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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

Can be interpreted as describing the interaction of two species with population densities \(x\) and \(y .\) In each of these problems carry out the following steps. (a) Draw a direction field and describe how solutions seem to behave. (b) Find the critical points. (c) For each critical point find the corresponding linear system. Find the eigenvalues and eigenvectors of the linear system; classify each critical point as to type, and determine whether it is asymptotically stable, or unstable. (d) Sketch the trajectories in the neighborhood of each critical point. (e) Draw a phase portrait for the system. (f) Determine the limiting behavior of \(x\) and \(y\) as \(t \rightarrow \infty\) and interpret the results in terms of the populations of the two species. $$ \begin{array}{l}{d x / d t=x(1-0.5 y)} \\ {d y / d t=y(-0.25+0.5 x)}\end{array} $$

Consider the system (3) in Example 1 of the text. Recall that this system has an asymptotically stable critical point at (0.5,0.5) , corresponding to the stable coexistence of the two population species. Now suppose that immigration or emigration occurs at the constant rates of \(\delta a\) and \(\delta b\) for the species \(x\) and \(y,\) respectively. In this case equations ( 3 ) are replaced by $$\frac{d x}{d t}=x(1-x-y)+\delta a, \quad \frac{d y}{d t}=\frac{y}{4}(3-4 y-2 x)+\delta b$$ The question is what effect this has on the location of the stable equilibrium point. a. To find the new critical point, we must solve the equations $$\begin{aligned} x(1-x-y)+\delta a &=0 \\ \frac{y}{4}(3-4 y-2 x)+\delta b &=0 \end{aligned}$$ One way to proceed is to assume that \(x\) and \(y\) are given by power series in the parameter \(\delta ;\) thus $$x=x_{0}+x_{1} \delta+\cdots, \quad y=y_{0}+y_{1} \delta+\cdots$$ Substitute equations (44) into equations (43) and collect terms according to powers of \(\delta\). b. From the constant terms (the terms not involving \(\delta\) ), show that \(x_{0}=0.5\) and \(y_{0}=0.5,\) thus confirming that in the absence of immigration or emigration, the critical point is (0.5,0.5) . c. From the terms that are linear in \(\delta,\) show that \\[ x_{1}=4 a-4 b, \quad y_{1}=-2 a+4 b \\] d. Suppose that \(a>0\) and \(b>0\) so that immigration occurs for both species. Show that the resulting equilibrium solution may represent an increase in both populations, or an increase in one but a decrease in the other. Explain intuitively why this is a reasonable result.

Each of Problems I through 6 can be interpreted as describing the interaction of two species with populations \(x\) and \(y .\) In each of these problems carry out the following steps. $$ \begin{array}{l}{\text { (a) Draw a direction field and describe how solutions seem to behave. }} \\ {\text { (b) Find the critical points. }} \\ {\text { (c) For each critical point find the corresponding linear system. Find the eigenvalues and }} \\ {\text { eigenvectors of the linear system; classify each critical point as to type, and determine }} \\ {\text { whether it is asymptotically stable, stable, or unstable. }}\end{array} $$ $$ \begin{array}{l}{\text { (d) Sketch the trajectories in the neighborhood of each critical point. }} \\ {\text { (c) Compute and plot enough trajectories of the given system to show clearly the behavior of }} \\ {\text { the solutions. }} \\ {\text { (f) Determine the limiting behavior of } x \text { and } y \text { as } t \rightarrow \infty \text { and interpret the results in terms of }} \\ {\text { the populations of the two species. }}\end{array} $$ $$ \begin{array}{l}{d x / d t=x(1.5-x-0.5 y)} \\ {d y / d t=y(2-0.5 y-1.5 x)}\end{array} $$

Consider the system $$ d x / d t=a x[1-(y / 2)], \quad d y / d t=b y[-1+(x / 3)] $$ where \(a\) and \(b\) are positive constants. Observe that this system is the same as in the example in the text if \(a=1\) and \(b=0.75 .\) Suppose the initial conditions are \(x(0)=5\) and \(y(0)=2\) (a) Let \(a=1\) and \(b=1 .\) Plot the trajectory in the phase plane and determine (or cstimate) the period of the oscillation. (b) Repeat part (a) for \(a=3\) and \(a=1 / 3,\) with \(b=1\) (c) Repeat part (a) for \(b=3\) and \(b=1 / 3,\) with \(a=1\) (d) Describe how the period and the shape of the trajectory depend on \(a\) and \(b\).

(a) Find the eigenvalues and eigenvectors. (b) Classify the critical point \((0,0)\) as to type and determine whether it is stable, asymptotically stable, or unstable. (c) Sketch several trajectories in the phase plane and also sketch some typical graphs of \(x_{1}\) versus \(t .\) (d) Use a computer to plot accurately the curves requested in part (c). \(\frac{d \mathbf{x}}{d t}=\left(\begin{array}{rr}{-1} & {0} \\ {0} & {-1}\end{array}\right) \mathbf{x}\)

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