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Chapter 8: Problem 11

\(x_{1}{ }^{\prime}=\left(x_{2}+x_{1} / 5\right)\left(1-x_{1}^{2}\right)\) and \(x_{2}^{\prime}=-x_{1}\left(1-x_{2}^{2}\right)\), with \(x_{1}(0)=0.8\) and \(x_{2}(0)=0\) on \([0,30]\).

Short Answer

Expert verified
Numerically solve the system with initial conditions using Runge-Kutta.

Step by step solution

01

Understand the Problem

We have a system of differential equations describing the dynamics of two variables \(x_1\) and \(x_2\). The goal is to solve these equations with specific initial conditions \(x_1(0) = 0.8\) and \(x_2(0) = 0\) over the interval \([0, 30]\).
02

Analyze the System of Equations

The equations given are:1. \(x_1' = (x_2 + x_1/5)(1-x_1^2)\)2. \(x_2' = -x_1(1-x_2^2)\)These are ordinary differential equations and they need to be solved simultaneously.
03

Use Initial Conditions

We start by acknowledging the initial conditions for integration: \(x_1(0) = 0.8\) and \(x_2(0) = 0\). This sets the starting values for solving the differential equations.
04

Choose a Numerical Method

A suitable method to solve initial value problems for systems of differential equations is the Euler method or more commonly, Runge-Kutta methods (such as the 4th order Runge-Kutta method). For highly accurate results, the 4th order Runge-Kutta method is preferred.
05

Implement the Numerical Solution

Implement a numerical solution (e.g., using a software package like MATLAB, Python with SciPy, or any numerical computing software) to solve the system of equations. Write the function defining the system and apply the chosen method with the initial conditions across the time span \([0, 30]\).
06

Interpret the Results

The numerical solution will provide values for \(x_1(t)\) and \(x_2(t)\) across the interval. Interpret these results to understand the behavior of the system over time.

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Key Concepts

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

Initial Value Problems
Initial value problems (IVPs) are a type of differential equation where the solution is required to satisfy certain conditions at the beginning of the interval. These conditions are called "initial values". In the context of solving differential equations, initial values specify the state of the system at the start of the time period of interest.
For our exercise, the system of equations is given initial values of \(x_1(0) = 0.8\) and \(x_2(0) = 0\). This means that at time \(t=0\), the variable \(x_1\) starts at 0.8, and \(x_2\) starts at 0.
Initial value problems are crucial in modeling scenarios where the initial state is known and the objective is to predict future behavior. Think of a rocket's trajectory starting from a launch pad, or the growth of a population starting from a known size.
Runge-Kutta Method
The Runge-Kutta methods are a family of iterative methods used to approximate solutions to ordinary differential equations. They are known for their accuracy and stability, especially the 4th-order Runge-Kutta method, which is one of the most commonly used.
In our problem, we use the 4th-order Runge-Kutta method to solve the initial value problem numerically. It approximates the solution by combining information from several points in each interval, rather than just the endpoints.
This approach improves the accuracy without significantly increasing the computational complexity. The Runge-Kutta method is favored because it balances efficiency and precision, making it especially useful for solving systems of differential equations like the one in our exercise.
Numerical Solutions
When analytical solutions to differential equations are difficult or impossible to find, numerical solutions provide an alternative. These are approximate solutions obtained using computational algorithms.
Numerical methods like the Euler method or Runge-Kutta can be implemented using software packages such as MATLAB, Python with SciPy, or similar tools. These programs numerically integrate the differential equations given their initial conditions across the desired time span.
For complex systems or equations, numerical solutions are invaluable, as they allow for the exploration and understanding of the system's behavior through simulated data. The beauty of numerical solutions is in their capacity to handle real-world complexities where exact solutions are not feasible.
System of Differential Equations
A system of differential equations involves multiple functions and their derivatives and are typically used to describe interconnected dynamic systems. In our exercise, we have two differential equations describing the relationship between variables \(x_1\) and \(x_2\).
Each equation in the system is dependent on the others, meaning the variables affect each other's rate of change. This makes the problem more complex compared to single differential equations, as all equations must be solved simultaneously.
Systems of differential equations are common in modeling situations such as ecosystem dynamics, economics, and engineering, where multiple factors interact continuously over time. Solving these systems helps in predicting how the system evolves, making them invaluable in scientific and applied research.

