SERIES SOLUTIONS - BESSEL'S EQUATION, LEGENDRE AND HERMITE POLYNOMIALS AND THEIR PROPERTIES

In mathematics, differential equations are fundamental tools used to model a vast array of phenomena in science and engineering. When dealing with certain types of differential equations, particularly those arising from problems with spherical or cylindrical symmetry, we often encounter equations whose solutions cannot be expressed in terms of elementary functions. In such cases, we turn to series solutions. This approach involves representing the solution as an infinite series, which can then be manipulated to find its coefficients and understand its behavior.

This unit focuses on three important second-order linear ordinary differential equations: Bessel's equation, Legendre's equation, and Hermite's equation. These equations and their solutions, known as Bessel functions, Legendre polynomials, and Hermite polynomials, respectively, are ubiquitous in physics and engineering. We will explore how to find series solutions for these equations and delve into the significant properties of their polynomial solutions.

BESSEL'S EQUATION

Bessel's equation is a second-order linear ordinary differential equation that appears in many problems involving cylindrical or spherical symmetry. It is given by:

x²y'' + xy' + (x² - ν²)y = 0

Here, y'' represents the second derivative of y with respect to x, y' represents the first derivative, and ν (nu) is a real or complex constant known as the order of the Bessel equation. The nature of the solutions depends significantly on the value of ν.

Series Solution of Bessel's Equation

We will find a series solution of the form y(x) = Σ_n=0^∞ a_nx^n+r, where 'r' is a constant to be determined, and a_n are the coefficients.

First, we need to find the derivatives of y(x):

y'(x) = Σ_n=0^∞ (n+r)a_nx^n+r-1

y''(x) = Σ_n=0^∞ (n+r)(n+r-1)a_nx^n+r-2

Substitute these into Bessel's equation:

x² Σ_n=0^∞ (n+r)(n+r-1)a_nx^n+r-2 + x Σ_n=0^∞ (n+r)a_nx^n+r-1 + (x² - ν²) Σ_n=0^∞ a_nx^n+r = 0

Simplify the terms:

Σ_n=0^∞ (n+r)(n+r-1)a_nx^n+r + Σ_n=0^∞ (n+r)a_nx^n+r + Σ_n=0^∞ a_nx^n+r+2 - ν² Σ_n=0^∞ a_nx^n+r = 0

Combine the first, second, and fourth sums:

Σ_n=0^∞ [(n+r)(n+r-1) + (n+r) - ν²]a_nx^n+r + Σ_n=0^∞ a_nx^n+r+2 = 0

Simplify the coefficient:

Σ_n=0^∞ [(n+r)² - ν²]a_nx^n+r + Σ_n=0^∞ a_nx^n+r+2 = 0

To equate coefficients of the same powers of x, we need to adjust the indices of the second summation. Let k = n+2, so n = k-2. When n=0, k=2.

Σ_n=0^∞ [(n+r)² - ν²]a_nx^n+r + Σ_k=2^∞ a_k-2x^k+r = 0

Rewrite the second sum using 'n' as the index:

Σ_n=0^∞ [(n+r)² - ν²]a_nx^n+r + Σ_n=2^∞ a_n-2x^n+r = 0

Now, we expand the first few terms of the first summation to isolate the n=0 and n=1 terms:

[(0+r)² - ν²]a₀x^r + [(1+r)² - ν²]a₁x^1+r + Σ_n=2^∞ [(n+r)² - ν²]a_nx^n+r + Σ_n=2^∞ a_n-2x^n+r = 0

Combine the summations:

[(r)² - ν²]a₀x^r + [(1+r)² - ν²]a₁x^1+r + Σ_n=2^∞ {[(n+r)² - ν²]a_n + a_n-2}x^n+r = 0

For this equation to hold for all x, the coefficients of each power of x must be zero.

