Constructing Lyapunov Functions Absence Of Closed Orbits In Dynamical Systems

July 7, 2025 by StackCamp Team 78 views

Constructing Lyapunov Functions to Prove the Absence of Closed Orbits in a Dynamical System

In the realm of dynamical systems and ordinary differential equations, understanding the stability of solutions is paramount. One powerful technique for analyzing stability, particularly the absence of closed orbits (limit cycles), involves constructing a Lyapunov function. This article delves into the process of constructing such a function for a given system, demonstrating how it can be used to prove the non-existence of closed orbits and, consequently, the absence of certain types of oscillatory behavior. We will specifically focus on the system:

dx/dt = x(y^2 + 1) + y
dy/dt = x^2y + x

and illustrate the steps involved in crafting a Lyapunov function to analyze its stability properties. This exploration is crucial for anyone studying dynamical systems, as Lyapunov functions provide a rigorous way to assess the long-term behavior of solutions without explicitly solving the differential equations.

Understanding Lyapunov Functions and Stability

Before diving into the specific example, let's establish a solid understanding of Lyapunov functions and their role in stability analysis. A Lyapunov function, often denoted as V(x, y) for a two-dimensional system, is a scalar function that provides a measure of the “energy” or “distance” from an equilibrium point of the system. The key idea is that if the system's trajectory consistently moves towards lower “energy” levels as time progresses, then the equilibrium point is stable. More formally, a Lyapunov function must satisfy the following conditions:

V(x, y) is continuously differentiable.
V(x, y) is positive definite in a neighborhood of the equilibrium point (i.e., V(x, y) > 0 for all (x, y) ≠ (0, 0) in the neighborhood, and V(0, 0) = 0).
The time derivative of V(x, y) along the system's trajectories, denoted as V̇(x, y), is negative semi-definite (i.e., V̇(x, y) ≤ 0) or negative definite (i.e., V̇(x, y) < 0 for all (x, y) ≠ (0, 0) in the neighborhood). If V̇(x, y) is negative definite, the equilibrium point is asymptotically stable.

In the context of proving the absence of closed orbits, we often seek a Lyapunov function whose derivative is strictly negative along any non-equilibrium trajectory. This implies that the system's trajectories are always moving inward, towards the equilibrium point, and therefore cannot form a closed loop. The existence of a suitable Lyapunov function is a powerful tool for demonstrating the absence of limit cycles, which are isolated closed trajectories in the phase plane. Limit cycles represent self-sustained oscillations, and their absence indicates that the system will eventually settle into a stable equilibrium state. The construction of a Lyapunov function is not always straightforward and often requires intuition and trial-and-error. There is no universal method for finding Lyapunov functions, but certain approaches and strategies can be employed, which we will explore in the context of our example system. The effectiveness of a Lyapunov function in proving stability or the absence of closed orbits stems from its ability to capture the overall trend of the system's dynamics. It provides a global perspective on the system's behavior, allowing us to infer stability properties without having to solve the differential equations explicitly. This is particularly useful for nonlinear systems, where analytical solutions are often difficult or impossible to obtain. Therefore, understanding Lyapunov stability theory and the techniques for constructing Lyapunov functions is essential for analyzing the behavior of a wide range of dynamical systems.

System Description and the Goal

Let's revisit the system we aim to analyze:

dx/dt = x(y^2 + 1) + y
dy/dt = x^2y + x

Our objective is to construct a Lyapunov function V(x, y) for this system and use it to demonstrate that the system has no closed orbits (limit cycles). This implies that any trajectory starting near the origin will either spiral into the origin or move away from it, but it will not form a closed loop. To achieve this, we will follow these steps:

Identify the Equilibrium Points: Determine the points (x, y) where both dx/dt and dy/dt are zero. These points represent the steady states of the system.
Propose a Candidate Lyapunov Function: Based on the system's structure, we will propose a function V(x, y) that is positive definite in a neighborhood of the equilibrium point. Common choices include quadratic forms like V(x, y) = ax^2 + by^2 or more complex expressions tailored to the system's specific form.
Compute the Time Derivative: Calculate the time derivative of V(x, y) along the system's trajectories, denoted as V̇(x, y). This involves using the chain rule and substituting the expressions for dx/dt and dy/dt from the system's equations.
Analyze the Sign of V̇(x, y): Examine the sign of V̇(x, y). If V̇(x, y) is negative semi-definite or negative definite, then V(x, y) is a Lyapunov function, and we can draw conclusions about the system's stability. In particular, if V̇(x, y) is strictly negative except possibly at the equilibrium point, then the system has no closed orbits.

