LEGENDRE’S EQUATION
2. LEGENDRE’S EQUATION
The Legendre differential equation is (2.1)
(1 − x 2 )y ′′ − 2xy ′ + l(l + 1)y = 0,
where l is a constant. This equation arises in the solution of partial differential equa- tions in spherical coordinates (see Problem 10.2 and Chapter 13, Section 7) and so in problems in mechanics, quantum mechanics, electromagnetic theory, heat, etc., with spherical symmetry. Also see an application in Section 5.
Although the most useful solutions of this equation are polynomials (called the Legendre polynomials ), one way to find them is to assume a series solution of the differential equation, and show that the series terminates after a finite number of terms. [There are other ways of finding the Legendre polynomials; see Sections 4 and 5, and Chapter 3, Section 14, Example 6.] We assume the series solution (1.2) for y and differentiate it term by term twice to get y ′ and y ′′ :
y=a 0 +a 1 x+a 2 x 2 +a 3 x 3 +a 4 x 4 +···+a n x n +···, (2.2)
y ′ =a 1 + 2a 2 x + 3a 3 x 2 + 4a 4 x 3 + · · · + na n x n−1 +···, y ′′ = 2a 2 + 6a 3 x + 12a 4 x 2 + 20a 5 x 3 + · · · + n(n − 1)a n x n−2 +···.
Section 2 Legendre’s Equation 565
We substitute (2.2) into (2.1) and collect the coefficients of the various powers of x; it is convenient to tabulate them as follows:
x 2 x 3 ··· x n ··· y ′′
const.
2a 2 6a 3 12a 4 20a 5 (n + 2)(n + 1)a n+2 −x 2 y ′′
−2a 2 −6a 3 −n(n − 1)a n −2xy ′
−2a 1 −4a 2 −6a 3 −2na n l(l + 1)y l(l + 1)a 0 l(l + 1)a 1 l(l + 1)a 2 l(l + 1)a 3 l(l + 1)a n
Next we set the total coefficient of each power of x equal to zero [because, as discussed in Section 1, y must satisfy (2.1) identically]. For the first few powers of x we get
(l − 1)(l + 2)
(l − 2)(l + 3)
= l(l + 1)(l − 2)(l + 3) a 0 ; 4!
and from the x n coefficient we get (2.4)
(n + 2)(n + 1)a n+2 + (l 2 +l−n 2 − n)a n = 0. The coefficient of a n in (2.4) can be factored to give (2.5)
l 2 −n 2 + l − n = (l + n)(l − n) + (l − n) = (l − n)(l + n + 1).
Then we can write a general formula for a n+2 in terms of a n . This formula (2.6) includes the formulas (2.3) for a 2 ,a 3 , and a 4 , and makes it possible for us to find any even coefficient as a multiple of a 0 , and any odd coefficient as a multiple of a 1 . Solving (2.4) for a n+2 and using (2.5), we have
(l − n)(l + n + 1)
a n+2 =−
(n + 2)(n + 1)
The general solution of (2.1) is then a sum of two series containing (as the solu- tion of a second-order differential equation should) two constants a 0 and a 1 to be determined by the given initial conditions:
y=a 0 1− x 2 + l(l + 1)(l − 2)(l + 3) x 4 −···
(2.7) (l − 1)(l + 2)
+a x 3 x− (l − 1)(l + 2)(l − 3)(l + 4) + x 1 5 −··· .
From equation (2.6) you can see by the ratio test that these series converge for x 2 < 1. It can be shown that, in general, they do not converge for x 2 = 1.
566 Series Solutions of Differential Equations Chapter 12
Example. Consider the a 1 series for l = 0. If x 2 = 1, this series is 1 + 1 3 + 1 5 + · · · , which is divergent by the integral test (Chapter 1, Section 6B). Now in many applications x is the cosine of an angle θ, and l is a (nonnegative) integer. We want a solution which converges for all θ, that is, a solution which converges at x = ±1 as well as for |x| < 1. We can always find one (but not two) such solutions when l is an integer; let us see how.
