Question

The Legendre equation of order \(\alpha\) is $$ \left(1-x^{2}\right) y^{\prime \prime}-2 x y^{\prime}+\alpha(\alpha+1) y=0 $$ The solution of this equation near the ordinary point \(x=0\) was discussed in Problems 22 and 23 of Section 5.3 . In Example 5 of Section 5.4 it was shown that \(x=\pm 1\) are regular singular points. Determine the indicial equation and its roots for the point \(x=1 .\) Find a series solution in powers of \(x-1\) for \(x-1>0 .\) Hint: Write \(1+x=2+(x-1)\) and \(x=1+(x-1) .\) Alternatively, make the change of variable \(x-1=t\) and determine a series solution in powers of \(t .\)

Short answer

Question: Find the indicial equation and its roots for the Legendre equation of order α at the regular singular point x=1. Also, find a series solution in powers of x-1 for x-1>0. Answer: The indicial equation for the Legendre equation of order α at the regular singular point x=1 is given by: $$r(r - 1) - r + \frac{\alpha(\alpha + 1)}{2} = 0$$ The roots of the indicial equation can be found by solving this quadratic equation. The series solution for the transformed Legendre equation is: $$y(t)= \sum_{n=0}^{\infty} a_n t^{n+r}$$ By determining the coefficients \(a_n\) for each root, we can then write the full series solution for the transformed Legendre equation and subsequently for the original Legendre equation (replacing \(t\) with \(x-1\)).

Step-by-step solution

Analyze the Legendre equation

The Legendre equation of order \(\alpha\) is given by: $$(1-x^{2})y^{\prime\prime}-2xy^{\prime}+\alpha(\alpha+1)y=0$$ We are asked to focus on the point \(x=1\), which is a regular singular point according to the exercise.

Change of variable

Using the hint provided, we can make the change of variable \(x-1=t\). So, we can rewrite the Legendre equation in terms of \(t\) as follows: $$(2-t) \frac{d^{2}y}{dt^{2}} - 2(1+t) \frac{dy}{dt} + \alpha(\alpha+1)y = 0$$

Assume the solution as a power series

We will assume the solution to the transformed Legendre equation has the form: $$y(t) = \sum_{n=0}^{\infty} a_n t^n$$ Taking the derivatives of \(y(t)\), we obtain: $$\frac{dy}{dt} = \sum_{n=1}^{\infty} n a_n t^{n-1}$$ $$\frac{d^{2}y}{dt^{2}} = \sum_{n=2}^{\infty} n(n-1) a_n t^{n-2}$$

Determine the indicial equation and its roots

Substitute the series expressions above into the transformed Legendre equation and collect terms with the same power of \(t\): $$\sum_{n=2}^{\infty} n(n-1) a_n (2-t)t^{n-2} - 2\sum_{n=1}^{\infty} n a_n (1+t)t^{n-1} + \alpha(\alpha+1)\sum_{n=0}^{\infty} a_n t^n = 0$$ For the indicial equation, focus on the terms with the lowest power of \(t\). Set the coefficient of the lowest power of \(t\) to zero, and solve for the roots \(r\). The indicial equation becomes: $$2r(r-1)a_r - 2ra_r + \alpha(\alpha + 1)a_r = 0$$ Divide by \(a_r\) (assuming \(a_r \neq 0\)) and solve for \(r\) to get the roots of the indicial equation: $$r(r - 1) - r + \frac{\alpha(\alpha + 1)}{2} = 0$$

Write the solution in the form of a series

The general series solution for the transformed Legendre equation is given by: $$y(t)= \sum_{n=0}^{\infty} a_n t^{n+r}$$ Since we have found the roots of the indicial equation, we can now substitute them into the series solution to get the particular solutions. By determining the coefficients \(a_n\) for each root, we can then write the full series solution for the transformed Legendre equation and subsequently for the original Legendre equation (replacing \(t\) with \(x-1\)).

Key concepts

Indicial Equation

Understanding the indicial equation is pivotal when solving differential equations with regular singular points, such as the Legendre equation. In our context, it's used to determine the roots that will form the basis of our series solution around a regular singular point—in this case, the point is at x=1.

The indicial equation emerges from the assumption that our solution can be represented as a power series, and it reveals the leading behavior of the solution near the singular point. When we plug our series assumption into the Legendre equation, we compare the coefficients of the lowest power of t (which is the transformed variable x-1 mentioned in the exercise). These coefficients must equate to zero because our series solution is required to satisfy the Legendre equation for all powers of t.

This process gives us an equation solely in terms of the indicial roots and the parameter α, which we solve to find the allowable values of r. These roots are crucial as they direct us in constructing a series solution that fits the original differential equation at a singular point. In summary, determining the indicial roots allows us to progress to finding the specific coefficients of our series and, consequently, the full series solution. Without this step, we would be at a loss when seeking an accurate solution near x=1.

Series Solution

The series solution is a method of resolving differential equations by expressing the solution as an infinite sum of terms, each multiplied by a power of the variable. This approach is especially useful for complex differential equations like the Legendre equation, where traditional methods (such as separation of variables) aren't applicable. After we've manipulated the Legendre equation by the change of variable x-1=t, as suggested in the problem's hint, we presume that our solution y(t) can be articulated as an infinite power series.

This presumption lets us transform the differential equation into a series where each coefficient has to be determined. Through alignment of the powers of t and equating them to zero, we systematically solve for the coefficients a_n, beginning with the indicial roots derived from the indicial equation. As we untangle the recurrence relations between the coefficients, we ultimately build the entire series solution, term by term. It's like constructing a complex puzzle where each piece must fit precisely to form the complete, correct picture of our solution. The series solution is essential for tackling differential equations with singular points, as it allows us to explore the nature of solutions in the often-tricky territory in the vicinity of these points.

Regular Singular Points

The concept of regular singular points is essential in the realm of differential equations. These points are locations in the equation where the usual analytical methods break down due to some form of singularity or peculiar behavior. However, contrary to what the name 'singular' might imply, 'regular' singular points still allow for the possibility of finding meaningful solutions.

In our case, the Legendre equation identifies x=±1 as regular singular points. This information is like a beacon, indicating that special techniques, such as the power series solution method, need to be applied to determine the behavior of the solution near these points. Interestingly, the way a differential equation behaves around regular singular points is quite structured—hence the 'regular' qualifier—and gives rise to specific solutions that can often be expressed as power series expansions.

The regular singular points crucially inform us about the structure of the equation's solutions. For instance, at a regular singular point, the solution to the differential equation may still exist and be uniquely defined, which wouldn't be the case for an 'irregular' singular point. Insights into the nature of these points guide us in selecting the right solution strategies, reinforce our understanding of the equation's behavior in complex cases, and ultimately enable us to describe physical phenomena accurately in regions that might otherwise be deemed troublesome to analyze.