By systematically following these steps, we can leverage the power of Lyapunov functions to gain insights into the qualitative behavior of dynamical systems. The absence of closed orbits is a significant property, as it rules out the possibility of sustained oscillations and simplifies the long-term dynamics of the system. In our specific example, the structure of the equations suggests that the origin (0, 0) is a critical point, and we will focus on constructing a Lyapunov function that demonstrates the stability properties around this point. The challenge lies in choosing a suitable candidate function V(x, y) that will yield a negative definite or negative semi-definite derivative V̇(x, y). This often requires some experimentation and a good understanding of the system's dynamics. However, once a suitable Lyapunov function is found, it provides a rigorous proof of the system's stability characteristics. The next sections will detail the actual construction and analysis of a Lyapunov function for the given system, illustrating the practical application of this powerful technique.

Constructing a Candidate Lyapunov Function

For the system at hand,

dx/dt = x(y^2 + 1) + y
dy/dt = x^2y + x

we aim to construct a Lyapunov function to prove the absence of closed orbits. A common starting point for such systems is to consider a quadratic form as a candidate Lyapunov function:

V(x, y) = ax^2 + by^2

where a and b are positive constants. This form is positive definite since V(x, y) > 0 for all (x, y) ≠ (0, 0) and V(0, 0) = 0, provided a and b are positive. However, the success of this choice depends on the resulting time derivative V̇(x, y). To improve our chances of finding a suitable Lyapunov function, we might consider a more general form that incorporates cross-terms:

V(x, y) = ax^2 + by^2 + cxy

where a, b, and c are constants. The inclusion of the cxy term allows for greater flexibility in shaping the Lyapunov function to match the system's dynamics. The key is to choose a, b, and c such that V(x, y) remains positive definite and its time derivative V̇(x, y) is negative semi-definite or negative definite. For V(x, y) = ax^2 + by^2 + cxy to be positive definite, we need to ensure that its level curves are ellipses centered at the origin. This requires the following conditions to be satisfied:

a > 0
b > 0
4ab > c^2

These conditions guarantee that V(x, y) represents a bowl-shaped surface, with a minimum at the origin. Now, the crucial step is to compute the time derivative V̇(x, y) and analyze its sign. This will involve substituting the expressions for dx/dt and dy/dt from the system's equations and carefully manipulating the resulting terms. The goal is to find values of a, b, and c that make V̇(x, y) negative semi-definite or, ideally, negative definite. If we can achieve this, then V(x, y) will be a Lyapunov function, and we can conclude that the system has no closed orbits. The process of choosing the coefficients a, b, and c often involves some trial and error, guided by the structure of the system's equations. For instance, terms in dx/dt and dy/dt that have opposite signs might suggest specific choices for a, b, and c that lead to cancellations in V̇(x, y). In our case, the presence of the +y term in dx/dt and the +x term in dy/dt hints that we might need a non-zero value for c to achieve a negative definite V̇(x, y). Therefore, the judicious selection of the candidate Lyapunov function is a critical step in this analysis. A well-chosen candidate can significantly simplify the subsequent calculations and increase the likelihood of finding a function that satisfies the Lyapunov criteria. The next step is to compute the time derivative of our chosen candidate and see if we can adjust the parameters to achieve the desired sign.

Computing the Time Derivative and Analyzing its Sign

Having proposed a candidate Lyapunov function of the form V(x, y) = ax^2 + by^2 + cxy, we now need to compute its time derivative V̇(x, y) along the trajectories of the system:

dx/dt = x(y^2 + 1) + y
dy/dt = x^2y + x

The time derivative V̇(x, y) is given by:

V̇(x, y) = (∂V/∂x)(dx/dt) + (∂V/∂y)(dy/dt)

First, we compute the partial derivatives of V(x, y):

∂V/∂x = 2ax + cy
∂V/∂y = 2by + cx

Next, we substitute these into the expression for V̇(x, y), along with the given expressions for dx/dt and dy/dt:

V̇(x, y) = (2ax + cy)(x(y^2 + 1) + y) + (2by + cx)(x^2y + x)

Expanding this expression, we get:

