Question

a. Sketch the nullclines and describe how the critical points move as \(\alpha\) increases. b. Find the critical points. c. Let \(\alpha=2\). Classify each critical point by investigating the corresponding approximate linear system. Draw a phase portrait in a rectangle containing the critical points. d. Find the bifurcation point \(\alpha_{0}\) at which the critical points coincide. Locate this critical point, and find the eigenvalues of the approximate linear system. Draw a phase portrait. e. For \(\alpha>\alpha_{0},\) there are no critical points. Choose such a value of \(\alpha\) and draw a phase portrait. $$x^{\prime}=-\alpha-x+y, \quad y^{\prime}=-4 x+y+x^{2}$$

Short answer

2. What are the critical points when \(\alpha=2\), and how can you classify them? 3. What is the bifurcation point \(\alpha_{0}\), and what is the critical point at this bifurcation? 4. How can you describe the phase portrait for \(\alpha>\alpha_{0}\)?

Step-by-step solution

Part a: Sketch the nullclines and describe how the critical points move as \(\alpha\) increases

To find the nullclines, we set each equation to zero and solve for \(x\) and \(y\). First, for the \(x\)-nullcline, set \(x^{\prime} = 0\): $$ -\alpha - x + y = 0 \Rightarrow y = x + \alpha $$ The \(x\)-nullcline is a straight line with slope 1 and the \(y\)-intercept at \(\alpha\). Next, for the \(y\)-nullcline, set \(y^{\prime} = 0\): $$ -4x + y + x^{2} = 0 \Rightarrow y = 4x - x^{2} $$ The \(y\)-nullcline is a downward-facing parabola with vertex at \((2, 4)\). As \(\alpha\) increases, the \(x\)-nullcline shifts vertically upwards. The critical points, which are the intersections of the nullclines, will move accordingly.

Part b: Find the critical points

To find the critical points, we solve the nullcline equations simultaneously: $$ x + \alpha = 4x - x^{2} $$ Rearrange the equation to find \(x\): $$ x^2 - 3x + \alpha = 0 $$ This is a quadratic equation for \(x\), and we can use the quadratic formula to find the values of \(x\) as a function of \(\alpha\): $$ x = \frac{3 \pm \sqrt{9 - 4\alpha}}{2} $$ Since for each value of x there is a corresponding y value, substitute x to find y as a function of \(\alpha\): $$ y = \frac{3 \pm \sqrt{9 - 4\alpha}}{2} + \alpha $$ Now we have the critical points \((x, y)\) as a function of \(\alpha\).

Part c: Classify each critical point with \(\alpha=2\) and draw a phase portrait

When \(\alpha=2\), we calculate the critical points by plugging it into our expressions for \(x\) and \(y\): $$ x = \frac{3 \pm \sqrt{1}}{2} = 1, 2 \\ y = \frac{3 \pm \sqrt{1}}{2} + 2 = 3, 4 $$ So, we have two critical points: \((1,3)\) and \((2,4)\). Next, we linearize the system by finding the Jacobian matrix: $$ J(x,y) = \begin{bmatrix} \frac{\partial x^{\prime}}{\partial x} & \frac{\partial x^{\prime}}{\partial y} \\ \frac{\partial y^{\prime}}{\partial x} & \frac{\partial y^{\prime}}{\partial y} \end{bmatrix} = \begin{bmatrix} -1 & 1 \\ 2x - 4 & 1 \end{bmatrix} $$ Now, evaluate the Jacobian matrix at the critical points \((1,3)\) and \((2,4)\): At \((1,3)\): $$ J(1,3) = \begin{bmatrix} -1 & 1 \\ -2 & 1 \end{bmatrix} $$ The eigenvalues of this matrix are \(-1 \pm \sqrt{2}\), which are both negative. Hence, the critical point \((1,3)\) is a stable node. At \((2,4)\): $$ J(2,4) = \begin{bmatrix} -1 & 1 \\ 0 & 1 \end{bmatrix} $$ The eigenvalues of this matrix are \(-1\) and \(1\), which indicates that the critical point \((2,4)\) is a saddle point. We can now draw the phase portrait of the system with nullclines, critical points, and their classifications.

Part d: Find the bifurcation point \(\alpha_{0}\) and draw a phase portrait

Bifurcation occurs when the discriminant of the quadratic equation for \(x\) is equal to zero: $$ 9 - 4\alpha_{0} = 0 \Rightarrow \alpha_{0} = \frac{9}{4} $$ At the bifurcation point \(\alpha_{0}=\frac{9}{4}\), there is only one critical point \((x,y)\). Plug in the value of \(\alpha\): $$ x = \frac{3}{2} \\ y = \frac{3}{2} + \frac{9}{4} = \frac{15}{4} $$ So the critical point at the bifurcation point is \((\frac{3}{2}, \frac{15}{4})\). Next, calculate the eigenvalues of the Jacobian matrix at the bifurcation point: $$ J(\frac{3}{2}, \frac{15}{4}) = \begin{bmatrix} -1 & 1 \\ -1 & 1 \end{bmatrix} $$ The eigenvalues are \(0\) and \(2\). Since one of the eigenvalues is zero, the linear approximation fails at the bifurcation point. We can now draw the phase portrait for the system with nullclines and the critical point at the bifurcation point.

