Spezielle Funktionen 3 Orthogonale Polynome

Transcription

1 Spezielle Funktionen 3 Orthogonale Polynome M. Gutting 28. Mai 2015

2 3.5 Application of the Legendre Polynomials in Electrostatics Let x R 3 and ρ(x) be a charge distribution with total charge R 3 ρ(x)dx. Then the fundamental equations of electrostatics are the electrostatic pre-maxwell equations. Let E denote the electric field, then for x R 3 div E(x) = 4πρ(x) curl E(x) = 0. We introduce the electric potential Φ and solve the equations by E(x) = Φ(x) which gives us the following Poisson equation Φ(x) = 4πρ(x). M. Gutting (Uni Siegen) Spezielle Funktionen 28. Mai / 16

3 Example 3.69 A point charge q at x 0 R 3 yields ρ(x) = qδ(x x 0 ) with the delta-distribution δ in R 3 (equality holds in the weak sense). We obtain Φ(x) = 0 for x R 3 \ {x 0 }. (1) The solution is given by Φ(x) = q x x 0 for x R3 \ {x 0 }. The corresponding electric field can be calculated to be E(x) = q x x 0 x x 0 3 for x R 3 \ {x 0 }. M. Gutting (Uni Siegen) Spezielle Funktionen 28. Mai / 16

4 Another approach to (1) introduces polar coordinates (r, ϕ, t) for x R 3 \ {0} by x 1 = r 1 t 2 cos ϕ, x 2 = r 1 t 2 sin ϕ, x 3 = rt where r = x > 0, ϕ [0, 2π), t [ 1, 1]. This gives us the representation of the Laplace equation in polar coordinates, i.e. [ ( ) r r r + 1 r 2 t (1 t2 ) t + 1 r t 2 ( ϕ ) 2 ] for x R 3 \ {0}. By separation of variables we set Φ(r, ϕ, t) = 0 Φ(r, ϕ, t) = U(r)V (ϕ)w (t). M. Gutting (Uni Siegen) Spezielle Funktionen 28. Mai / 16

5 We insert this in the equation above, i.e. ( ) 2 V (ϕ)w (t) U(r) + V (ϕ)w (t) 2 r r r U(r) + 1 r 2 U(r)V (ϕ) t (1 t2 ) t W (t) + 1 ( ) r 2 U(r)W (t) t 2 V (ϕ) = 0. ϕ Next we multiply by r 2 (1 t 2 ) and divide by U(r)V (ϕ)w (t): r 2 (1 t 2 ) U (r) U(r) + 2r(1 t2 ) U (r) U(r) + (1 t2 ) 2 W (t) W (t) 2t(1 t 2 ) W (t) W (t) + V (ϕ) V (ϕ) = 0. M. Gutting (Uni Siegen) Spezielle Funktionen 28. Mai / 16

6 This can be rewritten as r 2 (1 t 2 ) U (r) U(r) 2r(1 t2 ) U (r) U(r) (1 t2 ) 2 W (t) W (t) + 2t(1 t 2 ) W (t) W (t) = V (ϕ) V (ϕ). The left hand side only depends on r and t, the righthand side only on ϕ. Thus, the equation can only be fulfilled if both sides are equal to a constant λ R. Therefore, V (ϕ) V (ϕ) = λ V (ϕ) λv (ϕ) = 0, ϕ [0, 2π). M. Gutting (Uni Siegen) Spezielle Funktionen 28. Mai / 16

7 The non-trivial solutions of this differential equation are linear combinations of V 1 = exp(i λϕ) and V 2 = exp( i λϕ). V 1 and V 2 have to be 2π-periodic. This leads to a discretization of λ, i.e. λ = m 2, m N 0. The linear independent solutions are given by V m (ϕ) = exp(imϕ), ϕ [0, 2π), m Z. Now we consider the left hand side for these values of λ: m2 (1 t 2 ) = r 2 U (r) U(r) + 2r U (r) U(r) + (1 t2 ) W (t) W (t) 2t W (t) W (t) m2 (1 t 2 ) (1 t2 ) W (t) W (t) + 2t W (t) W (t) = r 2 U (r) U(r) + 2r U (r) U(r). M. Gutting (Uni Siegen) Spezielle Funktionen 28. Mai / 16

8 By the same argument as before we find that this equation can only hold if both sides are equal to a constant λ, i.e. m2 (1 t 2 ) (1 t2 ) W (t) W (t) + 2t W (t) W (t) = λ ) (1 t 2 )W (t) 2tW (t) + ( λ + m2 1 t 2 W (t) = 0 for t [ 1, 1] and for r > 0. λ = r 2 U (r) U(r) + 2r U (r) U(r) 0 = r 2 U (r) + 2rU (r) λu(r) M. Gutting (Uni Siegen) Spezielle Funktionen 28. Mai / 16

