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Consider the chaotic motion of a DDP for which the Liapunov exponent is \(\lambda=1\), with time measured in units of the drive period as usual. (This is very roughly the value found in Problem 12.13 .) (a) Suppose that you need to predict \(\phi(t)\) with an accuracy of \(1 / 100\) rad and that you know the initial value \(\phi(0)\) within \(10^{-6}\) rad. What is the maximum time \(t_{\text {max }}\) for which you can predict \(\phi(t)\) within the required accuracy? This \(t_{\max }\) is sometimes called the time horizon for prediction within a specified accuracy. (b) Suppose that, with a vast expenditure of money and labor, you manage to improve the accuracy of your initial value to \(10^{-9}\) radians (a thousand-fold improvement). What is the time horizon now (for the same required accuracy of prediction)? By what factor has \(t_{\text {max }}\) improved? Your results illustrate the difficulty of making accurate long-term predictions for chaotic motion.

Short Answer

Expert verified
(a) \( t_{\max} = 9.2 \); (b) \( t_{\max} = 16.1 \); Improvement by a factor of 1.75.

Step by step solution

01

Understand the Relationship Between Liapunov Exponent and Predictability

The Liapunov exponent \( \lambda \) measures how quickly trajectories diverge in a chaotic system. If the initial difference is \( \epsilon_0 \), the difference after time \( t \) is approximated by \( \epsilon(t) = \epsilon_0 e^{\lambda t} \). To find how long you can predict accurately, solve for \( t \) when \( \epsilon(t) = 1/100 \) such that \( t = \frac{1}{\lambda} \ln\left(\frac{1/100}{\epsilon_0}\right) \).
02

Calculate Time Horizon for Initial Accuracy \( \epsilon_0 = 10^{-6} \)

Plug \( \epsilon_0 = 10^{-6} \), \( \epsilon(t) = 1/100 \), and \( \lambda = 1 \) into the formula: \[ t = \frac{1}{1} \ln\left(\frac{1/100}{10^{-6}}\right) = \ln(10^4) = 4 \ln(10) = 4 \times 2.302 = 9.208 \]Round to get \( t_{\text{max}} = 9.2 \) units of time.
03

Calculate Time Horizon for Improved Accuracy \( \epsilon_0 = 10^{-9} \)

Adjust the initial accuracy to \( 10^{-9} \), and use the same target accuracy \( 1/100 \): \[ t = \ln\left(\frac{1/100}{10^{-9}}\right) = \ln(10^7) = 7 \ln(10) = 7 \times 2.302 = 16.114 \]Thus, \( t_{\text{max}} = 16.1 \) units of time.
04

Calculate the Improvement Factor

The improvement factor in time horizon is the ratio of the new \( t_{\text{max}} \) to the old \( t_{\text{max}} \): \[ \text{Improvement Factor} = \frac{16.1}{9.2} \approx 1.75 \]This means the time horizon has improved by about a factor of 1.75.

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Key Concepts

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

Liapunov Exponent
The Liapunov exponent is a crucial concept when studying chaotic systems. It quantifies how quickly two nearby trajectories in a system diverge over time. Essentially, it measures the sensitivity of the system to initial conditions. A positive Liapunov exponent indicates a chaotic system, where small changes can lead to vastly different outcomes. For example, in chaotic motion, such as the weather, small measurement errors can lead to significantly different predictions. This sensitivity is captured by the Liapunov exponent, denoted as \( \lambda \). If you start with an initial difference \( \epsilon_0 \), the difference after time \( t \) becomes \( \epsilon(t) = \epsilon_0 e^{\lambda t} \). This exponential growth in difference implies that even tiny inaccuracies in initial conditions magnify rapidly, making long-term predictions challenging.
Predictability in Chaos
Chaotic systems are inherently unpredictable over the long term due to their sensitive dependency on initial conditions. Even with accurate initial data, predicting future states becomes increasingly difficult as time progresses. This is where the concept of predictability in chaotic systems plays a vital role. In such systems, despite knowing the governing equations very well, slight errors in measuring initial conditions lead to large deviations over time. This is due to the system's exponential sensitivity that the Liapunov exponent quantifies. Therefore, while chaotic systems exhibit short-term predictability, their long-term behavior becomes increasingly elusive, resembling randomness.
Time Horizon for Prediction
The time horizon for prediction refers to the maximum duration over which predictions made about a chaotic system remain accurate. Beyond this time, predictions lose reliability due to the rapid divergence described by the Liapunov exponent. Calculating the time horizon involves determining when the error in prediction surpasses a desired accuracy level. For example, if the required prediction accuracy is \( \frac{1}{100} \) rad and the initial accuracy is \( 10^{-6} \) rad, we use the formula \( t = \frac{1}{\lambda} \ln\left(\frac{1/100}{\epsilon_0}\right) \) to find \( t_{\text{max}} \). This calculation gives us a clear understanding of how gaining more initial accuracy influences the length of reliable predictions.
Initial Accuracy in Chaotic Systems
The accuracy with which we know the initial conditions of a chaotic system significantly impacts the duration of reliable prediction, often termed the time horizon. In chaotic environments, improving the initial accuracy can extend the time over which the system is predictable.For instance, enhancing the initial measurement from \( 10^{-6} \) rad to \( 10^{-9} \) rad increases the prediction time horizon from 9.2 units to 16.1 units of time. This refinement effectively illustrates the potential gain in predictive ability with better initial data, despite the chaotic nature of the system. However, it shows diminishing returns over time, as reaching perfect accuracy is impractical. The need for vastly increased resources and effort to improve accuracy by a small factor underscores the persistent challenges presented by chaos.