Part e: Draw a phase portrait for \(\alpha>\alpha_{0}\)

For \(\alpha>\alpha_{0}\), no critical points exist. Choose a value for \(\alpha\) greater than \(\frac{9}{4}\), such as \(\alpha=3\). With this value, we can draw the phase portrait with the nullclines but no critical points. In conclusion, this exercise demonstrates how to analyze a given system of equations, find and classify critical points with different values of the parameter \(\alpha\), and investigate the bifurcation point. The phase portraits help visualize the system's behavior for various values of \(\alpha\).

Key concepts

Nullclines

Nullclines are a fundamental concept used to analyze systems of differential equations. These are the curves in the phase plane along which one of the variables in the system does not change - its derivative with respect to time is zero. Specifically, for a two-dimensional system like \[ x' = f(x, y) \]\[ y' = g(x, y) \],the x-nullcline is the set where \( x' = f(x, y) = 0 \) and the y-nullcline is where \( y' = g(x, y) = 0 \). They are invaluable for understanding the behavior of the system and predicting the locations of critical points, which are the points where the nullclines intersect.

As seen in the exercise, identifying the x- and y-nullclines can be done by setting \( x' \) and \( y' \) to zero and solving them as regular algebraic equations. The x-nullcline is a straight line, while the y-nullcline is a parabola; as parameters change, their relative positions shift, leading to movement of the critical points.

Bifurcation Point

Understanding Bifurcation

Bifurcation points are pivotal moments in the study of dynamical systems represented by differential equations. These are parameter values at which the qualitative nature of the system changes, leading to different long-term behaviors. Particularly, a bifurcation point is where the number or stability of critical points alters. For example, as a parameter crosses a bifurcation point, a stable point might become unstable, or two critical points may merge into one and then disappear.

In our exercise, by determining the bifurcation point \( \alpha_0 \), we see how the system's behavior transitions. At \( \alpha < \alpha_0 \), the system has distinct critical points, but at \( \alpha = \alpha_0 \), these points collide, and for \( \alpha > \alpha_0 \), there are no critical points. This observation is crucial for predicting system responses to parameter changes.

Phase Portrait

Visualizing System Dynamics

A phase portrait is a visual representation of a dynamical system in the phase plane. It consists of trajectories that represent possible states the system can be in over time. Phase portraits are incredibly useful for illustrating how a system's state evolves, especially near critical points.

In the context of our exercise, drawing phase portraits allows us to see not only the position of critical points for a given value of \( \alpha \) but also to observe the general flow of the system around these points. The portraits visually communicate stability (incoming trajectories) and instability (outgoing trajectories). Importantly, when \( \alpha > \alpha_0 \), the absence of critical points is clearly highlighted in the phase portrait, indicating a fundamentally different system behavior.

Jacobian Matrix

The Role of the Jacobian

In the realm of differential equations, the Jacobian matrix is instrumental in analyzing the local behavior of nonlinear systems around critical points. It is a matrix of first-order partial derivatives and represents the linear approximation of the system. For a two-dimensional system, it is a 2x2 matrix given by the formula:\[ J(x, y) = \begin{bmatrix} \frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y} \ \frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y} \end{bmatrix} \].

The exercise provides a clear application of the Jacobian to evaluate the stability of the critical points by investigating the eigenvalues of the matrix. It demonstrates how the Jacobian allows for the classification of critical points and guides the drawing of phase portraits by indicating the types of points we have (nodes, saddles, etc.).

Eigenvalues

Determining Stability and Behavior

Eigenvalues play a critical role in understanding the behavior of linearized systems near critical points. These values come from solving the characteristic equation of the Jacobian matrix. They offer insights into the nature and stability of critical points: positive eigenvalues suggest instability (sources), negative eigenvalues imply stability (sinks), and complex eigenvalues (with negative real parts) indicate a possibility of oscillatory behavior (spirals).

As shown in the exercise solution, the eigenvalues \( -1 \pm \sqrt{2} \) and \( -1 \) and \( 1 \) were used to classify critical points as stable nodes or saddle points, respectively. The occurrence of an eigenvalue of zero at the bifurcation point also highlights the limitation of the linear approximation in such cases and the need for further analysis.