Legendre's Equation and Orthogonal Polynomials

Definition Let $\lambda$ be a real constant. Legendre’s equation is

where $x$ is the independent variable and $y' = dy/dx$. The parameter is usually written $\lambda = n(n + 1)$, and the case of an integer $n \ge 0$ is the important one. Since $((1 - x^2)y')' = (1 - x^2)y'' - 2xy'$, the equation has the self-adjoint form

The leading coefficient vanishes only at $x = \pm 1$, so $(-1, 1)$ is a regular interval and $x = 0$ is an ordinary point. A power series solution about $0$ therefore converges at least for $\lvert x \rvert < 1$, the distance to the nearest singular point. The endpoints are regular singular points. Near $x = 1$ the normalized coefficients $p(x) = -2x/(1 - x^2)$ and $q(x) = \lambda/(1 - x^2)$ give $(x - 1)\,p(x) = 2x/(1 + x)$ and $(x - 1)^2q(x) = -\lambda(x - 1)/(1 + x)$, both analytic at $1$, and the indicial equation is $r^2 = 0$. Because of this repeated root, only the multiples of one solution stay bounded near $x = 1$, and every other solution grows like a logarithm there. Replacing $x$ by $-x$ leaves the equation unchanged, so the same holds at $x = -1$.

Definition Substitute a power series $y = \sum a_k x^k$ about the ordinary point $x = 0$. The coefficient of $x^k$ collects $(k + 2)(k + 1)a_{k+2}$ from $y''$, then $-k(k - 1)a_k$ from $-x^2y''$, $-2ka_k$ from $-2xy'$ and $\lambda a_k$, so

for $k \ge 0$, with $a_0$ and $a_1$ free. The even coefficients form one chain and the odd coefficients another. If $\lambda = n(n + 1)$ for an integer $n \ge 0$, the numerator is zero at $k = n$, so $a_{n+2} = 0$ and every later coefficient with the parity of $n$ vanishes. That chain terminates, and with a nonzero first coefficient it gives a polynomial of degree $n$. The other chain never terminates when its first coefficient is nonzero, because $k(k + 1) = n(n + 1)$ has no other solution with $k \ge 0$. The ratio $a_{k+2}/a_k$ along it tends to $1$, so it is an infinite series with radius of convergence exactly $1$. Its sum is not a multiple of the polynomial, so by the endpoint analysis above it is unbounded as $x \to 1$, and by its parity also as $x \to -1$. Hence for $\lambda = n(n + 1)$ the polynomial solutions are exactly the multiples of one polynomial of degree $n$. For any other $\lambda$ the numerator is never zero, so neither chain terminates and no nonzero solution is a polynomial. Here $n$ is the degree of a polynomial, while the differential equation has order $2$ for every $n$.

Definition For an integer $n \ge 0$, the Legendre polynomial $P_n$ is the polynomial solution of degree $n$ of Legendre’s equation with $\lambda = n(n + 1)$, scaled so that $P_n(1) = 1$. It is also given by Rodrigues’ formula

The first five are

Each $P_n$ contains only powers with the parity of $n$, so $P_n(-x) = (-1)^nP_n(x)$ and $P_n(-1) = (-1)^n$. Each $P_n$ has degree exactly $n$, so $P_0, \dots, P_n$ span all polynomials of degree at most $n$.

Definition The standard normalization of the Legendre polynomials is $P_n(1) = 1$. It is always possible, because a nonzero polynomial solution $p$ of the equation with $\lambda = n(n + 1)$ cannot vanish at $x = 1$. Differentiating the equation $m$ times and setting $x = 1$ gives

so $p(1) = 0$ would make every derivative of $p$ vanish at $1$, and the only polynomial with that property is zero. The case $m = 0$ also gives $P_n'(1) = n(n + 1)/2$. This normalization fixes a value, not a size. The squared integral norm of $P_n$ is

so the rescaled polynomials $\sqrt{(2n + 1)/2}\,P_n$ are the ones with norm $1$. Both scalings are in use, and a formula for expansion coefficients depends on which one is meant.

Definition Use the inner product $\langle f, g \rangle = \int_{-1}^{1} f(x)\,g(x)\,dx$, which has weight $1$. For integers $m, n \ge 0$ with $m \ne n$,

The reason is the self-adjoint form. Write $\lambda_k = k(k + 1)$ and $B(x) = (1 - x^2)(P_nP_m' - P_mP_n')$. Multiplying the equation for $P_m$ by $P_n$, the equation for $P_n$ by $P_m$, and subtracting shows that $B'$ equals $(\lambda_n - \lambda_m)P_mP_n$, so

Both $B(1)$ and $B(-1)$ are zero, because $1 - x^2$ vanishes at the ends and the polynomials and their derivatives are finite there. Since $k(k + 1)$ is strictly increasing for $k \ge 0$, the factor $\lambda_n - \lambda_m$ is not zero, and the integral must vanish. It follows that $P_n$ is orthogonal to every polynomial of degree less than $n$. Together with the norm $2/(2n + 1)$, orthogonality shows that if $f = \sum_{k=0}^{N} c_kP_k$, then for each $n \le N$

Definition The Legendre eigenvalue problem asks for the numbers $\lambda$ for which

has a nonzero solution $y$ on $-1 < x < 1$ that stays bounded as $x \to 1^-$ and as $x \to -1^+$. Such a $\lambda$ is an eigenvalue and such a $y$ an eigenfunction. The eigenvalues are exactly $\lambda = n(n + 1)$ with $n = 0, 1, 2, \dots$, and the eigenfunctions for $\lambda = n(n + 1)$ are the nonzero multiples of $P_n$. This is a singular Sturm-Liouville problem. The coefficient $1 - x^2$ of the derivative term vanishes at both endpoints, the weight is $1$, and the boundedness requirements take the place of boundary conditions. For $\lambda = n(n + 1)$, the solutions bounded near $x = 1$ are the multiples of $P_n$, and $P_n$ is bounded near $x = -1$ as well. For every other real $\lambda$, a standard result on the endpoint behavior of Legendre functions shows that every nonzero solution bounded near $x = 1$ is unbounded as $x \to -1^+$, so no eigenfunction exists. The orthogonality of $P_m$ and $P_n$ is the usual orthogonality of eigenfunctions with different eigenvalues, and the boundary term vanishes because $1 - x^2$ does.