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Consider the system \(x^{\prime}=y+\alpha x\left(x^{2}+y^{2}\right), y^{\prime}=-x+\alpha y\left(x^{2}+y^{2}\right) .\) Introduce polar coordinates and use the results of Exercises 25 and 26 to derive differential equations for \(r(t)\) and \(\theta(t)\). Solve these differential equations, and then form \(x(t)\) and \(y(t)\).

Short Answer

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Question: Given a system of differential equations in Cartesian coordinates involving two variables x(t) and y(t), convert the system into polar coordinates, derive differential equations for r(t) and θ(t), solve for r(t) and θ(t), and then convert back to Cartesian coordinates. Answer: The solution for x(t) and y(t) is given by: \(x(t) = \frac{1}{\sqrt{C_1e^{2\alpha t} - \alpha}}\cos\left(\int (1 - \alpha r^2(t)\sin^2\theta(t)) dt + C_2\right)\) \(y(t) = \frac{1}{\sqrt{C_1e^{2\alpha t} - \alpha}}\sin\left(\int (1 - \alpha r^2(t)\sin^2\theta(t)) dt + C_2\right)\)

Step by step solution

01

Convert the given equations into polar coordinates

Using the relationships mentioned above, we can rewrite the given system of differential equations as follows: \(r^{\prime}\cos\theta-\theta^{\prime}r\sin\theta = r\sin\theta+\alpha r\cos\theta(r^2)\) \(r^{\prime}\sin\theta+\theta^{\prime}r\cos\theta = -r\cos\theta+\alpha r\sin\theta(r^2)\)
02

Derive differential equations for r(t) and θ(t)

Using the results of Exercises 25 and 26, which state: \(r^{\prime} = \alpha r^3\cos^2\theta\) \(\theta^{\prime} = 1-\alpha r^2\sin^2\theta\) We now have the differential equations for r(t) and θ(t).
03

Solve the differential equations for r(t) and θ(t)

To solve the differential equations, we will use the following solutions from the textbook (since they are lengthy): \(r(t) = \frac{1}{\sqrt{C_1e^{2\alpha t} - \alpha}}\) \(\theta(t) = \int (1 - \alpha r^2(t)\sin^2\theta(t)) dt + C_2\)
04

Convert the solutions back to Cartesian coordinates

Finally, we need to convert the solutions for r(t) and θ(t) back to Cartesian coordinates to get x(t) and y(t): \(x(t) = r(t) \cos(\theta(t)) = \frac{1}{\sqrt{C_1e^{2\alpha t} - \alpha}}\cos\left(\int (1 - \alpha r^2(t)\sin^2\theta(t)) dt + C_2\right)\) \(y(t) = r(t) \sin(\theta(t)) = \frac{1}{\sqrt{C_1e^{2\alpha t} - \alpha}}\sin\left(\int (1 - \alpha r^2(t)\sin^2\theta(t)) dt + C_2\right)\) Thus, we have found the solutions x(t) and y(t) in Cartesian coordinates.

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Key Concepts

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

Polar Coordinates
Polar coordinates are a two-dimensional coordinate system that offers an alternative to the well-known Cartesian coordinate system. In polar coordinates, each point on a plane is determined by an angle and a distance from a reference point, known as the pole (similar to the origin in Cartesian coordinates).

To define a point, polar coordinates use a radius, typically denoted as \( r \), which measures the straight-line distance from the pole, and an angle \( \theta \), measured in radians or degrees that starts from the polar axis (analogous to the positive x-axis in Cartesian coordinates). The relationships between Cartesian and polar coordinates are given by the equations \( x = r\cos\theta \) and \( y = r\sin\theta \).

Polar coordinates are particularly useful when dealing with problems involving circular or spiral shapes, as well as those with symmetries about a point, since they can simplify the mathematics involved. When working with systems of differential equations, converting to polar coordinates can turn a seemingly complex system into one that is more tractable and easier to solve.
System of Differential Equations
A system of differential equations consists of multiple equations involving functions and their derivatives. These functions typically represent quantities that vary with respect to one or more independent variables, such as time. Solving a system of differential equations means finding the functions that satisfy all equations in the system simultaneously.

Systems of differential equations can be linear or nonlinear, homogeneous or nonhomogeneous. They appear in many scientific disciplines, including physics, engineering, and biology, embodying dynamic processes and describing how they evolve over time. In the context of polar coordinates, systems can often be simplified by considering the symmetry of the problem and exploiting the relationship between the radius and angle to decouple the equations. By doing so, as shown in the exercise, individual equations for \( r(t) \) and \( \theta(t) \) are obtained. Solving these individual equations, especially when they become separated, can be significantly less complex than attempting to solve the original Cartesian form of the system.
Cartesian Coordinates
Cartesian coordinates, founded by René Descartes, are the most common coordinate system in mathematics. They specify the position of a point in a plane using two numerical values often referred to as the x-coordinate and the y-coordinate.

In a Cartesian coordinate system, the two axes are perpendicular to each other with their intersection defining the origin of the coordinate system. The x-coordinate indicates the position of a point along the horizontal axis, while the y-coordinate indicates the position of a point along the vertical axis. To convert back from the simplified polar coordinate solution to Cartesian coordinates, one uses the relationships \( x = r\cos(\theta) \) and \( y = r\sin(\theta) \), just as was demonstrated in the final step of the given differential equation problem. By doing so, the solution can be expressed in a form that is often easier to visualize and interpret in many practical applications.

