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Chapter 2: Problem 2

Use linear stability analysis to classify the fixed points of the following systems. If linear stability analysis fails because \(f^{\prime}\left(x^{*}\right)=0\), use a graphical argument to decide the stability. $$ \bar{x}=x(1-x)(2-x) $$

Short Answer

Expert verified
In conclusion, using linear stability analysis and graphical arguments, we classify the fixed points of the given system as: \(x^* = 0\) (semi-stable), \(x^* = 1\) (unstable), and \(x^* = 2\) (stable).

Step by step solution

01

Find the fixed points

Find the fixed points by setting the equation equal to zero: $$ x(1-x)(2-x) = 0 $$ The fixed points are x = 0, x = 1, and x = 2.
02

Perform linear stability analysis

Differentiate the function with respect to x: $$ f^{\prime}\left(x\right) = \frac{d}{dx} \left[ x(1-x)(2-x) \right] $$ Use the product rule to differentiate: $$ f^{\prime}\left(x\right) = (1-x)(2-x) + x(-1)(2-x) + x(1-x)(-1) \\ f^{\prime}\left(x\right) = -3x^2 + 4x $$ Evaluate the derivative at each fixed point: 1. For \(x^* = 0\): $$ f^{\prime}(0) = -3(0)^2 + 4(0) = 0 $$ 2. For \(x^* = 1\): $$ f^{\prime}(1) = -3(1)^2 + 4(1) = 1 $$ 3. For \(x^* = 2\): $$ f^{\prime}(2) = -3(2)^2 + 4(2) = -8 $$
03

Classify the fixed points

Here's the stability of each fixed point based on the derivative values: 1. For \(x^* = 0\): As \(f^{\prime}(0) = 0\), linear stability analysis fails. We will use a graphical argument in the next step. 2. For \(x^* = 1\): As \(f^{\prime}(1) > 0\), the fixed point \(x^* = 1\) is an unstable fixed point. 3. For \(x^* = 2\): As \(f^{\prime}(2) < 0\), the fixed point \(x^* = 2\) is a stable fixed point.
04

Graphical argument for \(x^* = 0\)

To determine the stability of \(x^* = 0\) graphically, we analyze the original function \(f(x) = x(1-x)(2-x)\) around \(x^* = 0\). 1. For \(x < 0\): Since \(f(x) = x(1-x)(2-x) < 0\), we have negative feedback, so it will move towards 0. 2. For \(x > 0\): Since \(f(x) = x(1-x)(2-x) > 0\), we have positive feedback, so it will move away from 0. By analyzing the original function graphically, we can say that the fixed point \(x^* = 0\) is semi-stable because it shows stability for \(x < 0\) and instability for \(x > 0\). In conclusion, the fixed points of the given system are \(x^* = 0\) (semi-stable), \(x^* = 1\) (unstable), and \(x^* = 2\) (stable).

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Key Concepts

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

Fixed Points
In the study of dynamics systems, fixed points are like anchors within a system鈥檚 behavior; they are the values at which the system remains at rest if not perturbed. Formally, these are the points where the derivative of the system's function equals zero, signifying no change in the system's state over time.

To find fixed points, like in our exercise, we equate the system's function to zero. In our case, \( \bar{x} = x(1-x)(2-x) \) was set to zero to find the values of \( x \) that satisfy the condition. These points, in our exercise being \( x = 0 \), \( x = 1 \) and \( x = 2 \), represent the candidates for stability or instability within the system.

Understanding and identifying fixed points is crucial, as they reveal potential steady states of dynamic systems, around which we can examine the system's behavior.
Graphical Stability Argument
When linear stability analysis isn鈥檛 conclusive, as is the case with \( x^* = 0 \) where \( f'(0) = 0 \) because the first derivative also equates to zero, we turn to a graphical stability argument. By visualizing the original function around the questionable fixed point, we can see how the system behaves when it is near this point.

For instance, if the function's graph moves upward as we move away from the fixed point on either side, the fixed point is likely unstable. Conversely, if the graph moves downward, leading back toward the fixed point, it is stable. If stability is exhibited on one side but not the other, as demonstrated around \( x^* = 0 \), we classify that fixed point as semi-stable.