9 Symmetric Problems For the case of a point charge located at x 0 we can assume (after a certain rotation of the coordinate system) that x 0 is on the ε (3) -axis, i.e. x 0 = (0, 0, r 0 ) T where r 0 = x 0 > 0. In this case it is clear that the solution of (1) has a rotational symmetry, i.e. Φ does not dependent on ϕ. We can use Φ(r, ϕ, t) = U(r)W (t) with r > 0 and t [ 1, 1] (this corresponds to m = 0 before). Therefore, we find the equation for W : (1 t 2 )W (t) 2tW (t) + λw (t) = 0, t [ 1, 1]. This is the Legendre differential equation (or the Gegenbauer differential equation with λ = 1/2). M. Gutting (Uni Siegen) Spezielle Funktionen 28. Mai / 16

10 The ODE possesses a polynomial solution (in physical context: a solution with finite energy) if and only if λ = n(n + 1), n N 0. The solutions are the Legendre polynomials W n (t) = P n (t), t [ 1, 1], n N 0. M. Gutting (Uni Siegen) Spezielle Funktionen 28. Mai / 16

11 Now we consider the second equation for this values of λ: r 2 U (r) + 2rU (r) n(n + 1)U(r) = 0, r > 0. This ODE possesses two linearly independent solutions given by U 1,n (r) = r n, n N 0, U 2,n (r) = 1 r n+1, n N 0. Together we find the two linearly independent solutions of the Laplace equation in the rotational symmetric case to be for r > 0, t [ 1, 1], n N 0. Φ 1,n (r) = r n P n (t), Φ 2,n (r) = 1 r n+1 P n(t) M. Gutting (Uni Siegen) Spezielle Funktionen 28. Mai / 16

12 Every solution can be expressed as a linear combination of these two, i.e. there exist coefficients A n, B n such that Φ(r, t) = ( ) A n r n 1 + B n r n+1 P n (t). For the case of a point charge located in x 0 = (0, 0, r 0 ) T we know a solution, i.e. q Φ(x) = x x 0, x R3 \ {x 0 }. We use this solution to compute the coefficients A n and B n in the expansion. Restricting the solutions to the ε (3) -axis, the point x possesses the polar coordinates r = x, t = 1, ϕ = 0, i.e. Φ(x) = q r r 0, r > 0, r r 0. M. Gutting (Uni Siegen) Spezielle Funktionen 28. Mai / 16

13 We use the well-known geometric series and obtain q r r 0 = q 1 r 1 r 0 = q ( r0 ) n r r for r > r0, Φ(x) = r q r 0 r = q 1 ( ) n r 0 1 r = q r r 0 r0 for r < r0. r 0 Comparing this to the expansion for t = 1 (remember that P n (1) = 1), i.e. Φ(r, 1) = A n r n + B n, r > 0, r n+1 results in the following coefficients: { 0 if r > r0 A n = q if r < r r n { qr n B n = 0 if r > r 0 0 if r < r 0 M. Gutting (Uni Siegen) Spezielle Funktionen 28. Mai / 16

14 Thus, we finally obtain q Φ(x) = Φ(r, t) = x x 0 q r0 n P r = n+1 n (t) for x = r > r 0 = x 0, q r n P r n+1 n (t) for x = r < r 0 = x 0. 0 M. Gutting (Uni Siegen) Spezielle Funktionen 28. Mai / 16

15 From a mathematical point of view there is another interesting aspect. Canceling the charge q leads to (x x 0 ) 1 x x 0 = r n 0 r n+1 P n (t) for x = r > r 0 = x 0, r n r n+1 0 P n (t) for x = r < r 0 = x 0. The value x x 0 can be expressed using polar coordinates, i.e. x x 0 2 = (r 1 t 2 cos ϕ, r 1 t 2 sin ϕ, rt r 0 ) T 2 = r 2 (1 t 2 ) + (rt r 0 ) 2 = r (1 2 + r 0 2 ) r 2 2r 0 r t. M. Gutting (Uni Siegen) Spezielle Funktionen 28. Mai / 16

16 Take r > r 0 and we get for r > r 0 which is equivalent to Analogously, for r < r 0 1 = r 1 + r 0 2 r 2 2 r 0 r t 1 = 1 + r 0 2 r 2 2 r 0 r t 1 = 1 + r 2 2 r r0 2 r 0 t 1 x x 0 = r n 0 r n+1 P n(t) ( r0 ) n Pn (t). r ( ) r n P n (t). Substituting h = r 0 /r, respectively h = r/r 0, we obtain the result on the generating function of the Legendre polynomials. M. Gutting (Uni Siegen) Spezielle Funktionen 28. Mai / 16 r 0