Question

Show that the first derivatives of the Legendre polynomials satisfy a self-adjoint differential equation with eigenvalue: lamda = n(n+1)-2

Answer 1

The Legendre Differential Equation is

(3.1) (1 − x² )y'' − 2xy' + n(n + 1)y = 0, n ∈ R, x ∈ (−1, 1)

We know that x = 0 is an ordinary point of equation (3.1). We see that when we divide by the coefficient (1−x² ) that x ∈ (−1, 1). We will see later that the property of orthogonality falls out on the interval [−1, 1] by the Sturm-Liouville Theory. In order to find the series solution to this differential equation we will use the power series method.

Let y(x) = (k=0 to ∞ ) a_kx^k

y'(x) = (k=1 to ∞) a_kkx^k-1

y''(x) = (k=2 to ∞) a_kk(k − 1)x^k-2

Insert these terms into the original equation (3.1) to obtain

(1 − x² ) (k=2 to ∞) a_kk(k − 1)x^k-2 − 2x (k=1 to ∞) a_kkx^k-1 + n(n + 1) (k=0 to ∞) a_kx^k = 0.

which gives

( k=2 to ∞) a_kk(k − 1)x^k-2 − ( k=2 to ∞) a_kk(k − 1)x^k -2( k=1 to ∞) a_kkx^k + n(n + 1)( k=0 to ∞) a_kx^k = 0

making powers and indicies equal

( k=0 to ∞) a_k+2(k + 2)(k + 1)x^k − ( k=0 to ∞) a_kk(k − 1)x^k − 2( k=0 to ∞) a_k(k)x^k + ( k=0 to ∞) n(n + 1)a_kx^k = 0

simplify

( k=0 to ∞) [(k + 2)(k + 1)a_k+2 − (k)(k − 1)a_k − 2ka_k + n(n + 1)a_k]x^k = 0

equating coefficients

(k + 2)(k + 1)a_k+2 − (k)(k − 1)a_k − 2ka_k + n(n + 1)a_k = 0

solving for a_k+2 gives us a recurrence relation

(3.2) a_k+2 = (k(k + 1) − n(n + 1)/ (k + 2)(k + 1))a_k

Remark. We are looking for polynomial solutions. If we assume our solution has degree L then

( k=0 to L) a_kx^k = a₀ + a₁x + a₂x² + · · · + a_L + 0x^L+1 + 0x^L+2 + · · ·

Where all the terms following a_L will be zero, while a_L 0.

So we know,

a_L+2 = a_L (L(L + 1) − n(n − 1) /(L + 2)(L + 1)) = 0

L(L + 1) − n(n − 1)/ (L + 2)(L + 1) = 0

L(L + 1) − n(n + 1) = 0

L(L + 1) = n(n + 1)

L = n or L = −(n + 1)

So L = n is our solution because all terms after n+1 are zero. Therefore the degree of our polynomial solution is n where n is an integer. We get two linearly independent series solutions from the recurrence relation (3.2). The first solutions comes from the even values of k. While the second solution comes from the odds values of k. We assume a₀ 0 and a₁ 0.

y₁(x) = a₀[1 − n(n + 1)/ 2 x² + (n − 2)n(n + 1)(n + 3)/ 4! x⁴ − (n − 4)(n − 2)n(n + 1)(n + 3)(n + 5) /6! x⁶ + · · · ]

y₂(x) = a₁[x− (n − 1)(n + 2) /3! x³ (n − 3)(n − 1)(n + 2)(n + 4) /5! x⁵+· · · ]

Where both solutions are valid for x ∈ (−1, 1). Finding the Legendre polynomials can be very long and difficult. There are many methods including Rodrigue’s Formula that are useful in finding these polynomials. The first five Legendre Polynomials turn out to be

P₀(x) = 1

P₁(x) = x

P₂(x) = 1 /2 (3x² − 1)

P₃(x) = 1 /2 x(5x² − 3)

P₄(x) = 1/ 8 (35x⁴ − 30x² + 3)

·

·

·

By rewriting the Legendre Polynomial as a Sturm-Liouville problem, we can prove its orthgonality. We find that the operator can be written as

Ly = [(1 − x 2 )y' ]' .

where f, g ∈ C[−1, 1]. After imposing the conditions that any f, g ∈ BC² [−1, 1] whenever f, g meet the conditions. We want

<Lf|g> = <f|Lg>. That is we want L to be self-adjoint.

<Lf|g> − <f|Lg> = 0

(-1 to 1) Lf(x)g(x) − f(x)Lg(x)dx

= (-1 to 1) ((1 − x² )f ' ) ' g(x) − f(x)((1 − x² )g ' )' dx

= (-1 to 1) (−2xf ' + (1 − x² )f '')g − f(−2xg' − (1 − x² )g'')dx

= (-1 to 1) (1 − x² )f ''g − 2xf ' g + 2xfg ' − (1 − x² )fg ''dx

= (-1 to1 ) [(1 − x² )(f ' g − g ' f)]' dx

= [(1 − x² )(f ' g − g ' f)](-1 to 1 )

= 0.

Therefore L is self-adjoint with no imposed conditions.

Let y_n and y_m ,where n m, be polynomial solutions to the differential equation,

Ly_n = −n(n + 1)y.

− n(n + 1)<y_n|y_m>

= <Ly_n|y_m >

= <y_n|Ly_m >

= <y_n| -m(m+1)y_m >

= −m(m + 1)<y_n |y_m >

So, −n(n + 1)<y_n|y_m> = −m(m + 1)<y_n|y_m> since n m, <y_n|y_m> = 0. We could have also used Lemma (2.2) to say that the Legendre polynomials are orthogonal due to the Sturm-Liouville theory. The Legendre polynomaials are orthogonal on the interval [−1, 1] with respect to the the weight function r(x) = 1.