/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 21 Consider the differential equati... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider the differential equation \(d x / d t=k x-x^{3}\). (a) If \(k \leq 0\), show that the only critical value \(c=0\) of \(x\) is stable. (b) If \(k>0\), show that the critical point \(c=0\) is now unstable, but that the critical points \(c=\pm \sqrt{k}\) are stable. Thus the qualitative nature of the solutions changes at \(k=0\) as the parameter \(k\) increases, and so \(k=0\) is a bifurcation point for the differential equation with parameter \(k\). The plot of all points of the form \((k, c)\) where \(c\) is a critical point of the equation \(x^{\prime}=k x-x^{3}\) is the "pitchfork diagram" shown in Fig. \(2.2 .13 .\)

Short Answer

Expert verified
At \(k \leq 0\), \(x = 0\) is stable. At \(k > 0\), \(x = 0\) is unstable, and \(x = \pm \sqrt{k}\) are stable; \(k = 0\) is a bifurcation point.

Step by step solution

01

Find the Critical Points

First, identify the critical points of the differential equation \( \frac{dx}{dt} = kx - x^3 \). Critical points occur where \( \frac{dx}{dt} = 0 \), which gives the equation \( kx = x^3 \). Solving this equation by factoring gives \( x(x^2 - k) = 0 \), leading to critical points: \( x = 0 \) and \( x = \pm \sqrt{k} \).
02

Analyze Stability for \(k \leq 0\)

When \(k \leq 0\), the critical points \(x = \pm \sqrt{k}\) are not real numbers, leaving \(x = 0\) as the only critical point. For stability, consider the sign of the derivative of \(f(x) = kx - x^3\). The derivative is \(f'(x) = k - 3x^2\). At \(x = 0\), \(f'(0) = k \leq 0\). Since \(f'(0) \leq 0\), the critical point \(x = 0\) is stable when \(k \leq 0\).
03

Analyze Stability for \(k > 0\)

For \(k > 0\), there are three critical points: \(x = 0\) and \(x = \pm \sqrt{k}\). Consider \(f'(x) = k - 3x^2\) at each critical point. For \(x = 0\), \(f'(0) = k > 0\), indicating instability. For \(x = \pm \sqrt{k}\), \(f'(\pm \sqrt{k}) = k - 3k = -2k < 0\), indicating stability. Thus, \(x = \pm \sqrt{k}\) are stable.
04

Discuss Bifurcation

The stability analysis shows a change in behavior at \(k = 0\). For \(k \leq 0\), \(x = 0\) is stable, and there are no other critical points. For \(k > 0\), \(x = 0\) becomes unstable, and new stable critical points \(x = \pm \sqrt{k}\) appear. This change indicates a bifurcation at \(k = 0\), marked by the appearance of the 'pitchfork' structure in the bifurcation diagram as \(k\) increases past zero.

