LEGENDRE POLYNOMIALS AND APPLICATIONS We construct Legendre polynomials and apply them to solve Dirichlet problems in spherical coordinates.. Legendre equation: series solutions The Legendre equation is the second order differential equation ( ( x 2 y 2xy + λy = which can also be written in self-adjoint form as [ (2 ( x 2 y ] + λy =. This equation has regular singular points at x = and at x = while x = is an ordinary point. We can find power series solutions centered at x = (the radius of convergence will be at least. Now we construct such series solutions. Assume that y = c j x j be a solution of (. By substituting such a series in (, we get ( x 2 j(j c j x j 2 2x jc j x j + λ c j x j = j=2 j=2 j=2 j= j(j c j x j 2 j(j c j x j 2jc j x j + λ c j x j = After re-indexing the first series and grouping the other series, we get (j + 2(j + c j+2 x j (j 2 + j λc j x j = and then j= [ (j + 2(j + cj+2 (j 2 ] + j λc j x j =. By equating each coefficient to, we obtain the recurrence relations c j+2 = (j + j λ (j + 2(j + c j, j =,, 2, We can obtain two independent solutions as follows. For the first solution we make c and c =. In this case the recurrence relation gives c 3 = 2 λ c =, c 5 = 2 λ c 3 =,, c odd =. 6 2 The coefficients with even index can be written in terms of c : c 2 = λ 2 c, c 4 = (2 3 λ c 2 = 4 3 (2 3 λ( λ c 4! Date: April 9, 26.
2 LEGENDRE POLYNOMIALS AND APPLICATIONS we prove by induction that c 2k = [(2k (2k 2 λ][(2k 3(2k 4 λ] [3 2 λ][ λ] c (2k! We can write this in compact form as ( c 2k = c k [(2i + (2i λ] (2k! This give a solution y (x = c i= ( k k= i= [(2i + (2i λ] x 2k (2k! A second series solution (independent from the first can be obtained by making c = and c. In this case c even = and ( k c c 2k+ = [2i(2i λ] (2k +! The corresponding solution is y 2 (x = c k= i= i= ( k x 2k+ [(2i(2i λ] (2k +! Remark. It can be proved by using the ratio test that the series defining y and y 2 converge on the interval (, (check this as an exercise. 2. It is also proved that for every λ either y or y 2 is unbounded on (,. That is, as x or as x, one of the following holds, either y (x or y 2 (x. 3. The only case in which Legendre equation has a bounded solution on [, ] is when the parameter λ has the form λ = n(n + with n = or n Z +. In this case either y or y 2 is a polynomial (the series terminates. This case is considered below. 2. Legendre polynomials Consider the following problem Problem. Find the parameters λ R so that the Legendre equation [ (3 ( x 2 y ] + λy =, x. has a bounded solution. This is a singular Sturm-Liouville problem. It is singular because the function ( x 2 equals when x = ±. For such a problem, we don t need boundary conditions. The boundary conditions are replaced by the boundedness of the solution. As was pointed out in the above remark, the only values of λ for which we have bounded solutions are λ = n(n + with n =,, 2,. These values of λ are the eigenvalues of the SL-problem. To understand why this is so, we go back to the construction of the series solutions and look again at the recurrence relations giving the coefficients c j+2 = (j + j λ (j + 2(j + c j, j =,, 2,
LEGENDRE POLYNOMIALS AND APPLICATIONS 3 If λ = n(n +, then c n+2 = (n + n λ (n + 2(n + c n =. By repeating the argument, we get c n+4 = and in general c n+2k = for k. This means if n = 2p (even, the series for y terminates at c 2p and y is a polynomial of degree 2p. The series for y 2 is infinite and has radius of convergence equal to and y 2 is unbounded. If n = 2p + (odd, then the series for y 2 terminates at c 2p+ and y 2 is a polynomial of degree 2p + while the solution y is unbounded. For λ = n(n+, we can rewrite the recurrence relation for a polynomial solution in terms of c n. We have, (j + 2(j + c j = j(j + n(n + c (j + 2(j + j+2 = (n j(n + j + c j+2, for j = n 2, n 4, or. Equivalently, c n 2k = (n 2k + 2(n 2k + c n 2k+2, (2k(2n 2k + k =, 2,, [n/2]. I will leave it as an exercise to verify the following formula for c n 2k in terms of c n : c n 2k = ( k 2 k k! The polynomial solution is therefore y(x = [n/2] k= ( k 2 k k! n(n (n 2k + (2n (2n 3 (2n 2k + c n. n(n (n 2k + (2n (2n 3 (2n 2k + c nx n 2k where c n is an arbitrary constant. The n-th Legendre polynomial P n (x is the above polynomial of degree n for the particular value of c n c n = (2n! 2 n (n! 2. This particular value of c n is chosen to make P n ( =. simplification P n (x = [n/2] ( k (2n 2k! 2 n k!(n k!(n 2k! xn 2k. k= We have then (after Note that if n is even (resp. odd, then the only powers of x involved in P n are even (resp. odd and so P n is an even (resp. odd. The first six Legendre polynomials are. P (x = P (x = x P 2 (x = 2 (3x2 P 3 (x = 2 (5x3 3x P 4 (x = 8 (35x4 3x 2 + 3 P 5 (x = 8 (63x5 7x 3 + 5x We have the following proposition.
