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There are certain chemical reactions in which the constituent concentrations oscillate periodically over time. The system \(x^{\prime}=1-(b+1) x+\frac{1}{4} x^{2} y, \quad y^{\prime}=b x-\frac{1}{4} x^{2} y\) is a special case of a model, known as the Brusselator, of this kind of reaction. Assume that \(b\) is a positive parameter, and consider solutions in the first quadrant of the \(x y\) -plane. a. Show that the only critical point is \((1,4 b)\). b. Find the eigenvalues of the approximate linear system at the critical point. c. Classify the critical point as to type and stability. How does the classification depend on \(b ?\) d. As \(b\) increases through a certain value \(b_{0},\) the critical point changes from asymptotically stable to unstable. What is that value \(b_{0} ?\) e. Plot trajectories in the phase plane for values of \(b\) slightly less than and slightly greater than \(b_{0}\). Observe the limit cycle when \(b>b_{0} ;\) the Brusselator has a Hopf bifurcation point at \(b_{0}\). f. Plot trajectories for several values of \(b>b_{0}\) and observe how the limit cycle deforms as \(b\) increases.

Short Answer

Expert verified
Question: Determine the critical points, eigenvalues, stability, and bifurcations of the Brusselator dynamical system. Additionally, plot phase plane trajectories for different values of the parameter b and discuss its effects on the system. Answer: The critical point of the Brusselator is (1, 4b). The eigenvalues λ are obtained by solving the characteristic equation, which is given by λ^2 + λ - b^2 + 1/4 = 0. The real part of the eigenvalues (λ_r) determines the stability of the critical point: if b < 1/2, the critical point is an unstable spiral, while if b > 1/2, it is a stable spiral. The critical point changes its stability when b = 1/2. To plot phase plane trajectories, we need to numerically integrate the system for several values of b and visualize the outcomes. For b < 1/2, the system will exhibit a stable spiral. For b > 1/2, an unstable limit cycle occurs, indicating a Hopf bifurcation point at b = 1/2.

Step by step solution

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a. Finding the critical point

First, to find the critical point, we set the derivatives of x and y equal to zero and solve for x and y: \begin{align*} x^{\prime} = 0 &=1-(b+1)x+\frac{1}{4}x^{2}y\\ y^{\prime} = 0 &=bx- \frac{1}{4}x^{2}y \end{align*} We can rewrite the first equation as \(0 = x(\frac{1}{4}xy - (b+1)) + 1\). Since x=0 leads to a non-positive y, we must have \(\frac{1}{4}xy - (b+1) = 0\). Plugging this into the second equation, we find that \(y = 4b\). The critical point is \((1, 4b)\).
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b. Finding the eigenvalues of the linear system at the critical point

Now, we linearize the system by finding the Jacobian matrix evaluated at the critical point \((1, 4b)\): \[ J(x, y) = \begin{bmatrix} \frac{\partial x^{\prime}}{\partial x} & \frac{\partial x^{\prime}}{\partial y} \\ \frac{\partial y^{\prime}}{\partial x} & \frac{\partial y^{\prime}}{\partial y} \end{bmatrix} = \begin{bmatrix} -(b+1) + \frac{1}{4}2xy & \frac{1}{4}x^{2} \\ b - \frac{1}{4}2xy & -\frac{1}{4}x^{2} \end{bmatrix} \] At \((1,4b)\), the Jacobian matrix becomes: \[ J(1, 4b) = \begin{bmatrix} - b & \frac{1}{4} \\ b & -\frac{1}{4} \end{bmatrix} \] Now we find the eigenvalues by solving the characteristic equation \(\text{det}(J - \lambda I) = 0\): \[ \text{det}\begin{bmatrix} -b - \lambda & \frac{1}{4} \\ b & -\frac{1}{4} - \lambda \end{bmatrix} = \lambda^2 + \lambda - b^2 + \frac{1}{4} = 0 \]
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c. Classifying the critical point as to type and stability

To classify the critical point, we analyze the eigenvalues of the linearized system. Since both eigenvalues are complex, the critical point is a spiral point. Now we must determine whether it is stable or unstable by looking at the real part of both eigenvalues. This real part is given by: $$\lambda_r = -\frac{1}{2} + \frac{\sqrt{1-4(\frac{1}{4}-b^2)}}{2}$$ We see that the stability of the spiral point depends on the value of \(b\): when \(b < \frac{1}{2}\), \(\lambda_r > 0\), indicating the critical point is an unstable spiral; when \(b > \frac{1}{2}\), \(\lambda_r < 0\), indicating the critical point is a stable spiral.
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d. Finding the value of \(b_{0}\) at which the critical point changes its stability

Based on our analysis in part c, the critical point changes its stability when \(b\) crosses the value of \(\frac{1}{2}\), i.e., \(b_0 = \frac{1}{2}\).
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e. Plotting trajectories for the phase plane

To plot trajectories for the phase plane for values of \(b\) less than and greater than \(b_0\), we would first need to numerically integrate the system using a numerical solver such as Runge-Kutta. Then, we would plot the resulting trajectories in the \(xy\)-plane for values of \(b\) slightly less than and greater than \(\frac{1}{2}\). For \(b\) less than \(b_0\), we would observe a stable spiral, while for \(b\) greater than \(b_0\), we would observe an unstable limit cycle, indicating a Hopf bifurcation point at \(b_0\). (Plotting the trajectories is beyond the scope of this format and would require numerical methods and visualization software.)
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f. Plotting trajectories for several values of \(b > b_{0}\)

To plot trajectories for several values of \(b > \frac{1}{2}\), we again use a numerical solver such as Runge-Kutta to integrate the system for each chosen value of \(b\). Then, we plot the resulting trajectories in the \(xy\)-plane, observing how the limit cycle deforms as \(b\) increases. (Again, plotting the trajectories is beyond the scope of this format and would require numerical methods and visualization software.)

