Legendre differential equation

The equation $(1-x^2)y\prime \prime -2xy\prime +\ell (\ell +1)y=0$ has polynomial solutions for nonnegative integer $\ell$.

The equation $(1-x^2)y\prime \prime -2xy\prime +\ell (\ell +1)y=0$ has the Legendre polynomial of degree $\ell$ as its normalized regular polynomial solution when $\ell$ is a nonnegative integer.

For the Legendre differential equation $(1-x^2)y^{\prime \prime }-2x{y}^{\prime }+n(n+1)y=0$, put $t=1+x$. The equation becomes $t(2-t)y_{tt}+2(1-t)y_t+n(n+1)y=0$. Its indicial equation is $2r^2=0$. The exponent-zero power series $\sum a_k{t}^k$ obeys the displayed linear recurrence relation. For a nonnegative integer $n$, it terminates at degree $n$, giving a Legendre polynomial with $a_0\ne 0$. Every analytic solution at the endpoint is proportional to this one, since the recurrence determines all coefficients from $a_0$.

The second independent solution has the local form $P_n(x)\log (1+x)+B(x)$, where $B$ is analytic near $-1$. Indeed, reduction of order gives $P_n(x)\int dx/[(1-x^2)P_n(x{)}^2]$, whose integrand is $1/[2P_n(-1{)}^2(1+x)]$ plus an analytic function. Rescale the second solution to obtain the unit coefficient of $P_n\log (1+x)$. Reflection gives the corresponding expansion at $x=1$. This also proves the logarithmic divergence of the independent Legendre function of the second kind at each endpoint.

Substitute $y=x^{\sigma }{\sum }_{n\ge 0}{a}_n{x}^n$, $a_0\ne 0$, into the Legendre differential equation. The lowest powers give $\sigma (\sigma -1)=0$ and $\sigma (\sigma +1)a_1=0$; subsequent powers give the recurrence in the title. Thus exponent zero has independent even and odd coefficients, while exponent one requires $a_1=0$ and represents the odd solution. Zero is an ordinary point of the equation, so the exponent-one choice simply selects a solution that vanishes there.

The singular points at $\pm 1$ bound the generic radius of convergence about zero by one. Every nonterminating parity recurrence has successive-coefficient ratio tending to one in the variable $x^2$, giving radius exactly one. If $\lambda =N$ is a nonnegative integer, choosing the parity of $N$ makes the recurrence terminate at degree $N$, producing a Legendre polynomial with infinite radius. These are precisely the nonnegative parameters with a nonzero entire solution: even and odd parts of an entire solution would both have to be entire, so any nonzero nonterminating part is excluded.

For nonnegative integer $n$, a second solution of the Legendre differential equation is obtained by reduction of order from $P_n$ as $P_n(x)\int dx/[(1-x^2)P_n(x{)}^2]$. Normalize it as $Q_0=\frac{1}{2}\log |(1+x)/(1-x)|$ and $Q_1=x{Q}_0-1$, with the usual Legendre polynomial recurrence relation for higher degrees. Inside $(-1,1)$, $Q_n\sim -\frac{1}{2}\log (1-x)$ near $1$ and $Q_n\sim (-1{)}^n\frac{1}{2}\log (1+x)$ near $-1$. These follow from $P_n(1)=1$, $P_n(-1)=(-1{)}^n$ in the integral. Poles in the integral at interior polynomial zeros cancel in the product, by reduction of order across a zero of the known solution. On $x>1$, the integral from infinity defines the decaying normalization $Q_n(x)\sim 2^n(n!{)}^2{x}^{-n-1}/(2n+1)!$. For $n\ge 1$ all bounded exterior solutions decay this way; for $n=0$ an additional constant solution remains bounded.

This second-order linear differential equation extends the Legendre differential equation by its order parameter $m$. Its solutions are associated Legendre functions. In a degree-one potential problem, with $x=\tanh \eta$, two explicit solutions are $e^{m\eta }(m-\tanh \eta )$ and $e^{-m\eta }(m+\tanh \eta )$. Their Wronskian with respect to $\eta$ is $2m(1-m^2)$, so the two formulas are not a fundamental pair at $m=0$ or $m=\pm 1$.

Associated Legendre functions solve $(1-x^2)y^{\prime \prime }-2x{y}^{\prime }+[\nu (\nu +1)-{\mu }^2/(1-x^2)]y=0$. For integers $\ell \ge 0$ and $0\le m\le \ell$, the regular solution is $P_{\ell }^m(x)=(-1{)}^m(1-x^2{)}^{m/2}(d/dx{)}^m{P}_{\ell }(x)$, using a Legendre polynomial. It enters normalized spherical harmonics. Despite the traditional term associated Legendre polynomial, the factor $(1-x^2{)}^{m/2}$ is not a polynomial when $m$ is odd. Negative integer orders satisfy $P_{\ell }^{-m}=(-1{)}^m(\ell -m)!P_{\ell }^m/(\ell +m)!$.