4 LEGENDRE POLYNOMIALS AND APPLICATIONS P P 2 P 4 P 6 P P 3 P 5 P 7 Proposition. If y(x is a bounded solution on the interval (, of the Legendre equation ( with λ = n(n +, then there exists a constant K such that y(x = KP n (x where P n is the n-th Legendre polynomial. Remark. When λ = n(n + a second solution of the Legendre equation, independent from P n, can be found in the form Q n (x = 2 P n(x ln + x x + R n(x where R n is a polynomial of degree n. The construction of Q n can be achieved by the method of reduction of order. Note that Q n (x as x ±. The general solution of the Legendre equation is then y(x = AP n (x + B n Q n (x and such a function is bounded on the interval (, if and only if B =. 3. Rodrigues formula The Legendre polynomials can be expressed in a more compact form. Theorem. (Rodrigues Formula The n-th Legendre polynomial P n is given by the following (4 P n (x = [ (x 2 2 n n! dx n n] (thus expression (4 gives a solution of (3 with λ = n(n +. d n Proof. Let y = (x 2 n. We have following Claim. The k-th derivative y (k (x of y satisfies the following: (5 ( x 2 d2 y (k dx 2 + 2(n k x dy(k dx + (2n k(k + y(k =. Proof of the Claim. By induction. For k =, y = y (. We have and after differentiation, we get y = 2nx(x 2 n ( x 2 y + 2nxy = ( x 2 y + 2(n xy + 2ny =
LEGENDRE POLYNOMIALS AND APPLICATIONS 5 So formula (5 holds when k =. By induction suppose the (5 holds up to order k. We can rewrite (5 for k as ( x 2 dy(k dx + 2(n kxy(k + (2n k + ky (k =. We differentiate to obtain ( x 2 d2 y (k dx 2 + [2(n k 2]x dy(k dx + [2(n k + k(2n k + ]y(k =. which is precisely (5. Now if we let k = n in (5, we obtain ( x 2 d2 y (n dx 2 2x dy(n dx + n(n + y(k =. Hence y (n solves the Legendre equation with λ = n(n +. Since y (n is a polynomial of degree 2n, then by Proposition, it is a multiple of P n. There is a constant K such that P n (x = Ky (n (x. To complete the proof, we need to find K. For this notice that the coefficient of x n in P n is (2n!/(2 n (n! 2. The coefficient of x n in y (n is that of Hence d n (x 2n dx n = (2n(2n (2n n + x n = (2n! x n n! K (2n! = (2n! n! 2 n (n! 2 K = 2 n n!. This completes the proof of the Rodrigues formula. A consequence of this formula is the following property between three consecutive Legendre polynomials. Proposition. The Legendre polynomials satisfy the following (6 (2n + P n (x = P n+(x P n (x Proof. From Rodrigues formula we have P k(x = d ( d k dx 2 k k! dx k [(x2 k ] = 2k d k 2 k k! dx k [x(x2 k ] d k [ = ((2k x 2 2 k (k! dx k (x 2 k 2] For k = n +, we get d n P n+(x = [ ((2n + x 2 2 n n! dx n (x 2 n ] From Rodrigues formula at n, we get P n (x = d ( d n dx 2 n (n! dx n [(x2 n ] = 2n d n 2 n n! dx n [(x2 n ] As a consequence, we have P n+(x P n (x = 2n + 2 n n! d n dx n [(x2 n ] = (2n + P n (x.
