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

Answer: The value of R is \(\frac{5}{6}\).

Step-by-step solution

1. Identify the functions p(t) and q(t)

From the given ODE, we can identify the functions p(t) and q(t) as follows: $$ p(t) = \frac{1}{4 - 3t} $$ and $$ q(t) = \frac{3t}{5 + 30t} $$

2. Determine the intervals for p(t) and q(t) to be analytic

An analytic function is a function that can be represented by a converging power series. To find the intervals where p(t) and q(t) are analytic, we need to find the values of t for which the denominators of these functions are nonzero. For p(t), the denominator is: \(4 - 3t\). This is equal to 0 when: $$ 4 - 3t = 0 \implies t = \frac{4}{3} $$ So, for p(t), there is no issue of analyticity at \(t_0 = -1\), as \(t_0\) is not equal to \(\frac{4}{3}\). For q(t), the denominator is: \(5 + 30t\). This is equal to 0 when: $$ 5 + 30t = 0 \implies t = -\frac{1}{6} $$ So, for q(t), there is no issue of analyticity at \(t_0 = -1\), as \(t_0\) is not equal to \(-\frac{1}{6}\).

3. Find R, the distance to the closest singularity

Now, we need to find the value of R such that the initial point \(t_0 = -1\) is within the interval \(t_0 - R < t < t_0 + R\). To do this, we simply find the distance between \(t_0\) and the closest singularity: $$ R = \min\left\{ \left|t_0 - \frac{4}{3}\right|, \left|t_0 + \frac{1}{6}\right|\right\} $$

4. Calculate R

Now, we plug in the values of \(t_0\) and find R: $$ R = \min\left\{ \left|-1 - \frac{4}{3}\right|, \left|-1 + \frac{1}{6}\right|\right\} = \min\left\{ \frac{7}{3}, \frac{5}{6}\right\} $$ Therefore, R = \(\frac{5}{6}\), as it is the minimum value.

5. Conclusion

By applying Theorem 8.1, we have found that for the given ODE, there is a unique solution for the initial value problem that is analytic in the interval \(t_0 - \frac{5}{6} < t < t_0 + \frac{5}{6}\), or \(-\frac{11}{6} < t < \frac{1}{6}\).

Key concepts

Analytic Solutions

An analytic solution to a differential equation refers to a solution that can be expressed as a power series. This means it can be written as an infinite sum of terms of the form \( a_n (t - t_0)^n \), where \( a_n \) are coefficients and \( t_0 \) is the point of expansion. Analytic solutions are important because they give a precise mathematical representation of the behavior of the solution in a neighborhood of a point.
Analyzing the functions \( p(t) \) and \( q(t) \) in the original exercise is crucial to determine where these solutions exist. When we say a solution is analytic at \( t_0 \), it indicates the completeness and smoothness of the solution at that point.

• The existence of an analytic solution depends on the functions in the differential equation being analytic around the point \( t_0 \).
• For the provided differential equation, \( p(t) = \frac{1}{4 - 3t} \) and \( q(t) = \frac{3t}{5 + 30t} \) must be analytic around the prescribed point \( t_0 = -1 \).

Thus, ensuring both functions do not have singular points (points where they are undefined or not smooth) near \( t_0 \) is essential for determining where an analytic solution can be guaranteed.

Initial Value Problems

An initial value problem (IVP) in the context of ordinary differential equations involves finding a function \( y(t) \) that satisfies a differential equation and also meets the conditions at an initial point, typically expressed as \( y(t_0) = y_0 \) and \( y'(t_0) = y'_0 \). Solving an IVP ensures that the solution not only follows the dynamics described by the differential equation but also conforms to specific initial conditions.
In the given exercise, the initial value is provided at \( t_0 = -1 \), which is considered when determining the solution's behavior. The uniqueness and existence of the solution closely relate to the initial conditions and how they align with the entire equation setup.

• Having the initial values at \( t_0 \) offers a starting criterion to solve the ODE uniquely.
• Such problems often arise in real-world scenarios where the current state is known, and the system's future behavior is to be predicted.

By ensuring the functions are analytic, the initial value problem will have a unique and well-defined analytic solution over a certain interval.

Singular Points

Singular points in the context of differential equations are particular points at which the solution may fail to be analytic or even defined. Identifying singular points is crucial because they can prevent us from finding a power series solution around those points.
In the article exercise, singular points arise where the denominator of \( p(t) \) or \( q(t) \) becomes zero, making the functions undefined.

• For \( p(t) = \frac{1}{4 - 3t} \), a singularity occurs at \( t = \frac{4}{3} \).
• For \( q(t) = \frac{3t}{5 + 30t} \), a singularity occurs at \( t = -\frac{1}{6} \).

These singular points determine the boundaries beyond which the analytic nature of the solution cannot extend if they fall within the interval being considered. Hence, the distance of \( t_0 \) to the nearest singular point helps us establish the radius \( R \) within which an analytic solution is valid.

Interval of Convergence

The interval of convergence describes the specific range over which the power series solution of a differential equation remains valid and convergent. For an initial value problem, this interval is crucial as it provides the domain within which the solution behaves as expected and adheres to the initial conditions given.
In the exercise, determining this interval involves finding how far the initial point \( t_0 = -1 \) is from the nearest singular point.

• The interval of convergence is generally symmetric around \( t_0 \), determined by the distance \( R \) to the nearest point where the functions cease to remain analytic.
• As calculated, \( R = \frac{5}{6} \) gives us the interval \( -\frac{11}{6} < t < \frac{1}{6} \).

Within this interval, the solution is guaranteed to be both unique and analytic, allowing for accurate predictions and further mathematical manipulation. Outside this interval, the behavior of the solution may alter due to the presence of singularities.