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Chapter 6: Problem 3

Show that the given system is almost linear with \((0,0)\) as a critical point, and classify this critical point as to type and stability. Use a computer system or graphing calculator to construct a phase plane portrait that illustrates your conclusion. $$ \frac{d x}{d t}=e^{x}+2 y-1, \frac{d y}{d t}=8 x+e^{y}-1 $$

Short Answer

Expert verified
The system is almost linear at \((0,0)\), which is a saddle point and unstable.

Step by step solution

01

Identify the Critical Point

To find critical points of the system, set both \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) to zero:1. \(e^x + 2y - 1 = 0\)2. \(8x + e^y - 1 = 0\).Solve these equations simultaneously. For solving our equations with \(x = 0\) and \(y = 0\), substitute back into:- \(e^0 + 2(0) - 1 = 0\) holds true.- \(8(0) + e^0 - 1 = 0\) also holds true.Thus, \((0,0)\) is a critical point of the system.
02

Linearize the System

To linearize the system, find the Jacobian matrix at the critical point \((0,0)\). The Jacobian matrix \(J\) is given by:\[ J = \begin{bmatrix} \frac{\partial}{\partial x}(e^x + 2y - 1) & \frac{\partial}{\partial y}(e^x + 2y - 1) \ \frac{\partial}{\partial x}(8x + e^y - 1) & \frac{\partial}{\partial y}(8x + e^y - 1) \end{bmatrix} \]Evaluating partial derivatives at \((0,0)\):\[ J = \begin{bmatrix} e^0 & 2 \ 8 & e^0 \end{bmatrix} = \begin{bmatrix} 1 & 2 \ 8 & 1 \end{bmatrix} \]
03

Evaluate Eigenvalues

To classify the critical point, find the eigenvalues of the Jacobian \(J\), \[ \begin{vmatrix} J - \lambda I \end{vmatrix} = 0 \]where \(I\) is the identity matrix.The characteristic equation is:\[ \begin{vmatrix} 1 - \lambda & 2 \ 8 & 1 - \lambda \end{vmatrix} = (1-\lambda)^2 - 16 = 0 \]which simplifies to \(\lambda^2 - 2 \lambda - 15 = 0\).Solving the quadratic equation for \(\lambda\), we find:\(\lambda = \frac{2 \pm \sqrt{4 + 60}}{2} = \frac{2 \pm 8}{2}\).Thus, \(\lambda = 5\) and \(\lambda = -3\).
04

Classify the Critical Point

Since one eigenvalue \(\lambda = 5\) is positive while the other \(\lambda = -3\) is negative, the critical point \((0,0)\) is a saddle point. This implies the system is unstable at this point.
05

Construct Phase Plane Portrait

Using computer software or a graphing calculator, plot the direction field and integral curves around the point \((0,0)\).The saddle point is characterized by trajectories moving away in certain directions (unstable manifolds) and moving toward it in others (stable manifolds). This bifurcation clarifies that the system is unstable.