6 LEGENDRE POLYNOMIALS AND APPLICATIONS Generating function. It can be shown that the Legendre polynomials are generated by the function g(x, t =. 2xt + t 2 More precisely, if we extend g(x, t as a Taylor series in t, then the coefficient of t n is the polynomial P n (x: (7 = P n (xt n. 2xt + t 2 A consequence of (7 is the following relation between three consecutive Legendre polynomials. Proposition. The Legendre polynomials satisfy the following (8 (2n + xp n (x = (n + P n+ (x + np n (x Proof. We differentiate (7 with respect to t: (x t 2xt + t 2 3 = np n (xt n. n= We multiply by 2xt + t 2 and use (7 (x tp n (xt n = ( 2xt + t 2 np n (xt n. n= Equivalently, xp n (xt n P n (xt n+ = np n (xt n 2nxP n (xt n + np n (xt n+ n= and after grouping the series xp (x P (x + [(2n + xp n (x (n + P n+ (x np n (x] t n n= n= Property (8 is obtained by equating to the coefficient of t n. 4. Orthogonality of Legendre polynomials When the Legendre equation is considered as a (singular Sturm-Liouville problem on [, ], we get the following orthogonality theorem Theorem 2. Consider the singular SL-problem ( x 2 y 2xy + λy = < x <, with y bounded on (,. The eigenvalues are λ n = n(n + with corresponding eigenfunctions P n (x. Furthermore, the eigenfunctions corresponding to distinct eigenvalues are orthogonal. That is (9 < P n (x, P m (x >= n= P n (xp m (xdx =, n m.
LEGENDRE POLYNOMIALS AND APPLICATIONS 7 Proof. Recall that the self-adjoint form of the Legendre equation is [( x 2 y ] + λy =, (with p(x = x 2, r(x =, and q(x =. The corresponding weight function is r =. We have already seen that the eigenvalues and eigenfunctions are given by λ n = n(n + and P n (x. We are left to verify the orthogonality. We write the Legendre equation for P m and P n : λ n P n (x = [(x 2 P n(x] λ m P m (x = [(x 2 P m(x] Multiply the first equation by P m, the second by P n and subtract. We get, (λ n λ m P n (xp m (x = [(x 2 P n(x] P m (x [(x 2 P m(x] P n (x = [ (x 2 P n(xp m (x (x 2 P m(xp n (x ] Integrate from to (λ n λ m = [ (x 2 (P n(xp m (x P m(xp n (x ] P n (xp m (xdx = [ (x 2 (P n(xp m (x P m(xp n (x ] =. The square norms of the Legendre polynomials are given below. Theorem 3. We have the following ( P n (x 2 = P n (x 2 dx = Proof. We use generating function (7 to get ( 2 2xt + t 2 = P n (xt n = Now we integrate from to : dx 2xt + t 2 = n,m n,m 2 2n + P n (xp m (xt n+m ( P n (xp m (xdx By using the orthogonality of P n and P m (for n m, we get 2t [ ln 2xt + t 2 ] x= x= = P n (x 2 t 2n, and after simplifying the left side: t ln + t t = P n (x 2 t 2n. Recall that for s <, the Taylor series of ln( + s is ( j ln( + s = s j. j j= t n+m
8 LEGENDRE POLYNOMIALS AND APPLICATIONS Hence for t <, we have t ln + t t = (ln( + t ln( t t = ( j t j ( j ( t j t j j j= j= = ( j + t j t j j= 2 = 2n + t2n It follows that 2 2n + t2n = P n (x 2 t 2n. An identification of the coefficient of t 2n gives (. 5. Legendre series The collection of Legendre polynomials {P n (x} n forms a complete family in the space Cp[, ] of piecewise smooth functions on the interval [, ]. Any piecewise function f has then a generalized Fourier series representation in terms of these polynomials. The associated series is called the Legendre series of f. Hence, f(x c n P n (x where c n = < f(x, P n(x > P n (x 2 = 2n + f(xp n (xdx 2 Theorem 4. Let f be a piecewise smooth function on [, ]. Then, f av (x = f(x+ + f(x = c n P n (x. 2 In particular at the points x, where f is continuous, we have f(x = c n P n (x. Remark. For each m, the Legendre polynomials P, P,, P m forms a basis in the space of polynomials of degree m. Thus, if R(x is a polynomial of degree m, then the Legendre series of R terminates at the order m (i.e. c k = for k > m. In particular, x m = c P (x + c P (x + + c m P m (x = c + c x + c 2 2 (3x2 + For example, x 2 = 2 3 P (x + P (x + 2 3 P 2(x x 3 = P (x + 3 5 P (x + P 2 (x + 2 5 P 3(x
