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Chapter 4: Problem 4

Find the largest interval \(a

Short Answer

Expert verified
Based on the step-by-step solution provided, the largest interval for the existence of a unique solution for the given initial value problem is $-2 < t < 2$. This interval is determined by considering the points of discontinuity of the coefficients of the derivatives of the unknown functions and ensuring the continuity of the functions and their partial derivatives.

Step by step solution

01

Identify the Functions

We have a system of two first-order differential equations: \[ \begin{cases} (t+2) y_{1}^{\prime} = 3t y_{1} + 5 y_{2}, & y_{1}(1) = 0 \\ (t-2) y_{2}^{\prime} = 2 y_{1} + 4t y_{2}, & y_{2}(1) = 2 \end{cases} \]
02

Analyze the Continuity of the Functions and Their Partial Derivatives

The functions involved in the equations are: $$f_1(t, y_1, y_2) = 3t y_1 + 5 y_2$$ $$f_2(t, y_1, y_2) = 2 y_1 + 4t y_2$$ The partial derivatives are: $$\frac{\partial f_1}{\partial y_1} = 3t, \quad \frac{\partial f_1}{\partial y_2} = 5$$ $$\frac{\partial f_2}{\partial y_1} = 2, \quad \frac{\partial f_2}{\partial y_2} = 4t$$ All of these derivatives are continuous functions. However, we need to consider the coefficients of the derivatives of the unknown functions, which are \((t+2)\) for \(y_1'\) and \((t-2)\) for \(y_2'\).
03

Determine the Largest Interval for the Existence of a Unique Solution

From the analysis in Step 2, we notice that both \((t+2)\) and \((t-2)\) have discontinuities. To have a continuous interval for the unique solution, we need to avoid the points where \(t = -2\) and \(t = 2\). Since we are given \(y_1(1) = 0\) and \(y_2(1) = 2\), the initial point \((1, 0, 2)\) is inside the interval between the point of discontinuity \(t=-2\) and \(t=2\). Therefore, the largest interval for the existence of a unique solution is: $$-2 < t < 2$$

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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 relate some function with its derivatives. In real-world applications, these equations are used to model the behavior of complex systems, such as mechanical systems, populations in biology, or electric circuits.

In the case of the given exercise, we are dealing with a system of two first-order ordinary differential equations (ODEs). First-order means that the highest derivative in the equation is the first derivative. The 'initial value problem' requires finding a function that satisfies the ODEs for a given starting condition.

The initial conditions given are for the point where the independent variable, here denoted by 't', equals 1, specifically, \(y_{1}(1) = 0\) and \(y_{2}(1) = 2\). The goal is to determine where else, besides at \(t = 1\), we can expect the solution to be valid and unique.
Unique Solution Existence
The uniqueness of a solution to an initial value problem is an essential concept. There can be instances where no solution exists, or multiple solutions are possible. To ensure a unique solution, mathematicians often rely on the Picard-Lindel枚f theorem or its special case, the existence and uniqueness theorem.

This theorem requires certain conditions to be met. To have a unique solution on an interval, the functions described in the differential equations and their partial derivatives with respect to the dependent variables -- in this case, \(y_{1}\) and \(y_{2}\) -- must be continuous on that interval.

The existence of a unique solution is pretty straightforward for the given initial value problem because the partial derivatives do not involve any terms that could become undefined or infinite within the interval \( (-2, 2) \). Therefore, according to the theorem, a unique solution to the system of differential equations is guaranteed for any value of \(t\) within that interval.
Continuity and Differentiability
Continuity and differentiability are foundational concepts in calculus that are integral to solving differential equations. A function is considered continuous at a point if the left-hand limit, the right-hand limit, and the value of the function at that point all exist and are equal. Differentiability, on the other hand, requires that the function has a defined derivative at that point.

