Lefschetz (1977) discusses the following system of equations $$ \begin{aligned} &\frac{d x}{d t}=-y+x f\left(x^{2}+y^{2}\right) \\ &\frac{d y}{d t}=x+y f\left(x^{2}+y^{2}\right) \end{aligned} $$ Show that this system is equivalent to the polar equation $$ \frac{d r}{d \theta}=r f\left(r^{2}\right) \text {. } $$

Short Answer

Expert verified
Convert the original system into polar coordinates and simplify to obtain \( \frac{dr}{d\theta} = r f(r^2) \).

Step by step solution

01

Convert to Polar Coordinates

Use the transformations: \( x = r \cos(\theta) \) and \( y = r \sin(\theta) \). Then calculate the derivatives: \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).
02

Derivatives in Polar Coordinates

Write the derivatives in terms of \( r \) and \( \theta \): \[ \frac{dx}{dt} = \frac{dr}{dt} \cos(\theta) - r \sin(\theta) \frac{d\theta}{dt} \] and \[ \frac{dy}{dt} = \frac{dr}{dt} \sin(\theta) + r \cos(\theta) \frac{d\theta}{dt} \].
03

Substitute in Given Equations

Substitute \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) into the original system of equations: \[ \frac{dr}{dt} \cos(\theta) - r \sin(\theta) \frac{d\theta}{dt} = -r \sin(\theta) + r \cos(\theta) f(r^2) \] \[ \frac{dr}{dt} \sin(\theta) + r \cos(\theta) \frac{d\theta}{dt} = r \cos(\theta) + r \sin(\theta) f(r^2) \].
04

Simplify Equations

Simplify the above equations by factoring out common terms and separating the components involving \( \frac{dr}{dt} \) and \( \frac{d\theta}{dt} \). This yields: \( \frac{dr}{dt} \cos(\theta) - r \frac{d\theta}{dt} \sin(\theta) = -r \sin(\theta) + r \cos(\theta) f(r^2) \) and \( \frac{dr}{dt} \sin(\theta) + r \cos(\theta) \frac{d\theta}{dt} = r \cos(\theta) + r \sin(\theta) f(r^2) \).
05

Isolate \( \frac{dr}{dt} \)

Combine the obtained equations to isolate \( \frac{dr}{dt} \): \( \frac{dr}{dt} = r f(r^2) \).
06

Express \( \frac{d\theta}{dt} \)

By combining the simplified equations, obtain an expression for \( \frac{d\theta}{dt} \) (which should be a constant) and recognize that \( \frac{dr}{d\theta} = r f(r^2) \).
07

Final Form

Combine the results from previous steps to show the final polar equation form: \( \frac{dr}{d\theta} = r f(r^2) \).

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Key Concepts

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

polar coordinates
Polar coordinates provide an alternative way of representing points in the plane compared to Cartesian coordinates (x, y). Instead of using horizontal and vertical distances, we use:
- Radius (r): the distance from the origin to the point.
- Angle \( \theta \): the angle between the radius and the positive x-axis.

The conversion from Cartesian to polar coordinates involves:
\( x = r \cos(\theta) \) and \( y = r \sin(\theta) \).

This transformation can be utilized in various mathematical problems, especially in differential equations that exhibit radial symmetry.
differential equations
Differential equations involve derivatives of a function and can describe a multitude of physical and natural phenomena, such as motion, growth, and decay.
  • An ordinary differential equation (ODE) involves functions of a single variable and their derivatives.
  • Partial differential equations (PDEs) involve multiple variables and their partial derivatives.

In the provided exercise, we deal with an ODE system:\[\frac{d x}{d t}=-y+x f\left(x^{2}+y^{2}\right) \]
\[ \frac{d y}{d t}=x+y f\left(x^{2}+y^{2}\right) \]
These equations can be transformed into a more manageable form using polar coordinates, helping reveal properties of the system that are not easily noticeable in the Cartesian form.
coordinate transformation
Coordinate transformations change the perspective or basis of a mathematical problem. Here, we switch from Cartesian \( (x, y) \) to polar coordinates \( (r, \theta) \).