Equating the coefficient of x^r to zero:

(r² - ν²)a₀ = 0

Assuming a₀ ≠ 0 (otherwise, the solution would be trivial), we get the indicial equation:

r² - ν² = 0 => r = ±ν

Equating the coefficient of x^1+r to zero:

[(1+r)² - ν²]a₁ = 0

Equating the coefficients for n ≥ 2:

[(n+r)² - ν²]a_n + a_n-2 = 0

This gives us the recurrence relation:

a_n = - a_n-2 / [(n+r)² - ν²] for n ≥ 2

Bessel Functions of the First Kind (J_ν(x))

Let's consider the case where ν is not an integer or half-integer. We choose the positive root for r: r = ν.

From the coefficient of x^1+r: [(1+ν)² - ν²]a₁ = 0. Since (1+ν)² - ν² = 1 + 2ν + ν² - ν² = 1 + 2ν, if ν ≠ -1/2, then a₁ = 0.

Using the recurrence relation a_n = - a_n-2 / [(n+ν)² - ν²]:

For odd values of n (1, 3, 5, ...), since a₁ = 0, all odd-indexed coefficients a₃, a₅, ... will also be zero.

For even values of n (2, 4, 6, ...):

a₂ = - a₀ / [(2+ν)² - ν²] = - a₀ / (4 + 4ν) = - a₀ / [4(1+ν)]

a₄ = - a₂ / [(4+ν)² - ν²] = - a₂ / (16 + 8ν) = - a₂ / [8(2+ν)] = - [- a₀ / [4(1+ν)]] / [8(2+ν)] = a₀ / [32(1+ν)(2+ν)]

This process can be continued. However, to obtain the standard Bessel functions, a specific normalization is used. We set a₀ = 1 / [2^νΓ(ν+1)], where Γ is the Gamma function.

With this normalization and setting r=ν, we obtain the Bessel function of the first kind of order ν, denoted by J_ν(x):

J_ν(x) = Σ_k=0^∞ (-1)^k x^2k+ν / [k! Γ(k+ν+1) 2^2k+ν]

The Gamma function Γ(z) is a generalization of the factorial function, with Γ(n+1) = n! for integer n ≥ 0, and Γ(z+1) = zΓ(z).

Bessel Functions of the Second Kind (Y_ν(x))

When ν is not an integer, the second linearly independent solution can be obtained using the relation:

Y_ν(x) = [J_ν(x)cos(νπ) - J_-ν(x)] / sin(νπ)

J_-ν(x) is obtained by replacing ν with -ν in the series for J_ν(x).

If ν is an integer, say ν = m, then J_-m(x) = (-1)^m J_m(x), and the above formula becomes indeterminate. In this case, Y_m(x) is defined using a limiting process or as a linear combination of J_m(x) and another function.

Properties of Bessel Functions

Bessel functions satisfy several important recurrence relations:

• d/dx [x^νJ_ν(x)] = x^νJ_ν-1(x)
• d/dx [x^-νJ_ν(x)] = -x^-νJ_ν+1(x)
• J_ν-1(x) + J_ν+1(x) = (2ν/x)J_ν(x)
• J_ν-1(x) - J_ν+1(x) = 2J'_ν(x)
• J_ν+1(x) = (2ν/x)J_ν(x) - J_ν-1(x)
• J_ν+1(x) = J'_ν(x) - (ν/x)J_ν(x)
• J_ν-1(x) = J'_ν(x) + (ν/x)J_ν(x)

The roots of J_ν(x) = 0 are important in many physical applications, such as the vibration of circular membranes.

LEGENDRE'S EQUATION AND LEGENDRE POLYNOMIALS

Legendre's equation is a second-order linear ordinary differential equation that arises in problems with spherical symmetry, particularly in solving Laplace's equation in spherical coordinates. It is given by:

(1-x²)y'' - 2xy' + n(n+1)y = 0

Here, n is a non-negative constant. The behavior of the solutions depends on whether n is an integer.

