Question

Let $$ A=\left[\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right] $$ be a real invertible matrix, and consider the system \(\mathbf{y}^{\prime}=A \mathbf{y}\). (a) What conditions must the matrix entries \(a_{i j}\) satisfy to make the equilibrium point \(\mathbf{y}_{e}=\mathbf{0}\) a center? (b) Assume that the equilibrium point at the origin is a center. Show that the system \(\mathbf{y}^{\prime}=A \mathbf{y}\) is a Hamiltonian system. (c) Is the converse of the statement in part (b) true? In other words, if the system \(\mathbf{y}^{\prime}=A \mathbf{y}\) is a Hamiltonian system, does it necessarily follow that \(\mathbf{y}_{e}=\mathbf{0}\) is a center? Explain.

Short answer

Answer: The conditions for the matrix entries are \(a_{11} = -a_{22}, \quad a_{12}a_{21} > 0\).

Step-by-step solution

Conditions for a Center

To find the conditions under which the equilibrium point \(\mathbf{y}_{e} = \mathbf{0}\) is a center, we first recall the eigenvalues criterion. If all eigenvalues of the coefficient matrix \(A\) have negative real parts, the equilibrium point is considered a center. The characteristic equation of the matrix \(A\) is given by: $$ \det(A - \lambda I) = (a_{11} - \lambda)(a_{22} - \lambda) - a_{12}a_{21} = 0 $$ Solving for the eigenvalues, we have the equation: $$ \lambda^2 - (a_{11} + a_{22})\lambda + (a_{11}a_{22} - a_{12}a_{21}) = 0 $$ Let \(\lambda_1\) and \(\lambda_2\) be the eigenvalues of \(A\). We want the real parts of both eigenvalues to be zero; therefore, we require the discriminant of the equation to be less than zero, i.e., $$ (a_{11} + a_{22})^2 - 4(a_{11}a_{22} - a_{12}a_{21}) < 0 $$ By simplifying the inequality, we find that the conditions for the matrix entries are: $$ a_{11} = -a_{22}, \quad a_{12}a_{21} > 0 $$

Proving the System is Hamiltonian

To prove the system is Hamiltonian, we need to show that: $$ \frac{\partial H}{\partial y_1} = -\frac{dy_2}{dt}, \quad \frac{\partial H}{\partial y_2} = \frac{dy_1}{dt} $$ We are given that \(\mathbf{y}^{\prime} = A\mathbf{y}\). Since both eigenvalues have zero real parts, the matrix \(A\) can be written as: $$ A = \left[ \begin{array}{cc} 0 & a \\ -b & 0 \\ \end{array} \right], \quad a, b > 0 $$ The system of differential equations becomes: $$ \frac{dy_1}{dt} = ay_2, \quad \frac{dy_2}{dt} = -by_1 $$ Let's assume \(H(y_1, y_2)\) is a Hamiltonian function. By comparing the equations above with the conditions required for a system to be Hamiltonian, we find that: $$ \frac{\partial H}{\partial y_1} = -by_1, \quad \frac{\partial H}{\partial y_2} = ay_2 $$ From these equations, we can find the Hamiltonian function \(H(y_1, y_2)\) by solving the partial differential equations. Integrating both equations with respect to \(y_1\) and \(y_2\) respectively, we obtain: $$ H(y_1, y_2) = -\frac{1}{2}by_1^2 + f(y_2), \quad H(y_1, y_2) = \frac{1}{2}ay_2^2 + g(y_1) $$ Comparing the two expressions for \(H(y_1, y_2)\), we find that: $$ H(y_1, y_2) = \frac{1}{2}ay_2^2 - \frac{1}{2}by_1^2 $$ Since there exists a Hamiltonian function for the given system, it is Hamiltonian.

