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

Based on the given second-order linear differential equation, we determined the continuity of its coefficients around \(t_0 = 2\). Since both coefficients are continuous at \(t_0\), it guarantees the existence of a unique analytic solution in the interval \(2-R<t<2+R\) for some R. In this case, we found an appropriate R value of 1, leading to an interval of \(1<t<3\) where the unique analytic solution is guaranteed to exist.

Step-by-step solution

Identify and simplify the coefficients of the given differential equation

The given differential equation is $$ y^{\prime \prime}+\frac{1}{1+3(t-2)} y^{\prime}+(\sin t) y=0 $$ We can see that the coefficient of \(y^\prime\) is given by \(p(t)=\frac{1}{1+3(t-2)}\) and the coefficient of \(y\) is given by \(q(t) = \sin t\).

Check the continuity of the coefficients around \(t_0\)

In order to guarantee the existence of a unique analytic solution, the coefficients must be continuous around \(t_0 = 2\). Let's look at the coefficients. For \(p(t)=\frac{1}{1+3(t-2)}\), it is continuous for all \(t \neq \frac{1}{3}\) since the denominator does not equal zero for these values. Since \(t_0 = 2 \neq \frac{1}{3}\), it is continuous at \(t_0\). \(q(t)=\sin t\) is continuous for all real values of t because sine function is continuous everywhere. Therefore, it is continuous at \(t_0=2\).

Determine a value for R

Since both coefficients \(p(t)\) and \(q(t)\) are continuous at \(t_0=2\), we know that there exists a positive value R such that a unique solution exists in the interval \(t_0-R<t<t_0+R\). \(p(t)\) has a discontinuity at \(t=\frac{1}{3}\), which doesn't include \(t_0\). To find the smallest value of R considering both coefficients, we can find the distance between \(t_0\) and the nearest discontinuity, which is \(t=\frac{1}{3}\). The distance is \(|2-\frac{1}{3}|=\frac{5}{3}\). Thus, we can choose any positive value for R less than \(\frac{5}{3}\). For instance, we can choose \(R=1\). So, there is a guaranteed unique analytic solution in the interval, \(2-1<t<2+1\), or \(1<t<3\).

Key concepts

Understanding Initial Value Problems

An initial value problem (IVP) in the context of differential equations refers to finding a function that not only satisfies a differential equation, but also fulfills given conditions at a specific point, typically denoted as \( t_0 \). These conditions are usually related to the value of the unknown function and possibly its derivatives at \( t_0 \).

For instance, if we're given the differential equation \( y'' + p(t)y' + q(t)y = 0 \) with the initial conditions specified at \( t_0 \), the goal is to discover a function \( y(t) \) that solves this equation and also agrees with the initial conditions. In the exercise presented, \( t_0 \) is given as 2, and we are tasked to prove the unique existence of a solution that is analytic in an interval around this point.

It's crucial for students to grasp this concept as it serves as the foundation for understanding more advanced topics in differential equations.

Identifying an Ordinary Point

In differential equations, an 'ordinary point' is where the functions that make up the coefficients of the equation are analytic, indicating they can be expanded in a power series around this point. Analytically, this means the coefficients are not only continuous but also have continuous derivatives of all orders at that point.

In our example equation, \( t_0=2 \) was already identified as an ordinary point for the given second-order linear differential equation. This implies the coefficients \( p(t) \) and \( q(t) \) do not have any kind of singularity (like a division by zero, for example) at and around \( t_0=2 \).

This property is paramount because it influences the methods used to solve the differential equation and the behavior of the solution in the neighborhood of \( t_0 \).

Continuity of Coefficients

The continuity of coefficients in differential equations is a necessary condition for certain theorems that guarantee the existence and uniqueness of solutions. Specifically, coefficients \( p(t) \) and \( q(t) \) need to be continuous functions in an interval around \( t_0 \) to apply these theorems correctly.

In the given step by step solution, \( p(t) = \frac{1}{1+3(t-2)} \) and \( q(t) = \sin t \) are examined for continuity at \( t_0=2 \). Since neither function has any singularities or discontinuities at \( t_0 \)—with \( p(t) \) being continuous for all \( t eq \frac{1}{3} \) and \( q(t) \) being continuous everywhere due to the sine function’s inherent continuity—the resulting interpretation of their analytic properties is confirmed, thereby satisfying an important prerequisite for the existence of a solution.

This continuity ensures that nearby values of \( t \) will lead to small changes in \( p(t) \) and \( q(t) \)—a fact that's deeply connected with the predictability and stability of solutions to differential equations.

Unique Solution Existence

The existence of a unique solution to an initial value problem is a fundamental concern for those studying differential equations. Theorems such as the Picard-Lindelöf theorem (also known as the Existence and Uniqueness theorem) provide the conditions under which one can affirm that a given IVP will have a unique solution within a certain interval.

In our example, after verifying the continuity of the coefficients, we directly applied Theorem 8.1 to deduce the existence of a unique analytic solution around \( t_0=2 \) in a particular interval. Theorem 8.1 necessitates that the coefficients of the differential equation not only be continuous but also that they do not lead to singular behavior within the interval of interest.

By finding a value for \( R \) that keeps us away from the nearest discontinuity or singularity, we ensure that there’s an interval \( t_0-R < t < t_0+R \) within which the solution to the differential equation will be unique and analytic. It's insightful to students that a proper choice of \( R \) builds the bridge to this conclusion, emphasizing the interplay between the mathematical conditions and the physical interpretation of the problem at hand. Choosing \( R=1 \) in this case, provides a clear and comfortable interval \( 1 < t < 3 \) for students to work with when considering the unique solution.