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

Short Answer

Expert verified
The periodic solutions of the given ODE system are: r(t) = √2, θ(t) = -t + c.

Step by step solution

01

Rewrite the ODE system in polar coordinates

Using the polar coordinate transformation \(x = r\cos\theta\), and \(y = r\sin\theta\), rewrite the given ODEs in terms of \(r\) and \(\theta\). To do this, we first differentiate \(x\) and \(y\) with respect to time \(t\) and in terms of \(r\) and \(\theta\): $$ \frac{dx}{dt} = \frac{dr}{dt}\cos\theta - r\sin\theta\frac{d\theta}{dt} $$ $$ \frac{dy}{dt} = \frac{dr}{dt}\sin\theta + r\cos\theta\frac{d\theta}{dt} $$ Now, substitute the given ODEs into these equations to form a system of equations involving \(\frac{dr}{dt}\) and \(\frac{d\theta}{dt}\).
02

Isolate \(\frac{dr}{dt}\) and \(\frac{d\theta}{dt}\)

Solve for \(\frac{dr}{dt}\) and \(\frac{d\theta}{dt}\) to obtain the following ODE system in polar coordinates: $$ \frac{dr}{dt} = r\left(1 - \frac{2}{r^2}\right), $$ $$ \frac{d\theta}{dt} = -1 $$
03

Solve the ODE for \(r\)

The \(\frac{dr}{dt}\) ODE can be separated and solved for \(r(t)\): $$ \int\frac{dr}{r\left(1-\frac{2}{r^2}\right)} = \int dt. $$ By solving this integral, we find that: $$ r^2(t) = k e^{2t} + 2, $$ where \(k\) is an integration constant.
04

Solve the ODE for \(\theta\)

The \(\frac{d\theta}{dt}\) ODE is already separated, so it's easy to find the solution for \(\theta(t)\): $$ \int d\theta = -\int dt \quad \Rightarrow \quad \theta(t) = -t + c, $$ where \(c\) is another integration constant.
05

Identify periodic solutions

Since the ODEs are linear and autonomous, periodic solutions occur when \(\frac{dr}{dt} = 0\). This happens when the constant term in parentheses is zero: $$ 1-\frac{2}{r^2} = 0. $$ The solution to this equation is \(r^2 = 2\), meaning that \(r = \sqrt{2}\), which corresponds to a periodic solution. Therefore, the periodic solutions of the given ODE system are given by: $$ \boldsymbol{r(t) = \sqrt{2}, \quad \theta(t) = -t + c}. $$

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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 mathematical equations that involve an unknown function and its derivatives. They play a crucial role in expressing and investigating how different quantities change with respect to each other. The system in this exercise is a prime example where the dynamics of variables, such as position, are represented via differential equations. In particular, the given system shows what is known as a coupled differential equation. Instead of having a single differential equation, we have two that are interconnected through their variables. Differential equations can be tricky, but with practice, they become a powerful tool for modeling real-world phenomena.
Polar Coordinates
Polar coordinates are an alternative to the more familiar Cartesian coordinates, used for describing points in a plane. Instead of using horizontal and vertical measurements, polar coordinates use a distance from a reference point and an angle from a reference direction. In this exercise, the ODE system is rewritten using polar coordinates which substitute the Cartesian coordinates
  • With the transformation:
    • \( x = r \cos \theta \)
    • \( y = r \sin \theta \)
By doing this, we can simplify the complex interactions between the variables into radial and angular components. Polar coordinates can make certain problems easier to visualize and solve, especially those involving rotations or periodic behavior.
Autonomous Systems
An autonomous system is a type of differential equation where the system's rules do not explicitly depend on time. This means the system's behavior relies solely on the current state variables, not the temporal progression. The independence from time in autonomous systems simplifies analysis, especially when looking for equilibrium points and periodic solutions. In the given exercise, both
  • \( \frac{dr}{dt} = r\left(1 - \frac{2}{r^2}\right) \)
  • \( \frac{d\theta}{dt} = -1 \)
are autonomous. This allows the focus to be on the geometry of the solutions rather than their specific trajectory over time. Autonomous systems are especially powerful in identifying long-term behavior, such as stability and periodic orbits.
Polar Coordinate Transformation
Polar coordinate transformation involves converting a system given in Cartesian coordinates into a system expressed in polar coordinates. This is done by substituting
  • \( x = r \cos \theta \)
  • \( y = r \sin \theta \)
The transformation helps in solving certain types of differential equations that have rotational symmetry or cyclic behavior. By moving to polar coordinates, variables become easier to handle, such as moving from a system involving two variables
  • \( x \text{ and } y \)
  • to a system with
    • radius \( r \)
    • angle \( \theta \)
In this exercise, transforming to polar coordinates simplifies the process of finding periodic solutions, as seen in the expression
  • \( \frac{dr}{dt} \) and \( \frac{d\theta}{dt} \)
which are more straightforward to solve thanks to this transformation. Using polar methods can greatly assist in gaining insights into the system's overall dynamics.

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Most popular questions from this chapter

(a) Find all the critical points (equilibrium solutions). (b) Use a computer to draw a direction field and portrait for the system. (c) From the plot(s) in part (b) determine whether each critical point is asymptotically stable, stable, or unstable, and classify it as to type. $$ d x / d t=-x+2 x y, \quad d y / d t=y-x^{2}-y^{2} $$

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. $$ d x / d t=(1+x) \sin y, \quad d y / d t=1-x-\cos y $$

The motion of a certain undamped pendulum is described by the equations $$ d x / d t=y, \quad d y / d t=-4 \sin x $$ If the pendulum is set in motion with an angular displacement \(A\) and no initial velocity, then the initial conditions are \(x(0)=A, y(0)=0\) (a) Let \(A=0.25\) and plot \(x\) versus \(t\). From the graph estimate the amplitude \(R\) and period \(T\) of the resulting motion of the pendulum. (b) Repeat part (a) for \(A=0.5,1.0,1.5,\) and \(2.0 .\) (c) How do the amplitude and period of the pendulum's motion depend on the initial position \(A^{7}\) Draw a graph to show each of these relationships. Can you say anything about the limiting value of the period as \(A \rightarrow 0 ?\) (d) Let \(A=4\) and plot \(x\) versus \(t\) Explain why this graph differs from those in parts (a) and (b). For what value of \(A\) does the transition take place?

If \(x=r \cos \theta, y=r \sin \theta,\) show that \(y(d x / d t)-x(d y / d t)=-r^{2}(d \theta / d t)\)

Carry out the indicated investigations of the Lorenz equations. (a) For \(r=21\) plot \(x\) versus \(t\) for the solutions starting at the initial points \((3,8,0),\) \((5,5,5),\) and \((5,5,10) .\) Use a \(t\) interval of at least \(0 \leq t \leq 30 .\) Compare your graphs with those in Figure \(9.8 .4 .\) (b) Repeat the calculation in part (a) for \(r=22, r=23,\) and \(r=24 .\) Increase the \(t\) interval as necessary so that you can determine when each solution begins to converge to one of the critical points. Record the approximate duration of the chaotic transient in each case. Describe how this quantity depends on the value of \(r\). (c) Repeat the calculations in parts (a) and (b) for values of \(r\) slightly greater than 24 . Try to estimate the value of \(r\) for which the duration of the chaotic transient approaches infinity.

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