Thus, the graphical stability argument serves as a compelling tool, aiding in the comprehension of a system's trajectory near critical points where the mathematical method of linear stability falters.
Nonlinear Dynamics
The field of nonlinear dynamics delves into systems where changes are not directly proportional to inputs. Such systems often exhibit complicated and rich behaviors, including the possibility of multiple fixed points, as seen in the exercise. In linear systems, stability can be straightforwardly analyzed, but nonlinear systems require more nuanced approaches, such as the linear stability analysis discussed here.

In our exercise, we observed that the behavior of the system could not be fully predicted just by looking at the fixed points鈥攈ence the need for nonlinear analysis techniques. These include linear stability analysis, to discern whether small perturbations grow or decay, and graphical methods, for a more intuitive understanding of system behaviors around fixed points.

Nonlinear dynamics, with its fixed points and stability properties, plays a pivotal role in understanding complex systems ranging from populations in biology to orbits in celestial mechanics, showcasing the essentiality of mastery in these concepts for a wide array of scientific fields.

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Most popular questions from this chapter

\(\dot{x}=1-2 \cos x\)

Analyze the following equations graphically. In each case, sketch the vector field on the real line, find all the fixed points, classify their stability, and sketch the graph of \(x(t)\) for different initial conditions. Then try for a few minutes to obtain the analytical solution for \(x(t)\); if you get stuck, don't try for too long since in several cases it's impossible to solve the equation in closed form! \(\dot{x}=4 x^{2}-16\)

(The Allee effect) For certain species of organisms, the effective growth rate \(\dot{N} / N\) is highest at intermediate \(N\). This is the called the Allee effect (Edelstein-Keshet 1988 ). For example, imagine that it is too hard to find mates when \(N\). is very small, and there is too much competition for food and other resources when \(N\) is large. a) Show that \(\dot{N} / N=r-a(N-b)^{2}\) provides an example of Allee effect, if \(r, a\). and \(b\) satisfy certain constraints, to be determined. b) Find all the fixed points of the system and classify their stability. c) Sketch the solutions \(N(t)\) for different initial conditions. d) Compare the solutions \(N(t)\) to those found for the logistic equation. What are the qualitative differences, if any?

(Autocatalysis) Consider the model chemical reaction in which one molecule of \(X\) combines with one molecule of \(A\) to form two molecules of \(X\). This means that the chemical \(X\) stimulates its own production, a process called autocatalysis. This positive feedback process leads to a chain reaction, which eventually is limited by a "back reaction" in which \(2 X\) returns to \(A+X\). According to the law of mass action of chemical kinetics, the rate of an elementary reaction is proportional to the product of the concentrations of the reactants. We denote the concentrations by lowercase letters \(x=[X]\) and \(a=[A]\). Assume that there's an enormous surplus of chemical \(A\), so that its concentration \(a\) can be regarded as constant. Then the equation for the kinetics of \(x\) is $$ \dot{x}=k_{1} a x-k_{1} x^{2} $$ where \(k_{1}\) and \(k_{-1}\) are positive parameters called rate constants. a) Find all the fixed points of this equation and classify their stability. b) Sketch the graph of \(x(t)\) for various initial values \(x_{0}\).

(The leaky bucket) The following example (Hubbard and West 1991 , p. 159 ) shows that in some physical situations, non-uniqueness is natural and obvious, not pathological. Consider a water bucket with a hole in the bottom. If you see an empty bucket with a puddle beneath it, can you figure out when the bucket was full? No, of course not! It could have finished emptying a minute ago, ten minutes ago, or whatever. The solution to the corresponding differential equation must be nonunique when integrated backwards in time. Here's a crude model of the situation. Let \(h(t)=\) height of the water remaining in the bucket at time \(t ; a=\) area of the hole; \(A=\) cross- sectional area of the bucket (assumed constant); \(v(t)=\) velocity of the water passing through the hole. a) Show that \(a v(t)=A \hat{h}(t)\). What physical law are you invoking? b) To derive an additional equation, use conservation of energy. First, find the change in potential energy in the system, assuming that the height of the water in the bucket decreases by an amount \(\Delta h\) and that the water has density \(\rho\). Then find the kinetic energy transported out of the bucket by the escaping water. Finally, assuming all the potential energy is converted into kinetic energy, derive the equation \(v^{2}=2 g h\) c) Combining (b) and (c), show \(h=-C \sqrt{h}\), where \(C=\sqrt{2 g}\left(\frac{\alpha}{A}\right)\). d) Given \(h(0)=0\) (bucket empty at \(t=0\) ), show that the solution for \(h(t)\) is nonunique in backwards time, i.e., for \(t<0\).
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