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Chapter 9: Problem 25

Prove that if a trajectory starts at a noncritical point of the system $$ d x / d t=F(x, y), \quad d y / d t=G(x, y) $$ then it cannot reach a critical point \(\left(x_{0}, y_{0}\right)\) in a finite length of time. Hint: Assume the contrary, that is, assume that the solution \(x=\phi(t), y=\psi(t)\) satisfies \(\phi(a)=x_{0}, \psi(a)=y_{0}\). Then use the fact that \(x=x_{0}, y=y_{0}\) is a solution of the given system satisfying the initial condition \(x=x_{0}, y=y_{0}\) at \(t=a\).

Short Answer

Expert verified
Question: Prove that if a trajectory starts at a noncritical point of a system of first order 2D ordinary differential equations, it cannot reach a critical point in a finite length of time. Answer: We have proven that assuming a trajectory starting at a noncritical point can reach a critical point in a finite length of time leads to a contradiction, based on the Uniqueness Theorem for ODEs. Therefore, a trajectory starting at a noncritical point cannot reach a critical point in a finite length of time.

Step by step solution

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01

Understanding the given system

We are given a system of first order 2D ordinary differential equations as follows: $$ \frac{dx}{dt} = F(x, y), \quad \frac{dy}{dt} = G(x, y) $$ A critical point \(\left(x_{0}, y_{0}\right)\) is a point in the 2D plane \((x, y)\) such that \(F(x_0, y_0) = 0\) and \(G(x_0, y_0) = 0\).
02

Assuming the contrary

Let's now assume that the solution to the system \(x = \phi(t), y = \psi(t)\) satisfies $$ \phi(a) = x_{0}, \quad \psi(a) = y_{0} $$ for some finite time \(a\).
03

Defining the solution at the critical point

From the hint, we know that \(x = x_0, y = y_0\) is also a solution of the given system satisfying the initial condition \(x = x_0, y = y_0\) at \(t = a\): $$ x(t) = x_{0}, y(t) = y_{0}, \quad \text{for all } t \geq a $$
04

Using the uniqueness theorem

By the Uniqueness Theorem for ordinary differential equations, we know that the initial value problem has a unique solution for the given initial condition. This implies that the trajectory starting from the noncritical point must be identical to the solution of the critical point system for all \(t \geq a\): $$ \phi(t) = x_{0}, \quad \psi(t) = y_{0}, \quad \text{for all } t \geq a $$
05

Reaching a contradiction

However, this leads to a contradiction. The trajectory starting from a noncritical point must have \(F(x, y) \neq 0\) and/or \(G(x, y) \neq 0\). Meanwhile, at the critical point, we have \(F(x_0, y_0) = 0\) and \(G(x_0, y_0) = 0\), which means the trajectory cannot satisfy the condition of starting from a noncritical point. Thus, we've reached a contradiction.
06

Conclusion

Based on this contradiction, we can conclude that if a trajectory starts at a noncritical point of the system, it cannot reach a critical point in a finite length of time.

Key Concepts

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

Critical Points in Differential Equations
Ordinary differential equations (ODEs) are fundamental mathematical tools used to model continuous processes in nature and technology. In the realm of ODEs, critical points, also known as equilibrium points or stationary points, play a pivotal role in understanding the behavior of dynamical systems.

A critical point \( (x_{0}, y_{0}) \) is where the functions \( F(x, y) \) and \( G(x, y) \) representing the rates of change in the system both equal zero: \( F(x_0, y_0) = 0 \) and \( G(x_0, y_0) = 0 \). This signifies a state where the system's variables remain constant over time if the system is in that state. In analyzing critical points, one can determine the system's long-term behavior, such as stability and type of equilibrium.

When examining the system's behavior near these points, we often linearize the equations around them to obtain a clearer picture of the system's local dynamics. Tools like phase portraits and Jacobian matrices are valuable in assessing the nature of critical points—revealing whether they are nodes, saddle points, foci, or centers. By doing so, we gain insights into the trajectories of solutions and their tendency to converge to or diverge from these points.
Uniqueness Theorem
The Uniqueness Theorem in the context of ODEs is an essential aspect of both the theory and application of differential equations. It asserts that given an ordinary differential equation with certain conditions—specifically, if the function is continuous and satisfies a condition known as the Lipschitz condition—a unique solution exists that passes through a given point in the phase space, which is designated as the initial condition.

