Question

Show that the given differential equation has a regular singular point at \(x=0 .\) Determine the indicial equation, the recurrence relation, and the roots of the indicial equation. Find the series solution \((x>0)\) corresponding to the larger root. If the roots are unequal and do not differ by an integer, find the series solution corresponding to the smaller root also. \(x^{2} y^{\prime \prime}+x y^{\prime}+\left(x^{2}-\frac{1}{9}\right) y=0\)

Short answer

Question: Verify that x = 0 is a regular singular point of the given differential equation, and find the indicial equation, its roots, recurrence relation, and series solutions. Given differential equation: \(x^{2}y^{\prime\prime}+x y^{\prime}+\left(x^{2}-\frac{1}{9}\right) y = 0\) Answer: x = 0 is a regular singular point of the given differential equation. The indicial equation is \(r(r - 1) = 0\), with roots \(r_{1} = 0\) and \(r_{2} = 1\). The recurrence relation is given by \(a_n = -\frac{a_{n-1}}{((n + r)^{2} - \frac{8}{9})}\). The series solutions are: \(y_{1}(x) = a_0 + \sum_{n=1}^{\infty} a_n x^{n}\), \(y_{2}(x) = x + \sum_{n=1}^{\infty} a_n x^{n+1}\), and the general solution is: \(y(x) = y_{1}(x) + y_{2}(x) = a_0 (1 + x) + \sum_{n=1}^{\infty} a_n x^{n}(1+x)\).

Step-by-step solution

Verify x = 0 as a regular singular point

By definition, x = 0 is a regular singular point if both \(\lim_{x \to 0} xP(x)\) and \(\lim_{x \to 0} x^{2}Q(x)\) exist and are finite, where \(P(x)\) and \(Q(x)\) are given as below: \(Ly = x^{2}y^{\prime\prime}+x y^{\prime}+\left(x^{2}-\frac{1}{9}\right) y\) Comparing the given equation with \(Ly = y^{\prime\prime} + P(x) y^{\prime} + Q(x)y,\) we have \(P(x) = \frac{x}{x^{2}} = \frac{1}{x},\) and \(Q(x) = x^{2} - \frac{1}{9}\). Now, we need to find the limits: \(\lim_{x \to 0} xP(x) = \lim_{x \to 0} x \cdot \frac{1}{x} = 1,\) which exists and is finite. \(\lim_{x \to 0} x^{2} Q(x) = \lim_{x \to 0} x^{2} (x^{2}-\frac{1}{9}) = 0,\) which also exists and is finite. Since both limits exist and are finite, x = 0 is a regular singular point.

Write the differential equation in standard form

To obtain the standard form, we need to write the given equation as: \(y^{\prime\prime} + P(x)y^{\prime} + Q(x)y = 0\) Dividing the given equation by \(x^{2}\) gives: \(y^{\prime\prime} + \frac{1}{x}y^{\prime}+\left(1 - \frac{1}{9x^{2}}\right)y = 0\)

Find the indicial equation and its roots

For the standard form, assuming a series solution of the form \(y(x) = \sum_{n=0}^{\infty} a_{n}x^{n+r}\), we need to find the indicial equation: We have the equation: \(a_{0}x^{r-1} + a_{1}x^{r} + a_{2}x^{r+1} + a_{3}x^{r+2} + ... = \frac{1}{x}(a_{1}x^{r} + a_{2}x^{r+1} + a_{3}x^{r+2} + ...) + (1 - \frac{1}{9x^{2}})(a_{0}x^{r} + a_{1}x^{r+1} + a_{2}x^{r+2} + a_{3}x^{r+3} + ...)\) Now, let's find the coefficients by equating the exponents powers. First, we look at the lowest exponent of x, which is \(x^{r-1}.\) From this term, we can find the indicial equation: \(r(r-1)a_{0}x^{r-1} = 0\) Which implies \(r(r-1) = 0.\) The roots of the indicial equation are \(r_{1} = 0\) and \(r_{2} = 1.\)

Determine the recurrence relation

Using the Frobenius method, we can find the recurrence relation by comparing both sides of the original equation. For \(n>0\) we have the relation: \((n+r)(n+r-1)a_{n}x^{r+n-1} + (n+r)a_{n}x^{r+n-1} + \left( 1 - \frac{1}{9}x^{-2} \right)a_{n}x^{r+n} = 0.\) Rearranging and factoring out the common terms, we get: \(a_n[(n + r)^2+(\frac{1}{9}-1)]x^{r+n-1}=0.\) For the equation to hold true, either \(x^{r+n-1} = 0\) or \([(n + r)^2+(\frac{1}{9}-1)]= 0.\) Since x has no contribution to the recurrence relation, we consider the second part: \([(n + r)^2+(\frac{1}{9}-1)]= 0 \Longrightarrow a_n = -\frac{a_{n-1}}{((n + r)^{2} - \frac{8}{9})}\)

