Question

Find all the regular singular points of the given differential equation. Determine the indicial equation and the exponents at the singularity for each regular singular point. \(x^{2}(1-x) v^{\prime \prime}-(1+x) v^{\prime}+2 x v=0\)

Short answer

Answer: The regular singular points of the given differential equation are \(x=0\) and \(x=1\). The exponents at these singularities can be found by solving the indicial equation obtained from substituting a solution in the form of a Frobenius series into the given equation.

Step-by-step solution

Find the singular points

To find singular points, we need to find where the coefficient of the highest order derivative becomes singular. In our given equation, the highest order derivative is \(v^{\prime \prime}\), and the coefficient is \(x^{2}(1-x)\). We look for the values of \(x\) where this coefficient becomes zero or undefined: \(x^{2}(1-x)=0\) Here, we get two values of \(x\): \(x=0\) and \(x=1\). These are our potential singular points.

Check if the singular points are regular

For the singular points to be regular, the coefficients of the lower order derivatives (i.e., \(v^{\prime}\) and \(v\)) should be analytic at those points. Analytic functions are functions that can be represented by a convergent power series in a neighborhood of the point of interest, and their derivatives also remain analytic. The coefficient of the \(v^{\prime}\) is \(-(1+x)\), and it is clearly analytic at both \(x=0\) and \(x=1\) as it has no singularities. The coefficient of \(v\) is \(2x\), which is also analytic at both \(x=0\) and \(x=1\). Thus, both \(x=0\) and \(x=1\) are regular singular points.

Determine the indicial equation and exponents at the singularities

To find the indicial equation, we assume a solution in the form of a Frobenius series: \(v(x) = \sum_{n=0}^{\infty} a_n (x-x_0)^{n+r}\), where \(x_0\) is the regular singular point and \(r\) is the exponent. Let's first consider the singular point \(x=0\). Substitute the Frobenius series into the given equation and equate the powers of \((x-0)^n\) to get the recurrence relation for the coefficients \(a_n\). Now let's consider the singular point \(x=1\). For this, we substitute the Frobenius series into the given equation and replace all instances of \(x\) with \((x-1)\). Then equate the powers of \((x-1)^n\) to get the recurrence relation for the coefficients \(a_n\). In both cases, we will obtain an indicial equation that depends on the exponent \(r\). Solving the indicial equation for \(r\) will yield the exponents at each regular singular point.

Key concepts

Indicial Equation

Whenever a differential equation has regular singular points, it becomes crucial to find the so-called indicial equation. The indicial equation helps us determine the possible exponents related to the singularity of the solution, starting with an assumption involving a polynomial. Basically, to find the indicial equation, assume a solution in the form of a Frobenius series. Substitute this series into the differential equation. A common example might look like:

• Given a form: \( v(x) = \sum_{n=0}^{\infty} a_n (x-x_0)^{n+r} \)
• Where \( x_0 \) is the regular singular point and \( r \) is what we aim to find.

By balancing terms and focusing on the lowest power of \((x-x_0)\), we derive an equation solely dependent on \( r \). That equation we solve is our indicial equation. Solving it gives us the exponents that characterize solutions around the singular point.

Frobenius Series

A Frobenius series is an incredibly important concept when solving differential equations with regular singular points. It's a technique that enables us to handle these singularities elegantly. The basic idea of a Frobenius series is to express a solution as a power series—including a multiplicative factor with a non-integer exponent. Unlike a simple power series, a Frobenius series can handle solutions that aren't analytic at the point of interest.

• Possible form: \( v(x) = \sum_{n=0}^{\infty} a_n (x-x_0)^{n+r} \)
• Here, \( r \) is an exponent determined by solving the indicial equation.

Frobenius series are particularly useful because these series converge in a neighborhood around the singularity, allowing us to find solutions that might otherwise be impossible to determine through standard methods.

Exponents at Singularity

In the context of solving differential equations with regular singular points, exponents at singularity are critical to understanding the nature of possible solutions.When solving differential equations around a singular point, identifying these exponents indicates which types of terms need to be in our Frobenius series solution. Exponents \( r \) come from solving the indicial equation, and they inform us about the behavior of the function around this point. The nature of these exponents can tell us a lot:

• If \( r \) is a real number, often there will be more than one series that satisfies the equation.
• Complex or repeating exponents might require special handling to express the full solution set.

Finding these exponents provides insight into both the generality and the specific aspects of our series solutions.

Analytic Functions

In the study of differential equations, especially concerning regular singular points, understanding analytic functions is essential. An analytic function is one that can be expressed as a power series in the neighborhood of a point. For a singular point to be classified as 'regular,' certain criteria need to be met, specifically concerning the coefficients of the equation. It requires that:

• Lower order derivatives' coefficients must be analytic at the given points.
• This means being expressible as a power series with a radius of convergence greater than zero.

The coefficients integrating into our differential equations need to remain analytic at any singular points to ensure the point is indeed regular. This property allows us to apply the Frobenius method to construct solutions effectively.