How to Solve Legendre's Differential Equation

Last Updated: May 24, 2020

Legendre's differential equation

(1-x^{2}){\frac {\mathrm {d} ^{2}y}{\mathrm {d} x^{2}}}-2x{\frac {\mathrm {d} y}{\mathrm {d} x}}+l(l+1)y=0

is an important ordinary differential equation encountered in mathematics and physics. In particular, it occurs when solving Laplace's equation in spherical coordinates. Bounded solutions to this equation are called Legendre polynomials, an important orthogonal polynomial sequence seen in the multipole expansions of electrostatics. It is in this context that the argument of the solutions is $\cos \theta$ and therefore motivates us to look for solutions that are bounded by $[-1,1],$ so that every point is regular.

Because Legendre's equation contains variable coefficients and is not the Euler-Cauchy equation, we must resort to finding solutions using power series. Series methods usually involve a bit more algebra, but are still fairly straightforward.

Steps

1
Substitute the power series ansatz. This ansatz takes on the form $y(x)=\sum _{n=0}^{\infty }a_{n}x^{n},$ $y(x)=\sum _{{n=0}}^{{\infty }}a_{{n}}x^{{n}},$ where $a_{n}$ $a_{{n}}$ are coefficients to be determined. Its first and second derivatives are easily found as ${\frac {\mathrm {d} y}{\mathrm {d} x}}=\sum _{n=1}^{\infty }na_{n}x^{n-1}$ ${\frac {{\mathrm {d}}y}{{\mathrm {d}}x}}=\sum _{{n=1}}^{{\infty }}na_{{n}}x^{{n-1}}$ and ${\frac {\mathrm {d} ^{2}y}{\mathrm {d} x^{2}}}=\sum _{n=2}^{\infty }n(n-1)a_{n}x^{n-2}.$ ${\frac {{\mathrm {d}}^{{2}}y}{{\mathrm {d}}x^{{2}}}}=\sum _{{n=2}}^{{\infty }}n(n-1)a_{{n}}x^{{n-2}}.$
- ${\begin{aligned}\sum _{n=2}^{\infty }n(n-1)a_{n}x^{n-2}&-\sum _{n=2}^{\infty }n(n-1)a_{n}x^{n}\\&-\sum _{n=1}^{\infty }2na_{n}x^{n}+\sum _{n=0}^{\infty }l(l+1)a_{n}x^{n}=0\end{aligned}}$
2
Group all terms under a common sum. We proceed by first rewriting the first term so that there is an $x^{n}$ $x^{{n}}$ inside the summation (remember that $n$ $n$ is a dummy index). Then we explicitly write out all the $n=0$ $n=0$ and $n=1$ $n=1$ terms.
- $\sum _{n=2}^{\infty }n(n-1)a_{n}x^{n-2}=\sum _{n=0}^{\infty }(n+2)(n+1)a_{n+2}x^{n}$
- ${\begin{aligned}&2a_{2}+l(l+1)a_{0}+6a_{3}x-2a_{1}x+l(l+1)a_{1}x\\&\quad +\sum _{n=2}^{\infty }((n+2)(n+1)a_{n+2}-(n^{2}+n-l(l+1))a_{n})x^{n}=0\end{aligned}}$
- Notice the importance of the $l(l+1)$ constant, which has the same form as the $n^{2}+n$ contribution.
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3
Set the coefficients of each power to 0. In linear algebra, the sequence of powers can be thought of as linearly independent functions spanning a vector space. The linear independence demands that each coefficient of a power term has to vanish in order for the equality to hold true.
- $2a_{2}+l(l+1)a_{0}=0$
- $6a_{3}-2a_{1}+l(l+1)a_{1}=0$
- $(n+2)(n+1)a_{n+2}-(n(n+1)-l(l+1))a_{n}=0,\quad n=2,3,\cdots$
4
Obtain the recurrence relation. The recurrence relation is an important relation and is the goal of every power series solution method. The recurrence relation, together with limiting cases, gives the value of every coefficient in terms of $a_{0}$ $a_{{0}}$ and $a_{1}.$ $a_{{1}}.$
- $a_{2}=-{\frac {l(l+1)}{2}}a_{0},\quad a_{3}={\frac {2-l(l+1)}{6}}a_{1}$
- $a_{n+2}={\frac {n(n+1)-l(l+1)}{(n+1)(n+2)}}a_{n},\quad n=2,3,\cdots$
- Note that the first line is redundant - it came about from our handling of the series to begin at $n=2,$ so those coefficients are written out explicitly.
- The most important property in the recurrence is the fact that the even and odd contributions are decoupled - the $n\mathrm {th}$ coefficient is determined by the $(n-2)\mathrm {th}$ coefficient, which must be both even or both odd. This means that we may formulate our solution in terms of even and odd functions, which can be very useful.
5
Choose $l=n$ for certain values of $l$ . The coefficients $a_{0}$ $a_{{0}}$ and $a_{1}$ $a_{{1}}$ are the two constants resulting from the fact that Legendre's equation is a second-order differential equation. Because the recurrence relations give coefficients of the next order of the same parity, we are motivated to consider solutions where one of $a_{0}$ $a_{{0}}$ or $a_{1}$ $a_{{1}}$ is set to 0. For example, if $a_{1}=0,$ $a_{{1}}=0,$ then it follows that all odd terms vanish, and the solution is an even function; vice versa. The other important observation is the fact that the series can be bounded with a suitable choice of $l.$ $l.$ The obvious choice here is $l=n.$ $l=n.$ Then all terms $n>l$ $n>l$ vanish in the sum.
- For example, let's make a list of cases where $a_{1}=0.$ $a_{{1}}=0.$ Going through the possible values of $n,$ $n,$ the series truncates to the $n\mathrm {th}$ $n{\mathrm {th}}$ order term.
  - $n=0:y(x)=a_{0}$
  - $n=2:y(x)=a_{0}(1-3x^{2})$
  - $n=4:y(x)=a_{0}\left(1-10x^{2}+{\frac {35}{3}}x^{4}\right)$
- If $a_{0}=0,$ $a_{{0}}=0,$ we have the odd functions.
  - $n=1:y(x)=a_{1}x$
  - $n=3:y(x)=a_{1}\left(x-{\frac {5}{3}}x^{3}\right)$
- We could continue on like this to retain more terms.
6
Normalize the bounded solutions. By convention, the constants are set so that $y_{n}(1)=1$ $y_{{n}}(1)=1$ for all $n.$ $n.$ These constants are very easy to find, and this uniquely fixes each solution. The resulting polynomials are called the Legendre polynomials $P_{n}(x),$ $P_{{n}}(x),$ where $n$ $n$ is called the degree of the polynomial. Below, we list the first few Legendre polynomials.
- $P_{0}(x)=1$
- $P_{1}(x)=x$
- $P_{2}(x)={\frac {1}{2}}\left(3x^{2}-1\right)$
- $P_{3}(x)={\frac {1}{2}}\left(5x^{3}-3x\right)$
- $P_{4}(x)={\frac {1}{8}}\left(35x^{4}-30x^{2}+3\right)$