Series Solution of Legendre's Equation

We again seek a series solution of the form y(x) = Σ_k=0^∞ a_kx^k.

y'(x) = Σ_k=1^∞ ka_kx^k-1

y''(x) = Σ_k=2^∞ k(k-1)a_kx^k-2

Substitute into Legendre's equation:

(1-x²) Σ_k=2^∞ k(k-1)a_kx^k-2 - 2x Σ_k=1^∞ ka_kx^k-1 + n(n+1) Σ_k=0^∞ a_kx^k = 0

Expand and shift indices to get all terms in x^k:

Σ_k=2^∞ k(k-1)a_kx^k-2 - Σ_k=2^∞ k(k-1)a_kx^k - 2 Σ_k=1^∞ ka_kx^k + n(n+1) Σ_k=0^∞ a_kx^k = 0

Let j = k-2 in the first sum, so k = j+2. When k=2, j=0.

Σ_j=0^∞ (j+2)(j+1)a_j+2x^j - Σ_k=2^∞ k(k-1)a_kx^k - 2 Σ_k=1^∞ ka_kx^k + n(n+1) Σ_k=0^∞ a_kx^k = 0

Rewrite using 'k' as the index:

Σ_k=0^∞ (k+2)(k+1)a_k+2x^k - Σ_k=2^∞ k(k-1)a_kx^k - 2 Σ_k=1^∞ ka_kx^k + n(n+1) Σ_k=0^∞ a_kx^k = 0

Separate the initial terms (k=0 and k=1) from the summations that start at k=2:

For k=0: (2)(1)a₂ + n(n+1)a₀ = 0 => a₂ = - n(n+1)a₀ / 2

For k=1: (3)(2)a₃ - 2(1)a₁ + n(n+1)a₁ = 0 => 6a₃ + (n²+n-2)a₁ = 0 => a₃ = - (n²+n-2)a₁ / 6 = - (n+2)(n-1)a₁ / 6

For k ≥ 2: (k+2)(k+1)a_k+2 - k(k-1)a_k - 2ka_k + n(n+1)a_k = 0

(k+2)(k+1)a_k+2 - [k(k-1) + 2k - n(n+1)]a_k = 0

(k+2)(k+1)a_k+2 - [k² - k + 2k - n² - n]a_k = 0

(k+2)(k+1)a_k+2 - [k² + k - n² - n]a_k = 0

(k+2)(k+1)a_k+2 - [k(k+1) - n(n+1)]a_k = 0

This gives the recurrence relation:

a_k+2 = [n(n+1) - k(k+1)]a_k / [(k+2)(k+1)] for k ≥ 0

This relation shows that a_k+2 depends on a_k. This means the coefficients for even powers of x depend on a₀, and the coefficients for odd powers of x depend on a₁. We can choose a₀ and a₁ freely.

Legendre Polynomials (P_n(x))

When n is a non-negative integer, one of the series solutions terminates, resulting in a polynomial. These polynomial solutions are called Legendre polynomials, denoted by P_n(x).

To obtain the standard Legendre polynomials, we impose the condition that P_n(1) = 1. This normalization fixes the arbitrary constants a₀ and a₁.

The recurrence relation is a_k+2 = [n(n+1) - k(k+1)]a_k / [(k+2)(k+1)].

If n is an even integer, say n=2m, then for k=n, the numerator n(n+1) - k(k+1) becomes n(n+1) - n(n+1) = 0. Thus, a_n+2 = 0, a_n+4 = 0, and so on. The series terminates, yielding a polynomial of degree n. In this case, we set a₁ = 0.

If n is an odd integer, say n=2m+1, then for k=n-1, the numerator n(n+1) - k(k+1) becomes n(n+1) - (n-1)n = 0. Thus, a_n+1 = 0, a_n+3 = 0, and so on. The series terminates, yielding a polynomial of degree n. In this case, we set a₀ = 0.

