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Locate the unique equilibrium point of the given nonhomogeneous system, and determine the stability properties of this equilibrium point. Is it asymptotically stable, stable but not asymptotically stable, or unstable? $$ \begin{aligned} &x^{\prime}=-x+y+1 \\ &y^{\prime}=-10 x+5 y+2 \end{aligned} $$

Short Answer

Expert verified
Based on the steps above, we found out that the unique equilibrium point of the given system of first-order differential equations is \((\frac{7}{15}, -\frac{8}{15})\). After computing the eigenvalues of the Jacobian matrix and analyzing their real parts, we determined that this equilibrium point is unstable.

Step by step solution

01

Find the equilibrium point

To find the equilibrium point, set both derivatives equal to zero and solve for x and y: $$ \begin{aligned} -x+y+1 = 0\\ -10x+5y+2=0 \end{aligned} $$ Now we can solve this system of linear equations for x and y.
02

Solve the system of linear equations

Multiply the first equation by 5 to eliminate y, then add the two equations: $$ \begin{cases} -5x+5y+5=0 \\ -10x+5y+2=0 \end{cases} $$ Add them: $$ (-5x+5y+5)+(-10x+5y+2)=0 \\ -15x + 7=0 $$ Now, solve for x: $$ x=\frac{7}{15} $$ Plug x back into one of the original equations to find y: $$ - (\frac{7}{15}) + y + 1 = 0 $$ $$ y = -\frac{8}{15} $$ So, the equilibrium point is \((\frac{7}{15}, -\frac{8}{15})\).
03

Calculate the Jacobian matrix

To determine the stability properties of the equilibrium point, we need to find the Jacobian matrix and its eigenvalues. The Jacobian matrix for our system is: $$ J(x,y) = \begin{bmatrix} \frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y} \\ \frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y} \end{bmatrix} = \begin{bmatrix} -1 & 1 \\ -10 & 5 \end{bmatrix} $$
04

Calculate the eigenvalues of the Jacobian matrix

Next, we find the eigenvalues of the Jacobian matrix by calculating the characteristic equation: $$ \det(J - \lambda I) = \det \begin{bmatrix} -1-\lambda & 1 \\ -10 & 5-\lambda \end{bmatrix} = ( -1-\lambda)(5-\lambda) - (-10)=0 $$ Solve for lambda: $$ \lambda^2 -4\lambda + 15 = 0 $$ The eigenvalues are, \(\lambda_1 = 2 + 3i\) and \(\lambda_2 = 2 - 3i\).
05

Determine the stability properties

The stability properties of the equilibrium point depend on the real parts of the eigenvalues. If the real parts of both eigenvalues are negative, the point is asymptotically stable. If the real parts of at least one eigenvalue is positive, the point is unstable. If the real parts of both eigenvalues are zero and their imaginary parts are non-zero, the point is stable but not asymptotically stable. In our case, the real parts of both eigenvalues are 2, which are positive. Therefore, the equilibrium point \((\frac{7}{15}, -\frac{8}{15})\) is unstable.

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Key Concepts

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

Stability Analysis
In differential equations, stability analysis helps to understand the behavior of a system near its equilibrium points.
The goal is to determine whether small perturbations or changes in the initial conditions will die out over time or grow, causing the system to change significantly.
  • **Asymptotically stable**: Small disturbances in the system fade away, and the system returns to its equilibrium state.
  • **Stable but not asymptotically stable**: Small disturbances neither fade nor grow, so the system remains close to the equilibrium point.
  • **Unstable**: Small disturbances grow over time, pushing the system away from equilibrium.

