Question

In each exercise, \(t=t_{0}\) is an ordinary point of \(y^{\prime \prime}+p(t) y^{\prime}+q(t) y=0 .\) Apply Theorem \(8.1\) to determine a value \(R>0\) such that an initial value problem, with the initial conditions prescribed at \(t_{0}\), is guaranteed to have a unique solution that is analytic in the interval \(t_{0}-R

Short answer

For a given second-order linear differential equation, find a value \(R>0\) such that there exists a unique analytic solution in the interval \(t_0 - R < t < t_0 + R\), where \(t_0 = 0\), \(p(t) = \frac{1}{1+4t^2}\), and \(q(t) = \frac{l}{4+t}\). Answer: \(R = \min(1, \frac{4}{l})\)

Step-by-step solution

Identify the functions \(p(t)\) and \(q(t)\)

In the given equation, we see that \(p(t)=\frac{1}{1+4t^2}\) and \(q(t)=\frac{l}{4+t}\).

Find the radius of convergence for \(p(t)\) and \(q(t)\) at \(t_0=0\)

First, let's find the radius of convergence of \(p(t)\). Using the ratio test, we have $$ R_p = \lim_{t \to 0}\left|\frac{1+4t^2}{1}\right| = 1 $$ Now, let's find the radius of convergence of \(q(t)\). Again using the ratio test, we have $$ R_q = \lim_{t \to 0}\left|\frac{4+t}{l}\right| = \frac{4}{l} $$

Determine the minimum radius of convergence

The minimum radius of convergence is the smallest of \(R_p\) and \(R_q\). Here, the smallest of the two values \(R_p=1\) and \(R_q= \frac{4}{l}\) should be chosen. $$R = \min(R_p, R_q)$$

Verify the conditions of Theorem 8.1

Since \(p(t) = \frac{1}{1+4t^2}\) and \(q(t) = \frac{l}{4+t}\) are both analytic functions, and both have finite radii of convergence \(R_p=1\) and \(R_q=\frac{4}{l}\), Theorem 8.1 can be applied. This means there exists a unique solution to the initial value problem that is analytic in the interval \(t_0 - R < t < t_0 + R\).

Solution

The value of \(R > 0\) for which the initial value problem has a unique analytic solution in the interval \((t_0-R, t_0+R)\) is given by $$R = \min(R_p, R_q)$$

Key concepts

Initial Value Problem

An initial value problem (IVP) is a specific type of differential equation along with a specified value, known as the initial condition, at a given point. In the context of an ordinary differential equation (ODE), the initial value problem involves finding a function that not only satisfies the differential equation, but also meets the initial condition at a particular point.

In our exercise, the IVP is to find a solution to the second-order ODE \(y'' + p(t)y' + q(t)y = 0\), where the initial conditions are prescribed at \(t_0\). The solution to this problem provides a function \(y(t)\) that is defined and continuous within an interval around \(t_0\), and aligns with the given initial values. IVPs are fundamental in the study of differential equations as they simulate real-world phenomena where knowing the state at some starting point allows the prediction of future behavior.

Radius of Convergence

The radius of convergence is critical when dealing with power series solutions of ODEs. It indicates the distance within which a power series is guaranteed to converge, or in simpler terms, where the series 'makes sense' and doesn’t head off to infinity.

In our step-by-step solution, we calculated the radius of convergence for both \(p(t)\) and \(q(t)\). The smaller radius dictates the overall convergence radius for the solution series because if one of the series diverges (fails to converge), our overall solution won't hold. For the given exercise, the radius of convergence must be determined for both series to ensure that there is a guaranteed interval for our IVP solution to exist and be unique.

Theorem 8.1 Application

Theorem 8.1 pertains to the existence and uniqueness of solutions of IVPs for ODEs. It states that if the coefficients of the ODE are analytic (meaning they can be expressed as a power series) at a point \(t_0\), then there exists a unique analytic solution to the differential equation within an interval around \(t_0\), determined by the radius of convergence of the series.

In our step-by-step solution, we verified that \(p(t)\) and \(q(t)\) are both analytic at \(t_0=0\) and have finite radii of convergence. By applying Theorem 8.1, we confirmed there is a unique solution to the differential equation that is analytic in the interval \((t_0-R, t_0+R)\), where \(R\) is the minimum radius of convergence among those of \(p(t)\) and \(q(t)\).

Ordinary Differential Equations

Ordinary differential equations (ODEs) are equations involving derivatives of a function with respect to one independent variable. They serve as a foundational tool in modeling continuous dynamical systems such as mechanical systems, fluid dynamics, and electrical circuits.

In our case, the ODE \(y'' + p(t)y' + q(t)y = 0\) is second-order, meaning the highest derivative is the second derivative of \(y\). ODEs can be much more complex and come in various orders and forms, but they all share the common feature of relating some function with its derivatives. Understanding the solutions to ODEs is essential for predicting the behavior of physical systems and is a central topic in the study of applied mathematics, physics, and engineering.