Question

In this problem we indicate how to show that the trajectories are ellipses when the eigen- values are pure imaginary. Consider the system $$ \left(\begin{array}{l}{x} \\\ {y}\end{array}\right)^{\prime}=\left(\begin{array}{ll}{a_{11}} & {a_{12}} \\\ {a_{21}} & {a_{22}}\end{array}\right)\left(\begin{array}{l}{x} \\\ {y}\end{array}\right) $$ (a) Show that the eigenvalues of the coefficient matrix are pure imaginary if and only if $$ a_{11}+a_{22}=0, \quad a_{11} a_{22}-a_{12} a_{21}>0 $$ (b) The trajectories of the system (i) can be found by converting Eqs. (i) into the single equation $$ \frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{a_{21} x+a_{22} y}{a_{11} x+a_{12} y} $$ Use the first of Eqs. (ii) to show that Eq. (iii) is exact. (c) By integrating Eq. (iii) show that $$ a_{21} x^{2}+2 a_{22} x y-a_{12} y^{2}=k $$ where \(k\) is a constant. Use Eqs. (ii) to conclude that the graph of Eq. (iv) is always an ellipse. Hint: What is the discriminant of the quadratic form in Eq. (iv)?

Short answer

In conclusion, we went through the following steps to prove that the trajectories are ellipses when the eigenvalues are pure imaginary: (a) We found the conditions for the eigenvalues of the coefficient matrix to be pure imaginary, which are \(a_{11}+a_{22}=0\) and \(a_{11} a_{22}-a_{12} a_{21}>0\). (b) We showed that the equation obtained by converting the system of equations into a single equation is exact, using the conditions from part (a). Exact differential equations allow us to find potential functions. (c) We integrated the equation derived in part (b) and used the conditions from part (a) to show that the resulting equation represents an ellipse, as the discriminant of the quadratic form is positive.

Step-by-step solution

Part (a): Find the conditions for pure imaginary eigenvalues

Firstly, let's find the eigenvalues of the coefficient matrix $$ A = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} $$ The characteristic equation of matrix A is given by: $$ \det (A - \lambda I) = (a_{11}-\lambda)(a_{22}-\lambda) - a_{12}a_{21} = \lambda^2 - (a_{11} + a_{22})\lambda +(a_{11}a_{22}-a_{12}a_{21}) = 0 $$ Now, we have a quadratic equation in \(\lambda\). The eigenvalues are pure imaginary if the discriminant of the quadratic equation is negative. The discriminant \(\Delta\) is given by: $$ \Delta = (- (a_{11} + a_{22}))^2 - 4(1)(a_{11}a_{22} - a_{12}a_{21}) $$ To ensure that the discriminant is negative, we have: $$ (a_{11} + a_{22})^2 < 4(a_{11}a_{22} - a_{12}a_{21}) $$ This can be re-written as: $$ a_{11}+a_{22}=0, \quad a_{11} a_{22}-a_{12} a_{21}>0 $$

Part (b): Show that the equation is exact

The given equation is: $$ \frac{dy}{dx} = \frac{a_{21}x + a_{22}y}{a_{11}x + a_{12}y} $$ We can re-write this equation as: $$ (a_{11}x + a_{12}y)dy - (a_{21}x + a_{22}y)dx = 0 $$ Let's check the exactness condition: $$ \frac{\partial}{\partial y} (a_{11}x + a_{12}y) = a_{12} \\ \frac{\partial}{\partial x} (-a_{21}x - a_{22}y) = -a_{21} $$ From Part (a), we know that \(a_{12} = a_{21}\). Therefore, the exactness condition is satisfied. Hence, the given equation is exact.

Part (c): Integrate the equation and show that the graph represents an ellipse

We can write the exact differential equation as: $$ d(a_{21}x^2 + 2a_{22}xy - a_{12}y^2) = a_{21} x^{2}+2 a_{22} x y-a_{12} y^{2} = k $$ where \(k\) is an arbitrary constant. Now, let's rewrite the left side of the equation using the conditions from Part (a) where \(a_{11} = -a_{22}\) and \(a_{12} = a_{21}\): $$ a_{21} x^{2}-2 a_{11} x y-a_{12} y^{2}=k $$ To conclude that the graph represents an ellipse, we can look at the discriminant of the quadratic form: Discriminant \(= (-2a_{11})^2 - 4(a_{21})(-a_{12}) = 4a_{11}^2 + 4a_{21}a_{12}\) Since \(a_{11}a_{22}-a_{12}a_{21}>0\) from Part (a), the discriminant is positive. A positive discriminant indicates that the graph of the quadratic equation represents an ellipse. Therefore, we have shown that the trajectories are ellipses when the eigenvalues are pure imaginary.

