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Consider the nonlinear scalar differential equation \(x^{\prime \prime}=1-(1+x)^{3 / 2}\). An equation having this structure arises in modeling the bobbing motion of a floating parabolic trough. (a) Let \(y=x^{\prime}\) and rewrite the given scalar equation as an equivalent first order system. (b) Show that the system has a single equilibrium point at \(\left(x_{e}, y_{e}\right)=(0,0)\). (c) Determine the linearized system \(\mathbf{z}^{\prime}=A \mathbf{z}\), and analyze its stability properties. (d) Assume that the system is an almost linear system at equilibrium point \((0,0)\). Does Theorem \(6.4\) provide any information about the stability properties of the nonlinear system obtained in part (a)? Explain.

Short Answer

Expert verified
Based on the given step-by-step solution, the second-order differential equation was rewritten as a first-order system and an equilibrium point at (0, 0) was found. The linearized system revealed that the equilibrium point is a center equilibrium, but Theorem 6.4 doesn't give any information about the stability properties of the nonlinear system due to the eigenvalues having zero real parts.

Step by step solution

01

Rewrite the equation as an equivalent first-order system

Let y equal \(x'\), so we get the following system of first-order equations: \begin{align*} x' &= y \\ y' &= 1 - (1 + x)^{\frac{3}{2}} \end{align*} The first equation represents the relationship between x and y, while the second equation corresponds to the original nonlinear scalar differential equation.
02

Determine the equilibrium point

To find the equilibrium points, we look for the values of \(x_e\) and \(y_e\) that make both derivatives equal to zero: \begin{align*} x' &= y = 0 \\ y' &= 1 - (1 + x)^{\frac{3}{2}} = 0 \end{align*} Solving these equations simultaneously, we get \(x_e = 0\) and \(y_e = 0\). Hence, there is a single equilibrium point at \((x_e, y_e) = (0, 0)\).
03

Find the linearized system and analyze its stability properties

To find the linearized system, we need to find the Jacobian matrix A of the system at the equilibrium point \((0, 0)\). The Jacobian matrix has the following form: \[ A = \begin{bmatrix} \frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y} \\ \frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y} \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -\frac{3}{2}(1 + x)^{\frac{1}{2}} & 0 \end{bmatrix} \] Evaluating the matrix A at the equilibrium point \((0, 0)\), we get: \[ A = \begin{bmatrix} 0 & 1 \\ -\frac{3}{2} & 0 \end{bmatrix} \] To analyze the stability properties, we find the eigenvalues of the matrix A: The characteristic equation of the matrix A is given by: \[ \text{det}(A - \lambda I) = \begin{vmatrix} -\lambda & 1 \\ -\frac{3}{2} & -\lambda \end{vmatrix} = \lambda^2 + \frac{3}{2} = 0 \] Solving this equation for \(\lambda\), we find two imaginary eigenvalues \(\lambda = \pm i\sqrt{\frac{3}{2}}\). Since the real parts of both eigenvalues are zero, the linearized system has a center equilibrium at \((0, 0)\).
04

Discuss the applicability of Theorem 6.4

Theorem 6.4 states that for an almost linear system, if the linearized system at an equilibrium point has all eigenvalues with negative real parts, then the nonlinear system is locally asymptotically stable at that point. If the linearized system has at least one eigenvalue with a positive real part, then the nonlinear system is unstable at that point. In our case, the linearized system at the equilibrium point \((0,0)\) has two imaginary eigenvalues with zero real parts. Therefore, Theorem 6.4 does not provide any information about the stability properties of the nonlinear system obtained in part (a).

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Key Concepts

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

Equilibrium Points
Understanding the concept of equilibrium points in differential equations is essential in analyzing the behavior of dynamical systems. An equilibrium point, also known as a fixed point, is a solution to a system of differential equations where all derivatives equal zero. This implies that the system does not change at this point; it is in a state of balance.

When dealing with systems represented by a second-order differential equation, such as the one modeling the bobbing motion of a floating parabolic trough (\(x'' = 1 - (1 + x)^{3/2}\)), it is often helpful to first convert the system into a set of first-order equations. After this transformation, finding the equilibrium points involves determining the values of variables for which the derivatives (\(x'\) and \(y'\)) vanish. In this specific exercise, \(x_e = 0, y_e = 0)\) is the single equilibrium point, indicating that when the system reaches this state, it will remain unchanged if undisturbed.
Linearization of Differential Systems
Linearization is a method used to approximate the behavior of a nonlinear system near its equilibrium points. Because nonlinear differential equations can be difficult to solve analytically, linearizing the system around an equilibrium point allows us to apply linear algebra techniques to analyze the system's behavior.

The linearized system is obtained by computing the Jacobian matrix at the equilibrium point, which contains all the first-order partial derivatives of the system's equations. This matrix describes how the system behaves in the immediate vicinity of the equilibrium point. For the given problem, the Jacobian matrix A at the equilibrium point \((0, 0)\) reveals the nature of the system's dynamics around that point. Linearization gives us a powerful way to predict system behavior without having to solve the entire nonlinear problem directly.
Stability of Dynamical Systems
The analysis of a system's stability aims to understand how it reacts to small disturbances or changes from its equilibrium state. Does the system return to equilibrium, or does it diverge away? The stability of dynamical systems is a pivotal concept in various scientific fields, as it helps determine the long-term behavior of a system.

