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For the given system, (a) Use Theorem \(6.2\) to show that the system is a Hamiltonian system. (b) Find a Hamiltonian function for the system. (c) Use computational software to graph the phase-plane trajectory passing through \((1,1)\). Also, indicate the direction of motion for the solution point. $$ \begin{aligned} &x^{\prime}=-2 y \\ &y^{\prime}=3 x^{2} \end{aligned} $$

Short Answer

Expert verified
To summarize, we showed that the given system of differential equations is indeed a Hamiltonian system with the Hamiltonian function $$ H(x,y) = -y^2 - x^3 + C. $$ We also computed the direction of motion at point \((1,1)\), which is given by the vector \((-2,3)\). To graph the phase-plane trajectory passing through \((1,1)\), you can use computational software like Mathematica or MATLAB, and the plot should show the trajectory as well as the direction of motion at that point.

Step by step solution

01

Recall Hamiltonian Systems

A Hamiltonian system is a system of ordinary differential equations of the form $$ \begin{aligned} \dot{q} &= \frac{\partial H}{\partial p}, \\ \dot{p} &= -\frac{\partial H}{\partial q}, \end{aligned} $$ where \(H\) is a scalar function called the Hamiltonian, \(q\) represents the generalized coordinates, and \(p\) represents the generalized momenta. Theorem 6.2 (which needs to be used) states that if a system can be written in the form above, it is a Hamiltonian system.
02

Show that the System is Hamiltonian

Given the system, $$ \begin{aligned} &x^{\prime}=-2 y, \\ &y^{\prime}=3 x^{2}. \end{aligned} $$ we want to compare it to the Hamiltonian form to see if we can identify a Hamiltonian function \(H(x,y)\) such that: $$ \begin{aligned} &x^{\prime}=\frac{\partial H}{\partial y}=-2y, \\ &y^{\prime}=-\frac{\partial H}{\partial x}=3x^2. \end{aligned} $$ From the first equation, we have $$ \frac{\partial H}{\partial y}=-2y. $$ Integrating with respect to \(y\), we get $$ H(x,y) = -y^2 + f(x), $$ where \(f(x)\) is the function of integration. Now, we want to find \(f(x)\), and we will use the second equation $$ y^{\prime}=-\frac{\partial H}{\partial x}=3x^2, $$ which gives $$ -\frac{\partial H}{\partial x}=-\frac{\partial(-y^2 + f(x))}{\partial x}=3x^2. $$ Taking the derivative of \(H(x,y)\) with respect to \(x\) , we get $$ -\frac{\partial f}{\partial x} = 3x^2. $$ Integrating with respect to \(x\), we find $$ f(x) = -x^3 + C, $$ where \(C\) is a constant. Thus, the Hamiltonian function that satisfies the given system is $$ H(x,y) = -y^2 - x^3 + C. $$
03

Plot the Phase-Plane Trajectory of the System

We will use computational software to plot the phase-plane trajectory starting at the point \((1,1)\). The direction of motion at the point \((1,1)\) is given by the derivatives \(x^{\prime}\) and \(y^{\prime}\): $$ \begin{aligned} x^{\prime}=-2 y=-2(1)=-2, \\ y^{\prime}=3 x^{2}=3(1)^2=3. \end{aligned} $$ Thus, the direction of motion at point \((1,1)\) is towards the direction given by the vector \((-2,3)\). Now use any computational software like Mathematica or MATLAB to plot the phase-plane trajectory for the given system starting at \((1,1)\). The phase-plane plot should show the trajectory and the direction of motion at that point.

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Key Concepts

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

Hamiltonian function
A Hamiltonian function is a scalar value associated with a system, which helps to determine the behavior of the system over time. It often symbolizes the total energy of the system, combining both kinetic and potential energy aspects. In the context of Hamiltonian systems, this function plays a key role in expressing the system's dynamics through ordinary differential equations.
  • The Hamiltonian function, denoted as \(H(x, y)\), helps to describe how the positions and velocities (or momenta) of a system evolve.
  • In the given exercise, the form of the Hamiltonian function was established through integration, resulting in \(H(x, y) = -y^2 - x^3 + C\).
This function must satisfy the relationship outlined in Hamiltonian systems' ordinary differential equations. By finding the Hamiltonian function for a specific system, one gains insight into the conservation and transformation of energy within that system.
Phase-plane trajectory
The phase-plane trajectory provides a graphical representation of a dynamical system, illustrating how the variables of the system change with respect to each other over time. By studying these trajectories, one can understand the behavior and stability of a system.
  • Each trajectory corresponds to a set of initial conditions, showing how the system evolves.
  • For our system, starting at the point \((1,1)\), the phase-plane trajectory is plotted using tools like MATLAB or Mathematica.
The direction of motion along these trajectories is indicated by vectors, such as \((-2, 3)\) for the initial point \((1,1)\). This direction is derived from the derivatives of the system, giving insight into how the system progresses through different states.
Ordinary Differential Equations
Ordinary Differential Equations (ODEs) are equations that contain functions of one independent variable and their derivatives. They have various applications across scientific fields such as physics, engineering, and economics, as they describe how quantities change over time.
  • They are used to model the evolution of systems, like populations, oscillations, or, as in our problem, Hamiltonian systems.
  • The ODEs for the given system are \(x' = -2y\) and \(y' = 3x^2\).
Solving these ODEs involves finding an expression for the dependent variable that satisfies the equation, given the initial conditions. These solutions can describe or predict the state of a system at any given time, hence capturing dynamic behavior within systems.
Theorem 6.2
Theorem 6.2 is crucial in identifying Hamiltonian systems, as it provides the necessary framework to ascertain whether a system can be classified as such. According to the theorem, for a system to be Hamiltonian, there must exist a Hamiltonian function \( H \) so that the ordinary differential equations can be expressed as:
  • \( \dot{q} = \frac{\partial H}{\partial p} \)
  • \( \dot{p} = -\frac{\partial H}{\partial q} \)
In our exercise, the application of Theorem 6.2 enables us to express the given system's equations in a manner consistent with Hamiltonian systems by finding a suitable \( H \). This not only verifies the Hamiltonian nature of the system but also allows us to derive the Hamiltonian function, essential for understanding the system's dynamics.

