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Chapter 9: Problem 13

(a) Find all the critical points (equilibrium solutions). (b) Use a computer to draw a direction field and portrait for the system. (c) From the plot(s) in part (b) determine whether each critical point is asymptotically stable, stable, or unstable, and classify it as to type. $$ d x / d t=(2+x)(y-x), \quad d y / d t=(4-x)(y+x) $$

Short Answer

Expert verified
#Answer# The critical points are (-2, 2), (0, 0), and (4, 4). After analyzing the direction field and portrait of the system, classify the stability and type of each critical point as follows: 1. Critical point (-2, 2): If the trajectories are moving towards this point, it is asymptotically stable and can be classified as an attracting node or spiral. 2. Critical point (0, 0): If the trajectories are neither moving towards nor away from this point, it is stable, and can be classified as a saddle point. 3. Critical point (4, 4): If the trajectories are moving away from this point, it is unstable and can be classified as a repelling node or spiral. (Note: The specific type and stability of each critical point will depend on the direction field and portrait results, which are not provided here).

Step by step solution

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01

Find the critical points (equilibrium solutions)

To find the critical points, we set the right sides of both differential equations to zero and solve for x and y: $$ (2+x)(y-x) = 0, \quad (4-x)(y+x) = 0 $$ We have two possible cases: (i) Either \(2+x=0\) or \(y-x=0\) (ii) Either \(4-x=0\) or \(y+x=0\)
02

Step 1.1: Solve for x and y in case (i)

From the first case we have: (i) Either \(x=-2\) or \(x=y\) Now we substitute these values in the second equation: - If \(x=-2\): \((4-(-2))(y+(-2)) = 6(y-2) = 0 \Rightarrow y=2\) - If \(x=y\): \((4-y)(2y) = 0 \Rightarrow y=0\) or \(y=4\) In case (i), we found critical points as \((-2, 2)\), \((0, 0)\) and \((4, 4)\).
03

Step 1.2: Solve for x and y in case (ii)

From the second case we have: (ii) Either \(4-x=0\) or \(y+x=0\) Now we substitute these values in the first equation: - If \(x=4\): \((2+4)(y-4) = 6(y-4)=0 \Rightarrow y=4\) - If \(x=-y\): \((2-y)(2y)=0 \Rightarrow y=0\) or \(y=2\). But if \(y=2\), then \(x=-2\), which was already considered in case (i). In case (ii), we found critical points as \((4, 4)\) and \((0, 0)\).
04

Direction field and portrait (use a computer)

Using a computer software like Python or MATLAB, plot the direction field and portrait for the given system of differential equations.
05

Analyze the plot and determine stability of the critical points

By examining the direction field and portrait of the system, classify the stability and type of each critical point. 1. If the trajectories are moving towards the critical point, it is asymptotically stable. 2. If the trajectories are neither moving towards nor away from the critical point, it is stable. 3. If the trajectories are moving away from the critical point, it is unstable. Types of critical points include: - Nodes (attracting, repelling, or saddle) - Spirals (attracting or repelling) Using the stability analysis and classification of the critical points from the plot, determine the stability and type for each critical point found in step 1.

Key Concepts

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

Critical Points
In the study of differential equations, critical points, also known as equilibrium solutions, are vital for understanding the behavior of dynamic systems. These are the points where the system's variables do not change over time. To find these points, we set the derivatives equal to zero and solve the resulting equations. In our example, this involves solving the system:
  • \((2+x)(y-x) = 0\)
  • \((4-x)(y+x) = 0\)
By examining each case where either factor in each equation is zero, we discover the critical points: \((-2, 2)\), \(0, 0\), and \(4, 4\). These points are where the system balances itself, and knowing them helps us predict the future behavior of the system.
Stability Analysis
Stability analysis involves examining how a system behaves near its critical points. Specifically, whether the system returns to its equilibrium state or diverges away from it after a slight disturbance. In simple terms, stability tells us whether a point is:
  • Asymptotically stable, where trajectories converge to the point.
  • Stable, where trajectories neither attract nor repel.
  • Unstable, where trajectories diverge from the point.
In our exercise, after identifying the critical points, we evaluate their stability using a direction field plot. This visual tool illuminates the behavior of each nearby trajectory, offering insights into whether the system will settle, oscillate, or move away from these points.
Equilibrium Solutions
Equilibrium solutions are at the heart of many real-world systems, indicating states where the system's variables do not change over time. Such solutions offer vital insights into how systems naturally settle, balance, or evolve into stable forms. Once critical points are found, they are examined to determine if slight changes in the system return to equilibrium, stay the same, or diverge.
The equilibrium solutions, in our problem, found at \((-2, 2)\), \(0, 0\), and \(4, 4\), showcase different behaviors and stabilities that offer a glimpse into the overall system dynamics. Comprehending these points ensures that we can predict the system’s response to various initial conditions.
Direction Fields
A direction field, sometimes referred to as a slope field, is a graphical representation of a differential equation. It shows the trends and tendencies of the system's behavior without having to solve the equations explicitly. By plotting small line segments at various points, it provides an immediate visual clue to the trajectory of the system's solutions.
When approaching problems with differential equations, viewing the direction field can significantly aid in visualizing how solutions evolve over time. It captures the essence of the system's dynamics, offering a snapshot of how trajectories behave, converge, or diverge in relation to critical points. In our exercise, this involves using computer-generated plots to understand how each critical point behaves, thereby determining their stability classifications.

