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Chapter 13: Problem 5

Consider the gradient system (13.17) in the case \(U(x, y)=x^{2}(x-1)^{2}+y^{2}\). Find the critical points and their character. Sketch the phase portrait for the system.

Short Answer

Expert verified
Question: Determine the character of the critical points of the function \(U(x, y) = x^2 (x - 1)^2 + y^2\). Answer: The critical points are (0,0), (1,0), and \((\frac{1}{2},0)\). Point (0,0) and (1,0) are local minima, and point \((\frac{1}{2},0)\) is a saddle point.

Step by step solution

01

Identify the gradient vector

The gradient vector of the given function \(U(x, y) = x^2 (x - 1)^2 + y^2\) is obtained by calculating the partial derivatives of U with respect to x and y: \(\nabla U(x, y) = \left(\frac{\partial U}{\partial x}, \frac{\partial U}{\partial y}\right)\) So, we first find \(\frac{\partial U}{\partial x}\): \(\frac{\partial U}{\partial x} = 2x(x-1)(2x-1)\) Next, we find \(\frac{\partial U}{\partial y}\): \(\frac{\partial U}{\partial y} = 2y\) Hence, the gradient vector \(\nabla U(x, y) = \left(2x(x-1)(2x-1), 2y\right)\).
02

Find the critical points

Critical points are the points where the gradient vector is zero. So, we have to set both components of the gradient vector to zero: \(2x(x-1)(2x-1) = 0\) \(2y = 0\) Solving the above equations, we obtain 3 critical points: A: \((0,0)\) B: \((1,0)\) C: \((\frac{1}{2},0)\)
03

Analyze the character of the critical points

To determine the character of the critical points, we need to calculate the Hessian matrix of the function U(x, y) and compute its eigenvalues at these points. The Hessian matrix is given by: $H(x, y) = \begin{bmatrix} \frac{\partial^2 U}{\partial x^2} & \frac{\partial^2 U}{\partial x \partial y} \\ \frac{\partial^2 U}{\partial y \partial x} & \frac{\partial^2 U}{\partial y^2} \end{bmatrix}$ For our problem, we have: \(\frac{\partial^2 U}{\partial x^2} = 12x^2 - 12x + 2\) \(\frac{\partial^2 U}{\partial x \partial y} = 0\) \(\frac{\partial^2 U}{\partial y \partial x} = 0\) \(\frac{\partial^2 U}{\partial y^2} = 2\) So the Hessian matrix is: $H(x, y) = \begin{bmatrix} 12x^2 - 12x + 2 & 0 \\ 0 & 2 \end{bmatrix}$ Now, let's find the eigenvalues of the Hessian matrix at each critical point: 1. At point A: \((0,0)\) $H(0,0) = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$ Eigenvalues: \(\lambda_1 = 2\), \(\lambda_2 = 2\) Since both eigenvalues are positive, point A is a local minimum. 2. At point B: \((1,0)\) $H(1,0) = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$ Eigenvalues: \(\lambda_1 = 2\), \(\lambda_2 = 2\) Similar to point A, point B is a local minimum. 3. At point C: \((\frac{1}{2},0)\) $H(\frac{1}{2},0) = \begin{bmatrix} -3 & 0 \\ 0 & 2 \end{bmatrix}$ Eigenvalues: \(\lambda_1 = -3\), \(\lambda_2 = 2\) Since one eigenvalue is positive and the other is negative, point C is a saddle point.
04

Sketch the phase portrait

To sketch the phase portrait, we need to plot the critical points and provide the direction of the gradient vector field. 1. Points A and B are local minima and will have an inward spiral (attraction) on the phase portrait. 2. Point C is a saddle point and will have both an attraction and a repulsion on the phase portrait. Based on this information, we can now plot the phase portrait for the given system.

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Key Concepts

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

Phase Portrait
In understanding dynamical systems, the phase portrait offers a visual representation of the behavior of a system's trajectories over time. It is particularly useful in depicting the motion of a system in a phase plane, where each point represents a possible state of the system.