Let's compute the first few Legendre polynomials:

n=0: P₀(x). Here n=0 is even, so set a₁=0. The recurrence relation becomes a_k+2 = [-k(k+1)]a_k / [(k+2)(k+1)]. a₂ = [0(1) - 0(1)]a₀ / (2*1) = 0. So a_k = 0 for all k ≥ 2. We need P₀(1) = 1. If we set a₀ = 1, then y(x) = a₀ = 1. So, P₀(x) = 1.

n=1: P₁(x). Here n=1 is odd, so set a₀=0. The recurrence relation is a_k+2 = [1(2) - k(k+1)]a_k / [(k+2)(k+1)]. a₂ = [1(2) - 0(1)]a₀ / (2*1) = 0. So a_k = 0 for all k ≥ 2. We need P₁(1) = 1. y(x) = a₁x. Setting a₁ = 1 gives y(x) = x. So, P₁(x) = x.

n=2: P₂(x). Here n=2 is even, so set a₁=0. The recurrence relation is a_k+2 = [2(3) - k(k+1)]a_k / [(k+2)(k+1)]. a₂ = [2(3) - 0(1)]a₀ / (2*1) = 6a₀ / 2 = 3a₀. a₄ = [2(3) - 2(3)]a₂ / (4*3) = 0. So a_k = 0 for all k ≥ 4. y(x) = a₀ + 3a₀x². We need P₂(1) = 1. So, a₀ + 3a₀(1)² = 1 => 4a₀ = 1 => a₀ = 1/4. Thus, P₂(x) = (1/4) + 3(1/4)x² = (1/4)(3x² + 1). To match the standard convention, we often set a₀ = 1/2, which leads to P₂(x) = (1/2)(3x² - 1). Let's re-evaluate the normalization.

The standard normalization is P_n(1) = 1. Let's re-calculate P₂(x) properly. a₂ = [n(n+1) - 0(1)]a₀ / (2*1) = [2(3) - 0]a₀ / 2 = 3a₀. a₄ = [2(3) - 2(3)]a₂ / (4*3) = 0. All higher even terms are zero. Since n=2 is even, a₁=0. So, y(x) = a₀ + a₂x² = a₀ + 3a₀x² = a₀(1 + 3x²). We need P₂(1) = 1. So, a₀(1 + 3(1)²) = 1 => a₀(4) = 1 => a₀ = 1/4. This gives P₂(x) = (1/4)(1 + 3x²). The commonly accepted form is P₂(x) = (1/2)(3x² - 1). This implies a different choice of initial coefficients. Let's use the standard recurrence relation for Legendre polynomials directly.

Rodrigues' Formula and Properties of Legendre Polynomials

Legendre polynomials P_n(x) can be generated by Rodrigues' formula:

P_n(x) = (1 / 2ⁿ n!) * dⁿ/dxⁿ (x² - 1)ⁿ

The first few Legendre polynomials are:

P₀(x) = 1

P₁(x) = x

P₂(x) = (1/2)(3x² - 1)

P₃(x) = (1/2)(5x³ - 3x)

P₄(x) = (1/8)(35x⁴ - 30x² + 3)

They satisfy the following properties:

• Orthogonality: For n ≠ m, the Legendre polynomials are orthogonal over the interval [-1, 1] with respect to the weight function 1: ∫_-1¹ P_n(x)P_m(x) dx = 0 if n ≠ m ∫_-1¹ [P_n(x)]² dx = 2 / (2n+1)
• Recurrence Relation: (n+1)P_n+1(x) = (2n+1)xP_n(x) - nP_n-1(x)
• Symmetry: P_n(-x) = (-1)ⁿP_n(x) (i.e., they are even or odd functions)
• Values at x=1 and x=-1: P_n(1) = 1 and P_n(-1) = (-1)ⁿ
• Roots: The roots of P_n(x) = 0 are real, distinct, and lie strictly between -1 and 1.

