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Determine if the given system is Hamiltonian. If the system is Hamiltonian, find its Hamiltonian function. $$ \begin{aligned} &x^{\prime}=y \\ &y^{\prime}=-x+x^{3} \end{aligned} $$

Short Answer

Expert verified
Yes, the system is Hamiltonian with \( H(x, y) = (x^2/2) - (x^4/4) + (y^2/2) \).

Step by step solution

01

Identify the system of equations

The given system is \( x' = y \) and \( y' = -x + x^3 \). We need to determine if it is a Hamiltonian system.
02

Define Hamiltonian system criteria

A system is Hamiltonian if it can be written as \( x' = \, \frac{\partial H}{\partial y} \) and \( y' = -\frac{\partial H}{\partial x} \), where \( H \) is the Hamiltonian function.
03

Matching given equations with Hamiltonian form

From the given equations, equate \( x' = y \) as \( x' = \frac{\partial H}{\partial y} \) and \( y' = -x + x^3 \ \) as \( y' = -\frac{\partial H}{\partial x} \).
04

Finding \( \frac{\partial H}{\partial y} \)

From \( x' = y = \frac{\partial H}{\partial y} \), we deduce that \( H \) must be a function of \( y \) such that its derivative with respect to \( y \) equals \( y \). Thus, \( H(x, y) = f(x) + \frac{1}{2}y^2 \).
05

Finding \( \frac{\Refreshing as myselfpartial H}{\partial x} \)

For the second equation, \( y' = -x + x^3 = -\frac{\partial H}{\partial x} \), we deduce that \( \frac{\partial H}{\partial x} = x - x^3 \). Therefore, integrate this partially with respect to \( x \).
06

Integrating \( \frac{\partial H}{\partial x} \, term

Integrate \( x - x^3 \) to find m the function of \( x \): \( \int (x - x^3) \, dx = \frac{x^2}{2} - \frac{x^4}{4} \).
07

Combine results for complete Hamiltonian function

Combine the results from Steps 4 and 6: \( H(x, y) = \frac{x^2}{2} - \frac{x^4}{4} + \frac{1}{2}y^2 + C \), where \( C \) is a constant.
08

Verify Hamiltonian Equations

Verify inf Hmction_vsrsult satisfies \_HAVE'analyaan equations: \( \frac{\partial }{\partial y} (\frac{x^2}{2} - \frac{x^4}{4} + \frac{1}{2} AC + C) = y \) and \(-\frac{\partial}{\partial x}(...) = V F - x + x^3 \), which V __IlsiAMMtelliis.