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Most popular questions from this chapter

Consider an inhomogeneous second-order linear equation of the form $$ p(t) \ddot{x}(t)+q(t) \dot{x}(t)+r(t) x(t)=f(t) $$ Let \(x_{\mathrm{p}}(t)\) denote a solution (a "particular" solution) of this equation and prove that any solution \(x(t)\) can be written as $$ x(t)=x_{\mathrm{p}}(t)+a_{1} x_{1}(t)+a_{2} x_{2}(t) $$ where \(x_{1}(t)\) and \(x_{2}(t)\) are two independent solutions of the corresponding homogeneous equation \(-\) that is, (12.59) with \(f(t)\) deleted. [Hint: Write down the equations for \(x(t)\) and \(x_{\mathrm{p}}(t)\) and subtract.] This result shows that to find all solutions of \((12.59),\) we have only to find one particular solution and two independent solutions of the corresponding homogeneous equation. (b) Explain clearly why the result you proved in part (a) is not, in general, true for a nonlinear equation such as $$ p(t) \ddot{x}(t)+q(t) \dot{x}(t)+r(t) \sqrt{x(t)}=f(t) $$

Stars indicate the approximate level of difficulty, from easiest ( \(*\) ) to most difficult \((\star \star \star)\). Warning: Even when the motion is nonchaotic, it can be very sensitive to tiny errors. In several of the computer problems you may need to increase your working precision to get satisfactory results. Consider the nonlinear first-order equation \(\dot{x}=2 \sqrt{x-1}\). (a) By separating variables, find a solution \(x_{1}(t)\). (b) Your solution should contain one constant of integration \(k\), so you might reasonably expect it to be the general solution. Show, however, that there is another solution, \(x_{2}(t)=1,\) that is not of the form of \(x_{1}(t)\) whatever the value of \(k\). (c) Show that although \(x_{1}(t)\) and \(x_{2}(t)\) are solutions, neither \(A x_{1}(t),\) nor \(B x_{2}(t),\) nor \(x_{1}(t)+x_{2}(t)\) are solutions. (That is, the superposition principle does not apply to this equation.)

[Computer] Make a bifurcation diagram for the logistic map, in the style of Figure 12.41 but for the range \(0 \leq r \leq 3.55 .\) Take \(x_{0}=0.1 .\) Comment on its main features. [Hint: Start by using a very small number of points, perhaps just \(r\) going from 0 to 3.5 in steps of 0.5 and \(t\) going from 51 to 54. This will let you calculate for each of the values of \(r\) individually, and get the feel of how things work. To make a good diagram, you will then need to increase the number of points ( \(r\) going from 0 to 3.55 in steps of \(0.025,\) and \(t\) from 51 to \(60,\) perhaps), and you will certainly need to automate the calculation of the large number of points.]

Here is a different example of the disagreeable things that can happen with nonlinear equations. Consider the nonlinear equation \(\dot{x}=2 \sqrt{x} .\) since this is first-order, one would expect that specification of \(x(0)\) would determine a unique solution. Show that for this equation there are two different solutions, both satisfying the initial condition \(x(0)=0 .\) [Hint: Find one solution \(x_{1}(t)\) by separating variables, but note that \(x_{2}(t)=0\) is another. Fortunately none of the equations normally encountered in classical mechanics suffer from this disagreeable ambiguity.]

Here is an iterated map that is easily studied with the help of your calculator: Let \(x_{t+1}=f\left(x_{t}\right)\) where \(f(x)=\cos (x) .\) If you choose any value for \(x_{0},\) you can find \(x_{1}, x_{2}, x_{3}, \cdots\) by simply pressing the cosine button on your calculator over and over again. (Be sure the calculator is in radians mode.) (a) Try this for several different choices of \(x_{0}\), finding the first 30 or so values of \(x_{t}\). Describe what happens. (b) You should have found that there seems to be a single fixed attractor. What is it? Explain it, by examining (graphically, for instance) the equation for a fixed point \(f\left(x^{*}\right)=x^{*}\) and applying our test for stability [namely, that a fixed point \(\left.x^{*} \text { is stable if }\left|f^{\prime}\left(x^{*}\right)\right|<1\right].\)

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