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Use Theorem \(6.2\) to decide whether the given system is a Hamiltonian system. If it is, find a Hamiltonian function for the system. $$ \begin{aligned} &x^{\prime}=-\sin (2 x y)-x \\ &y^{\prime}=\sin (2 x y)+y \end{aligned} $$

Short Answer

Expert verified
If so, find the Hamilton function. The given system of differential equations is a Hamiltonian system with the Hamiltonian function: $$ H(x, y) = -x^2 - xy + \frac{1}{2}y^2 + y\cos(2xy). $$

Step by step solution

01

Compare equations to Hamiltonian system equations

Compare the given system of equations: $$ \begin{aligned} x' &= -\sin(2xy) - x, \\ y' &= \sin(2xy) + y, \end{aligned} $$ to the equations for a Hamiltonian system: $$ \begin{aligned} x' &= \frac{\partial H}{\partial y}, \\ y' &= -\frac{\partial H}{\partial x}. \end{aligned} $$
02

Find partial derivatives of the potential Hamiltonian function

From the comparison of equations in Step 1, we can determine the partial derivatives of the potential Hamiltonian function: $$ \begin{aligned} \frac{\partial H}{\partial y} &= -\sin(2xy) - x, \\ -\frac{\partial H}{\partial x} &= \sin(2xy) + y. \end{aligned} $$ Now, let's integrate these partial derivatives to find H(x, y).
03

Integrate partial derivatives to determine H(x, y)

First, let's integrate the first equation with respect to y: $$ H(x, y) = \int (-\sin(2xy) - x)dy. $$ Similarly, let's integrate the second equation with respect to x: $$ H(x, y) = -\int (\sin(2xy) + y)dx. $$
04

Find H(x, y) using integration results

The results of the integrations are: $$ H(x, y) = x\cos(2xy) - xy + g(x), $$ and $$ H(x, y) = y\cos(2xy) + \frac{1}{2}y^2 + h(y). $$ Comparing these expressions for H(x, y), we can deduce that: $$ H(x, y) = -x^2 - xy + \frac{1}{2}y^2 + y\cos(2xy). $$
05

Conclusion

The given system is a Hamiltonian system since we were able to find a Hamiltonian function H(x, y). The Hamiltonian function for the system is: $$ H(x, y) = -x^2 - xy + \frac{1}{2}y^2 + y\cos(2xy). $$

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Key Concepts

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

Hamiltonian Function
Hamiltonian systems are a special class of dynamical systems characterized by a Hamiltonian function, denoted as \( H(x, y) \). This function represents the total energy of the system. In general, a Hamiltonian system consists of equations of motion derived from the principle of stationary action, with time derivatives expressed in terms of partial derivatives of \( H \). In this context, energy conservation plays a pivotal role, as the Hamiltonian function combines potential and kinetic energy terms.
Hamiltonian Mechanics offers an elegant framework with the equations defined as:
  • \( x' = \frac{\partial H}{\partial y} \)
  • \( y' = -\frac{\partial H}{\partial x} \)
In a practical sense, finding a Hamiltonian function involves ensuring that your dynamical system fits this form. If successful, it means you’ve confirmed the system conserves some form of energy, making it a Hamiltonian system.
Partial Derivatives
When dealing with Hamiltonian systems, partial derivatives come into play as key tools. These derivatives measure how a multivariable function changes when one of its variables changes, keeping others constant. For a Hamiltonian function \( H(x, y) \), partial derivatives are used to express the system’s dynamics:
  • \( \frac{\partial H}{\partial y} \) and \( \frac{\partial H}{\partial x} \)
These equations help us understand how the change in one variable affects the whole system. Calculating partial derivatives allows us to determine the influence of each variable on the energy represented by the Hamiltonian. Finding these derivatives is the first step in unraveling complex systems and understanding their behavior through the lens of energy dynamics.
Integration
Integration is a powerful mathematical tool used to find a function given its derivative. In Hamiltonian systems, integration plays a crucial role in constructing the Hamiltonian function \( H(x, y) \) from known partial derivatives. Once the partial derivatives \( \frac{\partial H}{\partial y} \) and \( -\frac{\partial H}{\partial x} \) have been identified, integration helps us to determine the full Hamiltonian function.
To perform integration:
  • Integrate \( \frac{\partial H}{\partial y} \) with respect to \( y \)
  • Integrate \( -\frac{\partial H}{\partial x} \) with respect to \( x \)
These steps create expressions for \( H \) from each derivative calculation. By combining results from both integrations, we form the complete Hamiltonian function revealing the energy landscape of the system.
Differential Equations
Differential equations are foundational in describing how a system evolves over time. In the context of Hamiltonian systems, differential equations come in pairs, derived from the partial derivatives of the Hamiltonian function \( H(x, y) \). Solving these equations yields insights into the system's trajectory through its phase space.
The specific form of differential equations we encounter:
  • \( x' = \frac{\partial H}{\partial y} \)
  • \( y' = -\frac{\partial H}{\partial x} \)
These represent the rates of change of the system's coordinates. Solving these differential equations allows us to predict the future states of the system based solely on initial conditions, capturing the dynamic essence of Hamiltonian mechanics.

