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Chapter 4: Problem 16

16\. Examine the behaviour of the fixed points of the competing species model $$ \dot{x}_{1}=\left(1-x_{1}-x_{2}\right) x_{1}, \quad \dot{x}_{2}=\left(v-x_{2}-4 v^{2} x_{1}\right) x_{2}, \quad x_{1}, x_{2}>0 $$ as \(v\) varies through positive values. Show that changes in the number and the nature of the fixed points occur at \(v=\frac{1}{4}\) and \(v=1\). Sketch typical phase portraits for \(v\) in the intervals \(\left(0, \frac{1}{4}\right),\left(\frac{1}{4}, 1\right)\) and \((1, \infty)\).

Short Answer

Expert verified
Fixed points and their stability change at \(v=\frac{1}{4}\) and \(v=1\). Phase portraits differ across these intervals.

Step by step solution

01

Identify Fixed Points

Fixed points occur where \( \dot{x}_1 = 0 \) and \( \dot{x}_2 = 0 \). Solve: \((1-x_1-x_2)x_1 = 0\) and \((v-x_2-4v^2x_1)x_2 = 0\). This gives potential fixed points: (0,0), (1,0), and (0,v).
02

Analyze Fixed Points for v < 1/4

For \(v < \frac{1}{4}\), the stability of each fixed point needs analysis. Linearize and calculate the Jacobian matrix at each fixed point to determine stability. For example, at (1,0), compute Jacobian eigenvalues to conclude stability based on determined signs.
03

Analyze Fixed Points for v = 1/4

Find conditions at \(v = \frac{1}{4}\). Jacobian analysis reveals change in stability, indicating a bifurcation - nature of fixed points changes (e.g., from stable node to saddle).
04

Analyze Fixed Points for 1/4 < v < 1

Investigate any emerging new fixed points like the intersection of \(x_2 = v\) with \((1-x_1-x_2) = 0\) and continue stability analysis with the Jacobian matrix.
05

Analyze Fixed Points for v = 1

Determine effects at \(v = 1\). New bifurcation occurs with potential fixed points such as \((0,1)\). Continue Jacobian stability analysis at all fixed points.
06

Analyze Fixed Points for v > 1

Evaluate stability and existence of fixed points for \(v > 1\). Previously identified points may disappear, new nature of dynamically stable points identified.
07

Sketch Phase Portraits

For each interval \((0, \frac{1}{4})\), \((\frac{1}{4}, 1)\), and \((1, \infty)\), sketch typical phase portraits indicating direction of trajectories and stability around fixed points identified in earlier steps.

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Key Concepts

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

Fixed Points
In the context of a competing species model, fixed points represent the states where the populations remain constant over time. Finding these points involves setting both differential equations, \( \dot{x}_1 = 0 \) and \( \dot{x}_2 = 0 \), as it implies that at these points, there are no changes in \( x_1 \) and \( x_2 \). For the given system:
  • \( \dot{x}_1 = (1-x_1-x_2) x_1 = 0 \)
  • \( \dot{x}_2 = (v-x_2-4v^2x_1) x_2 = 0 \)
Solving these equations provides potential fixed points such as (0,0), (1,0), and (0,v). These are the baseline solutions that determine where the populations of the species do not change, assuming \( x_1 \) and \( x_2 \) are greater than zero.
Stability Analysis
Once fixed points are identified, the next step is to determine their stability. Stability analysis assesses whether small changes in the populations lead to the system returning to a fixed point (stable) or moving away (unstable). This is done through linearization and finding the Jacobian matrix:
  • A Jacobian matrix consists of partial derivatives of the system equations with respect to the variables \( x_1 \) and \( x_2 \).
  • The eigenvalues of the Jacobian at a fixed point determine its stability. If all eigenvalues have negative real parts, the fixed point is stable.
For \( v < \frac{1}{4} \), the computation of the Jacobian matrix at each fixed point reveals their stability. The stability may change as \( v \) increases, indicating potential bifurcations.
Phase Portraits
Phase portraits visually represent the dynamics of the system by showing trajectories in the phase space, usually defined by the axes \( x_1 \) and \( x_2 \). These diagrams help in understanding how populations evolve over time under different conditions for \( v \):
  • For \( 0 < v < \frac{1}{4} \), phase portraits typically illustrate trajectories converging to stable fixed points or diverging from unstable ones.
  • When \( \frac{1}{4} < v < 1 \), additional or modified trajectories reflect changes in stability or new fixed points.
  • Beyond \( v = 1 \), trajectories and stability may shift further, often showing more complex behavior and interactions between species.
By understanding how trajectories behave near fixed points, one can predict long-term population trends.
Bifurcation Analysis
Bifurcation analysis examines how a system's behavior changes as a parameter, in this case, \( v \), is varied. Bifurcations occur when a small change in \( v \) leads to a qualitative change in the number or nature of fixed points:
  • At \( v = \frac{1}{4} \), a bifurcation point may result in existing fixed points changing stability, indicated by a shift from a stable node to a saddle point.
  • At \( v = 1 \), another bifurcation can introduce new dynamics, such as the emergence or disappearance of fixed points.
Understanding bifurcations is crucial because it helps predict how the system transitions between different regimes, thereby aiding in the management and forecast of competing species interactions.