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Most popular questions from this chapter

The system $$ \begin{aligned} &m_{1} x^{\prime \prime}=-k_{1} x+k_{2}(y-x) \\ &m_{2} y^{\prime \prime}=-k_{2}(y-x) \end{aligned} $$ models a coupled oscillator. Imagine a spring (with spring constant \(k_{1}\) ) attached to a hook in the ceiling. Mass \(m_{1}\) is attached to the spring, and a second spring (with spring constant \(k_{2}\) ) is attached to the bottom of mass \(m_{1}\). If a second mass, \(m_{2}\), is attached to the second spring, you have a coupled oscillator, where \(x\) and \(y\) represent the displacements of masses \(m_{1}\) and \(m_{2}\) from their respective equilibrium positions. If you set \(x_{1}=x, x_{2}=x^{\prime}, x_{3}=y\), and \(x_{4}=y^{\prime}\), you can show that $$ \begin{aligned} x_{1}^{\prime} &=x_{2}, \\ x_{2}^{\prime} &=-\frac{k_{1}}{m_{1}} x_{1}+\frac{k_{2}}{m_{1}}\left(x_{3}-x_{1}\right) \\ x_{3}^{\prime} &=x_{4}, \\ x_{4}^{\prime} &=-\frac{k_{2}}{m_{2}}\left(x_{3}-x_{1}\right) \end{aligned} $$ Assume \(k_{1}=k_{2}=2\) and \(m_{1}=m_{2}=1\). Create an ODE file and name it couple.m. Suppose that the first mass is displaced upward two units, the second downward two units, and both masses are released from rest. Plot the position of each mass versus time.

In the 1920 's, the Italian mathematician Umberto Volterra proposed the following mathematical model of a predator-prey situation to explain why, during the first World War, a larger percentage of the catch of Italian fishermen consisted of sharks and other fish eating fish than was true both before and after the war. Let \(x(t)\) denote the population of the prey, and let \(y(t)\) denote the population of the predators. In the absence of the predators, the prey population would have a birth rate greater than its death rate, and consequently would grow according to the exponential model of population growth, i.e. the growth rate of the population would be proportional to the population itself. The presence of the predator population has the effect of reducing the growth rate, and this reduction depends on the number of encounters between individuals of the two species. Since it is reasonable to assume that the number of such encounters is proportional to the number of individuals of each population, the reduction in the growth rate is also proportional to the product of the two populations, i.e., there are constants \(a\) and \(b\) such that $$ x^{\prime}=a x-b x y $$ Since the predator population depends on the prey population for its food supply it is natural to assume that in the absence of the prey population, the predator population would actually decrease, i.e. the growth rate would be negative. Furthermore the (negative) growth rate is proportional to the population. The presence of the prey population would provide a source of food, so it would increase the growth rate of the predator species. By the same reasoning used for the prey species, this increase would be proportional to the product of the two populations. Thus, there are constants \(c\) and \(d\) such that $$ y^{\prime}=-c y+d x y . $$ a) A typical example would be with the constants given by \(a=0.4, b=0.01, c=0.3\), and \(d=0.005\). Start with initial conditions \(x_{1}(0)=50\) and \(x_{2}(0)=30\), and compute the solution to (8.22) and (8.23) over the interval \([0,100]\). Prepare both a time plot and a phase plane plot. After Volterra had obtained his model of the predator-prey populations, he improved it to include the effect of "fishing," or more generally of a removal of individuals of the two populations which does not discriminate between the two species. The effect would be a reduction in the growth rate for each of the populations by an amount which is proportional to the individual populations. Furthermore, if the removal is truly indiscriminate, the proportionality constant will be the same in each case. Thus, the model in equations \((8.22)\) and (8.23) must be changed to $$ \begin{aligned} &x^{\prime}=a x-b x y-e x \\ &y^{\prime}=-c y+d x y-e y \end{aligned} $$ where \(e\) is another constant. b) To see the effect of indiscriminate reduction, compute the solutions to the system in \((8.24)\) when \(e=0\), \(0.01,0.02,0.03\), and \(0.04\), and the other constants are the same as they were in part a). Plot the five solutions on the same phase plane, and label them properly. c) Can you use the plot you constructed in part b) to explain why the fishermen caught more sharks during World War I? You can assume that because of the war they did less fishing.

In Exercises \(4-7\), plot the solutions to all of the initial value problems on one figure. Choose a solution interval starting at \(t=0\) that well illustrates any steady-state phenomena that you see. Exercise 1 might be helpful. \(y^{\prime}+4 y=2 \cos t+\sin 4 t\), with \(y(0)=-5,-4, \ldots, 5\).

Separate variables and find an explicit solution of \(x^{\prime}=t /(1+x), x(0)=1\). a) Use ode45 to find the numerical solution of the initial value problem on the interval \([0,3]\). Store the solution in the variables \(t\) and \(x_{-}\)ode45. b) Use the explicit solution to find exact solution values at each point in the vector \(t\) obtained in part a). Store the results in the variable \(x_{-}\)exact. Compare the solutions graphically with plot \(\left(t\right.\), \(x_{-}\)exact, \(t\), x_ode45, 'r. '). Obtain a printout with a legend. c) For a finer graphical image of the error plot \(\mathrm{x}_{-}\)exact - \(\mathrm{x}_{-}\)ode45 versus \(\mathrm{t}\) in a new figure. Pay special attention to the dimensions on the \(y\)-axis.

\(y^{\prime}+4 y=t^{2}\), with \(y(0)=-9,-8,-7,-1,1,7,8,9\).
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