Question

The Legendre Equation. Problems 22 through 29 deal with the Legendre equation $$ \left(1-x^{2}\right) y^{\prime \prime}-2 x y^{\prime}+\alpha(\alpha+1) y=0 $$ As indicated in Example \(3,\) the point \(x=0\) is an ordinaty point of this equation, and the distance from the origin to the nearest zero of \(P(x)=1-x^{2}\) is 1 . Hence the radius of convergence of series solutions about \(x=0\) is at least 1 . Also notice that it is necessary to consider only \(\alpha>-1\) because if \(\alpha \leq-1\), then the substitution \(\alpha=-(1+\gamma)\) where \(\gamma \geq 0\) leads to the Legendre equation \(\left(1-x^{2}\right) y^{\prime \prime}-2 x y^{\prime}+\gamma(\gamma+1) y=0\) Show that, if \(\alpha\) is zero or a positive even integer \(2 n,\) the series solution \(y_{1}\) reduces to a polynomial of degree \(2 n\) containing only even powers of \(x\). Find the polynomials corresponding to \(\alpha=0,2,\) and \(4 .\) Show that, if \(\alpha\) is a positive odd integer \(2 n+1,\) the series solution \(y_{2}\) reduces to a polynomial of degree \(2 n+1\) containing only odd powers of \(x .\) Find the polynomials corresponding to \(\alpha=1,3,\) and \(5 .\)

Short answer

Question: Show that the series solution for the Legendre equation with given values of α reduces to a polynomial of degree 2n with only even powers of x for α as an even integer, and to a polynomial of degree 2n+1 with only odd powers of x for α as an odd integer. Find the polynomials for α = 1, 3, and 5. Answer: Using the Frobenius method to solve the Legendre equation and by setting the coefficients to zero, we can show that: For even α: 1) α = 0: y(x) = 1 2) α = 2: y(x) = 1 - (1/3)x^2 3) α = 4: y(x) = 1 - (3/5)x^2 + (1/14)x^4 For odd α: 1) α = 1: y(x) = x 2) α = 3: y(x) = x(1 - (1/3)x^2) 3) α = 5: y(x) = x(1 - (5/7)x^2 + (3/70)x^4) Hence, for even and odd values of α, the series solution reduces to the required polynomials with even and odd powers of x, respectively.

Step-by-step solution

The Frobenius method

To attempt to find the series solution, we will use the Frobenius method. We can start by assuming the series solution takes the form: $$y(x) = \sum_{k=0}^{\infty} a_k x^{k+r}$$

Plugging into the Legendre equation

Now, we will calculate the first and second derivatives of this generic series solution: 1) First derivative: $$y'(x) = \sum_{k=0}^{\infty} (k+r) a_k x^{k+r-1}$$ 2) Second derivative: $$y''(x) = \sum_{k=0}^{\infty} (k+r)(k+r-1) a_k x^{k+r-2}$$ After calculating both the first and second derivatives, we plug them back into the Legendre equation: $$\left[1 - x^2\right]\left[\sum_{k=0}^{\infty} (k+r)(k+r-1) a_k x^{k+r-2}\right] - 2x\left[\sum_{k=0}^{\infty} (k+r) a_k x^{k+r-1}\right] + \alpha(\alpha+1)\left[\sum_{k=0}^{\infty} a_k x^{k+r}\right] = 0$$

Simplifying the equation

Now, we will simplify the equation by multiplying, combining terms, and finally equating the coefficients to zero: $$\sum_{k=0}^{\infty} [(k+r)(k+r-1) - \alpha(\alpha+1)]a_k x^{k+r} - \sum_{k=0}^{\infty} 2(k+r)a_k x^{k+r} = 0$$ Setting the coefficients for each power of x to zero, we get: $$[(k+r)(k+r-1) - \alpha(\alpha+1)] a_k - 2(k+r) a_k = 0$$ With the simplified equation, we can now proceed to solve for the coefficients \(a_k\), considering the cases of α being even and odd, respectively.

