Problem 29 - Advanced Engineering Mathematics

Question

(a) Show that the plane autonomous system $$ \begin{aligned} &x^{\prime}=-x+y-x^{3} \\ &y^{\prime}=-x-y+y^{2} \end{aligned} $$ has two critical points by sketching the graphs of \(-x+y-x^{3}=0\) and \(-x-y+y^{2}=0\). Classify the critical point at \((0,0)\). (b) Show that the second critical point \(\mathbf{X}_{1}=(0.88054,1.56327)\) is a saddle point.

Step-by-Step Solution

Answer

(a) Two critical points are found: (0,0) and (0.88054, 1.56327). (0,0) is a spiral point. (b) (0.88054, 1.56327) is a saddle point.

Step 1: Identify Critical Points

Critical points occur where both derivatives are equal to zero. Set the equations \(-x+y-x^3 = 0\) and \(-x-y+y^2 = 0\) to find the values of \(x\) and \(y\) that satisfy both equations simultaneously.

Step 2: Solve the First Equation

For the equation \(-x+y-x^3 = 0\), rearrange it to get \(y = x + x^3\). This represents a curve that we will later sketch.

Step 3: Solve the Second Equation

For the equation \(-x-y+y^2 = 0\), rearrange it to get \(y = \frac{1 \pm \sqrt{1 + 4x}}{2}\). This represents two curves \((y = 1 + \sqrt{1 + 4x}/2\) and \(y = 1 - \sqrt{1 + 4x}/2)\) which we will sketch to find intersections with the first equation.

Step 4: Sketch Graphs and Find Intersections

Sketch the curves from both equations: \(y = x + x^3\) and \(y = 1 \pm \sqrt{1 + 4x}/2\). Identify the intersection points of these curves to find the critical points.

Step 5: Identify and Verify Critical Points

From the sketch, we find two intersection points: 1. \((0, 0)\)2. An approximate numerical solution which is given as \((0.88054, 1.56327)\). Verify these values satisfy both equations.

Step 6: Classify Critical Point at (0, 0)

Linearize the system around \((0, 0)\). Compute the Jacobian matrix \(J\) at this point and evaluate its eigenvalues to classify the stability of the point.

Step 7: Calculate Jacobian Matrix

The Jacobian matrix is: \[ J = \begin{pmatrix} \frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y} \ \frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y} \end{pmatrix} = \begin{pmatrix} -1 - 3x^2 & 1 \ -1 & -1 + 2y \end{pmatrix} \]. Evaluate \(J\) at \((0,0)\): \[ J = \begin{pmatrix} -1 & 1 \ -1 & -1 \end{pmatrix} \].

Step 8: Compute Eigenvalues

To classify the critical point, find the eigenvalues of \(J\). Solve \(\det(J - \lambda I) = 0\) where \(I\) is the identity matrix. This yields the characteristic equation \(\lambda^2 + 2\lambda + 2 = 0\). The eigenvalues are complex: \(\lambda = -1 \pm i\). This suggests that the point \((0, 0)\) is a spiral point (focus).

Step 9: Analyze Second Critical Point

For the critical point \((0.88054, 1.56327)\), calculate the Jacobian matrix similarly. Evaluate \(J\) at this point and compute eigenvalues.

Step 10: Show Saddle Point at (0.88054, 1.56327)

Calculate the Jacobian matrix: \[ J = \begin{pmatrix} -1 - 3(0.88054)^2 & 1 \ -1 & -1 + 2(1.56327) \end{pmatrix} \]. Find the eigenvalues. If they are real and of opposite sign, it confirms a saddle point. The calculations show one positive and one negative eigenvalue, confirming a saddle point.

Key Concepts

Critical Points

Critical points are vital in understanding the behavior of autonomous systems. These points are where the derivatives of the system are zero. Simply put, a system is "stationary" at a critical point. In the given example of the autonomous system, we find critical points by setting both equations,

• \(-x+y-x^3 = 0\)
• \(-x-y+y^2 = 0\)

to zero. Solving these equations simultaneously reveals the values of \(x\) and \(y\) where the system doesn't change. Graphically, critical points are represented by the intersection of the curves derived from these equations. In our case, the critical points are \((0,0)\) and approximately \((0.88054,1.56327)\). Verifying these ensures they satisfy both equations, proving their validity in describing the system's behavior.

Jacobian Matrix

To analyze the stability of critical points, we use the Jacobian matrix. This matrix arises from linearizing the system near the critical points and contains partial derivatives of the system's functions. The Jacobian matrix \(J\) for our autonomous system is:\[J = \begin{pmatrix} \frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y} \ \frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y} \end{pmatrix}\]Evaluating the partial derivatives for the given functions results in:\[J = \begin{pmatrix} -1 - 3x^2 & 1 \ -1 & -1 + 2y \end{pmatrix}\]This expression helps us assess the dynamics at the critical point by evaluating how small changes in \(x\) and \(y\) influence the system as a whole. Plugging in the values at specific critical points gives us crucial insights into the system's structure and behavior.

Eigenvalues

Eigenvalues of the Jacobian matrix play a crucial role in determining the behavior of critical points. They are obtained by solving the characteristic equation derived from the matrix. For our example, the Jacobian at \((0,0)\) makes this calculation straightforward:\[J = \begin{pmatrix} -1 & 1 \ -1 & -1 \end{pmatrix}\]From here, we find the eigenvalues by solving the equation:\[\det(J - \lambda I) = 0\]which leads to the characteristic equation:\[\lambda^2 + 2\lambda + 2 = 0\]The solutions, \(\lambda = -1 \pm i\), indicate complex eigenvalues, suggesting that \((0,0)\) is a focus or spiral point. This means the system exhibits oscillatory behavior at this critical point.

Saddle Point

A saddle point is a type of critical point characterized by one stable and one unstable direction. This occurs when the Jacobian matrix at the critical point has eigenvalues with opposite signs. Saddle points are significant because they indicate a change in stability, making them essential in understanding system behavior. In our example, at the second critical point \((0.88054, 1.56327)\), the Jacobian matrix yields real eigenvalues as calculated:\[J = \begin{pmatrix} -1 - 3(0.88054)^2 & 1 \ -1 & -1 + 2(1.56327) \end{pmatrix}\]Solving for eigenvalues, we find one positive and one negative value, confirming this is indeed a saddle point. Such points often separate regions of different dynamical behaviors, serving as partitions in the phase space.

Stability Classification

Stability classification involves determining the type of critical point based on the eigenvalues of the Jacobian matrix. We look at whether these eigenvalues are real or complex, and their sign, to classify points into categories:

• Stable Node: All real and negative eigenvalues.
• Unstable Node: All real and positive eigenvalues.
• Saddle Point: Real eigenvalues with different signs.
• Spiral (Focus): Complex eigenvalues; stability depends on the real part.

These classifications help predict how a system behaves near its critical points. In practical terms, they give insights into whether a system will settle into a steady state, spread outwards, or oscillate over time. For example, in our problem, the point \((0,0)\) is a spiral due to complex eigenvalues with negative real parts, suggesting relative stability with oscillations.