Types of Stability

  • Stable: The system returns to the equilibrium point after a small disturbance.
  • Unstable: The system moves away from the equilibrium point after a small disturbance.
  • Asymptotically stable: The system returns to equilibrium over time and remains stable.
  • Neutrally stable: The system remains in the new state but does not diverge.
A system like the floating parabolic trough, with its equilibrium at \((0, 0)\) analyzed in the exercise, exhibits what is known as 'neutral stability' in the linearized analysis, where any small disturbance would result in an oscillation around the equilibrium point.
Eigenvalues and Stability Analysis
Eigenvalues are at the core of determining the stability of a system once it has been linearized. They stem from the linearized system's characteristic equation, which is derived from the Jacobian matrix. The eigenvalues provide valuable insights into the system's behavior near equilibrium points.

For the linearized system obtained from the exercise, we calculated imaginary eigenvalues which suggest oscillatory behavior, but they cannot confirm stability on their own. As the real parts of these eigenvalues are zero, they indicate a center equilibrium. Hence, this system might be stable or unstable, depending on higher-order terms in the nonlinear system. Theorem 6.4, which aligns eigenvalues with system stability, does not apply here since it requires eigenvalues with non-zero real parts to make a definitive conclusion about stability. This limitation shows that eigenvalue analysis, while powerful, is not sufficient alone for nonlinear systems' stability analysis; comprehensive stability studies may require more advanced methods.

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Most popular questions from this chapter

Consider the system $$ \begin{aligned} &x^{\prime}=x+y^{2}-x y^{n} \\ &y^{\prime}=-x+y^{-1} . \end{aligned} $$ The slope of the phase-plane trajectory passing through the point \((x, y)=(1,2)\) is \(\frac{1}{6}\). Determine the exponent \(n\).

Let \(A\) be a \((2 \times 2)\) constant matrix, and let \((\lambda, \mathbf{u})\) be an eigenpair for \(A\). Assume that \(\lambda\) is real, \(\lambda \neq 0\), and $$ \mathbf{u}=\left[\begin{array}{l} u_{1} \\ u_{2} \end{array}\right] $$ Consider the phase plane for the autonomous linear system \(\mathbf{y}^{\prime}=A \mathbf{y}\). We can define a phase-plane line through the origin by the parametric equations \(x=\tau u_{1}, y=\tau u_{2}\), \(-\infty<\tau<\infty\). Let \(P\) be any point on this line, say \(P=\left(\tau_{0} u_{1}, \tau_{0} u_{2}\right)\) for some \(\tau_{0} \neq 0\). (a) Show that at the point \(P, x^{\prime}=\tau_{0} \lambda u_{1}\) and \(y^{\prime}=\tau_{0} \lambda u_{2}\). (b) How is the velocity vector \(\mathbf{v}(t)=x^{\prime}(t) \mathbf{i}+y^{\prime}(t) \mathbf{j}\) at point \(P\) oriented relative to the line?

Use the information given about the nature of the equilibrium point at the origin to determine the value or range of permissible values for the unspecified entry in the coefficient matrix. The origin is an asymptotically stable proper node of \(\mathbf{y}^{\prime}=\left[\begin{array}{rr}-2 & 0 \\ \alpha & -2\end{array}\right] \mathbf{y} ;\) determine the value(s) of \(\alpha\).

Consider the system encountered in the study of pendulum motion, $$ \begin{aligned} &x^{\prime}=y \\ &y^{\prime}=-\sin x \end{aligned} $$ at its equilibrium points \((0,0)\) and \((\pi, 0)\). (a) Let \(z_{1}=x, z_{2}=y\). Show that the system becomes $$ \mathbf{z}^{\prime}=\left[\begin{array}{rr} 0 & 1 \\ -1 & 0 \end{array}\right] \mathbf{z}+\left[\begin{array}{c} 0 \\ z_{1}-\sin z_{1} \end{array}\right] \text {. } $$ (b) Let \(z_{1}=x-\pi, z_{2}=y\). Show that the system becomes $$ \mathbf{z}^{\prime}=\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right] \mathbf{z}-\left[\begin{array}{c} 0 \\ z_{1}-\sin z_{1} \end{array}\right] $$ (c) Show that the system is almost linear at both equilibrium points.

These exercises explore the question "When one of two species in a colony is desirable and the other is undesirable, is it better to use resources to nurture the growth of the desirable species or to harvest the undesirable one?" Let \(x(t)\) and \(y(t)\) represent the populations of two competing species, with \(x(t)\) the desirable species. Assume that if resources are invested in promoting the growth of the desirable species, the population dynamics are given by $$ \begin{aligned} &x^{\prime}=r(1-\alpha x-\beta y) x+\mu x \\ &y^{\prime}=r(1-\alpha y-\beta x) y \end{aligned} $$ If resources are invested in harvesting the undesirable species, the dynamics are $$ \begin{aligned} &x^{\prime}=r(1-\alpha x-\beta y) x \\ &y^{\prime}=r(1-\alpha y-\beta x) y-\mu y \end{aligned} $$ In (10), \(r, \alpha, \beta\), and \(\mu\) are positive constants. For simplicity, we assume the same parameter values for both species. For definiteness, assume that \(\alpha>\beta>0\). Consider system (9), which describes the strategy in which resources are invested into nurturing the desirable species. (a) Determine the four equilibrium points for the system. (b) Show that it is possible, by investing sufficient resources (that is, by making \(\mu\) large enough), to prevent equilibrium coexistence of the two species. (c) Assume that \(\mu\) is large enough to preclude equilibrium coexistence of the two species. Compute the linearized system at each of the three physically relevant equilibrium points. Determine the stability characteristics of the linearized system at each of these equilibrium points. (d) System (9) can be shown to be an almost linear system at each of the equilibrium points. Use this fact and the results of part (c) to infer the stability properties of system (9) at each of the three equilibrium points of interest. (e) Sketch the direction field. Will a sufficiently aggressive nurturing of species \(x\) ultimately drive undesirable species \(y\) to extinction? If so, what is the limiting population of species \(x\) ?

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