HERMITE'S EQUATION AND HERMITE POLYNOMIALS

Hermite's equation is a second-order linear ordinary differential equation that appears in quantum mechanics, particularly in the study of the quantum harmonic oscillator. It is given by:

y'' - 2xy' + 2ny = 0

Here, n is a non-negative constant.

Series Solution of Hermite's Equation

We again seek a series solution of the form y(x) = Σ_k=0^∞ a_kx^k.

y'(x) = Σ_k=1^∞ ka_kx^k-1

y''(x) = Σ_k=2^∞ k(k-1)a_kx^k-2

Substitute into Hermite's equation:

Σ_k=2^∞ k(k-1)a_kx^k-2 - 2x Σ_k=1^∞ ka_kx^k-1 + 2n Σ_k=0^∞ a_kx^k = 0

Shift indices to get powers of x^k:

Σ_k=0^∞ (k+2)(k+1)a_k+2x^k - 2 Σ_k=1^∞ ka_kx^k + 2n Σ_k=0^∞ a_kx^k = 0

Separate terms for k=0:

k=0: (2)(1)a₂ + 2na₀ = 0 => a₂ = -2na₀ / 2 = -na₀

For k ≥ 1: (k+2)(k+1)a_k+2 - 2ka_k + 2na_k = 0

(k+2)(k+1)a_k+2 + (2n - 2k)a_k = 0

This gives the recurrence relation:

a_k+2 = - (2n - 2k)a_k / [(k+2)(k+1)] = 2(k-n)a_k / [(k+2)(k+1)] for k ≥ 0

As with Legendre's equation, a_k+2 depends on a_k. The even and odd terms are independent.

Hermite Polynomials (H_n(x))

When n is a non-negative integer, one of the series solutions terminates, resulting in a polynomial called the Hermite polynomial, denoted by H_n(x).

The recurrence relation is a_k+2 = 2(k-n)a_k / [(k+2)(k+1)].

If n is an even integer, say n=2m, then for k=n, the numerator 2(k-n) becomes 2(n-n) = 0. Thus, a_n+2 = 0, a_n+4 = 0, and so on. The series terminates, yielding a polynomial of degree n. We set a₁ = 0.

If n is an odd integer, say n=2m+1, then for k=n-1, the numerator 2(k-n) becomes 2(n-1-n) = -2. For k=n+1, the numerator 2(k-n) becomes 2(n+1-n) = 2. If n is odd, let's check k=n-1. a_n+1 = 2(n-1-n)a_n-1 / [(n+1)n] = -2a_n-1 / [n(n+1)]. If n is odd, k=n-1 is even. Let's check if the series terminates for odd n. If n is odd, then for k=n, the term a_n+2 depends on a_n. Let's re-examine. If n is an integer, then for k=n, the term a_n+2 = 2(n-n)a_n / [(n+2)(n+1)] = 0. This means all coefficients a_k for k > n and k have the same parity as n will be zero. So, if n is even, a_n+2=0, a_n+4=0, ... and we set a₁=0. If n is odd, a_n+2=0, a_n+4=0, ... and we set a₀=0.

The standard normalization for Hermite polynomials is that the coefficient of xⁿ is 2ⁿ.

Let's compute the first few Hermite polynomials:

n=0: H₀(x). n=0 is even. Set a₁=0. a₂ = 2(0-0)a₀ / (2*1) = 0. So a_k=0 for k ≥ 2. y(x) = a₀. We need the coefficient of x⁰ to be 2⁰ = 1. So a₀ = 1. H₀(x) = 1.

n=1: H₁(x). n=1 is odd. Set a₀=0. a₂ = 2(0-1)a₁ / (2*1) = -2a₁ / 2 = -a₁. a₃ = 2(1-1)a₁ / (3*2) = 0. So a_k=0 for k ≥ 3. y(x) = a₁x. We need the coefficient of x¹ to be 2¹ = 2. So a₁ = 2. H₁(x) = 2x.

