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Chapter 2: Problem 4

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. $$ \dot{x}=x^{2}(6-x) $$

Short Answer

Expert verified
The fixed points of the given system are \(x=0\) and \(x=6\). Using linear stability analysis, both fixed points have a derivative of zero. A graphical argument is utilized to analyze their stability. The fixed point \(x=0\) is unstable, as nearby values of \(x\) move away from it, and the fixed point \(x=6\) is stable, as nearby values of \(x\) move toward it.

Step by step solution

01

1. Find the fixed points

To find the fixed points, set the rate of change equal to zero: \[\dot{x} = x^{2}(6-x) = 0\] The fixed points are obtained by solving this equation: \[x^{2}(6-x) = 0\] The equation has 3 solutions: \(x=0\), \(x=6\). Therefore, these are the fixed points of the given system.
02

2. Linear stability analysis

To perform linear stability analysis, we need to find the derivative of \(\dot{x}\) with respect to x, and then evaluate it at each fixed point: \[\frac{d}{dx}\left(x^{2}(6-x)\right) = 12x - 6x^2\] Now, we will substitute the fixed points and check the results: At \(x=0\), the derivative is: \[\frac{d}{dx}\left(x^{2}(6-x)\right)\Bigg|_{x=0} = 0\] At \(x=6\) the derivative is: \[\frac{d}{dx}\left(x^{2}(6-x)\right)\Bigg|_{x=6} = 0\] Both fixed points have a derivative of zero, so linear stability analysis does not provide enough information to classify their stability. We will now use a graphical argument to classify the stability.
03

3. Graphical argument for fixed points

Since the analytic method doesn't work, we can use the reverse process to graph the function and verify the stability of the fixed points. The function is: \[\dot{x} = x^{2}(6-x)\] If we plot this function, we can analyze its behavior around the fixed points: - When \(00\) and \((6-x)>0\). So for values close to \(x=0\), 饾懃 will increase and go away from the fixed point. - When \(x>6\), the function is negative, as \(x^{2}>0\) and \((6-x)<0\). So for values close to \(x=6\), 饾懃 will decrease and go closer to the fixed point. Thus, from the graphical argument, we can determine: - The fixed point \(x=0\) is unstable because nearby values of x move away from it. - The fixed point \(x=6\) is stable because nearby values of x move toward it. In conclusion, using linear stability analysis and a graphical argument, we have classified the fixed points of the given system as \(x=0\) being unstable and \(x=6\) being stable.

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Key Concepts

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

Fixed Points
Fixed points in a dynamical system are pivotal for understanding its behavior. These are the points where the rate of change becomes zero, essentially where the system doesn't evolve further from. In the exercise, finding the fixed points involved solving the equation \(\dot{x} = x^{2}(6-x) = 0\), which gives us the values of \(x\) where the system is at rest.

Once identified, these fixed points act as critical markers. They can be thought of as 'equilibrium states'. If a system is at a fixed point, it will remain there unless perturbed by an external force. However, the nature of these fixed points鈥攚hether they attract or repel nearby points鈥攇ives profound insights into the long-term behavior of the system, making their study essential in dynamical analysis.
Dynamical Systems
Dynamical systems are mathematical objects used to model processes that evolve over time. These can be populations, chemical reactions, or even the orbit of planets. The defining characteristic of a dynamical system is how it changes with time, which is described by differential equations such as \(\dot{x}=x^{2}(6-x)\) in our exercise.

To analyze these systems, we look at the changes over tiny bits of time to predict future states from current conditions. Tools used in such analysis vary, but the main goal is always to understand the long-term behavior of the system and how sensitive it is to initial conditions. This feeds into forecasting and helps in designing controls or interventions for the system.
Stability Classification
Stability classification is vital in grasping the future behavior of fixed points in a dynamical system. Using techniques such as linear stability analysis, we can classify fixed points as stable, unstable, or neutrally stable. The results give us a clue about whether a system will return to equilibrium after a small disturbance or if it will diverge away.

In the context of the given exercise, we saw linear stability analysis fail as the derivative of \(\dot{x}\) at the fixed points was zero. This called for an alternative approach, using a graphical argument to ascertain stability. Through plotting and analyzing the behavior of \(\dot{x}\) around the fixed points, we concluded that \(x=0\) is unstable and \(x=6\) is stable. This highlights the importance of having multiple methods at our disposal for stability classification, as certain systems may defy straightforward analysis.

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Most popular questions from this chapter

(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 \(t<0\).

(A mechanical analog) a) Find a mechanical system that is approximately governed by \(\dot{x}=\sin x\). b) Using your physical intuition, explain why it now becomes obvious that \(x^{*}=0\) is an unstable fixed point and \(x^{*}=\pi\) is stable.

(Error estimate for Runge-Kutta) Show that the Runge-Kutta method produces a local errot of size \(O\left(\Delta t^{5}\right)\). (Warning: This calculation involves massive amounts of algebra, but if you do it correctly, you'll be rewarded by seeing many wonderful cancellations. Teach yourself Mathematica, Maple, or some other symbolic manipulation language, and do the problem on the computer.)

(Exact solution of \(\dot{x}=\sin x)\) As shown in the text, \(\dot{x}=\sin x\) has the solution \(t=\ln \left|\left(\csc x_{0}+\cot x_{0}\right) /(\csc x+\cot x)\right|\), where \(x_{0}=x(0)\) is the initial value of \(x\). a) Given the specific initial condition \(x_{0}=\pi / 4\), show that the solution above can be inverted to obtain $$ x(t)=2 \tan ^{-1}\left(\frac{e^{t}}{1+\sqrt{2}}\right) $$ Conclude that \(x(t) \rightarrow \pi\) as \(t \rightarrow \infty\), as claimed in Section 2.1. (You need to be good with trigonomctric identities to solve this problem.) b) Try to find the analytical solution for \(x(t)\), given an arbitrary initial condition \(x_{0}\)

(Analytically intractable problem) Consider the initial value problem \(\dot{x}=x+e^{-x}, x(0)=0 .\) In contrast to Exercise \(2.8 .3\), this problem can't be solved analytically. a) Sketch the solution \(x(t)\) for \(t \geq 0\). b) Using some analytical arguments, obtain rigorous bounds on the value of \(x\) at \(t=1 .\) In other words, prove that \(a
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