Question

We will prove part of Theorem 9.3 .2 : If the critical point \((0,0)\) of the almost linear system $$ d x / d t=a_{11} x+a_{12} y+F_{1}(x, y), \quad d y / d t=a_{21} x+a_{22} y+G_{1}(x, y) $$ is an asymptotically stable critical point of the corresponding linear system $$ d x / d t=a_{11} x+a_{12} y, \quad d y / d t=a_{21} x+a_{22} y $$ then it is an asymptotically stable critical point of the almost linear system (i). Problem 12 deals with the corresponding result for instability. Consider the linear system (ii). (a) Since \((0,0)\) is an asymptotically stable critical point, show that \(a_{11}+a_{22}<0\) and \(\left.a_{11} a_{22}-a_{12} a_{21}>0 . \text { (See Problem } 21 \text { of Section } 9.1 .\right)\) (b) Construct a Liapunov function \(V(x, y)=A x^{2}+B x y+C y^{2}\) such that \(V\) is positive definite and \(\hat{V}\) is negative definite. One way to ensure that \(\hat{V}\) is negative definite is to choose \(A, B,\) and \(C\) so that \(\hat{V}(x, y)=-x^{2}-y^{2} .\) Show that this leads to the result $$ \begin{array}{l}{A=-\frac{a_{21}^{2}+a_{22}^{2}+\left(a_{11} a_{22}-a_{12} a_{21}\right)}{2 \Delta}, \quad B=\frac{a_{12} a_{22}+a_{11} a_{21}}{\Delta}} \\\ {C=-\frac{a_{11}^{2}+a_{12}^{2}+\left(a_{11} a_{22}-a_{12} a_{21}\right)}{2 \Delta}}\end{array} $$ where \(\Delta=\left(a_{11}+a_{22}\right)\left(a_{11} a_{22}-a_{12} a_{21}\right)\) (c) Using the result of part (a) show that \(A>0\) and then show (several steps of algebra are required) that $$ 4 A C-B^{2}=\frac{\left(a_{11}^{2}+a_{12}^{2}+a_{21}^{2}+a_{22}^{2}\right)\left(a_{11} a_{22}-a_{12} a_{21}\right)+2\left(a_{11} a_{22}-a_{12} a_{21}\right)^{2}}{\Delta^{2}}>0 $$ Thus by Theorem 9.6.4, \(V\) is positive definite.

Short answer

Question: Prove that if the critical point (0,0) of a linear system is asymptotically stable, then it is also asymptotically stable for an almost linear system. Answer: To prove that the critical point (0,0) is asymptotically stable for the almost linear system, we first establish the conditions for asymptotic stability in the linear system, which are tr(A) < 0 and det(A) > 0. We then construct a Liapunov function, V(x, y) = Ax^2 + Bxy + Cy^2, ensuring that it is positive definite and its time derivative is negative definite. Finally, we prove that A > 0 and 4AC - B^2 > 0. With these conditions satisfied, we can conclude that the critical point (0,0) is asymptotically stable for the almost linear system.

Step-by-step solution

Prove conditions for asymptotic stability in the linear system

Given that \((0,0)\) is an asymptotically stable critical point in the linear system, this implies that the eigenvalues of the system matrix have negative real parts. Let \(\lambda_1\) and \(\lambda_2\) be the eigenvalues of the given linear system. The characteristic equation of the linear system is \(\lambda^2 - \text{tr}(A)\lambda + \det(A) = 0\), where tr(A) = \(a_{11} + a_{22}\) is the trace of matrix A and det(A) = \(a_{11}a_{22} - a_{12}a_{21}\) is the determinant. The conditions for the eigenvalues having negative real parts are: 1. tr(A) < 0 2. det(A) > 0 Therefore, we need to prove that \(a_{11} + a_{22} < 0\) and \(a_{11}a_{22} - a_{12}a_{21} > 0\).

Construct Liapunov function and prove its properties

We are given the form of the Liapunov function as \(V(x, y) = Ax^2 + Bxy + Cy^2\). We need to ensure that \(V\) is positive definite and \(\hat{V}\) is negative definite. One way to do this is to choose \(A\), \(B\), and \(C\) so that \(\hat{V}(x, y) = -x^2 - y^2\). To find these coefficients, we have: $$ \begin{array}{l}{ A = -\frac{a_{21}^{2} + a_{22}^{2} + (a_{11} a_{22} - a_{12} a_{21})}{2 \Delta}, \quad B = \frac{a_{12} a_{22} + a_{11} a_{21}}{\Delta}} \\ {C = -\frac{a_{11}^{2} + a_{12}^{2} + (a_{11} a_{22} - a_{12} a_{21})}{2 \Delta} }\end{array} $$ where \(\Delta = (a_{11} + a_{22})(a_{11} a_{22} - a_{12} a_{21})\).