n=2: H₂(x). n=2 is even. Set a₁=0. a₂ = 2(0-2)a₀ / (2*1) = -4a₀ / 2 = -2a₀. a₄ = 2(2-2)a₂ / (4*3) = 0. So a_k=0 for k ≥ 4. y(x) = a₀ + a₂x² = a₀ - 2a₀x² = a₀(1 - 2x²). We need the coefficient of x² to be 2² = 4. So -2a₀ = 4 => a₀ = -2. H₂(x) = -2(1 - 2x²) = 4x² - 2.

n=3: H₃(x). n=3 is odd. Set a₀=0. a₂ = 2(0-3)a₀ / (2*1) = -6a₀ / 2 = 0. So a_k=0 for k ≥ 2. (This is incorrect, a₂ depends on a₀, which is 0). Let's re-evaluate the recurrence for n=3. a_k+2 = 2(k-3)a_k / [(k+2)(k+1)] Since n=3 is odd, we set a₀=0. a₂ = 2(0-3)a₀ / (2*1) = 0. a₄ = 2(2-3)a₂ / (4*3) = 0. So all even terms are zero. We need to check odd terms. a₃ depends on a₁. a₃ = 2(1-3)a₁ / (3*2) = -4a₁ / 6 = -(2/3)a₁. a₅ = 2(3-3)a₃ / (5*4) = 0. So a_k=0 for k ≥ 5. y(x) = a₁x + a₃x³ = a₁x - (2/3)a₁x³ = a₁(x - (2/3)x³). We need the coefficient of x³ to be 2³ = 8. So, -(2/3)a₁ = 8 => a₁ = -12. H₃(x) = -12(x - (2/3)x³) = -12x + 8x³ = 8x³ - 12x.

The first few Hermite polynomials are:

H₀(x) = 1

H₁(x) = 2x

H₂(x) = 4x² - 2

H₃(x) = 8x³ - 12x

H₄(x) = 16x⁴ - 48x² + 12

Properties of Hermite Polynomials

Hermite polynomials satisfy the following properties:

• Orthogonality: Hermite polynomials are orthogonal over the interval (-∞, ∞) with respect to the weight function e^-x2: ∫_-∞^∞ e^-x2 H_n(x)H_m(x) dx = 0 if n ≠ m ∫_-∞^∞ e^-x2 [H_n(x)]² dx = 2ⁿ n! √π
• Recurrence Relation: H_n+1(x) = 2xH_n(x) - 2nH_n-1(x)
• Alternative Recurrence Relation: d/dx H_n(x) = 2nH_n-1(x)
• Symmetry: H_n(-x) = (-1)ⁿH_n(x) (i.e., they are even or odd functions)
• Values at x=0: H_n(0) = (-1)^n/2 (n! / ( (n/2)! )) if n is even, and H_n(0) = 0 if n is odd.

SUMMARY AND APPLICATIONS

Bessel's equation, Legendre's equation, and Hermite's equation are fundamental differential equations whose solutions are crucial in various fields of physics and engineering.

Bessel Functions are essential for problems with cylindrical symmetry, such as heat conduction in a cylinder, wave propagation in a circular waveguide, and the vibration of a circular membrane.

Legendre Polynomials are vital for problems with spherical symmetry, particularly in solving Laplace's equation and the wave equation in spherical coordinates. They are used in areas like electrostatics, gravitational potential, and quantum mechanics.

Hermite Polynomials are central to quantum mechanics, especially in describing the energy states of the quantum harmonic oscillator. They also appear in probability theory and statistics.

Understanding the series solution methods and the properties of these special functions (Bessel functions, Legendre polynomials, and Hermite polynomials) provides a powerful toolkit for solving a wide range of complex physical and mathematical problems. The orthogonality properties, in particular, are fundamental for developing solutions in terms of series expansions, analogous to Fourier series.