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Magazine Chapter 9: Problem 9
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) = 鈭�2, 胃(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:
- \( x = r \cos \theta \)
- \( y = r \sin \theta \)
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
- \( \frac{dr}{dt} = r\left(1 - \frac{2}{r^2}\right) \)
- \( \frac{d\theta}{dt} = -1 \)
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
- \( x = r \cos \theta \)
- \( y = r \sin \theta \)
- \( x \text{ and } y \)
- to a system with
- radius \( r \)
- angle \( \theta \)
- \( \frac{dr}{dt} \) and \( \frac{d\theta}{dt} \)