V̇(x, y) = 2ax^2(y^2 + 1) + 2axy + cxy(y^2 + 1) + cy^2 + 2bx^2y^2 + 2bxy + cx^3y + cx^2

Now, the critical step is to simplify this expression and see if we can choose the constants a, b, and c such that V̇(x, y) is negative semi-definite or negative definite. Rearranging the terms, we have:

V̇(x, y) = 2ax^2(y^2 + 1) + 2axy + cxy^3 + cxy + cy^2 + 2bx^2y^2 + 2bxy + cx^3y + cx^2

V̇(x, y) = 2ax^2y^2 + 2ax^2 + 2axy + cxy^3 + cxy + cy^2 + 2bx^2y^2 + 2bxy + cx^3y + cx^2

To make V̇(x, y) negative, we want to eliminate or combine terms with opposite signs. A careful observation reveals that we can potentially cancel the xy terms by choosing 2a + 2b = 0. However, since we require a and b to be positive for V(x, y) to be positive definite, this is not possible. Instead, let's try to simplify the expression by grouping terms and looking for cancellations. We can rewrite V̇(x, y) as:

V̇(x, y) = (2a + 2b)x^2y^2 + 2ax^2 + cy^2 + cx^2 + cxy^3 + cx^3y + (2a + 2b)xy

Now, let's try to choose the constants such that the coefficient of xy is zero, i.e., 2a + 2b = 0. However, as noted before, this is not feasible with positive a and b. Instead, let's consider setting c = -2a to eliminate some cross terms. This gives us:

V̇(x, y) = 2ax^2y^2 + 2ax^2 + cy^2 + cx^2 + cxy^3 + cx^3y + (2a + 2b)xy

Substituting c = -2a:

V̇(x, y) = 2ax^2y^2 + 2ax^2 - 2ay^2 - 2ax^2 - 2axy^3 - 2ax^3y + (2a + 2b)xy

V̇(x, y) = 2ax^2y^2 - 2ay^2 - 2axy^3 - 2ax^3y + (2a + 2b)xy

This expression still looks complicated, but we can see that the 2ax^2 term cancels. To simplify further, we can try to choose b such that the xy term vanishes. This requires setting 2a + 2b = 0, which again is not possible with positive a and b. However, we can try a different approach. Let's go back to the original expression for V̇(x, y) and try a simpler choice for V(x, y).

Revisiting the Candidate Lyapunov Function

Let's go back to the simpler candidate Lyapunov function:

V(x, y) = ax^2 + by^2

with a > 0 and b > 0. Then,

∂V/∂x = 2ax
∂V/∂y = 2by

And

V̇(x, y) = (2ax)(x(y^2 + 1) + y) + (2by)(x^2y + x)

V̇(x, y) = 2ax^2(y^2 + 1) + 2axy + 2bx^2y^2 + 2bxy

V̇(x, y) = 2ax^2y^2 + 2ax^2 + 2axy + 2bx^2y^2 + 2bxy

V̇(x, y) = (2a + 2b)xy + 2ax^2 + (2a + 2b)x^2y^2

We want V̇(x, y) to be negative. If we choose a = 1 and b = 1, we get:

V̇(x, y) = 4xy + 2x^2 + 4x^2y^2

This expression is not negative definite. Let's try to choose a and b such that the xy term cancels. This is not possible since the x^2 and x^2y2 terms are always non-negative. However, we can try a different approach.

A Modified Lyapunov Function

Let's consider a slightly modified Lyapunov function:

V(x, y) = x^2 + (y + f(x))^2

where f(x) is some function of x. This form ensures that V(x, y) is positive definite. Let's compute the derivative:

∂V/∂x = 2x + 2(y + f(x))f'(x)
∂V/∂y = 2(y + f(x))

V̇(x, y) = [2x + 2(y + f(x))f'(x)][x(y^2 + 1) + y] + [2(y + f(x))][x^2y + x]

V̇(x, y) = 2x^2(y^2 + 1) + 2xy + 2(y + f(x))f'(x)[x(y^2 + 1) + y] + 2(y + f(x))(x^2y + x)

V̇(x, y) = 2x^2y^2 + 2x^2 + 2xy + 2(y + f(x))f'(x)x(y^2 + 1) + 2(y + f(x))f'(x)y + 2(y + f(x))x^2y + 2(y + f(x))x

This expression is quite complex. Let's try a simpler form for f(x). If we choose f(x) = 0, then V(x, y) = x^2 + y^2, and

V̇(x, y) = 2x^2(y^2 + 1) + 2xy + 2y(x^2y + x)

V̇(x, y) = 2x^2y^2 + 2x^2 + 2xy + 2x^2y^2 + 2xy

V̇(x, y) = 4x^2y^2 + 2x^2 + 4xy

This is still not negative definite. Let's try f(x) = -x^3. Then V(x, y) = x^2 + (y - x³⁾2, and

f'(x) = -3x^2

This approach becomes very complex. Let's try a different strategy altogether.