LEGENDRE POLYNOMIALS AND APPLICATIONS 9 Remark 2. Suppose that f is an odd function. Since P n is odd when n is odd and P n is even when n is even, then the Legendre coefficients of f with even indices are all zero (c 2j =. The Legendre series of f contains only odd indexed polynomials. That is, f av (x = c 2j+ P 2j+ (x where c 2j+ = (2(2j + + f(xp 2j+ (xdx = (4j + 3 f(xp 2j+ (xdx. Similarly, if f is an even function, then its Legendre series contains only even indexed polynomials. f av (x = c 2j P 2j (x where c 2j = (2(2j + f(xp 2j (xdx = (4j + f(xp 2j (xdx. If a function f is defined on the interval [, ], then we can extend it as an even function f even to the interval [, ]. The Legendre series of f even contains only even-indexed polynomials. Similarly, if we extend f as an odd function f odd to [, ], then the Legendre series contains only odd-indexed polynomials. We have the following theorem. Theorem 5. Let f be a piecewise smooth function on [, ]. Then, f has an expansion into even Legendre polynomials f av (x = f(x+ + f(x 2 = c 2j P 2j (x. Similarly, f has an expansion into odd Legendre polynomials The coefficients are given by f av (x = f(x+ + f(x 2 c n = (2n + = c 2j+ P 2j+ (x. f(xp n (xdx. Example. Consider the function f(x = coefficient of f is { < x <, < x < The n-th Legendre c n = 2n + 2 f(xp n (xdx = 2n + 2 P n (xdx.
LEGENDRE POLYNOMIALS AND APPLICATIONS The first four coefficients are c = 2 dx = 2 Hence, c = 3 2 c 2 = 5 2 c 3 = 7 2 xdx = 3 4 2 (3x2 dx = 2 (5x3 3xdx = 7 6 = 2 P (x + 3 4 P (x 7 6 P 3(x + < x <, = 2 P (x + 3 4 P (x 7 6 P 3(x + < x < { < x <, Example 2. Let f(x =. Since f is odd, its Legendre series < x < contains only odd indexed polynomials. We have c 2n+ = (4n + 3 P 2n+ (xdx. By using the recurrence relation (6 ((2k + P k = P k+ P k with k = 2n +, we get c 2n+ = (P 2n+2(x P 2n(xdx = P 2n+2 ( P 2n+2 ( P 2n ( + P 2n (. Since P 2j ( = that P 2j ( = ( j (2j!/2 2j (j! 2 (see exercise, then it follows that ( ( n (4n + 3 (2n! c 2n+ = P 2n ( P 2n+2 ( = 2 n+ (n + (n! 2. We have then the expansion ( ( n (4n + 3 (2n! = 2 n+ (n + (n! 2 P 2n+(x, < x <. 6. Separation of variables for u = in spherical coordinates Recall that if (x, y, z and (ρ, θ, ϕ denote, respectively, the cartesian and the spherical coordinates in R 3 : x = ρ cos θ sin ϕ, y = ρ sin θ sin ϕ, z = ρ cos ϕ, then the Laplace operator has expression = 2 ρ 2 + 2 ρ ρ + 2 ρ 2 sin 2 ϕ θ 2 + 2 ρ 2 ϕ 2 + cot ϕ ρ 2 ϕ with ρ >, θ R, and ϕ (, π. Consider the problem of finding bounded solutions u = u(ρ, θ, ϕ of the Laplace equation u =. That is, u(ρ, θ, ϕ bounded inside the sphere ρ < A and satisfies ( 2 u ρ 2 + 2 u ρ ρ + 2 u ρ 2 sin 2 ϕ θ 2 + 2 u ρ 2 ϕ 2 + cot ϕ u ρ 2 ϕ =