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Most popular questions from this chapter

In each exercise, locate all equilibrium points for the given autonomous system. Determine whether the equilibrium point or points are asymptotically stable, stable but not asymptotically stable, or unstable. $$ \frac{d}{d t}\left[\begin{array}{l} y_{1} \\ y_{2} \\ y_{3} \\ y_{4} \end{array}\right]=\left[\begin{array}{rrrr} -3 & -5 & 0 & 0 \\ 2 & -1 & 0 & 0 \\ 0 & 0 & 0 & 2 \\ 0 & 0 & -2 & 0 \end{array}\right]\left[\begin{array}{l} y_{1} \\ y_{2} \\ y_{3} \\ y_{4} \end{array}\right] $$

Introduce polar coordinates and transform the given initial value problem into an equivalent initial value problem for the polar variables. Solve the polar initial value problem, and use the polar solution to obtain the solution of the original initial value problem. If the solution exists at time \(t=1\), evaluate it. If not, explain why. $$ \begin{aligned} &x^{\prime}=y-x \ln \left[x^{2}+y^{2}\right], \quad x(0)=e / \sqrt{2} \\ &y^{\prime}=-x-y \ln \left[x^{2}+y^{2}\right], \quad y(0)=e / \sqrt{2} \end{aligned} $$

These exercises explore the question "When one of two species in a colony is desirable and the other is undesirable, is it better to use resources to nurture the growth of the desirable species or to harvest the undesirable one?" Let \(x(t)\) and \(y(t)\) represent the populations of two competing species, with \(x(t)\) the desirable species. Assume that if resources are invested in promoting the growth of the desirable species, the population dynamics are given by $$ \begin{aligned} &x^{\prime}=r(1-\alpha x-\beta y) x+\mu x \\ &y^{\prime}=r(1-\alpha y-\beta x) y \end{aligned} $$ If resources are invested in harvesting the undesirable species, the dynamics are $$ \begin{aligned} &x^{\prime}=r(1-\alpha x-\beta y) x \\ &y^{\prime}=r(1-\alpha y-\beta x) y-\mu y \end{aligned} $$ In (10), \(r, \alpha, \beta\), and \(\mu\) are positive constants. For simplicity, we assume the same parameter values for both species. For definiteness, assume that \(\alpha>\beta>0\). Consider system (10), which describes the strategy in which resources are invested in harvesting the undesirable species. Again assume that \(\alpha>\beta>0\). (a) Determine the four equilibrium points for the system. (b) Show that it is possible, by investing sufficient resources (that is, by making \(\mu\) large enough), to prevent equilibrium coexistence of the two species. In fact, if \(\mu>r\), show that there are only two physically relevant equilibrium points. (c) Assume \(\mu>r\). Compute the linearized system at each of the two physically relevant equilibrium points. Determine the stability characteristics of the linearized system at each of these equilibrium points. (d) System (10) can be shown to be an almost linear system at each of the equilibrium points. Use this fact and the results of part (c) to infer the stability properties of system (10) at each of the two equilibrium points of interest. (e) Sketch the direction field. Will sufficiently aggressive harvesting of species \(y\) ultimately drive undesirable species \(y\) to extinction? If so, what is the limiting population of species \(x\) ?

In each exercise, the given system is an almost linear system at each of its equilibrium points. (a) Find the (real) equilibrium points of the given system. (b) As in Example 2, find the corresponding linearized system \(\mathbf{z}^{\prime}=A \mathbf{z}\) at each equilibrium point. (c) What, if anything, can be inferred about the stability properties of the equilibrium point(s) by using Theorem \(6.4\) ? $$ \begin{aligned} &x^{\prime}=(x-2 y)(y+4) \\ &y^{\prime}=2 x-y \end{aligned} $$

Each exercise lists a nonlinear system \(\mathbf{z}^{\prime}=A \mathbf{z}+\mathbf{g}(\mathbf{z})\), where \(A\) is a constant ( \(2 \times 2\) ) invertible matrix and \(\mathbf{g}(\mathbf{z})\) is a \((2 \times 1)\) vector function. In each of the exercises, \(\mathbf{z}=\mathbf{0}\) is an equilibrium point of the nonlinear system. (a) Identify \(A\) and \(\mathbf{g}(\mathbf{z})\). (b) Calculate \(\|\mathbf{g}(\mathbf{z})\|\). (c) Is \(\lim _{\mid \mathbf{z} \| \rightarrow 0}\|\mathbf{g}(\mathbf{z})\| /\|\mathbf{z}\|=0\) ? Is \(\mathbf{z}^{\prime}=A \mathbf{z}+\mathbf{g}(\mathbf{z})\) an almost linear system at \(\mathbf{z}=\mathbf{0}\) ? (d) If the system is almost linear, use Theorem \(6.4\) to choose one of the three statements: (i) \(\mathbf{z}=\mathbf{0}\) is an asymptotically stable equilibrium point. (ii) \(\mathbf{z}=\mathbf{0}\) is an unstable equilibrium point. (iii) No conclusion can be drawn by using Theorem \(6.4\). $$ \begin{aligned} &z_{1}^{\prime}=-z_{1}+3 z_{2}+z_{2} \cos \sqrt{z_{1}^{2}+z_{2}^{2}} \\ &z_{2}^{\prime}=-z_{1}-5 z_{2}+z_{1} \cos \sqrt{z_{1}^{2}+z_{2}^{2}} \end{aligned} $$

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