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Chapter 7: Problem 5

Show that the system \(\dot{r}=r\left(1-r^{2}\right)\), \(\dot{\theta}=1\) is equivalent to $$ \dot{x}=x-y-x\left(x^{2}+y^{2}\right), \quad \dot{y}=x+y-y\left(x^{2}+y^{2}\right) $$ where \(x=r \cos \theta, y=r \sin \theta\). (Hint: \(\dot{x}=\frac{\perp}{d i}(r \cos \theta)=\dot{r} \cos \theta-r \dot{\theta} \sin \theta .\) )

Short Answer

Expert verified
The given system of polar coordinate differential equations: \[ \dot{r} = r(1 - r^2), \quad \dot{\theta} = 1 \] is equivalent to the Cartesian coordinate system differential equations: \[ \dot{x} = x - y - x(x^2+y^2), \quad \dot{y} = x + y - y(x^2+y^2) \] by using the conversion formulas \(x = r \cos{\theta}\) and \(y = r \sin{\theta}\), differentiating both x and y with respect to time, and substituting the resulting expressions in terms of x and y.

Step by step solution

01

Write the system of differential equations in polar coordinates

We are given the following system: \[ \dot{r} = r(1 - r^2), \quad \dot{\theta} = 1 \]
02

Establish conversion formulas between polar and Cartesian coordinates

We have the following conversion formulas: \[ x = r\cos{\theta}, \quad y = r\sin{\theta} \]
03

Differentiate x and y with respect to time

We will need to find the derivatives \(\dot{x}\) and \(\dot{y}\) with respect to time, using the conversion formulas and the hint. Remember that we have \[ \dot{x}=\frac{d}{dt}(r\cos\theta), \quad \dot{y}=\frac{d}{dt}(r\sin\theta) \] Differentiating x with respect to time, we get: \[ \dot{x} = \dot{r} \cos \theta - r \dot{\theta} \sin \theta \] Now, substitute the given values of \(\dot{r}\) and \(\dot{\theta}\) in the above equation: \[ \dot{x} = r(1-r^2)\cos\theta - r\sin\theta \] Similarly, differentiating y with respect to time, we get: \[ \dot{y} = \dot{r} \sin \theta + r \dot{\theta} \cos \theta \] Now, substitute the given values of \(\dot{r}\) and \(\dot{\theta}\) in the above equation: \[ \dot{y} = r(1-r^2)\sin\theta + r\cos\theta \]
04

Express \(\dot{x}\) and \(\dot{y}\) in Cartesian form

Now we need to express the \(\dot{x}\) and \(\dot{y}\) in terms of x and y: For \(\dot{x}\): \[ \dot{x} = x(1-r^2) - y \] Now, express r^2 in terms of x and y using the conversion formulas: \[ r^2 = x^2+ y^2 \] So, we can rewrite \(\dot{x}\) as: \[ \dot{x} = x(1-(x^2+y^2)) - y = x - x(x^2+y^2) - y \] For \(\dot{y}\): \[ \dot{y} = y(1-r^2) + x \] Using the same conversion for r^2, we rewrite \(\dot{y}\) as: \[ \dot{y} = y - y(x^2+y^2) + x = x + y - y(x^2+y^2) \]
05

Conclude and write the final result

We have successfully shown that the given system of differential equations in polar coordinates is equivalent to the desired system in Cartesian coordinates. The resulting equations are: \[ \dot{x} = x - y - x(x^2+y^2), \quad \dot{y} = x + y - y(x^2+y^2) \]

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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 method to Cartesian coordinates for describing the position of points in a plane. While Cartesian coordinates use a grid like pattern with perpendicular x and y axes, polar coordinates specify each point based on its distance from a reference point (called the pole, analogous to the origin in Cartesian coordinates) and the angle relative to a reference direction (usually the positive x-axis).

In mathematical terms, a point's location in polar coordinates is given as \(r, \theta\), where \(r\) is the radius - the distance from the pole - and \(\theta\) is the angle, measured in radians, from the reference direction. These two values can be linked to Cartesian coordinates through simple equations: \(x = r \cos \theta\) and \(y = r \sin \theta\). This relationship allows for the conversion between coordinate systems, which is particularly useful in certain applications like complex analysis or when dealing with problems involving symmetry around a point.
Cartesian Coordinates
Cartesian coordinates are the most widely used coordinate system for representing geometric figures and spaces. It's composed of two perpendicular axes, usually labeled as the x-axis (horizontal) and the y-axis (vertical), that cross at a point called the origin. Every point in the plane can be described by an ordered pair \( (x,y) \) where x is the value on the horizontal axis and y is the value on the vertical axis.

The system is named after Ren茅 Descartes, who formulated its principles. It's highly valuable due to its simplicity and the straightforward way in which it can describe geometric shapes, handle algebraic equations, and perform calculus operations. In the context of differential equations, Cartesian coordinates allow us to express rates of change for variables x and y that are subject to changes over time or another variable.
System of Differential Equations
A system of differential equations consists of multiple equations that relate functions to their derivatives. These systems often describe complex phenomena where several changing quantities are related to each other through their rates of change. Solving these systems can give insights into the dynamics of the underlying problem, such as population growth, electrical circuits, or mechanical systems.

These systems can be linear or nonlinear and may require various methods to solve, such as separation of variables, linear algebra techniques, or numerical approximation methods. In the exercise provided, we are analyzing a nonlinear system of differential equations by leveraging the relationship between polar and Cartesian coordinates to translate the given polar equations into their Cartesian form, thus formulating the problem into a system that is often easier to visualize and solve.
Time Derivative
The time derivative is a measure of how a function changes as time progresses. It is a fundamental concept in physics and engineering, often denoted as \(\dot{x}\) for a function \(x(t)\) with respect to time \(t\). In mathematical terms, \(\dot{x} \equiv \frac{d}{dt} x(t)\), which tells us the instantaneous rate of change of \(x\) with time.