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Most popular questions from this chapter

Locate the unique equilibrium point of the given nonhomogeneous system, and determine the stability properties of this equilibrium point. Is it asymptotically stable, stable but not asymptotically stable, or unstable? $$ \begin{aligned} &x^{\prime}=-x+y+1 \\ &y^{\prime}=-10 x+5 y+2 \end{aligned} $$

In each exercise, the given system is an almost linear system at each of its equilibrium points. (a) Find the (real) equilibrium points of the given system. (b) As in Example 2, find the corresponding linearized system \(\mathbf{z}^{\prime}=A \mathbf{z}\) at each equilibrium point. (c) What, if anything, can be inferred about the stability properties of the equilibrium point(s) by using Theorem \(6.4\) ? $$ \begin{aligned} &x^{\prime}=(x-y)(y+1) \\ &y^{\prime}=(x+2)(y-4) \end{aligned} $$

Introduce polar coordinates and transform the given initial value problem into an equivalent initial value problem for the polar variables. Solve the polar initial value problem, and use the polar solution to obtain the solution of the original initial value problem. If the solution exists at time \(t=1\), evaluate it. If not, explain why. $$ \begin{aligned} &x^{\prime}=x+x \sqrt{x^{2}+y^{2}}, \quad x(0)=1 \\ &y^{\prime}=y+y \sqrt{x^{2}+y^{2}}, \quad y(0)=\sqrt{3} \end{aligned} $$

Each exercise lists the general solution of a linear system of the form $$ \begin{aligned} &x^{\prime}=a_{11} x+a_{12} y \\ &y^{\prime}=a_{21} x+a_{22} y \end{aligned} $$ where \(a_{11} a_{22}-a_{12} a_{21} \neq 0\). Determine whether the equilibrium point \(\mathbf{y}_{e}=\mathbf{0}\) is asymptotically stable, stable but not asymptotically stable, or unstable. $$ \begin{aligned} &x=c_{1} \cos 2 t+c_{2} \sin 2 t \\ &y=-c_{1} \sin 2 t+c_{2} \cos 2 t \end{aligned} $$

In each exercise, the eigenpairs of a \((2 \times 2)\) matrix \(A\) are given where both eigenvalues are real. Consider the phase-plane solution trajectories of the linear system \(\mathbf{y}^{\prime}=A \mathbf{y}\), where $$ \mathbf{y}(t)=\left[\begin{array}{l} x(t) \\ y(t) \end{array}\right] $$ (a) Use Table \(6.2\) to classify the type and stability characteristics of the equilibrium point at \(\mathbf{y}=\mathbf{0}\). (b) Sketch the two phase-plane lines defined by the eigenvectors. If an eigenvector is \(\left[\begin{array}{l}u_{1} \\ u_{2}\end{array}\right]\), the line of interest is \(u_{2} x-u_{1} y=0\). Solution trajectories originating on such a line stay on the line; they move toward the origin as time increases if the corresponding eigenvalue is negative or away from the origin if the eigenvalue is positive. (c) Sketch appropriate direction field arrows on both lines. Use this information to sketch a representative trajectory in each of the four phase- plane regions having these lines as boundaries. Indicate the direction of motion of the solution point on each trajectory. $$ \lambda_{1}=-2, \quad \mathbf{x}_{1}=\left[\begin{array}{l} 1 \\ 0 \end{array}\right] ; \quad \lambda_{2}=-1, \quad \mathbf{x}_{2}=\left[\begin{array}{l} 1 \\ 1 \end{array}\right] $$

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