The polynomial solution of polynomial degree $l$ normalized by $P_l(1)=1$ is the Legendre polynomial $P_l$.

For orthonormal Legendre polynomials on $[-1,1]$, the polynomial $q_n$ represents evaluation at zero on polynomials of degree at most $n$: $⟨r,q_n⟩=r(0)$. This follows by expanding $r$ in the orthonormal basis, or from the Riesz representation theorem. With $M=\lfloor n/2\rfloor$ and $c_m=\left(\genfrac{}{}{0ex}{}{2m}{m}\right)/4^m$, orthonormality givesThe last identity telescopes using $c_m/c_{m-1}=(2m-1)/(2m)$. The norms are unbounded: $c_m\ge 1/(2m+1)$ makes the sum dominate a divergent harmonic series. Thus the evaluation identities hold eventually for every fixed polynomial but cannot extend to convergence for every $L^2$ vector, by the Uniform boundedness principle. Point evaluation is also not defined on $L^2$ equivalence classes.

For normalized Legendre polynomials $p_n=\sqrt{(2n+1)/2}{P}_n$, parity makes the sign function orthogonal to every even index. Integrating the Legendre differential equation gives ${\int }_0^1{P}_n=P_n^{\prime }(0)/(n(n+1))$. The differentiated generating function $z(1+z^2{)}^{-3/2}$ gives the derivatives at zero and hence the displayed odd coefficients. The Parseval identity and squared norm two then evaluate ${\sum }_{m\ge 0}((2m)!{)}^2(4m+3)/[2^{4m}(m!(m+1)!{)}^2]=4$.

This instance of Rodrigues' formula constructs the Legendre polynomial for every nonnegative integer $l$. Put $v=(x^2-1{)}^l$; then $(x^2-1)v^{\prime }=2lxv$. Differentiating $l+1$ times by the Leibniz rule gives $(x^2-1)y^{\prime \prime }+2x{y}^{\prime }-l(l+1)y=0$ for $y=v^{(l)}$, which is the Legendre differential equation after reversing signs. The leading term of the normalized derivative is $(2l)!x^l/[2^l(l!{)}^2]$, proving its polynomial degree is exactly $l$. At $x=1$, the product $(x+1{)}^l(x-1{)}^l$ contributes only the term with all $l$ derivatives on the second factor, giving $v^{(l)}(1)=2^{l}l!$. Thus the normalization is $P_l(1)=1$.

Differentiating the Legendre differential equation gives $((1-x^2{)}^2{R}_n^{\prime }{)}^{\prime }+(n-1)(n+2)(1-x^2)R_n=0$ for $R_n=P_n^{\prime }$. Multiply the equations for two distinct indices by the opposite polynomial and subtract. The endpoint term vanishes because of $(1-x^2{)}^2$, giving the displayed weighted orthogonality. The zero polynomial $R_0$ is not a nonzero eigenfunction.

With $P_n(1)=1$, the generating function satisfies ${\partial }_x(x^2{\partial }_{x}G)+{\partial }_t((1-t^2){\partial }_{t}G)=0$. Expanding in Legendre polynomials gives $x^2{a}_n^{\prime \prime }+2x{a}_n^{\prime }-n(n+1)a_n=0$. Boundedness at zero removes $x^{-n-1}$, and the value at $t=1$ makes $a_n=x^n$. Orthogonality then gives ${\int }_{-1}^1{G}^{2}dt={\sum }_n{k}_n{x}^{2n}$. The integral is $x^{-1}\log ((1+x)/(1-x))$, so $k_n=2/(2n+1)$. Uniform analyticity on compact subdiscs $|x|<1$ justifies the generating expansion and termwise calculations.

The Legendre polynomials are orthogonal on $[-1,1]$ with unit weight:

The three-term recurrenceturns multiplication by $x$ into coupling between adjacent Legendre multipoles.

For a contour enclosing $t$,

For fixed $0<\theta <\pi$,

For fixed $\xi >0$,

The endpoint Laplace method for a Legendre polynomial ceases to localize when $n\sqrt{x-1}=O(1)$. On the scale $x=1+\nu /n^2$, its real integral representation givesThe Modified Bessel function of the first kind equals one at $\nu =0$ and has large-$\nu$ behavior $e^{\sqrt{2\nu }}/\sqrt{2\pi \sqrt{2\nu }}$, matching the Hyperbolic Debye asymptotic for Legendre polynomials. This boundary layer resolves an increasingly steep function at an endpoint where its value remains fixed.