LEGENDRE POLYNOMIALS AND APPLICATIONS The method at our disposal is that of separation of variables. Suppose that u(ρ, θ, ϕ = R(ρΘ(θΦ(ϕ solves the Laplace equation (we are assuming that Θ and Θ are 2π-periodic. After replacing u and its derivatives in terms of R, Θ, Φ and their derivatives, we can rewrite ( as (2 ρ 2 R (ρ R(ρ + (ρ 2ρR R(ρ + Θ (θ sin 2 ϕ Θ(θ + Φ (ϕ Φ(ϕ + cot ϕ Φ (ϕ = Φ(ϕ Separating the variable ρ from θ and ϕ leads to ρ 2 R + 2ρR λr =, Θ (θ sin 2 ϕ Θ(θ + Φ (ϕ Φ(ϕ + cot ϕ Φ (ϕ = λ Φ(ϕ where λ is the separation constant. A further separation of the second equation leads to the three ODEs (3 (4 (5 ρ 2 R (ρ + 2ρR (ρ λr(ρ = Φ (ϕ + cos ϕ ( sin ϕ Φ (ϕ + λ Θ (θ + αθ(θ = Φ(ϕ = α sin 2 ϕ with α and λ constants. The R-equation and Θ-equation are familiar and we know how to solve them: (3 is a Cauchy-Euler equation and (4 has constant coefficients. Furthermore, the periodicity of Θ implies that α = m 2 with m nonnegative integer and the eigenfunctions are cos(mθ and sin(mθ. A class of solutions of u =. Consider the case α = m =. In this case Θ(θ = and the function u is independent on θ. The R-equation and Φ-equations are (6 (7 ρ 2 R (ρ + 2ρR (ρ λr(ρ = Φ (ϕ + cos ϕ sin ϕ Φ (ϕ + λφ(ϕ = Equation (6 is a Cauchy-Euler equation with solutions R (ρ = ρ p, R 2 (ρ = ρ p2 where p,2 are the roots of p 2 + p λ =. To understand the Φ-equation, we need to make a change of variable. For ϕ (, π, consider the change of variable t = cos ϕ t (,, and let w(t = Φ(ϕ = Φ(arccos t Φ(ϕ = w(cos ϕ. We use equation (7 to write an ODE for w. We have Φ (ϕ = sin ϕ w (cos ϕ = t 2 w (t, Φ (ϕ = sin 2 ϕ w (cos ϕ cos ϕ w (cos ϕ = ( t 2 w (t tw (t By replacing these expressions of Φ and Φ in (7, we get (8 ( t 2 w (t 2tw (t + λw(t =. This is the Legendre equation. To get w a bounded solution, we need to have λ = n(n + with n a nonnegative integer. In this case w is a multiple of the
2 LEGENDRE POLYNOMIALS AND APPLICATIONS Legendre polynomial P n (t. By going back to the function Φ and variable ϕ, we get the following lemma. Lemma. The eigenvalues and eigenfunctions of equation (7 are λ = n(n + with n a nonnegative integer and Φ n (ϕ = P n (cos ϕ, where P n is Legendre polynomial of degree n. For λ = n(n + the exponents p,2 for the R-equation are p = n and p 2 = (n + and the independent solution of (6 are ρ n, ρ (n+ = ρ n+. Note that the second solution is unbounded. We have established the following. Proposition. If u(ρ, ϕ = R(ρΦ(ϕ is a bounded solution of the Laplace equation u = inside a sphere, then there exists a nonnegative integer n so that u(ρ, ϕ = Cρ n P n (cos(ϕ, (C constant. The following figure illustrates the graphs of the spherical polynomials P n (cos ϕ for n =,, 2, and 3. The surface has parametric equation ( P n (cos ϕ, θ, ϕ. 7. Dirichlet problem in the sphere with longitudinal symmetry Consider the following Dirichlet problem in the sphere of radius L: { u(ρ, θ, ϕ = (9 u(l, θ, ϕ = f(θ, ϕ Such a problem models the steady state heat distribution inside the sphere when the temperature on the surface is given by f. The case with longitudinal symmetry means that the problem is independent on the angle θ. Thus f = f(ϕ and the
LEGENDRE POLYNOMIALS AND APPLICATIONS 3 temperature u depends only on the radius ρ and the altitude ϕ ( u = u(ρ, ϕ. Problem (9 can be written as u (2 ρρ + 2 ρ u ρ + ρ 2 u ϕϕ + cos ϕ ρ 2 sin ϕ u ϕ = u(l, ϕ = f(ϕ Since u is assumed to be a bounded function, then the method of separation of variables for the PDE in (2 leads (as we have seen above to the following solutions with separated variables u n (ρ, ϕ = ρ n P n (cos ϕ, n =,, 2, The general series solution of the PDE in (2 is therefore u(r, ϕ = C n ρ n P n (cos ϕ. In order for