/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 8 Is the given function positive d... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 8

Is the given function positive definite in an open neighborhood containing \((0,0)\) ? Positive semidefinite? Negative definite? Negative semidefinite? None of these? Justify your answer in each case. $$ V(x, y)=2 x y-x^{2}-y^{2} $$

Short Answer

Expert verified
The function is negative semi-definite.

Step by step solution

01

Identify the Hessian Matrix

The Hessian matrix for a function \(V(x, y)\) is a square matrix of second-order mixed partial derivatives. For the given function, \( V(x, y) = 2xy - x^2 - y^2 \), the Hessian matrix \( H \) is:\[ H = \begin{bmatrix} \frac{\partial^2 V}{\partial x^2} & \frac{\partial^2 V}{\partial x \partial y} \ \frac{\partial^2 V}{\partial y \partial x} & \frac{\partial^2 V}{\partial y^2} \end{bmatrix} \]Calculate each element: \(\frac{\partial^2 V}{\partial x^2} = -2\), \(\frac{\partial^2 V}{\partial y^2} = -2\), and \(\frac{\partial^2 V}{\partial x \partial y} = 2\). Thus, \[ H = \begin{bmatrix} -2 & 2 \ 2 & -2 \end{bmatrix} \].
02

Determine the Definiteness Using Determinants

To determine definiteness, calculate the eigenvalues of the Hessian matrix or use leading principal minors. For matrix \( H \), the determinant of the top-left corner (\(-2\)) is \(-2\), which indicates typically negative definiteness could be possible. Check the determinant of the entire matrix: \[ \text{det}(H) = (-2)(-2) - (2)(2) = 4 - 4 = 0 \]. If the determinant is zero and the first minor determinant is negative, this matrix is not negative definite nor positive definite.
03

Analyze the Eigenvalues (Alternative Method)

For further verification, consider the eigenvalues of \( H \). The characteristic equation is obtained by \( \text{det}(H - \lambda I) = 0 \):\[ \text{det}\left(\begin{bmatrix} -2-\lambda & 2 \ 2 & -2-\lambda \end{bmatrix}\right) = (-2-\lambda)^2 - 4 = 0 \] This simplifies to \( \lambda^2 + 4\lambda = 0 \), giving eigenvalues \( \lambda = 0 \) and \( \lambda = -4 \). The presence of a non-positive and negative eigenvalue indicates that the function is negative semi-definite.
04

Conclusion on Definiteness

The function \( V(x, y) = 2xy - x^2 - y^2 \) has a Hessian matrix with determinant zero and a negative eigenvalue, which means it is neither positive definite nor negative definite. Since there is at least one negative eigenvalue, the function is negative semi-definite.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Positive Definite
A matrix is considered positive definite if every eigenvalue is positive and the leading principal minors are also positive. For our function, we determine its positive definiteness by examining its Hessian matrix. The general idea is that:
  • If all eigenvalues are positive, the matrix is positive definite.
  • If any eigenvalue is zero or negative, it cannot be positive definite.
When we apply this to the given Hessian matrix, calculated from the function's second derivatives, we find that the matrix has the eigenvalues: \( 0 \) and \(-4\). Since there is a zero eigenvalue, the matrix does not meet the criteria for positive definiteness. Therefore, this function is not positive definite.
Negative Definite
A matrix can be termed as negative definite if all its eigenvalues are negative. For the function we are analyzing, the Hessian matrix helps determine this by uncovering its eigenvalues. Here are the key points:
  • The Hessian matrix has eigenvalues of \( 0 \) and \(-4\).
  • To be negative definite, all eigenvalues should be strictly negative, which isn't the case here because one eigenvalue is zero.
Thus, despite having one negative eigenvalue, the presence of a zero eigenvalue implies it cannot be negative definite. Negative definiteness requires no zero values and all values strictly less than zero.
Eigenvalues
Eigenvalues are crucial in revealing the nature of a matrix, particularly when determining definiteness. For the Hessian matrix:
  • The characteristic equation is derived from \(\text{det}(H - \lambda I) = 0\).
  • Solving the equation yields the eigenvalues \(0\) and \(-4\).
This calculation suggests the behavior of the function's curvature at a point. The zero eigenvalue indicates a direction of constant change at that point. A negative eigenvalue indicates a direction of downward curvature. These eigenvalues are essential not just in pinpointing definiteness but also for understanding the function鈥檚 behavior.
Definiteness
Definiteness of a matrix is a key property that gives us insight into the behavior of multidimensional functions. With the given Hessian matrix, definiteness can be assessed using the leading principal minors or the eigenvalues. Key ideas include:
  • If all eigenvalues are positive, it is positive definite.
  • If all eigenvalues are negative, it is negative definite.
  • A zero eigenvalue combined with no positive ones often implies negative semidefinite.
The Hessian has a determinant of zero. The calculated eigenvalues \(0\) and \(-4\) show us that the matrix cannot be entirely positive or negative definite. Due to one being negative, the matrix is defined as negative semi-definite, meaning the function has some level of downward concavity but is not strictly decreasing in all directions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Find the equilibrium points and analyze their stability. $$ \begin{aligned} &x^{\prime}=y-z \\ &y^{\prime}=x-z \\ &z^{\prime}=x^{2}+y^{2}-2 z \end{aligned} $$

