Question

Find power series solutions in powers of \(x-1\) of each of the differential equations in Exercises. The differential equation $$ \left(1-x^{2}\right) y^{\prime \prime}-2 x y^{\prime}+n(n+1) y=0 $$ where \(n\) is a constant, is called Legendre's differential equation. (a) Show that \(x=0\) is an ordinary point of this differential equation, and find two linearly independent power series solutions in powers of \(x\). (b) Show that if \(n\) is a nonnegative integer, then one of the two solutions found in part (a) is a polynomial of degree \(n\).

Short answer

In summary, we showed that \(x=0\) is an ordinary point of Legendre's differential equation and found two linearly independent power series solutions: \[ y(x) = a_0\left(\sum_{k=0}^{\infty} y_k^{(0)} x^k\right) + a_1\left(\sum_{k=0}^{\infty} y_k^{(1)} x^k\right) \] We also proved that if \(n\) is a nonnegative integer, one of these two solutions becomes a polynomial of degree \(n\): \[ y^{(1)} = a_1\left(x^n + \text{lower-order terms}\right) \]

Step-by-step solution

(a) Proving ordinary point and finding power series solutions

First, we will rewrite the Legendre's differential equation by dividing all terms by \((1-x^2)\): \[ y'' - 2xy'+n(n+1)y = 0 \] To prove that \(x=0\) is an ordinary point, we should check if the coefficients of \(y'\) and \(y''\) are analytic at \(x=0\). The coefficients are: \(1\) (for \(y''\)) and \(-2x\) (for \(y'\)). Both coefficients are analytic at \(x=0\), so \(x=0\) is an ordinary point. Next, we will assume that the solution can be expressed as a power series: \[ y(x) = \sum_{k=0}^{\infty} a_kx^k \] Now, we will differentiate this solution once and twice: \[ y'(x) = \sum_{k=1}^{\infty} k a_kx^{k-1} \] \[ y''(x) = \sum_{k=2}^{\infty} k(k-1)a_kx^{k-2} \] Now, we plug the power series expressions for \(y(x)\), \(y'(x)\), and \(y''(x)\) into the differential equation: \[ \left(\sum_{k=2}^{\infty} k(k-1)a_kx^{k-2}\right) -2x\left(\sum_{k=1}^{\infty} k a_kx^{k-1}\right)+n(n+1)\left(\sum_{k=0}^{\infty} a_kx^k\right)=0 \] Now, we will simplify and compare the coefficients to find the recursion relationship.

(a) Finding recursion relationship

We will rewrite the equation with summations starting at \(k=0\), and for this, we need to consider the first terms in each series separately. Take the first terms outside each summation: \[ 2a_2 - 2a_1x + n(n+1)a_0x^0 + \sum_{k=0}^{\infty}[(k+2)(k+1)a_{k+2}x^k - 2(k+1)a_{k+1}x^k + n(n+1)a_kx^k] = 0 \] Now, gathering the constant terms and the terms involving powers of \(x\): \[ (2a_2 + n(n+1)a_0)+(n(n+1)a_1-2a_1)x+\sum_{k=0}^{\infty}[(k+2)(k+1)a_{k+2} - 2(k+1)a_{k+1} + n(n+1)a_k]x^k=0 \] For this equation to hold for all \(x\), the coefficients must be equal, which forms a recursion relationship: \[ 2a_2 + n(n+1)a_0 = 0 \] \[ n(n+1)a_1 - 2a_1 = 0 \] \[ (k+2)(k+1)a_{k+2} - 2(k+1)a_{k+1} + n(n+1)a_k = 0 \quad \text{for } k\geq0 \] Thus, the recursion relationship is given by: \[ a_{k+2} = \frac{2(k+1) a_{k+1} - n(n+1) a_k}{(k+2)(k+1)} \]

(a) Finding two independent solutions using the recursion relationship

Now, we will use the recursion to find two linearly independent power series solutions. We can express the power series solution in terms of \(a_0\) and \(a_1\): \[ y(x) = a_0\left(\sum_{k=0}^{\infty} y_k^{(0)} x^k\right) + a_1\left(\sum_{k=0}^{\infty} y_k^{(1)} x^k\right) \] Using the recursion relationship, we can find the coefficients of both series: \[ y_k^{(0)} = a_0, \quad y_k^{(1)} = 0 \quad \text{(for k=0)} \] \[ y_k^{(0)} = 0, \quad y_k^{(1)} = a_1x \quad \text{(for k=1)} \] For \(k\geq2\), we use the recursion relationship, and we obtain: \[ y_k^{(0)} = \frac{2(k-1) y_{k-1}^{(0)} - n(n+1) y_{k-2}^{(0)}}{k(k-1)}, \quad y_k^{(1)} = \frac{2(k-1) y_{k-1}^{(1)} - n(n+1) y_{k-2}^{(1)}}{k(k-1)} \] Now we have our power series solution: \[ y(x) = a_0\left(\sum_{k=0}^{\infty} y_k^{(0)} x^k\right) + a_1\left(\sum_{k=0}^{\infty} y_k^{(1)} x^k\right) \]