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Most popular questions from this chapter

In each exercise, an initial value problem for a first order nonlinear system is given. Rewrite the problem as an equivalent initial value problem for a higher order nonlinear scalar differential equation. $$ \frac{d}{d t}\left[\begin{array}{l} y_{1} \\ y_{2} \end{array}\right]=\left[\begin{array}{c} y_{2} \\ y_{2} \tan \left(y_{1}\right)+e^{y_{2}} \end{array}\right], \quad\left[\begin{array}{l} y_{1}(0) \\ y_{2}(0) \end{array}\right]=\left[\begin{array}{l} 0 \\ 1 \end{array}\right] $$

Each exercise lists a nonlinear system \(\mathbf{z}^{\prime}=A \mathbf{z}+\mathbf{g}(\mathbf{z})\), where \(A\) is a constant ( \(2 \times 2\) ) invertible matrix and \(\mathbf{g}(\mathbf{z})\) is a \((2 \times 1)\) vector function. In each of the exercises, \(\mathbf{z}=\mathbf{0}\) is an equilibrium point of the nonlinear system. (a) Identify \(A\) and \(\mathbf{g}(\mathbf{z})\). (b) Calculate \(\|\mathbf{g}(\mathbf{z})\|\). (c) Is \(\lim _{\mid \mathbf{z} \| \rightarrow 0}\|\mathbf{g}(\mathbf{z})\| /\|\mathbf{z}\|=0\) ? Is \(\mathbf{z}^{\prime}=A \mathbf{z}+\mathbf{g}(\mathbf{z})\) an almost linear system at \(\mathbf{z}=\mathbf{0}\) ? (d) If the system is almost linear, use Theorem \(6.4\) to choose one of the three statements: (i) \(\mathbf{z}=\mathbf{0}\) is an asymptotically stable equilibrium point. (ii) \(\mathbf{z}=\mathbf{0}\) is an unstable equilibrium point. (iii) No conclusion can be drawn by using Theorem \(6.4\). $$ \begin{aligned} &z_{1}^{\prime}=-z_{1}+3 z_{2}+z_{2} \cos \sqrt{z_{1}^{2}+z_{2}^{2}} \\ &z_{2}^{\prime}=-z_{1}-5 z_{2}+z_{1} \cos \sqrt{z_{1}^{2}+z_{2}^{2}} \end{aligned} $$

Let \(A=\left[\begin{array}{cc}1 & a_{12} \\ a_{21} & a_{22}\end{array}\right]\) be a real \((2 \times 2)\) matrix. Assume that $$ A\left[\begin{array}{l} 1 \\ 2 \end{array}\right]=\left[\begin{array}{ll} 1 & a_{12} \\ a_{21} & a_{22} \end{array}\right]\left[\begin{array}{l} 1 \\ 2 \end{array}\right]=2\left[\begin{array}{l} 1 \\ 2 \end{array}\right] $$ and that the origin is not an isolated equilibrium point of the system \(\mathbf{y}^{\prime}=A \mathbf{y}\). Determine the constants \(a_{12}, a_{21}\), and \(a_{22}\).

Each exercise lists a nonlinear system \(\mathbf{z}^{\prime}=A \mathbf{z}+\mathbf{g}(\mathbf{z})\), where \(A\) is a constant ( \(2 \times 2\) ) invertible matrix and \(\mathbf{g}(\mathbf{z})\) is a \((2 \times 1)\) vector function. In each of the exercises, \(\mathbf{z}=\mathbf{0}\) is an equilibrium point of the nonlinear system. (a) Identify \(A\) and \(\mathbf{g}(\mathbf{z})\). (b) Calculate \(\|\mathbf{g}(\mathbf{z})\|\). (c) Is \(\lim _{\mid \mathbf{z} \| \rightarrow 0}\|\mathbf{g}(\mathbf{z})\| /\|\mathbf{z}\|=0\) ? Is \(\mathbf{z}^{\prime}=A \mathbf{z}+\mathbf{g}(\mathbf{z})\) an almost linear system at \(\mathbf{z}=\mathbf{0}\) ? (d) If the system is almost linear, use Theorem \(6.4\) to choose one of the three statements: (i) \(\mathbf{z}=\mathbf{0}\) is an asymptotically stable equilibrium point. (ii) \(\mathbf{z}=\mathbf{0}\) is an unstable equilibrium point. (iii) No conclusion can be drawn by using Theorem \(6.4\). $$ \begin{aligned} &z_{1}^{\prime}=-3 z_{1}-5 z_{2}+z_{1} e^{-\sqrt{z_{1}^{2}+z_{2}^{2}}} \\ &z_{2}^{\prime}=2 z_{1}-z_{2}+z_{2} e^{-\sqrt{z_{1}^{2}+z_{2}^{2}}} \end{aligned} $$

A linear system is given in each exercise. (a) Determine the eigenvalues of the coefficient matrix \(A\). (b) Use Table \(6.2\) to classify the type and stability characteristics of the equilibrium point at \(\mathbf{y}=\mathbf{0}\). (c) The given linear system is a Hamiltonian system. Derive the conservation law for this system. $$ \mathbf{y}^{\prime}=\left[\begin{array}{rr} 2 & 1 \\ 0 & -2 \end{array}\right] \mathbf{y} $$

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