such a solution to satisfy the nonhomogeneous condition we need to have u(l, ϕ = f(ϕ = C n L n P n (cos ϕ. The last series is a Legendre series but it is expressed in terms of cos ϕ. To find the coefficients C n, we need to express the last series as a Legendre series in standard form. We resort again to the substitution t = cos ϕ with < t <. We rewrite f in terms of the variable t (for instance g(t = f(ϕ. The last series is g(t = C n L n P n (t. Now this is the usual Legendre series of the function g(t. Its n-th Legendre coefficients C n L n is given by C n L n = 2n + 2 g(tp n (tdt. We can rewrite C n this in terms of the variable ϕ as (2 C n = 2n + 2L n π sin ϕ f(ϕp n (cos ϕ dϕ. The solution of problem (2 is therefore u(ρ, ϕ = C n ρ n P n (cos ϕ, where C n is given by formula (2. Example. The following Dirichlet problem represents the steady-state temperature distribution inside a ball of radius assuming that the upper hemisphere is kept at constant temperature and the lower hemisphere is kept at temperature. cos ϕ u ρρ + 2 ρ u ρ + ρ 2 u ϕϕ + ρ 2 sin ϕ u ϕ =, f(ϕ = { if < ϕ < π/2, if (π/2 < ϕ < π, u(, ϕ = f(ϕ
4 LEGENDRE POLYNOMIALS AND APPLICATIONS The solution to this problem is where u(ρ, ϕ = C n ρ n P n (cos ϕ π C n = 2n + 2( n f(ϕ sin ϕ P n (cos ϕ dϕ 5(2n + π/2 = n sin ϕ P n (cos ϕ dϕ 5(2n + = n P n (t dt We can find a closed expression for C n. You will be asked in the exercises to establish the formula P n (x = 2n + [P n ( P n+ (] Thus, it follows from the fact that P odd ( = that C even = and from that P 2j ( = ( j (2j! 2 2j (j! 2 C 2j+ = 25( j (2j! 2j+ 2 2j (j + (j! 2 The solution to the problem ( j (2j! ( ρ 2j+ u(ρ, ϕ = 25 P2j+ 2 2j (j + (j! 2 (cos ϕ Example 2. (Dirichlet problem in a spherical shell. Consider the BVP u ρρ + 2 ρ u ρ + ρ 2 u ϕϕ + cos ϕ ρ 2 sin ϕ u ϕ =, < ρ < 2, < ϕ < (π/2, u(, ϕ = 5, u(2, ϕ =, < ϕ < (π/2, u(ρ, π/2 = < ρ < 2. Such a problem models the steady-state temperature distribution inside in a spherical shell when the temperature on the outer hemisphere is kept at degrees, that in the inner hemisphere is kept at temperature 5 degrees and the temperature at the base is kept at degree. The separation of variables for the homogeneous part leads to the ODE problems ρ 2 R + 2ρR λr =, sin ϕ Φ + cos ϕ Φ + λ sin ϕ Φ =, Φ(π/2 = As we have seen above, the eigenvalues of the Φ-equation are λ = n(n + and corresponding eigenfunctions P n (cos ϕ. This time however, we also need to have Φ n (π/2 = P n ( =. Therefore n must be odd. The solutions with separated variables of the homogeneous part are ρ n P n (cos ϕ and ρ (n+ P n (cos ϕ with n odd (note that since in this
LEGENDRE POLYNOMIALS AND APPLICATIONS 5 problem ρ >, the second solution ρ (n+ of the R-equation must be considered. The series solution is [ u(ρ, ϕ = A 2j+ ρ 2j+ + B ] 2j+ ρ 2j+2 P 2j+ (cos ϕ. Now we use the nonhomogeneous boundary condition to find the coefficients. [ = A 2j+ 2 2j+ + B ] 2j+ 2 2j+2 P 2j+ (cos ϕ 5 = [A 2j+ + B 2j+ ] P 2j+ (cos ϕ for < ϕ < (π/2. Equivalently, [ = A 2j+ 2 2j+ + B ] 2j+ 2 2j+2 P 2j+ (x, < x <, 5 = [A 2j+ + B 2j+ ] P 2j+ (x, < x <. By using the above series and the series expansion of over [, ] into odd Legendre polynomials (see previous examples we get = ( j (2j! 2 2j+2 (j + (j! 2 P 2j+(x < x <, A 2j+ + B 2j+ = 5( j (2j! 2 2j+2 (j + (j! 2 2 2j+ A 2j+ + B 2j+ 2 2j+2 = ( j (2j! 2 2j+2 (j + (j! 2 From these equations A 2j+ and B 2j+ can be explicitly found.