Legendre Polynomials We have seen that for l = 0 the a 1 series in (2.7) di- verges. But look at the a 0 series; it gives just y = a 0 for l = 0 since all the rest of the terms contain the factor l. If l = 1, the a 0 series is divergent at x 2 = 1, but the
a 1 series stops with y = a 1 x [since all the rest of the terms in the a 1 series contain the factor (l − 1)]. For any integral l, one series terminates giving a polynomial
solution; the other series is divergent at x 2 = 1. (Negative integral values of l would simply give solutions already obtained for positive l’s; for example, l = −2 gives the polynomial solution y = a 1 x which is the same as the l = 1 solution. Conse- quently, it is customary to restrict l to nonnegative values.) Thus we obtain a set of
polynomial solutions of the Legendre equation, one for each nonnegative integral l. Each solution contains an arbitrary constant factor (a 0 or a 1 ); for l = 0, y = a 0 ; for l = 1, y = a 1 x, and so on. If the value of a 0 or a 1 in each polynomial is selected so that y = 1 when x = 1, the resulting polynomials are called Legendre Polynomials, written P l (x). From (2.6) and (2.7) and the requirement P l (1) = 1, we find the following expressions for the first few Legendre polynomials:
(2.8) P 0 (x) = 1,
P 1 (x) = x,
P 2 (x) = 1 2 (3x 2 − 1).
Finding a few more Legendre polynomials by this method and other methods will
be left to the problems. Although P l (x) for any integral l may be found by this method, simpler ways of obtaining the Legendre polynomials for larger l will be outlined in Sections 4 and 5. Of course, if you just want the formula for a particular P l , you can find it by computer or in reference books.
Eigenvalue Problems In finding the Legendre polynomials as solutions of Leg- endre’s equation (2.1), we have solved an eigenvalue problem. (See Chapter 3, Sections 11 and 12.) Recall that in an eigenvalue problem we are given an equation or a set of equations containing a parameter, and we want solutions that satisfy some special requirement; in order to obtain such solutions we must choose partic- ular values (called eigenvalues) for the parameter in the problem. In finding the Legendre polynomials, we asked for series solutions of Legendre’s equation (2.1) which converged at x = ±1. We saw that we could obtain such solutions if the parameter took on any integral value. The values of l, namely 0, 1, 2, · · · , are called eigenvalues (or characteristic values); the corresponding solutions P l (x) are called eigenfunctions (or characteristic functions).
Note the parallel between the eigenvalue-eigenvector problems of Chapter 3 and the eigenvalue-eigenfunction problems of this chapter. Recall that in Chapter 3, we wrote an eigenvalue equation as Mr = λr where M was a matrix operator which operated on the eigenvector r to produce a multiple of r. The Legendre equation is of the form f (D)y(x) = l(l + 1)y(x) where f (D) is a differential operator which
Section 3 Leibniz’ Rule for Differentiating Products 567
operates on the eigenfunction y(x) to produce a multiple of y(x). See Section 22 and Chapter 13 for further examples of differential equations whose solutions are eigenfunctions.
The Legendre polynomials are also called Legendre functions of the first kind. The second solution for each l, which is an infinite series (convergent for x 2 < 1), is called a Legendre function of the second kind and is denoted by Q l (x) (See Problem 4.) The functions Q l (x) are not used as frequently as the polynomials P l (x). For fractional l both solutions are infinite series; these again occur less frequently in applications.
PROBLEMS, SECTION 2
1. Using (2.6) and (2.7) and the requirement that P l (l) = 1, find P 2 (x), P 3 (x), and P 4 (x). Check your results by computer.
2. Show that P l (−1) = (−1) l . Hint: When is P l (x) an even function and when is it an odd function? 3. Computer plot graphs of P l (x) for l = 0, 1, 2, 3, 4, and x from −1 to 1. 4. Use the method of reduction of order [Chapter 8, Section 7(e)] and the known
solution P l (x) of Legendre’s equation to find the second solution Q l (x) (in terms of an integral). Evaluate the integral for the cases l = 0 and l = 1 to find Q 0 and Q 1 . Note the divergence of the logarithms at x = ±1. Expand the logarithms in Q 0 to get the divergent series mentioned above [a 1 series in (2.7) with l = 0, x 2 = 1].