Find the series solutions

The series solutions corresponding to the roots of the indicial equation are given by: For \(r_{1} = 0:\) \(y_{1}(x) = a_0 + \sum_{n=1}^{\infty} a_n x^{n}\) For \(r_{2} = 1:\) \(y_{2}(x) = x + \sum_{n=1}^{\infty} a_n x^{n+1}\) Since the roots are unequal and do not differ by an integer, the general solution is given by the sum of these series: \(y(x) = y_{1}(x)+y_{2}(x)= a_0 (1 + x) + \sum_{n=1}^{\infty} a_n x^{n}(1+x)\)

Key concepts

Regular Singular Point

When tackling differential equations, recognizing the type of point we're dealing with is crucial for choosing the right solution method. A regular singular point is a specific kind of singular point where the solution to the differential equation behaves 'nicely' in the sense that, although the solution may become infinite at that point, it still has a predictable structure that we can work with.

For example, consider the differential equation given by
\(x^2 y'' + x y' + (x^2 - \frac{1}{9}) y = 0\).
To determine if \(x = 0\) is a regular singular point, we examine the limits of \(xP(x)\) and \(x^2 Q(x)\), where \(P(x)\) and \(Q(x)\) are coefficients of \(y'\) and \(y\) after dividing the equation by the coefficient of \(y''\). If both limits exist and are finite, as they do in our case, then \(x = 0\) is a regular singular point.

This has huge implications on how we approach finding a solution, because it allows us to use specialized techniques like the Frobenius method, which is designed specifically to tackle this type of problem.

Indicial Equation

Upon determining the nature of the singular point, our next step in the Frobenius method is to find the indicial equation. This is a crucial part of solving differential equations since it gives us information about the exponents of the power series solution.

For the differential equation
\(y'' + \frac{1}{x}y' + \left(1 - \frac{1}{9x^2}\right)y = 0\),
we propose a solution in the form of a power series multiplied by \(x^r\), where \(r\) is what we need to determine. By substituting this proposed solution into the differential equation and comparing coefficients for the lowest powers of \(x\), we arrive at the indicial equation. In this case, the indicial equation comes out to be \(r(r-1) = 0\), whose solutions tell us the possible exponents for the leading term in our series. These roots, \(r_1\) and \(r_2\), play a pivotal role in constructing the series solution to the differential equation.

Recurrence Relation

With the roots of the indicial equation in hand, we dive into the heart of the Frobenius method: deriving the recurrence relation. It's like a domino effect; knowing one coefficient can help us find the next, and so on, effectively building the solution piece by piece.

The recurrence relation for the differential equation
\(y'' + \frac{1}{x}y' + \left(1 - \frac{1}{9x^2}\right)y = 0\)
is derived by systematically comparing coefficients of equal powers of \(x\) after substituting the power series expression into the equation. What we end up with is a formula that relates each coefficient \(a_n\) to its predecessor. This is a powerful tool that allows us to calculate the coefficients of our series solution, one after another, starting from the lowest power.

In our example, the recurrence relation takes the form \(a_n = -\frac{a_{n-1}}{((n + r)^{2} - \frac{8}{9})}\), which clearly outlines how to get from one coefficient to the next.

Series Solution of Differential Equations

The culmination of our problem-solving journey arrives when we put together the series solution of the differential equation. Using the Frobenius method, we construct this solution as a power series that begins with the term dictated by the indicial equation and continues indefinitely, each term connected to the previous one by the recurrence relation.

With the roots \(r_1\) and \(r_2\) from our indicial equation, we find two series solutions:
\(y_1(x) = a_0 + \sum_{n=1}^{\infty} a_n x^{n}\) corresponding to \(r_1\),
and
\(y_2(x) = x + \sum_{n=1}^{\infty} a_n x^{n+1}\) corresponding to \(r_2\).

In cases where these roots do not differ by an integer, we have two distinct series solutions. These solutions express the behavior of our function around the singular point and allow us to predict its behavior in the vicinity of that point. The general solution is the sum of these series, each weighted by a constant coefficient.