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Chapter 9: Problem 3

Classify the equilibrium point of the system \(\mathbf{y}^{\prime}=A \mathbf{y}\) based on the position of \((T, D)\) in the trace-determinant plane. Sketch the phase portrait by hand. Verify your result by creating a phase portrait with your numerical solver. \(A=\left(\begin{array}{ll}2 & -4 \\ 8 & -6\end{array}\right)\)

Short Answer

Expert verified
The system has a stable spiral equilibrium point; all trajectories spiral towards the origin.

Step by step solution

01

Calculate the Trace

The trace \( T \) of a matrix \( A \) is the sum of its diagonal elements. For the matrix \( A = \begin{pmatrix} 2 & -4 \ 8 & -6 \end{pmatrix} \), calculate the trace as follows: \[ T = 2 + (-6) = -4 \] Thus, the trace of the matrix is \( T = -4 \).
02

Calculate the Determinant

The determinant \( D \) of a matrix \( A \) is given by \( D = ad - bc \), where \( a, b, c, \) and \( d \) are the elements of the matrix. For the matrix \( A = \begin{pmatrix} 2 & -4 \ 8 & -6 \end{pmatrix} \), - \( a = 2 \), \( b = -4 \), \( c = 8 \), \( d = -6 \).Calculate the determinant: \[ D = (2)(-6) - (8)(-4) = -12 + 32 = 20 \] Thus, the determinant of the matrix is \( D = 20 \).
03

Analyze the Trace-Determinant Plane

With \( T = -4 \) and \( D = 20 \), plot the point \((-4, 20)\) in the trace-determinant plane. The phase portrait's classification is dependent on the position of this point relative to the curves \( T^2 = 4D \) (parabola) and \( D = 0 \) (line). Calculate \( T^2 - 4D = (-4)^2 - 4(20) = 16 - 80 = -64 \), which is less than zero, indicating a spiral (as it's inside the parabola). Furthermore, since \( T < 0 \), it is a stable spiral point.
04

Sketch the Phase Portrait

Based on the analysis, the system demonstrates a stable spiral center. The eigenvalues of such a point will be complex with a negative real part, demonstrating trajectories that spiral inward towards the origin. On a sketch, draw trajectories that curve toward the equilibrium point (origin) in a spiral fashion, indicating the attracting nature of the fixed point.
05

Verify Using Numerical Solver

Use numerical software (e.g., Python, MATLAB) to solve the system \( \mathbf{y}' = A\mathbf{y} \) with \( A = \begin{pmatrix} 2 & -4 \ 8 & -6 \end{pmatrix} \). Simulate trajectories starting from various initial conditions and observe that they spiral inwards toward the origin. This will confirm that the equilibrium point is indeed a stable spiral as determined earlier in the Trace-Determinant analysis.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Trace-Determinant Plane
The trace-determinant plane is a powerful tool for analyzing the stability and type of equilibrium points in a dynamical system. In this system, the trace \( T \) is the sum of the diagonal elements of matrix \( A \), while the determinant \( D \) is calculated from the elements using \( ad - bc \). By plotting the point \((T, D)\) on this plane, you can determine the nature of the equilibrium point.
If \( T^2 - 4D < 0 \), the point lies inside the parabola \( T^2 = 4D \), indicating a spiral. Additionally, if \( T < 0 \), the spiral is stable, meaning trajectories spiral inwards.
Equilibrium Points
Equilibrium points are critical where the system's state doesn't change over time. For the system \( \mathbf{y}' = A \mathbf{y} \), the origin \((0, 0)\) serves as the equilibrium point when \( \mathbf{y} = 0 \). Identifying the stability of this point is essential.
In our example, the negative trace \( T \) and the position of \((-4, 20)\) in the trace-determinant plane help determine that the equilibrium point is a stable spiral. This means that any small perturbations will result in the system returning to this point over time.
Phase Portraits
Phase portraits offer a visual representation of a system's trajectories over time. They depict how solutions to differential equations behave, providing insight into the system's dynamics.
For a stable spiral, like in our example, phase portraits feature trajectories spiraling towards the equilibrium point. This visually confirms that disturbances in initial conditions will lead to solutions that converge towards the origin. Drawing these portraits allows us to interpret complex dynamical behavior in an intuitive way.
Matrix Analysis
Matrix analysis involves working with matrices to understand system properties. In the context of differential equations, a matrix \( A \) dictates how the system changes.
Calculating the trace and determinant, as well as using them in the trace-determinant plane, gives us valuable information about stability and type of equilibrium points. By doing these calculations, we know if a system is stable or unstable, and whether it spirals, forms a node, or acts like a saddle point.

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Most popular questions from this chapter

Each equation has a characteristic equation possessing distinct real roots. Find the general solution of each equation. \(y^{\prime \prime \prime}-2 y^{\prime \prime}-y^{\prime}+2 y=0\)

The matrix \(A\) has complex eigenvalues. Find a fundamental set of real solutions of the system \(\mathbf{y}^{\prime}=A \mathbf{y}\). \(A=\left(\begin{array}{rr}3 & -6 \\ 3 & 5\end{array}\right)\)

Provides a general solution of \(\mathbf{y}^{\prime}=A \mathbf{y}\), for some \(A\). Without the help of a computer or a calculator, sketch the half-line solutions generated by each exponential term of the solution. Then, sketch a rough approximation of a solution in each region determined by the half-line solutions. Use arrows to indicate the direction of motion on all solutions. Classify the equilibrium point as a saddle, a nodal sink, or a nodal source. \(\mathbf{y}(t)=C_{1} e^{-t}\left(\begin{array}{r}-5 \\\ 2\end{array}\right)+C_{2} e^{2 t}\left(\begin{array}{r}-1 \\\ 4\end{array}\right)\)

Find a fundamental set of solutions for the given system. Can be done by hand, but use a computer for the rest. \(\mathbf{x}^{\prime}=\left(\begin{array}{rrrrr}-2 & 2 & -2 & 0 & -3 \\ -1 & 0 & -1 & 0 & -3 \\ 15 & -16 & -1 & 10 & 33 \\ 12 & -13 & 1 & 6 & 26 \\ -5 & 5 & 0 & -3 & -12\end{array}\right) \mathbf{x}\)

Find the general solution of the given system. Write your solution in the form $$ \mathbf{y}(t)=e^{\lambda t}\left[\left(C_{1}+C_{2} t\right) \mathbf{v}_{1}+C_{2} \mathbf{v}_{2}\right], $$ where \(\mathbf{v}_{1}\) is an eigenvector and \(\mathbf{v}_{2}\) satisfies \((A-\lambda I) \mathbf{v}_{2}=\mathbf{v}_{1}\). Without the use of a computer or a calculator, sketch the half-line solutions. Sketch exactly one solution in each region separated by the half-line solutions. Use a numerical solver to verify your result when finished. Hint: The solutions in this case want desperately to spiral but are prevented from doing so by the half-line solutions (solutions cannot cross). However, the suggestions regarding clockwise or counterclockwise rotation in the subsection on spiral sources apply nicely in this situation. \(\mathbf{y}^{\prime}=\left(\begin{array}{rr}-4 & -4 \\ 1 & 0\end{array}\right) \mathbf{y}\)
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