Discuss the Converse of part (b)

We now examine whether the converse is true, i.e., if the system is Hamiltonian, does it necessarily imply that the equilibrium point \(\mathbf{y}_{e} = \mathbf{0}\) is a center? Consider the general Hamiltonian system with \(H(y_1, y_2) = \frac{1}{2}ay_2^2 - \frac{1}{2}by_1^2\). The corresponding equations are: $$ \frac{dy_1}{dt} = ay_2, \quad\frac{dy_2}{dt} = -by_1 $$ The eigenvalues for this system are \(\lambda = \pm i\sqrt{ab}\), which means their real parts are zero. As a result, the equilibrium point \(\mathbf{y}_{e} = \mathbf{0}\) can be classified as a center. Thus, the converse of the statement in part (b) is indeed true. If the system is Hamiltonian, it necessarily implies that \(\mathbf{y}_{e} = \mathbf{0}\) is a center.

Key concepts

Equilibrium Points

In the study of dynamical systems, an **equilibrium point** is where the system remains constant over time, given no external changes. For a system described by equations like \( \mathbf{y}' = A \mathbf{y} \), the equilibrium point is found by setting the derivative to zero, which makes any movement or change disappear.
In simple terms, it acts like a resting or stationary point where forces balance out. Here, we are particularly focused on the equilibrium point \( \mathbf{y}_e = \mathbf{0} \), which means that at zero, all configurations are calm and unchanged.

• Equilibrium points can be centers, nodes, or saddle points, depending on the nature of eigenvalues.
• For the origin to act as a center, certain conditions regarding eigenvalues and matrix properties need to be met.

Understanding equilibrium points is crucial because it helps predict the long-term behavior of a system, whether it will settle, oscillate, or diverge.

Eigenvalues

**Eigenvalues** are foundational in understanding the behavior of linear transformations. They come from solving the characteristic equation of a matrix, and they indicate how a system changes when influenced by the matrix.
In the context of our system \( \mathbf{y}' = A \mathbf{y} \), eigenvalues can inform us about stability and system dynamics.

• If all eigenvalues have a negative real part, the system tends towards stability.
• If eigenvalues are purely imaginary (zero real part), the equilibrium can be a center where trajectories neither converge nor diverge.

To determine these values, compute them through the determinant \( \det(A - \lambda I) \), leading to the characteristic polynomial. For complex systems, eigenvalues provide key insight into potential oscillations or stability tendencies.

Characteristic Equation

The **characteristic equation** is a crucial tool used to find the eigenvalues of a matrix. It's derived from setting the determinant of \( A - \lambda I \) to zero, where \( A \) is the matrix and \( I \) is the identity matrix. This equation often takes the form of a polynomial whose roots are the eigenvalues.
For a 2x2 matrix \( A \), the equation becomes:\[ (a_{11} - \lambda)(a_{22} - \lambda) - a_{12}a_{21} = 0 \]Solving this polynomial provides the eigenvalues \( \lambda_1 \) and \( \lambda_2 \). These are vital in determining system behavior at equilibrium points.

• The nature of these roots (real, complex, or repeated) helps conclude stability characteristics.
• A discriminant value less than zero indicates complex conjugates, suggesting possible oscillatory behavior.

Mastering the characteristic equation is essential for predicting and analyzing system dynamics in linear algebra and differential equations.

Real Invertible Matrix

A **real invertible matrix** is one that has an inverse, meaning there exists another matrix which multiplies with it to yield the identity matrix. For a matrix \( A \) to be invertible, its determinant \( \det(A) \) must not be zero.
Here are some important attributes:

• An invertible matrix implies that the linear system \( \mathbf{y}' = A \mathbf{y} \) has unique solutions.
• An inverse exists only when the matrix is square, and its rows (or columns) are linearly independent.

In the context of equilibrium and Hamiltonian systems, being real invertible facilitates analyzing system dynamics since the matrix directly influences the evolution and stability of trajectories.
Recognizing whether a matrix is invertible helps strengthen understanding of the potential reach and limitations of the system it represents.