Section 15.10 Series Solutions of Legendre’s Equation
Definition 15.15. Legendre’s Equation.
Legendre’s equation is
\begin{equation}
\left((1-z^2)\frac{\partial^2}{\partial z^2}
- 2z\frac{\partial}{\partial z} - A \right) P(z)
= 0\tag{15.10.1}
\end{equation}
It arises during the separation of variables procedure for any PDE involving the Laplacian in spherical coordinates, see
Section 19.5.
In this section, we will use series methods to find the general solution of
(15.10.1). Assume that the solution can be written as a power series
\begin{equation}
P(z) = \sum_{n=0}^{\infty} a_n\, z^n\tag{15.10.2}
\end{equation}
Then we have
\begin{align}
\frac{dP}{dz} \amp= \sum_{n=0}^{\infty} a_n\, n\, z^{n-1}\tag{15.10.3}\\
\frac{d^2 P}{dz^2} \amp= \sum_{n=0}^{\infty} a_n\, n(n-1)\, z^{n-2}\tag{15.10.4}
\end{align}
Our job is to solve for the unknown coefficients \(a_n\text{.}\)
\begin{align}
0
\amp = \sum_{n=0}^{\infty} a_n\, n(n-1)\, z^{n-2}
- z^2\sum_{n=0}^{\infty} a_n\, n(n-1)\, z^{n-2}\notag\\
\amp\qquad\quad - 2z\, \sum_{n=0}^{\infty} a_n\, n\, z^{n-1}
- A\sum_{n=0}^{\infty} a_n\, z^n\tag{15.10.5}
\end{align}
The next step is to get the power of the independent variable
\(z\) to be the same in each term of
(15.10.5). In the last three terms, the power is
\(z^n\text{.}\) (Don’t forget the powers of
\(z\) in front of the sum sign.) But in the first term, the power is
\(z^{n-2}\text{.}\) Since the summation variable
\(n\) is a dummy variable (just like a dummy variable of integration), we can shift
\(n\to n+2\) in the first term only, i.e.
\begin{align}
\textrm{First Term}\amp =\sum_{n=0}^{\infty} a_n\, n(n-1)\, z^{n-2} \\\notag\\
\amp= \sum_{n=-2}^{\infty} a_{n+2}\, (n+2)(n+1)\, z^n\notag\\
\amp= a_{0}\cancelto{0}{(-2+2)}(-2+1)\,z^{-2} \notag\\
\amp \qquad\quad + a_{1}(-1+2)\cancelto{0}{(-1+1)}\,z^{-1}\notag\\
\amp \qquad\quad + \sum_{n=0}^{\infty} a_{n+2}\, (n+2)(n+1)\, z^n\notag\\
\amp= \sum_{n=0}^{\infty} a_{n+2}\, (n+2)(n+1)\, z^n\tag{15.10.6}
\end{align}
Pay special attention to what happened to the lower limit of the sum. After the index is shifted, sum would start at \(n=-2\text{,}\) but the factor of \((n+2)\) in the first term and the factor of \((n+1)\) in the second term mean that these terms are zero and we can eliminate them from the sum. This cancellation doesn’t always happen, you should check. If it doesn’t happen, you just need to carry the extra terms through to the end of the calculation.
Plug
(15.10.6) into
(15.10.5). At the same time, bring any overall factors of
\(z\) into the corresponding sums. Finally, since each sum now has a factor of
\(z^n\) and runs over the same range, group the sums together.
\begin{align}
\amp\sum_{n=0}^{\infty} a_{n+2}\, (n+2)(n+1)\, z^{n}
- \sum_{n=0}^{\infty} a_n\, n(n-1)\, z^{n}\notag\\
\amp\qquad- 2\, \sum_{n=0}^{\infty} a_{n}\, n\, z^{n}
- A\sum_{n=0}^{\infty} a_n\, z^n\tag{15.10.7}\\
\amp = \sum_{n=0}^{\infty}
\left[ a_{n+2}\,(n+2)(n+1) - a_n\, (n(n-1)
+ 2\,n + A)\right] z^n\tag{15.10.8}
\end{align}
Now comes the MAGIC part. Since
(15.10.8) is true
for all values of \(z\), the coefficient of
\(z^n\) for each term in the sum (i.e. the quantity in square brackets) must separately be zero, i.e.
\begin{equation}
0=a_{n+2}\,(n+2)(n+1) - a_n\,(n(n-1) + 2\,n + A) \text{.}\tag{15.10.9}
\end{equation}
Definition 15.16. Recurrence Relation.