6 LEGENDRE POLYNOMIALS AND APPLICATIONS 8. More solutions of u = Recall, from the previous section, that if u = R(ρΘ(θΦ(ϕ satisfies u =, then the functions R, Θ, and Φ satisfy the ODEs (22 (23 (24 ρ 2 R (ρ + 2ρR (ρ λr(ρ = Φ (ϕ + cos ϕ ( sin ϕ Φ (ϕ + λ Θ (θ + αθ(θ = Φ(ϕ = α sin 2 ϕ The function Θ is 2π-periodic, R and Φ bounded. In the previous section we assumed u independent on θ (this is the case corresponding to the eigenvalue α =. Now suppose that u depends effectively on θ. The eigenvalues and eigenfunctions of the Θ-problem are { α = m 2 Θ, (θ = cos(mθ Θ 2, m Z + (θ = sin(mθ For α = m 2, the Φ-equation becomes (25 Φ (ϕ + cos ϕ sin ϕ Φ (ϕ + (λ m2 sin 2 ϕ Φ(ϕ = If we use the variable t = cos ϕ and let w(t = Φ(cos ϕ, then w solves (26 ( t 2 w (t 2tw (t + (λ m2 t 2 w(t =. or equivalently in self-adjoint form as [ (27 ( t 2 w (t ] + (λ m2 t 2 w(t =. Equation (26 (or (27 is called the generalized Legendre equation. Its solutions are related to those of the Legendre equation and are given by the following lemma. Lemma. Then the function Let y(t be a solution of the legendre equation ( t 2 d2 y dt 2 2tdy dt + λy =. w(t = ( t 2 m/2 dm y(t dt m solves the generalized Legendre equation ( t 2 w (t 2tw (t + (λ m2 t 2 w(t =. For λ = n(n +, the bounded solutions of (26 are Pn m (t = ( t 2 m/2 dm P n (t dt m these are called the associated Legendre polynomials (of degree n and order m. Note that since P n is a polynomial of degree n, then if m > n, the function Pn m is identically zero (Pn m (t =. Note also that Pn(x = P n (x.
LEGENDRE POLYNOMIALS AND APPLICATIONS 7 Example. P 2 (x = 3x x 2, P 3 (x = 3 2 (5x2 x 2 P 2 3 = 5x( x 2 P 2 4 (x = 5 2 (7x2 ( x 2 P 3 4 (x = 5x( x 2 3/2 P 3 5 (x = 5 2 (9x2 ( x 2 3/2 In summary we have constructed solutions of u = of the form The functions ρ n cos(mθp m n (cos ϕ (m n Y mn (θ, ϕ = P m n (cos ϕ cos(mθ are called spherical harmonics. Some of the surfaces given in spherical coordinates by ρ = P m n (θ are plotted in the figure Example: small vibrations of a spherical membrane. Consider the radial vibrations of a spherical membrane of radius L. Let u(θ, ϕ, t denotes the radial displacement at time t, from equilibrium, of the point on the L-sphere with coordinates (θ, ϕ. The function u satisfies the wave equation ( u tt = c 2 u ϕϕ + cot ϕ u ϕ + sin ϕ u θθ The method of separation of variables for the PDE leads to the following solutions P m n (cos ϕ cos(mθ cos(ω mn t,, with m n and ω mn = c n(n +.The solution u mn (θ, ϕ, t = L + ap m n (cos ϕ cos(mθ cos(ω mn t,,
8 LEGENDRE POLYNOMIALS AND APPLICATIONS are the (m, n-modes of vibration of the spherical membrane. Some of the profiles are illustrated in the following figure. m=, n=4 m=, n=4 m=2, n=4 m=3, n=4 m=3, n=7
LEGENDRE POLYNOMIALS AND APPLICATIONS 9 9. Exercises. Exercise. Use the recurrence relation that gives the coefficients of the Legendre polynomials to show that P 2n ( = ( n (2n! 2 2n (n! 2. Exercise 2. Use exercise to verify that ( 4n + 3 (2n! P 2n ( P 2n+2 ( = ( n 2n + 2 2 2n (n! 2. Exercise 3. Use P (x =, P (x = x and the recurrence relation (8: (2n + xp n (x = (n + P n+ (x + np n (x to find P 2 (x, P 3 (x, P 4 (x, and P 5 (x. Exercise 4. Use Rodrigues formula to find the Legendre polynomials P (x to P 5 (x. Exercise 5. Use Rodrigues formula to establish xp n(x = np n (x + P n (x. Exercise 6. Write x 2 as a linear combination of P (x, P (x, and P 2 (x. That is, find constants A, B, and C so that x 2 = AP (x + BP (x + CP 2 (x. Exercise 7. Write x 3 as a linear combination of P, P, P 2, and P 3. Exercise 8. Write x 4 as a linear combination of P, P, P 2, P 3 and P 4. Exercise 9. Use the results from exercises 6, 7, and 8 to find the integrals. x 2 P 2 (xdx, x 3 P (xdx, x 4 P 2 (xdx, x 2 P 3 (xdx x 3 P 4 (xdx x 4 P 4 (xdx Exercise. Use the fact that for m Z +, the function x m can written as a linear combination of P (x,, P m (x to show that x m P n (x =, for n > m.
