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. \(x y^{\prime \prime}+(1-x) y^{\prime}-y=0\)

Short answer

Short Answer: The given differential equation has a regular singular point at \(x=0\). By applying the Frobenius method, the indicial equation is found to be \(r(r-1)a_0=0\), which has roots \(r=0\) and \(r=1\). The series solutions corresponding to these roots are: \(y_1(x)=a_0\left(1 + \frac{1}{2}x + \frac{1}{3}x^2 + ...\right)\) and \(y_2(x)=a_0\left(1 + \frac{1}{2}x + \frac{1}{6}x^2 + ...\right)\). Thus, the general solution is given by the linear combination of these solutions: \(y(x)=C_1y_1(x)+C_2y_2(x)\), where \(C_1\) and \(C_2\) are constants.

Step-by-step solution

Determine if there is a regular singular point at \(x=0\)

To show that \(x=0\) is a regular singular point of the given equation, we need to confirm the following two conditions: 1. \(P(x)=\frac{(1-x)}{x}\) and \(Q(x)=-\frac{1}{x}\) should have a removable singularity or a pole of order 1 at \(x=0\). 2. \(xP(x)\) and \(x^2Q(x)\) should be analytic at \(x=0\). Let's check these conditions: 1. \(P(x)\) has a pole of order 1, and \(Q(x)\) has a pole of order 1 at \(x=0\). 2. \(xP(x) = (1-x)\) and \(x^2Q(x) = -x\) are both analytic at \(x=0\). Therefore, \(x=0\) is a regular singular point.

Applying the Frobenius method and finding the indicial equation

For the Frobenius method, we assume a solution of the form \(y(x)=\sum_{n=0}^{\infty}a_nx^{n+r}\), where \(r\) is the root of the indicial equation. Then, we have: \(y^{\prime}(x)=\sum_{n=0}^{\infty}(n+r)a_nx^{n+r-1}\) \(y^{\prime \prime}(x)=\sum_{n=0}^{\infty}(n+r)(n+r-1)a_nx^{n+r-2}\) Now, substitute these forms of \(y(x)\), \(y^{\prime}(x)\), and \(y^{\prime \prime}(x)\) into the given equation: \(x\left(\sum_{n=0}^{\infty}(n+r)(n+r-1)a_nx^{n+r-2}\right)+(1-x)\left(\sum_{n=0}^{\infty}(n+r)a_nx^{n+r-1}\right)-\left(\sum_{n=0}^{\infty}a_nx^{n+r}\right)=0\) Now, simplify and collect terms with the same power of \(x\) to rewrite and find the indicial equation: \(\sum_{n=0}^{\infty}(n+r)(n+r-1)a_nx^{n+r-1} +(1-x)\sum_{n=0}^{\infty}(n+r)a_nx^{n+r-1} -\sum_{n=0}^{\infty}a_nx^{n+r} = 0\) Consider the lowest power of \(x\), which is \(x^r\), corresponding to \(n=0\). This gives: \(r(r-1)a_0 = 0\) This equation is the indicial equation.

Finding the roots of the indicial equation

The indicial equation, \(r(r-1)a_0=0\), has roots \(r=0\) and \(r=1\). Since both roots \(\{0,1\}\) are unequal and do not differ by an integer, we need to find the series solution for both roots.

Finding the recurrence relation and series solution for the larger root

To find the recurrence relation, we equate coefficients of \(x^{n+r+1}\) to zero: \((n+1)(n+r)a_{n+1}=a_n - (n+r)a_n\) Now, solve for \(a_{n+1}\) and find the series solution for the larger root (\(r=1\)): \(a_{n+1}=\frac{(n-r)}{(n+1)(n+r)}a_n\) Using this recurrence relation, we can find the series coefficients: For \(r=1\), \(a_{n+1}=\frac{n}{(n+1)(n+1)}a_n\) Summing the power series gives: \(y_1(x)=a_0\left(1 + \frac{1}{2}x + \frac{1}{3}x^2 + ... \right)\)

Finding the series solution for the smaller root

Now, we find the series solution for the smaller root (\(r=0\)): For \(r=0\), \(a_{n+1}=\frac{n}{(n+1)(n+1)}a_n\) Summing the power series gives: \(y_2(x)=a_0\left(1 + \frac{1}{2}x + \frac{1}{6}x^2 + ... \right)\) Now, we can write the general solution as the linear combination of the individual solutions: \(y(x)=C_1y_1(x)+C_2y_2(x)\) Where \(C_1\) and \(C_2\) are constants.

Key concepts

Regular Singular Point

Understanding singular points in differential equations is pivotal for finding solutions. A regular singular point is a specific type where the equation's behavior is highly predictable. It's different from an ordinary point, where the equation behaves smoothly, or an irregular point, where things can get chaotic.

To confirm a regular singular point at \(x=0\) for a differential equation like the one presented, two conditions must be met:

• The coefficients \(P(x)=\frac{1-x}{x}\) and \(Q(x)=-\frac{1}{x}\) should either be removable singularities or have poles of order 1 at \(x=0\).
• The expressions \(xP(x)\) and \(x^2Q(x)\) must be analytic, or essentially smooth, at \(x=0\).

In simpler terms, a regular singular point ensures that the solution near \(x=0\) can be systematically approached using sophisticated techniques like the Frobenius method.

Frobenius Method

The Frobenius Method is a technique used to find power series solutions to differential equations near singular points. Unlike the standard power series method, it's suited for regular singular points, which exhibit more complex behavior.

The method involves assuming a solution in the form of a series:\(y(x) = \sum_{n=0}^{\infty} a_n x^{n+r}\), where \(r\) is determined by something called the indicial equation. Substituting this form into the differential equation allows us to obtain a recurrence relation, which is used to find the coefficients \(a_n\).

The beauty of the Frobenius Method lies in its ability to handle situations where ordinary power series methods fail, especially at singular points. By taking into account the singular nature of the point, it helps us construct a valid and convergent series solution.

Indicial Equation

The Indicial Equation is a key component when applying the Frobenius Method. It determines possible values of \(r\) in the assumed series solution. This equation is derived from finding the lowest power of \(x\) within the series substitution in the differential equation.

For our differential equation, the indicial equation takes the form \(r(r-1)a_0 = 0\). Solving this gives the roots \(r=0\) and \(r=1\), which become the possible exponents of \(x\) in our series solution. These roots are crucial because they dictate the specific structure of the solutions that will follow.

The indicial equation sets the stage by constraining \(r\), allowing us to start constructing our series solution. Both roots from the indicial equation, in this case, do not differ by an integer, which suggests unique solutions for each root.

Series Solution

The Series Solution is the actual solution we construct using the roots from the indicial equation and the recurrence relation. After identifying the roots \(r=0\) and \(r=1\), series solutions are built for each.

For the larger root \(r=1\), the series is:\(y_1(x)=a_0(1 + \frac{1}{2}x + \frac{1}{3}x^2 + \ldots )\)Similarly, for the smaller root \(r=0\), it is:\(y_2(x)=a_0(1 + \frac{1}{2}x + \frac{1}{6}x^2 + \ldots )\)
The coefficients \(a_n\) are determined using the recurrence relation:\(a_{n+1}=\frac{n}{(n+1)(n+1)}a_n\), specific for the chosen root. The final general solution is a linear combination of these individual series — combining them to capture all possible solutions of the differential equation.

Thus, the series solution offers a comprehensive view, illustrating how the equation behaves around the singular point \(x=0\), with the power series converging to represent the solution for \(x > 0\).