In the context of the exercise, time derivatives are used to describe the rate at which the radius \(r\) and the angle \(\theta\) change in polar coordinates, and equivalently, how \(x\) and \(y\) change in Cartesian coordinates. These derivatives give us insight into the movement and properties of a system over time. By converting the equations from polar to Cartesian, we were able to analyze the system's behavior in a different, often more intuitive coordinate system.

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Most popular questions from this chapter

Figure I shows the "tetrode multivibrator" circuit used in the carliest commercial radios and analyzed by van der Pol. In van der Pol's day, the active element was a vacuum tube; today it would be a semiconductor device. It acts like an ordinary resistor when \(I\) is high, but like a negative resistor (energy source) when \(I\) is low. Its current-voltage characteristic \(V=f(I)\) resembles a cubic function, as discussed below. Suppose a source of current is attached to the circuit and then withdrawn. What equations govern the subsequent evolution of the current and the various voltages? a) Let \(V=V_{12}=-V_{23}\) denote the voltage drop from point 3 to point 2 in the circuit, Show that \(\dot{V}=-I / C\) and \(V=L \dot{I}+f(I)\). b) Show that the equations in (a) are equivalent to $$ \frac{d w}{d \tau}=-x, \quad \frac{d x}{d \tau}=w-\mu F(x) $$ where \(x=L^{1 / 2} I, w=C^{1 / 2} V, \tau=(L C)^{-1 / 2} t\), and \(F(x)=f\left(L^{-1 / 2} x\right)\). In Section 7.5, we'll see that this system for \((w, x)\) is equivalent to the van der Pol equation, if \(F(x)=\frac{1}{3} x^{3}-x\). Thus the circuit produces self sustained oscillations.

Consider the initial value problem \(\ddot{x}+x+\varepsilon x=0\), with \(x(0)=1, \dot{x}(0)=0\). a) Obtain the exact solution to the problem. b) Using regular perturbation theory, find \(x_{0}, x_{1}\), and \(x_{2}\) in the series expansion \(x(t, \varepsilon)=x_{0}(t)+\varepsilon x_{1}(t)+\varepsilon^{2} x_{2}(t)+O\left(\varepsilon^{3}\right)\) c) Does the perturbation solution contain secular terms? Did you expect to see any? Why?

Consider the weakly nonlinear oscillator \(\ddot{x}+x+\operatorname{ch}(x, \dot{x}, t)=0 .\) Let \(x(t)=r(t) \cos (t+\phi(t)), \quad \dot{x}=-r(t) \sin (t+\phi(t))\). This change of variables should be regarded as a definition of \(r(t)\) and \(\phi(t)\). a) Show that \(\dot{r}=\varepsilon h \sin (t+\phi), r \dot{\phi}=\varepsilon h \cos (t+\phi)\). (Hence \(r\) and \(\phi\) are slowly varying for \(0<\varepsilon<<1\), and thus \(x(t)\) is a sinusoidal oscillation modulated by a slowly drifting amplitude and phase.) b) Let \(\langle r\rangle(t)=\bar{r}(t)=\frac{1}{2 \pi} \int_{1-\pi}^{t+\pi} r(\tau) d \tau\) denote the running average of \(r\) over one cycle of the sinusoidal oscillation. Show that \(d\langle r\rangle / d t=\langle d r / d t\rangle\), i.e., it doesn't matter whether we differentiate or time-average first. c) Show that \(d\langle r\rangle / d t=\varepsilon(h[r \cos (t+\phi),-r \sin (t+\phi), t] \sin (t+\phi)\rangle\) d) The result of part (c) is exact, but not helpful because the left-hand side involves \(\langle r\rangle\) whereas the right-hand side involves \(r\). Now comes the key approximation: replace \(r\) and \(\phi\) by their averages over one cycle. Show that \(r(t)=\bar{r}(t)+O(\varepsilon)\) and \(\phi(t)=\bar{\phi}(t)+O(\varepsilon)\), and therefore $$ \begin{gathered} d \bar{r} / d t=\varepsilon\langle h[\bar{r} \cos (t+\bar{\phi}),-\bar{r} \sin (t+\bar{\phi}), t] \sin (t+\bar{\phi})\rangle+O\left(\varepsilon^{2}\right) \\\ \bar{r} d \bar{\phi} / d t=\varepsilon\langle h[\bar{r} \cos (t+\bar{\phi}),-\bar{r} \sin (t+\bar{\phi}), t] \cos (t+\bar{\phi})\rangle+O\left(\varepsilon^{2}\right) \end{gathered} $$ where the barred quantities are to be treated as constants inside the averages. These equations are just the averaged equations (7.6.53), derived by a different approach in the text. It is customary to drop the overbars; one usually doesn't distinguish between slowly varying quantities and their averages.

Using the Poincar?-Lindstedt method, show that the frequency of the limit cycle for the van der Pol oscillator \(\ddot{x}+\varepsilon\left(x^{2}-1\right) \dot{x}+x=0\) is given by \(\omega=1-\frac{1}{16} \varepsilon^{2}+O\left(\varepsilon^{3}\right)\).

Sketch the phase portrait for each of the following systems. (As usual, \(r, \theta\) denote polar coordinates.) $$ \dot{r}=r\left(1-r^{2}\right)\left(4-r^{2}\right), \dot{\theta}=2-r^{2} $$
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