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Chapter 9: Problem 22

(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. $$ \text { Duffing's equation: } \quad d x / d t=y, \quad d y / d t=-x+\left(x^{3} / 6\right) $$

Short Answer

Expert verified
The function representing the trajectories of Duffing's Equation is: $$ H(x, y) = \frac{1}{2}y^{2} + \frac{1}{4} x^{2} - \frac{1}{20} x^{4} $$ To plot the trajectories, create level curves by setting \(H(x,y)=c\) and varying the value of \(c\). Additionally, evaluate the initial differential equations \(dx/dt=y\) and \(dy/dt=-x+x^3/6\) at different points on the level curves to determine and plot the direction of motion.

Step by step solution

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01

Find an expression for \(dy/dx\)

Given that \(dx/dt = y\) and \(dy/dt = -x + x^3/6\), we can find \(dy/dx\) by dividing both equations: $$ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-x + x^3/6}{y} $$
02

Solve the separable first-order differential equation

The equation we found in Step 1 is a separable first-order differential equation. To solve it, move \(dx\) to the right side and all \(x\) terms to the left side: $$ \frac{y \, dy}{-x + x^{3}/6} = dx $$ Integrate both sides of the equation: $$ \int \frac{y \, dy}{-x + x^{3}/6} = \int dx $$
03

Perform the integration

Integrate both sides, and don't forget the constant of integration (\(c\)): $$ \frac{1}{2}y^{2} + \frac{1}{4} x^{2}- \frac{1}{20} x^{4}=c $$ Now, we have our function \(H(x, y)\): $$ H(x, y) = \frac{1}{2}y^{2} + \frac{1}{4} x^{2} - \frac{1}{20} x^{4} $$
04

Plot level curves and indicate direction

To plot level curves, we set \(H(x,y)=c\) and vary the value of \(c\). For example, we can choose different constant values like \(c_1, c_2, c_3\), and solve the equation \(H(x,y)=c_i\) for each level curve. This can be done using a graphing tool or plotting software. The direction of motion is represented by the initial differential equations \(dx/dt=y\) and \(dy/dt=-x+x^3/6\). To find the direction at a particular point \((x, y)\) on the level curve, we can evaluate these two expressions and draw a small tangent line indicating the direction of motion at that point. To summarize, we found the function for the trajectories, which is given by: $$ H(x, y) = \frac{1}{2}y^{2} + \frac{1}{4} x^{2} - \frac{1}{20} x^{4} $$ Plot level curves of this function to visualize the trajectories of the system. Additionally, plot tangent lines on the level curves to represent the direction of motion at various points.

Key Concepts

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

First-Order Differential Equation
A first-order differential equation is a relation that involves functions, their derivatives, and constants. In our case, the Duffing's equation provided a set of first-order differential equations that describe the rate of change of a system over time.

Specifically, we have the equations \( dx/dt = y \) and \( dy/dt = -x + x^3/6 \). The process of solving these involves finding a function that satisfies this relationship. As seen in Step 1, we manipulate the two equations to express \( dy/dx \) in terms of \( x \) and \( y \) exclusively, which gives us the separable form. In essence, this method reduces the problem to a simpler one where we can integrate each side of the equation with respect to one variable, allowing us to find the hidden relationship between \( x \) and \( y \).

Through this method, we uncover the trajectory that a particle would follow under the influence of the forces described by the original set of differential equations.
Level Curves
Level curves, a fundamental concept in multivariable calculus, are crucial for understanding how a function behaves across two dimensions. In the context of Duffing's equation, level curves represent all the points \( (x, y) \) where the function \( H(x, y) \) has the same constant value, \( c \).

To visualize these curves, imagine a 3D landscape where altitude is dictated by the value of \( H(x, y) \)—each level curve then corresponds to a contour of constant height. By plotting several of these contours, as indicated in Step 2, we can see how the function's value changes across the plane. In terms of physics, each level curve represents a possible trajectory of an object described by the system—the path it would trace out over time if it had the specified energy level.
Trajectory Plotting
Trajectory plotting is a graphical representation of the path that a system will follow through its phase space.

After finding the function \( H(x, y) \) in Step 3, we plot its level curves to visualize these paths. These curves are significant in physics and engineering because they provide insight into the behavior of complex systems without solving the equations analytically at every point. \( H(x, y) = \frac{1}{2}y^{2} + \frac{1}{4} x^{2} - \frac{1}{20} x^{4} \) represents the energy of the system, and by plotting for different values of \( c \) in the equation \( H(x, y) = c \) we map out potential movements of the system.