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Key Concepts

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

Differential Equations
Differential equations are the heart of many dynamical systems in mathematics, modeling how things change over time. In a differential equation, we focus on understanding functions and their rates of change. The problem at hand revolves around a system of differential equations given by:
  • \( x' = y \)
  • \( y' = -x + x^3 \)
Differential equations can be of various types, and they play a crucial role in predicting behavior in engineering, physics, and beyond. The primary task when working with such equations is to identify the relationships between the rates of change of the system's variables. In this case, the equations represent the evolution of a system over time. Understanding the behavior of solutions to these equations provides insight into the system dynamics, potentially indicating stability or identifying periodic nature. Here, the goal is to determine if these differential equations represent a Hamiltonian system.
Hamiltonian Function
The Hamiltonian function is a powerful concept in classical mechanics, offering a comprehensive energy-based view of dynamical systems. It serves as an energy function combining potential and kinetic energies to describe the total energy of the system. A system of differential equations is Hamiltonian if it can be expressed as:
  • \( x' = \frac{\partial H}{\partial y} \)
  • \( y' = -\frac{\partial H}{\partial x} \)
Where \( H = H(x, y) \) is the Hamiltonian function. Identifying the appropriate Hamiltonian not only classifies the system but also simplifies studying its behavior, as it provides a conservation law studied over time.For our problem, we start by deriving the possible form of \( H \) by equating and integrating equations derived from the system's conditions. By fulfilling these conditions, you establish a connection between the given parts of the system and the potential Hamiltonian, leading to a differential equation that provides insights into its energy states.
Partial Derivatives
Partial derivatives involve differentiating a function with respect to one of its variables while keeping other variables constant. They are fundamental in formulating and solving Hamiltonian systems, as they enable us to express dynamic systems in terms of derivatives with respect to different variables. For our system, you use:
  • Partial derivative of \( H \) with respect to \( y \): \( \frac{\partial H}{\partial y} = y \)
  • Partial derivative of \( H \) with respect to \( x \): \( -\frac{\partial H}{\partial x} = -x + x^3 \)
By setting up these equations, the information allows us to identify or verify the exact nature of the function \( H(x, y) \). Calculating partial derivatives accurately is crucial for correctly interpreting the solutions of a differential system and ensuring the system's Hamiltonian nature.
Integral Calculus
Integral calculus complements differential calculus, focusing on accumulation functions such as areas under curves. When dealing with Hamiltonian systems, once we identify expressions for the partial derivatives as done in previous sections, we often need to integrate these expressions to derive the Hamiltonian's functional form.In this exercise, the task involves integrating the partial derivative with respect to \( x \):\[\int (x - x^3) \, dx = \frac{x^2}{2} - \frac{x^4}{4}\]This integral is central to forming the Hamiltonian:\( H(x, y) = \frac{x^2}{2} - \frac{x^4}{4} + \frac{1}{2} y^2 + C \)where \( C \) is an integration constant.Understanding and performing these integrals correctly is crucial for accurately constructing the Hamiltonian function. Practicing these techniques enhances skills in solving complex dynamic systems and interpreting the natural phenomena they model.

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Most popular questions from this chapter

Find the equilibrium points and analyze their stability. $$ \begin{aligned} &x^{\prime}=(x+9 y)(1-y) \\ &y^{\prime}=-x-5 y \end{aligned} $$

Perform each of the following tasks. (i) Sketch the nullclines for each equation. Use a distinctive marking for each nullcline so they can be distinguished. (ii) Use analysis to find the equilibrium points for the system. Label each equilibrium point on your sketch with its coordinates. (iii) Use the Jacobian to classify each equilibrium point (spiral source, nodal sink, etc.). \(x^{\prime}=y\) \(y^{\prime}=-\sin x-y\)

Create a model for a food chain consisting of three kinds of fish, where the second preys on the first and the third preys on the second. The first species thrives on its own but has limited resources. The second species will survive without the first but has limited resources. The third will die out if the second is not present.

Consider a food chain consisting of three species with populations \(x_{1}(t), x_{2}(t)\), and \(x_{3}(t)\), where \(x_{1}\) preys upon \(x_{2}\), which in turn preys upon \(x_{3}\). Assume that both predator populations will die out in the absence of their prey. Assume a logistic limit for \(x_{3}\) only. (a) Show that the food chain is modeled by $$ \begin{aligned} x_{1}^{\prime} &=\left(-a_{1}+b_{1} x_{2}\right) x_{1} \\ x_{2}^{\prime} &=\left(-a_{2}-c_{2} x_{1}+b_{2} x_{3}\right) x_{2} \\ x_{3}^{\prime} &=\left(a_{3}-b_{3} x_{2}-d x_{3}\right) x_{3} \end{aligned} $$ where all of the constants are positive. (b) Consider the special case $$ \begin{aligned} &x_{1}^{\prime}=\left(-1+2 x_{2}\right) x_{1} \\ &x_{2}^{\prime}=\left(-0.5-2 x_{1}+4 x_{3}\right) x_{2} \\ &x_{3}^{\prime}=\left(2-x_{2}-2 x_{3}\right) x_{3} \end{aligned} $$ Find and analyze the equilibrium point with all populations positive.

Perform each of the following tasks. (i) Sketch the nullclines for each equation. Use a distinctive marking for each nullcline so they can be distinguished. (ii) Use analysis to find the equilibrium points for the system. Label each equilibrium point on your sketch with its coordinates. (iii) Use the Jacobian to classify each equilibrium point (spiral source, nodal sink, etc.). \(x^{\prime}=x(6-2 x-3 y)\) \(y^{\prime}=y(1-x-y)\)

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