Problem 14
Question
In Problems \(11-20\), 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 node, or a saddle point. \(x^{\prime}=2 x-y^{2}\) \(y^{\prime}=-y+x y\)
Step-by-Step Solution
Verified Answer
Critical point (0,0) is a saddle point; (1, \(\sqrt{2}\)) and (1, \(-\sqrt{2}\)) are unstable spiral points.
1Step 1: Find the Critical Points
To find the critical points, set the derivatives equal to zero:1. \(x' = 2x - y^2 = 0\)2. \(y' = -y + xy = 0\)Solving these equations simultaneously:- From equation (2), \(y(-1 + x) = 0\), giving us critical points where either \(y = 0\) or \(x = 1\).For \(y = 0\): Substitute into equation (1):\(2x - 0 = 0 \rightarrow x = 0\).This gives a critical point at \((0,0)\).For \(x = 1\): Substitute into equation (1):\(2(1) - y^2 = 0\rightarrow y^2 = 2\rightarrow y = \pm \sqrt{2}\).This gives critical points at \((1, \sqrt{2})\) and \((1, -\sqrt{2})\).
2Step 2: Compute the Jacobian Matrix
Find the Jacobian matrix for the system:\[J(x, y) = \begin{bmatrix}\frac{\partial (2x - y^2)}{\partial x} & \frac{\partial (2x - y^2)}{\partial y} \\frac{\partial (-y + xy)}{\partial x} & \frac{\partial (-y + xy)}{\partial y} \\end{bmatrix}\]Calculate each partial derivative:1. \(\frac{\partial (2x - y^2)}{\partial x} = 2\)2. \(\frac{\partial (2x - y^2)}{\partial y} = -2y\)3. \(\frac{\partial (-y + xy)}{\partial x} = y\)4. \(\frac{\partial (-y + xy)}{\partial y} = -1 + x\)Thus the Jacobian is:\[J(x, y) = \begin{bmatrix}2 & -2y \y & -1 + x \\end{bmatrix}\]
3Step 3: Evaluate the Jacobian at Each Critical Point
Compute the Jacobian for each critical point. For \((0,0)\):\[J(0, 0) = \begin{bmatrix}2 & 0 \0 & -1 \\end{bmatrix}\]For \((1, \sqrt{2})\):\[J(1, \sqrt{2}) = \begin{bmatrix}2 & -2\sqrt{2} \\sqrt{2} & 0 \\end{bmatrix}\]For \((1, -\sqrt{2})\):\[J(1, -\sqrt{2}) = \begin{bmatrix}2 & 2\sqrt{2} \-\sqrt{2} & 0 \\end{bmatrix}\]
4Step 4: Analyze the Stability Using Eigenvalues
Calculate eigenvalues for each evaluated Jacobian, and use them to determine the nature of each critical point:For \((0,0)\):- Eigenvalues of \[\begin{bmatrix} 2 & 0 \ 0 & -1 \end{bmatrix}\] are \(\lambda_1 = 2\), \(\lambda_2 = -1\).- Since there's one positive and one negative eigenvalue, \((0,0)\) is a **saddle point**.For \((1, \sqrt{2})\) and \((1, -\sqrt{2})\):- The characteristic polynomial, \(\lambda^2 - (2 - 1 + 0)\lambda + (2 \times 0 - (-2\sqrt{2})(\sqrt{2})) = \lambda^2 - \lambda - 4 = 0\), has a negative discriminant, yielding complex eigenvalues with a positive real part. - Both points are **unstable spiral points**.
Key Concepts
Critical PointsJacobian MatrixStability AnalysisEigenvalues
Critical Points
In a dynamical system, critical points are locations in the phase space where the system's behavior changes significantly. These points occur where the derivative of the system's equations equals zero. This means the rates of change for the system variables temporarily halt, leading into a potential state transition. For our given system of equations:
Identifying these points is crucial since they often indicate where dynamic behaviors such as stability and convergence occur. They provide the anchors in the analysis, ventures from which reveal different phases of the system's behavior.
- \( x' = 2x - y^2 = 0 \)
- \( y' = -y + xy = 0 \)
Identifying these points is crucial since they often indicate where dynamic behaviors such as stability and convergence occur. They provide the anchors in the analysis, ventures from which reveal different phases of the system's behavior.
Jacobian Matrix
The Jacobian matrix is a fundamental tool in the analysis of nonlinear dynamical systems. It provides a linear approximation of a system around a critical point by considering all first-order partial derivatives of the system's functions. For our system:
The Jacobian not only aids in stability analysis but also simplifies solving complex systems through linearization, making it indispensable for analyzing nonlinear systems.
- \( 2x - y^2 \)
- \( -y + xy \)
The Jacobian not only aids in stability analysis but also simplifies solving complex systems through linearization, making it indispensable for analyzing nonlinear systems.
Stability Analysis
Stability analysis involves determining the behavior of solutions near the critical points. By analyzing the system's response to small disturbances, we determine if the system tends to return to the critical point (stable) or diverges away (unstable). The primary method for performing stability analysis is using the Jacobian matrix at critical points.
For the critical points in the example, evaluating the Jacobian helps determine their nature:
For the critical points in the example, evaluating the Jacobian helps determine their nature:
- At \((0, 0)\), the Jacobian reveals a saddle point due to the presence of both positive and negative eigenvalues.
- At \((1, \sqrt{2})\) and \((1, -\sqrt{2})\), the Jacobian indicates unstable spiral points due to complex eigenvalues with a positive real part.
Eigenvalues
Eigenvalues are pivotal in assessing the stability of a critical point. When derived from the Jacobian matrix, they indicate how perturbations behave. The real part of an eigenvalue determines stability:
Understanding eigenvalues is crucial as they directly express a system's tendency to either stabilize or diverge from its critical points, guiding both theoretical analysis and practical applications.
- Positive real part: The point is unstable; disturbances grow.
- Negative real part: The point is stable; disturbances shrink.
- Zero real part: Linear stability tests are inconclusive, requiring further analysis.
Understanding eigenvalues is crucial as they directly express a system's tendency to either stabilize or diverge from its critical points, guiding both theoretical analysis and practical applications.
Other exercises in this chapter
Problem 14
In Problems, find a circular invariant region for the given plane autonomous system. $$ \begin{aligned} &x^{\prime}=-y-x e^{x+y} \\ &y^{\prime}=x-y e^{x+y} \end
View solution Problem 14
Classify the critical point \((0,0)\) of the given linear system by computing the trace \(\tau\) and determinant \(\Delta\) and using Figure \(11.2 .12 .\) $$ \
View solution Problem 14
In Problems, find all critical points of the given plane autonomous system. $$ \begin{aligned} &x^{\prime}=\sin y \\ &y^{\prime}=e^{x-y}-1 \end{aligned} $$
View solution Problem 14
Find and classify (if possible) the critical points of the plane autonomous system $$ \begin{aligned} &x^{\prime}=x+x y-3 x^{2} \\ &y^{\prime}=4 y-2 x y-y^{2} \
View solution