Question

Use the results of Problem 21 to determine whether the point at infinity is an ordinary point, a regular singular point, or an irregular singular point of the given differential equation. \(x^{2} y^{\prime \prime}+x y^{\prime}+\left(x^{2}-v^{2}\right) y=0, \quad\) Bessel equation

Short answer

Based on the given Bessel equation and after transforming the point at infinity to a finite point, we found that the quotient of coefficients is analytic at the point \(z=0\), which corresponds to the point at infinity in the original variables. Therefore, we can conclude that the point at infinity for the given Bessel equation is an ordinary point.

Step-by-step solution

Recall definitions of point types

For a given differential equation, there are three types of singular points: 1. Ordinary point: A point at which all the coefficients of the equation are analytic. 2. Regular singular point: A point at which at least one coefficient has a singularity, but their quotient is analytic. 3. Irregular singular point: A point where the quotient of the coefficients has a singularity.

Analyze the Bessel equation

The Bessel equation given is: \(x^{2} y^{\prime \prime}+x y^{\prime}+\left(x^{2}-v^{2}\right) y=0\) First, let's divide the equation by \(x^2\): \(y^{\prime \prime}+\frac{1}{x}y^{\prime}+\left(1-\frac{v^2}{x^2}\right)y=0\) Now we have the following coefficients: \(p(x) = \frac{1}{x}\), \(q(x) = 1-\frac{v^2}{x^2}\) Notice that the point \(x = 0\) is a regular singular point for this equation. We want to analyze the point at infinity, so let's make a change of variable to transform the point at infinity to a finite point: \(z = \frac{1}{x}\), \(x = \frac{1}{z}\) Now, we have to differentiate twice: \(y' = \frac{dy}{dx} = \frac{dy}{dz} \cdot \frac{dz}{dx} = -\frac{1}{z^2} \frac{dy}{dz} = -z^2 y'_z\) \(y'' = \frac{d^2y}{dx^2} = \frac{d}{dx} (-z^2 y'_z) = \frac{d}{dz} (-z^2 y'_z) \cdot \frac{dz}{dx} = -\frac{1}{z^2} \frac{d}{dz} (-z^2 y'_z) = -2z y'_z + z^2 y''_z\) Let's substitute the new variables in the Bessel equation: \(-2z^3 y'_z + z^4 y''_z + \frac{1}{z} (-2z^3 y'_z + z^4 y''_z) + \left( 1 - v^2 z^2 \right) y = 0\) Finally, to get rid of the \(z^4\) term, let's multiply by \(z^2\): \(-2z^5 y'_z + z^6 y''_z -2z^4 y'_z + z^5 y''_z - v^2 z^4 y = 0\) Simplifying the equation, we have: \((z^6 - z^5) y''_z - (2 z^5 - 2 z^4) y'_z - v^2 z^4 y = 0\) Now, we have: \(p(z) = \frac{2 z^5 - 2 z^4}{z^6 - z^5}\), \(q(z) = \frac{-v^2 z^4}{z^6 - z^5}\) And we see that \(\lim_{z \to 0} p(z)\) and \(\lim_{z \to 0} q(z)\) both exist, meaning that \(z = 0\) is an ordinary point, which corresponds to the point at infinity in the original variables.

Determine the point type

Since we found that the point at infinity (after transforming to the finite point in our z variable) is an ordinary point in the transformed equation, we can conclude that the point at infinity is an ordinary point for the given Bessel equation.

Key concepts

Ordinary point

An ordinary point in the context of differential equations is a point where all the coefficients of the equation are well-behaved – specifically, they are analytic functions. This implies these coefficients can be expressed as a Taylor series in the vicinity of this point. In simple terms, there's nothing peculiar happening to the equation at this point.

For instance, in the Bessel equation \[x^{2} y^{\prime \prime}+x y^{\prime}+(x^{2}-v^{2}) y=0\], the first step involves normalizing the equation by dividing all terms by \(x^2\), resulting in \[y^{\prime \prime}+\frac{1}{x}y^{\prime}+(1-\frac{v^2}{x^2})y=0\]. At this stage, we express our coefficients in terms of \(x\):

• \(p(x) = \frac{1}{x}\)
• \(q(x) = 1-\frac{v^2}{x^2}\)

Now, an ordinary point would mean these coefficients do not have singularities or undefined values as \(x\) approaches any particular point of interest (except zero where \(p(x)\) becomes undefined).

However, to analyze the nature at infinity, we switch variables with \(z = \frac{1}{x}\) and transform it into a finite problem. This approach revealed that as \(z\rightarrow 0\), the new form of the equation becomes manageable, confirming the point at infinity is an Ordinary Point.

Regular singular point

A regular singular point is found where, after simplifying the equation, one or more of the coefficients might have singularities themselves, but when examined closer, their behavior suggests that the solutions should also still behave nicely - meaning the potential solutions remain analytic.

For example, when identifying singular points for our differential equation \[x^{2} y^{\prime \prime}+x y^{\prime}+(x^{2}-v^{2}) y=0\], by traditionally solving or simplifying it, the presence of \(x^2\) and \(\frac{1}{x}\) terms in the coefficients \(p(x)\) and \(q(x)\) respectively, suggest that \(x=0\) is a point of interest.

• \(p(x)\) has a singularity at \(x=0\) due to the \(\frac{1}{x}\) term
• Likewise, \(q(x)=1-\frac{v^2}{x^2}\) has a singularity at \(x=0\)

Despite these singularities, the Bessel equation is a regular type of differential equation commonly encountered in mathematical physics.

Still, because the behaviors of potential solutions at \(x=0\) can be mapped onto power series solutions — it is considered a regular singular point. Therefore, in practical application, we treat the equations in such a way that solutions can be obtained using methods like the Frobenius method or series expansion.

Irregular singular point

An irregular singular point is much more complex. It is a location in differential equations where the equations' coefficients and solutions become challenging to handle due to their drastic divergence. Unlike ordinary or regular singular points, here the behavior of the potential solutions could be chaotic and cannot be captured simply by power series.

This type of point can be identified when neither the coefficients of the differential equation nor their ratio is manageable or analytic in any simple form.

• Typically, this occurs when the order of the singularity is higher than allowed for regular.
• Solutions in this case may include terms that grow exponentially, oscillate wildly, or require more complex methods or special functions to understand directly.

In the original problem using the Bessel equation \[x^{2} y^{\prime \prime}+x y^{\prime}+(x^{2}-v^{2}) y=0\], the point at infinity was shown not to display the characteristics of an irregular singularity after transformation.

Consequently, it's important to differentiate between these points because the strategies for solving such differential equations extensively depend on the nature of the singular points present.