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Most popular questions from this chapter

By introducing suitable dimensionless variables, the system of nonlinear equations for the damped pendulum [Frqs. (8) of Section 9.3] can be written as $$ d x / d t=y, \quad d y / d t=-y-\sin x \text { . } $$ (a) Show that the origin is a critical point. (b) Show that while \(V(x, y)=x^{2}+y^{2}\) is positive definite, \(f(x, y)\) takes on both positive and negative values in any domain containing the origin, so that \(V\) is not a Liapunov function. Hint: \(x-\sin x>0\) for \(x>0\) and \(x-\sin x<0\) for \(x<0 .\) Consider these cases with \(y\) positive but \(y\) so small that \(y^{2}\) can be ignored compared to \(y .\) (c) Using the energy function \(V(x, y)=\frac{1}{2} y^{2}+(1-\cos x)\) mentioned in Problem \(6(b),\) show that the origin is a stable critical point. Note, however, that even though there is damping and we can epect that the origin is asymptotically stable, it is not possible to draw this conclusion using this Liapunov function. (d) To show asymptotic stability it is necessary to construct a better Liapunov function than the one used in part (c). Show that \(V(x, y)=\frac{1}{2}(x+y)^{2}+x^{2}+\frac{1}{2} y^{2}\) is such a Liapunov function, and conclude that the origin is an asymptotically stable critical point. Hint: From Taylor's formula with a remainder it follows that \(\sin x=x-\alpha x^{3} / 3 !,\) where \(\alpha\) depends on \(x\) but \(0<\alpha<1\) for \(-\pi / 2

(a) By solving Eq. (9) numerically show that the real part of the complex roots changes sign when \(r \cong 24.737\). (b) Show that a cubic polynomial \(x^{3}+A x^{2}+B x+C\) has one real zero and two pure imaginary zeros only if \(A B=C\). (c) By applying the result of part (b) to Eq. (9) show that the real part of the complex roots changes sign when \(r=470 / 19\).

Verify that \((0,0)\) is a critical point, show that the system is almost linear, and discuss the type and stability of the critical point \((0,0)\) by examining the corresponding linear system. $$ d x / d t=(1+x) \sin y, \quad d y / d t=1-x-\cos y $$

(a) A special case of the Lienard equation of Problem 8 is $$ \frac{d^{2} u}{d t^{2}}+\frac{d u}{d t}+g(u)=0 $$ where \(g\) satisfies the conditions of Problem 6 . Letting \(x=u, y=d u / d t,\) show that the origin is a critical point of the resulting system. This equation can be interpreted as describing the motion of a spring-mass system with damping proportional to the velocity and a nonlinear restoring force. Using the Liapunov function of Problem \(6,\) show that the origin is a stable critical point, but note that even with damping we cannot conclude asymptotic stability using this Liapunov function. (b) Asymptotic stability of the critical point \((0,0)\) can be shown by constructing a better Liapunov function as was done in part (d) of Problem 7 . However, the analysis for a general function \(g\) is somewhat sophisticated and we only mention that appropriate form for \(V\) is $$ V(x, y)=\frac{1}{2} y^{2}+A y g(x)+\int_{0}^{x} g(s) d s $$ where \(A\) is a positive constant to be chosen so that \(V\) is positive definite and \(\hat{V}\) is negative definite. For the pendulum problem \([g(x)=\sin x]\) use \(V\) as given by the preceding equation with \(A=\frac{1}{2}\) to show that the origin is asymptotically stable. Hint: Use \(\sin x=x-\alpha x^{3} / 3 !\) and \(\cos x=1-\beta x^{2} / 2 !\) where \(\alpha\) and \(\beta\) depend on \(x,\) but \(0<\alpha<1\) and \(0<\beta<1\) for \(-\pi / 2

Determine the critical point \(\mathbf{x}=\mathbf{x}^{0},\) and then classify its type and examine its stability by making the transformation \(\mathbf{x}=\mathbf{x}^{0}+\mathbf{u} .\) \(\frac{d \mathbf{x}}{d t}=\left(\begin{array}{rr}{0} & {-\beta} \\ {\delta} & {0}\end{array}\right) \mathbf{x}+\left(\begin{array}{r}{\alpha} \\\ {-\gamma}\end{array}\right) ; \quad \alpha, \beta, \gamma, \delta>0\)
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