Using Dulac's Criterion

Since finding a Lyapunov function directly is proving difficult, we can employ Dulac's Criterion to show the absence of closed orbits. Dulac's Criterion states that for a two-dimensional system:

dx/dt = P(x, y)
dy/dt = Q(x, y)

if there exists a continuously differentiable function B(x, y) such that:

∂(BP)/∂x + ∂(BQ)/∂y

has the same sign (excluding zero) in a region D, then there are no closed orbits in D. In our case:

P(x, y) = x(y^2 + 1) + y
Q(x, y) = x^2y + x

Let's choose B(x, y) = 1/x. Then:

BP = (x(y^2 + 1) + y)/x = y^2 + 1 + y/x
BQ = (x^2y + x)/x = xy + 1

Now we compute the partial derivatives:

∂(BP)/∂x = -y/x^2
∂(BQ)/∂y = x

So:

∂(BP)/∂x + ∂(BQ)/∂y = x - y/x^2

This expression does not have a definite sign. Let's try B(x, y) = 1/(x^2 + y^2). Then:

BP = (x(y^2 + 1) + y)/(x^2 + y^2)
BQ = (x^2y + x)/(x^2 + y^2)

∂(BP)/∂x = [(y^2 + 1) + x(-2x)(x(y^2 + 1) + y)/(x^2 + y^2)^2 - (x(y^2 + 1) + y)(2x)] / (x^2 + y^2)
∂(BQ)/∂y = [x^2(x^2 + y^2) - (x^2y + x)(2y)] / (x^2 + y^2)^2

These expressions are becoming extremely complex. Let's try a different B(x, y). Let's try B(x, y) = 1/x^2:

BP = (x(y^2 + 1) + y)/x^2 = (y^2 + 1)/x + y/x^2
BQ = (x^2y + x)/x^2 = y + 1/x

∂(BP)/∂x = -(y^2 + 1)/x^2 - y/x^3
∂(BQ)/∂y = 1

∂(BP)/∂x + ∂(BQ)/∂y = 1 - (y^2 + 1)/x^2 - y/x^3

This expression is also difficult to analyze. The choice of B(x, y) significantly impacts the complexity of the calculations, and there is no guarantee that a simple choice will work. However, the failure to find a suitable Dulac function does not necessarily imply the existence of closed orbits; it simply means that this particular method is inconclusive.

Conclusion

In conclusion, we explored the challenging task of constructing a Lyapunov function to prove the absence of closed orbits for the system:

dx/dt = x(y^2 + 1) + y
dy/dt = x^2y + x

We attempted various candidate Lyapunov functions, including quadratic forms and modified forms, but were unable to find a function that yielded a negative definite time derivative. This difficulty highlights the fact that constructing Lyapunov functions is not always a straightforward process and often requires significant insight and trial-and-error. Furthermore, we attempted to apply Dulac's Criterion with different choices of the function B(x, y), but the resulting expressions were complex and did not lead to a definitive conclusion. While we were unable to definitively prove the absence of closed orbits using these methods, it is important to note that the failure to find a suitable Lyapunov function or Dulac function does not necessarily imply the existence of closed orbits. It simply means that these particular techniques were inconclusive for this specific system. Other methods, such as phase plane analysis or numerical simulations, might provide further insights into the system's behavior. The exploration of stability in dynamical systems is a rich and complex field, and the techniques discussed here represent just a few of the tools available to researchers and engineers. The choice of method often depends on the specific characteristics of the system under consideration, and a combination of analytical and numerical approaches is often necessary to gain a comprehensive understanding of the system's dynamics. Despite the challenges encountered in this particular example, the principles and techniques of Lyapunov stability and Dulac's Criterion remain invaluable tools for analyzing the stability of solutions and the presence or absence of oscillatory behavior in a wide range of systems arising in physics, engineering, and other disciplines. Further investigation using other methods may be warranted to fully characterize the behavior of this system.