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Key Concepts

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

Stability Analysis
Stability analysis in the context of differential equations involves understanding how solutions behave when they are slightly perturbed. In the given differential equation \(\frac{dx}{dt} = kx - x^3\), stability can be determined by examining the derivative of the function at the critical points.
For a critical point \(x = c\), we analyze the derivative \(f'(x) = k - 3x^2\). If \(f'(c) < 0\), the point is stable, indicating that small perturbations will decay over time, leading the solution back to the critical point. Conversely, if \(f'(c) > 0\), the point is unstable, and small perturbations will grow over time.
  • When \(k \leq 0\): Only \(x = 0\) is considered, with \(f'(0) = k \leq 0\), confirming stability.
  • When \(k > 0\): The critical points \(x = \pm \sqrt{k}\) exhibit stability since \(f'(\pm \sqrt{k}) = -2k < 0\), while \(x = 0\) is unstable as \(f'(0) = k > 0\).
Stability analysis is crucial for predicting the long-term behavior of differential equations, allowing us to identify points of equilibrium and their nature.
Critical Points
Critical points occur where the derivative of a function equals zero, representing potential points of equilibrium in our differential equation. Solving \(kx - x^3 = 0\), involves factoring to find \(x(x^2 - k) = 0\), leading to the critical points: \(x = 0\) and \(x = \pm \sqrt{k}\). These solutions satisfy the condition \(\frac{dx}{dt} = 0\), where no change occurs over time.
For different parameters \(k\):
  • For \(k \leq 0\), only \(x = 0\) is a real critical point due to the non-reality of \(x = \pm \sqrt{k}\).
  • For \(k > 0\), both real and positive/negative \(x = \pm \sqrt{k}\) emerge, adding new dynamics to the system.
Identifying and analyzing critical points is an essential step in studying differential equations, as it highlights equilibrium solutions and potential bifurcations.
Differential Equation Analysis
Differential equation analysis involves examining how the equation \(\frac{dx}{dt} = kx - x^3\) behaves under different conditions. This helps us understand not just individual solutions, but the overall dynamics of the system.
In this situation:
  • For \(k \leq 0\): Only the critical point \(x = 0\) remains, and is stable, leading to a simple outcome where any initial condition tends towards zero.
  • For \(k > 0\): The introduction of \(x = \pm \sqrt{k}\) changes the behavior dramatically. The system can settle at one of these stable points, creating a more complex, multi-steady state system.
  • The concept of the bifurcation point \((k=0)\) arises, revealing shifts in stability and the emergence of new solutions as \(k\) changes.
Understanding differential equation dynamics through these analyses helps to capture the full spectrum of possible behaviors, guiding predictions and interpretations of long-term solution paths.

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Most popular questions from this chapter

A programmable calculator or a computer will be useful for Problems 11 through \(16 .\) In each problem find the exact solution of the given initial value problem. Then apply the RungeKutta method twice to approximate (to five decimal places) this solution on the given interval, first with step size \(h=0.2\), then with step size \(h=0.1 .\) Make a table showing the approximate values and the actual value, together with the percentage error in the more accurate approximation, for \(x\) an integral multiple of 0.2. Throughout, primes denote derivatives with respect to \(x\). \(y^{\prime}=\frac{1}{2}(y-1)^{2}, y(0)=2 ; 0 \leqq x \leqq 1\)

A hand-held calculator will suffice for Problems 1 through 10, where an initial value problem and its exact solution are given. Apply the improved Euler method to approximate this solution on the interval \([0,0.5]\) with step size \(h=0.1 .\) Construct a table showing four-decimal-place values of the approximate solution and actual solution at the points \(x=0.1,0.2,0.3\), \(0.4,0.5\) $$ y^{\prime}=y-x-1, y(0)=1 ; y(x)=2+x-e^{x} $$

A hand-held calculator will suffice for Problems 1 through 10, where an initial value problem and its exact solution are given. Apply the improved Euler method to approximate this solution on the interval \([0,0.5]\) with step size \(h=0.1 .\) Construct a table showing four-decimal-place values of the approximate solution and actual solution at the points \(x=0.1,0.2,0.3\), \(0.4,0.5\) $$ y^{\prime}=-2 x y, y(0)=2 ; y(x)=2 e^{-x^{2}} $$

In Problems, use a computer system or graphing calculator to plot a slope field and/or enough solution curves to indicate the stability or instability of each critical point of the given differential equation. (Some of these critical points may be semistable in the sense mentioned in Example 6.) $\$$ $$ \frac{d x}{d t}=\left(x^{2}-4\right)^{2} $$

In Problems, use a computer system or graphing calculator to plot a slope field and/or enough solution curves to indicate the stability or instability of each critical point of the given differential equation. (Some of these critical points may be semistable in the sense mentioned in Example 6.) $\$$ $$ \frac{d x}{d t}=\left(x^{2}-4\right)^{3} $$

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