For the unique solution to exist, it is not enough for the functions to be merely continuous; they must also be continuously differentiable across the interval in question. As shown in the exercise, while the functions \(f_{1}(t, y_{1}, y_{2})\) and \(f_{2}(t, y_{1}, y_{2})\) and their partial derivatives are continuous, we must be cautious of the coefficients \( (t+2) \) and \( (t-2) \) as they become zero at \( t=-2 \) and \( t=2 \) respectively. This makes the equations non-differentiable at these points, thus breaking the continuity of the interval.

In summary, it is the continuity and differentiability of the functions involved that dictate the interval where we can guarantee a unique solution for the given initial value problem. The exclusion of the points \( t=-2 \) and \( t=2 \) ensures that the criteria for the existence of a unique solution are satisfied.

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Most popular questions from this chapter

Write a program that applies the Runge-Kutta algorithm (12) to the given problem. Run the program on the interval given, with a constant step size of \(h=0.01\). Plot the components of the solution.\(\begin{array}{lll}y_{1}^{\prime} & =-y_{1}+y_{2}+2, & y_{1}(0) & =1, \\\ y_{2}^{\prime} & =-y_{1}-y_{2}, & y_{2}(0) & =0\end{array} \quad 0 \leq t \leq 2\)

For each of the exercises, (a) Rewrite the equations from the given exercise in vector form as \(\mathbf{y}^{\prime}(t)=A \mathbf{y}(t)\), identifying the constant matrix \(A\). (b) Rewrite the solution of the equations in part (a) in vector form as \(\mathbf{y}(t)=c_{1} \mathbf{y}_{1}(t)+\) $$ \mathbf{y}^{\prime}=A \mathbf{y}, \quad \mathbf{y}(0)=\left[\begin{array}{r} -1 \\ 8 \end{array}\right], \text { where } A=\left[\begin{array}{ll} 3 & 2 \\ 4 & 1 \end{array}\right] \text { and } \mathbf{y}(t)=c_{1} e^{5 t}\left[\begin{array}{l} 1 \\ 1 \end{array}\right]+c_{2} e^{-t}\left[\begin{array}{r} -1 \\ 2 \end{array}\right] $$

Consider the homogeneous linear system \(\mathbf{y}^{\prime}=A \mathbf{y} .\) Recall that any associated fundamental matrix satisfies the matrix differential equation \(\Psi^{\prime}=A \Psi\). In each exercise, construct a fundamental matrix that solves the matrix initial value problem \(\Psi^{\prime}=A \Psi, \Psi\left(t_{0}\right)=\Psi_{0}\).\(\Psi^{\prime}=\left[\begin{array}{rr}1 & -1 \\\ -1 & 1\end{array}\right] \Psi, \quad \Psi(1)=\left[\begin{array}{ll}1 & 0 \\\ 0 & 1\end{array}\right]\)

Exercises 1-5: For the given matrix functions \(A(t), B(t)\), and \(\mathbf{c}(t)\), make the indicated calculations $$ A(t)=\left[\begin{array}{cc} t-1 & t^{2} \\ 2 & 2 t+1 \end{array}\right], \quad B(t)=\left[\begin{array}{cc} t & -1 \\ 0 & t+2 \end{array}\right], \quad \mathbf{c}(t)=\left[\begin{array}{c} t+1 \\ -1 \end{array}\right] $$ $$ \mathrm{A}(t) B(t)-B(t) A(t) $$

In each exercise, the general solution of a \((2 \times 2)\) linear system \(\mathbf{y}^{\prime}=A \mathbf{y}\) is given, along with an initial condition. Sketch the phase plane solution trajectory that satisfies the given initial condition. $$ \mathbf{y}(t)=c_{1} e^{-t}\left[\begin{array}{r} 1 \\ -1 \end{array}\right]+c_{2} e^{-2 t}\left[\begin{array}{l} 1 \\ 1 \end{array}\right], \quad \mathbf{y}(0)=\left[\begin{array}{r} 3 \\ -1 \end{array}\right] $$
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