For the function given in the exercise, the phase portrait can illustrate how points move on the plane according to the gradient vector field. Critical points of a dynamical system, such as equilibrium points, are highlighted in a phase portrait. In this case, points A and B, both local minima, act as attractors—any initial condition close to these points will result in the system moving towards them over time. On the other hand, point C, a saddle point, is a type of unstable equilibrium where trajectories approach the point along certain directions but are repelled along others.

Overall, the phase portrait for this system would show spiraling motions inward towards the local minima and a mixed behavior at the saddle point, providing a comprehensive visualization of the system's dynamics.
Hessian Matrix Eigenvalues
The Hessian matrix, a square matrix of second-order partial derivatives, is a critical tool in multivariate calculus. It gives us valuable insights into the curvature of multivariable functions, aiding in identifying the nature of critical points—whether they are minima, maxima, or saddle points.

The Hessian matrix's eigenvalues are particularly telling. When all eigenvalues are positive, the critical point is a local minimum, as local changes around the point will increase the function's value. Conversely, if all eigenvalues are negative, we're dealing with a local maximum. A mixture of positive and negative eigenvalues indicates a saddle point, showing that the function's value increases in some directions but decreases in others.

In the given problem, the eigenvalues of the Hessian at points A and B being positive suggest that these are points of local minimum. However, at point C, the mix of positive and negative eigenvalues reveals its saddle point nature, signifying a strategic inflection point in the topography of the function.
Gradient Vector Calculus
The gradient vector calculus is foundational in field theory and optimization, representing the direction and rate of the steepest ascent of a function. Computed as the vector of partial derivatives, the gradient points towards the direction in which the function increases most rapidly.

In this exercise, we calculate the gradient vector to identify the critical points, which occur where this vector is zero—indicating points where the function does not increase in any direction, hence it's either at a peak, valley, or inflection point.

For the given potential function, the gradient vector is used to find the critical points by equating its components to zero. This allows us to identify where the behavior of the system changes, such as where a particle would come to rest in a potential field. Grasping the gradient vector calculus enables students to better understand these transition points, describing the topography of multivariable functions and the dynamics of systems described by them.

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Most popular questions from this chapter

For the pendulum equation \(\ddot{\theta}+(g / l) \sin \theta=0\) find the equations of the trajectories in the phase plane and sketch the phase portrait of the system. Show that on the separatrices $$ \theta(t)=(2 n+1) \pi \pm 4 \arctan [\exp (\omega t+\alpha)] $$ where \(n\) is an integer, \(\alpha=\) constant and \(\omega=\sqrt{g / l}\). [Hint: Use the substitution \(\left.u=\tan \frac{1}{4}\\{\theta-(2 n+1) \pi\\} .\right]\)

For the Rikitake dynamo system (13.35): (a) Show that there are two real critical points at \(\left(\pm k, \pm 1 / k, \mu k^{2}\right)\) in the \(\left(X_{1}, X_{2}, Y\right)\) phase space, where \(k\) is given by \(A=\mu\left(k^{2}-1 / k^{2}\right)\). (b) Show that the stability of these critical points is determined by eigenvalues \(\lambda\) satisfying the cubic $$ (\lambda+2 \mu)\left[\lambda^{2}+\left(k^{2}+\frac{1}{k^{2}}\right)\right]=0 $$ so that the points are not asymptotically stable in this approximation. (For the full system they are actually unstable.) (c) Show that the divergence of the phase-space flow velocity is negative, so that the flow causes volume to contract. (d) Given that \(\bar{Y}=\sqrt{2}(Y-A / 2)\), show that $$ \frac{1}{2} \frac{\mathrm{d}}{\mathrm{d} t}\left(X_{1}^{2}+X_{2}^{2}+\bar{Y}^{2}\right)=-\mu\left(X_{1}^{2}+X_{2}^{2}\right)+\sqrt{2} \bar{Y} $$ and use this result to determine in which region of the space the trajectories all have a positive inward component towards \(X_{1}=\) \(X_{2}=\bar{Y}=0\) on surfaces \(X_{1}^{2}+X_{2}^{2}+\bar{Y}^{2}=\) constant