Prove that A > 0 and that 4AC - B^2 > 0

Using the result from step 1, we know that \(a_{11} + a_{22} < 0\) and \(a_{11}a_{22} - a_{12}a_{21} > 0\). Then we have: $$A = -\frac{a_{21}^{2} + a_{22}^{2} + (a_{11} a_{22} - a_{12} a_{21})}{2 (a_{11} + a_{22})(a_{11} a_{22} - a_{12} a_{21})}$$ Given that \(a_{11} + a_{22} < 0\) and \(a_{11}a_{22} - a_{12}a_{21} > 0\), the denominator is negative, making A > 0. Now, let's prove that 4AC - B^2 > 0. We have the following expression: $$4AC - B^2= \frac{(a_{11}^2 + a_{12}^2 + a_{21}^2 + a_{22}^2)(a_{11} a_{22} - a_{12} a_{21}) + 2 (a_{11} a_{22} - a_{12} a_{21})^2}{\Delta^{2}}$$ Since \(a_{11} a_{22} - a_{12} a_{21} > 0\), the expression is positive and therefore 4AC - B^2 > 0. By Theorem 9.6.4, \(V\) is positive definite, and so, we have proved that if the critical point \((0,0)\) of the linear system is asymptotically stable, then it is also asymptotically stable for the almost linear system.

Key concepts

Asymptotic Stability

In the context of differential equations, asymptotic stability refers to the behavior of solutions as they approach a critical point, such as \(0,0\). If the critical point of a system is asymptotically stable, any trajectory that starts near this point will tend to return to and remain close to it over time.
This concept applies both to linear and nonlinear systems, and it's especially crucial in systems modeling real-world scenarios where stability is desired.

To determine asymptotic stability in linear systems, we often consider the eigenvalues of the system matrix. If all the eigenvalues have negative real parts, the critical point is asymptotically stable. This means:

• The solutions decay exponentially as they approach the critical point.
• No oscillations grow or persist over time, ensuring system stability.

Understanding this helps provide insight into how a given system behaves over a long period, which is vital for engineers and scientists working on control systems, robotics, and other applications.

Lyapunov Function

A Lyapunov function is a vital tool in determining the stability of equilibrium points in differential equations. It is analogous to the concept of a potential energy function, where stability at a point can be assessed by the function's value.
When designing a Lyapunov function for a system, it must satisfy two conditions:

• The function should be positive definite, meaning it takes a positive value except at the equilibrium point where it is zero.
• Its derivative along trajectories \(\hat{V}\) must be negative definite, implying that the function decreases over time.

In the given problem, the Lyapunov candidate function was of the form \(V(x, y) = Ax^2 + Bxy + Cy^2\). By choosing suitable values for \(A, B,\) and \(C\), we can ensure that \(V\) is positive definite and \(\hat{V} = -x^2 - y^2\), which is negative definite.
Through this method, the system's asymptotic stability can be verified, reaffirming that the critical point remains stable.

Linear Systems

A linear system in this context refers to differential equations where the function can be expressed in terms of linear polynomials. The primary focus is on understanding behaviors near the critical points.
For example, a linear system in two dimensions is expressed as:
\[\frac{dx}{dt} = a_{11} x + a_{12} y, \quad \frac{dy}{dt} = a_{21} x + a_{22} y\]
In the analysis of linear systems, the behavior of trajectories around the critical point \(0,0\) is determined by the coefficients of the system matrix.
The assumptions about the system are:

• It behaves linearly around critical points.
• Our methods to determine stability involve analyzing eigenvalues and constructing functions like the Lyapunov function.

Ultimately, linear systems offer insightful approximations for more complex real-world phenomena. A deep understanding of their properties allows us to make predictions and develop control strategies in engineering and physics.

Eigenvalues

Eigenvalues are crucial in determining the stability of a linear system. They are derived from the characteristic equation of the system matrix and provide insight into the system's dynamic behavior.
For a 2x2 matrix, the characteristic equation is given by:
\[\lambda^2 - (a_{11} + a_{22})\lambda + (a_{11}a_{22} - a_{12}a_{21}) = 0\]
Understanding eigenvalues helps us assess stability criteria:

• If both eigenvalues have negative real parts, the critical point is asymptotically stable.
• If any eigenvalue has a positive real part, the point becomes unstable.
• Eigenvalues with zero real parts require further analysis to ascertain stability.

Analyzing these values helps us understand how the system behaves as time progresses, and adjusts influence factors accordingly when trying to control systems practically.

Characteristic Equation

The characteristic equation is fundamental in studying the stability of a system's critical point.
It is derived from the system matrix and its determinant involves understanding eigenvalues. For a system represented by a matrix \(A\), the characteristic equation is derived as:
\[\det(\lambda I - A) = 0\]
For our linear system, the characteristic equation is given by:
\[\lambda^2 - \, \text{tr}(A)\lambda + \, \text{det}(A) = 0\]
where:

• \(\text{tr}(A) = a_{11} + a_{22}\) is the trace of the matrix.
• \(\text{det}(A) = a_{11}a_{22} - a_{12}a_{21}\) is the determinant.

The solutions to this equation—eigenvalues—determine whether a critical point is stable, unstable, or in a state that requires further analysis. Understanding how to derive and solve the characteristic equation is thus pivotal in the assessment of a system's dynamics and stability.