In addition, we use the original differential equations to determine the direction of this motion, as elaborated in Step 4. Knowing the direction at various points is invaluable for predicting future states of a dynamic system.
Integration
Integration is a fundamental operation in calculus that allows us to calculate the area under curves and the accumulated value of a function. In the context of solving Duffing's equation, integration is used to find a potential function \( H(x, y) \) that encapsulates the motion of the system.

When we integrate \( \frac{dy}{dx} \) with respect to \( x \) and \( y \) as seen in Step 2 and Step 3, we essentially sum up the small changes in \( y \) for tiny increments in \( x \) to reconstruct the overall change in \( y \) over the range of \( x \) considered. The result of this integration is a powerful function that binds \( x \) and \( y \) together and describes the potential energy landscape of the system, allowing us to understand and predict its movement through phase space. The constant of integration, \( c \)—often disregarded as a mere add-on—actually holds the key to understanding the complete family of possible trajectories.

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Most popular questions from this chapter

We will prove part of Theorem 9.3 .2 : If the critical point \((0,0)\) of the almost linear system $$ d x / d t=a_{11} x+a_{12} y+F_{1}(x, y), \quad d y / d t=a_{21} x+a_{22} y+G_{1}(x, y) $$ is an asymptotically stable critical point of the corresponding linear system $$ d x / d t=a_{11} x+a_{12} y, \quad d y / d t=a_{21} x+a_{22} y $$ then it is an asymptotically stable critical point of the almost linear system (i). Problem 12 deals with the corresponding result for instability. Consider the linear system (ii). (a) Since \((0,0)\) is an asymptotically stable critical point, show that \(a_{11}+a_{22}<0\) and \(\left.a_{11} a_{22}-a_{12} a_{21}>0 . \text { (See Problem } 21 \text { of Section } 9.1 .\right)\) (b) Construct a Liapunov function \(V(x, y)=A x^{2}+B x y+C y^{2}\) such that \(V\) is positive definite and \(\hat{V}\) is negative definite. One way to ensure that \(\hat{V}\) is negative definite is to choose \(A, B,\) and \(C\) so that \(\hat{V}(x, y)=-x^{2}-y^{2} .\) Show that this leads to the result $$ \begin{array}{l}{A=-\frac{a_{21}^{2}+a_{22}^{2}+\left(a_{11} a_{22}-a_{12} a_{21}\right)}{2 \Delta}, \quad B=\frac{a_{12} a_{22}+a_{11} a_{21}}{\Delta}} \\\ {C=-\frac{a_{11}^{2}+a_{12}^{2}+\left(a_{11} a_{22}-a_{12} a_{21}\right)}{2 \Delta}}\end{array} $$ where \(\Delta=\left(a_{11}+a_{22}\right)\left(a_{11} a_{22}-a_{12} a_{21}\right)\) (c) Using the result of part (a) show that \(A>0\) and then show (several steps of algebra are required) that $$ 4 A C-B^{2}=\frac{\left(a_{11}^{2}+a_{12}^{2}+a_{21}^{2}+a_{22}^{2}\right)\left(a_{11} a_{22}-a_{12} a_{21}\right)+2\left(a_{11} a_{22}-a_{12} a_{21}\right)^{2}}{\Delta^{2}}>0 $$ Thus by Theorem 9.6.4, \(V\) is positive definite.

an autonomous system is expressed in polar coordinates. Determine all periodic solutions, all limit cycles, and determine their stability characteristics. $$ d r / d t=r(r-1)(r-3), \quad d \theta / d t=1 $$

Verify that \((0,0)\) is a critical point, show that the system is almost linear, and discuss the type and stability of the critical point \((0,0)\) by examining the corresponding linear system. $$ d x / d t=x-y^{2}, \quad d y / d t=x-2 y+x^{2} $$

an autonomous system is expressed in polar coordinates. Determine all periodic solutions, all limit cycles, and determine their stability characteristics. $$ d r / d t=r(1-r)(r-2), \quad d \theta / d t=-1 $$

Construct a suitable Liapunov function of the form \(a x^{2}+c y^{2}\) where \(a\) and \(c\) are to be determined. Then show that the critical point at the origin is of the indicated type. $$ d x / d t=x^{3}-y^{3}, \quad d y / d t=2 x y^{2}+4 x^{2} y+2 y^{3} ; \quad \text { unstable } $$
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