Consider the Lorenz system \((13.33)\). (a) Show that the origin \(P_{1}(0,0,0)\) is a critical point and that its stability depends on eigenvalues \(\lambda\) satisfying the cubic $$ (\lambda+\beta)\left[\lambda^{2}+(\sigma+1) \lambda+\sigma(1-\rho)\right]=0 $$ Hence show that \(P_{1}\) is asymptotically stable only when \(0<\rho<1\) (b) Show that there are two further critical points $$ P_{2}, P_{3} \equiv[\pm \sqrt{\beta(\rho-1)}, \pm \sqrt{\beta(\rho-1)},(\rho-1)] $$ when \(\rho>1\), and that their stability depends on eigenvalues \(\lambda\) satisfying the cubic $$ \lambda^{3}+\lambda^{2}(\sigma+\beta+1)+\lambda \beta(\sigma+\rho)+2 \sigma \beta(\rho-1)=0 $$ (c) Show that when \(\rho=1\) the roots of the cubic in (b) are \(0,-\beta,-(1+\sigma)\) and that in order for the roots to have the form \(-\mu, \pm \mathrm{i} \nu\) (with \(\mu, \nu\) real) we must have $$ \rho=\rho_{\text {crit }}=\frac{\sigma(\sigma+\beta+3)}{(\sigma-\beta-1)}>0 $$ (d) By considering how the roots of the cubic in (b) change continuously with \(\rho\) (with \(\left(\sigma, \beta\right.\) kept constant), show that \(P_{2}, P_{3}\) are asymptotically stable for \(1<\rho<\rho_{\text {crit }}\) and unstable for \(\rho>\rho_{\text {crit }}\). (e) Show that if \(\bar{z}=z-\rho-\sigma\), then $$ \frac{1}{2} \frac{\mathrm{d}}{\mathrm{d} t}\left(x^{2}+y^{2}+\bar{z}^{2}\right)=-\sigma x^{2}-y^{2}-\beta\left[\bar{z}+\frac{1}{2}(\rho+\sigma)\right]^{2}+\frac{1}{4} \beta(\rho+\sigma)^{2} $$ so that \(\left(x^{2}+y^{2}+\bar{z}^{2}\right)^{1 / 2}\) decreases for all states outside any sphere which contains a particular ellipsoid (implying the existence of an attractor).

The simple SIR model equations for the transmission of a disease are $$ \begin{aligned} \dot{S} &=-a S I \\ \dot{I} &=a S I-b I \\ \dot{R} &=b I \end{aligned} $$ where \(S(t), I(t), R(t)\) are respectively susceptibles, infectives, removed/recovered and \(a, b\) are positive constants. (a) Show that the overall population \(N=S+I+R\) remains constant, so that we may consider \((S, I)\) in a projected phase plane. Hence show that a trajectory with initial values \(\left(S_{0}, I_{0}\right)\) has equation \(I(S)=\) \(I_{0}+S_{0}-S+(b / a) \ln \left(S / S_{0}\right)\) (b) Using the function \(I(S)\) show that an epidemic can occur only if the number of susceptibles \(S_{0}\) in the population exceeds the threshold level \(b / a\) and that the disease stops spreading through lack of infectives rather than through lack of susceptibles. (c) For the trajectory which corresponds to \(S_{0}=(b / a)+\delta, I_{0}=\epsilon\) with \(\delta, \epsilon\) small and positive, show that, to a good approximation, there are \((b / a)-\delta\) susceptibles who escape infection [the KermackMcKendrick theorem of epidemiology \((1926 / 27)]\).

Show that Rayleigh's equation in the form (13.28) has a single critical point at \((0,0)\) and that this is always unstable. Making use of substitution \(\dot{x}=v / \sqrt{3}\) show that \(v\) satisfies the Van der Pol equation (13.29).
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