Question

The Legendre differential equation \(\left(1-t^{2}\right) y^{\prime \prime}-2 t y^{\prime}+\alpha(\alpha+1) y=0\) has regular singular points at \(t=\pm 1\); all other points are ordinary points. (a) Determine the indicial equation and the exponent at the singularity \(t=1\). (b) Assume that \(\alpha \neq 0,1\). Find the first three nonzero terms of the series solution in powers of \(t-1\) for \(t-1>0\). [Hint: Rewrite the coefficient functions in powers of \(t-1\). For example, \(1-t^{2}=-(t-1)(t+1)=-(t-1)((t-1)+2)\).] (c) What is an exact solution of the differential equation when \(\alpha=1\) ?

Short answer

Question: Determine the exponent at the singularity \(t = 1\) in the series solution of the Legendre differential equation, along with the first three nonzero terms in the series solution when \(\alpha \neq 0,1\), and the exact solution when \(\alpha = 1\). Answer: The exponent at the singularity \(t = 1\) is given by \(1 + r_1\), where \(r_1\) is the smaller root of the indicial equation. The first three nonzero terms in the series solution when \(\alpha \neq 0,1\) are given by the expression \(y(r) = a_0\left[1 + \frac{\alpha(\alpha + 1)}{2} r + \frac{(\alpha - 1)(\alpha - 2)}{6} r^2 - \frac{(\alpha - 2)(\alpha - 3)(\alpha - 1)}{144} r^3\right]\). The exact solution when \(\alpha = 1\) is given by \(y(t) = \frac{e^{t^2}}{(1-t^2)^{1/2}}\).

Step-by-step solution

(a) Indicial equation and exponent at singularity \(t=1\)

To find the indicial equation, let us first manipulate the Legendre differential equation into the standard form of a Frobenius series solution. This requires a change of variables \(t - 1 = r\) (since we are considering the singularity at \(t = 1\)). Then \(t = r + 1\). Our Legendre differential equation is: \[\left(1-t^{2}\right) y^{\prime \prime}-2 t y^{\prime}+\alpha(\alpha+1) y=0\] Substitute \(t = r + 1\): \[\left(1-(r+1)^{2}\right) y^{\prime \prime}-2 (r+1) y^{\prime}+\alpha(\alpha+1) y=0\] Now, let \(y(r) = \sum_{n=0}^{\infty} a_n r^n\), and \(y'(r) = \sum_{n=1}^{\infty} n a_n r^{n-1}\), and \(y''(r) = \sum_{n=2}^{\infty} n(n-1) a_n r^{n-2}\). We can plug these into our differential equation and collect terms according to powers of \(r\): \[\sum_{n=2}^{\infty} n(n-1)a_n r^{n-2} - 2\sum_{n=1}^{\infty} n a_n r^{n-1} + \alpha(\alpha + 1)\sum_{n=0}^{\infty} a_n r^n = 0\] Looking at the coefficients of \(r^0\), we get the indicial equation: \[2a_2 - 2a_1 + \alpha(\alpha + 1)a_0 = 0\] Now, let \(r_1\) be the smaller root of the indicial equation. Then we can compute the exponent of the series solution at the singularity \(t = 1\) as \(1 + r_1\).

(b) Finding the first three nonzero terms of the series solution

First, we rewrite the coefficients in terms of powers of \(t - 1\). We already have: \[1 - t^2 = -(t - 1)((t - 1) + 2)\] Now for \(-2t\), we have: \[-2t = -2(r + 1)\] Substituting these back into the Legendre differential equation and equating the coefficients of equal powers of \(r\), we get a recurrence relation for the solution. The first few recurrence relations are: \[2a_2 - 2a_1 + \alpha(\alpha + 1)a_0 = 0\] \[3(2)a_3 - (2 + 2)a_2 + \left[1 + (\alpha - 1)(\alpha)\right]a_1 = 0\] \[4(3)a_4 - (4 + 2)a_3 + \left[2 + (\alpha - 2)(\alpha - 1)\right]a_2 = 0\] Now, we want to find the first three nonzero terms in the series solution. Let's solve the recurrence relations for \(a_1\), \(a_2\), \(a_3\) (given that \(\alpha \neq 0,1\)). \[a_1 = \frac{\alpha(\alpha + 1)}{2} a_0\] \[a_2 = \frac{(\alpha - 1)(\alpha - 2)}{6} a_0\] \[a_3 = -\frac{(\alpha - 2)(\alpha - 3)(\alpha - 1)}{144} a_0\] Thus, the first three nonzero terms are: \[y(r) = a_0\left[1 + \frac{\alpha(\alpha + 1)}{2} r + \frac{(\alpha - 1)(\alpha - 2)}{6} r^2 - \frac{(\alpha - 2)(\alpha - 3)(\alpha - 1)}{144} r^3\right]\]