In mathematical terms, if we are given \( \frac{dx}{dt} = F(x, y) \) and \( \frac{dy}{dt} = G(x, y) \) with initial conditions \( x(t_0) = x_0 \) and \( y(t_0) = y_0 \), and if \( F \) and \( G \) are Lipschitz continuous in a neighborhood of \( (x_0, y_0) \), then there exists a unique function \( x(t) \) and \( y(t) \) that satisfies the differential equation and the initial conditions. The implication of this theorem is significant; it helps ensure the predictability of the system modeled by ODEs and excludes the possibility of other solutions intersecting or diverging from the given trajectory at the initial condition.
Trajectory in Phase Space
The concept of a trajectory in phase space is central to visualizing and understanding the dynamic behavior of systems described by differential equations. Phase space is a multidimensional space in which each axis corresponds to one of the variables in the system. For a 2D system this is simply the \( (x, y) \) plane. A trajectory in this space represents the evolution of the system over time; it is the path that the system's state follows as dictated by the differential equations.

By plotting trajectories, one can see how a system moves from different initial conditions, how close trajectories come to critical points, and observe other qualitative behaviors such as periodic orbits or chaotic behavior. These visual insights supplement the analytical methods and help deepen the understanding of system dynamics. In phase space, critical points mark the locations where trajectories may either converge to a stable equilibrium, diverge away from an unstable one, or approach in the case of a center. Trajectories never cross in phase space due to the Uniqueness Theorem, and each trajectory is an individual path carved by a particular set of initial conditions.

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Most popular questions from this chapter

Each of Problems I through 6 can be interpreted as describing the interaction of two species with populations \(x\) and \(y .\) In each of these problems carry out the following steps. $$ \begin{array}{l}{\text { (a) Draw a direction field and describe how solutions seem to behave. }} \\ {\text { (b) Find the critical points. }} \\ {\text { (c) For each critical point find the corresponding linear system. Find the eigenvalues and }} \\ {\text { eigenvectors of the linear system; classify each critical point as to type, and determine }} \\ {\text { whether it is asymptotically stable, stable, or unstable. }}\end{array} $$ $$ \begin{array}{l}{\text { (d) Sketch the trajectories in the neighborhood of each critical point. }} \\ {\text { (c) Compute and plot enough trajectories of the given system to show clearly the behavior of }} \\ {\text { the solutions. }} \\ {\text { (f) Determine the limiting behavior of } x \text { and } y \text { as } t \rightarrow \infty \text { and interpret the results in terms of }} \\ {\text { the populations of the two species. }}\end{array} $$ $$ \begin{array}{l}{d x / d t=x(1.5-x-0.5 y)} \\ {d y / d t=y(2-y-0.75 x)}\end{array} $$

The motion of a certain undamped pendulum is described by the equations $$ d x / d t=y, \quad d y / d t=-4 \sin x $$ If the pendulum is set in motion with an angular displacement \(A\) and no initial velocity, then the initial conditions are \(x(0)=A, y(0)=0\) (a) Let \(A=0.25\) and plot \(x\) versus \(t\). From the graph estimate the amplitude \(R\) and period \(T\) of the resulting motion of the pendulum. (b) Repeat part (a) for \(A=0.5,1.0,1.5,\) and \(2.0 .\) (c) How do the amplitude and period of the pendulum's motion depend on the initial position \(A^{7}\) Draw a graph to show each of these relationships. Can you say anything about the limiting value of the period as \(A \rightarrow 0 ?\) (d) Let \(A=4\) and plot \(x\) versus \(t\) Explain why this graph differs from those in parts (a) and (b). For what value of \(A\) does the transition take place?

(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. $$ d x / d t=y, \quad d y / d t=x-\frac{1}{6} x^{3}-\frac{1}{5} y $$

(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

(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. $$ d x / d t=(2+x)(y-x), \quad d y / d t=y\left(2+x-x^{2}\right) $$
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