We can solve for \(a_{n+2}\) in terms of \(a_n\) to obtain the recurrence relation which relates coefficients in the power series to coefficients with smaller values of \(a_n\)
\begin{equation}
a_{n+2}=\frac{n(n+1)+A}{(n+2)(n+1)}\, a_n\text{.}\tag{15.10.10}
\end{equation}
By plugging successive even values of
\(n\) into the recurrence relation
(15.10.10), usining an iterative process, allows us to find
\(a_2\text{,}\) \(a_4\text{,}\) etc. in terms of the arbitrary constant
\(a_0\) and successive odd values of
\(n\) allow us to find
\(a_3\text{,}\) \(a_5\text{,}\) etc. in terms of the arbitrary constant
\(a_1\text{.}\) Thus, for the second-order differential equation
(15.10.1), we obtain two solutions as expected.
\(a_0\) becomes the normalization constant for a solution with only even powers of
\(z\) and
\(a_1\) becomes the normalization constant for a solution with only odd powers of
\(z\text{.}\) For example:
\begin{align}
a_2 \amp= \frac{A}{2} a_0\tag{15.10.11}\\
a_4 \amp= \frac{6+A}{12} a_2
= \left(\frac{6+A}{12}\right) \left(\frac{A}{2}\right) a_0\tag{15.10.12}\\
\amp\vdots\notag\\
a_3 \amp= \frac{2+A}{6} a_1\tag{15.10.13}\\
a_5 \amp= \frac{12+A}{20} a_3
= \left(\frac{12+A}{20}\right) \left(\frac{2+A}{6}\right) a_1\tag{15.10.14}\\
\amp\vdots\notag
\end{align}
Don’t forget to plug these coefficients back into the original power series anzatz
(15.10.2). It is rare to be able to find a closed form expression for general
\(n\text{,}\) so you may only be finding an approximation to a solution, but you can keep as many terms as you have patience to calculate.
\begin{align}
P(z) \amp= a_0 \left[ \frac{A}{2} z^0
+ \left(\frac{6+A}{12}\right) \left(\frac{A}{2}\right) z^2
+ ... \right]\notag\\
+\amp\qquad a_1 \left[ \frac{2+A}{6} z^1
+ \left(\frac{12+A}{20}\right) \left(\frac{2+A}{6}\right) z^3
+ ... \right]\tag{15.10.15}
\end{align}
In general, the solutions of an ordinary linear differential equation can blow-up only where the coefficients of the equation itself are singular, in this case at
\(z=\pm 1\) (which correspond to the north and south poles
\(\theta=0,\pi\) of spherical coordinates). But there is nothing special about physics at these points, only the choice of coordinates is special there. Therefore, in physics contexts, we may want to choose solutions of
(15.10.1) which are regular (non-infinite) at
\(z=\pm 1\text{.}\) This is an important example of a problem where the choice of coordinates for a partial differential equation end up imposing boundary conditions on the ordinary differential equation which comes from it. Therefore, the infinite series
(15.10.2) will typically blow up at the endpoints
\(z=\pm1\text{,}\) but a polynomial can not. So if we choose the special values for the separation constant
\(A\) to be
\(A=-\ell(\ell+1)\) where
\(\ell\) is a non-negative integer, we see from
(15.10.10) that for
\(n\ge\ell\) the coefficients become zero and one of the series terminates in a polynomial.
Definition 15.17. Legendre Polynomials.
The solutions of Legendre’s equation for \(A-\ell(\ell+1)\) are polynomials of degree \(\ell\text{,}\) denoted \(P_\ell\text{,}\) and called Legendre polynomials.