(b) Proving one solution is a polynomial of degree n

We now need to show that if \(n\) is a nonnegative integer, one of the two power series solutions becomes a polynomial of degree \(n\). Consider the series corresponding to \(a_1\), the coefficients \(y_k^{(1)}\): \[ y_k^{(1)} = \frac{2(k-1) y_{k-1}^{(1)} - n(n+1) y_{k-2}^{(1)}}{k(k-1)} \] If we have an integer value \(n\), and \(k=n+1\), we get: \[ y_{n+1}^{(1)} = \frac{2(n) y_{n}^{(1)} - n(n+1) y_{n-1}^{(1)}}{(n+1)(n)} \] The term \(n(n+1)\) will cancel out the term \(2n y_n^{(1)}\), making \(y_{n+1}^{(1)}=0\), and from the recursion relationship, all subsequent terms will also be zero. Thus, for a nonnegative integer value of \(n\), the power series solution corresponding to \(a_1\) becomes a polynomial of degree \(n\): \[ y^{(1)} = a_1\left(x^n + \text{lower-order terms}\right) \] Which completes the proof. So, we have shown that if \(n\) is a nonnegative integer, one of the two solutions found in part (a) is a polynomial of degree \(n\).

Key concepts

Legendre's Differential Equation

Legendre's differential equation is a second-order linear differential equation of the form: \[ (1-x^2)y'' - 2xy' + n(n+1)y = 0 \] where \(n\) is a constant. Its solutions, known as Legendre polynomials, are particularly important in physics and engineering, especially in fields like electrostatics and quantum mechanics. This equation is a type of Sturm-Liouville problem, which often arises in cases involving spherical coordinates. The presence of the term \((1-x^2)\) in the equation implies that solutions are typically found over the interval \(-1 \leq x \leq 1\). This restriction is key because it defines the region where Legendre polynomials remain stable and finite, ensuring valid and useful results.

Ordinary Differential Equations (ODEs)

Ordinary differential equations involve functions of a single variable and their derivatives. They are "ordinary" as compared to partial differential equations that involve partial derivatives of functions of multiple variables. Second-order ODEs, like Legendre's, consist of the second derivative of the unknown function, its first derivative, and the function itself. They frequently appear in various scientific disciplines:

• Physics, for modeling wave motion and dynamics.
• Biology, in population dynamics.
• Economics, for modeling growth processes.

The "order" of a differential equation indicates the highest derivative present. Solving an ODE may involve finding a function that satisfies the given equation over a certain interval.

Recursion Relationship

In the context of power series solutions to differential equations, a recursion relationship helps find successive coefficients in the series. This mathematical process converts a differential equation into a set of algebraic equations. For Legendre's differential equation, the recursion relationship is crucial to building solutions. It is established by equating coefficients of equal powers of \(x\) upon substituting the power series expansion back into the differential equation: \[ a_{k+2} = \frac{2(k+1) a_{k+1} - n(n+1) a_k}{(k+2)(k+1)} \] This formula allows us to compute subsequent terms \(a_{k+2}\), in terms of previous terms \(a_k\) and \(a_{k+1}\). Knowing initial conditions or base terms like \(a_0\) and \(a_1\) often completes the solution finding process.

Ordinary Point

An ordinary point in the context of differential equations is a point where the equation's coefficients, when divided individually by the leading coefficient of the highest derivative term, remain analytic (i.e., they can be expressed as a power series). Proving that a point is ordinary involves checking if functions representing coefficients of lower-order derivatives are analytic at that point. For the Legendre's equation: - At \(x=0\), the coefficients of \(y''\) and \(y'\) are 1 and \(-2x\), both analytic. An ordinary point is essential because it ensures that solutions can be expressed uniformly in a power series around that point, enabling a clear path to finding solutions via series expansions.

Polynomial Solution

A polynomial solution to a differential equation is a solution which takes the form of a polynomial, rather than an infinite series. For Legendre's differential equation, when \(n\) is a nonnegative integer, one of its solutions is a polynomial, specifically the Legendre polynomial of degree \(n\). This transformation from series to polynomial is due to the termination of the power series once a certain term (and all following it) become zero. This typically occurs when high enough powers render subsequent recursion steps multiples of zero due to cancellations in coefficients. Having a polynomial solution enhances computational simplicity and interpretability, especially in practical scenarios where closed-form solutions are preferable over infinite series.