Key concepts

Eigenvalues

Eigenvalues are a fundamental concept in linear algebra. They represent scalar values that provide insight into the properties of a matrix. When solving differential equations, particularly ones involving systems of linear equations, finding eigenvalues of the coefficient matrix helps understand the system's behavior.

For the given system, the coefficient matrix is \( A = \begin{pmatrix} a_{11} & a_{12} \ a_{21} & a_{22} \end{pmatrix} \). The eigenvalues are obtained by solving the characteristic equation given as \( \det(A - \lambda I) = 0 \). This simplifies to a quadratic form: \( \lambda^2 - (a_{11} + a_{22})\lambda + (a_{11}a_{22} - a_{12}a_{21}) = 0 \).

When we analyze the solutions for \( \lambda \), the eigenvalues are pure imaginary if the discriminant of this equation is negative. This implies:

• The sum of the diagonal elements \( a_{11} + a_{22} = 0 \).
• The product \( a_{11} a_{22} - a_{12} a_{21} > 0 \).

Therefore, these conditions help identify when the system's trajectories form ellipses, demonstrating a critical link between eigenvalues and the behavior of differential equations.

Ellipses

Ellipses are a type of conic section characterized by their closed, symmetric, oval shape. In differential equations, when the eigenvalues of a system's matrix are pure imaginary, the trajectories are ellipses. Here's why:

Considering the given system, where the differential equation is reduced to a form related to quadratic curves, the equation \( a_{21} x^{2} + 2 a_{22} xy - a_{12} y^{2} = k \) shows the nature of these trajectories. If the discriminant of this quadratic form is positive, it ensures the trajectory is an ellipse.

The discriminant of a general quadratic equation \( Ax^2 + 2Bxy + Cy^2 = D \) is given by \( B^2 - AC \). For an ellipse formation, the discriminant must be positive, signifying that the matrices properties (a combination of eigenvalues and **quadratic forms**) create elliptical paths.

Ellipses in this context mean that the solutions do not escape to infinity or collapse to a point but instead trace closed loops. This geometric interpretation aids in visualizing the system's stability and behavior.

Quadratic Forms

Quadratic forms arise in the study of polynomial equations where terms are quadratic expressions. In the context of the given problem, the quadratic form \( a_{21} x^2 + 2 a_{22} xy - a_{12} y^2 = k \) appears while converting the system of differential equations into a single second-degree equation.

These forms are key in analyzing and representing the properties of systems characterized by matrices. The specific form here helps in determining the type of conic section the equation represents: an ellipse, parabola, or hyperbola.

To determine which conic section it is, we look at the discriminant described by \( B^2 - AC \) where \( A = a_{21} \), \( B = a_{22} \), and \( C = -a_{12} \). For the system to represent an ellipse, the discriminant must be negative, under the condition \( a_{11} a_{22} - a_{12} a_{21} > 0 \) from our eigenvalue analysis.

This quadratic form not only helps in identifying the trajectory shape but also plays a role in simplifying the complexity of multidimensional systems into a manageable form, highlighting the interconnectivity of algebra and geometry in differential equations.

Exact Differential Equations

Exact differential equations represent a specific type where a direct integration method can solve them. For an equation to be exact, it must satisfy a certain equality involving partial derivatives.

The original exercise showed the differential equation given by \( \frac{dy}{dx} = \frac{a_{21}x + a_{22}y}{a_{11}x + a_{12}y} \). This equation can be expressed in the form \( (a_{11}x + a_{12}y)dy - (a_{21}x + a_{22}y)dx = 0 \), which resembles the standard form of an exact equation: \( M(x, y)dx + N(x, y)dy = 0 \).

For exactness, the condition \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \) must hold. By substituting and simplifying, we concluded that these conditions are satisfied, primarily when \( a_{12} = a_{21} \).

Exact differential equations are advantageous because they allow us to find an implicit solution directly through integration, making them a powerful tool in the analytical solution of differential equations. By recognizing and transforming our problem into an exact form, we simplify the work needed to solve for trajectories, such as ellipses.