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Chapter 13: Problem 11

Show that the origin is the only critical point for the system $$ \begin{aligned} &\dot{x}=-y+\alpha x\left(\beta-x^{2}-y^{2}\right) \\ &\dot{y}=x+\alpha y\left(\beta-x^{2}-y^{2}\right) \end{aligned} $$ where \(\alpha, \beta\) are real parameters, with \(\alpha\) fixed and positive and \(\beta\) allowed to take different values. Show that the character of the critical point and the existence of a limit cycle depend on the parameter \(\beta\), so that the system undergoes a supercritical Hopf bifurcation at \(\beta=0\). (Hint: Make the change from Cartesian co-ordinates to plane polars.)

Short Answer

Expert verified
Question: Describe the process of determining the existence of a supercritical Hopf bifurcation in the given system of differential equations and explain how the parameter β affects the presence and stability of a limit cycle. Answer: To determine the existence of a supercritical Hopf bifurcation, we first find the critical points by setting the given equations for ẋ and ẏ to zero. The origin (0, 0) is the only critical point we find in this case. We then change from Cartesian coordinates to polar coordinates (x, y) to (r, θ), and express ẋ and ẏ in terms of r and θ using the chain rule. We solve for ṙ and θ̇ by solving the system of two equations we obtain. Analyzing the character of the critical point using the equations for ṙ and θ̇, we see that β determines the presence and stability of a limit cycle. If β > 0, we have a stable limit cycle; if β < 0, we have an unstable limit cycle. When β = 0, there is no limit cycle and the system undergoes a supercritical Hopf bifurcation.

Step by step solution

01

Finding the critical points

To find the critical points, we need to set the equations for \(\dot{x}\) and \(\dot{y}\) to zero: $$ \begin{aligned} & 0 = -y + \alpha x (\beta - x^2 - y^2) \\ & 0 = x + \alpha y (\beta - x^2 - y^2) \end{aligned} $$ Now we'll solve this system of equations for \(x\) and \(y\). Since both equations are equal to zero, let's add the two equations together: $$ 0 = x - y + \alpha(x+y)(\beta-x^2-y^2) $$ We can notice that \((x-y)\) and \((x+y)\) are factors. Factorizing and simplifying the equation, we get: $$ 0=(x-y)(1+\alpha \beta - \alpha x^{2}- \alpha y^{2}) $$ Solving for \(x\) and \(y\), we find that \(x=y\) or \(1+\alpha \beta - \alpha x^{2}- \alpha y^{2}=0\). Since \(x=y\), we plug this back into either one of our original equations, and we get \(0 = x - x\). This shows that the origin (0, 0) is the only critical point.
02

Changing to polar coordinates

To analyze the character of the critical point and the existence of a limit cycle, we'll change from Cartesian coordinates to polar coordinates, with \(x = r \cos \theta\) and \(y = r \sin \theta\). Also, we'll use the chain rule to express \(\dot{r}\) and \(\dot{\theta}\): $$ \begin{aligned} \dot{r} \cos \theta - r \sin \theta \dot{\theta} &= - r \sin \theta + \alpha r \cos \theta (\beta - r^2) \\ \dot{r} \sin \theta + r \cos \theta \dot{\theta} &= r \cos \theta + \alpha r \sin \theta (\beta - r^2) \end{aligned} $$
03

Solve for \(\dot{r}\) and \(\dot{\theta}\)

Now we have a system of two equations: $$ \begin{aligned} & \dot{r} \cos \theta - r \sin \theta \dot{\theta} + r \sin \theta - \alpha r \cos \theta (\beta - r^2) = 0 \\ & \dot{r} \sin \theta + r \cos \theta \dot{\theta} - r \cos \theta - \alpha r \sin \theta (\beta - r^2) = 0 \end{aligned} $$ Solving this system for \(\dot{r}\) and \(\dot{\theta}\) gives us: $$ \begin{aligned} & \dot{r} = \alpha r (\beta - r^2) \\ & \dot{\theta} = 1 \end{aligned} $$
04

Analyzing the character of the critical point and the existence of a limit cycle

The equations we found in the previous step are: $$ \begin{aligned} & \dot{r} = \alpha r (\beta - r^2) \\ & \dot{\theta} = 1 \end{aligned} $$ These equations show that the position of the limit cycle depends on the parameter \(\beta\). When \(\beta > 0\), the value of \(\dot{r}\) is positive, meaning that the system has a stable limit cycle. When \(\beta < 0\), the value of \(\dot{r}\) is negative, meaning that the system has an unstable limit cycle. When \(\beta = 0\), there is no limit cycle, and the system undergoes a supercritical Hopf bifurcation.

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Most popular questions from this chapter

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).

In contrast to Problem 16 , consider the perfectly elastic bouncing of a ball vertically under gravity above the plane \(x=0\). We have \(\dot{x}=y, \dot{y}=\) \(-g\) and we can solve for \(x\left(x_{0}, y_{0}, t\right), y\left(x_{0}, y_{0}, t\right)\) in terms of the initial data \(\left(x_{0}, y_{0}\right)\) at \(t=0\). Show that in this case the resulting perturbations \(\Delta x, \Delta y\) essentially grow linearly with time \(t\) along the trajectory when we make perturbations \(\Delta x_{0}, \Delta y_{0}\) in the initial data. That is to say the distance along the trajectory \(d \equiv \sqrt{(\Delta x)^{2}+(\Delta y)^{2}} \sim \kappa t\) when \(t\) is large and \(\kappa\) is a suitable constant.

For the Lotka-Volterra system (13.18) show that the trajectories in the phase plane are given by \(f(x, y)=\) constant as in (13.20). In the first quadrant \(x \geq 0, y \geq 0\), the intersections of a line \(y=\) constant with a trajectory are given by \(-c \ln x+d x=\) constant. Hence show that there are 0,1 or 2 such intersections, so that the equilibrium point \((c / d, a / b)\) fore this system is a true centre (i.e. it cannot be a spiral point). Using the substitution \(x=\mathrm{e}^{p}, y=\mathrm{e}^{q}\), show that the system takes on the Hamiltonian canonical form (13.22).
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