Key Concepts

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

Differential Equations
Differential equations are powerful mathematical tools used to describe the rate at which quantities change. They are equations that relate a function to its derivatives, providing a dynamic system that unfolds over time or space. In the context of the Brusselator model, we encounter a system of nonlinear ordinary differential equations (ODEs) used to represent the evolution of chemical concentrations over time.

The Brusselator's ODEs capture the essence of feedback mechanisms in chemical reactions where the rate of change of one reactant influences, and is influenced by, the concentration of another reactant. Identifying the critical points—the conditions under which the system's state does not change over time—is the first step towards understanding the system's behavior. We do this by setting the derivatives equal to zero and solving for the relevant variables, which allows us to examine the system's dynamics near these points.
Linear Stability Analysis
Linear stability analysis is a method of studying the stability of equilibrium points within differential equations. It involves approximating the nonlinear system with a linear one, near the equilibrium point, by evaluating the Jacobian matrix there. This matrix contains the partial derivatives of the functions with respect to the state variables.

By examining the eigenvalues of the Jacobian, we gain insight into the behavior of the system around the critical point. If all eigenvalues have negative real parts, the system is stable, and any small disturbance will decay over time. On the other hand, if at least one eigenvalue has a positive real part, the equilibrium is unstable, and perturbations will grow. This gives us a clearer view of how the system might react to external influences and perturbations in the region near the equilibrium state.
Hopf Bifurcation
A Hopf bifurcation is a type of bifurcation, or structural change, in which an equilibrium point of a differential equation goes from stability to instability or vice versa as a parameter is varied. It typically leads to the creation or destruction of a periodic solution, which is the limit cycle.

In the Brusselator model, the parameter in question is the positive constant 'b'. As 'b' increases and passes through a critical threshold, the system transitions from having a stable equilibrium to having a stable oscillatory behavior. This Hopf bifurcation is fundamental to understanding how oscillatory chemical reactions can arise from changes in reaction rates or chemical concentrations.
Limit Cycle
A limit cycle is a closed trajectory in the phase space of a dynamical system that represents a periodic oscillation. Importantly, it is isolated, meaning that neighboring trajectories do not form closed loops. In chemical reactions, limit cycles depict the sustained periodic concentration changes of reactants over time.

Observing a limit cycle in the phase plane of the Brusselator model signifies that the system has experienced a Hopf bifurcation. For values of 'b' beyond the critical point, the system exhibits these periodic oscillations. The Brusselator's reaction transition from a non-oscillatory state to a limit cycle showcases the nonlinear behavior and multiple steady states that are characteristic of such chemical systems. Visualizing these oscillations for different values of 'b' highlights how the system's dynamics evolve with the change in the key parameter.

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Most popular questions from this chapter

Determine the periodic solutions, if any, of the system $$ \frac{d x}{d t}=y+\frac{x}{\sqrt{x^{2}+y^{2}}}\left(x^{2}+y^{2}-2\right), \quad \frac{d y}{d t}=-x+\frac{y}{\sqrt{x^{2}+y^{2}}}\left(x^{2}+y^{2}-2\right) $$

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For certain \(r\) intervals, or windows, the Lorenz equations exhibit a period- doubling property similar to that of the logistic difference equation discussed in Section \(2.9 .\) Careful calculations may reveal this phenomenon. (a) One period-doubling window contains the value \(r=100 .\) Let \(r=100\) and plot the trajectory starting at \((5,5,5)\) or some other initial point of your choice. Does the solution appear to be periodic? What is the period? (b) Repeat the calculation in part (a) for slightly smaller values of \(r .\) When \(r \cong 99.98\), you may be able to observe that the period of the solution doubles. Try to observe this result by performing calculations with nearby values of \(r\). (c) As \(r\) decreases further, the period of the solution doubles repeatedly. The next period doubling occurs at about \(r=99.629 .\) Try to observe this by plotting trajectories for nearby values of \(r .\)

(a) Find an equation of the form \(H(x, y)=c\) satisfied by the trajectories. (b) Plot several level curves of the function \(H\). These are trajectories of the given system. Indicate the direction of motion on each trajectory. $$ d x / d t=y, \quad d y / d t=2 x+y $$

(a) Determine all critical points of the given system of equations. (b) Find the corresponding linear system near each critical point. (c) Find the eigenalues of each linear system. What conclusions can you then draw about the nonlinear system? (d) Draw a phase portrait of the nonlinear system to confirm your conclusions, or to extend them in those cases where the linear system does not provide definite information about the nonlinear system. $$ d x / d t=-2 x-y-x\left(x^{2}+y^{2}\right), \quad d y / d t=x-y+y\left(x^{2}+y^{2}\right) $$

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