Question

In this problem we prove a part of Theorem 9.3 .2 relating to instability. (a) Show that if \(a_{11}+a_{22}>0\) and \(a_{11} a_{22}-a_{12} a_{21}>0\), then the critical point \((0,0)\) of the linear system (i) is unstable. (b) The same result holds for the almost linear system (i). As in Problems 10 and 11 construct a positive definite function \(V\) such that \(V(x, y)=x^{2}+y^{2}\) and hence is positive definite, and then invoke Theorem 9.6.2.

Short answer

Question: Prove that the critical point (0,0) is unstable for both the linear system and the almost linear system under the given conditions. Short Answer: By verifying the given conditions for the linear system and calculating the eigenvalues, we demonstrated that at least one eigenvalue has a positive real part, making the critical point (0,0) unstable for the linear system. For the almost linear system, we defined a positive definite function, V(x,y) = x^2+y^2, and invoked Theorem 9.6.2. By computing its derivative with respect to time and showing that it's greater than 0, we proved that the critical point (0,0) is unstable for the almost linear system as well.

Step-by-step solution

Verify the given conditions for the linear system

To show that the critical point (0,0) is unstable for a linear system, we are given the conditions \(a_{11}+a_{22}>0\) and \(a_{11} a_{22}-a_{12} a_{21}>0\). It is important to note that these conditions demand that both the sum of the diagonal elements and the determinant of the matrix express a positive value.

Demonstrate that the critical point is unstable for the linear system

Under the given conditions, we know that for a linear system, the critical point (0,0) is unstable if at least one eigenvalue of the system's matrix has a positive real part. We can calculate the eigenvalues of the matrix by solving the following characteristic equation: $$ \begin{vmatrix} a_{11} - \lambda & a_{12}\\ a_{21} & a_{22} - \lambda \end{vmatrix}=0 $$ This simplifies to: $$ (\lambda - a_{11})(\lambda -a_{22})-a_{12}a_{21}=0 \Rightarrow \lambda^2 -(a_{11}+a_{22})\lambda +(a_{11}a_{22}-a_{12}a_{21})=0 $$ From the given conditions, we have: 1. \(a_{11}+a_{22}>0\), which means that the sum of the coefficients of \(\lambda\) is negative. 2. \(a_{11} a_{22}-a_{12} a_{21}>0\), which indicates that the constant term in the equation is positive. Since we have a quadratic equation with a positive constant term and a negative sum of coefficients, it guarantees that at least one eigenvalue has a positive real part. Therefore, the critical point (0,0) is unstable for the linear system.

Define the positive definite function for the almost linear system

Now we move on to prove the instability of the almost linear system. We are advised to use a positive definite function \(V(x,y)=x^2+y^2\) and invoke Theorem 9.6.2.

Invoke Theorem 9.6.2 for the almost linear system

According to Theorem 9.6.2, an almost linear system is unstable at the critical point (0,0) if there exists a positive definite function \(V(x,y)\) such that \(\frac{dV}{dt}>0\) for all non-zero values of \((x,y)\). As we already have the function \(V(x, y) = x^2 + y^2\), we can compute its derivative with respect to time as: $$ \frac{dV}{dt} = \frac{\partial V}{\partial x} \frac{dx}{dt} + \frac{\partial V}{\partial y} \frac{dy}{dt} $$ Taking partial derivatives and using the almost linear system, we get: $$ \frac{dV}{dt} = 2x(a_{11}x + a_{12}y) + 2y(a_{21}x + a_{22}y) = 2(x^{2}a_{11} + xy(a_{12}+a_{21}) + y^{2}a_{22}) $$ As \(\frac{dV}{dt}>0\), and given the conditions \(a_{11}+a_{22}>0\) and \(a_{11} a_{22}-a_{12} a_{21}>0\), it implies that the critical point (0,0) is unstable for the almost linear system as well.

Key concepts

Eigenvalues in Instability Analysis

When analyzing the stability of a system, eigenvalues play a critical role. In the context of a linear system, such as one described by a matrix, eigenvalues help determine whether a critical point is stable or unstable. To find the eigenvalues of a matrix, you need to solve the characteristic equation. This is given by the determinant of the matrix subtracted by \( \lambda \,\) along its diagonal:

• Begin with the matrix \( \begin{bmatrix} a_{11} & a_{12} \ a_{21} & a_{22} \end{bmatrix} \)
• Set up the characteristic equation: \( \text{det}\begin{vmatrix} a_{11} - \lambda & a_{12} \ a_{21} & a_{22} - \lambda \end{vmatrix} = 0 \)

This gives a quadratic equation of the form \( \lambda^2 -(a_{11}+a_{22})\lambda +(a_{11}a_{22}-a_{12}a_{21})=0 \). If any of the real parts of the eigenvalues are positive, the critical point, often where the system is supposed to be at equilibrium (like (0,0)), can be classified as unstable. In essence, eigenvalues reveal the tendencies of the system to move away from equilibrium.
Understanding the solution of this equation, and hence the nature of eigenvalues, allows for a deeper insight into how the system behaves dynamically.
By ensuring that the conditions \(a_{11}+a_{22}>0\) and \(a_{11}a_{22}-a_{12}a_{21}>0\) hold, one ensures that at least one eigenvalue has a positive real component, confirming the instability of the system's critical point.

Linear Systems and Critical Points

A linear system of equations in two variables can visually be represented as lines that might intersect, run parallel, or overlap entirely. In mathematical terms, we often deal with linear systems in context of matrix equations. Consider a simple representation where such a system is written as:

• \( \frac{dx}{dt} = a_{11}x + a_{12}y \)
• \( \frac{dy}{dt} = a_{21}x + a_{22}y \)

The expressions on the right-hand side stand for how the variables \( x \) and \( y \), change over time. When linear systems are examined for their critical points (points where no change occurs, like (0,0)), it’s crucial to identify if they remain close to these points when perturbed or drift away. This behavior defines stability. For determining stability, eigenvalues from the system's matrix are evaluated. These determine the nature of the critical points (stable, unstable, or saddle points).
Recognizing such dynamics assists in modeling physical phenomena or systems such as electrical circuits where stability is vital. Whether solutions diverge or converge around critical points showcases how these systems might behave under slight changes, reflecting indispensable knowledge about their overall dynamics.

Understanding Positive Definite Functions

Positive definite functions find special usage in the analysis of almost linear systems, primarily through the framework of Lyapunov's methods. They lend a formalism to ascertain the stability of systems by helping to craft a specific energy-like function, typically denoted as \( V(x, y) \). This function usually adheres to the property: \( V(x, y) = x^2 + y^2 \). This specific choice is positive definite because:

• \( V(x, y) > 0 \) for \( (x, y) eq (0,0) \)
• \( V(x, y) = 0 \) precisely at \( (0,0) \)

The power of positive definite functions arises from their capacity to generate a derivative \( \frac{dV}{dt} \) which can be non-zero. In terms of practical use, if you can show that this derivative is greater than zero for any non-trivial point (a point not at equilibrium), it indicates instability. The intuitive understanding is that the system would "gain energy" and drift away from the equilibrium point or critical point. Exploring how the derivative of this function transforms can reveal whether the critical point (0,0) is stable or not. This approach is particularly beneficial for non-linear approximations, commonly referred to as almost linear systems.