2 LEGENDRE POLYNOMIALS AND APPLICATIONS Exercise. Use formula (8 and a property (even/odd of the Legendre polynomials to verify that h h P n (xdx = 2n + [P n+(h P n (h] P n (xdx = 2n + [P n (h P n+ (h] Exercise 2. Find the Legendre series of the functions f(x = 3, g(x = x 3, h(x = x 4, m(x = x. Exercise 3. Find the Legendre series of the function { for < x < f(x = x for < x < Exercise 4. Find the Legendre series of the function { for < x < h f(x = for h < x < (Use exercise. Exercise 5. Find the first three nonzero terms of the Legendre series of the functions f(x = sin x and g(x = cos x. In exercises 6 to 9 solve the following Dirichlet problem inside the sphere u ρρ + 2 ρ u ρ + ρ 2 u ϕϕ + cos ϕ ρ 2 sin ϕ u ϕ =, < ρ < L, < ϕ < π u(l, ϕ = f(ϕ < ϕ < π Assume u(ρ, ϕ is bounded. Exercise 6. L =, and f(ϕ = { 5 for < ϕ < (π/2, for (π/2 < ϕ < π. Exercise 7. L = and f(ϕ = cos ϕ. { 5 for < ϕ < (π/4, Exercise 8. L = 5 and f(ϕ = for (π/4 < ϕ < π. Exercise 9. L = 2 and f(ϕ = sin 2 ϕ = cos 2 ϕ. Exercise 2. Solve the following Dirichlet problem in a hemisphere u ρρ + 2 ρ u ρ + ρ 2 u ϕϕ + cos ϕ ρ 2 sin ϕ u ϕ =, < ρ <, < ϕ < (π/2 u(, ϕ = < ϕ < (π/2 u(ρ, π/2 = < ρ <. Exercise 2. Solve the following Dirichlet problem in a hemisphere u ρρ + 2 ρ u ρ + ρ 2 u ϕϕ + cos ϕ ρ 2 sin ϕ u ϕ =, < ρ <, < ϕ < (π/2 u(, ϕ = cos ϕ < ϕ < (π/2 u(ρ, π/2 = < ρ <.
LEGENDRE POLYNOMIALS AND APPLICATIONS 2 Exercise 22. Solve the following Dirichlet problem in a spherical shell u ρρ + 2 ρ u ρ + ρ 2 u ϕϕ + cos ϕ ρ 2 sin ϕ u ϕ =, < ρ < 2, < ϕ < π u(, ϕ = 5 < ϕ < π u(2, ϕ = < ϕ < π. Exercise 23. Solve the following Dirichlet problem in a spherical shell u ρρ + 2 ρ u ρ + ρ 2 u ϕϕ + cos ϕ ρ 2 sin ϕ u ϕ =, < ρ < 2, < ϕ < π u(, ϕ = cos ϕ < ϕ < π u(2, ϕ = sin 2 ϕ < ϕ < π. Exercise 24. Find the gravitational potential at any point outside the surface of the earth knowing that the radius of the earth is 64 km and that the gravitational potential on the earth surface is given by { 2 cos ϕ for < ϕ < (π/2, f(ϕ = 2 for (π/2 < ϕ < π. (This is an exterior Dirichlet problem Exercise 25. The sun has a diameter of.4 6 km. If the temperature on the sun s surface is 2, C, find the approximate temperature on the following planets. Planet Mean distance from sun (millions of kilometers Mercury 57.9 Venus 8.2 Earth 49.7 Mars 228. Jupiter 778.6 Saturn 429. Uranus 2839.6 Neptune 449.6 Pluto 588.2