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Chapter 3: Problem 14

Prove that the following systems have no fixed points (a) \(\dot{x}_{1}=\mathrm{e}^{x_{1}+x_{2}}, \quad \dot{x}_{2}=x_{1}+x_{2}\) (b) \(\quad \dot{x}_{1}=x_{1}+x_{2}+2, \quad \dot{x}_{2}=x_{1}+x_{2}+1\); (c) \(\dot{x}_{1}=x_{2}+2 x_{2}^{3}, \quad \dot{x}_{2}=1+x_{2}^{2}\) and sketch their phase portraits.

Short Answer

Expert verified
None of the systems (a), (b), or (c) have fixed points.

Step by step solution

01

Define Fixed Points for a System

A system has a fixed point at \((x_1, x_2)\) if both derivatives \(\dot{x}_1\) and \(\dot{x}_2\) are zero simultaneously. This means \[ \begin{align*} \dot{x}_1 &= 0, \\dot{x}_2 &= 0 \end{align*} \] must hold true.
02

Analyze System (a)

For the system \[ \begin{align*} \dot{x}_{1}&=\mathrm{e}^{x_{1}+x_{2}}, \\dot{x}_{2}&=x_{1}+x_{2} \end{align*} \] we have:1. \(\mathrm{e}^{x_{1}+x_{2}} = 0\) is impossible because the exponential function is never zero.2. Therefore, no solution exists for \(x_1\) and \(x_2\) that satisfies both equations being zero simultaneously.Thus, system (a) has no fixed points.
03

Analyze System (b)

For the system \[ \begin{align*} \dot{x}_{1} &= x_{1}+x_{2}+2, \\dot{x}_{2} &= x_{1}+x_{2}+1 \end{align*} \] we have:1. Subtract the second equation from the first: \((x_1 + x_2 + 2) - (x_1 + x_2 + 1) = 0\) simplifies to \(1 = 0\), which is a contradiction.2. Thus elements that make both \(\dot{x}_1 = 0\) and \(\dot{x}_2 = 0\) simultaneously do not exist, meaning no fixed points occur in this system.
04

Analyze System (c)

For the system \[ \begin{align*} \dot{x}_{1} &= x_{2}+2 x_{2}^{3}, \\dot{x}_{2} &= 1+x_{2}^{2} \end{align*} \] we have:1. \(1 + x_2^2 = 0\) is impossible since \(x_2^2\) is non-negative and \(1\) is a positive constant, hence cannot sum to zero.2. Therefore, there is no real \(x_2\) that can make \(\dot{x}_2 = 0\). Consequently, \(\dot{x}_1 = 0\) and \(\dot{x}_2 = 0\) are never both zero simultaneously, indicating system (c) has no fixed points.
05

Conclusion

Each system (a), (b), and (c) has non-zero conditions or impossibilities (like contradictions or summations always being positive) that prevent fixed points from existing. Thus, none of these systems can have fixed points, and their phase portraits would reflect trajectories not stabilizing to a fixed point.

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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 systems of differential equations, fixed points play a crucial role. A fixed point is a point in the phase space where the system does not change over time. In mathematical terms, for fixed points, the derivatives \( \dot{x}_1 \) and \( \dot{x}_2 \) equate to zero. This translates to the equations:
  • \( \dot{x}_1 = 0 \)
  • \( \dot{x}_2 = 0 \)
Fixed points are essential because they can indicate the stability or behavior of the system, showing points where trajectories remain constant. If a system has no fixed points, it often implies trajectories will not settle or stabilize into particular states. Instead, they will keep moving, possibly in repeating patterns or away from certain areas in phase space.
Systems of Differential Equations
Systems of differential equations describe how variables change together over time. They consist of multiple equations involving derivatives of related variables, showing their rates of change. Generally represented as:
  • \( \dot{x}_1 = f(x_1, x_2,...) \)
  • \( \dot{x}_2 = g(x_1, x_2,...) \)
These systems can model dynamic processes found in the real world, from ecosystems to electrical circuits. Key to analyzing these is understanding how variables interact, possibly leading to stability (fixed points) or ongoing change. Solutions explore how conditions evolve from initial states, examining whether paths converge, follow cycles, or diverge endlessly.
Phase Portraits
Phase portraits are graphical representations revealing the behvaior of systems of differential equations over time. They map the trajectories that solutions follow in the phase space, providing insights into system dynamics.
  • Each trajectory represents a possible behavior, showing how states evolve.
  • Fixed points, if any, appear as points where trajectories converge.
  • Absence of fixed points suggests trajectories will not settle.
Phase portraits help visualize not only stability and instability but also periodic behavior or chaotic paths. For systems lacking fixed points, the phase portrait may depict perpetual motion with trajectories filling certain areas or looping eternally without convergence.
Mathematical Analysis
Mathematical analysis uses logical reasoning and mathematical tools to explore and solve problems involving differential equations. It involves examining the behavior of solutions and understanding the qualitative aspects like stability and motion patterns.

