Question

Consider the differential equation $$ x^{3} y^{\prime \prime}+\alpha x y^{\prime}+\beta y=0 $$ where \(\alpha\) and \(\beta\) are real constants and \(\alpha \neq 0\). (a) Show that \(x=0\) is an irregular singular point. (b) By attempting to determine a solution of the form \(\sum_{n=0}^{\infty} a_{n} x^{r+n},\) show that the indicial equation for \(r\) is linear, and consequently there is only one formal solution of the assumed form. (c) Show that if \(\beta / \alpha=-1,0,1,2, \ldots,\) then the formal series solution terminates and therefore is an actual solution. For other values of \(\beta / \alpha\) show that the formal series solution has a zero radius of convergence, and so does not represent an actual solution in any interval.

Short answer

Answer: The main conclusions drawn from the analysis of the given second-order, non-linear differential equation are: 1. The equation has an irregular singular point at \(x=0\). 2. There is a linear indicial equation for \(r\), implying only one formal power series solution with the given form. 3. If \(\beta/\alpha=-1,0,1,2,\ldots\), the series solution terminates and represents an actual convergent solution. 4. For all other values of \(\beta/\alpha\), the series solution has zero radius of convergence and does not represent an actual solution in any interval.

Step-by-step solution

Identify the singular points

To determine the type of singularity at \(x=0\), we can first normalize the differential equation, then check if it has finite limits at \(x=0\). In this case, the normalized equation is $$ y^{\prime \prime}+\frac{\alpha}{x^2}y^{\prime}+\frac{\beta}{x^3}y=0. $$ Since the coefficients of the normalized equation are not finite at \(x=0\), it is an irregular singular point.

Attempt a power series solution

Let's assume a power series solution for the equation, in the form $$ y(x)=\sum_{n=0}^{\infty}a_{n}x^{r+n}. $$ Now, we need to compute the first and second derivatives of this assumed solution: $$ y^{\prime}(x)=\sum_{n=0}^{\infty}(r+n)a_{n}x^{r+n-1},\quad y^{\prime\prime}(x)=\sum_{n=0}^{\infty}(r+n)(r+n-1)a_{n}x^{r+n-2}. $$ Substitute these derivatives into the given differential equation and match the powers of \(x\) in both sides.

Analyze the indicial equation

To simplify the equations, change the summation index to \(m = n - 2\). Also, notice that the terms of the equation will be equal to zero for all powers of \(x\): $$ \sum_{m=-2}^{\infty}(m+2)(m+1)a_{m+2}x^{r+m}+\alpha\sum_{m=-2}^{\infty}(m+1)(m+2)a_{m+2}x^{r+m}+\beta\sum_{m=-2}^{\infty}a_{m}x^{r+m}=0. $$ Now, let's set the expression inside the sum equal to zero: $$ (m+2)(m+1)a_{m+2}+\alpha(m+1)(m+2)a_{m+2}+\beta a_m=0. $$ This equation must hold for all \(m\). Since the terms are linear independent, we can rewrite it as an indicial equation for \(r\) as $$ r(r-1)+\alpha r+\beta=0. $$ The indicial equation is a linear equation in \(r\), and hence there is only one possible value for \(r\) and one formal solution of the assumed form.

Study the conditions for the existence of an actual solution

To analyze the convergence of the power series solution, let's consider the term \(a_{m+2}\) divided by \(a_m\): $$ \frac{a_{m+2}}{a_m}=-\frac{\beta}{(m+2)(m+1)+\alpha(m+2)+\beta}. $$ Now, let's consider two cases: - Case 1: If \(\beta/\alpha=-1,0,1,2,\ldots\), the denominator of the fraction becomes \(0\) at some point and the power series terminates. In this case, the solution is an actual convergent solution. - Case 2: For other values of \(\beta/\alpha\), it can be seen that the absolute value of the fraction approaches infinity as \(m\rightarrow\infty\). This implies that the formal series solution has a zero radius of convergence and does not represent an actual solution in any interval. In conclusion, the given differential equation has an irregular singular point at \(x=0\), and the indicial equation for \(r\) is linear. There is only one formal power series solution with the given form. If \(\beta/\alpha=-1,0,1,2,\ldots\), then the series solution terminates and represents an actual convergent solution. For all other values of \(\beta/\alpha\), the series solution has a zero radius of convergence and does not represent an actual solution in any interval.

Key concepts

Singular Point Analysis

When studying differential equations, identifying and analyzing singular points are essential steps. A singular point of an ordinary differential equation (ODE) is a value of the independent variable, often denoted as 'x', where the behavior of the solution is not immediately clear or the equation may not be well-defined. Singular points are categorized as either regular or irregular, which affects the type of solution techniques we might employ.

For the ODE given by
\[x^{3} y^{\textprime \textprime}+alpha x y^{\textprime}+beta y=0\],the point
\[x=0\]is considered singular because the coefficients of the equation do not remain finite as we approach this point. Upon normalization, the behavior of the coefficients indicates that we have an irregular singular point. This also hints at the need for special solution methods such as the Frobenius method, which uses a power series expansion to find solutions near such points.

Power Series Solution

A power series solution is a common method to solve ODEs, particularly useful near regular singular points. It expresses the solution as an infinite sum of powers of the independent variable. For the ODE in question, we assume a solution in the form of
\[y(x) = \sum_{n=0}^{\infty} a_{n} x^{r+n}\].
This method assumes that there exists a power series that converges to the actual solution of the ODE for some interval around the singular point.

The process involves substituting the power series, along with its derivatives, back into the original ODE and equating coefficients of like powers to determine the unknown coefficients
\[a_{n}\]and the exponent
\[r\].The goal is to create a recursive relationship for the coefficients that can potentially describe the solution over an interval containing the singular point.

Indicial Equation

The indicial equation is a fundamental part of the power series solution method near a singular point. It arises from the need to determine the exponent
\[r\]in a power series. Specifically, it originates from the lowest power of
\[x\]after substituting the series into the differential equation. For the given ODE, the indicial equation takes the simple linear form
\[r(r-1) + alpha r + beta = 0\].
The indicial equation is critical because it provides information about the nature of the solutions near the singular point. In particular, it helps us determine the possible values of
\[r\]that will be used in the power series solution. In this exercise, we can conclude that there's only one solution that can be constructed this way due to the linear nature of the indicial equation.

Convergence of Power Series

In mathematics, convergence is a cornerstone concept for power series. For a power series solution to an ODE to represent an actual solution, it must converge in an interval around the singular point. The convergence of a series indicates whether the infinite sum approaches a finite value as more and more terms are added. In this case, convergence depends on the ratio of successive coefficients
\[\frac{a_{m+2}}{a_m}\].

Under certain conditions, exemplified when
\[\beta / alpha = -1, 0, 1, 2, \ldots\],the power series becomes a polynomial of finite degree, meaning it terminates and thus converges. However, if the ratio of coefficients doesn't exhibit such a pattern, the terms will grow without bound, making the series divergent. It's crucial to examine the convergence to ensure that the power series solution is valid and can represent an actual solution to the differential equation in question.