Solving for the case of α being zero or even

Taking α to be zero or a positive even integer 2n, solving for the coefficients of the equation: For the case \(k=2n\): $$[(k+r)(k+r-1) - 2n(2n+1)] a_k - 2(k+r) a_k = 0$$ For the case of \(\alpha = 0, 2, 4\), we can directly solve for the coefficients and form the respective polynomials: 1) \(\alpha = 0\): The polynomial is: \(y(x) = 1\) 2) \(\alpha = 2\): The polynomial is: \(y(x) = 1 - \frac{1}{3}x^2\) 3) \(\alpha = 4\): The polynomial is: \(y(x) = 1 - \frac{3}{5}x^2 + \frac{1}{14} x^4\)

Solving for the case of α being odd

Taking α to be a positive odd integer 2n+1, solving for coefficients: For the case \(k=2n+1\): $$[(k+r)(k+r-1) - (2n+1)(2n+2)] a_k - 2(k+r) a_k = 0$$ Finding the polynomial for the cases of \(\alpha = 1, 3, 5\): 1) \(\alpha = 1\): The polynomial is: \(y(x) = x\) 2) \(\alpha = 3\): The polynomial is: \(y(x) = x\left(1 - \frac{1}{3}x^2\right)\) 3) \(\alpha = 5\): The polynomial is: \(y(x) = x\left(1 - \frac{5}{7}x^2 + \frac{3}{70}x^4\right)\) Thus, we have shown that for even and odd values of α, the series solutions reduce to polynomials of the required degrees with the respective powers of x.

Key concepts

Frobenius Method

The Frobenius Method is a technique for finding series solutions to differential equations, especially when the solution involves a point where the equation is not analytic. Here's how it works:

• Assume the solution of the differential equation can be expressed as a power series: \(y(x) = \sum_{k=0}^{\infty} a_k x^{k+r}\).
• The value \(r\) is called the critical exponent and is determined by solving the indicial equation, which sets constraints for when the series solution is viable.
• Once \(r\) is found, the coefficients \(a_k\) are calculated by substituting the series and its derivatives back into the differential equation. This process involves simplifying terms and equating powers of \(x\) to solve for \(a_k\).

Using the Frobenius method with the Legendre equation helps to systematically determine solutions at the regular singular point, producing series that we often seek to simplify into polynomial forms.

Radius of Convergence

The radius of convergence is a critical aspect when dealing with series solutions of differential equations. It determines the extent over which the series solution is valid. Here's what you need to know:

• The radius of convergence is the distance from the point of expansion, usually denoted as \(x = 0\), to the nearest singularity of the differential equation's coefficients.
• For the Legendre differential equation, the polynomial \(P(x) = 1-x^2\) suggests that the singularities occur when \(x = \pm 1\). This indicates a radius of convergence of at least 1, as these are the boundary points.
• Within this radius, the series will converge, and the solution will be valid. Beyond that, there might be complexities or breakdowns in the solution's applicability.

Understanding the radius of convergence ensures that you know where series solutions apply accurately.

Polynomial Solutions

Polynomial solutions signal that a series solution terminates, producing a finite-degree polynomial instead of an infinite series. This is common in problems like the Legendre equation:

• Polynomials represent the exact solution within the limits of their derivation, making calculations simpler.
• For the Legendre equation, specific values of \(\alpha\), like zero or positive even integers, lead to these simplifications. The resulting polynomial is purely in terms of either even or odd powers of \(x\).
• Examples include \(\alpha = 0\), yielding the polynomial \(y(x) = 1\), and \(\alpha = 1\), resulting in \(y(x) = x\).

Such polynomial solutions often arise in physics and engineering, where they represent specific boundary conditions or special cases.

Differential Equations

Differential equations represent relationships involving functions and their derivatives. They play a foundational role in modeling real-world phenomena:

• Differential equations describe the change of systems, often used in physics, engineering, and other scientific domains.
• The Legendre equation is a form of linear second-order differential equation with variable coefficients, part of the family known as orthogonal polynomials, important in physics for solving problems involving spherical harmonics.
• Solving differential equations can yield a wealth of information, from dynamic system behaviors to steady-state conditions.

Understanding differential equations and their solutions, like those obtained through the Frobenius method, allows us to tackle complex problems in a methodical way.