Problem 11
Question
Classify the critical point \((0,0)\) of the given linear system by computing the trace \(\tau\) and determinant \(\Delta\) and using Figure 11.2.12. $$ \begin{aligned} &x^{\prime}=-5 x+3 y \\ &y^{\prime}=-2 x+5 y \end{aligned} $$
Step-by-Step Solution
Verified Answer
The critical point \((0,0)\) is a saddle point.
1Step 1: Identify the Linear System Coefficients
To classify the critical point, we start by identifying the coefficients from the given linear system. The system can be written in matrix form as \( \mathbf{X}' = A \mathbf{X} \), where \( A = \begin{pmatrix} a & b \ c & d \end{pmatrix} \). For the given system, \( a = -5 \), \( b = 3 \), \( c = -2 \), and \( d = 5 \).
2Step 2: Compute the Trace \(\tau\) of the Matrix
The trace \( \tau \) of the matrix \( A \) is the sum of the diagonal elements. Therefore, \( \tau = a + d = -5 + 5 = 0 \).
3Step 3: Compute the Determinant \(\Delta\) of the Matrix
The determinant \( \Delta \) of matrix \( A \) is calculated by \( \Delta = ad - bc \). Substitute the values we have: \[ \Delta = (-5)(5) - (3)(-2) = -25 + 6 = -19 \].
4Step 4: Classify the Critical Point Using \(\tau\) and \(\Delta\)
With the computed trace \( \tau = 0 \) and determinant \( \Delta = -19 \), we refer to Figure 11.2.12 in the textbook, which suggests that when \( \Delta < 0 \), the critical point \((0,0)\) is a saddle point.
Key Concepts
Critical Points in Differential EquationsMatrix Trace and DeterminantSaddle Point Classification
Critical Points in Differential Equations
In differential equations, critical points play a significant role in understanding the behavior of systems. A critical point, often referred to as an equilibrium point, is where the system does not change; that is, the derivatives are zero at these points. For a two-dimensional system of linear differential equations, such as \[\begin{aligned}&x' = ax + by \&y' = cx + dy\end{aligned}\]all critical points can be identified by solving the system of equations: \[x' = 0\] and \[y' = 0\]. This produces a straightforward result of \[(x, y) = (0, 0)\] for homogeneous linear systems.
Understanding the behavior at these critical points involves analyzing eigenvalues derived from the system's matrix. These values help determine whether a critical point is stable, unstable, or a saddle point.
Critical points are essential in predicting the long-term behavior of the system as it approaches stability under varying initial conditions.
Understanding the behavior at these critical points involves analyzing eigenvalues derived from the system's matrix. These values help determine whether a critical point is stable, unstable, or a saddle point.
Critical points are essential in predicting the long-term behavior of the system as it approaches stability under varying initial conditions.
Matrix Trace and Determinant
When analyzing a system of linear differential equations like the one given in the original exercise, the matrix trace (\(\tau\)) and determinant (\(\Delta\)) are crucial in classifying critical points. These two metrics are calculated from the coefficients of the system represented in matrix form. For a matrix \(A = \begin{pmatrix} a & b \ c & d \end{pmatrix}\):
In contrast, the determinant is related to the volume scaling factor, offering insight into rotational properties and stability.
By analyzing \(\tau\) and \(\Delta\) together through established mathematical criteria, we can predict the behavior of the differential equation near its critical points without solving the system entirely.
- The trace \(\tau\) is computed by summing the diagonal elements: \(\tau = a + d\).
- The determinant \(\Delta\) is found by the formula \(\Delta = ad - bc\).
In contrast, the determinant is related to the volume scaling factor, offering insight into rotational properties and stability.
By analyzing \(\tau\) and \(\Delta\) together through established mathematical criteria, we can predict the behavior of the differential equation near its critical points without solving the system entirely.
Saddle Point Classification
Saddle points in linear systems are a specific type of critical point characterized by certain trace and determinant values. A saddle point represents an instability within the system, where solutions diverge from the equilibrium point along certain paths while converging along others.
In the classification of critical points, if the determinant \(\Delta\) is less than zero, like in the given exercise with a determinant of \(-19\), this indicates the presence of a saddle point. The negative determinant suggests that one eigenvalue is positive while the other is negative, confirming an unstable system.
A saddle point is significant because it indicates that the system's trajectories exhibit different behaviors based on their initial conditions. This classification aids in understanding how systems transition through phases and approach equilibrium.
In the classification of critical points, if the determinant \(\Delta\) is less than zero, like in the given exercise with a determinant of \(-19\), this indicates the presence of a saddle point. The negative determinant suggests that one eigenvalue is positive while the other is negative, confirming an unstable system.
A saddle point is significant because it indicates that the system's trajectories exhibit different behaviors based on their initial conditions. This classification aids in understanding how systems transition through phases and approach equilibrium.
Other exercises in this chapter
Problem 11
Show that the plane autonomous system $$ \begin{aligned} &x^{\prime}=x\left(1-x^{2}-3 y^{2}\right) \\ &y^{\prime}=y\left(3-x^{2}-3 y^{2}\right) \end{aligned} $$
View solution Problem 11
Classify (if possible) each critical point of the given plane autonomous system as a stable node, a stable spiral point, an unstable spiral point, an unstable n
View solution Problem 11
Find all critical points of the given plane autonomous system. $$ \begin{aligned} &x^{\prime}=x\left(10-x-\frac{1}{2} y\right) \\ &y^{\prime}=y(16-y-x) \end{ali
View solution Problem 12
Classify the critical point \((0,0)\) of the given linear system by computing the trace \(\tau\) and determinant \(\Delta\) and using Figure \(11.2 .12 .\) $$ \
View solution