How to Solve Legendre's Differential Equation

Substitute the power series ansatz. This ansatz takes on the form y ( x ) = ∑ n = 0 ∞ a n x n , {\displaystyle y(x)=\sum _{n=0}^{\infty }a_{n}x^{n},} y(x)=\sum _{{n=0}}^{{\infty }}a_{{n}}x^{{n}}, where a n {\displaystyle a_{n}} a_{{n}} are coefficients to be determined. Its first and second derivatives are easily found as d y d x = ∑ n = 1 ∞ n a n x n − 1 {\displaystyle {\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 d 2 y d x 2 = ∑ n = 2 ∞ n ( n − 1 ) a n x n − 2 . {\displaystyle {\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}}. ∑ n = 2 ∞ n ( n − 1 ) a n x n − 2 − ∑ n = 2 ∞ n ( n − 1 ) a n x n − ∑ n = 1 ∞ 2 n a n x n + ∑ n = 0 ∞ l ( l + 1 ) a n x n = 0 {\displaystyle {\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}}} {\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}}

Group all terms under a common sum. We proceed by first rewriting the first term so that there is an x n {\displaystyle x^{n}} x^{{n}} inside the summation (remember that n {\displaystyle n} n is a dummy index). Then we explicitly write out all the n = 0 {\displaystyle n=0} n=0 and n = 1 {\displaystyle n=1} n=1 terms. ∑ n = 2 ∞ n ( n − 1 ) a n x n − 2 = ∑ n = 0 ∞ ( n + 2 ) ( n + 1 ) a n + 2 x n {\displaystyle \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}} \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}} 2 a 2 + l ( l + 1 ) a 0 + 6 a 3 x − 2 a 1 x + l ( l + 1 ) a 1 x + ∑ n = 2 ∞ ( ( n + 2 ) ( n + 1 ) a n + 2 − ( n 2 + n − l ( l + 1 ) ) a n ) x n = 0 {\displaystyle {\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}}} {\displaystyle {\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 ) {\displaystyle l(l+1)} l(l+1) constant, which has the same form as the n 2 + n {\displaystyle n^{2}+n} n^{{2}}+n contribution.

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. 2 a 2 + l ( l + 1 ) a 0 = 0 {\displaystyle 2a_{2}+l(l+1)a_{0}=0} 2a_{{2}}+l(l+1)a_{{0}}=0 6 a 3 − 2 a 1 + l ( l + 1 ) a 1 = 0 {\displaystyle 6a_{3}-2a_{1}+l(l+1)a_{1}=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 , n = 2 , 3 , ⋯ {\displaystyle (n+2)(n+1)a_{n+2}-(n(n+1)-l(l+1))a_{n}=0,\quad n=2,3,\cdots } {\displaystyle (n+2)(n+1)a_{n+2}-(n(n+1)-l(l+1))a_{n}=0,\quad n=2,3,\cdots }

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 {\displaystyle a_{0}} a_{{0}} and a 1 . {\displaystyle a_{1}.} a_{{1}}. a 2 = − l ( l + 1 ) 2 a 0 , a 3 = 2 − l ( l + 1 ) 6 a 1 {\displaystyle a_{2}=-{\frac {l(l+1)}{2}}a_{0},\quad a_{3}={\frac {2-l(l+1)}{6}}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 = n ( n + 1 ) − l ( l + 1 ) ( n + 1 ) ( n + 2 ) a n , n = 2 , 3 , ⋯ {\displaystyle a_{n+2}={\frac {n(n+1)-l(l+1)}{(n+1)(n+2)}}a_{n},\quad n=2,3,\cdots } 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 , {\displaystyle n=2,} 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 t h {\displaystyle n\mathrm {th} } n{\mathrm {th}} coefficient is determined by the ( n − 2 ) t h {\displaystyle (n-2)\mathrm {th} } (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.

Choose l = n {\displaystyle l=n} l=n for certain values of l {\displaystyle l} l. The coefficients a 0 {\displaystyle a_{0}} a_{{0}} and a 1 {\displaystyle 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 {\displaystyle a_{0}} a_{{0}} or a 1 {\displaystyle a_{1}} a_{{1}} is set to 0. For example, if a 1 = 0 , {\displaystyle 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 . {\displaystyle l.} l. The obvious choice here is l = n . {\displaystyle l=n.} l=n. Then all terms n > l {\displaystyle n>l} n>l vanish in the sum. For example, let's make a list of cases where a 1 = 0. {\displaystyle a_{1}=0.} a_{{1}}=0. Going through the possible values of n , {\displaystyle n,} n, the series truncates to the n t h {\displaystyle n\mathrm {th} } n{\mathrm {th}} order term. n = 0 : y ( x ) = a 0 {\displaystyle n=0:y(x)=a_{0}} n=0:y(x)=a_{{0}} n = 2 : y ( x ) = a 0 ( 1 − 3 x 2 ) {\displaystyle n=2:y(x)=a_{0}(1-3x^{2})} n=2:y(x)=a_{{0}}(1-3x^{{2}}) n = 4 : y ( x ) = a 0 ( 1 − 10 x 2 + 35 3 x 4 ) {\displaystyle n=4:y(x)=a_{0}\left(1-10x^{2}+{\frac {35}{3}}x^{4}\right)} n=4:y(x)=a_{{0}}\left(1-10x^{{2}}+{\frac {35}{3}}x^{{4}}\right) If a 0 = 0 , {\displaystyle a_{0}=0,} a_{{0}}=0, we have the odd functions. n = 1 : y ( x ) = a 1 x {\displaystyle n=1:y(x)=a_{1}x} n=1:y(x)=a_{{1}}x n = 3 : y ( x ) = a 1 ( x − 5 3 x 3 ) {\displaystyle n=3:y(x)=a_{1}\left(x-{\frac {5}{3}}x^{3}\right)} n=3:y(x)=a_{{1}}\left(x-{\frac {5}{3}}x^{{3}}\right) We could continue on like this to retain more terms.

Normalize the bounded solutions. By convention, the constants are set so that y n ( 1 ) = 1 {\displaystyle y_{n}(1)=1} y_{{n}}(1)=1 for all n . {\displaystyle 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 ) , {\displaystyle P_{n}(x),} P_{{n}}(x), where n {\displaystyle n} n is called the degree of the polynomial. Below, we list the first few Legendre polynomials. P 0 ( x ) = 1 {\displaystyle P_{0}(x)=1} P_{{0}}(x)=1 P 1 ( x ) = x {\displaystyle P_{1}(x)=x} P_{{1}}(x)=x P 2 ( x ) = 1 2 ( 3 x 2 − 1 ) {\displaystyle P_{2}(x)={\frac {1}{2}}\left(3x^{2}-1\right)} P_{{2}}(x)={\frac {1}{2}}\left(3x^{{2}}-1\right) P 3 ( x ) = 1 2 ( 5 x 3 − 3 x ) {\displaystyle P_{3}(x)={\frac {1}{2}}\left(5x^{3}-3x\right)} P_{{3}}(x)={\frac {1}{2}}\left(5x^{{3}}-3x\right) P 4 ( x ) = 1 8 ( 35 x 4 − 30 x 2 + 3 ) {\displaystyle P_{4}(x)={\frac {1}{8}}\left(35x^{4}-30x^{2}+3\right)} P_{{4}}(x)={\frac {1}{8}}\left(35x^{{4}}-30x^{{2}}+3\right)