(c) Exact solution for \(\alpha=1\)

Now we will find the exact solution for the Legendre differential equation when \(\alpha = 1\): \[(1 - t^2)y''(t) - 2ty'(t) + 2y(t) = 0\] This differential equation can be solved by using separation of variables. Let \(y(t) = v(t)u(t)\). Then we have: \[v(t)u''(t) + (1 - t^2)u'(t) + 2u(t) = 0\] Now we can separate the variables: \[\frac{v'(t)}{v(t)} = 2t\] Integrating both sides, we get: \[v(t) = e^{t^2}\] Now we substitute this back into our differential equation and we get: \[(1 - t^2)u'(t) + 2u(t) = 0\] We can rewrite this as \bigg[\frac{2}{1 - t^2} + (1 - t^2)\bigg]u'(t) = 0 Now if you integrate both sides with respect to t using (1-t^2) as the denominator on LHS, you will get: \[u(t) = \frac{1}{(1-t^2)^{1/2}}\] Thus, the exact solution for the Legendre differential equation when \(\alpha = 1\) is: \[y(t) = \frac{e^{t^2}}{(1-t^2)^{1/2}}\]

Key concepts

Singular Points in Legendre Differential Equation

The Legendre differential equation, written as \((1-t^{2}) y^{\prime \prime}-2 t y^{\prime}+\alpha(\alpha+1) y=0\), is an important type of linear differential equation. It has special points known as **singular points**, which differ from ordinary points. These are locations where the behavior of the equation changes significantly.

For the Legendre equation, the singular points are at \(t = \pm 1\). Here's why these points are singular:

• At these points, the coefficient \(1 - t^2\) becomes zero, potentially causing infinite behavior in the solutions.
• Understanding these points allows us to explore solutions using special techniques, like the Frobenius method, applicable near singularities.

In contrast, points other than \(t = \pm 1\) are ordinary, meaning regular solution methods apply and coefficients remain well-behaved.

Series Solution Approach

Finding solutions around singular points can be challenging. The **series solution** method helps find approximate solutions in the form of power series. Using this approach, we assume a solution \(y\) can be expanded as a series:
\[ y(r) = \sum_{n=0}^{\infty} a_n r^n \]
where \(r = t - 1\) since we're examining the point \(t = 1\).

To find the series solution:

• Substitute the series form into the differential equation.
• Match coefficients for each power of \(r\) to form equations that help find the series' terms \(a_n\).

For parameters \(\alpha eq 0,1\), you derive terms like:

• \(a_1 = \frac{\alpha(\alpha + 1)}{2} a_0\)
• \(a_2 = \frac{(\alpha - 1)(\alpha - 2)}{6} a_0\)
• \(a_3 = -\frac{(\alpha - 2)(\alpha - 3)(\alpha - 1)}{144} a_0\)

This showcases how a complex problem can be approached step-by-step, providing an approximate solution.

Finding an Exact Solution

Sometimes, an **exact solution** provides a precise answer to a differential equation. For specific values of \(\alpha\), the Legendre differential equation can be solved exactly.

For instance, when \(\alpha = 1\), the equation simplifies to:
\[ (1 - t^2)y''(t) - 2ty'(t) + 2y(t) = 0 \]
Solving this involves methods such as separation of variables or known solution formulas.

Here, if we let \(y(t) = v(t)u(t)\), we use integration and algebraic techniques to find:

• \(v(t) = e^{t^2}\)
• \(u(t) = \frac{1}{(1-t^2)^{1/2}}\)

Thus, the complete solution when \(\alpha = 1\) becomes:
\[ y(t) = \frac{e^{t^2}}{(1-t^2)^{1/2}} \]
This result highlights the power of exact solutions to give direct insights into the equation's behavior at particular parameter values.