The domain of R is the set of all first elements of the ordered pairs in the relation.
Hence, Domain of R = {1, 2, 3, 4}
The whole set A is the codomain of the relation R.
Hence, Codomain of R = A = {1, 2, 3, …, 14}
The range of R is the set of all second elements of the ordered pairs in the relation.
Hence, Range of R = {3, 6, 9, 12}
Ques. Let f, g: R → R be defined, respectively by f(x) = x + 1, g(x) = 2x – 3. Find f + g, f – g and f/g. (3 marks)
Ans. Given the functions f, g: R → R is defined as
f(x) = x + 1, g(x) = 2x – 3
Now,
(f + g) (x) = f(x) + g(x) = (x + 1) + (2x – 3) = 3x – 2
Thus, (f + g) (x) = 3x – 2
(f – g) (x) = f(x) – g(x) = (x + 1) – (2x – 3) = x + 1 – 2x + 3 = – x + 4
Thus, (f – g) (x) = –x + 4
f/g(x) = f(x)/g(x), g(x) ≠ 0, x ∈ R
f/g(x) = x + 1/ 2x – 3, 2x – 3 ≠ 0
Thus, f/g(x) = x + 1/ 2x – 3, x ≠ 3/2
For Latest Updates on Upcoming Board Exams, Click Here: https://t.me/class_10_12_board_updates
Check-Out:
Comments