A chemostat \({ }^{2}\) is a device used to maintain a constant supply of bacteria for biological studies in the laboratory. Think of a chamber containing bacteria and nutrients essential for the bacteria's growth. The bacteria eat the nutrient and multiply. A nutrient reservoir replenishes lost nutrients, and the experimentalist harvests available bacteria as needed for research. Let \(N\) and \(C\) represent the number of bacteria in the chamber and the concentration of the nutrient, respectively. One possible model for the chemostat (using dimensionless variables) is given by the system $$ \begin{aligned} \frac{d N}{d t} &=\alpha_{1}\left(\frac{C}{1+C}\right) N-N \\ \frac{d C}{d t} &=-\left(\frac{C}{1+C}\right) N-C+\alpha_{2} \end{aligned} $$ where \(\alpha_{1}, \alpha_{2}>0\). (a) Show that the system has two equilibrium points, $$ \left(N_{1}, C_{1}\right)=\left(\alpha_{1}\left(\alpha_{2}-\frac{1}{\alpha_{1}-1}\right), \frac{1}{\alpha_{1}-1}\right) $$ and \(\left(N_{2}, C_{2}\right)=\left(0, \alpha_{2}\right)\). (b) If we assume that \(\alpha_{1}>1\) and \(\alpha_{2}>1 /\left(\alpha_{1}-1\right)\), then the equilibrium point \(\left(N_{1}, C_{1}\right)\) lies in the first quadrant. Show that the Jacobian, evaluated at the first equilibrium point, is given by $$ J\left(N_{1}, C_{1}\right)=\left(\begin{array}{cc} 0 & \alpha_{1} A \\ -1 / \alpha_{1} & -(A+1) \end{array}\right) . $$ where \(A=N_{1} /\left(1+C_{1}\right)^{2}\). Use this result to show that the equilibrium point \(\left(N_{1}, C_{1}\right)\) is stable. (c) Show that $$ J\left(N_{2}, C_{2}\right)=\left(\begin{array}{cr} \alpha_{1} B-1 & 0 \\ -B & -1 \end{array}\right), $$ where \(B=\alpha_{2} /\left(1+\alpha_{2}\right)\). Assuming the same conditions as in part (b), show that the equilibrium point at \(\left(N_{2}, C_{2}\right)\) is a saddle. (d) Choose numerical values for \(\alpha_{1}\) and \(\alpha_{2}\) satisfying \(\alpha_{1}>1\) and \(\alpha_{2}>1 /\left(\alpha_{1}-1\right)\). Sketch the nullclines in the \(N C\) phase plane, label equilibrium points with their coordinates, and indicate the direction of solution trajectories across each nullcline. Then use all of this information to sketch the phase portrait in the first quadrant for the chemostat model. Check your results with your numerical solver.

Find and analyze the equilibrium points for the Lorenz system when \(a=10, b=8 / 3\), and \(r=18\). Numerically compute and plot a solution that starts near each of the equilibrium points.

Show that the \(x\) - and \(y\)-axes are invariant sets for the given system. Then explain why each of the four quadrants is invariant for the given system. $$ \begin{aligned} &x^{\prime}=4 x(3-x)-3 x y \\ &y^{\prime}=y(1-y)-x y \end{aligned} $$

Consider the system $$ \begin{aligned} &y^{\prime}=v_{+} \\ &v^{\prime}=f(y) \end{aligned} $$ In Exercises 13-16, use the given \(f(y)\) to aid in completing each of the following tasks: (i) Use your numerical solver to sketch the phase portrait of the system. Label and classify each equilibrium point: center, saddle, and so on. (ii) Use the relation $$ \frac{1}{2} v^{2}+U(y)=E $$ to find the equation of the separatrices. Sketch these separatrices on the phase portrait developed in part (i). $$ f(y)=y^{3}-y $$
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Logic and Functions

Read Explanation

Probability and Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

Read Explanation

Discrete Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.