The analysis involves evaluating the real parts of eigenvalues derived from the Jacobian matrix. If all have negative real parts, the system is asymptotically stable. If any have positive real parts, the system is unstable. Understanding these dynamics is crucial for predicting long-term behavior of systems in fields like physics, engineering, and biology.
Jacobian Matrix
The Jacobian matrix is a crucial tool in studying the behavior of systems of differential equations.
It is essentially a matrix of all first-order partial derivatives of a vector-valued function.
In the context of systems like the one given in the exercise, the Jacobian helps approximate the behavior of nonlinear systems near their equilibrium points.
  • Each element of the Jacobian matrix represents how one component of the system changes in response to another.
  • For a given system, such as \[x' = -x + y + 1, \quad y' = -10x + 5y + 2,\]the Jacobian matrix provides insights by showing immediate rate changes and interactions between variables.
The size of the Jacobian corresponds to the number of equations in the system, making it instrumental for performing linear stability analysis.
Eigenvalues
Eigenvalues are fundamental in the analysis of differential equations, especially in understanding stability.
They are essentially scalars associated with a matrix, indicating the factors by which a corresponding eigenvector is scaled during a transformation.
When analyzing a system like our given nonhomogeneous equations, eigenvalues of the Jacobian matrix offer direct insight into the system's stability.
  • **Positive real eigenvalue**: Indicates instability, leading the system state away from equilibrium.
  • **Negative real eigenvalue**: Suggests stability, as the system returns to equilibrium over time.
  • **Complex eigenvalues**: Often introduce oscillatory behavior and affect stability depending on the sign of their real parts.
Eigenvalues, especially when complex, provide essential information on whether small perturbations will grow, shrink, or oscillate.
Nonhomogeneous Systems
Nonhomogeneous systems of differential equations contain terms independent of variables themselves.
These additional constants or functions occur frequently in real-world scenarios, such as when external forces or inputs are present.
Our example system \[x' = -x + y + 1, \quad y' = -10x + 5y + 2,\]shows this trait with constant terms like 1 and 2 in each equation.
  • Such systems are more realistic, capturing effects that don't solely depend on system variables.
  • The equilibrium points in nonhomogeneous systems should be found by setting derivatives to zero, providing unique behavior insights not always present in homogeneous systems.
Understanding nonhomogeneous systems is key to solving many practical problems where variables and external inputs interact, making them crucial for engineers and scientists alike.

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Most popular questions from this chapter

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.

In each exercise, (a) Rewrite the given \(n\)th order scalar initial value problem as \(\mathbf{y}^{\prime}=\mathbf{f}(t, \mathbf{y}), \mathbf{y}\left(t_{0}\right)=\mathbf{y}_{0}\), by defining \(y_{1}(t)=y(t), y_{2}(t)=y^{\prime}(t), \ldots, y_{n}(t)=y^{(n-1)}(t)\) and defining \(y_{1}(t)=y(t), y_{2}(t)=y(t), \ldots, y_{n}(t)=y^{\prime(n-t)}(t)\) \(\mathbf{y}(t)=\left[\begin{array}{c}y_{1}(t) \\ y_{2}(t) \\ \vdots \\\ y_{n}(t)\end{array}\right]\) (b) Compute the \(n^{2}\) partial derivatives \(\partial f_{i}\left(t, y_{1}, \ldots, y_{n}\right) / \partial y_{j}, i, j=1, \ldots, n\). (c) For the system obtained in part (a), determine where in \((n+1)\)-dimensional \(t \mathbf{y}\)-space the hypotheses of Theorem \(6.1\) are not satisfied. In other words, at what points \(\left(t, y_{1}, \ldots, y_{n}\right)\), if any, does at least one component function \(f_{i}\left(t, y_{1}, \ldots, y_{n}\right)\) and/or at least one partial derivative function \(\partial f_{i}\left(t, y_{1}, \ldots, y_{n}\right) / \partial y_{i}, i, j=1, \ldots, n\) fail to be continuous? What is the largest open rectangular region \(R\) where the hypotheses of Theorem \(6.1\) hold? $$ y^{\prime \prime}+e^{t} y=\ln |t|, \quad y(-1)=0, \quad y^{\prime}(-1)=-1 $$

In each exercise, an initial value problem for a first order nonlinear system is given. Rewrite the problem as an equivalent initial value problem for a higher order nonlinear scalar differential equation. $$ \frac{d}{d t}\left[\begin{array}{l} y_{1} \\ y_{2} \\ y_{3} \end{array}\right]=\left[\begin{array}{c} y_{2} \\ y_{3} \\ \sqrt{y_{2} y_{3}+t^{2}} \end{array}\right], \quad\left[\begin{array}{l} y_{1}(1) \\ y_{2}(1) \\ y_{3}(1) \end{array}\right]=\left[\begin{array}{l} 1 \\ \frac{1}{2} \\ 3 \end{array}\right] $$

Consider the initial value problem \(y^{\prime \prime}+y^{2}=t, y(0)=y_{0}, y^{\prime}(0)=y_{0}^{\prime} .\) Can Laplace transforms be used to solve this initial value problem? Explain your answer.

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 \(\alpha\).

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