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Key Concepts

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

Critical Points
In the context of differential equations, finding critical points is essential for understanding system behavior. Critical points are where both derivatives of the system equations, \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \), equal zero. This leads to a state where the system does not change over time, indicating equilibrium. By setting the equations \( e^x + 2y - 1 = 0 \) and \( 8x + e^y - 1 = 0 \) to zero, we solve for the variables \( x \) and \( y \). In this exercise, substituting \( x = 0 \) and \( y = 0 \) satisfies both equations, confirming that \( (0,0) \) is a critical point.
  • Critical points reflect where the system experiences no change, or a steady state.
  • The algebraic solution shows how both derivatives equal zero at \( (0,0) \).
Understanding critical points is crucial as they help identify the dynamics around these equilibrium states.
Linearization
Linearization is a technique used to approximate a nonlinear system near a critical point using a linear one. The process involves finding the Jacobian matrix, which describes the system's behavior in terms of its variables. For our system, we calculate the Jacobian matrix by partially differentiating the system equations around the critical point \( (0,0) \). The Jacobian matrix, \( J \), is formulated as:\[J = \begin{bmatrix} \frac{\partial}{\partial x}(e^x + 2y - 1) & \frac{\partial}{\partial y}(e^x + 2y - 1) \ \frac{\partial}{\partial x}(8x + e^y - 1) & \frac{\partial}{\partial y}(8x + e^y - 1) \end{bmatrix} \]Evaluating this at \( (0,0) \) yields: \[J = \begin{bmatrix} 1 & 2 \ 8 & 1 \end{bmatrix} \]
  • Linearization provides insights into system behavior around equilibrium.
  • The Jacobian matrix captures the local linear behavior of the system.
Linearization is invaluable as it simplifies complex nonlinear dynamics into understandable linear variations near equilibrium points.
Stability Analysis
Stability analysis requires us to examine if small deviations from a critical point fade away or amplify with time. This is determined by the eigenvalues of the Jacobian matrix. Calculating eigenvalues involves solving the characteristic equation derived from \( J - \lambda I = 0 \).Given \( J = \begin{bmatrix} 1 & 2 \ 8 & 1 \end{bmatrix} \), the characteristic equation turns into:\[(1-\lambda)^2 - 16 = 0 \]This simplifies to:\[\lambda^2 - 2\lambda - 15 = 0 \]Solving this quadratic equation yields eigenvalues \( \lambda = 5 \) and \( \lambda = -3 \). The presence of one positive \( \lambda \) and one negative \( \lambda \) indicates a saddle point.
  • Positive eigenvalues hint at instability, suggesting growth of perturbations.
  • Negative eigenvalues imply stability, indicating decay of perturbations.
When a critical point is classified as a saddle point, it signifies that the system is unstable as trajectories move away in certain directions, confirming the complexity of equilibrium in dynamical systems.

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Most popular questions from this chapter

Solve each of the linear systems to determine whether the critical point \((0,0)\) is stable, asymptotically stable, or unstable. Use a computer system or graphing calculator to construct a phase portrait and direction field for the given system. Thereby ascertain the stability or instability of each critical point, and identify it visually as a node, a saddle point, a center; or a spiral point. \(\frac{d x}{d t}=x, \quad \frac{d y}{d t}=3 y\)

Find each equilibrium solution \(x(t)=x_{0}\) of the given second-order differential equation \(x^{\prime \prime}+f\left(x, x^{\prime}\right)=0 .\) Use a computer system or graphing cal culator to construct a phase portrait and direction field for the equivalent first-order system \(x^{\prime}=y, y^{\prime}=-f(x, y) .\) Thereby ascertain whether the critical point \(\left(x_{0}, 0\right)\) looks like a center, a saddle point, or a spiral point of this system. \(x^{\prime \prime}+\left(x^{2}-1\right) x^{\prime}+x=0\)

In Problems, investigate the type of the critical point \((0,0)\) of the given almost linear system. Verify your conclusion by using a computer system or graphing calculator to construct a phase portrait. Also, describe the approximate locations and apparent types of any other critical points that are visible in your figure. Feel free to investigate these additional critical points; you can use the computational methods discussed in the application material for this section. $$ \frac{d x}{d t}=x-2 y+3 x y, \quad \frac{d y}{d t}=2 x-3 y-x^{2}-y^{2} $$

Deal with the system $$ \frac{d x}{d t}=F(x, y), \quad \frac{d y}{d t}=G(x, y) $$ in a region where the functions \(F\) and \(G\) are continuously differentiable, so for each number a and point \(\left(x_{0}, y_{0}\right)\), there is a unique solution with \(x(a)=x_{0}\) and \(y(a)=y_{0}\). Suppose that \((x(t), y(t))\) is a solution of the autonomous system and that \(\gamma \neq 0\). Define \(\phi(t)=x(t+\gamma)\) and \(\psi(t)=y(t+\gamma)\). Then show (in contrast with the situation in Problem 27) that \((\phi(t), \psi(t))\) is also a solution of the system. Thus autonomous systems have the simple but important property that a "t-translate" of a solution is again a solution.

In Problems, apply Theorem I to determine the type of the critical point \((0,0)\) and whether it is asymptotically stable, stable, or unstable. Verify your conclusion by using a computer system or graphing calculator to construct a phase portrait for the given linear system. $$ \frac{d x}{d t}=x-2 y, \quad \frac{d y}{d t}=5 x-y $$
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