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Most popular questions from this chapter

Investigate the fixed points of the equations of motion $$ \dot{x}_{1}=x_{2}, \quad \dot{x}_{2}=-\omega_{0}^{2} \sin x_{1} $$ of the simple pendulum where \(\omega_{0}^{2}=g / l\). Here, \(q\) is the acceleration due to gravity and \(l\) is the length of the pendulum. Find a first integral and sketch the phase portrait. Suppose a damping term \(-2 k x_{2}\), \(k>0\), is added to \(\dot{x}_{2}\). Find the nature of the fixed points of the damped system when \(k\) is small and sketch its phase portrait. Interpret both of the above phase portraits in terms of the motions of the pendulum.

Consider the prey-predator equations with 'logistic' corrections $$ \dot{x}_{1}=x_{1}\left(1-x_{2}-\alpha x_{1}\right), \quad \dot{x}_{2}=-x_{2}\left(1-x_{1}+\alpha x_{2}\right) $$ where \(0 \leqslant \alpha<1\). Show that, at the non-trivial fixed point, the centre that exists for \(\alpha=0\) changes into a stable focus for \(0<\alpha<1\). Sketch the phase portrait.

11\. In a simple model of a national economy, \(I=I-\alpha C, C=\) \(\beta(I-C-G)\), where \(I\) is the national income, \(C\) is the rate of consumer spending and \(G\) is the rate of government expenditure. The model is restricted to its natural domain \(I \geqslant 0, C \geqslant 0, G \geqslant 0\) and the constants \(\alpha\) and \(\beta\) satisfy \(1<\alpha<\infty, 1 \leqslant \beta<\infty\) (a) Show that if the rate of government expenditure \(G=G_{0}\), a constant, then there is an equilibrium state. Classify the equilibrium state when \(\beta=1\) and show that then the economy oscillates. (b) Assume government expenditure is related to the national income by the rule \(G=G_{0}+k l\), where \(k>0\). Find the upper bound \(A\) on \(k\) for which an equilibrium state exists in the natural domain of \(t\) model. Describe both the position and the behaviour of this state \(\beta>1\), as \(k\) tends to the critical value \(A\).

An age-dependent population model is given by $$ \dot{P}=-\mu(P) P+B, \quad \dot{B}=[\gamma-\mu(P)] B, \quad \gamma>0 $$ where \(P\) is the total population and \(B\) is the birth rate. Prove that \(B=\gamma P\) is a union of trajectories for all choices of the function \(\mu(P)\). Investigate the phase portrait when \(\mu(P)=b+c P\) where \(b<0\) and \(c\) \(>0\). Show that for all positive initial values of the variables, both population and birth rate stabilize at non-zero values.

30\. A model for a population which becomes susceptible to epidemics is constructed as follows. The population is originally governed by $$ \dot{p}=a p-b p^{2} $$ and grow? to a certain value \(QA / B\). The population falls to the value \(q\), where \(A / B

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