To transform:
  • Use \( x = r \cos(\theta) \) and \( y = r \sin(\theta) \).
  • Calculate derivatives in polar form:
  • \( \frac{dx}{dt} = \frac{dr}{dt} \cos(\theta) - r \sin(\theta) \frac{d\theta}{dt} \)
  • \( \frac{dy}{dt} = \frac{dr}{dt} \sin(\theta) + r \cos(\theta) \frac{d\theta}{dt} \)


Substituting these into the original differential equations allows us to reframe the problem. This way, we obtain simpler, often more insightful forms of the equations that can be easier to solve or analyze.
Lephschetz system
The Lefschetz system, named after the mathematician Solomon Lefschetz, involves a set of differential equations that describe a particular dynamical system. In this exercise, the system is:

\[\frac{d x}{d t}=-y+x f\left(x^{2}+y^{2}\right) \]
\[ \frac{d y}{d t}=x+y f\left(x^{2}+y^{2}\right) \]
The goal is to show equivalence to a polar coordinate system equation:
\[ \frac{d r}{d \theta}=r f\left(r^{2}\right)\]

By transforming to polar coordinates, we discover that the behavior and evolution of the system can be expressed as a function of the radius and angle, simplifying the problem. The key steps include finding derivatives \( \frac{dr}{dt} \) and \( \frac{d\theta}{dt} \) in polar terms, highlighting underlying symmetries, and leading to simplified, insightful forms of the equations.

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Most popular questions from this chapter

Use Bendixson's negative criterion to show that if \(P\) is an isolated saddle point there cannot be a limit cycle in the neighborhood of \(P\) that contains only \(P\).

Show that the system $$ \frac{d x}{d t}=-2 y, \quad \frac{d y}{d t}=x-5, $$ has closed elliptical orbits. (Hint: Consider first transforming \(x\) to \(x-5\) and \(y\) to \(y / 2\) and then using a polar transformation.)

Consider the following system of equations (Odell, 1980), which are said to describe a predator-prey system: $$ \begin{aligned} &\frac{d x}{d t}=x[x(1-x)-y] \\ &\frac{d y}{d t}=k\left(x-\frac{1}{\mu}\right) y \end{aligned} $$ (a) Interpret the terms in these equations. (b) Sketch the nullclines in the \(x y\) plane and determine whether the PoincareBendixson theory can be applied. (c) Show that the steady states are located at $$ (0,0), \quad(1,0), \quad\left(\frac{1}{\mu}, \frac{1-1 / \mu}{\mu}\right) \text {. } $$ (d) Show that at the last of these steady states the linearized system is characterized by the matrix $$ \left(\begin{array}{lr} \frac{1}{\mu}\left(1-\frac{2}{\mu}\right) & -\frac{1}{\mu} \\ \frac{k}{\mu}\left(1-\frac{1}{\mu}\right) & 0 \end{array}\right) $$ (e) Can the Bendixson negative criterion be used to rule out limit-cycle oscillations? (f) If your results so far are not definitive, consider applying the Hopf bifurcation theorem. What is the stability of the strictly positive steady state? What is the bifurcation parameter, and at what value does the bifurcation occur? "(g) Show that at bifurcation the matrix in part (d) is not in the "normal" form required for Hopf stability calculations. Also show that if you transform the variables by defining $$ \hat{x}=x, \quad \hat{y}=\frac{2}{k^{1 / 2}} y, $$ you obtain a new system that is in normal form. (h) Find the transformed system, calculate \(V^{\prime \prime \prime}\) and show that it is negative. (i) What conclusions can be drawn about the system?

Consider the equations $$ \frac{d x}{d t}=y, \quad \frac{d y}{d t}=-x . $$ (b) By transforming variables, obtain $$ \frac{d r^{2}}{d t}=0, \quad \frac{d \theta}{d t}=-1 . $$ (b) Conclude that there are circular solutions. What is the direction of rotation? Are these cycles stable?

Consider the nonlinear system of equations $$ \begin{aligned} &\frac{d x}{d t}=-y+x\left(x^{2}+y^{2}-1\right)=f(x, y), \\ &\frac{d y}{d t}=x+y\left(x^{2}+y^{2}-1\right)=g(x, y) . \end{aligned} $$ Show that \(r=1\) is an unstable limit cycle of the equations.

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