Key Points in the Analysis

  • Identifying fixed points helps in understanding potential long-term behaviors.
  • Solving systems involves setting each equation to zero for theoretical explorations like finding fixed points.
  • Analysis includes proving or disproving scenarios, such as the impossibility of fixed points as seen in the given exercise.
This process involves algebraic manipulations, exploring function properties, including range and constraints. Successfully analyzing differential systems provides deeper insights into potential dynamics and behaviors evident in real-life analogues.

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Most popular questions from this chapter

Use the linearization theorem to classify, where possible, the fixed points of the systems: (a) \(\dot{x}_{1}=x_{2}^{2}-3 x_{1}+2, \quad \dot{x}_{2}=x_{1}^{2}-x_{2}^{2}\) (b) \(\dot{x}_{1}=x_{2}, \quad \dot{x}_{2}=-x_{1}+x_{1}^{3} ;\) (c) \(\dot{x}_{1}=\sin \left(x_{1}+x_{2}\right), \quad \dot{x}_{2}=x_{2}\); (d) \(\dot{x}_{1}=x_{1}-x_{2}-\mathrm{e}^{x_{1}}, \quad \dot{x}_{2}=x_{1}-x_{2}-1\) (e) \(\dot{x}_{1}=-x_{2}+x_{1}+x_{1} x_{2}, \quad \dot{x}_{2}=x_{1}-x_{2}-x_{2}^{2}\) (f) \(\quad \dot{x}_{1}=x_{2}, \quad \dot{x}_{2}=-\left(1+x_{1}^{2}+x_{1}^{4}\right) x_{2}-x_{1}\); (g) \(\dot{x}_{1}=-3 x_{2}+x_{1} x_{2}-4, \quad \dot{x}_{2}=x_{2}^{2}-x_{1}^{2}\).

Show that the non-linear change of coordinates $$ y_{1}=x_{1}+x_{2}^{3}, \quad y_{2}=x_{2}+x_{2}^{2} $$ satisfies the requirements of the flow box theorem for the system $$ \dot{x}_{1}=-\frac{3 x_{2}^{2}}{1+2 x_{2}}, \quad \dot{x}_{2}=\frac{1}{1+2 x_{2}} $$ in the neighbourhood of any point \(\left(x_{1}, x_{2}\right)\) with \(x_{2} \neq-\frac{1}{2}\).

Find the principal directions of the fixed points at the origin of the following systems: (a) \(\dot{x}_{1}=\mathrm{e}^{\mathrm{x}_{1}+x_{2}}-1, \quad \dot{x}_{2}=x_{2}\); (b) \(\dot{x}_{1}=-\sin x_{1}+x_{2}, \quad \dot{x}_{2}=\sin x_{2}\) (c) \(\dot{x}_{1}=\ln \left(x_{1}+x_{2}+1\right), \quad \dot{x}_{2}=\frac{1}{2} x_{1}+x_{2}, \quad x_{1}+x_{2}>-1 ;\) and use them to sketch local phase portraits. Sections 3.4-3.6

Find the local phase portraits at each of the fixed points of the system $$ \dot{x}_{1}=x_{1}\left(1-x_{1}^{2}\right), \quad \dot{x}_{2}=x_{2} $$ Use these results to suggest a global phase portrait. Check whether your suggestion is correct by using the method of isoclines. Section \(3.7\)

Prove that there exists a region \(R=\left\\{\left(x_{1}, x_{2}\right) \mid x_{1}^{2}+x_{2}^{2} \leqslant r^{2}\right\\}\) such that all trajectories of the system $$ \dot{x}_{1}=-w x_{2}+x_{1}\left(1-x_{1}^{2}-x_{2}^{2}\right), \quad \dot{x}_{2}=w x_{1}+x_{2}\left(1-x_{1}^{2}-x_{2}^{2}\right)-F \text {, } $$ where \(w\) and \(F\) are constants, eventually enter \(R\). Show that the system has a limit cycle when \(F=0\).
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